A pasting diagram is the analogue of composition of arrows in higher categories. The nlab page gives an example of a two dimensional diagram and how to compute it as a composite of two morphisms. I am looking for an analogue of this result in higher dimensions. i.e. two n-cubes sharing an n-1 dimensional face.

It's hard to draw the picture, but the guess for the answer, for example in 3 dimensions, is something like (1-morphism $\circ$ 3-morphism) +(2-morphism $\circ$ 2-morphism) +(3-morphism $\circ$ 1-morphism) where the first morphism is given by an edge of the first cube etc.

I didn't find the references on the nlab page really useful.

18823986863801739529In the following I will omit requirements of smoothness, extent of domain, finiteness, etc, both to simplify the exposition and because I don't know exactly what the requirements are. Please imagine that such requirements are stated correctly.

Let's say that a function $F : \mathbb{C}^n\to\mathbb{C}$ *separates* (where $\mathbb{C}$ is the complex numbers) if we can factorize it like
$F(z_1,\ldots,z_n)=F_1(z_1)\times\cdots\times F_n(z_n)$. In this case we can factorize the integral of $F$ like $\iiint F(z_1,\ldots,z_n) ~dz_1\cdots dz_n = \int F_1(z_1)dz_1\times \cdots \times\int F_n(z_n)dz_n$.

In case $F$ does not separate, we might be able to change variables to make it separate. If $g : \mathbb{C}^n\to\mathbb{C}^n$ is nice enough and $\Delta$ is its Jacobian determinant (or maybe its inverse depending on which way you like to define it), then $\iiint F(z_1,\ldots,z_n) ~dz_1\cdots dz_n = \iiint G(w_1,\ldots,w_n) ~dw_1\cdots dw_n$, where $G(w_1,\ldots,w_n)=F(g(w_1,\ldots,w_n))\Delta(w_1,\ldots,w_n)^{-1}$.

My question is: **for which $F$ can $g$ be chosen so that $G$ separates**?

An example everyone knows is $F(z_1,\ldots,z_n)=\exp(Q(z_1,\ldots,z_n))$, where $Q$ is a quadratic form. Then $g$ can be chosen to be a linear transformation that diagonalizes the quadratic form. In general, a non-linear transformation will be required.

215335811227603893672956284@MichaelAlbanese : You're right : I have written the argument for algebraic surfaces. On the other hand, it seems to me that on non-algebraic surfaces, there will not be high-dimensional linear systems of smooth curves (otherwise, you would get, after blow-up, a generically finite morphism to projective space). This should show that the space of smooth curves in $S$ is a countable union of spaces of bouded dimension. This cannot fill $M_g$ if $g$ is big. I do not have time now to write it down: does it sound sensible ?1085277782829907476639970 It depends what is meant by ``simple". One can view a holomorphic function just as the sum of a (convergent) power series, so it is quite easy (=simple) to give plenty of examples of such functions and manipulate them. But these functions satisfy many properties that plain continuous functions do not: the identity principle, the Cauchy- Riemann equations, they can be recovered through the integral formulas etc., so they are more elaborate (=non-simple). This refers to functions, but functions make up maps, give coordinates on manifolds, so they are a proxy for other holomorphic objects.According to the answers below, I think we should agree that $f$ is supposed to be surjective. Could you edit?22267962279505758414I agree with your example, thanks a lot ! I wonder if I assume that the component $Q$ of $p^{-1}(K)$ is bounded, is it true?For a layman (nay, this is already too generous, an ignoramus) as myself, this is a quite impressive answer.The idea is that if $S \setminus C$ is not disconnected, then you can write $S$ as a lower genus surface connect-sum a torus, and $C$ is an essential curve in that torus. On the other hand, I was kind of expecting that this problem would already be taken to pieces at some handbook. Because the question looks very natural to me.651877XDo half-sphere shaped domes need pentagons?695982420848lixindiyiAre the finite groups inclusions, almost all relatively cyclic?17583801687126lhttp://frothygirlz.com/2010/01/14/big-numbers-part-2/sutr902269409Andrej, thank you for your elaborations on constructive mathematics which I found to be quite interesting. I would have accepted your answer, but Tom was a little faster so I picked his.Thanks, but I don't think that it's worth it (posting this as an answer) :) Jayy2236658Let $f(n)=\sum\limits_{d \mid n}\frac{\log d}{d}.$

It is not hard to see that $f(n)\ll(\log\log n)^2$. Is there any reference for this inequality?

**EDT 1:** A possible answer is Analysis of the subtractive algorithm for greatest common divisors by Knuth D. E., Yao A. C. (Proc. Nat. Acad. Sci. USA
Vol. 72, No. 12, pp. 4720-4722).

**EDT 2:** Earlier this bound was proved by Ingham in Some Asymptotic Formulae in the Theory of Numbers (Journal of the London Mathematical Society,
V. 1-2, Issue 3, 1927, pp. 202-208).

could you tell me which page it is?

10449402235074tAlso, the $\leq$ in your bound should be a $\geq$, right?89161@MichaelBächtold As your first question, we do not require any relation with the usual Lie Bracket structure.But we hope that the $ad_X$ operator, would be an elliptic operator.Regarding the second question, I think it is included in the definition of Lie algebra that Lie bracket is $\mathbb{R}$-linear. right?355071You might consider changing the title of your question to mention Okounkov's paper explicitly. There are plenty of people here who know about homology of algebraic varieties, but unless one happens to have read the actual paper, this question is impossible to answer. On the plus side, there may well be people around who would be enticed to read the question if it mentioned the paper, but who would otherwise easily skip over such a generic title. 1252520|https://graph.facebook.com/100000760552999/picture?type=large1915895Let $A$ be a countable subset of the open unit disk $\mathbb D$ centered at 0. For a point $x\in\mathbb D$ denote by $\nu_x$ the associated harmonic measure on the boundary circle $\partial\mathbb D$.

Question: When does there exist a probability measure $\alpha$ on $A$ which satisfies the following

stationarity property: $ \nu_0 = \sum_{x\in A} \alpha(x) \nu_x ? $ If yes, what can one say about the collection of all measures $\alpha$ on $A$ with this property? - other than that it is a convex set :)

This question, of course, can be reformulated in terms of the hyperbolic plane $\mathbf H^2$: then $\nu_x$ is just the visual measure on $\partial \mathbf H^2$ as seen from the point $x$. In the same way one can ask it in numerous other situations when a natural boundary is present as well, for instance, for homogeneous trees, symmetric spaces, buildings, etc. Also, one can assume that $A$ is the orbit of a certain discrete group of isometries of the state space.

I am only aware of two results in this direction. One is a construction due to Furstenberg (later extended by Lyons and Sullivan). According to it, if $X$ is a Riemannian manifold of bounded geometry, and $A$ is an $r$-separated subset of $X$ such that the union of $r$-balls centered at points of $A$ is recurrent for the Brownian motion on $X$, then the Brownian motion can in some sense be approximated at infinity with a certain Markov chain on $X$. In particular, the transition probabilities of this chain satisfy the stationarity condition from my question.

The other result is due to Connell-Muchnik who proved that Gibbs measures on the boundary of a CAT(-1) space can always be stationarized in the above way with respect to the orbits of a discrete cocompact action.

1397749Is there anything known about the purely analytic aspects of the problem, just for the unit disk (aka the hyperbolic plane)?

One reference is: S. A. Amitsur, On the Characteristic Polynomial of a Sum of Matrices, Linear and Mult. Algebra 8 (1980), 177-182. (pp. 469-474 in Selected Papers of S. A. Amitsur, Part 2, AMS 2001.)

230751359931Thanks. That is an interesting way of demonstrating how rare such numbers must be.619370542509551543184522dEvery element of $A_k$ is a permutation of $S_n$.172643217873643818451802351Furthermore, we have $\mathcal{H}^{i}(F'') \simeq \mathcal{H}^{i}(F)$ for all $i>i_0$. By recursion, we know that $F'' = i_*G''$ and $\mathcal{H}^{i_0}(F') = i_* H$ for $H, G'' \in D^b(mE)$. Furthermore, by adjunction, the map $F'' \rightarrow \mathcal{H}^{i_0}(F')$ lifts to a map $G'' \rightarrow H$. Denote by $G'$ the cokernel of this map. Then, the five lemma shows that $F' \simeq i_* G'$ in $D^b(mE)$.1756761611031I study graduate mathematics in kerman university in iran , I work on noncommutative algebra and like division ringsHave you looked at Moriwaki's book from the Translations of Mathematical Monographs series? It has a focus on "birational" aspects (line bundles with appropriate analogues of "positivity), and generalizes to arbitrary arithmetic varieties, but certainly covers the curves/surfaces cases quite well, together with many of the results you mentioned. It doesn't explicitly give the classical algebraic results side-by-side, however.1290752@RobertIsrael: I can see your point, although I would have found it slightly more convincing if the person her/himself modifies the name, rather than other people misspelling or misidentifying it. This of course also happens a lot: there are a lot of Bobs out there whose legal name is Robert.713547398061884844Interesting! In the application of my argument, however, one could replace $\mathbb{Q}$ with the algebraic reals or any countable dense set of reals that can be finitely represented, and this may avoid such obstacles.2079560206505265309118574061145986Given $a,b\in\Bbb N$ of $n$ bits each and $c,d\in\Bbb N$ of $m$ bits each with $\frac{n}\beta<m<n$ with $\beta>1$ is there a better way to decide if $\exists e\in[c,d]\cap\Bbb N$ such that $e|a$ and $e|b$ than factoring $\mathsf{gcd}(a,b)$ that is of complexity $O((m + n)^\alpha)$ for some $\alpha\geq1$?

It is not obvious factorization reduces to this problem and so is there a possibility for a quicker path?

4863181081938:Yes, it is true over curves.Just to make sure I understand. The residue at $s=\frac{3}{2}$ is given by $I_f = \int_{\mathbb{R}_+^2} f(x,y) dx dy$, and at $s=\frac{1}{2}$ by $c_0=0$ since $f(0,0)=0$?Flach's article "Cohomology of topological groups ..." contains a definition of cohomology of a topological group $G$ with values in a sheaf on top. spaces (including the case where the sheaf is represented by a topological G-module) in terms of the topos $BGG.679116lsections of the cotangent bundle of elliptic surfaces1545883OK. Posted to soon earlier. The change from the T tetromino to the S tetromino actually allows going down to five pieces instead of six.@Victor: your guess is correct. The algebraicity of an abstract homomorphism follows from Borel Tits in the isotropic case for arbitrary fields but for global fields, in the anisotropic case, it is still true, and is a consequence of the Margulis super-rigidity (this consequence is deduced in Margulis' book, Chapter 8)159243dOh, good point; I'll make the corresponding edit.The square of the hypotenuse is equal to the sum of the squares of the other two sides.

414437Thank you for this answer! Yet could you give some more detail? Is there any substantial difference between $t$ and $\sum t^{p^{e_i}}$ in the field $\mathbb{F}_p((t))$?Thanks Tim. Actually, I was thinking in $X$ and $Y$ as symplectic varieties, I mean, as two Poisson varieties which are symplectic. In this case if the varieties are isomorphic as symplectic varieties their quantum cohomology rings are isomorphic. My question was really stupid. I apologise for that. Please, close it.

52551Given a commutative ring $A$ we say that a property P is *local* if

$A$ has P $\leftrightarrow$ $A_{p}$ has P for all prime ideals $p$ of $A$

It is usually the case that this requirement is equivalent to $A_{m}$ having P for all maximal ideals $m$ of $A$. I was wondering which (if any) are the strongest/most interesting local properties $P$ of a commutative ring that do not satisfy the second equivalence. Similarly, I would like to know the strongest/most interesting non-local properties P that are true at all localizations at $p$.

That is to say, what are the most interesting properties P of $A$ such that:

(1) $A_{p}$ has P for all prime ideals $p$ of $A$ but P is NOT local

or

(2) P is local BUT it is NOT true that if $A_m$ has P for all maximal ideals $m$ of $A$ then $A$ has P.

**EDIT**: After comments and answers received have edited (and expanded) the question. Hope it is clear and unambiguous now.

The book *A History of Mathematics: An Introduction*
by Victor J. Katz says:

"...probably the most famous mathematical technique coming from China is the technique long known as the Chinese remainder theorem. This result was so named after a description of some congruence problems appeared in one of the first reports in the West on Chinese mathematics, articles by Alexander Wylie published in 1852 in the North China Herald, which were soon translated into both German and French and republished in European journals..." (page 222)

This seems to suggest that the name "Chinese Remainder Theorem" was introduced soon after Wylie's 1852 article.

But the book *Historical Perspectives on East Asian Science,
Technology, and Medicine*, edited by Alan Kam-leung Chan,
Gregory K. Clancey and Hui-Chieh Loy says:

"A. Wylie introduced the solution of Sun Zi's remainder
problem (i.e. "Wu Bu Zhi Shu") and Da-Yan Shu to the
West in 1852, and L. Matthiessen pointed out the identity
of Qin Jiushao's solution with the rule given by C. F. Gauss
in his *Disquisitiones Arithmeticae* in 1874. Since then it
has been called the Chinese Remainder Theorem in Western
books on the history of mathematics."

This is ambiguous, as it is not clear whether the author means that the name "Chinese Remainder Theorem" came into use after 1852 or after 1874. But the phrasing does suggest that the name came into use before 1929.

In 1881, Matthiessen published the following article:

L. Matthiessen. "Le problème des restes dans l'ouvrage chinois Swang-King de Sum-Tzi et dans l'ouvrage Ta Sen Lei Schu de Yihhing." Comptes rendus de l'Académie de Paris, 92 :291-294, 1881.

But does the name "Chinese Remainder Theorem" ("le théorème chinois des restes") appear in this article?

5339461894239217865https://www.gravatar.com/avatar/09fa1ae2d20713a4308f83ed1e837689?s=128&d=identicon&r=PG&f=1The fixed-point theorem says that a group of isometries preserving a weakly compact convex subset of a Banach space must have a fixed point in the convex set.1932313796807612831274828271724268Rough paths can be thought of as taking values in a Lie group embedded in the tensor algebra of $\Bbb R^d$. See page 17/section 2.3. Lie groups represent the continuous symmetries of some object. That is, elements of the Lie group act on some other object in a way that preserves symmetry.

I am curious, since rough paths are elements of a Lie group, what exactly are they acting on? What symmetry do rough paths preserve? What do those actions look like?

6225200355637120490169845816816632290722A question on how polynomials split over $\mathbb{F}_p$ for large primes $p$MU is not an HZ-algebra because not all formal group laws are additive. I think that a similar approach will work for MGL1942254841478Lhttp://www.globesynctechnologies.com/2000082TDoes eigenvalue exist in a Hilbert space?My recollection is that the answer to 2) should be "no" but I will need to either think some more on this, or look it up in Bonsall and Duncan3388170Chapter 4 of Wells's "Differential Analysis on Complex Manifolds" is titled "Elliptic Operator Theory" and is, I think, close to what you want. It certainly explains why elliptic operators have finite-dimensional kernels and cokernels.

11831422946832... In a nutshell, "constructive proof" means a proof that gives an algorithm (which requires only finite time) which takes any $r\in R$ and gives back an $x\in R$ satisfying $rxr=r$; it can (and will have to) use a function (which is considered to be already implemented) that takes an increasing sequence $I_0\supseteq I_1\supseteq I_2\supseteq ...$ of ideals (i. e., an algorithm to construct $I_n$ for every $n$, a proof that it is indeed an ideal, and a proof that $I_n\supseteq I_{n+1}$) and gives you an $n$ such that $I_n=I_{n+1}$ (but no other oracles).275354Hi, my question is:

Is it acceptable to do a PhD in mathematics in a topic such as Number Theory and or Analysis and change subjects in postdoc to something different than these topics such as Geometry/Topology or something else?

I've looked on some pages of researchers in mathematics and most of them just continued their topic of dissertation after their graduation.

Thanks in advance.

When was the fact that a positive integers such that the sum of its decimal digits is divisible by 9 is itself divisible by 9 first recorded? As @AaronMeyerowitz implicitly suggests, this applies here339280265956See chapters II and III of the classic book of Vigneras. Beware however that in the later parts of chapter III she assumes widely that C.E. (the "Eichler condition") is verified, and C.E. is explicitly not verified for quaternion algebras $B$ over $\mathbf{Q}$ such that $B\otimes \mathbf{R} \cong \mathbb{H}$ (so called "definite" algebras). For lots of computation with definite algebras please see Chapter V of her book.

1377087Richard WongTdistribution of non-solvable group orders2109010pRemark 9 in v2 explains the error in the first version.What is significant about version 3.4.1, specifically? The numbering makes it appear to be a minor update to 3.4.31869811967141540116Thank you! Yet I am rather interested in weights defined in terms of eigenvalues of Frobenius actions; see the upd. to my question.33292715665161106451123234x@DenisNardin Edited. Thanks for the constructive criticism!I straightly say ordinary japanese.I say japanese honest opinion.I always fight with discussions on wikipedia.Also I recommend lang-8.com.I am on active in knowledge sites more than 4.

18414941692486527994661526 Anixx: I'm tempted to say that if your other formula gives the same values, then it's really the Fourier-analysis-related formula in disguise, so this is still a question about Fourier analysis. Nonetheless, if what you really want is a discussion about whether integration constants are useful in general, then I suppose it's best to leave the question as it is. Ultimately, my advice is to write your question so that you get answers which are satisfying to you. If you want answers like KConrad's below, you should write it one way; if you want answers like mine, you should write it another way.17402572289199https://www.gravatar.com/avatar/87799db0be2118c3a2b73e3fcf1e1583?s=128&d=identicon&r=PG&f=1branch locus of the discriminant map $\overline{\mathcal{H}}_{g',r} \to \overline{\mathcal{M}}_{g,n}$20400742581631103922697677Hurewicz versus Wadge hierarchy of zero-dimensional Borel sets?@A case of nested central limitsI asked a similar question at http://mathoverflow.net/questions/57766/why-are-there-no-wild-arcs-in-the-plane12600056273616148191009616thanks edgar. and all these definitions of dimension are invariant under isometry/hemeomorphism? https://graph.facebook.com/10155234999384865/picture?type=largedAustralia,Lifestyle Entrepreneur- Author- Speaker10387042948534Direct construction of the Stone-Čech compactification using ultrafilters?263471does the minimum number of bits in $x$ and $y$ always grow double exponential($2^{2^t}$) or faster in $t$? Could there be a smaller solution?11952942082098181938310921162306512003382For many purposes in category theory, we consider limit and colimits of diagrams $F\colon\mathsf{J\to C}$ where $\mathsf{J}$ is small category, that is, a category where the classes of objects and morphisms are actually sets.

However, what if instead of sets and classes we want to adopt the foundational system with Grothendieck universes? Do we even need the smallness condition anymore?

For instance, in the framework of classes (without universes) we say that a category $\mathsf{C}$ is complete (resp., cocomplete) if limit (resp., colimits) of all diagrams indexed by small categories exist. What are the appropriate versions of these notions in the framework of universes?

6587532200796117646438994933915228135154112431971631279Acknowledgment: *Let me acknowledge Theo Buehler from whom I have learnt about the theorem to be presented, however since Theo seems to be absent from MO, let me post it by myself.*

Lemma (Zabreiko, 1969)Let $X$ be a Banach space and let $p\colon X \to [0,\infty)$ be a seminorm. Suppose that for every absolutely convergent series $\sum_{n=1}^\infty x_n$ in $X$ we have $$ p\left(\sum_{n=1}^\infty x_n\right) \leqslant \sum_{n=1}^\infty p(x_n) \in [0,\infty]. $$ Then $p$ is continuous. That is, there is a constant $C\geqslant 0$ such that $p(x)\leqslant C\Vert x\Vert$ for all $x\in X$.

Now, using Zabreiko's lemma you may easily recover the open mapping theorem, Banach's bounded inverse theorem, the uniform boundedness principle, and closed graph theorem. For more details see this fantastic post.

Original references:

П. П. Забрейко, Об одной теореме для полуаддитивных функционалов, Функциональный анализ и его приложения, 3:1 (1969), 86–88.

P. P. Zabreiko, A theorem for semiadditive functionals, Functional analysis and its applications 3 (1), 1969, 70-72). (MathSciNet review.)

If $g(C) \geq 2$ then $\textrm{kod}(S)=1$.

The same holds also if $g(C)=1$ and $f$ is not locally trivial.

See [Barth-Hulek-Peters-Van de Ven, Compact Complex Surfaces], Proposition 12.5 page 215 (Chapter V).

532820I thought of algorithms given in the book, and implemented in "gap", which do work for real and complex Lie algebras.Does $$\|f\|:=\inf \left\{ \int_0^1 g(x) dx : |f|\le g \text{ everywhere}, g \text{ a step function} \right\} $$ work? Then $\|\cdot\|$ is a norm on step functions, Cauchy sequences for $\|\cdot\|$ are Cauchy for $L^1$, and using completeness of $L^1$, pointwise convergence for subsequences of $L^1$ convergent functions, and something like Egorov's theorem, the Riemann integrability of the $L^1$ limit should follow. But I haven't worked out all details :)

164878310768321633974Yes there are many counterexamples. The infinite dihedral group is solvable and so has no nontrivial perfect subgropus, but it has finite abelianization.Sorry I should have mentioned the result in my question. I am editing the question along with the definitions.Let $H=(V,E)$ be a hypergraph. We call it $T_0$ if for all $x\neq y \in V$ there is $e\in E$ with $\{x,y\}\not\subseteq E$ and $\{x,y\}\cap e\neq \emptyset$ (i.e., $e$ contains exactly one of $x,y$).

If $H=(V,E)$ is a $T_0$-hypergraph, it is possible that $|E|<|V|$: Let $V=\mathbb{R}$ and let $E = \{(-\infty, q):q\in\mathbb{Q}\}$.

**Question.** Is there a $T_0$-hypergraph $H=(V,E)$ such that $2^{|E|} < |V|$?

It is known that the distribution of the pair correlations of $\sqrt{n}$ mod 1 is asympotitically Poissonian (that is, "random"). See [1]. Note that having Poissonian pair correlations implies being equidistributed [2], so this is a stronger result than the equidistribution result.

I don't know if a similar result is known for $\sqrt{p}$. Also, I don't know about results for higher correlations or neighbor spacings.

You might also want to check the related paper [3].

[1] El-Baz, Daniel; Marklof, Jens; Vinogradov, Ilya. The two-point correlation function of the fractional parts of \sqrt{n} is Poisson. Proc. Amer. Math. Soc. 143 (2015), no. 7, 2815–2828.

[2] https://arxiv.org/abs/1612.05495

[3] https://www.maths.bris.ac.uk/~majm/bib/nato.pdf

464373The only other technique I know is to use existence of real analytic metrics and work with harmonic functions for these metrics .However all constructions I know of real analytic metrics are by complex analytic methods .7076452687533818202091525Since you mentioned locally presentable categories, I'll give one sufficient condition involving that.

Let $\mathcal{A}$ be a locally *finitely* presentable additive category. Then the canonical morphism $\sum_{i \in I} A_i \to \prod_{i \in I} A_i$ is a monomorphism. This is true when $I$ is finite because $\mathcal{A}$ is additive. Suppose $I$ is (possibly) infinite. Then $\sum_{i \in I} A_i$ is a filtered colimit of all its finite "sub-coproducts". But for any finite subset $I' \subseteq I$, the canonical morphism $\sum_{i \in I'} A_i \cong \prod_{i \in I'} A_i \to \prod_{i \in I} A_i$ is a (split!) monomorphism, and filtered colimits preserve monomorphisms, so the canonical morphism $\sum_{i \in I} A_i \to \prod_{i \in I} A_i$ is indeed a monomorphism.

In fact, all we really needed was Grothendieck's axiom AB5 (in addition to AB3 and AB3*, of course), i.e. that filtered colimits in $\mathcal{A}$ are exact.

VCollapsing the medial axis of a polytope 3487027223171188692I guess you mean weakly 1-convergent rather than weakly 1-summable. The answer is no; consider the summing basis for $c_0$.712020216147710638013842631458662nMechanical Engineer from Huntsville, AL.

The answer is sort of well known. A simplicial set $X$ is the nerve of a groupoid if and only if any $n$-horn has a unique filler for all $n\geq 2$. The horn $\Lambda^k[n]$ is the simplicial subset of $\Delta[n]$ obtained by removing the non-degenerate $n$-simplex of $\Delta[n]$ and it's $k^{\text{th}}$ face. The previous statement means that any map $\Lambda^k[n]\rightarrow X$ extends to $\Delta[n]$ in a unique way, for all $n\geq 2$ and $0\leq k\leq n$. ~~Alternatively, a simplicial set $X$ is the nerve of a groupoid if and only if it is a Kan complex and $\pi_n(X)=0$ for all $n\geq 2$.~~ The nerve of the category with only one object and endomorphism monoid $\mathbb N$ has trivial homotopy groups in dimensions $\geq 2$ but it is not the nerve of a groupoid (it is not a Kan complex).

The article linked by the OP proves that every degree $k$ spherical harmonic is a $k(k+1)$-eigenfunction of the Laplacian on $S^2$. The OP asks: is every eigenfunction of the Laplacian a spherical harmonic?

The answer is yes, though it seems that neither the linked article nor the Wikipedia page on spherical harmonics address this issue.

It is not hard to show that the spaces $H^k$ of degree $k$ spherical harmonics are mutually orthogonal, so it suffices to prove that the orthogonal direct sum $H$ of the $H^k$'s over all $k$ is dense in $L^2(S^2)$. (Proof: if $f \in L^2(S^2)$ is an eigenfunction not in $H$ then the span of $f$ is orthogonal to all $H^k$ which implies that $H$ is not dense.)

To prove that $H$ is dense in $L^2(S^2)$, note that every continuous function on $S^2$ can be expressed as the uniform limit of restrictions to $S^2$ of polynomials by the Stone-Weierstrass theorem. Every polynomial on $S^2$ is the sum of homogeneous polynomials and every homogeneous polynomial is the sum of harmonic polynomials, so since spherical harmonics are just the restriction of harmonic polynomials to $S^2$ it follows that $H$ is dense in $C(S^2)$. But $C(S^2)$ is dense in $L^2(S^2)$, so we're done.

There is probably a shorter and more sophisticated argument using representation theory (the Peter-Weyl theorem), but in the end it would probably be an elaborate reformulation of the argument above.

466553 It's a fair question, and I apologize for not answering sooner. Hard not to give an unduly long response. Even model categories are rarely directly at the heart of calculations, but they can streamline them, and they can set up structure that can lead to them. For example, localizations of spaces at homology theories are constructed model theoretically. Knowing they exist sets up a wealth of things calculated in modern homotopy theory. A not too sophisticated source leading up to that is ``More concise algebraic topology'', by Kate Ponto and myself. It is meant to introduce the general idea.1714517I would like to prove that for any family of balls $\{B(c_i,r_i)\}_i \subset \mathbb{R}^d$ such that $\{c_1, \dots, c_n\} \subset \bigcap_i B(c_i,r_i) $ and $\forall i, r_i \geq 1$, there exists a ball of radius $1-\frac{\theta_d}{2}$ included in the intersection $\bigcap_i B(c_i,r_i)$ where $\theta_d/2 = \frac{1}{2}\sqrt{\frac{2d}{d+1}}$ denotes the ratio between the diameter and the radius of the smallest enclosing ball of a regular simplex.

Intuitively, it seems that the way of making the smallest intersection is to assign all points $c_i$ to the vertices of a regular simplex of diameter $1$ and all $r_i$ to $1$. By doing so, one can check that the ball of radius $1-\frac{\theta_d}{2}$ centered at $x$ the barycenter of $\{c_1,\dots,c_n\}$ is included in the intersection of balls $ \bigcap_i B(c_i,r_i)$. Indeed, in the case of a simplex, the radius of the biggest ball centered at $x$ and included in the intersection of balls is $1-\text{Radius}(\sigma) = 1 - \frac{\theta_d}{2}$ (hence the constant is tight in this case).

I am having difficulties to prove that this case is indeed the worst case. I was just able to prove that the result holds when all balls have the same radius. Does this result seems familiar to someone? I would really appreciate any comment, idea or reference.

ps : The topology tag is here for several reason. One of them is that the biggest radius of the ball included in the intersection corresponds to the weak feature size of the complement of the intersection. Another one is that this result is linked to a collapsibily result.

893151Are $U,V$ orthonormal? If $U,V$ are orthonormal and $UV^T = I_n$ then isn't $U = V$? In that case $A$ is symmetric and $x,y$ originate from the same subspace right? In that case isn't this the basic Rayleigh inequality?2014836dI will verify it tomorrow but this seems to work.165074922654112147127|Ah that makes more sense. Thanks for clarifying the question.what is the status of this in 2015? (in december, mochizuki posted a progress report on the verification of universal teichmüller theory / his alleged proof: http://www.kurims.kyoto-u.ac.jp/~motizuki/IUTeich%20Verification%20Report%202014-12.pdf )The article mention by Martin in a comment (on the question) is very interesting and discusses among others some practical considerations in some detail.484460Thanks! a really enlightening answer, putting the problem into its correct abstract setting! Yet, to complete a few computations I'm currently fighting with, what I'd need is a completely explicit version of the isomorphism $Hom_{Rep(H)}(U,Res_H^G(W)) \stackrel{\sim}{\to} Hom_{Rep(G)}(Ind_H^G(U),W)$, with the explicit model for the induced representation mentioned above. I think I can work this out from the answer, but I'll wait still a couple of days to see if there's someone knowing this on the spot (and willing to post it here :) )8654811131334No. Saying that the point $x$ lies in the Shilov boundary of $\mathcal{M}(A)$ gives strong restrictions on the completed residue field $\mathscr{H}(x)$. In your case, its residue field $\widetilde{\mathscr{H}(x)}$ will be of transcendence $d$ over $\tilde{k}$, where $d$ is the dimension of the variety (see proposition 2.4.4 of Berkovich's book).

To handle the more general situation (with no assumptions on the field $k$), let me denote by $s(x)$ the previous transcendence degree and by $t(x)$ the dimension of the $\mathbb{Q}$-vector space $\sqrt{|\mathscr{H}(x)^\ast|}/\sqrt{|k^\ast|}$. Then you will find the condition $s(x)+t(x) = d$ (what is called an Abhyankar point).

If you want an explicit counter-example, pick any point of type 1 ($s+t=0$) when $d\ge 1$.

For a more interesting question, you could ask whether the result holds when you replace $X$ by the Zariski closure $Y$ of the point associated to $x$ in $X$. This is still not true, but you have to consider subtler things like points of type 4.

Let me also add that if the point is Abhyankar, then your result is true. You may find this in my paper "Les espaces de Berkovich sont angéliques" (Bulletin de la SMF 141 (2013), no. 2, 267–297).

1124204https://graph.facebook.com/10156555548034814/picture?type=large1541507705844115434800GrothendieckMainTheoremOne has a natural exact sequence: $$0 \rightarrow H^1(G_p, \mathbb{Q}_p(2)) \rightarrow H^1(G_p,U) \xrightarrow{s} H^1(G_p,\mathbb{Q}_p(1)) \rightarrow 0,$$ where $ H^1(G_p, \mathbb{Q}_p(2)) \cong H^1_{crys}(G_p,\mathbb{Q}_p(2))$, $\dim H^1(G_p,U)=\dim H^1_{crys}(G_p,U)+1$, $\dim H^1(G_p, \mathbb{Q}_p(1))=\dim H^1_{crys}(G_p, \mathbb{Q}_p(1))+1$. So $s^{-1}\big(H^1_{crys}(G_p, \mathbb{Q}_p(1))\big)=H^1_{crys}(G_p,U)$. In other words, if $U$ is crystalline, then $V$ is crystalline $\Leftrightarrow$ $W$ is crystalline.

7708944457291582725Hhttps://i.stack.imgur.com/s1NTa.jpg Keegan Landreth1299886One of my collegues, Lucien Guillou, told me that he was asked to give lessons in topology, especially knot theory, to psychanalysts of the Lacanian sort. One of the reason for their interest was the Borromean rings which they took for a illustration of the link between body, spirit and soul. Take one out and the remaining two fall apart.For the case with section and $q=0$ see http://www.math.colostate.edu/~miranda/preprints/weierstrassfibrations.pdf

A similar construction should work in the case (with section, $q$ fixed and $p_g$ sufficiently large) you should get a moduli space together with a morphism to $M_g$.

In the case (with section; $q=p_g=1$) then $p_g$ is ``sufficiently large" and you have that the Weierstrass equation of the elliptic surface depends on the base curve $C$, a choice of a line bundle $L$ of degree 1 on $C$ and two sections in $H^0(L^4)$ and $H^0(L^6)$. You easily can get the dimension of the corresponding moduli space from this. Also this moduli space is obviously connected. Moreover, I believe that the period map is locally an isomorphism in this case, hence different Picard numbers occur.

12948902625321415822There are many $ZFC$ results that their proof uses forcing: The idea is that we force the statement to be true, and then using absoluteness (or other reasons) we conclude that the result is true in $ZFC$.

**Question.** Are there any $ZFC$ results such that in their proof we use large cardinals (and then remove the use of that large cardinal by some arguments like absoluteness)?

Is there a known algorithm to compute the (generalized) Voronoi diagram of a set of points, line segments and triangles in $\mathbb{R}^3$? If yes, are there any available implementations?

I know that there are two methods with available code for line segments and points on the plane, described in the following papers:

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.112.8990 (CGAL) https://www.sciencedirect.com/science/article/pii/S0925772101000037 (VRONI)

I am also aware of a method that computes voxelizations for a closely related 3D problem, where the sites/objects are surfaces: http://sci.utah.edu/~jedwards/research/gvd/

But I'm not being able to find an algoithm for the case when the sites are vertices, line segments and triangles in space. The output should be the Voronoi diagram vertex coordinates and the (possibly curved) bisector descriptions.

2348540I had something like this in mind but couldn't make it work. Very nice answer!1416351486254854496594553@Mariano: That's obviously an understatement, but I wanted to emphasize that it isn't enough (for rationality questions) to limit everything to finite dimensional vector spaces `$V$`. At the same time, the full linear dual of an infinite dimensional space doesn't seem to come up in this context. Though it needs to be checked that Borel's later treatment doesn't rely on the overstatement in AG.11. https://www.gravatar.com/avatar/088fa9a14c66e1c352d6528372f8b26a?s=128&d=identicon&r=PG&f=11254941VRolling/width functions: Characterization?1385617526320dAs requested, here's a little more detail in the form of an answer.

**Disclaimer**: Let me start by saying that it'll be easier for me to talk about *stable* homotopy groups instead of homotopy groups, just because that's all I know about. By that I mean, I'm not gonna define, say, $\pi_{1/2}S^3$ but rather $\pi_{N+1/2}S^{3+N}$ when $N$ is really big. Throughout I will be lazy and write $\pi_kX$ for what should really be the stable value of $\pi_{k+N}\Sigma^NX$. Feel free to ignore this. Someone who knows more about this than I do can come around and fill in how to translate this back into something about ordinary homotopy groups. Also, unless I say otherwise, **my prime is odd**. (Contrary to normal practice here at Northwestern.)

As usual, it starts with $K$ theory. Except I said we're gonna $p$-adically interpolate, so we actually want $p$-complete $K$ theory, $K_p$. The nice thing about $p$-adic $K$-theory is that, for any number $m$ coprime to $p$, there's a natural transformation of (multiplicative) cohomology theories, called the Adams operations, $\psi^m: K_p \rightarrow K_p$. This is supposed to act like a linearization of taking exterior powers of vector bundles, and here's all you need to know about it:

- If $v \in K(S^2)$ denotes the Bott periodicity element, then $\psi^m(v) = mv$
- $\psi^m$ induces a graded ring endomorphism of $K^*(pt)$

Now, for various reasons, coming from manifold theory (if you're Sullivan) or just trying to understand some elements in the stable homotopy groups of spheres (if you're Adams), people got interested in understanding the cohomology theory you get by taking the 'fixed points' of $\psi^m$. That is, they wanted to understand the cohomology theory $J$ that has a natural transformation $J \rightarrow K_p$ and such that, for any space $X$, you get a long exact sequence

$$J^n(X)\rightarrow K^n_p(X) \stackrel{\psi^m -1}{\longrightarrow} K^n_p(X) \rightarrow J^{n+1}(X) \rightarrow \cdots$$

When you choose $m$ correctly (I'll say how in a second), this cohomology theory deserves the name $S_{K(1)}$ which stands for '$K(1)$-local sphere' because it sees a very special sector of the rainbow that makes up the sphere (or more accurately, the 'stable' phenomena of the sphere).

It is this object which we will now raise to the "1/2 th" power, and maps out of this will constitute elements of $\pi_{1/2}$. The key is to notice the following:

- $K^{2n}_p(pt)= \mathbb{Z}_p$
- The map $x \mapsto (m^{n}x - x)$ on the $p$-adics makes sense for (invertible) $p$-adic values of $m$.

Now I can tell you how to choose $m$ to get the $K(1)$-local sphere: it has to be a topological generator of the $p$-adic units. (Those following along at home just saw why my prime was odd.) If we think about Bott periodicity and stare at the formula a bit, it becomes clear that the $\lambda$th suspension, for $\lambda \in \mathbb{Z}_p$, should be obtained by taking the fiber of the map which, on homotopy groups, does $x \mapsto (m^nx - \lambda x)$. [Okay, it takes a little fussing to see this is the right thing to do- it helps to interpret $\lambda$ as a unit which is 1 mod $p$].

This fiber behaves like the $\lambda$th suspension of the $K(1)$-local sphere, and with it we may probe $K(1)$-local spaces and spectra $p$-adically. When $p=2$ there's a little more stuff you can probe with, but you get $2$-adic things as well.

All the stuff above is spelled out in Hopkins-Mahowald-Sadofsky's paper here. That's where they launched the program of computing Picard groups. As I said before, the *Picard group* of a symmetric monoidal category is the set of (isomorphism classes of) $\otimes$-invertible objects. These are good for indexing things like homotopy groups in all sorts of settings.

The most mysterious example, however, is the $K(n)$-local category (this is the thing that sees the $n$th layer of the rainbow of the sphere, we met $n=1$ earlier). We are still in the very early stages of understanding the Picard group of this category, but it's very desirable to do so. For one, there are mysterious duality phenomena in the $K(n)$-local category, analogous to the Poincaré duality example mentioned in the equivariant setting, which require exotic suspensions (see here). On the other hand, it can be useful to index the homotopy groups of something as a function of a $p$-adic number, say. Who knows- maybe the only way we'll ever get an answer to 'what are the ranks of the homotopy groups of spheres?' is to look for solutions like 'the special values of some horribly incomputable arithmetic-p-adic-y-L-type function'.

But that's speculation. For some recent calculations, see some of the papers here by Paul Goerss and his collaborators (Henn, Mahowald, Rezk).

2292039516315355361I am a little confused why you are asking this question, when it seems that you have already arrived at the correct conclusion. But, I will take it at face value. The category of primitive recursive functions you describe does not have pullbacks. Indeed, take your example of two different constant functions with the same codomain and just apply the definition of pullback. There is no way to complete the diagram to a square at all, let alone a universal way.

Let $G$ be a semisimple algebraic group over an algebraically closed field $k$ of characteristic 0 and let $B \subseteq G$ be a Borel subgroup with unipotent radical $U$. Let $P \supseteq B$ be a parabolic subgroup with unipotent radical $U_P \subseteq U$. Set $\mathfrak n := \textrm{Lie}(U)$ and $\mathfrak n_P := \textrm{Lie}(U_P)$. Then we have the enveloping algebras $U(\mathfrak n)$ and $U(\mathfrak n_P)$ associated to $\mathfrak n$ and $\mathfrak n_P$.

Let $\alpha_1, \ldots, \alpha_\ell$ denote the simple roots of $G$. It is well-known that $U(\mathfrak n)$ is generated as a $k$-algebra by generators $E_{\alpha_i}^{(n)}$ for $1 \leq i \leq \ell$ and $n \geq 0$. (Here we are using the divided-power notation $E_{\alpha_i}^{(n)} = \frac 1 {n!} E_{\alpha_i}^n$). Furthermore, the relations between these generators are given by the Serre relations $$ \sum_{a+b = -\alpha_j(\alpha_i^\vee)+1} (-1)^a E_{\alpha_i}^{(a)} E_{\alpha_j} E_{\alpha_i}^{(b)} = 0 . $$ Succinctly, using the adjoint action $*$ of $U(\mathfrak n)$ on itself, we can write the Serre relations as $$E_{\alpha_i}^{(c)}*E_{\alpha_j} = 0 \textrm{ for } c > -\alpha_j(\alpha_i^\vee) .$$

Now to my question: Is there a nice presentation of $U(\mathfrak n_P)$ by generators and relations given in a similar fashion?

10438321487347Why shouldn't the primordial example of a 2-topos be $\mathbf{Topos}$ (the 2-category of toposes, logical morphisms, and natural transformations)?You can find these details in Jech's "Set Theory" book in the chapter introducing forcing (Ch. 14).2095222VWriting down minimal Weierstrass equations183851353779Thank you for your answer and comment. This is a good starting point for me to study algebraic K-theory.1412354@Benjamin I'm not sure I follow... For a complete path algebra, the completion is always with respect to the radical, so this is respected by isomorphisms anyway. Is that what you meant?It also depends on what you mean by "moduli space". What is it you want to do with this concept in mathematical arguments?A direct construction of the H-space map is also given as thm 10.5.7 in Aguilar-Gitler-Prieto (algebraic topology from a homotopical viewpoint)117266821761041243349@Let me first note that there is a slight ambiguity when one says "ramified only at 2". Strictly speaking, that means that the extension is unramified at every place of $\mathbb Q$ except 2, including infinity. The latter mean that the extension is totally real. Often, however, "ramified only at 2" means "ramified only possibly at 2 and $\infty$", and it is probably what you mean. Here, to remove ambiguity, for $S$ a finite set of places of $\mathbb Q$, I will use "ramified only at $S$" in the strict sense.

That being said, the short answer is that whatever the finite set $S$, there are strong restrictions on the finite groups $G$ that can appear as the Galois group of an extension of $\mathbb Q$ unramified outside at $S$, but that in general, to "describe" all these restrictions (we may mean different things by that) is in general an open problem. To see what kind of restrictions can appear, let us consider several situation, from the very particular to the general.

If $S=\emptyset$ or $S=\{\infty\}$, then Minkowski's theorem tell that there is no nontrivial extension unramified outside $S$, so the only possible Galois group $G$ is the trivial one. A very strong restriction indeed.

If $S$ is any set, but you try to determine what **abelian** $G$ may appear, then the answer is given by class field theory. Precisely, if $S=\{\ell_1,\dots,\ell_k,\infty\}$, then the abelian group which appear are the ones that are quotient of $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$, and it is obvious that many abelian groups are not of this type (e.g. $\mathbb Z/\ell \mathbb Z$ for $\ell > \ell_1,\dots,\ell_k$). If $S=\{\ell_1,\dots,\ell_k\}$, replace $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$ by its quotient by the diagonal subgroup $\{1,-1\}$.

If $S$ is any set, and you're interested in groups $G$ that are $p$-groups (for a $p$ that may or may nor be in $S$), then again, class field theory can help you because of Frattini's argument saying that a $p$-group is generated by any set that maps subjectively to the maximal abelian $p$-torsion quotient of $G$. So by the above, the $p$-groups $G$ appearing this way have a system of generators with less elements than the dimension over $\mathbb F_p$ of the maximal $p$-torsion quotient of $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$. Is that all? No. We can describe in some cases all the $p$-group $G$ appearing, but not in all. For instance, with $S=\{2,\infty\}$, and $p=2$, the groups $G$ appearing are exactly the $2$-groups having a system of two generators, one of them being of square 1. For $S=\{2\}$, the only $2$-groups appearing are the cyclic groups. In general, the case $p \in S$ is better understood than the case $p \not \in S$. In the latter case, it is a folklore conjecture, on which not much is known, that only finitely many $p$-groups $G$ appearing (one can check readily that the conjecture is true for abelian $p$-groups, by the above paragraph).

If now we consider general finite groups $G$, well, the above shows that there are restrictions, but determining them all is largely open. For instance, a conjecture of Shafarevich states that there is an $n=n(S)$ such that every group $G$ which is the Galois group of an extension unramified outside $S$ has a system of generators with less than $n$ elements. But this is open in every non-trivial case.

Let me also mention a simple but too little known result of Serre, even if it is not strictly speaking part of your question: [Edit: Sorry, I messed up the statement of the result. Here is the right version]. If every finite group is a Galois group over $\mathbb Q$ (if the inverse galois problem has a positive solution) then every finite group is the Galois group of an extension of $\mathbb Q$ unramified at infinity -- that is, a real extension. In other words, according to the inverse Galois problem, there should need no restriction of $G$ in the case where $S$ is the set containing all finite places, but not the place at infinity.

546376Alladi and Vaaler are among the very few people who have Erdős number 1 as well as Erdös number 1.703486575691702143 After not thinking about this problem for awhile, I suddenly realized out of nowhere that my earlier statements about elements of order 2 are not always valid; they hold if the group is perfect (which is actually the mental picture I had in mind when I wrote them). I realized the assumption of perfectness was necessary when I realized that there are counterexamples to my assertion about having a unique conjugacy class of elements of order 2. Specifically, I realized that any nonabelian 2-group with exponent 4 and all irreducible characters of degree 1 or 2 must satisfy your condition.554783The proof of a more general fact can be found in Corollary 5.35 of the survey paper [S.P.Gulʹko, Semilattice of retractions and the properties of continuous function spaces of partial maps. Recent progress in function spaces, 93--155, Quad. Mat., 3, Dept. Math., Seconda Univ. Napoli, Caserta, 1998.]2497741158170@BACKGROUND: Over an algebraically closed field of arbitrary characteristic, most of the basic structure theory of affine (= linear) algebraic groups can be developed concretely without quoting difficult theorems from algebraic geometry, though the study of quotients $G/H$ gets more subtle and can benefit from the scheme setting. (This is even more so when one works over arbitrary fields.)

Similarly, basic finite dimensional (rational) representations of a connected semisimple group can be developed up to a point in down-to-earth ways, including the classification of simple modules by highest weights (though complete reducibility breaks down badly in prime characteristic). But already in characteristic 0 it is useful to work geometrically with the flag variety $G/B$ for a fixed Borel subgroup $B$. Here the classical theorems of Borel-Weil and Bott realize the sheaf cohomology groups $H^i(G/B, \mathcal{L}(\lambda))$ of a line bundle as simple modules or zero. The index $i$ ranges from 0 to $\dim G/B = $ number of positive roots.

More precisely, let $B$ correspond to the *negative* roots of $G$ relative to a maximal torus $T \subset B$. The characters $\lambda$ of $T$ determine homogeneous line bundles on the projective variety $G/B$. For $\lambda$ not in a root hyperplane after a shift by the half-sum of positive roots $\rho$, exactly one sheaf cohomology group $H^i$ is nonzero; it affords the simple module of dominant highest weight $w(\lambda + \rho)-\rho$ with $w$ of length *i* in the Weyl group $W=N_G(T)/T$.

In prime characteristic Kempf (1976) proved that higher cohomology all vanishes when $\lambda$ is dominant. But otherwise there can be multiple nonvanishing cohomology groups. (There are partial results by H.H. Andersen and others, though no definitive treatment; the vanishing patterns are conjectured to depend on Kazhdan-Lusztig theory for the dual affine Weyl group.) In any case, the nonzero cohomology groups seldom give simple modules; instead there are parallels with infinite dimensional Verma modules.

Even in this complicated setting, one can invoke the principle from algebraic geometry quoted in the header: *Euler characteristic is invariant under base change.* This requires suitable realization of the groups, weights, flag variety over $\mathbb{Z}$, using ideas which originated with Chevalley but have only recently been completed by Lusztig (*J. Amer. Math. Soc.*, 2008). The invariance principle should be applied to the individual weight spaces for each fixed weight, so that the alternating sum of dimensions of the weight spaces in the various cohomology groups recovers (up to sign) the classical weight multiplicity. As a result, $$\sum_i (-1)^i \dim H^i(G/B, \mathcal{L}(\lambda))$$ and the associated formal character of the (virtual) $G$-module are given (up to sign) by Weyl's formulas. Approaching this intrinsically in characteristic $p$ seems impossible.

Jantzen's book *Representations of Algebraic Groups* explains Kempf's theorem and its uses. But further treatment of higher cohomology groups is found mainly in research papers, where the invariance principle is quoted as "standard". It apparently originated in the work of Grothendieck and his school, only a small part of which should be needed as a prerequisite for the algebraic group application.

What source for the invariance of Euler characteristic under base change involves the fewest prerequisites for the limited application in modular representation theory?

ADDED: Concerning sources, it's just been pointed out to me that the last section of Chapter III in Hartshorne's *Algebraic Geometry* incorporates Mumford's proof of the Grothendieck theorem in question. So this is probably the canonical reference. I confess I haven't become familiar enough with this book, but in any case I still want to understand better how to minimize prerequisites for understanding the special application outlined above.

I'm a very very new user of Linux and Ubuntu. And the good thing is that I have started liking this. This is awesome especially for everything.

815580What is $E$? Of course $\det Rp_*(L\oplus E) = \det (Rp_*L \oplus Rp_*E) = \det Rp_*L \otimes \det Rp_*E$, so the formula that you like to have looks strange.60630120592401721152Yes, that Borcherds congruence would do it for rank $r=24m$ (and also say something about self-dual ternary codes that might already be nontrivial itself). But there'll still be the question of divisibility by 24 for rank $24m+12$. On further thought I can get a factor of 6, not just 2, but don't see how to go beyond that by elementary means.@MikeBenfield You are right. It is (if one can say) nonsmooth analysis on Carnot group. And we assume that functions could have only first derivatives (of derivatives in weak sense).1713749Let $f$ be a $C^2$ function defined on a neighborhood of $0$ in $\mathbb R^n$ such that $f(0)=0, df(0)\not=0$. By the Implicit Function Theorem, it is easy to get (after a rotation) that near 0 $$ f(x)=\bigl(x_n-\alpha(x')\bigr) e(x),\quad\text{$\alpha\in C^2$ near the origin in $\mathbb R^{n-1}$, $e\in C^1$.} $$ Moreover I claim that for $T=(t',t_n)\in \mathbb R^{n-1}\times \mathbb R$ $$ f''(x',\alpha(x'))T^2=-e(x', \alpha(x'))\alpha''(x')(t')^2 +2 (e'(x', \alpha(x'))\cdot T)(t_n-\alpha'(x')\cdot t'), $$ where $g''$ stands for the symmetric matrix of second derivatives. It would be a simple consequence of Leibniz formula if $e$ were $C^2$, but $e$ is no better than $C^1$. I guess a "direct" proof would do.

2170077547829229977782912@François Brunault：I use SAGE to calculate the integer points on $517c1$ and found no integer points on this curve. But my question is about a curve having many integer points.15782181041212In fact the root we want is the one which is about -10^{87} and within 10^{-6} of an integer. The other roots are "random" complex numbers (not near integers, or even near the reals in general) of size about 10^5.82360411879294Hhttps://alejandrohtadinom.github.ioThis page has a few of the equivalent definitions of incompressible surface: https://en.wikipedia.org/wiki/Incompressible_surface The one using the language of fundamental groups is the one you need for this. With that definition it's just the lifting property of a covering space that gives you the result.247486928469Obstruction to lifting of global sections of invertible sheavesjBy $[0;1]$ you mean $\left\lbrace0,1\right \rbrace$?311788192150922956631830212910015829501879161zRepresentability of the Weil restriction reference and proof1484682Regularity of solutions to $u' + Au = f$ for nonlinear monotone operator $A$507070 Sage21852422270961106854591414010Wasn't there another meta thread where we kept track of people popping in to answer questions about their work?|https://graph.facebook.com/101874644014961/picture?type=large44313322884291311449Martin, your question is fine, but the distinction between "algebraic" and "geometric" is a little subjective isn't it?1619911In addition, if you assume that polynomial root finding as a black box, you can write a rather short program to compute the diagonalization.496093TI use Maple software. What are you using?1743485dRequest for errata for Automorphic Forms on GL(2)The number of solutions is bounded from below by $cn^2$ (and from above too by obvious reasons). Indeed, let's choose odd $x$ and $y$ at random and independently from the segment $[1,2N]$ for $N$ about $n/100$ and put $z=n-x-y$. For any $d$, the probability that both $x$, $y$ are divisible by $d$ is at most $N^{-2}(N/d+2)^2=d^{-2}+4/Nd+4N^{-2}$. Analogously for $x$ and $z$. Summation over all $d=3,5,7,\dots,n$ gives the upper bound for the probability of at least one event of the form $2(1/9+1/25+\dots)+o(1)$. This is less then $0.8$, so with the probability at least $0.2$ both pairs $(x,y)$ and $(x,z)$ are coprime.

nCommon lower bounds for positive semidefinite matrices22787936785811319785https://www.gravatar.com/avatar/ab2e43b22cd13b2fbd7c14a9af7068a2?s=128&d=identicon&r=PG&f=1753158For every morphism $F\colon T\to U$ of $\mathrm{Vec}(C)$, consider the linear map $\hat F$ acting on the vector space generated by all morphisms of $C$ by $\hat F(g) = F(\mathrm{cod}(g))g$ if $\mathrm{cod}(g) \in T$ and $\hat F(g) = g$ otherwise. Then, one has $\widehat{FG} = \hat F\hat G$ (usual composition of linear maps) and associativity follows since $\hat F$ determines $F$ uniquely.@Scott: You are right! Added a corresponding image. @Ryan: Remaining inside $R$ eliminates many options. @Will: Thanks for the relevant link! Their focus is aesthetics, but still there are many ideas in there I might use. 1474844335906A comment on another question (linked below) states

"The group $PSL_2((\mathbb{Z}/p^n))$ is the automorphisms group of the $(p+1)$ regular tree of depth $n$, where at level $m$ of the tree you have the points of $\mathbb{P}(\mathbb{Z}/p^m)$."

I was unable to find a reference stating this. Is it true, and if so what is the correct reference?

Comment: Presentations of PSL(2, Z/p^n)

1867111760435https://www.gravatar.com/avatar/c78a66c1baa398549ac8f1e524ff363d?s=128&d=identicon&r=PG&f=1\https://i.stack.imgur.com/p9Qzu.jpg?s=128&g=1179985211014822073833368029297464134062580175931090296225343567531262791Indeed, the general theory is described best (imho) in the language of algebraic groups, so at some point you might want read a text dealing with these. However, maybe the first thing to do would be to work through a single illustrative example which you understand. I would start with $\text{SO}(3,1)$.1650851Let $X$ be a smooth algebraic curve over $\mathbb C$, and let $F$ be a vector bundle on it of degree $1$, take the dual of an extention $$0\rightarrow F^*\rightarrow E\rightarrow F\rightarrow0$$ is again an extension of $F$ by $F^*$, my questions are:

How can I explicitly describe this involution on vector space $Ext^1(F,F^*)$?

Are there any criterian for $E$ to be stable? ($F$ is taken to be semistable)

Thanks

2037978@GHfromMO: I think this is a combinatorics question. Cardinal number operations are defined combinatorially, so the relevant question is: *Is there a combinatorial interpretation of tetration*?14851986913482124676See my answer for an efficient conversion to CNF using auxiliary variables.I'm a first-year grad student at MIT working with Clark Barwick. I'm interested in algebraic *K*-theory, higher category theory, and homotopy theory.

A big fan of revisionist history.

Well what I was looking for is:

$d=\frac{ln(\frac{N(k-2)+2}{k})}{ln(k-1)}$

the Link helped, thx.

Nigar1481643178998This nice post of Terry Tao might help http://terrytao.wordpress.com/2007/08/23/random-matrices-the-circular-law/ 1445736hApplied math student at York University in Toronto.207797504418887936It seems I was a bit off on my initial go at showing $(.2)_{ccc}$ entails the failure of $\mathsf{CH}$. Here is the correct argument.

**Proposition 1**: Assuming $(.2)_{ccc}$.

- If $K\subset [\omega_1]^2$ is uncountable and c.c.c. then the graph $L = [\omega_1]^2 \backslash K$ is not powerfully c.c.c.

(in particular $\mathsf{CH}$ fails, see Galvin1977; also, if I understand correctly, this statement is an odd mix of some of the c.c.c. partition relations listed here LarTodo2001, p. 87.)

**Proof**: Consider the statement $\varphi(a) := (\exists H)(\vert H \vert > \omega$, and $[H]^2 \subset a)$, then for each uncountable c.c.c. graph $K$, we have $\diamondsuit \square \varphi(K)$ (as witnessed by the partial order $\mathbb{P}_{K}$ of finite $K$-cliques, which is assumed to be c.c.c.) Applying $(.2)_{ccc}$ we must conclude $\square \diamondsuit \varphi(K)$.

To establish the result, note that if the finite support product of countably many copies of $\mathbb{P}_L$ is c.c.c., then in the corresponding extension, $\omega_1$ is covered by countably many $L$-homogeneous sets. It follows that in no further $\omega_1$-preserving forcing extension, can the statement $\varphi(K)$ hold. $\square$

**Remark:** the use of "not powerfully c.c.c." in the above was a bit of a cop-out since I spent far to long treading water on simply "not c.c.c."; however it does seem to be important.

It turns out, under $(.2)_{ccc}$, there is a connection between being powerfully c.c.c. and productively c.c.c.; in particular

**Proposition 2:** Assuming $(.2)_{ccc}$:

- If $\mathbb{P}$ is powerfully c.c.c. (that is, each finite power of $\mathbb{P}$ is c.c.c.), then $\mathbb{P}$ is productively c.c.c. (i.e. the product of $\mathbb{P}$ with any other c.c.c. forcing, is again c.c.c.)

**Proof:** Given an order relation $a \subset \omega^{V}_1 \times \omega^{V}_1$ defined on $\omega^{V}_1$, let $\varphi(a)$ be the statement

$\varphi(a):= \left(\exists \mathcal{C} \subset \mathcal{P}(\omega^V_1)\right)(| \mathcal{C} | \le \omega$, each $C \in \mathcal{C}$ is centered with respect to $\le_a$, and $\omega^V_1 \subset \cup\mathcal{C})$

(it should be clear that $\varphi(a)$ holds, iff, the partial order $(\omega^V_1, \le_a)$ is $\sigma$-centered.)

Now, for each powerfully c.c.c. order relation $R \subset \omega^V_1 \times \omega^V_1$, the statement $\diamondsuit \square \varphi(R)$ holds (since by assumption, the finite support product of countably many copies of the partial order $(\omega^V_1, \le_R)$ is c.c.c. and in the extension generated by this product, the $\Sigma_1$ statement $\varphi(R)$ holds.) As such, from $(.2)_{ccc}$, we can conclude $\square \diamondsuit \varphi(R)$.

To establish the result, fix some c.c.c. partial order $\mathbb{P}$. Applying our interpretation of the modal operators $\square$ and $\diamondsuit$, we are guaranteed the existence of a $\mathbb{P}$-name $\dot{\mathbb{Q}}$, such that, the iteration $\mathbb{R}=\mathbb{P}\ast \dot{\mathbb{Q}}$ is c.c.c. and forces $\varphi(R)$.

Therefore, fixing an $\mathbb{R}$-name $\dot{\mathcal{C}}$ witnessing $\varphi(R)$ and arbitrary $\mathbb{P}$-name $\dot{A}$ forced by $1_\mathbb{P}$ to be an $R$ anti-chain, *the following hold*:

$\Vdash_\mathbb{R} (\forall C \in\dot{\mathcal{C}})( |C \cap \dot{A}| < \omega)$

$\Vdash_\mathbb{R} |\mathcal{\dot{C}}| < \dot{\omega}_1=\omega^V_1$, and

$\Vdash_\mathbb{R} \dot{A} \subset \bigcup \mathcal{\dot{C}}$

and combine to imply $\Vdash_\mathbb{P} | \dot{A} | < \dot{\omega}_1 = \omega^V_1$. Hence $\Vdash_{\mathbb{P}} (\omega^V_1, \le_\check{R}) \text{ is c.c.c. }$ and the result follows. $\square$

**Remark**: It's worth pointing out that under $\sf CH$, powerfully c.c.c. does not imply productively c.c.c. ( Galvin1977 )

The following dichotomy is an interesting corollary to Proposition 2.

**Product Dichotomy**: Assuming $(.2)_{ccc}$:
For every partial order $\mathbb{P}$: *Exactly one* of the following holds,

- The product of $\mathbb{P}$ with any c.c.c. partial order is c.c.c.

*Or,*

- Some finite power of $\mathbb{P}$ is not c.c.c.

**Proof:** Without loss of generality assume $\mathbb{P}$ is c.c.c. and apply Proposition 2.

In light of the previous corollary, the following seems like a natural question.

**Question:** Does $(.2)_{ccc}$ imply every c.c.c partial order is powerfully c.c.c.?

As pointed out by @JoelDavidHamkins, it was already enough to just consider the square of c.c.c. partial orders.

13385161755016735026887564Thanks, I had to add the properness. I do not quite understand your notation: what is $T_x(f)$? Also probably I should add that $X$ is not assumed to be smooth.Is $2D$ two or even? As far as I remember, $\operatorname{Spin}^{\mathbb{C}}$-structures are the integral lifts of $w_2$. If $2D=2$ and, hence, $w_2=0$, there is a distinguished lift, do the set is **naturally** isomorphic to $\mathbb{Z}$.2255129OK...I guess my main concern is that if the space of all extensions is an abelian group..65348422826641074981834560700501130985119979611644450593109911426911688867h@AlexandreEremenko I agree, but I still like it. :)731328226924JThis question has a solution presented in this paper even if with the jargon and notation of theoretical physics. So, I will use a somewhat different notation and I will change

$${\bf A}(t)\rightarrow -i{\bf A}(t).$$

Then, I will compute the eigenvalues and eigenvectors of ${\bf A}(t)$ through

$${A}(t)|n;t\rangle=\lambda_n(t)|n;t\rangle.$$

Now, you get a series with a leading order term

$${\bf B}(r)=\sum_n e^{i\gamma_n}e^{-ir\int_{-1}^1 dt\lambda_n(t)}|n;1\rangle\langle n;-1| \qquad r\rightarrow\infty$$

being $\gamma_n=\int_{-1}^1dt\langle n;t|i\partial_t|n;t\rangle$ known as *geometric phase*. Then, an expansion in the inverse of $r$ can be obtained with the matrix

$$\tilde {\bf A}(t)=-\sum_{n,m,n\ne m}e^{i(\gamma_n(t)-\gamma_m(t))}e^{-ir\int_{t_0}^tdt[\lambda_m(t)-\lambda_n(t)]}\langle m;t|i\partial_t|n;t\rangle|m;t_0\rangle\langle m;t_0|$$

being in this case

$$\tilde {\bf B}(r)=\prod_{-1}^1e^{-i\tilde {\bf A}(t)dt}$$

so that

$$B(r)=\sum_n e^{i\gamma_n}e^{-ir\int_{-1}^1 dt\lambda_n(t)}|n;1\rangle\langle n;-1|\tilde {\bf B}(r).$$

This represents a solution of the Schroedinger equation

$$-ir{\bf A}(t)B(r;t,t_0)=\partial_tB(r;t,t_0)$$

in the interval $t\in [-1,1]$ and $r\rightarrow\infty$.

**An example**:

$$ A(t) = \frac{1}{1+t^2} \begin{pmatrix} 2 & t\\ -t & -2 \end{pmatrix} $$

and one has to solve the problem $$ \dot U(t)=rA(t)U(t) $$ with $r\gg 1$. We want to apply the technique outlined above. We note that $A(t)$ is not Hermitian and so, solving the eigenvalue problem, we get $\lambda_{\pm}=\pm r\frac{\sqrt{4-t^2}}{1+t^2}$ and $$ v_+=\frac{1}{2}\begin{pmatrix} \sqrt{2+\sqrt{4-t^2}}\\ -\frac{t}{\sqrt{2+\sqrt{4-t^2}}}\end{pmatrix} \qquad v_-=\frac{1}{2}\begin{pmatrix}-\frac{t}{\sqrt{2+\sqrt{4-t^2}}} \\ \sqrt{2+\sqrt{4-t^2}}\end{pmatrix}. $$ But $v_+^Tv_-\ne 0$ and so these vectors are not orthogonal. We need to solve also the eigenvalue problem $u^T(A-\lambda I)=0$ producing the following eigenvectors $$ u_+=\frac{1}{2}\begin{pmatrix} \sqrt{2+\sqrt{4-t^2}}\\ \frac{t}{\sqrt{2+\sqrt{4-t^2}}}\end{pmatrix} \qquad u_-=\frac{1}{2}\begin{pmatrix} \frac{t}{\sqrt{2+\sqrt{4-t^2}}} \\ \sqrt{2+\sqrt{4-t^2}}\end{pmatrix}. $$ It is easy to see that $u_+^Tv_-=u_-^Tv_+=0$. It is important to note that $\lambda(t)=\lambda(-t)$ and $u_+(-t)=v_-(t)$ and $u_-(-t)=v_+(t)$ and so, these eigenvectors are just representing a backward evolution in time. Now, we want to study the time evolution of a generic eigenvector $$ \phi(t)=\begin{pmatrix}\phi_+(t) \\ \phi_-(t)\end{pmatrix} $$ and this can be done by putting $$ \phi(t)=c_+(t)e^{r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}v_+(t)+ c_-(t)e^{-r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}v_-(t) $$ that will produce the set of equations $$ \dot c_+=\gamma_+c_++e^{-2r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}\frac{u_+^T\frac{dv_-}{dt}}{u_+^Tv_+}c_- $$

$$ \dot c_-=\gamma_-c_-+e^{2r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}\frac{u_-^T\frac{dv_+}{dt}}{u_-^Tv_-}c_+ $$ having set $\gamma_+=\frac{u_+^T\frac{dv_+}{dt}}{u_+^Tv_+}$ and $\gamma_-=\frac{u_-^T\frac{dv_-}{dt}}{u_-^Tv_-}$. These equations are interesting because they provide the way time evolution is formed in a non-hermitian case. But this is also saying to us that each component may evolve in time differently: One can be really smaller than the other for $r\gg 1$. But we can also understand the form of the higher order corrections:

$$ c_+(t)=c_+(0)+\int_0^tdt'e^{\int_0^{t'}dt''(\gamma_+(t'')-\gamma_-(t''))}e^{-2r\int_0^{t'}dt''\frac{\sqrt{4-t^{''2}}}{1+t^{''2}}}\frac{u_+^T\frac{dv_-}{dt''}}{u_+^Tv_+}c_-(0)+\ldots. $$

Using a saddle point technique, we can uncover here that the correction is exponentially small and cannot be stated that is something like $e^{r}/r^k$ in the general case.

Now, we consider the simple case $c_+(0)=1$ and $c_-(0)=0$. The approximate solution will be

$$ \phi_+(t)=\frac{1}{2}\sqrt{2+\sqrt{4-t^2}}e^{r\int_0^{t}dt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}} \qquad \phi_-(t)=-\frac{1}{2}\frac{t}{\sqrt{2+\sqrt{4-t^2}}}e^{r\int_0^{t}dt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}} $$

and solving numerically the set of differential equations for $r=50$ we get the following

The agreement is strikingly good.

I always found it strange that, in the context of invariant and representation theory, averaging over the group is called the "Reynolds operator". As far as I know the work of Reynolds was in fluid mechanics. He introduced the idea of local averaging in order to distinguish slow and fast variables. As such it is a precursor of things such as the Lyapunov-Schmidt method and Wilson's renormalization group. How did this terminology end up in invariant theory?

112808Dear Dmitry, the (*analytic !*) isomorphism is proved by pulling back the projective $\mathbb P^1$ bundle $L\to E$ to the *analytic* covering space $\mathbb C \to E$ where it becomes trivial.Then you see that the pull back of $U$ is just $\mathbb C \times \mathbb C$ and $U$ is the quotient of that $\mathbb C \times \mathbb C$ by a suitable action of $\mathbb Z \times \mathbb Z$. The quotient is $\mathbb G_m \times \mathbb G_m$ and this is why $U$ is isomorphic to $\mathbb G_m \times \mathbb G_m$2776731182752right different ways to translate the same thing into English Find a generalized hypergeometric-based function yielding certain ratios of fifth-degree polynomials1530993To avoid choice in the ambient universe, you could also use the following argument: If $\varphi$ is a consequence of ZF+arb but not ZF, then there is a countable Henkin model of ZF+$\lnot \varphi$. This model can be expanded to a model of ZF+arb without choice. (But Shoenfield is fine, too.) 148329199032716845831089393660903Do you have a concrete example in mind where the existing theory is not applicable?19324801587211\Regarding representation of an outer functionCan you illustrate why you are interested in this, and what preliminary calculations you've done?Thanks a lot for the very helpful answer (and sorry if the question wasn't clearly phrased). I just have one more question. Isn't it normally assumed that f(1) = 0 for any f-divergence? This doesn't seem to be satisfied by the counterexample?17106062080850A round man cannot be expected to fit in a square hole right away. He must have time to modify his shape. (Mark Twain)

This epigraph is used in Bredon's book *Topology in Geometry*. It is a very nice illustration of the idea of homotopy theory. Also nice is the epigraph used in the preface:

The golden age of mathematics - that was not the age of Euclid, it is ours. (Keyser, 1916)

As I'm sure you must realize, it's unlikely that there exists a difference equation whose solutions are precisely the same as for a given differential equation. So I presume that by a difference equation you mean a numerical scheme to approximate solutions to Einstein Field Equations. There is an entire subfield of Relativity concerned with that: Numerical Relativity. It's already 50+ years old, has spawned hundreds of articles, also special conferences, textbooks and computer codes. Certainly, the numerical schemes that arise are not particularly simple, but they have been used to simulate situations like black hole collisions, that are not amenable to other approximation schemes. The linked Wikipedia article is a decent place to start reading.

2026130~Existence of $\kappa$-Suslin trees above a measurable cardinal2963102022331188191pIs there a n/2 version of the Erdős-Hanani conjecture?rWhen is the diagonal inclusion a $\Sigma_2$-cofibration?83494jAgain, you are neglecting unavoidable size issues...^Blow up along codimension one closed subschemeLet $\mathbb{N}=\{1,2,3,\ldots\}$ be the set of positive integers. For $n,k\in\mathbb{N}$ we define $$\text{Sol}(n,k) = \{(a,b,c)\in \mathbb{N}^3: |a^n + b^n - c^n| \leq k\}.$$ (The set $\text{Sol}(n,k)$ denotes the solutions of the inequality $|a^n + b^n - c^n| \leq k$ for fixed $n,k$.)

Moreover, for $j\in\mathbb{N}$ set $$\text{Inf}(j) =\{n\in \mathbb{N}: \text{Sol}(n,j) \text{ is infinite}\}.$$

Clearly, we have $\{1,2\}\subseteq \text{Inf}(j)$ for all $j\in\mathbb{N}$.

**Questions:**

- Is there $j\in\mathbb{N}$ such that $\text{Inf}(j)\neq \{1,2\}$?
- Is there $j\in\mathbb{N}$ such that $\text{Inf}(j)$ is infinite, and what is the smallest such $j$?

Hi,

Let $\Gamma$ be a free subgroup of rank 2 in $\mathbb{G}_m^2(\mathbb{Q})$. For all but finitely many primes p we can reduce $\Gamma$ modulo p. Let $S$ be the of primes for which $\Gamma$ does not reduce modulo p, and for any $p$ not in $S$, let $\gamma_p$ be the size of $\Gamma \mod p$. My question is what is known about the function

$f(x)= \sum_{p\not\in S,\ p\leq x}\frac{\log p }{\gamma_p}$

In particular what is the asymptotic behavior of $f$? Is the corresponding infinite series convergent whenever $\Gamma$ is *not* contained in an algebraic subgroup of $\mathbb{G}_m^2$? Do you know of any references that might be relevant to those questions?

Thanks in advance,

16040339914251512106194245314355126557318894691413710I honestly don't know. Maybe his point was that he valued intuition over logic.Illusie's thesis (1960's) Complexe cotangent et déformations I, Lec. Notes Math. 239, Springer 1971, xv+355 pp.; II, LNM 283, Springer 1972. vii+304 xv+355 pp.428601767840786640Can you also provide a link to the home-page from where you said that the book is available? I did quite a few Google searches but nothing came up except the flipkart and amazon and google books link for the book. 14418115384249390391476588What is the reference for some detailed discussion either lattice or only discrete subgroup? And what if n=4?See also "Simplexity of the $n$-cube" at The Open Problems Garden: http://garden.irmacs.sfu.ca/?q=op/simplexity_of_the_cube@joro Nice! So, in a sense, your latest solution has $40$ triples. And you get another form of the same by swapping $f,g$ and using $z \lt 0.$ How exactly are you getting an elliptic curve and what is the rank?7696775767585PThe Genus field and Hilbert class field3929383277242205771361940I have already asked this on stackexchange but did not get any answer.

Say I run a simple Bernoulli trial a number of times and compute the relative frequency for success. Clearly the relative frequency should represent the underlying probability for success better the more experiments I run.

My question is this: Is there any way to say how close the two values are given the number of times the experiment was run? For example if I run the experiment N times, what is the expected deviation of the relative frequency and the probability? Or turning the question around: If I want to know the probability up to an uncertainty of ε, how many experiments do I have to run?

Any pointers to results in this direction would be appreciated.

2065928@Jason Starr: I see what you meant. But the natural map should be $R^p f_*\mathbb{G}_{m,\text{et}} \to u_{Y*}R^p f_*\mathbb{G}_{m,\text{fppf}}$, where $u_Y:Y_{\text{fppf}}\rightarrow Y_{\text{et}}$. Then $u_{Y*}R^p f_*\mathbb{G}_{m,\text{fppf}}=0$ provided $R^p f_*\mathbb{G}_{m,\text{et}}$, But this doesn't imply that $R^p f_*\mathbb{G}_{m,\text{fppf}}=0$ 178150Seattle (UW)2022063How about fixing a non-zero element $g\in{\mathbb Z}_2^n$ and flipping a coin for each pair of elements that add up to $g$ to choose one for your set $X$? You will get a set of density $\alpha=1/2$ with $g\notin 2X$, and this set will typically be uniformly distributed in subspaces of co-dimension $1$.Sorry, I was describing the strict henselization there. Your description of the henselization is correct, I think. Why do you think it's not a DVR?4740911969463b@kenneth: You're welcome! I like that book, too.1742788376047If you write "surface of the polytope", do you assume that each point of the polytope is adjacent to an $n$-cell, and "surface" means boundary?This is not correct. If $M$ is a co-adjoint orbit, it is an homogenous space, and every invariant function is constant. It is however true if $s$ denotes the dimension of the polytope $\Phi(M)\cap C,$ if $C$ is a Weyl chamber.And not just semisimple, of course: there are Lie groups with bi-invariant metrics whose Lie algebras are not even reductive. 27418020564211528027413086933099Piotr: At the moment, the only candidate for $W$ I see is $\ell_\infty(G)$. (I think, if $b_g\in \ell_2(G)$, then $f$ has to be bounded, at least for finitely-generated groups, so $a_g$ is also bounded.) 8What do you mean by "rank"?Yes, it seems you are right. I have edited the post to point the GAP in my reasoning. 1070659938697I don't know the answer to this, but if you come to Cambridge the weekend after next then Marie Bjerrum might be able to tell you: https://www.dpmms.cam.ac.uk/~jg352/PSSL93.html1235760Schubfachprinzip ("drawer principle" or "shelf principle" or "Dirichlet's box principle"). It is now easy to guess we are talking about P-H P.

Many people know that there is a (3×3) nine lemma in category theory. There is also apparently a sixteen lemma, as used in a paper on the arXiv (see page 24). There might be a twenty-five lemma, as it's mentioned satirically on Wikipedia's nine lemma page.

Are the 4×4 and 5×5 lemmas true? Is there an n×n lemma? How about even more generally, if I have an infinity × infinity commutative diagram with all columns and all but one row exact, is the last row exact too? For all of these, if they are true, what are their exact statements, and if they are false, what are counterexamples?

*Note*: There are a few possibilities for what infinity × infinity means -- e.g., it could be **Z**×**Z** indexed or **N**×**N** indexed. Also, in the **N**×**N** case, there are some possibilities on which way arrows point and which row is concluded to be exact.

Let me flesh out my comment, and give some more details. I will denote $C_p$ the cyclic group of order $p$.

First, the structure of the group ring $\mathbb{Z}[C_p]$ is a little more complicated. If $\sigma$ denotes a choice of generator of $C_p$, then there is an isomorphism $\mathbb{Z}[C_p]/(\Phi_p(\sigma))\cong\mathbb{Z}[\zeta_p]$. This can be used to classify modules over $\mathbb{Z}[C_p]$, see this MO-question on representation theory over $\mathbb{Z}$ and in particular the paper of Reiner linked to there.

More relevant for the K-theory question is the fact that $\tilde{K_0}(\mathbb{Z}[C_p])$ is in fact isomorphic to the class group $\tilde{K_0}(\mathbb{Z}[\zeta_p])$ of the cyclotomic field $\mathbb{Q}(\zeta_p)$. This was shown in

- D.S. Rim. Modules over finite groups. Ann. of Math. 69 (1959), pp. 700-712,

based on the explicit description of modules over the group ring.

As a funny aside, you might want to have a look at

- M.A. Kervaire and M.P.Murthy. On the projective class group of cyclic groups or prime power order. Comment. Math. Helvetici 52 (1977), 415-452.

to see what can be said about class groups for $C_{p^n}$, $n>2$. As far as I know, these groups are not yet completely computed.

In any case, for prime order, you are then interested in the class groups of cyclotomic fields. The class numbers of cyclotomic fields for prime order roots of unity form the OEIS sequence A055513. For tables of class numbers as well as an extensive treatment on methods to understand class numbers, class groups and other facts about cyclotomic fields, see the book

- L.C. Washington. Inntroduction to cyclotomic fields, Springer, 1997 (for second edition).

For more precise statements, you better ask number theorists. As far as I understand, there are two conjectures for the $p$-part of the class group of $\mathbb{Q}(\zeta_p)$. The Kummer-Vandiver conjecture states that $p$ does not divide the +-part of the class group. A conjecture of Iwasawa states that the eigenspaces for the Galois action on the $p$-part of the class group are all cyclic groups (of order related to a special L-value). For more precise statements and references, see e.g.

- M. Kurihara. Some remarks on conjectures about cyclotomic fields and $K$-groups of $\mathbb{Z}$. Compositio Math. 81 (1992), pp. 223-236.

Summing up, an explicit list is probably only available for small primes, but there seems to be a (at least conjectural) conceptual understanding of the class groups of the group ring of cyclic groups of prime order.

2279974Is this the sort of idel curiosity that lead to the discovery of so many things, or a question motivated by a "real problem"? (I am asking out of idle curiosity because I can't image how and why anyone would think of these inequalities.)hPHP, Javascript, JAVA and NodeJS develoeper

@Agol : Your heuristic remark about the corresponding Teichmuller spaces is very interesting, thank you. 6394220374391567225https://lh3.googleusercontent.com/-4KCh2DKokzI/AAAAAAAAAAI/AAAAAAAAAAA/AAN31DXKuACEHABVih6IjwtALJDohTPLYg/mo/photo.jpg02009-12-08T22:49:37.6472006963^any knowledge about Hensel lifting complexity?29641502505472793461681622So I no longer need the result for the paper that I was working on, but I'm still interested in the question. I believe that, for example if $T$ is an Anosov automorphism of $\mathbb T^d$ (with a variety of Lyapunov exponents), then the answer to the question is `yes' (as the Bowen balls are translates of small ellipsoids around $x$).192994919522931217746606415It seems to me the answer should be no. Take a fg cancellative monoid $M$ not embeddable in a group. Enumerate the elements and invert them one by one. Then the original monoid would embed in the unit group of the direct limit.You have to reformulate the question as each family satisfying the Hall condition can be embedded into a family satisfying the condition in the question.1454988xUpper bound of $\frac{\sum_i c_ia_ie_i}{\sum_i d_ib_if_i}$?93519\https://i.stack.imgur.com/w4CSm.jpg?s=128&g=14771442183857Yes, that is the book I meant. I like the co-ordinate free vector field approach taken, and I like its brevity. By the way, a "vielbein" is just a "moving frame", i.e., a set of tangent vector fields on an open subset of the manifold that form an orthonormal basis of the tangent space at each point on which it is defined. Using such an object, you can also do the calculations using the dual 1-forms and the Maurer-Cartan equations (the equations that tell you what the exterior derivative of the left-invariant 1-forms of G are).1941404AI won't be able to give any references, so I hope some more experts can help me out, as there is much work on traces of functors. In general, there are two reasonable notions of "trace" of a functor, and they can be different. Throughout, I let $F$ denote an endofunctor of a nice-enough ($\mathbb C$-linear, etc.) category $\mathcal C$, and $\DeclareMathOperator\id{id}\id$ the identity functor.

Then one notion of "trace" is: $$ \operatorname{trace}(F) = \hom(\id,F)$$ where the $\hom$ is taken in the (monoidal) category $\operatorname{End}(\mathcal C)$ of endofunctors of $\mathcal C$, i.e. it is the space of natural transformations. For this definition to make sense, we need only that $\mathcal C$ is small enough for $\operatorname{End}(\mathcal C)$ to be locally small (otherwise, for generic categories, the hom spaces between functors can be proper classes), or at least for $\hom(\id,F)$ to be small. In the $\mathbb C$-linear setting, one expects that $\hom(\id,F) \in \mathrm{Vect}$, and in fact it is a $\hom(\id,\id)$-module. Note that $\hom(\id,\id)$ is always an algebra. In fact, since $\operatorname{End}(\mathcal C)$ is a monoidal category, $\hom(\id,\id)$ is always a commutative algebra. For example, when choose a $\mathbb C$-algebra $A$, and let $\mathcal C$ denote the category of left $A$-modules. Then $\hom(\id,\id)$ is the center of $A$.

There is an important generalization: work not with categories but $(\infty,1)$-categories. Then one can set $\mathcal C$ to be an appropriate "derived" category of chain complexes of $A$-modules, and $\hom(\id,\id)$ is then the Hochschild *cochain* complex of $A$.

There is another important notion of "trace", which is given by an end (or is it a coend?) of the functor $\hom(-,F-)$. This notion is slightly closer to the idea of "adding up the diagonal entries of a matrix for $F$". In the $A$-module case, this version gives $\operatorname{trace}(\id) = A / [A,A]$, where $[A,A]$ is the subvector space of commutators (and not an ideal or anything), so that the quotient is simply a vector space (with a distinugished element, namely the image of $1\in A$). In the derived setting, one gets the Hochschild *chains* of $A$.

The two constructions must give canonically-the-same answer if $\mathcal C$ is the image of an oriented but otherwise unframed 2-TQFT. But if you work with framed TQFTs, they can give different answers. Recall that a *2-framing* of a 1-manifold $S$ is a framing of $S\times \mathbb R$, and that a framing of an $n$-manifold is a collection of $n$ vector fields which are at *every* point linearly independent.
The first "trace" corresponds to the circle with "outward" framing, i.e. it has a "2-framing" inherited from embedding the circle as a simple closed curve in $\mathbb R^2$. The second trace corresponds to the "product" framing, i.e. the framed circle where one of the two vector fields is parallel to the circle and the other is orthogonal.

When thought of in this geometric picture, the "Deligne conjecture" that Hochschild *chains* has a homotopy-$S^1$-action becomes natural, and Hochschild *cochains* have their $E_2$-algebra structure coming from embedding two disks into a larger disk.

Actually, if you have a complete framed 2-dimensional TQFT which assigns $\mathcal C$ to a point, then the two notions of trace must agree for $\id$, at least in dimension. I mean, if you look at the torus (with its unique framing), the value of the torus must be the dimension of each $\operatorname{trace}(\id)$, by cutting the framed torus into an annulus in two different ways. A framed 2-TQFT does not pick out a *chosen* isomorphism between the two different traces, and I believe that any choice of such an isomorphism is pretty much enough to extend the framed TQFT to an unframed one. Algebras with such a choice are called "Calabi–Yau", at least by some people, because the data of such an isomorphism is roughly the same (when $A$ is commutative) as a trivialization of the canonical line of $\operatorname{Spec}(A)$.

Let's take a page from Silverman's book, VII.3. Let $\mathfrak{p}$ be one of the primes of good reduction of $A$. Let $K/k$ be any extension, and let $\mathfrak{P}$ be a prime of $K$ above $\mathfrak{p}$. The reduction map $A(K)\to A(\mathcal{O}_K/\mathfrak{P})$ becomes injective when you restrict to torsion points of order prime to the residue characteristic of $\mathfrak{p}$ -- this is proved using an appeal to formal groups.

Now choose two such primes $\mathfrak{p}$ and $\mathfrak{p}'$ with distinct residue characteristics. Convince yourself that there exists a $K/k$ and primes $\mathfrak{P},\mathfrak{P}'$ of $K$ for which $A(K)\to A(\mathcal{O}_K/\mathfrak{P})\times A(\mathcal{O}_K/\mathfrak{P}')$ has nontrivial kernel. Nontrivial points in the kernel must not be torsion.

tI think you mean "standard part" rather than "real part".2292441511108Jhttp://turing.plymouth.edu/~kgb1013/2091751https://www.gravatar.com/avatar/a2fcbfb3776f4ba3562b48bb7bce5fa8?s=128&d=identicon&r=PG&f=1445895~https://graph.facebook.com/2183731521845973/picture?type=large I've stated to write a detailed answer, but I'll just change it to a comment after seeing this beautiful thread - http://mathoverflow.net/questions/98405/fourier-decay-rate-of-cantor-measures , I think the correct notion for this question should be the H-dim of the support of $\mu$, and you can find relation like that for Bernoulli shifts (a.k.a diagonal actions in your phrasing, like times $2$). Notice that for the usual Kronecker system, the entropy is $0$ although the dimension is full, so you indeed need this encoding of your system to make sense out of your situation.1774714https://lh6.googleusercontent.com/-NERON8iJT84/AAAAAAAAAAI/AAAAAAAAACw/zIMh96xEN0A/photo.jpg?sz=128491444127541 OK thanks, this allows me to explain a bit. The thing is that without reading the linked question (at least to some level of detail) it might be very difficult to even relate where this particular answer is coming from (and that doesn't seem to have gotten any additional view ..... till yet ofc). As I understood the original question (I admit I am no logician), it was about developing a meaningful link between "infinite machines" and "formal Church's thesis". Here's the approach I took in my answer (certainly claiming that it is the only approach would be a bit too much so I don't claim that).Alexander: I do not understand your simpler example, I get $\frac{1}{\text{Tr}M}$. The role of the integer $p$ is as the dimension of the positive-definite matrix$U$ and $T$.For teams of 1, n=1, the results are trivial.

For n=2, and each match having an expected score for the weaker side of 0.3, elementary probability and contour maps can be used to illustrate that :

(I) the probability of the weaker side at least drawing is maximised if both x's are 0, (II)the probability of the weaker side scoring at least 1.5 is maximised if one x is 0 and the other is 0.3.

I know nothing about higher n, except n=12, when I suspect, in all practical cases, x(1)...x(12) equal zero: the "go for it" solution. In this seasons London League there are three "Drunken Knights" teams

In any event, I still believe and hope that the general n behaviour is an appropriate question for this site.

The question is pretty vague, but it sounds as if you might be interested in the work of Andreas Weiermann on phase transitions in logic.

715953Fair point. I described a simplified problem (just edited it to make it clear). It is not possible to do a Fourier transform for the real problem (the functions themselves are high-dimensional there), so I have to stick to Monte Carlo.2090375 I highly recommend asking professionals directly. If you write a summary that they can read in a few minutes, followed by a longer first draft (marked as such so they know what to spend time on), then you can reference these from your MathOverflow user page and in a brief request that you send them. If you were to send me (also a non professional) such a request, I might reply that I am not interested, but might make suggestions as to how to find interested people. If you can, get brief feedback from two sources or more. Gerhard "Get That Support Network Activated" Paseman, 2018.09.29.Employee of a tech SME in the UK. Willing to learn anything that can be taught, and teach anything that he can.

When I was a student in Lvov in 1970-s, I've heard many legends about Banach, so let me add a few points. Once Steinhaus was walking in a park, and he accidentally heard a conversation of two young people sitting on a bench. The words "Lebesgue integral" were pronounced. At that time very few people in Lvov had heard of the Lebesgue integral. So Steinhaus was curious, and introduced himself... Banach was an engineering student at that time. (Who was the other person on the bench, the story does not tell).

According to the legend, Banach worked most of his time in the Scottish cafe. Students and colleagues joined him for conversation. (One of the results of this was the famous "Scottish book" of unsolved problems. Prizes were offered sometimes and recorded to the book together with the problems. For example, in 1970-s, when Per Enflo solved the "basis problem" from the Scottish book, he won a prize, a live goose, which was delivered by Mazur). He used to write on the table cloth. The owner of the cafe never complained. At the end of the day, he changed the tablecloth for a new one. And the old one he sold to students.

Banach drunk a lot (and there are many stories about this, which I omit). Frequently he was short of money, and had to drink in credit. At some time, the debt grew large, and there was an argument with the owner of the Scottish cafe. Finally the owner proposed that Banach writes a calculus textbook to make money to pay for his drinks. (Some version of the legend says this was suggested by students). Indeed, he wrote a calculus textbook:-) But I have never seen his high school textbooks.

The Scottish cafe still existed in 1990-s, but under a different name, and in 1970-s this was a simple cantina. Then the rooms passed to some financial institution.

P.S. Wikipedia, http://en.wikipedia.org/wiki/Scottish_Caf%C3%A9 has somewhat different details of doing math in the Scottish cafe, based on Ulam's recollections.

1390724Thanks. I'm afraid I do not understand what Tachyon does in this scenario exactly. Is it even necessary? I checked the website but couldn't make heads or tails out of it.15845829182581593724That does sound like a great start, thanks! I've never looked at the second edition.162068519256461690818130566Sæsi4519902132682Yep, you are right. I have edited moments ago. The condition I gave may not be correct. But the usual criterion you raised is for continuous case not for discrete one. The discrete form is the eigenvalues of $A$ fall into the unit circle.@RazvanD: when fedja says "dimension" he means 3 dimensions as in 3D movies. He doesn't mean the actual side lengths of your box.431025You can obtain such a surjection for any finitely presented non-residually finite group $H$ using Daniel Wise's residually finite Rips construction, which is the main result of this paper: A Residually Finite Version of Rips's Construction.

18530261071359 Daniele PalombiYes, all I need is that the intervals $(\alpha_i,\alpha_{i+1})$ have length larger than $\alpha_i$.57670phttp://coupestan.com/pool_index.php?pa=5&lg=en_US.UTF-811069046570711927740Perturbing the constant term of a polynomial and implications to stability$TheoremOfMontague121904219149261672690296471But which case? Are you interested in the sequences of matrices which are not necessarily independent?88938724263837976I have an optimization problem with a semi-definiteness constraint: $$ N \preceq 0 $$ where the entries $N^{AB}$ of the matrix $N$ are defined through $$ N^{AB} = \sum_{i,j} x^i M_{ij}^{AB} x^j $$ The $x^i$ are my independent variables, whereas the $M_{ij}^{AB}$ are fixed and obey $M_{ij}^{AB}= M_{ij}^{BA}$ so $N$ is symmetric. The variable to optimize is a linear combination of the $x^i$.

When $N$ is a one-by-one matrix this is a simple quadratically constrained program. In this case the problem is convex if $M_{ij}$ is positive semidefinite, and the translation to a familiar semidefinite programming problem can be found, for example, in section 8.1 of these lecture notes.

When $N$ is of higher dimension the constraint is similar to a *bilinear matrix inequality*. In this tutorial it is simply mentioned that such inequalities do not carve out convex sets in general. However it seems to me that the problem can very well be convex for suitable $M^{AB}_{ij}$.

Question: can we think of constraints for $M^{AB}_{ij}$ so that the problem ends up being non-trivially convex? And what would then be the translation to a standard semidefinite programming problem?

1308226178399This question was apparently one of two challenges issued by Fermat in 1657, the other being the “inverse” equation $1+x+x^2+x^3=y^2$. Mahoney points out that Fermat didn't comment on the solution of this one (though he did solve the companion), but suggests both can be reduced to a Pell equation $u^2-dv^2=\pm k$. While I can see how the companion would be so transformed, can anyone tell me how this one would be turned into a Pell equation?105907157450As Hartshorne chapter III.9.2 claim，an Ox-module （need not be quasi coherent) F's flatness is stable under base change. But the stalks is not the tensor products, how can I prove the claim?

7054922151633318391651162392VCombination of $k$-powers and divisibility908375487630n(Why the downvote - did I misunderstand the question?)2253261143003877698719379463288711594798:DoobsOptionalSamplingTheoremWell, it's not really about the class being $G$-invariant, it's about some cocycle in the appropriate cohomology theory being $g$-invariant. You can see why it has to be that way by looking at the $H^1$ of the circle under the degree $2$ map to itself. It's obvious from any number of arguments that the map on $H^1$ has to be multiplication by $2$, so only even classes lie in the image. I don't really know what "is sensible to $E_2$ part" means.321610\https://i.stack.imgur.com/J1QBM.jpg?s=128&g=11540240663715Why do you attempt to perform an edit with a different username, given that you are a registered user of MO?1710422644249 @Geordie Williamson: I am slightly worried that the local to global degeneration for weight reasons that you mention uses that Hom between the ICs is pure. No? Let $X=X_0 \supset ⋯\supset X_1$ be the filtration by closed subspaces corresponding to the stratification. Let $v_k:X_k \to X$ be inclusion. The degeneration is obtained by looking $Hom(v^∗_k M,−)$ applied to $i_∗i^!v^!_k N \to v^!_k N \to j_∗j^!v^!_k N$ (I hope what my $i$ and $j$ are is clear). Now for degeneration we want the connecting map in the long exact to be zero. Without weights (or parity vanishing) I dont see how to get it450627NFourier Transform, for entire function208161684946816968861403336If oner works in a neighborhood of a point, there is no difference between smooth and analytic (or algebraic) case. The statements and proofs are the same. The space of derivations at a point is the tangent space, and has the dimension equal to the dimension of the manifold. On the other hand, derivations of the algebra of functions (without fixing the point) form an infinite dimensional space (Lie algebra) of vector fields. 1091720Is it provable that computers can't ask math questions? Strongly doubt it.Partial answer: if $X$ has underlying topological space of finite dimension $k$, then what you want is true, because $H^i(X, F)=0$ for any $O_X$-module $F$ and $i>k$. Let $F=\ker(f_{k+n})$, so that we have an injective resolution $$ 0\to F\to I_{n+k+1} \to I_{n+k} \to \ldots \to I_{n+1}\to I_n \to I_{n-1} \to\ldots $$ So if your $j$ didn't exist, $i$ would give a nonzero class in $H^{k+1}(X, F)=0$.

@Oleg567: Thanks. I had a feeling that the "smoothness" of small solutions might not remain if we remove the restriction that $F(s_m)$ be a perfect square.Let $T$ be a rooted tree with root $r$. Say an ordering $v_1,\ldots,v_n$ of the vertices of $T$ is a *search order* if $v_1=r$ and for all $2 \leq i \leq n$, there is $j < i$ such that $v_j$ is the parent of $v_i$. In other words, parents are explored before their children in the order.

For a given search order $v_1,\ldots,v_n$, let $w(v_i)=\max(j:v_iv_j \in E(T))-\min(j:v_jv_i \in E(T))$. The max is the time the last child of $v_i$ is explored, and the min is the time the parent of $v_i$ is explored. Say the width of the order is $\max(w(v_i):1 \leq i \leq n)$, and say the width of $T$ is the minimum width of an ordering of $T$.

Is anything known about the width? Is it a known concept under another name? Have any theorems been proved about it? Any equivalent characterizations/definitions? Any useful bounds~~, perhaps in terms of the maximum degree of $T$~~?

**Edit**: This is the directed bandwidth, as David Eppstein points out below. I'm still interested in any bounds -- perhaps some upper bound with a simple form, perhaps even with an approximation guarantee?

Not an answer, just
an illustration for $N=2$ policemen, starting at $x=2$, with the thief starting at $x=0$.
Time advances vertically. The thief (black) is caught (by the purple policeman) at $(x,t)=(-3,19)$:

The distribution of catching times is highly skewed, and so it is difficult to determine the
mean-time-to-catch from simulations. Here is a histogram for 100 random trials:

In one of my runs (not included above), it took 24,619 time steps to catch the thief (at $x=-49$)!

Just to add an illustration for $N=3$, as per Ori G.-G.'s latest estimation, here is an example where the thief is captured at $(t,x)=(912,2)$. Again the thief is black (the lower curve), but now time increases to the right:

nQuestion on resolutions for arbitrary chain complexes.330223tCorrection -- the $+$'s in my comment should be $\cap$'s.p@WillSawin Right. Sorry, I misunderstood your argument.1932668@VictoriaGitman I did not consider them when I wrote the question, but do not matter about it.2202820mcampoI've heard the name "ends" (of $E$) for the elements of your $E_\infty$, but I'm not sure how standard that name is.17525610371111878729575168https://www.gravatar.com/avatar/80a4403ef244f08631adef8ba6e206a1?s=128&d=identicon&r=PG&f=11358883Thanks, Georges, this is a great answer! As I am guessing you know, it is very closely related to a lot of things I have written about in various notes on my webpage, so it speaks deeply to me. 644990628018175646rThat is properly where the question have been motivated.14406001423895204632383448#

The standard weak formulation of the Neumann problem for the Poisson equation is to find $u \in H^1 ( \Omega)$ such that for every $v \in H^1 ( \Omega)$:

$$ \int_{\Omega} \nabla u \nabla v d x = \int_{\Omega} fv d x + \int_{\partial \Omega} gv d s $$

for given $f \in L^2 ( \Omega)$ and $g \in H^{- 1 / 2} ( \Omega)$.

My goal is to introduce an arbitrary coordinate change and arrive at an analogous weak formulation.

I start with the classical formulation $- \Delta u ( x) = f ( x)$ and $\partial_n u ( x) = g (x)$ with everything fine and smooth. Let now $\phi : \Omega \rightarrow \tilde{\Omega} : x \mapsto \phi ( x) = : \tilde{x}$ be $C^{\infty}$ and bijective and define $\tilde{u} \circ \phi = u$. Applying the chain rule to $ \Delta_x ( \tilde{u} \circ \phi)$ we find the differential operator $L$:

$$ L \tilde{u} ( \tilde{x}) = a_{i j} ( x) \partial_{i j} \tilde{u} ( \tilde{x}) + b_i ( x) \tilde{u} ( \tilde{x}), $$

where $a_{i j} ( x) = \partial_k \phi_i ( x) \partial_k \phi_j ( x)$, $b_i ( x) = \Delta \phi_i ( x)$ and we used the usual summation convention over repeated indices. Yes the coordinates are all mixed up: I left out a composition with $\phi^{- 1}$ because it's messy.

It now holds that $\Delta_x u ( x) = L \tilde{u} (\phi ( x))$ for all $x \in \Omega$.

**Question 1:** My operator is not in divergence form, so deriving the weak
formulation is going to be a real mess. What would be a better approach?

**Question 2:** How do I transform the boundary conditions? I'd like something
like

$$ \partial_{\nu} \tilde{u} ( \tilde{x}) = \tilde{g}( \tilde{x}) . $$

The chain rule applied to $\partial_n ( \tilde{u} \circ \phi)$ results in $\partial_{\nu} \tilde{u} ( x) = \nabla \tilde{u} ( x) \nu ( x) = \nabla \tilde{u} ( x) D \phi ( x) n ( x)$, which is partly natural since the normal vector field is transformed with the differential, but is far from what I'd need during the partial integration deriving the weak formulation, which would rather be (assuming $L$ were in divergence form)

$$ \partial_{\nu} \tilde{u} = a_{i j} \partial_i \tilde{u} \nu_j .$$

This must be pretty basic, but I'm a bit confused here...

xIf $\phi$ annihilates $L^1(G)$, then $\phi$ is zero itself.73115510725047005445Given any 4 positive numbers $p_{00}\,,p_{01}=p_{10}\,,p_{11}$,such that the sum of the 4 numbers is 1, now I want to find a sequence in $\{0\,,1\}^\mathbb{N}$ such that this sequence has uniform frequencies and recurrent properties. where $p_{ij}$ is the frequency of block $(ij)$ in the sequence. As we all known, uniform frequencies equivalent to unique ergodicity in dynamical systems, a good example is Sturmian sequence which has uniform frequencies and recurrent properties, but the frequencies are determined by the irrational rotation.

@martin: Yes, and that is how I understood it. The claimed equation for the zeros on the critical line would imply that only 0% of the nontrivial zeros are off the critical line. We certainly don't have a proof of that.Thank you! Shepherdson's construction of $R$ is given in this paper: http://plms.oxfordjournals.org/content/s3-1/1/71.full.pdf.@q-expansion principle for Γ(N)212546590722313336862192504147740413549961218737In general, when one says that $E$ is a multiplicative cohomology theory, that means that $E^*(X)$ is a ring for $X$ a space, not for $X$ a spectrum. For $X$ a spectrum we have that $E^*(X)$ is an $E^*E$-module.(I have asked this question at math stackexchange, it was upvoted but got no answers; maybe you can help.)

It is well-known that $\beta_1(R(f))\le\beta_1(X)$, where $\beta_1$ is the first Betti number (number of loops), $X$ is a topological space, $f:X\to\mathbb R$ a continuous funcion, and $R(f)$ its Reeb graph, i.e., the space of contours of $f$ (connected components of levels of $f$): see, e.g., Computational Topology: An Introduction by Edelsbrunner & Harer, page 141.

However, the commonly given argument looks flawed to me: it is said that the dimensionality of the space of **images** of all 1-cycles of $X$ in $R(f)$ does not exceed $\beta_1(X)$, but why cannot $R(f)$ have other 1-cycles not induced from any 1-cycle of $X$?

For example, the height function $f$ on a Warsaw circle (a topologist's sine curve including a segment at the "bad" place, not just a point; plus a piece of a circle) $X$:

is continuous and the $R(f)$ is a circle, $\beta_1(R(f))=1$, while $\beta_1(X)=0$ (here, Remark 2.7 on page 7). While it is true that the projection in $R(f)$ of any cycle in $X$ is zero, $R(f)$ has a cycle not projected from any cycle in $X$.

Where am I wrong?

And even if this particular counterexample is wrong, still how to prove that $R(f)$ does not have other cycles?

7512271794214casterbear1173379786315bIs there a k missing in your commutative square?10698841158894This seems way too hard. If G=SL2 and V=Sym^d(C^2) that's the ring of invariants for binary forms. I think it is only known up to d=10.If you want to consider the variation of Kahler Ricci flow on degeneration of Kahler-Einstein metrics $\pi:X\to \Delta$, then we need to resolve singularities by two times $s$ and $t$, i.e, $$\frac{\partial^2 \omega(s,t)}{\partial s\partial t}=-Ric_{X/\Delta}(s,t)-f(s)\omega(s,t)$$, which is hyperbolic PDE in this case we need to the positivity of initial metric and I am not sure the statement of Deane Yang is true in this case. where $f(s)$ is fiberwise constant. In KRF case we need to run the flow by one time $t$,9199847151647359151069511http://mathoverflow.net/questions/13171/how-many-trial-picks-expectedly-sufficient-to-cover-a-sample-space11964343572961179361I'm voting to close this question as off-topic because: I agree with Gerhard that it is inappropriate to discuss the flaws in a paper in a public forum.Thanks, but I am hoping somebody will read it critically, and highlight any dubious claims or possible inaccuracies!997112T@AJ not sure, I haven't thought about it.1294508By the way, do you think that my initial attempt could succeed, with some more work? or not?1837421(∞,1)-categories are a lot more like 1-categories than 2-categories are. Like Tom says you should start by trying to make sense of "2-topos".442273219087217120921423485It's not research, but I disagree it's quite elementary. There are a number of components to Paul's proof below that are not completely obvious, IMO. L(G) is the lattice of subgroups of G. Aut (G) is the group of automorphisms of G186273554831Jhttp://nagabhushansn95.wordpress.com204841542538When I attended a course by Jacek Brodzki (http://toknotes.mimuw.edu.pl/sem6/index.html) he indeed said that most of the time we'd be interested not in computing KK(A,B) but rather in understanding specific cycles. But I can't say whether he meant it generally, or only for the lecture (back then I though he meant it generally).It is easy to miss the point that in the second definition $\mathbb Q/\mathbb Z$ is required to be *discrete* in Hiro's question.

Hence even if a continuous morphism $f:G\to T$ has image in $\mathbb Q/\mathbb Z$ it cannot automatically be considered as a continuous map $f_0:G\to \mathbb Q/\mathbb Z$ and so might not be a character in the second sense.

An example for this failure is to take $G=T_{tors} =\mathbb Q/\mathbb Z\subset T$ , the torsion subgroup of $T$ with its induced topology from the circle and for the character $f $ (in the first sense) the inclusion $f:G\hookrightarrow T$.

Even though $G$ is torsion, the corestricted morphism $f_0:G\to \mathbb Q/\mathbb Z$ is not a character in the second sense, since it is not continuous.

However if a character on a compact group $G$ *happens* to have values in $\mathbb Q/\mathbb Z$,so that both definitions can be compared, then its image in the circle is finite and the two concepts coincide.

Xandi explains in his answer that this is always the case for profinite groups.

There is only one differentiable structure permitted in R^2, meaning, I think, that all atlases in R^2 are diffeomorphic to the Cartesian atlas. But, doesn't the polar coordinate system represent an atlas that is not truly diffeomorphic to the Cartesian atlas, due to the coordinate singularity it has at its origin?

17075041570198tSee Theorem 7.3.5(i) of the book by McConnell and Robson.@Emerton: I thought Leray invented sheaves during WWII. Am I (or the references on Wikipedia) mistaken?19525416456652223829Hmm? What about 5,14,7,20,10,5,... ? Another one is at 17 or 19 (if I recall right). But these two are then the only additional known ones besides the "trivial" 1,2,1,...The OP was asking about unitary matrices. Note that if $U$ is unitary and $|\omega|=1$ with $\omega \notin \sigma(U)$, $i (U+\omega I)(U-\omega I)^{-1}$ is hermitian, so solving the unitary problem is essentially like solving the hermitian problem.68835318704265293096416791786331How to define the orientation of a vector space over an arbitrary field?vDistribution of random Projections on the Stiefel Manifold1231957Haven't you just described a factorization of the functor through the category of monoids in C? So an alternative functor is just a functor F: C---> Mon(C)? I guess we have to encode the lax-monoidal part...27595263365416182187997812297088@FriederLadisch : The result was proved independently by Guralnick and Lucchini at around the same time, it was not a joint paper - I had forgotten that myself!I think it is straightforward to prove by induction that starting from the eight vertices $(\pm 1, \pm 1, \pm 1)$ of a cube in $\mathbb{R}^3$, the constructible points in the OP's extended sense all lie in $K^3$, where $K$ is the union of all Galois extensions of $2$-power degree over $\mathbb{Q}$. It follows that the possible distances determined by the so constructed points also lie in $K$, whence $\sqrt[3]{2}$ is not among them.

In short, the cube cannot be doubled even in the OP's extended sense. On the other hand, it will be hard to verify what the oracle of Delos had in mind.

https://graph.facebook.com/10213553414464106/picture?type=large15747945777349137829The following should be known, but I could not find an example.

Let $\kappa$ be an uncountable cardinal. Find a model $M$ of size $\kappa$ which has $\ge\kappa$ many automorphisms, but for some $m\in M$, $(M,m)$ has only the trivial automorphism (I call this element rigid on the title. Not sure if there is a standard name for it).

I as a Chemist by education find it extremely enlightening to read Mathematicians takes on this subject. Thank you very much!@Kris: the zero ring is not a zero object in the category of unital rings. It is only the terminal object, while the initial object is $\mathbb{Z}$.311974user601257999023436Hi, for someone less familiar with the theory, do you mean indicating (a reference perhaps) why the space of bounded harmonic functions on $M$ is two dimensional, and why this also carries over in the case of more than one tube?7422381287637811494437009bAssistant Professor at the University of Idaho.Hochschild homology gives invariants of (unital) $k$-algebras for $k$ a unital, commutative ring. If we let our algebra $A$ be the group ring $k[G]$ for $G$ a finite group, we get group homology. There are plenty of other connections to homological algebra. If we use cyclic homology, there are connections to geometry and topology involving the Chern character.

Von Neumann algebras are complex algebras, so we can take their Hochschild and cyclic homologies. When I have asked experts in the fields of von Neumann algebras and non-commutative geometry about what you get, I usually hear some approximation of the following: "There's also analysis in von Neumann algebras, so I wouldn't expect an algebraic invariant like Hochschild or cyclic homology to tell you anything useful."

Although this answer makes some sense, I find it very displeasing and cryptic. Why shouldn't it tell you something? Is there some way to make "it doesn't tell you anything" quantitative? Is there an example of a von Neumann algebra with nontrivial Hochschild or cyclic homology (different from that of the complex numbers)?

EDIT: After reading the responses so far, I should specify that I really want to know if there is a $II_1$-factor with nontrivial Hochschild or cyclic (co)homology.

826666413279This is a cute problem, but in most cases, you will be able to reach an arbitrarily high entropy.

Basically, for any $p(z)$ such that : $\int p(z)p(x|z)p(x|y) dxdydz $ is finite, you have a joint distribution that respects your condition.

Depending on your p(x|y) function, all p(z) could respect that condition (for example, if $\int p(x|y)dy \leq B$ for all $x$), in which case you can reach an abitrary entropy from p(z) alone

DNormal form in second-order logic1289295171790\@Donu: can it be proved 'purely algebraicly'?@Qiaochu: with all due respect (which is a tremendous amount!), I don't think questioners should be prompted to provide sufficient background for their question so as to make it comprehensible to a general mathematical audience (or to an undergraduate, no matter how brilliant and knowledgeable). That's what tags are for, and that's what experts are for. MO was all over the question as written, and to someone who knows about modular forms it was unimpeachably clear. If it is a formal group law associated to an elliptic curve, then it must be of height either 1 or 2. I don't believe that this elliptic curve corresponds to the height 3 formal group law I mention above.Do you know an old reference? Gromov's random group theorem is overkill, one could just use that a random relator will be small-cancellation, but was this observation made before Gromov? Anyway, an answer is more useful if it provides at least some reference. 126631In Mike Artin's book, some part of the Hilbert-Burch theorem is also attributed to Schaps.612456I'm no graph theorist, but if you're only interested in how many nodes have different degrees then what you want is surely just a statistical property of the frequency distribution. Your "hubbiness" property sounds like it corresponds to positive skewness, or perhaps to the ratio (standard deviation / mean) being large (edit: this ratio is apparently known as the coefficient of variation).

124511318061761704813https://www.gravatar.com/avatar/26e941e811df7ad3c2aaa03751e59f90?s=128&d=identicon&r=PG&f=11424723RStructural Mechanics Researcher.

I think this is the reference you mention http://www2.isye.gatech.edu/~nemirovs/Daureen.pdf Specifically Section 1.3156447For self-injective Koszul algebras of Loewy length greater than three such that the Yoneda algebra is Noetherian, all (stable) components of the Auslander-Reiten quiver for *graded* modules are of the form $\mathbb Z A_{\infty}$ (Martinez-Villa, Zacharia: Approximation with modules having linear resolution).

It shouldn't be too hard to find examples where all modules are gradeable, it is more difficult to keep track of graded shifts within each component. The easy way out is to say that each (ungraded) component is of the form $\mathbb Z A_{\infty}/G$ for some group of automorphisms $G$. (Edit: Also known as a tube when $G$ is non-trivial.)

14765459925918Vector bundle under Blow upPerfect! Thank you very much! \\ Thanks also for the comments --- concerning point 3., usually this theorem is used to isometrically embedded the universal cover of closed surfaces, that can be still very interesting.Sorry, I cant write a full example here in comments, it is too lengthy, but you can check how it works for example for the lowest root $\Lambda-\alpha_1-2\alpha_2-\alpha_3$ of (010) of $A_3$. All the scalar products $\langle E^+\dots E^+E^-\dots E^-\rangle$ which appear in $\langle v|\rangle$ here are equal to 1.1258939You can play around with this anyway. For example, draw the n lines, and place a circle somewhere around the concurrent point. The intersections determine a cyclic polygon, and there may be some relationship you can develop with opposing sides or pairs of sides as in cyclic quadrilaterals. Again, I don't see where you are going with this. Gerhard "But Have Some Fun Anyway" Paseman, 2016.12.07.207206121545172061176f- = 'tilde'. My apologizes for the bad formatting.4475621756621444826Yeah, if I had to guess I would assume that the original motivation was an attempt to generalize the [Gauss map](https://en.wikipedia.org/wiki/Gauss_map) beyond hypersurfaces1971591191807736774820052522085806Now, I see the deeper problem: the realization doesn't need to even be irreducible. If we take $S_n$ generic poly, this leads us to say that the reducible polynomials are thin between all degree n polys. This is even more shocking and counter-intuitive to prime number theorem.Impressed on two counts. a) That you managed to locate an undergraduate honors thesis, and b) this (the thesis) seems quite a sophisticated piece of work for an undergraduate.2953582747405In physics, I've been told that $\mu$ is a "mass", or perhaps "mass squared", but I don't claim to understand why. I've also been told that $\mu = \hbar$, a claim that I understand even less.I am a working mathematician/teacher in India, with interests in Operator Algebras and their K-theory.

Can someone help me prove the following identity? $$ \mathop{\mathrm{Tr}}\left(\prod_{j=0}^{n-1}\begin{pmatrix} 2\cos\frac{2j\pi}{n} & -m \\ 1 & 0 \end{pmatrix}\right)= \begin{cases} 2 & \text{if } n=1\pmod{2}\\ 2m^{\frac{n}{2}} & \text{if } n=0\pmod{4}\\ -2m^{\frac{n}{2}}-4 & \text{if } n=2\pmod{4}\\ \end{cases}, $$

For the 1st equality, since the product of $2\cos\frac{2j\pi}{n}$'s is 2 when $n$ is odd, we only need to prove that the trace is a constant polynomial in $m$. However, because of the cosine term, the approach of polynomial analysis given in my previous post does not seem to work here.

Situations are the same for the 2nd and 3rd equalities.

391319I don't really see how to get there from just compact groups, so in that sense this is not an answer. My take on the question is something like: how might one have guessed the existence of canonical basis theories*, knowing only the Lie theory of, say, 1975? (The type A case was well under way by then, motivating standard monomial theory.)

The standard way to make *the* canonical basis uses quantum groups, and uses $q$ in a fundamental way. The closest we can get to a quantum group in the classical theory is the semiclassical limit, the Poisson structure on the Lie group. (Still too recent.) One question that becomes meaningful now is to ask what subgroups are Poisson subgroups. There are very few: I'll focus on the Borel $B$ (in particular, this is about complex groups not compact), whose importance was well-appreciated by then.

Now we can go beyond representation theory of $G$, i.e. complete linear systems over $G/B$, to talk about Demazure modules, complete linear systems over Schubert varieties $\overline{BwB}/B$, and look at the surjections $H^0(G/B; \mathcal L) \twoheadrightarrow H^0(\overline{BwB}/B;\mathcal L)$. In particular, one can ask for a basis for the irrep such that each of these kernels is spanned by a subset of the basis. Moreover, you can filter your basis of the kernel according to the order of vanishing along Schubert varieties, which suggests a means of indexing the basis. This line of thinking leads to the Lakshmibai-Seshadri conjecture (about characters, but motivated by the putative existence of such a basis).

*There's a problem with this question in that there are several bases, at the very least the canonical basis and the (different) semicanonical basis, the latter of which may or may not be the MV basis you asked about. Most of the striking properties you ask about hold for all of these. As such it's hard to find overlooked smoking guns that would pick out e.g. the MV basis over the others.

193683116548226132jealcalat197317315964561674212433648246186dSecond order difference implies differentiability2034111h@Sasha at al., my construction can be still useful.201527441199 5453Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon1953994Although I think Hardy's book might be a bit too difficult for me at the moment...856251666597https://www.gravatar.com/avatar/3b77488573659e7331e9ce0311c3e305?s=128&d=identicon&r=PG&f=1It feels like I've seen the title "Estimating a sum involving binomial coefficients" on six other questions on this forum alone. In any case, break the trend! You are allowed to add things like ${\binom{n-k}{k}}^2$ to the title.38314126041111413311184341185906904400As well-known in toric geometry, we can define a toric divisor $D_i$ with respect to every face $F_i$ of the polytope. And it is well know that $-K_M=\sum D_i$. My question is that if we denote $D_i$ to be the toric divisor with respect to the face $F_i$, then is $-D_i$ also a toric divisor with respect to $F_i$?

The question comes from a represent of $-K_M$ while $M$ is the blow-up of $\mathbb{C}P^2$ at one point. We know that $-K_M=3H-E$, where $H$ is the hyperplane line bundle and $E$ is the exceptional divisor. So three of $D_i$s are $H$ and the other is $-E$. Are there some mistake?

20806561665476Exact ramification information of mod $p$ Galois representation223334537721137213353692I'll bet he means a subgroup of the ultraproduct; the square brackets in the definition surely indicate an equivalence class?@NeilStrickland I am afraid that the statement is incorrect, as one can see, for example, in the existence of non-orientable homogenous spaces. The problem in your argument is that G might act non trivially on LG hence the action might be orientation preserving on TM but not on V. The identification of U with LG is not canonical as well, since it depends on the choice of x.4168971309746PProperties of the function $\chi_{s,k}$2239959@GerryMyerson For (a) one doesn't have to be so fancy, but it could be the ring of bounded linear maps taking $l^2$ to itself, with multiplication given by composition. Here $y$ would be given by $T$, and $x$ by $e_0 \mapsto e_0$ and $e_{i+1} \mapsto e_i$.1729832The first functor inverts W so you get a (weak) retraction for the second one. This shows essential surjectivity.588719673704Also, the word "Scienceographic" is an abomination and whoever came up with it should be ashamed.Can you do time complexity? I would love to see the Time Hierarchy Theorem formalized (especially since the proof in the most well known textbook is wrong...). @TheMaskedAvenger You are of course right; my first comment above was unfortunately a case of "shoot first, ask questions later". The relation to Wilson was supposed to be (well, at least under the proviso that $p!$ is prime to $2p+1$) that $(p!)^2 \equiv (-1)^{p-1}$ modulo $2p+1$. (There's a nice pictorial proof of that based on the graph representing the reciprocation relation on $\{2, \ldots, p\} + \{-2, \ldots, -p\}$ modulo $2p+1$.) But of course that proviso is true only if $2p+1$ is prime, leading to the same *petitio principii* objection I had raised in response to GH's argument.This is exactly the type of answer I was hoping for. Thank you!1600902https://lh5.googleusercontent.com/-IOPY7NA7Igk/AAAAAAAAAAI/AAAAAAAAAAA/AAN31DVWkSBCPqQsCuR9AGOK-5jcASpJXg/mo/photo.jpgSuppose that $\lambda_1,\lambda_2,\lambda_3$ are partititions of $n$. When do there exist permutations $\sigma_1,\sigma_2,\sigma_3 \in S_n$ such that

(1) $\sigma_1\sigma_2\sigma_3$ is the identity;

(2) the $\sigma_i$ generate a transitive subgroup of $S_n$; and

(3) the cycle type of $\sigma_i$ is $\lambda_i$ for all $i$?

One can show that the total number of parts among the $\lambda_i$ must be an integer that is at most $n+2$ and that is congruent to $n$ modulo 2. Is this condition sufficient? If $\lambda_3 = (n)$ and the total number of parts of the $\lambda_i$ is exactly $n+2$, then the existence is known.

This problem is equivalent to the existence of an irreducible curve $X$ over the complex numbers and a nonconstant morphism $X \rightarrow \mathbb{P}^1$ whose branch locus is contained in $\{0,1,\infty\}$ (a Belyi map) and whose monodromy generators have cycle types $\lambda_1,\lambda_2,\lambda_3$.

@iPe : Why do you think this heavily depends on the geometry? I have now replaced the former ball $C$ by an appropriate compact set, which always exists. In the comment now appended at the end of my answer, it is explained that all we need are conditions (1), (3), (4), and it is also explained that for $B$ we can take a small enough symmetric neighborhood $B$ of $1_G$. Can you give a good example of a continuous locally compact group where some of the conditions (1), (3), (4) fail(s) to hold?1664511applied_math260063^Finite lattice representation problem checking"For the entire question, the interval I am integrating over is $[0,1]$.

Background: In order to exhibit an isometry from $L^2[0,1]$ into $l^2$, I need to either assume absolute continuity over some interval for a whole family of functions due to the technicality of integrating a derivative using the Lebesgue integral, or I can use the Gauge integral.

I have very little experience with the Gauge integral, but if what I understand is correct, I can integrate the derivative of a function which exists at all but countably many points, and it behaves just like the Fundamental Theorem of Calculus says. However, I need to integrate $G_n'(x)$, where $G_n$ are $L^{1}[0,1]$ functions, which Rudin says in Thm. 7.14 of "Real and Complex Analysis" exists only almost everywhere.

Almost everywhere $\ne$ countable unfortunately, so my question is whether one of these two statements is correct:

i) Because I am working on a fixed bounded interval rather than $R^1$, I can assume that $G_n'(x)$ exists at all but countably many places (unlikely, and I believe shown false by the "distance to the cantor set" function)

ii) The Gauge integral of $G_n'(x)$ behaves as expected even over uncountable sets of measure 0.

If both of these are false, then my question is what is the least stringent condition I can impose on $G_n$ to ensure that the integration of the derivative works as expected. The actual functions $G_n$ are defined recursively, and even "absolute continuity" provides a condition too difficult to describe for the input function $G_0$.

Thank you very much MathOverflowers,

Hunter Spink

Edit: After a good night's sleep, I realize that both i) and ii) are rendered moot by functions like the cantor staircase (and it doesn't even make sense to ask about countably many discontinuities if working in $L^p$ spaces because functions are equivalent mod sets of measure 0). However, I am still interested in what conditions one needs to impose on a function to be able to recover itself just from its derivative (which I thought was absolute continuity everywhere except a set of measure 0 until I learned about the Gauge integral).

|https://graph.facebook.com/956595697841477/picture?type=large276267659711hWhat are the reference(s) that says (b) is "exact"?677624$TheoremOfClifford1374382Fine work on the World Cup. http://www1.voanews.com/english/news/sports/Psychic-Octopus-Predicts-Spain-World-Cup-Win-98120159.htmlI assume that you are everywhere working up to multiplication by a power of $2$ (or else the result is trivially false if $p = 2$ for example) and assuming that your scale has some nontrivial interval ratio $n$ in it which is not a power of $2$. Then this is straightforward.

**Lemma:** A subgroup of $\mathbb{R}$ is dense if and only if it contains arbitrarily small elements.

*Proof.* The integer multiples of a small element $r$ are at most $|r|$ apart from each other.

We apply the lemma as follows. The subgroup generated by $1$ and $\log_2 n$ contains arbitrarily small elements because $\log_2 n$ is irrational by prime factorization, so it is dense in $\mathbb{R}$. Given $\alpha \in (0, 1]$, it then follows that we can find a sequence $a_k, b_k$ of integers such that

$$a_k + b_k \log_2 n \to \log_2 \alpha$$

hence such that

$$2^{a_k} n^{b_k} \to \alpha$$

and the conclusion follows.

21413817684211784205142932\Phd Student of Computer Science at VT

16238562963180931119@Kumar: What for? Why do you care about the native language of the book author?56447 In comments to the question @KevinCasto pointed out that for $M$ simply connected and $f$ with the property that all $f^k$ are pairwise non-homotopic we would obtain an example of $M$ with non-co-Hopfian fundamental group ($\mathbb Z$ in this case) that cannot cover itself nontrivially. However I could not find a rigorous argument for the reverse of your statement; more precisely, whether $M_f$ and $M_{f^k}$ are not homotopy equivalent if $f$ and $f^k$ are not homotopic. It might be not difficult to show this for some specifically chosen example, but is it true in general?The entries of $\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a'&b'\\c'&d'\end{bmatrix}=\begin{bmatrix}aa'+bc'&ab'+bd'\\ca'+dc'&cb'+dd'\end{bmatrix}$ are curiously given by the entries of the composition of rational functions $\frac{ar+b}{cr+d}$ and $\frac{a'r+b'}{c'r+d'}$ which yields $\frac{(a a' + b c')r + a b' + b d'}{(c a' + d c')r + c b' + d d'}$.

Does this have a generalization to $n\times n$ matrix multiplication?

1368375396614Nesting big-O with big-Omega $O(g(\Omega(h(n))))$: is it $O$ for all $\Omega$ or for one $\Omega$?5003722744775136641634103A continuous function is uniquely determined by its values on a dense subset, so continuous functions on Q extend uniquely to continuous functions on R. The function you wrote down isn't that unique extension.43189410665401867198823356Yes, there are lots of ways to be more concrete. Another would be to look at congruence covers.dSomeone, somewhere, looking at the sea...

66685918916261970804795326That is quite a complicated question and I have no time to think deeply enough about it. But just let me mention a more modern and more complete monograph: S. Benzoni-Gavage and D.S., *Multi-dimensional hyperbolic PDEs*, Oxford Univ. Press (2007).2577995Consider the elliptic equation $-\Delta u=\alpha f(u)$ in $R^n$ and assume that for some $\alpha^* \in R$ it has a bounded smooth solution $u^*$ in $R^n$ ($f$ is a nice smooth function). Under what conditoions one can guarantee that for $\alpha$ close to $\alpha^*$ this equation has a solution $u$ close to $u^*$?

Thank you for your answer and comments. when I read your writing, it implies that the analogous question for Maass forms is true.. Right?"This paper is worth publication, provided the authors add some words of explanation on their construction."4791242009390163297444758@Wojowu: the question in the title is my main question, although I am also curious as to whether this was the argument that was originally used.3099721186986551867620479Thank you for the helpful comment. I assumed that both $X,Y$ are smooth. The target space of the contraction have an ordinary double point. Is there a further contraction to a smooth point?146010810514982105106Thanks for the nice answer. Semisimple and simply connected is a start. Do you know any reference on this? Anyway i would be really interested why the situation in the non simply connected case is harder. Again, i would appreciate any reference. Best, Oliver.@LaurentMoret-Bailly: I don't think I follow. The OP wants to start with a purely insep. $\varphi$. In my comment $\varphi=i \circ \alpha$. Maybe I am being slow.570828Can anyone prove this statement? It seems true, but I'm finding it tricky to give a concise proof.

Fix $\alpha\in[0,1]$. Let $\mu$ be Lebesgue measure. Define $B(c,r)\equiv[c-r,c+r]$, where $[\cdot, \cdot]$ denotes an interval. For $i=1,\ldots,n$, fix $r_1,\ldots,r_n\in[0,\infty)$, and $c_1,\ldots,c_n\in\mathbb R$. Let $\bar c, \bar r, \tilde c$, and $\tilde r$ satisfy $B(\bar c, \bar r)=\bigcap_{i=1}^{n}B(c_{i},r_{i})$ and $B(\tilde c, \tilde r)=\bigcap_{i=1}^{n}B(c_{i},\alpha r_{i})$. Then, $\tilde r \le \alpha r$.

pWhy Does Induction Prove Multiplication is Commutative?314984101281lThe first eigenfunction of Dirac operator for surface1186358203685852209https://www.gravatar.com/avatar/cba1cf7251ead509e310809ffde2608e?s=128&d=identicon&r=PG&f=11884880Is there a hyperreal-valued finitely additive measure on all the subsets of [0,1), or at least the Borel ones, that

assigns $b-a$ to $[a,b)$ and to $(a,b]$ for all $a\lt b,$ and

assigns an infinitesimal--ideally, the same one--to each singleton?

It's (1) that's a problem. The Bernstein-Wattenberg construction yields a finitely-additive measure that gives (1) up to infinitesimals. But it would be nice to have (1) exactly.

8084782290890NVibration of point load on a halfspace756003human-computer interactive learning: the social and technical research which pursues building technologies to help people make decisions

2260336https://www.gravatar.com/avatar/c82d76b4ea24904ffc693dd561593242?s=128&d=identicon&r=PG&f=1356677288831262705Yeah, I was being too careless. The Amborse-Signer theorem really says that holonomy at a point is given by parallel transports of curvature from the whole connected component. So once you have a bump somewher, you get nontrivial holonomy everywhere. But in the real analytic category the local holonomy is the same as the infinitesimal one.11243177953262380112932931880357542419212685801123Tquestions on Néron-Tate canonical height8430972139895tGroup of homomorphisms with real target and circle target55203818052401725807Thanks abx, I will look for the Kobayashi-Nomizu book. Thanks Misha and Joseph.Thanks, I suppose I was too quick to assume that because these conditions look subtly different, they must actually be different!Sage has a library for this: http://www.sagemath.org/doc/thematic_tutorials/lie/branching_rules.html#branching-rules But since I don't understand much of branching rules yet, I couldn't appreciate this much.2277926mathstudent42656978717012171274440\https://i.stack.imgur.com/6rwnR.png?s=256&g=1nMathematics Ph.D. student at the University of Chicago1982135@Sebastian Thanks for your answer. but you don't defined the **connection map** and what is it **properties**.19555051321432(Let me work over an algebraically closed field $k$, let $X$ be a reduced $k$-scheme of finite type and let $G$ be a smooth connected $k$-group scheme acting on $X$, and acting trivially on an open dense subset $U$ of $X$. I will show that $G$ acts trivially on $X$.

I will only manipulate closed points. Since both $G$ and $X$ are reduced, it suffices to prove that for any $g\in G$ and $x\in X$, $gx=x$.

Let us fix $x\in X$. We may of course suppose that $x\notin U$. Let $V$ be an affine neighbourhood of $x$. By quasicompactness of $X$, the open subset $GV$ is covered by finitely many of the $gV$, say $GV=\cup_i g_iV$. Since $U$ is dense in $X$, it is possible to find a smooth curve $C$ and a morphism $C\to V$ whose image contains $x$ and intersects $U$. Up to removing points of $C$, we may suppose that $x$ is the only point of the image of $C$ not belonging to $U$.

Now, if $g\in G$, $gx\in g_i V$ for some $i$. Consider $g(C)$ (this is an abuse of notation for the composition of $C\to V$ and of the action of $g$) : its image contained in $g_iV$. It coincides generically with $g_i(C)$, hence, the morphisms $g(C)$ and $g_i(C)$ are equal by separation of $g_iV$ ($g_iV$ is even affine). This shows $gx=g_ix$. We have proved $Gx\subset ${$g_1x,\dots, g_rx$}. Since $G$ is connected, $Gx$ is also connected. This shows $Gx=${$x$} and finishes the proof.

Moreover, the hypothesis that $G$ is smooth may be removed as follows. First, we deal with the case where $X$ is separated and reduced. In this case, the arguments given by Anton in the question still work : the graphs of the action and of the projection are closed subschemes of $G\times X\times X$ isomorphic to $G\times X$ that coincide on an open dense subset $U$. Since $G\times X$ has no embedded point ($X$ is reduced and $G$ homogeneous), these two graphs need to coincide with the schematic closure of $U$ in $G\times X\times X$ and are thus equal.

Then we deal with the general case ($G$ connected, $X$ reduced) in the following way. Applying the result when the group is smooth connected and the space is reduced to $G^{red}$ acting on $X$ shows that $G^{red}$ acts trivially on $X$. In particular, $G^{red}$ stabilizes every affine open subset of $X$ ; this implies that $G$ also stabilizes every affine open subset of $X$. Applying the result when the group is connected and the space is separated and reduced to $G$ acting on any affine open subset of $X$ shows that $G$ acts trivially on any open subset of $X$, hence on $X$.

z@მამუკაჯიბლაძე: Thanks. Corrected.The idea (now well hidden I'm afraid) was really that this shouldn't hold because taking powers of functions can improve smoothness (for something like $x^{1/2}$, say), so makes it easier for the Fourier transform to be in $L^1$.On Benabou and Johnstone definition of "locally small" fibred (indexed) category198462122142951016103854439~https://graph.facebook.com/2130833303600779/picture?type=large49241215024221033261Let's denote $F_K$ the family of real-valued trigonometric polynomials corresponding to $K$, and assume that *$K$ has a point in the interior of its convex envelope.* Then, there is a function $f$ in $F_K$ for which $\{f\ge 0\}$ has a bounded component.

To show this we can freely apply a linear transformation to $K$, for $F_{LK}=\{f\circ L^T\, :\, f\in F_K \}$. In particular we can assume that $K$ includes the standard basis $ \{ e_1,\dots, e_n\}$, and there is in $K$ one more point $y $ with $y_j\ge0$ and $\|y\|_1:=\sum_{j=1}^n y_j <1$. Consider a trigonometric polynomial $$f(x)= \sum_{j=1}^n \lambda_j \cos(x_j) -\cos(y\cdot x)\, .$$ It belongs to $F_K$ and has a second-order expansion at $0$ $$f(x)= \sum_{j=1}^n\lambda_j - 1 - \frac{1}{2}\sum_{j=1}^n \lambda_j x_j^2 + \frac{1}{2}(y\cdot x)^2+o(\|x\|^2)$$ $$\le \Big(\sum_{j=1}^n\lambda_j - 1\Big) -\frac{1}{2}\sum_{j=1}^n (\lambda_j -\|y\|_1y_j)x_j^2 +o(\|x\|^2) $$ because by Cauchy-Schwarz, $(y\cdot x)^2 = \big( \sum_{j=1}^n y_j ^{1/2} y_j ^{1/2} x_j\big)^2\le \|y\|_1\sum_{j=1}^n y_j x_j^2 $.

We can now take e.g. $\lambda_j= \|y\|_1y_j +\frac{1}{ n}(1-\|y\|_1^2+\epsilon)$ with $\epsilon>0$ so that $f(0)=\epsilon$ and $f(x)\le\epsilon-\frac{1}{2n}(1-\|y\|_1^2)\|x\|^2+o(\|x\|^2)$ (unif. on $\epsilon$). So for $\epsilon$ small enough $f(x)<0$ on the boundary of a ball around $0$, meaning that the connected component of $0$ in $\{f\ge0\}$ is contained in the ball.

Not an answer, but if you want to look for different solutions in the future in the same vein as these you posed, this might be a general method.

For an alternate representation for $$:

$$\frac{1}{a-b}=\frac{1}{a}+\frac{b}{a^2}+\frac{a^2}{b^3}+\frac{b^3}{a^4}+\cdots\qquad(1)$$

Plugging in $a=(2k)^{5}$ and $b=(2k)^3$

$$\frac{1}{(2k)^3}+\frac{(2k)^3}{(2k)^{10}}+\frac{(2k)^6}{(2k)^{15}}+\cdots=\sum_{n\geqslant 1}\frac{1}{(2k)^{2n+1}}$$

And thus your sum can be evaluated as

$$\sum_{k\geqslant 1}\sum_{n\geqslant 1}\frac{1}{(2k)^{2n+1}}$$

I am not entirely convinced if this was helpful to you. I am sorry if it was not.

$(1)$ Reason

Thank you very much! This uniqueness of the extension is exactly what I needed. And now I also agree that there isn't "explicit" description. 17860091494153Primes dividing functions defined by linear recurrence relations with constant coefficients27710519201371072462That could happen from one very low probability outlier which is far above the mean for both random variables, so you still couldn't say much of anything. You are pushing on a rope. You need some other tools than the linear regression or correlation to say anything meaningful about either the density function or confidence regions.869697The OP actually specifies that both matrices are indefinite, but your example is easily modified to meet that.14140871230122Consider $x=(x_0,x_1,\dots)\in \ell^2$ as a function $F_x:=\sum_{k=0}^{\infty} x_k z^k\in L^2({\mathbb S}^1)$ (this is norm-preserving). Then shift operator multiples functions by $z$. If $x$ is so that $F_x(a)=0$ for some complex $a$, $|a|<1$, then also $a^k F_x(a)=F_{S^k x}(a)=0$, and since $$ f(z)\rightarrow f(a)=\frac1{2\pi i}\int \frac{f(z)dz}{z-a} $$ is a continuous functional on $L^2$ restricted to our space (well, not just our, but Hardy space), we see that orbit of $x$ under $S$ lies in a proper closed subspace. In Nik Weaver's example function is $1-2z$ and $a=1/2$.

1973831There is certainly no satisfactory answer to the question, one reason being that $S_{2n}$ has many solvable transitive subgroups, as pointed out by Walter Neff in the comments.

Another reason is that if there were a manageable criterion, then we could handle polynomials $g(x)$ of odd degree as well by considering $f(x)=g(x^2-a)$ with $a\in\mathbb Q$ such that $f(x)$ is irreducible. (The existence of such an element $a$ isn't hard to establish.)

672621Quantum invariants: the Jones polynomial started the field of quantum topology. See [Ed Witten's lecture](https://www.ias.edu/ideas/2011/witten-knots-quantum-theory)1934101fI think I can prove what I claimed in a few pages.2140128233147I defined / studied this function in a recent preprint (for computational purposes). The derivation is standard convex analysis, though slightly tedious to do explicitly.10204434701251614743266060123839818606372194204@Wolfgang I ve tried some examples. So far $N \ge \frac{n}{2}$ was sufficient.1085323342746599655LIsoperimetry and Poincaré InequalityI only ask for the subset of Reshetikhine-Turaev invariants for now.

Is the following sum up into cases a) and b) correct?

- Dots in Dynkin diagrams of Lie groups G come as a) symmetry-equivalent pairs
(most of $G=A_n$, some of $D_n$ and $E_6$), b) not.

- The irreps $R$ belonging to them are a) not self-adjunct b) self-adjunct.

- $R\otimes{R}$ contains a) not 1 b) 1

- The knot polynomial belonging to $R$ is defined for oriented knots anyway, and a) it doesn't work with unoriented ones b) you can as well drop the arrows.

At least this would explain why I can't find a R matrix for $A_2$(dim 3 irrep). (Or more precise, an R matrix for unoriented knots. I have exactly one R matrix that suggests itself, but it only works for oriented knots.)

182701118738481118937https://lh4.googleusercontent.com/-y1Zoij0OVJM/AAAAAAAAAAI/AAAAAAAAELI/bD05eNv1R50/photo.jpg?sz=1281009750765093bOn average length of sums of unit vectors in R^n2244769f${\mathbb R}(x)\not\subset C^\infty({\mathbb R})$.17528771647631054442533604148618018331232019670331644I never really thought of the Ackermann function as being *strange*, only *big*. But maybe that's just me.1082405278657655866I am not completely sure what the best tags are for this question. Feel free to add tags or retag as appropriate.This kind of issue is handled carefully in mathematical treatments of lambda calculus (which is just the abstract version of Lisp). The Wikipedia page on lambda calculus has some discussion of this under 'capture-avoiding substitutions'. I've not worked through it in detail but I don't think that anything very subtle comes up, you just need a rigorous formulation of what it means for a variable to be bound and the sense in which bound variables can harmlessly be renamed.

1144058769378X@SamHopkins: You just went ahead of me. :-)217355Yes, you are right. All that matters is that $\omega_1^{CK}$ itself is (countable and) indecomposable.Swift16047072980791782036Thank you all! Could you give a clue how to calculate the $map$, for example, $map(R^{0|d},M)$ for any supermanifold $M$?@Qiaochu: Yes, the matrix is not supposed to be symmetric: see JCollins' answer below. He didn't say that it was, but the language that he used might suggest that on a first reading. So I clarified it.180854743523For the equivalence of definitions of quasiconformal maps the reference is J. Heinonen, Lectures on analysis on metric spaces, Springer 2001. Notice that the $K$ in the definiton you cite is not the same $K$ as in the Ahlfors definitions. So your definition of quasiconformality is equivalent to the usual one, but with a different $K$. Another reference is Lehto and Virtanen, Quasiconformal mappings in the plane, Springer 1973. The equivalence of all these definitions is a non-trivial fact.

17362371091539Not quite: in characteristic $0$, $\mathbb{P}^1$ and $\mathbb{A}^1$ are both simply connected. \https://i.stack.imgur.com/Qq9Vb.jpg?s=128&g=1Hi guys, this was based on a false assumption. The proof I had in mind when writing that only works when the diagonal subsheaf $\Delta \to X \times X $ was the zero section of a vector bundle and that one could build a Koszul resolution for $\Delta$. I later realized this is wildly false. The fact about char(p)> dim(X) should be because when writing the HKR map to differential operators you need to divide by n! for n$0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$.

Of course $V_1$ and $V_3$ are nuclear if $V_2$ is. I recently asked myself if the converse might be true. I haven't found anything useful in the standard literature (Treves, Schaefer) but that might be just me being too ignorant to see the obvious. I'm grateful if someone could shed some light on this.

Cheers,

Ralf

1869221139843RAVI D PARIKHtLennart: Any place where this approach has been detailed?1827791fyes it does. Updated the question to reflect that.I'm not sure I understand your question : you want to extract a generating set from a big set of points? What if your big set of point is of the form P, 2P, 3P, 4P, etc?14583391184970Thanks for asking this question! I'm curious to read responses to this question. This is slightly tangential, but a (more general?) question that I have is whether higher-dimensional formal groups can influence homotopy theory in some way. For instance, the Gross-Hopkins period map is a special case of the crystalline period map (section 6 of the Scholze-Weinstein paper). The former has had enormous success in chromotopy --- can something in chromotopy be deduced from the latter?821287Which is a simultaneous generalization of the integer-valued functions "Euler characteristic" and "count points over finite fields"!75368411402391052327707947390340There's an article on the Kakeya problem that alludes to his work linking harmonic analysis and combinatorial geometry. http://www.ams.org/bull/2008-45-01/S0273-0979-07-01189-5/S0273-0979-07-01189-5.pdf1842879If $H$ is commutative and unit-cancellative, then so is the monoid of non-empty ideals of $H$1291471tAn upper bound on the Jordan condition number of a matrixLet me start by a very simple example; consider the following question:

"Let D1 be a square and D2 a rectangle (boundary included). View them as subsets of the complex plane. Does there exist a conformal map (extending to the boundary) taking D1 to D2?"

Of course, the answer is no, but I want to point out an unusual "proof" of this assertion. Suppose the answer was yes. I think we can assume that the center gets mapped to the center. Start a brownian motion from the center of D1. The probability that this brownian motion hits any of the four sides is equal. However the probability that a brownian motion hits any of the four sides starting from the center of of D2 is not equal. And this is a contradiction, because brownian motion is conformally invariant (which is a non trivial fact, but its true).

I believe this "proof" can be made rigorous. My question is the following:

Can this same idea be used to show for instance two complex manifolds are not biholomorphic to each other? Of course there maybe a simpler proof using more direct methods, but I am still curious to know if the idea of using brownian motion can be used to answer such a question (ie are two manifolds conformally equivalent).

2027166Like Stefan mentions, under AD every ultrafilter is $\omega_1$-complete. Then the proof that the c.u.b filter on $\omega_1$ is an ultrafilter proof goes through Solovay's game where players play codes for well-orderings. The players choose countable ordinals $\alpha_i$ for $i < \omega$. Player I wins if $sup${$\alpha_i:i<\omega$} $\in Y \subset \omega_1$ for some $Y \subset \omega_1$ over which the game is played.

https://www.gravatar.com/avatar/cfc42381f882128e1bcf692a9fed697c?s=128&d=identicon&r=PG&f=11274256363842If the initial integral $I$ is well defined it means there is absolute convergence, no ? Then when interchanging the integration the integral is not absolute convergent... Meaning one manipulation or one asumption is wrong...https://lh4.googleusercontent.com/-eDO_gCniT0g/AAAAAAAAAAI/AAAAAAAAAAA/APUIFaMKBA7ewak26rmBKaA2l6UniWw4vA/mo/photo.jpgA little while ago, I was reading Cathy O'Neil's post Why is math research important (subtext: why does Pure Math deserve funding), where she discusses 3 possible answers. One of these is the usual "because there may be applications some day." Of course, this is not too compelling, or an explanation of why people do Pure Math (nor does she claim it is).

I wondered if one can make a compelling argument that Pure Math is important because it binds all of Science together---it is, in some sense, the *most* interdisciplinary area of science because it is an exploration of ideas from all parts of Science and how they interact with each other. That is, Mathematics is not just the common language of Science, but a sort of playground where different Science ideas meet together, and sometimes (frequently?) copulate. If this is true, we should have a lot good examples where a question in Science A was turned into a Pure Math problem, and was solved based on idea from Science B. I'm not an expert on history, or applications, so I have no idea if there are many such examples, but this is my question.

What are examples of important problems in Science, which led to Pure Math research questions that were later solved by using an idea from a different area of science, which in turn resulted in a solution to or significant progress towards the original science problem?

Let me emphasize that I'm *not* looking for examples of solutions to pure math problems, which serendipitously found some important application later. Rather these should be pure math problems that originally studied because they were very clearly motivated from a "practical" question from field X. Preferably in situations where it is reasonable to posit that the other part(s) of science which led to the key insight(s) in solving this problem had little-to-no direct contact with field X, and Pure Math really was the essential conduit.

In many places the existence of automorphism is acknowledged as one of the reasons why fine moduli spaces cannot exist. A typical example is the following.

Consider a curve $C$ with a nontrivial automorphism, for instance a hyperelliptic curve with its involution $\phi$. Now let $B$ be any scheme with a free action of (in this case) $\mathbb{Z}/(2)$.

Let $D$ be the quotient of $B \times C$ by the diagonal action of $\mathbb{Z}/(2)$ and let $B'$ be the quotient of $B$ by the action of $\mathbb{Z}/(2)$.

Consider the map $f \colon D \rightarrow B'$. Then the fibers of $f$ are all isomorphic to $C$, but for a suitable choice of $B$ the family $f$ is not isomorphic to the product $C \times B'$.

How can I get an explicit example of the last assertion?

The number of orderings of elements with order-of-appearance constraints605399429665JExplicit families of elliptic curves13133171512060dThese comments don't really address the question.18212421128901321085Maybe you could include a sentence or two explaining *why* it is trivial? That way you can help your readers reconstruct the proof, without appearing patronizing.How can I produce $B$ as above such that $f$ is not isomorphic to the trivial family?

It is established in this post that you there is no computable model of ZFC, yet it can be computed in by a PA-degree oracle machine. Note that when we see "compute a model", we just mean that membership is decidable. My question is what Turing degree do you need if you want more than that?

In particular, we require the following:

- A computable set $S$, representing the sets.
- A decidable relation $\in_S$, such that $(S,\in_s)$ is a model for ZFC.
- That $=_s$ is computable (i.e. equality between sets in the model)
- Given a nonempty set $x$, we can compute a disjoint set $y$ such that $y \in_s x$ (axiom of regularity)
- For any formula $\phi(x)$ and given a set $x$, we may compute the set $\{x \in_s z : \phi(x)\}$ (axiom schema of specification)
- Given $x$ and $y$, we can compute $\{x,y\}$ (axiom of pairing)
- Given $x$, we can compute $\bigcup x$ (axiom of union)
- Given $x$, we can compute $P(x)$ (axiom of powerset)
- Given $x$, we can compute a well order for $x$ (Well-ordering theorem)

Note that we do need to worry about the axiom of infinity, or any of the axioms of replacement, since they assert that a certain set exists, and we can just "hard code" those constants as output.

Of course, this is based only just one axiomatization of ZFC. I could have asked about others, or I could just ask that for any provable statement $\exists x:\phi(x,y)$, given a $y$ we can compute a $x$ but this is just about getting a feel for what Turing degrees we will need for some axiomatization of ZFC.

What Turing degree is needed to compute all of the above. (My best guess is that any PA degree would be sufficient, since there is a PA degree computing a model of Morse-Kelly set theory, but I'm not sure.)

02012-07-05T08:24:57.280Here is a possible counterexample to the conjecture above: The real polynomial x^2 + (xy-1)^2 is bounded below but does not achieve its lower bound 0, which I think implies that it is not conjugate to a piecewise linear function of finite type. 724459Let $G$ be a graph with disjoint copies of $K_{1,3}$. Prove that if there are uncountably many copies of $K_{1,3}$ in $G$, then $G$ is not planar.

I have a proof of this statement by contradiction i.e. assuming it is planar with uncountably many copies of $K_{1,3}$ in $G$, but I am not satisfied. I am wondering if anyone has a proof that proves the contrapositive i.e. if $G$ is planar, then there is only countably many copies of $K_{1,3}$ in $G$.

My idea is to find an embedding of $G$ onto $\mathbb{R}^2$ (or $\mathbb{C}$), and use the fact that $K_{1,3}$ has $3$ edges, and hence if we find disjoint open sets which contain each $K_{1,3}$, they will have positive measure.

Thanks a lot!

583234 7509:@IgorBelegradek Good idea!:)573457525441689713Another possible approach is first read a paper or book about math you already know.I do not quite understand exactly what are you asking here. The question you refer to does not relate the sum $\sum_a r(a)$ to the size of the intersection $2A\cap A=S_1\cap A$. Exactly what line in my solution is unclear?12551508822205159316771991966144108814415157871696560your equation contains the Black-Scholes equation as a special case (if $b=-r$). the B-S equation can be solved explicitly*. why don't you test your conjecture on that special formula? *well, at least for *homogeneous* Dirichlet boundary conditions, whereas yours are inhomogeneous; but there are of course standard ways to derive a solution of the inhomogeneous problem from the solution of the homogeneous one.Say you have a surface of type $(g',\nu')$ and you put one or more disjoint closed loops on it such that when you cut along them, you cut your surface into two pieces of types $(g_1,\nu_1)$ and $(g_2,\nu_2)$. Then the boundary component of $M_{g',\nu'}$ corresponding to curves with the disjoint closed loops you chose shrunk to nodes is isomorphic to the product of moduli spaces $M_{g1,\nu_1}\otimes M_{g_2,\nu_2}$. And more generally this holds for any number of pieces that you cut your original surfaces into.$Yes, this is a field norm; it is the norm of $a + b \sqrt{3} + c \sqrt{5} + d \sqrt{15}$, from $K = \mathbb{Q}(\sqrt{3}, \sqrt{5})$ down to $\mathbb{Q}$. Note that $a+b \sqrt{3} + c \sqrt{5} + d \sqrt{15}$ acts on the basis $(1, \sqrt{3}, \sqrt{5}, \sqrt{15})$ by $$a \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} + b \begin{pmatrix} 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 \\ 0 & 0 & 1 & 0 \end{pmatrix} + c \begin{pmatrix} 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 & 0 & 15 \\ 0 & 0 & 5 & 0 \\ 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}.$$ Now conjugate by the matrix whose diagonal entries are $(1, \sqrt{3}, \sqrt{5}, \sqrt{15})$ to get your matrix. The entries are no longer rational, so I can't think of the result as describing the action on $K$, but the determinant is the same.

$\mathbb{Q}(\sqrt{15})$ has class number $2$ and $K$ is the class field. So, for a prime $p$ other than $2$, $3$, $5$, we have that $\pm p$ is a value of $x^2-15 y^2$ if and only if $p$ splits principally in $\mathbb{Q}(\sqrt{15})$ if and only if $p$ splits in $K$ if and only if $\pm p$ is a value of $f$. Also, neither $x^2-15 y^2$ nor $f$ can be $3 \bmod 4$, so the sign is the same in the two cases.

However, they don't take the same set of composite values. Look at $-119 = 7 \times 17$. We have $61^2 - 15 \cdot 16^2 = -119$, but, if $7 | f(a,b,c,d)$ then $7^2 | f(a,b,c,d)$.

I found this by hunting for two primes which are non-principally split in $\mathbb{Q}(\sqrt{15})$. In terms of quadratic forms, which I know you love, I needed primes of the form $3 x^2 - 5 y^2$, and I found $7=3 \cdot 2^2 - 5$ and $-17 = 3 \cdot 6^2 - 5 \cdot 5^2$. Then their product was of the form $x^2-15 y^2$.

Since these primes split non principally in $\mathbb{Q}(\sqrt{15})$, they don't split further in the class field. (We can also directly compute $\left( \frac{3}{7} \right) = \left( \frac{3}{17} \right) = -1$.) So things divisible by one power of $7$ or $17$ are not norms from $K$.

Recently I am considering a geometric question, which is reduced to the following problem.

Given $L<0$, let $a\in [L/2,0]$ and $b=L-a$. For any $c>0$, let $p,1-p$ solve $$x^2-x+c^2=0,$$ and $q,1-q$ solve $$x^2-x-c^2=0.$$ Using Gauss hypergeometric functions, we define \begin{align*} f_a(c)&=(a^2-a)\left(\frac{F(1+p,2-p;2;a)}{F(p,1-p;1;a)}+\frac{F(1+q,2-q;2;a)}{F(q,1-q;1;a)}\right)\\ &+(b^2-b)\left(\frac{F(1+p,2-p;2;b)}{F(p,1-p;1;b)}+\frac{F(1+q,2-q;2;b)}{F(q,1-q;1;b)}\right). \end{align*} Let $c_0>0$ be the first positive zero of $f_a(c)$. So $c_0$ depends on $a$, i.e. $c_0=c_0(a)$. To solve the geometric question I am considering, we need to prove $$\min_{a\in [L/2,0]} c_0(a)=c_0(0).$$ I am a beginner to learn Gauss hypergeometric function. I suppose I need to understand those Gauss hypergeomertic functions appearing in $f_a(c)$, including their monotonicity, zeros, etc. Using Matlab I can get some rough pictures. But that is not enough for a rigorous proof. Could anyone recommend some text books concerning the problem above?

lPath components of a monoidal category form a monoid?399629792553I've always wondered which descriptions of the Catalan numbers are actually isomorphic as species. That would be fun to figure out, maybe.372278n@Adam: write down a long exact sequence of cohomology.2191721110602994182ANRs are a key tool in this paper in geometric group theory: http://www.ams.org/journals/jams/1991-04-03/S0894-0347-1991-1096169-1/home.html6193981041397163574https://lh3.googleusercontent.com/-7AFmjz5EJRk/AAAAAAAAAAI/AAAAAAAAACY/zy1Dw1sseGc/photo.jpg`Homotopy colimits/limits using model categories256198Read chapter 1 of Goldfeld and Hundley. Anyway, the $p$-adic component of a Bruhat-Schwartz function is a locally constant compactly supported function, and is the indicator function of $\mathcal{O}_v$ for all but finitely many nonarchimedean places $v$.175348715708811111935That was silly of me! I corrected the original posting. Thanks for the tip.6171131077534In class field theory, the Takagi isomorphism says only that the abelian Galois group is isomorphic to the associated ideal class group. Keeping track of the isomorphism (the Artin reciprocity law) says that the L functions match, hence have analytic properties,... leading to the Langlands program, ...\https://i.stack.imgur.com/m8BWa.png?s=128&g=11395532BREAKING NEWS: a plane got blown up by a mathematician. At least five cups of coffee are thought to be dead.1839104829858466629210966811502142177629The theoretical implication is that this is really cool, and understanding why it is true is likely to require understanding something deep.For `$k < 2^32$` we have $|K_1| = 48717, |K_2| = 1657, |K| = 50374$. These were all verified to be unique, so Andrej's conjecture is confirmed for `$k < 2^32$`, given that we accept that the enumeration process finds all $k$. 10229198594692182704758411I think all that's happening here is that you're filtering your $R$ by powers of an ideal $I$. No? In which case $R$ is Noetherian so $gr_I\ R$ is too. Perhaps somebody else can say more about the relation to Pólya's Positivstellensatz and on whether the conjecture is true.>Let me give a few examples.

**Example 1.** Let us work in Gödel-Bernays set theory, and
assume that $T\subset {}^{<\text{Ord}}2$ is a proper class tree of
height Ord, but there is no cofinal branch.

(This theory is consistent relative to an inaccessible cardinal, because if $\kappa$ is inaccessible and not weakly compact, then there is a $\kappa$-Aronszajn tree $T\subset {}^{<\kappa}2$, and then $V_\kappa$ with all subsets is a model of GBC where $T$ has the desired property.)

In the logic $L_{\infty,\omega}$, which allows arbitrary sized conjunctions and disjunctions, with a constant for every element of $T$ and a unary predicate symbol $B$, consider the theory $P$ consisting of the assertions $\varphi_\alpha$ asserting first, that there is precisely one object $u$ on level $\alpha$ of the tree that satisfies $B$, and secondly, that in this case, every $v<_T u$ also has $B(v)$. These assertions can be made in the logic $L_{\infty,\omega}$ using constants for the elements of $T$. Thus, altogether, $P$ is the theory asserting that $B$ is a cofinal branch through the tree.

Every set-sized subtheory of $P$ mentions only a bounded number of levels, and so we can find a model by picking any node above that bound and using the predecessors of that node as the instantiation of $B$.

But under our assumptions that the tree $T$ is Ord-Aronszajn, there can be no model of all of $P$ or even of a proper class sized subtheory of $P$, because any such subtheory will involve the assertions concerning unboundedly many levels of $T$, and so the model of that subtheory will pick out a cofinal branch in $T$; but there is no such branch.

Meanwhile, there is a strong connection between your property and (non)weak compactness, because an inaccessible cardinal $\kappa$ is weakly compact just in case we have the $\kappa$-compactness property for $L_{\kappa,\kappa}$ theories of size $\kappa$. (And there are diverse variations on this.)

**Example 2.** Here is a different kind of related example using
only first-order logic.

**Theorem.** There is a proper class first-order theory $P$, such
that every set-sized subtheory of $P$ has a model, but no class is
a model of the whole of $P$.

**Proof.** We interpret this as a theorem scheme in ZFC, where by
"class" we mean a definable class (allowing parameters). Thus, I
shall provide a definition of a theory $P$, and then prove first,
that every set-sized subtheory of $P$ is satisfiable, and second,
that no definable class is a model of $P$.

Let $P$ be the theory in the language of set theory $\in$ augmented with a predicate $\newcommand\Tr{\text{Tr}}\Tr$, meant to serve as a truth-predicate, plus a constant for every object in the universe. The theory $P$ asserts that $\Tr$ obeys all instances of the recursive Tarskian truth definition:

- $\Tr(a\in b)$ just in case $a\in b$ holds.
- $\Tr(a=b)$ just in case $a=b$.
- $\Tr(\varphi\wedge\psi)$ just in case $\Tr(\varphi)$ and $\Tr(\psi)$.
- $\Tr(\neg\varphi)$ just in case $\Tr(\varphi)$ does not hold.
- $\Tr(\exists x\ \varphi)$ just in case there is $a$ such that $\Tr(\varphi(a))$.

For any set many such assertions, we can find a model, since we can find some $V_\theta$ large enough to contain all the parameters mentioned in the subtheory, and then use truth in $\langle V_\theta,\in\rangle$, which will satisfy all the assertions made in the subtheory.

But no definable class can satisfy $P$, because this is exactly
the content of Tarski's theorem on the non-definability of truth.
**QED**

Meanwhile, this theory $P$ of the theorem does has proper-class sized subtheories that are satisfiable, since we could, for example, restrict to the quantifier-free assertions; since there are so many constants, we can produce proper class trivially satisfiable subtheories.

1274320217468217739342226564 sztsbProbabilistic models and related issues.

7975581185768*TheoremOfSchoenflies$\def\F{\mathbb F}$ $\def\Z{\mathbb Z}$

One reason that Chevalley-Warning theorem is that amazingly useful is the fact that for a finite field $\F$, any function from $\F^n$ to $\F$ is a polynomial. This fails to hold in general for finite *rings*; say, it is easy to see that if $R$ is a finite (commutative) ring but not a field, then the delta-functions from $R^n$ to $R$ are not representable by polynomials. With this in mind, it would be very interesting to extend Chevalley-Warning onto the ring of all functions from $R^n$ to $R$. Confining specifically to the rings $R=\Z/m\Z$,

Can one define the "degrees" $\deg f$ of the functions $f\colon(\Z/m\Z)^n\to\Z/m\Z$ so that for any system of functions $f_1,\dotsc,f_k$, if $n>\deg f_1+\dotsb+\deg f_k$, then the system of equations $$ f_1(x)=0,\dotsc,f_k(x)=0 $$ cannot have a unique solution in the variable $x\in(\Z/m\Z)^n$?

The degrees $\deg f$ should not be excessively large, of course. Ideally, they should have properties like this:

- if $f$ is a polynomial, then $\deg f$ does not exceed its "polynomial degree";
- $\deg(fg)\le\deg f+\deg g$;
- $\deg(f+g)\le\max\{\deg f,\deg g\}$;
- $\deg f\le(m-1)n$ for any function $f$, and if $f$ is determined by just $s$ coordinates, then indeed $\deg f\le (m-1)s$.

Let $BMO$ the space of bounded mean oscillation functions on $\mathbb{R}^n$ equipped with the Lebesgue measure. If $Q\subset \mathbb{R}^n$ a cube, let $m_Q f$ the average of a function $f\in L^1_{loc}(\mathbb{R}^n)$ on $Q$. We say that $f\in L^1_{loc}(\mathbb{R}^n)$ is in $BMO$ if $$ \sup_{Q\in\Delta} m_Q(|f-m_Qf|)<\infty $$ where $\Delta$ is the set of the non-degenerate cubes in $\mathbb{R}^n$ (with sides parallel to the axis). The inequality of John-Nirenberg allows us to defined equivalent norms on $BMO$ for all $1<p<\infty$: $$ \Vert f\Vert_{\star,p}=\sup_{Q\in\Delta} m_Q(|f-m_Qf|^p)^{\frac{1}{p}}<\infty $$ Then let us defne the Hardy spaces, $H^{1,p}(\mathbb{R}^n)$, for $1<p\leq \infty$. Firstly, an p-atom on a cube $Q$ is a (Lebesgue)-measurable function $a:\mathbb{R}^n\rightarrow \mathbb{C}$ such that : $(i)$ $\text{supp}\, a\subset Q$, $$(ii)\Vert a\Vert_{L^p}\leq \frac{1}{\mathscr{L}^n(Q)^{1-\frac{1}{p}}}$$ $$ (iii)\int_{\mathbb{R}^n}a(x) d\mathscr{L}^n(x)=0$$ $H^{1,p}(\mathbb{R}^n)= \{f\in L^1(\mathbb{R}^n), \exists (\lambda_j)_{j\in\mathbb{N}}\;\text{and p-atoms}\; (a_j)_{j\in\mathbb{N}}\in l^1(\mathbb{N}), f=\sum_{j\in\mathbb{N}}\lambda_j\, a_j \}$. $H^{1,p}(\mathbb{R}^n)$ is a Banach space equiped with the norm $$ \Vert f\Vert_{H^{1,p}}=\inf\left\{\sum_{j\in\mathbb{N}}|\lambda_j|, \lambda\in l^1(\mathbb{N}), \exists (a_j)_{j\in\mathbb{N}}\, \text{p-atoms}, f=\sum_{j\in\mathbb{N}}\lambda_j\,a_j\right\}.$$ We can prove that all the spaces $H^{1,p}$ are the same with equivalent norms (recall that $1<p\leq\infty$), and we note the common space $H^1$.

My question concerns a precision about the theorem of Fefferman and Stein : the dual of $H^1$ is isomorphic to $BMO/\mathbb{R}$.

When we fix $1<p<\infty$ $\Vert\; \Vert_{\star,p'}$ and the norms $\Vert\;\Vert_{H^{1,p}}$ on $BMO/\mathbb{R}$ and $H^{1,p}$ respectively, can we construct an isometric isomorphism $BMO/\mathbb{R}\rightarrow (H^{1,p})^{\star}$?

Every (real) Lie group is a finite cover of a linear Lie group (*i.e.*, a closed subgroup of some $\mathrm{GL}_m(\mathbb R)$), and this can obviously be done for linear Lie groups; so maybe it's enough to show that this condition is preserved under covers? (Or, if it's false, then a non-linear group like a spin or metaplectic group is the place to look.)xHow networks with high largest eigenvalues are more robust?8725801315662http://projecteuclid.org/euclid.jdg/1175266207 , http://www.math.umn.edu/~akhmedov/FRG-Lec-2.pdf1318115The constraint equations are simply the expression for relationships that the metric and second fundamental form of a hypersurface inside a (Lorentzian) Einstein manifold must satisfy, according to the Gauss and Codazzi equations. So they are highly geometric (at least in the case of the vacuum Einstein constraints -- otherwise there is another term coming from the stress-energy tensor of the ambient manifold).

11067912175128While I agree with your interpretation of the question, I am very nervous about your answer. In particular, I don't like the idea that composing several poor sources of randomness is a good way to make a random number generator. As Donald Knuth says, "random numbers should not be generated with a method chosen at random". See the introduction to Chapter 3 in The Art of Computer Programming, Volume II, for an example of how this can fail.438828768741781995Check out Section 2 of Noohi's paper "Fundamental groups of topological stacks with slice property" http://front.math.ucdavis.edu/0710.2615.

rThe properness of the special singular simplicial spaces1729967414192168817519340852289922Software Engineer and Bachelor of Computer Science interested in all fields of CS(including computer architectures, digital systems and hardware), technology and the IT sector.

fUse the detexify webpage and you find code for it.~https://artelogic.net/industries/logistics-management-software359805239003818391685538The tensor product $F_1 \otimes_F F_2$ is not 0, hence it has a quotient which is a field. This contains the images of both $F_i$.

Let $f$ be a modular form -- more specifically, a normalized new eigenform which is not of CM type.

We say $f$ has **extra twists** if there exists some $\sigma \in \operatorname{Aut}( \mathbf{C})$ such that the Galois conjugate $f^\sigma$ is equal to the twist of $f$ by some non-trivial Dirichlet character, so $a_n(f)^\sigma = \chi(n) a_n(f)$ for all $n$ coprime to the level.

Any newform of nontrivial character has at least one extra twist (because the complex conjugate $\bar{f}$ is a twist of $f$.) But browsing some tables suggests that extra twists are pretty unusual for newforms of trivial character.

- Do "most" non-CM newforms of trivial character (in some precisely quantifiable sense) have no extra twists?
- Do "most" newforms of non-trivial character have no extra twists other than the obvious one?

(This question is prompted by this earlier question, which is subtle precisely when extra twists exist.)

1960501430115Suppose, for each $n>0$, I have two complexes of abelian groups $(A_{n}, d)$ and $(B_{n},d')$ and a quasi-isomorphism $$f_{n}: A_{n} \rightarrow B_{n}.$$ Furthermore, suppose I have maps of complexes $A_1 \rightarrow A_2 \rightarrow A_3 \dots$ and similarly for $B_n$, compatible with $f_n$.

Is it true that the map of direct limit complexes $$\lim_{n} f_{n}: \lim A \rightarrow \lim B$$ is also a quasi-isomorphism?

Complexes can mean unbounded or maybe just bounded above/below, depending on your preference.

Also: what if I replace the limit over natural numbers with a more general filtered colimit?

I would be surprised if the purported Grothendieck quote is really his. He does not lean to the short and sweet. It sounds more like an adaptation of another thing Deligne says in "Quelques idées maitresses de l'oeuvre de Grothendieck" (p. 13): "if the decision to let every commutative ring define a scheme gives standing to bizarre *schemes*, allowing it gives a *category of schemes* with nice properties."

Assume $K$ is a number field and $E$ is an elliptic curve defined over $K$.

Are there conditions under which the short exact sequence $$0\rightarrow E (K)/mE (K)\rightarrow H^1_{Sel}(K,E_m)\rightarrow Sha(E|K)_m\rightarrow 0$$ splits?

References?

2137045849817:@IanAgol, are you happy now?672348166051850757542096301097778I should clarify: as a candidate for an example of the kind you're after, you want a number field whose ring of integers has class number greater than one. Not being a PID is obviously a necessary condition, though not necessarily sufficient."Why use flat covers when etale covers are enough?" Finer topologies yield small toposes that are not functorial.The terminology is still not clear to me. What is the "unipotent radical" of an abstract group?110591$a(n)=\lfloor \alpha n\rfloor-\lfloor \alpha(n-1)\rfloor$, where $\alpha=\log_23$. Sequences of this type are called Sturmian and are well known (and easily shown) not to be periodic. Basically: for any $x,x'$ whose difference is not an integer, there is a multiple of $\alpha$ such that $\lfloor x+n\alpha\rfloor-\lfloor x\rfloor\ne \lfloor x'+n\alpha\rfloor-\lfloor x'\rfloor$ by density of multiples of $\alpha$ in the circle.

2583217925922All the prime numbers less than or equal to 2 are even, and all the prime numbers greater than or equal to 3 are odd :)Finding interesting correlations in seemingly trivial concepts. I love it.7389471554531686376dHow to check if a commutative ring is Gorenstein.ZIs the diagonal morphism a strong immersion?1485034hThe inverse image of a Noetherian topological spaceOne important case to test would be $D=15$, since it should result in a larger ratio yet.2157224789266105933Here's an idea. For any partition $(A,B)$ of $[2n]$, where $|A|=|B|=n$, we can ask each child if they prefer $A$ or $B$. If one prefers $A$ and the other prefers $B$, then we are done. Otherwise, they have the same preference function over all such partitions.

**Lemma.** There exists partitions $(A,B)$ and $(A',B')$ such that

both children prefer $A$ over $B$,

both children prefer $B'$ over $A'$, and

$(A',B')$ is obtained from $(A,B)$ by performing a single swap.

**Proof.** Perform swaps until $(A,B)$ becomes $(B,A)$. At some point, each child must switch preferences.

Given the assumption that the gifts are all roughly the same value, it seems fair to offer such an $(A,B)$ as a choice and to flip a coin to decide who gets $A$.

145506240888I need a reference (or a short proof) for the following statement:

Suppose a closed manifold $N$ is the result of a surgery (along an embedded sphere) on a closed manifold $M$. Then the difference $\sum dim H_i(N) - \sum dim H_i(M)$ (the homology is taken with coefficients in a field) is at most 2.

https://www.gravatar.com/avatar/12f18eaa7dbbc3ab8962df0cec786f41?s=128&d=identicon&r=PG&f=11060511This approach is very similar to what I have called the Naturalist Account of forcing, and written on here: http://jdh.hamkins.org/themultiverse/, and elsewhere.>DirichleteApproximationTheorem2127160ZI am a graduate student in Helsinki.

51742217141452280496I suspect Stirling's formula is involved, since $\pi$ and factorials appear. But I also suspect this is closer to being an exercise than a research question, especially since you seem sure of the formula. So this question might be more appropriate for stackexchange or maybe Wikipedia's mathematics reference desk than for mathoverflow.I too agree with this heresy. (Maybe enough people have already agreed that it's no longer heresy.) There are plenty of nice problems in algebraic geometry one can work on with a really minimal amount of commutative and homological algebra. I'm not recommending that approach, but I think the idea, which one might get from looking at Hartshorne, that you have to know that depth equals dimension in Cohen-Macaulay rings (or whatever) before you start thinking about curves on del Pezzo surfaces, is not accurate, and holds people back.For the special case where $X$ is projective, and reducing (by the "trivial" examples of surjections and normal quotients) to $G$ a semisimple group of adjoint type, with $H=P$ a parabolic, one version of the answer is given by Demazure in "Automorphismes et déformations des variétés de Borel", *Invent. Math.*, 1977. In a nutshell, the automorphism group of such an $X$ is the subgroup of $Aut(G)$ preserving the conjugacy class of $P$ --- unless the pair $(G,P)$ is one of several "exceptional" cases. (E.g., the $5$-dimensional quadric is both $G_2/P$ and $SO_7/P$.)

In other words, for the non-exceptional cases, you can recover $G$ as the connected component of the identity in $Aut(X)$.

149390291105813094059Floer cohomology for immersed Lagrangians is introduced by
*Akahi, Manabu; Joyce, Dominic*, **Immersed Lagrangian Floer theory**, J. Differ. Geom. 86, No. 3, 381-500 (2010) and its one-dimension version (in the absence of "teardrops")
is developed in *Abouzaid, Mohammed*, **On the Fukaya categories of higher genus surfaces**, Adv. Math. 217, No. 3, 1192-1235 (2008).
See also the embedded case described in *de Silva, Vin; Robbin, Joel W.; Salamon, Dietmar A.*, **Combinatorial Floer homology**. As in the embedded case, non-triviality of Floer cohomology of immersed Lagrangians obstructs Hamiltonian displaceability. For example, consider the following example of an immersion $f: L \to X $ in the two sphere $X$ (thought of as the one-point compactification the plane) dividing the two sphere into areas $A_0,A_1,A_2,A_3$ (smoothing used curve-shortening by A. Carapetis)

A computation shows that the Floer cohomology admits a weakly bounding cochain and is Floer-non-trivial if $A_2 < \min(A_1,A_3)$ and

$$A_0 = A_1 + A_3 - 3A_2.$$

Indeed in this case using the Morse model there are six generators in the Floer cochains, given by the max and min of the height function and four generators corresponding to the two self-intersection points; the zeroth Fukaya map is non-zero because of the two teardrops but the teardrops can be "killed" by taking a bounding cochain that is the sum of generators coming from the self-intersections with coefficients $q^{A_1 - A_2}$ and $q^{A_3 - A_2}$. On the other hand, Moser's principle *Moser, Jürgen*, **On the volume elements on a manifold**, Trans. Am. Math. Soc. 120, 286-294 (1965) implies that $f(L)$ is non-displaceable iff each of the areas is at most the sum of the other three:

$$ A_i \leq A_j + A_k + A_l, \quad i,j,k,l \ \text{distinct}. $$

Indeed, if one of these inequalities fails then there is a symplectomorphism that takes the complement of the interior of one of these regions into its interior, while if they all hold then such a symplectomorphism obviously does not exist. The conditions

$$A_0 + 3A_2 = A_1 + A_3, A_2 < A_1, A_2 < A_3$$

imply the "Moser conditions" (exercise), but are much stronger. So there seems to be a fairly big gap between Floer non-triviality and non-displaceability in this case.

Question: (1) Can anyone do better, i.e. is it possible to detect the non-displaceability of immersed curves in the two-sphere using some other version of Floer theory, or a different weakly bounding cochain?

(2) Suppose one takes the product of two such curves in $S^2 \times S^2$. Here Floer theory still works, but there is no analog of Moser's results although presumably one could use *McDuff, Dusa*, **Displacing Lagrangian toric fibers via probes**, Proceedings of Symposia in Pure Mathematics 82, 131-160 (2011). When is the image of an immersion of product form displaceable?

There is a long exact sequence in a paper of Hurewicz in the Bulletin of the AMS, 1941. The paper is called "On Duality Theorems". According to Weibel's History of Homological Algebra, this is the first appearance of an exact sequence.

876310224777514748405Do you intend or at least consider to combine this (potential) other PhD with a switch of fields (within maths)? 2052422f@AlexandreEremenko: thanks, that very interesting!2112928@NikWeaver But is that all such sets? What if $X$ is not locally compact?1416381262868@Keenan: Re LaTeX. Read the box entitled "how to write math" down there on the right hand side. You need to use more backticks. 206125112564991805217935093859715772841970575I'm a PhD candidate in mathematics at Yerevan State University (YSU). Obtained bachelor and master degrees in mathematics from Higher School of Economics (HSE) and Skolkovo Institute of Science and Technology (Skoltech) in 2016 and 2018, respectively.

@A hyperplane inside another one661529https://www.gravatar.com/avatar/bf168b79efd7caccf1a92d14a3020ed6?s=128&d=identicon&r=PG&f=112157082046850z@fedja How do you know that's optimal for the other problem?71642830941412755311057509Do you mean an n x n matrix? If so, the probability of being singular seems to be 1 as formulated in the first paragraph: Each row has a positive probability of being identically zero. Perhaps the diagonal changes this? 28015141646918549418842821162335dhttps://www.ps.uni-saarland.de/~forster/index.php1435993261517~I think $f(x)=\sum_{i=0}^{inf} (1/6*\sum_{j=1}^6 (x^{a_j}))^i$2770821638201852813This is not an answer, but rather a long comment (grad student level, so please don't take it seriously). I use surfaces for simplicity. The answer must yes in some form. My belief is from the moduli space theory. It is known that the normal stable surfaces admit at worst log canonical isolated singularities. This includes $xyz+x^p+y^r+z^q$ singularities. However, to have a complete moduli space of surfaces, we must include no isolated singularities of the form $xyz$, $xyz+x^p$, and $xyz+x^p+y^r$ (among others). The resemblance of the equations must be more than a coincidence. So, I can imagine we can have an isolated singularity and consider all the deformations from it to non isolated ones. Then to look for the minimal "complete" family of such degenerations.

I wish someone can say something more about all this.

tGlenn Green, blockchain... more data more blocks!

130315313302421775159Before I start, let me make a note on terminology. Curves are always smooth projective connected curves over an algebraically closed field of characteristic zero.

Let $\mathcal C$ be a class of curves. We say that $\mathcal C$ is dominant if, for all curves $X$, there exists a curve $Y$ in $\mathcal C$ and a finite morphism $Y\to X$.

Bogomolov and Tschinkel proved that the class of hyperelliptic curves and their unramified covers is dominant. Manin proved that the class of modular curves $X(n)$ and their unramified covers is dominant. Both proofs rely on Abhyankar's Lemma.

Let $k\geq 2$ be an integer. Let $\mathcal C_k$ be the class of $k$-gonal morphisms, i.e., the class of curves for which the gonality equals $k$.

Q1. Is $\mathcal{C}_2$ dominant?

Q2. Is $\mathcal{C}_k$ dominant?

Q3. Is $\cup_{2 \leq j \leq k} \mathcal{C}_j$ dominant if $k>>0$?

Let me repeat this in words. Let $X$ be a curve. Does there exist a $k$-gonal curve $Y$ and a finite morphism $Y\to X$?

I'm mainly interested in the case $k=2$. In this case, it suffices to answer the following question.

Q1b. Let $X$ be a curve. Does $\mathbf{P}^1$ admit a closed immersion into the symmetric product $X^{(2)}$?

93823211153751269234^What do you mean by "an ideal of $R$ and $S$"?269036Brian Borchers27416017880527180142116562Another nice property of your definition is that it is affine invariant. However, it seems hard to calculate it even for the simplest examples, say for a disk and its center.790434hHausdorff distance and sum of independent variables6618071957383lApply Krull's Hauptidealsatz to $C\cap D$ inside $D$.53197ZStable ∞-categories as spectral categories@S.Surace: It seems that you have read only the title of this post, but not its content. Indeed, the title suggests a completely diffferent question - the one that you have asked.36285996375Here, we are looking for examples which are not a product of lower dimensional manifolds, and is not a connected sum.2031748155247974996417789332171002JEntropy and total variation distance18786751762573https://lh4.googleusercontent.com/-cqGkX2ngzTQ/AAAAAAAAAAI/AAAAAAAAAQM/y9k8pJDMNF4/photo.jpg?sz=128997249 Fair enough. My question arose out of some textbook examples of probability spaces and random variables (e.g the interval $[0,1]$ with the Borel algebra and Lebesgue measure) in which the underlying space had some familiar topology and the $\sigma$-algebra was chosen to be the Borel algebra rather than its completion. Maybe this is more of a phenomenon of examples rather than of theorems in probability theory. Can you mention some basic theorem in probability requiring the space in question to be equipped with a complete probability measure? The CLT, for instance, does not, as far as I know.Oh, okay, I think I get it. $R(g)$ is made up of various derivatives of $g$. So we can view this instead as a (very nonlinear) PDE on $X \times \mathbb{R}$.Can you give a good example of this in another instance?

The Wikipedia definition seems to imply that if you do not have natural numbers defined yet, they follow as the lengths of the finite strings.

In particular, given any Free Monoid, the unique word length function maps elements of that monoid into the natural numbers.

2140500I'm not sure if this question is appropriate for MO. Add comments if it is not. Thanks.

How to explain/motivate the method of separation of variables for PDEs to undergraduates? What's the real math behind it? It's not just because the guy who fancied it is very smart, right? (Although I feel like it does give students this impression...)

(Background: At Berkeley there is a course called Math 54, in which students learn linear algebra, linear ODEs and then 1 or 2 weeks of PDEs. Teaching Separation of variables is always my nightmare...)

1671553Here's an answer which is a bit better than the one I suggested in the comment, but unfortunately still quite unnatural.

Define $U(x) := \forall y\;y\in x$. Then adjust the axioms of $\mathsf{ZF}$ (other than extensionality) so that they are "bounded over sets that aren't universal." That is, replace every universal quantifier $\forall x \;\ldots$ with $\forall x\;\neg U(x) \rightarrow \ldots$ and every existential quantifier $\exists x \; \ldots$ with $\exists x \; \neg U(x) \wedge \ldots$. Then keep the extensionality axiom the same and add the axiom $\exists x\;U(x)$. Call this theory $\mathsf{T}$. Note that we can easily produce models of $\mathsf{T}$ from models of $\mathsf{ZF}$ by adding one extra element for the universal set. We can produce models of $\mathsf{ZF}$ from models of $\mathsf{T}$ by removing the universal set.

This follows from the answer to question 1 by replacing $x \in y$ by $y \in x$ in every axiom of $\mathsf{T}$.

Define $Q(x) := \forall y\;y \in x \leftrightarrow y = x$. That is, $Q(x)$ says that $x$ is a *Quine atom*. Let $\mathsf{T}$ be the theory with the axiom $\exists ! x \;Q(x)$ and with the axioms of $\mathsf{ZF}$ adjusted as follows. Like in question 1, quantifiers should be bounded to non-Quine atoms. This time we also require that whenever $\mathsf{ZF}$ would assert the existence of a set, $\mathsf{T}$ asserts the existence of the same set but with a Quine atom added. So for example, empty set becomes
$$
\exists x \; \neg Q(x) \wedge (\exists z \; Q(z) \wedge z \in x) \wedge (\forall y \; \neg Q(y) \rightarrow \neg y \in x)
$$
Separation would become
$$
\forall x\,\neg Q(x) \rightarrow (\exists y\, \neg Q(y) \wedge (\exists z\; Q(z) \wedge z \in y) \wedge \forall z \; (\neg Q(z) \rightarrow (z \in y \leftrightarrow (\,z \in x \wedge \phi(z)\,))))
$$
Similarly to question 1, we can convert between models of $\mathsf{ZF}$ and models of $\mathsf{T}$ by just adding or removing a single element (this time corresponding to the Quine atom).

Here's another stab. It's based on the idea mentioned by Joel David Hamkins in his comments.

Let $a(x)$ and $b(x)$ be two such functions. We'll use the fact that deciding whether or not $a(x)$ and $b(x)$ are identically equal is undecidable. For this, we need a function $\Phi(a,b)$ that takes two functions and outputs $0$ if they are identically $0$ and some non-zero real otherwise.

One function we could take is simply $\Phi(a,b) = sup_{x} (a(x)-b(x))^2$. Perhaps this is not allowable for your class of functions. So I propose instead the function $$\Phi(a,b) = \int_{-\infty} ^{\infty} \frac{(a(x)-b(x))^2}{e^{|x|} (1+(a(x)-b(x))^2)} dx.$$

We then consider the integral of $e^{-t^2 \Phi}$, which is an elementary function iff $\Phi = 0$, which is iff $a(x)=b(x)$, which is undecidable.

A slightly different punchline might be to consider the double integral

$$ \int \int \frac{(a(x)-b(x))^2}{e^{|x|} (1+(a(x)-b(x))^2)} e^{-t^2} dx dt, $$ which is perhaps not great because it's a function of two variables, which isn't likely what you had in mind.

Or a third variation on this would be to define $\gamma(x) = |x|/x$ (and $\gamma(0) = 0$). Then consider the function $\gamma((a(x)-b(x))^2)e^{-x^2}$, which has an elementary antiderivative iff the leading coefficient is $0$ almost everywhere (which is undecidable).

A fourth variation is perhaps more satisfying. We may assume $a(x)$ and $b(x)$ are in the ring generated by $\mathbb{Z}[x, \sin(x^n), \sin(x \sin(x^n))]$. Let $C(x) = |a(x) - b(x)| - (a(x)-b(x))$. Then it is undecidable to determine if $C(x)$ is identically $0$. But for $C(x)$ of this form, it's all but certainly true that $e^{C(x) x^2}$ has an elementary antiderivative iff $C(x) = 0$. [I am unsure how to prove this claim, but it could probably be proven along the same lines that $e^{ax^2}$ has no elementary antiderivative]

2297159But this example is too easy: Maple's numerical integration says $\int_{-\infty}^\infty e^{-x^4-3x^2-1} dx = .3538037606$, and it would be straightforward if tedious to use estimates to prove rigorously that this integral is less than 0.4. Oh, and Maple can also get a closed form for the integral: it is $\frac{\sqrt{3}}{2} e^{1/8} K(1/4,9/8)$ where $K$ is a Bessel function of the second kind.492585436252278832993016A while ago over at our sister site, there was an interesting question [not by me] with next to no answers which I feel is, fleshed out in a more precise fashion, appropriate for MathOverflow.

**The question.** Which objects and concepts would you regard as *mathematical phantoms*? I'm especially interested in examples from applied mathematics, since I already know a couple of examples from pure mathematics (listed below).

**Mathematical phantoms.** Following John Baez, Gavin Wraith and surely others, a *mathematical phantom* is "an object that doesn't exist within a given mathematical framework, but nonetheless 'obtrudes its effects so convincingly that one is forced to concede a broader notion of existence'.
Like a genie that talks its way out of a bottle, a sufficiently powerful mathematical phantom can talk us into letting it exist by promising to work wonders for us. Great examples include the number zero, irrational numbers, negative numbers, imaginary numbers, and quaternions. At one point all these were considered highly dubious entities. Now they're widely accepted. They 'exist'."

**Being more precise.** The demarcation from what are merely very fruitful abstractions is obviously a bit blurry; perhaps useful criteria are:

- There should be a statement which was held to be obviously true before the discovery of the phantom, but which is false in view of the new concept. (Such as the statement "obviously no number squares to $-1$".)
- It should have required a nontrivial effort to make it precise (to "help it come into being").
- It should have great explanatory power and vast consequences.

**Examples.** Phantoms in this stricter sense could include:

- The irrational numbers. (Running counter to the basic tenet "all is number" by the Pythagorean school, where by "number" they meant "rational number".)
- The complex numbers.
- The $p$-adic numbers.
- Actual infinity, together with the flexible notion of sets we have nowadays (vastly surpassing recursive subsets of $\mathbb{N}$) and the axiom of choice.
- Sobolev function spaces.
- Infinitesimal numbers (as in the hyperreal numbers, where $\varepsilon$ is invertible, or as in synthetic differential geometry, where $\varepsilon^2 = 0$).

Phantoms in a broader sense (where I can't think of any held-to-be-obviously-true statement falsified by them) could include:

- Symmetries of zeros of polynomials, or more generally groups.
- The field with one element.
- Ideals in number theory.
- Motives.
- ∞-categories.
- Toposes. (Generalizing and unifying various cohomology theories.)
- Nonclassical logics. (Born in the foundational crisis, nowadays with applications in mainstream mathematics.)

For the purposes of this question, I'm interested in both kinds of phantoms.

8253171485844@DelioMugnolo Are there examples of domains which are isospectral to a disk (for Dirichlet problem with zero Cauchy data)?16879791972978428586I don't understand the votes to close; this question is reasonable, given how intricate forcing is.15019241723018> *Why it seems that mathematics is made of sheaves?* Maybe because every category of sheaves allows you to instantiate known mathematical structures inside it, thanks to its internal language?1890947lhttp://deliberatingsoftwaredevelopment.wordpress.com/1926876I just checked your example.. say $a = 0.8607538$ and $b = 0.017405778$. I took $p = [a,b,b,b,b,b,b,b,b]$ and $q = p \circledast p$. This turned out to be $q = [0.7433, 0.0321, 0.0321, 0.0321, 0.0321, 0.0321, 0.0321, 0.0321, 0.0321 ]$ whose entropy is pretty big, $1.5917$ to be exact. I hope I'm wrong, perhaps you could recheck your answer? Thanks for your comments.18977861964407497233Well, I mean, in many places such as textbooks or lectures, they merely tell us that they will or we should use Gaussian quadrature without explicitly mentioning the kind, and in other places they might also say that by doing this, we can get the exact solution if the integrand happens to be a (2N-1)-degree polynomial. So I wonder which of these methods they actually refer to by default when they say Gaussian quadrature.1849476464939This is really a striking-simple example. I wonder that I have not come across it so far. Morphisms in the arrow-category are commuting squares. I always imagined this was answered in Blasius-Harris-Ramakrishnan, MR1262930. Did you look there?Let $G$, $G_1$, $G_2$ be Hausdorff topological groups. I am mainly interested in the case when those groups are profinite.
Let $G$ act continuously on $G_1$ and $G_2$ via continuous automorphisms, i.e. we have **continuous** maps $G \rightarrow \mathrm{Aut}_\mathrm{cont}(G_i)$ (with compact-open topology on $\mathrm{Aut}$'s).

From this one can define a map of **abstract** groups $$G \rightarrow \mathrm{Aut}_\mathrm{cont}(G_1 *_\mathrm{top} G_2),$$
where $*_\mathrm{top}$ denotes the free topological product. This can be done using the universal property of the free topological products and is given explicitly via formula ${}^g(g_1g_2g_1'g_2' \ldots g_1^{(m)}g_2^{(m)})={}^gg_1{}^gg_2{}^gg_1'{}^gg_2' \ldots {}^gg_1^{(m)}{}^gg_2^{(m)}$, where $g_1g_2g_1'g_2' \ldots g_1^{(m)}g_2^{(m)}$ is a word in $G_1 *_\mathrm{top} G_2$ (note that this makes sense as the underlying abstract group of $G_1 *_\mathrm{top} G_2$ is the usual free product of $G_1$ and $G_2$ as abstract groups).

**Question**: Is the above map continuous? If not, what would be some additional assumptions to make it continuous?

Let $\ell$ be a prime number and let $f:X \to Y$ be a morphism of schemes of finite type over the complex numbers (or a regular scheme of dimension at most 1, in which $\ell$ is invertible). How to prove that the $\ell$-cohomological dimension of $Rf_\*$ in the etale topology is finite? I don't understand the proof in Kiehl and Weissauer's book "Weil Conjectures, Perverse Sheaves and l’adic Fourier Transform", Appendix D, since there is no base change for $f_*$. The crucial case seems to be when $f$ is an open immersion.

And what about the analytic version? $f:X \to Y$ is a morphism of complex analytic spaces, and $F$ is a sheaf of abelian groups on $X$ (in particular, no $\ell$).

11711401679844Now fedja's solution and this argument eventually produced a proof from the book ...1268477You need to assume at least enough regularity so that both side of the equation make sense. I believe that having the coefficients of $\vec{F}$ be in the Sobolev space $W^{1,1}$ suffices. The right side is bounded by the definition of the norm for $W^{1,1}$.. The left side requires a well-defined way to restrict $\vec{F}$ to the boundary. The operator that maps $\vec{F}$ to its restriction along the boundary is called the trace operator. See: https://en.wikipedia.org/wiki/Trace_operator (as well as the cited book by Evans ).590018348251~Implicit equation between rational function and its derivative1913991273669Good edits, but I think you missed one: the first time you say "my question is..." still needs to be reworded.dWhere is the dependence of the right-side on $R$?5390731492198What about replacing $\{0,1\}$ in Stone duality with another finite set?$A^flat(M)/G(M)≅Hom(π_1(M),SU(2))/SU(2)$ is easy to prove. Just use geometric definition of connection.Thanks, Sándor. Unfortunately, the zero set is difficult to determine, and fixing a variable, also. They're polynomials of high degree and of many (>10) variables.7120211537598Tsurprisingly difficult filtration problem577720Are you talking about Conjecture/Question: If F is a finite non-trivial union-closed family of finite sets, then some element appears in at least half the members of F. (from http://www.math.uiuc.edu/~west/openp/unionclos.html)3842656976796155822267778rWhat exactly is $\underset{n}{\varprojlim} \ \mu_{p^n}$?I usually email the author in case the paper is online in a place controlled by the author (i. e., his website or arXiv). The only downside of this approach (apart from authors not replying) is that some would just take down their papers (rather than correcting them) when they hear about a mistake in a part they consider substantial. This happened to me once (although fortunately the paper is still avaliable at other places in the net).

In case the paper is online but not in such a place, this is a more difficult question, but if the error is substantially confusing, I'd still mail the author. Unfortunately, this sometimes means notifying a clueless author that somebody else has put his paper online (most usually, a university workmate, for his students; I am not talking about pirate sites), and rather than correcting it tries to get it offline. I would be careful here.

In case the paper is offline or behind a paywall, I ignore it unless the mistake really destroys some results from the paper, in which case I rejoice about another little blow to copyrighted literature and the so-called peer review process.

But then again, I am mostly reading openly avaliable texts, so the first case is by far the most frequent.

I, personally, would prefer anyone telling me of any mistake, but I do not have (and do not plan to have) papers outside of open access, so there is no contradiction here.

,RiemannWeilConjecture Thank you very much for your reply. I knew the two first works. But let me insist on the "dynamical systems" aspect, thus reformulating a bit: Are there any other "natural" occurrences of packing dimension and/or packing "measure" in the context of dynamical systems? I tend to disagree with you that having positive packing measure is a good argument; even the Patterson-Sullivan procedure goes precisely by avoiding this problem, somehow in a blunt manner, but quite to the point, and so I also don't like my own example of Kleinian groups. By the way, I'm quite fond of your work.2214441809909169092JSorry, I had a bad reference

15071981444300When you say "normal affine domain over an algebraically closed field", do you mean that the ring is finitely generated over the field? For general rings containing a field, I am not sure this is true. However, in the case of finitely generated $k$-algebras, this is true. The simplest argument involves blowing up the point and passing to the stalk at each generic point of the exceptional divisor: this reduces your problem to a problem about extensions of DVRs.|$nw(X)$ denots the network weight of a topological space $X$.\https://i.stack.imgur.com/byW2M.jpg?s=128&g=1The archimedean component of the representation determines the type of the underlying automorphic form. Loosely speaking,

- if $\pi_\infty$ is a discrete series (of index $k$), the underlying form is an holomorphic cusp form (of weight $k$, and level the arithemtic conductor of $\pi_f$)
- if $\pi_\infty$ is a principal or complementary series (of index $\lambda$), the underlying form is a Hecke-Maass cusp form (of eigenvalue $\lambda$)

The correspondence is detailed and proved in Gelbart, Automorphic forms on Adele Groups, 5.C "Some explicit features of the correspondence between cusp forms and representations".

Maybe you should look at the more form-casted answer given here, and the associated reference in Bump.

2589PIP1418612Joël : the book you love is by S.D. Silvey, not Sylvey.

David Williams "Weighing the Odds" is a book about probability and statistics by a distinguished probability theorist. He has a great sense of humour and the book is a lot of fun. The exercises can be demanding, but they are also interesting. (I have latex solutions to many of them if you want to ask me after first trying an exercise.) Professional statisticians think this book is more maths than stats, but it does contain a lot of stats. It's an introduction that will hold your attention if you put effort into the exercises. It's aimed at advanced undergraduates, but in my opinion is also very suitable for research mathematicians (though I have my doubts as to its suitability for a professional probability theorist).

After reading and enjoying this book, a mathematician will find it much easier to get to grips with a good book on stats. But there's no substitute for trying to handle large amounts of real data, when nothing works out as it "ought" to. Learning stats without handling data is like learning maths without working out anything for yourself---it's too easy to kid yourself that you understand.

Yemon, I thought that Cauchy gave the epsilon-delta definition of a limit, but apparently this is not quite right; it was indeed Weierstrass. It seems that Cauchy gave a somewhat imprecise definition of a limit, but that when using this definition in proofs he used the epsilon-delta language (and maybe was the first to do so? I am unclear on this). nLab defines simple objects for categories with zero objects, and the considers the category of unital rings with $0\neq 1$, which has no zero object.835648319817bWhich limits does group cohomology commute with?lTietze's extension theorem for contractible manifolds1197955906616658781I become interested in this problem because $G\cong Aut(G)$ suggests a special symmetry in $G$ (This kind of group describes its own symmetry).

From $G/Z(G)\cong Inn(G)$ we know complete group is the answer for the simplest case, though this class of group itself is quite weird.

What about the case when $Z(G)$ or $Out(G)$ are non-trivial? $G\cong D_8$ is a good example for non-complete group satisfy $G\cong Aut(G)$.

Is there any method to find such non-complete $Aut$-invariant $G$? Maybe we can try apply $Aut$ iteratively to some G and look for fixed points (some interesting results for centerless or non-abelian simple groups can be found at this MO post).

And what other interesting properties does this class of group have?

1687616For the first point, note that by Sylvester's identity

$$\det(I_n + v^TA^{-1}v) = \det(I_n + (v^T)(A^{-1}v)) = \det(I_n+(A^{-1}v)(v^T)) = \det(I_n + A^{-1}vv^T),$$

so

$$\det(A)\det(I_n+v^TA^{-1}v) = \det(A)\det(I_n + A^{-1}vv^T) = \det(A + vv^T).$$

Given $u \in \mathbb{R}^m$ and $v \in \mathbb{R}^n$, their *outer product* is the $m\times n$ matrix $uv^T$. If $u$ and $v$ are non-zero, then their outer product has rank one. Conversely, a rank one $m\times n$ matrix can be written as the outer product of some non-zero $u$ and $v$.

Now suppose $C$ is an $n\times n$ matrix (not necessarily invertible) and $A$ is a rank one $n\times n$ matrix. By the above discussion, there are $u, v \in \mathbb{R}^n$ such that $A = uv^T$. So

\begin{align*} \det(I_n + tCA) &= \det(I_n + tCuv^T)\\ &= \det(I_n + (tCu)(v^T))\\ &= \det(I_1 + (v^T)(tCu))\\ &= \det(I_1 + tv^TCu)\\ &= 1 + tv^TCu \end{align*}

where the last equality follows because $I_1 + tv^TCu$ is a $1\times 1$ matrix.

A function of the form $f(t) = at + b$ is called *affine*. I'm guessing this is what is meant by affine-linear.

**Added Later:** Let me add a theorem which further demonstrates the relationship between the rank of a matrix and outer products. In the following statement, I am taking the definition of the rank of a matrix to be the dimension of its column space.

Theorem:An $m\times n$ matrix $A$ has rank $k$ if and only if the minimal number of outer products needed to express $A$ as a sum is $k$ (i.e. $A$ can be written as the sum of $k$ outer products, but not $k - 1$).

Proving this is a really nice exercise in elementary linear algebra. I would hate to rob you of the experience by posting the solution here.

I would use $\exists$ if I were writing in some setting where the expression using the symbol is being viewed as a mathematical object in its own right, as in logic. Otherwise I would avoid it in favor of "There exists".147423992304402277383https://www.gravatar.com/avatar/9e8188d784d1d6ec3d391288dedce228?s=128&d=identicon&r=PG&f=1Given a square matrix $M \in SO_n$ decomposed as illustrated with square blocks $A,D$ and rectangular blocks $B,C,$

$$M = \left( \begin{array}{cc} A & B \\\ C & D \end{array} \right) ,$$

then $\det A = \det D.$

What this says is that, in Riemannian geometry with an orientable manifold, the Hodge star operator is an isometry, a fact that has relevance for Poincare duality.

http://en.wikipedia.org/wiki/Hodge_duality

http://en.wikipedia.org/wiki/Poincar%C3%A9_duality

But the proof is a single line:

$$ \left( \begin{array}{cc} A & B \\\ 0 & I \end{array} \right) \left( \begin{array}{cc} A^t & C^t \\\ B^t & D^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\\ B^t & D^t \end{array} \right). $$

1864919575590516169104576414307671794001905577Practicing statistician with an interest in computer science.

Bill Murray is my spirit animal.

613733293473335264879798I'm trying to determine the state parameters for an hmm with a minimum of 9 states. I'm running chunks of my data through hmmtrain on MATLAB which I think uses the Baum/Welch aka forward/backward algorithm. This tends to be very slow. I have found several papers describing the segmental k-means algorithm for determining the hidden markov parameters. However, none of the papers make clear where in the algorithm the segmentation and k-means clustering occurs (see Rabiner 1990).

Any insight into this algorithm would be appreciated.

Thanks.

Yes, thanks for the additional information. Interestingly, it looks like $T(x + \pi) - T(x)$ is also periodic, with period 1.294514717145622014750333637148612526253410966501250078Certainly a pairing $\langle f, \mu\rangle = \int fd\mu$ is well defined for quite general vector spaces of functions and signed measures. It is still not clear to me what you are after. For example, do you want the vector spaces to have norms?199409246890112719228LThis answers expands on my comment on the original question.

**1.** *Given two circles in the plane, there is (at least) a line
which is tangent to both of them*:
this is not true unless we allow lines with complex coefficients
(and even then there's an exception, see below). In the real plane,
two circles have:

$\bullet$ no common tangents if one is contained in the other's interior,

$\bullet$ two common tangents if they intersect at two points, and

$\bullet$ four common tangents if they're disjoint and neither's interior contains the other.

At the boundaries between these cases, the circles can have 1 or 3 common tangents if they're tangent internally or externally. There's also a generate case where the two circles coincide and thus have infinitely many common tangents.

**2.** *Given three spheres in the space, there is a plane which is
tangent to all of them*: again not true, and and with a wider
variety of counterexamples, and a wider variety of degenerate cases where
there are infinitely many common tangents. Aside of these degenerate
examples, the maximum number of common tangent planes is $8$.
(See below for how to compute them.)
A simple configuration with infinitely many tangent planes is
three identical spheres with collinear centers.
More generally, three spheres with collinear centers that have
one tangent plane have infinitely many, because we can rotate
the plane about the line $l$ joining the centers. But make one sphere
a bit larger or smaller about the same center, and there's no common
tangent plane at all, and then this stays true if we move the centers
a bit off $l$.

**3.** *In general, given $n$ spheres in $n$-dimensional space, is
there a hyperplane which is tangent to all of them?* [The proposer
wrote "$n$ n-spheres", but current terminology uses "n-sphere" for
a sphere in ${\bf R}^{n+1}$.] Again, not necessarily; the number
can range from $0$ to $2^n$, or be infinitely large, and there are
various ways for the number to be zero, most simply when one sphere
is contained in another's interior.

The hyperplanes in ${\bf R}^n$ constitute an $n$-dimensional space, more precisely a projective space ${\bf RP}^n$ with a point removed (the hyperplane at infinity, which is not relevant to us because it is not tangent to any sphere). The hyperplanes tangent to a given sphere constitute a degree-$2$ hypersurface in this projective space. Generically $n$ such surfaces in $n$-dimensional space meet in $2^n$ points, but it is possible for none of them to have real coordinates; and there are also special configurations that intersect in a positive-dimensional variety, most simply when they are linearly dependent (but this is far from the only reason, as noted in this mathoverflow answer).

**4.** *What other generalizations does this problem admit?*
I wrote that one gets the same $2^n$ enumeration with $n$ arbitrary
quadrics in place of spheres, but Mariano Suárez-Alvarez
suggested a much more complete answer:

The ultimate generalization (or, at least, one of them) is, I guess, intersection theory and enumerative geometry.

Still, there are special features of the $n$-sphere problem that
do not survive generalization even to $n$ quadrics. Most notably,
the $2^n$ tangent planes can be given in closed form, each requiring
only the extraction of a single square root. Suppose the spheres have
centers $v_i \in {\bf R}^n$ and radii $r_i \in (0,\infty)$. A hyperplane
$a \cdot x = b$ is tangent to the sphere $|x-v_i|=r$ iff
$a \cdot v_i - b = \pm r_i |a|$. Now for each of $2^n$ choices of $\pm$ signs
we get $n$ simultaneous linear equations in $n+1$ variables: the coordinates
of $a$, and the length $|a|$. Generically there is a line of solutions,
and then the additional condition that $|a|$ really be the length of
the vector $a$ gives a quadratic equation on that line which usually has
$2$ solutions, either real or complex conjugate. [This gives $2^n$ hyperplanes,
not twice that number, because switching *all* the $\pm$ signs yields the same solution.]
But there are also non-generic cases where the equations are dependent, and so give
a higher-dimensional space of solutions (not necessarily defined
over ${\bf R}$), or inconsistent, and so give no solutions at all.
If I did this right, one of these happens **iff** the centers of
the spheres lie on a linear subspace of dimension at most $n-2$.
So for $n=2$ we're back to concentric circles, with infinitely many
solutions if they coincide, and none otherwise, not even over ${\bf C}$.
Likewise for $n=3$ we get the configurations of three spheres with
collinear centers.

Suppose a (linear algebraic) group $G$ acts on a variety $X$ and that $U$ is a $G$-invariant open subvariety. My question is: under what conditions is the restriction functor

$i^*: Vect^G(X) \rightarrow Vect^G(U)$

an equivalence of categories of $G$-equivariant vector bundles? Obviously not in general, but we do have: if $X$ is normal (I think Gorenstein or even less will do actually) and $U$ is dense with complement of codimension at least 2 then $i^*$ is fully faithful. What more is needed to make it surjective? What if the complement of $U$ is actually a fixed point of the $G$ action?

The example I have in my head here is that of $G$ being the product of a finite subgroup of $SL(2,\mathbb{C})$ and a $\mathbb{C}^*$, acting canonically on $X = \mathbb{C}^2$. Then the $G$-equivariant vector bundles on the punctured plane should be equivalent to those on the whole $\mathbb{C}^2$, that is to representations of $G$. So I guess I'm asking: why is this true?

Hi Steve! No, I think it is pretty much the same. Thank you for checking it out. The article has a survey section where I discuss many people effort to study $(2,2\times 2)$ generating sets in simple groups (this is based on classification - with few exceptions, almost all series and all sporadics have such generating triples). 1043509I will give an idea similar to that of Jaap Eldering. However, I find the "retraction to $[0,\infty)$" approach quite useful in general, so I decided to post this answer.

Consider the family $\{g_t\}_{t\in [2,\infty)}$ given by:

- $g_t(x)=f_1(x)$ for $t\leq 0.5-\frac{1}{t}$,
- $g_t(x)=f_2(x)$ for $t\geq 0.5+\frac{1}{t}$,
- $g_t(x)$ linear on $[0.5-\frac{1}{t},0.5+\frac{1}{t}]$.

This family should form (correct me if I am wrong) a closed subset in $C([0,1])$ that is homeomorphic to the half-line $[0,\infty)$. Since $[0,\infty)$ is an AR (absolute retract), the subset $\{g_t\}_{t\in [2,\infty)}$ is a retract of $C([0,1])$. But $[0,\infty)$ does not have the fixed point property, so $C([0,1])$ also does not have this property.

359867Some thoughts on your question are here: http://golem.ph.utexas.edu/category/2011/05/making_things_simpler_by_duali.htmlInjitea620140I've been stuck on this one for a while now.

Given an $\mathcal{H}^{s}$ measurable subset $E\subset \mathbb{R}^n$ with $0<\mathcal{H}^{s}(E)<\infty$, we let $\overline{D}^{s}(E,x)$ denote the upper spherical density of $E$, defined as $$\lim\sup_{r\rightarrow 0} \frac{\mathcal{H}^{s}(E\cap B_r(x))}{(2r)^s}$$

I'm trying to prove that given a $\mathcal{C}^1$ map $\phi$ with nonzero Jacobian, we have that $\overline{D}^{s}(E,x) = \overline{D}^{s}(\phi(E),\phi(x))$ for $\mathcal{H}^{s}$ almost all $x$.

I've been considering just linear maps so far. I tried isolating pieces of $E$ whose $\mathcal{H}^s$ measure could be approximated well using subsets $D\subset \mathbb{R}^{n}$ whose diameters changed by a factor of, say ($d\pm \varepsilon$) (for small $\varepsilon$) in an attempt to get estimates on $\mathcal{H}^s(\phi(E))$, but I've been unsuccessful so far. In the end the only estimates I've been able to obtain have been those which use the constants $M,m$ satisfying $m|x-y|\leq |\phi(x)-\phi(y)|\leq M|x-y|$, but this only takes me so far in proving the equality of the densities. I also attempted to relate the spherical density of $\phi(E)$ to the density of $E$ with respect to the family $\mathcal{A}$ consisting of those sets $A$ such that $\phi(A)$ is a ball, but for this to work I need estimates relating the latter density and $\overline{D}^{s}(E,x)$.

Any help would be greatly appreciated.

indeed this is part of a general picture where constructing new L-functions by finite-dimensional representations is usually pretty easy, but finding the correspond automorphic representations is usually incredibly hard.298992ni8mrThere is a notion of a closed subfunctor (introduced by Grothendieck), see e.g. [FGA]. Of course $X \to Y$ is a closed embedding if and only if $Mor(-,X)$ is a closed subfunctor in $Mor(-,Y)$.Malcev's theorem is equivalent to the statement that the kernel of each self-epimorphism of a fg group is contained in the intersection of all finite index subgroup. 16406702171660299Prove a function, defined by integration of a harmonic function, is log-convex15858283695171120304188211930236517163901211735The second 1. and 2. were typed as 3. and 4. (I just went back and checked). I'm not exactly sure why they're showing up as 1. and 2., but I imagine it has to do with my ineptitude as using computers ;-)6Yes, for the last comment.12880066834441936736PRegular holonomic $\mathcal{D}$-modules528750Ryes, A should be conencted. I edited it.1140180p@NikWeaver that’s why they’re comments, not answers@ Jeremy Rickard Since A can be seen as a projective A-module, so we can take a splitting map $\gamma : A \rightarrow G^{(I)}$, but if the index set I is infinite, why the image of $\gamma$ is contained in $G^{(J)}$ for some finite J?@JasonStarr: Thank you. I think that I will try to learn a bit about GIT and then consider this again.7034852014289175398612931844553vGeneralized cycle index polynomial for the symmetric group\https://i.stack.imgur.com/Pdfeh.jpg?s=128&g=16427482This has reminded of my own proof of a similar identity: http://artofproblemsolving.com/community/q1h352146p19156021318660206449010031723320061292834887695213828122572671319571635068vIs every distance-regular graph vertex-transitive?

In fact you know that the purported quotient has one cusp, at least one cone point of order $2k$, and at least one of order $3m$, right? This must make the arithmetic almost trivial.646077An abstract chain just has to have a $G$ action, not the ambient space. The idea I guess is that a chain in the Borel construction can be pulled back to $X\times EG$ and then I can approximate $EG$ by a smooth finite-dimensional manifold $E$ containing the pulled-back chain. Embed $X\times E$ in $\mathbb R^\infty$ and you have geometric chain (or rather its graph). Anyway the paper is nicely written if I remember, so probably it's better to read what they said!13729232693nIs a flop on Calabi-Yau threefolds always Atiyah flop?Perhaps the downvotes are for asking a well-known hard question without any sign of research effort.https://graph.facebook.com/10212574083362336/picture?type=large795192l@JimHumphreys: Thanks a lot for the nice references.1837830https://www.gravatar.com/avatar/05d4462e1512c36eafb7533f14abc31b?s=128&d=identicon&r=PG&f=11843429229679@AknazarKazhymurat. You will need to speak with Paul Seidel about that.&ConjectureOfRhodesOn math.stackexchange, there might be an even greater number of people who can benefit from an answer like this.5945981672001855539Harald Hanche-Olsen and coudy, thank you for your comments. I've been too hasty. 1540838136823@Peter Samuelson, The story is in Stanley's "Enumerative combinatorics" Vol.1. pp. 163-164.Let $M$ be a compact subset of $\mathbb R^2$ with smooth boundary, and let $g$ be a Riemannian metric on $M$. If $g'$ is another Riemannian metric which is "close" to $g$, then they should have almost identical curvature profiles. I would like to prove a concrete estimate on the total difference of their curvatures in terms of the distance of $g'$ to $g$. Before I state the question precisely, I need to introduce some notation.

Write $\operatorname{Sym}$ for the space of symmetric $2\times 2$ real matrices, and let $\operatorname{SPD} \subseteq \operatorname{Sym}$ be those matrices which are also positive-definite. Consider the function space $\Omega = C^2(M, \operatorname{SPD})$. Denote partial derivatives of $g_{ij} \in \Omega$ by additional subscripts following a comma, so that $\tfrac{\partial}{\partial x^k} g_{ij} = g_{ij,k}$, et cetera. Endow the space $\Omega$ with the norm $$\|g\| = \sup_{x \in M} \max_{i,j,k,l} \left\{|g_{ij}(x)|, |g_{ij,k}(x)|, |g_{ij,kl}(x)| \right\},$$ so that it has the structure of an open cone within the Banach space $C^2(M, \operatorname{Sym})$.

Each $g \in \Omega$ defines a Riemannian structure on $M$ via the inner product $\langle v, g(x) v' \rangle$ for $v, v' \in T_x M$. Let $K(g,x)$ be the scalar curvature of the metric $g$ at the point $x \in M$.

**What I want to prove:** For each $g \in \Omega$, there exist constants $C$ and $\epsilon$ so that if $g' \in \Omega$ with $\|g - g'\| < \epsilon$, then $$\sup_{x \in M} \left| K(g,x) - K(g',x) \right| \le C\|g- g'\|.$$

My current approach to this is quite clunky, and involves calculating everything directly from the Christoffel symbols of the metrics. Is there a better, more geometric approach to this than brute force calculations?

I'm sure this type of lemma is well known to geometric analysts. Is a proof of a similar result written down somewhere?

I thought it came from the theory of elliptic integrals / functions but I don't have a reference.31299617620071577879I'm going to say some things which might be either (a) obvious, (b) wrong, or (c) useless. (Or some combination!)

You could rephrase the question by asking that the map from SL_k x SL_k to SL_k x SL_k given by (x,y) -> (x,w(x,y)) is dominant. This seems like an improvement, because now you're talking about a map between two spaces of equal dimension.

If the corresponding map on the tangent spaces of Id x Id is an isomorphism, then certainly the map is dominant. (And this map is easy to compute, given w -- we just replace the product in w by the corresponding sum of tangent vectors.)

The map is dominant iff the map on tangent spaces is generically an isomorphism. I don't know how to check this, but my feeling is that it will be easier for this to fail than in the given conjecture.

206636313111120895418415978797501584192180763918424135258575326380 If you want a basis encoded by graph data, try searching for "cycle space basis" and/or "fundamental cycles". The null space of the constraint matrix is called the cycle space of the graph because it is spanned by the (signed) incidence vectors of the cycles. Although the cycles span the space, there are typically more than rank many of them. One way to choose a basis from among the cycles is to fix a spanning forest $F$ of $G$ and take, for each edge $e \in G \setminus F$, the unique cycle in $F \cup \{e\}$. These are the fundamental cycles of $G$ with respect to $F$ and they form a basis.799837919582267208a closed-form for mean/integral, but weighting positive differences between values and "mean" differently from negative differences?"Hilbert's goal was to use proof theory to show that from any (formal) proof of the continuum hypothesis, he could produce a disproof of the continuum hypothesis." - this should be the other way round.1248022Napoleon5478431619850A possible variant would be: - What about reading? Did you felt the same way about it? Because both are extremely important to understand the world.247747bAssistant Prof in Statistics at Rice University.615702522721875812Nb. for graphs (i.e. asking for the maximal number of $l$-cliques a graph can contain before it contains a $k$-clique) there are explicit bounds.198048819523341257274113813<Let me write $c=1/2+\delta$ and define $$P_n(\delta)\equiv \mathbb{P}\left[\sum_{i=1}^n X_i \leq n \cdot c\right].$$ Note that $P_n(0)=1/2$ and $P_n(1/2)=1$, irrespective of $n$.

It is convenient to work with the characteristic function of the Irwin-Hall distribution. I find the principal value integral $$\mathbb{P}\left[\sum_{i=1}^n X_i \leq n \cdot c\right]=\frac{1}{2}-\frac{1}{2\pi}\int_{-\infty}^\infty dt\,\frac{1}{(it)^{n+1}}\left(e^{it(1-c)}-e^{-ict}\right)^n,$$ which can be rewritten as $$P_n(\delta)=\frac{1}{2}+\frac{1}{\pi}\int_{0}^\infty dt\,\sin(2nt\delta)\,\frac{\sin^n t}{t^{n+1}}.$$

The plot shows $P_n(\delta)$ for $n=1,2,3,4$, from bottom curve to top curve. We need to demonstrate that this ordering of $P_n(\delta)$ with increasing $n$ holds for all $n$, so $P_n(\delta)$ increases with $n$ for all $\delta\in(0,1/2)$. I will attempt to prove this in several steps.

**(I) $P_n(\delta)$ increases with $n$ near $\delta=0$.**

For small $\delta$ the integral evaluates to

$$P_n(\delta)=\tfrac{1}{2}+2nC_n\delta+{\cal O}(\delta^3),$$

with the coefficient

$$C_n=\frac{1}{\pi}\int_0^\infty dt\,\frac{\sin^n t}{t^n}.$$

This integral over the $n$-th power of the sinc function is well-studied, we need to show that it decreases more slowly than $1/n$. For small $n$ this can be checked by explicit calculation, for large $n$ it follows from the asymptotic decay $C_n\propto 1/\sqrt n$. So the slope $P'_n(0)=2nC_n\propto \sqrt n$ is indeed an increasing function of $n$.

**(II) $P_n(\delta)$ increases with $n$ near $\delta=1/2$.**

This follows from a similar expansion around $\delta=1/2$, which shows that the first nonzero $p$-th order derivative $P_n^{(p)}(1/2)$ occurs for $p=n$. So near $\delta=1/2$ the function expands as

$$P_n(\delta)=1-(-1)^n A_n(\delta-1/2)^n+{\cal O}(\delta-1/2)^{n+1},\;\;A_n>0,$$

hence $P_n(\delta)$ increases with $n$ for $\delta$ just below $1/2$.

**(III) Large-$n$ asymptotics**

For $n\gg 1$ the sinc integral can be evaluated in closed form by means of the limit $$\lim_{n\rightarrow\infty}\left(\frac{\sin(t/\sqrt n)}{t/\sqrt n}\right)^n=\exp(-t^2/6)$$ so that $P_n(\delta)$ becomes asymptotically $$P_n(\delta)\rightarrow\frac{1}{2}+\frac{1}{\pi}\int_0^\infty dt\,\sin(2t\delta\sqrt n)\frac{1}{t}\exp(-t^2/6)=\frac{1}{2}+\frac{1}{2}\,{\rm Erf}\,(\delta\sqrt{6n}),$$ which has indeed a monotonically increasing $n$-dependence for each $\delta\in(0,1/2)$.

Note that we recover the $\sqrt n$ slope at $\delta=0$, $$P'_n(0)\rightarrow \sqrt{6n/\pi}\;\;\text{for}\;\;n\rightarrow\infty.$$ At $\delta=1$ the deviation from the exact limit $P_n(1/2)=1$ is exponentially small, $\propto\exp(-3n/2)/\sqrt n$.

The error function asymptotics is remarkably accurate already for small $n$, see the plot below for $n=3$ and $n=4$. For larger $n$ the exact and asymptotic curves are indistinguishable on the scale the figure.

**Bottom line.**

I would think that the finite-$n$ analytics near $\delta=0$ and $\delta=1/2$, together with the large-$n$ asymptotics for the whole interval $\delta\in(0,1/2)$, goes a long way towards a proof of the monotonic increase with $n$ of $P_n(\delta)$, although some further estimation of the error in the asymptotics is needed to complete the proof. I am surprised by the fact that the large-$n$ asymptotics is so accurate already for $n=4$.

It seems that it's certainly Hausdorff, as the topology of $k(X)$ is finer (if $U$ is open in $X$ then $U\cap K$ is open in $K$ for all compacta $K$, by definition of the subspace topology.) So the two separating sets that worked for $X$ still work for $k(X)$.1269177Not exactly a documentary, but appropriate for an undergrad math library: http://www.youtube.com/watch?v=wO61D9x6lNY21249701472321This is not really what the "derived" in "derived functor" means. A much closer analogy is Goodwillie calculus: https://ncatlab.org/nlab/show/Goodwillie+calculus1539908Try asking on math.stackexchange.com; here, your question is off-topic, as explained in the FAQ. Good luck!I think the clarification for $Map$ comes from different authors using it for different things, for example, simplicial enrichment, homotopy function complexes, etc etcNot every orientable 3-manifold is a double cover of $S^3$ branched over a link. For example, the 3-torus isn't. However, in 1975 Montesinos conjectured (Surjery on links and double branched covers of $S^3$, in: "Knots, groups and 3-manifolds", papers dedicated to the memory of R. Fox) that every orientable 3-manifold is a double branched cover of a sphere with handles i.e. the connected sum of a certain number of copies of $S^1\times S^2$ (this number can be zero, in which case we get $S^3$). Notice that this time $T^3$ does not provide a counter-example since if we take the quotient of $T^3$ by the involution $(x,y,z)\mapsto (x^{-1}, y^{-1},z)$ we get $S^2\times S^1$.

I was wondering what the status of this conjecture is.

Suppose we have a domain $\Omega\subset \mathbb{R}^n$ which is homeomorrphic to the unit ball $B(0,1)\subset \mathbb{R}^n$ and such that $\partial \Omega$ is of class $C^1$ (technically, this means that for every point in the domain we can give a $C^1$-diffeomorphism to the half-space in dimension $n$).

A map $f\colon (X,d_X)\to (Y,d_Y)$ between metric spaces is said to be bi-Lipschitz if there is a constant $K>0$ such that $$ \frac{1}{K} d_X(x_1,x_2) \leq d_Y(f(x_1), f(x_2)) \leq K d_X(x_1,x_2)$$ for all $x\in X$. The spaces $X$ and $Y$ are said to be Lipschitz equivalent if there is a surjective bi-Lipschitz map between them (any bi-Lipschitz map is necessarily injective).

*I wonder if in this case the domain $\Omega\subset \mathbb{R}^n$ is Lipschitz equivalent to the unit ball $B(0,1)\subset \mathbb{R}^n$. If true, is the condition $C^1$ necessary, sufficient?*

EDIT: My original question did not assume that the domain was homeomorphic to the ball. However, in this case it is clearly false, so I added this condition.

Robby: "I can't actually construct a situation where it doesn't seem somehow nicer to add "Now" or "Then" or "Consider when" or something similar" I think one reason for this is, that many mathematicians often insert a "Now" to avoid beginning a sentence with a symbol, so you are used to this kind of constructions. If you asked a non-mathematician, I think he might prefer a construction that began with a symbol. I agree that you shouldn't begin a sentence with a symbol, but I don't think that these constructions are "nicer" English. 19004841278051291218ZHow many of the true sentences are provable?@მამუკაჯიბლაძე Not really. If one of the $B[x]$ is empty, then the type $\prod(x:A).B[x]$ will also "be empty" in the sense of having no elements, but it will not be *definitionally* identical to "the empty type", so we cannot automatically conclude from this that it belongs to $U_i$.https://www.gravatar.com/avatar/adca2d099428f5452497c796ac810efe?s=128&d=identicon&r=PG&f=121017371495621608667This shows more generally that is $w(G)\ge d(G)-1$ then the answer to the question is: yes, there exists a generating set of minimal size, containing a subset realizing the weight.3820091663056672528998994Grad student at UC Berkeley, formerly an undergrad at Stanford. Interested in algebraic geometry, number theory, group theory, etc.

2268706184535285379pI would adopt it as a formula in combinatorics any day!TConformal Mappings for hyperbolic polygon961692256666216154032654014917291019902Serre's construction (or, I believe, a version of it which is enough here) takes a commutative ring $R$ and an $R$-category $C$ for which all idempotents have kernels. (An $R$-category is a category enriched in $R$-modules and which has direct sums.) Then for every finitely generated projected $R$-module $P$ and every object $A$ of $C$ we can define $A\bigotimes_RP$ characterised by the adjunction equality $\mathrm{Hom}_R(P,\mathrm{Hom}_C(A,B))=\mathrm{Hom}_A(A\bigotimes_RP,B)$. For $P=R^n$ we clearly can put $A\bigotimes_RP=A^n$ and for a general $P$ we write it as a summand of some $R^n$ and use the fact that idempotents have kernels in $C$. It is purely formal that this tensor product commutes with additive functors and $H_1(-,\mathbb Q)$ as a functor on the isogeny category has that property.

If I remember correctly Serre's construction is a version of this where $R$ is an arbitrary ring and we have a fixed ring homomorphism $R \rightarrow \mathrm{End}(A)$ and a right projective finitely generated $R$-module $P$ (and $C$ is an arbitrary additive category whose idempotents have kernels). We can then define $P\bigotimes_RA$ by the property that $\mathrm{Hom}_R(P,\mathrm{Hom}(A,B))=\mathrm{Hom}(P\bigotimes_RA,B)$. The proof of existence is almost identical to the one above.

455984"Any name deviating from a person's actual [legal] name can be considered a pseudonym" is setting the bar rather low. The French tend to spell my last name as "Israël", while Americans tend to render my first name as "Bob". I don't think those would be pseudonyms.20479801094087well, if we defined the sum over all $d\vert P_z$ we could bound it above nicely by $1/\log(z)^2$ using Euler products...it's the truncation of the sum that kills it. I was hoping that some work had been done on it.Most likely not. Why would one need such a table? And what is "every manifold"?1037151319593564071839798This post just received a downvote for some reason -- is there something I can do to clarify the question better? I've considered this for about a month on my own now and come up with no obvious generalizations, so prior to really digging in and trying to generalize it myself I was wondering if it had already been done.943424zAlgebraic planar curve with precisely $n$ closed components?146398017172164799741227105From the introduction: "A model-theoretic framework is employed, but I suggest that this is a consequence of the fact that intuitive inconsistent thinking is undeveloped (though not entirely absent) among mathematicians and logicians."676000lMaximal ideals of the algebra of measurable functionsFor an explicit example, you could use $A= \{ n: \lfloor \log_2 \log_2{n} \rfloor \text{ is even} \} $.

1654556dminimum of convex function in different variables363525772893406879tHard to say what it'll say without knowing what you want!149829618869712118623?In **physics**, the canonical example is the case of gauge theories. There are different levels of sophistication:

**Classical electromagnetism**: here the "physical" variables are the components of the two form $F$ (the electromagnetic field). In principle, one may formulate the theory purely in terms of $F$, with no extra camels. On the other hand, the general analysis greatly simplifies if one introduces extra variables in the form of a one form $A$, such that $F=\mathrm dA$. As usual, the differential form $A$ is defined up to an arbitrary exact form. General field configurations (the solutions of the equations of motion), when written in terms of $A$, depend on one arbitrary function, reflecting the presence of extra "unphysical" degrees of freedom.**Classical Yang-Mills**: here we have several two-forms $F$ (living in the adjoint representation of a certain Lie Group) such that $F=\mathrm dA+A\wedge A$. Here, and unlike before, it is impossible to formulate the theory purely in terms of $F$, and therefore one is forced to introduce the one forms $A$. As before, these are not uniquely determined given $F$, and therefore general field configurations depend on several arbitrary functions.**Quantum Yang-Mills**: in order to have a manageable and consistent theory, one is obliged to introduce more "unphysical variables", i.e., more extra fields. These are the so-called Faddeev-Popov ghosts, and they greatly simplify the general formalism but, once again, they are auxiliary, unobservable fields. They do not affect measurable predictions.**Other quantum gauge theories**: the most general and systematic approach to the quantisation of gauge theories that we know of as of today is the Batalin-Vilkovisky formalism. In this formalism, one introduces one auxiliary variable for each field that one is working with. That is, the first step towards the quantisation of the theory is to duplicate the number of variables (which include the gauge fields themselves, as well as any Faddeev-Popov ghost, or any other auxiliary field that is present). These variables are eventually eliminated, but their presence greatly simplifies the general analysis.

Other examples in physics:

**Stückelberg fields**, which are extra fields that are introduced to maintain gauge invariance in massive theories. In the same vein, the Goldstone-like components of the Higgs field.**Pauli-Villars fields**, which are introduced to ensure convergence of certain integrals. They are eventually eliminated (after making sure that there are no measurable divergences).**Inonu-Wigner contraction**: sometimes, one may understand a certain group as the contraction of a larger group. If the larger group is simpler than the contracted one, one may use this information to simplify the analysis of the contracted group. Such is the case of the Poincaré Group, which is the contraction of $\mathrm{SO}(4,1)$ (the latter being semi-simple, and the former not). The explicit contraction allows us to obtain, for example, the rank and Casimirs of Poincaré in terms of those of $\mathrm{SO}(4,1)$.The method of

**image charges**, which is a strategy to solve certain problems in electrostatics by introducing extra "fictitious" point-charges, and exploiting the symmetry of the resulting configuration. One should note that this technique can be used to obtain certain Green functions provided the domain is simple enough.

Finally, I'm surprised no one has mentioned the standard trick to calculate the **Gaussian integral**
$$
\int\mathrm e^{-x^2}\mathrm dx=\sqrt{\pi}
$$
by introducing a second coordinate $y$, such that
$$
\left(\int\mathrm e^{-x^2}\mathrm dx\right)^2=\int\mathrm e^{-r^2}\mathrm dx\,\mathrm dy
$$
and changing to polar coordinates.

other about me : http://orcid.org/0000-0001-8858-3789 https://www.reddit.com/user/nguyenthanhson9899/

Just FYI: I've just changed the nlab page to reflect the fact that $\|L_{(a,\lambda)}\|_{B(A)}$ does not equal $\|(a,\lambda)\|_{A^+}$ in the unital case.Although from 2009, the Kervaire invariant 1 problem is indeed worthy; the biggest breakthrough in algebraic topology in the last years.8398267 You can understand what is proved, because the definitions and theorems have to be understandable for someone to write and prove them in the first place. The theorems are the same you'd find in a book, only written with quantifiers more explicit. The formal proof objects are (unfortunately) not understandable, because they contain too much low-level detail. Development sizes vary a lot. Surprisingly, fancy stuff (eg, constructive topology) is often shorter than basic mathematics (eg, arithmetic or group theory), since "basic" things rely on a much wider set of obvious mathematical facts.@Tom Take $L= \mathcal{O}_{\mathbb{P}^1}(-2)$. Then $L \oplus L^*$ has three global sections, and we have a map $\mathcal{O}^3 \to L \oplus L^*$. Of course this map is not surjective, since the first summand in the right has no sections. However, we still have a map $\mathbb{P}^1 \to \mathbb{G}(1,3)$; the point is that this map is $degenerate$, in fact its image is a smooth conic contained in a $\mathbb{P}^2 \subset \mathbb{G}(1,3)$. Of course this $\mathbb{P}^2$ corresponds to the three sections of $L^*$.10187521689929.Here you have some of the coolest ones I have heard of:

**1)** Let $a$ be a positive integer. Then $a$ is a Fibonacci number if and only if at least one member of the set {$5a^{2}-4, 5a^{2}+4$} is a perfect square.

I think the result is original with Prof. Ira Gessel.

**2)** Let $\phi$ denote the Euler totient function. Prove that $\phi(F_{n}) \equiv 0 \pmod{4}$ if $n \geq 5$.

The proof consists of an unexpected application of Lagrange's theorem in Group Theory. Guess there are some other ways to prove it, but that approach will always remain my cup of tea. The problem was posed and solved in the Monthly in the 70's (if my memory serves me right). Look for all entries by Clark Kimberling in that magazine and you'll surely find it.

**3)** Can you find $(a,b,c) \in \mathbb{N}^{3}$ such that $ 2 < a < b < c$ and $F_{a} \cdot F_{b} = F_{c}$?

This problem would be trivial if instead of the $\cdot$ we had placed a plus sign there. In any case, there is no need to panic with this proposal. All you need to recall is the corresponding **primitive divisor theorem**.

**4)** Ben Linowitz mentioned above a beautiful result by Professor Florian Luca, namely:

**There aren't any perfect numbers in the Fibonacci sequence.**

I read the paper in my junior year and I didn't find it that hard to follow. The easy part of this cute note resides in the proof of the fact that there are no even perfect numbers in the Fibonacci sequence. Guess this result is interesting enough to deserve consideration in those lectures that you intend to give. If **this** proposal is not exactly your idea of excitement, you can take a look at some of the other papers by Professor Florian. He writes **a lot** about recurrence sequences. Another theorem of his, closely related with the subject matter of this discussion, ascertains that

**There is no non-abelian finite simple group whose order is a Fibonacci number.**

**5)** Last but not least... Prove that the sequence {$F_{n+1}/F_{n}$}$_{n \in \mathbb{N}}$ converges and use this fact to derive the continued fraction development for the golden ratio.

This one should be well-known, yet it would be nice to see what your students come up with...

**Added** (Nov 20/2010) I've just noticed that the **Fibonacci Assn.** has made available the articles published in *The Fibonacci Quarterly* between 1963 and 2003. I'm sure you will find plenty of additional material among those files that they have so generously released for our enjoyment. For instance, the seminal paper by J. H. E. Cohn that K. Buzzard mentions below can be found here.

Consider a non-compact complete Riemannian manifold $(M, g)$ with smooth compact boundary $\partial M$. Suppose also that $M \setminus \partial M$ has positive/non-negative Ricci curvature.

My question is, is it possible to remove $\partial M$, and replace it by another compact manifold $N$ with boundary $\partial N = \partial M$, and $M \cup N$ is a complete manifold (without boundary) of positive/non-negative Ricci curvature?

Let us assume the maximal volume growth on $M$, and let us also assume for convenience that $M \setminus K$, where $K$ is compact, has only one connected unbounded component. The naive mental picture I am having is "fitting a spherical cap to an end of a cylinder". But I am not sure how true such a heuristic is in general. Thanks in advance for any suggestions!

Yes, I managed to prove that it is possible to find three simple geodesic $\alpha,\beta$ and $\gamma$ that pairwise intersect only once and that cannot all intersect in the same point.@RW If you like, we could continue this discussion in chat or over email or, if anyone still uses it, on http://tea.mathoverflow.net However, as someone who has needed character theory of finite groups that is in Isaacs's book, apparently standard knowledge to various algebraists on MathOverflow, _but certainly not part of a functional analyst's toolkit_, I feel quite strongly that _one person's "obviously not research level" is another person's "exactly what I need to get me past a roadblock in my own research"._By maximal clique, do you mean a complete subgraph not contained in a larger complete subgraph or such a subgraph of maximal size?added an identity to clarify the relation between the fourth-root rate of growth of the Riesz function along the positive real axis, and the absence of zeroes of $\zeta(z)$ in the strip $1/2<{\rm Re}\,z<2$.92003378666567105421896411405787I suppose this is fair. As a category theorist, I generally treat preorders as categories, up to equivalence, so that they are equivalent to their preorder reflection. But I suppose if you're treating them differently then the difference matters. However, I do think this should have been a comment or chat discussion rather than an Answer.9049131544596Still learning to become a programmer... Interested in anything that pertains to computers.

Let $X$ be a singular irreducible projective curve over an algebraically closed field and $\pi : \widetilde{X} \to X$ the normalization morphism. In the book on Neron models by Bosch et al. (I have slightly simplified the setup), the Picard group of $X$ is computed by taking the long exact sequence associated to $$0 \to \mathcal{O}_X^* \to \pi_*\mathcal{O}_{\widetilde{X}}^* \to \pi_*\mathcal{O}_{\widetilde{X}}^*/\mathcal{O}_X^* \to 0.$$ This seems to depend on the assertion that $\text{Pic}(\widetilde{X}) \cong H^1(X,\pi_*\mathcal{O}_{\widetilde{X}}^*)$, which is not obvious to me. If $\mathcal{O}_{\widetilde{X}}^*$ were coherent, then this would be a consequence of the exactness of $\pi_*$ on coherent sheaves ($\pi$ is finite), but of course $\mathcal{O}_{\widetilde{X}}^*$ is not even a sheaf of $\mathcal{O}_{\widetilde{X}}$-modules. How can we resolve this difficulty?

An example where X is CM and X_{red} is not: Let k be a field of characteristic 2 and consider the subvariety of A^4 cut out by w^3*z=x^4 and w*z^3=y^4. This is a complete intersection, hence CM. On X_{red}, (wz)^4=(xy)^4 so, since char(k)=2, wz=xy. Similarly, x^{12}=w^9*z^3=w^8*y^4 so x^3=w^2 y and y^2=z^2 x. It is somewhat well known that k[w,x,y,z]/(wz-xy, x^3-w^2 y, y^3-z^2 x) is isomorphic to k[t^4, t^3 u, t u^3, u^4]. The latter ring is not normal (t^2 u^2 is missing), so X_{red} is not normal. Working a little harder, X_{red} is not S_2 so X_{red} is not CM.111159529872231552985Chintan Modi1783461h@YCor - The edit helps. Thanks for your assistance.1324377It is known that for $SU(N)$ $$ \int \chi_{\mu_1}(UV_1)\chi_{\mu_2}(U^{-1}V_2)\, dU = \delta_{\mu_1\mu_2}\frac{\chi_{\mu_1}(V_1V_2)}{\dim(\mu_1)} $$ where $dU$ is Haar measure on $SU(N)$ normalized such that with respect to it $\operatorname{Vol}(SU(N))=1$; and $\chi_{\mu}(U)$ means the trace of $U$ in the irreducible representation $\mu$.

I came across this integral in "On quantum gauge theories in two dimensions", Edward Witten, Commun. Math. Phys. 141, 153-209 (1991).

My question is simple. What else is known about integrals on $SU(N)$ of the form $$ \int \chi_{\mu_1}(U^{\pm 1}V_1)\dots\chi_{\mu_n}(U^{\pm 1}V_n)\, dU \ ? $$

1843388397509I don't understand. This question is the same as "classify all countable monoids", which is surely an unreasonable question. Equivariant quantum cohomology and K-theory of vector bundles over a Grassmannian1841219445389Algori, you're correct; I didn't see on a first look that you--and McGibbon, in his paper in the Handbook--are working with a different definition of "phantom" than I'm used to. Of course, your definition nixes anything I try to do with Eilenberg-Mac Lane spaces, and I don't have another example at hand.TInvariant two form under symplectic group1684639158106!First of all, this is probably best asked at stats.stackexchange.com

Aside from that, your question has a few distinct parts to it.

The calculated probability of 0.0159 is strictly the probability of there being an outage in any single random second in a year. If we make the simplifying assumption that the second to second probabilities of outage are not contingent on each other (which is a false assumption because outages are usually in blocks of time longer than a second) then the odds of experiencing an outage in any random 30 second session in a year is hyper-geometric:

$ \mathbb{P} \lbrace outage~in~30~seconds \rbrace = 1 - \mathbb{P} \lbrace no~outage~in~30~seconds \rbrace $

$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 1 - \frac{ \binom{31056926}{30} \binom{500000}{0} }{ \binom{31556926}{30} } $

This works out to 0.3806 or 38% chance of experiencing at least one second of outage in a 30 second session.

But this is perhaps not the question you are actually trying to address. One could also ask the question: given a random session then what are the odds of an outage in that session. To do this we need to make the simplifying assumption that the system load (number of concurrent sessions) does not affect system availability (this is also likely to be wrong). With this assumption, the probability can be found by marginalizing over the session statistics (we will denote the length of time of a single session as $r$ for residency, and the number of sessions that are resident for length $r$ as $n_r$):

$ \mathbb{P} \lbrace session~outage \rbrace = \frac{1}{16000000} \sum_{r=1}^{31056926} n_r (1 - \frac{ \binom{31056926}{r} \binom{500000}{0} }{ \binom{31556926}{r} }) $

One could introduce more nuanced Markov models to account for the affect of concurrent sessions and the dwell time of outages, but I'm not sure there would be a substantial gain in the predictive power and significance of the estimators, despite the clear inaccuracy of the simplifying assumptions that have been made.

See Flaxman et al, Strings with maximally many distinct subsequences and substrings, Electronic J Combinatorics 11 (2004) #R8, 10 pages, www.combinatorics.org/ojs/index.php/eljc/article/download/v11i1r8/pdf22415301565092Thanks to both of you: unknown and Greg. I think I understand. Your arguments sound convincing.8142381492491@GerryMyerson, you've written: "So far as I can see, ONE PERSON has downvoted, and you have no idea who it was, and it could have been an accidental slip of the finger for all you know, so, please, chill out" *** Gerry, young man, you act so ugly. Before your time such things among mathematicians I knew were unthinkable. Now, young man, that you think that you've provoked me again (:)), are you again eternally happy?But on this wikipedia page, "normed space" means normed space over the real or complex numbers, both complete fields. I'm not trying to be argumentative: I really don't know the answer myself (nor have I given myself a good chance to think about it; that's a little later in my course preparation), but I have concerns that the issue is more subtle than it appears. Can you give a reference that works with vector spaces over an arbitrary (not necessarily complete!) normed field?I have been working on characterizing the asymptotic behavior for large $n$ for the following sum:

$$\Gamma(1+t)\sum_{k=1}^n \frac{\binom{n}{k} (-1)^{k-1}}{k^t}$$

where $t(>0)$ is a given constant. The case for $t=1$ can be somewhat found here:

http://www.math.wvu.edu/~gould/Vol.4.PDF

in equation $8.25$. Using this equation, for large $n$, the sum behaves as $\log(n)$. I am trying to understand if similar behavior can be obtained for a general $t>0$.

Thanks in advance.

Hi!

Let f be a (continuous, $C^\infty$... whatever) function from $\mathbb{R}^n$ ($n \geq 2$) to $\mathbb{R}$. Assume that each connected component of $f^{-1} (0; \infty)$ and $f^{-1} (-\infty; 0)$ is unbounded. Also assume that $f$ changes sign so that neither $f^{-1} (0; \infty)$ nor $f^{-1} (-\infty; 0)$ is empty. Is it possible to prove that there exists at least one connected component of $f^{-1}(0)$ which is unbounded? Thanks.

424820631713https://www.gravatar.com/avatar/2ab9132a7101f2e65307bd1edeeefe5f?s=128&d=identicon&r=PG&f=1+Let $C$ be a convex shape in the plane. Your task is to cover the plane with copies of $C$, each under any rigid motion. My question is essentially: What is the worst $C$, the shape that forces the most wasteful overlap?

To be more precise, assume $C$ has unit area. Let $n_C(A)$ be the fewest copies of $C$ (under any rigid motions) that suffice to cover a disk of area $A$. I seek the $C$ that maximizes the "waste": $$ w(C) = \lim_{A \to \infty} n_C(A) / A \;.$$ So if $C$ is a perfect tiler of the plane, then $\lim_{A \to \infty} n_C(A) = A$ (because $C$ has area $1$) and $w(C)=1$, i.e., no waste.

Consider a regular pentagon $P$, which cannot tile the plane. Here is one way to cover the plane with regular pentagons:

If I've calculated correctly, this arrangement shows that $w(P) \le 1.510$. So one could cover an area $A=100$ with about $151$ unit-area regular pentagons, a $51$% waste. I doubt this is the best way to cover the plane with copies of $P$ (

Three questions.

. Is it known that the disk is the worst shape $C$ to cover the plane? My understanding is that L.F.Tóth's paper[1], which I have not accessed, establishes this forQ1latticetilings/coverings. Is it known for arbitrary coverings?

*Q1 Answered*. Thanks to several, and especially Yoav Kallus, for pointing me
in the right direction. Q1 remains an open problem. In [2,p.15], what I call
the waste of a convex body $C$ is called $\theta(C)$. It is about $1.209$ for a disk. The best upperbound is $\theta(C) \le 1.228$ due to Dan Ismailescu,
based on finding special tiling "p-hexagons" in $C$. A p-hexagon has two
opposite, parallel edges of the same length.

. Since every triangle, and every quadrilateral, tiles the plane, the first interesting polygonal shape is pentagons. What is the most wasteful pentagon?Q2

. More specifically, what is the waste $w(P)$ for the regular pentagon?Q3

[1] L. Fejes Tóth, "Lagerungen in der Ebene auf der Kugel und im Raum."

[2]
Brass, Peter, William OJ Moser, and János Pach. *Research Problems in Discrete Geometry*. Springer Science & Business Media, 2006.

10+ years experienced software engineer.

Dear Jim, I edited my answer to give the correct attribution for the construction of canonical models.2239386735028@Igor In fact, the math arXiv has had impressive growth and now receives more than 1/4 of total new submissions to the arXiv, so it is certainly not the case that physics has two orders of magnitude more volume. Also, as recounted elsewhere, Olga was quietly given the benefit of the doubt despite miscommunication with her. She is indeed distinguished, but that's not the reason that her case was reviewed last week. The ideal would be to give everyone the benefit of the doubt.3368705The point is the following. If one takes a reduced tensor product of a pair of $W^*$-algebras $A_1,A_2$ with states $\phi_1$, $\phi_2$, the resulting tensor product does not depend on the choice of $\phi_i$ as long as they give rise to equivalent GNS representations of $A_i$ (e.g. the states are faithful). Note, however, that it is not as straightforward as one might think to write an isomorphism between $(A_1,\phi_1)\otimes (A_2,\phi_2)$ and $(A_1,\phi'_1)\otimes (A_2,\phi'_2)$, since in general there is no unitary from $L^2(A_i,\phi_i)$ to $L^2(A_i,\phi'_i)$ which intertwines the actions of $A_i$. For $A_i$ finite-dimensional, there is an invertible linear map (similarity) $S_i : L^2(A_i,\phi_i)\to L^2(A_i,\phi'_i)$ with this property, but it is not isometric. There are analogs of this more general situations, too.

For infinite tensor products, the situation is even more delicate. Indeed, the naive way to prove isomorphism between $\bigotimes_i (A_i,\phi_i)$ and $\bigotimes_i (A_i,\phi'_i)$ even for finite-dimensional $A_i$ runs into the problem that similarities $S_i$ do not give a bounded map between the infinite tensor products of Hilbert spaces: $\bigotimes S_i$ may not exist.

It turns out, for example, that for $A_i\cong M_{2\times 2}$, the algebras of $2\times 2$ matrices and $\phi_i(x) = Tr(d x)$, with $d$ a positive operator with eigenvalues $\lambda_1,\lambda_2$ satisfying $\lambda_1+\lambda_2=1$, $\lambda_1/\lambda_2 = \lambda\in (0,1]$, the infinite von Neumann tensor product $\bigotimes (A_i,\phi_i)$ depends on $\lambda$. The resulting von Neumann algebra is called an ITPFI factor, and is the unique type $III_\lambda$ (if $\lambda\neq 1$) or type $II_1$ (if $\lambda=1$) hyperfinite factor.

8Buildings for Affine groups1817550221114584005710200522079965To give an extreme example, let's assume that the hat-numbering guy does not attempt to "choose a function at random" but simply always labels each hat with the number 0. Then it seems to me that precisely the same argument used to justify that the prisoners will win with prob 1 now also proves that they will win with prob 0, because what are the "chances" (whatever that means!) that they'll choose the zero function within its hugely uncountable equivalence class?? Zero! 1655998if the function $f$ is differentiable that is equivalent to $\frac{\partial f}{\partial x_i x_j x_m} \geq 0$186499342108120571491474719@JohnBaez $H^n(C_2\ast C_3,\mathbb{Z})\cong H^n(C_2,\mathbb{Z})\oplus H^n(C_3,\mathbb{Z})$, which is $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$ for $n=4$.1325013196725948813If you drop the positivity requirement then the result no longer holds (so it is not merely the case that the previous proof now fails, the desired conclusion is not true in general). Continuous bilinear forms on $C(X)\times C(Y)$ are sometimes known as _bimeasures_ and this is all related to various things concerning Grothendieck's inequality/theoremWith some preprocessing time, you should be able to do it in about N times n queries. I can't imagine it being done in much less time. For example, you need to distinguish between the case that you have N = 2^{n-1} - 1 disjoint two element intervals (when you are missing a two element interval from a complete cover) and the case with the same N where one of the intervals has four elements and is a complete cover. Here is how I would start, with the assumption that the N lattice intervals have been processed so that their top elements M_i and maximal elements (inside T_i and just below M_i) are accessible with a single query or at most n queries.

Start with the top level set which I will call n (avoiding square brackets). Preprocessing allows me to pick the set T that contains n, and the k many maximal elements of T just below n. If k=n (notation clash), then we are done as T covers the whole power set being an interval.

Otherwise there are n-k sets of size n-1 not covered by T, and we know which ones. For each of them, either there is another set T that covers that n-1 set, or we have found an uncovered set. Anyway, put the n-k sets in a list and go through them one by one.

Now if we are lucky, all the size n-2 sets under these (n-k) many size (n-1) sets are covered by n-k intervals T. Otherwise we have to start a list of size (n-2) sets that are not covered. (Also, if the first set T had k=1, that T is an interval containing n and a size (n-1) subset m, so we have to worry about all the size (n-2) subsets of m. So put those subsets of m in the n-2 list to check.)

In this way we descend level by level, checking uncovered sets of smaller size. However, because of the preprocessing, as we look at each potentially uncovered set, we either find a T which has that as its M, or we stop because the set M is not covered. For each such T, we check at most n subsets of its M to determine what sets one level down are not covered, which is at most nN queries.

Gerhard "Layer Cakes And Archaeological Excavations" Paseman, 2018.08.15.

EladjMalcev Lie algebra and associated graded Lie algebra1948137106069Thanks. I guess by degree I really meant the smallest possible degree of the polynomial representation of the algebraic map. Also in my application $M$ can be assumed to be compact and also equipped with a Riemannian metric.vWhat's the connection between the title and the questions?1815032hOK, an infinite graph... (I've edited the question))I'll try to help with intuition on why wavefunctions are complex-valued.

First, it's not actually true that they have to be complex. When you quantize electromagnetic waves, the wavefunction is simply the electric and magnetic fields, which are real. The correct statement is that a spin-1/2 particle's wavefunction has to be complex.

As a simple example of why this is, consider the case of two planar sine waves that are moving in antiparallel directions and then merge and superpose.

If the wavefunction is a real scalar, then at a time when the two superposed waves are 180 degrees out of phase, their sum is identically zero. This violates conservation of energy and, perhaps more importantly, conservation of probability.

If the wavefunction is a complex scalar, then one can prove that solutions of the Schrodinger equation always conserve probability. This is easy to check in an example like the superposition of $e^{i(kx+\omega t)}$ with $e^{i(-kx+\omega t)}$.

To see that this argument doesn't prove that wavefunctions are always complex-valued, consider electromagnetic waves in the same situation of superposition of antiparallel plane waves. Because there is a right-handed relationship among $\mathbf{E}$, $\mathbf{B}$, and $\mathbf{k}$, you can't make both $\mathbf{E}$ and $\mathbf{B}$ cancel. For example, you could choose the polarization so that $\mathbf{E}$ would cancel, but then $\mathbf{B}$ wouldn't.

To see that the fundamental issue is conservation of probability, so that this is really something specific to quantum mechanics rather than classical physics, consider the case of sound waves, which can be represented as real scalar functions $f$ measuring the pressure. Rerunning the same argument about superposing antiparallel plane waves, we find that it's possible for $f$ to cancel, but that's OK, because $f$ doesn't have a probability interpretation. We also still have conservation of energy, because the energy depends not just on $f$ (potential) but also on $\partial f/\partial t$ (kinetic), so $f$ can vanish without making the energy vanish.

Complex numbers come up in a lot of places in quantum mechanics, not just in wavefunctions, and it's not always obvious when they're just a notational convenience. For example, Pauli basically reinvented the quaternions in 1924. His spin matrices $\sigma_1$, $\sigma_2$, and $\sigma_3$ are equivalent to the quaternions i, j, and k if you multiply them by i.

Operators can be complex-valued, but expectation values are always supposed to be real, since they correspond to measurable quantities.

2204492135686523128The cocycle data which you review together is a map of 2-groupoids $B \Gamma \to B Aut(B G)$ to the delooping of the automorphism 2-group "of $G$" (really: of $BG$). As for any cocycle with coefficients in an automorphism group, there is the corresponding associatived 2-bundle, hence a $BG$-fiber bundle. That's the corresponding Giraud G-gerbe and that's essentially the group extension.

There are some details on this spelled out in the nLab entry *nonabelian group cohomology*, though maybe that needs another touch.

The general abstract story of nonabelian cocycle, infinity-gerbes and associated infinity-bundles is in section 4 of

- Thomas Nikolaus, Urs Schreiber, Danny Stevenson,
*Principal infinity-bundles - General theory*(arXiv:1207.0248, web)

The exponentials used in Fourier series are eigenvalues of shifts, and thus of any operator commuting with shifts, not just Laplacian. Similarly, spherical harmonics carry irreducible representations of $SO(3)$, and so they are eigenfunctions of any rotationally invariant operator.

If the underlying space has symmetries, it's no wonder that a basis respecting those symmetries has some nice properties.

Roots of a polynomial in a finite field related to Fermat's Last Theorem14055333722083302251021394http://www.nettv4u.com/celebrity/hindi/movie-actress/roma-manik26423755107422460671747762Depends on what you mean by "is"...In the sense that there exist higher homotopies that extend $(N(A), d, \Delta)$ to a full $E_\infty$ structure, yes. By itself though, $(N(A), d, \Delta)$ is not enough data to be $E_\infty$.60175355331931640773937361@darij At least the basic results in algebra need AC: the different definitions of noetherian aren't equivalent without DC. ( (http://mathoverflow.net/questions/53523/maximal-ideal-and-zorns-lemma). @SJR: What is the precise meaning of the statement "$\phi(t,x)$ exhibits exponential growth"? Thanks for your help.191926115814961977220@Donu: You're right, perhaps I mean a statement in algebra which becomes a surprising geometric interpretation.See the OP's update http://mathoverflow.net/questions/250266/average-distance-between-objects-in-a-cube#comment615365_2502661743547\Sato-Tate measure for GL(3) Automorphic forms10678221387154roghayeh abdi8188441222908bhttps://sites.google.com/csweb.haifa.ac.il/sajin179475020264878690571932406949286EFA can prove the exponential function to be total, but it cannot prove the superexponential function to be total. Is there an analog of Tennenbaum's theorem (which states the PA has no recursive non-standard models) for EFA which states that EFA has no "sub-superexponential" non-standard models? (Here "sub-superexponential" should mean something like "only finitely iterated exponential" space complexity.) Can one say even more, like that EFA has no primitive recursive non-standard models?

1362032881799It's possible for $a_n=n$ and probably most stepsizes without modular or growth obstructions.

We have covered some subset of an mxm square, are situated at the boundary, and want to visit a cell (x,y) in our square. Choose one of the x,y axes and move far away along it, (but not upon it), until stepsize s>>m and distance is some d from the axis. Then take either 1,2, or 4 more steps along the axis. Then alternately move away, and towards the axis, 2*d steps, until we land on it. Then by moving away and towards (x,y), n times, we can reach every point of the form j-1-3n on the axis, by just moving one more step towards (x,y) where j is our current coordinate, which we could shift to anything modulo 3 when we chose one of the 1,2,4 steps. And if 3n-n>m, we dont use any other squares within the the mxm square, to visit (x,y), and emerge on the opposit side. And since s>>m, if we take one more step we are at a boundary of a new square.

WLOG suppose we are at $(0,0)$, with stepsize $s(0)$, and want to visit $(x,y)$, $0\leq x\leq m$, $0\leq y \leq m$, The full path we take consist of these moves. We move south for $k$ squares (or $j(k)$ steps), then alternate west,east, $x$ times, now we are at $(x,-k)$ with stepsize $s(j(k)+2x)$. Then we alternate south, east, $n$ times, now we are at $(x,n-k)$ with stepsize $s(j(k)+2x+2n)$, select $n$ and $k$ such that $y=s(j(k)+2x+2n)+n-k$ and $s(j(k)+2x+2n)>m$. Then take two steps north, we are now at $(x,y+s(j(k)+2x+2n)+1)$. Move 1 step east, you are now at a corner of a square bounding all visited squares, define the new m to be the side of this new square, let (0,0) your position, pick a new point (x,y) and repeat.

PS. I asked a question about the less trivial 1-D version here: https://math.stackexchange.com/questions/111377/self-avoiding-walk-on-mathbbz

\https://i.stack.imgur.com/ejoji.jpg?s=128&g=112386201144739432907Good example -- I guess for k >> 1, there is no substantially better upper bound. I wonder if something is possible if $p(x)$ is a uniform distribution over $X$ and $|X|$ >> $e^k$...*I'll add an overlong comment to provide more perspective, in community-wiki format. The question is probably a bit misguided, even taken as a purely pedagogical one (the older mathematics involved having been developed over a century ago by Frobenius and Schur).

In the context of finite group representations, the groups $SL(2,q)$ provide an excellent example of how the classical theory works and are treated in a number of textbooks as well as in the notes by Mark Reeder linked by Peter. It's a good way to see how basic ideas in linear algebra, group theory, and field theory combine to produce some nontrivial results. On the other hand, the original computation of the character table (which encapsulates most of the essential information about the irreducible representations) involves an *ad hoc* step
after the study of the principal series. The problem is, as the question recognizes, that it's not immediately clear why you can't get all the desired representations via parabolic induction (here from the Borel subgroup of upper triangular matrices). The original calculations by Frobenius (over prime fields) and then by Schur (over arbitrary finite fields) are clever but don't provide a conceptual solution.

Some "high-tech" machinery may actually be essential here, to understand where the missing representations come from. The character table can be produced in this case by some clever moves involving the orthogonality relations, but beyond rank one progress gets very difficult. (J.A. Green used combinatorial and recursive methods for finite general linear groups, while his student Srinivasan pushed the ideas as far as $Sp(4,q)$.)

Only around 1976 did the much more sophisticated Deligne-Lusztig approach get developed, followed by Lusztig's extensive refinements; this is exposed in the rank one case in a concise text by Cedric Bonnafe (Springer, 2011). Much earlier, my 1975 expository article in the Math Monthly *here* laid out the Frobenius-Schur approach (working over a prime field, just for convenience), still with their *ad hoc* flavor.

Ultimately the pedagogy depends on what students actually know and where they intend to go next, but the question raised here about avoiding too much "high-tech" machinery probably has no good answer. The suggestions by Peter and Victor rely heavily on knowing the classical theory (relating representations and characters) as well as the main conclusions about characters of $SL(2,q)$, but they suggest no real explanation of what is going on.

2143112Ah, sorry, I was thinking of the other tensor product. The tensor product of a symmetric monoidal category $V$ lifts directly to the category ${\rm CMon}(V)$, where it is in fact the coproduct. But you're thinking of the tensor product induced by considering commutative monoids as a commutative theory, which represents 2-variable functors that *distribute* over the monoidal structure in each variable separately, which is different.1040401I don't know the original proof, but I heard that the trick of Rabinovich provided a drastic improvement of the proof of Hilbert's Nullstellensatz.You can do things this way. However, taking the Gromov closure only requires that the metric space in question be $\delta$-hyperbolic. It is an exercise to check that for any metric $\rho$ on $S$ (with $\chi(S) < 0$) the annular cover is $\delta$-hyperbolic. (However, it is true that $\delta$ depends $\rho$.)190342312315762044209I don't know how "different" you will consider this proof, but if you want a quick and efficient exposition I would suggest to have a look to Klaus Deimling's *Nonlinear Functional Analysis*.

I would like to share one picture which shows how flexibility of polyhedrons with faces removed can be tested. This is an illustration for the polytope proposed by Joseph:

The whole construction is made of plastic tubes that are hold by rubber bands.

If you can't get a position after a four years post-doc, I don't think that the problem can be where you got your PhD.111347263260017485265240131144670148668767586see http://smf4.emath.fr/Publications/Gazette/2010/125/smf_gazette_125_119-123.pdf1742694@Pierre: Thanks! @Péter: I've tracked down a copy and will be sure to take a look through it. (I also love that you included solutions!)7187211858788570961@Jim: To be honest my extremely limited understanding of the B-B result comes from the Morse theoretic description of it in section 2.4 of Chriss and Ginzburg's book on Complex Geometry and Representation Theory. However, I did originally look at the B-B Annals paper to see if my question was answered there. However, from my skimming of it I did not find anything helpful. But I do find that paper a tough read so could have easily missed a key point.Do you mean that the sum diverges almost surely? If so, then having constant expectation is not enough.If you wish, you may include the information that the lemma has been proved in your PhD dissertation, in the introduction, or in a line just before stating it.1811274@ Alex: For $p\geq 1$, a $p$-th order cone is defined as $\{(t,x) \in \mathbb{R} \times \mathbb{R}^{n} : ||x||_{p} \leq t \}$. The second-order cone is same as Lorentz cone, ak.a. ice-cream-cone. This is standard terminology in optimization-control.1636979jWhat are the minimal numbers of $(2\ 2)$-tetrahedra?15808661822027pParabolic cohomology of modular groups and cup-products496035As you say, the support of the measure is disjoint from the interior of the Reeb component.1148675508430168380813903662049554https://lh3.googleusercontent.com/-XkG88y387cM/AAAAAAAAAAI/AAAAAAAABIQ/HD4aeG4GDvo/photo.jpg1344978|It'd pretty natural, I have no reason to expect a connection.220207922343082259010@user74230 Thank you very much for your enlightening comment! Is the result true if we consider the cohomology on compact Stein manifold instead of arbitrary Stein manifold?1182816@DavidSpivak By the downset model I mean the one described in your post [Reference request: Heyting algebra structure on Catalan numbers](https://mathoverflow.net/q/272394/41291) that you link to. There you describe a bijection between downsets from $\Omega[m]$ and Dyck paths of length $m+2$. In $\Omega[m]$, the order is that of subset inclusion (of one downset into another), and I expressed doubt that this is the same order as transferred by that bijection from the height order on Dyck paths.22833491689275What's the definition of the scheme-theoretic complement for a non-reduced non-affine scheme anyway?Say $\rho$ is a tetrahedral representation of $G_F$, i.e., it is an irreducible 2-dimensional representation and the projection $\bar \rho$ to $PGL_2(\mathbb C)$ has image $A_4$. Then you can take $E/F$ to be a normal cubic extension such that $\bar \rho$ has image $V_4$ in $A_4$.

I guess you want to do the following: take a character $\xi$ of $G_E$ and induce to $G_F$ and say $Ind_E^F(\xi) \simeq \rho \oplus \psi$ for some character $\psi$. You cannot have such a decomposition because $E/F$ is normal, so this argument will not work.

To take an explicit example, suppose $\rho$ factors through an SL(2,3) extension, so you can view $\rho$ as an irreducible 2-dimensional representation of SL(2,3) (the double-cover of $A_4$). Then SL(2,3) has no index 2 subgroups so $\rho$ is not induced, and there is only one subgroup of index 3, a normal $Q_8$. The restriction of $\rho$ to this is irreducible, so you cannot construct $\rho$ as a (component of) induction of a character along a degree 2 or degree 3 extension. This is why Langlands' argument is so beautiful--it gives modularity for representations that are not easily constructed from characters.

11066731575320855332Yes it is : the kernel of $f^{\ast}$ is torsion, so that if you look at a "cohomology theory" without torsion, such as rational cohomology, $f^{\ast}$ will be injective.797312xOn the embedding of a function space $X$ into $L^2\cap L^4$Sorry, I believe it should be covered carefully at the undergraduate level, as expected preparation at the graduate level.2128627633098223559822681541292771ZDiscrete version of some topological object.vDecomposition of conic equation for two intersecting linesCool! If you find any bugs (and I'm sure there are plenty), I'd be much obliged if you send 'em my way.I'm with Reid. If G,H are finitely presented, and g,h elements, and we want to know if there's a morphism f:G->H with f(g)=h, wouldn't that restrict to a morphism on the cyclic group generated by g, and thus be immediately undecidable?

17310651015307467725I like to program Haskell and make things that do something cool.

I also like biking and building bikes. I am very proud of my America theme bike. I just need star-spangled handlebar tape now.

@FrançoisG.Dorais, my comment about the constructions of the integers in a set-theoretic context by von Neumann and Zermelo was precisely meant to point out that the Intended Interpretation *cannot* be understood in a set-theoretic context because that would make it vacuous: the theory is interpreted in a set-theoretic context by assigning to the symbols 1,2,3 their set-theoretic meaning. This is vacuous because this is true in any interpretation and not specifically the Intended Interpretation, so obviously this is not what Wang meant. Rather, what he seems to have meant is an informal...Is the superconformal algebra in five dimensions, $F(4)$, a subalgebra of the (maximal) six-dimensional superconformal algebra $OSp(8|4)$?

13824151443909411878292226893Well, the modern viewpoint relays on the interpretation of "fourier transform" (in any generalized fashion you like to define "fourier transform") in representation-theoretic language. As a consequence, there are several approaches today to get the trace formula (either more analytic by Green's functions or the more general representation theoretic manner).

A nice introductory account can be found here by Marklof - http://arxiv.org/pdf/math/0407288v2.pdf Another representation-free approach is done in Iwaniec's "spectral methods of automorphic forms".

The theorem that Hejhal mentioned is very well known for general (compact, closed) manifolds (follows from Poincare inequality), but in the (cocompact) homogeneous case, one can overcome many analytical complications by just mimicking the proof of the Peter-Weyl theorem in representation theory of compact groups (Hilbert-Schmidt operators and so on). In particular, no Sobolev computations whatsoever, that shows one simple example of the advantages of using representation theory.

A more advanced approach (which uses some representation theory) is found in Knapp's article - http://sporadic.stanford.edu/bump/match/trace.pdf Probably a good introduction to this article is Bump's book about automorphic representations (chapters 1-2 I guess, you only need the real part for this article).

\https://i.stack.imgur.com/xeh1I.jpg?s=128&g=11986019hWhen you say *limiting* do you refer to what limit?I edited again the question. As your example shown, I think that also some conditions on $z$ are necessary. Thanks.10269331607826I didn't understand the last argument - on what space does the group Mp(2) act?I was going to comment that the solution set condition from Giraud's theorem should ensure Grothendieck topoi satisfy WISC...229239918931971382086222633114748<@gordon,no it's not necessary1175542502236Also, just see the [equivalent definition (3)](https://en.wikipedia.org/wiki/Absolute_continuity#Equivalent_definitions) of absolute continuity on Wikipedia.1022318918242041785https://www.gravatar.com/avatar/88e8864f49b62a5024e39c6cfa4ea965?s=128&d=identicon&r=PG&f=1auceps18116201904696322443I case $n=2$, (it seems that) you are asking if any geodesic is stable. This is obviously not true for general $g$.@JamesSmith I don't remember now, but probably I found it here (I'm not sure about this -- maybe it was indeed Wikipedia): https://books.google.ru/books?id=hr2ZfC7upmgC&pg=PR37&lpg=PR37&dq=Tibetan+Dream+Yoga+ramanujan&source=bl&ots=LYhSjQUI36&sig=FGxVot-8n8w_En-5Cd9CJR12V5Q&hl=ru&sa=X&ved=0ahUKEwihiq2vvPLXAhUjYZoKHZZMAa0Q6AEIOjAH#v=onepage&q=Tibetan%20Dream%20Yoga%20ramanujan&f=false2695878677511553232243811719766fSuppose $\epsilon_{r} =(1-r)^{2}$ then it's false?1357969RynantI studied the inequality in case $n=3$, see

http://link.springer.com/article/10.1007%2Fs00010-012-0178-2

You can see from this paper that not every subadditive function ($n=2$) satisfies inequality for $n=3$. So (GH3) is not equivalent to (GH1) and (GH2). I don't know what happens for greater $n$.

15187631421390Todd, Rudy Rucker's book was what got me wondering about this question. He stops short of explaining measurable cardinals, except for saying "they are so big that comparing them to run-of-the-mill inaccessibles is like comparing $\omega$ to 2". But the definition of a measurable cardinal does little to convey WHY it is so large.2968411063392221728136171620026112109753255972@RajeshD: the Fourier transform **is** coordinate independent in the sense that the Euclidean group of symmetries augmented by the dilations and scaling act predictably on the Fourier transform. In particular, rotating your function in physical space can be balanced out by a rotation of your function in frequency space. Your characterisation, however, is not symmetric with respect to that.17901314375211298244941612126431896161902037bIs the Sorgenfrey Line monotonically monolithic?I use $\dim_H(E)$ to denote the Hausdorff dimension of a set $E \subseteq \mathbb{R}$ and $|E|$ to denote its Lebesgue measure. It is easy to see from the definition of Hausdorff dimension that if $\dim_H(E) < 1$, then $|E| = 0$. The converse is not true, and there are many cases where $\dim_H(E) = 1$ yet $|E| = 0$. So the question:

**What was the first (or most elementary) example of this phenomenon?**

After some looking around, I was able to prove that a central Cantor set $C$ with ratio of dissection $r_k = 1/(2+\frac{1}{k})$ satisfies the condition I want. It is easy to see that $|C| = 0$ since at step $n$ of the process to construct this Cantor set, it has measure $2^n(r_1 \cdots r_n)$ which in this case limits to 0, but for the Hausdorff dimension I required a non-trivial result from the paper *Sums of Cantor sets* (Cabrelli, Hare, Molter) that gave the formula

$\dim_H(C) = \liminf_n \frac{n \ln 2}{\ln r_1 \cdots r_n}$.

This result is fairly recent and sophisticated, and I feel that there should be older and simpler examples.

18974877808394OrentetfAre smooth functions dense in the *operator* norm?8127057379989502Yes, this is what is meant by the question. I was under that impression that such a function could not be Lipschitz continuous, for example. What's the argument that this is the support?Someone named "reference" attempted to add a comment as follows: you say in your answer that Pérez Marco's results are "not really so new". Could you give any references to the numerical analysis of sections 4 to 8? 309800531895Let $p$ and $q$ be probability densities on $\mathbb R$, with respect to the Lebesgue measure $dx$. The corresponding Hellinger integral is $H(p,q):=\int_{\mathbb R}\sqrt{pq}\,dx$.

Let now $p$ be the density of Student's distribution with $d$ degrees of freedom, so that $$p(x)=C_d\,(1+x^2/d)^{-(d+1)/2}$$ for real $x$, where $d\in(0,\infty)$ and $C_d$ does not depend on $x$.

At least for $d=1$, is there a closed form expression for $H(p,p_t)$ for real $t\ne0$, where $p_t(x):=p(x-t)$ for real $x,t$?

An obviously equivalent form of this question: Is there a closed form expression for the integral $$\int_{\mathbb R}\frac{dx}{(1+x^2)^a\,(1+(x-t)^2)^a} $$ for all real $t\ne0$ and all real $a>1/4$ (or at least for $a=1/2$)? Of course, this is not a problem for natural $a$.

113191718593731023928Thank you, that would answer the question. But, how would you apply Hartshorne's Connectedness Theorem to prove that? Also, I guess you mean the range where $X^{[n]}_{sm}$ is not an irreducible component of $X^{[n]}$, since $X^{[n]}$ is always connected.Here is a suggestion for a problem related to the one you are asking. I will give the two dimensional version; perhaps Joseph O'Rourke will be intrigued enough to illustrate this or a 3 or higher dimensional version for us.

I will suggest a coloring/labeling for an integer grid, with x and y coordinates being all integers greater than m respectively n. I recommend m=1 and n=1 ranging up to 40, but you can choose differently depending on your graph paper. Also, you can decide whether to take advantage of symmetry or not and restrict yourself (or not) to the region x <= y.

Each coordinate (x,y) will be assigned the integer xy+1, but do not label every such coordinate. Instead, use whatever color scheme to sieve out nontrivial multiples of 2 (meaning x and y are both odd, and xy>1), interesting multiples of 3 (those pairs where xy = 2 mod 3 and xy >2), big multiples of 5, and so on, not forgetting to mark the primes encountered. That is, label a coordinate (x,y) with xy + 1 only if that quantity is prime.

You will end up with a colored grid, with the colors having a pleasing pattern (I think) and a look up chart for small m and n of primes greater than mn, where you look at or above and to the right (if you use my orientation) of the coordinate (m,n) for such primes, and try to find the smallest one of the nearest such primes. From generating such a chart for various m,n, you may get a sense of how close heuristically your desired prime is. I predict the value of the smallest such prime p(mn)= 1 + (m+i)(n+j) for nonnegative i and j satisfies p(mn) < (n+m)*m*n for positive integers m,n.

Gerhard "Ask Me About Prime Guesses" Paseman, 2011.10.04

I am a little confused about 'to a unique solution for arbitrary k' since k is bounded by a function of q,n,m right?4126420272941890052$There is a rather obvious generalization of your example to cubical pushouts of any dimension. Let $I_n$ be the poset of **proper** subsets of $\{1, ..., n\}$, and $\chi\colon I \to $Sset a diagram indexed on $I$. Let $I_{n-1}$ be the subposet of sets not containing $n$, and $I^n$ the subposet of sets containing $n$ (you can think of these subposets as the back and front faces of a punctured cubical diagram). Suppose that the restrictions of $\chi$ to $I_{n-1}$ and $I^n$ are both cofibrant. Then the map $\mbox{ hocolim }\chi \to \mbox{colim }\chi$ is an equivalence. The reason is that $\mbox{hocolim } \chi$ is equivalent to the homotopy pushout
$$\mbox{hocolim}_{I_{n-1}}\,\, \chi\leftarrow \mbox{hocolim}_{I_{n-1}^1} \,\,\chi \rightarrow \mbox{hocolim}_{I^n} \,\,\chi $$
Here $I_{n-1}^1$ is $I_{n-1}$ minus its final object, which is the set $\{1, \ldots, n-1\}$. There is a similar decomposition of colim $\chi$. It is isomorphic to the strict pushout
$$\mbox{colim}_{I_{n-1}}\,\, \chi\leftarrow \mbox{colim}_{I_{n-1}^1} \,\,\chi \rightarrow \mbox{colim}_{I^n} \,\,\chi $$
Our assumptions guarantee that the left map in this pushout is a cofibration, so pushout=homotopy pushout. The assumptions also guarantee that the natural map from the first pushout diagram to the second is a pointwise equivalence, so it induces an equivalence of homotopy pushouts. So it induces an equivalence hocolim $\chi \to$ colim $\chi$.

Note that $I^n$ is isomorphic to $I_{n-1}$, so by induction you can replace the hypothesis that $\chi$ restricted to ${I^n}$ is cofibrant with a weaker one, so long as $n-1>1$.

The case $n=2$ is equivalent to your example.

Another example: Let $G$ be a finite group (probably can be a more general class of groups - how general?) acting on a **pointed** simplicial set $X$. You can think of this action as defining a diagram of pointed simplicial sets. The colimit of this diagram is the orbit space $X/G$. The homotopy colimit is the pointed homotopy orbit space $X\wedge_G EG_+$. $X$ is cofibrant if the action of $G$ is free except for the basepoint. However, the weaker assumption that $X^H$ is contractible for all non-trivial subgroups $H\subset G$ suffices to conclude that the map from the homotopy orbits space to the strict orbits space is an equivalence.

Instead of considering the *closed* stars, it is convenient to consider inscribed broken lines with constant leg lengths.

**Lemma.** Let $AOB$ be an angle, let the points $X$ and $Y$ move along $AO$ and $OB$ monotonically, so that $X$ moves with the constant speed, and $XY$ is constant. Then the coordinate of $Y$ changes concavely.

*Proof.* A direct computation via cosines rule.

**Corollary.** Let $X_0X_1\dots X_{2\ell+1}$ be a broken line with constant leg length inscribef into a regular polygon, whise vertices move monotonically along the sides of the polygon, $X_0$ with a constant speed. Then $X_{2\ell+1}$ moves concavely, until some vertex of the broken line meets a vertex of the polygon.

*Proof.* A composition of increasing concave functions is also concave.

Now return back to the original question. Let $X_0X_1\dots X_{2\ell+1}$ be an inscribed star (whose vertices are distinct from vertices of the polygon and modpoints of its sides), $X_{2\ell+1}=X_0$. Let its points move as in the corollary ($X_0$ and $X_{2\ell+1}$ become distinct). Either forward or backward, this broken line meets a configuration symmetric to the original one (which is also closed) before tracing a vertex of the polygon. By concavity, between these two positions $X_{2\ell+1}$ went ``ahead'' of $X_0$, which means that the inscribed stars starting at those positions of $X_0$ were shorter.

Notice that, due to the symmetry, between our two positions there was exactly one with a midpoint of a polygon's side on the star. This yields the required monotonicity.

71505966024Zcriterions for polar set of Feller processeszOne dimensional projection of complex analytic sets (Edited)641676477210It is not hard to show $(\Delta_2/\partial\Delta_2)^{\Delta_1}$ is finite (in fact, $4$-dimensional if I'm not mistaken). Just enumerate what a sequence $(x_0,\dots,x_n)$ can look like, and you'll only find $O(n^4)$ possibilities (this is a little messy but not hard, the point being that $x_i$ very nearly determines $x_{i+1}$).1710200(cont) I guess that the last section, comparing with the matrix models, concedes this point, but lacks lucidity as to wheher, let alone why, these two regimes can exhibit similar statistical properties.125697C & R looks like exactly the thing I've been looking for! Thanks for the input guys.167976514535578296712292324@FedorPetrov: That of course also immediately follows from the more complete result I quoted since any null set is contained in a $G_{\delta}$ null set by regularity.447735RUnderstanding the Weyl Character Formula20145421086874Exceptional isomorphisms between finite simple Chevalley groups@AlexeyUstinov Yes, I think they bear some links to algebraic topology. As I pointed out in an answer to your another MO question on the eigenvectors of these matrices, these polynomials are related to the generating function $\displaystyle\left(\frac{t}{\sinh t}\right)^y$. While I do not know precisely the link, but the similar function $\displaystyle \frac{\frac{\sqrt{t}}{2}}{\sinh \frac{\sqrt{t}}{2}}$ is used to define the $\widehat{A}$-genus of manifolds. See https://en.wikipedia.org/wiki/Genus_of_a_multiplicative_sequenceIts limit set should be homeomorphic to a circle. Do you want an algebraic or geometric criterion? Do you have an application in mind?100664317567899705111104760As for a result that was not simply incorrectly proved, but actually false, there is the case of the Severi bound(*) for the maximum number of singular double points of a surface in P^3. The prediction implies that there are no surfaces in P^3 **of degree 6** with more than 52 nodes, but in fact there are such surfaces in P^3 with 65 nodes such as the Barth sextic (and this is optimal by Jaffe--Ruberman).

(*) Francesco Severi; "Sul massimo numero di nodi di una superficie di dato ordine dello spazio ordinario o di una forma di un iperspazio." Ann. Mat. Pura Appl. (4) 25, (1946). 1--41, MR0025179, doi:10.1007/bf02418077.

|https://graph.facebook.com/143240243251983/picture?type=large17195351218439dUniversity of Illinois at Urbana Champaign (UIUC)1036192143873214822072107487538773ZansonOne general answer to this is to say: a generalized smooth thing is an oo-stack on some site of smooth test spaces like Diff or so. These are "smooth oo-groupoids" in a useful sense. There is some discussion aimed towards smooth classifying spaces from this perspective at motivation for sheaves, cohomology and higher stacks.

3266241754539You I right, I obviously missed point. So probably relation K should be king of "global relation" and not for particular vertexes, as Cayley graph is always self similar, although not always infinite...19902171116643Family of functions which satisfies $f(\boldsymbol{x}) = 0$ if $\nabla f(\boldsymbol{x})=0$?https://lh5.googleusercontent.com/-LvsX2h4HBLE/AAAAAAAAAAI/AAAAAAAAACs/DmEm0qAx1Iw/photo.jpg?sz=1289015171455869.. Thus I think the two orbits found by Stefan will fuse in $PGL_2(11)$. And, in general, there will be one orbit of $(2,3)$-pairs with product equal to $5$ in the automorphism group of $G$. For other product orders things will be different I guess.537254https://lh6.googleusercontent.com/-XNYZleX4FcI/AAAAAAAAAAI/AAAAAAAAAFE/OPf8xthL_sE/photo.jpgCS student Interesting in Computer Vision, Machine, Learning, Optimization, Python, C++, functional programming

88411733151921590631348558204224710540891491244310335lI've posted this as an answer with some explanations.2278486389741881044My unofficial best guess is that the length is some divisor of the product of the two lengths of the sequences, perhaps the minimal such length that has a whole number of periods from each sequence. Gerhard "Six Plus Three Equals Two?" Paseman, 2012.01.13What are the current views on consistency of Reinhardt cardinals without AC?294468019378831363203I'm not used to talking about species, bu the fact that L is a torsor for P should be restatable as saying that LxL is canonically isomorphic to LxP.You must be more precise in what you are willing to do. The answer to your question is depending on.1638092Hejhal in his book on the trace formula combined distinct $K$-types to obtain the Eichler-Selberg trace formula for modular forms. Hejhal attributes this idea to Selberg. I am not familiar with Selberg's original paper, but there is no argument with support on elliptic elements in Hejhal.76887113021761844908A general bit of advice for answering questions of this kind: calculate the first several values, and enter them into the box here: http://oeis.org/ 147073520773341314762147494720038051951439Could be because I'm in my last year of undergraduate and haven't taken an axiomatic set theory course19391101736924264358Consider two positive integers $x \ne y$ and let $n = max\{\lfloor \log_2{x} \rfloor +1 ,\lfloor \log_2{y} \rfloor +1 \}$. Choose a prime $p$ randomly from the first $3n$ primes. What is the probability that $x \bmod p = y \bmod p$?

I believe it is at most $1/3$. My reasoning is that there are only $n$ primes at most for which $x \bmod p = y \bmod p$. Does this make sense and is there a self contained proof?

I am interested in upper and lower bounds for this probability, especially those that hold for large $n$.

The same question is also at https://math.stackexchange.com/questions/567230/probability-of-equality-mod-p where an upper bound of $2/9$ is now conjectured.

2040183dContact structures on pseudo-riemannian manifoldsLooking at his webpage, he certainly could be, but I don't actually know -- this kind of theorem proving is (a) a huge and extremely active area, and (b) just outside my own expertise. SMT is a very applied field, and as a result they are happy to use any mathematics from any area to make their programs run faster. 1657008lLocus classicus for the Chevalley restriction theorem279164148489252418217704954168512250237Noah: yes but using $\liminf$ or $\limsup$ gives actually equivalent conditions in general, when stated for every sequence converging to $\gamma.$ Rinverse M-matrix times mixed-sign vector228906If $f(x,y)=0$ is a counterexample in two variables then $f(x_1+\ldots+x_n,y)=0$ is also.12853Indeed Will, there should be five: I conflated two classes when I should not have. I will edit later and include some more remarks. Gerhard "Ask Me About System Design" Paseman, 2011.12.221986290818342341165Your final formula doesn't involve T. But is this really of interest to research level mathematicians? It just seems like youre asking someone to check your calculations.160027819064491076560By the way, if $A$ is an algebra, there is a similar statement for strongly $G$-graded algebras with unit component $A$, and actions of $G$ on the category of $A$-modules. Do you know if this is also done in SGA 7 ? Thanks again.#Hi! I'm new here. It would be awesome if someone knows a good answer.

Is there a good lower bound for the tail of sums of binomial coefficients? I'm particularly interested in the simplest case $\sum_{i=0}^k {n \choose i}$. It would be extra good if the bound is general enough to apply to $\sum_{i=0}^k {n \choose i}(1-\epsilon)^{n-i}\epsilon^i$.

For the more commonly used upper bound, variants of Chernoff, Hoeffding, or the more general Bernstein inequalities are used. But for the lower bound, what can we do?

One could use Stirling to compute $n!$ and then ${n \choose k}$ and then take the sum:

${n \choose k} = \frac{n!}{k!}{(n-k)!}$, and Stirling's formula (a version due to Robbins) gives $$n! = \sqrt{2\pi}n^{-1/2}e^{n-r(n)}$$ with remainder $r(n)$ satisfying $\frac{1}{12n} \leq r(n) \leq \frac{1}{12n+1}$.

For the next step, it's easy to apply Stirling thrice. But, even better, I noticed that Stanica 2001 has a slight improvement to the lower bound that also is simpler to state (but more difficult to prove):

$${n \choose k} \geq \frac{1}{\sqrt{2\pi}}2^{nH(k/n)}n^{1/2}k^{-1/2}(n-k)^{-1/2}e^{-\frac{1}{8n}}$$

for $H(\delta) = -\delta \log \delta -(1-\delta)\log(1-\delta)$ being the entropy of a coin of probability $\delta$.

Now for step 3. If $k$ is small, it's reasonable to approximate the sum by its largest term, which should be the ${n \choose k}$ term unless $\epsilon$ is even smaller than $k/n$. So that's great, we're done!

But wait. This bound is off by a factor of at most $\sqrt{n}$. It would be better to be off by at most $1 + O(n^{-1})$, like we could get if we have the appropriate Taylor series. Is there a nice way to do the sum? Should I compute $\int_{0}^{k/n} 2^{nH(x)}\frac{1}{\sqrt{2\pi}}x^{-1/2}(1-x)^{-1/2}n^{1/2}e^{-1/8n} dx$ and compare that to the discrete sum, and try to bound the difference? (This technique has worked for Stirling-type bounds.) (The terms not dependent on $k$ or $x$ can be moved out of the integral.)

Another approach would be to start from Chernoff rather than Stirling (i.e. "How tight is Chernoff guaranteed to be, as a function of n and k/n?")

Any ideas or references? Thanks!

2287724\https://i.stack.imgur.com/goWbZ.jpg?s=128&g=16774181118156For a related question please see http://mathoverflow.net/questions/157420/a-converse-to-the-gauss-bonnet-theorem901179But is holomorphic scalar curvature necessarily equal to scalar curvature? It seems to me that scalar curvature has more terms in it.11550061893431128371I answer the second question myself, as I could find a reference for it: the indicated base change is indeed an iso. The reference is the article "Cohomology of Algebraic Varieties", paragraph 2, Thm. 2.2, "Algebraic Geometry II" (ed.: Shafarevich) where the claim is shown for proper morphisms and coherent sheaves.Say we have a map, C->D, of relative curves over a Dedekind scheme, S. What are some of the available methods for showing that this map has good reduction, or integral reduction, at some s∈S? By this I mean: what are some popular conditions that imply this? What are the tricks people usually use?

By a map having good reduction I mean that both C_{s} and D_{s} are regular integral curves. By integral reduction I mean that both C_{s} and D_{s} are integral curves.

You may assume whatever you want, this is part of the question. Assuming, for example, that C->D is generically Galois; or that D is smooth over S; is legitimate. This is pretty open-ended. Hence, community wiki.

6406632118568|I think I can do it; I'll write it up and post a bit later...346813If you neglect the third relation, the abstract group with the first two relations is the Artin braid group $B_N$ on $N$ letters. For $N=5$, there are certainly representations (The Burau representation) into $U(h)$ for suitable Hermitian forms $h$ in four variables, whose projection into $U(4)$ (a compact group) contains $SU(4)$ in their closure. See a recent paper by Curt Mcmullen (Braid groups and Hodge theory, Math Annalen ) where such constructions are given19905118908281381957It is true under large cardinals. So the question is asking if the assumption can be weakened to universal Baireness. Do I understand the question correctly?935034865174Neverender1922468j@ Andrés E. Caicedo : Yes, I mean Borel measurable.175565797268zCharacter table of $\mathrm{SL}_2(\mathbb{Z}/p^n\mathbb{Z})$Thanks for reminding me of the paper. The problem is just that scl=0 does not imply finite commutator width.1051054In case it helps people not native or not fluent enough in English (including myself): **bottleneck** = *a problem that delays progress*; from here: http://dictionary.cambridge.org/fr/dictionnaire/anglais/bottleneckI think the poster was asking about $O_P$, where $P$ is a prime ideal of $O$ above $p$, in which case the answer is different.1015957150816319408652282391vBirational automorphisms of varieties of Picard number one13421608756912087349682915@MarianoSuárez-Alvarez - Sorry I'm not really familiar with this book and can't find it. Could you tell me where exactly should i look.18875842264102Observe Sturm-Liouville problem on entire line $$-(p(x)y'(x))' + l(x)y(x)= \lambda r(x)y(x), \hspace{3mm} -\infty<x<\infty \tag{1} \label{1}$$ where $p(x)$ and $r(x)$ are positive on $\mathbb{R}$. Assume $p(x),r(x),l(x) \in C^{2}(\mathbb{R})$ (maybe it is enough $p(x)$ has a contionous first derivate and $p(x)r(x)$ has continous second derivate).

$\mathbf{Question:}$ How to reduce $\eqref{1}$ to the form
$$-u''(x) + q(x)u(x)= \mu u(x) $$
By Levitan, Sargasjan - Introduction to spectral theory,

substitution on finite interval $[a,b]$ is given
$$z=\frac{1}{c} \int_{a}^{x} \bigg(\frac{r(x)}{p(x)}\bigg)^{1/2} dx, \hspace{3mm}u=(r(x)p(x))^{1/4}y(x),\hspace{3mm} \mu=c \lambda,$$
where c is given by $$c=\frac{1}{\pi}\int_{a}^{b} \bigg(\frac{r(x)}{p(x)}\bigg)^{1/2}dx$$ and $$q(z)=\frac{\theta''(z)}{\theta(z)}-c^2 \frac{l(x)}{r(x)}, \hspace{3mm} \theta(z)=(r(x)p(x))^{1/4}$$
As the result of this change the interval $[a,b]$ is transformed to the interval $[0,\pi]$.

Is it same for whole line $\mathbb{R}$?

*Let me explain slightly modified example 4.6 by A'Campo-Neuen and Hausen*.

Suppose that $\mathbb C^*$ acts on $X=\mathbb P^1_{x_1:x_2} \times \mathbb A^2_{y, z}$ (indices denote corresponding coordinates): $$t \cdot ((x_1 \colon x_2); \; y; \; z)=((tx_1\colon x_2); \; ty; \; t^{-1}z).$$

There are two rational invariants, $x_1z/x_2$ and $yz$, and one can check that $X \to \mathbb A^1_{yz}$ is a categorical quotient. But the restriction $$Y=\pi^{-1}(\mathbb A^1_{yz}-0) \to (\mathbb A^1_{yz}-0)$$ is not a categorical quotient, because now there is $$Y \to \mathbb P^1_{x_1z:x_2} \times (A^1_{yz}-0).$$

4522502949188659196896160I would have thought this got mentioned in *some* book on either commutative algebra or differential geometry... I'm not an expert, but the universal property of Kahler differentials should be suggestive.143436612700672038093@LinusHamilton thank you for the comment. may you more explain about the picture?8779201415944292781fA Knowledge Enthusiast and a Learning Bug.

14741617680101154073I probably should have mentioned that the reduction theorem also holds with a single type $p$ in place of the sequence $\{p_n:n<\omega\}$, but the proof is slightly more complicated.383748584090210069Yes, @darijgrinberg that would be useful, especially if it included enough explanation for a non-combinatorialist about what a raising operator is to understand the issue.222132518115781169869I am a physicist and during some of my research I realized that $\log_2 \sec(\pi x)$ is *really well* approximated by $\frac{6x^2}{1-x}$ for small but positive $x$. Is there any reason this should be so? After all the former function looks quite unwieldy on first sight. In particular, I am wondering whether there is some context in mathematics where these functions would arise, with the hope of understanding why they should agree so well. (Note that if I instead simply Taylor expand the former function and keep the leading term, it performs much more poorly.)

EDIT: Here is a plot from WolframAlpha:

Note that at $x= \frac{1}{2}$, the former function has a singularity.

Edit: There were comments about the Taylor series and Padé approximants at $0$. The approximation is neither, and it is not exceptional at $0$, but it seems good on the interval $[0,1/3]$.

1024808130614@Ori: I guess I do that because most of my exposure to random walks came from Doyle and Snell, and from wikipedia (which was clearly written by someone who had just finished reading Doyle and Snell). Both of those sources define recurrence this way, but at least Doyle and Snell also discuss the original definition and prove the two are equivalent. I guess I'll stop referring to this as a definition and start calling it a theorem. This is the problem with learning math from wikipedia!Thanks for the translation Matthew, that's perfect. In some sens I am a little bit disappointed, since I though writing a text in French would have lead to protestations from some people on the forum and thus have lead to some kind of polemic...doing this was some kind of provocation from my part...but nothing.@EmilJeřábek, thanks. Free groups are very special and a lot is known about their first order theory. Especially the study of commutators (maybe not iterated commutators) has played an important role in the work of Zlil Sela.16089501372954846122And what about having more pieces in the partition? I guess in this case we don't want them to be intervals.'The following is not strictly speaking something that can be read off from the character table. However, it is an elementary combinatorial identity about partitions which one can deduce from understanding the character theory of symmetric groups well enough, and looking at the character table does play a central role:

For $\lambda \vdash n$ a partition of $n$ (i.e., $n = 1 \lambda_1 + 2 \lambda_2 + \cdots + n \lambda_n$) define
$$ A(\lambda) = \prod_{i=1}^{n} n^{\lambda_n}, \qquad B(\lambda) = \prod_{i=1}^{n} (\lambda_n)! $$
**Claim:** $$\prod_{\lambda \vdash n} A(\lambda) = \prod_{\lambda \vdash n} B(\lambda) $$

The character-theoretic proof proceeds as follows:

- For an element in the conjugacy class of $S_n$ indexed by the partition $\lambda$, it's centralizer has cardinality $A(\lambda) B(\lambda)$, i.e., the number of elements in the conjugacy class is $$\frac{n!}{A(\lambda) B(\lambda)}$$
- Take the character matrix $M$. The orthogonality relations tells us that a suitable rescaling of the character matrix is orthogonal, so has $\det = \pm 1$. From this, together with 1 to find the scaling factors for the columns, we obtain $$ (\det M)^2 = \prod_{\lambda} A(\lambda) B(\lambda) $$
- $M$ relates two bases for the spaces of class-functions: The characters of irreps (indexed in the Schur ordering by partitions), and the delta functions on conjugacy classes (indexed obviously by partitions). For symmetric groups, there is a third nice basis: For $\lambda \vdash n$, let $$ S_\lambda = \prod_{i=1}^{n} S_i^{\lambda_i} $$ and consider the characters of the induced reps $Ind_{S_\lambda}^{S_n} \mathbb{C}$.
- Consider the change of basis matrix relating characters of induced reps and the delta functions on conjugacy classes: Easy character theory shows that it is triangular with diagonal entries equal to $B(\lambda)$.
- Consider the change of basis matrix relating characters of induces reps and characters of irreps: Knowing how character theory for symmetric groups works over $\mathbb{Z}$ (i.e., that both span integrally), we can show that it has determinant $\pm 1$. More precisely, knowing the character theory well enough we can show that the change of basis matrix between them is upper triangular with ones on the diagonal.

Putting together 2, 4, 5 we obtain $$ \det(M)^2 = \prod_{\lambda \vdash n} A(\lambda) B(\lambda) = \left(\prod_{\lambda \vdash n} B(\lambda)\right)^2 $$ and so the claimed identity.

911662815262I was about to write a comment on the OP to this effect, but this answer is much more comprehensive, so +1.@i707107,those hold ,provided a series is convergent with radius r,otherwise,they may not hold,am I right?For instance,$$f(x)=\frac{1-\sqrt{1-4x^2}}{2x^2}$$392407^CLT for Bernoulli RV with negative correlation1484313nStrengthening the Compression Lemma in homotopy theory659906pPerhaps try http://math.stackexchange.com :-) Good luck517228145731Dima -- yes, "torsor" and "principal homogeneous space" are synonyms, at least in my usage.353689783367You posted on m.se and MO without linking each to the other. That's rude.413422168411314797281380401445896https://www.gravatar.com/avatar/ad6fab6bb69fcb628c09284a990ee7cb?s=128&d=identicon&r=PG&f=1In 1959, Lyndon showed that in a free group, the equation $u^2v^2=w^2$ has only commuting solutions: $uv=vu=w$. Is there in the litterature any information about the following "twisted" version of the above equation: $f(u)ug(v)v={fg}(w)w$, where $f$ and $g$ are (commuting) automorphisms of the free group? Any indication welcome.

1826771RGeneralization of Sprague-Grundy Theorem6737062013648Question: Can the Fell topology be expressed in terms of the distributions of the the tracial states of a unitary representations, that, is $\pi_j \rightarrow \pi$ if and only if $tr\; \pi_j \rightarrow tr\;\pi$?

This makes of course only sense in a more restricted setting, say reductive groups over a local field $F$. Otherwise, it isn't clear whether irreducible unitary reps have tracial states.

Examples: This is true for $GL(n,F_v)$ for $F_v$ local field and $n=1,2$. Also works well for $SL(2, \mathbb{R})$.

added to YCor : consequently, one must pronouce Yordan', while Camille Jordan is pronounced Geordã.9173119This seems to be a very difficult problem. You can think about the case of bounded degree graphs where it is known that graph isomorphism is in P. Still no set of invariants is known. (Do you have any suggestions for trees? for planar graphs?) For general graphs although there are good reasons to believe that graph isomorphism is in co NP there are also reasons to believe that showing it will be very hard. "Under deradomization" is not something to take lightly. (Here are some examples: http://gilkalai.wordpress.com/2009/12/06/four-derandomization-problems/ )Grunewald, Segal and Smith define in "Subgroups of finite index in nilpotent groups" a zetafunction associated to a finitely generated, torsion-free nilpotent group G by counting normal subgroups corresponding to their index: $$\zeta^\triangleleft_G(s)=\sum_{n} a_nn^{-s},$$ where $a_n=\{H\leq G: |G:H]=n\}$. This zetafunction decomposes as an Euler product by counting normal subgroups of p-powered index.

For such a group of nilpotency class 2 they define a Lie-ring by $L=G/Z(G)\oplus Z(G)$ with $Z(G)$ the center of $G$. Then they remark that it is easy to see that for every prime $p$ the number of normal subgroups of $G$ of index $p^k$ and of ideals of $L$ of index $p^k$ coincide. Christopher Voll states a similar result in "Zeta functions of groups and enumeration in Bruhat-Tits buildings" by considering $\mathfrak{L}=G/G'\oplus G'$ with $G'$ the commutator subgroup.

Can someone give me an idea how to prove one of these results?

47502Fibonacci numbers are defined by the recurrence relation $f_{n+2}=f_{n+1}+f_{n}$ and Tribonacci numbers by $f_{n+3}=f_{n+2}+f_{n+1}+f_{n}$

One can define, in general, K-Bonacci numbers as $f_{n+K}=f_{n+K-1}+...+f_{n+1}+f_{n}$

(they show up naturally if you consider the problem of counting binary strings of length n which do not contain sequences of K adjacent zeroes).

The characteristic polynomial associated to K-Bonacci numbers is $$P_K(t):=t^K-(t^{K-1}+t^{K-2}+...+t+1)$$ By the way, he same polynomial turns up when trying to calculate the asymptotic growth rate via generating functions and, as $K \to +\infty$, the biggest real root approaches 2.

Question: **do these polynomials have a already a name?**

Let $M$ be an $R$-module,where $R$ is a hereditary (or cohomological dimension less or equal to 1).Take $E(R)$ to be injective hull of $R$, then we have the essential extension $i:R^I\rightarrowtail E(R)^I$ (product $I$ times) and we also have $p:R^I\twoheadrightarrow M$ is epimorpshim. Then I take the push forward of these two morphisms $i$ and $p$, denote the push out by $(N,f,g)$, where $f:M\to N$ and $g:E(R)^I\twoheadrightarrow N$.

It is clear that $N$ is an injective module (because it is image of $E(R)^I$ and $R$ is hereditary) and $f:M\to N$ is injective morphism. However, $N$ is not necessarily injective hull of $M$ because in general, essential extension does not commutes with colimit.

My question is: can we give some conditions to $R$ or other extra conditions to make N is injective hull of $M$?

In general, I know it is not true, but it seems that it gives a approximation of "functor" $M \to E(M)$

19845801199007789444754009Oddly I find about zero resources talking about "complex symplectic reduction" upon a web search. Is there anything wrong with it?

I guess maybe there are two competing settings a priori: a complex symplectic manifold $(M, \omega)$ and an action of a Lie group $G$ preserving $\omega$, where either:

- $G$ is a complex Lie group and the action $G \times M \to M$ is holomorphic

or

- $G$ is merely a real Lie group acting on $M$ by biholomorphisms (actually this is automatic when the action preserves $\omega$)

I'm thinking that the two settings are more or less equivalent, but I'm not sure it's worth expanding my thoughts.

In any case, I think that one can define a complex (holomorphic) moment map and a complex symplectic quotient $M//G$ naturally equipped with a reduced complex symplectic structure (provided of course the appropriate conditions allowing symplectic reduction are satisfied, the action is Hamiltonian, etc). Right?

Any thoughts?

20671901711371758412If I understood correctly, this gives examples of manifolds that are stably equivalent but have different cup product structures. Doesn't an equivalence of the spectra imply an isomorphism of the cohomology rings?Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/73679/discussion-between-peter-heinig-and-lancejpollard). 4587I agree with @AndrésE.Caicedo's comment recommending interlibrary loan. Peter Hinman now has a smaller office than before he retired, so he may have discarded his copies of dissertations from long ago. Even if he still has his copy of Stan Stahl's thesis, I think it would be more appropriate to ask a librarian to scan a copy than to ask Peter to do it. (I might also have a hard copy of this thesis, but I'm far away from Ann Arbor until the end of the month.)John Asplund10613741770709|Question in the setting of generalized Diophantine $m$-tuplesIf we take $f(x,y)=2\sqrt x$, so that $g_n'(x)=1/\sqrt x$, then your integrals do not exist at all.

Existence of a canonical embedding from a quotient of $\mathscr{F}^\ast(\mathcal A(H))$ to another18408061324685148327fUniversity of Edinburgh, Edinburgh, United Kingdom934496413939673124554833125168922378851568435In that case,we need to put the kid under Secret Service protection on his way to the Pentagon.KConrad.6668542215808/I have the following **Question**:

1) Is it true that if $\Omega\subset\mathbb R^3$, $\Omega$ - bounded, $v\in H_0^1(\Omega)$ but $v\notin L_\infty(\Omega)$ implies $\int\limits_{\Omega}{e^vdx}=+\infty$ or $\int\limits_{\Omega}{e^{-v}dx}=+\infty$, (or both)? In other words, does it imply that $\int\limits_{\Omega}{cosh(v)dx}=+\infty$

Here is a **paper** which basically gives me that $\int\limits_{\Omega}{e^udx}< +\infty$ for functions $u\in W_0^{1,d}(\Omega)$. So, for $d=3$ it might be the case that: if a function $v$ is in $H_0^1(\Omega)$ but not in $L_\infty(\Omega)$, then no matter how weak singularity it has, the exponent is already not summable.

My question is motivated by the following problem that I am trying to solve:

Let $F: H_0^1(\Omega)\rightarrow \overline{\mathbb R}, $ $F(v)=\int\limits_{\Omega}{f(x,v(x))dx}=\int\limits_{\Omega}{k(x)cosh(v(x)+g(x))dx}$, where $k(x)\ge 0$ in $\Omega$, $k(x)$ is not identically 0 in $\Omega$ (it can be taken for positive constant) and $g\in S:=L_\infty(\Omega)\cap H_0^1(\Omega)$ (dense in $H_0^1(\Omega)$) is given. Let $V_\alpha :=\{v\in S | F(v)\leq \alpha\}$ for some arbitrary $\alpha\in\mathbb R$, is the lower level set of $F$ with elements in $S$. I would want to show that $V_\alpha$ is closed in the $H_0^1$ norm.
**Here is what I have done:**

First, note that $F$ is convex. Then, I show that $F$ is sequentially weakly lower semicontinuous in $H_0^1$ by using this **Theorem 3.20** and the fact that each weakly convergent sequence in $H_0^1$ is strongly convergent in $L_2$ and therefore also weakly convergent in $L_2$. Then proceed as usual: if $\{v_n\}_{n=1}^{\infty}\subset V_\alpha$ and $v_n\rightarrow v$ in $\|.\|_{H_0^1(\Omega)}$ norm, because $F$ is sequentially w.l.s.c $\Rightarrow$ $F(v)\leq \alpha \Rightarrow$ it remains to show that $v\in S=L_\infty(\Omega)\cap H_0^1(\Omega)$ (obviously $v\in H_0^1(\Omega)$). Now, I will be finished if I show one of the following

1) (the main question from above) - because if I assume $v\notin L_\infty\Rightarrow F(v)=+\infty$ and by w.l.s.c. of $F\Rightarrow F(v_n)\rightarrow +\infty$ which is a contradistion with $F(v_n)\leq \alpha$ for each $n$

or

2) (less restrictive than 1) ) Again assume $v\notin L_\infty(\Omega)$. We have $\{v_n\}\subset V_\alpha, \|v_n-v\|_{H_0^1(\Omega)}\rightarrow 0$. Is it true that $F(v_n)\rightarrow +\infty$ .

Also, as a consequence of $v_n\rightarrow v$ in $\|.\|_{H_0^1(\Omega)}$ and $v\notin L_\infty$, we have $\|v_n\|_{L_\infty(\Omega)}\rightarrow +\infty$ and 2) is like coercivity of $F$ w.r.t $\|.\|_{L_\infty}$ norm.

Any help or a reference on 1) or 2) will be very appreciated. Thanks.

zAlgorithm to generate hyperbolic metric on a compact surface1950519There are continuous functions with Fourier series divergent at a point, so your $L^p$ assumptions seems somewhat beside the point here. (Also, since you talk about Fourier series, presumably you're on the circle, so $L^{\infty}\subset L^p$.)1445854https://www.gravatar.com/avatar/32f65378137d03002e878890615aacc5?s=128&d=identicon&r=PG&f=11932942|https://graph.facebook.com/100002395968321/picture?type=large1202111180451213760491836736|**Question: Let $R$ be a reduced ring with all
non-prime ideals finitely generated. Then is $R$ Noetherian?**

The answer is Yes.

To lessen my typing, let me use the abbreviation
**NFG** to mean not-finitely-generated.
The result proved here is

**Theorem.**
If $R$ is a commutative unital ring whose
NFG ideals are prime, then $R$ has at most
one NFG ideal. If one exists, it
is a nonmaximal, nonzero prime which squares to zero.
[In particular, a reduced, commutative, unital
ring whose NFG ideals are prime is Noetherian.]

Throughout the rest of this post $R$ always denotes a commutative, unital ring whose NFG ideals are prime. I will also assume that $R$ has at least one NFG ideal. I now choose one and label it $\mathfrak p$. I choose $\mathfrak p$ arbitrarily, except that if there exists some NFG ideal that does not square to zero, I choose $\mathfrak p$ so that it does not square to zero. That way, by proving that, indeed, $\mathfrak p^2=(0)$, I will have established that all NFG ideals square to $(0)$.

**Lemma 1.**
$\mathfrak p$ is contained in the Jacobson radical, $J(R)$, of $R$

**Proof of Lemma.**
Let $\mathfrak m$ be a maximal ideal of $R$.
If $\mathfrak p\not\subseteq \mathfrak m$,
then $\mathfrak p\cdot \mathfrak m\subseteq
\mathfrak p\cap \mathfrak m$, and $\mathfrak p\not\subseteq \mathfrak p\cap\mathfrak m$, and $\mathfrak m\not\subseteq \mathfrak p\cap\mathfrak m$,
so $\mathfrak p\cap\mathfrak m$ is not prime.
The ideal $\mathfrak p\cap \mathfrak m$ is therefore finitely generated.

As $R$-modules, we have $\mathfrak p/(\mathfrak p\cap \mathfrak m)\cong (\mathfrak p+\mathfrak m)/\mathfrak m = R/\mathfrak m$. The latter is simple, hence $1$-generated. Altogether we have that, as $R$-modules, $\mathfrak p$ is $1$-generated over $\mathfrak p\cap \mathfrak m$, and that $\mathfrak p\cap \mathfrak m$ is finitely generated. This is enough to prove that $\mathfrak p$ is finitely generated as an $R$-module, hence as an ideal. This contradicts the assumption that $\mathfrak p$ is NFG.

We have shown that $\mathfrak p$ is contained in an arbitrarily chosen maximal ideal, hence it is contained in $J(R)$. (End proof of Lemma 1.)

**Lemma 2.**
If $a\in \mathfrak p$ and $a\cdot \mathfrak p\neq (0)$, then the annihilator
$\mathfrak q:=(0:a)$ is NFG and $\mathfrak q$ is a proper
subideal of $\mathfrak p$.
Moreover, $R/\mathfrak q$ is Noetherian.

**Proof of Lemma.**
We cannot have $a\in a\cdot \mathfrak p$ for the following reason.
It leads to $a = ab$ for some
$b\in \mathfrak p$, hence to $(1-b)a=0$ for some
$b\in\mathfrak p\subseteq J(R)$.
This yields $a=0$, since $1-J(R)$ consists of units.
But $a=0$ contradicts $a\cdot \mathfrak p\neq (0)$.

Therefore, we have $a\notin a\cdot \mathfrak p = (a)\mathfrak p$. In otherwords, the product $(a)\mathfrak p$ contains neither of the factors $(a)$ or $\mathfrak p$. (The factors each contain $a$ but the product does not.) Hence $(a)\mathfrak p=a\mathfrak p$ is not prime, and therefore $a\mathfrak p$ is finitely generated.

The map $x\mapsto ax$ is an $R$-module homomorphism of the nonfinitely generated $R$-module $\mathfrak p$ onto the finitely generated $R$-module $a\mathfrak p$. Necessarily this map has NFG kernel, which is $\mathfrak q:=(0:a)\cap \mathfrak p$.

The element $a$ does not annihilate $\mathfrak p$, i.e. $\mathfrak p\not\subseteq (0:a)$, so we derive that $\mathfrak p$ properly contains $(0:a)\cap \mathfrak p = \mathfrak q$. Now the fact that $(0:a)\mathfrak p\subseteq (0:a)\cap \mathfrak p\subseteq \mathfrak q$, implies that $(0:a)\subseteq \mathfrak q\subsetneq \mathfrak p$. Thus $\mathfrak q=(0:a)\cap \mathfrak p = (0:a)$.

To summarize the progress so far: if $a\in \mathfrak p$ and $a\mathfrak p\neq (0)$, then $\mathfrak q:= (0:a)$ is a proper NFG subideal of $\mathfrak p$.

To finish, we need to prove the last sentence of the lemma. Note that the class of rings whose NFG ideals are prime is closed under the formation of quotients. Moreover, it follows from the earlier part of this lemma that no non-Noetherian domain belongs to this class. (Starting with an NFG ideal $\mathfrak p$ we produced nontrivial zero divisors by establishing that for any $a\in\mathfrak p$ either $a\mathfrak p=(0)$ or $(0:a)$ is NFG. Either case produces nontrivial zero divisors.) Hence, the domain $R/\mathfrak q$ must be Noetherian. (End proof of Lemma 2.)

**Lemma 3.**
$\mathfrak p^2=(0)$ and $\mathfrak p$ is the unique
NFG ideal of $R$.

**Proof of Lemma.**
To obtain a contradiction to $\mathfrak p^2=(0)$,
suppose that $a, b\in \mathfrak p$ satisfy
$ab\neq 0$. Then $a\mathfrak p\neq (0)$,
so by Lemma 2 $\mathfrak q:=(0:a)$ is a proper NFG
subideal of $\mathfrak p$, which does not contain $b$.
Moreover, $R/\mathfrak q$ is Noetherian.

In particular, $\mathfrak p/\mathfrak q$ is a finitely generated ideal of $R/\mathfrak q$, so $\mathfrak p$ is finitely generated over the subideal $\mathfrak q$ in $R$. It follows that $\mathfrak p$ is also finitely generated over the larger subideal $(b^2)+\mathfrak q$. If this latter ideal $(b^2)+\mathfrak q$ was f.g. in $R$, then $\mathfrak p$ would also be f.g. in $R$, which is false. Hence $(b^2)+\mathfrak q$ is NFG, and therefore prime, in $R$. But $b^2\in (b^2)+\mathfrak q$, so by primality $b\in (b^2)+\mathfrak q$. It must be possible to express $b$ as $b=b^2c+q$ where $c\in R$ and $q\in\mathfrak q$. Rewritten, $b(1-bc)=q\in \mathfrak q$.

But $b\in\mathfrak p\subseteq J(R)$, so $(1-bc)$ is a unit, and we derive that $b=(1-bc)^{-1}q\in\mathfrak q$, a contradiction to $b\notin (0:a)=\mathfrak q$. What we have contradicted was the initial assumption that $\mathfrak p^2\neq (0)$.

Now suppose that $R$ has NFG ideals $\mathfrak p$ and $\mathfrak r$. By the first part of the lemma, $\mathfrak p^2=(0)=\mathfrak r^2$. Since $\mathfrak p^2=(0)\subseteq \mathfrak r$, and $\mathfrak r$ is prime, we get $\mathfrak p\subseteq \mathfrak r$. Similarly, $\mathfrak r\subseteq \mathfrak p$, so $\mathfrak p=\mathfrak r$. (End proof of Lemma 3.)

The theorem is essentially proved now, but let me write it out.

**Theorem.**
If $R$ is a commutative unital ring whose
NFG ideals are prime, then $R$ has at most
one NFG ideal. If one exists, it
is a nonmaximal, nonzero prime which squares to zero.
[In particular, a reduced, commutative, unital
ring whose NFG ideals are prime is Noetherian.]

**Proof of Theorem.**
The lemmas show that
if $R$ is a commutative unital ring whose
NFG ideals are prime, and $R$ has at least one
NFG ideal $\mathfrak p$, then $\mathfrak p$ is the unique NFG
ideal of $R$ and $\mathfrak p^2=(0)$. The ideal $\mathfrak p$
cannot be $(0)$, since $\mathfrak p$ is NFG and $(0)$ is not.
Thus $R/\mathfrak p$ is a domain that
acts on the module $\mathfrak p$
is such a way that this module is NFG, but all proper
submodules are f.g.
It follows that $\mathfrak p$ is not a maximal ideal of $R$,
since then $R/\mathfrak p$ would be a field,
and fields have no NFG modules
whose proper submodules are all f.g.
(This last observation also appears
in the answer to
this question.)

The square-bracketed claim at the end of the theorem statement is clear. (End proof of Theorem.)

**Comment.** There exists a ring $R$ whose NFG ideals are prime,
which actually contains an NFG ideal (i.e.,
there exists a non-Noetherian ring whose NFG ideals are prime). One is constructed in the answer to this question.

More generally, you could ask this for any irreducible Markov chain and any starting state. For each nonempty set S of vertices not containing the starting state $s_0$, let $T_S$ be the time (in steps) it takes to reach S (i.e. if $X_t$ is the state after $t$ steps, the least $t$ such that $X_t \in S$). Then the expected time to visit all states is $\sum_S (-1)^{|S|-1} E[T_S]$. Each $E[T_S]$ is straightforward to calculate: if $P$ is the transition matrix, $P_{S^c}$ the submatrix for rows and columns not in $S$, and $I_{S^c}$ the corresponding submatrix of the identity matrix, $E[T_S] = \sum_{j \in S^c} ((I_{S^c} - P_{S^c})^{-1})_{s_0,j}$. For your problem I get a final answer of 1996/95 = 21.01052632.

A real function $f:X\rightarrow \mathbb{R}$ is called **Baire-one function**, if there is a sequence $(f_n)_{n=1} ^\infty$ of continuous functions $f_n:X\rightarrow \mathbb{R}$ on $X$ so that for all $x\in X$ $$\lim_{n\rightarrow \infty}f_n(x)=f(x).$$

When $X$ is a Banach space, we have the following theorem referred to as **Baire factorization theorem**.

Theorem: The real function $f:X\rightarrow \mathbb{R}$ is in the class ofBaire-oneif and only if for all closed subset $K\subset X$, the restricted function $f|_K$ has a point of continuity with respect to $K$.

**Definition**: We denote the set of all Baire-one real functions on the space $X$ by $Ba_1(X)$.

As you could easily see, $Ba_1(X)$ forms a ring with pointwise addition and multiplication. For simplicity let me consider $X=[0 , 1]$.

Suppose $C[0 , 1]$ denotes the ring of all continuous real valued functions on the interval $[0 , 1]$. By the theorem of **Gelfond and Kolmogrov** we know that the set of all maximal ideals of the ring $C[0 , 1]$ is of the form $\{M_x: x\in X\}$, where $M_x=\{f\in C[0, 1]: f(x)=0\}$.

Compared with the ring $C[0 , 1]$ we could easily find that the sets of the form $M_x^1=\{f\in Ba_1[0 , 1]: f(x)=0\}$ are maximal ideals of the ring $Ba_1[0 , 1]$. From this property some questions came in my mind as follows:

1643052329006235621151986Are you sure the descriptions there are actually "citations" in any official sense? The one for Serre contains the word (if such it is) "exented", so I don't think they're written *that* carefully.127073938917420892181190994 Indeed it can't be done for $g = 1$ but can be for $g \geq 2$. Let $V$ be the $2g$-dimensional $\mathbb{F}_2$-vector whose elements are given by partitions of a set of $2g + 2$ elements into $2$ even-cardinality subsets. The addition law is induced by symmetric differences. The intersection pairing is defined using the parity of intersections between even-cardinality subsets. Then each permutation in $S_{2g + 2}$ induces an automorphism of $V$ which preserves the intersection pairing. The resulting homomorphism $S_{2g + 2} \to \mathrm{Sp}_{2g}(\mathbb{F}_2)$ is injective iff $g \geq 2$.177140022062131008357Van example of a strictly superstable field2263005695667@FrancescoPolizzi: The matrices $A$ and $B$ need not be symmetric, so the sums of the singular values are $\operatorname{trace}(\sqrt{A^*A})$ and $\operatorname{trace}(\sqrt{B^*B})$ rather than $\operatorname{trace}(A)$ and $\operatorname{trace}(B)$ .134163619733841915068

Question 1: Does there exist a maximal ideal in $Ba_1[0 , 1]$ other than maximal ideals of the form $M_x^1$ for $x\in X$?

Question 2: Is the ring $Ba_1[0 , 1]$ a $\mathbf{PM}$-ring? $($i.e. a ring in which each prime ideal is contained in a unique maximal ideal.$)$

Generally no. The Euler-Poincaré equation derives all of its structure from the Lie bracket, and an arbitrary constraint does not respect this structure. The most simple counter-example is probably a subspace constraint. Let $V \subset \mathfrak{g}$ be a subspace which is not a sub-algebra. This subspace generates a constraint distribution on $TG$ in the obvious way, which we'll denote by $VG$. The constrained Euler-Lagrange equations on $TG$ are given by

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{g}} \right) - \frac{\partial L}{\partial g} \in (VG)^\perp \\ \dot{g} \in VG$$

Here $(VG)^\perp \subset T^*G$ is the annihilator of $VG$. These equations can be reduced to $\mathfrak{g}$. Here they take the form

$$\frac{d}{dt} \left( \frac{\partial \ell}{\partial \xi} \right) \pm {\rm ad}^*_\xi \left( \frac{\partial \ell}{\partial \xi} \right) \in V^\perp \\ \xi \in V$$

Here $V^\perp \subset\mathfrak{g}^*$ is the annihilator of $V$. In any case, the above equation is not an Euler-Poincaré equation, and It can not be transformed into one (although, one may call it a "non-holonomic Euler-Poincar\'e equation).

%**Tldr: Is it possible to somehow correct a coordinate system by the maximization function when using the simplex algorithm so it would be possible to just always follow the steepest slope from point to point instead of plugging in the slope into the maximization function?**

Sidenote as to how I understand the simplex algorithm: When using the simplex algorithm to find the point on a body at which some function is maximized, we start at a random point on the body and calculate the slope to each adjacent point. We plug this slope into the maximization function. This results in a number that tells us how much we get into the direction of the optimal point on the body. When always following the biggest number, whe should eventually reach a point from where all adjacent points result in a negative number, i.e. a maximum.

Practical example: I tried to solve a production planning problem with three different products. The problem included 8 equations which served as boundaries and a function that should be maximized. (I left all these equations out of the question as I am not sure they provide more insight instead of just cluttering it.)

I visualized the 8 boundary equations as an 8 sided body in a three dimensional coordinate system (as each boundary equation results in a plane) in which the three coordinates represent the number of produced products respectively. The maximization function can be visualized as a plane in this coordinate system as well. My first instinct was to somehow "turn" the 8-sided body with the maximization function so that the three points at which the maximization function meets the zero points are at an equal distance (i.e. turn the plane and the body so that the plane meets the x, y and z axis where they are equal). This would basically eliminate the maximization function and we could just look for the point on the body which has the highest cumulative coordinates (I think).

So instead of calculating the slope to each adjacent point and plugging it into the maximization function to find the next point, we could just calculate the slope and always follow the steepest slope.

Is this possible? If so, how could it be achieved?

I am sorry if I used confusing terms as I learned math in another language and had to translate them into english to the best of my knowledge.

11433511889001bHow many unit balls can be put into a unit cube?In addition to the "canonical" references mentioned above, the book "A Basic Course in Probability Theory" by Bhattacharya and Waymire has a very nice treatment of this topic in Chapter 5.

added: also, as the boundary length goes off to infinity, all lengths of all closed curves would have a lower bound that goes off to infinity.1109956dYes, exactly. .\https://i.stack.imgur.com/rv4j0.jpg?s=128&g=1151920@jmc I know that Lean has a HoTT branch. Shouldn't the HoTT versions of Hom and $\otimes$ give you Ext and Tor?13003751845138881730Thanks to all of you. Ariyan, It seems that you gave me exactly the answer I need. Could you please post it as an answer so I can accept it?I think it exists in english as well, but my memory could be playing tricks on me.3518711824253532194473009317911020483324311423963Well I thought that the following was kinf of standard. If $F$ is quasi-isomorphic to a sheaf, then its true. Assume that it is true when there are at most $m$ non vanishing cohomology sheaves for $F$. Not let $F'$ with $m+1$ non-vanishing cohomology sheaves. There is an exact triangle $F'' \rightarrow \mathcal{H}^{i_0}(F') \rightarrow F'$ where $i_0$ is the smallest integer such that $\mathcal{H}^{i}(F') \neq 0$.2226310217685228139452210614281930372In general, no. Say with probability $1/2$, $X(t) = 1$ and with probability $1/2$, $X(t) = 0$. Then every realization of the process verifies the property "$X$ takes integer values", while $\mathbb{E}X(t) = 1/2$ doesn't verify that property.

If you specify what a "property" is, you can get such results. If your property says "$X(t)$ takes values in a convex set $A \subset \mathbb{R}^n$" then your assertion is true (see this question).

@Qfwfq, this may be what you mean, but to clarify: The cell decomposition coming from this answer computes Borel-Moore homology, which for a (compact or non-compact) manifold is isomorphic to singular cohomology.While I work at Amazon, my opinions and views here are completely and solely mine, not my employer's.

However, I think these are essentially the only restrictions, ie, for the full group of automorphisms of F_e there are exactly two orbits, the negative curve and the complement. Is this right? I don't have a reference at hand.@Aaron, I have edited my question. I had the implicit assumption that the largest pairwise sum from $G_{2n}$ to be $2n$. 217139414551076075352964375534152Professor of mathematics at Carnegie Mellon University. My main area of interest is Classificiation Theory for Abstract Elementary Classes.

2194447142979119288261030235http://math.stackexchange.com/questions/896661/cubic-polynomial-equal-to-a-cube7793604745980CeruleanBlue726948l@DanielLitt Yes, of course. Thanks for noticing that.6597091516848921718Lisl Novak published under that name and then as Lisl Gaal after she got married. And then as Lisl Gaal after her divorce.620142506961551546Unless I misunderstood your question, this can be entirely rephrased in terms of branching random walks. This goes as follows: at time 0 there is 1 individual at position 0. Each individual gives birth to two descendants, whose position is the position of the parent plus a jump, where all jumps are i.i.d. random variable. You are asking about the maximum position at time $k$, $M_k$.

This is a much studied problem, with deep links to traveling wave partial differential equations such as the Fisher-KPP equation. (Eg, in the space-time continuous case where branching random walk is replaced by branching Brownian motion, the function $u(t,x) = \mathbb{P}(M_t >x)$ solves the KPP equation with initial condition $u(0,x) = 1_{x<0}$.)

See this recent paper http://front.math.ucdavis.edu/1101.1810 by Elie Aidekon, which provides complete answers to your question under minimal assumptions on the jump distribution. The main result is then that $M_k - ck + (3/2) \log k$ converges to a random variable, where $c$ is a constant that is easy to compute. The distribution of the limiting random variable doesn't have to be either Gumbel or lognormal.

In the space-time continuous case where branching random walk is replaced by branching Brownian motion, this result is a famous result originally due to Maury Bramson (1983): Convergence of solutions of the Kolmogorov equation to travelling waves. Mem. Amer. Math. Soc. 44, no. 285.

Cut locus, conjugate points and smoothness of distance function1315971106355Thanks everyone for your help and insight! The answers are exactly what I was looking for. I would "accept" several answers if I could, but it looks like mathoverflow only allows one accepted answer per question.1837986Yes, exactly. I proved your conjecture with an explicit bound on the probability of exceptions.21293711921384Just some hints. The functions $f^{\,(k)}$ have the same zeros of the polynomials $P_k:=f^{\,(k)}\exp(-Q)$, that satisfy $P_0:=P$ and $P_{k+1}=P_k'+P_kQ'$. In particular $P_k$ has degree $\deg(P)+k\left(\deg(Q)-1\right)$, and this is also the total number of zeros of $f^{\,(k)}$. They may be all real: for instance if $Q:=-x^2$ and $P:=1$ one finds the Hermite polynomials, that are orthogonal, hence have all zeros real and simple.

282537https://www.gravatar.com/avatar/322a4ea21ca14b07213d10dc4e377dad?s=128&d=identicon&r=PG&f=173579665460732911414926841634366tadjacency matrix of a graph and lines on quartic surfacesGiven a projective smooth variety $X$ of dimension $n$. Pick a very ample divisor $D \in Div(X)$, which induces an inclusion $\phi_D : X \rightarrow \mathbb{P}^N $. Then we can define the degree of any $m$ dimensional irreducible subvariety $Y \subset X $ by $$\text{deg}_D(Y):=\text{deg}({\phi_D}_*(\langle Y\rangle ) \cdot H^m )$$ where $H$ is any hyperplane in $\mathbb{P}^n$ and $\langle Y\rangle$ is the $m$-cycle class of $Y$. We can extend this linearly to the Chow's group of $X$. My question is : when the $n-1$ dimension Chow's group $C_{n-1}(X) = Pic(X) \cong \mathbb{Z}/n\mathbb{Z} $, then the degree map $\text{deg}_D : \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}$ is the zero map, which means all $n-1$ dimensional subvariety of $X$ has degree zero. This is weird and counter intuitive, can someone shed some light? (When can $Pic(X) \cong \mathbb{Z}/n\mathbb{Z} $ and what does it mean geometrically) Thanks.

626420858721403249727443In the mixed strategy Nash equilibrium the column players will choose b with probability 1, thus there is never a mix that includes (15,1). However, the row player can mix with their weakly dominated strategy A. (probability A = 2/9, B = 7/9)

835507No, this is different. The definition you quote refers to normed spaces and adds a non-colinearity requirement.1171765|Just a math student passing by...

Teach me!!!

f(+1) Sounds correct! Thanks for posting, Mohammad!16755761675384https://lh5.googleusercontent.com/-F2wBCiUtBBo/AAAAAAAAAAI/AAAAAAAAAI4/6HyCuAX4rJk/photo.jpg?sz=1281784226I need a reference for the result which gives the number of solutions to the congruence $mx^2+ny^2 \equiv k\pmod{p}$. This result seems to be something that would be discussed in Gauss' Disquisitiones Arithmeticae, as it is proven from basic results in the theory of curves over finite fields.

Is anyone aware of a specific reference where the number of solutions to the above congruence is discussed?

The details you added are great! Unfortunately, I couldn't find what you wrote in the reference above. Could you give me another reference? I will need it for my work.As of the time I write this, there are a number of different answers (in two families) that have given me a great deal of enlightenment. Thank you Georges for this one!Thanks Anton, 1. I'm not familiar with Cantor torus, do you have a reference? 2. $K$ has zero measure, but $S$? Thanks, that looks useful. Ideally, I'd like a bound that doesn't tend to $0$ with large $n$ and constant $A$, but that may not be possible.@Joseph O'Rourke, I think we're more or less on the same page, except that I originally meant $P$ as the sum of the total number of repeated pairs, not the average (which is totally fine). Just to explain my original definition, on your 8 x 8 run, if an arbitrary pair has already occurred and you see another instance of it, you'd increment $P$ by one. In other words, if I see two pairs of, arbitrarily, "red" and "blue" vertices, both with distance one, and this is the only repeat I encounter on the grid, $P = 1$.235221RFair enough--I guess I'll read the book!880254.Here are some results concerning the degree of the polynomial $P$. We only treat the cases where $n$ is a positive, square-free integer congruent to $3$ modulo $4$, showing that the degree of $P$ is less than or equal to $2$.

We start by the weaker assumption that, given the positive integer $n$ and a solution $(a,b)$ of Pell's equation $a^2-nb^2=1$, there exist polynomials $D,P,Q\in\Bbb Q[X]$ and a rational number $k$ such that $P^2-DQ^2=1$, with $D(k)=n,P(k)=a$ and $Q(k)=b$. Let $d$ be the degree of $P$. As mentioned in my other post, and following the conventions and results in David Speyer's answer, there exist constants $\alpha\in\Bbb C$ and $\beta,\gamma\in\Bbb Q$, with $\alpha^2,\gamma^2\in\Bbb Q$ such that $$\begin{cases} P=\pm T_d\left(\alpha(X+\beta)\right),\\ Q=\gamma U_{d-1}\left(\alpha(X+\beta)\right),\\ D=\gamma^{-2}\left(\alpha^2(X+\beta)^2-1\right), \end{cases}$$ where $T_d$ (resp. $U_d$) denotes the degree $d$ Chebyshev polynomial of the first kind (resp. of the second kind). From the hypothesis on $P$, we can assume $\beta=0$. We have the identity $$T_d+U_{d-1}\sqrt{X^2-1}=\left(X+\sqrt{X^2-1}\right)^d,$$ which leads to $$P+Q\sqrt{D}=\left(\alpha X+\sqrt{\alpha^2X^2-1}\right)^d.$$ If $d$ is odd then the explicit expression of $T_d$ shows that $\alpha$ (and therefore $\gamma$) is rational. Evaluating at $k$, we then obtain the identity $$a+b\sqrt{n}=\left(u+v\sqrt{n}\right)^d,$$ with $u,v\in\Bbb Q$. It then follows that $u+v\sqrt{n}$ is a unit in the ring of integers of $\Bbb Q(\sqrt{n})$ and, since we are assuming $n\equiv3\pmod4$, the elements $u$ and $v$ are integers. In particular, for $d>1$, the couple $(a,b)$ cannot be a fundamental solution of Pell's equation.

From now on, we assume $d=2m$ even. For $\alpha\in\Bbb Q$, we proceed as above and deduce that $(a,b)$ cannot be a fundamental solution for $d>1$. Suppose then that $\alpha=\sqrt{w}$, we $w\in\Bbb Q$ not a square, so that $\gamma=t\alpha$, with $t\in\Bbb Q$. We have the relation $$T_d+U_{d-1}\sqrt{X^2-1}=\left(X^2-1+2X\sqrt{X^2-1}\right)^m$$ and thus $$P+Q\sqrt{D}=\left(\alpha^2X^2-1+2\alpha\sqrt{\alpha^2X^2-1}\right)^m.$$ Evaluating at $k$, we then easily obtain the identity $$a+b\sqrt{n}=\left(u+v\sqrt{n}\right)^m,$$ with $u,v\in\Bbb Q$. Once again, for $m>1$, it then follows that the couple $(a,b)$ cannot be a fundamental solution, leading to the result.

A final remark: for general $n$, if $(a,b)$ is a fundamental solution then $$a+b\sqrt{n}=\left(u+v\sqrt{n}\right)^r,$$ where $u+v\sqrt{n}$ is a fundamental unit of $\Bbb Q(\sqrt{n})$. For example, in Stefan Kohl's example for $n=13$, we have $$649+180\sqrt{13}=\left(\frac{11}2+\frac32\sqrt{13}\right)^3.$$ I'm definitely not an expert on this subject, but the above discussion combined with an explicit bound for the integer $r$ would lead to a bound for the degree of $P$ and therefore give a complete answer to the second question.

10163383542232517361024059213986018623612067290235078I'm an undergraduate math student in Germany, currently in my fifth semester. My favorite fields are algebra and topology. I also like coding although my skills are rather basic.

1520374XTranscendence of $\Gamma(1/3), \Gamma(1/4)$417097730825@VladimirSMatveev: My comment was only about the dimension $2$ case. I agree that the order of obstructing conditions is lower in higher dimensions, and it grows much faster than in dimension $2$. For example, in dimension $n>2$, the existence of a nonvanishing Killing vector field (i.e., $k=1$) implies at least $(n{-}2)$ conditions on the $3$-jet of the metric (and the actual number of third-order conditions probably grows something like $n^4/12$).2295416jA linear order obtained by forcing with P(omega)/fin5800381680536Can you fix the first link to point to the abstract rather than directly to the PDF? Thank you!1910929nPossible monotone decreasing sequence involving primes101811645376939267222628956911521172241912201171222https://www.gravatar.com/avatar/53174d6bb163a84ade71ff12481f4d3a?s=128&d=identicon&r=PG&f=11679509It's certainly true if C is abelian, but I guess you're looking for something a bit more interesting. 480202Ok. But my question was the following concerning the example, where $Y= \Delta^1 \times \Delta^1$: Let's given a square $ \alpha : p \to Hom(Y, F) = F_2 \times_{F_1} F_2 $ with $\alpha_1:= pr_1 \circ \alpha$ and $\alpha_2:= pr_2 \circ \alpha$ as defined above and $p: A \to A'$ a cofibration. Let $ q: A' \to dom( Hom(Y, F) ) = dom(F_2 \times_{F_1} F_2 ) $ a lift in the square $\alpha.$ Then $pr_1 \circ q \circ p= \alpha_1 $ and $pr_2 \circ q \circ p= \alpha_2. $2198801673584RoronoaIt is a theorem of Gopal Prasad (which I hope I am not misquoting...) that lattices in higher rank linear semi-simple Lie groups have finite outer automorphism groups. Is there some other reasonable class of groups with this finiteness property?

686377Nice. And this will also do for the proof that $\mathsf{GB}$ is conservative over $\mathsf{ZF}$, right?2236447272963One limitation of the von Neumann trick is that you don't know in advance how many samples will be needed.

If you limit the number of samples in advance, we can get a negative answer. That is, suppose for a certain bit in your construction you want to use a Boolean function $f(\vec X,\vec Y)$ of $n$ many $X$-values and $m$ many $Y$-values. Then $p:=\mathbb P(f(\vec X,\vec Y)=1)$ is a polynomial in $\alpha$ and $\beta$.

For instance, if $n=m=1$ and $f(X,Y)=X\vee Y$ then $p=\alpha\beta+\alpha(1-\beta)+(1-\alpha)\beta$.

Thus such a construction (with $\alpha/\beta<1$ and $\epsilon=0$) cannot work unless $\alpha/\beta\in\mathbb Z[\alpha,\beta]$.

Yes, certainly they have PL partitions of unity. For smooth manifolds all you need for the proof is small bump functions, which have nice piecewise-analytic explicit formulas. On a PL manifold you can manufacture your bump functions much more easily -- take a suitably-refined triangulation, force the function to be 1 on a top simplex, 0 away from a small neighbourhood and linearly interpolate the rest. The remainder of the proof is just like the smooth category.2208056327137I think that we have had a previous thread related idoneal numbers here. The known idoneal numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 16, 18, 21, 22, 24, 25, 28, 30, 33, 37, 40, 42, 45, 48, 57, 58, 60, 70, 72, 78, 85, 88, 93, 102, 105, 112, 120, 130, 133, 165, 168, 177, 190, 210, 232, 240, 253, 273, 280, 312, 330, 345, 357, 385, 408, 462, 520, 760, 840, 1320, 1365, and 1848 there is possibly one more. If there is an additional idoneal number it must be greater than 8436. If the generalized Riemann hypothesis holds the above sequence is complete. I got this number from the wikipedia article there is also a online integer sequence for these numbers at A000926.

1929875Some of the *good books* are:

Elements of the Theory of Functions and Functional Analysis by Kolmogorov, Fomin.Functional Analysis, by F.Riesz and Nagy.

Functional Analysis : Spectral Theory by V.Sunder. Freely availablehere.Analysis now. By Gert Kjeargård Pedersen. (As suggested by Theo Buehler at $\textbf{Math.SE}$

Stein, Elias; Shakarchi, R. (2011). Functional Analysis: An Introduction to Further Topics in Analysis. Princeton University Press. ISBN 0691113874.

Functional Analysis, Sobolev Spaces and Partial Differential Equations (Universitext) by Haim Brezis.

Elementary Functional Analysis by Georgi E. Shilov.

Introductory Functional Analysis with Applications by Erwin Kreyszig.

Notes on Functional Analysis by Rajendra Bhatia. (Hindustan Book Agency.)

Functional Analysis by S.Kesavan.( Hindustan Book Agency.)Elementary functional analysis By Barbara D. MacCluer

Functional analysis: an introduction By Yuli Eidelman, Vitali D. Milman, Antonis Tsolomitis (

AMS)Principles of functional analysis By Martin Schechter. (

AMS)You may also want to see this thread: Problem books in Functional Analysis

$\textbf{Note.}$ The books which are written in Italics are the ones which I have read partially. The ones which are not in Italics are the ones which I have come to know (by friends, teachers) are good books in Functional Analysis. Also, I really don't know which publisher actually publishes the book in foreign edition written by Kesavan and Bhatia.

15095931916Let $z$ be the $n$-tuple of complex variables $(z_1,\ldots,z_n)$ and define $H:{\bf C}^n\to{\bf R}$ by

$ H(z)=\sum_{i=1}^p |P_i(z)|^2 - \sum_{j=1}^q |Q_j(z)|^2, $

where $P_i,Q_j \in {\bf R} [z_1,\ldots,z_n]$ are homogeneous polynomials of the same degree, $d$.

For $\Omega=\{z\in {\bf C}^n: H(z)>0\}$, is it true that

$ \Omega \cap {\bf R}^n=\emptyset\quad \Longrightarrow\quad \Omega=\emptyset\;? $

Remarks: ${\bf R}$ real numbers, ${\bf C}$ complex numbers. All coefficients of the $P_i,Q_j$ are real. The degree $d$ is arbitrary.

4318168849094918261438988Can you not use the standard scattering theory introduced in Cazenave's book to solve this or what is the special difficulty here?How are you defining $\mathrm{Map}$? In general you'll need to take some kind of (co)fibrant replacement of the arguments to get the correct homotopy type of the mapping space...Presumably he means that regular languages are recognized by finite monoids and context-free languages are recognized by a product of a free group and a finite monoid and so what recognizes an arbitrary recursively enumerable language.How about using sheaves? Show that the sheaf of such forms is acyclic (no higher sheaf cohomology) and note that the usual Poincare lemma applies stalkwise.1544465Your sentence "we can homotope..." is totally confusing to me and I would strongly discourage you from inveting such phrases. I see no point in being ONE word, or in abbreviations which make things obscure in general. In geometry the deformations may not be only parametrized by time and continuous but also of some smoothness class. Homotopy theory has a variety of related more complicated notions e.g. deformation retract, deformed mod boundary, isotopic etc. and I see no need to add strange shortcuts to those. User31011017901211269Personally, I find formal groups much easier to understand than anything with the word etale in it. Though, while I am sure I am not the only one with this opinion, I am also quite sure this is uncalled for and self destructive...1993372639359bhttp://www.maths.ed.ac.uk/%7Epsamuels/index.htmlIf you don't know about $p$-adics, then use the field example in Todd's comment to his answer, which can be extended to ${\mathbf Q}(x_1,\dots,x_n)$ for any $n \geq 1$.1825237274478As in a previous incarnation of this problem, such numbers can be identified with multi sets of digits, and such multisets of d-self-contained numbers are then closed under union (for a fixed d). There may be finitely many primitive solutions as in the earlier problem, but this is less clear. You can try computer enumeration to find such solutions, but you may also find some theory to aid in the search (Tarry-Escott comes to mind). I don't think a quick analytic solution is available. Gerhard "If Base Is Large Enough" Paseman, 2016.06.27.147428221617701637634Structure of the unitary representation $L^2(N/M)$ when $N$ is a nilpotent Lie group1134853210248929643632There are a couple of subtle differences. Some of the concepts are only relevant when talking about geometry from an axiomatic perspective.

When one is talking about geometry from an axiomatic perspective ( you want to talk about points, lines, planes, angles etc.) you are really looking at a model for your axioms. Here we might talk about Euclidean, Riemannian, Hyperbolic, Projective, Spherical and (perhaps) Elliptic geometries. Usually the main difference is whether or not we choose to take the parallel postulate as an axiom or one of its negations. If we take the parallel postulate then we are working in a model of Euclidean geometry, this is sort of flat geometry where things are what you expect. In Spherical geometry (sometimes called Riemannian, and maybe elliptical but i am not sure about that) is where we have no parallel lines, a model of this is the sphere where we take lines to be great circles. Note though that typically Riemannian geometry is about manifolds with a Riemannian structure, but that we can save until later. Hyperbolic geometry is when we have infinitely many parallel lines through a given point, a model of this is the Poincare disk. Projective geometry has every two parallel lines meet... at the point at infinity. There are much better references for this stuff but so far we have just been looking at models of axiom systems that one could work with and "do geometry" in.

Algebraic, differential, and Riemannian geometry are more complicated. Here are some "one line slogans" which i am sure others can improve upon. Differential geometry is about curves, surfaces and homogeneous objects that you want to study via calculus, there is a priori some smooth structure on that object. Algebraic geometry wants to study similar objects but only when you are concerned with what you can tell about an object via rational functions. For me, this starts with the Gelfand-Naimark result. it gets much much richer. Riemannian geometry is differential geometry except you have a well behaved notion of distance between points, distance ON the hypersurface itself! All of these could be fixed by someone with more knowledge then I on the respective subjects.

I didn't mention affine geometry, but I will take a stab and say it is like a geometry in any one of the above models except you only care about "flat" things, lines, planes, etc.

The above comment suggests looking at some good books. It mentions Klein's Erlangen program, which is where Klein proposed studying a geometry by understanding the group of symmetries that preserve it. So you can think of the first big paragraph as looking at groups and the geometries you get from them. The second big paragraph you can get by looking at different types of sheaves on various topological spaces (I think) where the sheaf keeps track of the type of structure you care about. The two classes are sort of a bit different, but they are similar in that you can look at objects that act as receptacles for the geometric content: groups and sheaves.

Any suggestions are very welcome!

1281565710940561327The converse of 2 is very much false. For example, for $n = 1$ it would say that $X$ is affine if and only if $Y$ is affine. But Jouanolou's trick shows that the opposite is true: for every quasi-projective $k$-scheme $Y$, there exists an affine morphism $X \to Y$ which is a torsor under a vector bundle such that $X$ is affine.19373781673191I once spent some time contributing to a wikipedia article about a topic I felt confident about, and have spent my whole career studying, including all the historical papers by the great masters. Not long after, essentially all my content had been replaced by people who were parroting the usual abstract treatments they apparently copied from books. I never tried again. On topics where I am ignorant, I do learn a lot from wiki articles~Monoids (or semigroups) with a "finite decomposition" property102885917158061798536Hello,

I would like to know whether the only non trivial involutary group isomorphism $f:(\mathbb{C}^{*},\times)\to(\mathbb{C}^{*},\times)$ is the complex conjugation or not. Thank you in advance.

A commutative Banach algebra with an abundance of discountinuous functions1255480!Let $A \in M_n(\mathbb{Z}_p)$ be a nonsingular matrix which is nilpotent mod $p$, so $A^r \in pM_n(\mathbb{Z}_p)$ for some $r$. Then $\mathbb{Z}_p[[T]]$ acts on $\mathbb{Z}_p^n$ with $T$ acting by $A$. If I take $M/pM$, or equivalently tensor with $\mathbb{F}_p[[T]]$, then since $\mathbb{F}_p[[T]]$ is a local ring I get a decomposition of the form $\bigoplus_i \mathbb{F}_p[[T]]/(T^{n_i})$. However, the structure theorem for modules over the Iwasawa algebra tells us that there is a quasi-isomorphism (i.e. a $\mathbb{Z}_p[[T]]$-module map with finite kernel and cokernel) $$M \to \bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$$ where the $f_i$ are either distinguished polynomials (there is no free part and no $p$-torsion do to the initial assumptions on $M$). If I take the module $\bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$ and reduce mod $p$ to get an $\mathbb{F}_p[[T]]$-module, then it's already in the form $\bigoplus_i \mathbb{F}_p[[T]]/(\bar{f}_i(T)^{n_i})$ (where the $\bar{f}_i$'s are just powers of $T$ since the $f_i$ were distinguished polynomials).

My question is, is the $\mathbb{F}_p[[T]]$-module I get by reducing $\bigoplus_{i} \mathbb{Z}_p[[T]]/(f_i(T)^{n_i})$ mod $p$ the same as the one I got by reducing $M$ mod $p$, and then finding its decomposition? So in some sense I'm asking whether the operations "decompose my module" and "reduce mod $p$" commute. Note that this decomposition holds more information than the $\lambda$-invariant of $M$.

As Chris Wuthrich points out in a comment responding to an earlier version of this question where I only assumed $M$ is a finitely-generated torsion module which is free when restricted to a $\mathbb{Z}_p$-module, the answer is "no" under these conditions. However, the setup above gives some additional conditions, and hopefully these are enough to make the above true.

Note: I'm aware that the quasi-isomorphism from $M$ to the decomposed module will in general NOT become an isomorphism after tensoring the two modules with $\mathbb{F}_p[[T]]$, but I'm asking whether they will nonetheless be isomorphic by a different map.

I think I see... however, in the special case, you still need to know the max index for the summation (which happened to be $(3\delta-1)n/2=|A\cap 2A|/2$, right?). Does it also hold for "bigger" $\delta$'s, greater than $1/2$?18484532099844`codimension of unstable locus of vector bundles21002047534212nr is usually called the rank of the permutation group.205824512333291534442t"I don't know where that claim came from" -- BLAME CANADA735556857329527567Given a smooth algebraic variety $X$ and a Q-Cartier Q-divisor $D$ which is semiample and big. Assume that $Y=Spec ( \oplus_{m\geq 0}H^0 (X, mD))$ is Cohen-Macaulay. Do it follows that $H^i (X,mD)=0$ for $0 <i<\dim (X)$?

I know the answer is yes if $D$ is Q-ample and this characterize Cohen-Macaulayness for ring of sections of Q-ample divisors.

If this is true, what is a reference in the literature?

|https://graph.facebook.com/982898625200563/picture?type=large1496487\https://i.stack.imgur.com/yTxyT.png?s=128&g=1There is a part 2 of Guy's paper, Richard K Guy, The second strong law of small numbers, Math Mag 63 (1990) 3-20, MR 91a:11001 (and also by the same author, Graphs and the strong law of small numbers, in Graph Theory, Combinatorics, and Applications, Vol 2, Wiley, 1991, pages 597-614). 1123858663450Generically all matrices are diagonalizable, so I am not convinced that testing with random matrices tells us much...1702692When you say $x \in T$, do you mean, [generalized elements](http://ncatlab.org/nlab/show/generalized+element)? Also, $g(cod(f(x)))$ doesn't really seem to make sense in arbitrary (small or otherwise) categories. Let me ask you this. If C is, say, a [proset](http://ncatlab.org/nlab/show/preorder), does your definition make sense? (If not, you need to be more specific about your $C$.)6827498697241573135236052137801116046018294214648748Thank you for your answer! However, would you mind being more specific? I have tried this approach and I get stuck when trying to prove that $$\|A^{(N)}e_{j}-Ae_{j}\|_{2}\to 0$$ as $N \to \infty.$ It is clear to me that all the entries of the columns of $A^{(N)}$ converge to those of $A$, however convergence in $\ell^{2}$-norm doen't seem immediate. Could you give more details? I am very sorry for asking this dumb question.20668001351192hMaximally unipotent monodromy point of a K3 surface1242197We know that the Vandermonde determinant of order $n$ is the determinant defined as follows:

$$\begin{vmatrix} 1&x_1&x_1^2&\dots&x_1^{n-1}\\ 1&x_2&x_2^2&\dots&x_2^{n-1}\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 1&x_n&x_n^2&\dots&x_n^{n-1}\\ \end{vmatrix}=\prod\limits_{1\leq i<j\leq n}(x_j-x_i).$$

For any or some special nonempty subset $S\subseteq \{(i,j)\mid1\leq i<j\leq n\}$，does there exist some matrix $A_S$ which is similar to the Vandermonde matrix (for example, every element of $A_S$ is something just like ${x_i}^j$) such that $$\begin{vmatrix} A_S \end{vmatrix}=\prod\limits_{(i,j)\in S}(x_j-x_i)$$ $$\text{or}$$$$\begin{vmatrix} A_S \end{vmatrix}\neq 0 \quad\text{if and only if}\prod\limits_{(i,j)\in S}(x_j-x_i)\neq 0\quad?$$

dOh yes! I found the coefficients! Thank you much!73660https://www.gravatar.com/avatar/5630216c7ebca155bcba4a1c9f839ef9?s=128&d=identicon&r=PG&f=14112871626190But it is in French. I can't read this language. Non sérieusement, c'est très bien, merci, mais pour qppliquer tes resultats il me reste à savoir si dans mon cas B/A est syntomique. A part celle que tu dis (avec A et B regulier, ce qui est un peu trop fort pour moi), il y a d'autres conditions suffisantes simples de syntomicité ? There is continuum of Fuchsian groups, while there are only countable many triangle groups (up to conjugation). 12827257076421299943@Federico Poloni, yes, you are right! $(XX^\top)^{-1}$ really does not exist. Thank you!22267471. You would like to use a heavy machinery for an easy problem. 2. I do not see how you would like to use this here? My question was not what the theorem is (that was somehow clear), but how you would like to use it. PingI have several truly remarkable examples but they all concern holomorphic dynamics (not Hamiltonian or Reeb). If this is also allowed I will write about them.8931864771401320710362619 @Giulio: although the (usual) proof of Ehresmann's theorem doesn't work in the real-analytic case, and it is easy to make counterexamples to a complex-analytic analogue of the statement, what is a counterexample to the real-analytic analogue of the statement? (In various references it is asserted that the statement is false for the real-analytic setting, but counterexamples are not given.) Note that by work of Grauert and Morrey, real-analytic structures are more "floppy" than one might have expected from analogy with the complex-analytic case (due to a non-obvious link with Stein spaces).1945861BRTAnalytic continuation is much more general than Wick rotation. I was asking about the specific instances of $t \mapsto it$ substitution.28283970619463498741452161How related are 1/f noise and self-organized criticality? And separately. Does sandpile model exhibit 1/f noise of any kind?Joel: I don't see your argument. The fact that $\sum 1/n$ gets arbitrarily large only implies (to me) that the sets $nA$ are not all disjoint.342747619945127142920327276375301880706356678764231@DanielMoskovich, this serves the role of a disclaimer that wasn't provided by Lolli in his answer. There are some more difficulties, which the OP does not hurry to clarify. If the large field $K$ is arbitrary and complicated, we could expect to have other issues. Namely, fix $\mathfrak{h}$ to be the 2-dimensional non-abelian Lie algebra (over $K$). Let $\mathfrak{g}$ vary as the 3-dimensional Lie algebra $\mathfrak{o}(q)$, where $q$ is the quadratic form $x^2+y^2-tz^2$, and $t$ is a nonzero rational parameter. So there's an embedding iff $t$ is a sum of two squares. Possibly there are large fields $K$ for which the set of primes that are sum of 2 squares is not recursive.hello David Speyer, probably you're right, since over the $2$-adics the Hamiltonian quaternions have no zero divisors and thus reversibility is trivial. Probably this can help since $\mathbb H_{2^s}$, for every $s$, is an homomorphic image of the hamiltonian queternions over the $2$-adics. Also the quaternions over the $2$-adics is the inverse limite of $\mathbb H_{2^s}$, $s=1, 2, 3, \ldots$.1399142@Tyler: I think that in your example the groups do stabilize as usual, and in fact that for roughly $j>k$ the reduced $j$th suspension has the same $\pi_{j+k}$ as the infinite product of copies of $S^j$.105641I don't think this is really a research-level question,as the solution involves playing around with basic properties of permutations... One can write $x$ in cycle-notation, and then construct $y$ so that it `links together' different cycles of $x$ to give the right order for $xy$. Tricky, but elementary I guess. 400258111100256586692032955h@FedericoPoloni The natural measure _is_ normalized687568389361444760I am not sure that there is any example of this kind beside sorting, and the obvious bounds in terms of the size of the data or output.8817681552432This is not an exact answer, but if you take $k=\log n/\log |\mathcal A|$, then each string of length $(1+\epsilon)k$ should be unique. A substring is obtained by fixed left and right endpoints. Most of these substrings are much longer than $k$, so should be unique. This suggests that the number of distinct substrings should be approximately $\binom n2-kn\approx n^2/2-n\log n/\log|\mathcal A|$.772737Related: See the Appendix (pdf p. 14/16) in http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1986__27_2/CTGDC_1986__27_2_93_0/CTGDC_1986__27_2_93_0.pdf248465The quantity $\prod_{i=2}^n \prod_{j=1}^{i-1}(w_i-w_j)$ looks like the determinant of a Vandermonde matrix.12120611144453404173David's answer and its comments gives the necessary references for the necessary uniform distribution of the zeros.!Let $f$ be a monic univariate polynomial with real coefficients:

$$f_A(x) = x^n + a_{n-1}x^{n-1} + ... + a_{0}$$

The values of $A=(a_{n-1},...,a_0)$ are unknown, but are estimated as $B=(b_{n-1},...,b_0)$ with error $\epsilon$. Therefore, $b_i - \epsilon \leq a_i \leq b_i + \epsilon$. Furthermore, the value of $\epsilon$ can be reduced arbitrarily, but cannot be set to 0. (Changing $\epsilon$ will alter the $B$ estimates). $lim_{\epsilon\rightarrow 0} B = A$

**Question 1) Is there any method of determining if $f_A$ has a repeated root using $B$ and the ability to reduce $\epsilon$?**

I've worked out how to isolate the roots of $f_A$ if $f_A$ does not have repeated roots as follows:

$D(A)$, the discriminant of $f_A$, must be non-zero. As computing $D(B)$ from the Sylvester resultant of $f_B$ and $f_B'$ can be accomplished with only $+$ and $*$ operators, $g(\epsilon,B)$, the error propagated while calculating the resultant, must strictly increase. We can decrease the value of $\epsilon$ until $D(B) + g(\epsilon,B) < 0$ or $D(B) - g(\epsilon,B) > 0$. This gives us the bound $|D(A)| \geq \min(|D(B) + g(\epsilon,B)|,|D(B) - g(\epsilon,B)|)$. This lower bound for $|D(A)|$ can be plugged into Rump's inequality to generate a root separation bound. Then Sturm's theorem can be applied until the roots are sufficiently isolated (including the error generated from $\epsilon$).

**Question 2) If $f_A$ has multiple roots, is there any way of isolating these roots?**

In this case, Sturm's theorem can isolate the roots to an arbitrary degree of precision, but without a root separation bound, there is no way to disambiguate between repeated roots and close roots.

If we try to calculate the number of distinct zeros as $n - \gcd(f_B,f_B')$, there will always be a remainder term $r$. By reducing $\epsilon$, we might be able to make $r$ arbitrarily close to 0, but we run into the same problem as with Sturm's theorem in that we don't have an bound on $r$ which differentiates between $r=0$ and $r$ is just small.

1368440 @GilKalai: I do not think P versus NP is the formulation - even morally - of the idea that certain algorithms require exponential time. The latter question is essentially P vs EXP, where we've known the answer for half a century. I think P versus NP is much closer to the formalization of the question of whether some algorithmic problems require *brute force search* (which is, conjecturally, just a small subset of EXP). However, perhaps the more relevant point is that whatever P vs NP formalizes, it is "just" a flagship conjecture indicative of a huge number of related open conjectures.VConvexity of distance-to-boundary function21227741866148https://www.gravatar.com/avatar/278af9034cb73d36ec36ea4ec1e26704?s=128&d=identicon&r=PG&f=1Decomposition of spectrum of a (unbounded, non-self-adjoint) linear operator in two spatial dimensions131054414392312253847Actually, "controlling" that is a subtle point. The method I employ to extend the transformation from the two objects $\mathcal O, \mathcal O(1)$ involves dg-enhancements and functorial cones...224762118870121114461Regarding interpretations of eigenvalues, I recall that the spectra of a graph can tell you if it is bipartite. I think a connected graph is bipartite if and only if $-\lambda$ is one of its eigenvalues (here $\lambda$ is the largest eigenvalue). Please correct me if I've muddled the theorem.

Let $M$ be a complete Riemannian manifold of finite volume whose sectional curvatures $\kappa$ satisfy $a \leq \kappa \leq 0$ for some $a \leq 0$. Let $\tilde{M}$ be the universal cover of $M$. The space $\tilde{M}$ has a visual boundary $\tilde{M}(\infty)$ (homeomorphic to a sphere). Let $X \subset \tilde{M}(\infty)$ be the set of endpoints of axes of hyperbolic elements of $\pi_1(M)$.

Question : Must $X$ be dense in $\tilde{M}(\infty)$? I assume that the answer is no, but I'm having trouble coming up with counterexamples.

11393651683752rWhat is the analytic conductor of this Hecke L-function?\Prove or disprove a matrix logarithm equation1461754nBy Day: Android Custom ROM dev with bug fixer.

Even in Set, not all monics split: consider any function from an empty set to a nonempty one. (-:774927Aalborg2519848877822352851780366522957xCommutative algebraic version of algebraic geometric object2000511I would very much appreciate a good answer to this question, perhaps a follow up to Igor's answer as this is something that I have thought about before, but have not come across in the literature.

I quickly (perhaps too quickly) abandoned the discs model in favour of the little cubes model, or perhaps I should say the hard cubes model.

For hard 2-cubes and k=3 I think the homology groups are:

for r > 1/2: clearly 0

for 1/2 >= r > 1/3: H_0 = Z^6, H_1=Z^6 and H_i=0 for i>1

the three squares are effectively arranged in a circle which can be rotated, the order (and not just the cyclic order!) parametrises the 6 connected components.

for 1/3 >= r we get the usual configuration space: H_0 = Z, H_1 = Z^3, H_2 = Z^2 and H_i=0 for i>2.

1278565Sergei --- $f(x) = \langle x 1_{[0,1]}, 1_{[0,1]}\rangle$ for $x \in B(L^2[0,2])$.9284572739102064311Ed Kirkby1121110xAsymptotic expansion of a Gaussian integral and heat kernel1876167755781694357I'm interested in a theoretical tool. Sorry for making that point unclear.1047204No -- each high degree contains a cohesive set. And Cooper showed there is a high minimal degree. But no 1-generic has minimal degree.

bbest rank r approximation for non-Frobenius norm20705851484867Check out my twitter account: @ialuronico

nIs the completion of a CAT(0) open ball a closed ball?782251420250245162A related (and to me, when I first saw it, much more surprising) Fun Fact: Divide the n-dimensional cube in half in each of $n$ dimensions, to create $2^n$ smaller cubes of edge length 1/2. Inscribe a ball in each of these subcubes, and then construct the smallest ball tangent to each of those (and centered at the center of the original cube) like so:

What happens to the diameter of the central ball as $n$ gets large?

This question received much attention at an algebraic K-theory conference in Boulder in the early 1980s, where each new arrival was presented with a multiple choice problem: Without stopping to compute, is the limit $-1$, $0$, $1/2$, $1$, $10$ or $\infty$? You were allowed to choose any three answers out of six, and place a bet on whether the right answer was among them. I can report that an overwhelming majority of algebraic K-theorists reason thusly: the answer can't be negative and can't be greater than 1 (the ball, after all, is obviously contained inside a box of side 1!); therefore it's safe to bet on the set $\lbrace 0,1/2,1 \rbrace $. Feel free to make money off this.

7074771042586024311802455630087386857Regarding (3) in Dan Lee's answer, you don't need my work in the Ricci flow approach to the differential geometric version of the uniformization theorem. The reason is as follows. Take any metric $h$ on a closed orientable surface $M$ with $\chi (M) > 0$. Let $r$ denote the average scalar curvature of $h$, which is positive since $\chi$ is. Solve the equation $\Delta_h u = R_h - r$, where $R_h$ denotes the scalar curvature of $h$. This is possible since the integral of $R_h - r$ with respect to the measure induced by $h$ is zero. Define the pointwise conformally equivalent metric $g=e^uh$. Then we have $R_g=e^{-u}(-\Delta_h u + R_h) = e^{-u}r>0$. We can now use $g$ as an initial data for the normalized Ricci flow and apply Hamilton's theorem to obtain exponential convergence in any $C^k$-norm to a constant curvature metric in the same conformal class as $h$.

By now there are many approaches to the Ricci flow on closed surfaces, such as the Aleksandrov reflection method employed by Bartz--Struwe--Ye (inspired by Schoen's work on Yamabe (constant scalar curvature) metrics; in PDE see Serrin and Gidas--Ni--Nirenberg, etc.), Hamilton's isoperimetric monotonicity, application of Perelman's entropy formula, Andrews--Bryan's isoperimetric profile monotonicity, etc.

Regarding Igor Rivin's answer, it's been a long time since I've looked at this, but this is what I remember (please correct me if I am wrong). Osgood--Phillips--Sarnak looked at the Polykov action from the spectral point of view and also from the variational point of view. Aside: Ray--Singer's work on analytic torsion predates their work (I should also mention McKean, etal.). However they did not explicitly make the connection to Ricci flow, although they may have known this. They also did not reprove the differential geometric version of the uniformization in the positive Euler characteristic case since their proof of sequential convergence assumed conformality to the standard $S^2$. In my 2-sphere paper, I actually used the Polykov energy and the fact that it is bounded from below (I believe due to Onofri), which I learned from Osgood--Phillips--Sarnak (I am indebted to Richard Melrose for asking that I read this paper when I first arrived at MIT). I used the Polykov energy to control Hamilton' entropy in the variable signed curvature case. However, later in my entropy on 2-orbifolds paper, I found a way to avoid Polykov entropy to control the modified Hamilton's entropy. Yet another proof of the entropy bound, adapting the original Hamilton's contradiction argument, was in my paper with Lang-Fang Wu on 2-orbifolds with variable signed curvature.

Regarding some sort of convexity of the functional, a fancy way to interpret the energy functional is in terms of Bott--Chern secondary characteristic classes and this was originally used in Donaldson's work on Hermitian-Einstein metrics/Hermitian Yang-Mills connections on (semi-)stable vector bundles over algebraic surfaces (and later algebraic manifolds); Uhlenbeck--Yau had a different approach. In the case of closed Riemannian surfaces, the formula is: $$\ln \det \Delta_g - \ln \det \Delta_h = -\frac{1}{48\pi} E_h(g),$$ where the relative energy of two pointwise conformal metrics is defined by: $$E_h(g) = \int_M \ln (g/h)(R_g d\mu_g + R_h d\mu_h).$$ The Bott--Chern class is in effect the term $\ln (g/h)$ since $\partial \bar{\partial}$ of it is essentially the difference in the first Chern classes (Gauss--Bonnet integrands in this case) of $g$ and $h$.

19875982289387Just include the proof. If readers find it trivial then they will skip over it. Citing a dissertation in Ukrainian is worse than useless, it will just annoy people.1691553Topological classification of continuous real functions ($n=1$) may be obtained by considering their intervals of monotonicity. In general, differentiable and analytic classifications are harder than topological. A keyword is "singularity theory". 3211310The conjectured maximum of $N = \binom{\lfloor n/2\rfloor}{2}$ is correct except for $n=7$, when the maximum is $7$, and $8 \leq n \leq 11$, when the maximum is $14$. The maximal configuration is unique except for $n=12$, $13$, $15$, $16$, and $17$.

Let $L$ be the subgroup of ${\bf Z}^n$ generated by $(2{\bf Z})^n$
and the characteristic functions $e_i + e_j + e_k + e_l$ of each
4-set $\lbrace i,j,k,l \rbrace$ in our family $\cal F$ of subsets of
$\lbrace 1,2,\ldots,n \rbrace$. Give $L$ the structure of lattice
using the inner product
$$ \langle x, y \rangle = \frac12 \sum_{i=1}^n x_i y_i $$
(i.e. *half* the usual inner product). Then $L$ is generated by
vectors $2e_i$ and $e_i + e_j + e_k + e_l$ of norm $2$, any two of which
are either orthogonal or have inner product $1$. Hence $L$ is an
even integral lattice, with at least $2n+16|{\cal F}|$ roots
(vectors of norm 2), namely $\pm 2 e_i$ and $\pm e_i \pm e_j \pm e_k \pm e_l$
for $\lbrace i,j,k,l \rbrace \in \cal F$. Equality holds iff $\cal F$ contains every tetrad $\lbrace i,j,k,l \rbrace$ such that $e_i + e_j + e_k + e_l \in L$.

Now we can use the theory of *root systems* to partition the set of
roots of $L$ into mutually orthogonal simple root systems. Since
$L$ contains the root lattice $A_1^n = (2{\bf Z})^n$, the only possible
components of the root system of $L$ are $A_1$, $D_{2k}$ for $k \geq 2$,
and the exceptional systems $E_7$ and $E_8$. These contribute respectively
$0$, $\binom{k}{2}$, $7$ and $14$ tetrads to $\cal F$. Namely, each $A_1$
corresponds to a coordinate that does not appear in $\cal F$; each $D_{2k}$
corresponds to $k$ pairs of coordinates paired in each of $\binom{k}{2}$
possible ways; and $E_7$ and $E_8$ correspond to the tetrads of the
Hamming $[7,3,4]$ and extended Hamming $[8,4,4]$ codes respectively.

It is now elementary bookkeeping to obtain the maximum configuration.

$\circ$ Except for $7 \leq n \leq 11$, the maximal $|{\cal F}|$ is $\binom{k}{2}$ for $n = 2k$ or $n = 2k+1$, attained by the $D_{2k}$ configuration.

$\circ$ For $n=7$, the maximum of $7$ is attained by the $E_7$ (Hamming) configuration, and for $8 \leq n \leq 11$, by $E_8 \oplus A_1^{n-8}$ (extended Hamming).

$\circ$ For $n=12$ ($n=13$), the maximum of $15$ is attained by both $D_{12}$ ($D_{12} \oplus A_1$) and $E_8 \oplus D_4$ ($E_8 \oplus D_4 \oplus A_1$).

$\circ$ For $n=15$, the maximum of $21$ is attained by both $D_{14} \oplus A_1$ and $E_8 \oplus E_7$.

$\circ$ Finally, for $n=16$ ($n=17$), the maximum of $28$ is attained by both $D_{16}$ ($D_{16} \oplus A_1$) and $E_8 \oplus E_8$ ($E_8 \oplus E_8 \oplus A_1$).

[The lattice $L$ corresponds via "construction A" to a binary linear code generated by $\cal F$, which is doubly even by hypothesis. Koch developed a theory of "tetrad systems" of such codes that could be used to give a more direct but less familiar derivation of this answer.]

14928141358952Question: how does one enumerate all star-convex $2n$-vertex sublattices of the plane that have the unique domino-tiling property?

A subset S of the xy-plane is star-convex if there is a point s in S such that the line segment [st] is contained in S for every t in S. For example the Young diagram of a partition is star convex with s taken as the origin.

A sublattice G of the plane is a bipartite graph whose vertices are some points in the plane with integer coordinates. Two vertices in G whose x-coordinates differ by 1 or whose y-coordinates differ by 1 are joined by an edge.

A tiling of a graph by dominos (dimer covering or perfect matching) is a collection of edges such that every vertex belongs to exactly one edge in the tiling.

For $n\geq1$ there is a bijection between the Young diagrams of $2n$ which have a unique tiling by dominos and two copies of the set of partitions of $n$.

We see this as follows:

Represent the partition of $2n$ using beads on a $2$-runner abacus (as explained in James & Kerber's book The Representation Theory of the Symmetric Group). Moving a bead on a runner into a space immediately above it on the abacus correspond to the removal of a domino (2 adjacent boxes) from the diagram.

The Young diagram has a tiling by dominos if and only if there are an equal number of beads on each runner (the partition has an empty 2-core). The tiling is unique if one runner is the bead sequence of a partition of $n$ and the other runner represents the empty partition.

For $2n=2,4,6$ only the hook partitions $[k,1^{2n-k}]$ have a unique tiling by dominos. In addition to the $8$ hook partitions of $2n=8$, the partitions $[4,3,1]$ and $[3,2^2,1]$ have unique tilings. These correspond to the partition $[2^2]$ of $n=4$.

1379708I'm trying to figure out second moment of the following quantity

$$y = \frac{\langle x_1, x_2 \rangle}{\left\|x_1\right\|\left\|x_2\right\|}$$

Where $x_1$, $x_2$ are sampled independently from $\mathcal{N}(0, \Sigma)$

This can be solved exactly in 2-dimensions using algebraic manipulation: suppose eigenvalues of $\Sigma$ are $a$ and $b$, then

$$E[y^2] = \frac{a+b}{\left(\sqrt{a}+\sqrt{b}\right)^2}$$

Is there a similarly elegant expression for $n$ dimensions?

(update, I extended this formula to eigenvalues $a,b,c$ "by analogy" and it seems to hold numerically)

525421905370'$\newcommand{\bsV}{\boldsymbol{V}}$ $\newcommand{\bsE}{\boldsymbol{E}}$ $\newcommand{\bR}{\mathbb{R}}$ Suppose that $\bsV$ is an $N$-dimensional real Euclidean space. Denote by $\newcommand{\eA}{\mathscr{A}}$ $\eA$ the space of symmetric positive semidefinite operators $A:\bsV\to \bsV$. To each $A\in \eA$ we can associate in a canonical fashion a centered Gaussian measure $\gamma_A$ on $\bsV$ which is concentrated on $(\ker A)^\perp$. For example, if $A$ is nondegenerate, then

$$ \gamma_A(dv)= \frac{1}{\sqrt{\det 2\pi A}} e^{-\frac{1}{2} (A^{-1}v,v)}dv, $$

while if $A=0$, then $\gamma_0$ is the Dirac measure concentrated at the origin.

Fix a locally Lipschitz function $f:\bsV\to\bR$ which is positively homogeneous of degree $\alpha \geq 2$. For any $A\in \eA$ we denote by $\bsE_A(f)$ the expectation of $f$ with respect to the probability measure $\gamma_A$ on $\bsV$. Consider the function

$$ \eA\ni A\mapsto \bsE_A(f)\in \bR. $$

This function is continuous and positively homogeneous of degree $\frac{\alpha}{2}$, i.e.,

$$ \bsE_{tA}(f)=t^{\frac{\alpha}{2}} \bsE_A(f),\;\;\forall t>0,\;\;A\in\eA. $$

I am interested in its modulus of uniform continuity on the ball

$$\eA_1:=\bigl\lbrace A\in\eA;\;\;\Vert A\Vert\leq 1\bigr\rbrace, $$

I was able to prove that on this ball the above function is Holder continuous, with Holder exponent $\frac{1}{2N+3}$. This suffices for the applications I have in mind, but I strongly suspect that it is far from optimal. I believe that the Holder exponent $\frac{1}{2}$ is *uniformly* optimal in the following sense: there exist $C, r>0$ so that for any $A, B\in\eA_1$ satisfying

$$\Vert A- B\Vert \leq r, $$

we have

$$\bigl\vert \bsE_A(f)-\bsE_B(f)\bigr\vert\leq C\Vert A-B\Vert^{\frac{1}{2}}. \tag{1} $$

**Remark.** To see that the exponent $\frac{1}{2}$ is the best one can hope for consider the case $\bsV=\bR^2$, $f(x,y)=|xy|$ and $\newcommand{\ve}{{\varepsilon}}$ and the Gaussian measures

$$ \gamma_{A_\ve}=\frac{1}{2\pi\ve} e^{-\frac{1}{2\ve^2}x^2-\frac{1}{2}y^2} |dxdy| $$

Then $\Vert A_\ve-A_0\Vert =\ve^2$,

$$\bsE_{A_0}(f)=0,\;\; \bsE_{A_\ve}(f)=\left(\int_{\bR}|x|e^{-\frac{1}{2}x^2} |dx|\right)^2 \ve. $$

My question is the following: *have you encountered continuity results of this sort, and if so, can you indicate some references that deal with this?* Thanks!

The Riemann Hypothesis is also equivalent to $|\pi(x) - Li(x)| = O(x^{1/2 + \epsilon})$, so let's look at that instead. In other words, $\log$ of the error should be about $(1/2) \log x$.

The sequence of points plotted below is $( \log x,\ \log |\pi(x) - Li(x)|)$ for $x=10^k$, with $1 \leq k \leq 23$. The straight line has slope $1/2$, with constant term chosen by a least squares fit (specifically, the line is $x/2 -1.24878$). Interpreted in this way, you can definitely see the promised asymptotic behavior.

(Data set courtesy of Wikipedia)

Note: My $\log$'s are base $10$, since my data set was binned by powers of $10$ already. Of course, that doesn't effect the slope.

470128115555919021481578526hsystem administration, software development

2269140I was just going to comment about $z \bar z = 1$ but yours is even simpler. $$ $$ If there's no positive-dimensional component then Bézout gives a quadratic upper bound. Can this bound be attained with only real solutions?102303317727991881945@AliTaghavi: By the way, I don't know if you have noticed, but Daniele's observation that the $1$-form that you wrote down is degenerate along the whole line $y = z = x-w = 0$ shows that the problem is tricky near there. That is the locus of places where $\alpha\wedge \mathrm{d}\alpha$ vanishes, even though $\alpha$ itself only vanishes at the origin of $\mathbb{R}^4$. Are you sure that the $\alpha$ that you want to ask about is the one you wrote?21087901385236XBigalois Groupoid Of Drinfel'd Group Double441178Is think you want the [Thom-Mather stratification theorem](https://www.encyclopediaofmath.org/index.php/Thom-Mather_stratification) In particular, the first isotopy lemma seems like it could be the one referred to.given an instance of $\mathrm{SAT}(S)$, either the instance includes the empty relation, in which case it is unsatisfiable, or it does not, in which case it is satisfied by the assignment of all variables to 0 (or to 1, if you prefer).I still don't see any proof idea in the coloring; isn't it just a rewording of the problem?1481333166486514384504530321034997https://www.gravatar.com/avatar/7cf8b8d3dac9663ab01c6804fa807a94?s=128&d=identicon&r=PG&f=1\https://i.stack.imgur.com/lfMuz.jpg?s=128&g=11085635@Mehrdad - It's for anything. A PDF of the relevant paper is linked to in my answer to the question linked above.3859641018066533186Local spaces:finite subschemes::Factorization spaces:finite subsets.

A local space over X is a compatible collection of spaces over the Hilbert schemes of arbitrary numbers of points in X satisfying a factorization property for disjoint union. This is very close to the notion of a factorization space, in which we have a space over the Ran space Ran(X) parametrizing all finite subsets of X. The point is we're allowed (and encouraged!) to keep track of multiplicities. The idea is that in practice objects that don't care about multiplicities arise by allowing all multiplicities at once: vertex/factorization algebras/spaces naturally arise as objects living over formal discs in X, which can be filtered by objects living over increasing infinitesimal neighborhoods of points, which will form local spaces. This gives useful extra structure and allows the construction of interesting factorization spaces by "accumulation of dust" (to quote the creator, Mirkovic).

The amazing application for which local spaces were introduced is the geometric construction of the most interesting factorization space, the affine Grassmannian (and hence of reductive groups themselves) "out of combinatorics" - starting from a torus with an invariant bilinear form and a sequence of natural operations on local spaces Mirkovic builds the affine Grassmannian. This is a beautiful reformulation of the theory of "Zastava spaces" (also a cocreation of Mirkovic), which are models for transverse slices to semiinfinite orbits in the Grassmannian --- in retrospect their construcion fits into the language of local spaces and the idea is one can assemble these slices together with the orbits themselves to reconstruct the Grassmannian. Mirkovic has many exciting related ideas, including applications in higher dimensions. A recent talk on the subject is available here.

Let $\mathbf{v}:(a,b)\to\mathbb{R}^2$ be a continuous function, such that $||\mathbf{v}(t)||=1,\ \forall t\in (a,b)$. Find all continuous functions $\mathbf{r}:(a,b)\to\mathbb{R}^2$ so that:

$ \mathbf{v}(t_0)=\lim\limits_{t\searrow t_0} \displaystyle \frac{\mathbf{r}(t)-\mathbf{r}(t_0)}{||\mathbf{r}(t)-\mathbf{r}(t_0)||},\forall t_0\in (a,b). $

Is it true that that the only solutions are of the form: $\mathbf{r}(t)=\displaystyle\int f(\tau)\mathbf{v}(\tau)\ \mathrm{d}\tau +\mathbf{c}$, where $f:(a,b)\to (0,\infty)$ is continuous and $\mathbf{c}=(c_1,c_2)$ is a constant?

157618719680808927762017813Thanks Sam. What if I insist that the construction of $T^* C$ should not depend on any monoidal structure $\otimes$? For example it doesn't seem like such a structure is available in part 2.2281094dI am glad that my comment was useful after all :)1956415516041Oh, by larger max spectral norm, I meant the set of $x\in\{\pm\}^n$ satisfying $\|Dx\|_\infty \leq c\sqrt{n+1}$ for $c$ not necessarily 1, but potentially larger.2245179608761529124+Let $f: X \to Y$ be a morphism of schemes. Then $f$ is called (EGA IV.17) *formally smooth* if whenever $T$ is an affine $Y$-scheme and $T'$ a closed subscheme of $T$ defined by a nilpotent ideal (it's sufficient to consider ideals of square zero), any $Y$-morphism $T' \to X$ can be extended to $T \to X$. $f$ is called *formally étale* if such a lifting always exists and is unique. $f$ is called *formally unramified* if such a lifting is unique. I am trying to understand what this should mean geometrically. I don't have a specific question in mind, but I don't see much geometric intuition in the EGA definition, and would appreciate any explanation.

For the geometric intuition that I see, suppose $X,Y$ are algebraic varieties over an algebraically closed field $k$. Let $y \in Y$ be the image of $x \in X$ (consider only closed points for simplicity). Then, formal smoothness states that any tangent vector (i.e. a map $\mathrm{Spec} k[\epsilon]/\epsilon^2 \to Y$) to $y$ lifts to a tangent vector of $x$. Formal unramifiedness states that any tangent vector can lift in only one way to a tangent vector of $x$. Formal étaleness implies that the map on tangent spaces is an isomorphism.

I don't know whether these tangent space conditions are *equivalent* to the formal definitions for varieties. I believe it is true for formal unramifiedness, at least.

(For schemes of finite type over a noetherian scheme, formal unramifiedness is equivalent to the relative sheaf of differentials being zero. This can be checked on the fibers. So let $f: X \to Y$ be a morphism of $k$-varieties. Then $f$ is formally unramified if $\Omega_{X/Y} = 0$. The exact sequence $f^*{\Omega_{Y/k}} \to \Omega_{X/k} \to \Omega_{X/Y} \to 0$ shows that formal unramifiedness holds iff the map on the cotangent spaces is surjective, i.e. the map on the tangent spaces is injective.)

**Question:** Am I correct in thinking of a formally smooth map as a submersion, a formally unramified map as an immersion (in the sense of differential geometry), and a formally étale morphism as a local isomorphism (again, using neighborhoods smaller than the Zariski neighborhoods---I understand when the residue fields are the same étaleness implies that the map on the completions of the local rings are the same)? Is this geometric intuition reasonable?

(This intuition will, of course, ignore the distinction between plain unramifiedness/smoothness/étaleness and the formal analog, which only exists when one leaves noetherian and finite-type hypotheses.)

**Question'** Are the tangent space remarks sufficient for etaleness/smoothness/unramifiedness in the variety case?

No, generally they're not.

For example, there's only one homotopy class $S^2 \to \mathbb R^3$ but there's two isotopy classes of embeddings (given via how the embedding orients the compact 3-manifold it bounds).

edit: I think if your 3-manifold is irreducible and if your maps $S^2 \to M$ are not null homotopic then the answer is likely yes. But if your 3-manifold is say a connect sum of lens spaces then I suspect it's false but I haven't come up with a nice example yet. As Allen points out in the comments below, a connect-sum of lens spaces won't work, at least not when your surface is a sphere.

edit2: As Misha Kapovich points out, for irreducible 3-manifolds and incompressible surfaces homotopy implies isotopy. This is an old theorem of Waldhausen's. "On Irreducible 3-manifolds which are sufficiently large" Ann. of Math (2) 87 (1968) 56--88.

4919471235892https://lh4.googleusercontent.com/-fUL-_iAxyo0/AAAAAAAAAAI/AAAAAAAAAj0/uzeAJXrzaT8/photo.jpg?sz=128598477~Actually my own wolfy calculation suggests all roots are real!Thanks very much Martin, this answer is great! It answers everything I wanted, and more. I have an application in mind where the base is regular, so this is perfect for me.I have a matrix in which each element contains the coordinates of a 3D surface. Sometimes, some points will be "out of line" meaning that they will not conform to the general shape. For example you would have a slightly curved plane and all of a sudden a point with coordinates which are very different from the rest which produces a spike in the surface.

I was thinking of applying a smoothing filter or a lowpass filter.

Then I thought I might produce a distribution of the gradients of all points (w.r.t. the points around them) and only allow points with a gradient within two std. of the mean.

I'm really not sure what the best way would be (there might be others which I don't know about) so before I started implementing it into matlab I thought I'd ask.

2955698dGrothendieck monodromy theorem for l-adic sheaves1652807t$\mathbb{Q}$-forms of $\mathrm{SL}_2\times \mathrm{SL}_2$pYour definition needs to require $a$ and $b$ nonzero...1642498is there some background for this ? or it is just standing alone paper ? In this case I would be very much admired and surprised how can any one (except the authors) know about it :) 742853139944618589602044174377670847302https://www.gravatar.com/avatar/6cf1b35698ec15989db6c9be3ae967ac?s=128&d=identicon&r=PG&f=120988191283749514967521598207064520271041236993Could you put back the answer I will accept it. I just was not sure about re-posting the question.600638832788563321Thanks! However, I am mostly interested in decompositions in sets which are *cellular*, this imposes more restrictions. You can (and should) use $\TeX$ in your questions, answers and comments.8612101430382237991This sounds fascinating! I can't wait to find out where the number 576 comes from...201802482884You're just asking for square-free numbers below $x$ with no prime factor larger than $\sqrt{x}$. The answer is $\sim (6/\pi^2)x (1-\log 2)$, where $(6/\pi^2)$ comes from the square-free-ness, and $1-\log 2$ from having no prime factor larger than $\sqrt{x}$.18213Not sure why this isn't getting more upvotes. I think it works, and it is very different from my answer.See also http://mathoverflow.net/questions/17736/way-to-memorize-relations-between-the-sobolev-spaces1277218644641732766I am searching for a bijection between two types of bit strings (strings of 0's and 1's) both of even length (2n).

The restriction on the first type of bit string is that they must have the same number of 0's and 1's.

The restriction on the second type is that they can not have any prefix with the same number of 0's and 1's.

Note that I need the actual bijection, it is not enough to simply show that they are in bijection with one another. So I need both a transformation and it's inverse.

This is a sub-problem of a larger problem that I have been working on for some time now and it is the last piece that I have been unable to work out. Thanks in advance for any help.

If any additional clarification or if examples would be helpful, just let me know and I'd be happy to provide it

Hello Darij: the centre ${\cal z}\left({\bf g}\right)$ is an ideal - the formatting should be clearer now:Kolmogorov-Arnold-Moser (KAM) theorem in the study of perturbed Hamiltonian systems. This landmark theorem clarified a major point in the field, and answered in positive the question of existence of systems which are neither ergodic, nor integrable, but rather "near integrable" in some appropriate sense. It also dispelled the long held belief that arbitrary small perturbations can convert an integrable system into an ergodic one ("Ergodic Hypothesis"). The result was exactly the opposite of this belief.

The theorem spawned a whole subfield (KAM Theory) which still is very active.

182342019554001815994XReference for the Koszul--Malgrange TheoremTrigremmFor locally trivial bundles, it is also natural to ask the fibre to be homeomorphic (not just homotopy equivalent) to a compact CW-complex. I have also been looking at small values and have that $M(r,n,3)$ obtains the singelton bound for all odd $n \ge 3$. The even case has been much more difficult but I recently showed $M(4,3,3) = 16$, again the singelton bound. I am trying to see if I can generalize that construction to other evens61806hBy hand, I could also veryfy all assertions for k=199892213111422260465You probably want the slightly more sophisticated version of the Hairy Ball theorem which says that the sum of the indices of the zeros equals 2.Yes Pietro, it concerns only even $n$, but I thought that a more detailed presentation would cloud the essence of the question.111900910333689976911386376Thanks for your answers. As Liran Shaul noticed I was actually asking for rings on which every module is flat:), thus only the Von Neumann regular rings!1954988RHodge–Tate structures of modular forms22647812006462Well, I think there is an entire scale of obviousness, depending on the reader and his or her background. If the author intends to write for a certain audience, presuming a certain background, then certain statements would be obvious, which might not be so clear to readers not in that intended audience. So when one reads that something is obvious, but it doesn't seem so clear, then it needn't necessarily reflect poorly on the author, but rather simply be an indicator that the reader was not in the intended audience.*shrug* I share pretty much all my ideas, silly or not, with anyone who asks. In fact this is one of the reasons I've never worked for GCHQ (UK government crypto people---analogue of NSA)---I can't face the idea of having to work on a secret problem that I then can't talk about with other people. Sharing is exactly what supervising PhD students is about.469825When I first met the concept of "characters" of topological groups in Pontryagin's book "Topological groups", it was defined as follows:

Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the torus $T=\mathbb{R}/\mathbb{Z}$. Here, the torus $T$ has the induced topology from the usual topology of the real line $\mathbb{R}$.

I found the same definition on Wikipedia. So, I think this is a standard definition of character. But, in a lot of modern articles, it seems to me that characters are defined as follows:

Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the discrete additive group $\mathbb{Q}/\mathbb{Z}$.

Clearly, these two definitions cannot be the same for all topological groups. However, if $G$ is a (discrete) finite group, then the two definitions agree.

**Questions:**

What kind of conditions on a topological group $G$ one needs in order to identify the two definitions of character?

If $G$ is a "profinite group", then do the two definitions agree? If the answer is yes, then how can one prove it?

Please give me any advice.

I found a way to answer the second question. One only needs to show the following: for any profinite group $G$ and any continuous homomorphism $f:G \to T$, the image of $f$ is finite. This statement can be shown as follows. The torus $T$ has an open neighborhood $U$ of $0$ which contains no non-trivial subgroup of $T$. Since $G$ is profinite, there exists an open normal subgroup $H$ of $G$ satisfying $H \subset f^{-1}(U)$. This implies $H \subset \ker(f)$. So, the map $f$ factors through the finite group $G/H$. This implies that the image of $f$ is finite.

Hence the two definitions of character agree for any profinite group.

1362874131502The question is not well-posed. What are you assuming about $E$? There may not be a single generator. Or $E(\mathbb{Q})$ could be torsion, in which case finding a point of infinite order is going to be very hard indeed.153396117852141051816Information density is also part of a more general issue of mathematical style, which is to resort to symbols only when absolutely necessary. I've seen this in writing advice from Knuth, Halmos, Munkres...229003714484431682283209625mmm... Why are you saying that $X/X'$ is simple in the case that $X' \leq Z$? Wouldn't $Z/X'$ usually be a non-trivial normal subgroup of $X/X'$?1151153jIt is $(-J*u_1)u_2$ and * is the convolution in $r$.1957680Suppose $X$ is a singular variety over a field $k$, which admits an action by a finite group $G$. Suppose the quotient $X/G$ is also a variety over $k$. If $Y_G$ is a resolution of $X/G$, does there exist a variety $Y$ which is a resolution of $X$ such that the action on $X$ could be extended to an action on $Y$ and

$$Y/G \simeq Y_G$$

Put it in another way, is the resolution of the quotient "the same thing" as the quotient of (equivariant) resolution?

214558211829501389113832659This amounts to counting primitive integral vectors in a centered Euclidean ball. If you consider cubes (i.e., $\ell^\infty$-balls) instead of Euclidean balls, the answer (the asymptotics) is given in this MO answer: https://mathoverflow.net/a/303153/140941478715Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?16776121401926563509854571From the book of Boss "Classical and modern methods in summability":

"The class of Hausdorff methods includes the Hölder, Cesaro and Euler methods. A large number of other matrix methods which play an essential role in summability are Hausdorff methods too."

I wonder if there is any different known example of Hausdorff methods? For example, circle methods such as Meyer-Konig or Taylor is an Hausdorff method?

Thanks Geoff. What I had meant in my crossed-out question is exactly that it be constant on elements generating the same cyclic subgroup. I will reread your answer carefully and then most likely be satisfied and accept. 17393273274873780125Hi Guillermo,

If we mean an ideal generated by a regular sequence by "regular", then there is an example.

Let $R = k[x,y]$ be a polynomial ring over a field and $m = (x,y)R$. Let $J = (x^2, y^2)$ and $K = m^2$. Then one can check that $J K = m^4$. Notice that $JK : K = m^4 : m^2 = m^2$, but $xy$ is not in $J =(x^2, y^2)$. Hence $JK : K \neq J$.

I believe the questions you have might have a close connection to the integral dependence of ideals (or reduction of ideals). For instance, the condition (1) implies that the ideal $J$ is integral over $I$ if $K$ contains a regular element. The technique is exactly the same as in Steven's proof. Also, check out the m-full property which has to do cancellations in some cases. I recommend an excellent book by Huneke, Swanson on this subject for reference. Chapter 1 contains a good overview of the theory.

I would like to suggest you to move your answer to the question.

672306zWhy is the Euler-Mascheroni constant not a Liouville number?34869815275581686719@YemonChoi Given an arbitrary underground map as an input Mornington Crescent is probably also undecidable. The proof is trivial if going around the circle line more than once is not allowed. That should not arise in normal play, but I'm not sure how to prove it.106879397446196052hhttps://plus.google.com/117949022376731880906/posts"Hi

I'm not quite sure what title to give to this question or what tags to use, because this isn't really my area of expertise and I'm unfamiliar with the terminology. It is a problem that came up while trying to write a program to enumerate some graphs, however there is no graph theory left in this problem.

Let me first start by giving definitions for the different components. I have a set $C$ and two functions: $s:C\rightarrow \mathbb{N}$ and $r:C\rightarrow \mathbb{N}$. The values for these functions are completely known. It is also true that $\forall c \in C: r(c)=s(c) \vee r(c)=\frac{s(c)}{2}$. I don't know whether this will make a difference for the solution of my problem, but it might be relevant information.

I am now interested in finding all possible functions $m:C\rightarrow \mathbb{N}$ that satisfy the following conditions:

- $\sum_{c\in C}\frac{s(c)}{m(c)}=\sum_{c \in C}\frac{s(c)}{4}$
- $\forall c \in C: r(c) \mid m(c)$

I really need an algorithm to find all possible solution, but of course I'm also interested in any theoretical results that could help me find such an algorithm. As I said, I'm not really at home in this type of problems, so I have no idea what to look for and where to start. The first conditions makes it impossible to have an infinite number of solutions, but in some cases I have some better upper bounds and lower bounds available for the value of the function $m$ in certain points. It would be nice if this could be taken into account, but I think it would not be the bottle-neck if I just filtered the solutions to take these bounds into account.

Any help is greatly appreciated. And of course I will try my best to clarify anything in my explanation that isn't clear.

Edit: Let me first say that $C$ will always be finite.

A small example might also be more clarifying. Suppose we have $C=\{1,2\}$, the function $s:C\rightarrow\mathbb{N};c\mapsto 2$ and the function $r:C\rightarrow\mathbb{N};c\mapsto 1$.

We then have three possible solutions for $m$:

- $m(1)=3$, $m(2)=6$
- $m(1)=4$, $m(2)=4$
- $m(1)=6$, $m(2)=3$

I'm trying to pin down the various ways we can extend a property of commutative rings to a corresponding property for schemes. Let $P$ be a property of commutative rings. We could define a scheme $(X,\mathcal{O}_X)$ to have property $P$ in one of the following ways:

- $\mathcal{O}_X(U)$ has $P$ for every open subset $U\subset X$.
- $\mathcal{O}_X(U)$ has $P$ for every
*affine*open subset $U\subset X$. - there exists an affine open cover ${U_i}$ of $X$ such that $\mathcal{O}_X(U_i)$ has $P$ for all $i$.
- for each $x\in U\subset X$ with $U$ an open subset, there exists an affine open $V\subset X$ with $x\in V\subset U$ such that $\mathcal{O}_X(V)$ has $P$.
- $\mathcal{O}_{X,x}$ has $P$ for all $x\in X$.

Evidently, $(1)\Rightarrow (2)\Rightarrow (3)$. If the property $P$ is stable under inversion of single elements (that is, $A$ has $P$ $\Rightarrow$ $A[1/s]$ has $P$ for any element $s\in A$) then $(3)\Rightarrow (4)$. Furthermore, if the property $P$ is stable under arbitrary localizations (that is, $A$ has $P$ $\Rightarrow$ $S^{-1} A$ has $P$ for any multiplicative subset $S$ of $A$) then $(4)\Rightarrow (5)$.

Thus, for many properties of commutative rings $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4)\Rightarrow (5)$.

Now we need to consider going in the other direction. I believe it can be shown that if $P$ is a local property in the sense that $A$ has $P$ iff $A_{\mathfrak{p}}$ has $P$ for each $\mathfrak{p}\in$ Spec$(A)$ then $(5)\Rightarrow (2)$. Now, it seems to me that if $P$ is a local property in this sense then the property is stable under localization by arbitrary multiplicative subsets. Thus if $P$ is a local property in this sense then $(2) \Leftrightarrow (3) \Leftrightarrow (4) \Leftrightarrow (5)$.

Finally, here is another notion of a property being local. Suppose that $P$ is such that $A$ has $P$ implies that $A[1/s]$ has $P$ for each $s\in A$ and that on the other hand, if $s_1,\ldots,s_n \in A$ are such that Spec$(A)=D(s_1)\cup D(s_2) \cup \cdots \cup D(s_n)$ then $A[1/s_i]$ has $P$ for all $i=1,..,n$ implies that $A$ has $P$. Then I believe it can be shown that $(4)\Rightarrow (2)$ and thus for such a property we have that $(2) \Leftrightarrow (3) \Leftrightarrow (4)$.

Does this seem right to you? I haven't seen any books on algebraic geometry discuss this question to my satisfaction and I am nervous that there may be some holes in my proofs, so if anyone knows off the top of their head that what I have described seems right then I would be happy to hear from you. Do you have any further comments to make about this process of extending a property of commutative rings to schemes?

14568958989891741745196665793986https://www.gravatar.com/avatar/f765b1d56f521b2f4df390f655508e58?s=128&d=identicon&r=PG&f=1|https://graph.facebook.com/464896663984310/picture?type=large18730574961460616498668509Thanks for the comment. A sequence of balls in my phrasing was meant to be a subsequence of the sequence of the balls, indeed. I hope it is clear.204314612342914678441109834By an *arch diagram* of size $n$, I mean a diagram consisting of $n$ arches matching $2n$ points, where the points are ordered on a line running from left to right. An arch diagram is basically just a way of representing a fixed-point free involution in $S_{2n}$, and arch diagrams of size $n$ are counted by the double factorial numbers (A001147)
$$1, 1, 3, 15, 105, 945, 10395, ...$$
A special class of arch diagrams are the non-crossing (planar) arch diagrams, which are in bijection with (rooted planar) binary trees and are counted by the Catalan numbers (A000108)
$$1, 1, 2, 5, 14, 42, 132, ...$$
My question is about a family of arch diagrams living between these two extremes, or really about a family of *equivalence classes* of arch diagrams.

Let's say that two arcs of an arch diagram are *left-adjacent* if their left end points are adjacent in the linear order. Then say that two arch diagrams are *equivalent modulo left-adjacency* if one can be obtained from the other by successively swapping left endpoints of left-adjacent arches. For example, among the three arch diagrams of size 2,
diagrams (2) and (3) are equivalent modulo left-adjacency, but neither is equivalent to (1).

One (somewhat arbitrary) way of picking a canonical representative of each equivalence class is to enforce the condition that $$ i < \alpha_i \text{ and } i' < \alpha_{i'} \text{ implies } \alpha_i > \alpha_{i'} $$ for all $1 \le i\le 2n-1$ and $i' = i+1$, where $\alpha \in S_{2n}$ is the fixed-point free involution corresponding to the arch diagram. For example, this way of choosing representatives yields the following arch diagrams of sizes $1..4$: In any case, it appears that equivalence classes of arch diagrams modulo left-adjacency are counted by the factorial numbers (A000142) $$1, 1, 2, 6, 24, 120, 720, ...$$ I've managed to convince myself of this by considering a more general notion of arch diagram allowing unattached points (corresponding to fixed points of the associated involution), and defining a two-variable generating function $A(x,z)$ counting equivalence classes of arch diagrams by number of unattached points ($x$) and arches ($z$). An inductive decomposition of this family of equivalence classes of (generalized) arch diagrams implies that $$A(x,z) = 1 + \sum_k \frac{z^k}{k!} \cdot \frac{\partial^k x\cdot A(x,z)}{\partial x^k} = 1 + (x+z)\cdot A(x+z,z)$$ from which $A(0,z) = \sum_n n! \cdot z^n$ follows.

My questions are:

- Has this equivalence relation on arch diagrams been previously studied, and is there a standard name for arch diagrams modulo this equivalence relation?
- Is there a simple bijective proof of the fact that equivalence classes of arch diagrams modulo left-adjacency are counted by the factorial numbers?

I've been trying to determine the rationality of certain fields of invariants coming from G-lattices. More precisely, letting $G$ be a finite group, $L=\mathbb{Z}^n$ a free abelian group with a $G$ action, we set $F[L]$ be the group algebra (Laurent polynomials in n indeterminates) and $F(L)$ to be its field of fraction and then ask whether $F(L)^G$ is rational over $F^G$.

If $L$ is a permutation lattice then this is the usual Noether probelm. In my work I came across a family of lattices which are usually not permutation but are very close to being permutation. My question is if there is some way to move from this type of lattices to permutation lattices in a "nice" way.

To illustrate I added my smallest example below.

Let $L$ be the free abelian group on generators {$a,\hat{a},b,\hat{b},u$} (written multiplicativly) . Let $K=$ {$e,\sigma,\tau,\sigma\tau $} be the Klein four group and define a $K$ action on $L$ as follows:

$\sigma$: switches between $a$ and $\hat{a}$, fixes $b$ and $\hat{b}$ and $\sigma(u)=ab\hat{a}\hat{b}/u$.

$\tau$: switches between $b$ and $\hat{b}$, fixes $a$ and $\hat{a}$ and $\tau(u)=ab\hat{a}\hat{b}/u$.

The K-lattice $L_0$ generated by {$a,\hat{a},b,\hat{b}$} is easily seen to be a permutation lattice, so my hope is to find some sort of rationality condition (pure, stable, retract) of $F(L)^K$ over $F(L_0)^K$, so I can then use the solution to the usual Noether's problem (if it exists).

The only theorem that (I found and) is close to helping me in this case is a generalization of Luroth theorem which says that if the G action on $u$ in $F(L)$ is something like $g(u)=a_g u+b_g$ where $a_g,b_g \in F(L_0)$ then $F(L)^G$ is rational over $F(L_0)^G$. This is similar to the case above except that the action on $u$ is of the form $\sigma(u) = a_\sigma u^{-1}$.

In this particular example, we also have that $L$ is a permutation lattice over the subgroup {$\sigma\tau,e$}, because $\sigma\tau(u)=u$. I tried to first take the $\sigma\tau$ invariants and then the rest of the group but still didn't get anything helpful, though this way still seems promising.

Exponential Diophantine $\prod x_i^{e_{x_i}} + \prod y_i^{e_{y_i}} = \prod z_i^{e_{z_i}}$ ,$e_{x_i}>3,e_{y_i}>3,e_{z_i}>3$The case of $d = 0$ is even known in mixed characteristic and goes back at least to Lipman in his work in the 70s. I do have something to say about the question though which is probably too big to fit in a comment.1957992296155547845Arnav Tripathy\https://i.stack.imgur.com/vMT8h.jpg?s=128&g=1344831They show that $\DeclareMathOperator{\WKL}{WKL}\DeclareMathOperator{\RT}{RT}\DeclareMathOperator{\RCA}{RCA} \RT^2_2$ is $\Pi^0_3$-conservative over $\RCA_0$. Thus, there is no way that $\RT^2_2$ can be essential in a proof of simple **first-order** statements like for instance the *twin prime conjecture*, that have a $\Pi^0_3$ form:
$$
(\forall n)(\exists p>n)(p\text{ is prime and }p+2\text{ is prime.})
$$
(This is $\Pi^0_2$ which is, in particular, $\Pi^0_3$.)

**First-order** means you quantify over numbers only, whereas $\RT^2_2$ itself involves quantifying over sets of numbers, making it second order.

For comparison, Leo Harrington (1978) showed that $\WKL_0$ (weak König's lemma) is $\Pi^1_1$-conservative, hence in particular also $\Pi^0_3$-conservative, over $\RCA_0$.

hJust merged this with my long-lost account!

The question is whether there's an algorithm saying whether there's a positive root, not to compute this root.22717441185528I know that a contravariant functor is a covariant functor in the opposite category, but what is the relationship between the homology in the opposite category and the original homology?806268100087987717360022020348397152858The content of this note was the topic of a lecture by Günter Harder at the School on Automorphic Forms, Trieste 2000. The actual problem comes from the article A little bit of number theory by Langlands.

The problem is about a connection between two quite different objects. The first object is the following pair of positive definite quadratic forms: $$ P(x,y,u,v) = x^2 + xy + 3y^2 + u^2 + uv + 3v^2 $$ $$ Q(x,y,u,v) = 2(x^2 + y^2 + u^2 + v^2) + 2xu + xv + yu - 2yv $$ The second object is the elliptic curve $$ E: y^2 + y = x^3 - x^2 - 10x - 20. $$

To each of our objects we now associate a series of integers. For each integer $k \ge 0$ define $$ n(P,k) = | \{(a,b,c,d) \in {\mathbb Z}^4: P(a,b,c,d) = k\} |, $$ $$ n(Q,k) = | \{(a,b,c,d) \in {\mathbb Z}^4: Q(a,b,c,d) = k\} |. $$

As a matter of fact, these integers are divisible by $4$ for any $k \ge 1$ because of the transformations $(a,b,c,d) \to (-a,-b,-c,-d)$ and $(a,b,c,d) \to (c,d,a,b)$.

For any prime $p \ne 11$ we now put $$ a_p = |E({\mathbb F}_p)| - (p+1),$$ where $E({\mathbb F}_p)$ is the elliptic curve over ${\mathbb F}_p$ defined above.

Then Langlands claims

**For any prime $p \ne 11$, we have **
$ 4a_p = n(P,p) - n(Q,p).$

The "classical" explanation proceeds as follows: Given the series of integers $n(P,p)$ and $n(Q,p)$, we form the generating series $$ \Theta_P(q) = \sum \limits_{k=0}^\infty n(P,k) q^k = 1 + 4q + 4q^2 + 8q^3 + \ldots, $$ $$ \Theta_Q(q) = \sum \limits_{k=0}^\infty n(Q,k) q^k = 1 + 12q^2 + 12q^3 + \ldots. $$ If we put $q = e^{2\pi i z}$ for $z$ in the upper half plane, then $\Theta_P$ and $\Theta_Q$ become ${\mathbb Z}$-periodic holomorphic functions on the upper half plane. As a matter of fact, the classical theory of modular forms shows that the function $$ f(z) = \frac14 (\Theta_P(q) - \Theta_Q(q)) = q - 2q^2 - q^3 + 2q^4 + q^5 + 2q^6 - 2q^7 - 2q^9 - 2q^{10} + q^{11} - 2q^{12} \ldots $$ is a modular form (in fact a cusp form since it vanishes at $\infty$) of weight $2$ for $\Gamma_0(11)$. More precisely, we have $ f(z) = \eta(z)^2 \eta(11z)^2,$ where $\eta(z)$ is Dedekind's eta function, a modular form of weight $\frac12$.

Now we have connected the quadratic forms to a cusp form for $\Gamma_0(11)$. This group has two orbits on the projective line over the rationals, which means that the associated Riemann surface can be compactified by adding twocusps: the result is a compact Riemann surface $X_0(11)$ of genus $1$. Already Fricke has given a model for this Riemann surface: he found that $X_0(11) \simeq E$ for the elliptic curve defined above.

Now consider the space of cusp forms for $\Gamma_0(11)$. There are Hecke operators $T_p$ acting on it, and since it has dimension $1$, we must have $T_p f = \lambda_p f$ for certain eigenvalues $\lambda_p \in {\mathbb Z}$. A classical result due to Hecke then predicts that the eigenvalue $\lambda_p$ is the $p$-th coefficient in the $q$-expansion of $f(z)$. Eichler-Shimura finally tells us that $\lambda_p = a_p$. Putting everything together gives Langlands' claim.

Way back then I asked Harder how all this follows from the general Langlands conjecture, and he replied that he did not know. Langlands himself said his examples came "from 16 of Jacquet-Langlands". So here's my question:

* Does anyone here know how to dream up concrete results like the one above from Langlands' conjectures, or from "16 of Jacquet-Langlands"?*

Yes. The assumption implies that $\mathcal{B}$ is cofinal in the poset $[\kappa]^{<\kappa}$, and hence that this poset has cofinality $\le\kappa$. This implies that $\kappa$ is regular, by this answer.

Is there any result about the time complexity of finding a cycle of fixed length $k$ in a general graph? All I know is that Alon, Yuster and Zwick use a technique called "color-coding", which has a running time of $O(M(n))$, where $n$ is the number of vertices of the input graph and $M(n)$ is the time required to multiply two $n \times n$ matrices.

Is there any better result?

Note that $ker(f)$ is a sublattice of the integer lattice $\mathbb{Z}^r$. And conversely any such integer sublattice will give you such a homomorphism $f$. The index of this sublattice is exactly what you call $h$ (since the homomorphism is surjective). If $C$ is a symmetric convex body in $\mathbb{R}^r$ and $L$ a full rank lattice then the number of lattice points in $t C$ (the dilation of $C$ by a real factor of $t$) is asymptotic to $t^r vol(C)/D$, where $D$ is the volume of the fundamental domain of $L$. In your case $D=h$ and $C$ is the unit hypercube. Getting a good remainder is usually a much harder job. In general the remainder is bounded by something proportional to the area of $C$. When $C$ has a smooth boundary (which alas, the hypercube doesn't) you can get better estimates (but usually have to work quite hard). When you're working in a high dimension a large fraction of the volume is close to the vertices -- making things much more difficult.

In particular you might look at the work of Martin Huxley. For example his book "Lattice Points, Area and Exponential Sums" or a number of his papers on this subject.

1434079178993*I was reading "A Concise Course in Algebraic Topology" by J.P.May (page 52) and found the proof of the following theorem incomprehensible:

Let $p:E\rightarrow B$ be a map and let $\mathcal{O} $ be a numerable open cover of B. Then $p$ is a fibration if and only if $p:p^{-1} (U) \rightarrow U$ is a fibration for every $U\in \mathcal O$.

The only if part is clear. To check the if part, May tries to patch together path lifting functions of $p|p^{-1} (U)$ to construct a path lifting function. To do this, he defines some complicated functions, which I could not easily understand.

He starts by choosing maps $\lambda_U : B\rightarrow I$ such that $\lambda _U ^{-1} (0,1]=U$. For each finite ordered subset $T=\{U_1 , \cdots , U_n \}$ of $\mathcal O$, he defines $c(T)=n$, and $\lambda _T : B^I \rightarrow I$ as $$\lambda_T (\beta ) = \inf \{(\lambda_{U_i} \circ \beta ) (t) : (i-1)/n \le t \le i/n , 1\le i\le n \}$$ and let $W_T = \lambda _T ^{-1} (0,1]$. I understand that $\{W_T\}$ is an open cover of $B^I$, and that $\{W_T : c(T)<n\}$ is locally finite for each $n$. Then he defines $\lambda _T : B^I \rightarrow I $ as $$\lambda_T (B) = \max \{ 0,\lambda_T (B) - n \sum _{c(S)<c(T)} \lambda_S (\beta )\}, V_T = \{\beta : \gamma_T (\beta)>0\} \subset W_T.$$ I understand that $\{V_T\}$ is a locally finite open cover of $B^I$.

Now, he chooses path lifting functions $$s_U : p^{-1} (U) \times _p U^I \rightarrow p^{-1} (U) ^I $$ for each $U\in \mathcal O $, and he constructs a global path lifting function by the following steps.

First, choose a total order on the sets of finite ordered subsets of $\mathcal O $. For a given $T=\{U_1, \cdots , U_n \}$, he defines $s_T (e,\beta [u,v]):[u,v]\rightarrow E $ for each $\beta \in V_T$, $u,v\in [0,1]$, $e\in p^{-1} (\beta (u))$ canonically, by lifting $\beta[u,v]$ by letting its starting point be $e$. ($\beta [u,v]$ is defined as the restriction of $\beta$ to $[u,v]$.)

Now here is the part where I don't understand at all. For each $e,\beta$ such that $\beta(0)=e$, he defines $s(e,\beta)$ to be the concatenation of the parts $s_{T_j} (e_{j-1} , \beta [u_{j-1} ,u_j ])$, $1\le j \le q$ where the $T_i $ run through the set of all $T$ such that $\beta \in V_T$, and $$u_0 =0, \ u_j = \sum _{i=1} ^j \gamma _{T_i } (\beta ).$$

He asserts that $s$ is well defined and continuous, thus a path lifting function for $p$.

I could not understand at all why the definition of $s$ makes sense. For this to make sense, I believe that $u_q$ should be $1$, which I cannot get at all. For me, this definition seems completely random. Can someone please kindly explain why this definition makes sense?

zLooking for a limit related to the series in a previous postThank you, Dan. I am trying to get a pdf copy of the mentioned paper off the web, but it seems hard...17630051084374Do you mean Free Probability Theory (See e.g. Voiculescu, D. V.; Dykema, K. J.; Nica, A. Free random variables. A noncommutative probability approach to free products with applications to random matrices, operator algebras and harmonic analysis on free groups) ?

1203064710591694480%Here is my attempt at fleshing out Peter Mueller's answer. If anyone sees any inaccuracies (or worse) here, I would be grateful if you could let me know. If this answer seems correct to you, please be sure to upvote *Peter's answer*.
$\newcommand{\GL}{\operatorname{GL}}$
$\newcommand{\ra}{\rightarrow}$
$\newcommand{\Z}{\mathbb{Z}}$

If $G$ is a group, then we say a group $S$ is a **subquotient** of $G$ if
there are subgroups $N \subset H \subset G$ with $N$ normal in $H$ such that
$H/N \cong S$.

Step 1: Let $G$ be a finite group. We claim that every simple subquotient of $G$ is a subquotient of some Jordan-Holder factor of $G$. By induction it suffices to show that if $N$ is a normal subgroup of $G$, then every simple subquotient of $G$ is a subquotient of either $N$ or $G/N$. Let $G'$ be a subgroup of $G$, let $N'$ be a normal subgroup of $G'$, and let $\pi: G \ra G/N$ be the quotient map. Then $G'$ is an extension of $\pi(G')$ by $G' \cap N$, so the Jordan-Holder factor $G'/N'$ of $G'$ must appear up to isomorphism as a Jordan-Holder factor of either $G' \cap N$ or of $\pi(G')$, and it follows that $G'/N'$ is a subquotient of either $N$ or $G/N$.

Step 2: Let $m \in \Z^+$. Let $n \in \Z^+$ have prime power factorization $n = p_1^{a_1} \cdots p_r^{a_r}$. Then $\GL_m(\Z/n\Z) \cong \prod_{i=1}^r \GL_m(\Z/p_i^{a_i} \Z)$. Further, for $1 \leq i \leq r$, the kernel of the natural map $\GL_m(\Z/p_i^{a_i} \Z) \ra \GL_m(\Z/p_i \Z)$ is a $p_i$-group. Thus every noncyclic simple subquotient of $\GL_m(\Z/n\Z)$ is a simple subquotient of $\GL_m(\Z/p_i \Z)$ for some $i$. Thus we assume henceforth that $n = p$ is prime.

Step 3: We apply the following theorem of Larsen-Pink: there is a positive
integer $J'(m)$ such that every subgroup $\Gamma \subset \GL_m(\Z/p\Z)$
admits a subnormal series $\Gamma_3 \subset \Gamma_2 \subset \Gamma_1 \subset \Gamma$ such
that:

$\bullet$ $[\Gamma:\Gamma_1] \leq J'(m)$;

$\bullet$ $\Gamma_1/\Gamma_2$ is a direct product of finite simple groups of Lie type;

$\bullet$ $\Gamma_2/\Gamma_3$ is commutative; and

$\bullet$ $\Gamma_3$ is a $p$-group.

In view of Step 1, the result follows immediately from this and the fact that
for all $k \geq 9$, $A_k$ is *not* isomorphic to a finite simple group of Lie type.

It turns out that Beno Eckmann actually asked Hermann Weyl why he published this work in Spanish, as described here and here. The answer is remarkable and unexpected (and apparently does not involve his Spanish spouse).

I quote from the latter source (RK=Robert Kotiuga, BE=Beno Eckmann):

277081is there is some algebraic properties of the fundamental group that can garuentee the property without talking about curvature say for example if the fundamental group is torsion free168082ftriangles in a graph with specified clique number 283212vİt is too late to edit. But you are right, my question is2021526l(Nitpicking: The guy's name was Mertens, not Merten.)140044thogrhm5631521853360RK: In the hallway outside our offices was a high-tech espresso machine and every morning Beno would take a break to sit and enjoy an espresso outside our offices. The first day, I "coincidentally" joined him and he related wonderful anecdotes from 1950-1955, after Hermann Weyl retired from the IAS, resettled in Zürich, and frequented the department. The next day I resolved to ask Beno a question which I didn't think any living person could answer.

RK: Beno, there is something I really don't understand about Hermann Weyl.

BE: What is it?

RK: Well, in his collected works, there are are two papers about electrical circuit theory and topology dating from 1922/3. They are written in Spanish and published in an obscure Mexican mathematics journal. They are also the only papers he ever wrote in Spanish, the only papers published in a relatively obscure place, and just about the only expository papers he ever wrote on algebraic topology.It would seem that he didn't want his colleagues to read these papers.

BE:Exactly!

RK: What do you mean?

BE:Because topology was not respectable!

No.

Take a poset $P$ on the ground set $S=\{0,1,a,b,c,d\}$ determied by $a>c$, $a>d$, $b>c$, and $b>d$ (with $0$ the minimal and $1$ the maximal element). Let $L$ be a complete lattice on $S$ with $\{a,b,c,d\}$ being an antichain (and the same minimum and maximum). Then the identical map on $S$ is an order preserving map $L\to P$, but $P$ is not a lattice, so its completion contains more than 6 elements. THus there is no sujcection from $L$ to the completion at all.

15771741471452094350See also Lieven le Bruyn's blog http://www.neverendingbooks.org/index.php/who-dreamed-up-the-primesknots-analogy.html5263141737180116611Yes, there exists such a generalization. It was done by Marie-Helene Schwartz.

Formules apparentées à la formule de Gauss-Bonnet pour certaines applications d’une variété à n dimensions dans une autre. (French) Zbl 0057.38102 Acta Math. 91, 189-244 (1954).

Formules apparentées à celles de Nevanlinna-Ahlfors pour certaines applications d’une variété à n dimensions dans une autre. (French) Zbl 0057.31602 Bull. Soc. Math. Fr. 82, 317-360 (1954).

However, on my opinion, this generalization was much less successful than the original Ahlfors theory.

On the other hand, Ahlfors's one-dimensional theory has several interesting APPLICATIONS to several complex variables, the key word is "Ahlfors currents", see, for example,

Henry de Thélin, Ahlfors' currents in higher dimension. Ann. Fac. Sci. Toulouse Math. (6) 19 (2010), no. 1, 121–133.

Remark. In general, there is (at least) two very different generalizations of classical complex analysis to several dimensions. 1. Several complex variables, and 2. Quasiregular maps between real n-dimensional manifolds. The geometric part of the theory (where the main notion is conformity) better generalizes in the setting 2 (for analytic functions of SCV, conformality is meaningless). The work of M-H Schwartz belongs to the second direction.

|How is what you are asking for different from uniformization?https://lh4.googleusercontent.com/-2wKZaOXQKsk/AAAAAAAAAAI/AAAAAAAAAAA/AAnnY7pgoGyzGw0Wz9oZEOlwabxjjPJQjQ/mo/photo.jpg@WillieWong Changing dimension is definitely one case where the answer is "can't be done". Fortunately I can now show that my matrix avoids this problem.22962181513805ZConvergence of sequence of inverse functions1522247~https://graph.facebook.com/1668988663197705/picture?type=large14825651799466876025Also, if you want $\prod (A \coprod B) \cong \prod A \times \prod B$, you'd better take $\prod \emptyset$ to have one element.While you are correct regarding the relationship between the discriminant and the resultant, the OP assumed that the polynomial was monic (hence a_n=1) and that he was only interested in the discriminant up to sign.<While I don't know of any reference that answers this question explicitly, I have a few observations that might be helpful. For an upper-semicontinuous C$^*$-bundle $\pi : \mathcal{A} \to X$ and an open subset $U \subseteq X$, $\Gamma^b (U, \mathcal{A})$ denotes the C$^*$-algebras of continuous sections $s:U \to \mathcal{A}$ that are norm bounded. $\Gamma_0 (U, \mathcal{A} )$ denotes the C$^*$-subalgebra of $\Gamma^b(U,\mathcal{A})$ consisting of those sections $s$ that `vanish at infinity' on $U$ in the sense that $$ \{ x \in U : \| s(x) \| \geq \alpha \} $$ is compact for all $\alpha>0$.

Suppose that $A$ is a C*-algebra and $\phi: \mathrm{Prim}(A) \to X$ is continuous. Combining Remark 3.7.3 and Theorem 5.6 of the paper "Sheaves of C$^*$-algebras" by Ara and Mathieu, we do get an upper-semicontinuous C$^*$-bundle $\pi: \mathcal{B} \to X$ over $X$ together with a canonical embedding of the multiplier algebra $M(A)$ into the C$^*$-algebra $\Gamma^b (X, \mathcal{B} )$. If we then regard $A \subseteq \Gamma^b(X , \mathcal{B})$, one could then consider in each fibre $\mathcal{B}_x$ of $\mathcal{B}$ the subalgebra $$ \mathcal{A}_x:=\{ b \in \mathcal{B}_x : b=a(x)\mbox{ for some }a \in A \}. $$ It looks plausible that $A$ would then be isomorphic to (a subalgebra of) $\Gamma_0 (X, \mathcal{A} )$ where $\mathcal{A}$ is a sub-bundle of $\mathcal{B}$ with fibres $\mathcal{A}_x$, though I haven't checked the details.

As for the converse, suppose that $A = \Gamma_0 (X , \mathcal{A} )$ for some bundle $\pi : \mathcal{A} \to X$. I think that the existence of a map $\phi : \mathrm{Prim} (A) \to X$ can possibly be deduced from Lemma 2.25 of the paper "C*-algebras over topological spaces : the bootstrap class," by Meyer and Nest (here we need the assumption that $X$ is a sober space):

Let $\mathcal{O}(X)$ and $\mathcal{I} (A)$ denote the directed sets of open subsets of $X$ and closed-two-sided ideals of $A$ respectively (both ordered with respect to inclusion). Then there is a bijective correspondance between continuous maps $\mathrm{Prim}(A) \to X$ and maps $\mathcal{O} (X) \to \mathcal{I} (A)$ that commute with arbitrary suprema and finite infima.

Certainly, if $A$ is isomorphic to $\Gamma_0 (X, \mathcal{A} )$ for some u.s.c. C*-bundle $\pi:\mathcal{A} \to X$, then there is a natural map $\mathcal{O}(X) \to \mathcal{I} (A)$ where $U \subseteq X$ is identified with the ideal $$\Gamma_0(U,\mathcal{A})=\{ a \in A: a(x) = 0 \mbox{ for all } x \in X \backslash U \} $$ of $A$, which appears to have the required properties.

It is worth pointing out that when constructing the bundle in point 1. one cannot simply adapt the proof from the case of locally compact Hausdorff $X$ in general (the one used in Appendix C of "Crossed products of C$^*$-algebras" by Dana Williams for example).

Indeed, for $x \in X$ define the ideal $$ I_x : = \bigcap \{ P \in \mathrm{Prim} (A) : \phi (P) = x \} $$ and the quotient C*-algebra $A_x = A/I_x$. Then there is a natural way to regard each $a \in A$ as a cross section $X \to \coprod_{x \in X} A_x$, letting $a(x)$ be the image of $a$ under the quotient mapping $A \to A/I_x$. Indeed, for locally compact Hausdorff $X$, this identification gives the required $*$-isomorphism $A \to \Gamma_0(X,\mathcal{A})$.

Returning to the general case, if we take $X= \mathrm{Prim} (A)$ and $\phi$ to be the identity map, this construction yields an upper-semicontinuous C*-bundle with fibres $A_x$ and section algebra $\Gamma_0(X,\mathcal{A})\cong A$ if and only if $\mathrm{Prim} (A)$ is Hausdorff, in which case the bundle is in fact continuous. Thus it looks like the fibres of the bundle constructed in point 1. will not coincide with $A_x$ in general.

This paper might possibly be relevant:

http://www.ams.org/mathscinet-getitem?mr=2922602

It's not quite exactly what you are asking for. Instead of Gorenstein algebras of low rank, they are considering short algebras, i.e., ones with small socle degree. But hopefully it is related and helpful for you.

Some pointers to the literature that may help: The small-$R$ asymptotics of the Thermodynamic Bethe Ansatz equation can be expressed in terms of a Painleve III function with independent variable $R$ [1], and the small-$R$ asymptotics has been studied in that connection [2]. The $R\rightarrow 0$ limit (called the "massless" or "ultraviolet" limit in the physics literature) has the form of an Airy function [3].

[1] C.A. Tracy and H. Widom, Proofs of Two Conjectures Related to the Thermodynamic Bethe Ansatz, Commun.Math.Phys. 179 (1996) 667-680 [arXiv:solv-int/9509003].

[2] P. Fendley and H. Saleur, $N=2$ Supersymmetry, Painleve III and Exact Scaling Functions in 2D Polymers, Nucl.Phys.B 388 (1992) 609-626 [arXiv:hep-th/9204094].

[3] P. Fendley, Airy functions in the thermodynamic Bethe ansatz, Lett.Math.Phys. 49 (1999) 229-233 [arXiv:hep-th/9906114].

342117The proof given in that Monthly solution seems to me to be at best incomplete. I agree (in the notation of that problem) that the sets S_d for divisors d of k do not consume too much of the set G_k, but what about sets S_d where d < k but d is not a divisor. Perhaps they involve some elements of G_k. Does anyone see a way to repair this problem?224533534155178110bYou are a treasure to this community. Thank you.5810452179641255669The real question is why anyone thinks the phrase "natural numbers" is preferable to "nonnegative integers" or "positive integers." 997888445387142403707388Yes, the correct spelling is Zdzislaw. The "dz" here is pronounced by Poles roughly as "g" in "gym." Wikipedia has something to say on this name: http://en.wikipedia.org/wiki/Zdzis%C5%82aw2103638for this, even the simple Markov inequality is better than Hoeffding since one can carry out all the computations exactly. This gives something like: P[Y>X] < [ 2\sqrt{pq(1-p)(1-q)} + (1-p)(1-q) + pq ]^n 11516912142100The same argument works with ultrafilters on any set, not just ultrafilters on a countable set.^continued... do you suggest letting 2 variables, one for $e_1$ and $e_2$, say $t_1$ and $t_2$ and setting them simultaneously to 0 and 1? Thanks for your valuable suggestions!1513874Muirhead (1982, "Aspects of Multivariate Statistical Theory") references on page 59 a result (from MacDuffee, 1943, chap 3, "Vectors and Matrices") a book I cannot find):

" The only polynomials in the elements of a matrix satisfying $p(I)=1$ and $p(AB)=p(A)p(B)$ for all matrices, are the integer powers of det B: $p(B) = (\det B)^k$ for some integer $k$.

Where otherwise, can I find this result and discussions of it?

For completeness, I also post the public worksheet here: http://sage.lacim.uqam.ca/home/pub/14/18861531064437I'd guess you take the product of the cones over the varieties and divide out by the diagonal action of $\mathbb{G}_m$.Let's agree that if there is only one deformation available, then it's important, even if you didn't otherwise care about deformations. Actually, since $p$ is a critical point of $d(-,p)^2$, one can define the Hessian in $p$ without any choice of connection (just locally as the Hessian of that function in a chart).162783881017412749*The nLab page on partitions of unity mentions the application of partitions of unity as a way to construct continuous maps to geometric realizations of simplicial spaces. However I often feel uncomfortable with the continuity of maps constructed in the realm of this example.

Let me discuss my uncertainty with the help of an explicit example.

Let $X$ be a space with an open cover $X=\bigcup_{i\in I} U_i$ and a partition of unity $\{f_i\colon X\rightarrow\mathbb{R}\}_{i\in I}$ subordinate to this cover and $\{a_i\}_{i\in I}$ a sequence of arbitrary real numbers.

Consider the poset $(\mathbb{R},\le)$ as a category internal to Spaces with the usual Euclidian topology. The nerve of this is a simplicial space $N_\bullet(\mathbb{R},\le)$. It has a geometric realization $||N_\bullet(\mathbb{R},\le)||=B(\mathbb{R},\le)$, whereby I mean the so called "fat" geometric realization, i.e. I only factor out the face maps and not the degeneracies.

Now define a map $g\colon X\rightarrow||N_\bullet(\mathbb{R},\le)||$ by mapping $x$ to the residue class of $((a_{i_0}\le...\le a_{i_k}),(f_{i_0}(x),...,f_{i_k}))\in N_k(\mathbb{R},\le)$ where ${i_0,...,i_k}$ are exactly those indices where $f_i(x)\neq0$.

This gives a well-defined map. Is this map always continuous? The choice of the non-zero $f_i(x)$'s feels somehow uncomfortable to me. The values of $g$ seem to "jump" somehow.

However, my impression that many of the constructions summarized by the mentioned example of the nlab go along this example, so probably my feeling is wrong and I would be happy seeing somehow thin out my fog.

Edit: The comment of Omar Antolín-Camarena and the answer of Oscar Randal-Williams have shown, that the example is not suitable to address my concerns. Let my give it another try by means of a modified example using the notation of the example before.

Let $Y\subseteq \mathbb{R}\times X$ a subspace. View $Y$ as a topological preordered set by using the standard ordering in the first coordinate, i.e. $(a,x)\le(b,y):\iff a\le b\text{ and } x=y$. Assume there are $a_i\in \mathbb{R}$, such that $\{a_i\}\times U_i\subseteq Y$. Now define the map $$g\colon X\rightarrow ||N_\bullet(Y)||$$ analogously by $$g(x)=[((a_{i_0},x)\le...\le(a_{i_k},x)),(f_{i_0}(x),...,f_{i_k}(x)],$$ where the indices $i_0,...i_k$ are exactly those, where $f_i(x)\neq 0$. Now one does not have the freedom to take a bigger set of indices, because if $f_i(x)=0$ it could be possible that $(a_i,x)\notin Y$, so the suggestion of Oscar Randal-Williams and Omar Antolín-Camarena do not seem to work.

(3) The $8\times8$ square has two tilings by $8\times 1$ rectangles, but a region consisting of two $9\times 9$ squares with inner $7\times 7$ squares removed has four tilings. I don't think there is anything similar for $n>1$.Check that, that would be wrong. Factoring out by the span of the sums over all sets comprising a disjoint union of straight lines might work...@Fano blow ups of $\mathbb CP^n$15229247070081045143126472318241941125263jhttp://forum.xda-developers.com/member.php?u=184389918595836460021458960The key reference for this is

MR0799042 (87d:03141). Henle, J. M.; Mathias, A. R. D.; Woodin, W. Hugh.

A barren extension. InMethods in mathematical logic (Caracas, 1983), C. A. Di Prisco, editor, 195–207, Lecture Notes in Math., 1130, Springer, Berlin, 1985.

There, Henle, Mathias, and Woodin start with $L(\mathbb R)$ under the assumption of determinacy (and $\mathsf{DC}$), and force with $\mathcal P(\omega)/\mathrm{Fin}$; they refer to the resulting model as "the Hausdorff extension".

They use $\mathsf{ZF}+\omega\to(\omega)^\omega$ to prove that the Hausdorff extension is *barren*, meaning that every map from an ordinal into the ground model was already in the ground model. They also show (under $\mathsf{AD}+V=L(\mathbb R)$) that all strong partition cardinals below $\Theta$ are preserved in the extension. Easily, the extension also preserves $\mathsf{DC}$.

On the other hand, the forcing adds a Ramsey ultrafilter on $\omega$ (in particular, unboundedly many strong partition cardinals below $\Theta$ is not enough to ensure all sets all reals are Lebesgue measurable).

Curiously $4$ is the only dimension $d$ in which $R(C_p^d)$ approaches $1$ as $p \to \infty$, and the reason is something happening in the middle dimension. I wonder if this is related to the other reasons that $4$ is an exceptional dimension... 477818@YCor: Thanks for the comment. I've done a bit of editing but will also need to think over the strategy Ben uses when $\phi$ involves more than one summand. It seems awkward to have to consult tables like those of Lawther (or to make up one's own for classical types), but maybe there is no alternative.2300151129219089260I thought, I could turn the comments into an answer…

The approach by couplings does not work without modifications and the reason is that couplings do not exist if the measures have different total mass: If $P$ and $Q$ are two measures on $X$ with different total masses which were coupled by $\mu$, then $$ \int_x\int_y d\mu(x,y) = \int_x dP(x) \neq \int_y d\tilde P(y) = \int_y\int_xd\mu(x,y). $$ However, for a given metric $d$ for probability measures one can build a metric for measures with different masses as follows: If $P$ has total mass $P(X) = p$ and $Q$ has total mass $Q(X) = q$, define $$ D(P,Q) = |p-q| + d(\tfrac{P}{p},\tfrac{Q}{q}) $$ (see Gromov's "Metric structures on Riemanian and Non-Riemannian Spaces", Chapter $3\tfrac12$.B).

There are also other approaches to metrics on measure spaces such as the Kantorovich-Rubinstein norm $$ W(P,Q) = \sup\bigg\{\int f\, dP - \int f\, dQ\ :\ f\ \text{Lipschitz with constant}\ \leq 1\bigg\} $$ and others (cf. Villani's "Optimal Transport - Old and New", Chapter 6).

1561985Yes, G-S shows that there is no good voting system when you allow voters to state partial orders or preorders instead of total orders. You are adding possibilities to a set over which there is no good voting system, and this can't create a good voting system. The incoherent social choices guaranteed by G-S still exist in the enlarged set.1054050from topology point of view, you might want to disregard length 2 loops.30064164384343A beginner of developing.

Laugh to world, The world will laugh to you ;).

#SOReady2Help!

69196user829012088547This seems related: [Why is $\beta(\mathbb{N}\times\mathbb{N})\neq\beta(\mathbb{N})\times\beta(\mathbb{N})$?](https://math.stackexchange.com/q/1806304) (math.SE)Could you please teach me the genus of Y^3 = X^4 - 1 ?

1317920577722166347476454lCompact surface with genus$\geq 2$ with Killing field355643540474d@BenoîtKloeckner : Thank you for the suggestion.118471020338912212891@ToddTrimble: I found this after some googling: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=9&cad=rja&uact=8&ved=0CFoQFjAI&url=ftp%3A%2F%2Fftp.cis.upenn.edu%2Fpub%2Fcis610%2Fpublic_html%2Fconvex45.pdf&ei=s7iwU83bMJawyAS6soH4Dw&usg=AFQjCNGx8_2T9UdXGxBPK_Zn7dsxuDVNeg&bvm=bv.69837884,d.aWw This discusses the facts used by Daniel, but it seems to involve machinery.For $n=2$ you can try http://www.alpertron.com.ar/QUAD.HTM for various values of $x_0$.215582114139221690595Suvrit: It appears this paper too deals with the question of how concentrated $||X||$ is to its median. I am interested in showing it is close to $||\mathbb{E}X||.$ This will need some stringent constraint on the tails of the entries, and will not happen under such minimal assumptions as in the paper you provide.653094`References/Papers on analytic solutions to SDEshttps://www.gravatar.com/avatar/36e0f8049c8f84154ee832fb9bddac16?s=128&d=identicon&r=PG&f=1464943xDid you look in Steenrod's "The topology of fibre bundles"?472856Thanks for such a quick answer! Just one more question: I looked up the multiplicative ergodic theorem but I can only find its description in terms of real matrices instead of complex ones that we need. Is the extension to complex domain straightforward?703484936363For any $n\in\mathbb{N}$ let $S_n$ denote the set of all permutations (bijective maps) $\pi:\{1,\ldots, n\} \to \{1,\ldots,n\}$. For $\pi \in S_n$ we set $$\text{fix}(\pi) = \{x\in \{1,\ldots, n\}: \pi(x) = x\}.$$

For any $n\in \mathbb{N}$ the expected value of the number of fixed points of a randomly chosen element of $S_n$ is $$E_n := \frac{1}{n!}\sum_{\pi\in\S_n} |\text{fix}(\pi)|.$$

Is the real sequence $(E_n)_{n\in\mathbb{N}}$ bounded? If not, what are the values of $\lim_{n\to\infty}\frac{E_n}{n}$ and $\lim_{n\to\infty}\frac{E_n}{\log n}$?

https://lh6.googleusercontent.com/-Ol675BzDBHc/AAAAAAAAAAI/AAAAAAAAAEY/_RMlh9Ol_iA/photo.jpg?sz=1281671205 Ignore my above comment, I hadn't drunk enough coffee yet (I was thinking that bi-interpretability was a relationship between models, whereas in fact it's a relationship between theories). Am I right in thinking that ZF has an obvious interpretation in ZFC (just take the whole model)? So what we're looking for is an interpretation of ZFC in ZF with some nice properties. In this case (where one interpretation is trivial) the definition of bi-interpretation says that every model of ZF should be isomorphic to the model of ZFC it induces. Clearly this is stronger than what we want.7804521338186Vhttp://www.youtube.com/watch?v=d-sALU_hveA4160041426086For an alternate proof: it is easy to construct a diagonal matrix $D$ such that $DAD^{-1}$ is symmetric.1742552rSobolev Norm of distance function on Riemannian manifoldaxioman1367067225937915695558716221851238|https://graph.facebook.com/156481501915827/picture?type=large0The continued fraction length is usually a small constant factor away from the regulator. A more precise version can also be achieved, but I don't remember a reference, so if anyone does...

Then, we know the regulator times the class number is usually a small constant factor away from the discriminant (of the order, not necessarily the field).

In addition, if a discriminant has $n$ prime factors in its squarefree part, the class number will be divisible by $2^{n-1}$.

Finally, most positive numbers of some size don't have many prime factors, and we suspect the real quadratic fields composed from these to have relatively small class number (look up Cohen-Lenstra).

Combining these facts and heuristics we get that primorials, and even factorials (still have large squarefree part), will have larger class number, hence small regulator, and therefore smaller continued fraction period length.

That said, we can dig even deeper. For factorials, there is a hefty squareful part. When we go from the maximal order of the field $\mathbb{Q}(\sqrt{n!})$ to the order $\mathbb{Z}(\sqrt{n!})$, the discriminant is enlarged hugely. Each prime (even) power $p^{2m}\ ||\ n!$ adds $p^{m-1}(p-(squarefree(n!)/p))$ to the original $h_K\times R_K$. So for each such factor, something goes into the class number of the smaller order, and something goes into the regulator of the smaller order.

Here comes the interesting bit. The factors that make up the new regulator tell us how far the unit group in the small order is from the unit group of the maximal order. Since we are in a real quadratic field the unit groups are of rank 1, so this distance is just the exponent to which the fundamental unit is powered by in order to enter the smaller order. Say $p_1-1$ and $p_2-1$ (say Legendre symbol is 1) have a large gcd. Once we power the fundamental unit to, say, $p_1-1$ to get (locally) into the $p$-part of the smaller order, we can slack off when getting into the $p_2$-part of the smaller order, because we've already done some of the work.

So in the factorial case, or in any number with many square factors (not dividing the squarefree part), since many of the above mentioned factors will have a large gcd, most of the factors will have to go into the class number of the small order - hence the very small regulator and continued fraction expansion.

We can write the above explicitly, but we need another notion. When a prime divides the discriminant to an even power greater than 2, the power of $p$ from the above mentioned factor that goes into the regulator measures how the fundamental unit is far p-adically from being 1. For most $p$, the fundamental unit will not be $1\ (mod\ p^2)$, so pretty much all of the $p^{m-1}$ goes into the regulator.

Hence, we expect:

$$ \frac{period-length(n!)}{\sqrt{n!}} \sim \frac{lcm(\{\ p-(p/squarefree(n!))\ |\ p^{2m}\ ||\ n!\ \}}{\prod_{p^{2m}||n!} (p-(p/squarefree(n!))} $$

And how that factor behaves - I have no idea. Sounds like a combinatorical answer could exist.

46297998080Probably related: http://math.stackexchange.com/questions/1930461/find-an-integrable-dominated-function-for-a-convolution.The dihedral group of order $2n+2$ acts on $K_n$, the $n-2$-dimensional associahedron. Are there any other symmetries? References?

Does the answer to 1 change if we restrict to just the 1-skeleton of $K_n$? References?

It is "obvious" that any simple circuit (simple closed walk, simple closed path, whatever terminology you prefer) of length 4 or 5 is a 2-dimensional face of $K_n$. Is this true? Proof? Reference?

Sorry, I do not understand what you say in 5. What if A = k[x, y] and X_0 = Spec(A) and U_0 = D(x) and the effective Cartier divisor is given by the zero locus of y + t/x ? Then (y + t/x)(y - t/x) = y^2 and the scheme theoretic closure of Z_1 is given by an ideal containing both y^2 and xy + t which cannot be principal.

There are results in the literature concerning this question when one takes out something of sufficiently high codimension and X_0 is sufficiently nice. Look in Grothendieck's Cohomologie locale des faisceaux coh´erents et th´eoremes de Lefschetz locaux et globaux (SGA 2) for example.

Is it possible to check if a sequence of numbers (1 to 10) have been generated randomly or by a human?:While looking through old questions, I came accross this one, and decided to throw my hat into the "ring."

**If**the identity element has a countable neighborhood base,**then**trivially by the continuity of the operation of addition, every point has a countable neighborhood base (as continuous maps take a filter base at some point x, and map this onto a filter base of the image point this is a particular example of homogeneity exhibited by topologically equipped algebraic objects.).Provided, the topology defined on the ring is $T_0$, we have that the topology must also be $T_2$ and completely regular.

Moreover, since there is a countable dense subset $D$ by assumption, we have that given the above two assumptions, there exists a countable collection, of countable covers $U_n$, with the following property: Given any closed set $C$ from this space, and point $p\not\in C$ there exists an $n\in\omega$ such that for some $O\in U_n$, $p\in O$ and $O \cap C = \{\}$. (We get such a collection by ordering the neighborhood base of each point in $D$ by reverse inclusion, and taking fixed 'sections of fixed height' from each to form the open covers)

Putting (1), (2), and (3) together produces a topological space which is about as close to a normal Moore space as you can get without actually being one, that is to say, under these assumptions: $R$ is a separable completely regular developable Hausdorff space, (and we haven't even invoked the ACC yet)

Ignoring for the moment the previous assumptions, intuitively, the ACC should somehow produce a covering property for this particular space. However, there is an interesting problem when it comes to the definition of subgroup/subring (which is required to get to the notion of ideal needed to apply the ACC): will they be open, closed, neither? Because of this we cannot really apply the ACC, to produce a nice covering property that might have tied everything together (like Lindelöf.)

**Edit**: While poking around Wiki, I came across something I felt I needed to add

However, if you mean that the space is a Noetherian topological space, then we get some gnarlly consequences ( http://en.wikipedia.org/wiki/Noetherian_topological_space ), like the fact that the space is compact! Which is exactly the thread we would want to tie everything together, and produces a normal moore space.

The question is ill-posed, in that we do not have enough information to properly deduce a valid and fully general answer. My answer to this question has tried to highlight this point by giving you a case where, you can be about as close as you might ever want to be to something genuinely interesting, and then failing to make it interesting because of the incompatibility of the algebraic assumption with the topological ones. Even if we consider the other possibility we enter one of those strange and beautiful areas in topology where things being to become independent of $ZFC$.

Because of this freedom or lack of information, we are left with an answer of **Most Likely No**. (Weak answer I know) But I can make this claim, because we honestly do not completely understand the notion of hereditary separability (in fact it was only in 2006 that J T Moore was able to produce a ZFC example of an L-space Article)

The best you can hope for is $O(n \log n)$ plus a term dependent upon accuracy,
for finding an *approximate* Euclidean MST. The fastest known exact algorithm
is just a hair better than quadratic, $O(n^2)$. Below is a central paper on the topic. Note
especially: "The algorithm is deterministic and very simple."

Arya, Sunil, and David M. Mount. "A fast and simple algorithm for computing approximate Euclidean minimum spanning trees."

Proceedings 27th Annual ACM-SIAM Symposium Discrete Algorithms. SIAM, 2016. (PDF download.)

909257Is the compass with memory allowed? Or is there a clever trick to duplicate lengths on the sphere?312999842455

The answer is yes, it was pointed out to me by Vitaly Bergelson that it follows from Theorem 6.1 (d) of a paper by Bergelson, Hindman and Kra (Trans. Amer. Math. Soc. 348 (1996), no. 3, 893–912). The notation $g_{\alpha,\gamma}[A]$ is defined in the previous section and means $$g_{\alpha,\gamma}[A]:=\{\lfloor \alpha n+\gamma\rfloor:n\in A\}$$ where $\lfloor\cdot\rfloor$ represents the floor function. Note that $\lfloor x+1/2\rfloor=\lfloor x\rceil$

2955877133773718555001509360Sorry, can't answer, but the letter seems very interesting, not least because of the mixture of German and Latin. Not being a native English speaker, I'm aware of a similar phenomenon today: I often have conversations, nominally in Spanish, which are liberally sprinkled with many English words, due to their mathematical content. Although perhaps hardly surprising, I hadn't realised that this was the case also earlier with Latin.21662213108481668182The probability measure you're asking about is nonatomic, but your events of interest are countable. Therefore the unconditional probability of each is zero, and the conditional probabilities are undefined.

This is morally correct but technically wrong, as pointed out in Jose's answer and the comments to the question (including yours). But, yes, I do agree at a moral level that the bracket and the differential are dual. One way to make this precise, as you probably know but I'll mention it for other readers, is the Koszul duality between Lie algebra/oids and differential graded commutative algebras. When applied to tangent bundle, this Koszul duality returns the de Rham complex.204611919323711532313Assume that there is an smooth structure of the matrix algebra $M_{n}(\mathbb{R})$ on fibers of the tangent bundle of a $n^2$ dimensional manifold.

5751711689202b@T. Amdeberhan R symbolizes Ramanujan summation.2294421108341190700948027317200411377365I agree, I think it's two points inside the circle, not on the circumference. I also think 'piece of circumference' means [arc](https://en.wikipedia.org/wiki/Arc_(geometry)). The OP really needs to clear this confusion up!I have in mind the book of Lax and Philips who applied scattering theory to the study of automorphic forms. In that book I saw for the first time a truncated version of the hyperbolic Laplacian which has a compact resolvent. Then from a key observation of Colin de Verdiere, one may apply the Fredholm analyticity theorem to obtain the analytic continuation and the functional equation. In any case, thanks again for your answer.214803815675752004635Is there a Riemannian metric on $M$ such that all operator of parallel transports would be an algebra isomorphism?

Assume $F, G : \mathbf C \to \mathbf D$ be functors. Denote by $\widehat{\mathbf C} = \mathrm{Fun}(\mathbf{C}^{\mathrm{op}}, \mathbf{Set})$ the category of presheaves of sets on $\mathbf C$. Then, $F$ and $G$ induce "restriction" functors, obtained by composition with $F$ and $G$: \begin{align*} \mathrm{Res}_F, \mathrm{Res}_G : \widehat{\mathbf D} \to \widehat{\mathbf C}. \end{align*} If $F \cong G$ then clearly $\mathrm{Res}_F \cong \mathrm{Res}_G$. I believe that the converse is false; how can I find a counterexample? Or perhaps it is true under certain assumptions?

13503541665706140051114941931450406@Gerhard, than you. Now, it has finally occurred to me that there should be plenty of examples because $\ H(a\ b)\ $ is a 2-parameter set while $\ l\ $ is only a 1-parmeter set, i.e. it is expected that $\ H(a\ b) \mapsto l\ $ is not injective.Prabin Niroula2261968you kind of have to right? otherwise the original question is false too: e.g. if all the points are on a line11018481537517Let $H$ be a hypergraph of maximum vertex-degree $\Delta$. (That is, for all vertices $x$, we have $| \{ e \in H \mid x \in e \} | \leq \Delta$) Are there any bounds on the *chromatic index* $\chi_e(H)$ of the form
$$
\chi_e(H) \leq c \Delta
$$
for some constant $c$?

If not, are there any simple criteria on $H$ that can guarantee this?

Note that if $H$ is a multi-graph, then this follows from Shannon's theorem.

1148371Let $X$ be the real line with the usual topology. Then clearly $|C(X)| = c = |X|$ and on the other hand $|X| = 2^{\aleph_0}$.
Now my question is as in the title: **Is there a Tychonoff space $X$ of cardinality not of the form $2^\alpha$ such that $|C(X)| = |X|$** (where $C(X)$ is the set of all real valued continuous functions on $X$). Any reference or help would be appreciated.

I'm looking for a proof of the analytic implicit function theorem (IFT). The only related proof I could find was the holomorphic inverse function theorem (by Henri Cartan). On Wikipedia, the analytic IFT is mentioned casually in the general article "Implicit function theorem", saying that "Similarly, if f is analytic inside U×V, then the same holds true for the explicit function g inside U. This generalization is called the analytic implicit function theorem." Mmmh, that's fast...

A sketch of the proof may be the following :

- use analytic continuation to transform f into a holomorphic function
- use the holomorphic inverse function theorem (Cartan) to prove a holomorphic IFT
- restriction : g is holomorphic on $\mathbb{C}$, therefore analytic on $\mathbb{R}$.

But it seems weird and I don't think it would work (I have no idea whether a so-called holomorphic IFT exists or not). What would be an efficient proof of the theorem ? Thanks a lot by advance.

Rather than working with the image of $SL_2$, one can look at its image in the space of characteristic polynomials. This is necessarily $1$-dimensional and conjugacy-invariant, so taking the union of conjugates does not affect this. Because it is one-dimensional in a $g$-dimensional space, it is not dense. This works even if the embedding is only defined over $\mathbb C$, because any $2g$-dimensional algebraic representation of $SL_2$ is conjugate over $\mathbb C$ to one defined over $\mathbb Q$.1523498hNon-tame 3-manifolds covered by the Euclidean spaceTMaximum Number of Skew-Symmetric matrices1288596xHi I assume you mean the degree is equal to $d$, not $n$? 746132hhttp://www2.math.uu.se/~astrombe/emaass/emaass.htmlI would be happy if I could see some more details of the argument (which is rather sketchy)5075615802671878800Since roots give factorisation of polynomials even over the ring $\mathbf Z/n$, it is equivalent to search a characterisation of subsets $S \subset \mathbf Z/n$ s. t. for every $x$ not in $S$, $\prod_{s \in S} (x-s) \neq 0$.Yes, one reason you need to move to $E_\infty$ in characteristic $\neq 0$ is that you have no hope of having a model category on strictly commutative monoids unless the characteristic is 0. This means you need to use $E_\infty$ if you want homotopy theoretic meaning. See for instance: http://mathoverflow.net/questions/23269/non-examples-of-model-structures-that-fail-for-subtle-surprising-reasons/23885#23885108941312784822261193https://www.gravatar.com/avatar/19c87cb16e04f8fa5fc4f026fce486af?s=128&d=identicon&r=PG&f=16I recently stumbled upon the following identity, valid for any real numbers $\alpha_1,\dots,\alpha_n$ and $\lambda_{n1} \leq \dots \leq \lambda_{nn}$:

$$ \mathrm{det}( e^{\alpha_i \lambda_{nj}} )_{1 \leq i,j \leq n} = V(\alpha) \int_{GT_\lambda} \exp( \sum_{i=1}^n \sum_{j=1}^i \lambda_{ij} (\alpha_{n+1-i}-\alpha_{n-i}))$$ where $V(\alpha)$ is the Vandermonde determinant $$ V(\alpha) := \prod_{1 \leq i < j \leq n} (\alpha_j - \alpha_i),$$ $GT_\lambda$ is the Gelfand-Tsetlin polytope of tuples $(\lambda_{ij})_{1 \leq j \leq i < n}$ obeying the interlacing relations $\lambda_{i+1,j} \leq \lambda_{i,j} \leq \lambda_{i,j+1}$ and with the usual Lebesgue measure, and one has the convention $\alpha_0 := 0$. Thus for instance when $n=1$ one has $$ e^{\alpha_1 \lambda_{11}} = \exp( \lambda_{11} \alpha_1 )$$ when $n=2$ one has $$ \mathrm{det} \begin{pmatrix} e^{\alpha_1 \lambda_{21}} & e^{\alpha_1 \lambda_{22}} \\ e^{\alpha_2 \lambda_{21}} & e^{\alpha_2 \lambda_{22}} \end{pmatrix} $$ $$= (\alpha_2 - \alpha_1) \int_{\lambda_{21} \leq \lambda_{11} \leq \lambda_{22}} \exp( \lambda_{11} (\alpha_2-\alpha_1) + \lambda_{21} \alpha_1 + \lambda_{22} \alpha_1 )\ d\lambda_{11}$$ and when $n=3$ one has $$ \mathrm{det} \begin{pmatrix} e^{\alpha_1 \lambda_{31}} & e^{\alpha_1 \lambda_{32}} & e^{\alpha_1 \lambda_{33}} \\ e^{\alpha_2 \lambda_{31}} & e^{\alpha_2 \lambda_{32}} & e^{\alpha_2 \lambda_{33}} \\ e^{\alpha_3 \lambda_{31}} & e^{\alpha_3 \lambda_{32}} & e^{\alpha_3 \lambda_{33}} \end{pmatrix} $$ $$ = (\alpha_2 - \alpha_1) (\alpha_3 - \alpha_1) (\alpha_3 - \alpha_2) \int_{\lambda_{31} \leq \lambda_{21} \leq \lambda_{32}} \int_{\lambda_{32} \leq \lambda_{22} \leq \lambda_{33}} \int_{\lambda_{21} \leq \lambda_{11} \leq \lambda_{22}}$$ $$ \exp( \lambda_{11} (\alpha_3-\alpha_2) + \lambda_{21} (\alpha_2-\alpha_1) + \lambda_{22} (\alpha_2-\alpha_1) + \lambda_{31} \alpha_1 + \lambda_{32} \alpha_1 + \lambda_{33} \alpha_1)$$ $$ d \lambda_{11} d\lambda_{22} d\lambda_{21},$$ and so forth.

The identity can be proven easily by induction. I first discovered it by starting with the Schur polynomial identity $$ \mathrm{det}( x_j^{a_i} )_{1 \leq i,j \leq n} = V(x) \sum_T x^{|T|}$$ where $0 \leq a_1 < \dots < a_n$ are natural numbers in increasing order, $T$ ranges over column-strict Young tableaux of shape $a_n-n+1, \dots, a_2-1, a_1$ with entries in $1,\dots,n$, and $x^{|T|} := x_1^{c_1} \dots x_n^{c_n}$ where $c_i$ is the number of occurrences of $i$ in $T$, and taking a suitable "continuum limit" as the $a_i$ go to infinity and the $x_j$ go to one in a particular fashion. It can also be derived from the Duistermaat-Heckmann formula for the Fourier transform of Schur-Horn measure, combined with the fact that this measure is the pushforward of Lebesgue measure on the Gelfand-Tsetlin polytope under a certain linear map.

Note that the identity also provides an immediate proof that any $n$ distinct exponential functions on $n$ distinct real numbers are linearly independent.

I am certain that this formula already appears in the literature, and perhaps even has a standard name, but I was unable to locate it with standard searches. So my question here is if anyone recognizes the formula and can supply a reference for it.

356673923980\https://i.stack.imgur.com/oRQTd.jpg?s=128&g=16558691232442It sounds unlikely, unless maybe perhaps each $X_i$ has a very large kernel...I would like to start with considering the time-dependent heat equation on a connected graph and consider its Laplacian matrix. Suppose we have a connected graph with unknown temperature on vertices. Let's for simplicity assume that we have few heat sources and we want to transfer this heat through some edges of the graph. Deciding which edges will contribute in transferring the heat would be of interest. We are given a laplacian matrix and we want to find those edges that will appear in the transferring (I think binary variables will be best choice).

Another interesting idea would be as following. Suppose we have some sensors that we want to install these in some vertices to measure the heat. The question is to find the best place for installing these sensors on the vertices.

My idea: To start, I will need to model it respect to time discretization. something like this: $$A T^{k+1} = B T^k + Cf^k$$ where the initial time is given and the time step is assumed to be constant. First, I have to determine the matrix $A$, $B$, and $C$.

Would you please let me know if you know some references or have some comments on this topic?

18916461500766369630zFibered 3-manifolds over circle with harmonic projection map13642711784619906072In general, it is *false* that deforming equations produce birational varieties. Typical example: a nodal cubic curve is rational, but you can deform it to a smooth cubic curve, that has genus $1$. @DenisNardin Sure. You iteratively fill horns of the form HTT.3.1.1.1.(2), i.e. $(\Lambda^n_n,\mathcal{E} \cap (\Lambda^n_n)_1) \subseteq (\Delta^n,\mathcal{E})$, first for n=2, then for n=3, and so on. The vertices $n,n−1,n−2,n−3,\dots$ of $\Lambda^n_n$ are sent to the vertices $1,0,1,0,\dots$ of the simplicial set that you've constructed so far, starting with $(\Delta^1)^{\sharp}$. Finally you need to compose with a pushout of a map of the form HTT.3.1.1.1.(4) because you want the edge going from the vertex 1 to the vertex 0 to be marked, too. I think I have …Ah- apologies, I guess I spoke too quickly. So then I guess something like first $n^2-n$ will be nonzero and the rest will be zero?2142563The notion of nilpotency passes nicely from groups to semigroups. Define $q_1(x,y)=xy$ and $$q_{i+1}(x,y,z_1,\cdots,z_i)=q_i(x,y,z_1,\cdots,z_{i-1})z_iq_i(y,x,z_1,\cdots,z_{i-1})$$ inductively for all $x,y,z_1,z_2,\cdots$.

A. I. Malcev proved that nilpotent groups of class $c$ are described as the groups satisfying the law $q_c(x,y,z_1,\ldots,z_{c-1})=q_c(y,x,z_1,\ldots,z_{c-1})$. It is natural to define nilpotent semigroups of class $c$ as those satisfying this law.

Question: Do all f.g. nilpotent semigroups have polynomial growth?

Note that for cancellative semigroups the answer is "yes" since nilpotent cancellative semigroups satisfy the Ore's condition and so are group-embeddable.

(My guess is that in general the answer is "no" and there even perhaps exists a counter-example among matrix semigroups.)

21475802000153113037440845856231361944165992~Criterion for a equalizer to be a homotopy equalizer in spaces478763As for the algebraic geometry side, this would presumably mean that there is some sort of common generalization of Cartier and Weil divisors which gives "correct thing" for the singular case, is there such a thing?Let $G=(V,E)$ be a finite, simple, unconnected graph. We define the *total graph* $T(G)$ of $G$ as follows:

- $V(T(G)) = (V\times\{0\}) \cup (E\times\{1\})$,
- $E(T(G)) = E_v \cup E_e \cup E_{v+e}$, where
- $E_v = \big\{\{(v,0), (w,0)\}: \{v,w\}\in E\big\}$,
- $E_e = \big\{\{(e,1), (f,1)\}: (e,f\in E) \land (e\neq f)\land (e\cap f \neq \emptyset\big)\}$, and
- $E_{v+e} = \big\{\{(v,0), (e,1)\}: v\in e\big\}$.

I think that for the clique number we have $\omega(T(G)) = \Delta(G)+1$ (where $\Delta(G)$ is the maximum degree in $G$), except if $\Delta(G) = 1$, but I have to check this (it's probably an easy question).

**Question**: Is there a $G$ such that $\chi(T(G)) > \omega(T(G))$?

The book "A Panoramic View of Riemannian Geometry" by Marcel Berger includes a number of open problems.

~https://graph.facebook.com/1470438746425065/picture?type=large17935762092706h@AlexM. I disagree. The article is very accessible.2004864Le and Murakami (HERE and HERE) discovered several previously unknown relations between multiple zeta values through the study of quantum invariants of knots. Further relations were later discovered through knot theory by Takamuki, and by Ihara and Takamuki.

These relations stem from the fact that the Kontsevich invariant extends to an invariant not of tangles, but of bracketed tangles, also known as q-tangles or non-associative tangles (alternatively, to knotted trivalent graphs, e.g. HERE). An "associator" relates different bracketings, and the relations satisfied by the associator induce the relations between the multiple zeta values, upon choosing a weight system.

Now that these relations have been discovered, the scaffolding can be removed and they can be derived directly from the defining relations of the associator as shown by Furusho (see there for further references). The associator is an algebraic object rather than a topological one, so now the number theorist "no longer need the knot theorists" as far as the known relations between multiple zeta values are concerned.

391853Related: [this question](http://math.stackexchange.com/questions/110458/) on M.SE.Thanks! The way to get the constant is still a bit mysterious to me, so if you have time, an update would be much appreciated.2298833167402@Dirk not linear? The functional $L(\phi):=\int_I\phi'd\mu$ is linear right? Or am I make some trivial mistake here... I think the TGV definition is linear. What I expect is a kind of relation as I stated in equation $(1)$ above.1947004191415113423212645461608239Thank you for the correction: A semi-direct product needs of course divisibility by 3 of the number of invertible elements modulo $n$. I think you suggest that the answer is messy!@Christi Stoica: if you disagree with a closure the best thing to do is to start a thread on meta. (Link at the top; extra sign up necessary, but easy and instant. Sign-up, top right, then 'apply for membership' which is granted instantly.)Have a look at Whitney's "Geometric integration theory". It contains a proof of his embedding theorem which shows that the image of the embedding is closed.277935My understanding is that this started with the Dirac wave equation. This was a relativistic equation for an electron. However it happened to also introduce the idea that a point particle could have an internal state space. This was a successful theory and was taken up and imitated when it came to probing the structure of the nucleus.

For a description of the standard model a good place to start is: http://arxiv.org/abs/0904.1556

3589682128432https://www.gravatar.com/avatar/e9b313ea4d4e2c31dc90f2d8ee7897c0?s=128&d=identicon&r=PG&f=1128342414662491592654Suppose $K \subset \mathbb{C}$ is a Cantor set and let $u:\mathbb{C} \setminus K \to \mathbb{R}$ be the maximal smooth function such that the conformal metric $e^{2u}(\mathrm{d}x^2 + \mathrm{d}y^2)$ has constant curvature $-1$ on $\mathbb{C} \setminus K$.

Suppose we put the Hausdorff distance on the set of compact subsets of $\mathbb{C}$. I'm interested in how $u$ varies with respect to the Cantor set $K$.

**Question:** Does $u$ vary continuously in the smooth topology on compact subsets of $\mathbb{C} \setminus K$ with respect to $K$?

**Some more background:** The existence of a unique maximal hyperbolic metric (sometimes called the Poincaré metric of $\mathbb{C} \setminus K$) is proved for example in Ahlfors' book. I've looked around and found several references estimating the metric in terms of distance to the Cantor set (for example this one). I haven't found any result on how the metric varies with respect to the domain save the easy observation of domain monotonicity (i.e. $u$ grows if $K$ does).

Let $\gamma_+$, $\gamma_-:\mathbb{R}_+\to\mathbb{R}$ be two given functions. Assume that $\gamma_+$ ($\gamma_-$) is smooth, strictly increasing (decreasing) and $\gamma_{+}(+\infty)=+\infty$ ($\gamma_{-}(+\infty)=-\infty$).

Now consider the following system:

\begin{eqnarray} \lambda''(\gamma_+)\gamma_+'\gamma_+&=&\lambda''(\gamma_-)\gamma_-'\gamma_-,~ \forall l>0 \\ \lambda'(\gamma_+)-\lambda'(\gamma_-)&=&2g'(l)+2\lambda''(\gamma_-)\gamma_-'\gamma_-,~ \forall l>0, \end{eqnarray}

wherer $\gamma_+=\gamma_+(l)$, $\gamma_-=\gamma_-(l)$ and $g=g(l)$ is a given function regular enough. Now my question is how to obtain explicitly the solution $\lambda: \mathbb{R}\to\mathbb{R}$ and under which conditions $\lambda$ is convex. Thx a lot for the reply!

8316321603897Yang: I don't know how to do it, but I'm certain it should be true. In fact, in the book "Spin Geometry" by Lawson and Michelsohn, they assert that this is true, but their seem to assume $d = d_\infty$ in the setup I've given in a couple of comments below. I'm horrible at finding references and I was hoping someone would know it! 203554dCan you give the example or at least a reference?22055882194474@Dirk, but why do you point this out? I don't see how it contributes to answering my question...8663492304357704256Thank you very much for the clarification. But I am still confused about the third part. The Gromov clouser requires the cover to have a negatively curved metric right! So are we considering $\bar{Y}$ as a quotient of $\mathbb{H}^2?$ because then it is same as considering a hyperbolic metric in $S$.Let me add that the inverse function theorem of Nash developed for the purpose is quite amazing and is probably the best part of that results9895932175597809861024888In General Topology zdH-spaces are usually called totally disconnected spaces, see e.g., https://en.wikipedia.org/wiki/Totally_disconnected_space633651821203VHere, $t$ is the stable letter, of course.93320510370291619267138473210048597120191372460bAlternating series test for non-decreasing terms1056132$C(X)$ is the space of all continuous function from X to $\mathbb{C}$. This is a $C^{*}$ algebra.\https://i.stack.imgur.com/1cnAq.jpg?s=128&g=1pI didn't know that the category of bananas had kernels.216028410533532049772So far we have been able to show that the statement is true if matrix $A$ is normal or doubly stochastic.211446113904097982212003384Consider $f(x):=\sin(\frac1x)$ for $x\ne0$, $f(0):=-1$. Then clearly $f$ is lower semi-continuous but has neither a left nor a right limit at $0$.2221184vAssuming you are a student, what does your supervisor say?1980089What's "the real definition of a tensor"? Element of a tensor product? A section of a tensor bundle?10691371047336520327My previous comment is wrong, it is a one-sided double sum so is not directly related to Epstein zeta (I do not know how to delete a comment). I agree with Zhou's answer.13958861750009831871The answer depends on how smooth your domain is. In dimension 2, if very smooth, it is finite, but if it can have, say 90 degree corner, then the integral can be infinite.1098338133642+1 I have the App on my Galaxy Tab and enjoyed it. I guess that it is computationally hard (NP-complete).The only so-called "proof" that I can see is by just plugging in this big result, http://www.sciencedirect.com/science/article/pii/S0095895605000948. I don't seem to see any other first-principles way of seeing it!23229301898533935?Many of us presume that mathematics studies objects. In agreement with this, set theorists often say that they study the well founded hereditarily extensional objects generated *ex nihilo* by the "process" of repeatedly forming the powerset of what has already been generated and, when appropriate, forming the union of what preceded.

But the practice of set theorists belies this, since they tend—for instance, in the theories of inner models and large cardinal embeddings—to study classes that, on pain of contradicting the standard axioms, are never "generated" at any stage of this process. In particular, faced with the independence results, many set theorists suggest that each statement about sets—regardless of whether it be independent of the standard axioms, or indeed of whether it be formalizable in the first order language of set theory—is either true or false about the class $V$ of all objects formed by the above mentioned process. For them, set theory is the attempt to uncover the truth about $V$.

This tendency is at odds with what I said set theorists study, because proper classes, though well founded and hereditarily extensional, are not objects. I do not mean just that proper classes are not sets.

Rather, I suggest that no tenable distinction has been, nor can be, made between well founded hereditarily extensional objects that

aresets, and those that aren't.

Of course, this philosophical claim cannot be proved.

Instead, I offer a persuasion that I hope will provoke you to enlighten me with your thoughts.

Suppose the distinction were made. Then in particular, $V$ is an object but not a set. Prima facie, it makes sense to speak of the powerclass of $V$—that is, the collection of all hereditarily well founded objects that can be formed as "combinations" of objects in $V$. This specification should raise no more suspicions than the standard description of the powerset operation; the burden is on him who wishes to say otherwise.

With the powerclass of $V$ in hand, we may consider the collection of all hereditarily well founded objects included in *it,* and so on, imitating the process that formed $V$ itself. Let $W$ be the "hyperclass" of all well founded hereditarily extensional objects formed by this new process. Since we can distinguish between well founded hereditarily extensional objects that are sets and those that aren't, we should be able to mirror the distinction here, putting on the one hand the proper hyperclasses and on the other the sets and classes.

Continuing in this fashion, distinguish between sets, classes, hyperclasses, $n$-hyperclasses, $\alpha$-hyperclasses, $\Omega$-hyperclasses, and so on for as long as you can draw indices from the ordinals, hyperordinals, and other transitive hereditarily extensional objects well ordered by membership, hypermembership, or whatever. It seems that this process will continue without end: we will never reach a stage where it does not make sense to form the collection of all well founded hereditarily extensional objects whose extensions have already been generated. We will never obtain an object consisting in everything that can be formed in this fashion.

For me, this undermines the supposed distinction between well founded hereditarily extensional objects that are sets, and those that aren't. Having assumed the distinction made, we were led to the conclusion of the preceding paragraph. But that is no better than the conclusion that proper classes, including $V$ itself, are not objects. Indeed, it is worse, for in arriving at it we relegated set theory to the study of just the first two strata of a much richer universe. Would it not have been better to admit at the outset that proper classes are not objects? If we did that, would set theory suffer? In particular, how would it affect the idea that each statement about sets is either true or false?

1734105Dear Prof. Robinson, Thank you very much for your exact answer.919903So it isn't possible to define the relation $$e^x-1 \mid e^{kx}-1$$ in this language, right? To do that do we have to add the symbol $e^x$ in the language?I think there's still a confusion: "a la Gentzen" is just a reference to the fact that, for all natural theories (including $PA$), a very small finitary theory together with "enough" induction proves that theory's consistency. Now, what "enough" means here varies from theory to theory, and is known as the "proof-theoretic ordinal" of a theory. For example, the proof-theoretic ordinal of $PA$ is $\epsilon_0$, which means that "$\epsilon_0$ is well-founded" (together with a very weak - much weaker than $RCA_0$ - base theory) proves $Con(PA)$. (cont'd)The properties that different unitalizations may have is also contemplated in $K$-theory. A rng is $H$-unital iff it's $K$-theory doesn't depend on the unitalization chosen to construct it.MSH20882361770894@Nate: Not at all! I did a lot of simulations. With a Best-2-of-3 voting rule (which is something quite different!) the "lower" side of the distribution looks like an exponential distribution (which is quite useless, as those votes are on death row in the next steps anyway), so I won't exclude it for the $n=\infty$ case either.@anyone: I don't know about potential theory; Wikipedia says that a countable subset of $\mathbb{R}^n$ is polar. Is it really the case?I think not so much is known.

Conjecturally, elliptic curves (defined over $\mathbb{Q}$) of rank $0$ have density $1/2$, and since at any rate only a finite number of cases for the torsion part of $E(\mathbb{Q})$ are possible (by Mazur's celebrated result), I guess that elliptic curves with $E(\mathbb{Q}) \cong \mathbb{Z}/2 \mathbb{Z}$ should have density $1/2$ too.

For this reason, it seems to me that a complete classification is out of reach.

If you want to see an infinite family of cubic curves with this property, take

$y^2=x^3 + px$

where $p$ is a prime number such that $p \equiv 7$ (mod $16$), see [Silverman-Tate, Rational points on elliptic curves, p. 105].

9916363200102211499619280131674801lJust a curious geek interested in learning more math.2276965rlook up ambiguous class number formula and genus theory.Burn also has a book on introductory real analysis, which is quite good.5710646335021801553@Andej: I don’t think your P.S. is quite fair — “passing as research-level” is determined only by not getting close-votes, and we are the sort of people who should have been casting those close-votes, if anyone. I would have voted for migration to math.se, but the overlapping close-vote reasons mean that questions usually end up getting just closed not migrated, so I prefer to answer it here instead. Why did you choose not to close-vote?832882052026|https://graph.facebook.com/477290319337355/picture?type=large902564Have you thought about considering the following? Take the group structure of your totally ordered group, and as lattice structure take some non total order (compatible with the group operation). This has to work.19841512000170@Daniel: because it would distract from the point, I suppose. There's no curvature involved. 2059461Are there analogous statements for the number of zeros of a section in terms of the Euler class, even when the relevant spaces are not manifolds?The OP asks for steepest descent for a "stretched exponential" : if $M(z)=e^{z^\beta}$, with $\beta>0$, then $M(\lambda g(z))=e^{\lambda^\beta g^\beta}$ and there is no problem with steepest descent in the $\lambda\rightarrow\infty$ limit. This would apply, for example, to the Mittag-Leffler function $E_{2,1}$ (with $\beta=1/2$). If $M(z)$ is a rational function, for example $E_{0,1}$, the large-$\lambda$ asymptotics is not governed by steepest descent.

1974946https://www.gravatar.com/avatar/c06bd5ac140ad017a995d6aa83cd8aeb?s=128&d=identicon&r=PG&f=12955117183086277261928903948401drx16896354119911901930447737I think this is close to an answer but I do not understand why A+A containing an interval implies that A.fMost Asterisque volumes are not, in my experience.15026852057626Thank you, Matias; I did not know this reference and therefore had to obtain the same local description by hands. 2121077655700\Matrix inequality with arbitrary large ratios1601733@ThomasKlimpel What I didn't understand in your earlier comment was the claim that being conservative over $WKL_0$ is "stronger and more useful then merely being conservative over $RCA_0$." Being conservative over a stronger theory (like $WKL_0$) is a weaker property, not a stronger one.30237122119111701844@Richard Kent: I do not assume the surjectivity of f in the question. As you said, if we take a parametrization of a figure 8 lying in the plane by the circle, i.e., M = S^1 and N = R^2, then we can see it is false in general.5318961264670671058This is a standard ECC (http://en.wikipedia.org/wiki/Error-correcting_code) problem.6617163@Mark: The Zariski topology was invented substantially earlier by Zariski, not Grothendieck, for algebraic varieties and algebraic sets. It's more likely that Grothendieck noticed that since irreducible closed subsets are in canonical bijection with prime ideals, we can keep track of them as actual points. The fact that we can do this follows from the observation that while maximal ideals are not preserved under preimage, prime ideals are. This gives us an "enrichment" of the classical Zariski topology by adding generic points. 702905Percie Jackson897482^Philosophical Consistency Proof for Set Theory161229322758321730611202230761871413333591286984jWhat is the intuition inertia orbifold (or stack)? bOn the determinant of a class symmetric matrices10374631950285137626137249317652581715331244388https://www.gravatar.com/avatar/3ffdcd0968616723354678b888ebdaa9?s=128&d=identicon&r=PG&f=1182160115390206482621081717:Probably a finitist, Asaf...186292 Sorry, bootstrapping from the locus of smooth curves in $\overline{M}_{g,n}$ by using double and triple fiber powers hits a fatal snag over the issue you raised, so I have deleted my answer via that idea. Without loss of generality $g\ge 2$ (as $g \le 1$ is easy), so over an etale cover where we have an $n$-pointed stable structure we apply Knudsen's "contraction" of *enumerated* marked points one at a time, getting a stable genus-$g$ curve over $S'$ with the same Pic$^0$; alas, for non-smooth fibers it depends on the ordering (and position) of marked points, so descent is problematic. Hmm!1956755396608I’d prefer “Some Logicians Don’t Bother With Sophistication If There Is An Easy Way To Get Things Done”.xThe first few terms are $1,4,53,936,25325,933980,45504649$.204218014022381372931 Thank you, Tsuyoshi. I had forgotten the distinction. Your answer reminds me of precisely why the subgraph problem is harder.856542039391098841327366209664888097https://lh6.googleusercontent.com/-nJmWJ2qfow8/AAAAAAAAAAI/AAAAAAAAABU/3w3_AUikmLY/photo.jpg?sz=128215240811796406319451I have read in a number of places that the lower Bruhat interval $[e, w]$ is rank-symmetric if and only if the KL-polynomial $P_{e, w}(q) = 1$. All of the proofs I've come across use "rationally smooth Schubert varieties", which I don't really understand.

The KL polynomials can be defined purely in terms of the Iwahori-Hecke algebra of the Coxeter group, and satisfy a number of identities involving sums over Bruhat intervals. I would like to know then if there is a more direct way to prove that $[e, w]$ is rank symmetric iff $P_{e, w}(q) = 1$, using only the Hecke algebra (and Bruhat order).

The forcing also adds no functions from the ordinals to the ground model. So the extension won't be a counterexample to (1), unless the ground model is.This should now be the accepted answer. (@PierreSchroeder, it is possible and encouraged to unaccept an answer if a better one comes along). @NiclasBörlin, you might change the link to use the more stable DOI and add that the paper is open access.Dijon31191tUnfortunately, your bulleted question has no answer... :)786969bI meant algebraic closure* but can’t edit now.149046617360782034977Do you have a reference for the case where $X$ and $Y$ are smooth fourfolds and $S$ is still a surface contracting to a curve?34440592142 8501152122213108632289631278661|Formula for the Ordinal Number of k-Sets of Positive Integers346927 Indeed: in a nearby fibre, the vanishing cycles will be disjoint. Far from the critical points, the whole family over the disc is trivialized by radial parallel transport into the central fibre; near the critical points, one has a standard quadratic model. Show in this way that the smooth monodromy is a simultaneous Dehn twist along the vanishing spheres $V_i$. Then prove Picard-Lefschetz by observing that for any $n$-cycle $x$, $(S-1)(x)$ is alway homologous to a cycle supported in neighborhoods $T^*V_i$ of the vanishing cycles, hence to a linear combination of the $\delta_i$. @GerhardPaseman: no, sorry, it should be about a page. The point is that a student of mine has a bijection which solves a relatively old problem, but the algorithm is, to my eyes, terribly complicated. We checked it by computer (although the first interesting cases appear for standard Young tableaux with about 30 entries, so there are too many to check them all), and the proof looks correct. So the question is how to sell it. Thus I was hoping for an educational example, the focus being on the quality of the exposition.10065791867965When $X \times Y \cong X \times Z$ implies $Y \cong Z$ (in the category of finite topological spaces)tKernel of linear representation of Baumslag-Solitar group20230311630790353691Not using notes is very brave and respectful. But didn't your mind ever go blank? Or don't you fear it could? Anyway, I'll try to muster courage and emulate you by speaking without notes at least once in my life. 48172121600416777441566007RProjective line as a quotient by a torus2148382051983fhttps://www.math.cornell.edu/m/People/bynetid/zt66The structure sheaf is globally generated, but is not ample if $X$ is projective of positive dimension.Oh, I see now. Are you sure your $M,N$ are manifolds with boundary? (I think perhaps they are manifolds with corners, but I am not sure even that is true). Perhaps there is some singularity at the origin or (in $N$) in the "curve" you identify?16030463221109264711Cauchy-Schwarz proof of Sidorenko for 3-edge path (Blakley-Roy inequality)2181376425412\https://i.stack.imgur.com/YZOgi.jpg?s=128&g=1864454546597@AleksandarMilivojevic Can you suggest to me a reference where I can study these spectral sequences for fiber bundles?Perhaps the situation becomes clearer by looking at a similar problem. Replacing the circle by a hyperbola, we can look at rational points on $H: XY = R$. This conic has the obvious point $N = (1,R)$, and among the many rational points on $H$ there are a few integral points, each corresponding to a factorization of $R$. You can define a group law on $H$ with neutral element, but this does not help at all at finding "nontrivial" rational points (that is, integral points). Neither does it seem to help to replace the rationals by an algebraic number field. In fact the geometric picture does not seem to add to our understanding in this case (on this elementary level, at least).

2304223186775317298893099802212840Magnus Find185328359421815915031456846A student in the BeNeLux olympiad apparently proved that 56 is not a cube by observing that 56 = 4^3 - 2^3 and referring to Fermat's Last Theorem for the exponent 3. 1795776jThe unreasonable effectiveness of Pade approximation101772003277For $G$ finite, you can just take the set-theoretic quotient (and put an appropriate sheaf on this). See for example Section 3 of Conrad's notes on the Keel-Mori theorem: http://math.stanford.edu/~conrad/papers/coarsespace.pdfYes, I think the DJ definition is from the late 80s, and most of the existence and uniqueness results are from the late 90s and early 00s.h@ChristopherA.Wong Indeed. I searched with no luck.1634636@Dal There are two books on the mathematics of Paul Erdős: *Graham, Nešetřil and Butler - The Mathematics of Paul Erdős I/II* (http://www.springer.com/mathematics/book/978-1-4614-7257-5)713550 9328These could be called modified [Yamanouchi words](https://en.wikipedia.org/wiki/Lattice_word)Combinatorics: In my opinion, Discrete Mathematics is only a mediocre journal (I wouldn't consider this top journal). Yes, it contains good papers, but it contains a *lot* of papers... on average... it's average.

Some other ones worth a mention (on top of JCTA, JACO and EJoC mentioned earlier): Journal of Combinatorial Theory Series B, Journal of Combinatorial Designs, Annals of Combinatorics, Combinatorica.

The Electronic Journal of Combinatorics should probably go on the top list in combinatorics, but since it's a free, open access journal, it's usually assumed to be worse than it actually is.

Combinatorics, Probability, and Computing, the Journal of Graph Theory and the *Electronic* Journal of Combinatorics seem to be widely regarded as excellent journals, at the level of the ones mentioned above (except Discrete Mathematics).

Formerly, the "Journal of Combinatorics" referred to a printed version of the "Electronic Journal of Combinatorics" (which has led to some confusion, see e.g. http://symomega.wordpress.com/2010/03/30/the-arc-the-era-and-the-ejc/), although most people in combinatorics haven't even heard of it.

Joel Reyes Noche's comment points out that there is a new journal entitled "Journal of Combinatorics".

Yes Ramiro - chain is any linearly ordered subset. I was misled by a previous answer (already deleted) which assumed well ordering of the subspaces.for related beautiful Povray drawing, see [here](http://images.math.cnrs.fr/Une-chambre-hyperbolique.html).3456632216991655052196629327715https://www.gravatar.com/avatar/0f69272647ecc5035f13934bc3e55fe0?s=128&d=identicon&r=PG&f=12297730Note that you don't really need the explicit mapping $\phi$. What you want is a kernel $k(A, B)$ which has the property that $k(A, B) = \phi(A)^T \phi(B)$. Generally, you compute the kernel directly, instead of calculating the mapping and computing the dot product -- that's why it's called kernel `trick'. In your case, it seems you want the kernel to be $k(A, B) = tr(AB)$. So it seems you want to know how to compute the trace efficiently, rather than what the mapping is.181262418610062294044The Lewy example obeys the conditions of the Cauchy-Kovalewski theorem (it has analytic solutions from analytic data), so there are no formal obstructions to solvability. I mean that assume that we attach 2-cells to $K$, ecessarily with trivial attaching maps, to make $L$ and extend $\phi$ over these cells to get a map $\psi :L\longrightarrow X$ so that $\pi_1 (\psi )=\pi_2 (\psi )=0$ (as Wall has mentioned in his paper, p. 59). Now if $\pi_3 (\psi )$ is free, then how do I describe attaching maps?1433546Suppose that a solution $X$ exists. Taking the trace you get $tr(Q)=0$ and hence $Q=0$; similarly $tr(N^TXN)=0$ and hence $N^TXN=0$. Thus, you need to search for $X$ satisfying $XA-AX=0$ and $XN=0$. In any case, these equations are homogenous, so that if there is a solution, then there will be many solutions.155387166798121936101990333464367Of course, the analysis is more complex in comparison to the discrete state space case. Another approach is to avoid Markov analysis by using drift-based theorems and (super) martingale theorems that do not require Markov chains.So, to be clear, there's no requirement of uniqueness in your digit representations?1810586185832You prove that H^*(X;F) is the regular representation of Sigma_k, but this is false, at least for k=2, when this representation is trivial. 1624414tIs there an entire solution for the Van der pol equation?ashleym1771519736893758409http://sailormansoft.blogspot.com

Edited the post to try to clarify my intent. We do not need $|V_gf|$ to have a max at the origin, that was only a suggestion.LopesIf every Galois connection is a Lagois Connection and (you "expect") every Lagois Connection induces a topology, then are you suggesting that *every* Galois connection induces a topology?? This is certainly not true: every Moore closure operator on a set comes from a Galois connection and not every Moore closure operator is a Kuratowski closure operator. So could you clarify what you mean?$First of all, what I want to ask is slightly more elaborate than what stands in the title (hence the subquestion).

I am telling this since as it is, the title contains a meaningful question, but it sounds very much like a superficial blind shot, so I want to provide some more motivation for asking it.

However there is one problem: what I am going to describe I have seen somewhere stated in detail, but cannot remember where, and cannot remember the details. OK, here it goes.

A concise way to formulate the SW-duality (or maybe I should actually say generalized Alexander duality?) is that a map $\varepsilon:X\wedge Y\to S^n$ with certain properties produces isomorphisms$$h^k(X)\to h_{n-k}(Y)$$for every cohomology theory $h$.

What I have seen somewhere is a very interesting geometric description of this map. Suppose that $\varepsilon$ comes from some embedding $i:X\hookrightarrow S^n$, with $Y$ homotopy equivalent to $S^n\setminus X$. Then, given an $h$-cocycle on $X$ one may ask whether it is in the image of $h^*(i):h^*(S^n)\to h^*(X)$. One may try to extend the cocycle from $X$ to larger and larger neighborhoods of $X$ in $S^n$. And if it is not in the image of $h^*(i)$ then at some point one encounters certain obstructions - say, the spots where the cocycle has been extended start to overlap and the extensions do not match on the overlaps. So if one does this extension as far as one can, certain singular "obstruction set" is carved out in $S^n\setminus X$, and the upshot is that for some nice $h$ this "obstruction set" (where the extended cocycle fails to be a genuine cocycle as it has some sort of singularities along this set) can be interpreted as an $h$-cycle in $S^n\setminus X$ (which is $\simeq Y)$, with different attempts giving rise to homologous cycles. (I believe I saw it worked out in detail at least for de Rham cohomology and $K$-theory, maybe also for cobordisms.)

Now the subquestion: where could I see the above description?

So it is this description that inspired some vague analogies with things like branes, and hence the main question. I lost myself in zillion of dualities that physicists keep producing. Does the above extension obstruction interpretation of the duality resemble any of them?

827779(A very partial answer only, but too long for a comment.)

In light of your second comment under Monroe's answer, I interpret your question as follows:

Given a formula $\varphi(x)$, write $\kappa_\varphi$ for the least cardinal satisfying $\varphi$. ($\kappa_\varphi$ could be undefined, of course. But you are only concerned with formulas which are satisfied by a unique cardinal.)

Write $C(\kappa)$ to abbreviate $\kappa^{cf(\kappa)} < \kappa^{\kappa}$.

How can we find out if "ZFC + $\kappa_\varphi$ exists + $C(\kappa_\varphi)$" is consistent?

A sufficient condition is the following: $\varphi$ is absolute under cardinal-preserving extensions (so in particular, $\varphi$ may use ordinal arithmetic, cardinal successor, the $\aleph$-function, etc), ZFC proves that $\kappa_\varphi$ exists and is singular. (This is really just a reformulation of Monroe Eskew's answer.)

An example is the formula $\kappa=\aleph_\omega$.

10294287559151274546nhttps://sites.google.com/site/matroidintersection/homeAre you familiar with this paper : Relating first order set theories and elementary toposes from Awodey,Butz,Simpson and Streicher. I haven't read in detail yet, but it really seems to provide a machinery that answer your question.

I would also add that if such "general" theory as not been developed much is because (this is my personal opinion) it would be essentially useless:

For model theorist, because of the various representation theorems for toposes and boolean toposes, it is know that all the eventual model of set theory you could get this way can be obtained by first taking a permutation model and then taking a boolean valued model model inside of it. What I mean is that any boolean Grothendieck topos is localic over the classyfing group of a pro-discrete topological group, and even worst, any Grothendieck topos satisfying the axiom of choice admit an etale covering by a boolean locale.

And for topos theorist, well the main difference between a model of set theory and a topos is the possibility of comparing two arbitrary object for the membership relation, and I hardly see how this feature can be relevant for topos theory.

8013330909019293061033120$\mathbb{Q}$, as a $\mathbb{Z}$-module, has no minimal generating set.

By the way, in the paper "A characterization of left perfect rings" by Yiqiang Zhou, it is proven that a ring $R$ is left perfect if and only if every generating set of some $R$-module contains a minimal generating set.

184829Also related, it is difficult to give an example of a continuous real valued function on [0,1] that is not monotone on any interval, but it is not too difficult to prove using the Baire category theorem that almost all functions are like this.1158760~Is a 4-dimensional submanifold of a spin manifold always spin?In general this is sharp estimate for all $r,R$. I have to draw a picture, it requires some time for me.1963451196796103567818422026homotopy Sym and quotientsDear unknown, concatenation is non-assocative. Proof: Mädchen(handelsschule) $\neq$ (Mädchenhandels)schule797789XNonstandard analysis is one of these jokes.Addedhttp://mathoverflow.net/questions/191678/limiting-entropy-of-deterministic-sequences11200102045174393304313499There is an obvious inclusion but not equality in general. For a counter example take a quiver algebra and choose $e$ to correspond to a proper subset of the vertices $I$. Then if you have a directed path from a vertex in $I$ to a vertex not in $I$ and then to a vertex in $I$ this path is an element in one subspace but not the other.

~A formula for Frobenius number of certain numerical semigroups12150412032628 [we got the same answer for $d=4,5$ because these coefficients vanish]. I finally realised my error. I gave an argument in the main question that $b_p=1$ for $p=1951$ and it's not right. If $\chi_1$ and $\chi_2$ are two finite order characters of the absolute Galois group of $\mathbf{Q}_p$ and they both cut out totally ramified extensions of $\mathbf{Q}_p$ then, of course, their product might not. Hence my argument that $b_{1951}=1$ is not correct. In fact it's a random 10th root of unity. Thanks so much for enabling me to find my error! I'll edit my original post and finally acceptyouranswer!637473**Problem**

Visually, the "extreme" outliers in the following graph are somewhat obvious:

**Question**

Given:

- Set of all temperatures*T*- Set of all years*Y*- Σ
- Sum of temperatures.*T* - Σ
- Sum of years.*Y* - N - Number of elements
(n) - Temperature of the nth element in the temperature set*T*

How do you determine if ** T**(n) is an outlier?

**Related Sites**

The math on some of these sites is a bit above my understanding:

- Multiple Outliers Detection Procedures in Linear Regression (broken link)
- M-estimator
- Measure of Surprise for Outlier Detection
- Ordinary Least Squares Linear Regression

Many thanks!

11549681825013Okay, I remembered, but it doesn't answer this question. I was thinking of Schwede-Shipley Equivalences of Monoidal Model Categories, which lifts Dold-Kan to monoids, but not general diagram categories. Sorry that doesn't help.2298761Let's remark that sum(b_i)=0 in rectangle for example. Which is not regular. For more details,see our 4217 from "Crux Mathematicorum"-Marinescu and Giugiuc.https://graph.facebook.com/10207657559531763/picture?type=largefThanks! this looks great. Do you have a reference?1349562\https://i.stack.imgur.com/z2Xnl.png?s=128&g=13818381430830The uses of algebraic Galois theory today (e.g., in number theory) have nothing to do with the classical application to solvability of roots in terms of radicals. Even algebraists don't care about that application very much. So it is no surprise that analysts would also have little use for differential Galois theory to help them solve (linear) differential equations. The analysts have a much wider scope for what it means to solve a differential equation than anything that can be offered from differential Galois theory.@user46652 If I understand what you're asking, yes if $G$ has the homotopy type of a CW complex so does $BG$ (because after all it can be constructed using the bar construction).2151762Thank you for your answer. Could you give me an explanation for your last statement? I can't see why equality can't hold if $\rho(x)=\gamma+1$.103849075322829902@Greg: Is that correct that if an arXiv contributor receives strange automatic response from the arXiv, then he/she needs to contact you directly (as was suggested here)? 2164994637082260801194350220381616677651288918user970395499381990521812437In your first paragraph, didn't you mean to say "every Baire subset $A$"? (Or equivalently "Borel")?1630195Given a short exact sequence $0\to F_1 \to F \to F_2\to 0$, one has $pd(F)\leq \max \left( pd(F_1),pd(F_2) \right)$ with equality except when $pd(F_2)=pd(F_1)+1$. Suppose that $pd_B(M)<\infty$. Then $pd_B(N) = pd_B(M)$.

Now, $N$ is projective if and only if $pd_B(N)=0$. Therefore, one has to ask for $pd_B(M)=0$, which is the same as to say that $M$ is projective as a $B$-module.

For every vertex $x$ in $G$, have $\:\text{star}(x)\:$ denote the set whose members are $x$ and the vertices adjacent to $x$.

For every vertex $x$ in $G$, $\:\text{star}(x)\:$ is a biclique of size $\:\text{deg}(x)+1\;$. $\;\;$ The degrees of vertices in

$X$ are independent and distributed as $\:\text{Bin}(n,p)\;$. $\;\;$ From the central limit theorem, if $\: \alpha \leq p$

then for sufficiently large $n$, the probability that any particular vertex's star is not a biclique of

size greater than $\: \alpha \cdot n \:$ will be less than $\frac23$. $\;\;$ If $\: \alpha \leq p \;$ then as $n$ goes to infinity the probability

that no star centered on a vertex in $x$ is a biclique of size greater than $\: \alpha \cdot n \:$ converges to $0$.

Therefore, if $\: \alpha \leq p \;$ then $\;\;\;\; \displaystyle\lim_{n\to \infty} \: \text{Pr}\hspace{.01 in}(\text{the graph a a biclique with size greater than } \alpha \cdot n) \;\; = \;\; 1 \;\;\;\;\;$.

@Ori: Thanks! I did not know of Richardson's model. It does approach a convex limiting shape asymptotically, so maybe my process approaches a Voronoi diagram, but under a metric different from the Eulidean metric (as suggested by Edmund).3363285n special Lagrangian n-Torus has Tubular neighbourhood?24574I Recently finished the examination of Higher sec part 2nd and I took maths as main subject in s Govt. college.

!Let $\varphi\!:\!S\to R$ be a homomorphisms of $K$-algebras for some field $K$. Let $\{a_{\lambda}\}_{\lambda}$ be a family of ideals of $S$.

Is there some "natural" assumption on $\varphi$ to guaranty that $$ \cap_{\lambda} a_{\lambda}^e = (\cap_\lambda a_\lambda)^e, $$ where $\_^e$ denotes the extension to $R$.

The inclusion $\cap_{\lambda} a_{\lambda}^e \supseteq (\cap_\lambda a_\lambda)^e$ is always satisfied. Which conditions on $\varphi$ (as general as possible) guaranty that the converse inclusion is also satisfied?

For example, if $R= S\otimes_K T$ for some $K$-algebra $T$ and $\varphi(s)=s\otimes 1$. Let's check it.

The ring $R$ is a free $S$ module via $\varphi$. Moreover, given a $K$-basis $\{t_l\}_l$ of $T$, the set $\{1\otimes t_l\}_l$ is a $S$-base of $R$. So, given an ideal $I\subseteq S$ and $r\in R$ with $r=\sum_l s_l(1\otimes t_l)$, then $r\in I^e$ if and only if $s_l\in I$ for all $l$. Hence, if $r\in \cap_\lambda a_\lambda^e$, then $s_l\in a_\lambda$ for all $l$ and $\lambda$, that is $s_l\in\cap_\lambda a_\lambda$ for all $l$ and then $r\in (\cap_\lambda a_\lambda)^e$.

From the geometric point of view it is clear that the corresponding map $f\!:\! X\to Y$, where $X=Spec(R)$ and $Y=Spec(S)$, has to be surjective. Also flatness looks a reasonable assumption and then $\varphi$ is faithfully flat. But with this two assumptions on $\varphi$, I am not able to fine neither a proof that $\cap_{\lambda} a_{\lambda}^e \subseteq (\cap_\lambda a_\lambda)^e$ nor a counterexample. Hence, I am not sure whether "faithfully flat" is the assumption on $\varphi$ that I am looking for or not, but I think it is. (In the example $R=S\otimes T$, $\varphi$ is faithfully flat).

Facts about faithfully flat homomorphisms that could be useful are:

- $\varphi$ is injective, so $S$ is a subring of $R$.
- For every ideal $I\subseteq S$, $I=I^e\cap S$.
- For every prime ideal $p\subseteq R$ the ideal $p\cap S$ is a prime ideal of $S$.

Any suggestion or comment would be highly appreciated.

h@AlexDegtyarev Of course! I confused the two $f^*$.20980331530065As far as I understand it follows from the work of M. Freedman that there exist *locally flat* embeddings of two dimensional surfaces in $\mathbb R^4$ that can not be smoothed in the class of locally flat embedding. (For definition of locally flat see http://en.wikipedia.org/wiki/Local_flatness)

I am curious if one can draw a realistic picture of such a surface. At least is it possible to draw an intersection of such a surface with a (linear) hyperplane in $\mathbb R^4$? Is the Hausdorff dimension of such an intersection equals $1$?

1592609You are welcome. It happens that people keep a "wrong" belief for some time.768327(This was first posted to math.stackexchange but had no answers there after several days):

Let ${\mathbb R}$ be the set of real numbers in whatever is your favorite model of $ZFC$. Then (by Levy collapse) there is some larger model in which ${\mathbb R}$ is countable.

I wonder whether there are any mathematically interesting facts about ${\mathbb R}$ that are, on the one hand, entirely internal to the original model, but on the other hand, best proven (or perhaps best discovered) by reference to the countability in the larger model.

In other words, I am looking for arguments of the following form: "Choose an enumeration of ${\mathbb R}$ in some larger model $M'$. Then ..... and therefore $X$'', where $X$ is a statement about the real numbers that would both make sense and be interesting to a person who had never heard of model theory. Is there a reason to believe that no such arguments are likely to exist?

10104881091528I am confused, maybe i don't understand what the homotopy category is. I was thinking about working in the category where objects are spaces (CGH etc etc) and morphisms are homotopy classes of continuous maps. I would even be willing to restrict to CW complexes. Are the things you are thinking about zig-zags?17031311992378Hi! very new to this site, but have always loved mathematics and the opportunities it offers...

I will also apologize up front if some of my questions are a bit strange...

1732526834078528500Get in touch with me at prayag@broomble.com

Found my answers helpful - Buy me Coffee

918791I'm aware that mathematically speaking, Calabi-Yau manifolds are complex manifolds with vanishing first Chern number. However from a physics point of view, Calabi-Yau manifolds are related to the solution of Einstein's field equation in vacuum environment (i.e., with vanishing stress–energy tensor). Since Einstein's field equation is on a 4-dimensional real manifold, why Calabi-Yau manifolds are complex? Is there a "real version" of Calabi-Yau manifold?

11718111013037Crossposted from math.SE (http://math.stackexchange.com/q/607761/264) but seems appropriate here.I think the name **L** is chosen because it corresponds to the 'L operator' of 'old algebraic geometry' over the complex numbers (and that in turn is named L, I would agree, probably to honour Lefschetz). I don't know what notation Lefschetz used himself, but for example Griffiths-Harris book "Principles of Algebraic Geometry" introduces this L operator on page 111 in the deRham setting. Also as the key player for the Lefschetz decomposition/Hard Lefschetz.

The whole book makes no reference to motives or Weil conjecture stuff, just down to earth complex manifold stuff. So, just as a reading suggestion, if you want to read about it on an elementary level.

Maybe this is an easier starting point than things over finite fields etc., even though admittedly maybe not very sexy.

The L operator here wedges with the Kähler form, and that would be just an explicit (1,1)-form representing the point in P1.

(by representing I mean that it represents the cycle class Chow -> deRham of the point via an explicitly given differential form - and that should come down to representing the 'motive' of it)

158534552208414751There may be no precise connection. However, there seems to be difficulty in doing a loose packing even if you get to choose an order in which to pack, and are given a sequence ahead of time, and you have room to spare. That his problem is difficult suggests to me that the two dimensional version of your problem is even more so. Gerhard "It's How I See Things" Paseman, 2012.03.05@Deane: By _mean radius_, do you refer to $1/H(p)$ where $H(p)=\kappa_1(p)+\kappa_2(p)/2$ is the mean curvature at the point $p$.This looks to me like a question about philosophy rather than mathematics. On the other hand it wouldn't surprise me if the logicians who hang around here will have some interesting things to say about it.1628537211130920396384465721093629824407See also http://mathoverflow.net/questions/13320/cool-problems-to-impress-students-with-group-theory6124531952591218061 Are you considering the CIR case or the OU case ? For the CIR, of course, this is impossible, but for the OU, you can certainly write something, since you need to find the law of $ (G_1^2, G_2^4) $ where the $G_i$'s are correlated Gaussians (with explicit correlation function that you can write using the BM). And differentiating is simple. The computations that you have done are good ; I think nevertheless that your process is continuous in any case, CIR or OU, so you don't need the discontinuous bracket / the sum of jumps (there must be a theorem on that in the Revuz-Yor).2275630296843202695t@JoeSilverman: I changed the title to what you suggested.148660317644041310238dThe min in defining $V$ is over complex matrices?1433403508428In "my youth" I computed a finite presenation of the Poisson algebra on $S^2$ (finite presentation as a *Lie algebra*).
In what ways might this be useful? Does this allow you to extract information that would otherwise be harder to obtain?

(1) Is it possible to compute other invariants like Lie algebra cohomology from the finite presentation ... or is it much easier to compute it directly from the full Poisson algebra.

(2) I also do not fully understand how the topology on the algebra intervenes: My presentation is actually only a presentation for a dense sub-algebra of the full Poisson algebra. In what ways can it give information about the "full" algebra. Is the full-algebra already encoded in my finite presentation?

Thank you for any suggestion and help.

Klaus

we conclude that $\sigma$ acts trivially on global sections of $\omega_X^{\otimes2}$... Is there no easier way to see this?! 125481395116TConverse of Hamilton's Maximum Principle?1616551633046Can you say more about 2. - in particular what it means concretely, why it is satisfied for affine schemes, and what do you mean by $K=$"$e^{1/x}$ on $x \neq 0$"? I thought that $K$ is a set of global sections.3963841237188362796@Ashot: $i+1,i+2$ are missing on purpose? If you replace $i+3$ by $i+1$, the example exists?1689338Is there an example of a differential operator $A(z)$ with parameter $z \in \mathbb{R}^d$ and Frechet derivative $A_z(z)$ such that $\mathrm{im}(A_z(z)) \subseteq \mathrm{ker}(A^T(z))$. Can this still be done when $d = \mathrm{dim(\mathrm{ker}(A(z)))}$? I am interested in operators satisfying these properties due to their connection with a decomposition of the scores of multivariate normal distributions $N(\mu, 1)$ with $\mu = A(z)x$.

2096882https://www.gravatar.com/avatar/fe7cb00b0c39155261fe187f4cb53bd6?s=128&d=identicon&r=PG&f=12230969zThe asymptotic behavior of hypergeometric function around -1\https://i.stack.imgur.com/PO5f6.jpg?s=128&g=1322444What happens if $X=Spec(\mathbb{R})$, $Y=Spec(\mathbb{R}[x]/(x^2+1))$ and $f$ maps $X$ to $i\in Y(\mathbb{C})$?19970241907612Let $D$ be the unit $n$-ball (for concreteness). Let $\beta\in\Omega^1(D;R^n)$ be an $R^n$-valued one-form, having full rank (viewed as a section of $T^*D\otimes R^n$). Under what conditions on $\beta$, does there exist a section $Q$ of $SO(n,R)$ (over $D$), such that $Q\circ\beta$ is closed (hence exact)?

The question is non-trivial for the following reason: if there exist such $Q$ and an $f:D\to R^n$, such that $df = Q\circ\beta$, then $\beta^T\circ\beta = df^T\circ df$, and the latter is (up to a musical isomorphism) a flat metric on $D$, whose Riemann curvature tensor vanishes.

So in a sense, I have an answer to my question. What I am looking for is a more explicit condition; in particular, I wonder whether there exists a condition that is linear in $\beta$.

For the curious, this question came up twice in two different contexts in the theory of elasticity.

@Steve. I guess the Jordan curve theorem is a nice example of physical intuition.2282507194843219692813933539924443007630fWhat's your definition of Tschebyshev coordinates?666852514244tHistorical (personal) examples of teaching-based research701131Thanks! I knew the integral identity for $S-T$, but I used it for something very different and it somehow didn't occur to me that it would be useful here. Silly me.20848651822293https://graph.facebook.com/10214701016293111/picture?type=large14266746191431470423...thus get the same information you would get on the generic fibre (ie after base change from $\bf Z$ to $\bf Q$).4807098003907536881310961Susua Kionio142573823008949This is a sequel of this question where I asked for which positive integer $n$ the
set of primes of the former $x^2+ny^2$ was defined by congruences (a set of primes $P$ is *defined by congruences* if there is a positive integer $d$ and a subset $A$ of $\mathbb{Z}/d\mathbb{Z}$ such that a prime $p$ is in $P$ if and only if $p$ mod $d$ is in $A$, up to a finite number of exceptions). I was taught there that the answer was "exactly when $n$ is idoneal", that there is finitely many idoneal numbers, and that all are known but perhaps one.

When is the set of primes of the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ defined by congruences?

My motivation is not just an idle ternary generalization of the binary case. I really met this question while working on a problem concerning modular forms, and also the slightly more general question, given a fixed positive integer $a$: when is the set of primes $p$ such that $ap$ has the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ defined by congruences?

I am well aware that since the set of integers represented by a ternary quadratic form is not stable by multiplication, it is much less natural to ask the question for prime numbers instead of all positive integers than in the case of a binary quadratic form. Yet this is really the question for primes that appears in my study (for about a dozen specific ternary forms, actually).

I have found a very interesting paper by Dickson (Ternary quadratic forms and congruences. Ann. of Math. (2) 28 (1926/27), no. 1-4, 333–341.) which solves the question for the integers represented by $x^2+ny^2+mz^2$: there is only a finite explicit numbers of $(n,m)$ such that this set of integers is defined by congruences (in the obvious sense). But the proof does not seem (to me) to be easily generalizable to primes. Other mathscinet research did not give me any more informations.

When I try to think to the question, I meet an even more basic (if perhaps slightli more sophisticated) question that I can't answer:

When is the set of primes of the form $x^2+ny^2+mz^2$ ($x,y,z \in \mathbb{Z})$ Frobenian? (Is it "always"?)

A set of primes $P$ is called *Frobenian* (a terminology probably introduced by Serre) if there is a finite Galois extension $K/\mathbb{Q}$, and a subset $A$ of Gal$(K/\mathbb{Q})$ stable by conjugacy such that a prime $p$ is in $P$ if and only if Frob${}_p \in A$, except for a finite number of exceptions. A set determined by congruences is a Frobenian set for which
we can take $K$ cyclotomic over $\mathbb{Q}$, which is the same by Kronecker-Weber as abelian over $\mathbb{Q}$. For a quadratic binary
quadratic form (for example $x^2+ny^2$), the set of represented primes is always Frobenian ($K$ can be taken as the
ring class field of $\mathbb{Z}[\sqrt{-n}]$, and $A=\{1\}$, as explained in Cox's book). But I fail to see the reason
(which may nevertheless be trivial) for which the same result would be true for a general ternary quadratic form.
I should add that for my specific ternary forms, I can show that the set is Frobenian, but I am not sure how to extend the argument to all ternary quadratic forms.

Finally, let me say that I would be interested in any book, survey or references on this kind of question (which surely must have been studied), and that I am also interested in analog questions for quaternary quadratic forms (which might be easier, because of multiplicative properties related to quaternions).

18291311122093616801hClosure of tensor product /tensor product semigroup21784711777032I'm studying the Ramsey numbers, especially $R(3,6) = 18$

I understand that the proof using the theorem $R(m,n) <R(m-1,n)+R(m,n-1)$ can only prove that $R(3,6)<20$. However by Cariolaro's "On the Ramsey number $R(3,6) = 18$" I understand the proof for $R(3,6)<19$.

Now I try to understand the proof for $R(3,6)>17$, but the graph there is built with some program. I read that it is possible to build a graph to test $R(3,6)>17$ without programs.

In "thesis (Ph. D.)--University of Waterloo, 1966, Chromatic Graphs and Ramsey's Theorem" by J. G. Kalbfleisch this result supposedly exists, but unfortunately I can not find the document. Can someone give the idea of the construction or do you know of any document or article where you do it?

bI think you mean furthermore rather than however14146518013791619451If your field is CM or totally real and the involution is complex conjugation, the "unitary units" are exactly the roots of unity. Otherwise, if the involution is non-trivial there will be infinitely many such units.413938869751712291640668https://www.gravatar.com/avatar/d62ae7b017b6cde27caef3cf0e3b0f07?s=128&d=identicon&r=PG&f=119538662078597146683411684631419502The Jacobi sum of $n$ multiplicative character $\chi_1,\dots,\chi_n$ on a finite field $\mathbb F_q$ is defined as $$J(\chi_1,\dots,\chi_n) = \sum_{x_1,\dots,x_n \in \mathbb F_q, x_1+\dots+x_n=1} \chi_1(x_1) \chi_2(x_2) \dots \chi_n(x_n).$$ One also considers the variant: $$J_0 (\chi_1,\dots,\chi_n) = \sum_{x_1,\dots,x_n \in \mathbb F_q, x_1+\dots+x_n=0} \chi_1(x_1) \chi_2(x_2) \dots \chi_n(x_n)$$

It is well-known (see e.g. Ireland and Rosen) that one can compute the complex modulus of $J(\chi_1,\dots,\chi_n)$ and $J_0(\chi_1,\dots,\chi_n)$ (when the $\chi_i$ are "general" in some precise sense, $|J(\chi_1,\dots,\chi_n)|=q^{n-3/2}$, and the other cases are not hard to determine as well).

The definition of the Jacobi sum can be rewritten in a more compact way using the maximal diagonal torus $T$ of $Gl_n$: for $\chi=(\chi_1,\dots,\chi_n)$ a character of the maximal torus, $$J(\chi) = \sum_{x \in T(\mathbb F_q), \ tr\ x = 1} \chi(x)$$ $$J_0(\chi)= \sum_{x \in T(\mathbb F_q), \ tr\ x = 0} \chi(x)$$ Now it is clear that the definition above of $J(\chi)$ makes sense when $T$ is replaced by any subtorus of $Gl_n$ defined over $\mathbb F_q$, not necessarily maximal or split. My question is

Is it possible to determine, or at least estimate, $J(\chi)$ for general character of $T$ when $T$ is a general torus of $Gl_n$ defined over $\mathbb F_q$?

In particular, by how much (if anything) is it possible to improve on the trivial bound $|J(\chi)| \leq \sum_{x \in T(\mathbb F_q), \ tr\ x = 1} 1$?

If the general question is too hard, here is one case I am especially interested: $T$ is the torus of $Gl_4$ of diagonal matrices $(x,y,y^{-1},x^{-1})$. So in this case, $T$ is still
a split torus, but is *not* maximal (actually it it is the maximal torus of the symplectic group $Sp_4$ seen as a torus of $Gl_4$ through the natural inclusion). So in this case, in down-to-earth terms, $J_0(\chi)=\sum_{x,y \in \mathbb F_q, x + y + x^{-1} + y^{-1} =0} \chi_1(x) \chi_2(y)$.

Students learning about polar coordinates for the first time may investigate the "roses"

Maybe they will even discover "greatest common divisor" from these.

Historically, the result is due to Hervé Jacquet

MR0369624 (51 #5856) Jacquet, Hervé Sur les représentations des groupes réductifs p-adiques. C. R. Acad. Sci. Paris Sér. A-B 280 (1975), Aii, A1271–A1272. 22E50

(From MathSciNet)

"Let F denote a nonarchimedean nondiscrete locally compact field and G a reductive F-group. A representation r or G(F) in a complex vector space V is said to be "smooth'' if the stabilizer of each vector in V is open in G(F); r is said to be "admissible'' if it is smooth and the space of vectors in V fixed by any open compact subgroup of G(F) is finite-dimensional. The author proves that every irreducible smooth representation of G(F) is automatically admissible. The proof, like the result itself, is surprisingly simple and clever. "

22252619925292085765The definitions are not equivalent: consider $$\mathbb{A}^2 \times 0 \cup 0 \times \mathbb{P}^1 \subset \mathbb{A}^2 \times \mathbb{P}^1$$

Then you cannot see the line before you see the plane.

That is, you can't take any subvariety out of $\mathbb{A}^2$ first, since you can't stratify $\mathbb{A}^2$-minus-anything by cells. You can't take $0 \times (\mathbb{P}^1 \setminus 0)$ out first, because it's not closed. So there's no dimension one closed subvariety you can remove, leaving the complement a union of affines, so there's no Barry Mazur stratification. But you can take

$$\mathbb{A}^2 \times 0 \subset \mathbb{A}^2 \times 0 \cup 0 \times \mathbb{P}^1$$

which is a stratification in the other sense.

1086460@NikWeaver sure, you are right and in fact it is too esay to see it here: $\mathbb{Z}$ consists of the isolated points in $\beta\mathbb{Z}$...1133951@darijgrinberg There is a short Hall-avoiding proof by studying the Birkhoff polytope of doubly stochastic matrices. Namely, replace a regular multigraph to a graph with positive weights on edges such that the total weight of edges in any vertex equals 1. If the graph is a forest, it is already a perfect matching (look at isolated vertex and its neighbor). If it has a cycle, replace the weights along the cycle by $+t,-t$ alternatively so that the minimal weight becomes equal to 0. Proceed this way.66657805512061258866You mean if $G$ is a tree, then $\sum M_{G}$ is even? BTW, your construction works for every $m$-regular graph, $m>2$.214135@And now it is a community wiki.36293451701549Gauss--Lucas type theorem for tracts and higher derivatives of a polynomialDeep Learner at Heart. Data Scientist at Work. OpenCV Guru by Work. Image Processing fanatic at Heart.

Am confused by the apparent discrepancy in the above comments. @Goldstern: Is the result mentioned really well-known and, if so, could you give me a reference?I guess, judging by his name, that Sébastien is French. What we call "les entiers naturels" is the set of non negative integers.22223781086686VEuclidean symmetries of torus links in R^3328872fSt. Petersburg, Severo-Zapadniy Fed. Okrug, Russia504415907730570453488651153735767584211758I am a manager in the software development industry. I have long standing interests in computational linguistics and philosophy of language.

115933Intuitively, it seems clear that conditional on rows $1,2,\dots,n-1$, the $n$th row has zero probability to lie in the $n-1$-dimensional subspace spanned by the previous rows, since that's a finite-dimensional subspace in an infinite-dimensional space. By induction and countable additivity we should get that the matrix is a.s. injective.a limit by Gosper involving a product of arctan and $4^{1/\pi}$2030303\https://i.stack.imgur.com/COnFL.jpg?s=128&g=1https://lh6.googleusercontent.com/-bD9v8-Jnln0/AAAAAAAAAAI/AAAAAAAAApo/w6JpHtQ3px8/photo.jpg?sz=12817244231205926Which statement is usually called the Decomposition Theorem (for perverse sheaves)? Is this (roughly): a proper pushforward of an intersection complex could be decomposed into a direct sum of (shifted) intersection complexes (or is it something more general, or possibly something less general)? Which theorems of [BBD] should one combine to get the 'usual' formulation of the decomposition theorem?

34690512063861521950@Mohan: I do not think that all limits are trivial - there are non-trivial limits of trivial bundles over $\mathbb{P}^1$ and $\mathbb{P}^2$. Moreover, once you have a non-trivial limit, it may be possible to further deform the bundle to something that is no longer a limit of trivial bundles. This happens for $\mathbb{P}^2$, and this is the reason for stating explicitly that the parameter space does not need to be irreducible.169773162098 9660This seems to be a (sort of) 2-dimensional version of Golombs ruler, which itself is hard: http://en.wikipedia.org/wiki/Golomb_ruler

Morty574065@Thierry: the question of what it means to 'know' a prime is essentially at the heart of the question; does one actually have to have a decimal representation, or does an arbitrary description suffice? Or should there be some other requirement? As you point out, the limit of size of primes whose decimal representation is known is vastly different from the limit of size of primes known by some other description.41320412596531488r$M$ is assumed to be finitely generated in the question.1968625The nonabelian tensor product of groups which act on each other (in a compatible way) was defined by Jean-Louis Loday and myself in 1984 and the topic has been taken up by group theorists because of its relation to commutator theory. I have compiled a bibiliography on this topic which now runs to 131 items. The applications to homotopy theory do require tools of higher groupoids for the proofs, as well as advanced techniques from algebraic topology, and many useful aspects of category theory.

Just to give an idea of the subject, suppose $M,N$ are normal subgroups of a group $P$. Then the commutator map $[\;,\;]: M \times N \to P$ is not bimultiplicative, but is what may be called a *biderivation*. So it factors through a **universal biderivation** $M \times N \to M \otimes N$ giving rise to a morphism $\kappa: M \otimes N \to P$. Even the calculation of this tensor product is not so easy.

Although it seems like a textbook question, I was not able to find a textbook or even a research article answering the following question:

Let $M$, $N$ and $P$ be finite-dimensional smooth manifolds and let $f \in C^r(M \times N,P)$ for a given $r \in \mathbb{N}$.

Let $\hat{f}: M \to C^0(N,P)$ denote the adjoint map, given by $\hat{f}(x) = (y \mapsto f(x,y))$.

For which $k,l \in \mathbb{N}$ does it hold that $\hat{f} \in C^k(M,C^l(N,P))$?

10766251310066 4540It is better to write the symbol of the exponential function as $\exp$ `\exp` than as $exp$ `exp`.Nice answer! I think you have a typo in the first displayed equation: $\lambda_j-n$ should be $\lambda_j-z$.Averaging a Kaehler class over the involution, and taking the corresponding Ricci-flat metric, we may assume that the involution preserves a flat metric on a torus. At each fixed point, the eigenvalues of the involution are +1, -1, and the fixed point set is a subtorus $T_0\subset T$. This means that the involution acts as -1 on $T_1 :=T/T_0$, hence the quotient is a product $T_0\times T_1/\{\pm 1\}$, that is, $T_0 \times {\Bbb C} P^1$.

I just found a much earlier reference: it is Lemma 2.1 in: "Factoring Absolutely Convergent Series". William H. Ruckle; Robert E. Jamison. Mathematische Annalen (1976). Volume: 224, page 143-148.21512672263562@Chris: Are you talking about my comment? I asked the OP to clarify his / her question. Open-ended questions, such as "what is pretty", easily become argumentative. I think the OP's new, clarified title immediately makes clear the precise kind of question this is.80635~@abx Sorry for my typo (brain-o?), and thanks for catching it.1614519207317244706815912901840542Mahdi Ahmadi1315421This doesn't answer the main question, but I'm accepting it anyway since the argument is so beautiful!rHow is a descent datum the same as a comodule structure?Argh my original comment did not go through. I just chose my analysis class example because I wanted to point out that anecdotal evidence isn't adequate. I was wondering if there were better metrics one could use to determine the success of a particular method. The entire dept (which is quite small) is on board to trying new things so I'm not worried about that. Anyways, I think this entire study/analysis is geared towards lower level courses. 1631597604145It's mostly a surprisingly long collection of examples; I don't claim to have a structural "why". Heuristically what seems to happen is that — assuming the Diophantine equation is of the right dimension and complexity to yield a K3 surface in the first place — there's enough divisors coming from trivial solutions, and often enough symmetry, that there's barely enough room for the Néron-Severi lattice to accommodate them all under the constraint of rank at most $20$.589726Ok, thank you! I really appreciate your intuition and all the hints. This is not my field, so I was not sure if this was a feasible question. It was worth trying though :), and learning a small bit of the beautiful mathematics behind those questions..Dear Igor Rivin, I don't see the proof in Wikipedia. I don't believe any thing with out the proof on Wikipedia. Do you know how to prove it?1502040hWhat manifolds are boundaries of euclidian spaces ?122212020242821635911nKodaira embedding theorem for rigid analytic varietieshttps://www.gravatar.com/avatar/65f674c3b9b5e0802605bd0607579d19?s=128&d=identicon&r=PG&f=1659710988731@მამუკაჯიბლაძე, there's at least one way to do the coalgebraic reals constructively that ends up equivalent to the Dedekind reals; this is shown at the end of section D4.7 in *Sketches of an Elephant*. So that would certainly be acceptable. If there's a way to do coalgebraic reals constructively that doesn't end up equivalent to the Dedekind or Cauchy reals, then that wouldn't really be quite what I'm looking for, but it would nevertheless be interesting.1561865What is the status of the equidistribution root numbers of elliptic curves' L-functions140395581702https://www.gravatar.com/avatar/08eb09a321e372452bdfaf8c5571d3aa?s=128&d=identicon&r=PG&f=1377702216875That’s right. For a natural finitely axiomatizable theory, the answer is also positive for $\mathit{PV}_1$ (one can check that $\mathit{PV}_1+E$ proves EXP).216762712226201145681Consider the class of all Hausdorff compacts with distinct points (i.e. which have more than $1$ point) that are absolute retracts in the class of Hausdorff compacts. Then $[0,1]$ is up to homeomorphism the only member of this class that embeds into every other.

1717086898646622145The Newton Polygons of these polynomials are relatively easy to study, and by some inspection, proving the assertion needed, prime > n/2, seems quite hard... These are just my calculations on page. Someone else confirm98739431120101944536Actually, there is no need to take the Cesaro averages as the convolution powers themselves already converge to the stationary measure.45063019381712246544682044VIndecomposable quotient of Prüfer domains1577120Ugh, why aren't these posted yet:

Q: What's purple and commutes? A: An Abelian grape.

Q: What's sour, yellow, and equivalent to the axiom of choice? A: Zorn's lemon.

etc.

59379333771222085482000763First off, Prof. Figueroa-O'Farrill is correct in noting that you omitted the $G$-equivariance condition for the connection, which reduces to $G$-invariance in the case where $G$ is abelian (in particular, when $G=\mathbb{C}^*$).

The choice of normalization conventions really just comes down to how you identify the Lie algebra of $\mathbb{C}^* $ with $\mathbb{C}$. Śniatycki's convention amounts to identifying the $\mathrm{U}(1)$-generator (i.e. the vector field $2\pi\frac{\partial}{\partial \theta}$ in the Lie algebra of $\mathbb{C}^* $) with $1\in\mathbb{C}$, whereas the seemingly more logical choice would be to identify it with $2\pi i\in\mathbb{C}$. Śniatycki does this because it's also Kostant's convention (who got it from Weil I believe).

The relationship between your $\alpha$ and Śniatycki's $\alpha$ should be just $$ \alpha_{\mathrm{Blake}} = 2\pi i~\alpha_{\mathrm{Śniatycki}} $$

The condition for the existence of a line bundle $L$ over $X$ with connection $\alpha$ and $\alpha$-compatible pairing $\langle\cdot,\cdot\rangle$ can formulated in terms of the curvature $\omega$ of $\alpha$, defined by $d\alpha = \pi^*\omega$, where $\pi$ is the bundle projection. The big advantage of defining the connection $\alpha$ à la Kostant/Śniatycki is that the condition becomes that $\omega$ is

*integral*, i.e. gives an integer when integrated over closed two-cycles in the base manifold $X$. For this reason, the $2\pi i$ is a common normalization in the theory of Chern classes.

From personal experience, normalization conventions can drive you mad, especially when comparing results from different authors. For example, Guillemin and Sternberg seem to favor the convention that $\frac{\partial}{\partial \theta}\in\textrm{Lie algebra of }\mathbb{C}^* \leftrightarrow 1\in\mathbb{C}$. So best of luck :)

It is really cool to discuss with mathematicians! Thanks Jan for your reply. You did show that for the general case, the projection matrix is not non-negative definite.Thank you, Gjergji... I'd up-vote that beautiful old answer of yours, but it seems I've already done that time! My fault, I should have searched more carefully. So I voted to close my own question :(Let $S\subset M_{n}(\mathbb{R})$ be the singular points of the equation $Det=0$. That is $S$ is the critical points of the determinant function.

What matrices belongs to $S$, precisely?

Let $M=Det^{-1}\{0\}-S$ be the codimension one submanifold of $M_{n}(\mathbb{R})$ which has a natural Riemannian metric induced by the standard metric of $M_{n}(\mathbb{R})\simeq\mathbb{R}^{n^{2}}.$

What is a linear algebraic and matrix meaning for a matrix $A\in M$ with the following property:

**"The sectional curvature of $M$ at $A$ is independent of choosing a $2$-plane tangent to $M$ at $A$"**

153611258409214360046RMorphisms contracting a family of curves1888738fIf $\sigma(T)=\{0\}$, then, sure, $T$ is positive.The theory of operads covers lots of these structures -- though not all. [Loday and Vallette have a book on them](https://www.math.univ-paris13.fr/~vallette/Operads.pdf).Merci Mikael. It seems like I will need to do some "search and replace" tonight on something I have been typing...1635794What is a precise example of this situation, for $n=2$?

Let $\tau$ be $(-1+\sqrt{5})/2$, let $f(x)$ be $\lfloor (x+1)\tau \rfloor$, let $s_n$ be $\tau n (n+1) (n+2) / 6$, and let $S_n$ be $$\sum_{k=0}^{n} (n−2k)f(n−k) = n f(n) + (n-2) f(n-1) + (n-4) f(n-2) + \dots - (n-2) f(1) - n f(0).$$

Is $(S_n-s_n)/(n \log n)$ bounded for $n > 1$?
(It stays between -0.35 and +0.30 for all $n$ between 2

and $10^6$.)

This is a specific instance of the question Dedekind-esque sums that I posted a few weeks ago. It may be an atypical instance in some ways (since $\tau$ is a pretty atypical real number for Diophantine approximation problems) but it's the one that interests me most right now. An affirmative answer to my question would have implications concerning the "Goldbug machine" described in http://front.math.ucdavis.edu/0501.5497 .

The plot at the bottom of http://www.cs.uml.edu/~jpropp/Phi-short.pdf is a histogram of $(S_n - s_n)/n$ for $n$ going from 1 to a million. As you can see, it doesn't stray very far away from 0. So perhaps that $\log n$ in the denominator could be replaced by something smaller, like $\sqrt{\log n}$ or even $\log \log n$ (or maybe even 1, though I doubt it).

20741703070Did Bourbaki acknowledge *anyone* 's priority? I remember his thanking tits in a footnote for the exercises on Coxeter groups &c.No, what I meant is that there are varieties with no 1-dimensional deformations but with nontrivial first order deformations. But I agree that you didn't claim that, if you take $H^1(X,T_X)=0$ as definition of rigid.880558397832134966626933810344781968323033487355571088931719490https://www.gravatar.com/avatar/5c5a33246cdd789c6603532b734c2b5f?s=128&d=identicon&r=PG&f=1~@J.C. Ottem, How does Bertrand's postulate give us divergence?5228442453767148328I'm still looking forward to a detailed reference concerning Question 1.I am currently the Executive Administrator of The World Council for Gifted and Talented Children.

1660195!The generalization of Chebyshev series on $[-1,1]$ to an arbitrary connected and simply connected compact set $K$ of the complex plane were defined and studied by Faber in 1903. The series of Chebyshev polynomials corresponding to $[-1,1]$ are then replaced with series of Faber polynomials of the set $K$. These series play in $K$ the role of the Taylor series in the unit disk $\mathbb{D}$, and enjoy exactly the same convergence properties, which can be seen via the conformal map from $\overline{\mathbb{C}}\setminus\overline{\mathbb{D}}$ to $\overline{\mathbb{C}}\setminus K$. A standart reference for Faber polynomials is [1].

The case when $K$ has several components (like, for instance, the union of two intervals) was first considered by Walsh in 1958. Based on conformal mappings from multiply connected regions to lemniscates, that he also introduced, see [2], he defined in [3] the so-called Faber-Walsh polynomials which again have properties in $K$ similar to Taylor series. The papers [4] and [5] are recent references on that subject. In particular the case of two intervals is studied in detail in [5] along with numerical results.

[1] P. K. Suetin, Series of Faber Polynomials, Gordon and Breach, Amsterdam, 1998.

[2] J. L. Walsh, On the conformal mapping of multiply connected regions, Trans. Amer. Math. Soc., 82 (1956), pp. 128-146.

[3] J. L. Walsh, A generalization of Faber's polynomials, Math. Ann., 136 (1958), pp. 23-33.

[4] O. Sète, J. Liesen, On conformal maps from multiply connected domains onto lemniscatic domains. Electron. Trans. Numer. Anal. 45 (2016), 1-15.

[5] O. Sète, J. Liesen, Properties and examples of Faber-Walsh polynomials. Comput. Meth. Funct. Theory 17 (2017), 151-177.

143645913213471680524458942067719jBanja Luka, Republika Srpska, Bosnia and Herzegovina17931517109451143271496631108767rSome confusion regarding the definition of NPO reduction641678120067162459713946134646697905613026915@MattSamuel: But as you point out, the problem being GapP-complete is morally equivalent to it being impossible to find a Littlewood-Richardson rule. So presumably it would be big news if someone could show it is GapP-complete.1628422Yes. Majda's book is helpful. But it does not give a whole introduction to the standard methods to the mathematical analysis of hyperbolic equations. I hope to get a whole view of the modern theory.4213822704722062433You could see : https://www.youtube.com/watch?v=Vg7955TGp4U ...1832357217242GullitlCategorical representations of absolute Galois groupsI am wondering if there are some results about the depth of a diffeomorphism on a manifold.

More precisely, $(M,f)$ be a diffeomorphism. For each compact invariant subset $E$, let $\Omega(f, E)$ be the nonwandering subset of $f$ relative to $E$. Let $\Omega_1=\Omega(f,M)$, $\Omega_{n+1}=\Omega(f,\Omega_n)$, and $\Omega_a=\cap\Omega_b$ over $b < a$ for a limit cardinal $b$... etc

So my question is; are there some conditions under which the diffeo has a finite depth, that is, $\Omega_{n+1}=\Omega_n$ for some $n$?

There are examples of topological systems with countable depths. I do not know what can happen in the smooth category. I googled and found that the depths of circle maps or interval maps are less than 2.

Thanks!

To rpotrie: I am looking for sufficient conditions on the spaces (say, manifolds) and the maps (say, the regularity) such that $f$ has finite center depth. As rpotrie mentioned, Axiom A maps (hence all Anosov) always have center depth 1, the maps with $\Omega(f)$ hyperbolic have center depth less than 2.

For example partially hyperbolicity may not be a good candidate since the direct product $f\oplus g:M\times N\to M\times N$ has transfinite center depth if one of $f$ or $g$ has.

|https://graph.facebook.com/100000695860221/picture?type=largebAnswering 1), there is no such a correct notion.19931 9839130488411026934318541857802-On the basis of rather convincing numerical evidence (iterative optimisation that always converges to the same place regardless of starting point), I conjecture that the worst case is $$ a_i = \frac{i}{(n!)^{1/n}}.$$

I believe that gives $$C(n) = (n-1) \biggl( \frac{2}{n(n+1)!} \biggr)^{1/(n-1)} = e - \frac{e(7\ln n-\ln 2+\ln\pi)}{2n} + O((\ln n)^2/n^2).$$

The method is as boring as can be imagined. Start with a random vector **a** then make random changes to it, rejecting the change if the ratio of left side to right side gets greater. Stop when it hasn't changed for quite a while. Here is Maple code and sample output for $n=7$. Note that the vector is written with normalisation $a_1=1$ even though the function is calculated with $\prod a_i=1$; this is to make it easy to verify the conjecture by eye. Increase the value of *Digits* to get more precise verification.

```
# C(a) = left side divided by right side when a is normalised to product 1
C := proc(a::list) local b,s,k,n;
n := numelems(a);
s := mul(a)^(1/n);
add((1/k-2/n/(n+1))*a[k]/s,k=1..n) / add(k^2*s/a[k],k=1..n)^(1/(n-1));
evalf(%);
end proc:
eps := rand(-0.005..0.005):
r := rand(0.0..1.0):
n := 7; # Insert value of n here
p := rand(1..n):
a := [seq(r(),i=1..n)]:
Ca := C(a):
# Initial values
a,Ca;
while true do
currC := Ca:
for iter to 5000 do
a := a / a[1]; # Normalise to a[1]=1, note definition of C()
pos := p():
old := a[pos]:
a[pos] := max(old+eps(),0.0):
newCa := C(a):
if newCa < Ca then Ca := newCa: else a[pos] := old: end if:
end do:
print(a,Ca);
if Ca = currC then break end if:
end do:
# Compare to conjecture
evalf((n-1)*(2/(n*(n+1)!))^(1/(n-1)));
n := 7
[0.01506015221, 0.07126263187, 0.6247572599, 0.8230410086, 0.3693148049, 0.7607420702, 0.4647867183], 1.130896511
[1.000000000, 2.539500613, 16.40072680, 21.91302565, 10.73273281, 20.16793736, 13.31584408], 0.9921703822
[1.000000000, 1.724599348, 6.508732912, 8.844028091, 5.179715013, 8.535313648, 6.434865587], 0.8859967880
[1.000000000, 1.922974212, 3.388120139, 5.041522591, 4.236184244, 6.150065920, 4.972451482], 0.8403143690
[0.9999999998, 2.003100092, 3.006977468, 4.342457370, 5.023819919, 6.039027264, 6.158267516], 0.8324458625
[1.000000000, 2.000240008, 3.000252639, 4.000077928, 5.000611566, 6.000674528, 7.000510740], 0.8315464026
[1.000000000, 2.000204912, 3.000252639, 4.000077928, 5.000451793, 6.000674528, 7.000510740], 0.8315464022
[1.000000000, 2.000204912, 3.000252639, 4.000077928, 5.000451793, 6.000674528, 7.000510740], 0.8315464022
0.8315464026
```

11753942115210Could one perhaps make a heuristic argument saying that if the constant term was "random" (in some sense) then there would only be finitely many pairs $(p,h)$ satisfying the OP's constraints? I'm not sure what "random" should mean here.There is a weak choice principle called $DC_\lambda$ which holds in $L(V_{\lambda+1})$ under the assumption of a non-trivial elementary embedding $$j:L(V_{\lambda+1})\prec L(V_{\lambda+1})$$ and it is known that this choice principle is not sufficient to split the ordinals below $\lambda^+$ (a regular cardinal) which have cofinality $\omega$ into disjoint stationary sets.

Is there a choice principle $\Phi$, which, when augmented with $DC_\lambda$ and strictly weaker than full AC that suffices to prove Solovay's Theorem on Partition of Stationary Sets? A little more specifically, what is the minimal $\Phi$ such that $T=ZF + DC_\lambda + \Phi$ where $$T\vdash \text{All stationary subsets of }\lambda^+\text{ have a disjoint partition into stationary sets}?$$ One such candidate could be $Unif(V_{\lambda+1}\times V_{\lambda+1})$: given any $R\subseteq V_{\lambda +1}\times V_{\lambda +1}$ there exists some function $f\subset R$ with the same domain as $R$. (This question is related to both A proposed axiom of Laver (updated) and Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? .)

Are there other weak choice principles $\Phi$ that could be considered?

Are there (perhaps) some partition properties with infinite exponents that would prohibit Solovay's Theorem?

EDIT: While I am interested in the more general question (which I believe Asaf Karagila has addressed in the comments and chat), I am really specifically interested in the context that Woodin's Axiom $I_0$ holds. Specifically, any assertion $\Phi$ I'm looking for can't imply that $[\lambda]^\omega$ is well-ordered (in conjunction with $DC_\lambda$).

If $R$ is Artinian, any such function is uniquely and freely determined by what it does to simple modules, isn't it? 621211As Fernando says: there's literally a proof of the result *in the link you provide* to Borel's paper.1620235356676Not sure if this question really qualifies for MO.

Anyway, the answer very much depends on the group $G$. In most cases $n_p(G)\ne n_q(G)$ for distinct prime divisors of the group order. However, there are infinitely many examples where equality occurs: If $r$ is an odd prime, then $n_p(\text{PSL}(2,r))=r(r+1)/2$ for each odd prime divisor $p$ of $r-1$.

But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizers of the $3$-Sylows and $5$-Sylows have order $60$.

You could also use Hurewicz to say $H_1 = 0$ and $H_2 = \mathbb{Z}^2$. The rest follows from this. (Here I'm using the fact that the $S^2$ bundle has a section.)1814691@David: Have you heard from the authors? I can't find any signs that the paper has been corrected or withdrawn. This answer reminded me of a lot of things that I knew about but had forgotten. But now I'm not sure I believe it. Actually, the problem is probably my fault: either I am simply confused, or I chose bad notation. My notation "Vect_SuperVect" is not intended to mean "categories that are direct sums of copies of SuperVect", but rather "module categories for SuperVect" (so perhaps I should have used "Mod_SuperVect"), with the balanced tensor product. Now, surely the Morita category of superalgebras embeds in here just as in the usual case? E.g. Cliff(1) corresponds to Vect as a SuperVect-module.Consider a large, fixed $M>2$. For each $n$, let $\alpha_n$ denote the smallest algebraic integer of degree at most $n$, all of whose Galois conjugates lie in the real interval $(0,M)$.

Is there anything known on the rate at which $\alpha_n$ decays to $0$ as $n\to\infty$ ?

The exponential lower-bound $\alpha_n>\frac1 {M^{n}}$ is not hard to prove, but is it reasonable to expect that $\alpha_n$ will actually decay much more slowly, i.e. like an inverse power of $n$ ?

4315102Ok, so here is how I like to think of it from a geometric perspective.

**Definition:** (Under moderate hypotheses) *analytic spread* of an ideal $J$ is the minimum number of generators of an ideal with the same integral closure as $J$.

**Definition:** Two ideals $J, J'$ have *same integral integral closure* if and only if their normalized blowups $X = X'$ are the same with equal pullbacks $J \cdot O_X = J' \cdot O_X$.

The number of generators of an ideal of course gives a bound on the number of affine charts you need to cover a blowup.

Putting these together you obtain that:

**Fact:** Analytic spread of $J$ is a canonical upper bound on the number of affine charts you need to cover the blowup of $J$. In other words, it measures the complexity of a blowup of $J$.

Let's look at the usual:

**Example:** Consider $J = \langle x^3, x^2y, xy^2, y^3 \rangle \subseteq k[x,y]$. Then the analytic spread is two, corresponding to the ideal $J' = \langle x^3, y^3 \rangle$. The normalized blowup of $J'$ is the same as the blowup of $J$, which is just blowing up the origin in $\mathbb{A}^2$. On the other hand, you only need two charts to cover the blowup of $J$, the chart corresponding to inverting $x^3$, and the chart corresponding to inverting $y^3$.

Now, it could be that the analytic spread is *exactly* the minimum number of charts needed to cover the blowup, I don't know (if it's important I could think of a couple people I would ask).

Finally, if you were looking for more algebraic interpretations, I would see if anything in the Swanson-Huneke book on integral closure helps.

Did you try:

XWhy are they called "screen" distributions?20123722224126124945052543its good to keep in the mind the following example: take $N \ge 3$ and $ x_0 \in \partial \Omega$ and $ x_m $ not be in $ \overline{\Omega}$ and consider $u_m(x) = |x-x_m|^{2-N}$ with $x_m \rightarrow x_0$.(KotschickConjecture371635613273199451017228091833048^I realy thank you very much for your answer.2990212101156541057I believe this question is too basic to be appropriate for MO, but it is very well written and I will give a short answer in any case.36093415444022176734Could you flesh this out? Or give a link to a reference? Why is finding super-linear lower bounds hard? If it is obvious, I apologize in advance.@user76758: I did not claim that the action of $\mathfrak S_2$ on $L$ or $U$ is by swapping rows.2064532292887dComputer science student and indie game developer1035438Yes, Tor will be zero away from $Y$, but it won't be zero on $Y$. It's important that you are taking $\mathrm{Tor}_X$ and not $\mathrm{Tor}_Y$.I have added back some details which were in the comments but not in the answer.915391157313623017541790272I edited the question in view of several helpful replies (thanks).

When we define ind-objects in a category, we use in general filtered diagrams in a category, not just sequences $A_1 \rightarrow A_2 \rightarrow A_3 \dots$ indexed by the integers. More generally, if one wants a "bigger" limit, we could look at diagrams indexed by an ordinal.

Is there a simple example of an ind-object that can't be indexed by an ordinal?

14477751114076"(assuming X is projective)" But of course X is projective - via L^k (for k sufficiently big) :-)Also FYI for sample paths: "Semimartingales are "good integrators", forming the largest class of processes with respect to which the Itō integral can be defined." http://en.wikipedia.org/wiki/Semimartingale19083642121212https://www.gravatar.com/avatar/7df580c0d3b153423b3a91d232ab4b78?s=128&d=identicon&r=PG&f=1I have the following problem:

Given a matrix with n rows and m columns. Some elements of the matrix are unavailable. For each column, you have a set containing a number of zeros and ones which must be distributed over the available elements in that column. The total number of zeros and ones in a set is equal to the number of available elements in the corresponding column. The question is: How should the values of the sets be distributed over the elements in the corresponding colum, such that the number of rows for which the total number of ones is less than 20% of the available elements in that row, is minimized.

Does someone knows a similar problem which has a polynomial time algorithm solving the problem to optimality, or knows a polynomial time algorithm solving the problem to optimality (if it exists)?

256336115448310933136481033I know that the asymptotic for the sum of all the primes up to n is $n^2/2\log n$. But I'm trying to find the formula (an estimate) for when $n^2$ is divided by each of the primes up to $n$, in turn -- and then summing the results.

1613217I think Fourier analysis / transforms / linear algebra / Finite Element Method don't really fit so well in Rings and Algebras (*although, obviously, you can put them there if you really insist*) - obviously they use algebraic techniques, but their main core is elsewhere (analysis, numerical analysis, etc.) in my opinion.https://www.gravatar.com/avatar/e339869339bde538587091585cadd8db?s=128&d=identicon&r=PG&f=1Currently I am a postdoctoral researcher, with my interests mainly in operator K-theory and noncommutative geometry.Let $\Gamma$ be the category of finite pointed sets. The abelian category $\mathrm{Mod-}\Gamma$ is the category of functors $\Gamma^{\mathrm{op}} \to \mathrm{Vect}_k$, where $k$ is a field (see Pirashvili's paper).

Is every object of $\mathrm{Mod-}\Gamma$ the direct sum of simple objects? (If necessary, add finiteness assumptions to the modules.)

If yes: Is every simple object a direct summand of the "regular representation" $\displaystyle\bigoplus_{n \geq 0} K\bigl[\mathrm{Hom}(-,[n])\bigr]$?

I am aware that Peter Webb and others have studied representation theory of categories, but I couldn't locate such a result so far.

Actually I am not so much interested in $\Gamma$, but rather the category of finite sets (i.e. without base points), but I thought it would be wise to ask the question for $\Gamma$ first since this category seems to appear more frequently in the literature, especially in algebraic topology.

I highly appreciate references to the literature, because I am not really not sure where to look for more properties of these module categories. Google almost only gives me results on $(\phi,\Gamma)$-modules in algebraic number theory.

13806141648831907565209677820325523320071966257Dear Robert Haslhoer, you are right. I need pairing of $f$ with $\delta_x$ for all $x\in R$, which equivalent to saying the convolution. :-)1154895vI think you must assume the algebra $A$ is self-injective?lCan you say a little about the context or motivation?37716721526619226923Senior software engineer with over 7 years of professional experience developing object-oriented frameworks, libraries, and applications in .NET and C++; over 4 years managing small groups of developers through product releases.

Are you familiar with Kato's Perturbation Theory for Linear Operators (Ch 2), [downloadable from U. of Edinburgh](www.maths.ed.ac.uk/~aar/papers/kato1.pdf) .1284391For $C^{\star}$-algebras I would suggest in addition the lovely book of Gerard Murphy "$C^\star$-algebras and operator theory".

If you are interested in K-Theory Blackadar's "K-Theory for Operator algebras" might be a good choice after Murphy's book.

For Noncommutative Geometry, there are nice little introductory books by Varilly "An Introduction to Noncommutative Geometry" and Khalkhali "Basic Noncommutative Geometry". They are maybe more "beginner friendly" than Connes' master piece.

156176236773793395771772094710536The kernel of the $G$-action on $G/P$ is contained in $P$. So, if you find any Lie subalgebra $\mathfrak{h}\subseteq\mathfrak{g}$ meeting $\mathfrak{p}$ trivially and integrate it to a closed subgroup $H$, then the action map gives you a subgroup of $Aut(G/P)$ with dimension equal to that of $H$. I imagine you can choose $H$ to have dimension greater than $1$ in most examples. In these cases, you really have a subgroup of $Aut(G/P,L)$ of dimension greater than $1$, meaning your quotient cannot be discrete.13130231175753 Thank you for pointing this out. I agree that this should come from five-term relations, but I thought Wojtkowiak’s theorem is the case with rational functions as arguments whereas the above relation involves square roots. Also, what I really wanted to do is to show the above from known relations (five-term relations etc), rather than taking a derivative. The square root above comes from extremizing w.r.t. a certain variable, and there seems to be generalizations to the case with more variables where we cannot use eliminate variables. Let me see if I can make the question sharper.1441356504615I designed the math T-shirts for http://thenerdiestshirts.com, including one on the Calkin-Wilf tree, a proof of the Fundamental Theorem of Algebra, a truncated icosahedron, and others. Perhaps most mathematicians wouldn't call these high brow, but they are supposed to be a big step above the usual math shirts.Dritan Skarra942301zHow hard is it to tell when a finite set tiles the integers?2269965I am not really a MathOverflow reader, but I just came across this discussion. I first saw the fracture square that Neil describes (in the classic case of interest as above) in a (handwritten) letter to me from Pete Bousfield dated January 22, 1987. It is in the midst of a paragraph that begins with " ... I'll make some little comments which may be well known to you.", and describes how to (easily) construct distinct nice spectra X and Y whose K(n)-localizations agree for all n. (His letter was part of a correspondence we had around then about how one could generalize his telescopic functor for n=1 to all n.)

Very possibly Pete knew the fracture square result in the late 1970's, when he was thinking about the Boolean algebra of localization functors and such. But it doesn't have a lot of meat until one has some naturally arising smashing localizations, which needed developments in the 1980's.

207273810565231309485@FredDashiell: You certainly don't need "full" AC; an Ulam matrix is enough.In other words: the tangent vector space to $G/H$ at $eH$ is the quotient of the Lie algebra of $G$ by the Lie algebra of $H$. $G$ acts on its Lie algebra by the adjoint action and the restriction of this to $H$ descends to act on the quotient vector space. The vectors fixed under this action are exactly the ones that extend to invariant vector fields.10185871538101lHomogeneity degree one functions of a matrix argument1425352jInteger matrices whose determinant equals their norm10241622123316141634~Generalizing a pattern for the Diophantine $m$-tuples problem?1875115Such sections are tantamount to solutions of the unit equation $u + u' = 1$ in $O_K^*$. This is indeed impossible for $K = {\bf Q}$, when $O_K = {\bf Z}$ and the only units are $\pm 1$; but there can be such solutions for other number fields $K$, though it is known that in each $K$ there are only finitely many solutions.

A "section of ${\bf P}^1_{O_K}$ over $O_K^{\phantom|}$" is a $K$-point of the projective line, i.e. either $\infty$ or a field element. The "sections" $u$ disjoint from $\infty$ are precisely the algebraic integers, because $u$ "intersects $\infty$ at the prime $\wp$" **iff** $u$ has negative valuation at $\wp$, and the algebraic integers are precisely the field elements none of whose valuations are negative. Likewise $u$ is disjoint from $0$ **iff** $u$ has no positive valuation at any $\wp$, and disjoint from $1$ **iff** $u-1$ has no positive valuations. Therefore, $u$ is disjoint from $0$, $1$, and $\infty$ **iff** both $u$ and $u-1$ are units, which is to say **iff** $(u,1-u)$ is a solution of the unit equation.

One easy way to get such $(K,u)$ is to make $u(1-u)$ a unit in some number field $F$, say $\epsilon$, because then $u$ and $1-u$ are themselves algebraic integers in a number field containing $F$ with degree at most $2$ (they're the roots of the monic quadratic polynomial $x^2-x+\epsilon$ over $O_F$), and thus units because they divide the unit $\epsilon$. For example, taking $F={\bf Q}$ and $\epsilon = 1$ or $-1$ we recover the simplest solutions of the unit equation: the sixth roots of unity $(1 \pm \sqrt{-3})/2$, and the golden ratio $(1 \pm \sqrt{5})/2$. More generally, if $x^{m+n} - x^m = \epsilon$ for some unit $\epsilon$ and positive integers $m,n$ then both $x$ and $1-x^n$ are units, so we may take $u = x^n$.

13721691279667RUnderstanding a formula in Ozsvath-Szabo859573https://www.gravatar.com/avatar/b7f01b027e23bc786b64b3380d180186?s=128&d=identicon&r=PG&f=1447402~https://graph.facebook.com/1881632725476544/picture?type=large20106631743951254465One can get a full asymptotic expansion as an application of Watson's Lemma. One need only observe that the integrand is maximized at $x_0 = \left(\frac{Kb}{ac}\right)^{1/(a-b)}.$

Substituting $x = x_0 u,$ one gets(where $I$ is the original integral) $I = x_0 \int_0^\infty \exp(-c^{b/(a-b)} K^{a/(a-b)}((b/a)^{a/(a-b)} u^a - (b/a)^{b/(a-b)} u^b)) d u.$ Letting $t = c^{b/(a-b)} K^{a/(a-b)},$ the integral breaks up into two Watson Lemma integrals, one from $0$ to $1,$ the second from $1$ to $\infty.$ I leave the final computation of the asymptotics to the interested reader.

**EDIT** Actually, this is not quite Watson's lemma. You approximate the function $\phi(u) = (b/a)^{a/(a-b)} u^a - (b/a)^{b/(a-b)} u^b)$ by its Taylor series (at the maximum point $1).$ Since it is the maximum, this will look like $\phi(1) - (u-1)^2 \phi^{\prime\prime}(1)/2.$ This means that the integral is asymptotically approximated by a Gaussian integral (I am too lazy to compute $\phi^{\prime\prime}(1)$...$)

"A short tale of hybrid mice", by Grigor Sargsyan.

6373161732976I'm a LaTeX enthusiast.

I'm the maintainer of a few LaTeX packages such as `chemnum`

, `chemformula`

or `translations`

.

**Any code of mine that I publish or have published previously on the TeX.SE main site I hereby relicense under the WTFPL.**

There are some special cases where the CSM classes of singular varieties can be computed easily.

For example there is Ehler's formula for $c_{SM}(X)$ where $X$ is any complete toric variety. Let $\Sigma$ be the fan of $X$ with torus orbits $B_\sigma$ for $\sigma \in \Sigma$, then the CSM class of $X$ is given by

$$ c_{SM}(X) = \sum_{\sigma \in \Sigma}[\overline{B}_\sigma] \in A_*(X) $$

where $\overline{B}_\sigma$ are closures of the torus orbits.

Paolo Aluffi has done a lot of work on computing these things. For example (see here) he has constructed a generalization that lives in proChow groups instead of the usual chow groups. The proChow group of a variety $X$ agrees with the Chow group when $X$ is proper and the proCSM class of a not necessarily complete toric variety still satisfies the same formula as above. This also generalizes to other varieties with stratification by smooth locally closed subvarieties but the formula gets more complicated.

In another direction, if we have an embedding $i : X \subset M$ where $M$ is smooth, then we can compute $i_*c_{SM}(X)$ in $M$ by the following formula.

$$ i_*c_{SM}(X) = c(TM) \cap [M] - \pi_*\left(c(\Omega^1_{\tilde{M}}(log X')^\vee)\cap [\widetilde{M}]\right) \in A_*(M) $$

where $\pi : \widetilde{M} \to M$ is an isomorphism away from $X$ and $X' = \pi^{-1}(X)_{red}$ is an snc divisor. See this paper.

When $X$ is itself a hypersurface in $M$ there is a much more computable formula:

$$ c_{SM}(X) = c_F(X) + c(TM) \cap \left(\frac{1}{c(\mathcal{O}(X))}\cdot (s(Y,M)^\vee \otimes \mathcal{O}(X))\right) $$

where $c_F$ is the Chern-Fulton class and $s$ is the Segre class and $Y$ is the singular locus of $X$. Each of the pieces of this formula are in general much more computable then $c_{SM}$ itself in the sense that there are algorithms and Macaulay2 codes for doing this. For more on this aspect and especially techniques and examples of these computations, check out this survey.

As for your second question, it is true in general that the degree of the zero dimensional piece of $c_{SM}(X)$ is the Euler characteristic,

$$ \int c_{SM}(X) = \chi(X). $$

So the answer to your second question is yes.

\https://i.stack.imgur.com/nXwCf.png?s=128&g=1No, since your topology is stronger than the $C^1$ topology... I assume here $M$ is supposed to be a surface.111164216921301578703 Dear Professor Dalawat - thanks for the links! This is addressing the last question I presume? If I understand you, you are saying that the "1-motive" should be an abelian variety over $\mathbf{Z}$ which can't exist. But I was thinking along the following lines: is there some integral "Motivic like" object $[\mathcal{M}]$ for which "$H^1$" returned $\mathbf{Z}/\mathbf{Z}12$. I don't even know if Motives are supposed to form a category with "integral" properties... and maybe this is related to "torsion" automorphic forms... perhaps your remarks merely expose my ignorance, but I like to dream!If you're *maximizing*, I can't see how you'd hit a `-Inf` unless your starting values for your iteration are *really* bad...if $\lambda$, $\mu$, $\nu$ are the three fixed points of a quadratic rational map, and $\sigma_i$ are the symetric elementary functions, the map is $\bar{f} \mapsto (\sigma_(\lambda,\mu,\nu), \sigma_2(\lambda,\mu,\nu)$55911765762019368613234061274675534461049240474491388188Thanks, I see for the case finitely many variables. So you think it is also true for the case infinitely many variables? Could you be more precise for the inclusion. Thanks alot.20298552048470417842b@ArunDebray Yes $M$ is assumed non zero operator828398827108At $n=5$ you say 'so the corresponding abelianisation corresponds to something 2-dimensional geometrically' could you say whether at $n=2^k+1$ where $k\in\Bbb N_{>0}$ this is something $k$-dimensional geometrically and for other $n=2t+1$ where $t\in\Bbb N_{>0}$ and not of form $n=2^k+1$ where $k\in\Bbb N_{>0}$ can we tell anything at all?$TheoremOfWodzicki@Ralph: You are right. In fact, I proved it but I am not sure about it, the graded structure confuses me.1042421According to https://www.intlpress.com/site/pub/files/_fulltext/journals/cag/2001/0009/0002/CAG-2001-0009-0002-a001.pdf

it was observed by Epstein, Gersten, and Mess that any extension of a Fuchsian group by ${\mathbf Z}$ is quasi-isometric to ${\mathbf H}^2 \times {\mathbf R}$, and such extensions are typically not finite extensions of lattices in $Isom({\mathbf H}^2 \times {\mathbf R})$.

Of course that does not yet prove that these groups do not act simultaneously on any other space...

7896271148716Perhaps I am not understanding something. Given a graph which I would like to see as the 1-skeleton of a polytope, if possible, how do I write down the equations for the facets of this polytope?Are we assuming X is chosen from the union of A_i with uniform probability? Even if so, we can only put bounds on the probability, since we don't seem to know the sizes of the intersections between the other sets.193958318878641643743153628910138791260903hSwitch to chat. I feel it is not going to be short.1491940https://lh3.googleusercontent.com/-sPL4bZXU7W4/AAAAAAAAAAI/AAAAAAAAAWg/FOG3peQI48A/photo.jpg?sz=128240828198535772236686375tOr $e^x$. Homology and Morse theory hardly seem relevant.209447522401705167462134637444757Continuing with this Socratic approach, is there a countable $\mathbb{R}$-vector space?85110713484841799244Rum Raisin173819490296I'm selecting this answer because, after graphing $\sqrt{\frac{1}{12t}}$, it appears very close to the answer!7322971594979https://lh5.googleusercontent.com/-s3SX3w8Gzw0/AAAAAAAAAAI/AAAAAAAAAAA/AGDgw-jriHFNBT9-gWeItsUsSQCsQQI2Ig/mo/photo.jpgWrite $H = \mathbb{R}^4$. Call the standard basis elements $\{1,i,j,k\}$. Then $X,L,M$ and $N$ are $4\times 4$ matrices, and therefore can be considered as linear maps $H\to H$.833329You do not even need lex order, just look at this product as a polynomial in, say, $x_1$. It must be linear, thus one factor is linear and other factors are constant. So, each factor $p_i$ is linear with respect to the variables from some set $A_i$, and these $A_i$ must be disjoint. But by symmetry, if $|A_j|=k$, all subsets of size $k$ must appear between $A_i$'s. This all is possible only if $A_i$'s have size 1.`Dimension of Affine Bundles on Projective Space1015090nMinimal steps of construction for constructible number444194|Sorry, I forgot to mention the matrices should be symmetric. Does each discrete solvable group admit an injective homomorphism to a compact topological group?fConsecutive means that O' is infinitely near to O.Paper by Moser on commuting circle diffeomorphisms and simultaneous Diophantine approximations@Dietrich: I think you mean *indecomposable*. Otherwise you also have the abelian one, as well as the direct product of the indecomposable 5-dimensional one by the 1-dimensional one.15205641032926194044920348566199471869644I’m a freelance writer. I write almost all topics .I really loves the freelance writer job. Actually I was born in Belgium. I do all the works in my home itself. And at the present of working college essay writing services. Lot of articles and essay writings has published in online services. Custom essays which is very helpful for them.lts not easy to write essay. It should learn more books and other articles and write.

2285739171779https://www.gravatar.com/avatar/9fe7050a1a202996cdedd62466bb424a?s=128&d=identicon&r=PG&f=1425973Well, that was easy (perhaps I shouldn't ask people interested only in finite-dimensional algebras...)RFilter on $\omega$ or filter in general?17824775783743277079v@ChristianRemling I think you could post this as an answer15311031289798|https://graph.facebook.com/100001075155914/picture?type=large113474622543501722978Thanks Emil!! I was just about to make a more complicated revision ... :)15323918611621725325449135 @Rbega: I assume that $\mathbb{D}$ means the Poincaré disk. Actually, you will have an even worse problem in that you don't have compactness in any sense. For example, $f(z) = \epsilon z$ will have $S(f)\equiv0$, but, as $\epsilon$ goes to zero, $f$ converges to something that is not a linear fractional transformation (and for which $S(f)$ is not even defined). There is a way to 'break' the $\mathrm{SL}(2,\mathbb{C})$ symmetries that cause noncompactness and get a quantitative statement of the kind that you want, but you may not regard the resulting norms as 'natural'.74801921498968956442046509Dear David, you may find the article "Integrable systems, toric degenerations and Okounkov bodies", arXiv:1205.5249, useful.

Maybe I misunderstand the setup, but at least in the case of a finite-dimensional simplicial (or regular CW) complex $X$, wouldn't one be able to reduce everything to a Markov process on the Hasse graph?19459241068724862145zeromorphismI think you are looking at the radial part of a 2D Brownian motion, which is well-known to be a Bessel process. So you are really asking for the expected value of the Bessel process at time t. This is also well-known.One way of finding which is the best journal is look at the number of citations the journal gets.

157224905506lhttp://math.stackexchange.com/users/1284/dan-brumleve392762Thanks for the examples. Would you know if there are physical systems that can implement these different discrete process ? For our device, we are counting the number of electrons so the underlying stochastic process is measuring the total counts \sum_{n \in N} I(n) where I(.) is an indicator function and n is time instant representing the occurrence of an event.6834491350730Wlerin19496281453401Probabilistic heuristics suggest that the set of such Fourier coefficients (or equivalently, the set of non-trivial integer solutions to $m_1^4 + n m_2^4 = m_3^4 + n m_4^4$) is very sparse (only about $O(\log X)$ such solutions up to height $X$). So it is unlikely that analytic methods will be of much help here. Maybe there is some algebraic number theory approach but it doesn't look too promising (e.g. I don't see a norm form or other obviously algebraic structure here).208684415383482110859That's why many disciplines have, in effect, switched to a global language of science - bad english. And those who don't switch, don't get read nor cited anymore.3I am currently reading the famous article "Universal Properties of Maps on an Interval" by Collet, Eckmann and Lanford related to the Feigenbaum-Coullet-Tresser universality. I am in particular interested in Theorem 6.3 page 236 in that article. See the article for the precise statement, but very roughly the theorem says:

Consider a transformation $T$ on an infinite dimensional space which has a hyperbolic fixed point with one unstable direction with eigenvalue $\delta$ and a codimension one stable manifold. Then there is a change of coordinates to a new system (x,y) where:

- the stable manifold is given by $y=0$
- the unstable manifold is given by $x=0$
- the transformation T takes the form $$ (x,y)\longmapsto (M(x,y), \delta \ y) $$ in this new coordinate system.

In other words, this realizes a linearization in the unstable direction only.

I would like to know about similar/related theorems, follow-ups, improvements, etc. that exist in the literature.

Using keyword searches etc. has been quite disappointing and I can definitely use the help of people with expertise in the area. For instance, I did not know the above paper contained such a theorem until a chance discussion with one of the authors.

Edit with some context:

The method used in the CEL article is as follows. They first do some prep work in order to have a coordinate system $(x,y)$ satisfying the first two properties, i.e., such that the stable and unstable manifolds are straight. Then they construct the partial conjugation as $$ z(x,y)=\lim_{n\rightarrow \infty} \delta^{-n} y_n(x,y) $$ where $y_n$ denotes the $y$ coordinate of the $n$-th iterate of the point $(x,y)$ by a suitable cut-off modifcation of $T$.

This is very similar to the construction of wave operators in scattering theory.

The reason I am interested in this is because in recent joint work (see this paper) we proved the following:

Assume $T$ is analytic and has a hyperbolic fixed point $v_{*}$ with only one expanding direction with eigenvalue $\delta$. Then $$ \Psi(v,w)=\lim_{n\rightarrow \infty} T^n(v+\delta^{-n}w) $$ exists and is analytic (jointly in $w$ and the component of $v$ along the stable tangent space used in the analytic parametrization of the stable manifold). Here $v$ belongs to the stable manifold and $w$ is arbitrary but not too big. This function $\Psi$ is not a true linearization, not even a partial one such as the $z$ function of CEL but it shares some of that flavor. Namely, it satisfies the properties:

- $T\circ\Psi(v,w)=\Psi(v,\delta\ w)$.
- $\Psi$ takes its values in the unstable manifold.
- $\Psi(v,w)=\Psi(v_{*},L_{v}(w))$ where $L_v$ is a $v$-dependent linear map onto the unstable tangent space.

The $\Psi$ function can be seen as the $w$ directional derivative of the $z$ function on the stable manifold. It is a "true linearization" on the unstable manifold only. I would like to know if similar results exist in the literature.

1422914Yes, this is correct. Actually, "self-adjoint trace-class" is more than you need; any Hilbert-Schmidt operator can be represented as an integral operator. The Hilbert-Schmidt operators from $L^2(X)$ to $L^2(Y)$ are precisely the integral operators with kernel in $L^2(X\times Y)$ (at least for $\sigma$-finite $X$ and $Y$). This should be in a standard reference, probably Dunford-Schwartz, for example.

I would be surprised if one could prove without using the classification that there is a universal constant $N$ such that for each integer $g$, there are at most $N$ simple groups of order $g$. However, I would be very happy to be proven wrong!3476901998162How do I go about checking if the graph with the given paramaters is a Cayley graph?

2119824Exactly I want to work in the non-DG case ("classical deformation theory" à la Schlessinger) and it seems that to get this theorem, one **has to** go through higher deformation theories (DG Artin rings, coalgebras, simplicial sets...) That's why I'm having such a hard time coming back to the classical case !@Kevin: Could you explain your statement "because any move will reduce the situation to a direct sum of rings of the form $k[x]/(x^n)$"? 2322542043022344948942429The answer is false due to torsion $\chi$, such as for PGL$_2$ (center trivial, hence connected!) with $\chi({\rm{diag}}(x,1)) = (-1)^{{\rm{ord}}_p(x)}$. [EDIT: This is wrong, as the OP notes below.] But if we ignore torsion, which amounts to considering continuous $\chi:T(\mathbf{Q}_p) \rightarrow \mathbf{Q}_p$, then the answer is affirmative. Indeed, this reduces [EDIT: not quite, but see comments below] to an analogous question at the level of the dual space ${\rm{Lie}}(T(\mathbf{Q}_p))^{\ast} = {\rm{X}}(T) \otimes_{\mathbf{Q}} \mathbf{Q}_p$, for which it suffices to show that the subspace of $w$-fixed points for any $w \ne 1$ is contained in some root hyperplane. This in turn is a statement of purely linear algebraic nature, so the problem over $\mathbf{Q}_p$ is equivalent to the one over $\mathbf{Q}$, which in turn is equivalent to the one over $\mathbf{R}$, where we may use considerations with Weyl chambers to conclude.

Any textbook in discrete mathematics. Or try [Wikipedia](https://en.wikipedia.org/wiki/Recurrence_relation) if you only need statements.@Matthew: Yes, my student Francois Lemeux wrote this argument up in a preprint on the Haagerup property for the complex reflection quantum groups. But it only works in the Kac case. I haven't seen any progress in the general case.349651169790436202952278A second order differential equation on a manifold $M$ is a vector field $X$ on $TM$ which is not only a section of the vector bundle $T(T(M)) \to TM $ with the obvious structure, but also is a section of another bundle structure $(T(T(M)), TM, D\pi)$ where $\pi:TM \to M$ is the standard map and $D\pi$ is its differentiation.

I don't think you'll find any non-finitely generated examples; see Theorem 4.3 of http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.ijm/1255637479b@Henry Yes. $N=1$ is standard. What about $N>1$?290247019133351417814713929What is an example of a real analytic vector field $ X$ on $TM$, the tangent bundle of a manifold $M$, such that its set of singularities is a discret set and is topological equivalent to NO second order vector field?

In particular, is there a polynomial vector field $X$ on $\mathbb{R}^2\approx T\mathbb{R}$ such that $X$ has a finite number of singular points(a generic case) but $X$ is not topological equivalent to any vector field in the following form? $$\begin{cases} x'=y \\ y'=g(x,y) \end{cases}$$

Well, this *particular* strategy generalizes for finding the k best horses when the track size is $n = (k-1)(k+2)/2$ and the number of horses is $n^2$, and it takes n+2 races as in your example:

Split them into n groups of size n, race them in those sets, and label as $a_{11}, a_{12}, \dots, a_{1n}, a_{21}, a_{22}, \dots, a_{2n}, \dots, a_{nn}$ as before (so the horse who came in $j$th place in the $i$th race has label $a_{ij}$. Then race $a_{11}, a_{21},\dots,a_{n1}$, and relabel the first subscripts of all horses using the results of this race. The winner of that race is the best horse. To determine the other k-1 best horses, race the n other horses who have fewer than k horses that are better than them (directly or by transitivity): $a_{12}, a_{13}, \dots, a_{1k}, a_{21}, a_{22}, \dots, a_{2(k-1)}, a_{31}, a_{32}, \dots, a_{3(k-2)},\dots, a_{k1}$. (Note here that conveniently $n = (k-1) + (k-1) + (k-2) + (k-3) + \dots + 2 + 1 = (k-1)(k+2)/2$.)

But this still leaves open the question of what to do for other cases.

It might be useful to translate @MikeMiller's comment into the fact that codimension (defined as $n-k$ for a $k$-dimensional submanifold in an $n$-dimensional manifold) is additive under transverse intersection. Heuristically, the codimension of a submanifold can be thought of as the "number of equations" that it satisfies, so the intersection of two (transverse) submanifolds of codimension $m_1$ and $m_2$ should be the solution to a system of $m_1+m_2$ equations (in this heuristic, transversality guarantees that the equations are independent).2152211587720They would correspond to equilateral paths or cycles of size at most $n$, together with a closed walk of length $n$ on this graph (a graph homomorphism from the cycle $C_n$ of length $n$ to the equilateral thing). Or the same thing modulo rotations/reflections of the homomorphed $C_n$ graph.842213318670214348816Measures on Young tableaux31957918372371165781791093XCan the Turing degrees be linearly ordered?I had some trouble with LaTeXing the inequality in the penultimate paragraph. How do you get { and } to show up properly in "math mode"?2191280xThis answer is incredibly deep and well-written, thank you.118416Thank! However I don't understand now what was my mistake in https://wwwsnd.inp.nsk.su/~silagadz/I4.pdf In fact I'm interesting to evaluate (1) from this manuscript in direct way without physics input. Maybe you can tackle this integral too?844454194833669288611693281087907bDecomposing maximal compact subgroups of SO(n,1)2951051945783\https://i.stack.imgur.com/t2CQE.jpg?s=128&g=1fFoliations, von Neumann algebras and measurability14173942076873I accept your answer temporarily for now, but may award to some other people if there are better answers!1764829542613In numerical mathematics there is a recent new grand theme called **randomized numerical linear algebra** (RandNLA). One example (probably even a paradigm) is the "randomized range finder" from the paper "Finding structure with randomness". In a nutshell, you hit a (probably very large) matrix $A$ from the right with a random matrix $\Omega$ (which should be a short matrix) and then use a traditional numerical algorithm to find an orthonormal base of the range of $A\Omega$. By hitting the matrix from the right, one can reduce the number of columns of the matrix and hence, the potential computational effort can be reduced. On the other hand, one loses some dimensions of the range of the matrix but one hopes that, due to the randomization, the most important dimensions are kept.

The general idea is that in most cases the interesting quantities are not such "high-dimensional" as they look at first glance. In the case of a high dimensional range of a matrix $A$, it may be that the "usual element" $Ax$ lives in a space of lower dimension.

Interesting questions in this area include: What guarantees can be given for the output of a randomized algorithm? To what extend does the distribution from which the random object in the algorithm is drawn influence the quality of the output? Under what circumstances does randomization pay off (e.g. in terms of computational effort or storage)?

18050181719892348260Of course, your class probably don't all have this form. But if the relators are all of length four (as is the case here) then there's a good chance that there's some sort of uniform geometric solution to the word problem.2011933rWave front set of vector-valued Dirac delta distribution4336941707007102529Well, one mistake in your proof is that the final inequality in Step 3 only holds for $n$ sufficiently large, where "sufficiently large" depends on $x$.469239Suppose I have some nicely defined "fractal" subset of (to make life simpler) Euclidean space $\mathbb{E}^n,$ of some arbitrary Hausdorff dimension $s,$ such that the corresponding Hausdorff measure $H_s$ has positive mass. The first question is whether it makes sense to talk of a "uniform sample" (or Poisson point process) with respect to $H_s$ (since it is only an outer measure), and second question is: if it makes sense, have people figured out how to generate uniform $H_s$ variates (in simple cases, like the von Koch snowflake, or the more interesting cases, like limit sets of Kleinian groups)?

2248931496703573802706197386593rRenorming a Banach space to make projections contractive2. follows from the Uniformization theorem + louville, right? The holomorphic universal cover is the disk, and so any such $(f,g)$ would need to factor through the disk, which would then be constant by Louville.Btw, the superseeker couldn't figure out the (in hindsight) rather simple rule for the sequence 2, 4, 9, 16, 27, 38.If you just care about multiplicities, you might want to consider restrictions instead of inductions (Frobenius equivalence).69057820080011877090@LisaS. Thank you for catching the mistake. I have now corrected the exponent.20245881853957You’re right! And inflation does not exist on homology. I think this boils down to an issue with the map on group rings $\mathbb{Z}[G]\to\mathbb{Z}[G/H]$. All the other maps use functoriality of (co)homology (in both variables: groups and modules) and Shapiro’s lemma associated to (co)extension of scalars coming from inclusion of H into G.|Thanks. This makes the motivation for the question clearer. 7372301357111068029Yes, with all the same eigenvalues as $\Delta(z)$ (perhaps except for the Hecke operator $T_p$).There's some information on the groups $H^q(S,\mathbb{G}_m)$ in Grothendieck's *Le groupe de Brauer, II* in *Dix Exposés*. Proposition 1.4 says that if $S$ is regular then these groups are torsion for $q \ge 2$. Corollaire 3.2 says that, under suitable finiteness hypotheses, the Kummer sequence gives isomorphisms $H^q(S,\mathbb{G}_m)[\ell^\infty] \cong H^q(S,\mu_{\ell^\infty})$ for $q \ge 3$ and $\ell$ invertible on $S$. These finiteness hypotheses are satisfied, for example, if $S$ is either proper or smooth over a field that is either separably closed or finite.

Using this you can find plenty of examples with $H^3(S,\mathbb{G}_m)=0$, such as any projective space over an algebraically closed field of characteristic zero: the standard calculation of the cohomology of projective space gives $H^3(\mathbb{P}^m,\mu_n)=0$ for all $n$.

I don't think Mathematica is needed. Consider the zero set of $x+x^2y+z^2+t^3$. Note that for $x\ne 0$ we have the unique value for $y$, so this gives $(p-1)p^2$ points ($p-1$ choice for $x$, $p$ choices for $z$, $p$ choices for $t$). For $x=0$, the polynomial becomes $z^2+t^3$, so there are $p$ choices for $y$ and a choice of a zero $(z,t)$ of that polynomial. However, it clealy has $p$ zeros by the usual parametrisation of a singular cubic curve: for $t=0$ there is just $z=0$, and for $t\ne 0$, we have $t=-(z/t)^2$ so denoting $z/t=u$, we have $(p-1)$ solutions $(u^3,-u^2)$.2047630I believe the question is asking for the smallest size of a set $A\subset [0,m]$ such that $[0,m]\subset 2A\cup A$, not $[0,m]\subset 2A$.10060068027111451013jYes. The $\alpha^\vee$'s also define a root system.174079619664481566591+1 for correct answer; this is something that's addressed in basic physics classes, I'm surprised this question is still open...22899981505973295316111391501263868Please read FAQ. This site is for research level questions. Which means, in particular, that BEFORE you ask a question on complex analysis here, you are supposed to take at least an undergraduate course on it.417104I cannot fix the latex for displaying the matrices properly. Someone in charge please help!1198052A numerical check shows that $a^b+b^a$ is unique for $2\leq aTo supplement Angelo's answer, you should be able construct it as analytic space by taking the quotient of the Siegel upper half plane $H_g$ by the subgroup of matrices $M\in Sp_{2g}(\mathbb{Z})$ with first column congruent to $e_1=(1,0,\ldots 0)^T\mod N$. Given $\Omega\in H_g$, its image corresponds to the abelian variety with $N$-torsion point $(\mathbb{C}^g/\mathbb{Z}^g+\Omega\mathbb{Z}^g, \frac{1}{N}e_1)$.

But I think that your first couple of paragraphs give me the answer I want, because I really do want $z$ to be a constant. I had assumed that even in order for $z$ to be a declared constant of type $B$, the type $B$ would have to be a well-formed type in some context---which I guess $A(z)$ is since $z$ is a constant and all, but I had somehow felt that $B$ ought to be well-formed "before" $z$ is declared. But I guess that sort of "before" is not even applicable here.11792852002368https://www.gravatar.com/avatar/a58998b066a7e1d475a7c99924397d83?s=128&d=identicon&r=PG&f=1Semiabelian actions appearing in the toroidal campactification of a degenearting abelian varieties6005111034697198397470748018325772941179157051@MaxAlekseyev Yes, this results are asymptotic. OP din't ask for exact expression, just for nice formula. And existence of exact expression is a big question. The error term here is combined from complicated expressions with Kloosterman sums.521107It is indeed true that when the sequence $(x_d)_{d\mid n}$ comes from a finite simple group $G$, then $G$ is the only group (up to isomorphism) producing that sequence. In fact, $G$ is determined by its order $\lvert G \rvert$ and the set of its element orders, that is, the set of $d$'s such that $x_d \neq 0$. This has been proved using the classification of finite simple groups (Vasilʹev, A. V.; Grechkoseeva, M. A.; Mazurov, V. D.: *Characterization of finite simple groups by spectrum and order*, MR2640961 (2011b:20073)).

Since the finite simple groups are classified, you have in principle a parametrization of the possible "{simple $G$}-solutions", but I doubt if there is any useful number-theoretic characterization of these solutions, other than the parametrization itself. (In fact, I don't know whether there is any number-theoretic characterization of the orders of finite simple groups other than the list coming from the classification itself). Of course, there are various necessary conditions coming from various theorems in group theory. For example, when $(x_d)$ is a $G$-solution, then $(1/d)\sum_{e\mid d} x_e\phi(e)$ is an integer by a theorem of Frobenius, and by a conjecture of Frobenius which has been proved in the meantime (using CFSG), this integer is $\geq 2$ for all proper divisors $d$ of $\lvert G \rvert$ when $G$ is simple.

A famous problem of Thompson (still open, to the best of my knowledge) is whether the sequence $(x_d)$ (equivalently, the *order type* of a group) determines whether this group is solvable (Kourovka notebook, Problem 12.37). In general, it is a difficult problem to characterize solutions $(x_d)$ coming from (certain classes of) groups.

Text Mining And Search Engine Expert In CRCIS

19774974863172954887**The Shoenfield Absoluteness Theorem**

$\Pi^1_2$ and $\Sigma^1_2$ statements are absolute for transitive models of set theory.

Thus, a $\Pi^1_2$ statement is true (in $V$) if and only if it is true in $L$ (the constructible universe). Since $L$ satisfies the Axiom of Choice, every true $\Pi^1_2$ statement is consistent with the Axiom of Choice. This fact is often used to show that assuming the Axiom of Choice is often completely harmless in mathematical practice. Examples of $\Pi^1_2$ theorems include

- The (classical) Brouwer Fixed Point Theorem
- The separable Hanh-Banach Theorem
- Ascoli's Theorem
- The existence of algebraic closures for countable fields
- The existence of prime or maximal ideals for countable rings
- König's Tree Theorem
- Ramsey's Theorem
- The Completeness Theorem for countable first-order languages

and many, many more...

The Shoenfield Absoluteness Theorem is often used in conjunction with forcing. Indeed, if a $\Pi^1_2$ or $\Sigma^1_2$ statement can be forced in an extension, then it must have already been true in the ground model. Other absoluteness results are used in this way too. For example, the original proof of the Baumgartner-Hajnal Theorem (see here and here) combines forcing and the absoluteness of well-founded relations.

Sorry for my typo, but it should be $G_1\cap yG_2\cap zG_3$, and that's quite a difference to $G_1\cap G_2y\cap G_3z$.As far as I know the question of outliers is a very difficult one. You need a model of the process that produces the data, and argue that the underlying process is better behaved than the data, so there is a rare mistake in some outlying point. Given this it is a mixture model to be fit. There are issues about biases introduced through this method, and fitting distributions is very annoying from a statistics perspective due to measure theory issues.

You might try robust statistics to avoid being sensitive to outliers, without the issues of chopping data after looking at it. You could also determine the sensitivity of your procedure to each piece of data, and see if excluding 54 gets significantly different results from excluding other entries.

235222ZWhy limit of discrete series representation?206188219044571943641227113453358827395316240091069702146848241275753Here the subscheme $S$ is a finite coproduct of copies of $\mathrm{Spec}(\mathbb{Z})$, so its Picard group is actually trivial (so lemma $7.3$ is not needed here).`What are typical examples of u.p. (semi)groups?2973232186560502231Visualizing this in my head, for n = 3, 4, 5, 6 the numbers appear to be 16, 19, 24, and 20. To see this (even if my mental picture turns out to be flawed), imagine bending your 2-torus, then adding 3 cubes to get a 3-torus, and the same again for a 4-torus. Adding one more cube can get a 6-torus (the 3x3x3 cube with face and body centers removed), whereas a 5-torus requires adding 5 cubes along the periphery of the 4-torus. OEIS comes up dry on the initial sequence 8, 13, 16, 19, 24, 20, though there are 4 matches for the first 4 terms.Any of the three following references should give you enough background to work this out, though the answer to your question may not be spelled out in detail: [1] "Complex differential geometry" by F. Zheng, [2] "Lectures on Kahler geometry" by A. Moroianu, [3] "Complex analytic and differential geometry" by J-P. Demailly (http://tinyurl.com/2dcnycm).It may be helpful, if you could provide more background about your question. In particular, why do you need p>3 and what's the reason for the condition involving r(n,p) ?Say $k=np$ (imagine $p=1/2$), then that is about $C/\sqrt{n}$ for some constant $C$. This is by (for example) a normal approximation to the underlying binomial random variable, or (perhaps more to the point) recognizing that the binomial random variable has standard deviation about $C\sqrt{n}.$

This ($1/\sqrt{n}$) is the biggest the sum gets. If $|k-np|/\sqrt{n}$ is much larger than 1, then this expression really really quickly falls to $0$. (See many different bounds for this, but perhaps just bound it by the normal approximation.)

You may find useful the bounds $(n/k)^k \leq {n \choose k} \leq (en/k)^k,$ which almost always help clear things like this up considerably.

135468292550hHalin Graphs with Highest Number of Hamilton Cycles840698rCan you include what Atiyah-Hirzebruch gives you so far?jI know this one under the name *Hofmeister's bound*.236395226141121685315115497880971255928Let $X$ be a separable Banach space and $T_1:D(T_1) \subset X \rightarrow X$ and $T_2:D(T_2) \subset X \rightarrow X$ two closed operators with $D(T_2)\subset D(T_1)$ and $D(T_2^*) \subset D(T_1^*).$

We say that $T_1$ is relatively $T_2$ bounded if $D(T_2) \subset D(T_1)$ and for all $x \in D(T_2)$

$$\left\lVert T_1x \right\rVert \le \alpha \left\lVert T_2 x \right\rVert + \beta \left\lVert x \right\rVert.$$

I am interested in the following question:

Are there sufficient conditions such that $T_1^*$ being relatively $T_2^*$ bounded implies that $T_1$ is $T_2$ bounded with the same relative bound?

What about the converse: Does $T_1$ being relatively $T_2$ bounded imply that $T_1^*$ is relatively $T_2^*$ bounded under the above conditions?

163580https://www.gravatar.com/avatar/fe029b87616766431a5b85b5a809bef0?s=128&d=identicon&r=PG&f=185083471256800118557211183841627133Here are two very elementary examples.

It's not 100% different from your Cauchy's inequality example, but the fact that if X is a random variable, then $(\mathbb{E}X)^2\leq\mathbb{E}X^2$ is very useful and follows from the fact that the difference equals the variance of X.

The fact that $|\cos x|\leq 1$ and $|\sin x|\leq 1$ follows from the fact that $\cos^2x+\sin^2x=1$.

1038329Regarding Coret, he is not explicit about his formulation of the Axiom of Infinity. Perhaps none is needed - after all, if no infinite set exists then every set is hereditarily finite, though to deduce this in the sense as posted seemd to require Foundation.1352135x@Eugene Lerman: In what sense? Could you give some details?Is the Bruns-Vetter book ( http://www.home.uni-osnabrueck.de/wbruns/brunsw/detrings.pdf ) of any relevance?131352513963020264451799111999187Thank you for your comment. As far as I know, the concept of slices only appears in the context of Tychonoff spaces throughout the literature (Bredon, Palais, tom Dieck …). One of the key steps in Bredon's result is the existence of slices for actions on a Tychonoff space. Manifolds are Tychonoff (no smoothness required), so, unfortunately, your condition does not add anything to my list of known results above.10018831189438181298550823421628041181219rCould you please clarify what you mean by $\|\cdot\|_2$?Yes. Joe Roberts book on Elementary Number Theory has some. There is also Carmichael's lambda function. You may find some more with a well crafted search term and an Internet search. Gerhard "Ask Me About System Design" Paseman, 2012.01.13Thank you! I am familiar with model categories and the modes structure on simplicial sets, but didn't think to look at it that was.145298414772061698364|https://graph.facebook.com/459314984586524/picture?type=largelhttp://research.microsoft.com/en-us/um/people/mohits/60072Maybe it would have been clearer to use different symbols, but yes, I'm asking about a fixed probability distribution that has compact support, and letting the barriers go to infinity.7089732266346It depends on your semiring. In some cases, semirings do not obey the distributive law, and adding elements will not fix it, and often quotients also will not. Gerhard "Ask Me About System Design" Paseman, 2012.03.07Dear Felipe, as long as we restrict ourselves to elliptic curves over $\mathbb{Q}$, the only results known in this direction are that the 2, 3, 5, 7 and 13 primary torsion of Tate-Shafarevich groups can be arbitrarily large. These are due to Kramer, Cassels, Fisher, Matsuno and Matsuno, respectively. It is still unknown whether it is true that for any $p$, the $p$ primary torsion of sha can be arbitrarily large. It is not even known whether the $p$-torsion of sha can be non-trivial for arbitrary $p$.

As far as I know, almost everybody nowadays believes that ranks of elliptic curves over $\mathbb{Q}$ can be arbitrarily large, but when pressed for evidence, most people point to the function field case. In fact, it is not even known, whether the rank can get arbitrarily large over number fields of uniformly bounded degree.

151033582165019604496216691991728Let $u\in C^\infty(\mathbb{R}^n\times\mathbb{R}^n)$ be symmetric and of strictly positive type on some hypersurface $S \subset \mathbb{R}^n$ diffeomorphic to $\{0\}\times\mathbb{R}^{n-1}$. This means there is some $c>0$, s.t. $$\forall N>0, \enspace x_1,...,x_N \in S,\enspace v\in\mathbb{R}^N: \quad \sum_{i=1}^N \sum_{j=1}^N u(x_i,x_j)v_iv_j \geq c\left(\sum_{i=1}^N v_i\right)^2. \quad (*)$$ Of course this is satisfied for $u$ being uniformly bounded from below $u\geq c>0$. I would guess ($*$) is merely necessary for boundedness from below but only sufficient on the diagonal ($u(x,x)\geq c$ f.a. $x\in S$) and not on whole $S\times S$. Now I have two questions concerning this kind of functions:

Is there some symmetric and smooth $u$ fulfilling ($*$), that is not bounded from below?

Assuming ($*$) is it possible to show positivity on some open neighborhood $O$ of $S$, i.e. $$\forall N>0, \enspace z_1,...,z_N \in O,\enspace v\in\mathbb{R}^N: \quad \sum_{i=1}^N \sum_{j=1}^N u(z_i,z_j)v_iv_j \geq 0$$ or are there maybe counterexamples?

Thanks a lot!

13764243506641897687465702This is wonderful confirmation of the start @$ 90^\circ, \Omega=0.59\pi \approx 1.85$ to @$0^\circ, \Omega = \pi$. Great that you could compute the solid angle explicitly!If the motivation is to use spectral methods, then there is no need to interpret things as a graph. To count walks in a graph one gets a recurrence relation (given by the adjacency matrix) and proceeds using spectral methods. If $A$ is an $n \times n$ matrix and $x_0$ a column vector then setting $x_{m+1}=Ax_m$ leads to a system of $n$ first order linear recurrences in the $n$ positions. In most non-degenerate cases one can get a single $nth$ order recurrence for a particular entry (or linear combination of entries). Similarly for the $r,c$ entry of $A^m$, the trace and other linear combinations. Really one is studying the entries of $A^m$ since $A^mx_0=x_m$. In the case that one starts with an $nth$ order linear recurrence one just has a rather special kind of graph.

That said, if the entries of $A$ are non-negative integers then one can naturally interpret $A^m$ as counting length $m$ walks in a certain $n$-vertex directed graph with multiple edges and loops allowed. For an arbitrary $n \times n$ matrix with entries from a ring one could consider the entries as multiplicitive edge weights on the complete $n$-vertex directed graph (with loops) and $A^m$ as recording in position $u,v$ the total weight of the length $m$ paths starting at $u$ and ending at $v$.

But again, once one enjoys the fact that every linear recurrence can be interpreted as a weighted path enumeration problem, there is usual not much motivation to actually do so. I suppose that the initial conditions are not really accounted for in this sketch, but they don't really come into solving recurrence relations until the very end.

2269870I think this is an important point. The story (i.e., motivation) behind a piece of mathematics is very important but usually easy to remember. What is hard to remember are the rigorous details. So we want a book to remind us of the hard details. A book that is too wordy is in the long run just something that is too big and heavy to keep around. Now that we have the web, maybe that's a better place to maintain and organize the stories?1881896821362049289Simon: With this "grading", $(x^2-x)x$ has degree 3, which means its square should have degree 6. But in fact the square is $(x^2-x)(x^4-x^3)$, which is not even homogeneous. 892397You may want to see this paper as well by Berndt:

1512177Equivalently, let $X = \left\{n \geq 1: \frac{x_{mn}}{mn} \geq \frac{1}{m}\right\}$.1582583 I don’t know of a reference, but since this is an easy basic result, it should be somewhere. One way to prove this is to use the following variant of the ordered Mostowski model (this is a permutation model, i.e., a model of ZFA, but it can be made a model of ZF using the Jech–Sochor embedding theorem): assume that the set $A$ of atoms is in 1–1 correspondence with $[0,1]$, and consider the permutation model $M$ determined by the group of order-preserving permutations of $A$ and the normal filter of its subgroups with finite support. Then $M$ contains $A$ and its order, which induces ...1670502ZProving $\omega_1$ is inaccessible in $L(r)$525861156272bI've made edits based on the two comments above.1513209322678286247111459520026881063679The space of stochastic maps on a finite probability space is a convex polytope. So you can choose one using standard Euclidean measure.A ternary self-distributive algebra is an algebra $(X,t)$ that satisfies the identity $$t(u,v,t(x,y,z))=t(t(u,v,x),t(u,v,y),t(u,v,z)).$$

Is the equational theory of the variety of ternary self-distributive algebras decidable? If so, then when are two terms equivalent in the variety of ternary self-distributive algebras. It may be easier to prove the decidability of this variety if one assumes strong large cardinal hypotheses.

\https://i.stack.imgur.com/NkO9k.jpg?s=128&g=1437013657070I guess there's nothing irrational about calling a function that's not rational "irrational," but it seems unusual.IlanaThis question has been answered by Victor Ostrik in the comments.

Ah, that's right! I was being cavalier as usual. I think the existence of $u_s$ should force it to actually lie in $Hom(A_{\eta},B_{\eta})$.2121862158226619993292096211262737Doug Fir2246345https://lh4.googleusercontent.com/-w_RKA9Ee5g0/AAAAAAAAAAI/AAAAAAAAVmI/W3-6ECRJVZs/photo.jpg?sz=128Let $F$ be a non Archimedean local field with ring of integers $\mathcal{O}_F$. Let $P$ be a standard proper parabolic subgroup of $GL_n(F)$. Let $M$ be the standard Levi subgroup of $P$ and $\sigma$ a supercuspidal representation of $M$ and $N$ be the unipotent radical of $P$. Let $s=[M,\sigma]$ be a Bernstein component of $GL_n(F)$. Is it possible to chose a semi-simple type $(J_s,\lambda_s)$ for $s$ such that $J_s\cap N=N(\mathcal{O}_F)$?

22247692201495739322In the case that $H$ is time-independent (so independent of $x$), I think your formula is the same as Theorem 2.19 in section IX.2.6 in Kato's _Perturbation theory_.2100514In logic, the terminology seems to have been influenced by two factors. The very early development of various deductive systems was done by people who were more philosophers than mathematicians and who seem to have used "calculus" to refer to anything that looked mathematical. Also, that development took place before "algebra" had acquired all of its current meanings.

My impression is that the use of "calculus" in logic is restricted to the meaning of "formal deductive system" --- and usually rather old systems. As for the SKI system of combinators, I would call it a calculus if you're talking about rules of inference. But if you mean the system of all combinators, with the operation of application, generated by S and K (I is redundant), then this is an algebra.

dRoots of lacunary polynomials over a finite field@Pasten Thanks for pointing out that the conjecture is well-known . My formulation looks different but no doubt is equivalent to Dressler's conjecture. I think the way I wrote the problem in terms of sets made it difficult to find references or literature already published pertaining to the problem.I recently came upon this condition for a vector field X on M to be affine $$[L_{X},\nabla_{Y}]-\nabla_{[X,Y]}=0$$ Operating this on another vector Z gives me ${\nabla}^{2}_{Y,Z}X+R(X,Y)Z=0$

Where $\nabla^{2}_{Y,Z}X=\nabla_Y \nabla_Z X-\nabla_{\nabla_YZ}X$. Now if I choose $Y=Z=T$ some tangent vector to a geodesics in M then I get back the geodesics deviation or Jacobi equation. I know that affine transformations maps geodesics to geodesics hence the deviation vector is also an affine vector field. Is this observation correct?. Further I have seen this affine condition also written as $$(L_X\nabla)_{Y}Z=\lim_{s\to 0}{1\over s}\left((\phi^{*}_s\nabla)_Y Z-\nabla_Y Z\right)=0$$ If I break this down I get $$\lim_{s\to 0}\lim_{t\to 0}{1\over s}{1\over t}\left((\phi^{-1}_{*s}{\bar{\tau}}^{t}_{0}\phi_{*s})Z_{x_t}-\tau^{t}_{0}Z_{x_t}\right)=0$$ Where $t$ is the parameter along the curve whose tangent is $Y$ and similarly $s$ is along $X$. $\phi_s$ are the flow maps along $X$ and ${\tau}$ are the parallel transport map. The above condition is equivalent to the following condition $${\bar{\tau}}^{t}_{0}\phi_{*s}=\phi_{*s}\tau^{t}_{0}$$ i.e, they commute. Further I have noticed that the transformations $C_t=\tau^{t}_{0}\phi_{*t}$ is a local $1$ parameter group of linear transformations of $T_x(M)$ from this one can see that there is a linear endomorphism A of $T_x(M)$ such that $C_t=\exp{-t(A_X)_x}$. Using this one can again check that $A_X=L_X-\nabla_X$ and one can write the affine condition as $$(\nabla_Y A_X)Z=R(X,Y)Z$$ I have tried to summarize here what I have read but somehow I am not able to make a concise understanding of what is going on?. What is the connection between $(A_X)$ and affine transformations and how is this whole thing connected to geodesic deviations?. Can anyone summarize this in a proper clear order?. Most of this material can be found in "Foundations of Differential geometry volume one" by Kobayashi and Nomizu. Please forgive me if the arguments are not clear as I have just started studying this topic.

194968159242lIf you don't mind it would be appreciated. Thank you.1088909Do you want a pair $(G,v)$, or a single $G$ that works when any of its vertices $v$ is deleted?908580@YemonChoi: I agree. Anyway, I’m trying hard not to rely too much on knowledge about $ C^{*} $-unitizations.1925721!One can certainly make the basic definitions, and the real issue is to show that the definition "works" using any etale map from a scheme. More precisely, the real work is to show that a weaker definition actually gives a good notion: rather than assume representability of the diagonal, it suffices that $R := U \times_X U$ is a scheme for *some* scheme $U$ equipped with an etale representable map $U \rightarrow X$. Or put in other terms, we have to show that if $U$ is any scheme and $R \subset U \times U$ is an 'etale subsheaf which is an etale equivalence relation then the quotient sheaf $U/R$ for the big etale site actually has diagonal representable in schemes. Indeed, sometimes we want to *construct* an algebraic space as simply a $U/R$, so we don't want to have to check "by hand" the representability of the diagonal each time.

That being done, then the question is: which objects give rise to a theory with nice theorems? For example, can we always define an associated topological space whose generic points and so forth give good notions of connectedness, open behavior with respect to fppf maps, etc.? (The definition of "point" needs to be modified from what Knutson does, though equivalent to his definition in the q-s case.) The truth is that once the theory is shown to "make sense" without q-s, it turns out that to prove interesting results one has to assume something extra, the simplest version being the following weaker version of q-s: $X$ is "Zariski-locally q-s" in the sense that $X$ is covered by "open subspaces" which are themselves q-s. This is satisfied for all schemes, which is nice. (There are other variants as well.)

In the stacks project of deJong, as well as the appendix to my paper with Lieblich and Olsson on Nagata compactification for algebraic spaces, some of the weird surprises coming out of the general case (allowing objects not Zariski-locally q-s) are presented. (In that appendix we also explain why the weaker definition given above actually implies the stronger definition as in Chris' answer. This was surely known to certain experts for a long time.)

12484131282643I'm a master student of science and technology - applied physics and mathematics with discipline in machine learning and statistics.

2149996@André: Could you explain the definition of the attaching map? Your function is not periodic.zIf Lie algebras are far away your area, why do you need it ?12684008858164771361690965You can cook up lots of normal subgroups by looking at any characteristic subgroup of the free group. For example, if $F^{(k)}$ is the $k$th term of the lower central series, there is a surjection $$Out(F_n)\twoheadrightarrow Out(F_n/F_n^{(k)}).$$ The kernel of this surjection is an interesting normal subgroup. Oscar's construction is a special case of this.

(Edit: I misread "perfect" as "simple," so this is not an answer to the question.)

2514391647886326357Jonathan Lubin's "Torsion in the Nottingham group" discusses the structure of finite order $\sigma$.57613415168901468919The puzzle in your answer is different from the one on the page you linked to. So I solved it: http://i.imgur.com/jTbwfnm.png7168721149665342745The post of GH from MO reminds me of oscillation results of Chebyshev's first function (log of primordial), so I now no longer believe there are finitely many exceptions. I suspect only finitely many of them are not primorials, however. Gerhard "Always Have A Fallback Position" Paseman, 2017.07.22.1769925256201~But what if $v$ is not an eigenvector? I intend it to be free.1454166+1 for self-reflection and enthusiastic punctuation. I think you're being a little hard on yourself.328432702242Does this the sequence go to zero?

$\Pi_{n=1}^{N}\text{sin}(2\pi n\omega)$ as N $\rightarrow \infty$ for any $\omega \in (0,1)?$

I can see this sequence is always decreasing for general $\omega$. And for some specific $\omega$ (in $\mathbb{Q}$ I believe), sin($2\pi n_0\omega$) becomes 0 for some specific $n_0$ so making the sequence to be 0 there after. But how to show such sequence goes to zero for general $\omega \in (0,1)$ though?

1408989@Will That sheafification is exact is proved [here](http://stacks.math.columbia.edu/tag/00WJ) in the Stacks Project. (Since links in comments are hard to read - it's here: http://stacks.math.columbia.edu/tag/00WJ) $r$-numbers by real numbers in all claims and be free in their reasoning). But if one's choice of set $A$ is merely taken because of convenience then we have some trouble, because we can't replace $r$-numbers by real numbers in our claims. In this case, we can still freely reason when talking "only" about $r$-numbers. But when we do try to reason about real numbers we will have to be quite careful (and probably very conservative) due to possibility that $\mathbb{R} \supset R$. Secondly clearly there must be some countable ordinal $\alpha$ whose well-order relation isn't contained in $A$ (cont.8276087638741737896h@LSpice You are right. It should be "a". Thank you.1879698The case for odd primes being slightly more complicated to write down due to the Bockstein, we deal with the case $p=2$ here, although the odd prime cases work essentially in the same way. Since it is a Hopf algebra, we have $$\Delta (x\cdot y) =\Delta (x)\cdot \Delta (y).$$ Thus it suffices to know the coproduct on algebra generators. However, the algebra generators are of the form $Q^{I_1}Q^{I_2}\cdots Q^{I_r}([1])$, and $[-1]$. The coproduct on $[\pm1]$ is given by $\Delta ([\pm 1])=[\pm 1]\otimes [\pm 1]$, and the repeated use of Cartan formula gives $$\Delta (Q^{I_1}Q^{I_2}\cdots Q^{I_r}([1]))=\Sigma _{J_1,J_2,\ldots J_r} Q^{J_1}Q^{J_2}\cdots Q^{J_r}([1])\otimes Q^{I_1-J_1}Q^{I_2-J_2}\cdots Q^{I_r-J_r}([1]).$$

What is the cardinality of $U$? Is it $n$? Or greater than $n$?542047boy! if there's one thing I would like more than anything else, it would be to read that paper by Grothendieck. Poincare also liked to study the mind (his mind!). As someone who thinks about the craft more than one practices it (but hardly a philosopher!) I would say one of the greatest services a mathematician can do besides mathematics is to analyze their own mind. See also for example the thread here: https://mathoverflow.net/questions/38639/thinking-and-explaining1840983There was a Gibbs lecture "Integrable Systems: A modern View": http://jointmathematicsmeetings.org/meetings/national/jmm/deift528189+1. For the converse, if we know that a program runs in polynomial time then we can use an oracle for the halting problem to find a polynomial upper bound. Just check all the polynomials, one after another, until you find one that is an upper bound. The property that a given total program does not run in a fixed time bound is Sigma^0_1 and so it can be answered by a query to the halting problem. This shows that the lower bound you give is sharp. (P.S. I doubt that the Turing reduction I describe here could be improved to a stronger reduction.)11995461656108180870.. resonate as something fundamental and deep about natural numbers, while modular arithmetic and quadratic residues would probably seem as something interesting, but not very deep or general.2251138@Robin: Right I saw potential problems only because for some strange reason I had confused what was subgroup of what.|https://graph.facebook.com/846640465535896/picture?type=large5144571614923https://lh3.googleusercontent.com/-fzBkFpaRF8M/AAAAAAAAAAI/AAAAAAAAAAA/AGDgw-h0zL2Uf0cmsWWzIbfuauFJDdNcvQ/mo/photo.jpgOne should add that the result for sofic groups was first shown by Elek and Szabo in "Sofic groups and direct finiteness", Journal of Algebra Vol 280, Issue 202012-10-12T13:40:53.707194622Thank you, Igor! Is "Closed geodesics on surfaces" (M. Freedman, J. Hass and P. Scott) the reference for this? I don't have access to that paper.23031701724207XSquare root for Hamiltonian diffeomorphismshttps://www.gravatar.com/avatar/53dc2d3cb253ee39eea8a81ba2d2fdd0?s=128&d=identicon&r=PG&f=1I personally feel that almost every sentence in this solution could be improved by including the justifications for claims made, starting with the sentence beginning "Clearly...". I hope that Anton will choose to expand it!1681384512491074881There's something about the notation you should know before you get confused when trying to do non-abelian gauge theory. The second term in the field strength should involve a combination of the wedge product of forms and the Lie bracket: the field strength (in the case of an arbitrary gauge group $G$ with Lie algebra $\mathfrak{g}$) should actually be $$F = dA + \tfrac{1}{2}[A \wedge A],$$ where if $\omega$ is a $\mathfrak{g}$-valued $k$-form and $\eta$ is a $\mathfrak{g}$-valued $p$-form, then $$[\omega \wedge \eta](X_1, \dots, X_{k+p}) = \sum_{\sigma \in S_{k+p}} (-1)^{\text{sgn}(\sigma)} [\omega(X_{\sigma(1)}, \dots, X_{\sigma(k)}), \eta(X_{\sigma(k+1)}, \dots, X_{\sigma(k+p)})]$$ for any $k + p$ vector fields $X_1, \dots, X_{k+p}$. In particular, if $A$ is a $\mathfrak{g}$-valued $1$-form, then $$[A \wedge A](X_1, X_2) = [A(X_1), A(X_2)] - [A(X_2), A(X_1)] = 2[A(X_1), A(X_2)].$$ So in components, the field strength is given by $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu, A_\nu],$$ which is the form you'll see most frequently in the physics literature.

When the gauge group $G$ is abelian (*e.g.* in ${\rm U}(1)$ gauge theory), the Lie bracket on $\mathfrak{g}$ is trivial so that $[A \wedge A] \equiv 0$ and the field strength is just the exterior derivative of the gauge potential: $F = dA$.

I'd like to know the largest almost quasisimple group $G$ which acts faithfully and irreducibly on the spin module for an odd dimensional orthogonal group $\Omega(2k+1,q)$, or on each of the two half-spin modules for an even dimensional orthogonal group $\Omega^\epsilon(2k,q)$.

I've convinced myself using the weights of the representation in the corresponding algebraic group that $G/Z(G)$ contains no field or graph automorphisms.

I'm also half convinced that the correct answer is the Clifford group or the spin group $\mathrm{Spin}^\epsilon(n,q)$, but I'm not sure.

Any help or guidance would be much appreciated!

573958359930Thank you for your patient!!! I believe I understand it now: any Cartier divisor on $X$ can be extends to a Cartier divisor on $P$ (or any meromorphic function on $X$ can be extends to a meromorphic function on $P$).18304981438990Sorry, where did you repeatedly say this? (And please: don't get testy -- I am trying to understand.) Anyway, the question as stated is explicitly about Piaget's postulation as you call it, so I must stand by my previous comment. Your last comment also refers to Piaget's famous "structures", right?386161474167161636121332791308609577728661525$I am trying to compute thousands of integrals of the below type, that comes up in a conformal mapping problem, to as many accurate digits as possible (preferably 50+):

$$ \int_{-1}^1\textrm{d}t \frac{\mathcal{Re}\{\log[(\cos{(\pi/130)} - t)]\}}{\sqrt{1 - t^2}} $$

The results from PARI/GP, Sage and Python's mpmath library respectively are:

```
-2.1770705767584673426214016567105099553,
(-2.1775860588840983, 1.2746272565903925e-05),
-2.1774410877577223893132496923831596284
```

Clearly $-2.177$ is correct, but what's the best way to find a more accurate answer, accurate to 50+ digits?

I've tried splitting intervals from $[-1, \tau] \cup [\tau, 1]$, where $\tau = \cos{(\pi/130)}$; that doesn't improve things but actually makes it worse. I am working at much higher decimal precision than 3.

**UPDATE**: Following the suggestion of IgorRivin, the below is the Mathematica attempt:

```
tau = N[Cos[Pi/130], 50];
epsilon = 2^-10;
limit = N[ArcCos[1 - epsilon], 50];
T1 = NIntegrate[Re[Log[(tau - t)]]/(Sqrt[1 - t^2]), {t, -1, 1 - epsilon},
WorkingPrecision -> 50, AccuracyGoal -> 50]
T2 = NIntegrate[Re[Log[tau - Cos[theta]]], {theta, 0, limit},
WorkingPrecision -> 50, AccuracyGoal -> 50]
answer = T1 + T2
```

gives the results:

```
Out[1]= -1.7968036050143567231750633164742621583459497767361
During evaluation of In[881]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed
accuracy after 9 recursive bisections in theta near
{theta} = {0.024170580099064656300274166159163078224443688377727}.
NIntegrate obtained -0.38078136551592029350719560689304843811203900465893
and 6.9050153103011684224977203192777088512613553501207`50.*^-7 for the
integral and error estimates. >>
Out[2]= -0.38078136551592029350719560689304843811203900465893
Out[3]= -2.1775849705302770166822589233673105964579887813950
```

As may be seen the error is still in the seventh decimal place; increasing the working precision or accuracy does not really improve things.

**UPDATE**: The integral is exactly soluble; the result is given in my comment to the accepted answer of Emil Jeřábek, and compares well to the exact value.

Entrepreneur, physics enthusiast, excited to build stuff and learn new things with great people.

fSubstructural types, the lambda calculus, and CCCs14172626774229224697781476387gionniIs every functor inducing a homotopy equivalence a composition of adjoint functors?2222646-*This question was the motivation behind an earlier question of mine; having thought about it some more, that question seems nontrivial and the connection is actually pretty tenuous anyways, so I've decided to ask this directly:*

For $\mathcal{K}\subseteq\omega_1$, we let $G_{CD}(\mathcal{K})$ be the game defined as follows *(which is a special case of a broad class of games defined by Montalban)*: players $1$ and $2$ alternate building objects by finite extensions (with passing allowed) - in player $1$'s case, a single linear order $L^1$, and in player $2$'s case an array of linear orders $L^2_i$ ($i\in\omega$). Player $2$ wins iff $(i)$ each $L_i^2$ is in $\mathcal{K}$, and $(ii)$ if $L^1\in\mathcal{K}$ then $L^1\cong L^2_i$ for some $i\in\omega$. Put another way, player $1$ is trying to *either* build an element of $\mathcal{K}$ which player $2$ doesn't build, *or* trick player $2$ into building something not in $\mathcal{K}$.

If $\mathcal{K}$ is unbounded, then $G_{CD}(\mathcal{K})$ can't be a win for player $2$ by $\Sigma^1_1$-bounding. Conversely, if CH holds it's easy to build an unbounded $\mathcal{K}\subseteq\omega_1$ such that $G_{CD}(\mathcal{K})$ isn't won by player $1$ either:

Fix an enumeration $\{\Sigma_\alpha:\alpha<\omega_1\}$ of all strategies for player $2$.

Let $\mathcal{K}_0=\emptyset$ and let $\mathcal{K}_\lambda=\bigcup_{\alpha<\lambda} \mathcal{K}_\alpha$ for $\lambda$ limit.

For successor stages, having constructed $\mathcal{K}_\alpha$ for $\alpha\in\omega_1$ we fix some enumeration $\Pi_\alpha$ of $\mathcal{K}_\alpha$; we then let $\mathcal{K}_{\alpha+1}=\mathcal{K}_\alpha\cup\{\omega_1^{\Sigma_\alpha\oplus\Pi_\alpha}\}$.

Now let $\mathcal{K}=\bigcup_{\alpha<\omega_1}\mathcal{K}_\alpha$. For each $\alpha\in\omega_1$, the strategy for player $1$ which simply enumerates $\mathcal{K}_\alpha$ according to $\Pi_\alpha$ beats $\Sigma_\alpha$ in $G_{CD}(\mathcal{K})$. At the same time, $\mathcal{K}$ is unbounded in $\omega_1$. So $G_{CD}(\mathcal{K})$ is undetermined.

However, in lieu of CH things seem much harder. Intuitively, there are continuum-many strategies but only $\omega_1$-many "choices" involved in creating $\mathcal{K}$. "Obviously" ZFC alone should prove that there ar many $\mathcal{K}\subseteq\omega_1$ such that $G_{CD}(\mathcal{K})$ is undetermined, but I don't see how to do that. So my question is:

670322Does ZFC

aloneprove that there is some $\mathcal{K}\subseteq\omega_1$ such that $G_{CD}(\mathcal{K})$ is undetermined?

It's fairly easy to do this for finite groups. In fact, the functor $R \mapsto R[G]$ is naturally representable by a ring scheme: the underlying set functor is represented by $\mathbb A^n$ where $n = |G|$, and the ring structure comes from the functor of points $R \mapsto R[G]$. Write $Y$ for this ring scheme (say over $\operatorname{Spec} \mathbb Z$).

Now the unit group can be constructed as the closed subset $V \subseteq Y \times Y$ of pairs $(x,y)$ such that $xy = 1$. It is closed because it is the pullback of the diagram $$\begin{array}{ccc}V & \to & Y \times Y\\\downarrow & & \downarrow \\ 1 & \hookrightarrow & Y\end{array},$$ where the right vertical map is the multiplication morphism on $Y$. This shows that $R \mapsto R[G]^\times$ is representable. It naturally becomes a group scheme, again by the functor of points point of view. $\square$

In the infinite case, this construction doesn't work, because the functor $R \mapsto R[G]$ is not represented by $\mathbb A^G$ (the latter represents the infinite direct product $R \mapsto R^G$, not the direct sum $R \mapsto R^{(G)}$). I have no idea whether the functor $R \mapsto R^{(G)}$ (equivalently, the sheaf $\mathcal O^{(G)}$) is representable, but I think it might not be.

On the other hand, in the example you give of $G = \mathbb Z$, the functor on fields $$K \mapsto K[x,x^{-1}]^\times = K^\times \times \mathbb Z$$ is representable by $\coprod_{i \in \mathbb Z} \mathbb G_m$, but this does not represent the functor $R \mapsto R[x,x^{-1}]^\times$ on rings for multiple reasons. Indeed, it is no longer true that $R[x,x^{-1}]^\times = R^\times \times \mathbb Z$ if $R$ is non-reduced, nor does $\coprod \mathbb G_m$ represent $R \mapsto R^\times \times \mathbb Z$ if $\operatorname{Spec} R$ is disconnected. These problems do not cancel out, as can already be seen by taking $R = k[\varepsilon]/(\varepsilon^2)$.

47157117465932625311709547I think this is still a perfectly fine question, and should remain open.Hi Joël, I'm having trouble with the toy model. If the Hecke eigenvalues of the forms $e_1$ and $e_2$ agree modulo $p$, then the forms themselves must agree modulo $p$. But then $(e_1-e_2)/p$ has integral coefficients, so should also belong to the module $M$.10579061756226If $n=18m+1$ then $(10^n-1)/9$ is $1 \bmod 7$, which is a cube.648441772884So F(X), besides being a container of X's, contains a rule for picking out an element of Y from F*(Y) and vice versa. So F(X) (or even just F(1)) acts like a pointer into the data structure F*(Y), and vice versa. This is really neat! Pointers into structures are usually thought of in terms of derivatives of functors so this is a nice twist.2261371I am trying to understand theorem 7.7.1 in Hormander's Analysis of linear partial differential operators, vol.1.

Let $K \subset \mathbb{R}^n$ be a compact set, $X$ an open neighborhood of $K$ and $j, k$ non-negative integers. If $u \in C^k_0(K), f \in C^{k+1}(X)$ and $\text{Im} f \geq 0$ in $X$, then

$\omega^{j+k}\lvert\int u(x)(\text{Im} f(x))^je^{i \omega f(x)}dx\rvert \leq C\sum_{|\alpha|\leq k} \text{sup}|D^{\alpha}u|(|f'|^2+\text{Im}f)^{|\alpha|/2-k}, \quad \omega>0$

Here $C$ is bounded when $f$ stays in a bounded set in $C^{k+1}(X)$.

It's the "$f$ stays in a bounded set" part that is unclear. The way I understand it is that if $f$ depends on a parameter and $f$ and its derivatives are *uniformly* bounded then $C$ is bounded. Is that correct?

If it is, then I think this ($f$ and its $k+1$ derivatives are uniformly bounded) is *only* a sufficient condition for $C$ to be bounded. What is the necessary condition for $C$ to be bounded?

First, anything that is proved using the Frobenius theorem can also be proved using the existence and uniqueness theorem for ODE's and the fact that partials commute. The theorem is used in differential geometry to show that local geometric assumptions imply global ones. Here are a few examples that come to mind:

1) If a submanifold of $R^n$ has zero second fundamental form, it is an affine subspace of $R^n$. A similar statement holds if the second fundamental form is a constant nonzero multiple of the metric.

2) More generally, if you have an abstract Riemannian $n$-manifold with a symmetric $2$-tensor that satisfies both the Gauss and Codazzi-Mainardi equations, then it determines an isometric immersion of the manifold into $(n+1)$-dimensional Euclidean space unique up to rigid motion. In an elementary course, you might present the $2$-dimensional case.

3) A Riemannian manifold with vanishing sectional curvature is isometric to the standard flat metric on $R^n$. Again, the $2$-dimensional case can be presented in an elementary course.

4) A Riemannian manifold with constant sectional curvature $1$ ($-1$) is locally isometric to $S^n$ ($H^n$).

Whether you prefer to use ODE's or Frobenius when proving these statements depends on how you set everything up. In particular, if you set everything up using differential forms ("moving frames"), then it is natural to use Frobenius. If you set things up in co-ordinates, it is perhaps more direct to use ODE's instead.

433806You are absolutely right. Indeed, within Lehmer's paper, it's only in the title that an umlaut is used, within the text proper it's always "Størmer". Apologies to Wiki and anybody else I might have inadvertently offended! :-)Correction: "Are you assuming that $N_1$ and $N_2$ are conjugate subgroups of $G$?" --> "Are you assuming that $H_1$ and $H_2$ are conjugate subgroups of $G$?"13849712000381582873401222Can you explain how we guarantee that if one of them is finite dimensional then the projective tensor product space is complete?2007191180229132429tHow are constants/functions named after their discoverer?2172628697676dhttp://www.math.rochester.edu/people/grads/yzou7/638823153466267017839327@AbdelmalekAbdesselam : having a "good" set of generators for such a ring (or only a system of parameters for such a ring) would allow one to solve the isomorphism problem for bipartite graphs (which is as hard as for general graphs -- cf. e.g. https://en.wikipedia.org/wiki/Graph_isomorphism_problem#GI-complete_and_GI-hard_problems). How a similar idea works for general graphs, is explained in the book I cited in the answer.6486732958980XSimilarly, depends on the mathematician....gI will have another go at arguing that dagger-categories are *not* evil.

Let’s look at a simpler case first. Consider the property “$1 \in X$” on sets. As a property of abstract sets, this is evil: it’s not invariant under isomorphism, e.g. any iso $\{1,2\} \cong \{2,3\}$.

But it is manifestly non-evil as a property of, say, “sets equipped with an injection to $\mathbb{N}$”. Looking at this gives a non-evil structure on abstract sets: “an injection $i : X \to \mathbb{N}$, such that $1 \in \mathrm{im}\ i$”.

Ah (you may say) so if this is non-evil, it can’t reflect our original idea correctly: it would have to transfer along that $\{1,2\} \cong \{2,3\}$. But that’s not such a clear-cut complaint. As a structure on abstract sets, it does transfer along that isomorphism. But we were thinking of them from the start not just as abstract sets, but as subsets of $\mathbb{N}$, i.e. as *already* equipped implicitly with injections to $\mathbb{N}$. Considered as such, one of them contains $1$ and the other doesn’t; and there’s no isomorphism between them which commutes with those injections.

Summing up: “containing 1” is certainly non-evil as a property of “sets with a mono to $\mathbb{N}$”. This induces a non-evil property/structure on abstract sets which you may or may not agree matches our original idea, because it requires considering different monos to $\mathbb{N}$ besides the ones we were already (implicitly) thinking of.

Now, back to dagger-categories. It seems reasonable that dagger-categories are evil *when regarded as structure on categories*. The post by Peter Selinger linked by Simon Henry argues this quite persuasively: proving a specific no-go theorem, showing there can exist no notion of dagger-structure satisfying certain desirable properties.

However, *structure on categories* is not the only way to look at dagger-categories. They can instead be seen as structure on “pairs of categories connected by a faithful and essentially surjective functor $i : \mathbf{C}_u \to \mathbf{C}$”. (Full definition below.)

You may be thinking: this can’t be right, because it induces a non-evil structure on categories (“equip with a fully faithful inclusion from some other category, and then the dagger-structure”) which would violate Selinger’s no-go argument. However, it doesn’t: this structure doesn’t allow a definition of unitary maps in the sense Selinger’s argument assumes. Given $A, B \in \mathbf{C}_u$, we can say a map $iA \to iB$ is unitary if it’s the image of some map $A \to B$. But given just $A, B \in \mathbf{C}$, this unitariness isn’t well-defined for $f : A \to B$; different ways of expressing $A$ as $iA'$ and $B$ as $iB'$ might give different answers as to whether $f$ is unitary.

Expressed in this form, transferring the “weak dagger structure” on $\mathbf{fdHilb}$ along the equivalence to $\mathbf{fdVect}$ yields a weak dagger structure where the functor $i$ is not injective on objects. Selinger’s argument shows that something like this is unavoidable.

Summing up again: dagger-structure is certainly not evil when viewed as a structure on “categories with a distinguished faithful, ess. surj. inclusion”. This gives a definition of weak dagger structure as a non-evil structure on categories, which you may or may not accept, because it requires us to loosen up our original expectation that the “subcategory” of unitary maps should be literally bijective on objects, i.e. to go beyond the kind of “subcategories” we were originally thinking of.

Full definition: a *weak dagger category* may be taken to consist of:

- a category $\newcommand{\C}{\mathbf{C}}\C$;
- a groupoid $\C_u$, with a faithful and essentially surjective functor $i : \C_u \to \C$;
- a functor $\dagger : \newcommand{\op}{\mathrm{op}}\C^\op \to \C$;
- a natural isomorphism $\varphi : \dagger \cdot i^\op \cong i \cdot (-)^{-1} : \C_u^\op \to \C$;
- a natural isomorphism $\psi : \dagger \cdot \dagger^\op \cong 1_\C$;
- such that for all $A \in \C_u$, $\psi_{iA} = \varphi_A (\varphi_{A}^\dagger)^{-1} : (iA)^{\dagger \dagger} \to iA$, or equivalently, as natural transformations, $\psi \cdot i = (\varphi \cdot (-)^{-1})(\dagger \cdot \varphi^\op)^{-1} : \dagger \cdot \dagger^\op \cdot i \to i$;
- and such that for any $A, B \in \C_u$, the image of $\C_u(A,B) \to \C(iA,iB)$ consists of
*all*$u$ such that $\varphi_A \cdot u^\dagger \cdot \varphi_B^{-1}$ is a 2-sided inverse for $u$.

Given this, I claim:

- each component is “non-evil” as structure on the earlier components (this can be made precise as a lifting property for forgetful functors between 2-categories);
- strict dagger categories are precisely weak dagger categories such that $i$ is bijective on objects and $\varphi_A = 1_{iA}$ for all $A \in \C_u$;
- given any weak dagger cat with $i$ bijective on objects, one can modify $\dagger$, $\varphi$, and $\psi$ to obtain a strict dagger-structure, equivalent to the original, with the equivalence acting trivially on $\C$, $\C_u$, and $i$;
- therefore, given a category $\C$ equipped with a distinguished all-objects subgroupoid $\C_u$, “strict dagger structures on $\C$ with unitaries $\C_u$” correspond to “weak dagger structures on $\C_u \to \C$”; so strict dagger structure is non-evil as a structure on “cats with a distinguished all-objects subgroupoid”;
- if we drop the last component of the definition, we similarly get a non-evil structure of “strict dagger structures on $\C$ with unitaries including at least $\C_u$”;
- given any weak dagger cat $\C$, there is an equivalent strict one, with the same unitary category $\C_u$, but with new ambient category given by the objects of $\C_u$ with the morphisms of $\C$. So overall, weak dagger categories do not give us anything essentially different from strict ones.

(I’m pretty sure that I’m remembering most of the ideas here from somewhere, but I can’t find where. The nearest I can find is this post by Mike Shulman on that same categories list thread. Better references very welcome.)

@Faisal: Thanks for calling attention to this earlier reference for this use of ampleness. (Kempf's 1976 theorem in prime characteristic essentially carries over the conclusion of Borel-Weil without appealing to Kodaira vanishing or the like.)@PeterMcNamara Right, thanks. The correct argument is that $Z_{\mathfrak{g}}(X)$ is *contained* in the generalized $0$-weight space of $\mathrm{ad}(X)$ which is a Cartan subalgebra since $X$ is regular.1322496Yes, I forgot to say that $\widetilde{\mathfrak{X}}$ has to be nontrivial, I've edited the question.202748656547https://www.gravatar.com/avatar/eefdcf66547f4bde3750175e6611a2ea?s=128&d=identicon&r=PG&f=1zThanks, I edited my question to add your answer in the list.839266Is it obvious that the equation from this example is formally integrable? I guess that equations without solutions abound if one allows "hidden" integrability conditions.2291447330222https://www.gravatar.com/avatar/f3f91acd2c902a15dcd5efe413d2979c?s=128&d=identicon&r=PG&f=1xI believe the answer Tim is looking for is "Dedekind cuts".\And are you also aware of any nonsmooth ones?1806958Well, I don't know if it's utterly standard, but I say "preserves", "reflects", "preserves and reflects" respectively.279349156223479176321124836985110104251303083f@MaxAlekseyev I had a look and added an item (+1).14911302017510233146712017I have a question about fibrations on Irreducible log holomorphic symplectic manifolds. Lets give some introduction

**Motivation;** A holomorphic symplectic manifold (HSM) is a $2n$-dimensional compact K\"ahler manifold $X$ which admits a non-degenerate holomorphic $2$-form $\omega\in H^0(X,\Omega^2)$. One of the examples is $K3$ surfaces. The non-degeneracy here means that $\omega^{\wedge n}$ trivialises $\Omega^{2n}=K_X$ , i.e.
$c_1=0$. A holomorphic symplectic manifold $(X,\omega)$ is irreducible (IHSM) if $H^0(X,\Omega^2)=\mathbb C\omega$. Matsushita showed that for $X$ be an IHSM, and $\pi:X\to B$ be a proper surjective morphism with connected fibres, and such that $B$ is smooth with $0<\dim B<\dim X=2n$, then the generic fibre is a complex torus (Abelian variety which is Calabi-Yau variety) and $\dim B=n$ which is the same as the dimension of the fibres. Moreover the restriction of $\omega$ on each fiber is zero, i.e. every fibre of $\pi$ is Lagrangian

816673661802Let $F:\mathscr{A}\to\mathscr{B}$ be an equivalence of Abelian categories. Must $F$ be additive?2251454Now if we consider the pair $(X,D)$ where $D$ is a snc divisor on $X$. Then is there Matsushita like theorem on pair $(X,D)$ where fibers are log Abelian varieties?

Etienne Ghys, Pierre de la Harpe, "Sur les Groupes Hyperboliques d’après Mikhael Gromov"

A detailed exposition of Gromov's ideas, outlined in his "Hyperbolic groups" article.

29557692294989The term "US-space" appears to go back to Albert Wilansky (1967) - http://www.ams.org/mathscinet-getitem?mr=2085571316345135843612865641806739 @Kimball : Yes.7144988VERY good choice,Slobadan-and a book that's strangely little known on this side of the Atlantic. 952461117184193529120696132173939320255jReducing system of polynomials with symbolic factors1831136you may want to email Lizhen Qin about this question ( qinl@math.purdue.edu ). He has though a lot about Morse theory on Hilbert manifolds. If the $X_i$ are all mutually nonisomorphic and the $Y_j$ as well, then it is, I believe, the product in the category of subobjects of the coend $\Omega_{\mathcal{C}}$, which is the direct sum of all simples. But I'm unsure about this, and I don't know how to get rid of the additional assumption.hGreat Ref!!! Thanks Matthias, very very helpful....342919202937419028761362993@PerAlexandersson do you mean chromatic symmetric function or there is symmetric chromatic function? can you plz suggest me some references?127260922366161312148https://www.gravatar.com/avatar/89ba1fbeb81cf036ddd327b3bd974940?s=128&d=identicon&r=PG&f=1620482161645119780741422323I am interested to know, why is this tagged under [set-theory]? I mean, what does this has to do with set theory that it hasn't got to do with functional-analysis or measure-theory?3553603550531025117I found this: Aharoni: Tutte for countable https://www.sciencedirect.com/science/article/pii/0095895684900522 Also the general case https://core.ac.uk/download/pdf/82044729.pdf220536716349591843163The mathematics was clear (of the question and of Anton's answer--I believe). It was hard only (so-to-speak) linguistically. The English was fine but the style was, well, next to impossible :-),I apologise if I am being thick. Can you tell me where I can read about how to associate a tree to a continuous function?2157366173936880782412602842072005No. Let $A=\mathbb{Z}[1/2]/\mathbb{Z}$. Let $R\to S$ be $\mathbb{Z}\to \mathbb{Z}_2\times \mathbb{Z}_3$, where $\mathbb{Z}_n$ denotes $\mathbb{Z}[1/n]$.

Then $Ann_{\mathbb{Z}}A=0$. But $(1,0)\in S$ kills $A\otimes_{\mathbb{Z}}S$.

I wanted to use $\delta_{i,j}$ for the Kronecker delta (instead of $\delta(i,j)$ in the text) but mysteriously that doesn't work. Any idea ? 4ReidemeisterSingerTheorem~If $h$ is the Hermitian form, what is the meaning of $\log h$?I just realized this is Exercise 1(a) from Section 9.1.1 in Montgomery-Vaughan: Multiplicative Number Theory I.18693631990491250872479963The Betti numbers are 1, 0, 2, 0, 1 for all n, and there is no torsion. Probably the reference you have contains this. Are you asking you do you actually check this?@Micheal Lugo: I assume the second German should be English. That is, "band" is a perfectly reasonable English word.I have seen a couple of questions related to the Eichler-Shimura Isomorphism, but almost all of them have to do with hodge theory (things I am unfamiliar with) and seem, to me, different/unrelated.

Let $S_k(\Gamma)$ denote the space of modular cusp forms of level $\Gamma \subset SL_2(Z)$ and let $V_{k-2} \subset \mathbf{C}[X,Y]$ be the homogenous polynomials of degree $k-2$ with coefficients complex numbers. $SL_2(Z)$ acts on $V_{K-2}$ as follows. If $\gamma \in SL_2(Z)$ has first row $(a_1 b_1)$ and second row $(a_2 b_2)$, then $\gamma( P(X,Y)) := P(a_1X + b_1Y, a_2X + b_2Y)$

Define a map from the cusp forms to the first group cohomology of $\Gamma$ with coefficients in $V_{k-2}$

$\phi: S_k(\Gamma) \to H^1(\Gamma,V_{k-2})$

$\phi(f)[g] := c_f(g) = \int_0^{g(0)}f(z)(X-zY)^{k-2}dz$

I have already verified that $c_f$ is a cocycle, hence embedded in $Z^1(\Gamma, V_{k-2})$ but what remains is my question:

**How do I prove that if $f\neq0$ then $c_f$ is not a boundary?**

The category of all groups contains all groups. There are no others. Yet other than triangles in constant curvature spaces, there are going to be other triangles in non-constant curvature spaces. I think this post doesn't fully convey what your original question in the other post wants. XAlmost convex combinations in $\mathbb R^n$795311307982882184hThanks a lot! Which book on CM would you recommend?1140681677623What is Known About the Complexity of Calculating Minimal Surface Polyhedra?1187064@FedorPetrov and AmirSagiv and everybody, sorry. I missed word *convex*, of course.SandeepJ, thanks. I will see if the bibliography holds some clues as well. Gerhard "Ask Me About System Design" Paseman, 2010.07.1120650127831998584261343765xOops, complex tori don't work: $h^{0,1}$ is always nonzero.\https://i.stack.imgur.com/dPy20.jpg?s=128&g=1170302316460682780332289260For every integer $k$, the derivation action of $H^0(G,T_G)=\mathfrak{sl}_n(\mathbb{C})$ on $H^0(G,\mathcal{O}_G(k))$ is equivalent to the irreducible $\mathfrak{sl}_n(\mathbb{C})$-module $\mathbb{S}^{(k,k)}((\mathbb{C}^n)^\vee)$, where $(k,k)$ is the partition of $2k$ into two parts of size $k$, and where $\mathbb{S}^{(k,k)}(-)$ is the corresponding Schur functor. (I am assuming that $n$ is at least $3$.)99086251663280*

I can give individual answers to a lot of your questions, but I can't answer any of them completely, nor can I fit all these answers together into a coherent whole.

For string theory, there does seem to be something special about the Deligne-Mumford compactification. Morally, what's going on is this: string theorists are allowing cylinder-shaped submanifolds of their Riemann surfaces to become infinitely long. The only finite energy fields on such infinitely long submanifolds are constant, so you can replace the long cylinder with a node. (Likewise, morally, if you allow vertex operators at two marked points to come together, you should take their operator product. This is what bubbling when marked points collide does for you.)

Somewhat more technically: The first step in (bosonic) string theory is to compute the partition function of the nonlinear sigma model as a function on the space of metrics on your worldsheet. This function on metrics descends to a section of some line bundle on the moduli stack of complex structures on the worldsheet. When you can compute it at all, you can show that this section has exactly the right pole structure it needs to be a section of the 13th power of the canonical bundle tensored with the 2nd power of the dual of the line bundle corresponding to the boundary divisor of the Deligne-Mumford compactification. (There's an old Physics Report by Phil Nelson that explains this pretty well, although not with anything you'd call a proof. Should also credit Belavin & Knizhik, who did the initial calculations.)

There's a somewhat more modern perspective on this (Zwiebach, Sullivan, Costello,...) that says that the generating function of string theory correlation functions for smooth Riemann surfaces satisfies a certain equation (a "quantum master equation"), which gives instructions for how to extend the theory to nodal Riemann surfaces. Different master equations give different recipes for extending to the boundary, if I understand your advisor correctly.

People have played around with other compactifications. There are a lot of different compactifications of the stack of smooth marked curves. People have already mentioned a few of them. David Smyth has some cool results which classify the "stable modular" compactifications of the stack of curves (http://arxiv.org/abs/0902.3690 ). For compactifications of the moduli of maps, the only one that comes immediatley to mind is Losev, Nekrasov, and Shatashvili's "freckled instanton compactification", in which IIRC, you allow zeros and poles to collide and cancel each other out.

78121113081461257986rI forgot to mention that $k$ have to be bigger than $1$Uniform continuity is another possibility between continuity and Lipschitz continuity. It's much weaker that Lipschitz, and in particular boundedness would no longer be preserved, but Cauchy sequences would be.@kjetil b halvorsen: That it might be on-topic on stats.stackexchange.com does not automatically imply that it is off-topic here. I thought about trying to redirect it to stats.stackexchange.com, but decided against it. Since the question had an upvote and was about probability, and the law of total variance is not as well known as it should be, I figured that it would be fine here.I do not understand why you believe that the lower bound is not true. Assume that the sup's are max's, so $\mathbf{P}(X_i = x_i) = p_i$. Then, the probability that the sum is equal to $\sum x_i$ is obviously at least $\prod p_i$.2055719979291728514637131186931|https://graph.facebook.com/217398395604400/picture?type=large892306phttps://graph.facebook.com/739522728/picture?type=large2546067350252045434“*Kindergarten Quantum Mechanics*” by Bob Coecke / arXiv:quant-ph/0510032v1

This is true. There is a very thorough discussion of this result in Mumford's book on abelian varieties.

910429165364241194250842616564771319255574437crazy muman180756120756164109462219556What is $Y$? You need some sort of convergence in $Y$ for the question even to make sense. Do you want an "order" convergence as you wrote in the real-valued case?02010-09-27T00:14:29.800I didn't check it, but I think the standard technique should do the job: if there were a fixed point for each $t$, those points would converge as $t\to0$, and something bad would happen at the limit point (the velocity vector would vanish?)369432I guess I'll add that one thing that impressed me enormously when I was a child was the film "Donald Duck in Mathmagicland" -- the excitement I felt watching it is a memory I carry to this day. Ostensibly, I was sitting passively and doing nothing except watching, but I was very excited by what I had seen -- I'm not sure I would have been able to articulate that excitement to my parents, but that film (with things like Fibonacci numbers, logarithmic spirals, music and the Pythagoreans) -- it struck a deep chord. 1332149lQuantity of partition sets intersecting a compact set396610fSmooth manifold with affine structure: aspherical?4155522094918In other words, is there a reason it is defined this way as opposed to something else? 229680I recommend the following for a breadth first search or a constraint solver with n=24. Enumerate all 132 legal triples. Start by picking out one of the 11 triples that has the number 24. Make a copy of the 132 triples minus those that share a number with the picked triple. Recurse until an exact cover is found or no exact cover is possible. My guess is that you can go five triples deep, cover less than a million cases this way, and find most of them can't be continued. Gerhard "Might Be Short In Python" Paseman, 2017.08.10.117343215795627070615867442032494https://www.gravatar.com/avatar/064e95e6a156cbf991ff5a815bec3e01?s=128&d=identicon&r=PG&f=1Alan Turing, widely regarded as the father of Computer Science, is most widely-cited for his paper *The Chemical Basis of Morphogenesis* (wikipedia entry). The paper is integral in the development of theoretical biology and chaos theory.

Consider a finite measure space $(X,\Sigma,\mu)$. Consider the function $d:\Sigma \times \Sigma \to [0,1]$ given by $$d(\sigma_1,\sigma_2) = \mu \left\{ (\sigma_1^c \cap \sigma_2) \cup (\sigma_1 \cap \sigma^c_2) \right\}.$$ One can verify that $d$ is a pseudometric (where $d(\sigma_1,\sigma_2) = 0$ means that $\sigma_1$ and $\sigma_2$ differ by a set of measure zero). Does $d$ have a name?

1900015 Laurent - thanks for a very enlightening answer! Certainly I wouldn't expect a geometric classification of supercuspidals.. but one can still ask for a geometric (categorical?) description one Bernstein component at a time (as Marty suggests), like we have in the tamely ramified case.. for GL_n I guess this follows from type theory as you explained? There is now a local geometric Langlands program for arbitrary depth.. it's rather formal right now except in some special cases, but there is a precise general conjecture (the tamely ramified case is the theorem of Bezrukavnikov McGerty mentions)33921522375695804234960381634916114031620746572095116MankkaXCentralizer of a generator in a braid group!Okay, so here is what we *can* do, and this I'mma put in an answer, because it's something like an answer. If we assume something stronger than P \neq NP, like that the polynomial hierarchy doesn't collapse, then we can prove that some things aren't NP-complete.

The best example of such a result (actually, it's the first one I thought of, although in retrospect I've seen the ones other people mentioned too) is that if graph isomorphism is NP-complete, the polynomial hierarchy collapses to the second level. I can't seem to find an online version of the paper in which this is proved (it's by Boppana et al.) but I believe the argument goes more or less as follows: Graph isomorphism is in NP, easily. Graph isomorphism is also in co-AM; if you don't know what this means or haven't seen the proof, basically it means that if you have two graphs and I tell you they're *not* isomorphic, you can check that I'm telling the truth probabilistically. How: ou secretly choose one of the two graphs, randomly permute its vertices, and send it to me; I tell you which one it is. If the graphs are isomorphic, I won't be able to give you the right answer more than half the time, but if they aren't isomorphic, I can always give you the right answer. (We're assuming here that I have unbounded computational power.)

So we apply a derandomization result, to show that co-AM is contained in some level of the polynomial hierarchy (I think the second level), and then we're done.

As far as I know this is the only method that gives these kinds of conditional results. The reason it works is that PH is robust against things like randomization, but of course we have to assume stronger conjectures in complexity theory, so there's definitely a tradeoff.

Here's something else I just thought of: Is there anything known about what a conditional proof that a certain problem isn't NP-hard *can't* be? (Along the lines of relativization, etc.) It's not as straightforward as relativization, since all we want is a conditional proof, but it seems like an interesting question.

Mathematics PhD Student at the University of Liverpool. Mainly working on applied Algebraic Topology (stratified spaces), and I completed my MSc. dissertation on embeddings of Cantor sets.

194076213507941509487Another example of the equivalence of a ratio of fluxions to a derivative and its source in implicitly defined functions is given in the footnote 42 on pg. 32 of "From velocities to fluxions" by Panza."e.g." means "for example". But no, you cannot insert $1^{(4)}$ if you're talking about norms in $\mathbb R^3$. Why would you?9913516549042295241@FrançoisG.Dorais basically I agree with you, but this doesn't apply to typos/very minor corrections.rInverse of a correlation matrix that has arcsin elementsRsure is that np complete is the problem.196504bhttp://mathoverflow.net/users/1732/mahesh-velaga17416061419851Just to simplify, replace $h(v):=g(v)+p$. It's still continuous and still as arbitrary as $g$. Also, say $y:=c+w>0$.122757223269510658701288466138818Very astute comments! Many answers to "best textbook" questions are way off the deep (macho) end.1571173I'm reading James Maynard's paper "Small gaps between primes". Lemma 6.1 (p.14) in this paper confused me. This lemma was taken from Goldston-Graham-Pintz-Yildirim's paper "Small gaps between products of two primes". In GGPY's paper, this lemma is referred to as lemma 4.

The point what I'm confused is the discontinuity of weighting function G. If we follow GGPY's proof of this lemma, I think Riemann-Stieltjes(?) integrals in their proof need not always exist because of discontinuity of G. If this was not problem, in the integration by parts at the last of this proof, we have perhaps some more terms arising from discontinuity of G. My question is following one related to this.

I think we need the total variation or the number of discontinuities of G as a factor in the error term of

$$\sum_{d<z}\mu(d)^2g(d)G\left(\frac{\log d}{\log z}\right)\\ =\mathfrak{S}\frac{\log^\kappa z}{\Gamma(\kappa)} \int_0^1G(x)x^{\kappa-1}dx+O_{A_1,A_2,\kappa}\left(\mathfrak{S}LG_{\max}(\log z)^{\kappa-1}\right).$$

Do we need this factor or not?

Certainly, this is not influence on Maynard's proof because G has only one discontinuity for application.

666853183248316903381094979Added http://mathoverflow.net/questions/191683/limiting-entropy-of-deterministic-sequences-21480820Intuitionistic consistency of surjection from naturals to reals78423156446 96572186938Hi! I'm trying to understand why a hyperbolic n-manifold has finite mapping class group if $n \geq 3 $. In books I'm reading it's said it's a consequence of Mostow's rigidity theorem: "If M and N are complete hyperbolic manifolds with finite total volume, any isomorphism of fundamental groups is realized by a unique isometry."

A corollary of this is that: " If M is hyperbolic (complete, with finite total volume) and $n \geq 3 $, then Out($\pi_{1}(M)$) is a finite group, isomorphic to the group of isometries of M ".

But how could this could solve my problem? I mean, I know there is Dehn-Nielsen Theorem which states that Out($\pi_{1}(M)$) is isomorphic to MCG(M), but I know this to be true only in dimension 2...what can I say in dimension (at least) 3? Thank you.

This game is addictive. (Here is the link to the app: play.google.com/store/apps/details?id=jp.co.global.regionselect )The Wikipedia page for L-systems links to the Thue-Morse sequence for which there are a large number of applications. In particular it can be interpreted as the sequence of distances between atoms in a 1D quasicrystal. Hence I suppose you might think about the work on substitution tilings and higher dimensional quasicrystals as an application of similar ideas.

Thinking even more broadly, the recent work by Borrelli, Jabrane, Lazarus and Thibert on implementing convex integration for a $C^1$ embedding of a flat torus in $R^3$ uses explicitly an L-system-like rewriting process to add corrugations.

18983235339371974633376894I guess it's worth noting even _with_ RH you can't push it much further - at best to $O(x^{1/2}\log x)$, so even $O(x^{1/2})$ would be be a significant achievement!dYou're right. I've edited the answer accordingly.177784932958758407xHow to find a closed form of following matrix's determinantThis the best I can come up with (using Morlet's reconstruction formula), four integrals instead of three: $$\int_{-\infty}^\infty x(t)^3\,dt=\frac{1}{D^3_\psi}\int_{-\infty}^\infty dt\int_0^\infty d\xi_1\int_0^\infty d\xi_2\int_0^\infty d\xi_3$$ $$\xi_1^{-3/2}\xi_2^{-3/2}\xi_3^{-3/2}\mathring{x}(\xi_1,t)\mathring{x}(\xi_2,t)\mathring{x}(\xi_3,t)$$This is a question originally asked in https://math.stackexchange.com/questions/2535604/dirichlet-problem-for-div-textbff-0, but I there has been no answer yet so I'm seeking some ideas or comments here.

Consider $\Omega \subset \mathbb{R}^n$ to be a given bounded smooth domain and $\textbf{G}$ to be a given smooth vector field defined in $\mathbb{R}^n$. I'm curious about the existence of smooth solutions to the following equation with Dirichlet boundary data:

\begin{cases} div \, \textbf{F}=0 \quad \text{in $\Omega$}\\ \textbf{F}=\textbf{G} \quad \quad \text{on $\partial \Omega$} \end{cases}

My first try is to consider $n=3$, then for any $\textbf{F}$ with divergence free condition, $\textbf{F}$ can be written as $\textbf{f} \times \textbf{g}$, and then the problem reduces to finding two smooth vector fields $\textbf{f}$ and $\textbf{g}$ such that $$\textbf{f} \times \textbf{g}=\textbf{G} \quad \textbf{on $\partial \Omega$}.$$ I guess there are many solutions to the equation, but I just cannot give a specific construction for those solutions. Can anyone help solve this problem? Any ideas and comment will be really appreciated.

oops, forgot a critical condition: I want the answer for the 2-generator case.295947As stated above, you want to start with:

- D. Mumford and P.E. Newstead, Periods of a moduli space of bundles on curves, Amer. J. Math. 90 (1968), 1200–1208
- M.S. Narasimhan and S. Ramanan, Deformations of the moduli space of vector bundles over an algebraic curve, Ann. Math. (2) 101 (1975), 391–417

Here is a very naive idea (hopefully without mistakes, but sometimes I make really stupid ones when I am brainstorming like this), and it is probably not even in the category you want. Here it goes.

Take the Cartesian product of the (twisted) $G$-character variety of $C$, in your notation $M(C,G)$, with the moduli space of complex structures on $C$, in your notation $M_g$, and define an equivalence relation as follows: $([\rho_1],[C_1])\sim ([\rho_2],[C_2])$ if and only if $M_{hol}(G,C_1)\cong M_{hol}(G,C_2)$ and $[\rho_1]=[\rho_2]$.

Why not let $\mathcal{M}$ be $(M(C,G)\times M_g)/\sim$?

I am not sure what kind of space you get (identification space of a real algebraic space), but you do get maps $M_g\to \mathcal{M}$ just by inclusion (pair with any fixed representation) followed by projection. Suppose these maps were injective, then: $M_{hol}(C_1,G)\cong M_{hol}(C_2,G)$ implies $C_1\cong C_2$; as is the case in the above references.

You also get a projection $\mathcal{M}\to M(G,C)$, since the equivalence relation is trivial on $M(G,C)$; its fibers are exactly the complex structures $[C]$ that have $M_{hol}(G,C)$ isomorphic (as varieties).

Your vague question was "How does this structure changes when the complex structure on $C$ changes?"

Even if this $\mathcal{M}$ is not what you had in mind, everything in the construction and the maps have concrete descriptions. Perhaps it can be analyzed.

11294172948993@Dustin G. Mixon: I can just repeat my previous comment. The way you formulated the problem it only asks about distributions of $\overline X$ and $\overline Y$ separately, so that their joint distribution is irrelevant.~I am an IT student and a beginner programmer hobbyist.

1014670353688385901633505I have edited the statement of the question to reflect my motivation more accurately.531894Given a set $A\subset \mathbb{R}^n$ such that $A\cap (x+\mathbb{Z}^n)\ne \emptyset$ for any $x\in \mathbb{R}^n$ (that is, $p(A)=\mathbb{T}^n$ for the projection $p:\mathbb{R}^n\rightarrow \mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n$). Is it true that supremum of Eucledian distances between points of $A$ is not less then $\sqrt{n}$? (equality holds for the unit cube)

Or maybe even two points with $x=(x_1,\dots,x_n)$, $y=(y_1,\dots,y_n)$ with $|x_i-y_i|\geq 1-\epsilon$ for each coordinate?

It is not hard to check both claims for $n=2$, but already for $n=3$ I do not know.

1690841039862509134 Mathieu Baillif720037I mostly agree with Joe Silverman. You certainly need some commutative algebra first. But I'd start with Mumford's red book rather than Hartshorne. The first part is a very clear presentation of a (somewhat) more classical approach so you understand what schemes are intended to generalize, and the second part is a very clear presentation of scheme theory with emphasis on why it's the right generalization. There's no cohomology, but first things should come first.3299843610427969361413092mrpyopOrder of magnitude of extremely abundant numbers and RHdSo $C{}^\psi\times H$ does not depend on $\beta$?140082440397834285622233A cryptographical application that generated random primes in this way would have a far bigger problem, namely that the resulting distribution would be heavily nonuniform (skewed towards primes that follow larger gaps).82608721025462257365301130Sorry, a graphic (not "graphical") sequence is a sequence of numbers which can be the degree sequence of some graph.16088523302161750501https://lh4.googleusercontent.com/-yL7wJOidtrw/AAAAAAAAAAI/AAAAAAAAAAA/AB6qoq3nDa44b9oy-Reqod_j0TMFQ4jFOw/mo/photo.jpgȭSpent a month checking, this is what I suspect is the complete list of 'sporadic' or 'exceptional' pairs. No restriction that they be in the same genus or have the same discriminant. I was able to check discriminant ratio $4$ and discriminant ratio $1$ very high. The other ones just seem sort of random little sets, two quadruples (see the repeated forms). Also some patterns that dry up, discriminants $111,333,999$ but not $2997,$ also on the left $24, 72, 216, 648, 1944,$ but not $5832.$ This last begins with three pair of regular forms, I typed those in. The list of pairs of regular forms that agree is enormous.

```
---------------------------------------------------------------
54 : 2 2 4 1 2 0 216 : 2 4 8 4 1 1
75 : 1 4 5 1 1 0 111 : 1 4 7 1 0 0
78 : 3 3 3 1 1 3 142 : 3 3 5 2 3 1
78 : 3 3 3 1 1 3 158 : 3 3 5 -1 2 1
78 : 3 3 3 1 1 3 190 : 3 5 5 5 2 3
142 : 3 3 5 2 3 1 158 : 3 3 5 -1 2 1
142 : 3 3 5 2 3 1 190 : 3 5 5 5 2 3
156 : 3 3 5 2 2 0 284 : 3 5 6 4 2 2
156 : 3 3 5 2 2 0 316 : 3 5 6 0 2 2
156 : 3 3 5 2 2 0 380 : 3 5 7 2 0 2
158 : 3 3 5 -1 2 1 190 : 3 5 5 5 2 3
162 : 2 2 14 1 2 2 648 : 2 6 14 3 1 0
177 : 2 4 7 4 2 1 213 : 2 4 7 0 1 1
225 : 3 4 7 4 3 3 333 : 3 4 7 1 0 0
232 : 3 3 7 1 2 1 232 : 3 5 5 3 1 3
284 : 3 5 6 4 2 2 316 : 3 5 6 0 2 2
284 : 3 5 6 4 2 2 380 : 3 5 7 2 0 2
316 : 3 5 6 0 2 2 380 : 3 5 7 2 0 2
324 : 4 4 6 0 3 2 567 : 4 6 7 3 2 3
486 : 2 2 41 1 2 2 1944 : 2 6 41 3 1 0
531 : 5 5 6 0 3 2 639 : 5 5 8 -1 2 4
648 : 4 7 7 5 2 2 2592 : 4 7 25 -4 2 2
648 : 5 5 8 0 4 3 648 : 5 7 7 6 1 5
675 : 5 5 8 -1 4 2 999 : 5 8 8 -5 1 4
These are pairs of positive quadratic forms that represent the
same numbers, and violate a Kaplansky conjecture.
Delta : A B C R S T means
f(x,y,z) = A x^2 + B y^2 + C z^2 + R y z + S z x + T x y,
and Delta = 4ABC + RST - A R^2 - B S^2 - C T^2.
The two pair within a genus each are
232 : 3 5 5 3 1 3 232 : 3 3 7 1 2 1
648 : 5 7 7 6 1 5 648 : 5 5 8 0 4 3
The most productive discriminant ratio is 4,
which includes Kap's two infinite families, also
24 : 1 2 4 2 1 1 6 : 1 1 2 1 1 0
72 : 2 2 5 1 1 1 18 : 2 2 2 1 2 2
216 : 2 5 6 3 0 1 54 : 2 2 5 1 2 2
648 : 2 6 14 3 1 0 162 : 2 2 14 1 2 2
1944 : 2 6 41 3 1 0 486 : 2 2 41 1 2 2
or
48N-24: 2 6 N 3 1 0 12N-6: 2 2 N 1 2 2
where N = (1+ 3^k)/2, and the pairs for N = 1,2,5 are regular
and have been Schiemann reduced.
------------------------------------------------------------
Reminder: Kap's two infinite families are equivalent to those
below, which need not be "reduced." For the first, require
gcd(A,C) = 1 and 0 <A,C. For the second, gcd(A,R) = 1, with
A > 0 and -A < R < 2 A.
4D : A 3A C 0 0 0 D : A A C 0 0 A
4D: A 2A-R 2A+R 0 2R 0 D : A A A R R R
For the first, D = 3 A^2 C, for the second D = (A+R)(2A-R)^2 .
```

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

Why not. The complete list of regular pairs, evidently 182 pairs among which various triples, quadruples and so on may be found. Smaller discriminant put first on the line.

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

```
2 : 1 1 1 1 1 1 18 : 1 2 3 2 1 0
2 : 1 1 1 1 1 1 32 : 1 2 4 0 0 0
2 : 1 1 1 1 1 1 8 : 1 1 2 0 0 0
3 : 1 1 1 0 0 1 12 : 1 1 3 0 0 0
3 : 1 1 1 0 0 1 12 : 1 2 2 1 1 1
3 : 1 1 1 0 0 1 21 : 1 2 3 0 0 1
4 : 1 1 1 0 0 0 16 : 1 2 2 0 0 0
4 : 1 1 1 0 0 0 36 : 1 2 5 2 0 0
5 : 1 1 2 1 1 1 20 : 1 2 3 1 0 1
5 : 1 1 2 1 1 1 20 : 1 2 3 2 0 0
6 : 1 1 2 0 0 1 22 : 1 2 3 0 1 0
6 : 1 1 2 0 0 1 24 : 1 2 3 0 0 0
6 : 1 1 2 1 1 0 15 : 1 1 4 0 1 0
6 : 1 1 2 1 1 0 24 : 1 1 6 0 0 0
6 : 1 1 2 1 1 0 24 : 1 2 4 2 1 1
6 : 1 1 2 1 1 0 33 : 1 2 5 1 1 1
8 : 1 1 2 0 0 0 18 : 1 2 3 2 1 0
8 : 1 1 2 0 0 0 32 : 1 2 4 0 0 0
8 : 1 1 3 1 1 1 32 : 1 3 3 2 0 0
8 : 1 1 3 1 1 1 72 : 1 3 7 2 1 1
9 : 1 1 3 0 0 1 36 : 1 3 3 0 0 0
9 : 1 1 3 0 0 1 36 : 1 3 4 3 1 0
9 : 1 1 3 0 0 1 63 : 1 3 6 3 0 0
10 : 1 2 2 2 1 1 15 : 1 2 2 1 0 0
10 : 1 2 2 2 1 1 40 : 1 2 5 0 0 0
12 : 1 1 3 0 0 0 12 : 1 2 2 1 1 1
12 : 1 1 3 0 0 0 21 : 1 2 3 0 0 1
12 : 1 1 4 0 0 1 48 : 1 3 4 0 0 0
12 : 1 2 2 1 1 1 21 : 1 2 3 0 0 1
12 : 1 2 2 2 0 0 44 : 1 2 6 2 0 0
12 : 1 2 2 2 0 0 48 : 1 2 6 0 0 0
14 : 1 1 5 1 1 1 30 : 1 3 3 1 1 1
14 : 1 1 5 1 1 1 46 : 1 3 5 3 1 1
14 : 1 1 5 1 1 1 56 : 1 3 5 2 0 0
15 : 1 1 4 0 1 0 24 : 1 1 6 0 0 0
15 : 1 1 4 0 1 0 24 : 1 2 4 2 1 1
15 : 1 1 4 0 1 0 33 : 1 2 5 1 1 1
15 : 1 2 2 1 0 0 40 : 1 2 5 0 0 0
16 : 1 2 2 0 0 0 36 : 1 2 5 2 0 0
18 : 1 1 6 0 0 1 72 : 1 3 6 0 0 0
18 : 1 2 3 2 1 0 32 : 1 2 4 0 0 0
18 : 2 2 2 1 2 2 45 : 2 2 3 0 0 1
18 : 2 2 2 1 2 2 72 : 2 2 5 1 1 1
18 : 2 2 2 1 2 2 72 : 2 3 3 0 0 0
18 : 2 2 2 1 2 2 99 : 2 3 5 3 1 0
20 : 1 1 7 1 1 1 80 : 1 3 7 2 0 0
20 : 1 2 3 1 0 1 20 : 1 2 3 2 0 0
22 : 1 2 3 0 1 0 24 : 1 2 3 0 0 0
24 : 1 1 6 0 0 0 24 : 1 2 4 2 1 1
24 : 1 1 6 0 0 0 33 : 1 2 5 1 1 1
24 : 1 2 4 2 1 1 33 : 1 2 5 1 1 1
25 : 2 2 2 -1 1 1 100 : 2 2 7 -1 1 1
25 : 2 2 2 -1 1 1 100 : 2 3 5 0 0 2
27 : 1 1 7 0 1 0 27 : 1 2 4 1 0 1
27 : 1 1 9 0 0 1 108 : 1 3 10 3 1 0
27 : 1 1 9 0 0 1 108 : 1 3 9 0 0 0
27 : 1 1 9 0 0 1 27 : 1 3 3 3 0 0
27 : 1 3 3 3 0 0 108 : 1 3 10 3 1 0
27 : 1 3 3 3 0 0 108 : 1 3 9 0 0 0
27 : 2 2 2 1 1 1 108 : 2 2 8 2 2 1
27 : 2 2 2 1 1 1 108 : 2 3 5 0 2 0
27 : 2 2 2 1 1 1 189 : 2 3 8 0 1 0
28 : 2 2 3 2 2 2 112 : 2 3 5 2 0 0
28 : 2 2 3 2 2 2 60 : 2 3 3 0 0 2
28 : 2 2 3 2 2 2 92 : 2 3 5 2 0 2
30 : 1 1 10 0 0 1 120 : 1 3 10 0 0 0
30 : 1 3 3 1 1 1 46 : 1 3 5 3 1 1
30 : 1 3 3 1 1 1 56 : 1 3 5 2 0 0
32 : 1 3 3 2 0 0 72 : 1 3 7 2 1 1
36 : 1 1 12 0 0 1 144 : 1 3 12 0 0 0
36 : 1 3 3 0 0 0 36 : 1 3 4 3 1 0
36 : 1 3 3 0 0 0 63 : 1 3 6 3 0 0
36 : 1 3 4 3 1 0 63 : 1 3 6 3 0 0
36 : 2 2 3 0 0 2 144 : 2 3 6 0 0 0
44 : 1 2 6 2 0 0 48 : 1 2 6 0 0 0
45 : 2 2 3 0 0 1 72 : 2 2 5 1 1 1
45 : 2 2 3 0 0 1 72 : 2 3 3 0 0 0
45 : 2 2 3 0 0 1 99 : 2 3 5 3 1 0
46 : 1 3 5 3 1 1 56 : 1 3 5 2 0 0
48 : 1 4 4 4 0 0 192 : 1 4 12 0 0 0
50 : 1 4 4 3 1 1 200 : 1 5 10 0 0 0
50 : 1 4 4 3 1 1 75 : 1 4 5 0 0 1
54 : 1 1 18 0 0 1 216 : 1 3 18 0 0 0
54 : 1 4 4 2 1 1 216 : 1 6 9 0 0 0
54 : 1 4 4 2 1 1 297 : 1 6 13 3 1 0
54 : 2 2 5 1 2 2 135 : 2 5 5 5 1 2
54 : 2 2 5 1 2 2 216 : 2 5 6 0 0 2
54 : 2 2 5 1 2 2 216 : 2 5 6 3 0 1
54 : 2 2 5 1 2 2 297 : 2 5 8 -2 1 1
54 : 2 3 3 3 0 0 216 : 2 3 9 0 0 0
60 : 1 4 5 4 1 0 132 : 1 5 7 1 0 1
60 : 1 4 5 4 1 0 96 : 1 4 7 4 0 0
60 : 2 2 5 0 0 2 240 : 2 5 6 0 0 0
60 : 2 3 3 0 0 2 112 : 2 3 5 2 0 0
60 : 2 3 3 0 0 2 92 : 2 3 5 2 0 2
64 : 3 3 3 -2 2 2 256 : 3 3 8 0 0 2
64 : 3 3 3 -2 2 2 576 : 3 3 19 -2 2 2
72 : 2 2 5 1 1 1 72 : 2 3 3 0 0 0
72 : 2 2 5 1 1 1 99 : 2 3 5 3 1 0
72 : 2 3 3 0 0 0 99 : 2 3 5 3 1 0
75 : 1 4 5 0 0 1 200 : 1 5 10 0 0 0
80 : 3 3 3 2 2 2 320 : 3 4 7 0 2 0
90 : 1 1 30 0 0 1 360 : 1 3 30 0 0 0
92 : 2 3 5 2 0 2 112 : 2 3 5 2 0 0
96 : 1 4 7 4 0 0 132 : 1 5 7 1 0 1
98 : 3 3 3 -1 1 1 392 : 3 5 7 0 0 2
100 : 2 2 7 -1 1 1 100 : 2 3 5 0 0 2
100 : 3 3 3 1 1 1 400 : 3 5 7 0 2 0
108 : 1 1 36 0 0 1 432 : 1 3 36 0 0 0
108 : 1 3 9 0 0 0 108 : 1 3 10 3 1 0
108 : 1 4 7 0 1 0 108 : 1 5 7 5 1 1
108 : 1 6 6 6 0 0 432 : 1 6 18 0 0 0
108 : 2 2 8 2 2 1 108 : 2 3 5 0 2 0
108 : 2 2 8 2 2 1 189 : 2 3 8 0 1 0
108 : 2 2 9 0 0 2 432 : 2 6 9 0 0 0
108 : 2 3 5 0 2 0 189 : 2 3 8 0 1 0
108 : 3 3 5 3 3 3 432 : 3 5 8 4 0 0
135 : 2 5 5 5 1 2 216 : 2 5 6 0 0 2
135 : 2 5 5 5 1 2 216 : 2 5 6 3 0 1
135 : 2 5 5 5 1 2 297 : 2 5 8 -2 1 1
144 : 3 4 4 4 0 0 576 : 3 4 12 0 0 0
150 : 2 5 5 5 0 0 600 : 2 5 15 0 0 0
180 : 2 2 15 0 0 2 720 : 2 6 15 0 0 0
180 : 3 5 5 5 3 3 288 : 3 5 5 2 0 0
180 : 3 5 5 5 3 3 396 : 3 5 8 2 0 3
192 : 1 8 8 8 0 0 704 : 1 8 24 8 0 0
192 : 1 8 8 8 0 0 768 : 1 8 24 0 0 0
196 : 3 5 5 -4 2 2 784 : 3 5 14 0 0 2
216 : 1 6 9 0 0 0 297 : 1 6 13 3 1 0
216 : 2 5 6 0 0 2 216 : 2 5 6 3 0 1
216 : 2 5 6 0 0 2 297 : 2 5 8 -2 1 1
216 : 2 5 6 3 0 1 297 : 2 5 8 -2 1 1
240 : 1 4 16 4 0 0 384 : 1 4 24 0 0 0
256 : 3 3 8 0 0 2 576 : 3 3 19 -2 2 2
270 : 3 3 11 3 3 3 1080 : 3 9 11 6 0 0
288 : 3 5 5 2 0 0 396 : 3 5 8 2 0 3
300 : 1 10 10 10 0 0 1200 : 1 10 30 0 0 0
384 : 4 4 7 0 4 0 528 : 4 7 7 6 0 4
400 : 3 7 7 -6 2 2 1600 : 3 7 20 0 0 2
432 : 1 12 12 12 0 0 1728 : 1 12 36 0 0 0
432 : 5 5 5 -2 2 2 1728 : 5 8 12 0 0 4
448 : 5 5 5 2 2 2 1472 : 5 8 12 8 4 0
448 : 5 5 5 2 2 2 1792 : 5 8 12 0 4 0
448 : 5 5 5 2 2 2 960 : 5 5 12 -4 4 2
450 : 5 5 6 0 0 5 1800 : 5 6 15 0 0 0
540 : 5 5 8 2 4 5 1188 : 5 8 9 6 3 2
540 : 5 5 8 2 4 5 864 : 5 8 8 8 2 4
540 : 6 6 7 6 6 6 2160 : 6 7 13 2 0 0
576 : 3 8 8 8 0 0 2304 : 3 8 24 0 0 0
704 : 1 8 24 8 0 0 768 : 1 8 24 0 0 0
720 : 3 8 8 4 0 0 1152 : 3 8 12 0 0 0
768 : 1 16 16 16 0 0 3072 : 1 16 48 0 0 0
864 : 5 8 8 8 2 4 1188 : 5 8 9 6 3 2
900 : 3 10 10 10 0 0 3600 : 3 10 30 0 0 0
960 : 5 5 12 -4 4 2 1472 : 5 8 12 8 4 0
960 : 5 5 12 -4 4 2 1792 : 5 8 12 0 4 0
960 : 5 8 8 8 0 0 3840 : 5 8 24 0 0 0
1024 : 3 11 11 -10 2 2 4096 : 3 11 32 0 0 2
1152 : 5 5 12 0 0 2 1584 : 5 5 17 -2 2 2
1280 : 7 7 7 -2 2 2 5120 : 7 12 16 0 0 4
1350 : 7 7 7 -1 1 1 5400 : 7 13 15 0 0 2
1472 : 5 8 12 8 4 0 1792 : 5 8 12 0 4 0
1728 : 1 24 24 24 0 0 6912 : 1 24 72 0 0 0
1728 : 8 8 9 0 0 8 6912 : 8 9 24 0 0 0
2160 : 5 8 17 8 2 4 3456 : 5 8 24 0 0 4
2304 : 3 16 16 16 0 0 9216 : 3 16 48 0 0 0
2700 : 9 11 11 -8 6 6 10800 : 9 11 30 0 0 6
2880 : 8 8 15 0 0 8 11520 : 8 15 24 0 0 0
3136 : 3 19 19 -18 2 2 12544 : 3 19 56 0 0 2
3456 : 8 11 11 -2 4 4 4752 : 8 11 15 6 0 4
4800 : 1 40 40 40 0 0 19200 : 1 40 120 0 0 0
6144 : 11 11 16 8 8 6 8448 : 11 11 19 2 2 6
6144 : 7 15 16 0 0 6 8448 : 7 15 23 -6 2 6
6400 : 3 27 27 -26 2 2 25600 : 3 27 80 0 0 2
6912 : 1 48 48 48 0 0 27648 : 1 48 144 0 0 0
6912 : 9 17 17 -14 6 6 27648 : 9 17 48 0 0 6
8640 : 13 13 13 2 2 2 34560 : 13 24 28 0 4 0
14400 : 3 40 40 40 0 0 57600 : 3 40 120 0 0 0
18432 : 17 17 20 -4 4 14 25344 : 17 20 20 -8 4 4
18432 : 5 20 48 0 0 4 25344 : 5 20 68 -8 4 4
43200 : 9 41 41 -38 6 6 172800 : 9 41 120 0 0 6
55296 : 11 32 44 -16 4 8 76032 : 11 32 59 8 10 8
```

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

@royalalexander: o(1) denotes a quantity that goes to zero as $n$ goes to infinity. You may check http://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation, if you will,Cloven Kan fibrations115035Assume the polynomials here are dense.

In here it was asked the difficulty of counting prime factors of an integer.

We know for the cases of primitive polynomials in $\Bbb Z[x]$ and any polynomial in $\Bbb F_q[x]$ counting irreducible factors is easy since for primitive polynomials in $\Bbb Z[x]$ we have deterministic LLL algorithm and for any polynomial in $\Bbb F_q[x]$ we have randomized algorithms.

$1$ My first question is whether it would be possible to have an algorithm to count number of irreducible factors for either the case of primitive polynomials in $\Bbb Z[x]$ or the case of any polynomial in $\Bbb F_q[x]$ in at least randomized polynomial time or even in randomized subexponential time without factoring?

$2$ Given that we know how to test primality without factoring my second question is whether it would be possible to test for irreducibility without factoring polynomials?

Query $2$ is affirmative if we are in $\Bbb F_q[x]$ but unclear for primitive polynomials in $\Bbb Z[x]$.

What is know about these in sparse polynomial case?

747104XUniformization theorem in higher dimensions396121Reid: if you change 1/2 to 1/2+epsilon does "FALSE" become "TRUE"? That would be a check to see if you'd done the translation to coordinates correctly.45457316262322280785 I think you are missing the point, so I will try to spell it out one last time. Sets in axiomatic set theory have structure, which is called $\in$. In a lot of applications of category theory the morphisms of our category preserve structure. If we apply this to set theory, our morphisms should preserve $\in$. However, this notion of morphism is very different to the notion of set function. All this implies that the standard technique of morphisms preserving structure does not seem to apply to standard set theory, and why category theory has not been too successful.What is the "action" of $g$ on $T(M) \times T(M)$? If you just mean applying $g$ to two vector fields, in what sense is this an action?Actually, complexity-wise, for a polynomial of degree d, approximating the roots and then calculating $Res(P,P')=\prod P'(\alpha_i)$ directly, would be a soft-oh of d, rather than $d^3$ for the Sylvester.@Dmitri: If you *fix* the lines, you just get a ${\bf P}^4$ of quartic curves, which contains no double conic. Do you want the lines to vary too? 787121361501184606Artin is going to be rough going for undergraduates who are not well versed in basic geometry and linear algebra,fpqc.But you can't help but love the infectious passion with which Artin weaves his craft in front of the students.He loves algebra and he's trying to prosyletize his students to it. A book with a similar geometric bent,level and also by a master that students will probably find easier going is E.B.Vinberg's A COURSE IN ALGEBRA. But Artin's book is very good and it's good news for all of us that Artin is revising it.7369932186806\Currenty I'm doing PhD in mathematics

867330562272213991848541231301898967Floquet theory applies to linear systems with periodic matrices, and can be sometimes used to deduce the stability of the origin. In this case all we can say is that the Floquet exponents sum to zero, but that does not decisively rule out stability. I was thinking a perturbation approach might work.1823460Let me address your question in the comments.

If one starts with a weakly compact cardinal $\kappa,$ then there exists a forcing extension in which all cardinals and cofinalities are preserved (in particular, $\kappa$ remains a regular limit cardinal) and in which the tree property holds at $\kappa$. In fact $\kappa$ can become the least weakly Mahlo cardinal.

This result is due to Boos, see Boolean extensions which efface the Mahlo property.

As far as I know the question of whether the least weakly inaccessible cardinal can have the tree property is open (it is asked in Boos paper).

2236970618895fSo, was there anything useful in those references?52541533029Possible duplicate: http://mathoverflow.net/questions/195/is-there-an-example-of-a-formally-smooth-morphism-which-is-not-smooth445683211957Derek, thanks a lot! This is precisely what I need. I didn't expect $H$ to be all of $GL(2,\mathbb Z)$ but in fact it suits me perfectly. 2125568I think the natural level of generalization is a group extension - or possibly a quotient group extension of a dynamical system. A key property here is that the convolution of Haar measure with anything is Haar measure.1552350@Jan-Christoph Schlage-Puchta. Thanks, but $A$ can be infinite.1092920+1. That's exactly what I love about lectures, in particular those where the lecturer makes a lot of throwaway remarks, which start making perfect sense and give me perspective when I go home and read a text.https://www.gravatar.com/avatar/5fc6ba04864f38011bf84e698b1a5ad5?s=128&d=identicon&r=PG&f=134980614120722198484See the discussion on FOM mailing list. As far as I remember, according to some members of the prize committee, Wolfram announced it without proper contact with them. There was also discussion about what was Pratt's objection to the proof. See this:

http://cs.nyu.edu/pipermail/fom/2007-October/012132.html

and the other posts in that thread.

733339830606https://www.gravatar.com/avatar/4968488f16a73c88f6c5ed08c3877643?s=128&d=identicon&r=PG&f=1Also used all over the place in various reaults about Banach algebras, especially in questions regarding derivations1910289615579 9329I believe that all examples must come from finite quotients:

A collection of subgroups is necessarily good if the element $1\in K_0(G)$ is in the subgroup generated by the images of the induction maps $K_0(H)\to K_0(G)$. That's because these images are ideals, using the formula $V\otimes Ind(W)=Ind(Res(V)\otimes W)$.

Therefore every good collection contains a finite good collection. And of course for any collection of finite index subgroups $H_i$ there is a finite index normal subgroup $N$ of $G$ that is contained in them all.

And then the collection ${H_i}$ will be good in $G$ if and only if the collection $H_i/N$ is good in the finite group $G/N$. (This last requires a little argument splitting representations of $G$, or $H_i$, into the part fixed by $N$ and its orthogonal complement and noting that this splitting is compatible with induction.)

No. Let $R'$ be any non-excellent dvr whatsoever, hence of equicharacteristic $p > 0$, and let $t \in R'$ be a uniformizer. Let $R = \mathbf{F}_p[t]_{(t)}$. The local inclusion $R \hookrightarrow R'$ has induced residue field extension that is separable since $\mathbf{F}_p$ is perfect. And of course $R$ is excellent for any of a million reasons. So this seems to be a counterexample.

1595564I just need to state that, in reality, all I did was point out that this result was a consequence of the theorems in the linked paper of Mathew-Naumann-Noel..I see, Thanks a lot！I think you missed his point. In the paper of HMX they actually used the fact that the automorphism group is finte, which goes back to Matsumura. Hanamura’s paper also proved this.Unfortunately, the simple framework I described of thinking about extensions along a map $A \hookrightarrow B$ doesn't include things like the limit of a diagram (except a few cases like split idempotents). But I think this is still good to know.The Bollobas'1965 theorem is the following:

If $A_1,...,A_n$ and $B_1,...,B_n$ are two sequences of subsets of $X=\{1,...,r\}$ such that $A_i\cap B_j = \emptyset$ if and only if $i=j$, then $$\sum_{i=1}^n\binom{|A_i|+|B_i|}{|A_i|}^{-1}\leq 1.$$

I am interested in a generalization of this theorem where for each set $A_i$, there are more than one set $B_j$ for which we have $A_i\cap B_j = \emptyset$.

More formally, the assumptions of my problem can be expressed as:

**Let $A_1,...,A_n$ and $B_1,...,B_n$ be two sequences of subsets of $X=\{1,...,r\}$. For each $i=1,...,n$, we have a subset $J_i=\{i,i+1,...,i+t\}$(mod n) with $|J_i|=t+1$. $A_i\cap B_j = \emptyset$ if and only if $j\in J_i$.**

If $t=0$, we obtain the basic version of the Bollobas theorem.

How can I adapt the result of the Bollobas theorem taking into account these new assumptions for $t\geq 1$?

Could You consider the conjecture as follows to make additional proposals for polymath project?

Let $A, B, C$ be three positive integer numbers such that $A+B=C$ with $\gcd(A, B, C) = 1$. By Fundamental theorem of arithmetic we write:

$A=a_1^{x_1}a_2^{x_2}...a_n^{x_n}$,

$B=b_1^{y_1}b_2^{y_2}...b_m^{y_m}$,

$C=c_1^{z_1}c_2^{z_2}...c_k^{z_k}$

Let $d=\min\{x_i, y_j, z_h \}$ where $1 \le i \le n, 1\ \le j \le m, 1\le h \le k$ then:

My conjecture:$$d \le 5$$

The conjecture in here

PS: I researched about one hundred papers, in any case $h \le 3$?

18726001225568403015Thanks. It does seem common for Phd students to fall into the habit of reading books and papers in isolation, and not "getting out" and talking to other students and collegues.1814452rDid you mean to write $P(S)\times \beta S$ in the title?74246478614I didn't really like Simon's link. The tutorial there seems to be based solely on drawing the curves by specifying coordinates. I find that a more, say, tabular or rigidly based approach is much easier to work with.

There's a good xy-pic tutorial for creating knots at http://lf.starlogik.de/res/XYPic-Knot-Intro.pdf

There's a great xymatrix tutorial at http://www.ctan.org/tex-archive/macros/generic/diagrams/xypic/xy/doc/xyguide.pdf

There are only 10 pages to read in each manual and you'll have mastered them both. You could get away without reading the xymatrix tutorial, but I'd reccommend reading it because it goes more in-depth into some of the parameter modifiers that aren't actually explained in the knot manual.

814211254747Maybe you already know about this, but you might want to look at the 1974 'Stable Mappings and their Singularities', by M. Golubitsky and V. Guillemin. That book contains an extensive discussion (and motivation) of the kinds of properties that you'd want for the `manifold' of smooth maps between two finite dimensional manifolds, and it contains a very nice treatment of Mather's fundamental work in this area. 2251235Carlos_San1308248552037zYou probably want to square the $d$ appearing in $\phi$, no?19290883818351408113@StephanMüller I am saying the same thing. One way to show the coherence theorem is to consider the Yoneda functor, which is an embedding.746959I think this actually should be much easier than the other problem, just because $\frac{\pi^2}{6}-1<\frac{3}{4},$ which gives you a chance of reducing this to a finite computation. In particular if there is an $n$ so that you can pack squares of side lengths $1/2\cdots, 1/(n-1)$ and $n$ right triangles of side length $2/n$, then you can also pack the squares in the question. This is because you can pack squares of side length $1,1/2,\cdots$ in a right triangle of side length $2$.Maybe you should index the rows and the columns of $A_n$ not by numbers from $1$ to $2^{n-1}$, but by length-$\left(n-1\right)$ bitstrings (which correspond to these numbers via base-$2$ representation). The entry in row $x_1x_2\cdots x_{n-1}$ and column $y_1y_2\cdots y_{n-1}$ is then obtained as follows: Let $k$ be the largest index such that $x_k = y_k$. (This is set to be $0$ if no such $k$ exists.) If some $i < k$ satisfies $x_i \neq y_i$, then the entry is $0$. If not, then the entry is $\left(-1\right)^{x_{k+1}} a_{n-k}$, where $x_n = 0$.1036597bInjective morphism from curves to $\mathbb CP^2$1348770In general, Soviet Doklady for those years are not available on the web. They were not translated into English at that time, and the Russian original has not been digitalized (unlike ALMOST ALL other Russian journals which are available free on mathnet.ru (one has to know some Russian to use this site)).

However, if you have an access to a university library, even one which does not have Doklady, you can use Interlibrary Loan. Just ask the librarian. I worked in several countries, and everywhere this option was avialable in research libraries. In any case, much more libraries have old Russian Doklady than Collected papers of Gelfand.

Re-interpreting vector spaces in a choice-less model of ZF as modules over a regular ring in ZFCpMedian and mean of the sample mean of i.i.d. log-normal293009The map $I \mapsto \chi_k(N(I))$ defines a character of the group of fractional ideals prime to $k$, so I think what you have is the $L$-function of a Hecke character of the quadratic field.69032762215577It is not *really* different, but I would say that it is less mysterious.351179176925510836361366207https://www.gravatar.com/avatar/4802fc608c0335295342fdf008e990c4?s=128&d=identicon&r=PG&f=1VDefining Massey products as transgressions1688120At first let us note that, we focused on the Baer*-rings not Baer rings. In a Baer*-ring $A$, an element $a$ is called positive if there are $x_1,\cdots,x_n$ in $A$ with $a=\sum x^*_ix_i$. As David Handelman determined just right now!221978Stuff that is indented by 4 spaces gets put into a block. I've edited your question to do that.13227361748949454669419934864470771693231VAre spherical harmonics uniformly bounded?1643607@YCor: Don't you mean that the answer to my question is in fact "yes"? Which book of Margulis are you referring to?16594961208023@EmilJeřábek which results of Hesse, Allender, and Barrington are you quoting?12140471172631@JeanMarieBecker You mean $t\in\mathbb{Q}$, presumably? Yes, but it doesn’t lead to anything any more obviously tractable. I’m really wondering, since I‘m pretty ignorant of modern number theory, whether there are any techniques I might not know about that are applicable to this sort of question.2207877707933On the one hand, given an extended TFT you can certainly compute its values using a triangulation. However, this process is a bit more complicated than you might expect (see pps. 4-5 of http://www.math.utexas.edu/users/odavidovich/Talks/BerkleyNov2010 for some discussion of these issues). On the other hand, I don't think you get the state sum model directly from an extended TFT. Instead to get the state sum model directly you really want an open-closed theory. In 2d you remove a little window around every vertex of your triangulation, compute that and then normalize.193098018375421831606444389For hyperbolic space, I suggest trying either the unit sphere of radius -1 in Minkowski space model or the Poincare disk model.Sylow 2's are elementary abelian so Walter's theorem applies and gives some examples with simple $G$ -- for instance $PSL(2,q)$ for certain values of $q$ and $J_1$.81925142806257374wThe following is what I wrote about six months ago during a private discussion (posted on request :-)

Let me start with a positive (obvious :-) result. $\newcommand{\word}[1] { \mathit{#1} } \newcommand{\catl}[1] { \mathbb{#1} } \newcommand{\catw}[1] { \mathbf{#1} } \newcommand{\mor}[3] { {#1 \colon #2 \rightarrow #3} }$

1) Let $\catl{C}$ be a small category. Denote by $\word{Cont}(\catl{C}^{op}, \catw{Set})$ the full subcategory of presheaves $\catw{Set}^{\catl{C}^{op}}$ that consists of such $\mor{H}{\catl{C}^{op}}{\catw{Set}}$ that preserve small limits that exist in $\catl{C}^{op}$ (i.e. map existing colimits from $\catl{C}$ to limits in $\catw{Set}$). It follows from abstract nonsense that $\word{Cont}(\catl{C}^{op}, \catw{Set})$ has small (co)limits and, in fact, is a reflective subcategory of $\catw{Set}^{\catl{C}^{op}}$.

Of course, every presheaf preserves colimits, therefore the restricted Yoneda embedding $A \mapsto \hom(-, A)$ gives functor: $$\mor{y}{\catl{C}}{\word{Cont}(\catl{C}^{op}, \catw{Set})}$$ This functor almost by definition preserves all limits that exist in $\catl{C}$. Observe, that it also preserves colimits. Let $\mor{F}{\catl{J}}{\catl{C}}$ be a functor from a small category, and assume that the colimit $\mathit{colim}(F)$ exists in $\catl{C}$. Consider any $H \in \word{Cont}(\catl{C}^{op}, \catw{Set})$. Then morphisms: $$y(\mathit{colim}(F)) = \hom(-, \mathit{colim}(F)) \rightarrow H(-)$$ by Yoneda, are tantamount to elements: $$H(\mathit{colim}(F)) \approx \mathit{lim}(H \circ F)$$ Similarly, morphisms: $$y(F(J)) = \hom(-, F(J)) \rightarrow H(-)$$ are tantamount to elements: $$H(F(J))$$ and one may easily verify, that the above exhibits $\hom(-, \mathit{colim}(F))$ as the limit $\mathit{lim}(y \circ F)$.

The above construction is described, for example, in "Basic concepts of enriched categories" of Max Kelly (Section 3.12). Moreover, as pointed out there, $\word{Cont}(\catl{C}^{op}, \catw{Set})$ inherits any monoidal (closed) structure from $\catl{C}$ via (restricted) convolution.

On the other hand, if $\catl{C}$ is not (monoidal) closed, then $\word{Cont}(\catl{C}^{op}, \catw{Set})$ generally will not be (monoidal) closed, due to the following fact.

2) There is no universal limit and colimit preserving fully faithful embedding of a small category to a cartesian closed complete and cocomplete category.

In fact, there is no universal embedding into cartesian closed category that preserves terminal object and binary coproducts.

For let us assume that $\catl{C}$ is non-degenerated and has a costrict terminal object (terminal object $1$ is costrict if whenever there exists a morphism $1 \rightarrow X$ then $X \approx 1$) and binary coproducts (you may take $\catl{C} = \catw{FinSet}^{op}$), and there is such an embedding $\mor{E}{\catl{C}}{\overline{\catl{C}}}$.

Because $1$ is costrict in $\catl{C}$ we have that $1 \sqcup 1 \approx 1$ in $\catl{C}$ and since coproduct and the terminal object are preserved by $E$, we have also $1 \sqcup 1 \approx 1$ in $\overline{\catl{C}}$. Let $\mor{f,g}{1}{A}$ be two morphisms in $\overline{\catl{C}}$. By the universal property of coproducts they induce the copairing morphism $\mor{[f,g]}{1 \sqcup 1}{A}$ that commutes with the coproduct's injections. However, since $1\approx 1 \sqcup 1$, the coproducts injections are identities and so $f = [f,g] = g$. Therefore, for every $A \in \overline{\catl{C}}$ there is at most one arrow $1 \rightarrow A$. If we take for $A$ an exponent $E(Y)^{E(X)}$, then by cartesian clossedness of $\overline{\catl{C}}$ and faithfulness of $E$:

$$\hom_\catl{C}(X, Y) \leq \hom_\overline{\catl{C}}(E(X), E(Y)) \approx \hom_\overline{\catl{C}}(1, E(Y)^{E(X)}) \leq 1$$

which contradicts the fact that $\catl{C}$ is non-degenerated.

3) Generally, the completions obtained in any of the mentioned ways will not be the smallest completion. In fact, for a general $\catl{C}$, the smallest reflective subcategory of $\catw{Set}^{\catl{C}^{op}}$ containing representables may not be a smallest limit and colimit completion of $\catl{C}$.

Let us take for example $\mathcal{Z}_3$ group thought as of a category with a single object. Actually one may easily compute the category $\overline{\mathcal{Z}_3}$ induced by the idempotent monad associated to the monad $T$ on the Isbell cojugation for $\mathcal{Z}_3$ (by the theorem of Fakir this category can be computed by taking the objects that respects $T$-weak equivalences). Explicitly, category $\overline{\mathcal{Z}_3}$ is the full subcategory of $\catw{Set}^{\mathcal{Z}_3^{op}}$ on free permutations, together with the terminal permutation (up to the terminal permutation it is equivalent to the Kleisli resolution of the monad $\mor{(-) \times \mathcal{Z}_3}{\catw{Set}}{\catw{Set}}$, but I'm not sure if this is deep or meaningless...).

One may easily see that $\overline{\mathcal{Z}_3}$ is not self-dual, thus cannot be the smallest limit and colimit completion (since $\mathcal{Z}_3$ is self-dual, if $\overline{\mathcal{Z}_3}$ was the smallest, then $(\overline{\mathcal{Z}_3})^{op}$ would be the smallest). Moreover, $\overline{\mathcal{Z}_3}$ satisfies the following properties:

- its every object is a colimit over $\mathcal{Z}_3$,
- its every object is a double-limit over $\mathcal{Z}_3$

Therefore (by the second property) $\overline{\mathcal{Z}_3}$ is the smallest reflective subcategory of $\catw{Set}^{\mathcal{Z}_3^{op}}$ containing representables.

4) The Dedekind-MacNeille construction as described in Todd's answer works when the monad induced by the Isbell conjugation is itself idempotent. For example, this is always true in the world of posets, in the world of Lawvere metric spaces, and more generally in the world of categories enriched over affine quantales (because in such a case every enriched monad is idempotent).

I actually think that only Dedekind-MacNeille completitions for categories enriched over (co)complete posets deserve the name.

The reason is that there is a subtlety here (which may look minor at first, but I think is crucial for the whole construction) that makes that the direct categorification of the original Dedekind-MacNeille requirements for completion have no sense --- the completion should be complete and cocomplete, which means that it should be closed under *all* limits/colimits. On the other hand, the term "a category is complete and cocomplete" means that the category is closed under *small* limits and *small* colimits (obviously, there is no (co)completion under *all* (co)limits in a $\catw{Set}$-enriched world). Therefore, the direct categorification of the requirement is possible only when the base of the enrichment has *all* limits/colimits, which in turn, implies that the base of the enrichment is a poset (at least in the classical mathematics).

As I said, the distinction between *small* and *all* may look minor at first, but in fact the distinction is not about *quantity*, but about *quality*. This distinction shows up on many occasions. For example, constructively if a category is complete (with respect to all cones) then it is also cocomplete (with respect to all cocones), and every colimit can be expressed as a limit; however, this is no longer true if the category is only small complete --- the cone from the canonical representation of a small cocone may be large, and there may be no reason for its limit to exists.

Postdoc in applied (algebraic) topology.

389165838915I roughly mean if $g'$ is "close" to $g$ then the minimal surface with respect to to $g'$ is "close" to S.1717902168576569189216612131004688202031817830301404326335822The only association I get for it is some sort of boundary value problem: if you can view $X$ as the boundary of some kind of compactification of its complement in the sphere, then, given a distribution (~cycle) on this boundary one seeks for a function (~cocycle) on the complement which limits (in some sense?) to this given distribution "at infinity"...114825A Hint/ very rough sketch: The definition of "Locally small" should says that given two objects $X,Y$ in a same fiber over $Z$, you have a certain objects in the codomain "representing morphisms between $X,Y$". You need to use the condition that the map your taking base change along is an epi with respect to morphisms to this specific objects and this will give you immediately that the base change is invective on morphisms between $X$ and $Y$.1800608~Calculate channel capacity of general channel under constraint1942564138990199308The Cayley plane $\mathbb{O}P^2$ is an example. See this question:

What is the Cayley projective plane?

1506054<There are at least two different $\sigma$-algebras that Lebesgue measure can be defined on:

- The (concrete) $\sigma$-algebra ${\mathcal L}$ of Lebesgue-measurable subsets of ${\bf R}^d$.
- The (abstract) $\sigma$-algebra ${\mathcal L}/\sim$ of Lebesgue-measurable subsets of ${\bf R}^d$, up to almost everywhere equivalence.

(There is also the Borel $\sigma$-algebra ${\mathcal B}$, but I will not discuss this third $\sigma$-algebra here, as its construction involves the first uncountable ordinal, and one has to first decide whether that ordinal is physically "permissible" in one's concept of an approximation. But if one is only interested in describing sets up to almost everywhere equivalence, one can content oneself with the $F_\delta$ and $G_\sigma$ levels of the Borel hierarchy, which can be viewed as "sets approximable by sets approximable by" physically measurable sets, if one wishes; one can then decide whether this is enough to qualify such sets as "physical".)

The $\sigma$-algebra ${\mathcal L}$ is very large - it contains all the subsets of the Cantor set, and so must have cardinality $2^{\mathfrak c}$. In particular, one cannot hope to distinguish all of these sets from each other using at most countably many measurements, so I would argue that this $\sigma$-algebra does not have a meaningful interpretation in terms of idealised physical observables (limits of certain sequences of approximate physical observations).

However, the $\sigma$-algebra ${\mathcal L}/\sim$ is separable, and thus not subject to this obstruction. And indeed one has the following analogy: ${\mathcal L}/\sim$ is to the Boolean algebra ${\mathcal E}$ of rational elementary sets (finite Boolean combinations of boxes with rational coordinates) as the reals ${\bf R}$ are to the rationals ${\bf Q}$. Indeed, just as ${\bf R}$ can be viewed as the metric completion of ${\bf Q}$ (so that a real number can be viewed as a sequence of approximations by rationals), an element of ${\mathcal L}/\sim$ can be viewed (locally, at least) as the metric completion of ${\mathcal E}$ (with metric $d(E,F)$ between two rational elementary sets $E,F$ defined as the elementary measure (or Jordan measure, if one wishes) of the symmetric difference of $E$ and $F$). The Lebesgue measure of a set in ${\mathcal L}/\sim$ is then the limit of the elementary measures of the approximating elementary sets. If one grants rational elementary sets and their elementary measures as having a physical interpretation, then one can view an element of ${\mathcal L}/\sim$ and its Lebesgue measure as having an idealised physical interpretation as being approximable by rational elementary sets and their elementary measures, in much the same way that one can view a real number as having idealised physical significance.

Many of the applications of Lebesgue measure actually implicitly use ${\mathcal L}/\sim$ rather than ${\mathcal L}$; for instance, to make $L^2({\bf R}^d)$ a Hilbert space one needs to identify functions that agree almost everywhere, and so one is implicitly really using the $\sigma$-algebra ${\mathcal L}/\sim$ rather than ${\mathcal L}$. So I would argue that Lebesgue measure as it is actually used in practice has an idealised physical interpretation, although the full Lebesgue measure on ${\mathcal L}$ rather than ${\mathcal L}/\sim$ does not. Not coincidentally, it is in the full $\sigma$-algebra ${\mathcal L}$ that the truth value of various set theoretic axioms of little physical significance (e.g. the continuum hypothesis, or the axiom of choice) become relevant.

1888381163938013251221959303lI got that paper you mentioned, I will go through it.6689181552049726125380378 705610810942151701The regularity mentionned in the theorem is not accurate, for two reasons. The first is that the spaces ${\cal C}^k$ don't behave well with PDEs. Often, it is better to work with ${\cal C}^{k,\alpha}$ with $\alpha\in(0,1)$.

The second and deeper reason is that the solutions have what is called a co-normal regularity. Physicists say that the solution is polarized. Let me explain this with the one-dimensional case ($x\in\mathbb R$). Then the equation reads $$(\partial_t+\partial_x)(\partial_t-\partial_x)u=F(u).$$ from this, you deduce that the quantity $w:=(\partial_t-\partial_x)u$ has a better regularity in the direction of $\partial_t+\partial_x$ than in other directions. In other words, $(\partial_t-\partial_x)w$ is more regular than $\partial_tw$ and $\partial_x w$ separately.

In three space dimensions, this can be quantified using pseudo-differential operators. But again, appropriate regularity results are directional, or polarized. The important theorems for propagation of regularity are due to Egorov and Taylor.

|and then there are all those variants of Priestley duality...Let $m$ be the permutation order.

It's easy to compute $n$-th inversion table by representing $n$ in the mixed radix system with bases from $1$ to $m$, and then turn it into a permutation (e.g., see this discussion).

522051819175Thanks everyone. I think I have solved it, and it doesn't require the inverse after all. @Robert 's answer led me to the solution by showing that I could construct the integral from a to b rather than some function of $u$ or its inverse. Once in that space, it is not too hard.2270906I don't really understand your comment because we still get to make the conclusion about properties of the ground model reals in the presence of a Woodin cardinal. And the countability of "old" sets of reals is key to this.212650221439006541871700387@abx I said "(rational) primes". In algebraic number theory, this is the way one refers to the primes in (to distinguish them from prime elements or prime ideals in other number fields). So it's not "prime in F". It's the usual prime integers.125615user98517A bit late, but the introduction of http://arxiv.org/abs/1206.4001 mentions two distinct "Erdos-Szekeres" theorems. 108253769175719391Zeta is not entire, but you can look at any non-polar L-function and you can subtract off the polar parts. ie: $zeta(s)+\frac1{1-s}$. Isn't there a selberg zeta function that, once you compensate for the scattering matrix, you morally know where the zeros are? I heard it given as an example but haven't worked it out firsthand. You can also remove finitely many zeros by dividing by a polynomial. A simpler example in my case is $\exp(L(s))$.z@HJRW: O.k., point taken -- I have taken back my close vote.1958689I was reading the abstract of a recent preprint (Division Algebras and Supersymmetry III by Juhn Huerta), and I wondered if something much simpler than what he was talking about had been worked on: have Lie $2$-groups been applied to the resolution of differential equations, in the same manner that Lie groups originated from the study of differential equations?

In other words, do Lie 2-groups arise as *symmetries* for (certain kinds of) differential equations, and can these in turn be used for the integration/resolution of those same differential equations? If they do not, then in what setting can a 2-group be understood as a symmetry (if any), and to what 'use' can this information be put?

My motivation here is to expand the toolset I can use to solve problems in classical analysis (like differential equations), and **not** to explore the other areas where Lie groups have developed in to (like Lie algebras and their classification, etc). For the purposes of this question, these issues are out-of-scope. In-scope are applications (to classical analysis) of generalizations going all the way to $\infty$-Lie groupoids.

I don't know if you're there anymore or interested in this question still, but perhaps these two resources could provide some juice for your investigation into the intuitive logic aspect of your question.

"Proofs are Programs: 19th Century Logic and 21st Century Computing" Philip Wadler, 2000. http://homepages.inf.ed.ac.uk/wadler/papers/frege/frege.pdf

"The modern development of the foundations of mathematics in the light of philosophy" Kurt Gödel, Collected Works, Volume III (1961) publ. Oxford University Press, 1981. https://www.marxists.org/reference/subject/philosophy/works/at/godel.htm (forgive the domain, I couldn't find it elsewhere)

I'm currently experimenting with a combination of proof diagrams, axiomatic proofs, and phenomenological reduction, in a very amateurish way.

Maybe these topics would be of use to you too. For more technical information on the combination of intuitive mathematical knowledge and axiomatic proof maybe this would be instructive: https://philarchive.org/archive/HIPPPAv1

17514010415831210133154834https://www.gravatar.com/avatar/a4323e2156183749a6e2ff21bc12911e?s=128&d=identicon&r=PG&f=1717144Dear Fernando: Thanks for you answer. I especially enjoyed the answer http://mathoverflow.net/a/49572/461 of Todd Trimble's to the question you linked to!If it's homework then it's not appropriate for MO (which is for research-level questions). You might get a better response at math.stackexchange.com which is for questions of all levels.1764548XIt is Theorem 2.4 in S, Krein's little bookaslan19057722068620Oh, I had just meant that when it comes to decidability questions about general diophantine sets, most of what we know comes from computability theory and the characterization of the diophantine sets as the c.e. sets, rather than from the general theory of diophantine sets. I think of the diophantine characterization as just another characterization on the list, like (many versions of) Turing machines, register machines, numerous other machine characterizations, group presentations, (failures of) tiling problems, game of life, and so on.1931645An interesting example involving self-intersection, but the valence is 4 is given on page 158 of Wenninger's *Polyhedron Models* (Model #102), this is Great Dodecahemicosahedron, with 10 hexagons and 12 pentagons. You can check it out as well on wikipedia. A similar example is the small demidecahemicosahedron. This provides a partial answer to Pisanski's follow up question.

Software Engineer with a keen skill for C# and Devexpress. Did a lot of PHP, HTML, CSS and Wordpress at the Jönköping School of Engineering. Recently started to refresh C++ by experimenting with OpenCV.

The claim in the question that maximal tori are the same as inclusion-maximal abelian subgroups is not correct. For example, the diagonal matrices with +1 or -1 on the diagonal form a maximal abelian subgroup of SO(n) that is not a torus.

4775691334085While reading this paper, the author provides an alternative definition of the Lagrange inversion formula. Call me crazy, but my intuition tells me that there's something wrong with his derivation. Can anyone verify if his argument is correct or not?

You should have love for technologies, love the work you are doing.Interest in something what all matters.

Oh yes!! This is an awesome book. The short version actually exists also in Russian and is very good as well but the long version is so much better.11023751598408hCould you remind us what the density of a graph is?124364459963@MatthiasWendt well, it's got to be more complicated than that, otherwise Kuiper's theorem would be a corollary of the contractibility of S^ooThanks, Qiaochu! You are encyclopedic. The reference to Math Intelligencer in the answer is exactly the one I've seen some years ago.391765253738178182215825069179522242544840898849537I don't understand your example $\phi_1$. If $G$ has an even number of vertices, then $\phi_1(G-v)=-1$, right? So the deck ratio is just minus $\phi_1$. So how can one be NP-hard and the other NP-easy? I think it might be sufficient to mandate nonzero values.Dear Pete, What Jim means is that H^1(X,G) should equal continuous homs. from pi_1 to G for any G (not just finite G). For the nodal curve X, H^1(X,Z) is non-zero, so there there should be a hom. from pi_1 to Z. The fact the dual graph is making this non-zero contribution to the H^1 is not coincidental. It is related to the fact that this part of the H^1 is of motivic weight 0. (See the discussion at the secret blogging seminar for more on this.) 1853093In the 19th century, a lot of efforts were made in order to solve the general quintic equation $x^5+a_4x^4 +a_3x^3 +a_2x^2 +a_1x +a_0$ using special functions. It turns out that the roots of this equation are expressible in terms of hypergeometric series. To wit, one possibility is by first reducing the number of parameters, to the form $x^5-x-t=0$. Then a Lagrange inversion argument essentially gives a root $$ z=t {}_4 F_3(\frac15,\frac25,\frac35,\frac45,\frac12,\frac34,\frac54,\frac{5^5}{4^4}t^4)=t+t^5+10\frac{ t^9}{2!}+15\cdot 14 \frac{t^{13}}{3!}+\ldots $$

1859444https://www.gravatar.com/avatar/99bcbdfb92ea453926ed592a7b660924?s=128&d=identicon&r=PG&f=1282530Olivier, thanks. The book is Hilbert Modular Forms and Iwasawa Theory (Oxford Mathematical Monographs).713228141469206563681570918460781335030Is this long list of two word answers OK? This almost feels like some sort of spam.1266232From: Twin Cities

Fractal_Love.gif: http://i.imgur.com/Nmv0tMG.gif

1725272xAdd "is of codimension 1" to your conditions and it's okay.610307dBrgy Sta. Teresita, Quezon City, NCR, Philippines@Scott: Oops, fixed. @Petya: yes, which was the comment that Deane made in my more trivial question.1107707194646I have to agree. Also, the construction of the Hilbert Scheme is barely scheched on page 8. That's a pretty huge shortcut for someone who has just finished reading Hartshorne. I'd suggest learning about this first for example in Mumford or FGA explained. Maybe it can be a good book to look at from time to time to measure how familiar you are getting with the general picture. Plus I find the font used in "Moduli of Curves" really bloated and this doesn't help reading it. 1003862So, in $R-Mod$, we have the rather short sequence

$\mathrm{Ext}^0(A,B)\cong Hom_R(A,B) $

$\mathrm{Ext}^1(A,B)\cong \mathrm{ShortExact}(A,B)\mod \equiv $, equivalence classes of "good" factorizations of $0\in Hom_R(A,B)\cong\mathrm{Ext}^0(A,B)$, with the Baer sum.

Question:

- $\mathrm{Ext}^{2+n}(A,B) \cong\ ??? $

While I suppose one could pose a conjugate question in algebraic topology/geometry, where the answer might look "simpler", I'm asking for a more directly algebraic/diagramatic understanding of the higher $\mathrm{Ext}$ functors. For instance, I'd expect $\mathrm{Ext}^2(A,B)$ to involve diagrams extending the split exact sequence $A\rightarrow A\oplus B\rightarrow B$, but precisely *what sort* of extension? Or is that already completely wrong?

This question is perhaps somewhat soft, but I'm hoping that someone could provide a useful heuristic. My interest in this question mainly concerns various derived equivalences arising in geometric representation theory.

**Background**

For example, Bezrukavnikov, Mirkovic, and Rumynin have proved the following: Let $G$ be a reductive algebraic group over an algebraically closed field of positive characteristic. Then there is an equivalence between the bounded derived category of modules for the sheaf $\cal D$ of crystalline (divided-power) differential operators on the flag variety, and the bounded derived category of modules with certain central character for the enveloping algebra $\cal U$ of Lie($G$). What is interesting is that it is *not* true that this equivalence holds on the non-derived level: The category of $\cal D$-modules is not equivalent to the category of $\cal U$-modules with the appropriate central character. This is true in characteristic 0 (this is the Beilinson-Bernstein correspondence), but something is broken in positive characteristic: there are certain "bad" sheaves that are $\cal D$-modules which make the correspondence not hold.

There are other results in geometric representation theory of this form. For example, Arkhipov, Bezrukavnikov, and Ginzburg have proved that there is an equivalence (in characteristic 0) between the bounded derived category of a certain block of representations for the quantum group associated to $G$, and the bounded derived category of $G \times \mathbb C^*$-equivariant sheaves on the cotangent bundle of the flag variety of $G$. Again, this equivalence does not hold on the non-derived level.

In general, there are a number of results in geometric representation theory that hold on the derived level, but not the non-derived level.

**Question**

Here's my question: Why would one be led to expect that a derived equivalence holds, when the non-derived equivalence does not? It seems as though the passage to the derived level in some sense fixes something that was broken on the non-derived level; how does it do that?

fDid you leave out the word "symmetric" on purpose?VX/S flat, X, S regular, but X/S not smooth20402951399489Are you familiar with the sparse fourier transform (e.g. https://people.csail.mit.edu/ludwigs/papers/sp14_sfft.pdf) ?1854773tWhen is a commutative ring the limit of its factor rings?1419772154903417446193758411697331595999https://graph.facebook.com/10211269888919155/picture?type=large890521979601 1452Yes, I was talking about closed manifolds. In section 2.2 of Aschenbrenner-Friedl-Wilton you find a discussion of $3$-manifolds with incompressible boundary.$\Gamma(k+1)=k!$ cancels,right?Makes me think you have a misprint there. Also, are $x$, $n$, and $t$ independent real (complex) variables or what?1357614922721208132717292502307461927459Have you looked at the coefficient of $t^{n-m}$ in the characteristic polynomial of the obvious square matrix?The dialogue is a very old (but funny!) routine, dating back at least to Abbott and Costello. See, e.g., http://freeclownskits.info/abbott-and-costello-skits/i-bet-you-that-youre-not-here-abbott-and-costello/1457944Since surface groups are also hyperbolic, this also shows that surface subgroups are never of finite index, as asked in the body. However, as Misha says, there are much more elementary ways to see either of these facts.309277203489814527321045217197944512745615417471013758@JasonStarr What is a reference for this? I will give my references. Believe they definitely mean _geometric_ genus by "genus".When people ask how to visualize things in higher dimensions, I think it is good to mention problems like this showing that we struggle to visualize things in $3$ dimensions.BobbyConsider the subset $\Omega\subset \mathbb{R}^N$ with boundary $\partial\Omega$ sufficiently regular and let $\Gamma\subset\partial\Omega$ be a $(N-1)$- dimensional submanifold of $\partial\Omega$. Consider the following problem \begin{cases} -\Delta u = \lambda u & \mbox{in }\Omega\\ u=0 & \mbox{in }\Gamma^c\\ \partial_\nu u=0 & \mbox{in }\Gamma \end{cases} where obviously $\Gamma^c$ is the complementar of $\Gamma$ in $\partial\Omega$, s.t. $\partial\Omega = \bar\Gamma \cup \bar\Gamma^c$ and $\Gamma \cap \Gamma^c =\emptyset$. What is the better functional space to define a variational formulation for this Mixed Dirichlet-Neumann problem? More precisely the space $$ H^1_0(\Omega\cup \Gamma) = \mbox{closure of } \mathcal{C}^1_0(\Omega\cup\Gamma) \mbox{ with respect to the norm of } H^1(\Omega) $$ can be studied and considered as an usual $H^1_0(D)$ with $D\subset\mathbb{R}^N$, in order to use the general results associated to the eigenvalue problem with Dirichlet boundary condition, or the fact $\Omega\cup\Gamma$ is not open can obstruct this idea?

20800110022161037848If $A$ is non-ordinary, then $T_p A$ is irreducible as a Galois representation. The submodule you are defining is Galois stable, isn't it? So it must be either zero or everything.nhttp://www.maths.ox.ac.uk/people/profiles/minhyong.kim5265238034076550551414134345215Ngo Bao Chau has recently proved the so called "Fundamental Lemma" (he is very likely to get the Fields medal for his work!) and this will have many dramatic applications to the Langlands program, though the full functoriality conjecture of Langlands is still far from being proved (but I am not an expert). James Arthur gave a talk at UBC on applications of the FL to the Langlands program but I have not been able to find a writeup of the talk.

12198902099791My apologies! I wasn't trying to rack up points but was concerned about the apparent size limit on comments. Eg: my discussion of the polynomials in the answer immediately above, would surely not fit?3671971802482701949https://www.gravatar.com/avatar/1e82fe8166ba5cccde3ce8b478820653?s=128&d=identicon&r=PG&f=1223234022432021189766374789400249https://www.gravatar.com/avatar/ea93710811be94ff1e1753dcee37b94b?s=128&d=identicon&r=PG&f=192022111729786597082267799What abut the subgroup generated by the matrices [[2,1],[1,1]], [[1,1],[1,2]] ?10019861656035I am currently trying to understand the general Green-Julg theorem, where $G$ is a compact group, $A$ and $B$ are $G$-$C^*$-algebras, and where $G$ acts trivially on $A$. The Green-Julg theorem states that there is an isomorphism $$ \mathrm{KK}^G(A,B) \rightarrow \mathrm{KK}(A, B\rtimes G).$$ Unfortunately, in all of the papers I can find, it's always the special case $A=\mathbb{C}$. Does anybody know a good paper where a proof is given for the general case?

Thank you.

2125415 Forming the set S is much the same kind of abuse as thinking of N as a subset of Z. You should never form the union of two sets that are not already given as subsets of a fixed ambient set; instead, you should form their disjoint union (which has a couple of common encodings in ZFC). Then of course 3 is not (barring the possibility of mistyped junk) an element of the disjoint union; but when considering whether 3 is a root, you really want to know whether i(3) is an element of the disjoint union (where i and j are the inclusion maps into a disjoint union). And this is true iff 3 is a root.551538The matrix $A = E -\epsilon I$, is a positive matrix for any $0<\epsilon <1$ and its eigenvalues are $n-\epsilon >0$ (simple Perron eigenvalue) and $-\epsilon$ with multiplicity $n-1$. So $A$ has $n-1$ negative eigenvalues.18304231588392pSorry for the silly question. What does rad mean here?Thanks for your answer. I will see this talk of Mandelbrot. But in fact we can choose whatever distribution to start.21554058657071776821This kind of issue may be one reason that we relax the notion of "efficient" to "polynomial-time computable"; it seems to generally be nicely model-independent.Isn't that just good ol' boring English? (It was probably Germanic a few hundred years ago, of course :) )2090835ZDerivative with multiple summation operators881053430956v@Zhi-WeiSun Can I publish this conjecture on your journal?I don't think there's any problem here. The question you want to ask is: Given a section $u$ of $E$, whether you get a consistent answer for $\nabla u$, using either frame. In other words, does $\nabla (u^ie_i) = \nabla (\hat{u}^i\hat{e}_i)$ using your change of frame formulas? I think you'll see what's going on if you work this out.

Let me elaborate a little bit. I never write change of frame formulas for the connection $1$-forms in isolation, because I am always confused by whether I should be acting on the frame by $g$ or $g^{-1}$. And whether $G$ is supposed to be acting on the right or left. But I know I will always get the right answer if I think about applying the connection to local section $u$ and expand it using two different local frames. Everything always automatically works out correctly when I do it this way.

2954570700552324318Would $P(-2)=1/3$, $P(1)=2/3$ be an asymmetric probability distribution? Or by asymmetry do you mean a distribution where the *expected* amount of movement on the first step is (say) to the right?Not an answer, but in the general context of 2-monads, a (strict) algebra $A$ with the property that every pseudo morphism with domain $A$ (and strict codomain) is isomorphic to a strict one is called *semi-flexible*. See Blackwell-Kelly-Power, *2-dimensional monad theory*, Theorem 4.7. In general, not every algebra is semi-flexible, and that is probably also the case here.Your alternate definition is ok if you restrict property 4 to level $N$ forms $g(z)$, where $N$ is the level of the newform $f(z)$. Actually it is sufficient to restrict property 4 to forms $g(z)$ of level at most $N$.

The restriction is necessary, because for any $d\in\mathbb{N}$ the oldform $f(dz)$ of level $dN$ has the same Hecke eigenvalues $a_n$ as $f(z)$ for $(n,dN)=1$. In particular, $f(Nz)$ has the same Hecke eigenvalues as $f(z)$ for $(n,N)=1$, rendering property 4 invalid in its original version.

In fact the name "newform of level $N$" comes from the fact that its Hecke eigenvalues $(a_n)$ are "new" for level $N$, i.e. they do not occur for smaller levels. In addition, these Hecke eigenvalues $(a_n)$ determine a 1-dimensional subspace within the space of level $N$ forms.

**Remark 1.** Note that the same concept makes sense for Maass forms where the Hecke eigenvalues $a_n$ are not expected to be algebraic numbers. So I suggest that in property 3 you allow $a_n\in\mathbb{C}$.

**Remark 2.** You quote the original definition of a newform incorrectly. The correct definition is that $f(z)$ is orthogonal to any oldform $g(dz)$, where $g(z)$ is a Hecke eigenform of level $M\mid N$ such that $M<N$ and $dM\mid N$. See Section 4 in Atkin-Lehner's paper "Hecke operators on $\Gamma_0(m)$" (Math. Ann. 185 (1970), 134-160).

The question I have arose while reading Waterhouse's Thesis (Abelian varieties over finite fields. Ann. Sci. École Norm. Sup. (4) 2 1969 521–560.), and motivates another question I recently asked.

Let $p$ be a prime number, and $\mathcal{C}$ an $\mathbf{F}_p$-isogeny class of $\mathbf{F}_p$-simple abelian varieties of dimension $g$ associated with a certain Weil $p$-number $\pi$ which will be assumed not real, i.e., $\pi\neq\pm\sqrt{p}$. For any $A$ in $\mathcal{C}$, the $\mathbf{Q}$-algebra $\textrm{End}_{\mathbf{F}_p}(A)\otimes\mathbf{Q}$ is commutative and equal to the non-real CM number field $\mathbf{Q}(\pi)$, which has degree $2g$. The endomorphism ring of any object of $\mathcal{C}$ is an order of this field containing $\pi$ and $p/\pi$. I would like to assume, for simplicity, that

(*) the order $\mathbf{Z}[\pi, p/\pi]$ is maximal at $p$.

Let $A$ be a fixed object of $\mathcal{C}$, and let $R_A$ be its endomorphism ring. To any nonzero ideal $I$ of $R_A$ we can attach, following Waterhouse, the finite $\mathbf{F}_p$-subgroup $H(I)$ of $A$ consisting of the (subgroup-scheme theoretic) intersection of all kernels of the nonzero elements of $I$.

On the other hand, associated to any finite $\mathbf{F}_p$-subgroup $N$ of $A$ there is the ideal $J(N)$ of $R_A$ consisting of all morphisms $\varphi:A\to A$ vanishing on $N$.

Waterhouse calls a nonzero ideal $I$ of $R_A$ a *kernel ideal for $A$* if the equality $I=J(H(I))$ holds.

The question I have asks for a correspondence between ideals of $R_A$ and finite subgroups of $A$ defined over $\mathbf{F}_p$:

**Q: Can we find $A$ such that for all nonzero ideals $I$ of $R_A$ and for all finite $\mathbf{F}_p$-subgroups $N$ of $A$ we have $I=J(H(I))$ and $N=H(J(N))$?**

The idea behind the question is to ask for an abelian variety $A$ such that the category $\mathcal{C}$ can be reconstructed starting from the contravariant functor $T\mapsto h_A(T)=\mathrm{Hom}_\mathcal{C}(T,A)$ going from $\mathcal{C}$ to rank-one, torsion free $R_A$-modules.

If I got it right, question Q is equivalent to ask $h_A$ be an anti-equivalence of categories. (At least if Q holds for a certain object $A$ then the equality $I=J(H(I))$ ensures essential surjectivity of $h_A$ and the assignment $I\mapsto A/H(I)$ gives an inverse functor defined on nonzero ideals of $R_A$).

Waterhouse proves several facts related to Q (which led to me ask the question to begin with):

1) Let $I$ and $J$ be kernel ideals for $A$. Then $A/H(I)\simeq A/H(J)$ in $\mathcal{C}$ if and only if $I\simeq J$ as $R_A$-modules;

2) If $I$ is a kernel ideal, then $\textrm{End}_{\mathbf{F}_p}(A/H(I))$ is the order of $I$ (inside $\mathbf{Q}(\pi)$);

3) If the $\ell$-adic Tate module $T_\ell(A)$ is free (of rank one) over $R_A\otimes\mathbf{Z}_\ell$ for all primes $\neq\ell$ and the Dieudonn\'e module $T_p(A)$ is free over $R_A\otimes\mathbf{Z}_p$, then every nonzero ideal of $R_A$ is a kernel ideal for $A$;

(the assumption (*) above ensure freeness of $T_p(A)$, if I'm right)

4) Any $A'$ in $\mathcal{C}$ is isogenous to an $A$ as in 3) in such a way that $R_{A'}\simeq R_A$ (the isomorphism between the two rings being definable from the isogeny from $A'$ to $A$);

5) Any order of $\mathbf{Q}(\pi)$ containing $\pi$ and $p/\pi$ arises as endomorphism ring $R_A$, for some $A$.

These results suggest that in order to find a good candidate for question Q we should start with an object $A$ such that $R_A$ is $\mathbf{Z}[\pi,p/\pi]$ and such that $T_\ell(A)$ is free over $R_A\otimes\mathbf{Z}_\ell$ for any $\ell$. It follows from 3), 4) and 5) that such an $A$ exists.

Since by 3) we have $I=J(H(I))$, we see that Q has a positive answer if for any $\mathbf{F}_p$-subgroup $N$ of $A$ we have $N=H(J(N))$. This equality holds if and only if for all primes $\ell$ the $\ell$-primary parts $N_\ell$ and $H(J(N))_\ell$ of the two subgroups coincide. If $\ell\neq p$, then the equality $N_\ell=H(J(N))_\ell$ amounts to the following:

Let $\Lambda$ be the $R_A\otimes\mathbf{Z}_\ell$-stable lattice of $V_\ell(A):=T_\ell(A)\otimes\mathbf{Q}_\ell$ containing $T_\ell(A)$ and corresponding to $N_\ell$.

Q($\ell$): Is it true that $\Lambda$ is equal to the intersection of all pre-images of $T_\ell(A)$ under the $R_A\otimes\mathbf{Z}_\ell$-endomorphisms of $V_\ell(A)$ sending $\Lambda$ to $T_\ell(A)$? Equivalently, is it true that for any $v\in V_\ell(A)$ with $v\notin\Lambda$ there exists an $R_A\otimes\mathbf{Z}_\ell$ lattice $\Lambda'\subset V_\ell(A)$ containing $\Lambda$, not containing $v$, and isomorphic to $T_\ell(A)$?

Since the assumption (*) ensures that $N_p=H(J(N))_p$, Q has a positive answer if and only if Q($\ell$) does for a single object $A$ for all $\ell\neq p$.

Question Q can be tested on *ordinary* $\mathcal{C}$ using a result of Deligne (Variétés abéliennes ordinaires sur un corps fini. Invent. Math. 8 1969 238–243) which says, when applied to the isogeny class $\mathcal{C}$ (and using the maximality assumption (*)), that $\mathcal{C}$ is equivalent to the category of rank one, torsion free $\mathbf{Z}[\pi,p/\pi]$-modules. The proof of this elegant result uses Serre-Tate canonical lifting of ordinary AVs. Using this theorem, Q translates into this other question, where $R$ should be taken equal to $\mathbf{Z}[\pi,p/\pi]$.

If you have anything to say about Q or Q($\ell$), or both, I would be happy to hear it.

Heard in high-school History class, no reference unfortunately: in the early 20th century, someone published a monograph on the dimensions of a certain small building; derived many important constants from basic operations on said dimensions, showing the intent of the architects. The building in question was... ...a public urinal!I have asked this question at MSE but got no response. I have rephrased it so that anyone who knows operator theory and integral equations could help me out... I faced a problem in physics which is a non-linear Fredholm integral equation (Hammerstein). It is of the form: \begin{equation} y(p)=f(p)+\int_0 ^{\infty}\frac{e^{ik∧p}}{y(k)}dk \end{equation} There are two subtleties:

- The kernel is non-degenerate or non-separable.
- The non-linearity is reciprocal.

The kernel is symmetric as the wedge product is anti-symmetric,i.e. $\overline{K(p,k)}=K(k,p)$. I tried solving it using the collocation method but got divergence due to it's severe bound on the upper limit. I am now starting to work with Monte-Carlo integration and it's too complicated. That's why I want to know if any of you could help me.

\https://i.stack.imgur.com/ESFqL.jpg?s=128&g=1795818This is a special case of Remark 5.33 in my book with Adámek which deals with small cone injectivity.4265846737031310840905278fElementary quantum scattering problem on the line.1127387rHow to ensure a clean conservative extension in general?1126605jAn identity of operator norms and de Leeuw's theorem222528863423859764444754It is said from http://www.artofproblemsolving.com/community/c6h1289401_inequality7916792053305532283https://lh4.googleusercontent.com/-GDcri2FdbqQ/AAAAAAAAAAI/AAAAAAAAAIo/Ze5eyHGkUec/photo.jpg665052The book *Conformal Geometry of Discrete Groups and Manifolds* by Boris N. Apanasov contains a detailed description of what it means for a group to be arithmetic, and why the definition manifests as it does for topology applications. He includes examples that illustrate the ideas behind statements 1 and 2, and discusses Vinberg's "simplest type" groups separately, calling them "real arithmetic" groups. He does not address the characterization using quaternion algebras, as far as I can tell. See, the book is $234.40 so I'm looking at the Google preview. Searching the word "quaternion" there yields no results. This use of quaternions does not apply in arbitrary dimensions, and that is the author's topic.

The use of quaternion algebras in characterizing arithmeticity is described in the Hilden-Lozano-Montesinos chapter of *Topology 90*. There's also a good explanation of it, with examples, in Kate Peterson's article "Arithmetic groups and Lehmer's conjecture." I have seen noting in print proving the 3rd statement, but it's a good exercise to work it out for oneself.

Cubical type theory is a variant of type theory which has all the usual (and some unusal) computational properties, and the Univalence Axiom is a theorem of cubical type theory. As was already pointed out in the comments, there are implementations of cubical type theory that one can play with (even for 90 hours).

237124467061Presumably "excessive" should be "excisive". Or perhaps topologists have a better sense of humour than algebraists.Hecke-module structure implicit in definition of automorphic forms in Borel-Jacquet's Corvallis articleI would like to add another geometric description of $\mathbb C\mathbb P^n \#\overline{\mathbb C\mathbb P^n}$ as sphere bundle (same works for $\mathbb C\mathbb P^n \# \mathbb C\mathbb P^n$):

Let $S^1 \to S^{2n+1} \to \mathbb C\mathbb P^n$ be the Hopf fibration. Then $S^1$ acts on $\mathbb R^2$ by rotations and the associated vector bundle $E$ is the normal bundle of $\mathbb C\mathbb P^{n-1}$ in $\mathbb C\mathbb P^n$. Thus the total space of $E$ is diffeomorphic to $\mathbb C\mathbb P^n$ with a disc removed. Hence if we glue the disc bundles of $E$ with $\overline E$ (which denotes here the the bundle with the reversed orientation) along their common boundary one obtains $\mathbb C\mathbb P^n\# \overline{\mathbb C\mathbb P^n}$.

Moreover one deduces from this description that the connected sum is given as the quotient $(S^{2n-1}\times S^2)/S^1$, where $S^1$ acts on $S^{2n-1}$ such that it induces the Hopf fibrations and on $S^2$ by rotations. This quotient has a projection map to $\mathbb C\mathbb P^{n-1}$ with fibre $S^2$. (This is just the associated $S^2$-bundle to the Hopf fibration)

zhttp://ambassadorconstruction.com.au/home-renovations-perth/119364919274942261141(er, make that "when x + 355 and x + 113 pi are not close to odd multiples of pi/2, that is, the points of discontinuity of tan x").413934146692822865278910799215562284944No, but the question had the word "stack" in it, and for most people, this means "stack of groupoids".542843129951771136Gil, as you said, this is one of those typical FPP problems which seems obvious but is hard to prove. What have you tried already? It'd be helpful to know of some naïve attempts which didn't work.

Here are my thoughts:

Claim: There exists non-random $\lambda$ such that, with probability one, for large n, all shortest paths between $0$ and $(n,0)$ meet $\lambda n + o(n)$ edges. (this is a LLN-type theorem so it shouldn't be hard to prove; e.g., via energy-entropy methods, since your passage time distributions are bounded)

Thus one can consider the probability space $\Omega_n$ consisting of all paths between $0$ and $(n,0)$ which meet $\lambda n + o(n)$ edges. A shortest path is a random variable $X_n$ on this space with a certain probability distribution.

Claim: There exists $\sigma$ such that $|\Omega_n| \approx \sigma^n$. (should be easy: $\log|\Omega_n|$ is probably subadditive)

Let $\Omega_n'$ be the subspace of paths which meet the middle edge, so that $|\Omega_n'| \approx \sigma^{n/2} + \sigma^{n/2}$.

Suppose that there exists $p > 0$ such that the shortest path between $0$ and $(n,0)$ meets the middle edge with probability at least $p$. (*)

Here is the part which I'm struggling to quantify. Intuitively, the distribution of $X_n$ should be smeared smoothly over $\Omega_n$. Certainly the mean is a horizontal line segment, but even paths which veer quite far away aren't unreasonable. However, if (*) holds, with probability at least $p$, $X_n$ concentrates on the much smaller subspace $\Omega_n'$. This seems wrong.

Perhaps all I've done is to translate one "obvious" statement into another. Hopefully this helps a bit. Good luck!

:Given $M$, minimize $|Mx|_0$ @MikeBattaglia: If I remember right, Cantor introduces cardinality (which he called *Potenz*, i.e. *power*) by saying two sets have the same power if they can be put in 1-to-1 correspondence. So in modern terms, he’s implicitly defining cardinalities as the quotient of the class of sets by the “equipotence” relation; and in a modern treatment, general class quotients constructed by Scott’s trick do indeed take a bit of machinery. But historically, it’s not about a set theory or a fragment of one in the modern sense; it’s that he was admitting such quotients as a principle of reasoning.Dear Barbara, thanks for your reply. I understand, that foliations are in general only Hölder and that I cannot expect $\delta_x(y)$ to be smooth in $x$. My question is rather if for given $x$ the conditional measure is smooth along the sheat $W_x$kwejhr11919771296928lThis is exactly what I was looking for, thanks a lot! 1017103617130825To prove that $\mathbb P^1\times E$ is not rational, one can also use the fact that its $\pi_1$ is obviously non-trivial. 1141695141500117265371742009Dear Agno, I was looking for this question and had trouble finding it because of the (lack of) tags. Perhaps you could add a gamma-function tag. Regards, Matthew 1705981623052284859As with Jim's answer also the comment I added there vanished, I'll repeat the simple counterexample here: take the embedding $SL_2(q) \to GL_{2e}(p)$ by simply considering a $2$-dimensional $F_q$-vector space as $2e$-dimensional $F_p$-vector space.1893011@NikWeaver Usually, I personally flag my question for the attention of the moderator to make it CW. In this particular case, I didn't, since I was under the impression that though there could be multiple answers, none of them would be opinion based, and each one could be only correct or wrong, like almost every other math question on MO.The chart function is a diffeomorphism, which suffices. You can take any oriented orthonormal basis of the tangent space of $\partial \Omega$ at a point, and wedge it together to give a volume form. That is the volume form whose associated measure we want to integrate, and that is the volume form whose integrals agree with the integrals I have described by extending functions to be constant in perpendicular directions.362612Another important example is given by groups acting on graphs, especially (but not only) in finite group theory. Quite a few of the sporadic finite simple groups have actually been discovered as automorphism groups of graphs, e.g. the Hall-Janko group $J_2$ or the Higman-Sims group $HS$.

A related class of examples with more "structure" are so-called incidence geometries (combinatorial objects with geometric structure), and the most prominent example of those are the (Tits) buildings, introduced by Jacques Tits in the early 70's. (In fact, the example of Bruhat-Tits trees mentioned by Qiaochu Yuan is a very specific example of this situation; these are buildings of type $\tilde A_1$.)

183229429132Reid, it should be possible to prove by explicit constructions like the ones I mentioned that ratios strictly between 1/2 and 3/4 are attainable.3835761719591766563I still not quite understand. The assumptions in the CLT are that that X_n are square integrable (loosely speaking) which more or less is eqivalent to the fact that their tails decays like o(t^{-2}). @Felipe Voloch, thanks for your comment. Perhaps Selberg's collected works (+Iwaniec-Kowalski perhaps) would be the closest to what I look for now.436894033856237675812296339XPresentability of the source of a fibration1593593890566pThe proper place to ask: http://math.stackexchange.com/I'm guessing you are using Sylvester's construction S for a Hadamard matrix . What happens if you use PS instead of S for P a random or clever choice of signed permutation matrix? Gerhard "Irregularity May Lead To Density" Paseman, 2016.09.11.120168018863471869815As you say, I expect that this has been known to experts for a long time, but it follows as a special case of Theorem 11.22 in my paper Enriched indexed categories, since when $V$ is the self-indexing of $S$, small $V$-categories include internal $S$-categories, and large $V$-categories are essentially fibrations over $S$.

2152054It's been a while since I looked (and I don't have a copy handy), but doesn't Mumford give a construction of the dual isogeny over arbitrary fields in his book on *Abelian Varieties*? Probably Lang does, too, in his *Abelian Varieties" book, but that's in the older Weil-style language, so probably harder to read.285943^Sperner's theorem and "pushing shadows around"18178722157945This question may be elementary, I have asked it on math.stackexchange.com but have not received any answer yet. Note that I am not an expert on theta/elliptic functions.

Define Jacobi's theta function with argument zero and nome $q$ (often also denoted by one of the following: $\theta_0$, $\theta_4$, $\vartheta$, $\vartheta_4$ or $\vartheta_0$):

$$\theta(q) = 1+2\sum_{n=1}^\infty (-1)^n q^{n^2}$$

plot of the function via Wolfram|Alpha

I am looking for a simple/standard/illuminating proof of the fact that $\theta(q)$ is convex for $q\in[0,1]$. The proof I found goes like this: We have

$$\theta'(q) = 2\sum_{n=1}^\infty (-1)^n n^2 q^{n^2-1}$$

and one can show that for some $q_0\in(0,1)$, $n^2q^{n^2-1} - (n+1)^2q^{(n+1)^2-1}$ is increasing in $[0,q_0]$ for any $n\ge 2$. This gives convexity of $\theta(q)$ in $[0,q_0]$. For the remaining values of $q$, one uses the representation of $\theta$ as a sum over Gaussian kernels:

$$\theta(e^{-\pi^2t/2}) = 2 \sqrt{\frac{2}{\pi t}}\sum_{n=1}^\infty \exp\left(-\frac{(2n-1)^2}{2t}\right)$$

With this representation, one can show that the second derivative (wrt $q$) of each summand is positive for $q \ge q_1$, with $q_1 < q_0$. This yields convexity of theta.

I don't like this proof, because it requires calculating $q_1$ and $q_0$ explicitly and it is not very illuminating. I tried playing around with the representation of $\theta(q)$ as the infinite product

$$\theta(q) = \prod_{n=1}^\infty (1-q^{2n-1})^2(1-q^{2n}),$$

but didn't manage to find anything, except that the partial products

$$\prod_{n=1}^N (1-q^{2n-1})^2(1-q^{2n})$$

all seem to be convex in $[0,1]$, which would prove the statement.

All suggestions are very welcome!

Ah, that's a nice fact to have around. Thanks for the insight! (and consider writing it into the answer).179942312726831048018508821You only have a mild solution for z0∈H. A solution for the differential equation exists for $z_0\in D(C)=D(A)$ only in the classical sense.105308810553577~$C_n^{n/2}$ is old notation still common in some countries... 826004They have one of these at the nearby Queens Hall of Science. I wonder if MoMath can just purchase the Hall of Science's math exhibits. They're nice but completed ignored by most visitors.1543384828266867652182819Note that my earlier comment referred to an earlier version of the question.22054582083453nHow would you call a subscheme of a smooth $S$-scheme?2156680pYou could check out $p=239$ when $\frac{p^2+1}{2}=13^4$1465744You probably mean to say "...decomposition of $T(M)\otimes_{\mathbb{R}} \mathbb{C}$ into..."2110640It is a (folklore?) fact that if $\kappa$ is a regular cardinal, and $\mathbb{P}$ is a $\kappa$-closed poset such that $\Vdash_\mathbb{P} |\mathbb{P}| = \kappa$, then $\mathbb{P}$ is equivalent to $Col(\kappa,\mathbb{P})$, the collection of partial functions from $\kappa$ to $\mathbb{P}$ of size $<\kappa$, ordered by inclusion. Hence if $\mathbb{P}$ is any $\kappa$-closed forcing, then $\mathbb{P} \times Col(\kappa,\mathbb{P}) \sim Col(\kappa,\mathbb{P})$, so $\mathbb{P}$ is completely embeddable into $\mathcal{B}(Col(\kappa,\mathbb{P}))$.

If $\mathbb{P}$ is $\kappa$-strategically closed (following the notation from Cummings' chapter of the *Handbook of Set Theory*, not Jech's book), then one can show that $\mathbb{P}$ is completely embeddable into $\mathcal{B}(Col(\kappa,\mathcal{P}(\mathbb{P})))$, since after forcing with this collapse we can use the strategy to build a filter on $\mathbb{P}$ that is generic over the ground model. It is not clear to me whether we can always embed such $\mathbb{P}$ into the smaller forcing $Col(\kappa,\mathbb{P})$.

**Question:** Are there counterexamples to the statement, "If $\kappa$ is regular and $\mathbb{P}$ is $\kappa$-strategically closed, then there is a complete embedding of $\mathbb{P}$ into $\mathcal{B}(Col(\kappa,\mathbb{P}))$?

student at bvicam currently pursuing mca,certified ethical hacker,ccna,mcitp,java and c developer.

It also helps Google index the site. I changed the title 16 hours ago, but this thread is already the top result for the search "connectedness fibers" (without quotes): http://www.google.com/search?q=connectedness+fibers0AbelianConvexityTheorem79060620419111728771676291Two resources that haven't been mentioned are Cornell University's digital collection of historical math monographs and the University of Michigan's historical mathematics collection.

327717https://www.gravatar.com/avatar/358114a66e6e6adbd67ac902c05f03d6?s=128&d=identicon&r=PG&f=1Prove the functor $[n]\to [n]\star [n]$ preserves inner anodyne198717614707541631411593805$There are two simple and classic enumerations that still I'm puzzled about. Let's start with a simple counting problem from a well-known dynamical system.

**fact 1** Consider the "tent map" f:[0,1]→[0,1] with parameter 2, that is

f(x):=2min(x,1-x).

Clearly, it has 2 fixed points, and more generally, for any positive integer *n*, there are 2^{n} periodic points of period *n* (it's easy to count them as they are fixed points of the *n*-fold iteration of *f*, which is a piecewise linear function oscillating up and down between 0 and 1 the proper number of times). To count the number of periodic orbits of *minimal* period *n*, a plain and standard application of the Moebius inversion formula gives

Number of n-orbits of (I,f) = $\frac{1}{n}\sum_{d|n} \mu(d)2^{n/d}.$

(**rmk**: any function with similar behaviour would give the same result, e.g. *f(x)=4x(1-x)*,...&c.)

Now let's leave for a moment dynamical systems and consider the following enumeration in the theory of finite fields.

**fact 2** Clearly, there are 2^{n} polynomials of degree *n* in $\mathbb{F}_2[x]$. With a bit of field algebra it is not hard to compute the number *I(n)* of the irreducible ones. One can even make a completely combinatorial computation, just exploiting the unique factorization, expressed in the form:

$\frac{1}{1-2x}=\prod_{n=1}^\infty (1-x^n)^{-I(n)}.$

One finds:

Number of irreducible polynomials of degree

nin $\mathbb{F} _ 2[x]$ = $\frac{1}{n}\sum_{d|n} \mu(d)2^{n/d}.$

**Question**: it's obvious by now: is there a natural and structured bijection between periodic orbits of *f* and irreducible polinomials in $\mathbb{F}_2[x]$? How is interpreted the structure of one context when transported ni the other?

(**rmk**: of course, analogous identities hold for any p > 2)

The method explained in Hausemöller's book on elliptic curves is as follows:

Take a general quartic $v^2=f_4(u)=a_ou^4+a_1u^3+a_2u^2+a_3u+a_4$, and let

$$u=\frac{ax+b}{cx+d}\qquad v=\frac{ad-bc}{(cx+d)^2} y$$

Substituting you get:

$$v^2=\frac{(ad-bc)^2}{(cx+d)^4}y^2=f_4\bigg(\frac{ax+b}{cx+d}\bigg)$$

which implies

$$(ad-bc)^2y^2=f_4\bigg(\frac{ax+b}{cx+d}\bigg)(cx+d)^4=\sum_{i=0}^4a_i(ax+b)^{4-i}(cx+d)^i=$$ $$=c^4f_4\bigg(\frac{a}{c}\bigg)x^4+f_3(x)$$

where $f_3(x)$ is a cubic polynomial whose coefficient of $x^3$ is $c^3f'_4(a/c)$. For $a/c$ a simple root of $f_4$ and $ad-bc=1$, this leaves the cubic equation $y^2=f_3(x)$.

From here you can use the tools you mention in the question to take care of the cubic.

226186960338564167569521116171~Who first considered constructibility of simplicial complexes?1063470414191477651193108814369191690718~https://graph.facebook.com/1669025039851384/picture?type=large1698576138136685874Is there literature about bands (semigroups which have these relations as identities) which would talk about these sets in things that are not bands? Gerhard "Looking For The Binding Tie" Paseman, 2018.11.24.This is also in Kodaira's "Complex Manifolds and Deformation of Complex Structures" for general complex tori (not necessarily algebraic). See pages 216-218 for the calculation.In the Schwartz Christoffel differential vector equation, just use higher derivatives instead of first derivatives.

527637**!**

(Title of a talk about the factorial function by Manjul Bhargava at the Clay conference in Paris in the year 2000.)

3677431659550hOn the Internet, nobody knows you're a dog.

17822562076790792563600205150336886058I don't know what you mean by "the morphisms ... over X are uniquely determined"1111040The number of $K$-permutations of the numtiset $\{ 1^D, 2^D, \ldots, N^D \}$ is $$\sum_{j_1+j+2+\dots+j_N=K\atop 0 \leq j_i \leq D} \binom{K}{j_1,j_2,\dots,j_N}.$$ (summands here are multinomial coefficients)

Alternatively, denoting by $m_i$ the number of $j$'s equal $i$, we get a formula as the sum over restricted partitions: $$\sum_{0m_0+1m_1 + \dots + Dm_D = K\atop m_0 + m_1 + \dots + m_D = N,\quad m_i\geq 0} \binom{N}{m_0,\dots,m_D} \frac{K!}{1!^{m_1} 2!^{m_2} \cdots D!^{m_D}}$$

For the example with $N=9$, $D=3$, $K=4$, the latter formula consists of four summands and gives: $$\binom{9}{5,4} \frac{4!}{1!^4} + \binom{9}{6,2,1}\frac{4!}{1!^2 2!^1} + \binom{9}{7,1,1}\frac{4!}{1!^1 3!^1} + \binom{9}{7,2}\frac{4!}{2!^2}$$ $$ = 3024 + 3024 + 288 + 216 = 6552$$ as expected.

1842355Ayman, I have just erased a the alluded comment, as it was not justified, sorry. One more remark is that my earlier comment regarding maximality of $G$ in $O(d)$ assumed $d>2$. It is possible that things get more rigid in higher dimension.Do you know how to construct a variety from the embedding of $\mathbb{P}^1$ in $\mathbb{P}^3$ using $\mathcal{O}(3)$?1738452356674895951https://lh3.googleusercontent.com/-VPT66UiThc8/AAAAAAAAAAI/AAAAAAAAUSY/wBLajtpUyzw/photo.jpg?sz=128643106Why do you think completeness is an issue? It's actually not that much of a problem with Lean because it has actual quotients and it's not that hard to work with them!I am a PhD student at University of Toronto. I am moving to Washington University in Saint Louis this fall.Dear Jack -- Precisely $R_d$ parameterizes first order deformations of $X$ as an abstract projective manifold which come from first order deformations of $X$ as a closed submanifold of $\mathbb{P}^n$. This part can be seen directly: given a base space / parameter space $B$ for deformations of $X$ with local coordinates $t_1,\dots,t_d$, and given a deformation of $X$ with defining equation $F(t_1,...,t_n;x)$, then the image in $R_d$ of the generator $\partial/\partial t_i$ of $T_0 B$ is just $\partial F/\partial t_i$. 36201I code in C,C++,JS and R. My core interests are algorithms, Machine learning and Math.

RIs there a weak strong regularity lemma?198520322965521566976Yes, this is true as soon as you assume that the isomorphism of sheaves $\mathcal O(E)\cong\mathcal O(F)$ is $\mathcal O_X$-linear.

To prove, pick a cover $(U_i)_{i\in I}$ of $X$ be opens which are trivializing both for $E$ and $F$. Now if you have an isomorphism of sheaves $\phi:\mathcal O(E)\to\mathcal O(F)$ then you can restrict it to each of these opens and get holomorphic functions $\phi_i:U_i\to GL_n(\mathbb{C})$. If you write carefully the gluing property you'll get something like: on intersections, $f_{ij}\phi_i=\phi_je_ {ij}$, where $e_{ij}$ and $f_{ij}$ are transition functions for $E$ and $F$.

This defines a bundle map.

203385113084201660143So now we deduce thanks to your wonderfull answer following result, because we can reduce to affine case. If $f:X \to Y$ is faithfully flat morphism of integral schemes and $X$ is normal, then also $Y$ is normal.( Maybe should add $f$ locally of finite presentation to be sure $f$ is open?). Is that true, Professor?1086381186191513902881527908lI enhanced my answer to cover the case $k=\mathbb Q$.Vdo they have a trivial automorphism group?10976582250694426515635232812843281532608(What you get is that if j:V to M is the ultrapower by any ultrafilter U on any set X, then every element of M has the form j(f)([id]). You can prove this by building an isomorphism from the ultrapower to the sets of this form. This way of thinking is also known as "seed theory".

**Theorem.** Suppose that j:V to M is an elementary embedding of the universe V into M. Then j is the ultrapower map by a measure on a set if and only if there is some s in M such that every element of M has the form j(f)(s).

That is, the ultrapower embeddings are precisely the embeddings whose target is generated by a single element.

Proof. If j is the ultrapower by U on X, then let s=[id], and argue that [f]_U is j(f)(s). Conversely, if M = { j(f)(s) | f in V }, where we assume that f is a function on some set X such that a ∈ j(X), then define the measure U by A ∈ U iff s ∈ j(A). This is a κ-complete ultrafilter on P(X). One can show that Ult(V,U) is isomorphic to M, by mapping [f]_U to j(f)(s). QED

**Theorem.** If U is a κ complete ultrafilter on κ with ultrapower embedding j:V to M, then every element of M has the form j(f)(κ) if and only if U is isomorphic to a normal measure.

Proof. You know the backwards implication. For the forward implication, suppose that every element of M has form j(f)(κ). In particular, β = j(f)(κ), where β = [id]_U is the seed for U. But also, κ = j(g)(β), where g(α) is the smallest ξ for which f(ξ)=α. Let μ = { X subset κ | κ ∈ j(X) } be the induced normal measure. Note that X in μ iff j(g)(β) in j(X) iff β in j(g^{-1}X) iff g^{-1}X in U. So μ is Rudin-Kiesler below U. Also, U is Rudin-Kiesler below μ since X in U iff f^{-1}X in μ. So μ and U are isomorphic.QED

One may illustrate the situation with product measures. Suppose that U is normal. The product measure UxU is isomorphic to the two-step iteration, where j_0:V to M is the ultrapower by U, and h:M to N is the ultrapower in M by j_0(U). Every element of M has the form j_0(f)(κ), and every element of N has the form h(g)(κ_{1}), where κ_{1} = j_0(κ). If j is the composition of j_0 and h, then j:V to N and every element of N has the form j(f)(κ, κ_{1}). If one only looks at j(f)(κ) inside N, then you will only get ran(h), which is isomorphic to M, but not all of N. So this would be a counterexample to what you asked about.

Researcher in theoretical computer science focusing on circuit complexity and communication complexity.

|Any reason to look for induced trees and not induced forests?2292598dSelf-equivalences of the stable homotopy category681798Is the following fact true?

Let $v_1,\ldots, v_k \in \mathbb{R}^2$, $\|v_i\|\leq 1$, be vectors that add up to zero. Does there exist a permutation $\sigma\in S_k$ and vectors $w_1,\ldots, w_k \in \mathbb{R}^2$, $\|w_i\|\leq 1$, such that $v_{\sigma(i)}=w_i-w_{i-1}$? (here, I assume $w_0=w_k$)

**Edit. Related (known) facts:**

1. The same fact in $\mathbb{R}^1$ is true. (can be easily proven by choosing $w_0=0$ and $\sigma(i)$ such that $\|w_i\|\leq 1$ for $w_i:=w_{i-1}+v_{\sigma(i)}$)

2. For each $\epsilon>0$ there exists a family of vectors $v_i$, such that $\max_i\|w_i\|>1-\epsilon$. See my comment below for the proof.

Based from Harminc and Sotak's result, www.fq.math.ca/Scanned/36-3/harminc.pdf

We know that under certain condition, an arithmetic progression can contain an infinitely many palindromes.

My question will be, if I have a system of Arithmetic Progression such as

\begin{equation*} 3t+2\tag{1} \end{equation*} \begin{equation*} 4t+1\tag{2} \end{equation*} can I always find an integer $t$ such that $(1)$ and (2) are both palindromes? I know in my example that the answer is yes. But in general, if I have the system

\begin{equation*} pt+j\tag{1} \end{equation*} \begin{equation*} qt+k\tag{2} \end{equation*}

with $\text{gcd}(p,q)=1$, can I find a $t$ for all $j<p$ and for all $k<q$ such that $(1)$ and $(2)$ are both palindrome? I am curious about this but I do not know how to answer it.

Or are there any reading materials that can help me on answering my query? Kindly help me.

Thanks for your help.

7630166226461876316617282274571196812819260481969647Not in general. There's a spectral sequence $H^i(G,Ext^j(C,A))\implies Ext^n(C,A)$ where the left hand Ext is as abelian groups (if $C$ is finitely generated). See, for example, the first section of Milne's book on Arithmetic Duality Theorems.@Martin: I don't think that the OP is positing infinite formulas. But only countably many of the maps from $\omega$ to $\omega$ (for example) are defined by formulas.553392Consider a bi-coloring of $\mathbb{N}^2$, (black and white), where we wish to maximize the limit (limsup) of the density of black squares in $[n] \times [n]$ as $n \to \infty$. Here, we identify each point with a square in the obvious way.

However, we have specified a finite set of black sub-patterns that are not allowed.
That is, a *pattern* is a subset of $[k]\times [k]$, and any translation of this subset is not allowed in the coloring as all black squares.

For example, perhaps we do not allow two adjacent black squares. Then, a checkerboard coloring has limit density $1/2$.

*Updated after answer*

**Q1:** For any finite set of forbidden patterns, will the coloring(s) maximizing the density always be periodic? **NO**

**Q2:** Is there a finite set of forbidden patterns, such that the maximal limit density can only be obtained by a non-periodic coloring? **YES**

**Q2b:** Is there a finite set of forbidden patterns, such that the maximal limit density is an irrational number? Note that this requires a non-periodic pattern.

**Q3:** Is there a limit density that is not realizable, that is, there is a sequence of colorings, $c_1,c_2,\dots$ with increasing limit density, but the limit density is not realizable by any coloring?

**Q4:** Since this feels closely related to tiling problems and permutation patterns, might it be that the question "Does the set $F$ allow a coloring with limit density $\geq d$?" is undecidable? **YES**

According to bijection with Wang tiles, where all tiles having equal density. Deciding if a finite set of Wang tiles, tile the plane, is undecidable.

*Note that for every set of forbidden patterns, coloring all squares white is a valid coloring. *

Note: We only consider a finite set of forbidden patterns in all questions.

Let $\hat{G}$ be the set of irreducible representations of a group $G$. Kirillov's method provides a geometric method for understanding $\hat{G}$ as the orbits of $G$ in the dual of the Lie algebra. It is also a means to introducing non-commutative harmonic analysis to engineering.

Tony Dooley from the university of Bath in the UK gave a talk on it and the Kirillov character formula as part of a seminar at my university so it may be productive to check out his homepage.

137291527664@paulgarrett: $SU(3,1)$ that you ask about maps to $SO^*(6)$ (a.k.a. $SO(3,\mathbb H)$), according to Helgason (§X.6.4) or Besse (Einstein Manifolds, p. 201). You may need to add a third section "Over $\mathbb H$"!Do compact complex manifolds fall into countably many families?)Some "equidistributions $\mod{p}$" are discussed in Washington's "Introduction to Cyclotomic Fields". I adress here only your first question, and have nothing to say about questions b) and c).

The first equidistribution is about irregular primes, and is discussed after Theorem 5.17 (I refer, here and henceforth, to the second edition), on page 62-63. The idea is to assume that the classes of Bernoulli numbers are random $\mod{p}$, in the sense that the probability that $p\mid B_j$ for some $j\in[2,4,6,\dots,p-3]$ is $1/p$. In this case one coops with a Poisson distribution and the deduced heuristic is that approximately $60\%$ of primes are regular, which agrees with computations.

A second distribution argument is briefly touched upon after Theorem 5.37 and concerns the possibility that the Bernoulli number $B_{(p-1)/2}$ be random $\mod{p}$. This would have consequences on the residue $\mod{p}$ of the class number of $\mathbb{Q}(\sqrt{p})$: as it is explained in Exercice 5.9, though, one can show that $B_{(p+1)/2}$ is *not random* $\mod{p}$, as a consequence of Brauer-Siegel (at least for $p\equiv 3\pmod{4})$.

A third one concerns Vandiver's conjecture and is discussed in the Remark on page 158 ($\S$ 8.4). It shows that the naive approach of assuming that the $i$-th cyclotomic unit be a $p$-th power with probability $1/p$ fails dramatically, while there is a more refined example that combines the above assumption with the needed hypothesis that $p$ be irregular; the very first heuristics I mentioned, about equidistribution of the $j$-th Bernoulli number $B_j\pmod{p}$, implies that possible exceptions to Vandiver's up to $p\leq x$ go like $1/2\log\log(x)$, and is therefore of no significance that we haven't found any counterexample so far. The very same section of Washington's book has a final remark on Vandiver based on the equidistribution assumption that $p\mid h^+$ with probability $1/p$ (where $h^+$ is the class number of the totally real field $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$). In this case again, Vandiver could be false for a prime $p\leq x$ with probability
$$
\log\log(x)
$$
which is bigger than before but roughly analogous. In both cases, since computations (at the time of Washingtons' writing, namely 1996) can go up to $p\sim 4.000.000$ and $\log\log(4.000.000)=2.72...$, that no exception has been found is "not *too* strange".

I should perhaps add that it seems that P. Mihailescu has some argument for turning Washington's heuristic "in favor" of Vandiver here.

105466012356651449195Let $u \in H^{\frac 12}(\Omega)$ with $\int_\Omega u = 0$ and consider the solution $v \in H^1(C)$ where $C=\Omega \times (0,\infty)$ of $$-\Delta v(x,y) = 0$$ $$\partial_\nu v = 0$$ $$v(x,0) = u(x)$$ in the weak sense. We can write $v(x,y) = \sum_{k=1}^\infty e^{-\lambda_k^{\frac 12}y}(u,\varphi_k)_{L^2}\varphi_k$ where $\lambda_k$ and $\varphi_k$ are the eigenelements corresponding to the Neumann Laplacian.

If $u \in L^\infty(\Omega)$, does it follow that $v \in L^\infty(C)$?

And also by Siksek, see J. Théor. Nombres Bordeaux 15 (2003), 839–846.I think you referring to reverse mathematics. This science investigates which axioms are required for a certain theorem.I added a definition of extremely amenable group, as I suppose that it is not something that everybody knows. I also tried to create a corresponding tag, but it is forgotten from mobile version (why?!)1115417101906411121821033473560018It seems the only development so far is a response from Bender, Brody and Müller from 18 May 2017 to Bellissard's criticism of their work, also to be found on arxiv:

Comment on 'Comment on "Hamiltonian for the zeros of the Riemann zeta function" '

They seem to be addressing all the issues pointed out by Bellissard, although I don't know if it's to a satisfactory level.

940933805646486653There is no upper bound for finite index subgroups of $SL_n({\mathbb Z})$. That is, given any integer $k$, there exists a finite index subgroup of $SL_n({\mathbb Z})$ which needs at least $k$ generators. I do not know to whom this is originally due, but it is a remark of Rapinchuk that if you pull the centre of, say $SL_{2n}({\mathbb Z}/m {\mathbb Z})$ where $m$ is the product of $k$ distinct primes back to $SL_{2n}({\mathbb Z}) $, you get a finite index subgroup of $SL_{2n}({\mathbb Z})$ whose abelianisation has $({\mathbb Z}/2{\mathbb Z})^k$ as a quotient and hence is at least $k$ generated.

On the other hand, a finite index subgroup of $SL_n({\mathbb Z})$, contains a smaller finite index subgroup which is generated by 3 elements; this result is true for any higher rank non-uniform arithmetic group in a semi-simple linear group (and is due to Ritumoni Sarma and Venkataramana)

vholomorphic sectional curvature and total scalar curvatureHas anyone (using whatever language) spelled out the formalism of the four operations for $\mathcal{O}$-modules? Or better yet, for quasi-coherent sheaves? I'd be especially interested in coherence for base-change diagrams. (the only reference I could find is buried in one of Lurie's DAG's and for some reason there is a quasi-affineness condition which I still haven't figured out why appears)200477Thanks very much for Viterbi, it seems I understand what you mean if have time will try to check. But still, I am puzzled can the answer be expressed in terms of polynoms - the question seems quite belong to the realm of algebra, while Viterbi is something outside. Or may be Viterbi have some interpretation in terms of algebra ? 1073016Most of the existing queueing theory are based on models like $M/M/1$ or $M/G/1$ or $G/G/1$, where the server service rate is normalized to one. However, in practice, suppose we have a queueing model with Poisson packet arrival $A$, packet size distribution $B$, and the service rate distribution $C$. Does this model corresponds to the $M/G/1$ model with $G$ corresponding to the distribution obtained by analyzing $B/C$?

2220861|https://graph.facebook.com/448018982378880/picture?type=large153487fThe manifold is Riemannian and dimension is three.6048461578177The real question is: are $X = (X_t)$ and $Y = (Y_t)$ indeed $K$-valued random variables? This is not so obvious in continuous time, and in fact the answer may depend on the choice of the $\sigma$-field in $K$.

If $K$ is equipped with the product $\sigma$-field (generated by cylinder sets), then the answer is clearly "yes": $X = (X_t)$ and $Y = (Y_t)$ are $K$-valued random variables, and they are independent.

On the other hand, if $K$ is equipped with the Borel $\sigma$-field associated with the norm in $K$, then the answer can be "no". For example, take $\Omega = [0, 1]$ with the Lebesgue $\sigma$-field, consider a non-measurable set $E \subset \Omega$, and define the process $X_t$ so that $X_t(\omega) = 1$ if $t = \omega$ and $\omega \in E$, $X_t(\omega) = 0$ otherwise. Then $X_t$ has bounded, Borel-measurable paths, but $$\{\omega \in \Omega : \|X(\omega)\| < 1\} = E$$ is not measurable (here $\|X(\omega)\| = \sup \{|X_t(\omega)| : t \in [0, 1]\}$).

What happens if we know *a priori* that $X$ and $Y$ are measurable with respect to the Borel $\sigma$-field? I guess it is still possible to cook up an example in which $X$ and $Y$ may fail to be independent as $K$-valued variables, but I do not know that.

If, however, your processes are regular enough (right-continuous with left limits should do), then independence over the cylinder $\sigma$-field should imply independence over the Borel $\sigma$-field.

865256**Claim:**
Let $T$ be an ideal in a commutative ring. If $T$ contains
a unique maximal proper subideal, then $T$ is principal.

*Proof*: Indeed take any element outside the unique maximal proper subideal. The ideal generated by that cannot be a proper subideal and hence has to be $T$.

So, this suggests that this is a very special property. For instance, that maximal subideal would have to contain the square of any generator of $T$ and if those generators are not zero-divisors, then they are associate (i.e. they can only differ by a unit multiple). If any generator is a zero-divisor then every element in $T$ is a zero-divisor. I think you can continue this line of thought to derive more special properties when this holds.

There is also a partial converse:

**Claim:**
Let $T$ be a non-zero principal ideal in a commutative ring. Then every proper subideal of $T$ is contained in an ideal that is maximal among proper subideals in $T$.

(By the usual proof of existence of maximal ideals)

*Proof*: Let $T\subseteq R$ be a principal ideal generated by $t\in T$, $T'\subsetneq T$ a proper subideal, and consider the set of ideals $I\subseteq T$ such that $T'\subseteq I$, $t\not\in I$ which is the same as the set of proper subideals of $T$ containing $T'$. If there is a chain of such ideals $\ldots\subseteq I_\lambda\subseteq \dots$ with $\lambda\in \Lambda$, then $t\not\in J:=\cup_{\lambda\in\Lambda}I_\lambda$ is an upper bound of the chain and hence the statement follows by Zorn's lemma.

However, of course, this does not mean that there is a unique one...

On the positive side, here is a situation where what you are hoping for holds:

**Claim:** If $R$ is a DVR, then the ideals in $R$ form a totally ordered set.

*Proof*: This follows essentially from the definition of a DVR.

So, I suppose one can conclude that this holds sometime, but for most ideals in most rings it does not.

I know this method under the name *lagged diffusivity*. I learned it from the paper

Vogel, Curtis R., and Mary E. Oman. "Iterative methods for total variation denoising." SIAM Journal on Scientific Computing 17.1 (1996): 227-238.

and the paper

Chan, Tony F., and Pep Mulet. "On the convergence of the lagged diffusivity fixed point method in total variation image restoration." SIAM Journal on Numerical Analysis 36.2 (1999): 354-367.

analyzes the case $p=1$ (with additional regularization). There they write that "This idea is quite commonly used in other PDE applications, e.g., CFD, and our theory may have applications there as well. " but I don't know more about the background.

There are some newer paper which you find under the same buzzword, but I do not know of any papers that treat other $p$-Laplacians than $p=1$.

Yes, I agree with $(a,b,c)^2\cap(c)=(ac,bc,c^2)$ but I think an error has occurred before that. Do you agree $(a,b,c)^2\cap(c)\cap(c,d)^2=(c^2,bcd,acd)$? If you agree with what I just wrote then you will see the problem.(Please tell me if you think I'm totally off base here and people would be offended by this behavior... but it seems reasonable to me.)1139229Having a compact manifold helps, since then (iirc) the $L^2$ structure is essentially independent of a chosen Riemannian metric that defines it. So for a compact $M$ you can take any Riemannian metric on $M$ and then define $\mathcal{F}(M,\mathbb{R})$ to be the $L^2$ functions with respect to this metric. I think that the mapping $F_T$ can be always written as an infinite matrix once you choose bases.2018850`functions from Q to itself with derivative zero1751214184521Indeed, it does not appear to mention a restriction on $f$ in the definition. It seems to me that theorem 7.1.6 claimed in the book is not correct, in light of my argument above.The following standard argument works if $k$ is assumed bounded. I don't think that assumption should be necessary, but maybe this is at least a helpful start.

If $T$ is symmetric then for every $f,g \in C^\infty(M)$ we have $$\int \int k(x,y) f(x) g(y)\,dx \,dy = \int \int k(x,y) g(x) f(y)\,dy\,dx$$ or changing variables and using Fubini's theorem (here we use the assumption that $k$ is bounded), $$\iint k(x,y) f(x) g(y) \,dx\,dy = \iint k(y,x) f(x) g(y)\,dx\,dy.$$ In other words, for every $F : M \times M \to \mathbb{R}$ which is of the form $F(x,y) = \sum_{i=1}^n f_i(x) g_i(y)$ where $f_i, g_i \in C^\infty(M)$, we have $$\iint (k(x,y) - k(y,x)) F(x,y) \,dx\,dy = 0.$$ Now using a monotone class argument, show that the same holds for all bounded measurable $F: M \times M \to \mathbb{R}$. Taking $F(x,y) = k(x,y) - k(y,x)$ you get $$\iint |k(x,y) - k(y,x)|^2 \,dx\,dy = 0.$$ (If $k$ is assumed continuous you can instead use Stone-Weierstrass in the last step.)

575587@T.Amdeberhan, from what is written in 1), $n$ is the index of the components in the vector $Y_N(t)$ and the successive quotients (or their inverses) of these components are decreasing (or increasing) with $n$.10962912073759You may wish to make your comment an actual answer, and, after the enforced delay, accept it.823339Let $\mathbf{S}$ be an excellent model category in which all objects are cofibrant, viewed as an $\mathbf{S}$-enriched category by its canonical self-enrichment. Then we know that there is an obvious enriched full embedding of the subcategory of fibrant objects $\mathbf{S}^\circ \hookrightarrow \mathbf{S}$.

Since $\mathbf{S}$ is a combinatorial model category, the small object argument provides us with a fibrant replacement functor $\mathbf{S}\to \mathbf{S}^\circ$ that is *not necessarily enriched*.

First question: If $\mathcal{A}$ is an $\mathbf{S}$-enriched combinatorial model category (maybe in which all objects are cofibrant), is there an enriched version of the small object argument that provides us with a fibrant replacement functor $\mathcal{A}\to \mathcal{A}^\circ$ that is an *enriched* functor?

Second question: If the first question is not true in general for $\mathbf{S}$-enriched combinatorial model categories, is it at least true for the special case $\mathcal{A}=\mathbf{S}$ itself?

1403203This is false already for ordinary fibrations of 1-categories. It is a theorem that if $D$ has pullbacks and $F:C\to D$ is a fibration, then $C$ has and $F$ preserves pullbacks if and only if each fiber of $F$ has pullbacks and the reindexing functors preserve them. Thus, the Grothendieck construction of any functor $D^{\mathrm{op}}\to \mathrm{Cat}$ which does not factor through $\mathrm{Cat}_{\mathrm{pullbacks}}$ will give a counterexample.

16499561244786Is there any specific reason why you assume $n>15$? Is the result wrong for smaller $n$?\https://i.stack.imgur.com/6DsXG.png?s=128&g=1@TomekKania is correct in this case an arbitrary $h$ may fail to satisfy a (local) extension. After a discussion, we are still unable to claim how/whether such an extension will meet the requirement.32005313679Oh right, in fact we have $H_n(X;\mathbb{Z}_{2k})\cong\mathbb{Z}_2$ for all integers $k$, which forbids your option.11983454580401813098In my paper, found at arxiv or AoP, the proof of an exact Rosenthal-type bound on absolute moments of sums of independent zero-mean random variables relied to a large extent on identities of "a calculus of variations", given by Lemma 2.5 there, of moments of infinitely divisible distributions with respect to variations of the corresponding Lévy characteristics (as well as on a number of other identities, such as the ones in display (71) there).

My apologies for giving another example related to my own work. I can try to justify this by saying that (i) my work is what I know the best, (ii) the question itself was rooted in my experience, and (iii) there have been not many answers so far.

554830141459515430791480413blearning probability theory and analysis

6526813716968794042163350(In addition to Anton Petrunin's answer, here is a trick to simplify (and in some sense solve) the geodesic equation.

Since the metric has three-dimensional group of isometries (generated by rigid motions of the plane), the corresponding Noether's integrals make the geodesic flow completely integrable. More precisely, the first derivative is uniquely determined by the position and the values of the integrals. Actually a two-dimensional group is sufficient because for the third integral one can always take the speed (= the norm of the velocity vector) of a geodesic.

So let's use invariance under parallel translations only. We have a manifold of all unit segments in the plane, and a tangent vector to that manifold can be thought of as the vector field along a segment (representing velocities of all its points). This vector field has a form $$ v(t) = tx + (1-t)y, \qquad t\in[0,1] $$ where $x,y\in\mathbb R^2$ are such that the vector $x-y$ is orthogonal to the segment. The norm of this tangent vector is $\int |v(t)| dt$ (as in Anton's answer), and the Noether integral corresponding to translations (if my quick computation is correct) boils down to the following: if $v$ is a velocity vector of a geodesic represented in the above form, then $$ \int_0^1 \frac{v(t)}{|v(t)|} \ dt = V_0 $$ where $V_0$ is a constant (for each geodesic) vector in the plane. This vector $V_0$ (and, say, the assumption that the geodesic is unit-speed) determines the parameters $x$ and $y$ in the formula for $v(t)$ uniquely (as a function of the segment's current direction), so the geodesic equation is reduced to a 1st order ODE. Furthermore, the above integral can be found as an explicit (but weird) function of $x$ and $y$, so the equation becomes easy to solve, at least numerically.

By the way, I second Jean-Marc Schlenker's proposal to consider quadratic mean of the distances (i.e. $L^2$ norm rather than $L^1$), especially if you have any physics-related application in mind. In this case the metric is Riemannian, the energy is just the standard kinetic energy of the moving segment, so the Noether intergrals are just the standard conservation laws: the linear momentum and the angular momentum. And the geodesics are very simple: the segment rotates at a fixed angular speed while its midpoint (the barycenter) moves along a straight line with a constant speed.

12908617761181278062@მამუკაჯიბლაძე Non-empty Stein manifolds contain Floer theoretically non-trivial immersed Lagrangian submanifolds.@Elliot Most bounds for codes $C$, even if they are stated and proven for linear codes, work for non-linear codes over finite alphabets of size $q$ as well, except that one has to replace $\dim C$ with $\log|C|/\log q$. A notable exception is the Griesmer bound, which truly uses the linearity of the code.62339121030271838532"Natural" metrics on the Grassmannian should be invariant under isometries of the undelying Euclidian space. Actually there are many such metrics, and this nice paper gives a general method for constructing them: Qiu, Zhang, Li, "Unitarily invariant metrics on the grassmann space". SIAM J. Matrix. Anal. Appl. vol 27, no 2, 507-531.1063920&TheoremOfChenevierIt is indeed easy to find a description of the pencil method. It is however not easy to find a description where numerical stability is considered.http://mathoverflow.net/questions/80482/if-x-is-an-affine-variety-is-x-one-component-of-a-complete-intersection-withWorst case, you'll need $n$ powers (vs $O(1)$ random characteristic polynomial checks).Here are two examples. I describe it as Lie algebras (over any field $K$).

(1) The 7-dimensional, 3-step nilpotent Lie algebra with basis $(X_1,\dots,X_7)$ and nonzero brackets

$$ [X_1,X_2]=X_4,[X_1,X_3]=X_5,[X_2,X_3]=X_6,[X_1,X_4]=[X_1,X_5]=[X_2,X_4]=[X_3,X_6]=X_7$$

(2) The 6-dimensional, 4-step nilpotent Lie algebra with basis $(X_1,\dots,X_6)$ and nonzero brackets

$$[X_1,X_2]=X_3, [X_1,X_3]=X_4, [X_2,X_3]=X_5, [X_1,X_5]=[X_2,X_4]=X_6.$$

Here's a common proof. In (1), let $V$ be the subspace generated by $X_1,X_2,X_3$, $W$ the subspace generated by $X_4,X_5,X_6$, and $Z$ the subspace generated by $X_7$. In (2), let $V$ be the subspace generated by $X_1,X_2$, $W$ the subspace generated by $X_4,X_5$, and $Z$ the subspace generated by $X_6$.

Then in each $\mathfrak{g}$ of these two: $Z$ is the 1-dimensional center, and the bracket $V\times W\to Z$ defines a non-degenerate pairing. Thus for each hyperplane $X$ in $V$, its centralizer in $W$ is equal to its orthogonal $X^\bot$ with respect to this pairing.

Each codimension 1 ideal of $\mathfrak{g}$ has the form $I_X=X\oplus \mathfrak{g}$ where $X$ is some hyperplane, and the center of $I_X$ is equal to the 2-dimensional ideal $X^\bot\oplus Z$.

Both Lie algebra being defined over $\mathbf{Q}$, we can consider the corresponding lattice, which therefore answers your question.

15730891603954If $\Gamma \subset SL_2({\mathbb C})$ is a Kleinian subgroup, will you please define what the trace field and the invariant trace fields are? One can then think about your question (since I have not seen Maclachlan Reid) .80871iluvatar1888304That's not nonsense; it is management jargon adopted to mathematics! @Nate: I suspect you have written a similar program for management training courses.213725759292839936015512041015166470124fWhich topological spaces are (topological) groups?547841h$\sin(x-y)=s_xc_y-c_xs_y$, which is at most rank-2.9236422149074148326206129314463091921202083308https://www.gravatar.com/avatar/3b27f01cca088f8757cc0aa120689c78?s=128&d=identicon&r=PG&f=1I was talking about the sesquilinear inner product, not the bilinear inner product. But things like this should be cleared up by the OP, in my opinion.1199488rAh yes, should have mentioned I'm working in real space.387407A sequence $(x_{n})_{n}$ in a Banach space $X$ is said to be unconditionally $p$-summable if $$\sup_{x^{*}\in B_{X^{*}}}\Bigl(\sum_{n=m}^{\infty}\lvert\langle x^{*},x_{n}\rangle\rvert^{p}\Bigr)^{1/p}\rightarrow 0\qquad(m\rightarrow \infty).$$ We say that an operator $T:X\rightarrow Y$ is unconditionally $p$-summing if $(Tx_{n})_{n}$ is unconditionally $p$-summable in $Y$ whenever $(x_{n})_{n}$ is weakly $p$-summable in $X$. We can prove that $T$ is unconditionally $p$-summing whenever $T^{**}$ is unconditionally $p$-summing. Is the converse true?

1863232001837Thanks Brendan! - some less precise calculations are also getting me to $H''(0)=\infty$ in that case.275355I'm pretty sure there is a couple of neat PDEs that the solution satisfies, though likely there is also a direct approach - not that I know of unfortunately. Let's take a look at the integrals. If $p(x,y,t)$ is a density of lending at $y$ from $x$ in time $t$, then the last integral is not even a probability. $$ \int_C p(x,y,t)\mathrm dy = P(x,C,t) $$ that is a probability of landing in a set $C$ from $x$ in time $t$. Your latter integral is integrating this probability over $x\in \Bbb R$ and $t\in \Bbb R_+$ - that's **not** a probability of hitting $C$.71594817839618814251811031678536267699193954313894841955088633943@JasonStarr Ah right I misunderstood the meaning of zero locus.One of the first things to learn about category theory is that not everything in mathematics is a category. (-:1656235229478017912182245900Shouldn't this question be community-wiki, since the goal is to have a list rather than to find a definitive answer?**The main interest in the solution is that all pairs $(a,b)$ are integers** and $f(a,b)$ is a perfect square rational number."The answer is yes. Let $[E^{-1} \rightarrow E^0]$ be a perfect obstruction theory on $M$. After localizing in $M$ we can assume that the map $E^0 \rightarrow \Omega_M$ is induced as $\mathcal{O}_M \otimes \Omega_{\mathbf{A}^n} \rightarrow \Omega_M$ for some map $M \rightarrow \mathbf{A}^n$.(*) As the map on differentials is surjective the relative differentials vanish and the map is unramified. By EGA IV.18.4.7, we can (at least after localizing further in $M$) factor the map $M \rightarrow \mathbf{A}^n$ as a closed embedding $M \rightarrow V$ followed by an étale map $V \rightarrow \mathbf{A}^n$.

Let $I$ be the ideal of $M$ inside $V$. Then we have an exact sequence

$I/I^2 \rightarrow \mathcal{O}_M \otimes \Omega_V \rightarrow \Omega_M \rightarrow 0$

where the first two terms are $\tau_{\geq -1} L_M$. By the definition of an obstruction theory, we are given a map $E^{-1} \rightarrow I/I^2$ inducing a surjection $H^{-1}(E) \rightarrow H^{-1} L_M$. By the 5-lemma, $E^{-1} \rightarrow I/I^2$ is surjective.

Localizing further in $V$ and $M$, we can assume that the map $E^{-1} \rightarrow I/I^2$ is induced from a map $F \rightarrow I$ of coherent sheaves on $V$ with $F$ locally free. By Nakayama's lemma we can assume, possibly after further localization, that $F \rightarrow I$ is surjective. Then consider the section of the vector bundle $F^\vee$ dual to the composition $F \rightarrow I \subset \mathcal{O}_V$. The vanishing locus of this section is $M$ and the induced complex $F \rightarrow \mathcal{O}_M \otimes \Omega_V$ is the obstruction theory we started with.

(*) We can represent $\tau_{\geq -1} L_M$ as $[ J / J^2 \rightarrow \Omega_W ]$ for some closed embedding $M \subset W$ and we are given a map of complexes $E^\bullet \rightarrow \tau_{\geq -1} L_M$. Replacing $E^\bullet$ with a quasi-isomorphic complex, we can assume that the map $E^0 \rightarrow \Omega_W$ is surjective. But $E^0$ and $\Omega_W$ are vector bundles, so after localizing in $M$ we can assume that $\Omega_W$ is a direct summand of $E^0$. Choosing a basis for the complementary summand, we can identify $E^0 \cong \Omega_{W \times \mathbf{A}^n}$.

469808Many years ago, a professor of mine in a graduate algebra course wrote down a totally impenetrable statement and then added:

"Of course, it's obvious. It may not be *clear* that it's obvious...but it's obvious."

The words "it's obvious", etc., should be discarded just as the words "We have..." should be discarded. Also, the words "Consider the following function...." As Estermann comented..."I do not know what that means."

1483367RNonpathological nonnormal covering space97149General solution of the Multiplicative symmetry equation $f(xf(y))=f(f(x)y)$ in nonabelian groups305744dAssistant prof at University of Kentucky.

21501941618950vQuestion about Notation for Spaces of $n$-ary $k$-ic Forms6075442600231738811In this blog post you'll find a computation of the cohomology ring of a hypersurface of degree $d$ in $\mathbb{CP}^3$ using characteristic classes. This turns out to be a weirdly good exercise in using characteristic classes: the computation invokes, in order, Euler classes, Chern classes, Pontryagin classes, Wu classes, and Stiefel-Whitney classes, and doing it is what made me comfortable with characteristic class computations.

82021445892zhttps://www.southampton.ac.uk/maths/about/staff/ijl1y09.page2301398@Scott: Thanks, fixed. @Akela: since C(X,R)oC = C(X,C), you get it from the same place as normally.XWhich Banach spaces have categorical duals?101896215244141254804Expressing a value related to an infinitary relation through ultrafiltersThe coherence theorem for bicategories states that every bicategory is equivalent to a strict one. What do you mean by embedding into a strict 2-cat?1626708518853695648Thanks! So this covariance operator is of the ``Wiener distribution" on the space of $\mathbb{R} \rightarrow \mathbb{R}$, right? Here the "stochastic process" is just an intermediary and not really the important point - am I right? And the $W_t$ whose spectral decomposition is given here, https://en.wikipedia.org/wiki/Wiener_process#Wiener_representation is really the value of a randomly sampled function using the the Wiener distribution. Right?149583110601831694889Hi, is there a "Galois" theory for rational functions of real algebraic sets similiar to the one relating coverings of riemanien surfaces with Galois Extension of meromofphic functions?

Thanks

163947662013758236422811932136037I would be interested in something which is *not* of this sort. Anyway, all of Harvey's examples are true in the standard model. (Under appropriate background assumptions on large cardinals.)n@DavidSteinberg I think that is a nice option as well.15695471022120149870880359721522733025331110667Dear Paul, thanks for portrait that you have brushed. Regarding your comment about Kubota's book I get your point but my feeling is that his book was probably the first comprehensive and accessible reference on $GL_2$-real analytic Eisenstein series for non-experts. Also in the nice portait that you just depicted, one should also probably mention the substantial contributions which came from physics and functional analysis.If one wants to work with $Q_l$-adic sheaves, should the scheme be of finite type over a 1-dimensional one?142338173526114099012177409142901 @Rhett Butler: You are confused. As has been pointed out much earlier here, the payoff in roulette is not the proportion $B/(B+G)$. So, if you think there is an application to roulette of a method for changing the expected value of $B/(B+G)$, the burden is on you to show it. Your claims that my correct statements are equivalent to claiming to have a winning roulette strategy are wrong and insulting. I tried to help you to understand some math, and you lie and ridicule me. Any more lies claiming that I am trying to invent a winning roulette strategy will be flagged as spam. 14003051708701André Belottoremote 34420353524910186But it would be nice if this was true whenever out original model obeyed choice. Do you know of an interpretation of ZFC in ZF that does this? i.e. Is there an interpretation of ZFC in ZF such that whenever you apply it to a model of ZF that happens to obey Choice the induced model is definably isomorphic to the original? That question is probably what I would have asked if I was trained to think in terms of logic rather than category theory.342121A lot of neuron behaviors are modeled by dynamical systems. A nice reference on this topic is this book: http://mitpress.mit.edu/books/dynamical-systems-neuroscience&I would like to know a natural categorification of the rig of integers $\mathbb{Z}$. This should be a $2$-rig. Among the various notions of 2-rigs, we obviously have to exclude those where $+$ is a coproduct (since otherwise $a+b \cong 0 \Rightarrow a \cong b \cong 0$). So let's take rig categories instead.

**Question.** What is a natural and non-discrete rig category whose rig of isomorphism classes is $\mathbb{Z}$?

I've read MO/3476, but the answers are not really satisfactory. Both the category of tangles and Schnauel's categories of polyhedra don't qualify.

Here is my approach. Notice that $\mathbb{N}$ is the initial rig, so its categorification should be the initial rig category, which turns out to be the groupoid of finite sets and bijections, which is equivalent to the permutation groupoid $\mathbb{P}$. The rig $\mathbb{Z}$ is the free rig on one generator $x$ subject to the relations $x^2=1$ and $x+1=0$. For short, $\mathbb{Z} = \mathbb{N}[x]/(x^2=1,x+1=0)$. This suggests that our categorification is $\mathbb{P}[X]/(X^2 \cong 1,X+1 \cong 0)$, where these isomorphisms probably should satisfy some coherence condition (which are not visible in the decategorification $\mathbb{Z}$) in order to "flatten" the rig category. Namely, if $e : X^2 \to 1$ and $f : X+1 \to 0$ are the isomorphisms, we could require (here I omit the coherence isomorphisms of the rig category) that $eX = Xe : X^3 \to X$ and $fX = Xf =f \circ (e+X): X^2+X \to 0$.

I think that $\mathbb{P}[X,X^{-1}]$ should be the rig category of vector bundles on the projective space $\mathbb{P}^1_{\mathbb{F}_0}$, where $\mathbb{F}_0$ is the "field with no element" in the sense of Durov. Here, the Serre twist $\mathcal{O}(1)$ is not inverse to itsself, but this changes when we consider the Möbius strip $H$ on $S^1$. But vector bundles in topology have too much morphisms. For the same reason we cannot take $\mathbb{P}^1_{\mathbb{C}}$.

**Question.** What is a natural realization of the rig category $\mathbb{P}[X]/(X^2 \cong 1,X+1 \cong 0)$?

I have looked through the answers but haven't come across Cantor's theorem, that the there is no surjection from a set $M$ on its power set $P(M)$.

I don't know whether the proof should be considered trivial, but it is short and easy to understand. The assumption of a surjection $f\colon M\rightarrow P(M)$ leads to the contradictory set $A_f:=\{m\in M\colon m\not\in f(m)\}$, the contradiction being $A_f\in A_f\longleftrightarrow A_f\not\in A_f$.

The implication of the theorem is that there is no "largest" set/infinitude!

I advert to some StackExchange proposals, success of which I yen:

♦ Art History & Theory ♦ Canada ♦ Greek ♦ Rhetoric ♦ UK

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(:Em Space) ✘ ✓ ¿ ● ■ « » ⟨ ⟩ ← ⟵ ⟽ ⟸ ⟶ → ⟹ ⟾ ⟷⟺

≈ ≡ ∨ (`∨`

) ∧ (`∧`

) ¬ ¶ ∃ ∀ ∴ ∵

You should definitely take a look at Lang's "Topics in Cohomology of Groups", chapter 4. There a general notion of cup-products on delta-functors is introduced. This may look like a lot of abstract nonsense to most people, but I like it :)

3916417436812123582@abx: it does not quite identify one space with the dual of the other, e.g. as $\text{GL}(V)$-representations we do not get duality because the action of $\text{GL}(V)$ on $\Lambda^n(V)$ is nontrivial.1180241592072x(I mean, with $n/2$ as the OP asked, not with $O(\log n)$.)VThis question looks like à blind shot ...1756821@KConrad I’m studying Artin and Tate’s book on class field theory. I’m trying myself on this kind of questionsddon't see a problem defining volume forms thought148315421501521767631I assume you know about the book Kollár is writing? http://www.math.princeton.edu/~kollar/ Maybe that's what you meant though?18915313274912952601154890213541877322962246422510742x@ToddTrimble I was thinking specifically in $\mathbf{Top}$.https://mathoverflow.net/questions/184216/how-to-change-the-successor-of-a-singular-with-a-woodin https://mathoverflow.net/questions/178112/inaccessible-becomes-successor-of-singularI guess this was in the 20th century. These days you can check quickly with a computer that the sum of the first five million primes is already over pi.627618Geometric interpretation of the exact sequence for the cotangent bundle of the projective space1354621280351https://graph.facebook.com/10154780099329327/picture?type=large-In

- Wang, Shuzhou. Quantum symmetry groups of finite spaces. Comm. Math. Phys. 195 (1998), no. 1, 195--211. MR1637425 (99h:58014), link

a quantum version of the symmetric group $\mathbb{S}_n$ is defined.

Let me sketch Wang's construction.

Let $u_{ij}$ be the characteristic function of the set of $\sigma\in\mathbb{S}_n$ such that $\sigma(j)=i$.

Assume that all entries $u_{ij}$ are projections, and on each row and column of $u=(u_{ij})$ these projections are orthogonal, and sum up to $1$. Then the commutative $C^*$-algebra generated by these $u$ is $C(\mathbb{S}_n)$.

Now drop the commutativity condition and let $A_s(n)$ be the $C^*$-algebra generated by all the $u_{ij}$. Then we have a *quantum* analogue of $\mathbb{S}_n$.

It turns out that $A_s(n)$ is a finitely generated Hopf algebra.

The group $\mathbb{S}_n$ acts on an set $X=[1,2,...,n]$ with $|X|=n$. The corresponding action map $(i,\sigma)\mapsto \sigma(i)$ gives by transposition a certain morphism $\alpha$ ($\alpha$ is called *coaction*). This coaction can be expressed as
$\alpha(\delta_i)=\sum\delta_j\otimes u_{ji}$. Furthermore, $\alpha$ is a sort of universal coaction.

It is possible to prove that the following diagram is commutative $$ \begin{array}{ccc} C(X) & \to & C(X)\otimes A_s(n)\\\\ \downarrow & & \downarrow\\\\ C(X) & \to & C(X)\otimes C(\mathbb{S}_n) \end{array} $$

Furthermore, $C(\mathbb{S}_n)=A_s(n)$ if $n=1,2,3$. For $n\geq4$, $A_s(n)$ is not commutative and infinite dimensional.

For a nice survey about quantum permutation groups and some applications see the following paper:

- Banica, Teodor; Bichon, Julien; Collins, Benoît. Quantum permutation groups: a survey. Noncommutative harmonic analysis with applications to probability, 13--34, Banach Center Publ., 78, Polish Acad. Sci. Inst. Math., Warsaw, 2007. MR2402345 (2009f:46094), link

For a quantum version of the automorphism group of finite graphs (and a quantum version of the dihedral group $\mathbb{D}_4$):

- Bichon, Julien. Quantum automorphism groups of finite graphs. Proc. Amer. Math. Soc. 131 (2003), no. 3, 665--673 (electronic). MR1937403 (2003j:16049), link

A complete classification of quantum permutation groups acting on 4 points was given in:

- Banica, Teodor; Bichon, Julien. Quantum groups acting on 4 points. J. Reine Angew. Math. 626 (2009), 75--114. MR2492990 (2010c:46153), link

On the one hand, the comments explain that a low-growth-rate tree can still have continuum many branches. Indeed, the growth rate can be extremely slow, with most levels having no splitting at all, but then every once in a very long while, a single node splits. Just make sure at the $k^{th}$ such splitting that you are splitting at a node above the $k^{th}$ binary sequence of the tree (in some fixed standard enumeration of all finite binary sequences), and then the resulting tree $T$ will have the property that every node lies below a splitting node. This property will ensure a subtree of type $2^{<\omega}$, and hence give you continuum many branches.

In contrast, it is the converse question that I find extremely interesting, and which could be considered the real content of your question. Namely,

**Question.** What is a sufficient growth rate on the tree $T$ to
ensure that it has continuum many branches?

The answer is that the growth rate must be essentially close to $2^n$, in the precise senses described by the following theorems. On the one hand, such a high growth rate suffices for continuum many paths:

**Theorem.** If $T\subset 2^{<\omega}$ is a binary tree with
growth rate $\Omega(2^n)$ (in the Knuth sense big-Omega notation), then $T$
has continuum many branches.

Proof. What I intend to assume on the growth rate is that is that there is a positive real number $r>0$ such that the number of nodes on the $n^{th}$ level of $T$ is at least $r2^n$. Thus, the complement of the set of paths through $T$ is approximated by the unions of the cones above the omitted nodes on this level, and this has measure at most $1-r$ with respect to the usual coin-flipping probability measure on Cantor space. Since this is true at each level, it follows that the measure of the paths through $T$ is at least $r$, which is positive, and so there must be uncountably many paths through $T$. Since this is a closed set, it follows by the Cantor-Bendixson theorem that $T$ has continuum many paths. QED

On the other hand, any lower growth rate does not ensure continuum many paths.

**Theorem.** If $f:\mathbb{N}\to\mathbb{N}$ is in $o(2^n)$ (see
little-o notation),
then there is a tree $T$ with growth rate
exceeding $f$, but having only countably many paths.

Proof. The growth rate assumption is that for every $r>0$, it will be true for large enough $n$ that $f(n)<r2^n$.

Let me describe a general procedure for building a tree $T$ with a very high growth rate, but with only countably many branches. We build $T$ in stages. First, we let $T$ use the full binary branching up to some level $n_0$. Then, at this stage, we kill off the future growth of half the branches, the ones beginning with $0$, forcing them to become isolated branches in the tree starting at level $n_0$, continued only with more $0$s after $n_0$. Meanwhile, we let the other "live" nodes, those having $1$ in their first bit, to continue splitting above $n_0$ until some much larger stage $n_1$. At that level, we again kill off half of them. Namely, any node beginning with $10$ will become isolated at $n_1$, and the others, beginning with $11$, will continue splitting up until the very much later stage $n_2$. And so on.

The general procedure is: at level $n_k$, the nodes beginning with $1^k0$ will become isolated at level $n_k$ (which is much larger than $k$), and the nodes beginning with $1^k$ are allowed to continue branching up to level $n_{k+1}$.

The resulting tree $T$ will have only countably many branches, since the only branches will be the isolated branches that we forced to occur, plus the all $11111\cdots$ branch, since whenever a sequence has its first $0$ in position $k$, it will become isolated at level $n_k$.

But now, the main point, is that by choosing the levels $n_0<n_1<n_2<\cdots$ to be very fast growing, we can ensure a growth rate exceeding $f$. Specifically, since $f$ is $o(2^n)$, there is for each $k$ a number $n_k$ such that $f(n)<2^{n-k}$ for all $n\geq n_k$, and we may also assume $n_0<n_1<n_2<\cdots$. Now, with the $T$ as defined above, it follows that the number of nodes of $T$ on level $n_k$ is at least $2^{n_k-k}$, and it remains at least $2^{n-k}$ for $n_k\leq n<n_{k+1}$, and this is larger than $f(n)$, as desired. So the growth rate of $T$ is at least $f$, even though $T$ has only countably many paths. QED

14895322201281Dear Qph: Chapter 6 of "Neron Models" is a good place from which to learn descent theory. The natural map from $A_a[[t]] \otimes_{A[[t]]} A_{a'}[[t]]$ to $A_{aa'}[[t]]$ induces an isomorphism modulo $t^n$ for each $n \ge 1$, so the $t$-adic separatedness is equivalent to this natural map being injective. But such injectivity is unclear to me. bI may have missed this somewhere, but what is W?2291541https://lh3.googleusercontent.com/-iI5ZgjkwvFU/AAAAAAAAAAI/AAAAAAAAABM/uhPzDzbRjdw/photo.jpg?sz=1281990471317340205498383751851855I'm afraid this question might be too localized, but I have no better place to ask it:

In the section Classes which interpret any structure of his *Model Theory* Hodges shows how each $L$-structure $B$ can be converted into a graph $A = \Delta(B)$, $L$ a first order language with finite signature.

He associates every element of $B$ with a so-called 5-tagged element $a$ of $A$. Likewise he associates the *i*-th symbol of the signature $S_i$ with a (*i*+6)-tagged element of $A$. For further details see Hodges, p. 228 ff.

The only question I have at this point is: Why isn't it enough to associate all the relation symbols with different 6-tagged (and 6-tagged *only*) elements of $A$? Why is it necessary - for his argument's sake - to distinguish them further by their "tag count"?

http://www.jwz.org/doc/markup.html

21633751221496Both of these examples fall into the case of "paving by affines" (or affines minus several points) that Jonathan mentions in his question. Your note is a nice point---this reduction is essential to Dwork's proof of rationality. I see this as an indication that the hypersurface case is quite hard, though.XRuelle-Perron-Frobenius for continuous time86185415371801680217Of the modern proofs, I like the Jacobi field approach the best. You start by using the distance from a given point as one co-ordinate function and showing that you can extend the angular variables from the tangent space at the origin. This is most easily done using Jacobi fields $J_1, \dots, J_n$,. The metric in these co-ordinates is given by $g_{ij} = J_i\cdot J_j$, so its Taylor expansion is easily calculated using the formal solution to the Jacobi equation.38744This is quite related to another MathOverflow question from some years back: https://mathoverflow.net/questions/187116/what-is-the-expected-dimension-of-the-zariski-closure-of-the-rational-points-on?rq=120563871964362992512Hi,

I'm a new Java programmer who's learning and want to give some of the learned things from SO back to the SO-Community ;)

According to Greg O'Keefe on fa.isabelle,

\https://i.stack.imgur.com/kkbjg.png?s=128&g=1205553620059772636522109496I am saying the spectrum is the categorification of the spectrum in the classical case. By considering the main object of study R-mod(corresponding to QCsheaves on a scheme) we make the transition from sets to abelian categories. We replaces our spaces with 'spaces' which are simply categories. In this sense, it is a categorification.Just to bring out the main theme of my answer, if you fix a smooth structure (as you can in many cases, for instance in low dimensions), you are asking for conditions that ensure that a homeomorphism is topologically conjugate to a diffeomorphism. There is no easy answer to this -- it depends in a delicate way on the dynamics of the homeomorphism. For example, the $5$-sphere has a unique smooth structure, and I have no idea what possible form an answer to this question would take there.@A.Gonus: I was recently told by someone knowing more than me about this that the A.C.C. book has quite a few unusual definitions.181230816734611114533M.BigdeliHi Nik! Yes, in this particular case it is the Sturmian system, and on balance the way to go is probably to use the fact that we know explicitly what the natural extension is. I'd still like to know the general answer though!1479544782242pNo work since 1990 it seems. I haven't seen the sources, but [3] claims that adjunctions, triples and Cartesian closed categories were all defined. Adjoint functor theorem proved.

Lecturer at Xi'an Jiaotong-Liverpool University

572579Yes. The induced graph would have to have exactly two connected components. Pick a component of $V \backslash S$ to be a component of $V \backslash S_0$ and figure out what $S_0$ should be.415730331947936856See for instance section 7 of http://arxiv.org/pdf/0902.0648.pdf I guess they are assuming also twisting $\omega_Y$ by high powers of relatively ample line bundles, so that probably is non-optimal for you... The case where that's not needed is if you have a family of Calabi-Yaus (see the above reference).@Zuhair I don't understand most of what you wrote, but let me comment on the little bit where I can say something. My platonc perspective is that there are sets and some of them are members of each other. That membership is what I mean by "actual elementhood relation". The "pseudo-elementhood" that I referred to is some other relation. Like any relation, it can be represented as a class of ordered pairs. [continued in next comment]2230204814117rLoop space functor and sequential colimits of inclusions1588926@Andrej: Read some of Zeilberger's own work, where he himself makes similar claims. He's got papers which were (essentially) written by his computer; he puts his computer as co-author on his papers often enough.6164971493680https://www.gravatar.com/avatar/af95b0737e9b572ffc90e42dff3b5f07?s=128&d=identicon&r=PG&f=1479966263030My guess is that this not possible. In fact I think if you look at the Serre spectral sequence for the fibration you mention you can see that if the cohomology of G is bounded, then the cohomology of BG must be unbounded (hence it can't be finite complex).263285https://www.gravatar.com/avatar/8823395c150508912a3c87cc4f19c029?s=128&d=identicon&r=PG&f=1sorry, the link doesn't seem to work. The paper is "On a question of Remeslennikov" by James McCool. 147790119172791164816There is a related problem in my current work: to find the residue of the following function at any negative integer $s=-n$: $$f(s)=\frac{\Gamma^3(s)}{\Gamma(3s)(e^{2\pi is}-1)}$$ It seems to be a tedious calculation. As far as I know, the first few terms of Laurent series of $\Gamma(s)$ around $s=-n$ are given by $$\Gamma(s)=\frac{(-1)^n}{n!}\left(\frac 1{s+n}+\psi(n+1)+\frac 16(3\psi(n+1)^2+\pi^2-3\psi'(n+1))(s+n))+\frac16(\psi(n+1)^3+(\pi^2-3\psi'(n+1))\psi(n+1)+\psi''(n+1))(s+n)^2+O((s+n)^3)\right)$$ where $\psi(s):=\frac{\Gamma'(s)}{\Gamma(s)}$ is the digamma function. So we might expect that the residue formula of $f(s)$ at $s=-n$ involves the values of digamma function and/or its derivatives. However, numerical values indicate that the formula for $Res (f,-n)$ might only involve $\pi$, $i$, and rational numbers, which is quite unexpected. Here are a few examples: $$Res (f,-3)=7527 + \frac{4299 i}{4 \pi} + 2100i\pi$$ $$Res (f,-10)=\frac {1065144125784453}{40} - \frac {17136782690536253 i}{11200 \pi} + 6938745989175 i \pi$$ I have verified some other integers as well. Can we find any reason for this pattern or any counterexample? It also seems that $f(s)$ is very special, for instance, if we multiply another $\Gamma(s)$ to the numerator, i.e. $g(s)=\frac{\Gamma^4(s)}{\Gamma(3s)(e^{2\pi is}-1)}$, then the formula for $Res(g,-n)$ does not have such property.

572560zConnected subgroup of $\mathbb G^n_m$ of Zariski dimension 11210207Hint: if $\lambda^+$ embeds into ${}^\lambda 2$, then fix an embedding; eventually, the first digits of the targets stabilize, and then the next, and so on. But $\lambda^+$ is regular, so every $\lambda$ sequence is bounded.Yes to both questions, because both $E$ and $X \times X$ are flat over $X \setminus B$.129184321296101092511339901872377hWhat about constant weights $v=\infty$ and $v_n=n$?rNTT Assistant Professor at Vanderbilt University

207789510882691353333I am a *novice* and a simpleton in most all respects. I am only just now beginning to program (in Python) and I hope to get better at it as time goes by. My primary interest is to learn how to code solutions to math problems.

I did a little work on this problem in Bielefeld in 1991, based on a suggestion by Waldhausen.

Let $G$ be a finite group. The algebraic $K$-theory $A^G(X)$ of the category of finite retractive $G$-spaces and $G$-maps over and under $X$, with $G$-homotopy equivalences as the weak equivalences, does indeed have a Segal-tom Dieck style factorization in terms of non-equivariant $A$-theory, indexed over the conjugacy classes of subgroups $H$ of $G$, as you describe. This follows from the additivity theorem. However, I was unable to realize this as the $G$-fixed points of a $G$-spectrum $A_G(X)$.

If you naively define the latter as the algebraic $K$-theory of the $G$-category with the same objects as above, but consider all (not necessarily $G$-equivariant) maps under and over $X$ as morphisms, and take the non-equivariant homotopy equivalences as the weak equivalences, then its $G$-fixed part $A_G(X)^G$ is instead the algebraic $K$-theory of the category of finite retractive $G$-spaces and $G$-maps, but with respect to a coarser notion of weak equivalence, namely a $G$-equivarant map whose underlying map is a homotopy equivalence

There was some interest in trying to prove a version of the Segal conjecture for this theory, i.e., to see if $$ A_G(X)^G \to A_G(X)^{hG} $$ is an equivalence after suitable completion. This could then contain the original Segal conjecture/Carlson's theorem about $Q_G(X_+)^G \to Q_G(X_+)^{hG}$ as a retract. I observed that in general there could not be such an equivalence for a version of $A_G(X)$ satisfying the Segal-tom Dieck splitting $$ A_G(X)^G \simeq \prod_{(H)} A(X^H_{hW_GH}) $$ and the other expected formula $$ A_G(X)^{hG} \simeq F(BG_+, A(X)) $$ (for $X = X^G$, probably). The argument used your (Goodwillie's) computation of the derivative of $A \colon X \mapsto A(X)$ in terms of $\Sigma^\infty(\Lambda X)_+$, to see that the derivatives of the two right hand sides above were not equivalent. Basically free loop spaces do not commute with Borel constructions. So my conclusion was while the Segal conjecture could hold for one space, it would then mostly fail in a neighborhood of that space.

Let $X$ be a Riemann surface of genus $g \geq 2$ or in other words a complex curve.

Let $P_1, \ldots, P_m$ be points in $X$ and $E \rightarrow X$ surjective map such that is is a complex $n$-dimensional vector bundle map on $X-\{P_1, \ldots, P_m\}$ since in $P_1, \ldots, P_m$ the local triviality condition is not satisfied.

I am looking for a general notion of this phenomena. In the spirit of, "locally free sheaves correspond to vector bundles". I am not sure if coherent sheaves are the right objects because I do not want the fibre dimension to vary but only to have no local triviality at a finite number of points.

As a cute "observation", we can use the topology on the unit interval as a "seed" to produce all the standard n-simplices. The topologically-enriched free commutative monoid on the unit interval is the disjoint union all the standard n-simplices. If this somehow fit in, it would be nice.168485612771736434191372148202066969114619042521556522738674945451819388848057307362lWhy is it important that partial derivatives commute?14115857389161747391fComputing distribution of non-identical coin flips13015921974806Given an $n\times n$ invertible matrix $\mathbf A$ and two column vectors $\mathbf u$, $\mathbf v\in\mathbb R^n$, suppose that $1 + {\mathbf v}^T {\mathbf A}^{-1}\mathbf u \neq 0$. Then the Sherman-Morrison formula states that \begin{equation*} (\mathbf A + \mathbf u \mathbf v^T)^{-1} = \mathbf A^{-1} - {\mathbf A^{-1}\mathbf u\mathbf v^T \mathbf A^{-1} \over 1 + \mathbf v^T \mathbf A^{-1}\mathbf u}. \end{equation*}

**Question:** I'm wondering whether we have a similar formula when the inverse in the Sherman-Morrison formula is replaced by the Moore-Penrose Pseudoinverse in case that $\mathbf A$ is singular matrix.

Alexandrov manifold means Alexandrov space which happens to be a manifold, i.e. the space of directions is homeomorphic to shpere. Sorry for introducing this new term.

For such a open manifold does the soul theorem holds? i.e. the manifold is homeomorphic to the disk bundle over it's soul?

XGluing of Quotient Stacks of vector bundles2285635https://lh3.googleusercontent.com/-fKOMzxaaBUI/AAAAAAAAAAI/AAAAAAAAAq8/RAlGoolXYak/photo.jpgJudging by the first expression for $\omega$, I conclude that you are speaking about tensor rather than direct product. (Also, that's what they usually do in physics, and I conclude physics judging by the name "state" :)84592)This is an interesting question, and Greg's answer is excellent. Thinking about this was very nice!

Regarding the Pontryagin-Thom construction, as opposed to the Pontryagin construction which other comments and answers have addressed, I think Thom's original approach, whose goal was to solve Steenrod's realization problem, was quite geometric if you think of it the right way (see this question and this question, where people ask Steenrod's realization problem as an MO question).

Going one way, our initial data is a bordism class, *i.e.* a closed manifold M mapping into a polyhedron X, modulo cobordism. Steenrod's realization problem, by the way, is to determine which homology classes in X are realized this way for some M. Well, what Thom does first is to embed X in some $\mathbb{R}^n$ for n large enough, and take a regular neighbourhood N of X, which, as Greg points out, is diffeomorphic to the normal bundle of X. You now collapse the boundary of N to a point, and the quotient $N/\partial N$ to the Thom space of the regular neighbourhood (normal bundle) of the image of M in X. This is a map into a Thom space, so you can compose with a map into the universal Thom space. As I understand it, the point of this whole exercise is to leverage the nice properties of Euclidean space- namely the existence of the universal Thom space. So taking stock, starting with a bordism class, I have given you a map from $N/\partial N$ to the universal Thom space, whose image is a homotopy class of the right dimension. This is totally geometric- you can "see" the class as the image of $N/\partial N$ in the universal Thom space, where everything outside the image of M is mapped to the basepoint. Thom proved that this is well-defined, and that it's a surjection.

To go the other way is even easier. Starting from a class in the universal Thom space, you realize it as a map g from $N/\partial N$ to the universal Thom space, and apply the Thom isomorphism. What is the Thom isomorphism? It's the intersection of the (relative) cycle in the Thom space of $N/\partial N$ corresponding to g with the zero section (algebraically, you're multiplying by the Thom class), which gives you a closed manifold $M^\prime$ in N. Retract N to X, and you are done, modulo transversality and other technical issues.

Here is a solution that I have found while working on other lattice sums. It utilizes a very simple result:

Define $f$ by $$f(x)=\sum_{n=0}^{\infty} (-1)^n (2n+1) e^{-\pi x (n+\frac12)^2}.$$ Then $$\int_0^{\infty} e^{-y x} f(x)dx=\operatorname{sech}\sqrt{\pi y}.\tag{$\star$}$$

The proof is simple - just integrate term by term, and the result follows from the series $$\sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)}{(2n+1)^2+(2x)^2}=\frac{\pi}{4} \operatorname{sech} \pi x.$$

Therefore, $$\large \operatorname{sech} \frac{\pi}{2} \sqrt{n^2+m^2} = \int_0^{\infty} e^{-\frac{\pi}{4} x(n^2+m^2)} f(x)dx.$$ Now put $(2n+1)$ and $(2m+1)$ instead of $n$ and $m$, multiply by $\displaystyle \, \frac{(-1)^{n+m}}{(2n+1)(2m+1)},$ and sum both $n$ and $m$ over the non-negative integers:

$$ 2 S=\sum_{n,m=0}^{\infty} \frac{(-1)^{n+m}}{(2n+1)(2m+1)} \operatorname{sech}\left(\frac{\pi}{2} \sqrt{(2n+1)^2+(2m+1)^2}\right) = \int_0^{\infty} f(x) \left(\sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} \right)^2 dx$$

Here comes the neat part: we notice that $$\frac{d}{dx} \sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} = -\frac{\pi}{4} f(x),$$ and conclude that $$ 2 S = \frac13 \left(-\frac{4}{\pi}\right) \left(\sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} \right)^3\Biggr{|}_0^\infty \\= \frac13 \left(\frac{4}{\pi}\right) \left(\frac{\pi}{4}\right)^3=\frac{\pi^2}{48}.$$

Can you see how nicely it generalizes to higher dimension versions of this sum?

**Note**:
The function $f(x)$ is actually $\eta^3(i x)$ where $\eta$ is the Dedekind eta function (This is a consequence of Jacobi's triple product identity). The result $(\star)$
was obtained by Glasser, along with other beautiful formulas for integrals involving $\eta(i x)$, in his article
*Some Integrals of the Dedekind Eta-function* (2008) (arxiv link). I omitted this detail simply becuase it is not important for
this purpose.

Otherwise, if all the elements in a set can be represented by a at most n symbols (finite Kolmogorov complexity), I could count them by creating a n dimensional pairing function. Or atleast, that is my assumption.

Any thoughts?

Edit: as @Carl has pointed out the correct term for sequences that have no finite representation is Kolmogorov Random not infinite as I used in my title.

finding permutation matrix I which minimizes TRACE( I* C*( I^T)* D) matrix1686453 But you do not use anywhere sequential continuesness which could help us to approach points in various ways in any way desired? I mean, with the help of that property of continuous functions could it be that we can conjecture the existence of continuous f that sends rationals to irrationals and irrationals to irrationals and then approach some point (or points) with some special sequences to show that such an f cannot exist? What result(s) from analysis do you think we need to prove that such an f cannot exist, and please, choose some result(s) that is(are) as weak as possible.16854742038976X@NickS thanks I have deleted it from there.I would imagine the key advantage of the first definition is not having to think about whether the set is measurable.6208881720985600641@Todd, I get it now. I had a dig and found the argument in a topos theory book. I had a look at Day-Kelly, and it may be enough.85114792670342742375301077906370275I couldn't find anything in your link that such moduli space may not exists, can you explain for me?666851vatna556327Yes.

Assume we have two distinct functions $f$ and $g$ such that $f^{-1}-f\equiv g^{-1}-g$. Take a sequence $x_n=f(x_{n-1})$. Clearly $f(x_n)-g(x_n)=0$ or $(-1)^n[f(x_n)-g(x_n)]$ has the same sign for all $n$.

Sinse $\int_0^1f=\int_0^1g$, there are two sequences $x_n$ and $y_n$ as above such that $f(x_n)=g(x_n)$, $f(y_n)=g(y_n)$ and say $(-1)^n[f(x)-g(x)]>0$ for any $x\in(x_n,y_n)$. Note that $x_n,y_n\to 0$ and $\int_{x_n}^{y_n}|f-g|=const>0$. It follows that $\limsup_{x\to0} |f(x)-g(x)|\to\infty$, a contradiction.

No, this is not the case. Indeed, there is exactly one type of singular fibre on the list for which the scheme structure is not unique up to isomorphism, namely type $I_0^\ast$ in Kodaira's notation. The point, of course, is that here we have a $\mathbf P^1$ component which intersects 4 other componenents, and automorphisms of $\mathbf P^1$ don't act transitively on sets of 4 points.

In fact ${\rm SL}(n,q^h)$ and ${\rm PGL}(n,q^h)$ are btoh simple groups in this case. I think it might be enough to have $n$ coprime to $(q^h-1)/(q-1)$.1825331170989921647To state the obvious, one possible reason is that it's false...7878491465411bNow fixed (after 5 years of intense thinking...)1354466@Lin: In fact I noticed on page 2 of the paper, that their proof is not constructive (see the last paragarph).1678284The condition that the inverse is the transpose is precisely the condition that P is orthogonal, not that it is positive-definite. 4453862088623575165Zone: Hadoop, Spark, Bash Programming, Python & Hive & Pig Latin, R & SPSS, Machine learning

My favor editor: Sublime Text.

Macbook pro with Retina, #15, 2012Mid

137633244773494326419807261746311I don't really understand your request. My description of $f$ is as the unique map satisfying an equation. In which way is this not the kind of structural you're looking for? 1309469 Well, I'm actually trying to figure out an exact formula. The recurrence there represents the probability that a specific Galton-Watson process will have terminated by a given depth of its tree. knowing this, the expected height of the tree can be approximated. Your approximation indicates ways to generally approximate a much broader variety of Galton-Watson processes, and could have actual application in modeling population dynamics. Your method is also generally really interesting to me. I'd have been floudering with much cruder methods for a long time without this. Thank you! 196049101113zColin: did you mean 'don't get a simplicial group' in line 4The equivalence you write, $\pi^\ast\mathcal{K}\simeq\mathcal{T}$, is this a homotopy equivalence?23640111164221215083DC's answer is very good and I'm going to accept it, but I don't yet understand why there might not be a stronger result. I would welcome further answers which could enlighten me and the rest of MathOverflow.99915068212527704213963511893140lNothing about me is of any interest whatever.

2162295It does the trick. The Jordan curve in that paper is a "thinned out" variant of the Peano curve.2173672827981zDoes the Fano plane "embed" in the complex projective plane?1617248186396721998661887178There is of course no such name, as this wouldn't bear much information about the semigroup.

Now, some beautiful paper, where the global idempotency really appears, and really means something:

Robertson, E. F.; Ruškuc, N.; Wiegold, J. Generators and relations of direct products of semigroups. Trans. Amer. Math. Soc. 350 (1998), no. 7, 2665–2685.

Also, if you are interested in power semigroups or finitary power semigroups, there is a whole thesis by Peter Gallagher (with very nice results about finitary power semigroups of groups, and some nice chacracterisation theorems and some really difficult to come up with examples!) from St Andrews.

Also, there is a series of papers by Volodymyr Mazorchuk on various power semigroups of finite transformation semigroups. Also check on the web some results by Mykola Rybak (I think from the journal "Algebra and Discrete Mathematics")

There is some work by Ash -- which is great if you like finite semigroups, you can find a lot about this in the recent book by Benjamin.

There is the most important result in this topic that there are two infinite non-isomorphic semigroups S and T, such that their finitary power semigroups are isomorphic.

Finally, the most important question in the topic is whether there are two non-isomorphic FINITE semigroups whose power semigroups would be isomorphic.

Also, just in case, my e-mail is: victor.maltcev@gmail.com

|https://graph.facebook.com/315539418859777/picture?type=large629502216060013822671508240I recommend the book Geometry and Spectra of Compact Riemann Surfaces from Peter Buser. The great pedagogy of the author is reflected in his book and many results are illustrated. I particularly like his approach to the famous problem "Can we hear the shape of a drum?" using the Sunada theorem. As a further motivation, note that while Jeff Cheeger proved the upper bound on the smallest nonzero eigenvalue of the Laplace operator (Cheeger's inequality), it is Peter Buser who proved the lower bound (Buser's inequality).

1287966Awesome answer :). Thank you so much Prof. Bryant. You read me correctly :). I was trying out a "dead alley" so to speak. I had realized that early on, though I did not have a proof that it was a "dead alley", but I had moved on, and succeeded in doing what I wanted to do. You are so smart, not just in Mathematics, but also in Psychology :). I will of course accept this brilliant answer, but please fix the small typo: you wrote $\mathbb{C}P^4$ instead of $\mathbb{C}P^2$ in the definition of $M^7$. Thank you.4913501644235K. Thuong808355The concept of conjugate points is meaningless once you leave the setting of (smooth) Finsler manifolds.5517531370242330511389313thank you! Yes, for residually finite groups it is trivial. Are there some known examples of classes of groups for which finite FC(G) implies virtually ICC?Ah. I was composing the sum with C. Got it now, thanks. Gerhard "Ask Me About System Design" Paseman, 2011.04.19 1434142XGrowth of stable homotopy groups of spheres2252071nMonoidal structures on modules over derived coalgebras873923l@MTS, Maybe not, but I am not sure. What about Magma? 8293 5489https://www.gravatar.com/avatar/7e33c384ce2407a3629ab467cd800fa1?s=128&d=identicon&r=PG&f=120226049679129319584363209I don't know which mono you mean, but nowhere uniqueness of something is required.13628981808124https://www.gravatar.com/avatar/30a507aa3b9cb553ebe9c2bd35f48c52?s=128&d=identicon&r=PG&f=19287351838291\https://i.stack.imgur.com/q3JiY.jpg?s=128&g=12253037998207568423the equation is correct and is a necessary condition for every solution but it isn't sufficient (every $w(t)$ that satisfy the equation can not be a solution). @AlexandreEremenkoIt was proved by Birget, Margolis, Meakin and Weil that deciding whether a finitely generated subgroup $H$ of a free group is pure (meaning $H$ is closed under taking roots) is PSPACE-complete using semigroup theory.

The main idea is that to each Stallings graph of a finitely generated subgroup of a free group, you can associate a finite inverse semigroup given by generators. They proved that the subgroup is pure if and only if the inverse semigroup is aperiodic (meaning any subsemigroup which is a group is a trivial group). This can be checked in PSPACE. Then, they modified a classical construction in automata and finite semigroup theory showing that it is PSPACE-complete to decide if an automaton has an aperiodic transition semigroup, to prove completeness for purity.

I advice to not train too much stuff that is also done in elementary school because otherwise the child might get bored a lot there. So nothing with basic arithmetic operations or simple geometrical shapes. Do the stuff, they don't do there (actually a pity that it's not done) in a playful way: logic, minimax strategies, graphs.5893971909455I guess the question is "Do there exist C, 0 < C\_1 < C\_2 such that for every l, there exists n < Cl and C_1 < x_i, y_i < C_2 such that ..."Yes. On $\mathbb{R}^n$, a differential form is, more or less, a vector-valued function, and your definition works for vector-valued functions. You just need to show that the definition is preserved under sufficiently smooth changes of variable. The second derivative exists in the weak sense and, using that, $d^2 = 0$.1307223Also @Joro, note that $\zeta(s;6,2)$ is not primitive in the sense that $gcd(6,2)=2$ and therefore $\zeta(s;6,2)=2^{-s}\zeta(s;3,1)$ where $\zeta(s;3,1)$ is now primitive.15063582049733Classically, in the setting of commutative Banach algebras, people speak of the "maximal ideal space". **No one** says "maximal dual space"!163013171539212182313542891312906/Note that because $T_k$ are disjoint, it is trivial to **check** whether they cover $[n]$:
just add up their sizes ($|T_k| = 2^{|M_k| - |m_k|}$) and compare the sum with $2^n$: if
it is smaller than $2^n$, then there exists a uncovered subset of $[n]$, otherwise there is no such
subset.

Quite often, if there exists a polynomial time algorithm for decision problem, there also is a polynomial time algorithm for finding a certificate. In the case of this problem it is true as well. Indeed, consider the following procedures:

do_cover($n, T$):

$~~~~$Returns whether $T$ covers $[n]$. Checks this by simply summing up $|T_k|$.

reduce($n, T, b$):

$~~~~$Reduces each $T_k$ by leaving only sets that contain $n$ or not

$~~~~$if $b = 1$ or $b = 0$ respectively. In case of $b = 1$ element $n$

$~~~~$is also deleted from all sets that belong to $T_k$.

$~~~~$Returns reduced $T$ (and does not modify $T$)

$~~~~$This reduction preserves disjointness of $T_k$ and the

$~~~~$fact that $T_k = [m_k, M_k]$ for some $m_k, M_k \subset [n]$.

$~~~~$Reduced $T_k$ may become empty, but that does not matter much.

find_uncovered($n, T$):

$~~~~$if do_cover($n, T$):

$~~~~~~~~$no such element then

$~~~~$if n == 0:

$~~~~~~~~$just try both possibilities

$~~~~$Now, we know for sure that there exists a subset of $[n]$

$~~~~$that is uncovered and want to find this subset.

$~~~~$There are 2 possibilities: either it contains element $n$,

$~~~~$or it does not. We will just iterate over them.

$~~~~$if do_cover($n - 1$, reduce($n, T, 0$)):

$~~~~~~~~$return find_uncovered($n - 1$, reduce($n, T, 0$))

$~~~~$else if do_cover($n - 1$, reduce($n, T, 1$)):

$~~~~~~~~$return find_uncovered($n - 1$, reduce($n, T, 1$)) $\cup \{ n \}$

$~~~~$else:

$~~~~~~~~$can't happen, because there is an uncovered set

Clearly, find_uncovered runs in polynomial time of $n$ and $N$ (because do_cover does).

Now, note that the same problem is NP-hard if there is no restriction of $T_k$ being disjoint. Indeed, consider some 3-SAT instance with $n$ variables $x_1, x_2, \ldots, x_n$ and $N$ clauses. $T_k$ for $k = 1, 2, \ldots, N$ will consist exactly from such subsets $S$ of $[n]$ such that setting $x_i := (i \in S)$ for all $i = 1, 2, \ldots, n$ will make $k$-th clause false. Clearly, a solution of original problems for such $T_k$ would yield a solution to our 3-SAT instance.

P. S. The solution above for the case of disjoint $T_k$ can be implemented quite efficiently if you forego the immutability of $T$ that I used for clarity. Indeed, you only need to keep $m_k, M_k$ and sum of $|T_k|$. All queries and changes that you'll do all end up very small: you just ask whether some element belongs to some set $m_k$ or $M_k$, do similar minor modifications to $m_k$'s and $M_k$'s (don't forget to update the sum of $|T_k|$, when doing so) and check whether the sum of $T_k$ is equal to $2^x$ for some $x$.

198913240679420019301553472@TheoJohnson-Freyd, when I said I'd have to take some time to think about your comment, I didn't realize it would be five years! Now I see why what you said is true, though, and it's very cool. Thanks!The referenced results (including Lemma 8) are asymptotic, aren't they? But OP asks for an exact expression, as I understand.@@YangMills, Interesting comment1466702577247792022Markus Krötzsch, Professor for Knowledge-Based Systems, Faculty of Computer Science, TU Dresden

1267258bPostdoc working on quantum information theory. 14334481934744@peng you are right about $Q$ and the case $N=3$, $K=2$. I corrected it.@Bora: My problem is that the expression $\prod_{i=1}^n L_i(x)^{-s}$ is not theta-like. To me a theta-like function is an infinite series such as $\sum_{n\in\mathbb{Z}} e^{-n^2 x}$ whose Mellin-transform is a zeta-like function.186503416764001055663209622I ran a quick check for Q3, iterating over all abelian groups of order at most $10^6$. The only integers I got were 1,2,4,6,7,14, and 21. 255521xok, yes, thank you, But why? It doesn't seem trivial to me!210800821141018094091418292127302A trivial observation about this duality, yet somewhat unexpected for me: the dual module doesn't depend on the field used to construct it. At least, under certain finiteness conditions. $\mathrm{Hom}$ cancels the field out, just like $\otimes$ would do...617110Understanding the completed stalk of the dualizing sheaf of a family of nodal curves at a node425415The use of the term "spectrum" to denote the prime ideals of a ring originates from the case that the ring is, say, $\mathbb{C}[T]$ where $T$ is a linear operator on a finite-dimensional vector space; then the prime spectrum (which is equal to the maximal spectrum) is precisely the set of eigenvalues of $T$. The use of the term "spectrum" in the operator sense, in turn, seems to have originated with Hilbert, and was apparently **not** inspired by the connection to atomic spectra. (This appears to have been a coincidence.)

A cursory Google search indicates that Hilbert may have been inspired by the significance of the eigenvalues of Laplacians, but I don't understand what this has to do with non-mathematical uses of the word "spectrum." Does anyone know the full story here?

12165162283866Given two positive $\sigma$-finite measures $\mu_{1/2}$ on the spaces $X_{1/2}$ one can define the product measure $\mu_1\otimes\mu_2$ on the product space $X_1\times X_2$. It can be proved that the following representation holds $$ \mu_1\otimes\mu_2(A)=\int_{X_2}\mu_1(A^{x_2})d\mu_2(x_2)=\int_{X_1}\mu_2(A_{x_1})d\mu_1(x_1) $$ where $A_{x_1}$ and $A^{x_2}$ are the usual slices of $A$ in the two "directions" for fixed points $x_1,x_2$.

Now suppose to have two commuting projection valued measures $E_{1/2}:\mathcal{B}(X_{1/2})\rightarrow P(H)$ where $H$ is a complex separable Hilbert space, then there exists a unique joint projection valued measure $E:\mathcal{B}(X_1\times X_2)\rightarrow P(H)$ such that $$ E(A_1\times A_2)=E_1(A_1)E_2(A_2). $$

My question is: does there exist some "integral representation" of $E$ in terms of $E_1$ and $E_2$ similarly to the $\mu_1\otimes\mu_2$ case?

I would like to have some relation like $$ E(A)=\int_{X_2} E_1(A^{x_2})d E_2(x_2)=\int_{X_1} E_2(A_{x_1})d E_1(x_1) $$ although I don't know how to define this rigorously.

1439271I should have asked: Would you consider Euclid's Elements as a textbook? Wikipedia writes: "Euclid's Elements is the most successful and influential textbook ever written": http://en.wikipedia.org/wiki/Euclid%27s_Elements - And also "The Disquisitiones Arithmeticae is a textbook ...": http://en.wikipedia.org/wiki/Disquisitiones_Arithmeticae15005381854013 So far, we managed to find a pair `$(1111, 11010)$` as a candidate of the minimal length. For this pair, the possible states not necessarily leading to a dead-end state break into three types: 1) `$f_1(f_1(1)) = 1$`, 2) `$f_1(f_1(f_1(1))) = 1$`, and 3) `$f_1(f_1(f_1(f_1(1)))) = 1$`. In a state of type (2), a node `$f_1(1)$` changes; let us call it "write". In turn, (1) and (3) change the root node; let us call them "next". (1) and (2) leave a node whose arrow labeled 1 points to the root node, while (3) leaves a linked list of the three nodes. We have rewriting scenarios for an arbitrary node.82160818449621984502At the top, we have $X_k$ is non-negative; in the middle, you added that $\mathbb{E}[X_k]=0$, so $X_k$ would be identically zero; is this what you meant?tWell, there is your #1. You can use that as a definition.OEIS points to "D. H. Greene and D. E. Knuth, Mathematics for the Analysis of Algorithms; see pp. 95-98." I don't have the book handy to see whether it answers the questions.n"the kinetic energy overwhelms": Ah, that makes sense!842967105093Bochner method was introduced by Bochner in 1946-48 and used by numerous authors to prove vanishing theorems for cohomology groups of vector bundles ever since. This remark, of course, is not meant to diminish Kobayashi-Wu's 1970 paper. 514960278654167525415857671813928197166918906001422615zthe relation between projective and quasi-projective modules77703715893118320351053937VL^1-convergence of convolution exponential144365111254659116414858890~Is there an integer a such that f(X)+a is irreducible in Z[X]?I don't think "exercise-level" has much meaning. In Bourbaki or in Serre's books there are exercises of research level (at least the sense which makes it appropriate for this site). "Exercise" doesn't refer to the level, but rather to the origin of the problemhttps://lh5.googleusercontent.com/-RZDFB2xWAH0/AAAAAAAAAAI/AAAAAAAAAgY/s6l_ZSj3usg/photo.jpg?sz=12814279741173541According to Dickson's History of the Theory of Numbers, this proof was first found by J. Petersen, Tidsskrift for Mathematik (3), 2, 1872, 64-5. (Petersen divides everything by 2, but the idea is the same.) Dickson gives a number of references to rediscoveries of Petersen's proof in the 19th century. Petersen was also apparently the first to discover the combinatorial proof of Fermat's theorem.

1778668What is the relation between conductor of a multiplicative character of a local field and the conductor of square of the character?

774118pCohomology of configuration space of a compact manifold79535XWhat if we take the lcm instead of product?jThis is upper semi-continuity in multi-valued lingo.647547I'm voting to close this question as off-topic because it sounds like it has been assigned as an exercise772887271558359929239691274198I know of no example of a non-finitely-generated algebra over a field which becomes finitely-generated when you tensor it up to a bigger field (one might exist, but none spring to mind, and it might never happen). You know that if both X and Y happen to be subsets of Z, then X tensor Y isn't just the subthing of Z generated by X and Y, right? It can be much bigger, and just surject onto it (like your $F[x_i]$ above).151667187337020195675I am considering the following problem:

Suppose you are given a function $u: C \rightarrow C$, find a function $g$ such that $g(g) = u$ (Let's assume that such a function exists). And by "find", I mean give a series representation of $g$ in terms of known special functions as well as values of $u$.

This is a special case of (the Inverse Operator Problem):

Suppose you are given a function $u: C \rightarrow C$, find a function $g$ such that for some metafunctional $L$ $L(g) = u$. Which is solved by giving a series representation of $L^{(-1)}$

Which is a generalization of (the Inverse Function Problem):

Suppose you are given a number $x \in C$, find a number $t$ such that for some function $f$ we have that $f(t) = C$. This is solved by giving a series representation of $f^{-1}$ which can be done by the Lagrange inversion theorem given knowledge of the values of $f$ and it's derivatives.

So now I was wondering, how does one generalize the techniques of Lagrange to the operator problem I had given?

Here's the meta-plan I set up:

```
create: Linear Meta-Meta Functionals L
Express some Meta-Functional M as a series using L and a suitable basis
Then attempt lagrange Inversion Theorem trick, by determining a chain rule for L
```

However I've gotten stuck on point 2.

Some notation:

functions will be of the form $\text{symbol}(x)$ example $f(x)$, $A(x)$, $e^x+2^x$

meta-functions will be of the form $\text{symbol}(f)$ example $L(f)$, $f'+\frac{1}{f^2 + f(x+1)} $

meta-meta-functionls will be of the form $\text{symbol}(L)$ example $O(L)$, $L(f(x+1)-f(x)) + L^2$

Binding notation: The expression $(U)_{\alpha \leftarrow \beta}$ indicates to evaluate the expression U and then substitute every instance of $\alpha$ with $\beta$. This will be used to avoid ambiguity on operators.

Consider $O(L) = \frac{\partial L}{\partial f}$ this isn't well defined for all meta functions but it doesn't happen to be defined for $L(f) = f(f)$ for which it takes on the value $$O(L) = O(f(f)) = f'(f)$$

Furthermore its null-space is the set of all functions, and its linear which gives rise to the following

$$ O(A(x)) = 0$$ $$ O(A(x)f) = A(x)$$ $$ O(\frac{1}{2}A(x)f^2) = A(x)f$$

etc... which is the naturally way a Taylor series is generated. Thus we have that:

$$ f(f) = f(f)_{f \leftarrow g} + O[f(f)]_{f \leftarrow g} (f - g) + \frac{1}{2}O^2[f(f)]_{f \leftarrow g}(f - g)^2+ ... \frac{1}{n!} O^n[f(f)]_{f \leftarrow g}(f- g)^n + ... $$

Over some radius of convergence. This becomes:

$$ f(f) = g(g) + g'(g)(f - g) + \frac{1}{2}g''(g)(f-g)^2 + \frac{1}{6}g'''(g)(f-g)^3 + ... $$

Except there's a slight problem. It's obvious that we already know

$$g(f) = g(g) + g'(g)(f - g) + \frac{1}{2}g''(g)(f-g)^2 + \frac{1}{6}g'''(g)(f-g)^3 + ... $$

What has happened here is we have generated a meta-taylor series for the meta function $f(f)$ with radius of convergence 0.

I want another operator that actually gives me some non-zero radius of convergence. Because only once I have such a series representation, can I then progress to create the machinery for a generalized Lagrange inversion theorem.

1184639758408675762712018355633478511Integral estimate in the Levitan's paper "On expansion in eigenfunctions of the Laplace operator"Responding to D.Z., calling it r suggests the more general problem of using an arbitrary algebraic number (such as the golden ratio). Then the main question is this: Is $G(n)$ linearly recurrent.254364353686912011One reason might be that there are some numerical quantities that imply amenability of a group and some of the experimental approximations of these quantities for F seem to hang right near the border.The question is not appropriate. Because Selberg's trace formula is just a link between geometry and analysis.Thank you. I'll need some time to digest the references, and not citing articles behind the paywall is definitely welcome :) 13218491898367180028153551912100752145893472839There's now a question on this topic here http://mathoverflow.net/questions/25974/calculus-on-rationals303310422249621983627 Good question! Making $f\in\mathbb{Z}[[x]]$ an univalent (slicht) map on the unit disk seems hard to me (but it should be known, and it is worth posting a new question). The sufficient condition on coefficients for univalence I know is Alexander's one: $n|a_n|$ *be decreasing*, but for $f\in\mathbb{Z}[[x]]$ it implies $f$ is a polynomial. On the other hand, de Branges' necessary condition $|a_n|\le n|a_1|$ leaves some hope. If univalence proves hard or impossible, one should try to build an $f$ depending on both $\alpha$ and $\beta$, not only on $\beta$, possibly not slicht.126080320684281714748http://www.texample.net/tikz/examples/spherical-and-cartesian-grids/

https://www.gravatar.com/avatar/21634b7b3bfe9d8021a384cc8fc587c3?s=128&d=identicon&r=PG&f=120406841602549Non-Haken hyperbolic 3-manifolds without nonorientable surfacesThis question is not appropriate for MathOverflow. In addition to asking this question on math.stackexchange, you might find a web search to be quite fruitful. Gerhard "Ask Me About System Design" Paseman, 2012.01.273891681766437229285In principle, we could have a recursively axiomatized theory for which the property numbers-an-axiom (even relative to some routine Gödel numbering scheme) is recursive but not primitive recursive. But are there any natural examples?

Of course, any such theory can be primitive-recursively reaxiomatized using Craig's trick. So we know that there can't be theories which are recursively axiomatizABLE but not primitive-recursively aziomatizABLE. But that's not the issue. The question is whether there is a theory *T* which *when presented in a natural way* requires open-ended searches to check whether a purported *T*-proof is indeed a proof according to that specification.

[I couldn't think of one when I wrote the first edition of my Gödel book, and I still can't as I work on the second edition. But maybe I'm just being dim/ignorant!]

1233817884260788096624365Seems that you were led astray by notation. The exponential notation for cardinal numbers has nothing to do with asymptotics of the exponential function, or with exponentiation as a series of multiplications, or anything else in arithmetics. alyscia5449931027552In your "we know that" sentence, you really need the fields of transcendence degree 1 to be finitely generated.hPerhaps some people from Deepmind could be helpful!$\textbf{Hey everybody}$: please upvote Zev Chonoles' response, especially if you've already upvoted David's or mine. Reasons: 1) His response came at the same time as David's and before mine (even though all three were written independently); 2) his response -- unlike David's and mine -- was 100% correct on the first pass; and 3) he's a second-year undergraduate (unlike David and I, who are each several years post-PhD). Show some encouragement for the new generation!19518621723506I am certainly not an expert on these things, but I think we already have those. There is a formalisation of the Halting Theorem in the math library of Lean.1392054Can you give some examples and non-examples to help find reasonable conditions?mainly interested in financial mathematics; probability theory, stochastic analysis, extreme value theory...A simplicial category $\mathcal{C}$ is an $\mathbf{sSet}$-enriched category which is bitensored, in that the functors $\mathbf{Map}(-,X)$ and $\mathbf{Map}(X,-)$ admit left adjoints for every $X\in \mathcal{C}$.

It is mentioned on page 114 in Model Categories that the category of topological spaces does not admit a simplicial model structure, with respect to the above definition. On the other hand, Example 9.1.45 in Model Categories and Their Localizations states that it does, and it defines the simplicial structure.

Which statement is correct?

I think the obstruction to the existence of such structure is the existence of exponentials $X^{\mid K\mid}$ for every $X\in \mathbf{Top}$ and $K\in \mathbf{sSet}$. So,

2013950is there an example of a simplicial set whose geometric realisation is not locally compact?

In general, if $V = \bigoplus_{\mu \in \Lambda} V_\mu$ and $W = \bigoplus_{\nu\in \Lambda} W_\nu$ are $\Lambda$-graded vector spaces (for some abelian group $\Lambda$), then

$V \otimes W = (\bigoplus_\mu V_\mu) \otimes (\bigoplus_\nu W_\nu) = \bigoplus_{\mu,\nu} V_\mu \otimes W_\nu = \bigoplus_\lambda \left( \bigoplus_{\mu + \nu = \lambda} V_\mu \otimes W_\nu\right)$

shows that $V \otimes W$ is again a $\Lambda$-graded vector space, and that its $\lambda$-component is $\bigoplus_{\mu + \nu = \lambda} V_\mu \otimes W_\nu$.

`Sarrus determinant rule: references, extensions1661340|volume over a hypercube, over simplex: twist by Euler numbers46524317023846900603397508$f=x^4+100x^2y^2+y^4$ and $g=x^3y+y^3x$. $f/g$ is unbounded (when $y=1$, $x$ is large), so maximum is not attained for $x=y$. On the other hand, $417/10=f(2,1)/g(2,1)```
<!DOCTYPE html>
<html>
<head>
<title>Web Developer</title>
</head>
<body>
<article>
</article>
</body>
</html>
```

@Qiaochu: thank you! I did not know this. So this is a way to model any arbitrary finite set, correct? http://ncatlab.org/nlab/show/hereditarily+finite+setNo problem it's common, but I don't know all common notation of all fields... and $\Omega^2$ evoked me a module of differentials.1332213Let $R$ be a commutative ring with identity. There are so many ways to associate a graph to $R$. Consider this: take the elements of $R$ (All elements including zero) as vertices an two distinct vertices are connected if their product is zero. Lets call this graph, $G_R$.

Suppose that the chromatic number of $G_R$ is finite. I am interested in minimal prime ideals of $R$. My guess is that for any minimal prime ideal $P$ the ring $R_P$ is either a field or finite. Any suggestion or references would be helpful.

1862611282532One interesting way to do this is to set $\mu(x_i,y_i)=m$ for $i\le m$, $\mu(x_i,y_j)=0$ for $j\le m$ with $i\ne j$ , and $\mu(x_i,y_j)=1$ for $j>m$.1758609phttp://www.azimuthproject.org/azimuth/show/Tim+van+Beek230804338387Extension of $X$ by $Y$ has 2 usual meanings, the group theorists's convention being that $X$ is the kernel and the algebraic geometer's convention being that $Y$ is the kernel. In your claim that "$A$ is extension of $\mathbb{Z}^\omega$ by the group of bounded integral sequences" if I'm correct you use the second convention. Here you can define the quotient map to be $(a_n+b_n\sqrt{2})_{n\ge 0}\mapsto (b_n)_{n\ge 0}$ (the image is the set of all integral sequences, the kernel is the set of bounded integral sequences).Sorry, I perhaps was a bit too unspecific there. $P_{Y|X}{Y=1|X=x}=x$ and $P_{Y|X}(Y=0|X=x)=(1-X)$ where $Y\in\{0,1\}$ and $X\in[0;1]$. 7080421728927For a better understanding of this question, please see the question and answer here.

In $Spin(8)$ there are plenty of copies of $Spin(7)$; consider, for instance, the antiimage of $SO(7)<SO(8)$ by the double cover for any $SO(7)<SO(8)$ obtained by fixing some direction in $\mathbb{R}^8$. So it is true, there are plenty of $Spin(7)<Spin(8)$ (let's call this subgroups *primitives*).

By the triality automorphism of $Spin(8)$, the $Spin(7)$ subgroups go to other $Spin(7)$ subgroups. But the hot point is that it is not a permutation between primtives subgroups: that is, if we perform triality on a $Spin(7)<Spin(8)$ obtained by fixing $e_1\in\mathbb{R}^8$, then we get a copy $Spin(7)<Spin(8)$ whose proyection onto $SO(8)$ is not a double cover of some subgroup $SO(7)<SO(8)$, but an isomorphism over some $Spin(7)<SO(8)$. And it happens that this $Spin(7)<SO(8)$ may be described as the stabilizer of a 4-form, as described in the link above.

So the corner question is: **Why this $Spin(7)$ obtained by performing triality to a primitive $Spin(7)$ should in fact be so strangely described, as the stabilizer of some 4-form? Is there a nice way to understand this fact?**

Any idea or suggestion is welcome.

2002412892962464938I did some calculations: For the special case $r=2, s =1, b=-1, c=-1$ (i.e. the fibonacci quadratic polynomial) one gets $\mathrm{Res}(x^n +a, x^2-x-1)=a^2 + \mathrm{Lucas}(n) \cdot a +(-1)^n$, with $(\mathrm{Lucas}(n)) = (1,3,4,7,\ldots)$. Should not be too difficult to prove and could be a starting point for more general cases.1565735156131575754358380@LaurentBerger: That looks interesting. I haven't looked at it, but I will surely do. Thank you!106902Let $(R, \mathfrak{m})$ be a Noetherian normal local domain and $I$ an $R$-ideal. Write $X$ for the normalization of $\mathrm{Proj}(R[It])$ and $E$ for the effective Cartier divisor defined by the invertible ideal sheaf $I\mathcal{O}_X$. We may place suitable hypothesis on $R$ (excellent / essentially of finite type over a field / that completion has no nilpotents) to guarantee that $X$ is of finite type over $\mathrm{Spec}(R)$.

One sees from the exact sequence $0 \to I\mathcal{O}_X \to \mathcal{O}_X \to \mathcal{O}_E \to 0$ that $\mathrm{H}^0(E, \mathcal{O}_E) \supseteq R/\overline{I}$, where
$\overline{I}$ denotes the integral closure of $I$.
Is it true that $\mathrm{H}^0(E, \mathcal{O}_E) = R/\overline{I}$? I am specifically interested in the case: $\dim R = 2$, $I$ is
$\mathfrak{m}$-primary and $R$ is *not* pseudorational. (If $R$ is pseudorational, then the answer is positive.)

Any help/reference will be appreciated.

Thanks; Manoj.

1166341The definition of $p_{A,B}$ is asymmetric in $A$ and $B$, unlike Jaccard distance. The error in the proof occurs in "we look at the first element which in in $A\cup B$" since the first element which is in $A\cup B$ might be in $B$, but not in $A$; even though the first element in $A$ is also in $B$.1946024"if $x$ is a real number of $L$, then it appears at some countable stage $L_\alpha$ for a countable ordinal $\alpha$" So thaaat's how it works. (Could I find a proof that $\alpha$ must be countable somewhere?) As far as verifying that the structure thinks V=L, how would you define "the nth formula with the mth tuple of parameters constructs set x in the $\beta$th level" arithmetically? Or would you not actually need to do that?41603314502221825196Hi, are limitations on the fundamental group for compact complex manifolds known?

Can an arbitrary (finite represantable) group be the fundamental group of a compact complex manifold?

Thanks

1764289Let $i:U\rightarrow X$ be an open immersion and let ${\cal F}$ be a coherent sheaf on $U$.

Let $z$ be a point in $X$. For any affine open set $V\subset X$ containing $V$, you can compute $H^n(V\cap U,{\cal F})$. The fiber of $R^ni_*{\cal F}$ is a limit of these cohomology groups over all such $V$.

Note that if $z$ is contained in $U$, then it has an affine open neighborhood $V$ contained in $U$, and $H^n(V,{\cal F})$ must be zero; therefore the fiber at $z$ is zero.

More generally, the group $H^n(V\cap U,{\cal F})$ can be non-zero only if $V\cap U$ is non-affine, which can happen only if $V\cap U\neq V$, which can happen only if $U\neq X$. So it's the fact that $U$ is ``missing some points'' that allows these cohomology groups to be non-zero. The fiber at $z$ is, therefore (and speaking very roughly) a measure of the extent to which $z$'s absence from $U$ is allowing ${\cal F}$ to have non-zero cohomology on the restrictions of affine open sets.

15213061062810\Adjoint of an operator-valued linear operatorThank you very much so far! An additional question: If B happens to be a 3-torus, what might I try to calculate the "$H^3$-part" of $K^1(B)$ ?834314There was indeed a typo in the Zarankiewicz problem. Thank you.619348nI don't understand what you mean. I made it an answer.1060404102380917690412119094By the way, after a little digging, I found a SIAM review on "semidefinite programming" that might interest you at this address http://stanford.edu/~boyd/papers/sdp.html201309For any hessian matrix, of say 300 by 300, and may or may not necessarily be positive semi-definite, thus algorithms such as Cholesky decomposition may not be used.

I've found that some algorithms are not very accurate enough, and sometimes accuracy is traded with efficiency either.

For non-sparse Hessian matrix, what would be the best algorithm for computing an inverse? (the specific application of it is for online newton step)

Not necessarily. In fact, if the answer was yes, there would be an easy (yet time consuming) algorithm for computing Mordell-Weil ranks, which we don't have.

The lowest conductor example with positive rank and no integral points is this elliptic curve.

```
sage: E = EllipticCurve('189b3')
sage: E.rank()
1
sage: E.gens()
[(-143/4 : -3/8 : 1)]
sage: E.integral_points()
[]
```

I looked it up by typing this into sage:

```
sage: for E in cremona_curves(1..1000):
if E.rank() > 0:
if len(E.integral_points())<3:
print E.cremona_label(), E.rank(), E.integral_points()
```

This functionality should be helpful in answering a lot of other questions of this type you might have about elliptic curves. Some can even by answered simply by using the search feature on LMFDB, but this isn't one of them. You can read more about Cremona's database (now up to conductor at least 300,000) in this report.

1864274500808An example of a similar type is Gromov's Theorem that a finitely generated group has polynomial growth if and only if it is virtually nilpotent (i.e. it has a normal nilpotent subgroup of finite index).

Or Mal`cev's theorem that a polycyclic group is nilpotent by abelian by finite. So a group is polycyclic-by-finite if and only if it is (finitely generated nilpotent)-by-(finitely generated abelian)-by-finite.

Partially Observable Markov Decision Process - finding a hidden object with some positive probability^@BorisBukh Great! Why not make this an answer?I added a rmk for (1). The fact (2) about traces comes from the embedding $W^{1,1}(I\times J)$ in the Bochner space $L^1(I, W^{1,1}(J))$ which is essentially Fubini-Tonelli theorem.15515521156427Take a look at Bob Easton's book ```
Geometric Methods for Discrete Dynamical Systems'.
He was a student of Conley's, who took a very topological/geometric perspective. Conley's tiny monograph,
```

Isolated Invariant Sets and the Morse Index' is also a gem
but perhaps too dense.

(Cross posted from math.SE)

I'm looking for a modern reference to the subject line, preferably one that doesn't use Ore's generalizations to modular lattices.

To clarify terminology...

Suppose we have a group $G$ satisfying the descending and ascending chain conditions on normal subgroups. Then we can find indecomposable subgroups $H_1,...H_n$ of $G$ such that $G=H_1\times \cdots \times H_n$. This is often known as the Krull-Schmidt theorem, though Wedderburn, Remak, and Ore are also commonly associated to it. Moreover, $n$ is the same for any such decomposition, and terms in any two such decompositions are pairwise isomorphic and replaceable. This is found in most (graduate) texts for an introduction to group theory.

The proof is accompanied by Fitting's lemma and the concept of normal endomorphisms. A normal automorphism is also called a central isomorphism. They can all be expressed as the convolution of a morphism $G\to Z(G)$ with the identity on $G$, hence the name. They have several other characterizations as well. These I've rarely seen mentioned in texts, but they are simple and common enough.

However, the subject line is a fact I've never seen mentioned even as an exercise in modern texts: given any two decompositions there is a central automorphism taking one to the other. This is not terribly difficult to work out by playing with the Krull-Schmidt theorem and its proof, but I would like a modern reference if possible.

Wikipedia only mentions it was proved in Remak's thesis from 1911. His thesis is actually available online, and I can confirm that these results are in there (though mostly as an excuse to say that Wedderburn did a poor job proving them). The closest I have to a modern reference is Hall's book on group theory, but this has several shortcomings. It is over 40 years old; he uses Ore's modular lattice proof (which won't quite work for what I want); and he only states that individual terms in two decompositions are pairwise centrally isomorphic, which is not as strong a statement as the central automorphisms acting transitively on the decompositions.

Does a reference better suiting my needs exist? I'd very much like one. Even a book that leaves these to the exercises would be better than what I currently have.

A quick Google search brings up http://plato.stanford.edu/entries/type-theory/ .156265010708731731948If you're looking for a wiki that can handle LaTeX-style equations, then you should take a look at instiki. Not only does it display mathematics properly, it can also export pages to LaTeX.

http://www.instiki.org/show/HomePage

1860957Dan KneezelF*This is a problem which has been bothering me for a while now; it doesn't seem inherently too hard, but I haven't been able to make any real headway, so I'm putting it out in the open since at this point I just want to know the answer. I don't think it has any deep value, but it's a natural question (at least to me) which I haven't seen addressed in the literature.*

In brief: is there a single game $G$, defined by a formula in the language of second-order arithmetic, such that the statement "$G$ is determined" isn't true in any Turing ideal with a maximal element (viewed as an $\omega$-model of RCA$_0$)? Say in this case that $G$ has **boldface strength** (as opposed to the merely *lightface* strength that we're used to individual instances of principles having, in reverse mathematics).

*(Note,of course,that because a model $M$ thinks $G$ is determined doesn't mean $M$ contains an actual winning strategy, or even that $M$ is correct about who wins.)*

Let me say a bit about my thoughts so far:

**Why the answer might be yes**: the statement "$G$ is determined" is $\Sigma^1_2$ on top of $G$; so it easily has enough quantifier complexity to have boldface strength, even if $G$ is low in the projective hierarchy. Now of course $G$ can't be *too* low in the projective hierarchy - after all, we need the Turing ideal generated by a winning strategy for $G$ to not satisfy "$G$ is determined," so membership in $G$ has to be sufficiently non-absolute. But once $G$ gets complicated enough, all bets seem off. That is: if $G$ is sufficiently complicated, then merely having a winning strategy doesn't matter, the model needs enough structure to be able to verify that the strategy always wins, and there's no obvious upper bound to how complicated this extra structure could be.

**An obstacle**: A natural guess for a game with boldface strength is the following: Player $1$ plays a second-order sentence $\varphi$ of some bounded complexity (say, $\Pi^1_7$), and player $2$ plays "true" or "false" then player $2$ wins iff $\varphi$'s truth value is player $2$'s play. (The bound on the complexity of $\varphi$ is needed for this game to be definable.) However, this doesn't work, since by Martin's cone theorem the models corresponding to all sufficiently large degrees will yield the same theory, so once our topped models get big enough they'll see how to win this game.

**A response**: More generally, the obstacle above suggests that the game needs to involve playing a *specific* element of the top degree of the model (if there is one). This suggests a game like the following: player $2$ builds a real of maximal Turing degree, and also responds to player $1$'s queries about that real. To give player $2$ a chance, we should demand that player $1$ only gets to ask finitely many questions. Meanwhile, we have a real synergy between this game and the existence of a maximally complicated real: on the one hand, in a model with a maximally complicated real player $2$'s strategy must build something as complicated as itself, while in a model with no maximally complicated real player $1$ wins vacuously (e.g. the classical analysis has no relevance to the case we're interested in).

In particular, a game of the following form seems promising: Player $2$ builds a real $r$, while player $1$ plays finitely many second-order sentences of some fixed complexity (say, $\Pi^1_7$), and player $2$ responds to player $1$'s queries with "True" or "False" as above; player $2$ then wins iff $r$ has maximal Turing degree and they never guessed incorrectly about the theory of $(\omega\cup\mathcal{P}(\omega); +,\times, r)$. (Note that all of this is defined *internally*.)

**Another obstacle**: This suggests some recursion theorem trickery, where player $1$ manages to ask player $2$ about their own strategy in such a way that player $2$ is trapped. In particular, the real that player $2$ builds is guaranteed to compute all of their strategy, if player $2$ is playing according to a winning strategy. However, I haven't managed to get this line of attack to pan out, the main problem being that player $2$'s real (and hence the way player $2$'s real computes player $2$'s strategy) depends on player $1$'s queries.

So I'm asking now:

509133xA characterization of flat metrics via global vector fieldsSure, but the question was about p-adic L-functions in general, not specifically for elliptic curves. 143686722292055469171671907.Is there a single game with boldface strength?

(I assume that by

linear functionsyou meanaffine functions, i.e. that you are not assuming $\ L_*(0) = 0$).

First of all, the class of your functions is contained in the class of all convex functions. Each convex function $\ Ł : \mathbb R^n\rightarrow\mathbb R\ $ (including your functions) admits a unique representation as follows:

$$ Ł\ = \ \max_{s\in S} Ł_s $$

where $\ S\subseteq \mathbb R^n\ $ is the set of all smooth arguments
$\ s\ $ (see below), and $\ Ł_s\ $ is the **unique** affine function such that
$\ Ł_s(s) = Ł(s),\ $ and
$$ \forall_{x\in\mathbb R^n}\quad Ł(x)\ge Ł_s(x) $$

The set $\ S\ $ is defined as the set of points $\ s\ $ for which such a (supporting) function is indeed unique.

Thus, your sum has got (or *gotten*) replaced by a single summand.

Some different $\ s\in S\ $ may lead to the same $\ Ł_s,\ $ especially in the case of your finite expressions. The $\ \max\ $ function doesn't care. In the case of a finite expression, the number of the

differentfunctions $Ł_s$ will be finite.

In the case of a finite expression, take just one $\ s\ $ for each $n$-domain around which the function is affine (i.e. $\ s\ $ should belong to that domain).

The question seemed a little bit vague to me, hence I didn't attempt any algorithmic considerations.

**EDIT** (*more details*)**:**

The function (a finite sum) considered in the *Question*

$$ Ł\ :=\ \sum_i \max \{ L_{i1},L_{i2},..,L_{ik} \}$$

is convex, hence the set:

$$ C\ :=\ \{(x\ t)\in\mathbb R^n\times\mathbb R: Ł(x)\le t \} $$

is convex and closed. Thus the topological boundary

$$\ B\ :=\ \mathcal B (C)\ =\ \{(x\ t)\in\mathbb R^n\times\mathbb R: t=Ł(x) \} $$

(i.e. the graph of $Ł$) contains a dense subset $S\subseteq B$--dense in the boundary $B$--such that for every $\ (x\ t)\in S\ $ there exists exactly one $n$-plane $\ H\ $ such that $\ (x\ t)\in H\ $ and $\ C\ $ is on the one side of $\ H.\ $ Of course $\ t=Ł(x)\ $, and $\ H\ $ is the graph of a sum of the original affine functions $\ L_*.\ $ We see that there are only finitely many such affine functions simply because there are finitely many of the original affine functions to start from (the same affine plane will appear $\infty$-many times but it doesn't matter when they appear unde $\max$). Thus function $\ Ł\ $ is just one $\max$ of finitely many affine functions.

A further detail: each said $H$ is a graph of the sum of functions $L_*$ taken one from each summand (of expressions $\max$) of the original expression.

I truly believe that all this is basic and true. I am sorry if I am somehow blind or worse.

Quadratic twists of elliptic curves (or, more generally, abelian varieties) are familiar objects in arithmetic geometry. I would like to extend that definition to the category of 1-motives over global fields, but I have no idea where to begin. Unfortunately, a google search didn't help much. Can anyone suggest literature that might be related to this matter or pitch in with some ideas? Thanks a lot in advance and Happy New Year 2016!

Let's begin with a few observations. Suppose we consider the set of $N\times N$ matrices and consider the matrices with positive determinant. There are several connected components in this set; let $S$ be the component that contains the identity matrix. Observe that $S$ corresponds to the set of positive definite matrices. This fact (and the convexity of $S$) is important for interior-point techniques in semi-definite programming.

Suppose we replace "determinant" with "permanent". What happens? In other words, suppose we consider matrices with positive permanent. Let $C$ be the connected component that contains the identity matrix. The closure $\overline{C}$ is a pointed cone.

- Is $C$ convex?
- Does $C$ or $\overline{C}$ have a name, or can anyone point me to literature discussing its properties?

I am somewhat doubtful that the question as posed as any sort of reasonable answer. (Also, I don't really see how the Lorentzian metric even enter into the problem.)

(a) ANY hyperbolic PDE in (1+3) dimensions will, by definition, not have unique solution for your problem.

(b) Even elliptic PDEs are not guaranteed to have unique solutions, even for nice ones that come from minimization of a strictly convex energy functional: Euclidean space simply has too large a symmetry group, and if the equation itself is invariant under translations, any finite translate of a solution that decays at spatial infinity is another solution.

(c) If you restrict yourself to linear scalar PDEs $L[u] = 0$, by linearity, $u \equiv 0$ is necessarily a solution that satisfy your conditions. So your question as posed reduces to "which linear operator admits only the trivial solution". For constant coefficient $L$'s you can simply take the Fourier transform of both sides and study the zero-set of the PDO.

If you clarify your motivation for considering the question your are asking, maybe more reasonable answers can be given.

Edit: To say a bit more about the Fourier transform method. For example, if you assume that your solution is in Schwarz space on spatial slices, and assume the the PDO $L$ does not have $\partial_t$ terms, then you can ignore the time component. Restricting to the spatial slice and taking Fourier transform, you see that $P(\xi)\hat{u}(\xi) = 0$ where $P$ is the symbol for $L$, and is a polynomial in $\xi$. Using that the Fourier transform is injective on $\mathcal{S}$, you immediately have that $\hat{u}$ can only be supported on the zero set of $P$, but as $P$ is a polynomial, its zero set has vanishing measure unless $P\equiv 0$. Thus you see that in Schwarz space any non-trivial constant coefficient $L[u] = 0$ only has trivial solutions. Now, it is well known that the smoothness of $\hat{u}$ is related to the decay properties of $u$. By considering restriction theorems/Strichartz type estimates, there can be solutions for certain PDOs in $L^p$. (As an example, let the space-dimension be 4 [only because I remember the Lebesgue exponent explicitly in this case], Then the PDO on $(t,w,x,y,z)$ given by $-\partial_w^2 + \partial_x^2 + \partial_y^2 + \partial_z^2$ admits infinitely many solutions in $L^4(\mathbb{R}^4)$, with Fourier transform supported on the set $\hat{w}^2 = \hat{x}^2 + \hat{y}^2 + \hat{z}^2$. You can even ask that the solutions be smooth: they just cannot decay too fast in the $w$ direction.)

Edit 2 [After the updated question, one should probably read the comments below first before reading this]: A few things

(i) Intuition from 1+1 dimension can be misleading. Waves don't decay there. In the 1+3 case, solutions to the wave equation disperses. So for the equation give by $L = -c^{-2}\partial_t^2 + \triangle$, even when the wave speed is bigger than the speed of gravity, the solution is not incompatible with mere decay at spatial infinity. (Though of course, since the equation is no-longer Lorentz invariant, speaking of Spatial Infinity is somewhat of a red-herring: the conformal compactification of the space-time does not give a compatible compactification of the solution to the equation.) If to rule out such cases you need to also impose a rate of decay.

(ii) When I said local well-posedness of $L$, implicitly I mean on a suitable function space on space-like slices. For (strictly/symmetric/regularly) hyperbolic equations, you can of course study the characteristic cones and compare against the back ground null cone to have the decay automatically also satisfied for boosted slices.

(iii) As a trivial example, also note that in the $\partial_t v + Av$ formulation, you can take the Fourier transform to get the solution to be $\hat{v}(t,\xi) = e^{-A(\xi)t}\hat{v}_0(\xi)$. If the matrix $A(\xi)$ has only eigenvalues with negative real parts, then Schwarz data will lead to Schwarz solutions, and lead to non-uniqueness of the solutions. This is sort of the explicit version of the Hille-Yoshida type theorems.

Here is simplified and compact version of sage worksheet: http://sagenb.com/home/pub/4603/|persistence of regularity for nonlinear Klein-Gordon equation1321390Ryan, a translation of $\mathbb R^2$ extends to a diffeomorphism of $S^2$ with one fixed point; it is hardly exotic.Is there a known primitive recursive upper bound on the nth "Zhang prime"227384017288868916172231850https://www.gravatar.com/avatar/77710382ebcd47aaa6b8e5c4a0905d8b?s=128&d=identicon&r=PG&f=1509523@user40276 I'll try that, but I do not know how to define a Pfaffian, or even a volume form on a sheaf. I don't even know if it's been done before. Has it?2267932~https://graph.facebook.com/2166509826951964/picture?type=large20985161954377871268The space $U$ in Ariyan's comment is just $M_{0,2g+2}$. So it is a $K(\pi,1)$ because its universal cover is a Teichmuller space. One can also use this description to see that the fundamental group is the hyperelliptic mapping class group, given that the latter is a Z/2 central extension of the mapping class group of a sphere with 2g+2 unordered punctures.1056098I am reading about Lyapunov functions for Markov processes, and I am having trouble thinking of examples to keep in mind as I read. If $X_t$ is a continuous-time Markov process with generator $L$, a Lyapunov function is supposed to be a function $V$, in the domain of $L$, with $V \ge 1$ such that $LV \le -aV + b 1_C$, where $a,b$ are constants and $C$ is a "petite" set. It seems that the existence of a Lyapunov function leads to good results on the rate of convergence of $X_t$ to a stationary distribution.

What are some simple examples of processes with explicit Lyapunov functions? Continuous processes would be best. I was trying to think about something like Brownian motion on the circle, but got stuck.

Perhaps it is worth reading the review on MR that R. Fintushel wrote about the paper quoted above: http://www.ams.org/mathscinet-getitem?mr=28231133249738521797$\newcommand{\RR}{\mathbb{R}}\newcommand{\calF}{\mathcal{F}}\newcommand{\diam}{\mathrm{diam}}$ In geometric measure theory there are various notions of $m$-dimensional measure for sets $A\subset \RR^n$ for $m\leq n$ (some of them also for non-integer $m$, but this is not the point here). They all build on Carathéodory's general construction which works by covering $A$ with countably many sets $E_i$ from a specific base family $\calF$, measuring the sets $E_i$ with a function $\zeta$ and then building the infimum over all these coverings that are $\delta$-fine (i.e. $\zeta(E_i)\leq \delta$), and letting $\delta\to 0$.

Among these specific measures are

**Hausdorff measure**for $s>0$: This uses all sets for covering and takes $\zeta = \diam^s$, the diameter of the set raised to the $s$th power.**Spherical measure**is the same but restricts the family $\calF$ to consist only of balls.**Dyadic net measure**is again similar but uses cubes with dyadic corner points as the family $\calF$.**Gross measure**for $m=0,1,\dots$ is a bit different: It uses the Borel sets for $\calF$ and reuses the $m$-dimensional Lebesgue measure as follows. For some $E$ define $\zeta(E)$ as the largest $m$-dimensional Lebesgue measure you get by projecting $E$ onto any $m$-dimensional subspace.**Carathéodory measure**is similar to Gross measure but takes as $\calF$ only closed and convex subsets.

(There are more, e.g. Federer measure or Gillespie measure…)

Somehow, the Gross measure seems most natural to me as the method of covering really drives the minimizing covers to follow the set as close as possible and also directly counts the size of the covering sets in the way one wants to have in the end (slight drawback is that it only works for integer $m$). However, the Hausdorff measure seems to produce a very reasonable definition as illustrated by various examples (e.g. rectifiable curves, Cantor dust in two dimensions with Hausdorff dimension 1).

In the books on geometric measure theory I considered (like Federer, Mattila, Krantz & Parks, Morgan) they describe the construction and basic estimates between them quite detailed and mainly use the Hausdorff measure afterwards. Especially for the notion of dimension it turns out that Hausdorff measure, spherical measure and the dyadic net measure all give the same notion of dimension.

However, I could not find answers to these questions:

Are there sets for which the Gross measure gives something unreasonable (in comparison to intuition or Hausdorff measure), e.g. a totally "wrong" size or even a "wrong" dimension?

Are there sets for which the spherical or dyadic net measure gives some intuitively wrong size (or some size different from the Hausdorff measure)?

I would also love to have some examples of the usefulness of the measures different from the Hausdorff measure (which appear to be handy for the analysis of self-similar sets). Hence, a third question is

${}$3. What are the spherical, dyadic net, Gross or Carathéodory measure good for?

Somehow I am most interested in Gross and Carathéodory measure - they are defined in several books but basically not used. Also it is daunting to search the web for "Gross measure" and totally not helpful to search for "Carathéodory measure".

16246387135422226141Let $K$ be a number field, and let $$\zeta_{K}(s):= \sum_{0 \neq I \text{ ideal of }O_K} \frac{1}{N_{K/\mathbb{Q}}(I)^s} = \sum_{n \ge 1} \frac{a_n}{n^s}$$ be the Dedekind zeta function of $K$. The quantity $s_K(x):=\sum_{n \le x} a_n$ counts ideals of $O_K$ of norm up to $x$.

$\zeta_K$ is analytic in $s\ge 1$ apart from a simple pole at $s=1$, with residue given by the class number formula. The Wiener-Ikehara theorem implies: $$s_K(x) \sim c_{K} x$$ as $x\to \infty$, where $c_K$ is given by the class number formula.

Let $E_K(x):=s_K(x) - c_K x$.

When $K=\mathbb{Q}(i)$, the problem of studying $E_K$ is known as the Gauss circle problem. Proving $E_K(x) = O(x^{1/2})$ is easy, but it is believed that $E_K(x) = O_{\varepsilon}(x^{1/4 + \varepsilon})$.

What is known about $E_K(x)$ for general number fields (conditionally and unconditionally)? What is the heuristic for that?

What distinguishes the case $K=\mathbb{Q}(i)$ from other cases? (Apart from the elementary geometric interpretation)

No. Let $a,b,c,d$ be complex numbers of modulus $1$, chosen so that $a\neq c$, $b\neq d$, and $ab^*\neq cd^*$. Form the column vectors $v=(1,a,b)^T$ and $w=(1,c,d)^T$. Then the matrix $$ vv^* +ww^*=\begin{pmatrix} 2 & a^*+c^* & b^*+d^* \\ a+c & 2 & ab^*+cd^* \\ b+d & a^*b+c^*d & 2 \end{pmatrix} $$ has rank two, but all the off-diagonal entries have modulus strictly less than $2$.

*EDIT*: A simple example would be
$$
\begin{pmatrix} 1 & \frac35 & \frac45 \\ \frac35 & 1 & 0 \\ \frac45 & 0 & 1\end{pmatrix}
$$
where one can replace $(\frac35, \frac45)$ with any pair $(a,b)$ such that $|a|^2+|b|^2=1$.

Let me denote $u_i = x^{i+1} + 1$. As I learned from the previous question, under the "matrix system" OP understands the following matrix equation (up to a factor $\frac{1}{x}$): $$My = b,$$ where $$M:=\begin{bmatrix} u_1^0 + x - 1 & u_1^1 + x - 1 & \cdots & u_1^{n-1} + x - 1 \\ \vdots & \vdots & \ddots & \vdots \\ u_n^0 + x - 1 & u_n^1 + x - 1 & \cdots & u_n^{n-1} + x - 1 \end{bmatrix} \quad\text{and}\quad b := \begin{bmatrix} u_1^n + x - 1 \\ \vdots \\ u_n^n + x - 1 \end{bmatrix}. $$

First, it is easy to see that it always has a unique solution: $$y = M^{-1} b = AA^{-1}M^{-1} b = A (MA)^{-1} b,$$ where $$A := \begin{bmatrix} \frac{1}{x} & - \frac{x-1}{x} & - \frac{x-1}{x} & \cdots & - \frac{x-1}{x} \\ 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$ so that $$MA = V = V(u_1,\dots,u_n)$$ is a Vandermonde matrix.

Second, let us find the value of $V^{-1}b$.

The Vandermonde matrix inverse $V^{-1}$ has elements $$(V^{-1})_{i,j} = [z^{i-1}]\ \frac{F(z)}{F'(u_j)(z-u_j)},$$ where $$F(z) := (z-u_1)\cdots (z-u_n)$$ and $[z^k]$ is the operator taking the coefficient of $z^k$. Hence, $V^{-1} b$ is composed of the coefficients of $$G(z):=\sum_{j=1}^n \frac{F(z)}{F'(u_j)(z-u_j)} (u_j^n+x-1)=z^n - F(z) + x-1,$$ where the latter equality holds since the right-hand and left-hand sides as polynomials in $z$ both have degree $\leq n-1$ and at $z=u_j$ evaluate to $u_j^n+x-1$ (i.e., they have equal values at $n$ distinct points).

Hence, $$(V^{-1}b)_k = [z^{k-1}]\ G(z) = \begin{cases} (-1)^{n-1} u_1\cdots u_n + x-1, & k=1;\\ (-1)^{n-k} e_{n+1-k}(u_1,\dots,u_n), & k>1; \end{cases}$$ where $e_k()$ are elementary symmetric polynomials.

Finally, from $y=A(V^{-1}b)$ we get $$y_1 = (-1)^{n+1} u_1\cdots u_n + x-1 - \frac{x-1}{x}G(1)$$ and $$y_k = (-1)^{n+k} e_{n+1-k}(u_1,\dots,u_n),\quad k>1.$$ Recalling the definition of $u_i$, we conclude that the free term of $G(1) = x -(1-u_1)\cdots(1-u_n)$ as polynomial in $x$ is zero, and thus $\frac{x-1}{x}G(1)$ is a polynomial in $x$ with integer coefficients. Then so are all $y_1,\dots,y_n$.

Sorry, I should say not the closest, but the closest Gaussian integer to the lower-left; i.e., take the floor of each component.255971291401@Yemon: I think you forgot finite additivity in your list of conditions. `:-)` Anyway, Terry Tao gave a good description of the difference between the dimensions on his blog: http://terrytao.wordpress.com/2009/01/08/245b-notes-2-amenability-the-ping-pong-lemma-and-the-banach-tarski-paradox-optional/551541288751@Raphael: you may get more answers at the sister site devoted to math finance http://quant.stackexchange.com/18477921383049429171170919Many thanks! This is a perfect reference (and I am happy to discover the nice and simple idea of the proof).1129480350819ZTerminology generalizing "quasi-isomorphism"722231$1+{n+m-1\choose m}$ is an easy (and probably not too bad) upper bound to your first question.1792374522071237480There is the ring of diferential operators à la Grothendieck, which is infinite dimensional and has elements of all orders. It is an increasing union of matrix rings in the case of the line, though, so it is a bit strange.68409422720862016338Here is an example from probability theory. Let $X_i$ be a sequence of independent identically distributed random variables in $L^1$. The strong law of large numbers asserts that the mean converges to the expectation.

$$a.e. \quad {1\over n}\ \sum_{k=0}^{n-1} X_k \rightarrow E(X_0).$$

What can be said about the speed of convergence ? If we assume that the $X_i$ are in $L^p$ for some $p\in ]1,2[$, then we have:

$$a.e. \quad {1\over n}\ \sum_{k=0}^{n-1} X_k = E(X_0) + o(n^{1/p-1}).$$

18746881525454203557445547dCompetitive programmer and problem solver

RAh! Thank you, that answers my question.13570336193550782112019412281499584089~Wild guess: the question is "Does Witten's claim hold water?".746135126966922320352Yes,coudy,you are right.1862328617861097832177162812726438ruakh88286444439518027201408757186300587628bWhen does $A^A=2^A$ without the axiom of choice?1903033*I tried to comment on this question since it is rather badly posed. Since I do not have enough credits yet on mathoverflow I was not allowed to comment. Unfortunately, it appears that nobody with enough computational matrix theory knowledge has edited it or commented on it.

(a) You have not said why you want to do this decomposition. Numerical computational experts have design many alternative solutions depending on the usage. As a start, please read "Matrix Computations" by Gene Golub and Charles Van Loan before trying to reinvent the wheel. If software is required then google "LAPACK". The original package is in Fortran but has been translated to almost all popular high level languages.

(b) You have not defined what is meant by $E'$. Is it the standard transpose operation on $E$ or the complex conjugate transpose operation?

(c) If it is the transpose operation, then the product $E'E$, in general, is not real if $E$ is complex.

(d) Is the symmetric matrix $A$ positive definite or at least positive semi-definite? If $A$ is positive definite, then $E$ can be taken as a real upper triangular matrix and the decomposition $A = E'E$ is known as the Cholesky factorisation. If $A$ is positive semi-definite, then a permutation matrix $P$ can be used for the decomposition $PAP' = E'E$ where $E$ is upper triangular. Equivalently, $A = P'E'EP$.

(e) If the matrix $A$ is sign indefinite, then the matrix $E$ cannot be taken as real. If this is the case, then we use the $E'DE$ decomposition (sometimes known as the $RDR'$ decomposition) where $E$ is upper triangular and $D$ is a block diagonal matrix where the blocks are $1 \times 1$ or $2 \times 2$. Note that a blocks of the form $$ \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right] $$ cannot be diagonalised without complex arithmetic. Normal convention is to avoid complex arithmetic, as much as possible, for real matrix problems. Thus, $2 \times 2$ blocks which cannot be diagonalised without complex arithmetic are left alone. Of course, if you really want to find the square-roots then operate only on these blocks.

(f) As far as I know, this problem cannot be solved using the SVD. Please indicate the steps if this problem can be solved using the SVD.

(g) The complexity of the problem (based on multiplication counts) does not always reflect the running time. There are very efficient programs based on level 3 BLAS (Basic Linear Algebra Subroutines) operations which optimally utilises modern hardware. Usually, cache management is more important than basic multiplication counts since certain arithmetic operations can be done in parallel.

tTrace of a weighted composition operator on Bergman spaceYes, I should have said canonical. Universal has a categorical meaning which is not true here. Thank you!jGot the book, and it seems to have the same problem.86604921598591825983There is an explanation of sorts in Section 1.4 of Elkies's "The Klein quartic in number theory". There is a three-dimensional lattice $L$ over the cyclotomic field $k=\mathbf Q(\zeta_7)$, and $G$ can be defined as its group of isometries. The resulting three-dimensional representation of $G$ has the unusual property of remaining irreducible when reduced modulo every prime of $\mathcal O_k$. Its reduction modulo a prime over $2$ turns out to be $\mathrm{GL}(3,\mathbf F_2)$ acting on $\mathbf F_2^3$, and its reduction modulo a prime over $7$ is $\mathrm{PSL}(2,\mathbf F_7)$ acting on $\mathbf F_7^3$ as the symmetric square of the two-dimensional representation of $\mathrm{SL}(2,\mathbf F_7)$. (Note that since $-1$ acts trivially on the symmetric square, the symmetric square really is a projective representation.)

if you phrase it as "how many objects can I fit in the circle", the "field of mathematics" you are looking at is that of [packing problems](https://en.wikipedia.org/wiki/Packing_problems)9Another question that's getting no answers on stackexchange:

Once upon a time, when Wikipedia was only three-and-a-half years old and most people didn't know what it was, the article titled functional equation gave the identity $$ \sin^2\theta+\cos^2\theta = 1 $$ as an example of a functional equation. In this edit in July 2004, my summary said "I think the example recently put here is really lousy, because it's essentially just an algebraic equation in two variables." (Then some subsequent edits I did the same day brought the article to this state, and much further development of the article has happened since then.)

The fact that it's really only an algebraic equation in two variables, $x^2+y^2=1$, makes it a lousy example of a functional equation. It doesn't really involve $x$ and $y$ as functions of $\theta$, since any other parametrization of the circle would have satisfied the same equation. In a sense, that explains why someone like Norman Wildberger can do all sorts of elaborate things with trigonometry without ever using trigonometric functions.

**But some** trigonometric identities do involve trigonometric functions, e.g.
$$
\sin(\theta_1+\theta_2)=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2
$$
$$
\sec(\theta_1+\cdots+\theta_n) = \frac{\sec\theta_1\cdots\sec\theta_n}{e_0-e_2+e_4-e_6+\cdots}
$$
where $e_k$ is the $k$th-degree elementary symmetric polynomial in $\tan\theta_1,\ldots,\tan\theta_n$. These are good examples of satisfaction of functional equations.

So at this point I wonder whether **all** trigonometric identities that do seem to depend on which parametrization of the circle is chosen involve adding or subtracting the arguments and no other operations. In some cases the addition or subtraction is written as a condition on which the identity depends, e.g.
$$
\text{If }x+y+z=\pi\text{ then }\tan x+\tan y+\tan z = \tan x\tan y\tan z.
$$

**QUESTION:** Do **all** trigonometric identities that **do** involve trigonometric functions, in the sense that they are good examples of satisfaction of functional equations by trigonometric functions, get their non-triviality as such examples only from the addition or subtraction of arguments? Or is there some other kind? And if there is no other kind, can that be proved?

In comments below the stackexchange posting, Gerry Myerson mentioned the identities $$ \cos \frac x2=\sqrt{\frac{1+\cos x}{2}} $$ and $$ \prod_{k=1}^\infty \cos\left(\frac{x}{2^k}\right)= \frac{\sin x}{x} $$ The latter is somewhat like the one involving tangents above: One can say that if $x_n = x_{n+1}+x_{n+1}$ for $n=1,2,3,\ldots$ then $$ \prod_{k=1}^\infty \cos \left(\frac{x_k}{2^k}\right) = \frac{\sin x_1}{x_1}. $$ A similar but simpler thing applies to the half-angle formula.

**Postscript:** Wikipedia's list of trigonometic identities is more interesting reading than you might think. It has not only the routine stuff that you learned in 10th grade, but also some exotic things that probably most mathematicians don't know about. It was initially created in September 2001 by Axel Boldt, who was for more than a year the principal author of nearly all of Wikipedia's mathematics articles---several hundred of them.

I'm looking for a result of the form: Let $B_\epsilon$ denote a "natural" smooth $\epsilon$-approximation to an $n$-dimensional Brownian motion $B$ (e.g. by mollification or simply piecewise linear) with $\sigma$ and $\mu$ smooth coefficients, and denote by $X_\epsilon$ the solution to the random ODE $$dX_\epsilon = \mu(X_\epsilon)dt + \sigma(X_\epsilon)dB_\epsilon.$$

Then $X_\epsilon$ converges in probability, as $\epsilon \to 0$, so the solution of the SDE $$dX=\mu(X)dt+\sigma(X)\circ dB.$$

In particular I'm seeking results with no ellipticity requirement on $\sigma$. Many thanks.

https://lh6.googleusercontent.com/-bK7Mz4b1nUE/AAAAAAAAAAI/AAAAAAAAAAA/oh9lH8zJbXg/photo.jpg?sz=128zThe minimal growth rate of the countable family of sequences1517248Bounded analytic functions are not "completely regular", though; the idea behind my example was basically to enlarge them to make them completely regular.@AlexanderChervov Hard to say exactly. Some combination of 1) computational feasibility of the implementation (definitely not true in de Finetti's time), 2) good marketing/branding, 3) reaching a new audience (computer scientists reading JMLR) and 4) the contemporary relevance of the applied problem (text classification in the internet age). Regarding branding, LDA was actually a "taken" term in traditional statistics, referring to a classification method invented by Fisher. https://en.wikipedia.org/wiki/Linear_discriminant_analysis ("Not to be confused with...")However, one more important property (which I did not explicitly mention in my question) we have not considered yet is connectedness of $X$. The non-existence of non-trivial clopen subsets makes life more difficult (or easier, depending on the perspective).480726878448,But is it true that, for V a commutative theory, TW-V-monoids form a variety again, with operations given by those of V (applied pointwise on endomorphisms) plus the monoid structure coming from composition - i.e. the structure of the set endomorphisms?

Yes, this is true. And the simplest example of a TW-V-monoid is where V is abelian groups whereupon TW-V-monoids are simply rings (unital, not necessarily commutative).

What led us to write our paper is that we couldn't find an accurate description of what we wanted (the full structure on unstable operations in a "generator-relation" type description). After a paper chase through near rings and composition rings, we finally came across Tall and Wraith's 1970 paper in PLMS and the Borger-Wieland paper. Neither was quite what we wanted, though, since the algebras from cohomology theories are more complicated than "mere" algebras. The Hunting of the Hopf Ring paper (we were sure that a referee was going to complain about the title, by the way) concentrates on those technicalities rather than trying to explain the simple details of TW V-monoids. We're currently writing another paper which tries to lay out the simpler ideas and which will include a proof of the commutative case (not that this is difficult).

The slogan "TW V-monoids are that-which-acts-on-V-monoids" is intended as an adaptation of the slogan "Rings are that-which-acts-on-abelian-groups". The story is more complicated because it is not automatic that Hom(V,V) is a TW V-monoid. One simple case where it is is when V is a finite ring, but in general one needs a Kunneth-type formula to hold. Nonetheless, it is quite easy to find lots of examples of TW V-monoids and the Tall-Wraith paper, and the Borger-Wieland paper contain plenty of examples (as will ours). The point about TW V-monoids being *the* thing that acts is that they are the *representable* things that act. The general idea is that whenever you have a **set-with-structure** acting on a V-algebra then there is an associated Tall-Wraith V-monoid. Of course, one might wonder why bother with that bloated gadget, but then one might wonder why bother with group rings when you already have the group.

I feel that this isn't really a sufficient answer, though. My problem is that we came at this with examples in hand that we already knew about and wanted some algebraic way of encoding the structure that we already had. As we couldn't find such a description, we invented one (and we hope that Tall and Wraith don't object to the approbation of their names!). So for us, the intuition is all in cohomology operations and not in "bare algebra". If you can expand on what you are looking for a little, I may be able to refine my answer somewhat.

~https://graph.facebook.com/1452583041488356/picture?type=large9918188Ah, so it was quite easy...493036417712Indeed, since "$\alpha$ is memorable" is $\Sigma_2$ expressible in $L_{\omega_1}$, they will all be below the first $\delta$ where $L_\delta\prec_{\Sigma_2} L_{\omega_1}$. Perhaps this will be a characterization?3001434Isoperimetric inequality from Cartan-Hadamard manifold extend to Alexandrov space14872610030908855470793711599301243825349690I'm a C++ developer at Yandex.Maps (Moscow, Russia). I'm mostly interested in algorithms and competitive programming and have spent several years mastering it, having been awarded two ACM ICPC World Finals medals at the top of my competitive career.

2244584@paulgarrett: I agree, but of course the epsilon factors combine nicely!In your example, $u_{xx}=0$ for both solutions. So you might start by asking about the ODE $u''(x) = f(u)$ instead of the PDE above.RThe question can be restated as follows: let $F$ be a field $T$ be an endomorphism of a vector space over $F$. Suppose that it decomposes as a direct sum of finite-dimensional $T$-stable subspaces. Does the same property hold for every $T$-stable subspace?

In turn, it can be formalized as follows, in a pure language of commutative algebra: let $V$ be a $F[t]$-module. Suppose that $V$ is a direct sum of finite length submodules. Does every submodule share the same property?

It's true and follows as a particular case of the following

TheoremLet $R$ be a PID. Let $\mathcal{C}_R$ be the class of $R$-modules that decompose as a (possibly infinite) direct sum of finite length submodules (and hence of finite length cyclic submodules). Then $\mathcal{C}_R$ is stable under taking submodules.

This is a theorem of Kulikov (1941) when $R=\mathbf{Z}$, in which case the proof can be found in L. Fuchs' book "Infinite abelian groups, Vol 1", §17-18.

When $R=k[t]$, polynomial ring in one indeterminate over a field $k$, this exactly yields your claim.

I don't know if anybody wrote the result for arbitrary PID (even for such polynomial rings, for which it should not be easier), but the proof in Fuchs' book extends with essentially no change to arbitrary modules over PIDs.

The main lemma in Fuchs' book (Theorem 17.1) is the following (in the case of $\mathbf{Z}$). For an irreducible element $r$ in $R$, an $R$-module is called a $p$-module $M$ if for every $x\in M$ there exists $n$ such that $p^nx=0$. Also, the (possibly infinite) number $h(x)=\sup\{n:x\in p^nM\}$ is called the height of $x$.

LemmaLet $R$ be a PID and $p$ an irreducible element. A $p$-module $M$ is a direct sum of cyclic modules if and only if one can write $M$ as union of an ascending chain of submodules $M_1\le M_2\le\dots$, $\bigcup_nM_n=M$, such for every $n$, we have $\sup_{x\in M\smallsetminus\{0\}}h(x)<\infty$.

It clearly implies, for a given PID $R$, the above theorem.

So, for completeness, let me write the proof of the lemma; it essentially consists in a copy of the proof in Fuchs' book.

Up to reindexing (adding redundancies if necessary), we can first suppose that $M_n\cap p^n M=\{0\}$ for all $n$. Second, we can suppose that $(M_n)$ is maximal for this property (namely, among sequences $(N_n)_{n\ge 1}$ satisfying $N_n\cap p^n M=\{0\}$ for all $n$ with the ordering $(N_n)\le(M_n)$ if $N_n\le M_n$ for all $n$).

Denote by $M[p]$ the kernel of $m\mapsto pm$. For every $n\ge 1$, we choose a maximal independent subset $J_n$ in the subgroup $M_n[p]\cap p^{n-1}M$, and define $J=\bigsqcup_{n\ge 1} J_n$. For every $c\in J$, choose $a_c\in M$ with $p^{h(c)}a_c=c$.

We claim that $M=\bigoplus_{c\in J}Ra_c$.

The first part is that this is indeed a direct sum. First, since all nonzero elements of the $R$-module $\langle J_n\rangle$ generated by $J_n$ have height $n-1$, we see that the $\langle J_n\rangle$ generate their direct sum, so $J$ is a $R/pR$-free family in $M[p]$.

Next suppose that we have a nontrivial combination, namely $\sum_{i\in I} r_ca_c=0$, with $I$ a nonempty finite subset of $J$ and $r_c\notin p^{h(c)+1}R$ for all $i$. If $r_c\notin p^{h(c)}R$ for some $c\in I$, we multiply everything by $p$. So we can suppose that $r_c=p^{h(c)}R$ for all $i$, namely $r_c=p^{h(c)}s_c$ with $s_c\notin pR$. Hence $\sum_{c\in I}s_cc=0$, contradicting that $J$ is a $R/pR$-free family in $M[p]$.

Now let us show that $M=\sum_{c\in J}Ra_c$. The first step is to show that $\langle J\rangle=M[p]$. Clearly $\langle J_n\rangle=M_n[p]\cap p^{n-1}M$. Suppose by induction that every element in $M_r[p]$ belongs to $\langle J\rangle$ (which is clear for $r=1$), and consider $b\in M_{r+1}[p]\smallsetminus M_r$. The maximality of $(M_n)$ and $b\in M_r$ implies that $(M_r+Rb)\cap p^rM\neq\{0\}$. So we can find $0\neq c=g+kb\in p^rM$, with $g\in M_r$ and $k\in R$. Then $kb\neq 0$; hence $k\notin pR$. Multiplying by $k'$ with $kk'-1 \mod pR$, we can suppose that $k=1$, so we now assume that $c=g+b$. We have $c\in M_{r+1}$ and $h(c)\ge r$, and thus $h(c)=r$ since $M_{r+1}\cap p^{r+1}M=\{0\}$. Also $h(pc)\ge r+1$, and $pc=pg\in M_r$; if nonzero it would be a contradictio to $M_r\cap p^rM=\{0\}$, so $pc=0$. So $pg=0$ as well. By induction, $g\in\langle J\rangle$, and hence $b=c-g$ as well. Thus $\langle J\rangle=M[p]$.

By another induction, we prove that every $m\in M$ with $p^nm=0$ belongs to $M'=\sum_{c\in J}Ra_c$; this holds for $n=1$ and we can now consider the case of $b$ with $p^{n+1}b=0$, with $n\ge 1$; we have to show that $b'\in M'$. By the previous paragraph, we have $p^nb\in\langle J\rangle$. We write $$p^nb=m_1c_1+\dots +m_jc_j+n_1d_1+\dots +n_kd_k,$$ with $c_1,\dots,d_k\in J$, $h(c_i)\ge n$, $h(d_i)\le n-1$. We can write $m_ic_i=p^nm'_ia_{c_i}$, and hence we have $$p^n(b-m'_1a_{c_1}-\dots -m'_ja_{c_j})=n_1d_1+\dots +n_kd_k\in M_{n}.$$ Since $M_{n}\cap p^nM=\{0\}$, we deduce that $p^nb'=0$, with $b'=b-m'_1a_{c_1}-\dots -m'_ja_{c_j}$. By induction, we have $b'\in M'$, and $b\in M'$ follows.

Oh yes. Thanks a lot. So I see. When we have given a finite product of arbitrary representable simplicial sets $Y$ can we also find a functor $H: B \to sSet$ with colimit Y such for every generating cofibration $ i:A→A′$ of $C$ and every map $ i \to Hom(Y,F) $ there is a common lift of all squares $i \to Hom(Y,F) \to Hom(H(b), F) $ for all $b \in B$ that promotes to a map $A' \to \dom( Hom(Y,F))?All the standard examples for model categories are large categories. Is it possible to have a small model category? Are there any interesting examples?

EDIT:

Since a complete small category is a preorder (proposition V.2.3 in MacLane's Categories), I'd be glad to compromise the limit axioms to be as in Quillen's original definition, demanding only finite limits and colimits. In particular, I don't consider a trivial model structure to be interesting.

2183445Yes. This is classical, maybe originally in Milnor and Moore's paper on Hopf algebras. For a recent exposition see for example "More concise algebraic topology" by Kate Ponto and myself. If $X$ is a connected $H$-space (say with finitely generated rational homology groups), then the rationalized Hurewicz homomorphism is a monomorphism with image the primitive elements of $H_*(X;\mathbf Q)$. The essential point is that the rationalization of $X$ is equivalent to a product of Eilenberg-MacLane spaces.

14729131458163FloatingForestIn the first paragraph, does $Diff(N)$ mean diffeomorphisms fixing the boundary pointwise or not?12350731623726lDeveloper, Tinkerer, Father, Human (not sure)

16662751991887205328520607221446985@AlexeyUstinov The epigraph is a quotation from a book by French sociologist Pierre Bourdieu reading "...qui a le loisir de s’arracher aux évidences de l’existence ordinaire pour se poser des questions extra-ordinaires ou pour poser de manière extra-ordinaire des questions ordinaires". As ACL writes above, it might be necessary to read the acknowledgement section and the title or introduction of the thesis to get why I chose it.XMathematicians who were late learners?-list14738011172500Mac Lane and Moerdijk's book "Sheaves in Geometry and Logic" spells out the connections very nicely.Cheever's result need the estimate $d_{C^1}(H, id)< \epsilon$, where $H$ is the extension of $\varphi$, under the assumption $d_{C^1}(\varphi, id)< \epsilon$. I know Hirsch's book, it seems not enough for this case.9629817This is **very** computable, using several methods. I assume you are over $\mathbb{C}$ and
that by cohomology
you mean singular cohomology, but other choices are also just as straight forward.
Let $P=\mathbb{P}^n$.
Then by the Gysin sequence
$$ \ldots H^i(P)\to H^i(X)\to H^{i-1}(Y)\to H^{i+1}(P)\ldots $$
you can basically reduce it to the computation of $H^i(P)$, which is standard, and
$H^i(Y)$.
There are a number of explicit formulas for the latter. See, for example,
http://www.math.purdue.edu/~dvb/preprints/book-chap17.pdf

**Notes**: The connecting map $H^{i-1}(Y)\to H^{i+1}(P)$ is the Gysin map, which is Poincaré dual
to the restriction $H^{2n-i-1}(P)\to H^{2n-i-1}(Y)$. Alternately, choose a tubular neighbourhood
$Y\subset T\subset P$ and
identify
$$H^{i-1}(Y)= H^{i-1}(T) = H^i(T,\partial T)=H^{i}(P,X)$$
by excision and the Thom isomorphism. Then this is the usual connecting map, and the
sequence is long exact sequence for the pair $(P,X)$. There other ways to understand
this as well.

I have the data of frequences of specific representatives in population. I have these data for 3 populations and 6 years. The frequence is usually decreases in every population with the timeflow, ecxept one case. for example:

year population A population B Population C

2001 42 35 36

2002 **54** 33 35

2003 37 30 34

2004 35 29 32

2005 31 29 31

2006 30 26 30

The question is, whether there is any statistical procedure, which can show the probability, that the outlier 54 is really different from the distribution? Thanks in advance.

1068178104035https://www.gravatar.com/avatar/7313bf68274f026c8864fb6b928b66ae?s=128&d=identicon&r=PG&f=1here is a link to notes by Milne, especially if you do not have access to the volume Donu referenced. http://www.jmilne.org/math/CourseNotes/AV.pdfLet $G$ be a group, either a Lie group or a discrete group. Let a principal $G$-bundle $$ G\to E\to B,$$ then $B=E/G$, the orbit space under action of $G$.

Let $BG$ be the classifying space of $G$.

My question:

How to get the fiber sequence

$G\simeq \Omega BG\to E\to B\to BG$?

1046237\https://i.stack.imgur.com/2dDlu.jpg?s=128&g=1Section 6.3 of Henri Cohen's "A Course in Computational Algebraic Number Theory" is about computing Galois groups using resolvents. Subsection 6.3.4 is specifically about the quintic case. Cohen gives an algorithm which addresses the case of $\mathbb{Z}_{5}$ versus $D_{5}$ in step 6 (once a 5-cycle contained in the Galois group is known). This algorithm is implemented in PARI/GP (see this page).

In light of David Speyer's remarks, the difficulty is in proving the Galois group is $\mathbb{Z}_{5}$ when it appears that the only possibilities for the factorization of $f(x)$ modulo $p$ are irreducible, and a product of $5$ linear factors.

230543So far we only discussed the latter. It is very well possible that the result is still true by a different argument.907678@David: You use the property of the hom