Example can be found, for instance, in Boris Weisfeiler's paper "Abstract homomorphisms of big subgroups of algebraic groups", pages 149-150, see

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ndml/1175197662

His example of a discontinuous representation $\rho$ of $SO(n, {\mathbb R})$ to a semidirect product $H$ of $SO(n, {\mathbb R})$ with the abelian group ${\mathbb R}^N$ (the Lie algebra of $SO(n)$), works for $SL(2, {\mathbb R})$ as well. Actually, Weisfeiler's example is even more dramatic: The image of the compact group $SO(n)$ under $\rho$ is *dense* in the noncompact Lie group $H$. Weisfeiler's paper also lists many positive results on rigidity of abstract homomorphisms of Lie groups.

As a general idea, without any other information on a sequence, it seems rather unlikely to find a natural metric in which it converges. Yet a metric in which it has a convergent subsequence is a more reasonable task; then you may prove that your sequence actually converges there, by means of additional informations on the particular sequence.

Here, if these compact sets $C_n$ are included in a closed ball $B$ of $L^2(\Omega)$, you may use the Hausdorff distance induced from the weak topology, which makes $B$ a metric compact space. So $\mathcal{H}(B)$ is a compact metric space, and your sequence has a convergent subsequence there, that converges to a weakly compact set. Otherwise you may just consider the trace on each ball, $C_n\cap \bar B( 0,r)$, and get by a standard diagonal argument a subsequence $C_ {n _ k}$ that converges on the Hausdorff distance of the weak topology on bounded sets.

8440761316568172410I wanted to say that most likely your requirement forces to have all kind of things like $\{a,\{b,\{c,d\}\},\{e,\{f,\{g,\{h,k\}\}\}\},...\}$ so that you will need all iterates of the powerset1658001@Adam: First, I should correct the wording to Strong Weil Curve: a Weil curve is any modular curve, and a strong Weil curve one such that any modular parameterization of an isogneous curve factors through the parameterization of this one. So, since the strong Weil curve in an isogeny class is a canonical choice for an identity, I guessed it would in fact have j-invariant $j(O)$. But I can't seem to find a reference for this. Hopefully someone else will intervene. In any case, the strong Weil curve provides for a canonical choice of identity.3170091554188224812118179901604480107739467852875500665288274881221732539439102873061726542A theory containing the axiom of arithmetics without the multiplication operator is complete. In some sense, a recursive theory containing the axiom of arithmetics (with the multiplication operator) may be considered as the minimum that is required for incompleteness. You ask whether any incomplete theory with one recursive model has to have self-reference property. Well it seems that we can embed this theory into the recursive theory of PA. Recursive theory that contains PA with PM as its inference rule is the minimum required for achieving incompleteness.

57311098006553801\Separating the spectrum of a Hermitian matrixThe adjoint rep is always bounded away from $0$. Let $\mathfrak{g}_0$ be a simple quotient of $\mathfrak{g}$. (I consider the $1$-dimensional Lie algebra to be simple, so there is always a simple quotient.) Let $\mathfrak{h}$ be the kernel of $\mathfrak{g} \to \mathfrak{g}_0$ and let $H = \exp(\mathfrak{h})$.

The adjoint action preserves $\mathfrak{h}$, so the adjoint representation is block upper triangular. The upper left block is the adjoint action of $G/H$ on its Lie algebra, which is $\mathfrak{g}_0$. Since $\mathfrak{g}_0$ is simple, it is unimodular, meaning that the adjoint rep has determinant $1$. This idea is taken from anton's answer.

In summary, the adjoint rep of $G$ can be put in block upper triangular form with the upper left block a matrix of determinant $1$ (and not a $0 \times 0$ matrix).

14050192072533217432877619735962011248Let $T^d$ be the standard simplex, $$ T^d = \left\{(t_1,\cdots,t_d)\in\mathbb{R}^{d}\mid\sum_{i = 1}^{d}{t_i} = 1 \mbox{ and } t_i \ge 0 \mbox{ for all } i\right\} $$

For any partition $\lambda\vdash n$,The Schur function is defined

$$ \displaystyle s_\lambda(x_1, \ldots, x_d) = \frac{\det\Bigl(x_i^{d + \lambda_j -j}\Bigr)_{ij}}{\det\Bigl(x_i^{d-j}\Bigr)_{ij}}. $$

I would like to ask the value of the following integration, and the asymptotic behaviour of the integration,

$$ \int _{T^d} ~~s_\lambda(t) dt $$

It's not clear in the question what kind of "quantum group" is wanted, since they come in different flavors. Apart from that, Harish-Chandra and many others have contributed to the classical theory for noncompact groups. The questions here need a more specific context. viratWilliam TozierLet $R_n$ be checkered rectangle sized $n \times 4, n \ge 1$.

Let $a_n$ be number of different $R_n$ tiling with rectangles sized $1 \times 3$.

$\ \ \ $ $\ \ \ $ $\ \ \ $ $\ \ \ $ $\ \ \ $ $\ \ \ $ $\ \ \ $ $\ \ \ $ Example tiling for $R_9$.

We assume the same tilings $A'$ and $A''$ if $A' = G_1 \circ \ldots \circ G_k \circ A''$, $G_1,\ldots,G_k$ are reflections about both axes of symmetry $R_n$.

I want to find generating function for $a_n$, $a_0 = 1, a_1 = 0$.

Also posted in https://math.stackexchange.com/questions/1533637/generating-function-for-number-of-different-tessellation-checkered-rectangle.

Thank you for any help!

1207701879125FQuite clean and simple, excellent!@AryehKontorovich I believe that I was just reacting to the placement of a mathematical statement and calling it a theorem, without any indication of the source. I see, the person posting it had a likely proof.1941165@FedorPetrov Suppose I wanted to calculatee the number of residues $a$ for which $a$,$a+1$ and $a+2$ are coprime with $n$. Does that mean that you would have $q \prod (1-3/p_i)?$6870261950690277104dSubprojectivity of the spaces $B_{p}(1One knows that graded ideals in polynomial rings over a field are primary iff they are graded-primary. What about the irreducible ideals?

Let $I$ be a graded ideal in a polynomial ring over a field. Is it true that $I$ is irreducible if and only if it can't be written as an intersection of two graded ideals (properly containing $I$)?

If the answer is positive in this case, then what's going on if change the ring by an arbitrary $\Bbb N$ (or $\Bbb Z$)-graded ring?

https://www.gravatar.com/avatar/3520b98c01c24e25389017d9d53419a0?s=128&d=identicon&r=PG&f=1344750Let $M^3$ denote the Poincare homology sphere. I am wondering what the possible orders of (smooth) automorphisms of $M$ are (I'm not sure if allowing arbitrary homeomorphisms changes things?). By presenting the space as $-1$-surgery on the trefoil, there is a automorphism of order 3 given by the 3-fold symmetry of the trefoil. Similarly, the framed link description of $M$ obtained by presenting $M$ as the 5-fold cyclic branched cover of $S^3$ branched over the trefoil has a 5-fold symmetry. Both of these symmetries can also be seen from the dodecahedron/icosahedron definition of $M$.

I believe that the $5k$-fold cyclic branched covers of $S^3$ branched over the trefoil are all $M$ for any $k \geq 1$ and therefore, by drawing the resulting framed link representations, we obtain automorphisms of order $5k$ for any $k$. I am not sure if any of these automorphisms are isotopic to the identity.

What are the possible orders of automorphisms of $M$? Which possible orders have an automorphism of that order that is not isotopic to the identity?

9259881261705@MikeShulman: Suppose I have a sheaf $V$ over a space $X$. Then surely the analogous operation is to take $\Gamma(X, V)=\pi_*V$, where $\pi: X\to *$ is the constant map? This does have the structure of a $\Gamma(X, O_X)$-module, where $O_X$ is the ring of your favorite kind of functions, but one can simply forget this to obtain an object in $Sh(*)$. So surely the correct thing to do here is to take Uri's operation and forget the module structure, no?930794|Isolated hypersurface singularities, Chow groups and D-branes24862512319511866534205752415833692951307020654060871106456121541026997Why did you say that for nonarchimedean valuations over number fields this is trivial?4moduli space of polytopes477159I was more interested in the standard metric, but if there are answers for arbitrary metrics that may be interesting too.74441521441862961276139036How about $X=\Bbb RP^2\times L^2_3$, where $L^2_3$ is the cone of the $3$-fold cover $S^1\to S^1$. Here $H^4(X;\Bbb Z)\simeq\Bbb Z/2\otimes\Bbb Z/3=0$, but $H^4(X;\mathcal L)\simeq\Bbb Z\otimes\Bbb Z/3$, where $\mathcal L$ is the pullback of the orientation sheaf of $\Bbb RP^2$.

As a side remark, "singular cohomological dimension" is not something that people normally do, perhaps because singular cohomology is not Brown representable, or because of the Barratt-Milnor example of a compact subset of $\Bbb R^3$ (in fact it is just the one-point compactification of $\Bbb R^2\times\Bbb Z$) which has infinite "singular cohomological dimension". If you do care about spaces not homotopy equivalent to CW-complexes, you can look up some books on traditional dimension theory, which deal with usual (that is, Cech) cohomological dimension. For spaces homotopy equivalent to CW-complexes, there's no distinction because all ordinary cohomology theories coincide.

146724515405181274394Here is the author:https://webvpn.ucr.edu/+CSCO+0h756767633A2F2F6A6A6A2E6E7A662E626574++/mathscinet/search/publdoc.html?pg1=IID&s1=194474&vfpref=html&r=1&mx-pid=1307945Indeed, I would not tag this as convex-optimization or linear-programming. It is most certainly neither. This is quite a difficult problem and I am unaware of any good heuristics (not that I would know them all!)1094507In this formula that I'm trying to use (actually, I think it's just the definition of integration over manifolds), the determinant of the metric tensor shows up. I thought that this can be called 'Jabcobian' in the special case that we're talking about.77259320675981192860119755961470544AswathDear Philippe, thanks. I could do with some help in understanding that paper in terms of growth diagrams.0Yes, there is a simple reason why this happens, but instead of giving the "magic" bijection, I will describe how to construct it. I will need to express this in the language of partitions, where I'm more familiar, although it shouldn't be hard to switch back to the dominoes interpretation. Below, I assume the weight of a partition of $n$ is $q^n$, and the weight of a pair of partitions is the product of their weights.

**Proposition A**: There is a weight preserving bijection between tilings of an $n\times 1$ rectangle with $k$ dominoes and $a$ black squares and pairs of partitions $(P,Q)$ where $P$ has $k+a$ parts each of size $\le n-2k-a$, and $Q$ has $k+a$ parts $q_1\geq q_2\geq \cdots \geq q_{k+a}=1$ and $q_{i}-q_{i+1}\in \lbrace 1,2\rbrace$, where these differences are $=2$ exactly $k$ or $k-1$ times.

Proof: Let $p_{k+a}$ denote the number of white squares before the first black square or domino, $p_{k+a-1}$ denote the number of white squares before the second black square or domino etc. We will let $P$ be the partition $p_1,p_{2},\dots$. We let $q_{k+a-i}-q_{k+a-i+1}=2$ if the $i$'th black square or domino is a domiono, and $q_{k+a-i}-q_{k+a-i+1}=1$ otherwise. It should be clear that this is a bijection. Proving that it is weight preserving is also easy and is left as an exercise.

**Proposition B**: The partitions $Q$ from the previous theorem are in a weight preserving bijection (up to a power of $q$ which depends only on $k$ and $a$) with partitions which have $k$ parts which are $\le a$.

Proof: Take the partition $Q$ and subtract the partition $(k+a,k+a-1,\dots,1)$, so we obtain $Q'=(q_1-k-a,q_2-k-a+1,\dots,q_{k+a}-1)$. Now let $\hat{Q}$ be the transpose partition of $Q'$. Check that $Q'$ has $k$ distinct parts which are $\in [0,k+a-1]$. Therefore we make a fourth partition $R=\hat{Q}-(k-1,\dots,1)$. Check that $R$ has $k$ parts which are $\le a$.

Now let's denote by $B(x,y)$ the set of partitions contained in an $x\times y$ rectangle (so partitions that have at most $x$ parts and each part is at most $y$). It is well known that the generating function for these is ${x+y\brack x}$.

**Proposition C** There is a weight preserving bijection between pairs of partitions $B(k+a,n-2k-a)\times B(k,a)$ and pairs of partitions $B(n-2k,k)\times B(n-2k-a,a)$.

Proof: Look up your favourite bijective proof of $${n-k \brack k+a}{k+a\brack a}={n-k\brack k}{n-2k\brack a}.$$ (Note that this is the part which messes things up a bit, so that the resulting final bijection won't be "obvious".)

So propositions A,B give a bijection between tilings with $k$ dominos and $a$ black squares and $B(k+a,n-2k-a)\times B(k,a)$. Applying proposition C, we get a bijection with $B(n-2k,k)\times B(n-2k-a,a)$. Finally since $B(n-2k,k)$ is in a weight preserving bijection (up to an inconsequential power of $q$) with domino tilings (no black squares) and $B(n-2k-a,a)$ is up to a power of $q$ the coefficient of $z^a$ in $(1+zq^{k+1})\cdots(1+zq^{n-k})$, we get the desired result. $\square$

5146211096257For a random $d$-regular graph, where $d$ can be fixed or can grow slowly with the size of the graph $n$, what can we say about its spectrum - Do you believe it has simple spectrum?

Thank you,

3046062Indrava Roy21842251460468192759714957141057614168271831955524236@verret Oh, I'm relatively unfamiliar with StackExchange - I'm not even sure whose answer would be most appropriate to formally Accept in this instance, since I think most of the answers (and even the comments here) have been valuable?67777489141@Ludolila: well then $C(X)$ is a vector space and thus contractible.1553359320564091389087551931093I think something you should look at is Johnson graphs. They are extremely easy to optimize in your scheme. The Petersen graph is the complement of a Johnson graph. Not sure if that helps.The critical values of $t$ comprise the support of the relative sheaf of differentials for $X$ over `$\mathbb{P}^1_{\mathbb{C}}$` This is the (slightly more global) function field analogue of the co-different ideal (that is, the inverse of the different). It's just the co-kernel of the map `$t^*:\Omega^1_{\mathbb{P}^1}\to\Omega^1_X$`. If $t$ is defined over $\overline{\mathbb{Q}}$, then so is its relative sheaf of differentials, and so also is its support.I also should say $t \in [0,T]$ so the $\|u\|_{L^2(x,t)}$ is boundedGI'll post an answer to what I think is a reasonable question, hoping that someone more expert will improve on this answer. My apologies if this answer is too chatty.

First, a very simple reason you might hope that there is something like a cotangent complex and that it should be an object in the derived category of quasi-coherent sheaves. Given a morphism $f: X \rightarrow Y$ of schemes (over some base, which I leave implicit), you get a *right exact sequence* of quasi-coherent sheaves on $X$:

$$f^{*}\Omega^{1}_{Y} \rightarrow \Omega^{1}_{X} \rightarrow \Omega^{1}_{X/Y} \rightarrow 0.$$

Experience has taught us that when we have a functorial half-exact sequence, it can often be completed functorially to a long exact sequence involving 'derived functors', and in abelian contexts such a long exact sequence is usually associated to a short exact sequence (or exact triangle) of 'total derived functor' complexes, the complexes being well-defined up to quasi-isomorphism and hence objects in a derived category. This is the semi-modern point of view on derived functors, which one can learn for instance from Gelfand-Manin's Methods of Homological Algebra. Experience has also shown that not only is the cohomology of the total derived functor complex important in computations, but that the complex itself, up to quasi-isomorphism, contains strictly more information and is often easier to work with, until the very last moment when you want to compute some cohomology.

Once you've seen this work a number of times (say for global sections, for $Hom$, and for $\otimes$), one might ask if there is a total derived functor of $\Omega^{1}$, call it $\mathbb{L}$, which among other things produces an exact triangle

$$f^{*}\mathbb{L}_{Y} \rightarrow \mathbb{L}_{X} \rightarrow \mathbb{L}_{X/Y}$$

such that the long exact sequence of cohomology sheaves begins with the original right exact sequence.

If you believe that such a thing should be useful, then you might go about trying to construct it. One way to do this involves interpreting $\Omega^{1}$ as representing derivations which in turn correspond to square-zero extensions, and this is where deformation theory comes in. So one might begin to ask what 'derived square-zero extensions' should be, and you might guess that you should try to extend not just by modules but by bounded above complexes of modules. When you do this, such a square-zero extension becomes not just a commutative algebra but some kind of derived version thereof, such as a simplicial commutative algebra. In these terms, the cotangent complex $\mathbb{L}$ of a commutative algebra turns out to be nothing but K\"ahler differentials of an appropriate resolution of our commutative algebra in the world of simplicial commutative algebras. Following this idea through and figuring out how descent should work leads, after a long song and dance, to the desired theory of the cotangent complex.

Once you set this all up, it becomes clear that there was no reason to restrict oneself to classical commutative algebras in setting up algebraic geometry, but that one could have worked with simplicial commutative algebras to begin with, and this leads to `derived algebraic geometry'. In some sense, this is the natural place in which to understand the cotangent complex, and here the higher cohomology of the cotangent complex has a natural geometric interpretation.

One should also point out that Quillen's point of view on the cotangent complex was as a homology theory for commutative algebras (search for Andre-Quillen homology), which in a precise sense is an analogue of the usual homology of a topological space. This is described in the last chapter of Quillen's Homotopical Algebra and is also discussed in Goerss-Schemmerhorn's Model Categories and Simplicial Methods.

*To summarise.* Deformation theory is about square-zero extensions (as well as about other more general infinitesimal extensions). K\"ahler differentials corepresent derivations which in turn correspond to square-zero extensions. For each morphism of schemes $X \rightarrow Y$, there is a natural right exact sequence involving K\"ahler differentials, which it would be useful to complete to a long exact sequence. Even better,
we'd like this long exact sequence to come from an exact triangle of objects in the derived category of quasi-coherent sheaves. Realising this goal naturally leads to derived or homotopical algebraic geometry.

In this context, a remarkable phenomenon to consider is also lack of *uniqueness*, that may be considered an instance of bifurcation.

The following concept seems to be useful:

Definition.Let $\mathbf{J}$ and $\mathbf{C}$ denote categories, and suppose we're given a functor $F:\mathbf{J} \rightarrow \mathrm{End}(\mathbf{C}).$ Ageneralized coconefrom $F$ is an object $X$ of $\mathbf{C}$ together with, for each object $Y$ of $\mathbf{J}$, a morphism $\varphi_Y :F(Y)(X) \rightarrow X$, such that for each $f : Y \rightarrow Y'$ in $\mathbf{J}$, we have: $$\varphi_Y=\varphi_{Y'} \circ F(f)(X).$$ Given cocones $(X,\varphi)$ and $(X',\varphi')$, amorphism$g : (X,\varphi) \rightarrow (X',\varphi')$ is just a morphism $g : X \rightarrow X'$ such that for all $Y \in \mathbf{J}$, we have $g \circ \varphi_Y = \varphi'_Y\circ F(Y)(g)$.A

generalized colimitis an initial generalized cocone.

The main examples are:

Given a functor $F:\mathbf{J} \rightarrow \mathbf{C}$, we can get a functor $G:\mathbf{J} \rightarrow \mathrm{End}(\mathbf{C})$ by assigning to every object $Y$ of $\mathbf{J}$ the constant functor whose value is $F(Y)$. A cocone from $F$ is the same thing as a generalized cocone from $G$.

Let $\mathbf{J} =\{Y_0,Y_1\}$ denote the discrete category with two objects. Let $F : \mathbf{J} \rightarrow \mathrm{End}(\mathbf{Set})$ denote a functor such that $F(Y_0)$ is a constant endofunctor with value $S \in \mathbf{Set}$. Then a generalized colimit of $F$ is just an $F(Y_1)$ algebra freely generated by $S$. For example, we can define $\mathbb{N}$ this way, by taking $S = \mathbb{1}$ and $F(Y_1) = \mathrm{id}_\mathbf{Set}$.

8183111058603853007Sorry for a vague comment but somehow it reminded me of universal spaces used in definitions of the generalized Lusternik-Schnirelman category and dual cocategory by Clapp&Puppe, Ganea, Hopkins and others. There is a [recent paper](http://arxiv.org/pdf/1209.2384v3.pdf) where these are put in the context of Goodwillie calculus. I wonder if all this might be relevant here...210318313701912918890120380363621@

Question.What are generalized cocones/colimits really called, and where can I learn more about them?

je pense, donc je suis.

160016822665604http://www.perfectlens.ca6852621817004Jerry Shurman1125669183091716650583494391329514Tsuyoshi Ito14427967979253I propose to collect here open problems from the theory of continued fractions. Any types of continued fractions are welcome.

Does Hartogs's Theorem for complex-analytic functions hold for real-analytic functions?129001014337831572708 You might be interested in questions 19074 and 22601 as well as some of the answers given to them even though the sort of minimizing curve sought for in these questions is of the "y=f(x)" type. With regard to your question, it occurs to me that if your curve C was the arc of a spiral enclosing a sufficiently large portion of the Euclidean plane, the maximum absolute value of its curvature at any point might be made arbitrarily small while all your n points could still be points of C. Of course this would violate your condition (c) unless your rectangle R was large enough. jExample is now included. Hopefully without mistakes.561456176434020459471487442301200846568I've been thinking about this for a while, and I'm a little stumped trying to rectify this with Sandor's response. The special fiber should indeed just be the smooth quartic rational curve in $\mathbb{P}^3$, and I don't see where any nilpotents could come from. I think the issue was in fact due to failing to saturate the ideal, so that your description of the ideal didn't actually provide generators for the ideal in the local charts he was working. Playing around in Macaulay, I can't get any of the cohomology groups to jump whatsoever. 602208Igor, x is fairly huge (the determinant of an n by n random 0,1 matrix) and then your upper bound is superexponential in n. It does not look very good.7367461702344Majoring in Mathematics and Computing at IIT Kharagpur, India.

Prefers to write clean, Pythonic code | GSoC '16 with SymPy

I don't follow your comments about the projective plane. Surely the geometric realization of the simplicial complex consisting of $$\{a,b,c\},\{a,c,d\},\{a,d,e\},\{a,e,f\},\{a,f,b\},$$ $$\{b,c,e\},\{c,d,f\},\{d,e,b\},\{e,f,c\},\{f,b,d\}$$ and their subsets is the real projective plane (I got these by identifying antipodal faces of a regular icosahdron).

1286729A better bound is $\dim W\geq 2\dim V-3$. This is obtained as follows. In the projective space $\mathbb{P}(\wedge^2V)$, the set of decomposable bivectors is the Grassmannian $\mathbb{G}(2,V)$, of dimension $2\dim V-4$. The bilinear map $B$ induces a linear map $b:\wedge^2V\rightarrow W$, and the hypothesis is $\mathbb{P}(\mathrm{Ker}\, b)\cap \mathbb{G}(2,V)=\varnothing$. This implies $2\dim V-4<\mathrm{codim}\, \mathrm{Ker}\, b=\dim \mathrm{Im}\, b$, hence the result.

494955PI have often heard the slogan that "a matrix algebra has no deformations," and I am trying to understand precisely what that means. While I would be happy with more general statements about finite-dimensional semisimple algebras over non-necessarily algebraically closed fields, I am mainly interested in the case when the base field is $\mathbb{C}$, so the question is phrased in terms of matrix algebras over $\mathbb{C}$. I will state the question in three different contexts, in increasing order of interest to me, and say what I know about each one.

A formal deformation of an associative $k$-algebra $A$ is a $k[[h]]$-bilinear multiplication on $A[[h]]$ of the form $$ a \cdot b = ab + m_1(a,b)h + m_2(a,b)h^2 + \dots $$ where the first term is given by the original multiplication on $A$. This type of deformation is connected with the Hochschild cohomology $HH^\bullet(A)$. As I understand it, for a semisimple algebra $A$, the Hochschild cohomology vanishes in degree $\ge 1$, and this implies that a matrix algebra has no deformations in the formal sense. So I guess there isn't much of a question so far.

In this context, we fix a natural number $n$ and a vector space $V$ with a fixed basis $a_1, \dots, a_n$, and we consider associative algebra structures on $V$. The *structure constants* of an algebra $A$ with underlying vector space $V$ with respect to this basis are the complex numbers $c_{ij}^k$ such that
$$ a_i a_j = \sum_{k}c_{ij}^k a_k, $$
and these numbers clearly determine the algebra. This gives us a point in $\mathbb{C}^{n^3}$. Associativity gives us some polynomial constraints on the structure constants, so we can think of the collection of $n$-dimensional associative algebras as some kind of algebraic variety in $\mathbb{C}^{n^3}$. (I know very little algebraic geometry, so please excuse/correct me if am using the wrong terminology.) Since one can take many different bases for an algebra, this variety overcounts associative algebras.

I don't know how this is proved, but as I understand things, semisimplicity is a Zariski-open condition in the variety of associative algebra structures, so semisimplicity is preserved under small deformations, and so a small deformation of $M_n(\mathbb{C})$ is still semisimple.

Is the property of being a matrix algebra preserved under deformation of structure constants?

This is the most flexible setting, and the one in which I am most interested. Here we consider algebras with a fixed collection of generators and a fixed number of relations, and we allow the relations to vary smoothly. This allows the dimension of the algebra to vary as well. However, this is also the hardest notion to formalize.

An example will illustrate what I mean. Consider the family of algebras $$ A_t = \mathbb{C}[x]/(tx^2 -x),$$ where $t$ varies in $\mathbb{C}$. Clearly there is a drop in dimension at $t=0$, although generically the dimension is constant. A slightly more complicated example is given by the family $B_t$, where $$ B_t = \mathbb{C} \langle x,y \mid x^2 = y^2 = 0, \quad xy + yx = 2t \rangle.$$ In this example the dimension is constant, but at $t=0$ one has the exterior algebra $\Lambda(\mathbb{C}^2)$, while for $t \neq 0$ this is the Clifford algebra of $\mathbb{C}^2$ with respect to a nondegenerate bilinear form, and hence $B_t \simeq M_2(\mathbb{C})$ for $t \neq 0$.

What is a good way to formalize this notion of deformation?

A first guess would be something like this: fix a space of generators $V$, a number of relations $k$, and a base manifold $M$ (in my above examples $M=\mathbb{C}$). To limit things a bit, I guess I would like to consider only quadratic-linear relations. Allowing these relations to vary means giving a smooth function $$F : M \to \mathrm{Gr}_k(\mathbb{C} \oplus V \oplus V^{\otimes 2})$$ and considering the "bundle" (sheaf?) of algebras $$ A_p = T(V)/ \langle F(p)\rangle, \quad p \in M $$ over $M$.

However, I'm not sure if that completely captures what I want. I might also insist that the projection of each $F(p)$ onto $V^{\otimes 2}$ has no kernel (which is not the case for the first example I described, but is the case for the second).

Given the setup just described (or an appropriately modified version of your choosing), what are some conditions on $M$, $F$ such that the property "$A_p$ is a matrix algebra" is local on $M$?

For this question I am mainly interested in the cases when $M = \mathbb{R}$ or $M = \mathbb{C}$, although more general base spaces are of course interesting as well.

I have heard about something called the *Azumaya locus* which may be related to this, but I don't really know anything beyond the name.

I didn't realize this would be so long! If you're still with me, thanks for reading.

PThis is what I was looking for. Thanks.77750751827353https://www.gravatar.com/avatar/8d582189fbba39915e019d1243338c78?s=128&d=identicon&r=PG&f=12046633Maybe you could get it started with a couple of examples of what you have in mind?1312366@Yes, of course, you are correctRWhat do you mean by the standard $U(1)$?102029518941ncr13818711711235432721920851pOpenness of stability in the set of flat vector bundles1646468650210213242518472192152550460248No: the first uncountable ordinal (non paracompact) is a subspace17826694264791689702261529269033@BenMcKay : you are right. The dimension is 21. Sorry for the mistake.163722215360223881695986143907951801651249691720738I believe the Lemma is sometimes referred to as the Harris Kleitman inequality. Probabilistically, you can view it as saying the events that a random subset belong to F' and G are negatively correlated.@PiotrAchinger Besides, it seems that I confused with the concepts "ramified cover along $D$" and "a morphism is ramified over $D$". In view of your second comment, a ramified cover could be a unramified morphism? (In the curve case, over $char=0$, this is true.)999994196065Ron Linton- Unless I am not thinking straight, it seem that the curvature of the pull back metric from the sphere is constant. - The other metric I could not show has positive curvature. If there were an infinite sequence of metrics produced by repeated choice of this principal curvature metric then would there have to be a limiting geometry? On the surface (so to speak) this seems impossible since it would have to be the constant curvature metric on the sphere. So maybe after a while the metric can not be embedded in 3 space. 684699475382(Edit: added solution that actually works, distinct from Scott's.) ChinaAnhuiHefei212699415010836249651519122bHistorical reference request on Nilpotent groups@john Good point. An accessible endofunctor of Set, however, *can* be described by a "very small" amount of data, since $Set_\lambda$ can be regarded as an *internal* category in Set and a functor $Set_\lambda \to Set$ as an internal diagram thereon. This suggests to me that maybe the answer I'm looking for might be something like "a monad obtained by internal left Kan extension from an internal diagram on an internal category". I wonder whether that can be made to work.210526213750591285645It is extremely well-known that $H^*(BQ_8;\mathbb{Z})=\mathbb{Z}[\alpha,\beta,\gamma]$ with relations $2\alpha=2\beta=8\gamma=\alpha^2=\beta^2=\alpha\beta-4\gamma=0$, $|\alpha|=|\beta|=2$ and $|\gamma|=4$. I was wondering if anyone can given an explicit reference showing that $c_2(\sigma)$ generates $H^4(BQ_8)\cong \mathbb{Z}_8$, where $\sigma: Q_8 \hookrightarrow SU(2)$ is the standard representation (i.e., $c_2(\sigma)=c_2(E_\sigma)$, $E_\sigma= S^\infty\times_{Q_8}\mathbb{C}^2)$.

Any help would be greatly appreciated, as I would like to avoid the following reasoning if possible:

$E_\sigma$ is the pullback of the canonical quaternionic line bundle $E=S^\infty\times_{SU(2)}\mathbb{C}^2\rightarrow \mathbb{H}P^\infty$ by the projection $S^3/Q_8\hookrightarrow BQ_8 \stackrel{\pi}{\rightarrow}\mathbb{H}P^\infty$, and Leray-Hirsch gives $H^4(\mathbb{H}P^\infty;\mathbb{Z}_2)\cong H^4(BQ_8;\mathbb{Z}_2)\cong\mathbb{Z}_2$. Looking at $H^4(\mathbb{H}P^\infty;\mathbb{Z})\cong \mathbb{Z}\rightarrow H^4(BQ_8;\mathbb{Z})\cong \mathbb{Z}_8$ then shows that $c_2(\rho)=\pi^*(c_2(E))$ must therefore represent an odd number in $\mathbb{Z}_8$.

15954830Let $W$ be a compact manifold with boundary and $D^W$ a graded Dirac type operator on $W$, of product type near the boundary acting on a vector bundle $E\to W$. One obtains a graded Fredholm operator $D_{\mathrm{APS}}^W$ once equipping $D^W$ with the APS-boundary condition. The index formula from Atiyah-Patodi-Singer's paper "Spectral asymmetry and Riemannian geometry. I." states that $$\mathrm{ind}(D_{\mathrm{APS}}^W)=\int_W \alpha_D-\overline{\eta}(D^{\partial W}),$$ where $\alpha_D$ is a local term determined by the principal symbol of $D^W$, $D^{\partial W}$ is the boundary operator and $\overline{\eta}(D^{\partial W})$ its reduced eta invariant. See the paper of Atiyah-Patodi-Singer for details. In analogy with the Atiyah-Singer index theorem on closed manifolds, this index theorem can be viewed as an “even-dimensional” index theorem. The “odd-dimensional” version was considered by Dai-Zhang in “An index theorem for Toeplitz operators on odd-dimensional manifolds with boundary“ (see http://arxiv.org/abs/math/0103230).

The paper by Dai-Zhang considers $W$ and $D^W$ as above, but $D^W$ is not necessarily graded, and a smooth function $g:W\to U(N)$. The product structure near the boundary guarantees that $D^{\partial W}$ is graded and after a choice of Lagrangian subspace $L\subseteq \ker(D^{\partial W})$, one can define an APS-type projection on the boundary $P(L)$. The projection $P(L)$ gives rise to self-adjoint boundary conditions on $D^W$. Denote the associated self-adjoint Fredholm operator $D^W_{P(L)}$. The details of this construction can be found in the paper of Dai-Zhang. It is shown by Dai-Zhang that for $P_{P(L)}:=\chi_{[0,\infty)}(D^W_{P(L)})$ and $P_{gP(L)g^{-1}}:=\chi_{[0,\infty)}(D^W_{gP(L)g^{-1}})$, the operator $$T_g(L):= P_{gP(L)g^{-1}}g P_{P(L)}: P_{P(L)}L^2(W,E) \to P_{gP(L)g^{-1}}L^2(W,E)$$ is Fredholm. They compute the index by an APS-type formula: $$\mathrm{ind}(T_g(L))=-\int_W \alpha_D\wedge \mathrm{ch}(g)-\overline{\eta}(D^{\partial W},g)+\tau_\mu(gP(L)g^{-1}, P(L),P_W).$$ Here $\overline{\eta}(D^{\partial W},g)$ is an $\eta$-type invariant defined from $D^{\partial W}$ and $g$. The operator $P_W$ is the Calderon projection, a pseudo-differential operator of order $0$ on $\partial W$ differing from $gP(L)g^{-1}$ and $P(L)$ by a lower order pseudo-differential operator. The term $\tau_\mu(gP(L)g^{-1}, P(L),P_W)$ is the Maslov triple index, considered in greater detail in the paper “The $\eta$-invariant, Maslov index, and spectral flow for Dirac–type operators on manifolds with boundary” by Kirk-Lesch (https://arxiv.org/abs/math/0012123).

Structurally, the difference between the two index theorems is the appearance of the Maslov triple index in the Toeplitz case.

After this lengthy pre-amble, we arrive at a question. Is there an explicit example where the Maslov triple index $\tau_\mu(gP(L)g^{-1}, P(L),P_W)$ has been computed to be non-zero?

2090469You might look at the papers of Rade T. Zivaljevic, particularly

Živaljević, Rade T. Combinatorial groupoids, cubical complexes, and the Lovász conjecture. Discrete Comput. Geom. 41 (2009), no. 1, 135–161.

and the references there, including a number of homotopical ones. There is an arXiv version of this paper.

850261580436@Marty: Right you are. It's considerations of exactly that sort that made me delay my comment. But the exponential has, in some sense, no properties, at least in comparison to the log, which has loads of structure (I'm thinking of how the roots are placed.) Maybe, like Abel, you'd get a handle on things by looking at the inverse of any function you got as a solution of a DE?!Surprisingly, this was the only instance I heard the voice of Grisha Perelman. I thought his lectures in Stony Brook in April 2003 were videotaped and asked Dr.Christina Sormani whether that was so---she attended the lectures and had also posted her transparencies here. She told me that she believed he refused his lectures being videotaped. I still do not know what happened to his lectures in MIT…

**Have you come across any other?**

So, coming to the transcription, here is the translation of Perelman’s replies copied verbatim from Science 2.0 There you can find the whole story.

Grisha has picked up the phone. Here is approximately the dialog which followed:

Teacher: "Hello, can you hear me, do you recognize me ?"

Grisha: "Sure, of course, nice to hear from you."

Teacher: "Grisha, I am sorry but I am really overwhelmed by all these journalists who would like to know why you have refused to accept your prizes. May be you can tell me ?"

Grisha: "But I have already told all of them that I would not like to give any kind of interview."

Teacher: "But you know, I am not journalist !"

Grisha: "I will not tell you as well."

Teacher: "It is OK, sure I will conform to the common rule. But, please, you'd definitely benefit a lot, if you'd try not to be so adversarial to the people surrounding you ... Well, OK, tell me please, what are you doing right now ?"

Grisha: "Couldn't you please call me any other time ? I have a feeling that our present conversation is being recorded somehow ...".

Teacher: "No problem, I will definitely call you again soon and possibly even come to see you. Is it OK ?"

Grisha: "That's fine, thank you. Bye."

Teacher: "Yes, never mind, Grisha ! Bye !"

I admit that the above answer remains the best.

1161286340865I have spotted the mistake in my proof. So here is the "wrong" proof:

Let $v_{1},\ldots,v_{m}$ be linearly independent elements in $A^{n}$, where $m\lt n$. Write them as $n$-tuples of elements in $A$, thereby forming an $n$-by-$m$ matrix. Linear independence of $v_{i}\ $s means that the rank of this matrix is $m$. So there is an $m$-by-$m$ minor with non-zero determinant. By exchanging rows if necessary, bring these $m$ rows to the top part of the matrix. Now add a colomn to the right side of the matrix whose entries are $0$ except at the $n+1$ th position, where the entry is $1$. Then the new $n$-by-$(m+1)$ matrix has rank $(m+1)$ and hence the $(m+1)$ columns, the first $m$ of which are the $v_{i}\ $s, are linearly independent.

The mistake was the notion of rank of a matrix. When the entries are not from integral domain, the proper definition should be the largest integer $m$ such that there is no nonzero element in the ring annihilating the determinant of every $m$-by-$m$ minor. In the above example, I can't conclude that the rank of the $n$-by-$(m+1)$ matrix is $(m+1)$.

With this, I can now exhale a sigh of relief and continue believing that this is not true in general. (By the way I also know that it is true for free modules of infinite rank)

I suppose I should have asked two questions 1)Are all planar Hamiltonian graphs 4 colorable ? 2)If so, then are does this imply all planar graphs are 4 colorable? Then the first paragraph is about reducing Hamiltonian planar graphs to compositions of trees which reduces the first question to the mutual edge colorability of any two trees. (This presupposes you know Taits theorem that 4 coloring faces is equivalent to 3 coloring the edges.) The second question I have not made any real headway on, so I have no discussion , only the question.1526869@Conifold You're quite right: the Gödelian formulation of our result requires that the "consistent, recursive axiomatisation of mathematics" include a sufficient portion of arithmetic, as usual. The statement of this theorem in the long version of the paper was rather sloppy on our part - I'll make sure to clarify this when we revise the paper. Thanks!1961520531510124936248552522585414954646925792168688291640The "experts" also more or less assumed that mortgage failures were events independent from each other...101681890202"If we suppose that $F$ is stable then $E$ is stable for all non trivial extension ..." Unfortunately this is not correct, for instance, on $\mathbb{P}^1$. The semistable bundle $\mathcal{O}^{\oplus 2}$ is a nontrivial extension of $\mathcal{O}(+1)$ by $\mathcal{O}(-1)$.If you go that high up with $n$, you can trivially have $P\le 2e^{-\frac 12\varepsilon^2d^{-2}n}$. On the other hand, I'm not sure myself whether $n\ll d^2/\varepsilon^2$ can give you anything interesting here.Relevant: https://mathoverflow.net/questions/54621/what-is-the-best-lower-bound-for-the-domination-number-in-regular-graphs-of-girt7334091945677This is in answer to your second question. There is a note by Gustafson:

- MR0327901 (48 #6243) Gustafson, W. H. What is the probability that two group elements commute? Amer. Math. Monthly 80 (1973), 1031–1034.

where he proves the result Ben mentions, viz. if $G$ is a finite nonabelian group, then $d(G) \leq 5/8$. He goes on to prove that the same result for the case where $G$ is a compact, Hausdorff topological group (endowed with the Haar measure).

While $d(G)$ has received some attention over the years (I think it was first mentioned in a paper of Erdos in the late 60s and there have been sporadic papers since then) very little seems to have been said about $d(G)$ where $G$ is an infinite group until recently. The basic results (most of which are analogous to the finite case) are proved in

- MR2558527 (2010m:22003) Rezaei, Rashid; Erfanian, Ahmad(IR-MASHM) On the commutativity degree of compact groups. (English summary) Arch. Math. (Basel) 93 (2009), no. 4, 345–356.

Ben has already mentioned the nice paper of Levai and Pyber where it is proved that if $G$ is a profinite group and $d(G) > 0$, then $G$ is abelian-by-finite. This result is extended to all compact groups in a recent preprint by Hofmann and Russo. There is much more besides in this preprint, I'm still digesting it myself!

I want a good introduction to localization in equivariant $K$-theory. This introduction can be simple in several ways:

- I only care about torus actions.
- I only care about $K^0$.
- I only care about very nice spaces. I would be fine if the only spaces considered were $G/P$'s and smooth projective toric varieties.

However, I want this exposition to include the following:

- How to compute a $K$-class from a Hilbert series.
- Given $X \to Y$ nice spaces, how describe push back and pull forward in terms of the map $X^T \to Y^T$.
- Ideally, this reference would also give the generators and relations presentation of $K$-theory for the sort of examples mentioned above. I mean formulas like $$K^0(\mathbb{P}^{n-1}) = K^0(\mathrm{pt})[t, t^{-1}]/(t-\chi\_1)(t-\chi\_2) ... (t - \chi_n)$$ where $\chi_i$ are the characters of the torus action. But I can find other references for this sort of thing.

The best reference I currently know is the appendix to Knutson-Rosu.

137992918950678613166274561493748Yes, that's what she means.833891@A question on (1,1) bridge KnottNon alternative $k$-linear maps vanishing on $\sum x_i=0$313806Dyes m* is Lebesgue outer measure.I was to start with, but now I'm not sure. I'd like any references between 1975/83 (Grothendieck) and 2000 (Stevenson), apart from Baez-Dolan. I was wondering how much it was 'in the air', given the progress on TQFTs and the like in the 1980s-90s.If you are willing to accept an approximation, there is quite a bit of work on finding shortest paths on convex polyhedra, much of it implemented. For example, here are two images of shortest paths from one point to all vertices of a polyhedron inscribed in an ellipsoid:

There is a nice Stanford presentation on the topic:

"Approximating Shortest Paths on a Convex Polytope in Three Dimensions." (PDF download.)

Exact shortest paths can even be computed in optimal $O(n \log n)$ time now, but the algorithm is quite complicated:

43765969829vonPetrushev116690But how does this unify the approach for archimedean and non archimedean fields?Your "lattice model" is a description of the torus as a quotient of a simpler manifold, the plane, by a discrete group action without stabilizers, translation by a lattice. The E8 manifold has no such description, because it it simply connected. One can describe it as a 4-cube with the boundary glued together a certain way, but describing that way is difficult and probably can't be done combinatorially.8862386277941837999768646PYevgeny Schreiber, Micha Sharir. "An Optimal-Time Algorithm for Shortest Paths on a Convex Polytope in Three Dimensions."

Discrete & Computational Geometry, March 2008, Volume 39, Issue 1–3, pp 500–579. (Springer link.)

*Edit: I've updated this answer to reflect the helpful comments made by Andres Koropecki and Ian Morris.*

As the other answers mentioned, the first crucial distinction you must make is that some properties refer to a **topological** dynamical system $(X,T)$, while others refer to a **measure-preserving** dynamical system $(X,T,\mu)$. Thus there are two different sets of definitions. Let me attempt a sketch at some of the relationships within each set.

First suppose you have a topological dynamical system $(X,T)$. Then four of the key properties are topological transitivity, topological mixing, minimality, and unique ergodicity. The first three are related by

- topologically mixing $\Rightarrow$ topologically transitive;
- minimal $\Rightarrow$ topologically transitive.

Unique ergodicity is independent of those three properties. The picture is the following.

Counterexamples **1-9** illustrating the strict containments are as follows. *(These may not be the simplest or the earliest counterexamples in each case, and I welcome corrections or improvements. This is based on some quick googling for things not already in my memory, plus the helpful additions offered by commenters.)*

**1.** $X = \Sigma_2 \times \{a,b\}$, the direct product of a full two-shift with a period-two orbit, where the dynamics is $\sigma\times S$, with $\sigma$ the shift map and $S$ the map interchanging $a$ and $b$.

**2.** $X=\Sigma_2$.

**3.** Constructed by Bassam Fayad, Topologically mixing and minimal but not ergodic, analytic transformation on $\mathbb{T}^5$, 2000.

**4.** Constructed by Furstenberg, Strict ergodicity and transformation of the torus, 1961.

**5.** An irrational flow on the torus, slowed down near a single point: see the comment below by Andres Koropecki.

**6.** As Ian Morris points out in the comments, the identity map on a singleton set works here. A less trivial example was given by Karl Petersen, A topologically strongly mixing symbolic minimal set, 1970.

**7.** Rotation of the circle by an irrational angle.

**8.** Direct product of the example from **5** with a periodic orbit. (Again as suggested by Andres in the comments.)

**9.** North-south map: a map $T\colon [0,1]\to [0,1]$ with fixed points at $0,1$ and such that $T(x) < x$ for all $x\in (0,1)$. Identify the endpoints $0$ and $1$ so that this is a uniquely ergodic circle map.

A couple things are probably worth pointing out.

- Terminology is not always uniform. For example one of the papers I referenced (I think Petersen's) uses "ergodic" in place of "topologically transitive", to highlight the analogy with the measure-preserving case. So people may sometimes use different words for the same thing.
- Conversely the same word may mean different things. There are two definitions of topological transitivity, one involving open sets ($f^n(U) \cap V \neq \emptyset$ for some large $n$) and the other involving existence of a dense orbit. The definition involving open sets more closely mirrors the definition of topological mixing (non-empty intersection for
**every**large $n$), while the definition with a dense orbit more closely mirrors minimality (denseness of**every**orbit). The definitions are equivalent if $X$ is separable, second category, and has no isolated points.

All of the above is for topological dynamical systems, where no invariant measure is specified. Then there are the ergodic properties: those that depend on a system preserving an measure $\mu$. For these one has the ergodic hierarchy.

It is very often the case that one wishes to study a topological dynamical system as a measure-preserving system by equipping it with an invariant measure, and in this case it is quite reasonable to ask about the relationships between the two different classes of properties. But this depends on which invariant measure you choose, because in general there may be very many of them. One may ask what properties of $(X,T)$ let you pick invariant measures $\mu$ with certain nice properties, and this is a whole different story which would expand this answer far beyond the bounds of propriety.

The base is $k=n+1$, as he states. By property 1.2 there are no regular points, and therefore every statement about regular points is true. Some meaningless things are better than others :)@John Thanks. I agree about the parabola. The wedge is the limit of one branch of a hyperbola, but with both foci limiting to the corner, not to infinity. So I can't see how it follows as a limit of a conic with Newtonian gravity.792804329831Let me remind some of notations for readers: $$\mathcal{L}=\log x,$$ $$k_0=3.5\times 10^6, \varpi=\frac{1}{1168},$$ $$D_0=\exp(\mathcal{L}^{\frac{1}{k_0}}), P_0=\prod_{p < D_0}p,$$ $$D_1=x^{\varpi},P=\prod_{p < D_1}p,$$

$$D=x^{\frac{1}{4}+\varpi}, D_2 = x^{\frac{1}{2}-\epsilon}$$

Here, the effect of replacing $\theta$ by $\Lambda$ produces an error of $$O(x^{\frac{1}{2}}\mathcal{L}^B),$$ for some positive $B$. So, this change is negligible compared to $O(x\mathcal{L}^{-A})$.

For imposing $(d,P_0)< D_1$, we claim that the following sum is also negligible compared to $O(x\mathcal{L}^{-A})$: $$ \sum_{\substack{{D_2 < d {<} D^2} \\\ {d|P} \\\ {(d,P_0)\geq D_1}}} \sum_{c\in C_i(d)} |\Delta(\Lambda,d,c)| $$ Note that the trivial bound for $|\Delta(\Lambda,d,c)|$ is: $$ |\Delta(\Lambda,d,c)|=O(\frac{x\mathcal{L}}{d})$$

Since we have $(d,P_0)\geq D_1$, the number $w(d)$ of distinct prime divisors of $d$, must satisfy $$ w(d )\geq \mathcal{L}^{\varpi(1-\frac{1}{k_0})}.$$

The sum is bounded by: $$ x\mathcal{L} \sum_{\substack{{ d {<} D^2} \\\ {w(d )\geq \mathcal{L}^{\varpi(1-\frac{1}{k_0})} } }} \frac{\tau_{k_0}(d)}{d }$$

Standard argument now applies, and the sum above is bounded by: $$ \frac{1}{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}}}\sum_{d < D^2} \frac{2^{w(d)}\tau_{k_0}(d)}{d}=O(\frac{\mathcal{L}^B }{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}}}),$$ for some positive constant $B$.

Combining all together, we obtain an upper bound: $$ \sum_{\substack{{D_2 < d {<} D^2} \\\ {d|P} \\\ {(d,P_0)\geq D_1}}} \sum_{c\in C_i(d)} |\Delta(\Lambda,d,c)| =O(\frac{x\mathcal{L}^{B+1}}{2^{\mathcal{L}^{\varpi(1-\frac{1}{k_0})}} })$$

Hence, our claim follows.

If the uncountable number of sets that you're intersecting is small enough, you *might* be able to guarantee that the intersection is measurable (and in fact of measure 1) --- it depends on information about the set-theoretic universe that the usual axioms (ZFC) don't decide. Specifically, the *additivity of measure* is defined to be the smallest cardinal $\kappa$ such that some $\kappa$ sets of Lebesgue measure 0 have a union that is not of Lebesgue measure 0. Obviously, this cardinal is at least $\aleph_1$ and at most the cardinality $\mathfrak c$ of the continuum. So if the continuum hypothesis holds, there's nothing more to be said about it. But in the absence of the continuum hypothesis, the additivity of measure is as small as $\aleph_1$ in some models of set theory, as large as $\mathfrak c$ in other models, and in between in still other models. If you intersect strictly fewer than this many sets of measure 1, the intersection will have measure 1. Whether that fact covers any uncountable famlilies of sets is, as indicated above, not decided by ZFC.

I'm reading up on maximal sets and the word "coinfinite" pops up in the first sentence. I tried searching on Wolfram Mathworld as well as Google, but nothing concrete has come up. What does it mean and in what context can it be used?

251676Thanks Joel for this beautiful example and for suggesting having a look at a stronger notion of homgeneity!334578818652135445910489091648231 moep1772838834211447044858613@Qiaochu: The uniqueness is a direct consequence of existence. In any representation, the last digit should be the same as in the canonical one, thus the step of induction.I am pretty sure Shannon assumes the $t_i$ to be commensurable (not "commensurate"), and actually positive integers. He wants $N$ to be a sequence, not a function.

If not, then we've got two possible interpretations of the situation:

1st interpretation: $N(t)$ denotes the number of all transmissions containing only the symbols, with NO pauses inbetween. Then $N(t)$ is a very discontinuous function, being zero at all points which cannot be written as sums of some $t_i$'s, and the asymptotics is toast.

2nd interpretation: $N(t)$ denotes the number of all transmissions containing symbols and pauses, where pauses are ignored on decryption. This makes $N(t)$ not a continuous, but at least a monotonically increasing function. However, in this case the formula $N(t)=N(t-t_1)+N(t-t_2)+...+N(t-t_n)$ should be replaced by something more complicated, and the asymptotic is wrong as well (check $n=1,\ t_1=1$, in which case $N(t)$ should be the floor function of $t$, which is hardly asymptotic to $1^t$).

7124191534663149574130242417629621953502204884711362338577723711061813011434075jBest examples of physics providing insight into mathLet's also assume that $l$ is too large for us to want to try $l,l-1,l-2,\dots$ until we find a divisor.

Let's also assume that we know the prime factorization of $n$, $n=\prod_1^mp_i^{r_i}$.

Then we want to maximize $\sum a_i\log p_i$, subject to $0\le a_i\le r_i$ and $\sum a_i\log p_i\le\log d$. Looks like a problem in integer programming, a topic on which there is a considerable literature available.

839325156596&**Edit.** I had posted this answer to complement Eric's original answer, which showed that the number of classes was at least ${\frak c}^+$, since at that time we didn't quite yet know whether there were $2^{\frak c}$ classes. Afterwards, however, Eric improved his answer to get $2^{\frak c}$ directly. Following the comments, though, I have left this answer up.

Let me complement Eric's answer by showing that it is relatively consistent to have strictly more than ${\frak c}^+$ many equivalence classes. Indeed, it is relatively consistent with ZFC to have $2^{\frak c}$ many equivalence class, in a case where this is larger than ${\frak c}^+$.

Specifically, I claim that if the continuum hypothesis holds and there is a thick Kurepa tree (an $\omega_1$ tree with $2^{\omega_1}$ many branches), then there are $2^{\omega_1}=2^{\frak c}$ many equivalence classes. Indeed, I shall construct an almost-disjoint family of $2^{\omega_1}$ many Vitali sets.

To see this, let $T$ be a thick Kurepa tree, and let $\langle A_\alpha\mid\alpha<\omega_1\rangle$ enumerate the equivalence classes of reals under translation-by-a-rational. Label the $\alpha^{th}$ level of $T$ with the countably many elements of $A_\alpha$. For any path $s$ through $T$, the set $A_s$ of labels appearing on the nodes of $s$ will be a Vitali set, and therefore non-measurable. Further, any two distinct paths $s\neq t$ will have $A_s\cap A_t$ being countable, and so $A_s\not\sim A_t$. Since $T$ is a thick Kurepa tree, we therefore have $2^{\omega_1}$ many branches and thus this many equivalence classes modulo your relation. The collection $\{\ A_s\mid s\in[T]\ \}$ is an almost-disjoint family of $2^{\omega_1}$ many Vitali sets.

Finally, let me explain that it is relatively consistent from an inaccessible cardinal that there is a thick Kurepa tree, yet CH holds and $2^{\omega_1}$ is very large. One way to do this is as follows. Start with $\kappa$ inaccessible in $V$ and $2^\kappa$ very large (by forcing if necessary). Let $V[G]$ be the forcing extension by the Levy collapse, so that $\kappa=\omega_1^{V[G]}$. Consider the tree $T=(2^{<\kappa})^V$ in the model $V[G]$. Since every ordinal less than $\kappa$ was made countable, this has become an $\omega_1$-tree. Yet, since $2^\kappa$ was very large and cardinals $\kappa$ and above were preserved, we have $(2^\kappa)^V$ many branches through this tree. So it is thick.

5762161275086207345072906the method for trees should generalize to graphs of bounded treewidth, where the formula might get exponentially long in the treewidth itself. Also, do you care how (algorithmically) complicated the formula is, because it's probably possible to write the general formula as some kind of sum over spanning trees of the graph.

176552015820171140014This will happen iff $\pi_1(S)$, seen as a subgroup of the isometry group of $\tilde{S}\simeq \mathbb{H}^2$ (via the deck transformation action), has a discrete cocompact torsion free group which normalizes it. In this case, take $G$ to be the quotient group. Said differently, this is the case iff $\pi_1(S)$ is a normal subgroup in a surface subgroup of $\text{PSL}_2(\mathbb{R})$.10258092058341200734142707851437076Since edits were made to the English in the original post I thought I might as well join in. However, like @GerryMyerson I am confused as to what your first question is asking for35521717637157601611382260112493811085162Of course a special case of a Fourier series in complex notation, the one for the theta function, the fundamental solution of the heat equation, was given by Jacobi in 1829 and generalized by Riemann in the 1850's.A deterministic and explicitly described walk which is like random onestokenarky140730419360781427161Bib21691842281198(To clarify, my last comment refers to the current version of the question, which is fine. The original question was incomprehensible.)Could you explain it more slowly? I don't really understand that.141982567148@Olaf: I meant the particular P in my answer, I thought that was obvious from contextp@Gerald: Yes, I should ask about *nonempty* sets X, Y. 8http://dineshsonachalam.me/745967As far as you question 1 goes, the better way to think about why the connection between K-groups and L-functions exists is by viewing the K-groups as "almost" being etale cohomology groups (i.e., identical up to the 2-torsion part).275310190902946569215948923513231000967781494https://www.gravatar.com/avatar/85d7f4844a729bf5a19dbc550be8b01f?s=128&d=identicon&r=PG&f=11733542I posted the research level original question of the question in the followin address. http://mathoverflow.net/questions/155542/energy-of-repeated-filter245653I am reading part of Dipendra Prasad's paper found here: http://arxiv.org/pdf/1306.2729v1.pdf.

In it (in the middle of page 8) he writes that compactly induced representations are projective. Why is this true? In general, what objects are known to be projective in this category?

1530025What prevents you from projecting everything to the (affine) plane generated by your $N$ points and working in $\mathbb R^N$?Interior of fundamental domains of lattices in locally compact groupsB$\mathbb M_m=\mathbb N_m$, sorry1533803723778998064Too lazy to check up details to write an answer, but wasn't some of the early Coates-Wiles work published in Journal of Aus Math Soc? (Coates being Australian) and weren't some of Hodge's breakthrough papers (and Atiyah's) published in the LMS journals (which are Ok but not super-super-top; Hodge and Atiyah both British though).30034910045392957440100466 @Kevin: My comment wasn't a criticism -- just a refinement. As for converse theorems and L-functions and epsilon-factors, I was referring to "Local Converse Theorems". By results of Henniart, and later results of J. Chen, the epsilon factors of twists are enough to characterize irreps of GL_n over a p-adic field. So knowledge of L- and epsilon- factors $L(\pi, r,s)$ would suffice to "nail down" functoriality purely locally. Here I require "lots" of L- and epsilon- factors, corresponding to lots of "twists" of $\pi$ by irreps of $GL_m$ (with $m \leq n-1$ or $n-2$ maybe). 117094304293The Grothendieck ring of $M$ has two elements iff on the one hand, for every definable subset $A\subseteq M^n$, there is a definable bijection from $A$ onto some $A'\subseteq M^m$, a definable set $B\subseteq M^m$ such that $|A'\cap B|\le1$, and a definable bijection from $B$ onto $B\cup A'$, and on the other hand, there is no definable bijection from some $A\subseteq M^n$ onto $A-\{a\}$, where $a\in A$. I don’t see how this implies conditions 1 and 2, or vice versa.17881111894512I also stumbled upon http://www.jstor.org/stable/119840?seq=1 where they also state that nonamenability is "sort of" necessary in the analogous problem of existence of non-constant harmonic functions.@Koose: I am a bit confused about your second paragraph. If you already know that $I$ is a intersection of primes, then it is radical, right? 16706651877572Is this some kind of extension of the kuramoto model ? Center manifold theory was used in the analyzing the original kuromoto model by many folks.,..you can start there 97051242696156366114809252003920132735616930781563332952605I do not understand the bit about a client "choosing one item among $n$ items". Are you saying that for every $(x_1,\ldots,x_n)$ there exists an $i$ such that $f(x_1,\ldots,x_n)=x_i$ ?What I meant is that it sounded too much like "The only correct method is". As if every other way (like writing to the author of the review first) would be considered an ignorance of proper channels. But then again I am hypersensitive to bureaucracy, so probably the meaning I read into your answer was never intended.nonlinearality18354353548158XRational singularities under flat morphisms1104390zWhat is the history of the name "Chinese remainder theorem"?Take $V=W=A$, so you get the map $A \to A\otimes A$, $a \mapsto a(1\otimes 1)$. I'm guessing the symmetric monoidal axioms are going to tell you that's a ring homomorphism (but haven't checked at all).6This answer is a follow-up to the other answers (particularly to Carlo Beenakker's answer from September 4).

First off, Carlo's conjecture is indeed true. That is:

Theorem.If $n \geq k$ then $$a_{2k}=\pi^{-1/2}2^{k}\Gamma(k+1/2).\quad [*]$$

*Idea of proof:* Recall that the Brauer algebra describes the set of invariants of $2k$-fold tensor copies of orthogonal matrices, which the following operator projects onto:
$$
\int_{O_n}X^{\otimes 2k} dX. \quad [\dagger]
$$
If we let $\{e_i\}$ be the standard basis of $\mathbb{R}^n$, then after you unwrap everything, you find that $[\dagger]$ is the orthogonal projection onto the span of the $\pi^{-1/2}2^{k}\Gamma(k+1/2)$ different vectors of the form
$$
W_\pi\sum_{i_1,i_2,\ldots,i_k=1}^n e_{i_1} \otimes e_{i_1} \otimes e_{i_2} \otimes e_{i_2} \otimes \cdots \otimes e_{i_k} \otimes e_{i_k}, \quad [**]
$$
where $W_\pi$ is a unitary operator that permutes the $2k$ different tensor factors according to an arbitrary permutation $\pi$ (this is essentially Theorem 3.1 in [1]). The reason that there are exactly $\pi^{-1/2}2^{k}\Gamma(k+1/2)$ vectors of the form $[**]$ is that this is the number of perfect matchings of $2k$ objects (which in turn is equal to the dimension of the Brauer algebra, as expected).

So far, this shows that $a_{2k} \leq \pi^{-1/2}2^{k}\Gamma(k+1/2)$ always (regardless of $n$ and $k$). To show that the other inequality holds when $n \geq k$, it suffices to show that the vectors $[**]$ are linearly independent. I will just refer to Theorem 3.4 of [1] for this claim, but it's not difficult to prove directly using standard linear algebra tools.

This still leaves the question of what $a_{2k}$ equals when $n < k$. Here are some random observations and conjectures based on the numerical evidence found by Carlo and that I've dug up elsewhere:

Conjecture.If $n = k-1$ then $a_{2k} = \pi^{-1/2}2^{k}\Gamma(k+1/2) - \frac{(2k)!}{k!(k+1)!}$. This is the number of perfect matchings of $2k$ objects that have at least one "crossing" (in the same sense that the Catalan numbers give the number of perfect matchings of $2k$ objects with no "crossings").

Conjecture.If $n = k-2$ then $a_{2k}$ seems to be the number of perfect matchings of $2k$ objects with at least three crossings (there is a known formula for this quantity, but it's messy so I won't write it here).

The other answers gave formulas for when $n = 2$ or $n = 3$. It turns out that there is also already a known formula for the $n = 5$ case (see OEIS A095922: the even-indexed terms are the values of $a_{2k}$ that we want). That OEIS entry cites the book [2], but I haven't been able to track down a copy of it. Numerical evidence also suggests the following:

Conjecture.If $n = 4$ then $$ a_{2k} = \frac{1}{2}\frac{(2k)!}{k!(k+1)!}\left(\frac{(2k)!}{k!(k+1)!}+1\right).$$

**References:**

- G. Lehrer and R. Zhang, The second fundamental theorem of invariant theory for the orthogonal group, Ann. of Math. 176:2031-2054 (2012).
- A. Mihailovs, A Combinatorial Approach to Representations of Lie Groups and Algebras, Springer-Verlag New York (2004).

Some notation : $i:E\hookrightarrow X$ the inclusion, $b: X\rightarrow \mathbb{P}^n$ the blowing-up, $p:E\rightarrow W$ the projection, $N$ the normal bundle of $W$ in $\mathbb{P}^n$. The map $A^1(W)\rightarrow A^2(X)$ that you mention is $i_*p^*$. Now $$H\cdot E=H\cdot i_*1=i_*i^*H= i_*p^*\mathscr{O}_{\mathbb{P}^n}(1)_{|W}\ .$$ The computation of $E^2$ is trickier. We have $E^2=i_*i ^*E$; by construction of the blow up $i^*E =-h$, where $h$ is the class of the tautological line bundle $\mathscr{O}_E(1)$. The so-called "key formula" gives $i_*(h+p^*c_1(N))=b^*[W]=\deg(W)\,H^2$, so $$E^2= i_*p^*c_1(N)-\deg(W)\,H^2\ .$$ For the "key formula", see the classical paper of Lascu-Mumford-Scott.

2133600752174755600Whenever you take $m+1$ columns in your matrix, there are $m+1$ different minors you can form from those columns. If $m$ of those vanish, the $(m+1)$-st does too, and more generally you can compute the $(m+1)$-st from the other $m$.

In order to see this, add one of the rows you already have to form a square matrix. It's determinant is 0, and knowing this gives you a relation between those minors by expanding along the row you added. If the rank of your matrix is exactly $m$, I don't think you can improve this, or not by much anyway. If the rank is smaller, of course, that's another story.

This is related also to Determinantal Varieties, though you would need to ask someone more knowledgeable than me about the details.

1269047Given a dense matrix $A \in \mathbb{R}^{n \times m}$, with $n < m$, I am interested in finding a good approximation by choosing $s$ rows and zeroing the rest. This leads me to the following optimization problem

$$\underset{A_I}{\min}\left\lVert A - A_I \right\rVert_F^2$$

where $I$ is the index set of $s$ selected rows and $A_I$ is the restriction to this set (with rows $I^c$ set to zero). Are there any known results in this direction?

Smola's paper seems to be close to what I want, but I can not follow its notation when performing column selection. Is $K_i$ a matrix or a column in Equation (11)? At first I thought it was about columns, but then when discussing selection strategies in Equation (25) it seems that there is a Gram-Schmidt-like linear relationship between $K_i$'s even though the previous selected columns are already orthogonal.

100374817501061529584124542lHodge numbers of diffeomorphic complete intersectionsAlready for three-term progressions it's somewhat surprising that there are infinitely many solutions, because the usual probabilistic guess for the expected number of solutions leads to a convergent sum: a random number of size $M$ is a square with probability about $M^{-1/2}$, so we're summing something like $1/(abc)$ over all three-term progressions $(a,b,c)$, etc. To be sure such a guess cannot account for non-random patterns arising from polynomial identities, but it does suggest that past a certain point such identities will be the only source of solutions.

Now a mindless exhaustive search over progressions $(x,x+y,x+2y)$ with $0 < x,y < 10^4$ finds only the first six examples $$ (1,7),\phantom+ (4,26),\phantom+ (15,97),\phantom+ (56,362),\phantom+ (209,1351),\phantom+ (780,5042) $$ of an infinite family associated with the solutions $(2,1)$, $(7,4)$, $(26,15)$, $(97,56)$, $(362,209)$, $(1351,780)$, etc. of the Pell equation $x^2-3y^2=1$. If it can be proved that these are the only solutions then it will immediately follow that there are no four-term arithmetic progressions with the same property. But that seems like a very hard problem.

Here's the **gp** code; with a bound of $10^4$ it takes only
a few minutes. One can surely do better with a more intelligent
search procedure (e.g. start by finding all solutions of $ab+1=r^2$
by factoring $r^2-1$).

```
H = 10^4
progsq(x,y,n) = sum(i=0,n-2,sum(j=i+1,n-1,issquare((x+i*y)*(x+j*y)+1)))
for(x=1,H,for(y=1,H,if(progsq(x,y,3)==3,print([x,y]))))
```

152894244384848Mobile and Web apps

38938819499761216052170520816572787289231397086IMHO $y=0+$ means that we are in the asymptotic case $n>> 1$, and this is apparently possible to solve in the affirmative by different means, see the answer by Matt Young. (But certainly, thanks, I stand corrected)1873140347462A complicated integral / a complicated Laplace transform involving the error function15977787156861655875Textbooks talk at length of the modular properties of $\theta(z)$ or $\tau(z)$ and the prominent role of $SL(2,\mathbb{Z})$ or one of the congruence groups.

In that case, aren't the basic objects in elementary number theory such as the set of squares $\square = \{ n^2 : n \in \mathbb{Z}\}$ also related to the geometry of $\mathbb{H}/\Gamma_0(4)$ ?

Texts on modular forms say nothing about continued fractions, symbolic coding or the geodesic flow on the modular surface. Am I missing something here?

"Is there a topological groupoid structure on the pair $(Gl(n,\mathbb{R}), O(n))$, with their standard topologies?"

This is already asked here but this linked question is a very general question, so we consider its special case about general linear group, as an independent question.

A topological groupoid structure on a pair $(X,A)$

69656313168832113161402743HMPanzo1248554My personal point of view is that once you know that there is a constructive proof, then it shouldn't be too hard to find it, for example in this case it is very reasonable that the 'sperner lemma' argument can be transposed to give a constructive proof of this version of the theorem...7903611197594355060More scaleable meaning applicable in situations where you're not dealing with topological spaces -- applicable in a wider-variety of contexts. It isn't a question of being "dated" or not, it's a question of breadth of applicability. 1440648@Mark Why a comment and not an answer ? Anyways "then if you are outside you will come to a point on the convex hull from which you can see that the whole fence is on one side of you and there is no fence on the other side" - at that point, looking at any other angle, you'd see the sea which is simpler right ? This is the algorithm I originally thought about - walk the fence and look at all angles for the sea. Obviously this would work, but the "1 meter" solution seems faster if you don't have line of sight and have to walk every timeDCan someone edit the tex, please?1122942562072421933992144389Wow, this implies there are no points at a rational distance on perpendicular bisector of a side of the square!1340326hComputing transition operators for Markov processes4774211697218187422But I don't see the point of the downvote. Wouldn't it be more helpful to other people to correct the mistake instead of downvoting the answer? Is there still anything wrong with it?Quick commentary on how to find this example (using natural log instead of log base 2): Note that $\|f\|_p$ is an increasing function of $p$ and its limit as $p\rightarrow 0$ is $\exp -H(f)$. So $f$ must be in $L^1$ but not $L^p$ for any $p > 1$. $f = x^{-\alpha}$, $\alpha > 0$, does not work, so the next thing to try is $x^{-1}(1-\log x)^{-\beta}$, where we add 1 to $-\log x$ to prevent the function from blowing up at $x = 1$. This blows up at $0$ slower than $x^{-1}$ but faster than $x^{-\alpha}$ for any $\alpha < 1$. Testing this, it works if $1 < \beta \le 2$. 21691704935393287The integral $M_2$ was studied in a 1961 US military report, to calculate the probabability of a missile hitting an elliptical target. See equation 31, where the integral $M_2=p(a,b)$ is given as a sum over Bessel functions $I_n$. (There does not seem to be a closed form solution.)

The answer can be seen as a two-dimensional analogue of the error-function, defined as an integral over $I_0$,

$$E(R,r)=e^{-r^2/2}\int_0^R e^{-t^2/2}I_0(rt)tdt$$

Then the desired integral over the ellipse is

$$M_2=\frac{1}{2\pi}\int\int_{x^2/a^2+y^2/b^2\leq 1} e^{-(x^2+y^2)/2}dxdy$$ $$=E[(a+b)/2,(a-b)/2]-E[(a-b)/2,(a+b)/2]$$

77757422470731899506@optima: And I think that locality cannot occur. Suppose two different assignments connect $s$ and $t$, through paths $p$ and $q$ respectively. Look at the last place in $p$ where it goes through a switch that is in the opposite position for the $q$ path. You can safely switch the $q$ path there, because no later switches in $p$'s tail disagree. Repeat until the $q$ path is identical with the $p$ path and then switch any irrelevant switches.Consider a graph $G$ where each vertex has degree at least two. Prove that this graph has a cycle.

One could prove this by arguing that we can start at a point, and just follow edges, until we meet an already visited vertex, but I think the following argument is slicker:

$G$ belongs to the complement of the set of forests, since every tree has a leaf, but $G$ does not. Forests are the only graphs that do not have cycles. Hence, $G$ has a cycle.

525098Yes; see Smale, On the structure of manifolds (Amer. J. Math. 84 1962 387–399) where it's shown that in high dimensions, you can eliminate handles under various connectivity assumptions. The h-cobordism theorem is a special case. For index greater than 1, one usually doesn't do handle-trading, because you can in fact just do handle cancellation, which is simpler. The reason for doing handle trading for 1-handles (and dually, for (n-1)-handles) instead of cancellations is to avoid tricky issues related to presentations of the fundamental group, related to the Andrews-Curtis conjecture.

3563411654856In any case, there should be *one* way of thinking of *all* categories, and if (2) is only possible for *some* categories, we have to take (1) for all, so even if (2) is appealing for concrete categories, we have to avoid it, right?1137635Thank you for the explanation! This is something nice to know about.10407655428221667824417131993782@Jeremy: Actually the formulation I suggested doesn't use choice. :-)Let $X$ be a smooth projective variety with polyhedral finitely generated effective cone $Eff(X)$. Let $f:X\dashrightarrow X$ be a birational automorphism of $X$ that is an isomorphism in codimension one, that is neither $f$ nor $f^{-1}$ contracts any divisor.

Does $f$ necessarily map a divisor generating an extremal ray of $Eff(X)$ to a divisor with the same property? In other words does the automorphism induced by $f$ on $Pic(X)$ preserve the set of the extremal rays of $Eff(X)$?

145610221382161929086513772See also https://mathoverflow.net/questions/301022/determining-if-a-rational-function-has-a-subtraction-free-expression/301468#3014684890532069231101313411310702271912Ah, very true, I'll head back and delete those comments. If $k$ is fixed, then this is a notoriously difficult problem.695987160543Harry, since you seem to like this Morita category (as do I), do you happen to know whether there is a good notion of whether two morphisms from $A$ to $B$ are EQUAL? Equal, not isomorphic or something like that. Because the best I can do is requiring the two bimodules to be isomorphic, but restricting to classical homomorphisms this would mean that the homomorphisms differ from each other by an inner automorphism of the target ring, which is somewhat weaker than equality. Or is equality lost when we pass to the Morita category?191521@Piertro: yes, E.T. Bell was an amateur historian of mathematics. :-)1935607@AbdelmalekAbdesselam Even if it is not a complete answer, surely the definition of nuclear space, restated as a theorem, is one example of a theorem satisfied by finite-dimensional spaces and Schwartz spaces but not typical infinite-dimensional vector spaces. EquationDude99918148601858052I think this would be true if the inclusion from $V$ into $H$ is a compact linear map645333@http://nicolabernini.github.io/1170560620278341473642777050102002722446325733432221224082541008571tIs there a percolation threshold in the hard discs model?526052061740209441314971501173406103900336381574878525765It would probably help if you explained **exactly** what you mean by «homogeneous space».22888962057345125978014089542147552796491198557418651331570987355658You can use sage to get two points you are looking for sage: E = EllipticCurve([0,0,0,-1563056672958141,0]) sage: G = E.gens() sage: print G [(48408867 : 194361588954 : 1), (260509445493025 : -4204701905638250451710 : 1)]bIrrationality of generalized continued fractions22185461930291BYou can check with Finch's book.1084224560592501690293446Qiaochu Yuan : this is the unique answer which satisfies this properties when we restrict to groupoid such that $\pi_i = \{1\}$ for $i$ large enough. One can imagine class of higher groupoid which does not satisfy this condition and for which there is a notion of "cardinal" satisfying the condition you state and which cannot be expressed under this form, at least not without some sort of renormalized product. (for example, using the Euler carateristic)51508422210361779909425184I'm aware of the standard results about Fourier Mukai on Weierstrass elliptic fibration.

What I need is the references about Fourier Mukai on elliptic fibration which can have reducible fibers and/or nontrivial Mordel Weil groups, or even genus one fibration.

All I know is the Caldararu's thesis and the paper by Donagi/Pantev on the spectral data on genus one fibrations. I appreciate if you let me know something other than this...

424986I used to work in information theory and haven't had formal math training beyond the basic undergraduate level. I still retain a lot of interest in mathematics and try to work on my skill set whenever possible.151287zoe_tang640927177815311263061321380189852Kelvyn Welsch14627011373308464292489275621872285010Well... Yes, to keep it simple let's say I'm just looking for a considerably shorter proof than the one in Schensted's paper.142401Can someone suggest some sources/references dealing with the Perron-Frobenius theory for nonnegative matrices that are reducible?

Specifically, if $A\ge 0$ is a $d\times d$ matrix with no assumptions about irreducibility, I am interested the precise description of

a) the set of all non-negative eigenvectors $v\ge 0$ in $R^d$ for $A$; and b) the dynamics of the action of $A$ on the non-negative quadrant $W=\{w\in R^d | w\ge 0\}$, particularly the limiting behavior (up to projectivization) of the sequences $A^nw$ where $w\in W$ and $n=1,2,3,...$.

The standard sources on the Perron-Frobenius theory only deal with primitive and irreducible nonnegative matrices, and I could not find any sources that treat in detail the case of an arbitrary $d\times d$ matrix $A\ge 0$.

[grr I want to make longer comments!]. For example the existence of the global Langlands group is a conjecture that, it seems to me, is almost unfalsifiable. Langlands makes some conjecture in Corvallis of the form "this set (iso classes of reps of GL_n(adeles) for all n at once) should have the structure of a Tannakian category in some natural way" for example. Is that really a conjecture or just a really good idea?830247281388722bHmm, that Otal book seems to be out of print. :(The heuristic from circle method for integral points on diagonal cubic surfaces $x^3+y^3+z^3=a$ ($a$ is a cubic-free integer) seems to fit well with numerical computations by ANDREAS-STEPHAN ELSENHANS AND JORG JAHNEL. The only known exception is the surface $x^3+y^3+z^3=2$. Circle method predicts that the number of integral points $(x,y,z)$ with $\max(\vert x\vert,\vert y \vert,\vert z\vert)<N$ is $\approx 0.16\log N$. But parametric solutions $(1+6t^3,1-6t^3,-6t^2)$ are missing.

**Question:** Why the heuristic fails for $x^3+y^3+z^3=2$?

Let $\xi_t$ be a zero-mean gaussian process on $[0,1]$ with covariance operator $C$. I would like to better understand the relation between the covariance operator and the regularity of the trajectories.

I already know that

Theorem(Kolmogorov) If there exists $\alpha>1,C\geq 0$ and $\epsilon>0$ such that $$E[|\xi_t-\xi_s|^{\alpha}] \leq C > |t-s|^{1+\epsilon}$$ then there exists a modification of the process that is almost surely Hölder-$\delta$ for $\delta\in ]0,\epsilon/\alpha[$.

I would like to know if there are other results in this direction (with other spaces than Holder ?) and especially one that relates directly a norm (spectral norm under stationnarity assumption ?) of the covariance operator and the regularities of the trajectories.

note: if I remember there is a link between the closure of the cameron martin space and the support of the gaussian measure associated to the process... how can I reformulate this to answer my question ? )

5348526995715857608965388I've been guilty of that...230548Oh, sorry. Finally you were managed to explain the trouble to (stupid) myself, thanks; now I see that the array is rearranged.1647344Alexander's Horned Sphere is not a submanifold of R^{3} is a manifold on its own but not a submanifold. Perhaps I dont understand the notion of locally flat submanifold.6079531771037$S_n=S_0\cap B[-e_1,\pi/2]\cap\dots\cap B[-e_n,\pi/2]$, so it is a ball-intersection. 1720084This is true for large $d$, and probably for all $d$. I'll prove that the sum is $$e^d(1-1/d+1/d^2+O(1/d^{2.5+\epsilon}))$$ and leave the explicit bounds to you.

Set $k=d+\ell$. For $|\ell| \leq d^{0.5+\epsilon}$, we have $$1-1/(2d+2-k) = 1-\frac{1}{d} \frac{1}{1-(\ell-2)/d} = 1-\frac{1}{d} - \frac{\ell-2}{d^2} - \frac{(\ell-2)^2}{d^3} + O(d^{-2.5+3 \epsilon})$$ $$=1-\frac{1}{d} - \frac{\ell}{d^2} + \frac{2 d - \ell^2}{d^3} + O(d^{-2.5+\epsilon}).$$

We will show later that the difference between your sum and
$$\sum_{k=0}^{\infty} \frac{d^k}{k!} \left( 1-\frac{1}{d} - \frac{\ell}{d^2} + \frac{2 d - \ell^2}{d^3} \right)$$
is very small, where $\ell = k-d$. Assuming that, let's compute the new sum. With the aid of *Mathematica*,
$$\sum_{k=0}^{\infty} \frac{x^k}{k!} \left( 1-\frac{1}{d} - \frac{k-d}{d^2} + \frac{2 d - (k-d)^2}{d^3} \right) = e^x \left( 1-\frac{1}{d}+\frac{x}{d^2}-\frac{x^2}{d^3} + \frac{2}{d^2} - \frac{x}{d^3} \right). $$

Plugging in $x=d$, $$\sum_{k=0}^{\infty} \frac{d^k}{k!} \left( 1-\frac{1}{d} - \frac{k-d}{d^2} + \frac{2 d - (k-d)^2}{d^3} \right) = e^d \left( 1-\frac{1}{d} + \frac{1}{d^2} \right).$$ The error coming from $O(d^{-2.5+\epsilon}) \sum d^k/k!$ is $e^d O(d^{-2.5+\epsilon})$.

We now just need to think about the error coming from discarding the terms with $|\ell|>d^{0.5+\epsilon}$. No matter what $\ell$ is, that error is no worse than $d^k/k! \cdot ( O(\ell^2/d^3) + O(1))$. But, for $|\ell| > d^{0.5 + \epsilon}$, we have $d^k/k! \leq e^{-d^{2 \epsilon}}$ as I pointed out on math.SE, so the contribution from these terms is exponentially small.

I have not found a slick way to give a proof for all $d$, rather than an asymptotic result.

RStrong ideals that are not pre-saturated21529132023655@Peter, You said "For other groups you have to enrich the coadjoint orbit to a bundle over them to get more representations". Can you explain the applications of non-positive line bundles for such representations?1389283nowadays it also holds in positive characteristic:

Higher direct images of the structure sheaf in positive characteristic. Algebra & Number Theory, vol 5, No. 6 (2011), 693-775

294981020667141065105167624230543249635318251891348342150296013009131027617*Dee Dee Bar* sounds like a nice place where to enjoy a drink...432696640373There's another volume conjecture formulated by Chen and Yang for Turaev-Viro invariants of closed manifolds. They present some evidence for the conjecture in the paper. In a second paper, Yang and collaborators formulate another volume conjecture based on the colored Jones polynomial, but evaluated at different roots of unity: see Question 1.7. They prove this version for the figure eight knot.

114141011975977908380The knowledge allowed to be shared is truth of the statement `a = b`.This is not an answer to your question about Zariski-local projectivity, but it is relevant to being locally free and you might be interested.

One can get away with finitely generated rather than finitely presented if one has a little more to work with. In particular, if $M$ is finitely generated and flat over $R$ and either

(i) $S$ is a multiplicative set consisting of non-zero divisors such that $S^{-1}M$ is projective over $S^{-1}R$

or

(ii) $M/rad(R)M$ is $R/rad(R)$-projective

then $M$ is projective.

This first result is due to Endo and the second is not so hard. More details as well as more of these types of results can be found in Vasconcelos' paper "On Finitely Generated Flat Modules".

163288612217201372413481201basil160264319336451400610If you do it, make sure to keep a count of the group sizes so that the labelled count can be computed as well. Also, you know I don't think the method you describe should be called "orderly". I think only the method of Read and Faradzev should be called that.R S1056134@StevenStadnicki Your point is exactly what i meant. The "applied mathematicians" believe that NS are well-posed for conditions that occur in applications.@Kevin. Ok. That's probably the thing for me to look at, then. No reason to continue this conversation here till I do. Thanks for your help.Ben, thank you very much! In your answer, you used `` homeomorphism" several times. I would like to know that '${exp}_{p}: U\in {T}_{p}M \rightarrow M$´ is really a homeomorphism?56279I happen to have been thinking about this question recently. The proof I like uses the fact that a nested sequence of *open* intervals has non-empty intersection provided neither end point is eventually constant. Now one inductively constructs a sequence of such intervals as follows. Each interval is a component of the complement of the union of the first n closed sets, for some n. Then wait till the next closed set intersects that interval. (If it never does, then we're trivially done.) It cannot fill the whole interval, and indeed must miss out an interval at the left and an interval at the right. So pass to one of those subintervals in such a way that your left-right choices alternate. Done.

PS The question (with closed intervals instead of closed sets) was an exercise on the first sheet of Cambridge's Analysis I course last year.

660023PYes, apologies for not pointing it out.I cannot find it there. Are these categories algebraic structures?183828348691930110It is NOT an isomorphism, by dimension count: the LHS has dimension $g(g+1)/2$, whereas the RHS has dimension $g^2$.

It is injective though. A (polarized) abelian variety is the same as a polarized weight 1 Hodge structure. Such a Hodge structure on $H^1(X_b,\mathbb C)$ is determined by the subspace $H^{1,0}(X_b)$, and the RHS is the tangent space to the Grassmannian of $g$-subspaces of $H^1(X_b,\mathbb C)$. The LHS is smaller because $H^{1,0}(X_b)$ must be Lagrangian with respect to the alternating form on $H^1(X_b,\mathbb C)$ for it to yield a polarized Hodge structure.

1792174zThat gives you *a* prime; he is looking for *all* the primes22752165778859432331946118734082Hi Gabriel, your question is a great one: well written, with good motivation and comments on your own thoughts. But I agree with the other comments that it is standard material in many linear algebra classes. I think thus question would be excellent at math.stackexchange. Perhaps part of what's causing an objection is the opening paragraph, which is worded like a homework question. I should say, this is a nontrivial result that you are asking about: why should the group of invertible matrices have _precisely_ two components, and not more? 1466140417841nWeak and Strong Integration of vector-valued functions18545321840709@Tom: I'm not one of the downvotes, but I don't think phrasing a question so provocatively when there are obvious alternatives is particularly professional. (Fine in a cocktail party or a casual conversation, not so much for a paper, except perhaps as a quick aside.) I will say that I clicked through to the question with some trepidation, since "shakehands/condom puzzle" sounds like it could be considerably dirtier than what's actually here... It's possible I'm just being prudish, though.49307251050808218266014035271492186856102217139111767838620The paper by de Medts-Tărnăuceanu, Finite groups determined by an inequality of the orders of their subgroups, Bull. Belg. Math. Soc. Simon Stevin 15 (2008), 699-704 (preprint at http://cage.ugent.be/~tdemedts/preprints/ineqgroups.pdf) might be relevant, as it considers two closely related notions.Lars: No. What is needed is this property: the normalization of $R$ is a finite $R$-module. Japanese implies that.@nfdc23 (cont'd) that the usual $j$-invariant map from $X \to \Bbb{A}^1$ is ramified.455684999858Smooth proper scheme over Z15034025563622259976>I'm still not sure about that detail in the proof in Lang, but here's a slightly different proof that $\exp_{NY}:NY\rightarrow X$ is a diffeomorphism.

*Step one*: $\exp_{NY}$ is surjective.
To see this, let $x\in X\setminus Y$. Let $B$ be a closed metric ball centered at $x$ containing some point of $Y$.
Since $X$ is complete, $B$ is compact, and since $Y$ is closed in $X$, $B\cap Y$ is compact.
Thus, there is $y_0\in B\cap Y$ which is closest to $x$, and so we have $d(x,y_0) = d(x, B\cap Y) = d(x, Y)$.
By differentiating $y\mapsto d(x,y)$ along any curve in $Y$ through $y_0$, we find that the geodesic connecting $y_0$
to $x$ is normal to $Y$ at $y_0$. In other words, if $\gamma:[0,1]\rightarrow X$ is the geodesic with $\gamma(0)=y_0$, $\gamma(1)=x$,
then $\gamma'(0)\in NY_{y_0}$. So $\exp_{NY}(\gamma'(0))=x$, and we have surjectivity.

*Step two*: $\exp_{NY}$ is injective.
Suppose there were distinct $v_0, v_1 \in NY$ such that $\exp_{NY}(v_0) = \exp_{NY}(v_1)=x$.
If $v_0,v_1$ were based at the same point of $Y$, then the exponential map based at that point would fail to be injective,
contradicting the Cartan-Hadamard theorem. So $v_1, v_2$ are based at distinct points $y_1,y_2\in Y$.
Let $\gamma:[0,1]\rightarrow Y$ be the geodesic with $\gamma(0)=y_0, \gamma(1)=y_1$.
Then we have $\frac{d}{dt}|_{t=0}d(x,\gamma(t)) = \frac{d}{dt}|_{t=1}d(x, \gamma(t))=0$,
because the geodesic connecting $y_0$ to $x$ is normal to $\gamma'(0)$ at $y_0$ and
the geodesic connecting $y_1$ to $x$ is normal to $\gamma'(1)$ at $y_1$. This contradicts
the convexity of the distance function of $X$.

So $\exp_NY$ is bijective. As Deane pointed out in the comments, it now suffices to show

*Step three*: The differential of $\exp_{NY}$ is everywhere injective.
For this, fix $y_0\in Y$, and for each $y\in Y$, let $P_{y_0}^{y}$ denote parallel transport
from $y_0$ to $y$. Then the map $Y\times NY_{y_0} \rightarrow NY$ given by $(y, v)\mapsto P_{y_0}^y v$
is a diffeomorphism, so it suffices to show that the map $E$ defined as the composition $Y\times NY_{y_0} \rightarrow NY \stackrel{\exp_{NY}}{\rightarrow}X$
has everywhere injective differential.
To this end, choose $(y,v)\in Y\times NY_{y_0}$ and nonzero $(z,w)\in T_yY\times NY_{y_0}$.
We need to show $dE_{(y,v)}(z,w)\neq 0$. Let $\gamma(t)$ be the geodesic passing through $y$ at time 0 with velocity $z$.
We need to show that $\frac{d}{dt}|_{t=0} \exp_{\gamma(t)} P_{y_0}^{\exp(\gamma(t))} (v+tw)$ is not zero. Let
$\alpha(s)= \exp_y P^{y}_{y_0}(sv)$.
Let $\Gamma(s,t) = \exp_{\gamma(t)} P_{y_0}^{\exp(\gamma(t))} s(v+tw)$, and
$J(s) = \frac{\partial}{\partial t} \Gamma(s,0)$. $J$ is a Jacobi field along the geodesic $\alpha$. Set $f(s) = \langle J(s), J(s) \rangle$. We will show $f(s)>0$ for $s>0$.
Setting $s=1$ will then yield the desired result.

Compute \begin{align*} f'(0) &= 2\langle \frac{D}{\partial s}\frac{\partial}{\partial t}\Gamma(0,0), \frac{\partial}{\partial t}\Gamma(0,0) \rangle\\ &= 2\langle \frac{D}{\partial t}\frac{\partial}{\partial s}\Gamma(0,0), \frac{\partial}{\partial t}\Gamma(0,0) \rangle\\ &= 2\langle \frac{D}{\partial t}|_{t=0}P_{y_0}^{\exp(\gamma(t))} (v+tw), \frac{d}{dt}|_{t=0}\gamma(t) \rangle\\ &=2 \frac{\partial}{\partial t}|_{t=0} \langle P_{y_0}^{\exp(\gamma(t))} (v+tw), \frac{d}{dt}\gamma(t) \rangle\\ &=0, \end{align*} since $P_{y_0}^{\exp(\gamma(t))} (v+tw) \in NY$ and $\frac{d}{dt}\gamma(t)\in TY$ for all $t$. Also, \begin{align*} f''(s) &= 2\langle J'(s), J'(s) \rangle + 2\langle J''(s), J(s) \rangle\\ &= 2\langle J'(s), J'(s) \rangle - 2\langle R(\alpha'(s), J(s))\alpha'(s), J(s) \rangle\\&\geq 0. \end{align*}

Now, $J(0) = z$, so if $z\neq 0$, then $f(0)>0$, which combined with the two computations above, proves $f(s)>0$ for all $s>0$. If $z=0$, then $w\neq 0$ and $J'(0)=w$, so $f''(0)>0$, and again we conclude $f(s)>0$ for all $s>0$.

28214889437204364Mable1858606204067912943306I am reading the paper "AN ARITHMETIC FORMULA FOR THE PARTITION FUNCTION" by Kathrin Bringmann and Ken Ono. In the proof of Proposition 4.2 they make reference to the existence of a bijection between the solutions to the Diophantine equation with $k,c>0$

$b^2-24kc=1-24n$

and the points of the orbits

$\displaystyle \bigcup_{Q\in Q_{24n-1}^p/\Gamma_0(6)}\lbrace{ A\tau_{Q}:A\in \Gamma_0(6)/\Gamma_{\tau_Q}\rbrace}.$

They do give some reasons why this exists but I am having trouble filling in the details. Does anyone know how to fill in the details that I think I am missing?

665418No, I'm just using set stability to get finiteness of the group. The difference between set stability and point stability is at worst $S_n$, of size $n!$, which is finite.273109One thing to consider is that [Chaitin's constant](https://en.wikipedia.org/wiki/Chaitin%27s_constant) might be measurable, potentially. Then we would need only a finite number of bits to measure the consistency of ZFC, for example.Ah, good, I didn't know a reference for this. Looking in MathSciNet for a few more details: it is Über die Erzeugung von $(F)$-Räumen durch selbstadjungierte Operatoren", Math. Ann. 164 1966 219–224. 616170885324335726770766**EDIT: this was for a different problem, but it has now been changed; so ignore this!**

Too long for a comment...

If the integral is over $D$, then $f'(z)$ is (after changing variable $w$ to $\bar{w}$ in the measure $\mu$) \[ \frac{d}{dz} \int_D \frac{d\mu(w)}{1-\bar{w}z} = \int_D \frac{\bar{w} \, d\mu(w)}{(1-\bar{w}z)^2} = P_{L^2_a}(\bar{w} \mu), \] which is (formally) the $L^2_a(D)$ Bergman space projection operator applied to the measure $\bar{w} \mu$; but I don't know if this function of $z$ necessarily does lie in $L^2_a$.

So it's closely connected to Bergman space Toeplitz operators, which are known to be very tricky (much worse than on the Hardy space). Even worse, there is usually no nice way to characterise various function spaces which arise in terms of Taylor coefficients.

So your question really combines *two* interesting and highly non-trivial problems! I doubt there is any really simple answer.

The answer is $K_t(H)$, where $H$ is a handlebody with boundary $F$. If $t$ is a root of 1 and we are taking the usual semisimple quotient, then $K_t(F\times I)$ is isomorphic to a matrix algebra and $K_t(H)$ isomorphic to the standard representation. Also, in this case we can let $H$ be *any* 3-manifold with boundary $F$ -- it doesn't even need to be connected.

In order to get an upper bound on $\alpha_4$ (which is probably far from being a tight bound), you can use the Kahn-Lovász Theorem (a generalisation of Brégman's Theorem). The Kahn-Lovász Theorem says that the number of perfect matchings in any graph $G$ is at most $$\prod_{v\in V(G)}(d(v)!)^{1/2d(v)}.$$ Its not immediately clear to me how to obtain an upper bound of the form $2^{c|V(G)|}$ from this, but it looks like it might be possible since planar graphs have at most $3|V(G)|-6$ edges (and therefore the sum of the vertex degrees is at most linear in $|V(G)|$).

Like I said, this is unlikely to be tight, but at least it might give you a bound. The Kahn-Lovász Theorem is tight for *general* graphs, but I think that the unique tight example is a disjoint union of balanced complete bipartite graphs, which is not planar unless every component is isomorphic to $K_2$ or $K_{2,2}$.

Every $A \in \text{GL}_n(\mathbb{R})$ has a unique orthogonal polar factor $O_A=A(\sqrt{A^TA})^{-1}$,

( $A=O_AP_A$, $O \in \operatorname{O}_n, P \in \operatorname{Psym}_n$see Polar decomposition).

I am interseted in characterizing the set $W:=\{(A,B)\in \text{GL}_n^+ \times \text{GL}_n^+|\,O_{AB}=O_AO_B \}$. I have made some progress, but am still lacking a unifying necessary and sufficient criterion. Even the special case where $A=B$ does not seem trivial. Any advancement would be welcome.

*Edit:* Another natural question is: How rare is it for a pair $(A,B)$ to be in $W$? (we can take the uniform distribution on $\text{GL}_n^+ \times \text{GL}_n^+$ and ask what is the probability of a random pair to be in $W$).

**Motivation:**

This question has a *geometric* nature, since $O_A$ is the closest matrix in $\text{SO}_n$ to $A$ (w.r.t to the Frobenius norm). Thus, it reflects the "rotational part" of $A$, after the scaling of the different directions is removed; This is very clear when looking at the SVD of $A$, $$A=U\Sigma V^T \Rightarrow O_A=UV^T$$ So, we can think of $O_A$ as the best "isometric approximation" (or projection) of $A$. If we know the two approximations for $A,B$ it is natural to compose them and to ask if we obtained in this way the best approximation to $AB$. (This comes up
in some applications of geometry and analysis).

**What I have so far:**

**(0)** The relation is not symmetric, i.e $(A,B) \in W \nRightarrow (B,A) \in W$.

**(1)** A *sufficient* condition for $(A,B) \in W$ is that $P_A$ commutes with $O_B,P_B$. (This is far from being a necessary condition, see below).

**(2)** $\forall A \in \text{GL}_n^+ \,: \, (A,A^{-1}) \in W$.

**(3)** If $B \in \operatorname{Psym}_n$, then $$(A,B) \in W \iff [P_A,B]=[P_A,P_B] =0$$ (see here for details )

In particular, for $A,B \in \operatorname{Psym}_n$: $$(A,B) \in W \iff [A,B]=0$$

**(4)** None of these implications hold in the general (proofs at the end):

(a) $(A,B) \in W \nRightarrow [A,B]=0.$

(b) $[A,B]=0 \nRightarrow (A,B) \in W.$

(c) $(A,B) \in W \nRightarrow [P_A,P_B]=0.$

(d) $[P_A,P_B]=0 \nRightarrow (A,B) \in W.$

**(5)** If $A$ is a normal matrix, then $(A,A) \in W$.

(This is easy to see since $A$ is normal $\iff$ $[O_A,P_A]=0$, via the uniqueness of the positive square root).

**Is it true that $(A,A) \in W$ implies $A$ is normal ?** (I suspect the answer is negative)

**Proofs (for the statements in (4)):**

(a) Take $B \in \operatorname{Psym}_n,O \in \operatorname{SO}_n$ such that $[B,O] \neq 0$, and define $A=OB$. Then $(A,B) \in W$ but $\, AB=OB^2 \neq BOB=BA$.

(b) Assume by contradiction $[A,B]=0 \Rightarrow (A,B) \in W$. Then $[A,B]=0 \Rightarrow [O_A,O_B]=0$; Indeed, $$[A,B]=0 \Rightarrow [B,A] =0 \Rightarrow (B,A) \in W,$$ so $$O_BO_A=O_{BA}=O_{AB}=O_AO_B \Rightarrow [O_A,O_B]=0.$$ However, this does not always hold:

Let $A \in \text{GL}_2^+,B \in \text{GL}_2^-$ be commuting matrices. Note that $[O_A,O_B] \neq 0$ since for any two elements $X \in \operatorname{O}_2\setminus \operatorname{SO}_2,Y \in \operatorname{SO}_2\setminus\{\pm Id\}$, $[X,Y] \neq 0$. However, I want counter-examples where $A,B$ are both in $\text{GL}_n^+$. This is impossible to find for $n=2$ since $\operatorname{SO}_2$ is abelian. However, we can take $4 \times 4$ matrices, by "doubling" $A,B$ in two diagonal blocks.

(For a concrete example see here)

(c) See counter-example here.

(d) See counter-example here.

@Ruberman Thank you for your helpful comment. May I ask you that why any self-diffeomorphism of $\Sigma_g\times S^1$ extends to a self diffeomorphism of $\Sigma_g\times D^2$ for $g\geq2$ ?57514430677512491831549253I don't quite get what do you mean by "chosen unifromly randomly". But if one multiplies a matrix with iid standard Gaussian entries by a non-random vector $h$ from $H$, the result is a well defined vector of iid centered Gaussians with variance $\|h\|^2$. Not sure if this helps you.127544Maybe it would be a good idea to share with us your attempted proof and tell us which parts thereof seem doubtful to you.14265604917519724201898471364211247955@NeilStrickland. Thanks, this fully answers my question. If you want to write it as an answer, I will accept it. Are there easy to verify conditions on $X$ that insure that $H_*F(X,Y)\cong Hom(H_*X,H_*Y)$ ?18017412009352JDieudonné's is quite idiosyncratic!Very nice, thank you. I have accepted this answer. I observe that the non-trivial extensions between the associated quotients of the filtration on $V_{l,N}$ that your example construct are not $f$ at $p$ in the sense of Bloch-Kato: that is, this is an extension of two unramified representations at $p$, that is itself ramified at $p$. Do you know if one can produce some non-trivial extensions that are $f$ in the sense of Bloch-Kato? And in general, do you know other type of examples of non semi-simple $V_{l,N}$?1890455155360038149514270799Here is a suggestion that may not be liked much, but has worked well for strict versions of $\infty$-groupoids and categories, namely to use the cubical model.

In

R. Brown, F.A. Al-Agl, R. Steiner, `Multiple categories: the
equivalence between a globular and cubical approach', *Advances in
Mathematics*, 170 (2002) 71-118.

we showed that strict globular $\omega$-categories are equivalent to strict cubical $\omega$-categories with connections. Now (see section 10) the monoidal closed structure on the latter category is not so hard to write down, following the groupoid version given in 1987 by myself and Higgins.

R. Brown and P.J. Higgins, ``Tensor products and homotopies for
$\omega$-groupoids and crossed complexes'', *J. Pure Appl.
Alg.* 47 (1987) 1-33.

Basically this uses the simple idea that for the $n$-cube $I^n$ we have the formula $I^m \times I^n \cong I^{m+n}$.

So the above equivalence of categories allows one to translate the closed monoidal structure in the cubical case to one on strict globular $\omega$-categories. This is related to earlier versions by Street and by Crans.

So the question is: can one use similar cubical methods for weak $\infty$-structures?

The other main advantage of cubical methods for the work with Philip Higgins was the easy description of multiple compositions as arrays of the form $[\alpha_{(r)}]$ where $(r)$ is a multi-index, so allowing *algebraic inverses to subdivision*.

A student who has looked something up on google and copied it out has probably learned something from doing this, so why worry? In fact, may be better for them than copying the answer from a friend.

2152071225670133971513590521652074What do you mean by a "good description"? For a $3$-dimensional vector space $V$, the kernel $K(V)$ of the map $\text{Sym}^2(\text{Sym}^2(V))\to \text{Sym}^4(V)$ is a representation of $\textbf{GL}(V)$ that can be explicitly described via Schur functors. For a $4$-dimensional subspace $U$ of $\text{Sym}^2(V)$ with $2$-dimensional quotient $W$, there is an induced surjection $T_W:K(V)\to \text{Sym}^2(W)$. The $3$-dimensional space you describe is the kernel of $T_W$.I think of the axiom of regularity along with the axiom of extensionality as formalizing what I mean by "set". Once upon a time, before paradoxes, one could think of sets as just any collection of things. Unfortunately, axioms based on that picture, in particular the unrestricted comprehension axiom, led to contradictions, so it became clear that the original, contradictory notion of "set" must be replaced by something clearer. (People might have thought the original notion was perfectly clear, but the paradoxes show that it isn't.) The clearer picture that emerged (in a development beginning with Russell's type theory, and continuing through simple type theory) is of a cumulative hierarchy, in which sets are obtained as follows.

Begin with some non-set entities called atoms ("some" could be "none" if you want a world consisting exclusively of sets), then form all sets of these, then all sets whose elements are atoms or sets of atoms, etc. This "etc." means to build more and more levels of sets, where a set at any level has elements only from earlier levels (and the atoms constitute the lowest level). This iterative construction can be continued transfinitely, through arbitrarily long well-ordered sequences of levels.

This so-called cumulative hierarchy is what I (and most set theorists) mean when we talk about sets. A set is anything that is formed at some level of this hierarchy. This meaning of "set" has replaced older meanings.

The axiom of regularity is clearly true with this understanding of what a set is. It expresses the idea that the stages of the cumulative hierarchy come in a well-ordered sequence. (Without well-ordering, the instructions for each level, namely "form all sets whose elements are at earlier levels," would not be an inductive definition but a circularity.)

Although there are set theories that contradict regularity, I would say that any such theory (and also any theory that contradicts extensionality) is not about sets but about some different (though presumably similar) entities.

1515307@Equivariant singular cohomology223202712488411569095Right - the cube is actually a simpler example than the simplex...1672749152404094311410346311533757Consider the $n$-dimensional Hamming cube, $C = \{-1,1\}^n$. Given an $n \times n$ orthogonal matrix $O$, I'll measure "how close $O$ is to being an isometry of $C$" by the following scoring function:

$s(O) = \frac{1}{2^n}\sum_{x\in C}\left| \prod_{i=1}^{n} (Ox)_i \right|.$

Intuitively, $s(O)$ measures how close the image of each $x\in C$ is to belonging in $C$, averaged over all $x\in C$. It's not hard to see that $s(O)\le 1$ for all $O$, with equality if and only if $O$ is just a product of reflections and permutations of the $n$ coordinates (i.e., "trivial"). My question is the following:

**If $s(O)$ is non-negligibly large (say, $\ge n^{-O(1)}$), then must $O$ be "close" to a product of reflections and permutations of the coordinates?**

Here, by "close," I mean that $O$ contains some diagonal whose product is at least $n^{-O(1)}$ in absolute value.

I should mention that my summer student Sumegha Garg did a computer search for interesting examples of such $O$, for $n$ up to about $10$, but didn't find much. When $n=4$, the Hadamard matrix does fairly well, achieving $s(O)=1/2$, but we've checked that $s(O)$ decreases exponentially with $n$ for $n\times n$ Fourier and Hadamard matrices and tensor products thereof.

A few words about where this question came from: it's not hard to show that the *permanent* of an $n\times n$ matrix $A$ can be expressed in the following form (the so-called "Glynn formula").

$\operatorname{Per}(A) = \frac{1}{2^n}\sum_{x\in C} \prod_{i=1}^{n} x_i (Ax)_i $.

So in particular, we have $\left|\operatorname{Per}(A)\right| \le s(A)$, and if $O$ is orthogonal, then $\left|\operatorname{Per}(O)\right| \le s(O) \le 1$, with equality achieved if and only if $O$ is a product of reflections and permutations of the coordinates. Now, for reasons arising from linear-optical quantum computing, my student Alex Arkhipov and I would like to know whether there are any $n\times n$ orthogonal matrices $O$ satisfying $\left|\operatorname{Per}(O)\right| \ge n^{-O(1)}$, which are far from the "trivial" examples---i.e., reflections and permutations of the coordinates. (Actually, we'd like to know the answer for unitary matrices, but if we don't even know it for real orthogonal matrices, then we might as well start there.) And this naturally led to the stronger conjecture that there isn't even any nontrivial rotation $O$ of the Hamming cube that produces a large value of $s(O)$.

As a final remark, Arkhipov managed to show that, if $O$ is orthogonal and $\left|\operatorname{Per}(O)\right| \ge \sqrt{1-1/e}$, then $O$ is close to a product of reflections and permutations of the coordinates. I'm not sure whether his proof generalizes to all $O$ such that $s(O) \ge \sqrt{1-1/e}$. In any case, the techniques don't seem to generalize to $s(O) \ge n^{-O(1)}$.

**Update:** Sumegha Garg notes that, if we consider $n\times n$ unitary matrices $U$ rather than just real orthogonal matrices $O$, then there *is* a way to get $s(U)=1$, despite $U$ not containing any diagonal whose product is $n^{-O(1)}$ in absolute value. The way to do it is to let $U$ be block-diagonal, with $2\times 2$ blocks of the form

$\frac{1}{\sqrt{2}}\left( \begin{array} [c]{cc}% 1 & i\\ 1 & -i \end{array} \right) $.

On the other hand, this is essentially the *only* counterexample we know, and even in the complex case, the conjecture could easily be reformulated to account for it.

I don't know if you are still interested, but here you go:

- Write $f(x)=\Pi(n,u,m)$ as a series $f(x)=S(u)$ in $u$ around $u_0=0$.
- Inverse/reverse that series $S(u)$ to the reverse series $T(v)$ with the property that $T(S(u))=u$. Then insert $f(x)$ into $T(v)$ to get a series representation $u=T(f(x))$ of $u$.

This can be done with help of a computer algebra system like Mathematica or Maple. A numerical approach is given here: T. Fukushima: Numerical Inversion of General Incomplete Elliptic Integral

18026883283562072132221837In his answer to user33038's mathoverflow question "What axioms are stronger than the Axiom of choice?", Prof. Hamkins writes:

"What's more, the axiom of choice is equivalent over $ZF$ to the assertion "there are unboundedly many strong limit cardinals."

My question is simply this: Is there an analogous assertion (meaning stated in terms of strong limit cardinals) to "there are unboundedly many strong limit cardinals" that is equivalent to Global Choice over $ZF$?

536870Thank you. On the second thought I am not sure I understand completely your argument. If $Z_0$ contains an entire component of $Z$, that probably means that the map $g$ is not flat. Also if $Z_0$ is smooth and $g$ is flat, then $g$ is smooth. Does it imply that nearby cycles of $Z$ and $Z_0\times S$ are the same?7474900paradoja1904552Here is an easy example which illustrates Francesco Polizzi's answer: Consider the morphisms $\mathbb{Z} \rightarrow\mathbb{Z}/12$ and $\mathbb{Z} \rightarrow\mathbb{Z}/6$, sending 1 to the class of 3. These are objects in the comma category. Now the canonical surjection $\mathbb{Z}/12 \rightarrow \mathbb{Z}/6$ induces a morphism in the comma category (it sends the class of 3 to the class of 3).

91303617635291654175@Harry. You could carry it out in your mind. Read a book like Demazure and Gabriel for the basic machinery, and read an article at Romagny's, with the constant aim in mind to see whether each statement can be proved without the crutches of schemes. Maybe you can write down your conclusions; or maybe it is sufficient to carry it out in your mind. 3308479356092038833user627316156118993251594052555047If you mean for all positive numbers $x>0$ (not just integers), then this surely fails, because the two sides jump at different values.Expanding my comment.

Notice that $\frac{\Gamma(n+2\alpha+1)}{\Gamma(n+1)\Gamma(2\alpha+1)}=\binom{n+2\alpha}{n}=(-1)^n\binom{-2\alpha-1}{n}$ and thus the sum can be rewritten as $$\frac{1}{[(s+1)(t+1)]^{\alpha+1}}\sum_{n=0}^{\infty} \binom{-2\alpha-1}{n} \left(-\frac{st}{(s+1)(t+1)}\right)^n$$ $$ = \frac{1}{[(s+1)(t+1)]^{\alpha+1}}\left(1-\frac{st}{(s+1)(t+1)}\right)^{-2\alpha-1} = \frac{[(s+1)(t+1)]^{\alpha}}{(s+t+1)^{2\alpha+1}}$$ as expected.

156231920305504288921407437X$x\geq 1465993117$. I think I'll stop here.264835631517:I meant $l\ne j$, of course.1919708Thanks for all the interesting perspectives on this question. It seems that there are two quite distinct reasons for focusing on expected value. The first is based on the law of large numbers, and works fine for common events. This does not require any theoretical apparatus of utility functions or choice under uncertainty. This type of reasoning is on very firm ground.

There is a completely distinct approach, based on expected utility. Expected value is often used, sometimes ignorantly, as a proxy for expected utility. This approach seems more treacherous to me, as it requires postulating a specific utility function, and even more so, it requires making some very strong moral or psychological assumptions.

In many cases, it seems, the difference between the two approaches is silently ignored.

1111349514493129900818626052906261211397Jack of all trades, master of none.

P...but upon reflection, also difficult.17584588815173436051247347Hmm. Thanks for pointing this out! I may have to reword the relevant paragraph.@Stromquist's 3 knives procedure625589156023515570714373553374823Just a slight further clarification. When I wrote something like "chain of length $n$ is present in every segment in $L[s_n]$" ..... obviously only those segments are meant which have links of form $(a_0,b)$ (where $b \ne P_{12}(a_0)$). I forgot to mention that. Other than that, I probably should have written the definition of a chain (in given context) more precisely. Anyway, there doesn't seem to be much interest in the main question anyway so I will just update the document(i) at suitable time (without bumping the question).16119751635086There is also a GAP package called Quagroup which for the quantized envelopping algebra of any simple Lie algebra can construct its highest weight modules and compute explicitely the corresponding image of the $R$ matrix. In fact it can even compute the isomorphism $V\otimes W \rightarrow W\otimes V$ for different modules $V,W$. http://www.science.unitn.it/~degraaf/quaman.html374800948497*"By teaching, we learn." - Seneca the Younger

This question arose from the recent one, roots of a polynomial linked to mock theta function?. Let $$ g(x):=\sum_{k=0}^\infty x^k\prod_{j=1}^{k-1}(1 + x^j)^2\\=1+x+x^2+3 x^3+4 x^4+6 x^5+10 x^6+15 x^7+21 x^8+30 x^9+43 x^{10}+59 x^{11}+...; $$ the sequence $1,1,1,3,4,6,10,15,21,30,43,59,...$ with the generating function $g(x)$ is A059618 on OEIS, it is the sequence of numbers of strongly unimodal partitions.

Now let $$ f(q):=g(q)\prod_{n=1}^\infty(1-q^n), $$ and let $a_k$ be the $k$th coefficient in the Maclaurin series for $f$, $$ f(x)=\sum_{k=0}^\infty a_kx^k\\=1-x^2+x^3+x^6+x^7-x^9+x^{10}-x^{14}+x^{18}-x^{20}+x^{21}+x^{25}+x^{26}-x^{27}\\+x^{28}-x^{30}+x^{33}-x^{35}+x^{36}-x^{39}-x^{40}+x^{42}-x^{44}+2x^{45}-x^{49}+x^{52}-x^{54}\\+x^{55}+x^{56}+x^{57}-x^{60}-x^{65}+... $$ The sequence of $a_k$, starting with

1,0,-1,1,0,0,1,1,0,-1,1,0,0,0,-1,0,0,0,1,0,-1,1,0,0,0,1,1,-1,1,0,-1,0,0,1,0,-1,1,0,0,-1,-1,0,1,0,-1,2,0,0,...

is not on OEIS. Among the first 1000 terms of the sequence, there are 609 zeroes, 182 ones, 161 -1s, 19 of them are 2 ($a_{45},a_{150},a_{210},a_{221},a_{273},a_{300},...$), 22 are -2 ($a_{77},a_{90},a_{165},a_{225},...$), and two of them ($a_{525}$ and $a_{825}$) are 3; seems like $a_k$ are zero for $k=2^j$ ($j>0$), for $k=p$ or $k=2p$, with $p$ prime $>7$, $k=3p$ and $k=4p$ with $p$ prime $\geqslant23$, $k=5p$ with $p$ prime $>31$, $6p$ for $p>37$, $7p$ and $8p$ for $p>43$, $9p$ for $p>47$, $10p$ for $p>61$, $11p$ for $p>67$,...

What may (or may not) be relevant is another sequence obtained from introducing new variable in the way I learned from a paper by Rhoades linked to from the above OEIS page for $g$.

Let $$ g_t(q):=\sum_{k=0}^\infty q^k\prod_{j=1}^{k-1}(1 + q^jt)(1+q^j/t), $$ and let $$ f_t(q)=g_t(q)\prod_{n=1}^\infty(1-q^n), $$ so that $g_1(q)=g(q)$ and $f_1(q)=f(q)$. Then $$ f_t(q)=1-q^2+\frac{1+t^3}{(1+t)t}q^3+\frac{1+t^5}{(1+t)t^2}q^6+q^7-\frac{1+t^3}{(1+t)t}q^9+\frac{1+t^7}{(1+t)t^3}q^{10}+...; $$ most coefficients have form $\pm\frac{1+t^{2j+1}}{(1+t)t^j}$, except that I cannot figure out how $j$ depends on the number of the coefficient. Exceptions here start from the $15$th coefficient, which is $\frac{1+t^9}{(1+t)t^4}-1$ and the $45$th one which is $\frac{1+t^{17}}{(1+t)t^8}+\frac{1+t^3}{(1+t)t}$.

Despite all these clues, to my shame I've given up searching for an explicit formula for $a_k$. Is there one? I am pretty sure there is, but what is it?

8168147748305Suppose $N$ is closed. Let $p:U\to N$ be a tubular neighborhood of $N$ in $M$. So $p:U\to N$ is a vector bundle. Now let $g_N$ be a Riemannian metric on $N$ and $g_U$ be a fiber metric on $U$, and let $\nabla$ be a linear connection on the vector bundle $p:U\to N$. Split $TU= V\oplus H$ into the vertical and horizontal bundle over $U$. Put $T(p)^*g_N$ on $H$, Put $g_U$ on $V$ via the affine structure, and declare $V$ an $H$ to be orthogonal everywhere. This gives you metric $\tilde g$ on $U$ so that $p$ is a Riemannian submersion by construction. Thus horizontal geodesics in $U$ project to geodesics of the same length in $N$, so $N$ is totally geodesic in $U$. Now use a partition of unity argument to get a metric on $M$ which equals $\tilde g$ near $N$.

1255588425273148551715838892212201556788237749user12078576480734957This is why I use the terms "Proof sketch" or "Proof idea" in doing write-ups. I think professionals need to make more explicit the fact that they are communicating ideas for recipes and not recipes themselves (unless they are providing a mechanically checkable proof). This would help people outside the group understand and appreciate (or be dismayed at) how the professionals "do their thing". Also, creating a class of "mechanics" who can test and fix a proof sounds like a good thing. Gerhard "'Let's Look Under The Hood...'" Paseman, 2018.01.24.825315184768530781218418221971322105322817290321956517Regarding the status of variables, you probably want to look at Chung-Kil Hur's PhD thesis "Categorical equational systems: algebraic models and equational reasoning". Roughly speaking, he extends the notion of formal (as in formal polynomials) to signatures with binding structure and equations. He was a student of Fiore's, and I think they've been interested in giving better models (inspired by the nominal sets approach) to things like higher-order abstract syntax. I've been meaning to read his thesis for a while, to see if his treatment of variables can suggest techniques that could be used for writing reflective decision procedures which work over formulas with quantifiers.

For schematic variables or metavariables, there's a formal treatment of them in MJ Gabbay's (excellently-titled) paper "One and a Halfth-Order Logic"

1677913@Hurkyl, the issue here was the length of an existing proof. That's a pretty clear sufficient condition for being *explicitly written*. Do you have an example of a published proof about which it is unclear whether its length is a standard integer?192010062993I'm probably showing my ignorance, but is the first step involving Chebotarev effective?19993761422446Not crucial, but the word 'covered' is usually used for other stuff, imho. I would say 'hit' or 'pierced' or 'stabbed'.1950879@HJRW Well, if that's the case, then the question would have been better left unanswered. Answering a question can be seen as tacit approval of the question as appropriate for this site. We should not be sending mixed messages. (That is to say: if a question is to be answered at all, it should be answered seriously, not flippantly.)6768428173360817981321522059277508Following Lam's notation, a ring (with identity) $R$ is called dedekind-finite if $ab=1\iff ba=1$ in $R$.

There are a lot of result about left invertible implies right invertible. But the results all require some finiteness property on the ring or the matrix ring. I am asking a proof or a couterexample of that that $R$ is dedekind-finite impies that the matrix ring $\mathbb{M}_n(R)$ is dedekind-finite.

352691158016bCohomology of the Moduli of G-bundles on a Curve"Extending scalars" for (motivic) ring spectra and for modules over them: are the corresponding Moore spectra highly structured ring objects?134739310036699076711130718@Igor: Your angry birds example does not have any intersection with MO. My post is not a math research question but chances that someone in MO community might have encountered this are higher than in SO. Anyways, feel feel to vote it down - let the votes decide.925362126768222607337213067I think $\omega_i[dim(i)]$ is isomorphic to $\mathcal{O}_E(-1)[-1]$.1240004This has relation with my another question in [http://mathoverflow.net/questions/259865/the-expectation-of-two-sides-of-rectangle-is-equal-can-we-deduce-that-in-the-ex].2144707You are right, Jason, already in genus 1! If the generic fiber of f is an elliptic curve without complex multiplication, then the Picard rank of any (good) reduction will be larger because elliptic curves over finite fields always have complex multiplication. See Milldenhall's paper in DMJ 67 (1992).1516676Let $f_1, f_2, \ldots f_k$ be a set of polynomials in $n$ variables, with integer coefficients. These define an affine scheme $X$ of finite type over $Spec \mathbb{Z}$. (We could also consider homogeneous polynomials, and the associated projective scheme.)

Question: How do I check (by hand or by computer) whether this scheme is regular or not?

A necessary but not sufficient condition is that the generic fiber $X_\mathbb{Q}$ is smooth as a variety over $\mathbb{Q}$, which is something we can check by the Jacobian criterion.

$t$ is not a closed term unless you are using $w$ as a meta-variable.117737010759033871255t@Gjergji, what do you mean exactly by "This should work"?19586551464649211715015088311172772@M.FarrokhiD.G. So, just to input data, you will need $O(n^2)$ memory. The input yields the value of $n=i+p$. After that you would like to make $o(n)$ operations and obtain the values of the index and period? Right?1470354206015This seems like much more than a mathematical mistake - really it's a cognitive bias which incidentally can be turned into a mathematically false statement. @Charles: The 2-category of (1-)toposes looks much more like a category of locales (0-toposes) to me, and that one is certainly not a 1-topos, although many 1-toposes can be extracted from it, as reflective subcategories in overcategories (etale spaces). Also, $\mathbf{Topos}$ (and even $\mathbf{Topos}/S$) lack any good cartesian properties: most objects are not exponentiable, limits and colimits are generally difficult to describe explicitly, there is no way to classify subobjects and no good substitute for their class. It is not even well-powered! $\mathbf{Cat}$ is so much nicer.229100@MartinBrandenburg: I think Wikipedia's choice to call this the change-of-variables-formula is somewhat misguided. It's a useful formula, I give you that, but "change of variables" usually refers to the theorem that actually *computes* the push-forward measure in terms of the Jacobian.user205567That's very interesting. An immediate consequence is that the eigenvectors are orthogonal, since $D^{1/2}$ is diagonal.And, of course, one also adds structure to the moduli space itself …. (I'd also like to take this-comment to re-register my objection to the term "moduli space" by anyone who doesn't also say/write "vectors space".)6Andy: see my answer below.9858932Isomorphic simple groupsThis should really be a comment to the original question, and/or to Finn Lawler's answer, complementing John Bourke's comment. I apologise for posting as an answer, but I don't yet have comment privileges here.

As Tom believes above, there is a monad with two different enrichments. See Example 4.1 in John Power's "Unicity of enrichment over Cat or Gpd" which John Bourke mentions above.

164649Existence of unique critical points to second order elliptic PDEsThank you for your reply. I am aware that the answer is "yes" when $\mathcal{E}$ is countably generated (see the first comment below the question). As for the ergodic decomposition: the reference you gave uses the fact that $\mathcal{E}$ is countably generated $\bmod P$ **specifically** to prove the "invariance" part [the easy part], *not* the "ergodicity" part [the part that I'm concerned about]. For the "ergodicity" part, the reference you gave still uses Birkhoff's ergodic theorem.JSure, for example let $g(x)=x^2-2x$.TongchengToo many questions, let me just answer the first one. The original reference is Grothendieck's Bourbaki lecture no. 221 (May 1961), *Techniques ... IV: les schemas de Hilbert*, §4 c. Note that Grothendieck denotes the Weil restriction by $\prod_{S'/S}(X'/S')$.192845439682852515Consider the parameterization of (hypercubic) grid graphs: $(x_1,\ldots,x_n)$ specifies a subgraph of the lattice $\mathbb{Z}^n$ which looks like a box with edge lengths $x_i$ and has $P=\prod x_i$-many vertices. Such a graph is Hamiltonian iff $P$ is even. Once one leaves bipartite graphs, such characterizations seem hard to generalize.How does $\prod_{\zeta}\varphi(q^{1/5}\zeta) = \varphi^6(q)/\varphi(q^5) $ hold?Technically, the question is not properly posed, since the isomorphism class of any single group (even the trivial group) is itself a proper class, and not a set. Thus, you have a "collection" of proper classes already, and this is neither a set nor a class. Rather, what you want to ask about is: Is there a proper class of pairwise non-isomorphic groups? And this is the question that the answerers answered.89035123869Gerhard: Valuation has at least three different meanings (at least one each in logic, measure theory and algebra according to wikipedia), none of which appear to be what the OP wants. In light of your comment, I'd recommend "weighted relation". For bonus points, a map $A \times B \to W$ for any $W$, ordered or otherwise, could be called a $W$-weighted relation.15599761752426Using advanced algorithms, Wilson primes have been searched up to $2 \cdot 10^{13}$, see https://arxiv.org/abs/1209.3436 (A search for Wilson primes, Mathematics of Computation 83 (2014), 3071-3091)69679118590427695951478133@GunnarMagnusson Perhaps the OP is looking for connections to other parts of mathematics? The mirror symmetry B-field is a cohomological shadow of more complicated objects.124237321807661185718114550583686In a pre-order $\prec$ (or a category) one can speak of *initial* objects $0$, or *terminal* objects $1$, meaning that $0\prec x$ for all $x$ --- (or $0\rightarrow^! x$ ) --- which also gives the notion of a universal object under several. E.g., among objects preceding both of $b_1,b_2$, with the restricted relation $\{(a_1,a_2)|a_1\prec a_2 ,a_i\prec b_j\}$ one can talk again about maximal objects and terminal objects, either of which notions might make a sensible candidate for "greatest lower bound" in this setting.

If you're not assuming the relation is transitive, you might want to take a (possibly graded category) transitive closure, or look at "transitive neighborhoods", or even just immediate neighborhoods as suggesed by Joel David Hamkins.

Of course, this is all quite speculative; I've not done any work where this notion was wanted.

1535595Wilem2219032618672541481008You should look at Positivity and canonical bases in rank 2 cluster algebras of finite and affine types by Sherman and Zelevinsky where they are able to construct a canonical basis explicitly for *finte type* and *affine type* (i.e. $bc < 4$ and $bc = 4$ respectively). Your $B$ is a basis for $A$ called the *standard monomial basis*. However it is not canonical as you can choose any four consecutive $x_i, x_{i+1}, x_{i+2}, x_{i + 3}$. Also it does not have the following notion of positivity.

Given $y \in A$ we call $y$ *positive* if the expansion of $y$ as a Laurent polynomial in every cluster $x_i, x_{i+1}$ has only positive coefficients. We want a basis so that the positive elements of $A$ are positive combinations of the basis elements.

For $(b,c) = (2,2)$ consider $z = x_0x_3 - x_1x_2$, then $$z = \frac{x_1^2 + x_2^2 + 1}{x_1 x_2},$$ and by symmetry is also a Laurent polynomial with positive coefficients for any other cluster.

For finite type the canonical basis is all cluster monomials and for affine type the basis is all cluster monomials along with some additional elements.

1354294594489435159831077Let $f:X \to Y$ be a smooth morphism between projective varieties. Suppose $Y$ is a homogeneous space. Under what additional condition on $f$, can we conclude that every fibers of $f$ are isomorphic?

I think the question is vague and probably does not have a unique answer.

I would says that this kind of concern is actually not so different from the "size issue " generally presented by the set of objects, it's just that because this problem only arise under very weak set theoretical foundation this question is not mentioned in books.

In the same way that you don't care whether the "set of object" is actually a set or not, you don't really care whether the "sets of morphisms" are sets or not. what you need is "a notion of object" and a "a notion of morphism" that you can compose, but if you want to devise categories as algebraic structures then it's convenient to say that you have "a set of objects" and "a set of morphisms" but those "sets" of objects and morphisms does not really have to be elements of what you call the category of 'sets'. Although you would have to be careful at some point: certain theorem of category theory require to take limit and co-limit indexed by hom sets and might become false if you are not assuming that "hom-set" are actually sets. (for example the special adjoint theorem)

Now, any category theory book I've ever opened was working in ZFC and hence had no reason to care about this kind of questions. In fact, if we just want to drop the axiom of choice then a good thing to do would be to replace the notion of "functor" by the notion of "anafunctor" (which is a generalization of functor where for each object $X$, '$F(X)$' is well defined only up to canonical isomorphism) for which most theorems of category theory, like the fact that a fully faithfull and essentially surjective functor is an equivalence of category, remains true without the axiom of choice. And I don't know of any book of category theory which follows this point of view.

10258607095864I'm looking for an elegant proof that any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$.

503249049149176539921486321743481714964@Henry Wilton. Thanks. I think one needs to consider also quotients of S3. Apparently there are also some quotients of the 3 -sphere that are circle bundles over $RP^2$. For example, I read in a paper that $S3^/Q$, where $Q=A_3/Z_2$ is the quaternion group of order 8, and $S^3/Z_4$ , the lens space $L(4,1)$, are such circle bundles, although there are no references given for this.1458303First, some terminology: if $G\cong\mathrm{Aut}(\Gamma)$ then $\Gamma$ is often called a GRR (for graphical regular representation). This may help in looking for references. Determining whether a Cayley graph is a GRR given $G$ and $S$ is very difficult in general.

One necessary criterion is the following: let $\mathrm{Aut}(G,S)$ be the group of automorphisms of $G$ that preserve $S$. If $\Gamma$ is a GRR then $\mathrm{Aut}(G,S)=1$.

Sometimes, this is also sufficient, see for example:

-Godsil, C. D.; On the full automorphism group of a graph. Combinatorica 1 (1981), 243–256.

-Li, Cai Heng; Sim, Hyo-Seob; The graphical regular representations of finite metacyclic p-groups. European J. Combin. 21 (2000), 917–925.

In fact, the semidirect product of $G$ and $\mathrm{Aut}(G,S)$ is exactly the normaliser of $G$ in $\mathrm{Aut}(\Gamma)$, so that part is somewhat under control, but the part of $\mathrm{Aut}(\Gamma)$ that does not normalise $G$ is very difficult to deal with in general.

2901537573401148542 Oliver Borchert494258I tried to find Killing vector fields with this property on surfaces of revolution but in the examples that build the liminf was zero. The motivation of this question is to find a vector field X on a complete manifold with finite volume such that $\rm{div} X$ is integrable and $$ \displaystyle\int_{M} (\rm{div} X) d\nu_{g} =0 $$ but the vector field X does not satisfy the hypothesis of the Karp's theorem in On Stokes’ Theorem for noncompact manifolds , 1981.1592890986875203049aidan shafer422833121375317422141626351339657192082) should be formulated in a way to make sense, what's the input? Right now it sounds like "given a non-empty subset of positive integers, can we exhibit an element in this set?"1480887141660The precise question is this: given a irrational $r\in \mathbb{R}$, is it true that $r\mathbb{Z}$ is dense (topologically) in $\mathbb{R}/\mathbb{Z}$?

The reason this came up was actually a teaching moment for Calc II; I wanted to use $sin(n)$ as an example of a bounded divergent sequence, but I was actually not sure (and certainly not sure how to explain to Calc II students) that it doesn't converge to some limit. I suspect the question above is true, which would imply that the example works (among other, more interesting, things).

Edit: The question was originally phrased in terms of transcendentals, but given the answer below, this is overkill and deceptive.

5330877382594503885486016527345168122076591The tangles should have two further properties:

The six ends of the ropes should align with the x, y and z axes. Ropes can also start along one axis and end along another.

The region where the three ropes tangle up should be as simple as possible, but at the same time, the three ropes should be tangled or braided up, so as to keep the tangle of the three ropes bound up together.

What would the simplest tangles look like?

Remark: I have found one simple candidate tangle. Two of the ropes look like the two parentheses ) and ( shifted over each other, with a third rope going vertically through the "hole" created by the first two ropes. This "tangle", however, is not really bound up together. Therefore I pose the above question.

I have found one complex candidate tangle. The three ropes are braided like a girl's hair braid, with a total of six braiding motions.

Are there other possible tangles, maybe with intermediate complexity?

jHow to make a sandwich from just one piece of bread?1370588You can take the Hochschild homology of any algebra or category. For algebras, see the first few sections of Loday's book "Cyclic Homology". There is a simple construction using the standard bar resolution. For categories, there is a similar "bar complex" construction, you can find it for example in Costello's paper on Calabi-Yau categories. For a more abstract approach, see Toën's "The homotopy theory of dg categories..." 379156 Daniel, thank you for feedback. (I'm still puzzled by Grothendieck's contraction-gluing; the result of gluing two PL manifolds along a PL homeomorphism of their boundaries doesn't need any extra information.) There are, of course, things like geometric structures on cell complexes (popular in geometric group theory, see e.g. the Bridson-Haefliger book), harmonic functions on simplicial complexes (see R.Forman's 1989 paper in Topology), combinatorial Gauss-Bonnet formula (see Yu Yan-Lin's 1983 paper in Topology), connections and parallel transport on PL manifolds (see M.A.Penna's 1978 paper... FHow to proof this subject?

Theorem of Narasimhan and Seshadri is a special case of what Carlos Simpson calls *nonabelian Hodge theory* developed by Hitchin and Simpson. This theory was generalized to the characteristic $p$ case in the paper of Ogus and Vologodsky, Nonabelian Hodge Theory in Characteristic p. Hope this helps.

Update: Below, on Alexander's request, is a brief explanation of relation between nonabelian Hodge theory and NS theorem. Consider vector bundles with vanishing 1st and 2nd Chern classes (I will call this condition ($\star$)), then the story in higher dimensions is exactly the same as for complex curves. The detailed explanation is in the pages 12-19 of Simpson's 1992 paper [S1992]. From there, it follows that flat unitary connections correspond exactly to vanishing Higgs fields (subject to ($\star$)). Briefly, every semistable Higgs bundle $E=(V, \bar\partial, \theta)$ has a hermitian YM metric $K$. Define the connection $D_K$ (as in [S1992], page 13), then then $D_K$ is flat (subject to ($\star$), page 17 of [S1992]).

If Higgs field $\theta$ vanishes then $D_K=\partial_K+ \bar\partial$ and, hence, by definition of $\partial_K$, connection $D_K$ preserves the metric $K$. Thus, our bundle reduces to a flat unitary bundle. Conversely, if bundle is flat unitary (with unitary metric denoted $K$) then the associated (multivalued) map $\Phi_K$ defined on page 16 of [S1992], is constant, so it has zero derivative. But its derivative is $0=d\Phi_K=\theta+ \bar\theta$ (here $\theta$ is the Higgs field determined by $K$). Since $\theta, \bar\theta$ have different types, the only way we can have $\theta+ \bar\theta=0$ is that $\theta=0$.

181973019103663800011182143`Iwasawa and $KAK$ Decomposition for Diff$(S^1)$1882954Yes, 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.

719730154886Ricci20442681268696388665Pholochtairze6499891947634~vertex arboricity and chromatic number of triangle-free graphs117113419430531035575Thanks! Just what I was looking for. Skimming through their paper, the examples are for n>=8, right? If so, what can be said about n=3 and n=4 for my simpler question, about fields with equal regulators?758912Distribution of the RKHS norm of the posterior of a Gaussian process1641290320136131337214723437940394@BigM If you remove the condition that the function is positive, then its derivative is a function satisfying my stated condition, so you get only a slight variation of Bernstein's theorem.1383332The abstract of the paper you link to refers to "mixed Weil cohomology theories" -- is their definition not what you want?The book and the references therein were very helpful. Thanks again.656282726582155403 In fpqc's formula the ring O_M(U) must be replaced with the restriction sheaf (O_M)|_U, just like definition of Hom sheaf. Otherwise it doesn't even make sense as a presheaf. As for the cotangent sheaf, a definition is no harder: (Delta_M)*(I/I^2) on M akin to algebraic case. Strictly speaking, modules of differentials in the algebraic sense are not logically relevant at this step since M x M is not built via tensor products (but the theories behave similarly, and for completed stalk calculations for analytic M the link with modules of continuous differentials is very useful)1048277589734850186139039309318Ok for the unramified case. When you say "on any unramified representation, the trace of this operator is zero", should it read "ramified" instead? Also, that would mean that ramified representations always have zero Hecke eigenvalues?303941Homogeneous space for what sort of groups? The elliptic curve iis homogeneous over itself! :-)1945087Suppose $M$ is a complete Riemannian manifold with very large injectivity radius (say larger than $100$) and $\left\lbrace x_i: i \in I\right\rbrace$ is a maximal $1$-separated subset of $M$.

Is diffeomorphism class of $M$ determined by the (possibly infinite) distance matrix $(d(x_i,x_j))_{i,j \in I}$?

Suppose now that we are only given the information of which points $x_i,x_j$ are at distance less than $2$. Is this enough to determine $M$ up to diffeomorphism?

7661511625404For an element $T\in B(F)$, $\sigma(T)=\{0\}$ iff $T=0$, this is because the identity function on $\{0\}$ is the zero function.Just add some other text to the ink, then - e.g. "The book A=B".179222820531516252231544468382951034395834083your terminology is strange; typically, "simple" means that there are no multiple arcs. If we assume that, then the digraphs in question are called tournaments, and are well-studied.2182871481706Given an irreducible solvable equation $P(x)=0$ of prime degree $p>2$ with rational coefficients and $\zeta^p=1$, define the usual *Lagrange resolvents* of the roots $x_i$ as,

$$R_n = \big(x_1+x_2\zeta^n+x_3\zeta^{2n}+\dots+x_p\zeta^{(p-1)n}\big)^p$$

where the $R_n$ are the roots of an equation of deg $p-1$ with rational coefficients.

* Question*: For any prime $p>2$, can half of the resolvents be $R_n = 0$ while the other half yields an

For example, the case $p=3$ is trivial, the *DeMoivre quintic* is for $p=5$, while the septic quadrinomial,

$$7x^7+14x^4+7x^3-1=0\tag{1}$$

can be solved as,

$$x = 7^{-2/7}\,\big(y_1^{1/7}+y_2^{1/7}+\dots+y_6^{1/7}\big)=0.423604\dots$$

and the $y_i$ are the roots of the "*sextic*",

$$y^3(7y^3-7y+1)=0\tag{2}$$

If it can be done for $p=11$ and higher, can one give an example? (P.S. The cubic in (2) looks so familiar.)

648190654711I am aware of that paper. It's an excellent reference but containing only Hasse diagrams of weights. I was actually looking for Hasse diagrams which draw positive roots. That helps me easier to read from.6http://stas.kolenikov.name1005291868241For a pleasant introduction that includes many beautiful pictures, I suggest the book "Indra's Pearls" by Mumford, Series, and Wright. They also give examples of explicit computation of the relevant matrices. Here's a link to the copy at Google books.

6Co-induction understanding34468021468551056257188936415633441229907163363380331106812943645The idea works for closed geodesic convex subsets of $X$ with diameter $\le k<\pi/2$.for positive $a,b$, $\max(a,b)\geq (a+b)/2$. Now reread Thomas' answer, or your reply. 2 is absorbed in C.155853115225331538500user33618The paper you mention has all sorts of interesting goodies, but from a quick skimming it doesn't seem to contain the sort of topological information I am after. It's a helpful starting point though. Thanks!I do not know from what angle you're coming, but "really exist" might mean "exists constructively". In this case you should look at Stefano Berardi, Silvio Valentini: *Krivine's intuitionistic proof of classical completeness (for countable languages)* Ann. Pure Appl. Logic 129(1-3): 93-106 (2004). Even though existence of the usual Tarski models for consistent theories cannot be proved construtively, one can still prove a slightly weaker version of completeness.

It is known that any inner automorphism of a unital $C^{\ast}$-algebra $A$ induces the identity map on $K_{0}(A)$ because unitary equivalence implies Murray-von Neumann equivalence. What is known about the induced map on $K_{1}(A)$?

1247408846912131879255791714571271787335For a closed plane curve $C$, define its sequence of winding numbers to be the sorted list of the winding numbers of each of the distinct regions of the plane demarcated by $C$. For example, this curve (if I've calculated correctly) has sequence $001111223 = 0^2 1^4 2^2 3$.

A winding number sequence must include $0$ for the unbounded region of the plane. I am wondering if there are any other restrictions:

24526Is it at least true that every discretely Lindelof P-space is Lindelof? "Discretely Lindelof" means "the closure of every discrete set is Lindelof". It's not too hard to see that every discretely Lindelof space is linearly Lindelof.20485672026045

. Can any winding sequence of consecutive integers that includes $0$ be realized by some curve $C$?Q

Perhaps it will be useful. https://core.ac.uk/download/pdf/11423881.pdf I did not go into details, my hypothesis is if we take a constant $c< \frac{131}{416}$, then the weakened condition in this article will still be preserved, and moreover $k$ can be limited, i.e. condition $\int_{X}^{X+Y}(\Delta (x+h(x))-\Delta (x))^{k}dx$ is valid for $k< C $ ($C$ - some constant).

Shouldn't there be a regulator factor in the Lichtenbaum conjecture formula (even when F is totally real)?961291484991fQuantum cohomology of isomorphic Poisson varieties167936618244072064484PyRulez2149452175000 I'm still a little confused as to how $O_X$ could be non-finitely presented...It has an obvious finite presentation and if a module over a ring is finitely presented then it is always finitely presented right???12603501472184#Hi, David. This is roughly Robin's comment, just with longer discussions, and, I am afraid, links to my own answers. That's life.

I think you would probably enjoy J.H. Conway, "The Sensual Quadratic Form," especially PSL_2(Z) on pages 27-33.

I always like "The Markoff and Lagrange Spectra" by Thomas W. Cusick and Mary E. Flahive and "Binary Quadratic Forms" by Duncan A. Buell.

On particular issues I think you are raising, answers by me and by Dror in:

Reasons for switching from simple continued fractions for $\sqrt D$ to reduced forms: Upper bound of period length of continued fraction representation of very composite number square root

Numbers (here primes) occurring as "diagonal" coefficients: Primes as the first coefficient of a reduced indefinite quadratic form

Other: Numbers characterized by extremal properties

In my language, I think you are asking mostly about small numbers occurring as diagonal coefficients, just not necessarily prime. For me, such numbers need also to be *primitively* represented by the forms in the cycle (for you I guess it is by $ x^2 - D y^2 ,$ using
$a_0 = \lfloor \sqrt D \rfloor$ your quadratic form is equivalent to the reduced form $\langle 1,\; 2 a_0, \; a_0^2 - D \rangle.$ )

```
phoebus:~/Cplusplus> ./Pell
Input n for Pell
67
0 form 1 16 -3 delta -5
1 form -3 14 6 delta 2
2 form 6 10 -7 delta -1
3 form -7 4 9 delta 1
4 form 9 14 -2 delta -7
5 form -2 14 9 delta 1
6 form 9 4 -7 delta -1
7 form -7 10 6 delta 2
8 form 6 14 -3 delta -5
9 form -3 16 1 delta 16
10 form 1 16 -3
disc 268
Automorph, written on right of Gram matrix:
-1106 -17901
-5967 -96578
Pell automorph
-48842 -399789
-5967 -48842
Pell unit
-48842^2 - 67 * -5967^2 = 1
=========================================
phoebus:~/Cplusplus>
```

Anyway, let me know if you want to see any more worked examples. Or the computer program. It is C++ so the numbers are bounded. Easy enough in other languages, of course.

221780682868816244191681999854676880685110094272391494481180163209461114642051671704kaffeeauf@HJRW, After thinking of it a bit I was not able to figure it out. It can have a different answer in the nonabelian case in which I am interested most.112850517407584382021485963For conjugacy classes of integral matrices there is no straightforward description of what happens, but if you focus on integral matrices with a fixed *irreducible* characteristic polynomial (OK, that won't have an analogue over C[z] very often) then there is a bijection between the conjugacy classes of such matrices and ideal classes in an order in a number field. This is a theorem of Latimer and MacDuffee. See http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/matrixconj.pdf.4792081435556Are you picking $i$ and $j$ or is that supposed to be the determinant of the Hessian?:Grothendieck group of curves|When is the canonical divisor of an algebraic surface smooth?127240512555291145649167619910753915147371907742546350941826@David: I think the answer is negative unless you assume that $Z$ is aspherical (and even in this case I am not sure). Take Whitehead manifold $W$, a contractible 3-manifold which is not homeomorphic to ${\mathbb R}^3$ and let $Z$ be the 2-dimensional skeleton of a triangulation of $W$. I think this would be a counter-example. The reason is that $W$ does not admit an exhaustion by simply-connected compacts, so the same should apply to $Z$. 18949114068731897835This function does not verify the conditions stated in the question.67462114758651180875217107215988981196537I see 3 reopen votes, but no related (or any other recent) comment.Given a variety V and a locally free (coherent) sheaf $\mathcal{F}$ of rank 1 (equivalently a line bundle $L$), I can do a Cech cohomology on it. Then $H^0(V; \mathcal{F})$ are just global sections. Is there a similarly understandable meaning to elements of $H^1(V; \mathcal{F})$?

Thanks!

19990142148258The multiplicativity of $\tau$ was conjectured by Ramanujan, and eventually proved by Mordell using the operators he discovered (now called Hecke operators). This is not hard if you consider this technology "well-known", but required non-trivial innovation from Mordell.Thank you for your comments. But I still do not know how to obtain $M$ even for the the example. All I have is the congruence conditions from the fact that $f$ is isotropic. @grghxy, could you please be more specific about your reference to Gauss' Disquisitiones?48283510302011132698Yeah, the arithmetic-geometric mean inequality you mentioned is clear. As to the independent case, it boils down to showing the inequality for just one random variable (because the expectation of the product is the product of expectations then). But then it reduces to the observation that any $C^2$ function $F$ defined on $(-a,+\infty)$ for some $a>0$ that is $1$ at $0$ and grows not faster than $e^{Ct^2}$ at $+\infty$ satisfies $F(t)\le e^{tF'(0)+Ct^2}$ for $t>0$ with some $C>0$.@guest Regarding this construction, see Section VIII of https://www.laurentlafforgue.org/math/NoriMotivesInformation.pdf1059746885514726721993321735811thank you Michael and Sux, Actually modular game and fixed point theorem on lattices are usefull however I am looking for non-lattice set that has fixed point.2950661783782I've heard of this paper, but not read it. What is the answer ;-) ?DChapter V, section 1 of Vigneras?I have a simple follow up question, which you may already know the answer to - is there a model where $2^\kappa$ is a limit cardinal $\iff$ $\kappa$ is a limit cardinal.1884572841038142708995558202301206167542254725536579118480622059376211367Odessa, Texas129938321918411435991007320On further thought, in reaction to comments posted by Tyson, I think the idea of learning differential geometry in a coordinate-free framework from the start is misguided. My advice now is to learn everything from a good textbook, even if it uses coordinates or at least orthonormal bases of vector fields ("moving frames"). Once you understand it thoroughly from this point of view, making the transition to a coordinate-free framework is straightforward.1843381584638294692343927970991629041439349307919Thanks for your answer. I actually need both. Generating all digits and finding nth digit.149261518764901889199Given a matrix $A$, each element $A_{i,j} \geq 0$, find the vector $\vec x$ that maximizes the minimum element in $\vec b$ ($\vec b = A \vec x$). Note that this is not a linear equation system as I don't know $\vec b$.

Extra contraints on the solution are $x_i \geq 0$, and $\sum x_i = 1$.

Is this possible to solve, and if so, how? Can it have 0 or more than one solution?

`Intuition for holomorphic bisectional curvature13195695614582071624595274349510721180421833264704539bWhat is the fundamental group of a hypersurface?:http://mathcs.slu.edu/~clair@Timkinsella: Dear Tim, If you decide to move this quetion to MO, don't cross-post; rather, flag for moderator attention and ask them about migrating your question. Regards,4499064Galois twist of a variety29652In my proof that mapping class groups are automatic, Ann. of Math. (2) 142 (1995), no. 2, 303–384, I used a theorem from ECHLPT "Word Processing in Groups" which says that if a groupoid is automatic then the corresponding group is automatic.

That theorem was applied in the situation of a finite type surface $S$ with one or more punctures, using the groupoid mentioned in Bruno Martelli's answer which has come to be called the "Ptolemy groupoid" of $S$, due to connections with work of Robert Penner. That groupoid needs to be altered slightly for purposes of my proof, by adding data which breaks the finite symmetry group of an ideal triangulation. The data I added was an enumeration of the prongs of the triangulation, so the objects of the resulting groupoid are "ideal triangulations with enumerated prongs". The generating morphisms of this groupoid are of two types: permutations of the enumeration; and the flip relators mentioned by Bruno Martelli, called "elementary moves" in my paper, together with some rule for enumerating the prongs of the new ideal triangulation resulting from the elementary move.

The group corresponding to this groupoid turns out to be the mapping class group of $S$, and hence the theorem from ECHLPT is applicable.

119348813208451529285Why did Dedekind claim that $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$ hadn't been proved before?105254330435971864683175132@MannyReyes Thanks for the reminder. Once upon a time I used to remember Harrison's paper and all that Barr-Gerstenhaber-Schack jazz: http://arxiv.org/abs/0709.3325 (see the list of references)15369411730527@Marguax: thanks for all three of your comments! Your last one helped me understand your first one. I may have one or two further questions later on, but what you've said is already extremely helpful.647194778950henfiber50057847738323571720010821859773295132146248622632592122029287942A reductive group is an algebraic group $G$ over an algebraically closed field such that the unipotent radical of $G$ is trivial 66284744013110781472189804Re. the question of "why not use some nonabelian coefficients like all $G$-spaces", the point for me is that my access to the category of $G$-modules comes via constructible sheaves on a space with fundamental group $G$. Of course, one can take constructible sheaves with nonabelian coefficients. However, I need to use microlocal techniques in sheaf theory, which fundamentally depend on the ability to take cones, shifts, etc. Thus, I need to use coefficients which allow this; ergo, spectra rather than spaces.This is a largely a question of pedagogy/references, though I may have overlooked some nuance of actual mathematics.

I am planning to introduce the concept of Turing machines and the halting problem to utter novices.

The proofs (or rather proof sketches) of undecidability for the halting problem that I have seen fix a universal Turing machine and vary a program and an input that both reside on the TM's tape. In effect an infinite 0-1 array is formed which is indexed by programs and inputs, with entries indicating whether or not the TM supposedly halts on the corresponding program/input pair. One then employs a diagonal argument.

But it seems more concrete to me to ditch the stored program idiom, let the TM vary instead, and use a TM/input pair for the diagonal argument. It is very easy to produce an explicit enumeration of (binary) TMs, and far less so to detail an encoding scheme for a universal TM's tape that allows one to sensibly distinguish the program from the input (of course I know this *can* be done, but an existence proof is not a great primary approach for very inexperienced students, I think).

So, are there any references that discuss a proof of the undecidability of the halting problem along these lines? (A little part of me actually wonders whether or not I have missed some trivia that obstruct such a proof, because I haven't seen one.)

A website ( http://www.math.unicaen.fr/~nitaj/abc.html#Consequences ) says that the $abc$ conjecture implies that there are only **finitely** many solutions to the equation $x^n+y^n=z^n$ with $\gcd(x,y,z)=1$ and $n\ge 4$. This one I have proven.

Lang's Algebra (p. 196) says that the $abc$ conjecture implies that for all $n$ sufficiently large, there are **no** solutions to the equation $x^n+y^n=z^n$ with $x,y,z\ne 0$. This one I have not proven (is it true?).

What are the strongest known assertions about Fermat's Last Theorem that follows from the $abc$ conjecture, and how are they proven? I searched the web but people tend to just say vague things like "implies asymptotic version of FLT" and such.

1012324728324118139993859014149792204663681867@Sally: Will isn't claiming that the kernel is $(x^p - 1)$. He only needs that the map is surjective, from which it follows that $R$ is a quotient of $\mathbb{F}_2[x]/(x^p - 1)$, and then he classifies all such quotients.499942t@IgorRivin : I have edited and explained in more details.For instance, in the setting of the logarithmic potential theory, what you "measure" are complex moments, aka harmonic moments. You can find more references e.g. here.

I hope you don't mind I edited your answer! There's a "revert" link if you don't like it. I plan to rewrite some parts a bit more as I understand them better, is it ok?1817422Suppose $A$ is a real $n\times n$ matrix with real eigenvalues: $$ 1=\lambda_1>|\lambda_2|\ge \ldots\ge |\lambda_n|>0. $$ Suppose $B$ is an involution, for simplicity let us assume that $B$ is diagonal, has $k$ ones on the diagonal, and $n-k$ minus ones. Suppose I also know that $AB$ has an eigenvector $ABx=x$. Thus I know that $$ ABx=x,\quad Ay=y $$ for some $y$.

Can I conclude that $A(I+B)/2$ has eigenvalue $1$ as well, ie. $$ A\cdot \frac {I+B}{2} z=z $$ for some $z$?

@JoshuaGrochow I don't think I do know a published reference. I could try to write details here if you really need a reference.Nisogeny clases of CM abelian varieties1319632123345888935Oh no, I meant just the germ of an irreducible one dim. family, i.e. (C1,0). So, no global complicationsDeane, if you have time, could you check out the constant in front the $f$? Basically in the end I get $C^{1-\frac 1p}p^{1-\frac 1p}\left(\int_0^T \left(\int_\Omega |f|^{(np)/(n+2p-2)}\right)^{(n+2p-2)/(np)}\right)^{\frac 1p}$ which doesn't seem to converge thanks to the $p^1$ factor out front. (We may as well asume $f \in L^\infty(0,T;L^2)$.) Here $C$ is the constant from the Sobolev inequality, and the $p$ coefficient terms come in when using Young's inequality with epsilon.185243642852821023371494859I've read somewhere that every monoidal category with a well behaved diagonal is allready cartesian monoidal. And in this case 'somewhere' is the nLab: http://ncatlab.org/nlab/show/cartesian+monoidal+categoryTopologyLet $k>7$, $B=1.5^k$, $A=\lfloor B \rfloor $ and $C=A+1=\lceil B \rceil$ then $A \lt B-0.75^k \lt B \lt B+0.75^k \lt C $ is an **open question** occuring in the Waring problem. An "inner" inequality $ B-0.75^k \lt {3^k+1\over 2^k+1} \lt {3^k-1\over 2^k-1} \lt B+0.75^k$ can be shown by some algebraical reworking. That ${3^k-1\over 2^k-1} \lt C$ might be a consequence of the Steiner-proof for the 1-cycle problem in the Collatz-problem but I don't remember exactly from the top of my head. All those are compatible with your second *floor* inequality.Have a look at the knot program by Kodama:

http://www.math.kobe-u.ac.jp/~kodama/knot.html

You can draw and manipulate knot diagrams using the mouse and there is an option to export the figures to pstricks. It also comes with all the prime knots of up to 10 crossings.

(1) I think it's better to overcount and explicitly prune, that is, to add an extra instance to construction 2, the same as the $D_4$ invariants, and to add an extra instance to construction 3, which appears in construction 1; (2) you can identify these seven examples with the seven reflection groups in $GL_2(\mathbb Z)$ by saying that the ring of invariants is a localization of the lambda-ring. But that isn't a construction, so you still need all three constructions.1391966852919@Wolfgang: The thing with odd powers is it is better to move them all to one side. Thus I prefer $\sum\limits^8 u_i^k = 0$ versus $\sum\limits^4 x_i^k = \sum\limits^4 y_i^k$. The latter can be mis-leading. For example, one of the seven in the table actually is a $\sum\limits^3 x_i^k = \sum\limits^5 y_i^k$ in positive terms.564666535862119680124250857739919328071727426I do not know why you think Cech cohomology stopped being interesting! For one thing, every time you construct something locally you are in fact doing it à la Cech. Also, Cech cohomology (or the Cech `idea') is one of the reasons cohomologies are computable... Combinatorics-the maximum number of subsets with a given property1341828130113219023012885308609162231856iwillnotOne can use the lemma: Let $R$ be a ring finitely generated over $\mathbb Z$. Then a maximal ideal of $R[x]$ is a maximal ideal of $R$ plus a polynomial in $R[x]$ that is irreducible modulo that maximal ideal. Apply to $R=\mathbb Z$, then to $R=\mathbb Z[x]$, etc. Proof: The quotient is some finitely generated field, thus some finite field, so the image of $R$ is a subfield, thus the quotient by a maximal ideal $I$. The field extension is generated by $x$, so it is the quotient of $F[x]$ by the minimal polynomial $m_x$, so it is $R[x]/(I,m_x)$.Yes, this is a simple application of the Chebotarev density theorem.7241981095213It's probably very standard, but what's a good reference for your first sentence?1609880>Residues of Zeta-like FunctionDrJokepuLet me give you the inner forms of $\mathrm{GSp}(4)$ with compact adelic quotient.

Every nonsplit inner form of $\mathrm{GSp}(4)$ is obtained in the following way: take $D$ a division quaternion algebra over $F$, let $V$ be a $D$-Hermitian space of $D$-dimension $2$, and construct $G = \mathrm{GU}(V)$.

By a theorem of Borel and Harish-Chandra, the adelic quotient of $G$ is compact if and only if $G$ modulo its center is anisotropic (in the algebraic group sense: it contains no nontrivial split torus).

This turns out to be equivalent to $V$ being anisotropic (in the quadratic/hermitian form sense: it has no nonzero isotropic vector). Over a number field, by the local-global principle for quadratic forms we can test anisotropy locally: over $p$-adic places, $V$ has $8$ variables as a quadratic form and is therefore isotropic, and over real places, $V$ is anisotropic if and only if it is positive definite or negative definite.

In summary, $G = \mathrm{GU}(V)$ has compact adelic quotient if and only if $F$ admits a real place at which $D$ is definite and $V$ is positive definite or negative definite.

For instance, the $D$-Hermitian form $x\bar{x}+y\bar{y}$ works over a totally definite quaternion algebra, but the one you wrote down is $x\bar{y}+y\bar{x}$.

8086842066768Ali Enayat7761093253614536911997223Existence of connections in a vector bundle whose parallel transport preserves a function on a total space756231915573Let me address just Q2. The set of points at distance $D$ from a point $p_1$ in $C_1$ form an annulus
centered on the center of $C_1$,
of outer radius $D+r_1$ and inner radius $\max \{ D-r_1, 0 \}$. So you just intersect this annulus with $C_2$,
obtaining (in general) zero, one, or two arcs of $C_2$
for the locus of $p_2$ points that are distance $D$
from some point $p_1$.
**Edit**. See Sergei's comment below; I ~~likely~~ misinterpreted "p1 \in C1," which I read
as "in" rather than $\in$.

In Colmez's work on the p-adic local Langlands correspondence for ${\rm GL}_2(\mathbb{Q}_p)$, he works with ${\rm GL}_2(\mathbb{Q}_p)$-representations on $p$-adic Banach spaces which admit an invariant norm, so the reduction modulo $p$ makes sense. To each irreducible admissible representation of this kind (let's call these "unitary" Banach representations), he attaches a rank 2 $(\varphi, \Gamma)$-module, and hence a 2-dimensional p-adic representation of ${\rm Gal}(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$.

The unitary condition is quite strict -- it rules out all nontrivial finite-dimensional algebraic representations of ${\rm GL}_2$, for instance. Is there any natural way to extend the correspondence to *non-unitary* admissible Banach space representations of ${\rm GL}_2(\mathbb{Q}_p)$, and what sort of Galois-theoretic objects would these match up with?

I guess you have seen sheaf cohomology as being the right derived functor of the global section functor, taking a sheaf $\mathcal{F}$ on a space $X$ to the abelian group $\Gamma(X,\mathcal{F})$. Suppose $X$ is a $k$-scheme, where $k$ is any field, with structural morphism $f:X\to\mathrm{Spec}(k)$. Then you can consider, on $\mathrm{Spec}(k)$, the sheaf $f_* \mathcal{F}$. Since sheaves on the spectrum of a field are not terribly sexy, you see that this guy is defined by its global sections, which by definition coincide with global sections of $\mathcal{F}$ over $X$: in other words, the functor $\Gamma(X,-)$ "coincides" with the functor $f_*$ (the reason for my quotes is that the first functor takes values in **Ab** while the second takes values in **Sh($\mathrm{Spec}(k)$)** but you can figure out the point, I guess).

Then, in general, given any map of schemes $f:X\to Y$ you can define for any sheaf $\mathcal{F}$ on $X$ its direct image $f_* \mathcal{F}$ getting a functor from **Sh($X$)** to **Sh($Y$)** who is left exact. Its right derived functors $R^if_*$ now produce sheaves on $Y$ and the $R^if_*\mathcal{F}$ can be thought of as the relative cohomology of $\mathcal{F}$, precisely as before. This is indeed done in Hartshorne, see Section 8 of chapter $III$ and self-references therein. You can also find something on this point of view in Weibel's *Homological Algebra*. Note that what I have said above does not need $f$ to really be a map between schemes, it works in a more general setting once you have a formalism taking "sheaves over somebody to sheaves over somebody else" – and this is the starting point of many cohomology theories you might encounter, like étale cohomology.

Although the answer is provided by Igor's pointers to the *Triangulations* book, it might be useful
to supplement those pointers with the explicit bounds.
The lower bound is due to Gil Kalai, and the upper bound to Tamal Dey.
For fixed dimension $d$, the cyclic polytope has at least $\Omega( 2^{n^{ \lfloor d/2 \rfloor }})$
triangulations, and for $d$ odd, at most $2^{ O( n^ {\lceil d/2 \rceil} ) }$ triangulations.
So, for $d=3$, the case posed in the question, the bounds are between $c^n$ and $c^{n^2}$.
See Section 8.4 (pp. 396-398) of *Triangulations*.

The naming comes from elliptic surfaces. http://en.wikipedia.org/wiki/Elliptic_surface

My recollection, III and III* are quadratic twists $v(\Delta)=3,9$,

as are II and IV* $v(\Delta)=2,8$ and are IV and II* $v(\Delta)=4,10$.

Note that $v(\Delta)=2,3,4$ for II,III,IV, and it goes up/down by 6 when twisting.

I think everything you have said is correct. The idea with "exceptional" good reduction, is that you acquire good reduction after a field extension, unlike [potentially] multiplicative case. The field to do this is easy for $p\ge 5$, but harder for primes above 2,3. See the paper of Kraus (in French, abstract in English). You can work locally, or also get a global field if wanted.

http://link.springer.com/article/10.1007%2FBF02567933

20990692031029I disagree with your comment about higher sheaves not being motivated well. As soon as you start talking about functor of points and moduli problems, stacks / groupoid valued functors pop up pretty fastMy question refers to the paper http://arxiv.org/pdf/alg-geom/9710020.pdf where Batyrev proves that birational Calabi-Yau algebraic varieties have equal Betti numbers by counting points over finite fields using p-adic integration and so computes the Betti numbers using the Weil conjectures.

It seems that he is doing the following. Given a variety $X$ over $\mathbb{C}$, we can actually write it (and all the associated data we care about) as a variety $\mathcal{X}$ over $\mathcal{R}$, some finite-type $\mathbb{Z}$-algebra: i.e. such that $\mathcal{X} \otimes_\mathcal{R} \mathbb{C} = X$. We then fix an approriate maximal ideal $J(\pi)$ of $\mathcal{R}$ which lies above $p \in \mathbb{Z}$. I think we then turn our attention to the variety $\mathcal{X}\otimes_\mathcal{R} (\mathcal{R}/J(\pi))$, and using a ring of integers $R$ of a local number field with this as special fibre, we can count the number of points this variety has over every finite field extension of $\mathbb{F}_q = \mathcal{R}/J(\pi)$ using p-adic integration.

So the Weil conjectures give us the Betti numbers of this variety, and by proper smooth base change these Betti numbers are the same as those of $\mathcal{X} \otimes_\mathcal{R} \mathbb{C}$, but where the map $\mathcal{R} \rightarrow \mathbb{C}$ is not the natural inclusion but rather $\mathcal{R} \rightarrow R \hookrightarrow \mathbb{C}$. Since he is trying to compute the cohomology of the former, this doesn't make sense to me.

Can anyone see if I'm making a mistake somewhere? (or how my issue can be resolved?)

Thanks, Tom.

1929331109568145860855436Lake Como865845841974RUnsurprisingly, Chris is faster than me.In $GL(2n,\mathbb{R})$, if you intersect $Sp(2n,\mathbb{R})$ with $O(2n)$ you'll get $U(n)$. If you intersect $Sp(2n,\mathbb{R})$ with some conjugate $g O(2n)g^{-1}$ (i.e., the orthogonal group of an arbitrary inner product on $\mathbb{R}^{2n})$ you'll get ... nothing that deserves a name besides $g O(2n)g^{-1}\cap Sp(2n,\mathbb{R})$, in general.1146329174550Take all planes of facets of your polyhedron. They divide the space onto convex sets. Each of them either is contained to your polyhedron, or does not meet it at all. Take those which are contained.

1556375The matrix $M = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ need not be in $\operatorname{GL}_2(\mathbb{Z})$. For example, take the quartic form

$$\displaystyle F(x,y) = x^4 + 3x^3 y + 5x^2 y^2 - 21xy^3 + 49y^4.$$

One checks that $F$ is fixed by the matrix given by $\frac{1}{\sqrt{7}} \left(\begin{smallmatrix} 0 & 7 \\ -1 & 0 \end{smallmatrix} \right)$ under substitution, and this implies that the roots of $F(x,1)$ are permuted by the Mobius transformation

$$\displaystyle x \mapsto \frac{7}{-x}.$$

If you assume a priori that $M = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) \in \operatorname{SL}_2(\mathbb{Z})$ and $F$ is irreducible, then $M$ must be an automorphism. Indeed, since the Galois group of $F(x,1)$ acts transitively on the roots of $F(x,1)$ since $F$ is assumed to be irreducible, it follows that you can apply Galois action on $\beta, \alpha$ to obtain an equation of the shape

$$\displaystyle \theta_1 = \frac{a \theta_2 + b}{c \theta_2 + d}$$

for any root $\theta_1$ of $F(x,1)$, so that all of the roots of $F(x,1)$ are related to another root by the same Mobius transformation (note that the Galois group of $F$ acts trivially on $a,b,c,d$, since we assumed that these are rational integers). Thus the Mobius transformation given by the matrix $M = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ permutes the roots of $F(x,1)$. Since $M \in \operatorname{SL}_2(\mathbb{Z})$, it is discriminant preserving and one can check that it fixes the leading coefficient of $F(x,y)$ as well, and since it permutes the roots and fixes the leading coefficient, it must be an automorphism.

22749121490084Suppose $X$ is a projective, connected, nodal curve (can be reducible) over an algebraically closed field $k$ of arbitrary characteristic. Let $F$ be a pure sheaf on $X$ and denote by $\pi^{*}(F)$ its pullback to the normalization of $X$. Suppose $\pi^{*}(F)$ is Gieseker semistable, then is $F$ Gieseker semistable?

Is it true when $X$ is irreducible?

4241435291Is it possible to have an even probability measure $\mu$ (that is $\mu(A)=\mu(-A)$ for any set $A\subset \mathbb{R}^d$) supported on the unit sphere $S^{d-1}$ such that its Fourier Transform $$ \widehat{\mu}(\xi) = \int_{S^{d-1}} e^{-2\pi i x\cdot \xi} d\mu(x) $$ is non-negative, that is $\widehat{\mu}(\xi)\geq 0$. If not, then how small $\left|\inf_{\xi \in \mathbb{R}^d} \{\widehat{\mu}(\xi)\}\right|$ can be?

650208955122157769248460881001OK I see. It probably does what I said in my previous comment...39449316295031929658463540@L.Guetta First question: no, I mean that, for each $n$, in degree $n+1$ the chain map $\alpha$ has component $\alpha_{n+1}$ that maps into the summand $(\sigma_{\leq n})[1]$ of $Y$. This makes sense as a chain map, since in each degree it maps into only one summand, but is not null homotopic, even though each $\alpha_n$ is, since a chain homotopy would have to involve a non-zero map in degree zero into every summand of $Y$, but a map $R\to\oplus_nR$ can only map into a finite subsum.2287499551516Lattice in the demotic sense --- in type A it's half a square lattice.22433199460901565090(Consider two doubly-connected open subsets $A$ and $A'$ of the Riemann sphere. We assume these two domains to be of same modulus (the moduli space being one real parameter), i.e. we assume that there exists a holomorphic bijection $\phi:A\rightarrow A'$. Note that the map $\phi$ is then unique up to precomposition by the automorphisms of the annulus $A$, loosely speaking the choice of whether we switch the two boundary components, plus a rotation.

**Question**

Is it possible to factor $\phi$ as a composition of holomorphic maps $\phi=f_1\circ g_1 \circ \cdots \circ f_n \circ g_n$, where the maps $f_i$ and $g_i$ are defined on simply-connected domains?

If yes, can we find such a factorization with $n=1$?

**Motivation: case $\phi=f_1$.**

To fix ideas, suppose the annuli $A$ and $A'$ separate $0$ and $\infty$, and that the map $\phi$ sends outer boundary to outer boundary. Assume that the map $\phi$ can be extended to the inside of $A$, i.e. $\phi$ is the restriction to $A$ of a holomorphic function defined on a simply-connected domain containing $0$ and bounded by the outer boundary of $A$. I claim that this gives no information on the outer boundary of the domain $A'$. However, once we fix the outer boundary of $A'$, its inner boundary has to live in a (small) family of three real parameters. Inded, fix the outer boundary of $A'$ to what you wish, and see by Riemann uniformization that the possible maps $\phi$ form a three real parameters family.

Thinking in terms of degrees of freedom, it seems it could be possible, for any couple of annular domains $A$ and $A'$, to write $\phi$ as a composition $f\circ g$, where $f$ extends to the inside, and $g$ extends to the outside: either one of these maps $f$ and $g$ allows complete freedom on either the exterior or the interior boundary of its image.

**Note: uniqueness if $n=1$.**

In the $n=1$ case, the factorization, if it exists is unique up to Moebius transformation.

Indeed, suppose $\phi = f\circ g = \tilde f \circ \tilde g$. Let us consider the holomorphic function $\tilde f^{-1} \circ f = \tilde g \circ g^{-1}$, a priori a holomorphic bijection $g(A)\rightarrow \tilde g(A)$. However, $\tilde f^{-1} \circ f$ extends inside $g(A)$ i.e. it is actually defined on a simply-connected domain bounded by the outer boundary of $g(A)$. Whereas $\tilde g \circ g^{-1}$ extends outside of $g(A)$. The function $\tilde f^{-1} \circ f$ can thus be extended to the whole sphere, and hence is a Moebius transformation.

73413635279537337631507317Nicolas, yes you are right. I misunderstood Ben's answer and the comment following it. I have deleted my incorrect "contribution" to the conversation. So the answer to your question, combining Ben's first sentence, with Mariano's answer is, in fact, yes?55440801610106426Also I think $G$ and $H$ must be reductive groups, with $H$ closed subgroup of $G$ (so that $G/H$ is a reductive space means $(G,H)$ is a reductive pair).Rather than suggest specific topics, which must be your personal choice, let me recommend three collections as possible sources:

(1) Brian Hayes' collection, *Group Theory in the Bedroom and Other Mathematical Diversions*. Every chapter is quite good,
and the title chapter is a gem.

(2) *The Best Writing on Mathematics 2010* and *2011*. The Forwards (by William Thurston
and Freeman Dyson respectively) are already worth the price of admission. Thurston:
"I have decided that daydreaming is not a bug but a feature." :-)

(3) The AMS series, *What's Happening in the Mathematical Sciences*, with many articles by Barry Cipra and Dana Mackenzie (some reprinted in *The Best Writing*).

There's unpublished work by Gay-Balmaz and Pavlov, *Variational Discretization of Compressible Fluids*, described here, with an instructive summary of the difficulties involved in extending the discrete theory from incompressible to compressible fluids.

Consider the following pair of principal bundle descriptions of $\mathbb{CP}^2$: $$ \mathbb{CP}^2 \simeq SU(3)/U(2) \simeq S^5/U(1). $$ If I have a principal $U(2)$-bundle connection for $\mathbb{CP}^2$, will that correspond to a principal $U(1)$-bundle connection? (Where by correspond I suppose I mean give the same covariant derivative.)

I am also interested in how this generalises to the $n$-case $$ \mathbb{CP}^{n} \simeq SU(n+1)/U(n) \simeq S^{2n+1}/U(1). $$

It seems one can obtain the required estimate by a straightforward application of Dynkin's formula. In particular, Dynkin's formula allows you to expand $\partial_x P_h f(x)$ about $h=0$, $$ \partial_x P_h f(x) = \partial_x f(x) + \int_0^h \partial_x \mathbb{E}_x L f(X_s) ds $$ where $L$ is the infinitesimal generator of $X$. You can iterate this formula to obtain a better rate of convergence, meaning reapply it to $Lf(x)$ in place of $f(x)$.

1148668103952mathFB2015251986Jin Xiao1874091723051296395112362701207450As Andrew says, it is not possible to construct such an operator. However, if you have some additional information (closability, symmetry, etc.), then you get automatic continuity.Thank you for your contribution. PLease note however that this software was already mentioned http://mathoverflow.net/a/161805/1893896Thanks for looking that up, but it only deals with Minkowsky (http://en.wikipedia.org/wiki/M-matrix) matrices!11490477812411114252172599491381713901203585111904864"Immediately less good than normal schemes" probably does not make much sense, since "goodness" depends on what you want to do.

For instance, often one likes normal varieties since it is possible to define on them a canonical (Weil) divisor $K$. This does not depend really on normality, but only on the fact that any normal variety $X$ has a singular locus of codimension at least $2$: one considers the canonical divisor on the smooth locus $X^{0}$and then push it forward on $X$.

Thus, one of the (many) possible answers to your question could be "the class of schemes whose singular locus has codimension at least $2$".

The two conditions "normal" and "singular locus of codimension at least $2$" are equivalent for hypersurfaces of $\mathbb{P}^n$, but in general the second is weaker.

For instence, the union of two $2$-planes in $\mathbb{P}^4$ intersecting in a single point is not normal, as follows immediately from Zariski's Main Theorem.

10118731138383574200660557I have a perpetual open invitation to do an AIME with me while running a half marathon or similarly long race. runbobby@mit.edu

Have you looked at Andy Baker's papers on TAQ? He approaches TAQ by thinking of it as a sort of cellular homology theory (which is an awesome perspective). See the paper by Baker Gilmour and Reinhard first, then the computational approach ones on the arxiv.118256810700758076312461622550531I'd say the answer to the first question is no, by the same Kolmogorov's counterexample quoted by coudy in the linked answer. Since $|S_{n_k}|\to \infty$ in a set $E$ of positive measure, by Severini-Egorov (applied to $1/|S_{n_k}|$) we also have that $|S_{n_k}|$ converges to infinity *uniformly* on a subset of positive measure, thus it is unbounded in $L^1$.

*Any sufficiently complex deterministicity is indistinguishable from stochasticity.*

Postdoc of physics dealing with complex systems, chaos theory, scientific computing, networks, analysis methods, and theoretical ecology. To get an idea what I am doing, take a look at my publications or this journalistic report on my work.

Blackletter enthusiast. Here is an interview with me on my work on Unifraktur Maguntia.

My username is pronounced [ˈvʀ̩ʦl̩ˌpʀm̩ft] and comes from a list of names you should not give to your child, which in turn got it from this guy.

1990046160284914692768roots of analytic functionsLet $V$ and $W$ be topological vector spaces over $\mathbb{F}$ (with $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$), and let $T:V \to W$ be a linear transformation. It is well-known that $T$ is not necessarily continuous. But is $T$ necessarily measurable (with respect to the Borel structures of $V$ and $W$)? Does the answer change when we specifically consider $W=\mathbb{F}$?

800116This is a sort of a follow-up to this question, and especially to Sean Lawton's answer: The book Fundamental Groups of compact Kähler manifolds (which, in my opinion, is one of the best mathematics books on any subject) is fantastic, but is over twenty years old now. Can someone summarize how the state of the art in that field has advanced in the intervening period?

DInverse of matrix-valued function5384322287262142472https://www.gravatar.com/avatar/d3c2d82d7a80f0f8db4f92351b0371e2?s=128&d=identicon&r=PG&f=1Thank you very much, it is clear to me now. Also thank you for mentioning theses books.@Starr: Thank you for the answer. I may be wrong, but I thought local complete intersection curves $X$ in $\mathbb{P}^3$ are Gorenstein, since for any closed point $x \in X$, $\mathcal{O}_{X,x}$ is the quotient of a regular ring by a regular sequence, hence a Gorenstein ring (several texts mention that l.c.i. is Gorenstein without giving proofs, including E. Sernesi's books on Deformation of Algebraic schemes).Lovas and Andai--in the cited reference (Conclusion section)--state that they desire to compute ("a very challenging problem") the volume of the set given in the first formula of my question as a function of $\varepsilon \in [0,1]$. In their Table 2 they give the value of the function at $\varepsilon=1$ for $\mathbb{R}$ as $\frac{2}{3} \pi^2$ and for $\mathbb{C}$ as $\frac{\pi^4}{6}$. The value of $\frac{2}{3} \pi^2$ agrees with that in the (much-viewed) https://mathoverflow.net/questions/1464/euclidean-volume-of-the-unit-ball-of-matrices-under-the-matrix-norm373776138508518608321721122Let G be a word-hyperbolic group with torsion and let ∂G be its boundary. Do there exist criteria that imply that all non-trivial finite order elements of G act fixed-point freely on ∂G?

1722458181597917936692012581Let $R$ be a (not necessarily commutative) ring.

If $M$ is a finitely generated submodule of a projective module $P$, is there a finitely generated projective submodule $P'$ such that

$M \subseteq P' \subseteq P$ ?

Kaplansky's "Projective modules" (Ann. of Math. Vol. 68, No. 2, pp. 372-377), Section 5, Lemma 3, answers this affirmatively in the situation $R$ commutative semi-hereditary. A few more positive results are known, e.g. work of Bass, but also negative ones and it appears in general one should expect this to fail.

Nonetheless, I wonder is there a clear criterion / philosophy / guideline what to expect for a given ring, e.g., are there sufficient criteria for the answer being negative?

fLinks between tight closure and deformation theory666557334573compare the two linearly independent solutions of Gauss' hypergeometric equation https://dlmf.nist.gov/15.10.E2 . They seemingly belong to different hypergeometric ODE's.20026146852691589371853218type theory that does not treat the terms of $\mathrm{Prop}$ as types33628382070511185811246299594338(Above comments refer to the question as it was before being edited)20310751314001174407As mathwonk said, Friedman and Smith has what you want although is still over C. For a more general result, stated algebraically and a connection with Serre-Tate deformation theory, see my paper with Coleman, also in Inv. Math. (1992).

138908024142728911313239465036841317754For what it's worth, I see that recent versions of Maple, especially Maple 16 onwards, have expanded abilities in finding real solutions to polynomial systems. See, for instance, the RegularChains package.user706844120046952161204231It is a real shame if the one involving skeleta got its title changed!Santa Ana, CA177193320931121684048The answer is "no" in general.

By the definition of the Selmer group, you can replace the target of the map by the product of $E(K_q)/p^n E(K_q)$. Now $E(K)/p^n E(K)$ is a subgroup of the Selmer group. So if your map were always injective, this would imply that a global point in $E(K)$ is divisible by $p^n$ if and only if it is divisible by $p^n$ in all localisations at finite places. However there are known counter-examples for $K=\mathbb{Q}$ and $p^n$ with $p=2$ and $3$ and $n>1$. The first was found by Dvornicich and Zannier.

On the positive side, one can show that it is injective for $p>2$ and $n=1$ by a lemma of Tate. Also if $K=\mathbb{Q}$, $p>3$ and $Ш(E/\mathbb{Q})$ has no $p$-torsion, then it is injective for all $n$, because the local-global divisibility holds.

In the limit, however, the $p$-primary Selmer group $\varinjlim \operatorname{Sel}_{p^n}(E/\mathbb{Q})$ can only injects into $\prod_{v\nmid \infty} H^1(\mathbb{Q}_v, E[p^{\infty}])$ if the rank of $E(\mathbb{Q})$ is zero or one. This is because the image of this map is contained in $E(\mathbb{Q}_p)\otimes \mathbb{Q}_p/\mathbb{Z}_p$, which is of corank $1$.

2036653164913416387991209915(This question has been killed in the comments, but it is still lacking the useful pointers.)

This is a weak form of the Hardy-Littlewood Conjecture which moreover predicts an asymptotic density for the number of such prime k-tuplets. Special cases of this include: twin primes, cousin primes, sexy primes, prime quadruplets, quintuplets, and sextuplets. While your conjecture is much weaker than Hardy-Littlewood, Kevin Buzzard's trick in the comments shows that it globally implies the infinitude of prime k-tuplets for any admissible pattern.

As far as I know, the infinitude of prime k-tuplets is an open problem for all fixed admissible patterns with k ≥ 2. Note that the Green-Tao Theorem falls short of proving any instance of this since the step size of the arithmetic progressions is not fixed. (Even the Tao-Ziegler Theorem falls short since the polynomials are required to have vanishing constant term.)

1877591072078Kolmogorov’s complexity for positive definite matrices

1471492158844812962001383636324884102692716720495Based on Kolmogorov’s idea, complexity of positive definite matrices with respect to a unit vector is defined. We show that the range of the complexity coincides with the logarithm of its spectrum and the order induced by the complexity is equivalent to the spectral one. This order implies the reversed one induced by the operator entropy.

**A.** To the extent that you think of Brownian motion as a random walk, the natural quantum extension is the *quantum random walk*. For a physics perspective, see Quantum random walks - an introductory overview, but you might prefer the more math-oriented exposition of Martin boundary theory of some quantum random walks and On algebraic and quantum random walks.

We give a concise prescription of the concept of a quantum random walk (QRW), using the example of QRW on integers as paradigm. It briefly explains the notion of quantum coin system and the coin tossing map, and summarizes two emblematic properties of that walk, namely the quadratic enhancement of its diffusion rate due to quantum entanglement between the walker and the entropy increase without majorization effect of its probability distributions. We conclude with a group theoretical scheme of classification of various known QRW's.

**B.** Concerning the relation between Wiener processes and quantum Brownian motion: A quantum version of the wavelet expansion of a Wiener process has been developed in A Levy-Cielsielski expansion for quantum Brownian motion and the construction of quantum Brownian bridges.

Classical Brownian motion has a delightful wavelet expansion obtained by combining the Schauder system with a sequence of i.i.d. standard normals. Our main technical result is to obtain a quantum version of this expansion and so construct quantum Brownian motion in Fock space. Consequently, only the discrete skeleton provided by a "quantum random walk" is required to generate the continuous time process. Our result seems easier to establish than the classical one of Lévy-Cielsielski as we don’t require logarithmic growth estimates on the squares of i.i.d. Gaussians, thanks to the nice action of annihilation operators on exponential vectors.

**C.** Concerning a mathematical description of the *physical phenomenon* of Brownian motion: We are then concerned with the effect of an environment having a large (infinite) number of degrees of freedom on the dynamics of a particle with a few degrees of freedom. So we are seeking a quantum theory of friction, diffusion, and thermalization. The seminal paper here is the path integral theory of Caldeira and Leggett. The literature is very extensive, an older but still relevant review is Quantum Brownian Motion: The Functional Integral Approach.

191975923164137900519032511284191647798I would say that 'topos' means 'elementary topos' rather than 'sheaf topos' by default.16130571164935The quantum mechanical dynamics of a particle coupled to a heat bath is treated by functional integral methods and a generalization of the Feynman-Vernon influence functional is derived. The extended theory describes the time evolution of nonfactorizing initial states and of equilibrium correlation functions. The theory is illuminated through exactly solvable models.

There is also a nice book by Sturmfels called Algorithms in Invariant Theory that discusses the problem of finding invariants. It may be worth a look.

@Wonderful, thank you very much!1648496Arthur RainbowYour argument for the $\sigma$-compactness is very nice and clear. Thank you. Concerning the density of $\varepsilon$-connected components I am still not convinced. Could you provide a more precise argument?6296222134296Very interesting question.If I remember correctly a related problem is given in the book of Herstein, Topics in algebra.402658160225429622110854271347017Regarding ghosts: for each $n$, there are maps $f$ of finite complexes such that $\Sigma ^{2k} f$ induces $0$ on $\pi_*$ for $2k\leq n$.It is true that $f$ restricted to $U$ has no critical values in its image. But I am actually interested in the complement of this Zariski open subset $N \setminus U$, and how its tubular neighborhood grows in volume. IXNThe function $C_l(i) =_{\text{def }} i^{l/i}\Gamma(1 + l/i)$ for large enough $l$ decreases as $i$ increases past 3. This says to me that the values of $i$ to be concerned with in general are at most $4$. For $n > 15$, this should break down into few enough cases to prove that the inequality does hold, or at least holds in all but finitely many cases, even if $c_1$ is allowed to be less than $n/2$.

Gerhard "Not Ready For The Details" Paseman, 2011.12.24

773042NHmm. I should have thought of that :/ 8090031053913117342993109841238184tOf course, thanks! I must have slept upside down tonight.13772601562872694933Frank MorganI haven't thought very hard about this, but it seems to me the answer is the sigma algebra is non trivial. You might want to look at section 2 of http://arxiv.org/abs/1407.5605%If $(x_0, x_1, \dots)$ is a strictly increasing, unbounded sequence of positive real numbers, then there exist fixed $M,N \geq 1$ such that the sequence $(x_0, x_1, \dots)$ contains an ($M,N$)-expander of length $k$ for every $k\in \mathbb{N}$.

If $M,N \geq 1$ are integers, then an $(M, N)$-expander of length $k$ of $(x_0, x_1, \dots)$ is a subsequence $(x_{i[1]},x_{i[2]},\cdots,x_{i[k]})$ of $\mathfrak{X}$ such that $i[j+1]-i[j]\leq M$ for all $1\leq j\leq k-1$ and either \begin{equation} \frac{x_{i[n+1]}-x_{i[n]}}{x_{i[m+1]}-x_{i[m]}}\leq N \textrm{ for all }1\leq m\leq n\leq k-1 \end{equation} or \begin{equation} \frac{x_{i[m+1]}-x_{i[m]}}{x_{i[n+1]}-x_{i[n]}}\leq N \textrm{ for all }1\leq m\leq n\leq k-1 \end{equation}

This is a question that was asked (formulated a little differently) in the following paper, of which I am one of the authors:

‘Relative ranks of Lipschitz mappings on countable discrete metric spaces’, Topology and its Applications 158 (2011) 412-423;

In that sense, it is an open problem. However, as far as I know, this question has been not been widely considered, and so it is not a well-known open problem that is known to be difficult. If, nevertheless, this question is inappropriate for this forum, then I appologise.

If the answer is "yes, it is true", then the results in the the paper mentioned above prove the following conjecture about the semigroup $\mathfrak{L}_{\mathfrak{X}}$ of all Lipschitz functions from a countable subset $\mathfrak{X}$ of $\mathbb{R}$ to itself (where the semigroup operation is composition of functions):

Conjecture:If $\mathfrak{X}$ is any countable subset of the real numbers, then

either $\mathfrak{X}$ contains a Cauchy sequence and there exists a single function from $\mathfrak{X}$ to $\mathfrak{X}$ that together with $\mathfrak{L}_{\mathfrak{X}}$ generates all functions from $\mathfrak{X}$ to $\mathfrak{X}$;

or $\mathfrak{X}$ contains no Cauchy sequences and the least number of functions from $\mathfrak{X}$ to $\mathfrak{X}$ that together with $\mathfrak{L}_{\mathfrak{X}}$ generate all functions from $\mathfrak{X}$ to $\mathfrak{X}$ is the uncountable cardinal $\mathfrak{d}$ (the dominating number).

18421161268978There are a number of different ways to show that a local diffeomorphism $f: X \rightarrow Y$ is a global diffeomorphism:

For example, this follows if $X$ and $f(X)$ are both connected and simply connected. (this appears to be incorrect based on the comments below)

Or if $X$ and $Y$ are both simply connected compact manifolds without boundary.

I think but am not sure that the statement also follows if $X$ and $Y$ are both simply connected manifolds without boundary and the map $f$ is proper.

45200274574410446826885061199718174132912523863193741941723249390748814610962419158Hmm, what do you mean by $\empty$? there is always a 0 in the intersection of two subspaces. For your example, $S_1 \cap T_4 = 0 = S_1 \cap T_3 = 0$, but $M(1,3) = 0$ and $M(1,4) = 1$. I don't see why any of the actions need to be transitive. The left action on $X \times_G Y$ is given by the left action of $G$ on $X$. (I forgot to mention the action in the question; I'll fix this.) Even if the action is not transitive on $X$ or $Y$, that does not a priori mean that it's not transitive on $X \times_G Y$. We could get from $(x, y)$ to $(x', y')$, for instance, by taking $x' = x \cdot g$ for some $g \in G$, whence $(x, y) = (x, g \cdot y)$, and then taking some $g'$ such that $(g \cdot y) \cdot g' = y'$. So it seems that the only (obvious) requirement is that...1869589Re topologically finitely generated -- I'm not sure what to expect! If I knew that there were inertial elements at each prime which generated a dense subgroup, I would just ask that the discrete group they generate has tau with respect to the finite quotients of G, which is a more "customary" definition.2012978914551123729We begin with a lemma. Let $F$ be an increasing function tending to infty as $x\to\infty$. Then $F'(x)\leq F^2(x)$, for $x\not\in E$, where $E$ is a set of finite measure.

Proof. Let $E$ be the set where $F'(x)\geq F^2(x)$. Then $$|E|\leq\int_{E\cap[1,\infty)}\frac{F'}{F^2}dx<\int_1^\infty\frac{F'}{F^2}dx<\infty,$$ by the change of the variable $y=F(x)$.

Now your conditions imply that $g(x)\to\infty$; $g$ and $g'$ are increasing.

Suppose first that $g'(x)\to\infty$. Your relation between $g$ and $f$ means that $g^{\prime\prime}=f$. Applying the lemma first to $g$, and then to $g'$, we obtain the result $\log f(x)=O(\log g(x))$. The case when $g'$ is bounded I leave as an exercise.

87936765216020032312442446194791399760Let $H \subset G$ closed subgroup of an algebraic group. We want to prove the existence of the quotient $G/H$ which is a quasi-projective variety and homogeneous G-space.

We can find a vector $0 \ne y_0 = [v] \in \mathbb{P}^n$ such that $\forall h \in H: h\cdot y_0 = y_0$ and thus $H = \mathrm{Stab}_G(y_0) =G_{y_0}$ (stabilizer) and $G/H =$ the orbit of $y_0$ under $G$.

We define a group action G-morphism $\varphi: G \to Y$ by $g \mapsto g\cdot y_0$. We want to prove that the differential $(d \varphi)_e$ is separable and surjective.

The tangent space of an alebraic group is a derivative module $\mathrm{Der}_k(k[G],k)$ where $D \in \mathrm{Der}_k(k[G],k)$ is a k-linear derivation (i.e. $\forall x \in k: D(x)=Dx = 0$). We also define a derivation $\delta \in \mathrm{Der}_k(k[G],k[G])$ which returns a derivative function. We call $$\mathcal{L}(G) = \left( \{ \delta \in \mathrm{Der}_k(k[G],k[G]) | \delta \mbox{ is left invariant}, \delta \lambda_g(f(x)) = \lambda_g \delta f(x) . \} \right) $$ There is isomorphism between these spaces, $D \mapsto \delta_D$.

According to [Springer, "Linear Algebraic Group", 4.4.7, p.72] we deine $J$ to be the ideal in $k[G]$ of functions that vanish on $H$. We want to prove $$T_eH = ( \{ D \in T_eG | \delta_D I \subseteq I \} ) $$ (this is a set and a tagent space).

After that we want to show that $T_eH = \ker (d \varphi)_e$ and by a theorem it is equivalent that $\varphi$ is separable.

I tried but didn't managed to prove the last two assertions (the two equalities about $T_eH$), so I need some help with this.

2219467lAre these systems of linear equations always solvableThe smooth points on a cuspidal cubic curve in $\mathbb P^2$ can be given the structure of an algebraic group, just as is done for smooth cubic curves. If the tangent directions of the cuspidal point are defined over your ground field $K$, then you get a copy of $\mathbb G_m$. If not, the tangent directions generate a quadratic extension $L/K$, and the group you get is the $L/K$ twist of $\mathbb G_m$. See for example, *The Arithmetic of Elliptic Curves*, Proposition III.2.5 for the split case. As usual, the group law is determined by the condition that three points add to 0 if and only if they are colinear.

If X is not smooth, then it is possible for the Chow groups and the A^1-represented motivic cohomology theory to disagree.

For instance, if we take X to be two copies of A^1 identified at a point then CH^0(X) has rank 2, but the sheaf represented by X in the A^1 category is contractible, so H^0(X,Z(0)) has rank 1.

To see this last point, we consider X as the colimit of a diagram A^1 <- * -> A^1. Since the maps in this diagrams are monomorphisms of schemes, they are cofibrations. The colimit of the diagram is therefore equivalent to the homotopy colimit, which is invariant under pointwise equivalences of diagrams. Since A^1 is contractible, we are left with the colimit of * <- * -> *, a point.

Morally, A^1 is contractible, so we can shrink down the A^1s without changing the A^1 homotopy type.

@Matt F.: The trivial example is uniform on the set of landmarks, but not on the set of all vertices of the graph.552616Oh, now this is a completely different question! I would expect that if your domain has sufficiently narrow (infinite) "horns", you cannot hope for the Feller property. Indeed, the first coordinate of a reflecting BM in a horn $\{(x,y):x>0,|y|The exercise in question is actually a theorem of Kaplansky. It appears as Theorem 5 on page 4 of his *Commutative Rings*. [I was not able to tell easily whether the result appears for the first time in this book.] The proof is reproduced in Section 10 of an expository article I have written [but probably not yet finished] on factorization in integral domains:

http://math.uga.edu/~pete/factorization.pdf

Regarding your second question, there has been some work on understanding Euclidean domains from more intrinsic perspectives. Two fundamental articles are:

Motzkin, Th. The Euclidean algorithm. Bull. Amer. Math. Soc. 55, (1949). 1142--1146.

http://math.uga.edu/~pete/Motzkin49.pdf

Samuel, Pierre About Euclidean rings. J. Algebra 19 1971 282--301.

http://math.uga.edu/~pete/Samuel-Euclidean.pdf

I have not had the chance to digest these papers, so I'm not sure if they directly answer your question (maybe not, but I think they will be helpful).

627797121844Let $A$ be a commutative ring with unity. Denote $X:=Spec(A)$.

- $X$ is connected if and only if $A$ cannot be written as a Cartesian product $A_1\times A_2$ for nontrivial $A_1$ and $A_2$. So, that gives you some description of $A$ (but I don't know if it is any useful to you).

Now, the nilradical of $A$ is prime if and only if $X$ is irreducible (aka. hyperconnected).

- Relation between irreducible and connected: hyperconnected always implies connected for set-theoretical topological reasons. Since you've imposed the Noetherian condition in your edit, for the inverse implication it suffices to additioinally assume that $X$ is smooth, that is, $A$ is a regular ring. I don't know if there is a more general condition that yields the inverse implication.

Jonah's question makes me wonder: What is with uniformization in algebraic/arithmetic geometry? E.g. this article by Faltings seems to be about that, the Shimura-Taniyama statement too, Mochizuki discusses a p-adic version of Fuchsian uniformization of hyperbolic curves. Do you know surveys or expositions of the theme? Or is 'uniformization' a mistaken concept in the context of arithmetic geometry (a remark in Faltings article sounds as if saying that)?

764928150023196721357679215148731785064150812333068146894As for black hole thermodynamics, I've never looked into that but it seems that I should.1215704In an additional complication, the word "champ" (field) in French also refers to the algebraic structure English speakers call a "stack." Worse and worse...I think it's the ring of complex numbers that Adam wants to replace, rather than the domain of the functions.1367715`@JohnPardon Sorry, meant to write affine scheme496179Ok, I see where I made a mistake. Because of simplified notation, I got $0$ at $C_4/e$ level, which is obviously wrong. Thank you!I would have said "the free symmetric monoidal 1-category on a dualizable object", but you seem to think that this isn't quite right? How's your site-specific google-fu? Have you tried searching the TWF archives? See also http://ncatlab.org/nlab/show/cobordism+hypothesis . For the unoriented category, you might say it's "the free symmetric monoidal 1-category on a self-dual object". As you move up the ladder, you'll also have to start thinking about other questions, like framed versus oriented versus ...Explicit data for $E_n$-monoidal model and simplicial categoriesB@FP: Thank you for your answers!Let $X$ be a smooth projective variety and $K_X$ the canonical line bundle. If $K_X$ is nef, then the abundance conjecture predicts that it is semiample, so in particular a multiple $mK_X$ has many sections. If $-K_X$ is nef, then it does not have to be semiample (see examples in reference below). I have the following weaker questions:

**Question 1:** If $-K_X$ nef, does there exist an integer $m>0$ such that $-mK_X$ effective?

**Question 2:** If $-K_X$ nef, does there exist $m$ as above, and $H$ ample divisor, such that $-mK_X|_H$ effective?

Of course, if $-K_X$ is big - in the interior of the effective cone, both questions are true, so the issue is really on the boundary. From the paper ''Nef Reduction and Anticanonical Bundles'' I can extract that Bauer-Peternell claim that the result is true in dimensions $\leq3$: see 1.5-1.7 for surfaces, and the various theorems in the case of threefolds. An idea for a counterexample for Q1 would be to cook up a semistable projective bundle, perhaps over something $K$-trivial, and follow something like the Hartshorne-Mumford construction for non-closed effective cone, but it does not seem clear how to make this work in higher dimension or for $-K_X$. Of course, in higher dimension there is more space for a complicated base locus for $-K_X$, which an ample divisor $H$ could perhaps not avoid. Note that one could also ask Q2 for $K_X$ as a weak version of abundance, but even here I am not sure what is known.

1090562Suppose that $(A,m)$ is a Noetherian local ring, $M$ is an $A$-finite module. Assume that $x_1, ..., x_n$ are elements in $m$. Is the following equality true:

$$ \mbox{ann}(M/(x_1, ..., x_n)M) = (x_1, ..., x_n) + \mbox{ann}(M). $$

133493817913464784544507767172041021401641576231050282080143991949668808I just thought of another possibility, which seems likely after reading Anton's post again. Is $\mathbb{G}_M / \mathbb{Z}$ a scheme? if so what scheme? It seems like it would have to be a single point.8128549908715739841676850787853569373143618694161115429155207941562073Rio de JaneiroI don't know an explicit reference for you, but I can tell you how these things are done and point you to some related reference. These groups are closely related to general linear groups via accidental isomorphism, so one can write down the Langlands parameters in terms of those for general linear groups and this should also give preservation of epsilon and L-factors.

SO(4) is essentially GL(2) $\times$ GL(2). See, e.g., the last section of Gross and Prasad's original paper (Can. J, 1992).

SO(6) is essentially GL(4). Both of the relations are essentially described in Gan and Takeda's paper

*Theta correspondences for GSp(4)*(Rep Thy 2011). (Technically they work with describe slightly related groups, GSO, which contain the SO's.)

During a lecture I gave on Catalan numbers, I pointed out that that it
is possible to give a continuum number of combinatorial
interpretations of these numbers. See the solution to (f$^5$) on
page 54 of http://math.mit.edu/~rstan/ec/catadd.pdf. After the lecture
someone from the audience (I don't know who) asked me if one can give
*more* than a continuum number of combinatorial interpretations. For
instance, for each subset of $\mathbb{R}$ can one give a different
combinatorial interpretation? Can anyone shed some light on this
question?

For the scope of the four color problem and without lack of generality, maps can be represented in different ways. This is generally done to have a different perspective on the problem.

For example, the graph-theoretic representation of maps has become so common and important that generally the four color problem is stated and analyzed directly in terms of graph theory: http://en.wikipedia.org/wiki/Four_color_theorem.

I am trying to collect other representations that may in some way help to get a different point of view on the problem. If you know one of these representations that is not listed and wish to share, report it here. If you also have a web reference that explains or shows the representation, it would be great.

The representations have to be general and applicable to all maps with the simplification that only regular maps (no exclaves or enclaves, 3 edges meeting at each vertex, etc.) can be considered.

These are some classic representations:

- Natural: As a 3-regular planar graph (boundaries = edges)
- Canonical: As the dual graph of the "natural" representation (region = vertex, neighbors = edges)
- As a straight line drawing graph (Fáry's theorem)
- As a graph with vertices arranged on a grid
- As a rectilinear cartogram
- As circle packing

Plus, I found these:

- As a circular map
- As a rectangular map
- As clefs (derived from rectangular maps)
- As pipes map (derived from the clefs representation)
- ...

Here is an example of some of these representations for the original map shown:

And here are other representations after the comments received:

UPDATE: 19/May/2011 - Added other representations of graphs

2071849196051214136795895131338284No, there is no ellipsoid satisfying equivariance and monotonicity.

Suppose we have such an ellipsoid. Consider the square whose corners are $(\pm 1, \pm 1)$. Since this is invariant under a quarter-rotation, so is its ellipsoid, and the ellipsoid must be a circle of radius $r$.

By equivariance, we get the following table of shapes: \begin{matrix} \text{Quadrilateral with Corners at} & \text{Semi-axes of corresponding ellipsoid}\\ (\pm 1, \pm 1) & r,\ r \\ (\pm \cos(t), \pm \sin(t)) & r \cos(t),\ r \sin(t) \\ (\pm \sqrt{2}, 0) \text{ and } (0, \pm \sqrt{2}) & r,\ r \\ (\pm \sec(t), 0) \text{ and } (0, \pm \csc(t)) & r\sec(t)/\sqrt{2},\ r\csc(t)/\sqrt{2} \\ \end{matrix}

But the second quadrilateral is contained in the fourth. So by monotonicity, $r \cos(t) < r \sec(t) / \sqrt{2}$, and $\cos^2(t) < 1/\sqrt{2}$. We can choose $t$ for which this is false, showing the impossibility.

**EDIT** (by Jairo):

I'll illustrate Matt's solution (or a minor variation of it) with a figure.

As Matt explained, $\pi$(square) = circle. Multiplying $\pi(.)$ be a constant if necessary, we can assume that $\pi$(square) = inscribed circle. By equivariance, $\pi$(rectangle) = inscribed ellipse. But then monotonicity is violated: the red rectangle is contained in the blue square, but the red ellipse is **not** contained in the blue circle.

My category theory is a bit rusty, but my copy of Basi Category Theory seems to say that $ (-)\times G $ functor does not preserve the initial object, it cannot have a right adjoint. Actually, it says this for Grp and Abgp, but then proceeds to mention that the same argument holds for Mon (and I see no reason why this should not hold for commutative monoids).

1500212390223200763417073011827051Wilhelm Magnus (W. Magnus, Über $n$-dimensionale Gittertransformationen. Acta Math. 64 (1935), no. 1, 353--367.) seems to have shown that the free metabelian group on $n$ generators has a faithful representation of degree $2$. It follows that there is a copy of the free solvable group on $n$ generators and of derived length $2$ inside $GL_2(\mathbb{C})$.

That answers your question affirmatively in a special case.

9225691433397410939225773113321@PeterDukes: For the version restricted to translations - perhaps so.57557118885803689211212170@L.Spice. http://www.tandfonline.com/doi/abs/10.1142/S1402925110000635241021714615I'm not sure whether that's necessarily the question, but it might be sufficient. The motivation behind my question is to understand if there is a natural map $C_r^*(G)\rightarrow C^*(G)$ that induces a map on the level of $K$-theory. In particular, if $C^*(G)=C_r^*(G)\oplus D$ for some $C^*$-algebra $D$, this implies the $K$-theory of one is a product of the $K$-theories of the other two.732309205545@stupid_question_bot The free abelian group on the right cosets, or equivalently $\Bbb Z \otimes_{\Bbb Z[G]} \Bbb Z[G']$.197404013038191765592Yes @DominicvanderZypen , this has the same flavor as a star-finite covering but quite stronger. Note that if $\varphi$ takes the constant value $X$ then it produces the star-finite covering $\{X\}$ but $\varphi$ does not have property 2 (unless $X$ is finite!).7207391692301355660I have been looking at categories of sup semilattices and sup preserving maps. If $A$ and $B$ are two such, the set I denote $[A,B]$ is sup preserving homomorphisms between is also a sup semilattice with pointwise sup. Even if $A$ and $B$ are lattices, $[A,B]$ might not be. It is, however, if $B$ is complete or $A$ is finite. The question I asking is when is it a distributive lattice, assuming both $A$ and $B$ are. It is if $A$ is finite, but I can get no handle on it otherwise. Does anyone know?

1233979Let $X$ be a smooth projective variety with Picard number 1 over $\mathbb{C}$. Let $F$ be a coherent sheaf on $X$ such that $c_1(F)$ is algebraically trivial, and hence numerically trivial. Also rank $F=0$.So $F$ is supported on a proper closed subscheme of $X$. Can $F$ be supported on a divisor, or is the codimension of Support $ F\geq 2$.

Since the Picard number of $X$ is 1, any non-zero effective divisor is ample. Since $c_1(F)$ is numerically trivial, if $F$ is supported on a divisor, it is non-effective. Is this possible?

Note, I posted this question on MathSE, but have not received any answers.

20547307908461824039How to compute KO characteristic classes/numbers?

They were introduced by Anderson/Brown/Peterson to study the structure of the spin cobordism ring. I looked through the literature but I did not find a nice example of computation. For instance, I would like to know how to determine $\pi^J(\mathbb{H}P^2)$. Many thanks for any comments in advance.

@Victor $\mathfrak d<2^\omega$ is consistent relative to ZFC. It holds, for example, in models obtained by adjoining a lot of random reals to increase the cardinality of the continuum.8294131999744116272953031462887But over the complex numbers your example is not a counterexample anymore, so I wouldn't say that we "need" positive characteristic for that :)1497519883132Let $X$ be a suspension spectra whose $BP$-homology is infinitely generated ($BP_*(X) = \Sigma^d BP_*/I$, where $I$ has the form $I=(v_0^{i_0}, \dots , v_n^{i_n})$ such that the homology is a $BP_*(BP)$ coalgebra).

Let $C_nX$ the fiber of the map $ X \to L_nX$ and let $\Sigma C_n X$ be its cofiber.

What can be said about the natural map $$BP_*(\Sigma C_n X) \to BP_*(\Sigma C_{n-1} X) ? $$

My guess would be that it's always injective, but i'm not entirely sure about it.

Thanks

Thank you for your reply! But I can not really understand the last sentence "(in particular you can do it if your vector field X has a zero, in which case X has a holomorphy potential)". Would you like to give some details? In my opinion, the difficulty of this problem is that we get less information from $g=\triangle f$, in particular, the Hessian of $f$.1535680130213020781633502482718377720861475011548953which one: http://www.springer.com/gb/search?query=Linear+algebraic+groups&submit=Submit ?1160041@Dan, thank you for pointing that out. Of course we shouldn't be taking values in the empty set. Sorry! By the way, had I of used the correct notion of skyscraper sheaf (replace $\emptyset$ with {*}) then, as written, this would have been the constant sheaf sending everything to the terminal object in $Ens$.1468327@OP: There's a way to make all of this function business precise using a construction called the espace étalé. 1703330A question about fibrations of simplicial sets and their fibers1284765Yes, those are great pictures! The view from above would also be good, for thinking about knot types. I suggest experimenting with combinations of things such as ViewPoint -> {1,2,20.}, BoxRatios -> {1,1,.1} or so to see what looks good. I'd also suggest --- now adding an arbitrary moderately high vertical component (maybe ~2 or 3 vertical bounce per horizontal traversal, so the simulation doesn't slow too much) to get something that looks nicely knotted, viewed from above./It is easy to see that the disjoint union $\bigsqcup_i X_i$ of a collection of metric spaces is metrizable, simply by rescaling or chopping off the individual metrics to have diameter at most one, and then making points in the various components distance two apart. This produces a metric on the disjoint union giving the disjoint union topology.

A subtle point arises, however, if we are given not a collection of metric spaces, but instead merely a collection of metrizable topological spaces $X_i$, where a particular metric has not yet been chosen. In this case, in order to carry out the construction above, we would need first to choose a particular metric realizing the metrizability of each space, and this would seem to require the axiom of choice.

**Question.** Without assuming the axiom of choice, when is the
disjoint union of a set of metrizable spaces metrizable?

One can see that some choice is definitely required by the following argument. Suppose that we are given a collection of countable sets $S_i$. Let's regard these sets as discrete topological spaces and let $S_i^+$ be the corresponding one-point compactification, which is metrizable, since each $S_i$ is countable. If we had a metric on the disjoint union $\bigcup S_i^{+}$, however, we could consider the largest ball at the new point in each $S_i^+$ to omit some elements of $S_i$, and this would provide us with a canonical choice of non-empty finite sets $T_i\subset S_i$. But the existence of such a kind of choice surely requires some form of choice. Thus, in general, we cannot prove without any AC that the disjoin union of metrizable spaces is metrizable.

So what I seek is a nice sufficient condition for making this conclusion. Here is one answer: given a family of metrizable spaces $X_i$ suppose there exists a finite family of distinct metrizable spaces $M_j$ such that

(1) each $M_j$ has a metric that makes every topological automorphism of $M_j$ an isometry; and

(2) but for finitely many $i$, each $X_i$ is homeomorphic to some $M_j$.

(Given an $X_i$ and a homeomorphic $M_j$, avoid choice by working
with *all* the homeomorphisms from $X_i$ to $M_j$.)

This condition seems to me far too restrictive. For example one could have infinite sets of such $M_j$s as long as they came pre-equipped with a selection of appropriate metrics.

So what I seek are general conditions on such $M_j$'s (or even for say just rigid metrizable spaces) that permit the selection of a canonical metric based solely on knowledge of the topology.

If the general question is still too vague, I don't even know the answer to this: does a compact, rigid space homeomorphic to some subset of the unit disk in ${\Bbb R}^2$ admit a distinguished metric (perhaps optimal for some objective function)?

Contrarywise, I'd be interested in models where a countable disjoint union of rigid metrizable spaces might have no compatible metric.

48119986140623005164315419768925947271713684fBasis for the Algebraic numbers over the rationals841649This question has a negative answer (given by Gregorz Plebanek), which follows from the following theorem of Gitik and Shelah.

**Theorem (Gitik-Shelah, 1989):** If a set $X$ admits an atomless probability $\sigma$-additive measure $\mu:\mathcal P(X)\to[0,1]$ defined on the algebra of all subsets of $X$, then the Banach space $L_1(\mu)$ has density $>|X|$.

On the other hand, if for some $X\subset\mathbb R$ with $\nu=\mu^*(X)$ the restriction $\mu^*|\mathcal P(X)$ is a measure, then $L_1(\nu)$ is separable (since the $\sigma$-algebra of Borel subsets of $X$ is countably generated), which contradicts Gitik-Shelah Theorem.

A combinatorial proof of Gitik-Shelah Theorem was given in
[A. Kamburelis, *A new proof of the Gitik-Shelah theorem.*
Israel J. Math. 72:3 (1990) 373–380].

Results on the subject seem spotty, but the last of them seems to be:

```
MR3271270 Reviewed
Donald, Andrew(4-GLAS-SMS)
Embedding Seifert manifolds in S4. (English summary)
Trans. Amer. Math. Soc. 367 (2015), no. 1, 559–595.
```

PInteresting. Thanks for the reference.20771341232568220255254729616529851573363417242I suppose you are right. The approach I give shows (inter alia) that if A+B is a complete set of residues mod N=#A#B and both A and B contain 0 (no loss of generality in assuming that) then at least one of the two contains no elements relatively prime to N.1410109432217Local questions:

1) Given a commutative ring $A,$ is $Sh_\infty\left(Spec(A)\right)$ hypercomplete?

2) Given a commutative ring $A,$ is $Sh_\infty\left(Et\left(A\right)\right)$ hypercomplete, where $Et\left(A\right)$ is the small etale site?

3) and 4) The same questions but for $A$ a simplicial commutative ring

5) and 6) The same questions but for $A$ and $E_\infty$-ring spectrum

Or perhaps something general is known about when $Spec^{\mathcal{G}}\left(X\right)$ is hypercomplete, where $\mathcal{G}$ is a geometry in the sense of Lurie's DAG V?

"Large" Global questions:

Ignoring size issues (e.g. by using universes) are infinity sheaves on any of the following hypercomplete?

A) Affine schemes with the Zariski topology

B) Affine schemes with the Etale topology (or flat, etc.)

C) and D) same question for simplicial affine schemes

E) and F) same question for spectral schemes

(I have a feeling that the answer might depend on things being Noetherian)

Any references would be great also, thanks!

Denis, I think you are right. I am looking at my quadratic forms books, it seems Hermite reduction is far more specific thing. I saw your sum as part of that and incorrectly put his name on this general method.In a Polish space that is not locally compact, there may be no nonzero continuous functions with compact support.7869531027096math201418362541301318408265I think I've spotted you -- maybe on the categories mailing list? -- also decrying the (perhaps all too typical?) approaches to fixed point theorems based on transfinite iteration. Is it correct to say, however, that big chunks of theory that you worked out on developing transfinite induction and recursion *constructively*, and that appear in your book, might not have been developed at all had you not missed Pataraia's proof at the time? (Not sure I've phrased all this optimally.)67764710293691661821106317317621991359959937655726481561247Given your two axioms, the relation is determined for all cycles by its value on cycles of the form $(0,1,z)$, on which it should divide $\mathbb C$ into the upper half plane, the real line, and the lower half plane. For $x \in \mathbb R$ you get the order relation on $\mathbb R$ by considering $(0,1,ix)$. So one way to produce such an axiomatization would be to list all the axioms of the real closed field in this way and then the additional axiom $\mathbb C = \mathbb R + \mathbb R i$. This would no doubt be an ugly axiomatization, but at least it shows that more is possible.

But we can use this as a guide by writing reasonable axioms for the orientation and making sure they imply all the real closed field axioms. For instance if $(z_1,z_2,z_3)$ is oriented and $(z_1,z_4,z_2)$ is not oriented, then $(z_1,z_2,z_3+z_4-z_1)$ is oriented. This implies that the reals are closed under subtraction and hence are a subgroup and that the positive reals are closed under addition.

@paul garrett: you are right. I would appreciate any information, topological representations are of course more relevant.1481831349417In version 1, we can prove g is an open mapping. However, g maybe not a closed mapping.The all-ones vector is always an eigenvector of $(P^{-1})^\mathrm{T}\circ P$ with eigenvalue $1$. To see this, note that the $i$th entry of $((P^{-1})^\mathrm{T}\circ P)\mathbf{1}=(P\circ (P^{-1})^\mathrm{T})\mathbf{1}$ is precisely the $(i,i)$th entry of $PP^{-1}=I$ by the definition of matrix multiplication.

**UPDATE:** In hindsight, this is a special case of Theorem DMHP from Dietrich Burde's link.

I want to know is there any interesting properties of this approach or generalization to find $S_k(n)=1^k+2^k+3^k+\cdots+n^k$ by using Pick's Theorem $S=i+\tfrac{b}{2}-1$, where $i$-number of interior points and $b$ - number of boundary points and the following geometric interpretation:

$(\textbf{1})$ - Let consider $S_1(n)$ then for first steps:

\begin{array}{|c|c|c|} \hline n& b_n & i_n \\ \hline 1& 4& 0\\ \hline 2& 8& 0\\ \hline 3& 12& 1\\ \hline 4& 16& 3\\ \hline 5& 20& 6\\ \hline ...& ...& ...\\ \hline n& 4n& \tfrac{(n-1)(n-2)}{2}\\ \hline \end{array}

$(\textbf{2})$ - Let consider $S_2(n)$ then for first steps:

\begin{array}{|c|c|c|} \hline n& b_n & i_n \\ \hline 1& 4& 0\\ \hline 2& 10& 1\\ \hline 3& 18& 6\\ \hline 4& 28& 17\\ \hline 5& 40& 36\\ \hline ...& ...& ...\\ \hline n& n(n+3)& \tfrac{(n-1)(n^2+n-3)}{3}\\ \hline \end{array}

$(\textbf{3})$ - Let consider $S_3(n)$ then for first steps:

\begin{array}{|c|c|c|} \hline n& b_n & i_n \\ \hline 1& 4& 0\\ \hline 2& 14& 3\\ \hline 3& 30& 22\\ \hline 4& 52& 75\\ \hline ...& ...& ...\\ \hline n& n(3n+1)& \tfrac{(n-1)(n+1)(n^2+2n-4)}{4}\\ \hline \end{array} $...$

$(\textbf{k})$ - Let consider $S_k(n)$ :$\textbf{???}$ So, $\textbf {the questions}$ are: How to find general formulas for $b_n$ and $i_n$ for any step $k$ when we depict our sums as follows: $S_k(n)=1^2\cdot1^{k-2}+2^2\cdot2^{k-2}+3^2\cdot3^{k-2}+...+n^2\cdot n^{k-2}$ - so $n^{k-2}$ times we take $n^2$ and add belows the previous squares? And, the second question: are such constructions involved in any math research or have some interesting properties?

Consider a function $f: \{0,1\}^n \to \{0,1\}^{cn}$, where $c>1$.

A random $f$ with high probability generates optimal covering of $\{0,1\}^{cn}$, i. e.:

$\forall x \in \{0,1\}^{cn}$ $\exists y \in Im(f): |x - y| < r + O(\log n)$,

where || - Hamming norm and $r$ is number such that volume of a full-sphere with radius $r$ is equal to $|\{0,1\}^{cn}| : |\{0,1\}^n| = 2^{cn} : 2^n = 2^{(c-1)n}$.

We don't know any $f$ such that can be calculated in polynomial of $n$ time and generates optimal covering.

But what is about Cryptographically secure pseudorandom generator(PG, https://en.wikipedia.org/wiki/Pseudorandom_generator) $f$?

In this case $f$ can be calculated in polynomial of $n$ time. Unfortunately from the definition of PG it is not follows that $f$ generates optimal covering (as in the case of Nisan-Wigderson generator). But anyway if $f$ doesn't generate optimal covering it may be used for some attack on a conjecturally PG function (i. e. for proving that some conjecturally PG are not PG really), I think.

Are there a results about functions that might be PG, but don't generate optimal covering?

202968716750110707321633532121009I was discussing the tree picture of ${\bf Z}_p$ and ${\bf Q}_p$ and mentioned that the idea can be extended to ${\bf C}_p$, with the caveat that the tree is no longer locally finite (as the value group $\bf Q$ is not discrete like ${\bf Z}$ is), and that there is no longer a root. My professor asked a question I hadn't considered: when we look at the extended picture with ${\bf C}_p$, is there a way to distinguish between algebraic and transcendental values simply by how they *look* as leaves of the tree? By extension we can also consider the difference between general elements of ${\bf C}_p$ and (say) the spherical completion $\Omega_p$ (see SBSeminar for a definition).

There might be necessary and sufficient conditions (for e.g. being in ${\bf C}_p\setminus \overline{{\bf Q}_p},\Omega_p\setminus{\bf C}_p$) on the set of exponents of $p$ in a number's $p$-adic expansion. This might be a stretch, though; it's hard to imagine there being conditions on the decimal expansion of a *real* number that determines if it is transcendental.

However, if $x=\sum_{\ell\ge u/v}a(\ell)p^\ell$ and the support of $a(\cdot)$ additively generates a discrete and hence cyclic subgroup of ${\bf Q}$ generated by say $r/s$ then we can write $x$ as a ${\bf Q}_p$-linear combination of the numbers $1,p^{1/s},p^{2/s},\cdots,p^{1-1/s}$ and hence $x\in{\bf Q}_p(p^{1/s})$ is algebraic. Is there a converse?

I have approximately zero familiarity with the field structure of extensions of ${\bf Q}_p$ unfortunately (this discussion occurred in what is an introductory class in the $p$-adic numbers following Gouvea, in fact), so I'd appreciate input from anyone with more background. Also, perhaps highlighting the differences would be easier if we use Teichmüller representatives?

Suppose $x,y \ne 0, 1$. Then by nondegeneracy of the Vandermonde determinant, any remaining point $(w, w^2, w^3, z, z^2, z^3)$ must have $(w, z) \in \{0,1,x\} \times \{0,1,y\}$. If $x = 0, 1$, then in fact $(w,w^2,w^3)$ has to be a scalar multiple of $(1,1,1)$; this only happens $w = 0, 1$. Thus in any case your plane hits $V$ in at most 9 points and in particular contains no curves.1226701178714Stephane Louboutin has worked on related problems for cubic fields; see e.g. Proc. Amer. Math. Soc. 140 (2012), 429-436.It's known to be consistent with ZF+DC that every subset of $\mathbb{R}$ has the Baire property (BP). (E.g. Shelah's model). If so, then every subset of every complete separable metric space has the BP.

Can we drop the word "separable" here?

Is it consistent with ZF+DC that every subset of every complete metric space has the BP?

In other words, working in ZF+DC, can we prove there exists a complete (non-separable) metric space $X$ and a subset $E \subset X$ without the BP?

I'm not sure which way my intuition goes. On the one hand, non-separable metric spaces are big and maybe they can be weird, even without AC. On the other hand, a counterexample would have the property that the intersection of $E$ with every separable $S \subset X$ would (consistently) have the BP in $S$. That seems unlikely but I don't see how to disprove it.

As one possible approach, a nonmeager proper linear subspace of any Banach space $B$ must not have the BP (by the Pettis lemma). Can we create such a beast in ZF+DC, for some non-separable $B$? We can't do it by taking the kernel of a discontinuous linear functional $f$, because it's consistent with ZF+DC that there aren't any (by considering sequences, the restriction of $f$ to some closed separable subspace $S$ would be discontinuous, making its kernel nonmeager in $S$).

7857392128642501190>I don't have an answer to the one-variable decision question that you asked.

But you did introduce multi-variable
expressions, and there are some fascinating results on the
multi-variable analogue of your domination problem.
(Perhaps you know all this already...) That is, given two rational
exponential expressions f(x_{1},...,x_{k})
and
f(x_{1},...,x_{k}), the problem is to decide whether
f(n_{1},...,n_{k}) <=
g(n_{1},...,n_{k}), for all positive integers (n_{1},...,n_{k}) except at most
finitely often.

The answer is that this decision problem is undecidable, and it is undecidable even for polynomial expressions. There is no algorithm that will tell you when one multi-variable (positive) polynomial expression eventually dominates another.

The reason is that the decision problem of
Diophantine equations
is encodable into this domination problem. That work
famously shows that the problem to decide if a given
integer polynomial p(x_{1},...,x_{k}) has
a zero in the integers is undecidable. This is the famous
MRDP solution to Hilbert's 10th
problem.

It is easy to reduce the Diophantine problem to the
domination problem, as follows. First, let us restrict to
non-negative integers, for which the MRDP results still
apply. Suppose we are given a polynomial
p(x_{1},...,x_{k}) expression over the
integers, and want to decide if it has a solution in the
natural numbers. This expression may involve some minus
signs, which your expressions do not allow, but we will
take care of that by moving all the minus signs to one
side. Introduce a new variable x_{0} and consider
the domination problem:

- Does 1 <=
(1+n
_{0})p(n_{1},...,n_{k})^{2}for all natural numbers except finitely often?

We can expand the right hand side, and move the negative
signs to the left, to arrive at an instance of your
domination problem, using only positive polynomials. Now,
if p(n_{1},...,n_{k}) is never 0, then the
answer to the stated domination problem is Yes, since the
right hand side will always be at least 1 in this case.
Conversely, if p(n_{1},...,n_{k}) = 0 has
a solution, then we arrive at infinitely many violations of
domination by using any choice of n_{0}. Thus, if
we could decide the domination problem, then we could
decide whether p(n_{1},...,n_{k}) = 0 has
solutions in the natural numbers, which we cannot do by the
MRDP theorem. QED

I think the best situation with the Diophantine equations is that it remains undecidable with nine variables, and so the domination problem I described above is undecidable with ten variables. (Perhaps Bjorn Poonen will show up here and tell us a better answer?)

Of course, this doesn't answer the one-variable question that you asked, and probably you know all this already.

My final remark is that if one can somehow represent the
inverse pairing function, then one will get undecidability
even in the one variable case. That is, let f(n,m) =
(n+m)(n+m+1)/2 + m be one of the usual pairing functions,
which is bijective between ω^{2} and ω.
Let p be the function such that p(f(n,m)) = n, the
projection of the pairs onto the first coordinate. If the
expressions are enriched to allow p, then one can in effect
work with several variables by coding them all via pairing
into one variable, and in this case, the domination problem
in the one-variable case, for rational exponential
expressions also allowing the function p, will be
undecidable. It would seem speculative to suppose that p is
itself equivalent to a rational exponential expression, but
do you know this?

I need a help:

What will be the distribution of sum of $n$ independent F distributed random variables with parameters 1 and 1 (i.e., $F(x;1,1)$?

Formally, say $x_1,\ldots,x_n$ are i.i.d. as F(1,1), what is the distribution of $\sum x_i$?

Great if you can suggest some references too, Thanks in advance.

4774221607009Random matrices are completely unrelated to the trivial zeros. Likewise the conjecture that there exists a chaotic system which, when quantized, provides the zeta zeros only applies to the nontrivial ones.5065621111853141168086455412910392556012527071536392 288316122885036This looks like a good approach to the "translations only" version.261751400586169365964044522234881876320The functions must have zeros at the same place; can you show the zeros have the same order? If so, then the ratio $f/g$ should also be entire, hence (because it's bounded) be constant be LiouvilleJack SmithOn second thought I should be able to make all weights positive!418129431984417115Hardy-Littlewood-Polya's "Inequalities" is an excellent source for proving these types of statements. You will want to check out sections 3.4 and 3.15 on mean values of arbitrary functions. Be forewarned, the book has more theorems than pages. I've now verified the whole proof and accept the answer. To see that the expected value of the square of the number of bisecting subspaces is asympt. equal to $E[X]^2$ more explicitly one can write $X=\sum_{L} X_L$ where $X_L:=[L\text{ bisects }A]\in\{0,1\}$. Note that Grassmann's equation $dim(L_1)+dim(L_2)=dim(L_1\cap L_2)+dim(\leftYour question originates in a confusion about the words *Cauchy problem*.

When you have a linear PDE of order $k$, and a hypersurface $S$, it seems natural to try to solve the Cauchy problem with data posed on $S$. *Cauchy problem* means here that you give the unknown $u$ and the $k-1$ first normal derivatives (whence $k$ scalar data). When you do that, you immediately have all the derivatives of $u$ over $S$ up to order $k-1$. Then you ask yourself whether the PDE gives you the rest of the jet, by induction on the order of derivatives. When the answer to this question is positive, you say that $S$ is *non-characteristic*.

When the answer is negative, you realize that the PDE actually imposes non-trivial linear relations between $u,Du,\ldots,D^{k-1}u$ over $S$. In other words, you are not free to choose the Cauchy data over $S$. This means that you have to start with less many data (less than $k$).

This is what happens for the Heat equation with $S$ the set $t=0$. The Cauchy problem, in the original sense of these words, would be to give both $u$ and $u_t$ at initial time, because $k=2$. But $S$ is characteristic and therefore one may not give two scalar data. In the case of the Heat equation, it is not hard to see that what is important is the partial order of the equation with respect to the direction normal to $S$ (the time). This order is one, whence only one initial data. This is a general principle.

4Thomason-Trobaugh TheoremzComputing Mertens' function in time O(sqrt(x)) - in practiceRe question 2. The answer is provided by the Ambrose Singer theorem. You can see how it connects with your question in my book *A Tour of sub-Riemannian Geometry*. When the connection form is analytic then it is easy to state the Ambrose-Singer theorem.
The curvature of the connection, together with all its covariant derivatives, are forms with values in the Lie algebra of the structure group G of the principal bundle. Take the Lie algebra generated by all of these values and then
take the subgroup of the structure group having this Lie algebra. [This is the "holonomy group" of the bundle-with-connection.] The accessible set (points you can get to by horizontal paths) is a subbundle of the original bundle
whose fiber is this (possibly) smaller connected Lie group.

Is it something known about $L^2$-Betti numbers for Golod-Shafarevich groups?

20091391235398186586`Invariance of reduced trace of Azumaya algebras785968663111065656230033235711185428830601 Rasoul Ahangari1449951519706@Buschi: Gray uses "pseudo-functor" to mean what nowadays is called a "lax functor". As far as I can tell, his discussion of why the tensor product doesn't work for lax functors has no bearing on pseudo-functors.20568714822501940292- If $R$ is a field of characteristic $0$, then the differentials of a transcendence basis of $R((x))$ over $R$ constitute a basis of $\Omega^1_{R((x))/R}$ over $R((x))$.

In this sense this module is big: If $d$ denotes the transcendence degree, then we have $|R|^{\aleph_0} = max(|R|,\aleph_0,d)$. Thus if $R$ is countable, we have $d = |R|^{\aleph_0}$. But I don't know $d$ if $|R| = |R|^{\aleph_0}$. Since $\Omega^1$ commutes with localization, this also shows that $\Omega^1_{R[[x]]/R}$ is very big.

- There is an exact sequence of $R[[x]]$-modules $0 \to \cap_ {n \geq 1} x^n \Omega^1_{R[[x]]/R} \to \Omega^1_{R[[x]]/R} \to R[[x]] \to 0$. Here the map on the right is induced by the formal derivative $R[[x]] \to R[[x]]$.

For a proof (thanks owk!), we use the exact sequence of $A/I$-modules $I/I^2 \to \Omega^1_{A/R} \otimes A/I \to \Omega^1_{(A/I)/R} \to 0$, where $A$ is a $R$-algebra and $I \subseteq A$ an ideal.

If we put $A=R[[x]]$ and $I=(x^n)$, it follows that $0 \to x^n \Omega^1_{R[[x]]/R} \to \Omega^1_{R[[x]]/R} \to \Omega^1_{R[[x]]/x^n / R} \to 0 (*)$ is exact. If we put $A = R[x], I = (x^n)$, we get a exact sequence $0 \to x^n R[x] \to R[x] \to \Omega^1_{R[[x]]/x^n / R} \to 0$. Using this, take the inverse limit of $(*)$ and get the desired sequence.

- If $M$ is a $R[[x]]$-module which is complete and separated with respect to the $x$-adic topology, then the map $Hom(\Omega^1_{R[[x]]/R},M)=Der_R(R[[x]],M) \to M, \delta \mapsto \delta(x)$ is bijective. For a proof, remark that the leibniz equality ensures that $\delta$ is continuous.

I think that $\Omega^1_{R[[x]]/R}$ is a rather pathological object. But the last remark shows that it is simple free of rank $1$ (as in the polynomial case) if we restrict to complete and separated modules, which also fits to $R[[x]]$ as $R$-module.

770767173744774388995087Proving such results falls into three parts. First you take an effective version of the prime number theorem, which implies all your desired bound for sufficient. Second you write a computer program (or use computations done by somebody else) to check your inequalities for small $x$. Finaly you bridge the gap between the two ranges by improving the general bounds for large $x$ with more specialized arguments using numerical information on zeros of the Riemann $\zeta$-function.

The first part is easy, since you can just cite the error terms of the effective PNT. As far as I know, the bet results are due to Trudgian ( https://arxiv.org/pdf/1401.2689.pdf ). The difficult part is making the different ranges overlap. Note that as your inequalities are rather weak, and $\pi(x)$ can be computed in sublinear time, you should compute $\pi(x_0)$, use an upper bound sieve to compute some $x_1$, such that your inequality holds in $[x_0, x_1]$. Since $x_1$ will be of magnitude $x_0+\frac{x_0}{\log^C x_0}$, the savings are substantial.

170786I'm not sure the approximation above $(LOG_2n)*(LOG_2LOG_2n)$ is correct.

Let's formulate the task in the folllowing way:

```
We have some number N>2,and we can divide it by $LOG_2 (N)$ while we get a number that is smaller or equal to 2. How many iterations do we need?
```

For $N>4$ $LOG_2(N)>2$,so,if we have sufficiently large N-- let's divide it by log(N) each time while we get a number which is <= 4. The number of iterations we will apply before we get a number that is smaller than 4 is < $LOG_2(N)$(because we divide by a number which is larger than 2 each time),while for N<4 the number of iterations is constant (Actually it is 1 ) and and we can ignore it for sufficiently large N.

It means we have O(LogN) as an upper bound.

So,the exercise from the book is solved. But, just out of curiosity, I'm interested more in some method to solve the equation and to get the **exact** solution. I'm not sure we can replace the difference equation with the differential equation here.

*Definitions*: Let $W$ be a representation of a group $G$, $K$ a subgroup of $G$, and $X$ a subspace of $W$.

Let the *fixed-point subspace* $W^{K}:=\{w \in W \ \vert \ kw=w \ , \forall k \in K \}$.

Let the *pointwise stabilizer subgroup* $G_{(X)}:=\{ g \in G \ \vert \ gx=x \ , \forall x \in X \}$.

Let $G$ be a finite group, $H$ a core-free subgroup, $U$ and $V$ two irred. complex representations of $G$.

Let the subgroup $L := G_{(U^H)} \cap G_{(V^H)}$.

*Question*: Is there an irr. complex rep. $W$ of $G$ such that $G_{(W^H)} \cap G_{(V^H)}$ and $G_{(U^H)} \cap G_{(W^H)} \subset L$?

*Motivation*: This would be very helpful for proving the dual version of a theorem of Øystein Ore.

We do not know that $\mathsf{PFA}$ and supercompactness are equiconsistent. In a sense, we are far from knowing, but in another sense, most set theorists who have worked on the problem are almost certain.

The problem is that the tools we have for deriving lower bounds in consistency strength are not strong enough to reach supercompactness (or even $\kappa^+$-strong compactness). As far as the current tools can reach, we know that $\mathsf{PFA}$ is at least as strong as that.

Typically, lower bounds are established not by using the full power of $\mathsf{PFA}$ but instead by verifying that some combinatorial consequence of $\mathsf{PFA}$ already requires at least that lower bound. For years, the only combinatorial consequence of $\mathsf{PFA}$ that we could extract such strength from was the failure of square principles, first established by Todorcevic, in

Stevo Todorcevic.

A note on the proper forcing axiom, inAxiomatic Set Theory (Boulder, Colorado, 1983), Contemporary Mathematics, vol. 31, American Mathematical Society, Providence, RI, 1984, 209-218. MR0763902 (86f:03089).

Using this, Sargsyan proved in *On the strength of PFA. I* that if $\mathsf{PFA}$ holds, then there is an inner model containing all the reals and satisfying $\mathsf{AD}_{\mathbb R}+$"$\Theta$ is regular". This is stronger than the existence of a proper class of Woodin cardinals and a proper class of strong cardinals. Sargsyan's techniques already give us (a bit) more than this, and the bounds should increase as descriptive inner model theory reaches stronger determinacy models. This is the current best bound we have.

Recently, alternatives to the approach via square have also been established. Neeman (see also his work with Schimmerling) proved in

Itay Neeman.

Hierarchies of forcing axioms II, J. of Symbolic Logic, vol. 73 (2008), pp. 522–542. MR2414463 (2009d:03123),

that the strength of $\mathsf{PFA}$ restricted to $\mathfrak c$-linked posets is that of a $\Sigma^2_1$-indescribable cardinal, and the restriction to $\mathfrak c^+$-linked posets should be what he calls a $\Sigma^2_1$-indescribable $1$-gap $[\kappa,\kappa^+]$. There is a natural hierarchy associated with the notion of $\Sigma^2_1$-indescribability, that ends up in supercompactness, so this gives us a new route to attempt establishing lower bounds. Already at the level described in the paper, though, the fine structural models required for the results are beyond what we can reach currently.

Another alternative was suggested in

Matteo Viale, and Christoph Weiß.

On the consistency strength of the proper forcing axioms, Advances in Mathematics, vol. 228 (2011), no. 5, 2672-2687. MR2838054 (2012m:03131).

Their results show that the usual iteration techniques for forcing $\mathsf{PFA}$ require at least a strongly compact cardinal, and if the forcing used is proper, then a supercompact is indeed needed. This is not the same as saying that we indeed need supercompactness in strength, but it goes to explain the "almost certainty" I mentioned above.

Two remarks need to be added, both related to the strength of failures of square.

**I.**

There are many combinatorial variants of square principles, of varying strength, for instance the family of principles $\square_\kappa,\square_\kappa^2,\dots,\square_\kappa^{<\omega},\square_\kappa^\omega,\dots,\square_\kappa^\kappa=\square_\kappa^*$, and also $\square(\lambda)$ and its variants. I will not discuss here all known results; they are due to a number of authors, including (in rough historical order) Todorcevic, Magidor, Cummings, and Strullu. (The list is grossly incomplete, and I apologize for this.)

We have that $\mathsf{PFA}$ implies not just the negation of $\square_\kappa$ for $\kappa\ge\omega_1$ but, in fact, of $\square_\kappa^{\omega_1}$. It actually implies the failure of $\square(\lambda,\omega_1)$ for $\lambda\ge\omega_2$. The stronger principle $\mathsf{MM}$ reaches farther. For instance, $\mathsf{PFA}$ is consistent with $\square_{\kappa,\omega_2}$ holding for all $\kappa\ge\omega_2$, while $\mathsf{MM}$ implies $\square_\kappa^*$ fails whenever $\kappa$ has cofinality $\omega$, etc.

A number of set theorists (notably, Magidor) have taken this as partial evidence that in fact strong compactness and supercompactness have different consistency strength, with strong compactness sufficing for the failures of square visible to $\mathsf{PFA}$, and supercompactness being responsible for the additional failures coming from $\mathsf{MM}$.

Together with $\mathsf{PFA}$ and $\mathsf{MM}$, in the last 30 years or so, a variety of reflection principles have been identified, some incompatible with $\mathsf{MA}$ (such as Todorcevic's Rado conjecture), some being a consequence of $\mathsf{PFA}$ (say, Moore's Mapping reflection principle), and some being consequences of $\mathsf{MM}$ (the Strong reflection principle, for example).

In spite of their obvious differences and mutual incomparability, all these principles imply failures of square, and so they have significant consistency strength (the bounds discussed by Sargsyan). Unless all these principles turn out to be equiconsistent with supercompactness, a finer analysis than we can currently anticipate will be needed to separate their strength.

The reason for the last remark is that the way strength is derived via failures of square can be understood as a proof by contradiction: If the principle under consideration does not imply the existence of inner models with certain large cardinals, then we can use this smallness assumption to build appropriate local core models $K$ (which are fine structural models and therefore satisfy square principles at suitable cardinals). We can then argue that the failure of square implies that these models must compute successors incorrectly. The point is that $\square_\kappa$ is upwards absolute between models where $\kappa$ and $\kappa^+$ remain cardinals. This is a violation of *weak covering* (one of the key properties of $K$), which implies that $K$ cannot actually exist, meaning that the large cardinals we were after were indeed reached during the attempted construction of $K$.

An entirely different approach seems needed if (as Magidor suspects) these reflection principles end up having different consistency strength. In particular, weak versions of square are equivalent to the strongest known versions in fine structural models, so weak covering or its relatives cannot be enough to distinguish lower bounds between these principles.

**II.**

Recently, we have begun to understand how single failures of square principles are significantly weaker than *consecutive* failures. In particular, the strength of $\lnot\square(\kappa)+\lnot\square_\kappa$ has been investigated. This started with work of Schimmerling and continued in

Ronald Jensen, Ernest Schimmerling, Ralf Schindler, and John Steel.

Stacking mice, J. Symbolic Logic, vol. 74 (2009), no. 1, 315–335. MR2499432 (2010d:03087).

Independently of the study of square principles, the "stacking" technique described in this paper has been significant in modern investigations in inner model theory. The strength reached from the failure of both $\square_\kappa$ and $\square(\kappa)$ for $\kappa\ge\omega_3$ is comparable to that reached through Sargsyan's method -- the goals and techniques are different: Sargsyan looks at the failure of $\square_\kappa$ for a singular strong limit cardinal $\kappa$, and is directly developing the core model induction, while the techniques in the stacking paper are purely inner model theoretic and pursue showing the existence of non-domestic mice.

See here for *Square principles in $\mathbb P_{max}$ extensions*, work involving me, Larson, Sargsyan, Schindler, Steel, and Zeman. We look at the failure of $\square(\omega_2)$ and $\square_{\omega_2}$, in ($\mathsf{ZFC}$) forcing extensions of models of determinacy. Looking at strengthenings of this failure led to the isolation of the notion of $\Pi^2_1$ subcompactness as playing a key role. That this is indeed the appropriate large cardinal has been essentially confirmed in *Equiconsistencies at subcompact cardinals*, by Neeman and Steel. The goal here has not been to identify the strength of $\mathsf{PFA}$ or $\mathsf{MM}$, but rather of some of their fragments, below $\mathsf{MM}(\mathfrak c^+)$. We expect that a good understanding at this level will be key for an eventual understanding of the strength of $\mathsf{PFA}$ via an appropriate stratification (in this respect, the goal is similar to that pursued in the *hierarchies* paper by Neeman mentioned above).

I was reading about the passing of Alexander Grothendieck, and something caught my interest:

Mr. Grothendieck was able to answer concrete questions about these relationships by finding universal mathematical principles that could shed unexpected light on them. Applications of his work are evident in fields as diverse as genetics, cryptography and robotics. New York Times

After extensive googling, I haven't been able to find examples. **Has Grothendieck's mathematical work been applied to robotics, cryptography, or genetics, and if so, how?**

**EDIT** : In what follows the weight $k$ is assumed to be an integer.

The matrix $\begin{pmatrix} 1 & 1/2 \\ 0 & 1 \end{pmatrix}$ normalizes the group $\Gamma_0(4)$, so in your case $g$ is still a modular form on $\Gamma_0(4)$ with trivial Nebentypus.

More generally if you start with $f$ on $\Gamma_0(N)$ and twist it by $1/m$ then you get a form on $\Gamma_0(N') \cap \Gamma_1(m)$ with $N'=\operatorname{lcm}(N,m^2)$.

In general, when you consider $g(z)=f(z+1/m)$, you are twisting $f$ by the *additive* character $\alpha(n)=\exp(2\pi i n/m)$. You can always write $\alpha$ as a linear combination of (not necessarily primitive) Dirichlet characters $\chi$ of level dividing $m$. Thus you can write $g$ as a linear combination of twists $f \otimes \chi$ for such $\chi$ (up to bad Euler factors). If you want to do this completely explicitly then the formulas are quite complicated in general, see Merel, Symboles de Manin et valeurs de fonctions L, Section 2.5. At some point, it may be useful to switch to the adelic language and work with the automorphic representation associated to the newform $f$.

The distribution of the maximal distance between a pair of random points on the circle is known - when you scale it by $n/\log n$ you get a Gumbel distribution with scale 1, location 1., see, e.g.,

*Schlemm, Eckhard*, **Limiting distribution of the maximal distance between random points on a circle: a moments approach**, Stat. Probab. Lett. 92, 132-136 (2014). ZBL1294.60045.

so it is quite obvious the the probability of maximal distance being $O(1)$ goes to zero exponentially fast (it is obvious that it goes to zero about as fast as $(3/4)^{2n}$, in fact.

1354041161407070529614947311491725I've been slightly vague about my "definable" well-orderings: I think everything I said is true whether we take "definable" to mean with or without parameters. But I welcome corrections on this point!#Let $G$ be a simple Lie group and $P$ its parabolic subgroup such that on the level of Lie algebras we have $\mathfrak{p} = \mathfrak{g}_0 \oplus \mathfrak{n}$. The dual $\mathfrak{n}^{*}$ is identified with the nilradical of the opposite Lie algebra via the Killing form of $\mathfrak{g}$. Now for any representation $V$ of $\mathfrak{g}$ one can define Lie algebra cohomology of $\mathfrak{n}$ with values in $V$ and Lie algebra homology of $\mathfrak{n}^*$ with values in $V$. The cochain and chain spaces are the same: $C^k(\mathfrak{n},V) = C_k(\mathfrak{n}^*,V) = \Lambda^k \mathfrak{n^*}\otimes V$.

All details can be found in the paper "Lie Algebra Cohomology and the Generalized Borel-Weil Theorem" by Kostant, where it is proved that for finite dimensional $V$, the Lie algebra homology differential $\delta$ is adjoint to the Lie algebra codifferential $d$ with respect to some invariant positive definite inner product on (co)chain spaces.

From the fact that these operators are adjoint follows that there is a sort of Hodge theory on the (co)chain complex for the operator $\square = d\delta + \delta d$. In other words, there is a unique representative in each (co)homology class that is harmonic; i.e. it lies in the kernel of $\square$. The induced isomorphism of (co)homology groups with $\ker \square$ lies at the heart of the famous Kostant's theorem on the structure of $H^*(\mathfrak{n},V)$.

One actually doesn't need $d$ and $\delta$ to be adjoint wrt an invariant inner product in order to get such a result. As Kostant actually proved in the paper, it is sufficient to prove that these operators are disjoint: $\delta d u = 0 \implies du=0\quad \&\quad d\delta u = 0 \implies \delta u = 0$.

Let $G$ be a simple complex Lie group and let $P$ be a parabolic subgroup. For which representations $V$ of $(\mathfrak{g},P)$ are the Lie algebra differential and codifferential acting on $C(\mathfrak{n},V)$ disjoint?

Remark: (Co)Chain complexes with values in Segal-Shale-Weil representation are an example that these operators are not disjoint in general. This in particular means that there is no $\mathfrak{sp}$-invariant Hermitian product on the Segal-Shale-Weil representation.

253375@FelipeVoloch Same mix-up. Here is an example. $N=221$ then $G=14+5i$. Since $13$ divides $N$ and $13=(2+3i)(2-3i)$, one of $2\pm 3i$ divides $G$. Check that $G(2+3i)=13+52i$ is divisible by $13$. Conclude that $2-3i$ divides $G$ and $G=(2-3i)(1+4i)$.@DavidSpeyer. Yes. By Lazard's correspondence there is a bijective map $\mathcal{L}$ from an arbitrary nilpotent finite dimensional Lie algebra $L$ over a finite field of characteristic $p$ of nilpotent class less than $p$ to a finite $p$-group $G$ (of the same nilpotency class as $L$) such that $[x, y]^\mathcal{L} = [x^{\mathcal{L}}, y^{\mathcal{L}}]$ for all $x, y \in L$, where the right hand side is the usual group commutator $[a, b] = a^{-1}b^{-1}ab$. This implies that the answer to the question is negative for all such Lie algebras $L$.230891N"numerical space" -> "Euclidean space"436560Madeleine607561698799That there are two different notions that are called "locally compact" does not help.(in fact it is not difficult to list these $2^n$ explicitly; I just do not see how to show that there are no others)5259124Forcing is intuitionisticIntegral Lie algebras are also important in homotopy theory, if $X$ is a simply connected space and $\pi_*(X)=\bigoplus_{n\geq 1}\pi_n(X)$ are its homotopy groups, $\pi_*(X)[-1]$ is a graded Lie algebra. The Lie bracket is the Whitehead product. This Lie algebras satisfy the feature that $[x,x]\neq 0$ in general since over the integers this is not equivalent to antisymmetry.

For a large number x>0, how many Fibonacci numbers are there in the interval [1,x]? I have saw the corresponding results in certain places but I have forgotten now. Can anyone help me? Thanks!

6261465150863722071Phil Launt1940653424988 19916133801958224I think Gabriel Paternain proved that a real analytic Finsler metric on a surface of genus g > 1 cannot be integrable. That makes me quite curious about this example of Bangert !!147760044043329487072208050)You can derive this identity from boson-fermion correspondence, once you correct the bounds on the sum, and fiddle with the right side. If you multiply both sides by $\prod_{j = 1}^\infty (1-q^{2j})^{-1}$, you get the character of the lattice vertex operator algebra on the left, and the character of the charged free fermions on the right. These VOAs are isomorphic, so their characters are equal. There is a treatment in chapter 5 of Frenkel and Ben-Zvi's *Vertex algebras and algebraic curves*, but you have to fiddle with the grading a bit to get the identity you want instead of the Jacobi triple product formula.

**Edit:** Okay, here is how it works: On the bosonic side, you have a Heisenberg VOA $\pi_0$, which as a graded vector space is isomorphic to the polynomial ring $\mathbb{C}[b_{-1}, b_{-2}, \ldots]$, where $b_{-i}$ has degree $i$ - it is an induced module of the Heisenberg Lie algebra. If we choose the convention of $Tr(q^{2L_0})$ for characters, the VOA has character $\prod_{j > 0} (1-q^{2j})^{-1}$. For each complex number $\lambda$, you have an irreducible module $\pi_\lambda$ of the VOA, also an induced module of the Lie algebra, and it has character $q^{\lambda^2} \prod_{j > 0} (1-q^{2j})^{-1}$. If you assemble modules for $\lambda \in \mathbb{Z}$, you get a bosonic Fock space that is a super VOA whose character is $\prod_{j = 1}^\infty (1-q^{2j})^{-1} \sum_{n \in \mathbb{Z}} q^{n^2}$. If you take the trace of parity, you get:
$$\prod_{j = 1}^\infty (1-q^{2j})^{-1} \sum_{n \in \mathbb{Z}} (-1)^n q^{n^2}.$$
If you multiply by $\prod_{j = 1}^\infty 1-q^{2j}$, you get $\sum_{n \in \mathbb{Z}} (-1)^n q^{n^2}$, which is what the left side of your equation ought to be.

On the fermionic side, the Fock space is a Clifford representation generated by vectors $\psi_n$ and $\psi^\dagger_n$ for $n \in \mathbb{Z}_{<0} + \frac12$. You can think of it as a polynomial ring in odd (i.e., anticommuting, square zero) variables. For each half-integer $n$, the operator $\psi_n$ contributes weight $n$. The character is then $\prod_{j > 0} (1+q^{2j+1})^2$. The trace of parity is: $$\prod_{j > 0} (1-q^{2j+1})^2 = \prod_{j > 0} \frac{(1-q^j)^2}{(1-q^{2j})^2} = \prod_{j > 0} \frac{1}{(1+q^j)^2}.$$ If you multiply by $\prod_{j = 1}^\infty 1-q^{2j}$, you get $\prod_{j = 1}^\infty \frac{1-q^j}{1+q^j}$, which is the right side of your equation.

You can also see this identity (with signs on the left) as part 2 of exercise 12.4 in Kac's *Infinite Dimensional Lie algebras*. This is no coincidence - I think it follows from twisting the Frenkel-Kac construction.

This was proven by Hatcher in 1983. It states that the diffeomorphism group $\mathrm{Diff}(S^3)$ of the $3$-sphere has the homotopy type of the orthogonal group $O(4)$, which in particular implies that $\pi_0\,\mathrm{Diff}(S^3)= \pi_0 (O(4))$, or equivalently that $\Gamma_4=\pi_0\,\mathrm{Diff}(D^3\mathrm{rel}\,\partial)=0$ (this latter result, due originally to Cerf, was simplified here). The case of the $2$-sphere is even more famous and much easier, but the Smale Conjecture is a major foundational result, which implies for example that ``the space of smooth unknotted curves retracts to the space of great circles, *i.e.* there exists a way to isotope smooth unknotted curves to round circles that is continuous as a function of the curve'' (quoted from here).

Hatcher's proof is considered to be very hard, and I have heard experts say that there might be only a handful of people in the world who truly understand it. I am not aware of the proof having been substantially simplified.

13794466Netz has emphasized that the Greeks of antiquity used their diagrams in a much more *topological* manner than one might expect, and perspective seemed limited. Archimedes was silent on perspective in spheres and liked to refer to the great circle therein; instead of a sphere in a cylinder, Archimedes' tombstone might have merely been a circle in a square!

In more detail, although the scholars of the 19th and early 20th century were really good at reconstructing authoritative *texts* of the ancients, along with doing really boss things like dating the time of an author's flourishing to some throwaway line the author might make about an eclipse or comet, these same scholars had the tendency to amend the figures of Euclid/Archimedes/etc. to fit the standards of rigor of the 19th/early 20th century.

Netz is one of the first to note that figures are amenable to the critical, philological method, and thus to answer @Quuxplusone's and @Matt F's questions, even if there are no extant manuscripts (on papyri) in the hand of, say, Archimedes, by comparing and contrasting the manuscripts that *are* available, one may go back centuries, maybe a millennia, to Archimedes himself (or at least to the scribe at the library of Alexandria). One such manuscript was famously hidden behind a prayer book and lost to the world until rediscovered in a monastery outside of Jerusalem.

As an example of a critical reconstruction of a diagram, consider the figure below from *On the Sphere and the Cylinder*. $K$ is a cone and each of the segments $AZ$, $ZH$, $HB$ etc are *lines* (here showed as arcs). In one branch of the manuscript tree, the bottom $A$ is missing, most likely due to scribal error. Because the text refers to line $AB$, one of the copies on that branch *added* a line between the left $A$ and the top $B$, while also fixing the segments straight. But upon reading the proof, the line $AB$ is intended to be the *diameter*.

Because the manuscript tree most likely branched off relatively early, e.g. in the early 1st millennium, Netz concludes from the study that, going back at least to antiquity and maybe to Archimedes himself, the figure had an $A$ on the left and a $A$ or maybe a $\Lambda$ on the bottom, along with arcs for line segments.

Netz discusses these in *The Works of Archimedes: Volume 1, The Two Books On the Sphere and the Cylinder: Translation and Commentary 1st Edition*, which is quite scholarly; he also has a popular book, *The Archimedes Codex*, with his coauthor William Noel, which provides figures available in the extant codices for the image above.

A scanned copy of the thesis is available here.

fQuestion about the square of the Jacobson radical2169252@Mark Meckes Many thanks, Mark. In retrospect, if I had been more adept with my Googling I might have discovered that the Gaussian RV approach was a "well-known" answer to my question, and would not have needed to ask it on MO. But as usual, I have come away very impressed with the utility of this great website and learned a lot from the answers I received.Felix: Every point in the cone is contained in the interior of a closed box $B\subset C$ (where $C$ is the open cone of positive-definite matrices). If you consider the closed cone of PSD matrices, then there is no countable covering by closed boxes (see the example of $2\times 2$ matrices): You cannot cover a closed round disk $D$ by countably many rectangular boxes contained in the disk. (Since every such box intersects the boundary circle in at most 4 points.) the case of $n\times n$ matrices reduces to the $2\times 2$ case. 660021Reference for Kakutani result on power sum bases of symmetric functions@Tom: but you can run it forward; the yoga of Lawvere theories implies that you can define commutative algebras as precisely those things whose elements you can apply polynomials to. So everything comes full circle.616172@ThomasKojar: I spent sometime reading through his papers. However it is not entirely clear to me how the projecting procedure works even for 2D, where he used TG domains and did impose boundary conditions. Do you mind to turn your comment into an answer?4435765057214924321The maximum may not exist in general. Take X=Spec k[T,U] the affine plane, A the complement of the vertical line L passing through the origin (A=Spec k[T,U,1/T]) and B the origin (Spec k[T,U]/(T,U)). Then, the maximum C of A and B in the ordered set of subschemes of X does not exist. If it was the case, then C should be some subscheme of X. Such are usually described as closed subschemes in some open subset of X, but can also be described as open subsets in some closed subschemes Z of X (I cannot find the reference in general, but it is certainly true when the schemes are noetherian). As U is (schematically)-dense in X, this Z can be nothing else as X itself, then C should be an open subset of X containing A and B, that is the open complement of a finite set of closed points of the line L except the origin, in which case it is clear that we could find an open subset of X containing A and B and strictly contained in C, which would give a contradiction.

zOh yes, I can see that. I was being very naive: no surprise.153996243835618038616What is a monotonic curve?5616702035509The Talbot workshops are mostly aimed at graduate students and early professionals. The topics and plenary speakers vary year-to-year, but the ones I have attended were all very illuminating. The 2009 workshop, on the Fukaya category, was especially good, and there are notes from each talk on the web page linked above.

17645111083839jtbandes493787Feu2158476also $c_n$ non zero and not $w_s = k w$ - edit: well actually why not $w_s = k w$...thanks for making me think about this, I think I have something that might work. Cheers.20949261230169531508 Antonino Travia13326631604303172966718333221951008Oops, I apologize - I didn't realize that the question has already been answered with *specific* references to Egyptian fractions.348226"...the chances that you will find a book that seems well-motivated for you are low." That really is an important point! And if it's a choice between a) a book with lots of motivation which makes no sense to me; and b) a book with no motivation, but which is consequently shorter, I'll pick (b) every time.12629215039501144451103992350778984632311476831316451141940517957877661522Completion of an algebra1775890Research Student : https://github.com/syedalamabbas

H@Joseph: I changed the references. 91736521408821991827Michael: subjective and argumentative is not bad *per se*, but I don't think it's a good use of MO (I don't think it should be a forum or an "online place to hang out"). On the other hand, your question strikes me as eminently worthwhile -- I'll soon be teaching caculus and analysis courses, and the kind of issue you raise is the sort I want ultimately to think about more. Voting to reopen.Noam, what privileged role does computing with radicals have within linear algebra?2197820294568436727928502I also want to mention Kosniowski: Equivariant Cohomology and Stable Cohomotopy - this paper from 1974 also introduces RO(G)-graded cohomology theories and attributes the idea to Graeme Segal. I just now looked through EGA2 chapter 1, and it only seems to discuss vector bundles associated to a quasicoherent sheaf. Your proof uses the construction of the associated $\mathbb{G}_a^n$-bundle $E$, which I understand is essentially a vector bundle in the fpqc topology, but I don't see why you can find a locally free sheaf $\mathcal{F}$ such that $E$ is the vector bundle associated to $\mathcal{F}$, because the usual correspondence between locally free sheaves and vector bundles uses a definition of vector bundles which require them to be Zariski locally trivial...2481316914706795302221251118709It seems that Bertrand Toen's thesis gives an answer to this question. It's long but appears to be quite thorough.

Toen also has an earlier paper where he does Grothendieck-Riemann-Roch for Deligne-Mumford stacks. This material seems to have been subsumed in the paper above, but perhaps it is easier to read here.

118715616420552031991981593491633694310895934761981581414654153376142481602077356100626634955477157DUNKIRKRUNAWAY12187041282222139035**Yes** they are simple. This is a peculiarity of the 1-dimensional eigenvalue problem. By shiftin $q$, we may always consider the eigenvalue $\lambda=0$.

Geometric simplicity : as mentionned by Remling, two eigenfunctions $u_1$ and $u_2$ must be linearly dependent. For we may find a constant $a$ such that $u_2'(0)=au_1'(0)$, and then $u:=u_2-au_1$ is a solution of a linear second order ODE, with homogeneous initial data $u(0)=u'(0)=0$. By Cauchy-Lipschitz, we have $u\equiv0$.

Algebraic simplicity : the operator $A$ is symmetric in $L^2(0,1)$ (this is already used to prove that the eigenvalues are real), hence the algebraic multiplicity equals the geometric multiplicity (in other words, $A$ is semi-simple). Proof: if $Av=0$ and $Au=v$, then $$0=\int_0^1uAv\,dx=\int_0^1vAu\,dx=\int_0^1v^2dx$$ hence $v\equiv0$.

Remark that the result still hold true if $A=-pD^2_x+bD_x+q$ is not symmetric, because it can always be symmetrized. This is false in higher dimension.

10920131063431571991148164517449048633722I retagged the question.714126276165Bill, that idea with facets, did it get any formalization in the question about $\ell_\infty$?18516131280566X2D Problems Which are Easier to Solve in 3D625586214821996583I dont think a closed form of the Green's function exists. But the closed form of heat kernel do exists. Green function is the integral from $0$ to $\infty$ of heat kernel.134983416614376769036409241898015pArveson's extension for normal completely positive maps18465151444521To extend my comment and emphasize the self-similarity of Julia sets and the Dragon curve, here is an interpolation between the two.

Each frame is generated by two complex functions,

```
f1[z_, t_] := ((1.0 + I) z/2) t + (1 - t) (Sqrt[z + 0.9 I]);
f2[z_, t_] := (1 - (1.0 - I) z/2) t + (1 - t) (-Sqrt[ z + 0.9 I]);
```

where $t$ goes from $0$ to $1$ in the animation frames. For $t=0$, we have the classical Julia fractal for $c=-0.9i$, and at $t=1$, we have the two generators for the Dragon curve.

So what about the colors? Let $J$ be attracting set. Then $f_1(C)$ is the black set, and $f_2(C)$ is the blue set, and $C=f_1(C) \cup f_2(C)$. This puts emphasis of the self-similar nature.

So, note that in the Dragon curve case, since $f_1$ and $f_2$ are analytic and affine, they do not distort the picture at all, so you'll see *exact* copies at smaller levels. In the Julia case, we only have analytic maps, so there is some distortion caused by the square root, but the picture is more or less preserved (this is the nature of analytic maps).

In **Berkeley, California**, Moe's Books, just north of Dwight Way on Telegraph Avenue, has a nice selection of used math books on the third floor. I've found a few gems there.

i like mostly discrete mathematics.

i am currently studying a bachelor of medicine and surgery at the same institution.

Let me comment on points 4) and 5). The problem with infinities in QFT or traditional equilibrium statistical field theory is related to the one addressed by Martin's theory but there are some differences. For concreteness let me talk about the $\phi^4$ model only. Mathematically, the problem it poses is to make sense of the probability measure $$ \frac{1}{\mathcal{Z}}\exp\left( -\int_{\mathbb{R}^d}\{ \frac{1}{2} (\nabla\phi)^2(x)+\mu \phi(x)^2+g\phi(x)^{4} \} d^dx \right)\ D\phi $$ on the "space of all functions" $\phi:\mathbb{R}^d\rightarrow\mathbb{R}$. This is the kind of heuristic formulas one finds in physics QFT textbooks. The symbol $D\phi$ stands for Lebesgue measure on this space of functions and $\mathcal{Z}$ is a normalization constant so the full space has measure one as befits a probability measure. Now let's turn this into a well posed mathematical question.

First remove the $\phi^2$ and $\phi^4$ terms, i.e., consider the case $\mu=g=0$. Then this measure $\mu_{C_{-\infty}}$
makes perfect sense. It is the centered Gaussian measure on the space of temperate distributions $S'(\mathbb{R}^d)$
and with covariance $C_{-\infty}$ given by
$$
C_{-\infty}(f,g)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^d}\frac{\overline{\widehat{f}(\xi)}
\widehat{g}(\xi)}{|\xi|^{2}} d^d\xi
$$
for all test functions $f$ and $g$ in $S(\mathbb{R}^d)$. Using this first rigorous step, one can reformulate
the problem as that of making sense of
$$
\frac{1}{\mathcal{Z}}\exp\left(
-\int_{\mathbb{R}^d}\{
\alpha (\nabla\phi)^2(x)+\mu \phi(x)^2+g\phi(x)^{4}
\} d^dx
\right)\ d\mu_{C_{-\infty}}(\phi)
$$
with a new normalization constant $\mathcal{Z}$ that I will still keep calling $\mathcal{Z}$.
I also introduced the "wave function renormalization coupling constant" $\alpha$ for more generality.
We made a bit of progress (we avoided the problematic Lebesgue measure $D\phi$),
but this still does not make mathematical sense because $\mu_{C_{-\infty}}$ is supported on nasty Schwartz distributions
and pointwise powers like $\phi^2$ and $\phi^4$ are ill-defined, just like $\Phi^3$ in Martin's answer. This is the source
of the UV (ultraviolet) infinities. There are also IR (infrared) problems due to the integration inside the exponential
being over $\mathbb{R}^d$ instead of a compact set. To address these issues, we need what the French call *troncature et
régularisation*. Let $\rho_{\rm UV}$ be a mollifier, i.e., a compactly supported $C^{\infty}$ function
$\mathbb{R}^d\rightarrow\mathbb{R}$ with $\int \rho_{\rm UV}=1$.
Let $\rho_{\rm IR}$ be a cut-off function, i.e., a nonnegative compactly supported $C^{\infty}$ function
$\mathbb{R}^d\rightarrow\mathbb{R}$ which is equal to 1 in a neighborhood of the origin. To slice Fourier momenta
into (Littlewood-Paley) shells we introduce an integer $L>1$, not necessarily equal to 2 as is customary in
harmonic analysis. For $r,s\in\mathbb{Z}$, define the rescaled functions $\rho_{{\rm UV},r}(x)=L^{-dr}\rho_{\rm UV}(L^{-r}x)$
and $\rho_{{\rm IR},s}(x)=\rho_{\rm IR}(L^{-s}x)$, and consider the probability measure $\nu_{r,s}$ given by
$$
\frac{1}{\mathcal{Z}}\exp\left(
-\int_{\mathbb{R}^d}\rho_{{\rm IR},s}(x)\{
\alpha (\nabla\phi)^2(x)+\mu \phi(x)^2+g\phi(x)^{4}
\} d^dx
\right)\ d\mu_{C_{r}}(\phi)
$$
where $\mu_{C_r}$, or regularized Gaussian measure, is the direct image of $\mu_{C_{-\infty}}$
by the convolution map $\phi\mapsto \rho_{{\rm UV},r}\ast\phi$.
In other words, $\mu_{C_r}$ is the centered Gaussian measure with covariance
$$
C_{r}(f,g)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^d}\frac{|\widehat{\rho_{{\rm UV},r}}(\xi)|^2\ \overline{\widehat{f}(\xi)}
\widehat{g}(\xi)}{|\xi|^{2}} d^d\xi\ .
$$
A good metaphor would be to say that your orginal flat-screen TV was too smart. The linear size of the
screen was $L^s=\infty$ and that of a pixel was $L^r=0$. Instead one should make $r$ and $s$ finite so that $\nu_{r,s}$
is mathematically well defined, and then study the limit where $r\rightarrow-\infty$ and $s\rightarrow\infty$
in the sense of
weak convergence of probability measures on the topological space $S'(\mathbb{R}^d)$.
Now renormalization theory in physics tells us that unless we allow the couplings $(\alpha,\mu,g)$ to depend on the
UV cut-off scale $r$, the following is more likely to happen: 1) we don't converge (e.g., loss of tightness), 2)
we converge to something utterly uninteresting like the atomic measure on the singleton $\{\phi=0\}$, 3) we
converge to something less trivial but still uninteresting, namely, a Gaussian measure like the GFF $\mu_{C_{-\infty}}$
or white noise, or massive free fields interpolating between the two.
Therefore the weak limit we need to study depends on the choice of bare ansatz $(\alpha_r,\mu_r,g_r)_{r\in\mathbb{Z}}$
(or rather the germ of this sequence at $r=-\infty$). Finally, the well posed mathematical question I promised,
regarding trying to make sense of the original $\phi^4$ functional integral is the following.

**Problem:** Find an explicit parametrization of all weak limits (of probability measures on $S'(\mathbb{R}^d)$)
given by $\lim_{r\rightarrow-\infty}\lim_{s\rightarrow\infty}\nu_{r,s}$ for all possible choices of bare ansatz
$(\alpha_r,\mu_r,g_r)_{r\in\mathbb{Z}}$.

What renormalization theory in physics also tells us is that although it seems one has a hugely infinite-dimensional amount of freedom in choosing the sequence $(\alpha_r,\mu_r,g_r)_{r\in\mathbb{Z}}$, the set $\mathscr{T}$ of weak limit points is a finite-dimensional variety. For $d=3$, one expects three parameters or "renormalized coupling constants" $(\alpha_{\rm R},\mu_{\rm R},g_{\rm R})$ suffice. One can even get rid of $\alpha_{\rm R}$ if one quotients by taking constant multiples of the random field $\phi$.

There are rigorous renormalization group techniques for constructing portions of $\mathscr{T}$. Kupiainen's work mentioned
by Ofer is an adaptation of these techniques to the time-dependent rough SPDE setting.
The above is what I call the *non-anchored* Gibbsian way
of trying to construct elements $\nu\in\mathscr{T}$. There is a completely different approach which I call the
*anchored* stochastic quantization approach. There one need to make sense of the SPDE in Martin's answer, which he
did locally in time. Then one needs to understand this SPDE globally in time and construct an invariant
measure which gives $\nu\in\mathscr{T}$. This is also fraught with difficulties but there has been quite a bit of progress in this direction
(see, e.g., this article related to invariant measures and Martin's comment below).
The key difference between the two approches is that in the non-anchored setting one does not have a fixed probability space
to work with. In the second anchored situation one does since all fields are functionals of the driving noise. In the anchored setting,
$L^2$ estimates involving second moments only are enough to prove convergence in probability and thus in
law for the random fields of interest. In the non-anchored situation one needs to control all moments
(correlation functions) with uniform $n!$ bounds on these moments.

It is hard to say more as an MO answer, but you can see these preprints for further explanations:

- "A second-quantized Kolmogorov-Chentsov theorem"
- "Towards three-dimensional conformal probability"
- "QFT, RG, and all that, for mathematicians, in eleven pages"

Before reading these articles it may helpful to consult the slides of my recent Colloquium talk "A Toy Model for Three-Dimensional Conformal Probability". They should be much easier to follow since they contain lots of pictures.

The article in bullet point 1) provides an alternate way of defining pointwise products of random distributions (like $\phi^2$ and $\phi^4$ above) using Wilson's operator product expansion (OPE). Regarding 5) in the OP's question, I believe it would be of great interest to prove that the moments of the solution $\Phi$ constructed by Martin satisfy the (dynamical version) of Wilson's OPE, and then compare $\Phi^3$ obtained by the theory of regularity structures with the one constructed (from the OPE) in my 2nd quantized KC paper.

Update: The article by Mourrat and Weber mentioned by Martin in his comment below has now appeared in CMP, see here. It provides a new construction of the scalar $\phi^4$ model in three dimensions, *in finite volume*.

(This is a bit late, but I hope you find it interesting!)

Here's smooth representation of the generator of $\pi_4(Sp(1))$ (and so the same homotopy group of $S^3$ and $SU(2)$). Consider $S^4 = \mathbb{HP}^1$, and $Sp(1)$ the unit sphere in $\mathbb{H}$. Then the following function $t\colon \mathbb{HP}^4 \to Sp(1)$ represents the nontrivial homotopy class $S^4 \to S^3$: $$ t[p;q] = \frac{2p\bar{q}i\bar{p}q - |p|^4 + |q|^4}{|p|^4 + |q|^4} $$ where [p;q] are homogeneous coordinates on $\mathbb{HP}^4$. I don't know if this has appeared previously (I would love to know!), but I presented this as part of some slides at the Australian Mathematical Society's annual conference last year (see slide 6), and originally worked it out with a pointer from Michael Murray to the Hopf fibration described using quaternions (that is, $Sp(1) \to S(Im\mathbb{H})$, the unit sphere in the pure imaginaries). That this map is the generator (i.e. is not null-homotopic) I calculated following the answer at my question Detecting homotopy nontriviality of an element in a torsion homotopy group.

Note that this function followed by the inclusion $Sp(1) \hookrightarrow Sp(2)$ (as the top left entry) is the generator of $\pi_4(Sp(2))$ (by results of Mimura and Toda). And thus we also get a representative for the generator of $\pi_4$ of $Spin(5) = Sp(2)$.

JThis is very interesting, thank you!1331601691539803277RGenerally speaking a unique lifting does not exist and I believe it is open as to what the possible liftings can be.

As an example of the non-uniqueness consider a slight variant of the particular category you asked about - namely $K^b(\mathbb{Z})$ the homotopy category of bounded complexes of finitely generated abelian groups. Considering this as a graded category it has at least 2 different structures of triangulated category. The usual one and one we denote $K^b(\mathbb{Z})^{-}$ where we declare that

$$X \stackrel{u}{\to} V \stackrel{v}{\to} Y \stackrel{w}{\to} \Sigma X$$
is a triangle if and only if

$$X \stackrel{-u}{\to} V \stackrel{-v}{\to} Y \stackrel{-w}{\to} \Sigma X$$
is a triangle with respect to the usual triangulation. The point that was implicit (before this edit) is that these triangulations do not agree. This also works for $D^b(\mathbb{Z})$, the bounded derived category of finitely generated abelian groups, indeed one just needs to suppose the two categories agreed and consider [TR3] applied to a diagram obtained by mapping
$$\mathbb{Z} \stackrel{3}{\to} \mathbb{Z} \to K(3) \to \Sigma \mathbb{Z}$$
to
$$\mathbb{Z} \stackrel{-3}{\to} \mathbb{Z} \to K(3) \to \Sigma \mathbb{Z}$$
by $(1,-1,h,1)$ where $h$ is some map provided by [TR3] but one can check this is ridiculous.

If you want more lifts of the suspended category with the usual suspension to a triangulated category I feel like the answer should be there aren't any but I don't know a proof or where this is written down if it is true. I think I know how to make some headway on the problem more generally if one rigidifies the situation a little but this is still work in progress.

More generally one can play this game with any unit in the degree zero part of the central ring of our suspended category (in plainer language we can twist by automorphisms of the identity functor commuting with suspension). In this note Balmer shows that this leads to suspended categories with infinitely many triangulations. I believe that it is unknown if there can be other sorts of triangulations at least on an indecomposable category.

I currently can't think of any examples that arise naturally and I am not familiar enough with the one you mention to be able to say at the moment whether or not the triangulations agree. If the suspensions agree then I'd guess they are equivalent up to something of the form above. I'd be interested to know if they weren't.

**Update** Here is the sort of thing I now have in mind for the bounded derived category of finitely generated abelian groups. As a warning I haven't really thought this through too thoroughly.

The statement I think one should be after is that no matter the triangulation there is no choice involved in the isomorphism class of the cones - more precisely if one is given $f\colon X\to Y$ then $\mathrm{cone}(f)$ is up to isomorphism independent of the triangulation. It should follow from this that the sign trick that gives you the second triangulation is all the wiggle room one has.

Denote by $\#X$ the non-negative integer $\lvert \{i\in \mathbb{Z} \;\vert\; H^i(X)\neq 0\} \rvert$, and try to run an induction on this quantity. If $X$ is the suspension of a group and we are given $f\colon X\to Y$ we can desuspend and assume it is in degree zero since this will just rotate the triangle. Complete $f$ to a triangle $$ X \stackrel{f}{\to} Y \to Z \to \Sigma X$$ Applying $\mathrm{Hom}(\mathbb{Z},-)$ gives an exact sequence of abelian groups which tells us that (where since we are not assuming the standard triangulation the use of cohomological notation is slightly abusive but just denotes the corresponding graded piece) $H^i(Z) \cong H^i(Y)$ for $i\leq -2$ and $i\geq 1$, that $H^0(Z) \cong \mathrm{coker} \; H^0(f)$, and that $H^{-1}(Z)$ is determined up to extension by $$0 \to H^{-1}(Y) \to H^{-1}(Z) \to \mathrm{ker}\;H^0(f) \to 0$$ but there is still no choice involved here since the element of $\mathrm{Ext}^1$ giving this extension is determined by the "degree 1 part" (what I mean by this is if $X$ has torsion it might map up to the piece of $Y$ in degree 1) of $f$ (I think - this is the bit I didn't really check). Thus the graded pieces of $Z$ are determined up to isomorphism and hence so is $Z$ since everything is a sum of its cohomology.

Now if we assume we have uniqueness for maps out of guys with less than $n$ pieces and let $\#X = n$ and suppose we are given $f\colon X\to Y$ which completes to a triangle with cone $Z$. Write $X$ as $X' \oplus X''$ where $\#X' < n$. Using the octahedral axiom we get a triangle (we can actually cook up 4 in this way but we only need 1) $$W \to \Sigma^{-1}Z \to X' \to \Sigma W$$ and since $\#X' < n$ the inductive hypothesis implies $Z$ is uniquely determined up to isomorphism.

I'd be interested if anyone has any ideas for an invariant that would detect the difference between these two triangulations on $D^b(\mathbb{Z})$ - they are really almost the same in every way I can think of except that the mapping axiom can fail between the triangles in each of them.

314160Create a doubly-linked list $L$, initially empty. Scan a given permutation $p=(p_1,p_2,\dots,p_n)$ from left to right, and for each element $p_i$ perform the following operations:

(loop) While $L$ is not empty and $p_i$ is smaller than the last element of $L$, remove this last element from $L$.

Append $p_i$ to $L$.

At the end, $L$ will contain a required subpermutation of $p$. Notice that visiting (i.e., comparison to) every element of $L$ results in either appending or removal of an element to/from $L$. The total number of such operations does not exceed $2n$, since every element of $p$ may be appended to $L$ only once and removed from $L$ only once. So, this is a linear-time algorithm.

14745231557721862029 I apology to those who find the comment offensive. @ Joel David Hamkins: For the resemblance between prime numbers, Turing degrees, Large cardinal, they are all object that can not be reached from below by certain operation. This resemblance is rather superficial, but I think we can argue that it seems like the latter might be just as interesting as the former. This point is, however, beside the main one. The point main point is that if Turing degrees and large cardinal has some universal characteristic and not merely artifacts of the language, they should be considered a natural maths object.6431891187929Making video games, read and love to discover the beauty of mathematics. I don't have any knowledge of math except K12 education. Interested in algebra from basic rules that we learn in primary school to abstract algebra. Self-learning most of the time (in some cases i have my friends helping me), love good music too. Leaving in Serbia, Belgrade.

Main goal is to really master algebra as much as i can as i see it as most interesting part of mathematics for sure.

Always ready to help if can. :)

7430479I assume that the output is continious. Whatever you do, don't try calculating the correlation between the methods. I think that the paper "Statistical methods for assessing agreement between two methods of clinical measurement" is what you need

1767351114316211404021168710A p-adically closed field is a field with the same first order theory that Qp. In terms less model theoretist, it is an henselian field with residual field Fp and a value group that is discrete, and such that [G:G^k] = k (these are called Z-groups as they have the same first order theory as Z). I had a glance at Prestel-Roquette, but they don't do what I am interested in. In fact they mainly proved what I just stated above (the explicit axiomtisation of the theory of Qp), but I may have read too fast...Probably I picked up the use of `$\delta$` from Jacobson's book on Lie algebras, but conventions have varied for a long time. The names of root systems or simple roots also vary from one source to another. After 1968 Bourbaki standardized usage somewhat, but I think made a mistake in switching to roman letters `$R$` and `$B$` (the former common in English for rings and the latter ubiquitous for Borel subgroups). I've pretty much followed Borel and Tits, using `$\Phi$` and `$\Delta$` (the latter often being replaced by a numbered list of simple roots). 1578001603819I'm going to do research on dynamics of master equation of $n$ states $$\dot p_i=A_{ij}p_j\qquad i=1\ldots n$$ where $p_i$ is the $i$-th component of probability vector and $A_{ij}$ is transition rate matrix (Q matrix, i.e. each column adds up to zero). For one thing, I'd like to regard master equation as linear dynamical system. But I've no idea how to introduce the property $\sum_ip_i=1$ properly and also that $p_i$ has unique fixed point, i.e. stationary distribution $\pi$ provided $A$ is irreducible. I know the system is restricted on standard $(n-1)$-simplex.

So my question is, are there any suggestions or references about general theory of dynamic systems restricted on simplex? This is different from LDS because such systems have fixed structures and dynamical variable is self correlated. I want directions to help involve structures. I'll appreciate it.

16858133492292236291223903122242501342388If you're saying that this is a puzzle, and someone somewhere knows the answer, then some people would say that this question isn't appropriate for this site and you should take it to MSE.525828RToo long for a comment, so I will add an answer.

I heard often about this, but so far I have yet to find an actual proof that RH implies that the zeroes of the RZF form a 1-dimensional quasicrystal. Actually there is no formal definition of quasicrystal yet, the new definition (1992) of a crystal is intentionally vague and we will probably not have a better formal definition until we understand them better.

Also note that the zeroes of RZF have arbitrarily large gaps, which is not something the quasicrystal community accepts. And there is a reason for that: The diffraction of an infinite quasicrystal is formally defined as the "limit" of diffraction of larger samples of the material. In general this limit might not exist, which happens when large samples of the solid have completely different diffraction, but it can easily be proven that all those finite diffractions are inside some compact spaces, thus we always have cluster points and we can get a limit by simply going to subsequences/subnets (depending if we work in $\mathbb{R}^d$ or arbitrary lcag). This means that some solids have different diffractions depending on how we average.

While this looks clumsy, in reality is not. It only happens if different larger and larger samples of our solid have completely different properties, which means that different large pieces of the solid are basically different materials.... Imagine finding a new material and drilling some samples to diffract. Unknown to us, the rock we find has two halves, one material $A$ and one material $B$. What will our diffraction be? Well the answer is: depends where we drill the sample. We could get the diffraction of $A$, of $B$ or many different diffractions of mixtures of $A$ and $B$. Those are exactly the cluster points we get.

We typically ask that the solid is nice (typically an uniquely ergodicity assumption), which leads to many nice properties and unique diffraction, but this is most of the times not a needed assumption.

So back to zeroes of RZF. The fact that the zeroes of RZF have arbitrarily large holes imply that $0$ is a diffraction measure of this system (i.e. we could drill by chance all our samples just from those holes). The system definitely has multiple diffractions, which makes the problem harder: deciding if this is a quasicrystal depends on which one we pick.

Since we are in $\mathbb{R}$, there are some choices which are more natural, so lets ignore this first issue... And lets simplify things, lets forget the measure approach and work directly with distributions.

If the tempered distribution $ \sum_{j = 1}^\infty e^{ik_j x}$ is a **translation bounded** discrete measure, then I think Hof/Lagarias proved that the set has pure point diffraction, which is understood to imply quasicrystal. This seems to be the case here.

**But** there is also to consider the following Theorem by Cordoba([1]), which I only know from a paper of Lagarias (I didn't read the original yet):

**Theorem**: Let $S \subset \mathbb{R}^d$ be an uniformly discrete set. If the Fourier transform of $\delta_S= \sum_{x \in S} \delta_x$ is a tempered distribution which is a translation bounded discrete measure, then $S$ is a finite union of translates of full rank lattices in $\mathbb{R}^d $.

This shows that the zeroes of RZF cannot fit this description. The only possibility left is that Fourier transform of $\delta_S= \sum_{x \in S} \delta_x$ is a distribution which is a sum of Dirac deltas, but is not a translation bounded measure. But then, as far as I know, there is nothing done in the quasycrystal community in this direction, and I don't know how this fits within our Theory (Hof might have done something in this direction though).

Last but not least, classifying all one dimensional quasicrystals seems like a problem which is impossible to solve, at least in a reasonable way. The issue comes from the fact that at any point in $\chi \in \hat{\mathbb{R^d} } $ we lose in the diffraction the phase information, which is a complex number on the unit circle. There are $c^c$ potential values the phase could take, not all of the potential values would work, but it is easy to show that in general, for a given diffraction, there are at least $c$ values which can work and I suspect that there are more than $c$ good values.

Because of this, each diffraction corresponds to infinitely many solids, some which are related and some which are not. If somehow we manage to get the classification of 1-D quasicrystals, we still have to face two huge issues :

- The classification will be an uncountable list of classes, each with uncountable elements. And given the many cases of models with the same diffraction, the classification is highly likely to be not nice..
- If we have a list of all quasicrystals, how do we check if the zeroes of RZF are or are not in the list, without knowing already the zeroes?

To me, this approach seems similar to trying to classify the zeroes of all analytic functions, and then check which ones correspond to the RZF....

([1]) A. Cordoba, Dirac Combs, Letter Math Physics, 17 (1989), 191-196

1909661The antipode is an anti-algebra homomorphism. Is the opposite of an algebra well-defined in the category of algebras?20674732023685@David Speyer : I believe that Bredon's book on algebraic topology does it too. Neither Hatcher nor May discuss smooth manifolds at all, so they really can't discuss this interpretation of cup products.Johannes Trost14372112294011Let A be a self-injective connected Nakayama algebra. What is the Loewy length of any indecomposable projective A-module?

@When Max(R) is Hausdorff space?2185962102063878130318535019970473This is by far not a complete solution, but at least a partial one, plus a potentially useful interpretation and, perhaps, a direction to pursue.

The way I understand the problem, given a $K$-element set $X$, we call its $k$-element subsets *blocks*, and we say that a block $B\subset X$ *separates* the elements $x_1,x_2\in X$ if exactly one of these elements is contained in $B$. We want to estimate the smallest possible number of blocks, say $n=n(K,k)$, separating every pair of elements of $X$. We can in fact confine to the range $k\le K/2$ since $n(K,K-k)=n(K,k)$, as it follows by looking at the complements of the blocks.

For $k=1$, one needs at least $K-1$ blocks (otherwise there will be at least two elements not contained in any block, hence not separated by any block). Considering, on the other hand, the system of $K-1$ pairwise disjoint, one-element blocks, we get $$ n(K,1) = K-1. $$

For $k=2$, one can identify $X$ with the set of vertices, and blocks with the edges of a graph. Two vertices $x_1,x_2\in X$ are separated whenever at least one of them is adjacent to a vertex other than another one. Thus, we want to minimize the number of edges subject to the condition that the graph does not have isolated edges, and has at most one isolated vertex; in other words, every connected component of the graph has at least three vertices, except that there can be one component containing an isolated vertex. In the optimal case, every component is a tree (take a spanning tree and remove all other edges), and then the number of edges of the graph is $K$ less the number of components. Thus, minimizing the number of edges is equivalent to maximizing the number of components. Consequently, all components must be paths of length $2$, with an obvious adjustment in one component if $K$ is not divisible by $3$. This gives $$ n(K,2) = \begin{cases} 2K/3 \ &\text{if}\ K\equiv 0\pmod 3, \\ 2(K-1)/3 &\text{if}\ K\equiv 1\pmod 3, \\ 2(K-2)/3 + 1 &\text{if}\ K\equiv 2\pmod 3. \end{cases} $$

In the general case, the following approach can be useful. Identifying subsets of $X$ with the vectors of $\mathbb F_2^K$, consider the bipartite graph on the vertex set $V_k\cup V_2$, where $V_k$ is the set of all vectors of weight $k$, and $V_2$ is the set of all vectors of weight $2$, with $b\in V_k$ adjacent to $x\in V_2$ if and only if $b$ is **not** orthogonal to $x$. For any $x_1,x_2\in X$ with $x_1\ne x_2$, the vector in $V_2$ corresponding to the subset $\{x_1,x_2\}\subset X$ is adjacent to a vector $b\in V_k$ corresponding to the block $B\subset X$ if and only if $x_1$ and $x_2$ are separated by $B$. It follows that $n(K,k)$ is the smallest possible number of vertices from $V_k$ dominating any vertex from $V_2$.

In principle, you can now take any of the numerous estimates for the domination number of a graph and try to adapt it to the settings described. The simplest estimate is as follows: since any $b\in V_k$ dominates $k(K-k)$ vertices from $V_2$, to dominate all $\binom K2$ vertices of $V_2$ one needs at least $K(K-1)/(2k(K-k))$ vertices $b\in V_k$; that is, $$ n(K,k) \ge \frac{K(K-1)}{2k(K-k)}. $$

1410245852700Ironically the original sense of "theorem" allows for a theorem to be false, since it just meant "assertion". So Euler could say that a theorem must be true even though he cannot prove it. But yes, modern usage doesn't allow for this.1547534657696user6617436200154447Arif Ullah197372216125471288771them decide whether research is for them. On the other hand, I totally agree with you that the emphasis on originality is misguided -- doing original work on a boring question isn't as good for the student as re-discovering interesting things.5988281008125Oh, now the definition of $L^*$ is finally clear (since your last edit). So I agree that $L^*$ is normal.1798696221739420314294654532187394It is well known that for a number of structures, their existence is equivalent to the existence of a projective plane for a given order. Some of them depend on more than one parameters, which means that there can be several ways of measuring "how close" one can approach a (conjecturally non existing) projective plane for a given order $n$ with more than one prime divisor.

To illustrate this, take e.g. the biggest number $a(n)$ of mutually orthogonal latin squares (MOLS) of order $n$. For prime order, we have trivially $a(n)=n-1$, and moreover non isomorphic sets of $n-1$ MOLS correspond to non isomorphic projective planes of order $n$. On the other hand, it is well known that for $n=6$, there are not even two MOLS, i.e. $a(6)=1$, further that $2\le a(10)\le 6$. This survey is not very recent but gives a lot of lower bounds. The well-known fact that $a(12)\ge5$ means e.g. that for $n=12$, one can come "much closer" (in a certain sense) to a projective plane than for $n=6$.

A different way of measuring is invoked in this recent thread.

Which other structures allow to "measure" how close one can get?

(I would have thought that such a collection already exists in MO, but I couldn't find any.)

15456911727602The problem you are running into is due to the fact that the Hurwitz zeta-function does not, in general, have an Euler product representation.142113314360213994371552840I think that this is a great and very helpful answer, and I regret that it will be the last of this thread. However, as I said, I perfectly understand if this question is not considered appropriate for this site.1302687Sufficient criteria for a nerve of a topological category to be good86730862906Qlzqlzuup20105122108850>Stable Dold-Kan correspondence566070!Probably worth pointing out a certain distinction. The quoted property is Pete L. Clark's ADC property, see:

Must a ring which admits a Euclidean quadratic form be Euclidean?

For positive forms in three variables with integer coefficients, this is stronger than L. E. Dickson's "regularity," which means that the form integrally represents any number represented by some form in the same genus. The ADC property is not implied by class number one here, nor does ADC guarantee class number one. I sent Pete the 103 positive ternary ADC forms; there are 794 positive ternaries alone in their genera. Also, thanks to Pete for actually checking something I foolishly assumed, there are several ADC forms not alone in their genera, such as $\langle 1,1,3,0,1,0 \rangle$ or $g(x,y,z) = x^2 + y^2 + 3 z^2 + z x,$ of discriminant 11. I also gave an example for the 810 failures, each such regular form (including 14 not proved yet, after B-K Oh proved 8) represented some $p^2 n$ without representing $n,$ where the prime $p$ divides the Brandt-Intrau-Watson discriminant.

So the subtle point is that the ADC property in "Cassels's lemma" includes primes that divide the discriminant, which in the case of three squares is $-4,$ or taken as 4 by me and J. L. Lehman. The fact that the sum of three squares represents $n$ if and only if it represents $4n$ is, well, among regular forms, an artifact of a fortunate arrangement of local conditions, in this case the single "congruence obstruction" $4^k(8 m + 7).$

For example, the regular form $\langle 1, 1, 3, 1, 1, 1 \rangle$ or $g(x,y,z) = x^2 + y^2 + 3 z^2 + y z + z x + x y,$ integrally represents 8 but not 2. Oh, the discriminant of this form is 8, so the only prime available to distinguish between regularity and the stronger ADC property is 2.

Here is an easier one: $x^2 + y^2 + 4 z^2$ is regular, related in an evident way to the sum of three squares, and is not ADC as it integrally represents 12 but not 3.

575146Actually, I think there is a simpler derivation of the other bound. There is a best linear lower bound for this problem, since $\mu_k$ is clearly a convex (concave?) function for $k$. It's the kind where if you interpolate linearly between $k=1$ and $k=3^n$, that's a lower bound for $\mu_k$. $\mu_1=2^k$, $\mu_{3^k}=4^k$.Motivation of my question: Does conjugate Bailey pair have conjugate Bailey lemma like Bailey pair have Bailey lemma?

There is no convergence - $\liminf a_n=0$ while $\limsup a_n\ge 1$.17718111320438One possible way to show that a word w gives non-surjective map is to show that the derivative of the map y -> w(x,yz) at the origin is singular for x and z generic. In fact, I think this is an if-and-only if condition. Now, as to how to show that...1443767~Is there triangulated category version of Barr-Beck's theorem?1656679163016421758221801612209571402905https://www.gravatar.com/avatar/0d45f5504d8314e2806caa2e4d00499e?s=128&d=identicon&r=PG&f=1329028P.S. Given the new formulation, complete reducibility makes it easy to control the highest weight occurring in such a tensor product. So there seems to be no real problem here.499941379159@dhy, thank you very much. The only property I need is the image of the projection $G \to L_J$ is $L_J$. Does such projection exist?The truth about an an analogy between prime ideals and prime geodesics170934160074002047163:@AD1984: You're welcome! :-)1937304That last part is easy: The thought that $x$ is correct is formalized as $\{x\}$.@Strongly rigid Hausdorff spacesFor a fixed $k \geq 2$, are there infinitely many *non-trivial* coprime integer pairs $(x,y)$ for which $xy$ divides $(x+y)^k-1$? By *trivial* I mean parametrized pairs $(x,y)$ of the form $(-1,-1),(x,1),(1,y),(x,1-x),(x,1 - x^n),(1-y^n,y),(x,p(x)),(p(y),y)$, where $p(x)$ is a polynomial factor of $\frac{x^k-1}{x-1}$. When $k$ is even there are a few more trivial pairs, namely $(x,-1),(-1,y),(x,-1-x),(x,x^n-1),(y^n-1,y)$; where $n>0$ is a multiple of $k$.

As explained in Answer 3, the shift can be explained by the inverse power iteration. $x'=(A-rI)^{-1}x$ has larger component in $v_1$ if $r$ is closer to $\lambda_1$.

However in QR iteration, we don't use the inverse iteration, but the power iteration.

In the power iteration, $x'=(A-rI)x$ has smaller component in $v_1$ if $r$ is closer to $\lambda_1$. It seems the conclusion is opposite. I don't know what my misunderstanding is. I appreciated if anyone can help. Thanks.

6420104625091372874740502687440Yet (in another universe), Walter Pitts notified Russell of several mistakes in the Principia (when Pitts was 12!). http://nautil.us/issue/21/information/the-man-who-tried-to-redeem-the-world-with-logic12693992210094Another thing that doesn't convince me is that you claim to find a closed path terminating at $V_i$ for every $i$, which is stronger than the statement in the problem (for at least one $i$). I actually think this is false and one of the sets mentioned in the comments above is probably a counterexample, but I haven't checked carefully.39838520902401514150228540212889551922886135005518321611294529218513453894131437832539825V@StevenGubkin - This comment made my week.@Lucia's edits had significant improvements and did not change the meaning of the question. I restored them. n40886: This is how MO works. If you have questions or concerns about this, please ask on meta.22840682302823189238516567511027370156274811219512081831184245Please add a backslash before each underscore. The first compiler, which italicizes, etc., will remove each backslash but leave the underscore, and then JSMath will do its thing.2036164LOops sorry, I had missed the comment.1235514322749596340 @GarrettErvin My argument was terse, perhaps to the point of being cryptic. The intervals are defined inductively, and at each step we make sure that the newly chosen interval has room on both sides, the o.t. of the gap being a "nonzero right multiple of $\alpha$". Note that any nonzero right multiple of $\alpha$ can be expressed in the form $\alpha\theta$ where $\theta$ has no first or last element; this is because $\alpha\theta=\alpha\xi\theta$ and $\xi\theta$ has no first or last element. This is why the new interval can be chosen without crowding the old intervals.828544110971929060560104I could be wrong but doesn't Burnside's Basis Theorem only deal with the $p'$-automorphisms? (I think David hints at this in his question.) Also, I am slightly unsure that the result which David states is true! Are we sure that the given bound holds for $|Aut P|$ as opposed to $|Out P|$ (see the paper by Neumann "Proof of a conjecture by Garrett Birkhoff...")?Suppose $M$ is a 1-connected closed manifold with sectional curvature $\ge 1$. So the diameter $D$ of $M$ satisfies $$ D \le \pi $$ When equality holds $M$ is isometric to round sphere. In fact this rigidity holds under the assumption of $Ric \ge n-1$. Hence it is natural to ask what happens to almost extreme case. i.e. when $$D \ge \pi -\epsilon $$Under the Ricci assumption only, the manifold is not necessary sphere by a counter example of M. Andersen. However with extra assumption: sectional curvature is bounded from below, Perelman proved it is homeomorphic to a twist sphere. (Is Perelman the first one prove this?)

By Grove-Shiohama's Diameter Sphere Thereom, under the assumption sectional curvature $\ge 1$ and $D\ge \pi-\epsilon$, $M$ is a twist sphere. (It also follows from Perelman's theorem above)

*My question is: Is $M$ diffeomorphic to the standard sphere?*

Either Perelman's proof or Grove-Shiohama's Diameter sphere theorem uses the 'soft' approach, i.e. It is not by convergence argument, nor a Lipschitz distance between $M$ and $S^n$ is derived.

Actually, I suspect that it is not Gromov-Hausdorff close to the round sphere, as one might round off two tips of $S^2/\mathbb Z_p$, where $\mathbb Z_p$ acts on $S^2$ by rotation along the $z$-axis.

What if one assume there is also an upper bound $K$ on sectional curvatures, i.e. $$1\le sec(M) \le K$$

I'm looking for examples of the following phenomena. Let $X$ be a reasonable space (say, a CW complex) and $G$ be a finite group acting on $X$. For all $k \geq 1$, the projection map $X \rightarrow X/G$ induces a map $H_k(X;\mathbb{Q}) \rightarrow H_k(X/G;\mathbb{Q})$ which factors through the $G$-coinvariants $(H_k(X;\mathbb{Q}))_G$; let $\psi_k : (H_k(X;\mathbb{Q}))_G \rightarrow H_k(X/G;\mathbb{Q})$ be the resulting map. I want examples of $X$ and $G$ and $k$ such that $\psi_k$ is not an isomorphism.

If $G$ acts freely, then the map $X \rightarrow X/G$ is a finite regular covering map and $\psi_k$ is an isomorphism by (for instance) the Cartan-Leray spectral sequence (Theorem VII.7.9 in Brown's book on group cohomology). But I have no idea what happens for non-free actions. My guess is that if it were true that $\psi_k$ were always an isomorphism, then I would have seen it somewhere, so I expect that there is a counterexample. However, I have not managed to come up with one.

user212006554072No. The category of models of an equational theory (i.e. a variety in the sense of universal algebra) is always a regular category, but the category of posets is not regular.

:bounded homogeneous quarticsHi Pietro. I tried to investigate on how to find $U$ such that $U^T U = I$, when $U$ is $n \times m$ with $m > n$ but that is impossible no? For it to be possible you must have that $n > m$ which is not my case. Am I right?1401889Totsuka, Japan5051581884669757338Characterisation of a class of group homomorphisms related to a central extension2058549@Eric Naslund: a nit-pick, there is no Omega *plus/minus*, but of course you are still right regarding the main issue.Let $A$ be a bounded linear operator from a Banach space $M$ to itself. Suppose that $\rho(A)<1$ where $\rho(A)$ is the spectral radius of $A$. For any $\varepsilon>0$, does there exist an open neighborhood $U$ of the origin such that $U\subset B(0,\varepsilon)$ and $A(U)\subset U$, where $B(0,\varepsilon)=\{x\in M | \|x\|<\varepsilon\}$?

155036719732308790A recent good book is Algebraic Operads by Jean-Louis Loday and Bruno Vallette.

121195119330371340367175691176097822117441170326208271Let $G$ be a compact subgroup of $O(n)$. Let $\rho$ be a continuous finite dimensional representation of $G$.

QuestionIs it true that there exists a continuous finite dimensional representation $\pi$ of $O(n)$ such that $\rho$ is a direct summand of the restriction of $\pi$ to $G$?

I am almost sure that this is true. A reference would be very helpful.

|"Dva okoshka" has more charm than the plain "two windows. :-)344760221100724867It means what it says: one object in a category instead of every object. But I think the Cayley example is legitimate: we do in fact consider all Homs to the object because in this case the category is the group G considered as a one-object category! 1831499Is geometric formality preserved under rational homotopy equivalence?747010912621120253413678831974638@ChristianRemling: Isn't it obvious from the text between parantheses?21627521488991658416Bill, it is independent of ZFC. See this answer of Joel David Hamkins: http://mathoverflow.net/questions/8972/do-sets-with-positive-lebesgue-measure-have-same-cardinality-as-r/9027#9027 or this expository note by Briggs and Schaffter: http://www.jstor.org/stable/2320153.Your question is essentially about surjective stability for *relative* symplectic $K_1$. The latter follows from the usual (absolute) surjective stability for $K_1$, which in symplectic case starts at $2n\geq \mathop{\mathrm{sr}}(R)$. To prove this, one can use so-called "Stein relativization", as described in M. Stein, "Relativizing Functors on Rings and Algebraic K-Theory", J.Algebra, 1971. See Corollary 1.7 therein or the remark after Theorem 4.2 in Stein's other paper, "Stability theorems for $K_1$, $K_2$ and related functors modelled on Chevalley groups", Japan J. Math, 1978.

It is also possible to prove it directly under somewhat weaker assumprion on a ring by explicit calculations with generators, but since you are interested in Dedekind domains, the usual stable rank condition should suffice.

For the case of a square see http://mathoverflow.net/questions/124579/mean-minimum-distance-for-n-random-points-on-a-unit-square-plane.Finding such cliques is problem D15 in Richard Guy´s Unsolved Problems in Number Theory (3rd edition). Many examples are known of K4. Stan Wagon apparently found examples of K5.1017683163614220061275405741862833738592Pete is also implicitly using the odd order theorem, to know that the nonabelian simple factor cannot be odd.2287989@ Jim Thanks for that reference. My library does not seem to have the book but I will keep looking. It seems like the place to look for my obscurely worded question.I guess the point is supposed to be that the space of subspaces of corank exactly n (which you say is homotopy equivalent to X) models BU(n) because one can build a vector bundle over it where the fibers are V/W? And they you're trying to argue that when you include the subspaces of smaller rank, you're not changing the homotopy type. I guess I see better the strategy now.1658503<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.

The question is imprecise concerning the topology involved and quasi-compactness conditions on the group schemes, so in view of the motivation let's first stick to the affine case before we venture beyond that.

Rather generally, consider any left-exact sequence $$1 \rightarrow G' \rightarrow G \rightarrow G''$$ of affine group schemes over a field $k$; i.e., $f:G \rightarrow G''$ is a homomorphism with $G' = \ker f$. For instance, if we begin with such a left exact sequence that is equivariant for the action of a group $\Gamma$ then the induced diagram between closed subgroup schemes of $\Gamma$-invariants is also left-exact.

Hence, the question posed is a special case of the general claim that for *any* left-exact sequence of affine $k$-group schemes as above, there is a unique factorization of $G \rightarrow G''$ as the composition of a faithfully flat homomorphism onto a closed $k$-subgroup scheme $\overline{G} \subset G''$ (and hence by fpqc descent, $G \rightarrow \overline{G}$ represents an fpqc sheaf quotient for $G$ modulo $G'$).

But this is "easy" in terms of coordinate rings: define $\overline{G}$ to correspond to the image of $f^{\ast}:k[G''] \rightarrow k[G]$. This is visibly a closed $k$-subgroup scheme of $G''$ through which $f$ factors, with the resulting map $G \rightarrow \overline{G}$ visibly having the same kernel $G'$ as $f$. Moreover, the map $G \rightarrow \overline{G}$ is faithfully flat because the corresponding map on coordinate rings is injective and *any* injective homomorphism between Hopf algebras over a field is always faithfully flat (this is proved in Waterhouse's book on affine group schemes, for example).

Of course, this argument was all rather specific to the affine case. But one can do much better: any quasi-compact homomorphism between arbitrary group schemes over a field factors as a faithfully flat homomorphism onto a closed subgroup scheme of the target (so that closed subgroup represents the fpqc sheaf quotient by the kernel); i.e., the conclusions above hold with affineness removed, provided we just assume $G \rightarrow G''$ is quasi-compact. For a proof, see Corollary 6.7 in Expose VI$_{\rm{A}}$ of the new edition of SGA3 (it is not in the original edition). The key input is a theorem of Perrin not proved there (but recorded as Theorem 6.5 there), concerning approximation of a general quasi-compact $k$-group scheme by finite type $k$-group schemes; in effect, every such $k$-group is an extension of a finite type one by an affine group scheme.

118662random_person1155881@DelioMugnolo Indeeed, I didn't noticed that theorem. I didn't read the chapter yet. Is the theorem applicable for $Au:=\Delta u$, $u\in\mathcal D(A):=\left\{u\in H_0^1(\Lambda):u\text{ admits a weak Laplacian }\Delta u\in L^2(\Lambda)\right\}$, and $\Lambda$ being just bounded, convex and open (without any assumption on the boundary)?1761547157697115597896992962570693052For now, I just can remember Monique Hakim thesis, but maybe other standard books on elementary topos will have something, like Johnstone's book "Topos theory" or "Sketches of an elephant". Actually, for any topos the abelian group objects will form an abelian category and you have the global sections functor (the unique morphism to the terminal topos of sets), then you can just apply the usual machinery of cohomological functors and pick the derived functor of global sections.645057176035229854433965965020670466991825213454717245713604522206991909846652888419162How much is known about unstable homotopy of truncated projective spaces? - Reference request I'm not sure if this works (I'm too lazy to do the tensor calculation), but if you differentiate $\Delta f = 0$ twice, you should get something like $\Delta \nabla^2f + A\nabla^2 f = \nabla\cdot(B\nabla f)$, where $A$ and $B$ depend on the curvature tensor (and maybve its covariant derivative). If so, then a straightforward Moser iteration argument gives an $L^p$ bound on $\nabla^2f$ in terms of the Sobolev constant, curvature (and maybe its covariant derivative). This in turn implies a Holder bound on $\nabla f$, where the constant can also be bounded in terms of geometric invariants.50835314609062185426For My Applications, I am happy with the existence of such an averaging map indeed.>Minimal Hausdorffness reversed197590516160181718080175775119333376333815980086997342005512p"Orthogonal complement" of a subspace of a Banach space996088 Re rock: it splits into a physics problem to determine an adequate mathematical model of the physical reality in question (Newton’s laws, gravity, friction forces, ...), and a mathematical problem to figure out the trajectory in said model. Likewise for Zeno’s paradoxes, finding a mathematical model of the situation is a problem of physics, and working out the solution of the paradox in such a model is a problem of mathematics. Each model yields a different mathematical problem. In the recent literature you mentionn, if someone proposes a novel model of motion and immediately solves ...To see why Morse theory gives more info in the variational case consider system of equations on a torus $T^n$:

$$f_1(\theta_1,\dotsc, \theta_n)=\cdots=f_n(\theta_1,\dotsc, \theta_n)=0.$$

where $f_,\dotsc, f_n$ are smooth functions. Degree theory cannot say much about this system. However if

$$ (f_1,\dotsc, f_n)= \nabla g (\theta_1,\dotsc, \theta_n) $$

for some smooth function $g$, then Morse theory predicts that system of equations

$$ \nabla g=0 $$

has at least $n+1$ solutions. If we a priori know that all the solutions are nondegenerate, then we can conclude that that are at least $2^n$ solutions.

Look at the simplest case when $f(\theta)$ is a smooth $2\pi$-periodic function. The equation $f(\theta)=0$ may have no solutions, but the equation $f'(\theta)=0$ has at least two solutions.

The variational nature of the equation $f'(\theta)=0$ adds some subtle features which Morse theory speculates to produce more refined results.

112271721160871754152You can identify $\operatorname{Sym}^2(\mathfrak{h})$ with the space of symmetric bilinear forms on $\mathfrak{h}$. For each root $\alpha$, you can define a symmetric bilinear form $\varphi_\alpha$ on $\mathfrak{h}$ via $\varphi_{\alpha}(h, h') = \alpha(h)\alpha(h')$ for all $h, h' \in \mathfrak{h}$. Then it seems in the case of $\mathfrak{sl}_n$ that $\{ \varphi_\alpha : \alpha \text { positive root} \}$ is linearly independent in $\operatorname{Sym}^2(\mathfrak{h})$. ...9927652884784653252043331203312499921847393Not only are those monotonic sequences of length $n = 2k(k+2)$ as fair as a sequence can ever get, but the score both players expect with them is asymptotically $n-o(n)$, leaving them both quite satisfied !87072183414271265Think of it locally. $\mathbf{Q}_2(\sqrt 5)$ is the unramified quadratic extension of $\mathbf{Q}_2$, and $\mathbf{Q}_2(\sqrt{-1})$ and $\mathbf{Q}_2(\sqrt{-5})$ are ramified quadratic extensions, so the compositum $\mathbf{Q}_2(\sqrt 5, \sqrt{-1})=\mathbf{Q}_2(\sqrt{-5}, \sqrt{-1})$ is an unramified extension of $\mathbf{Q}_2(\sqrt{-1})$ and of $\mathbf{Q}_2(\sqrt{-5})$.682448129882792182136824016816177483081466972153266867461814979881292037147244151496257To get yours from mine, set $B = A + D$. To get mine from yours, set $A = B$ and $D = 0$. Setting $D = 0$ is harmless, since positive semidefinite matrices are limits of positive definite matrices.885934@Turbo Since I am interested in real multiplications, the conjectured lower bound is $7$ (from the article you referenced).720373889071343462270505The reference to Dijkstra's transformation rules indicates what I want to do. More explicitly I want to transform expressions in general, not just predicates. Also it 'hides states under the carpet' as I want. My interest is in programming language semantics and design. Model-based semantics strives to explain states, transformation-based strives to eliminate states. The latter potentially allows programs (including imperatives ones) to be transformed to pure mathematics.4787298Discrete logs vs. factoring1967333203936017364871259081367758Thanks Daniel. I've had a look and section 12.2 (The curvature tensor in local coordinates) has been helpful, but it only deals with Kähler metrics. 60849168322:Cohomology of quotient space@MarianoSuárez-Alvarez: Sorry for the vagueness, I simply mean an extension such that $g_1^2+\dots+g_n^2=z_1^2+\dots+z_n^2$.H@Michael, I like my version better.117749213404327306921293902\Discrete Laplace operator and its eigenvalueszReally? I appreciate the use of funny words in mathematics.15492251586505The hit-and-run algorithm and variants are popular choices. These are Monte Carlo methods but should be much better than rejection sampling. Unfortunately I don't know of a canonical reference.

444508145913018007002236538The second claim is false for $n=3$. Choose $\varepsilon$ small and $\delta\ll\varepsilon$. Let $A$ be the set of all points $(x,y,z)\in\mathbb R^3$ satisfying the following inequalities: $$ \begin{cases} -1.5+\varepsilon+\delta &\le x+y+z &\le 1.5+\varepsilon \\ -1.5+\delta &\le x+y-z &\le 1.5 \\ -1.5+\delta &\le x-y+z &\le 1.5 \\ -1.5+\delta &\le -x+y+z &\le 1.5 \\ \end{cases} $$ The integer translates of this set cover the space, but its $\ell_1$-diameter is no greater than $3-\delta$.

**Added.** The first claim is false too. In the above example, fix $\delta=\varepsilon/10$ and add the inequality
$$\max\{|x|,|y|,|z|\}\le 0.5+10\varepsilon$$
to the system.

To the best of my understanding an exceptional collection of elements in $\mathcal{D}(X)$ determines a quiver - and quiver mutations come to represent corresponding mutations among the elements of the collection, as developed by Rudakov et. al. (vertices of the quiver). In this context, there is a beautiful construction due to P. Seidel - on the mirror analog of such mutations in the Fukaya category

18825195802613011901158011A natural Lascoux-Schützenberger involutions on plane partitions29512938092210167631186136@Name Likely you will need to find a library holding the journal [Sphinx-Oedipe](http://www.worldcat.org/title/sphinx-oedipe/oclc/9034835). Alas, I no longer have a copy.2251240user131379Let $\Gamma$ be a discrete subgroup of $SL_2(\mathbb{R})$ which has a cusp at $\infty.$ suppose that $\mu(\Gamma\setminus\mathbb{H})<\infty,$ consider the Eisenstein series :$$E(z,s,\Gamma)=\sum_{\gamma\in\Gamma_\infty\setminus\Gamma}\dfrac{y^s}{|cz+d|^{2s}}$$ what is the analytic properties of $E(z,s,\Gamma)$ ?

If I remember right, LISP was invented in the late 50's. Something like 1958 or so.see my answer. still, AFAIK, the transformations needed are explicit.20865981556763208856 Hm. I suppose you are right. I should have been more careful. Well, here's at least an example of a valuation ring where there *exists* a non-maximal nonzero prime $\mathfrak p$ with no proper non-primary subideals. Let's take a valuation ring with value group $\mathbb{Z} \oplus \mathbb{Z} \oplus \cdots \oplus \mathbb Z$, with lex order. That is, you have countably many copies of $\mathbb Z$ with order type $\mathbb N$, plus one at the end. Then let $\mathfrak p$ be the ideal of elements whose value has some nonzero entry *before* the last entry. Then $\mathfrak p$ is the union of ...143515430781122933822197091The object of the algorithm is to find $S$. I would like $S$ to have the property such that, for every other vertex-set $T\subseteq V(G)$, the cut associated with $S$ contains at least an $\epsilon$-fraction of the edges in the cut associated with $T$.2015067168524511353552277858Take an organism that ordinarily migrates south, put it in an artificial environment, and reset its circadian clock by $x$ hours (by gradually shifting the hours of daylight). Release it into the wild, and instead of migrating south, it will migrate in a new direction $y$. Thinking of $x$ and $y$ as elements of $S^1$ and $y$ as a function of $x$, this gives a map $S^1\rightarrow S^1$. Apparently the winding number of this map is characteristic of the species (e.g. $0$ for the pond skater, $1$ for the sunfish). (Source: Arthur Winfree's Geometry of Biological Time.)

I want to point out that the recursive subdivision algorithm can be made a little faster, and the infinite recursion problem can be avoided in the following way: As a first step iterate over $C$ and discard any members, $C_j$, where the sum of the radius of $C_j$ and the radius of $C_{n+1}$ is less than or equal to the distance from the center of $C_j$ to the center of $C_{n+1}$. You could also save yourself a few recursions by making the initial rectangle a square bounding only $C_{n+1}$ instead of a rectangle bounding all of the circles.4032031060145But here the characteristic, for example, is one of the unknowns :)57499722379001122229This doesn't answer the question of why Fourier analysis works, but it certainly is an answer to how one might think "Hmm ... perhaps we're in the domain of Fourier analysis here." It's that the surface area of a shape X is defined in terms of the volume of X+B when B is a small ball. There is a close relationship between sumsets and convolutions (the sumset is precisely the set of points where the convolution of the characteristic functions of the two sets is non-zero), and every time you have a convolution, thoughts of Fourier analysis should be triggered.

The reason I say this doesn't answer the question of why Fourier analysis works is that there is a difference between the convolution and the support of the convolution, and the latter does not transform nicely. But that shouldn't stop one thinking of Fourier analysis and attempting to find some way of relating surface area to convolutions.

214143611573511714113Yes, though personally I find "vertex (up to a small ball)" more intuitive.17997But that does not answer the question. For, on $\mathbb{H}^2$ there are indeed groups acting isometrically and properly discontinuously so that $\mathbb{H}^2/G$ is compact. Any compact, hyperbolic surface arises in this way. The question is asking, given $S$, does there exist a $G$ making $S/G$ compact. So exhibiting a $G$ such that $S/G$ is not compact does not answer the question.576981258577This has been in my mind for quite some time. Looking at Artin Schreier Theorem for fields:

If L is a field and K its algebraic closure and if 1< [K:L] < infinity then L=K[i] and L is a real closed field (Thus L has characteristic 0. Here i is just the square root of -1).

I was wondering if a "generalized" Artin Schreier exist or if someone could refer to me to some paper that attempts this. There is a concept of real closedness and "algebraic closedness" of reduced commutative rings, but I doubt that the statement would hold.

So one has the following conjecture:

If L is a reduced commutative ring and K is its total integral closure (this is an equivalent notion of algebraic closure if K and L were fields) and if 1<[K:L]< infinity (here I mean that K is a finite L-module that is not the same as L) then L is real (thus its characteristic is 0.. and one can add that L is real closed in the sense of reduced commutative rings).

Can one easily show this, even at least prove that L has characteristic 0?

3469873724@FilippoAlbertoEdoardo: Well, I actually meant $p^n$-cyclotomic extensions. Yup, I knew about the result about unramifiedness (?), but I'm actually interested in wild ramification.560231$\newcommand{\End}{\operatorname{End}}$

let $R$ be a local ring, $\varphi\in \End(R_{R}^{2})$, $\overline{\varphi}\in \End(\overline{R}_{\overline{R}}^{2})$, $\overline{R} =R/J(R)$ , $J(R)$= Jacobson radical $R$. where neither $\varphi$ nor $1-\varphi$ is invertible. Why neither $\overline{\varphi}$ nor 1- $\overline{\varphi}$ is invertible in $\End(\overline{R}_{\overline{R}}^{2})$ ?

jHow do we even define Banach manifolds with corners?22748306904271244650Set $\tilde{C}:=\overline{C\times_{\mathbb{P}^1}C\smallsetminus\mathrm{diagonal}}$, then $J\tilde{C}$ is a quotient of $JC\times JC$, and $JC$ maps into $J\tilde{C}$, where the kernel is a subgroup of the $2$-torsion points. Moreover, $J\tilde{C}/\mathbb{P}^1$ is Galois (with Galois group $S_3$). Now let $C'$ be the unique intermediate cover of $\tilde{C}/\mathbb{P}^1$ which is degree $2$ over the $\mathbb{P}^1$, then $\tilde{C}/C'$ is a cyclic degree $3$ cover, so you get some $3$ torsion data.

31728453117308869Is there a similar statement for symmetric bihomomorphisms? I'm looking for something of the form "Every symmetric bihomomorphism on $G$ is of the form $\kappa \cdot \kappa^T$ for a 2-cochain $\kappa$ with property X and two 2-cochains with property X give the same bihomomorphism if they differ by Y"

What are $\lambda$ and $\rho$ in this question? Parameters? Variables? I find the question hard to understand.13401831387431214012918047135028114450081334101171800790803651866144user159888I have lecture notes that I'd like to turn into a book one day. I have not yet had time to adapt to my new TeX system, and the drawings done via ps-tricks do not yet come out as planned. In addition, there are gaps and mistakes that I have not yet had time to fill and correct (the chapter on composition should be essentially correct and complete, however). In any case, the present set of notes, for the time being, can be found here. Comments and corrections are welcome.

Also, this idea of proof is given in some detail in "Fundamentals of the Theory of Groups", by Kargapolov and Merzljakov. There it is shown any free group is residually a finite p-group for all primes p, which resolves both residual finiteness and residual nilpotence.Hi, 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?1558863I actually did intend for independence to be "neither provable nor refutable in $T$". For example, assuming that $ZFC$ is consistient, then ZFC $\vdash 0=1$ is neither provable nor refutable from ZFC. Its obvious that if $ZFC \vdash \phi$ is true, then even PA can prove that fact. I guess I'm more interested in when $ZFC \nvdash \phi$ is true, but not provable in strong theories.227085Thanks donovan, you should post that as an answer so I can give you some points =). Also, thanks to Will for reminding me of the term "Arc Length" which is a good keword to use when searching for this problem.$G({\mathbb Z}_p)$ is compact, hence the only discrete subgroups are finite. On the other hand, $G({\mathbb Q}_p)$ doe have discrete subgroups, which, if the ${\mathbb Q}_p$ rank is high, are arithmetic, if they are of finite covolume.FI see. Thanks for clarifying that.Tom Jalinki2178295219139515398201207470Note that the notion of a (not necessarily chiral) genus zero CFT is not the same as the notion of a full CFT. Every full CFT yields an example of a genus zero CFT, but there are more genus zero CFTs than full CFTs. Your Theorem 3 does not hold for genus zero CFTs. See the first 10 pages of my course notes for some clarifications about various notions of CFT: http://www.staff.science.uu.nl/~henri105/Teaching/CFT-2014.pdf (in there, I called genus zero CFTs "weak CFTs").Ooh, very nice --- a classic "check your implicit assumptions" example. Good point!:So thanks, Pro. Dr. Y. Choi.3803486608481456817Steven Sivek's answer is much more elementary than the reference I gave, in fact torus knots have bridge numbers $\to \infty$, yet are tunnel number 1. The point of Jesse's construction is that he gave hyperbolic examples with this property.This is a follow-up to an earlier question.

The answer to that question was found on this page. The discussion on OEIS seems to suggest that, for any prime $p$, there should exist a $p$-length arithmetic progression of primes beginning with $p$.

Is this known?

212662082192You might want to look at the paper "Spectral Analysis for Adjacency Operators on Graphs" by Mantoiu et al.https://www.gravatar.com/avatar/bab5b20c0d9bed604b5c27897376904c?s=128&d=identicon&r=PG&f=180128175286112512091157007334574884798@user17597 My question aimed pretty much at the cases where the values of $n,p$ is such that Sobolev embedding does not hold. How would we use Poincare inequality in those settings?200485617046492121351486610183663219498903486244981You might have a look at Hatcher and Thurston's paper "A presentation for the mapping class group of a closed orientable surface". They use Morse functions on the surface to facilitate the derivation of their presentation. On page 223 and following, they discuss how to form a graph from a Morse function on a surface, which is the leaf space of the level sets of the Morse function. This graph, together with its map to $\mathbb{R}$, uniquely determines the Morse function on the surface (up to graph isomorphism preserving the function). In particular, one need only know the values of the Morse function at the vertices of the graph (corresponding to the critical points of the Morse function on the surface) since it is monotonic on each of the edges. The space of such functions will therefore be parameterized by a subset of a vector space, satisfying certain inequalities. As they indicate on p. 224, the surface with the Morse function may be recovered from the graph (with its Morse function) by embedding it in $\mathbb{R}^3$ and taking the boundary of a regular neighborhood (after a small perturbation).

21333819This property is called "coherence", and no, it doesn't always hold.

Establishing this property holds for a given semantics of proofs is a proof obligation. An example of when it doesn't arises with coercive subtyping -- if the diagrams corresponding to possible coercions do not all commute, then the semantics is not coherent, in that the meaning of a term depends critically on its typing derivation. If all you have is the term and knowledge that it has a typing derivation, you can't tell what it means, necessarily.

For a proof-theoretic characterization, the thing to look for is a cut-elimination theorem. One way of reading the cut-elimination theorem is precisely that it says that eliminating intro/elim and elim/intro pairs (i.e., $\beta$ and $\eta$ equations) induces a genuine equivalence relation on proofs.

However, we can still have essentially different proofs. For example, $x : A \land A \vdash \pi_1(x) : A$ and $x : A \land A \vdash \pi_2(x) : A$ are two different proofs that $A$ and $A$ entails $A$. However, cut-elimination guarantees that the addition of proof terms does make the typing derivations unique -- any proof of this entailment must be equal to one or the other, since every proof is equivalent to a cut-free proof, and these are the only two cut-free proofs, and the proof terms let us distinguish the two possibilities.

EDIT: There must be a treatment of this subject in somebody's book (Barendregt?), but I don't know off-hand, since I learned it via osmosis. Probably the best reference I can point you at is Jan Schwinghammer's paper "Coherence of Subsumption for Monadic Types", which nicely illustrates the idea, and whose references also contain pointers to the best available literature.

Noam quite rightly points out that you might actually be interested in the case where all derivations of a judgement are equivalent. (This possibility honestly hadn't even occurred to me, since I'm too intuitionistically contaminated.) The keyword to search for in this case is "proof irrelevance", and I recommend Awodey and Bauer's paper "Propositions as [Types]".

EDIT 2: When cut-elimination holds, then the derivations of a calculus *with proof terms* become equivalent. If you don't have proof terms, then you can't guarantee equivalence of derivations.

For example, suppose we erased the proof terms from the derivation above, so that we simply had the judgment $A \land A \vdash A$. In this case, even if we know that this judgment holds, we still don't know whether the proof used the left or the right $A$, which are the two intrinsically different derivations of the same judgment. On the other hand, if we have proof terms, then we have a judgment $x : A \land A \vdash e : A$. In this case, the derivation does bebgcome unique, because now we can $\beta\eta$-normalize $e$ and just check whether it is equal to $\pi_1(x)$ or $\pi_2(x)$.

This is why lambda-calculus expressions are called "proof terms": these terms are evidence to establish which proof you meant.

To connect this back with proof irrelevance, a type is "proof irrelevant" when all the proof terms of that type are equivalent. That is, if $\Gamma \vdash e : A$ and $\Gamma \vdash e' : A$, then we know that the two derivations are equivalent, regardless of $e$ and $e'$. For example, the unit type $1$ is proof irrelevant, since every term in it is equivalent to the unit value $\left<\right>$.

22051045794112088234Given an undirected connected graph on $n$ nodes, let $S$ be the subspace of vectors $x \in \mathbb{R}^n$ which satisfy $$\sum_{j \in N(i)} x_j = 0,$$ for all $i=1, \ldots, n$. Here $N(i)$ is the set of neighbors of node $i$; the graph may have self-loops so that it is possible that $i \in N(i)$. I am interested in understanding the dimension of $S$.

Some examples:

- On the complete graph with self-loops at every node, ${\rm dim}(S)=n-1$.
- On the ring with four nodes, ${\rm dim}(S)=2$.
- On the line with three nodes and self-loops at every node, ${\rm dim}(S)=0$.

My question: can any connections between ${\rm dim}(S)$ and any combinatorial graph quantities be made?

1422883no because in the infinite case the result says "there exists some C'..." I need a bound on C' in terms of the parameters b and C.74002049496115462647267266148381020455We seem to be talking at cross purposes. The dual of $M(K)$ with the weak star topology is $C(K)$, just as the dual of any dual Banach space with the weak star topology coincides with the original space.67222vSecond order Taylor expansion to solve system of equationsFrpzzd213512011888291321756Simon Scarfe17522332044049434343I'm reading a paper and here the authors say that a connected 4-manifold with zero rational top homology has a homotopy type of 3-dimensional CW-structure. I can't figure out how it can be done.

1317109genonymous2203674922839:@Ralph: no, see the picture.Hi, I am wondering what the difference between 'generalized gradient' and 'subgradient' of a (potentially non-differentiable) convex function 'f' is.

The generalized gradient I am interested in is meant in the sense of the paper http://www.is.tuebingen.mpg.de/fileadmin/user_upload/files/publications/ICML2010-Kim_6519[0].pdf (see page 3, footnote 2).

many thx for any help.

Let $g$ be a semi-simple Lie algebra, and $0 \to I \to h \to g \to 0$ an Abelian extension of $g$. Then $g$ acts on $I$. Considering $g$ under the adjoint action, when is there a $g$-module isomorphism between $g$ and the k-th exterior power $\Lambda^k(I)$ for some $k$? Only when $g = so(n)$, $k=2$, $k = n-2$?

9346331450968263774Fernando's answer is excellent, but I can't resist mentioning what is perhaps the simplest counterexample to a generalization to your question. As Fernando says, there are counterexamples if you generalize from rings concentrated in degree 0 to dg-algebras. These examples are somewhat complicated, but if you generalize further to $A_\infty$ ring spectra, there is a very easy example.

Fix a prime $p$ and an integer $n>0$ and consider the Morava K-theory spectrum $K(n)$, which can be given an $A_\infty$ structure. The homotopy groups $\pi_*K(n)=\mathbb{F}_p[v_n^{\pm1}]$ are a graded field, and it follows that every $K(n)$-module is free (a wedge of suspensions of $K(n)$). In particular, the homotopy category of $K(n)$-modules is semisimple, and is actually equivalent (as a triangulated category) to the category of graded $\mathbb{F}_p[v_n^{\pm1}]$-vector spaces. This category is also equivalent to the homotopy category of $H\mathbb{F}_p[v_n^{\pm1}]$-modules (or equivalently, dg-modules over the graded ring $\mathbb{F}_p[v_n^{\pm1}]$). However, the corresponding $(\infty,1)$-categories are not equivalent. Indeed, the space of endomorphisms of a simple $K(n)$-module is $\Omega^\infty K(n)$, while the space of endomorphisms of a simple $H\mathbb{F}_p[v_n^{\pm1}]$-module is $\Omega^\infty H\mathbb{F}_p[v_n^{\pm1}]\simeq \prod_i K(\mathbb{F}_p,2i(p^n-1))$. These spaces have isomorphic homotopy groups, but are otherwise quite far from being homotopy equivalent.

20435619360571866024426101184825312078766When did you "meet Polya"?991658After correcting the sign issue the numerical test also checks out.15026161296651801339140265711066I am not an expert on this topic, so someone please correct me if I'm wrong, but I believe the answer to this question is yes.

The stack BG (resp. BH) is represented by the simplicial scheme also usually denoted BG (resp. BH) which is obtained by covering BG (resp. BH) by a point and then taking the nerve of this covering. Then a map from BG \to BH should just be given by a map of the corresponding simplicial schemes, which in particular includes a map G \to H (these are the 1-simplices). However, I think that this map completely determines the map BG \to BH (this should have something to do with the fact that BG and BH have no nontrivial homotopy groups beyond \pi_1, so we really only need to work with groupoids).

Please delete your incorrect answer because it is generating downvotes.7839003940422956778Great! Thanks a lot for the clarification. I indeed liked the interesting fact of the non-amalgamable extensions. By the way, where could I find the theorem of yours regarding the positive side of amalgamating extensions? Thanks again!2038275203343019072011377464947420902350the reason for my question is the following: the two-dimensional canonical singularities are the ADE-singularities, which all are quotients of either affine space or another ADE-singularity by finite abelian Groups and even more, quotients of affine 2-space by finite solvable Groups but one - namely $E_8$, which is factorial and a quotient of $A_1$ by the perfect Group $A_5$.

Now a generalization of the ADE-singularities in Dimension 3 are the compound du Val singularities, which are canonical and are formally equivalent to $f(x,y,z)+tg(x,y,z,t)$, where f is the polynomial of an ADE-singularity.

Now some of these are again quotients of affine 3-space, while a lot more than the one before ($E_8$) are factorial. For example the one in the heading. Which means in particular, that they cannot be a Quotient of affine space by a solvable Group.

As you may see now, it would be very nice, if these singularities were also some quotients by any Groups. Any idea? I am quite sure that it can't be a finite Group and affine 3-space.

Thank you and looking forward to your suggestions, Lukas

1696492The latter classifies orientation-preserving actions of finite groups in genus 2 and 3: http://www.sciencedirect.com/science/article/pii/002240499190021S#bHolomorphic functions in almost-complex geometry1422961One important example of a monadic category over $C$ is one obtained by adjoining to $C$ a system of $n$-ary operations satisfying some relations. For example, the category of rings with involution is monadic over sets: it can be given by two binary operations ($+,\times$), two $0$-ary operations ($0,1$), and two $1$-ary operations (negation and the involution). Perhaps if you adopt a suitably enlightened definition of operations with relations then all monadic functors would arise in this way.

Now a $1$-ary operation is just a morphism, so you can also view it as a $1$-ary operation in the opposite category. So if you're adding only $1$-ary operations, then you should get a category which is both monadic and comonadic over the original category.

Indeed, the usual way of defining lambda-rings is by adjoining a bunch of $1$-ary operations to the category of commutative rings. The monad is the free lambda-ring functor, and the comonad is the big Witt vector functor (also called the co-free lambda-ring functor). Another example is the category formed from objects of the original category $C$ together with an action of your favorite group (or monoid) $G$. In representation theory, the monad is often called the induced representation functor, and the comonad is called the co-induced representation functor.

I'd expect that, as above, with suitably enlightened definitions, all examples would arise in this way.

Some things that follow formally from this set up are that the new category has all the same kinds of limits and colimits that the original category $C$ has, and the forgetful functor preserves them. Beck's theorem says that some kind of converse is also true.

22368771013728825278@მამუკაჯიბლაძე: (1) Of course it does. It's a vector space, $\alpha$ is unique once you've fixed $v$. (2) No, $0$ is always sent to $0$. So you cannot be zero everywhere.Certainly there are more: any polynomial in $H_e$'s. They are not covered by the Hadwiger's theorem. Thus the question is if there are other invariants independent of $H_e$'s.739499525721>Here's my attempt to rigorously show that it cannot generalize. I originally wrote this for my blog, using the following formulation of the puzzle:

- All islanders are immortal, except as explained below.
- All islanders know the color of every other islander's eyes, but not their own.
- If an islander has enough information to logically deduce the color of their own eyes, they will perish at midnight of the following day, which will be noticed by the other islanders.
- Any message washed ashore in a bottle can be trusted.

Now as it happens, at 12:01am on Day 0, a message washes ashore in a bottle, and is read by all the islanders:

There is at least one islander with blue eyes.

What happens?

The following solution works for infinitely many islanders, as long as only finitely many have blue eyes. I go into more information on my blog post, but since you already know the blue eyes puzzle you can probably prove the following.

**Lemma 1** If an islander sees at least $N$ other islanders with blue
eyes, she knows on Day $0$ that she will not be able to deduce her eye
color before Day $N$.

**Lemma 2** If an islander sees $N$ other islanders with blue eyes who
are alive on Day $N$, then she can deduce her eye color by Day $N+1$.

**Theorem 3** If there are $N+1$ islanders with blue eyes, they will
perish at midnight on Day $N+1$, and all others will survive.

So, we've handled the case where there are an arbitrary number of islanders (even infinite!), but only finitely many islanders with blue eyes. So, what if there are infinitely many islanders with blue eyes?

Well, we already have a very nice lemma above, **Lemma 1**. That allows us
to prove that, since all islanders see infinitely many pairs of blue eyes,
no islander can deduce their eye color by Day $N$ for all
$N\in\{0,1,2,\dots\}$.

So, assuming that our days are limited to the finite ordinals, the case is closed. But why should we stop there? What if there exists a day for every ordinal number?

We should consider what happens on Day $\omega$. Will anyone wake up dead? The answer is of course, no. All islanders were able to deduce from the beginning that no one could perish after a finite number of days, so by the time Day $\omega$ begins, no one has gained information they didn't already have. We shall be able to extend this argument by transfinite induction to handle every ordinal (regardless of the cardinality of the ordinal or the set of blue-eyed islanders).

**Theorem 4** If there infinitely many islanders with blue eyes, then no islanders
can deduce their eye color by Day $\alpha$, where $\alpha$ is any ordinal.

*Proof.* If $\alpha$ is finite, we are done by Lemma 1. By transfinite induction,
we may assume the theorem holds for all $\beta\lt\alpha$. In order to
deduce her eye color by Day $\alpha$, $Alice$ requires new information on
a previous day. However, by the induction hypothesis, $Alice$ knows from
the beginning that every islander cannot deduce their eye color by a previous
day. So, if $\alpha$ is a limit ordinal, we are done. The other case
is when $\alpha=\beta+1$ for some $\beta\lt\alpha$. Alice wakes up
on Day $\beta$, and is immediately bored. She already knew that everyone
sees infinitely many islanders with blue eyes, so she already knew that
everyone would be alive. Indeed, they are. So, she has gained no new information
that morning, and therefore cannot deduce anything new by the end of the day,
the beginning of Day $\alpha$. $\square$

In a commutative field $K$, the Zariski dimension of an algebraic subset of $K^n$ over $K$ does not vary if one enlarges $K$ if I understood well. In particular, for two Zariski-closed vector spaces $V(K)\subset W(K)\subset K^n$, if one considers an extension field $L/K$, one has $$[V(K):W(K)]=[V(L):W(L)].$$

Let $D$ be a division ring with infinite centre $k$, infinite dimentionnal over $k$. Let $a$ be a non-central element of $D$. Its centraliser $C_D(a)$ in $D$ is defined by the simple linear equation $$ax-xa=0.$$ The ring $D$ is a right vector space over $C_D(a)$ having dimension $[D:C_D(a)]_{right}$.

Let $L/D$ be a division ring extending $D$ and still having $k$ for centre.

Question 1: Is it possible to have $[D:C_D(a)]_{right}$ finite and $[L:C_L(a)]_{right}$ infinite ? And vice-versa ?

119611817629501680180"Can we bring some order to this situation?" Please formulate a proper question. While likely not a propblem in itself and isolation, such formulations from established users set a bad example. Also, you should use a top-level tag.Question 2: Is it possible to have $[C_D(a):k]$ finite and $[C_L(a):k]$ infinite ?

Programmers who trade abstraction for minor conveniences deserve neither and will lose both.

Too long for a comment.

I note first that I made an attempt to reduce the problem to the case where $K$ is normal, but it turned out to be false; I'm thankful to Ian Agol for his discussion. The case where $K$ is normal follows at once from a theorem of Zelmanov stating that every periodic torsion group is locally finite.

This second attempt is not a complete answer; however, it reduces the problem considerably:

*Let $G$ be as above, then, virtiually, $L_p(G)$ satisfies a PI, for every prime $p$; with $L_p(G)$ denotes the Lie algebra associated to the dimension subgroups (over the field of $p$ elements).*

Indeed, let $K_n$ denote the set of elements of $G$ satisfying $x^n\in K$. As $K$ is closed, each $K_n$ is closed. By assumption, $\cup_{n\geq1} K_n=G$, so by the standard Baire category theorem, there exists $n$ such that $K_n$ contains an open subset, so there is an open normal subgroup $N$ and $t\in G$ such that $tN \subseteq K_n$.

Let $H=\langle t \rangle N$, then $H$ is open. Consider the subgroup generated by $X$, the set of the elements $(tx)^n$, with $x\in N$. Then $X$ is a normal subset of $M$, and $X \subseteq K$ by the above paragraph. It follows that the closed subgroup $L$ generated by $X$ lies $K$.

Let us work now in the finitely generated profinite group $M/L$ ($K$ may be identified with $K/L$. We have $M/L$ satisfies the coset identity $X^n=1$ with respect to $N/L$ (see Wilson and Zelmanov's http://www.sciencedirect.com/science/article/pii/0022404992901386), by the main result in the previous paper, the Lie algebra $L_p(M/L)$ satisfies a polynomial identity. This proves the claim.

**Remark.** If $G$ is a pro-$p$ group, then we can find a finite generating set of $M/L$ in which every element satisfies the identity $X^n=1$ (take $t$ together with $tx_1,..,tx_s$, where $x_1,..,x_s$ generate $N$; or actually thier images in $M/L$). I wished to deduce from this (using the remark by Professor Yiftach Barnea in his answer Elements of infinite order in a profinite group) that $M/L$ is finite, from which it follows that $K$ has a finite index in $G$. Unfortunately, it seems that this remark is incorrect.

Suppose we have pair $(X,Y)\sim Normal([\mu_x,\mu_y],{{\sigma_x^2\atop\rho \sigma_x\sigma_y } {\rho \sigma_x\sigma_y \atop \sigma_y^2} }] $ How is $U=X\cdot Y$ distributed? I've tried to compute this by substituting y=u/x in the bivariate normal pdf and taking integral(from $-\infty$ to $\infty$ with respect to x. I find the pdf of U as the sum of two exponential distributions(one for U<0 and one for U>0) that are weighted unequally. Is my method valid, or do I have to deal with cumulative distribution functions instead?

2079122Q*[Follow-up to Can every finite graph be represented by one prescribed sequence of natural numbers?, reformulated thanks to a hint from Jacques Carette*]

Let $V(n,\nu)$ and $E(n,m,\mu)$ be computable predicates with parameters $\nu, \mu$.

Consider the class $\Gamma(V,E)$ of finite graphs $G$ for which there are parameters $\nu, \mu$ such that

the vertex set of $G$ is in bijection with $\lbrace n \ |\ > V(n,\nu)\rbrace$ and

$x_i$ and $x_j$ are adjacent iff $E(n_i,n_j,\mu)$.

**Motivation**

The existence or non-existence of computable predicates $V, E$ such that $\Gamma(V,E)$ coincides with a class $\Gamma$ of graphs characterized in the language of graph theory might reveal a "hidden" structure of the natural numbers, but to be honest, balanced strictly $k$-ary trees and hypercubes are the most interesting structures I did "discover" so far.

**Problems**:

Given a pair of predicates $V,E$ as above $\Rightarrow$ characterize $\Gamma(V,E)$ in the language of graph theory.

Given a class $\Gamma$ of graphs characterized in the language of graph theory $\Rightarrow$ find $V,E$ with $\Gamma = \Gamma(V,E)$.

Given $\Gamma$ as above $\Rightarrow$ find $V,E$ with $\Gamma = \Gamma(V,E)$ and minimal complexity.

Given $\Gamma$ as above $\Rightarrow$ find $V,E$ such that $\Gamma(V,E)$ contains almost all graphs in $\Gamma$ and no or almost no others. ("Almost" in the sense of Erdos–Rényi.)

**Examples ($\Gamma \Leftrightarrow V$ # $\ E$)**

complete graphs $ \Leftrightarrow n < \nu$ # $0 = 0$

empty graphs $ \Leftrightarrow n < \nu$ # $0 = 1$

path graphs $ \Leftrightarrow n < \nu$ # $ n = m + 1$

cycle graphs $ \Leftrightarrow n < \nu$ # $ n = m + 1$ mod $\mu $

balanced strictly $\mu$-ary trees $ \Leftrightarrow 0 < n < \nu$ # $ 0 \leq n - \mu\cdot m \leq \mu-1 $

hypercube graphs $ \Leftrightarrow n < 2^\nu$ # $ (\exists k < \nu) | n-m | = 2^k $

empty graphs $ \Leftrightarrow n < \nu$ is even # $n$ and $m$ are coprime (CoP for short)

complete graphs $ \Leftrightarrow n < \nu$ is prime # CoP

**???**$ \Leftrightarrow n < \nu$ is odd # CoP**???**$ \Leftrightarrow n < \nu$ # CoP**???**$ \Leftrightarrow \nu_0 \leq n < \nu_1$ # CoP**???**$ \Leftrightarrow (\exists k < \nu_2)\ n = \nu_0 + k\cdot \nu_1$ # CoP (see another question of mine)

**Observation**

The graphs in $\Gamma(V,E)$ are necessarily the more symmetric the more $V$ and $E$ are *regular* (in an admittedly vague sense). To get classes of more *irregular* graphs one needs at least one *irregular* predicate, e.g. coprimeness^{1}, even though this is not sufficient, see examples 7 and 8. In general, $\Gamma(V,$CoP$)$ seems hard to be characterized in graph theoretic terms, if at all.

^{1} Note that every finite graph is isomorphic to an induced subgraph of the coprimeness graph.

*One for all*

Niel's elaborate construction yields a predicate $V_0(n,\nu)$ – with the parameter $\nu$ being interpreted as the Gödel number of a graph – such that $\Gamma(V_0,$CoP$)$ is the class of *all* finite graphs. This fact doesn't really reveal a "hidden" structure of the natural numbers. It just uses the fact that graphs are gödelizable and shows – as Niel points out – how we can do "packing and unpacking of data structures in the integers". This comes along with the fact, that the sets $\lbrace n \ |\ V_0(n,\nu)\rbrace$ have a high level in the arithmetical hierarchy.

4obviously not. Thank you!zCoordinates of Dirichlet Distribution Negatively Associated?I don't think this is about real vs $Z_2$, but more about dropping signs that are irrelevant for $Z_2$ but should not have been dropped in other characteristics. To define a real cycle space I would assign an (arbitrary) orientation to each edge, and then define the characteristic vector of an (oriented) cycle to have $+1$ in the coordinate for an edge when the orientation of the cycle is consistent with that of the edge, and $-1$ when it's inconsistent. I'm pretty sure that if you do it that way you still get the same dimension, $E-V+1$, as in the $Z_2$ case.27722541799513144754027463039392027710"

Questions:

Can anyone imagine (or even give) really astonishing, non-trivial predicates $V,E$, yielding for example the class of all (unbalanced and/or not strictly $k$-ary) trees?

Does anyone have an idea how the graphs in the

???examples might be characterized in the language of graph theory?Can anyone imagine predicates $V$ significantly less complex than $V_0$ and predicates $E$ not significantly more complex than coprimeness such that $\Gamma(V,E)$ is the class of (almost)

allgraphs?

This can always be done:

Given nontrivial groups $A_i$ for $0 \le i \le n$, there exists a group $G$ and a subnormal series $H = H_0 < \cdots < H_n = G$ such that $H_i/H_{i-1} \cong A_i$ for $0 \le i < n$ and such that no shorter subnormal series from $H$ to $G$ exists.

Here is my proof:

We can assume $n > 1$, and we induct on $n$. By the inductive hypothesis, let $W$ be a group with subnormal series $V = V_1 < \cdots < V_n$, such that $V_i/V_{i-1} \cong A_i$ for $1 \le i < n$, and such that there exists no shorter subnormal series for $V$ in $W$. Write $A = A_0$ and let $G$ be the wreath product of $A$ with $W$ corresponding to the action of $W$ on the right cosets in $V$. In other words, $G = BW$ is a semidirect product, where $B \triangleleft G$ and $B$ is the direct product of $|W:V|$ copies of $A$. Also, $W$ acts to permute these direct factors of $B$, and this action is permutation isomorphic to the action of $W$ on the cosets of $V$ in $W$. (In fact, we assume that we are given a specific bijection from the set of cosets of $V$ onto the set of direct factors of $B$.)

Now let $C$ be the product of all of the direct factors of $B$ that correspond to nontrivial cosets of $V$, and note that ${\bf N}_W(C) = V$. Let $H = H_0$ be the group $CV$, and for $i > 0$, let $H_i = BV_i$. It is easy to see that $H_0 < H_1 < \cdots < H_n = G$ is a subnormal series with factors $A_i$ as wanted. We must show that no shorter subnormal series for $H$ exists. Note that the subnormal depth of $H_1$ is exactly $n - 1$. (This can be seen by intersecting a subnornal series for $H_1$ in $G$ with $W$. This yields a subnormal series for $V$ in $W$.)

Suppose $H \triangleleft K$. We argue that $BK = BV$. Otherwise, $BK > BV$, so $BK \cap W > V$. But $BK$ normalizes $C$ since $C = B \cap H$, and this contradicts the fact that $V$ is the full normalizer of $C$ in $W$. Now if $H = K_0 < K_1 < \cdots < K_m = G$ is a subnormal series for $H$, then $H_1 = BV = BK_1 \subseteq \cdots \subseteq BK_m = G$ is a subnormal series for $H_1$ with length at most $m-1$, and thus $m \ge n$, as wanted.

92569Suppose $G = F/R$ is a finitely presented group with $F = F_n$ a free group and $R$ the normal closure of words $W_1, \dots, W_p$, not all of which are products of commutators. An obvious necessary condition for $G$ to admit a presentation with relators contained in the commutator subgroup of $F$ is that the abelianization be torsion-free; then the number of generators in a commutator-relators presentation will be equal to the rank of $G^{ab}$. This certainly isn't sufficient, as a nontrivial perfect group shows. Is there a criterion for deciding whether a given presentation can be changed by Tietze moves into one with relators in the commutator subgroup? Is there an algorithm for doing so?

EDIT: Andy Putman observes that a general algorithm would solve the problem of identifying the trivial group, so there can't be such a criterion. Given this, are there any known results for special classes of presentations?

881183A compact Hausdorff space has the property you want if and only if it is totally disconnected. See theorem II.16.21 Bredon, "Sheaf theory" (second edition, GTM 170). Examples of compact totally disconnected spaces are the Cantor set and $\mathbb{Z}_p$.Yes, you are right. So both of your solutions now rely on existance of a well-behaving cross-section. The existance of such a plane is equivalent to existance of strict supporting plane (just shift it to the origin), which is in turn equivalent to the original question (just as it is for convex sets). Consider the cone $A - B$ (obviously closed and convex without zero). If there is a strict supporting plane for it, then it is a strict separating plane for $A$ and $B$. So you have not solved the main difficulty of the problem.1164778BNot for 64 bit, 63 is composite.11603465202542179245Fair enough. If you get a chance, take a look at Fremlin's statement. I think you'll like it.2When Russell discovered his paradox, two ways were invented to avoid Russell's paradox.

For logic with sets, ZFC was developed which restricts the creation of the definition of sets. Only sets that are subset of an existing set can be defined with the axiom of separation. And in some case you first need to to use the axiom of power set.

In type theory a different approach was taken. By giving the functions types, it is not possible to express something as element of itself.

My question is, did the logicians in the 20th century consider other ways to avoid paradoxes (only serious attempts)?

I am not completely satisfied with this way of paradox avoiding. Not because it doesn't work (I think it does work), but it forces me to accept the whole system. It is not an universal trick to avoid paradoxes, but inseparable linked to the system.

So, are there ways to avoid paradoxes that for instance also apply to the Liar's paradox?

I think that in general the following scheme is followed to obtain a contradiction from a paradox (whether it is Russell's paradox or the liar paradox, I don't think it is much different):

1) Assume the paradox.

2) Apply the paradox on itself to obtain the contradiction.

3) Express this in one statement, by discharging the assumption.

4) Reformulate previous result such that becomes exactly the paradox.

5) Apply the paradox on itself to obtain the contradiction.

So, to avoid paradoxes, one or more of the above steps can not be fully unlimited.

Note, that applying a rule on itself in general should not be forbidden, because this happens often in a harmless way (for instance when you instantiate an universal quantifier over a predicate variable, which may result again in something with a similar quantifier).

I am experimenting if you can do the trick with levels. Suppose we have a sentence $C$ with definition: $$ C := A \rightarrow B $$

Here $C$ is not just a propositional parameter, but a type of sentence that can produce one, more or infinite other logical sentences, if a logical sentence is given as input. In case of the Liar's paradox it takes itself as input and will produce $\bot$.

To avoid the paradox, each sentence is given a level and in the case above $A$ and $B$ must be of a lower level than $C$. Since, $A$ and $C$ can not be the same anymore, due to the level restriction, the contradiction can not be concluded anymore.

The level is taken from $\mathbb Z$, rather than $\mathbb N$. This prevents that you run out of levels.

If you have a theorem on a certain level, and the theorem is not under assumption, you may replace the level by the "any" level, since you can do a similar reasoning with any other level. The idea is that most statements gets the "any" level and you don't have to bother about it.

You are not allowed to do this under assumption. So, reasoning under assumption is limited in two ways:

1) You are not allowed to assume the "any" level.

2) You are not allowed to generalize to the "any" level under assumption.

Has anything attempted like this before and is this doomed to fail?

supplement: [**28** proofs of the irrationality of $\sqrt{2}$](http://www.cut-the-knot.org/proofs/sq_root.shtml)4307021822469198793011941071409191481703140977629785bForcing, cuts, and Dedekind-finite cardinalities11498941316236Yes, this is very open-ended. A 'big picture' answer to your question is that the closed geodesics on the surface, and the eigenvalues of the Laplacian (corresponding to Maass forms) are on opposite sides of the Selberg trace formula. The geodesics are analogous to primes, and the eigenvalues analogous to the Riemann zeros. The Selberg trace formula is the analogy of Riemann's explicit formula.

Of course, there's lots of other possible answers...

66678646914410698781898820718951Indeed my question contains many (maybe trivial) subquestions: 1) Does a 4-manifold with amenable fundamental group have any characterizing properties? 2) Is there an *explicit* way to construct a 4-manifold whose fundamental group is the Thompson group?@GHfromMO Yes, I know it is cruder, the question is how much cruder...20318621830699174858118053218901701781827Fedja, thank you for the example. But I don't think it works - it seems to me that on some of these vertical spirals $x$ takes minimum value at the top or at the bottom of the cylinder?A question fairly close to Question B was studied in [1] (JSTOR link here), which finds the choices of $n$ such that some volume-$n$ integer-side-length rectangular prism minimizes the volume/surface area ratio among *all* integer-side-length rectangular prisms with volume *at most* $n$.

Note, however, that on p. 194, Alspaugh notes that his/her method doesn't actually find the factorization $n = xyz$ that is optimal, and gives good examples of where the easy conjectures fail.
Still, this should give you some bounds on the optimal volume/surface area ratio.

[1] S. Alspaugh, "Farmer Ted goes 3D," *Math. Magazine* **78**(3), June 2005, pp. 192-204. DOI: 10.2307/30044156

For a concrete example, take $U_r$ to be the closed interval $[r^{-1},2]$ (for $1\leq r\leq\infty$), which clearly satisfies the conditions of the question. Now define the extension operator $m_r$ as follows: given $f$ continuous and real-valued on $U_r$, extend it to all of $[0,2]$ by putting $m_r(f)(0)=0$ and interpolating linearly, i.e.

$$ m_r(f)(t) = r^{-1}t f(1/r) \hbox{ if } 0\leq t\leq r \hbox{ and } m_r(f)(t)=f(t) \hbox{ if } 1/r\leq t \leq 2.$$ Clearly each $m_r$ is a linear extension operator with norm $1$.

Now let ${\bf 1}$ be the function on $[0,2]$ with constant function $1$. Then

$$ \Vert m_r\circ\phi_r({\bf 1}) - {\bf 1} \Vert \geq \vert m_r\circ\phi_r({\bf 1})(0) - {\bf 1}(0) \vert = 1 $$ for all $r<\infty$.

2060425Ayan Biswas@Hailong, I fixed the strike tag: closing the tag with `` won't work in HTML documents in most browsers.872540I'd say the number can be arbitrary large if $X$ is not assumed smooth along $Z$. Take the simplest example: hypersurface singularity $\{f_p+f_{p+1}=0\}\subset\Bbb C^n$, where $f_p$ is totally reducible and $f_{p+1}$ is generic enough. So that the singularity is isolated. Blowup the origin. The exceptional divisor (=the projectivization of the tangent cone) is the hyperplane arrangement, $\{f_p=0\}\subset\Bbb P^{n-1}$.

(Of course, you can compactify the hypersurface, if you insist on projective varieties)

2961762The expected value of common neighbors on a random regular graphFor some classes of varieties one can prove that this cannot happen. For example, if a smooth cubic hypersurface over $\mathbb{Q}$ has a point over a *single* quadratic extension of $\mathbb{Q}$, then it already has a point over $\mathbb{Q}$.1250506You seem to be actually answering https://mathoverflow.net/questions/31846/is-the-riemann-hypothesis-equivalent-to-a-pi-1-sentence rather than the original question here.581905Let $R=\oplus_{n\geq 0}R_n$ be a standard Noetherian commuative graded ring over a local ring $(A,m)$ where $R_0=A.$ Put $R_+=\oplus_{n\geq 1}R_n.$ Let $M$ be a finitely generated $\mathbb Z$-graded $R$-module. Then Castelnuovo-Mumford regularity of $M$ is defined here. My question is

What is the definition of Castelnuovo-Mumford regularity when we consider $R=\bigoplus\limits_{(r,s)\geq (0,0)}R_{(r,s)},$ $R_+=\bigoplus\limits_{(r,s)\geq (1,1)}R_{(r,s)}$ and $M=\bigoplus\limits_{(r,s)\in \mathbb Z^2}M_{(r,s)}$ a finitely generated $\mathbb Z^2$-graded $R$-module.

I have seen the algebraic geometric version of the definition of multigraded Castelnuovo-Mumford regularity. But I am interested to know the definition with respect to local cohomology and bound of the regularity .

433092419751Every graph is an induced subgraph of some vertex transitive graph (a result I first learned from Chris). In fact Fink and Ruiz showed in 1984 that for any graph $H$, there is a circulant graph $G~$ whose edge set can be partitioned into copies of $H~$ that are induced subgraphs of $G$. The second question is more difficult, but is there a reason (such as an example) to believe that some graphs cannot be the neighbourhood graphs of Cayley graphs? The neighbourhood graphs of circulant graphs always have an automorphism of order 2, so cyclic groups are not sufficient in this case.

10457801112563Let $X\neq \emptyset$ be a set and let ${\cal J} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ be a collection of non-empty subsets of $X$. We say that a topology $\tau$ on $X$ is ${\cal J}$-*compatible* if for every $J\in {\cal J}$ there is a continuous surjective map $f:X\to J$, where $J$ carries the subset topology inherited from $(X,\tau)$.

Let $\text{Cptb}(X, {\cal J})$ denote the collection of ${\cal J}$-compatible topologies, ordered by $\subseteq$.

Note that ${\cal P}(X)$ is always the greatest element of $\text{Cptb}(X,{\cal J})$, and the indiscrete topology $\{\emptyset, X\}$ is always the smallest element.

If $\tau\in \text{Cptb}(X, {\cal J})$ and $\tau\neq{\cal P}(X)$, is there a minimal element in $\text{Cptb}(X, {\cal J})$ properly containing $\tau$ (= "above" $\tau$, in the poset we are considering)?

872942416684688085646639986434101421537385169853117117646btw, what is the question?887880This is very far from true it seems to me. For example, why not take a smooth hypersurface $D$ in $P^n$ of very high degree and remove it? I've never read Van den Bergh's paper but presumably it is a precursor to more modern categorical notions of Calabi-Yau which reduce to the ordinary commutative notions in the case of affine varieties.18284081565369164291483508620640871989279If we change the setup and require that each line have infinitely many points of every color, we can extend Anthony Quas's idea to show that you can add any two patterns. That is, if there is a $C_d(n_1,k_1)$ and a $C_d(n_2, k_2)$, there is a $C_d(n_1 + n_2, k_1 + k_2)$. Overlay the two patterns, and at each lattice point, pick randomly which pattern to display. With probability 1, every line will contain infinitely many points of each color it contained in the original patterns. In particular, there is a $C_d(1,1)$, so given a $C_d(n,k)$, we can find a $C_d(n+1,k+1)$.

If we only want periodic patterns, you can replace the random mask we chose with another periodic pattern. Pick a prime which doesn't divide the periods of the two patterns, and construct a $C_d(2,2)$ with that period, then overlay the three patterns and choose which of the two original ones to display based on the color of the third. (Note that this doesn't require the periods of the original patterns to be relatively prime.)

If we do have relatively prime periods for our $C_d(n_1,k_1)$ and $C_d(n_2, k_2)$, then we can produce a $C_d(n_1 n_2, k_1 k_2)$, just by overlaying the two and taking the new colors to be ordered pairs of colors from the two patterns.

Like I said in the comments, I don't think your simplex construction actually works for $d>2$, but it's fine for $d=2$. You can make a slightly more general construction for a $C_2(2k-1,k)$ with period $p^{k-1}$: Given a lattice point $(x,y)$, let $n_x$ be the number of times $p$ divides $x$, up to a maximum of $k-1$, and similarly for $n_y$. Then color the point $(x,y)$ with the number $n_x + n_y$.

Since we're free to add $1$ to both $n$ and $k$, this means we can construct every $C_2(n,k)$ for $k \leq n < 2k$. By adding these to the rainbow constructions which give us $C_2(k^2,k)$, we can also fill chunks of the range $2k \leq n \leq k^2$, in particular, something like the range $k \leq n < C k^2$ for some $C < 1$. There are no $C_2(n,k)$ for $n > k^2$, so we have the asymptotic answer.

944036I am relatively new to complexity and computability theory. I just came across the concept of Permanent of a matrix and read that it is NP hard problem to compute the permanent of 0-1 matrix. Of course it struck me with surprise as it would have to anyone new that it is NP hard where determinent is polynomial. Could any one give an easy proof of this by reducing any other NP complete problem into this?

1986395I meant efficiently computable $G(n)$, not $\infty$. Probably I will edit the question.220690920665941145677Is it true that if $(R,m)$ is a (not necessary commutative) local ring then $R$ and $m$ have the same cardinal ? (Exclude TWO trivial cases: when $R$ is finite and when $R$ is a division ring)

On the other hand if the ring is Artinian then the above claim is true.

@Mohammad: I have added a bibliography. Point d) is proved in Fischer's book. Tyler, I have the feeling that the Hochschild cohomology spectral sequence that you write down is going to have bad convergence properties when $E$ is not connective. Do you know any ways around this for K-theory?1891873@vap Indeed, however I don't see why the proof for cyclic homology would be easier (it is rather the same computional problem).4052529150691535831133190814743951768958Thanks Mark, that's exactly what I was after. I was actually reading Friedman's paper, but with my poor grasp of deformation theory, I hadn't managed to extract what I needed!6952152160560NYes, it is. Thanks for the supplement.437654Kevin Brown16947215585107717317240415113481744761This 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.

1111509198529612102137It seems the following remarks in M. Kapranov's paper http://arxiv.org/abs/alg-geom/9604018 page 64 bottom, has not been mentioned so far

According to the point of view going back to Y.I. Manin and B. Mazur, one should visualize any 1-dimensional arithmetic scheme X as a kind of 3-manifold and closed points x ∈ X as oriented circles in this 3-manifold. Thus the Frobenius element (which is only a conjugacy class in the fundamental group) is visualized as the monodromy around the circle (which, as an element of the fundamental group, is also deﬁned only up to conjugacy since no base point is chosen on the circle), Legendre symbols as linking numbers and so on. From this point of view, it is natural to think of the operators (algebra elements) af,x,d for ﬁxed f and varying x, d as forming a free boson ﬁeld Af on the “3-manifold” X; more precisely, for ±d > 0, the operator a ± f,x,d is the dth Fourier component of Af along the “circle” Spec(Fq(x)). The bosons a ± f,x,d and their sums over x ∈ X (i.e., the Taylor components of log Φ ± f (t)) will be used in a subsequent paper to construct representations of U in the spirit of [FJ].

It might be that recent paper by Kapranov and coauthors: http://arxiv.org/abs/1202.4073 The spherical Hall algebra of Spec(Z)

is somehow developing ideas quoted above.

The question which I heard from V. Golyshev and others many years ago is the following: if Spec (Z) is analogous to 3-fold, what should be arithmetic analog of Chern-Simons theory ?

10875641216609530110340332416107271655658~How do I check if an abelian variety is principally polarized?215014615663481038851@You meant "even worse behaved"?195151016129911895669Let us work over the etale site $\mbox{Aff}/S$ (for the sake of definiteness) for some fixed base scheme $S$, where the covers are jointly surjective etale maps $\{ U_i \rightarrow U\}_{i\in I}$ (and $I$ is finite if you like). Let us also consider a prestack $F$ fibred in groupoids. Recall the definition of a category of descent data $F(\{ U_i \rightarrow U\}_{i \in I})$ associated to a cover $\{ U_i \rightarrow U\}_{i \in I}$. The objects of this category are collections of elements $\xi_i\in F(U_i)$ together with morphisms $\phi_{ij}$ between their appropriate pullbacks satisfying the cocycle condition. This definition is the one appearing on p. 15 of "Champs algebriques" by Laumon & Moret-Bailly (among other sources, e.g., Vistoli's notes in "FGA explained").

However, one can also consider coverings $U^\prime \rightarrow U$ consisting of one element only. In $\mbox{Aff}/S$ starting with any covering one can obtain one of such form by taking $U^\prime := \bigsqcup U_i$. However, it is not clear to me how this passage to a cover with a single morphism interacts with the associated categories of descent data. I suspect, they should be equivalent (in "Champs algebriques" for instance, the authors switch to the latter when exhibiting the stackification of a prestack in (3.2)) but on the other hand I don't see how is one supposed to get a single $\xi \in F(U^\prime)$ starting off with the $\xi_i$ as above, let alone an equivalence of categories between $F(\{ U_i \rightarrow U\}_{i \in I})$ and $F(\{ U^\prime \rightarrow U\})$. Are these categories equivalent and what is a functor exhibiting this equivalence? And if not, why is one allowed to consider only coverings of the form $U^\prime \rightarrow U$ when constructing the stackification?

4738912998257718557241961651042482837206310929548721608426874044 753719512571975462bMarkov chain approximates a fractional diffusion@ Frank The infinitesimal action on $\mathbb{R}^n$ of a matrix in the Lie algebra of $O(n)$ is the usual action of a $n\times n$ matrix on $\mathbb{R}^n$. The Lie algebra of $O(3)$ is $3$-dimensional and can be canonically identified with $\mathbb{R}^3$ using the Pauli matrices. Under this identification, the Lie algebra action becomes the cross product. Also, I think that Robert Bryant makes a very good point.429835On the level of functions it is the "constant term map" , see e.g. http://www.math.ucdavis.edu/~kapovich/EPR/HKM.pdf Somebody probably worked this out on the level of sheaves too.150632018766181518930634769478256Ok, I see, I was already expecting that this is automatic. Is $G$ necessarily abelian?The divergence operator is simply the exterior derivative operator acting on $(n-1)$-forms, and the divergence theorem is Stokes theorem. This can be seen by just writing everything in local coordinates. Any other version is just using an additional geometric structure to identify an $(n-1)$-form with a vector field or 1-form.

2Oriented Cobordism Rings`It's not a closed set, so not a pi 0 1 class...2831757261401525986A small addition : in the special case of Shimura varieties, this Lefschetz $\mathfrak{sl}_2(\mathbb{C})$-action admits an automorphic interpretation which lead Arthur to his conjectures about which local representations occur in automorphic ones. This is rather vague, so here is a wonderful explanation of this by Arthur : http://www.claymath.org/cw/arthur/pdf/L2andautomorphic.pdf127250165290515973861902123Inspired_Blue215271813329312763841243843524229537779when I say something appears to be true, then it implies that I don't know how to prove it! In this case it is just evidence based on lots of computer calculations. I would guess that, if true, then it is provable.1945033704626HAHA I get that, but if there is even one is enough to me! thanks12812021058901665785Among the concrete examples of a non-borel subset of $\mathbb{R}$, I know only the Lusin example.

This is the set $L$ of all irrational numbers whose continued fraction representation $(a_0,a_1,\cdots)$ is such that there exist an infinite subsequence $(a_{k_0},a_{k_1},\cdots)$ such that each for all $j$, $a_{k_j}$ divides $a_{k_{j+1}}$.

Knowing this, I am totally unable to see why this set $L$ would not be Borel, is there some intuition behind this ? What is the idea behind the proof of this fact ?

1758152301380_These problems can be viewed as a "toy versions" of the string theory landscape or multiverse concept_ <- what is this I don't evenBut there is an arbitrarily short path to a diagonalizable matrix.Bob CLet $C_2$ be a smooth genus $2$ curve and $J(C_2)$ its Jacobian. It is well known that the blow-up of $J(C_2)$ at the origin $o$ is isomorphic to the second symmetric product $\textrm{Sym}^2(C_2)$, and that the blow-up morphism $$\pi \colon \textrm{Sym}^2(C_2) \to J(C_2)$$ coincides with the Abel-Jacobi map.

The exceptional divisor $E \subset \textrm{Sym}^2(C_2)$ corresponds to the unique $g^1_2$ of $C_2$, so it intersects transversally the diagonal $\Delta \subset \textrm{Sym}^2(C_2)$ at six points, corresponding to the six Weierstrass points of $C_2$. Hence the image $\pi(\Delta) \subset J(C_2)$ is a curve with an ordinary singular point of multiplicity six at $o$.

Finally, the curve $\Delta$ is the branch locus of the double cover $C_2 \times C_2 \to \textrm{Sym}^2(C_2)$. This implies that its class is $2$-divisible in the Néron-Severi group $\textrm{NS}(\textrm{Sym}^2(C_2))$, and so the same holds for the class of $\pi(\Delta)$ in $\textrm{NS}(J(C_2))$.

More precisely, it is possible to show that $\pi(\Delta)$ is algebraically equivalent to $4 \Theta$, where $\Theta$ is the theta divisor, namely the principal polarization of the Jacobian (see for instance this MO question).

In particular, we have $$\pi(\Delta) \cdot \pi(\Delta) = 16 \Theta^2 = 32.$$

I would like to know whether this is the only possible occurrence for this situation, so let me ask the following

14519481692484739395184796721099841029528Hmmm... Your first suggestion (mapping each ball onto $R^n$) certainly sounds plausible, and the second one might work too; but at the moment I don't see how to prove either one or to find a counterexample. Let me think about it a little more.22993261619618@Timothy Chow, I spent the first 20 years of my career in Soviet Union, where all journals had strong length restrictions. Splitting a long paper always worked for me.counterexample: $\phi(x)=\cos(x)$ $a(x)= 2+\sin(Nx)$ gives $$ 2\pi > N^2\frac{\pi}{2}.$$1847113If 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...1046797@DavidLoeffler: I can explain the proof in terms of coroots (this is the same as Knop writes). Let $\alpha_1^\vee,\alpha_2^\vee,\alpha_3^\vee$ be the simple coroots of $G=SU(2,2)$ with respect to $T$ and $B$. Since $G$ is simply connected, these simple coroots consitute a basis of the cocharacter group of $T$. Since $B$ is defined over $\mathbb{R}$, the complex conjugation permutes the simple coroots. Thus $T\simeq \mathbb{R}^*\times\mathbb{C}^*$, hence it is quasi-split.1025248You might want to check out openmathematics.org, which doens't have GitHub support for files and comments yet, but soon will have.Have you looked at the exponential generating function of $A(j,k)$?LInteger point in a non-empty polytope94621520162421109385@IoannisSouldatos: Yes, I agree that the construction is a bit ad hoc. However, do notice that, while we have introduced $5$ new colors, our cycle is in an old color. This is why a transfinite iteration allows you to get odd cycles in every color eventually. I agree, though, that a nice, symmetric example would be more satisfying.197877448295814374341545129

Question.Let $A$ be an abelian surface and $D$ be an effective divisor on it such that

- $D$ has an ordinary point of multiplicity $6$ at the origin $o \in A$ and no other singularities;
- $D^2=32;$
- the class of $D$ is $2$-divisible in $\textrm{NS}(A)$.
Is the pair $(A, \, D)$ necessarily of the form $(J(C_2), \, \pi(\Delta))$ for some genus $2$ curve $C_2$?

finding the set of such $P$ (not necessarily unique)

Just a remark, not an answer. For $H$ a $4$-cycle, there are three partitions edit-distance $2$ from $H$:

So if $G$ is composed of $k$ disjoint copies of $H$, there are $3^k$ equally minimal partitions. So the cardinality of the set you seek can be exponential in the number of vertices. 7459366471003>Help with a double sum, pleaseTypos corrected:It might help to note that the binary octahedral group is isoclinic to ${\rm GL}(2,3)$, which is clear since they are both double covers of $S_{4}$, and both genuine 2-dimensional representations of these groups yield the same projective (in Schur's sense) representation of $S_{4}$. To be specific, ${\rm SL}(2,3)$ embeds the same way in each of these groups, and if we take an involution $t \in {\rm GL}(2,3) \backslash {\rm SL}(2,3)$, we may replace $t$ by $it$, and we now generate the binary octahedral group. 10379682171830If $f_n \to f$ a.e. and $\|f_n\|_p \to \|f\|_p$ then $\|f-f_n\|_p \to 0$ for any $p<\infty$. This is problem 6.10 in Folland's Real Analysis and is in many other RA books. Clarkson is not needed for the proof.7210938639771109334A knot in a $3$-manifold is called nonlocal if it is not contained in any closed ball. An open $3$-manifold containing no nonlocal knots is homeomorphic to $\Bbb R^3$, http://www.ams.org/journals/proc/1971-028-01/S0002-9939-1971-0271919-1/ so Zack's remark suffices to answer the 3-dimensional case of the question. To see an explicit nonlocal knot in the Whitehead manifold $W$ it suffices to know that the $n$th Whitehead link is nontrivial (and this is how Whitehead originally proved that $W\not\cong\Bbb R^3$).9301858464119903342178839120275312245267444831687523746479490748211261306

I am using Anthony Quas's reformulation to restate the problem. Letting the $n$th prime gap be given by $d_n= p_{n+1} - p_n$, and given $n$ we relabel $a=d_n$ and $b=d_{n+1}$, we look at $L$ as the set of $p_{n+1}$ in which $n$ satisfies $b-a\gt 0$ and we look at $Q$ as the set of $p_{n+1}$ in which $0 \lt (b-a)p_{n+1} -ab$. As noted in the post, $p_{n+1}$ not in $L$ readily implies $p_{n+1}$ not in $Q$; it is natural to ask if $p_{n+1} \in L$ implies $p_{n+1} \in Q$. The question further notes that when $p_{n+1}$ is observed to be in $L$ one also has $ ab - p_{n+1}(b-a) \gt (b-a)$, which I think should be reversed as $7=p_4 \in L$ but $49 - 55 \lt 11 - 14 + 5$.

If the last inequality is reversed, it says $ab \lt ( p_{n+1} + 1)(b-a)$. If $p_{n+1} \in Q$, then clearly this last inequality holds. Finally as Anthony Quas observes, if $p_{n+1} \in L\setminus Q$ then $ab \geq (b-a)p_{n+1}$ and $b \gt a$, so one would have $ab \geq 2p_{n+1}$ if $n \gt 1$.

The formulation shows that the basic question is about consecutive prime gaps, and that $L$ is different from $Q$ only when a large gap $d_n$ is greater than the square root of an adjacent prime $p_n$. Such large gaps have not been observed for $n \gt 30$ (so $p_n \gt 113$), and the stronger inequality $ab \geq 2p_{n+1}$ is also not observed for $1 \lt n \leq 30$. The case $n=1$ is left to the reader, as is the conclusion that $L$ properly containing $Q$ would violate expectations and many conjectures in prime number theory.

the link doesn't seem to be working right. The full url is here: http://en.wikipedia.org/wiki/Series_acceleration11131321507497From the comments above I feel there might be a confusion about what is a "_constructive_ algorithm", I don't really know what the OP means by putting "constructive" in front of "algorithm", I guess a typical interpretation would be the code of the algorithm for $Halt_n$ is uniformly computable from $n$. But then the OP goes on to say that "for _some_ $n$" that contradicts that interpretation.1351539To n40886: I'm sorry you don't like the edits I made, but you can simply revert back to your original version. In my opinion, to withhold relevant information that may help the reader is not such a great idea.@ Leandro Vendramin Leandro, thank you for nice examples. It is really Yang–Baxter for Dummies. Are there any simple examples connecting Yang–Baxter with special functions?40469218644617848672231064 Atomic Overflow181964555438661442You can call it an $\ell$-field, which is a topological field that is an $\ell$-space (locally compact, Hausdorff, and totally disconnected). More generally, the eigenvalues of an irreducible Hessenberg matrix are geometrically simple (the argument is that described by Federico). This statement, applied to Hermitian matrix tells you that if $H$ is tridiagonal and irreducible (the $h_{ij}\ne0$ for $|j-i|=1$), then it has simple eigenvalues. Therefore, no matter what is the diagonal $D$, the rank of $H+D$ is always $\ge n-1$.736745@Can you give an example? ThanksAlso, approximation within $\epsilon$ does not require matching derivative.1969481'Fix $(a,b)=1$, $a<b<2a$ and $a,b>n^{1/(2t)}$ and fix a prime $T\approx n^{\tau+\frac1k}$ where $\tau\geq1$ and $k=2(t-1)$. We can show using exponential sums there is an $m_{_T}$ such that $T/a^{2(t-1)}<m_{_T}<T-T/a^{2(t-1)}$ such that each of $c_{i,T}=m_{_T}a^{2(t-1)-i}b^i\bmod T$ is bound by $T^{\frac{2t-2+\kappa}{2t-1}}$ for some $\kappa=\frac{k(\tau-1)+2}{k\tau+1}>0$. Example if $t=2$ and $\tau=1$ we can get $\kappa=2/3$ and $T^{\frac{2t-2+\kappa}{2t-1}}=T^{8/9}$.

$\underline{\mbox{Problem} 1}$: Assume we have such $c_{0,T},...c_{k,T}$ that are proved to exist using exponential sums. Fix distinct primes $q_j>T^4>\max(c_{i,T})$ for $j\in\{0,\dots,k\}$. Are there integers $p_j$ such that $q_j/\min(c_{i,T})<p_j<q_j-q_j/\min(c_{i,T})$ such that each of $p_jc_{i,T}\bmod q_j$ at $i\neq j$ is bound in interval $[0,q_j^{r+\epsilon}]$ while $p_jc_{j,T}\bmod q_j$ is bound in $[k\cdot q_j^{r+\epsilon}+1,(k+1)\cdot q_j^{r+\epsilon}]$ for some $r\in(0,1)$?

Can we get $r<\frac{k^\alpha}{k^\alpha+1}$ for any fixed $\alpha\geq1$ in this case? If $c_{i,T}$s were uniform then $\alpha=1$ is possible.

Note that for every $\alpha_i\in\Bbb Z$ with $|\alpha_i|<n^\frac1{2t}$ and $(\alpha_1,\dots,\alpha_k)\neq(\underbrace{0,\dots,0}_{k\mbox{ times}})$ we have $$|\sum_{i=0}^k\alpha_ia^{2(t-1)-i}b^i|\leq(k+1)n^{\frac{2t-1}{2t}}$$ and if $(k+1)<n^{\frac1{2t}+\tau}$ then $$|\sum_{i=0}^k\alpha_ia^{2(t-1)-i}b^i|<T$$ holds. Moreover since $|\alpha_i|<\min(a,b)$ and $(\alpha_1,\dots,\alpha_k)\neq(\underbrace{0,\dots,0}_{k\mbox{ times}})$ $$0<|\sum_{i=0}^k\alpha_ia^{2(t-1)-i}b^i|<T$$ holds.

This implies $$\sum_{i=0}^k\alpha_i[m_{_T}a^{2(t-1)-i}b^i\bmod T]=\sum_{i=0}^k\alpha_ic_{i,T}\neq0\bmod T$$ as well. It means the discrepancy of the derived sequence $(c_{0,T},\dots,c_{k,T})\bmod T$ is at most $n^{-\frac1{2t}}$.

$\underline{\mbox{Problem} 2}$: Is it possible to get a prime $q>T^4$ such that discrepancy of $(c_{0,T},\dots,c_{k,T})\bmod q$ is at most $n^{-\frac1{t}}$? If so there is an $m'$ such that $$(m'c_{0,T},\dots,m'c_{k,T})\bmod q\in\underbrace{[0,n^{-\frac1{t}}q]\times\dots\times[0,n^{-\frac1{t}}q]}_{k+1\mbox{ times}}$$ holds?

Is it always true that if for a given sequence we can exclude a linear relation with coefficients $<n^\delta$ then we can always achieve $r=1-\delta$ for that sequence? Is there a proof or good reference? Is there a converse?

If $A\subseteq B(H)$ is hereditary and has unit $p$, then we must have $A=pB(H)p\cong B(pH)$. So the answer is, such an embedding exists if and only if $A\cong B(H)$ for some $H$. In particular, the only commutative example is $A=\mathbb C$.

!The answer for $p>2$ is NO:

Suppose $M \to L$ is a permutation covering with kernel $K\leq M$, and the rank of $M$ smaller than $p^2$. The permutation lattice $M$ decomposes into a direct sum

$$ M = M_1 \oplus \dotsb \oplus M_r, $$
where each summand corresponds to an orbit of the underlying $G$-set. These orbits can have sizes $1$ or $p$, as summands of rank $p^2$ are forbidden.
But summands of rank $1$ are in the kernel $K$, since $L$ contains no elements that are fixed by $G$. Thus we can omit these from the beginning, and assume that each $M_i$ has rank $p$.

Each $M_i$ has a (unique) submodule $S_i$ of rank $1$ on which $G$ acts trivially. Again, we must have $S_i \leq K$, as no element of $L$ is fixed by $G$. So each $M_i$ contributes at most rank $p-1$ to $L$. Thus we must have $r\geq p-1$. Since we assume that $M$ has rank $<p^2$, we have $r=p-1$.

Now both sublattices
$$ S_1 \oplus \dotsb \oplus S_{p-1} \subseteq K $$
are *pure* sublattices of $M$, and by comparing ranks we can conclude that actually
$$ S_1 \oplus \dotsb \oplus S_{p-1} = K.$$
Thus we would have
$$ L \cong (M_1/S_1) \oplus \dotsb \oplus (M_{p-1}/S_{p-1}). $$
But I claim that $L$ is indecomposable. (Edited later, in view of comments, and since the first proof was somewhat unclear:) The $P$ of the question is clearly the group ring $\mathbf{Z}G$, and $L = \mathbf{Z}G/I$, where $I$ is some ideal of $\mathbf{Z}G$.
It is, however, well known that the group ring of a $p$-group over a field of characteristic $p$ or over a local ring with residue field of characteristic $p$ is local. Thus $L\otimes \mathbf{F}_p$ or $L\otimes \mathbf{Z}_p$ (localisation at $p$) has a unique maximal submodule, and is indecomposable. Therefore, $L$ must also be indecomposable. So when $p-1>1$, we get a contradiction.

(On the other hand, $L \otimes \mathbf{Q}$ has a decomposition into $p-1$ irreducible $\mathbf{Q}G$-modules. An alternative proof would be to compute matrices of the projections to these subspaces in terms of a $\mathbf{Z}$-basis of $L$, and see that they do not have integer entries.)

Recall that a topological space is called $A$-affine for a sheaf of algebras $A$ if taking global sections of coherent sheaves of $A$-modules is an equivalence of categories to finitely generated $\Gamma(A)$-modules. (for example, an affine scheme is one which is affine for the structure sheaf).

It seems to be a "well-known fact" that the variety $G/P$ for any simple complex algebraic group $G$ and parabolic $P$ is $D$-affine where $D$ is the sheaf of differential operators (and that more generally, one can quite explicitly describe the set of TDO's which are affine). I've found this stated in several books and papers (Beilinson and Bernstein's original paper, "Algebra V: homological algebra", this paper of Alexander Samokhin) but have yet to find an actual proof. One place one might guess it would be that it seems to not be is the book of Hotta, Takeuchi and Tanisaki.

Does anyone know a published source where this is proved?

I'll emphasize, what I want is not a proof of the theorem in the answers here; that's easy once you understand Beilinson and Bernstein's original argument. What I'm looking for in a place in the literature where this result is clearly and precisely stated, with a proof or clear reference to a proof.

The authors are requiring more than just the sentence in question. The whole paragraph is needed. The lemmata proven say a minimal counter example $M$ must have no parallel elements and corank at most 2. The authors then note this also holds when replacing $M$ with its dual. So, the claim really is that if $M$ is connected, does not have any parallel nor any coparallel elements, and both the rank and corank are at most 2. Then $M$ is isomorphic to $U_2(4)$.

~In 1D, is a $W^{1,p}$ function always Lipschitz, for $p\ge1$? 4864899195512025724(There are other ways to relay questions from MSE to MO, by the way. For example: http://meta.mathoverflow.net/questions/1967/interesting-and-not-sufficiently-answered-questions-on-math-se)No. that is definitely false. Have you tried to read Brown's original paper? It does the case of principal bundles in section 5.1 (the trick is to classify principal bundles with a chosen basepoint). If you prefer I can try to write down an answer after dinner.14829681877616471961691659does there always exists a path $g:[0,1] \rightarrow X$ from $f(0)$ to $f(1)$ that has the same image as $f$ and ..?jendomorphisms of modules over symmetric ring spectra2144120122422421382517688251108622572557The paper DESIGNING RANDOM ALLOCATION MECHANISMS: THEORY AND APPLICATIONS by ERIC BUDISH, YEON-KOO CHE, FUHITO KOJIMA, AND PAUL MILGROM Also describes an extension of the Birkhoff-von Neumann theorem.

This is related to the following: There is some relation between the Birkhoff-von Neumann theorem and Scarf's theorem that asserts that balanced games have non empty core. The fact that that the houses allocation game is balanced follows directly from B-vN theorem. The fact that the game describing the stable-marriage problem requires a certain generalization.

2097217460887For another family of examples accessible to students, see also puzzle 12 at http://www.math.harvard.edu/~elkies/Misc/index.html#puzzles and its solution at http://www.math.harvard.edu/~elkies/Misc/sol12.html .5174913413158133361630708472586Yes, it's true, and at least morally I think it's equivalent to the statement about the natural map on $\pi_1$'s being an isomorphism. Someone else will probably come along with a more detailed answer, but in the meantime have you tried looking in Szamuely's book *Galois groups and fundamental groups*?617768670814819679116436829460646724491653188As long as the action of $G$ on $A$ is not trivial, this is not going to work, because $X/G$ will not be an $A$-scheme.

Here is a concrete, but somewhat general example: Let $Z$ be an arbitrary affine scheme and $X=A\times Z$. Define the $G$-action by acting on $A$ with the given free action and act trivially on $Z$. This gives a free $G$-action on $X$. The quotient $X/G$ is just $(A/G)\times Z$, which will not map to $A$ in general.

The context you may have a chance of making this work is if you want $X/G$ to be a relative $\mathrm{Spec}$ over $A/G$. That actually seems to work. Since $G$ is finite, $X$ is also affine over $A/G$, so you may actually assume that $G$ acts trivially on $A$. In that case, you can just look at the $G$-invariant part of $R$ and take the relative $\mathrm{Spec}$ according to that. This should give you what you want, since locally you just have $\mathrm{Spec}$'s and in that case you already know what you want and the base is fixed, so everything is dandy.

21108026196286846732289195PierreBdRThe standard Sage implementation relies on Cremona's mwrank, and mwrank "searches" for appropriate two-covers, rather than directly computing them. I think Magma is better for things like this, and the [online calculator](http://magma.maths.usyd.edu.au/calc/) may handle most of the OP's cases.1988409Sums like

$f_n = \sum_{k \ge 0} (1 - (1-2^{-k})^n)$

(and even much more terribly looking ones) have been considered at many places in the literature. There are methods for evaluating them asymptotically at arbitrary precision (i.e. up to any error of order $n^{-c}$). One such method is the **Mellin Transform**, see

http://algo.inria.fr/flajolet/Publications/mellin-harm.pdf

for an excellent survey by Flajolet et al. There, $f_n$ serves as one of several "running examples", and the authors show that

$f_n \sim \log_2 n + \gamma/\ln 2 + 1/2 + Q(\log_2 n) + O(n^{-1/2})$

where $Q$ is an explicitly given Fourier series. Interestingly, this series remains bounded, but oscillates forever.

773044N@Wadim: (-: That's a good use of MO. @ViniciusdosSantos The vertices of planar graph can be partitioned in two trees if and only if its planar dual is hamiltonian. It is known that a hamiltonian planar triangulation has at least four hamilton cycles and I think that the result is true for other planar graphs (perhaps some connectivity or minimum degree constraint). The existing proof for triangulations fundamentally relies on it being a triangulation, so I wanted to change the domain and see if techniques from the "vertex arboricity" literature may help. But it must be trees.118995611645942223746725897247651601551007002 The operators $Z_{1,l}$ act in the Macdonald basis as $Z_{1,l}\cdot P_\lambda = \sum_{\mu = \lambda + 1} a(i,\mu-\lambda) P_\mu$, where $\lambda + 1$ is the set of partitions obtained from $\lambda$ by adding one box, and the $a(i,\mu-\lambda)$ are some constants. I'm pretty sure the Schiffmann-Vasserot paper has an explicit expression in the DAHA for $Z_{1,l}$, it's one of their hard computational lemmas, I think. If you want some operator $F_i$ that acts on $P_\lambda$ by adding a box of content $i$, if possible, and by 0 otherwise, I'm not sure if this is inside the DAHA.jth22238692300950The related equation $$\sigma(n) = An + B(n)$$ where $B(n)$ "is a function that may depend on properties of $n$" is considered in the paper Variations on Euclid’s Formula for Perfect Numbers by Farideh Firoozbakht and Maximilian F. Hasler, published in the Journal of Integer Sequences.

182029613979185069723021792223156This problem is a lot like pricing a perpetual american option (google should give plenty of hits). The main difference that I see here is that the payoff depends on the *proportion* of heads, so that you are dividing by the number of trials (which translates to the time at which the option is exercised). In option pricing you would discount at some interest rate, which is a factor exponentially decaying in the exercise time.I have been led to believe that there is a result giving a description of the quotient of a Bruhat-Tits building $\Delta(G,k)$, for a semisimple algebraic group $G$ over a non-archimedean local field of positive characteristic $k$, by a non-uniform arithmetic lattice $\Gamma$. I believe it says that such a quotient is a union of a finite simplicial complex with finitely many cusps which are a product of $\mathbb{R}$ with a spherical building. I have had some difficulty in locating a reference for this and would be grateful if anyone could point me in the right direction.

The Cayley-Hamilton theorem is usually presented in standard undergraduate courses in linear algebra as an important result. Recall that it says that any square matrix is a "root" of its own characteristic polynomial.

**Question. Does this theorem have important applications?**

Being not an algebraist, I am aware of only one application of this result which I would call important; it is very basic for commutative algebra, algebraic geometry, and number theory. It is as follows. Let a commutative unital ring $A$ be imbedded into a field $K$. Consider the set of elements of $K$ which are integral over $A$, i.e. are roots of a polynomial with coefficients in $A$ with the leading coefficient equal to 1. Then this set is a subring of $K$.

1215035172974919339381158591@Spock: Neither could I. But it's in Hurewicz's collected works.19140701757447223226279774537773And also $3^{15} = 14348907$ which is not a very large set to process.14060901845503What you describe is not $\sf AC_\omega$. It's $\sf DC$, which is strictly stronger.Let $f(x,y)$ be a polynomial with integer coefficients, and let $\alpha=(\alpha_1,\alpha_2)\in \mathbb{C}^2$ be a complex point. I want to show that $f$ cannot vanish at $\alpha$ to high order unless $\alpha$ is ``simple''. More precisely, suppose that

- f(x,y) vanishes to order at least $0.99\cdot \deg f$ at $(\alpha_1,\alpha_2)$, and
- f(1,1)=1

Does it imply existence of an integer polynomial $g(x,y)$ satisfying $g(\alpha_1,\alpha_2)=0$ and $g(1,1)=1$ that is

linear, i.e., of degree $1$?

``Vanishing to order $m$'' at $\alpha$ means that all possible partial derivatives of order at most $m-1$ vanish at point $\alpha$.

Note that the single-variable version is a simple consequence of Gauss's lemma.

I previously asked an analogous question over $\mathbb{C}$, but the answer turned out to be contrary to what I expected. This is the original question that I am interested in.

I do not know how to answer this question even if one replaces $\mathbb{Z}$ by $\mathbb{Q}$ throughout.

20269235573801879205user375756191693615266241155213637681Dmitry F1958602The $k=0$ and $k=1$ case are drawn up nicely in the second appendix of Freed and Uhlenbeck's classic book *Instantons and Four-Manifolds*. It's entitled the Pontrjagin-Thom Construction, and is motivated by wanting to compute $[M,S^3]$ (for any compact simply-connected 4-manifold) whose nontriviality depends on the parity of the natural intersection form.

The best part: there is a cool picture of a dinosaur (that is, a framed cobordism) being cut open (that is, by two homotopy-equivalent framings).

Let $ c_{0}:=\lbrace x:\mathbb{N}\rightarrow \mathbb{R} :\lim_{j\rightarrow\infty} x_{j}=0 \rbrace$ denote the usual Banach sequence spaces. Given Banach spaces $X,Y$ let $L(X,Y)$ denote the Banach space of bounded, linear operators from $X$ to $Y$ . Also let $ \ell_{1} :=\lbrace x:\mathbb{N}\rightarrow \mathbb{R} :\Sigma_{j=1}^{\infty}\vert x_{j}\vert <\infty \rbrace$.

is $ c_{0} $ isometrically embedding in $ L(c_{0},\ell_{1}) $ ?

Could this be a particular case of Theorem 2.18 in [William Kuszmaul, *New Results on Doubly Adjacent Pattern-Replacement Equivalences*, arXiv:1402.3881v2](https://arxiv.org/abs/1402.3881v2)? This is **not** a rhetorical question -- I haven't read that paper -- but it sounds at least seriously relevant to the question.22106102258537Raa AnandI meant for $d=3$. What do you mean by "there's no nonconstant solution in Q(t) to x^3+y^3+z^3=d unless d is a cube or twice a cube"? Isn't rational parametrization for $d=3$ in Q(t)?1073712848808immibis662360>@SteveHuntsman: okay, and ...?147042420122389369064I am studying the basics of constructible and lisse sheaves, and am trying to understand SGA 4, IX. As Grothendieck himself observes at the beginning of the chapter, one is forced to work with *torsion* sheaves, and there are several "moral" explanations for this: Serre's example of an elliptic curve with quaternionic multiplication which cannot provide a $\mathbb{R}$-valued representation of the quaternions, the fact that constructible/lisse sheaves should correspond to continuous representations of $\pi_1^\mathrm{et}$ and these are of pro-finite image, and so on. But I am trying to understand a more technical detail, basically around Lemma 2.1 of SGA 4, Chap. IX.

Grothendieck fixes a topos $T$ and goes on to find sufficient conditions for a subsheaf $\mathcal{F}$ of a constant sheaf $S_T$ to be in turn (locally) constant: this depends a bit on the kind of object $S$ is. If $S$ is an abelian group, and we look at $S_T$ and $\mathcal{F}$ as being $\textbf{Ab}$-valued, then we need $S$ to be finite. If we have fixed a (noetherian) ring $A$, $S$ is an $A$-module and $\mathcal{F}$ to be $\textbf{Mod-A}$-valued, then we need $S$ to be finitely generated. These conditions turn out to be precisely those needed to make the category of locally constant, and hence of constructible, sheaves abelian. As a consequence (see Remark 2.3.1 *ibid.*) a constructible sheaf of $\mathbb{Z}$-modules is not the same thing as a constructible sheaf of abelian groups. I have tried to come up with examples to see that these finiteness assumptions are really needed, but did not succeed.

- Is there an example of a subsheaf of a locally constant sheaf with stalks (say) $\mathbb{Z}^n$, which is not locally constant as $\textbf{Ab}$-valued étale sheaf?
- Being a noetherian object in the category of $\mathbb{Z}$-modules or in that of abelian groups is the same thing: what is the reason for imposing that constructible $\mathbf{Ab}$-sheaves be valued in finite groups (which, in passing, kills injective objects)?

Let me add that I'd prefer to see examples in an algebraic setting, so working with either étale or Zariski topology: I was able to cook up an example of an infinitely-generated locally constant sheaf and a non-locally constant subsheaf on the one-point compactification of $\mathbb{Z}$ (with discrete topology) but this is not the kind of topological space which look natural to me.

** Edit**: In a first version, I asked also about $\mathbb{Z}_\ell$-sheaves and Nicolás adressed this in his very correct answer; but I am more interested in the true need for the finiteness conditions mentioned above. Here is the rest of the old question.

*In the literature, a "constructible $\mathbb{Z}_\ell$-sheaf" is usually defined as a projective system of constructible $\mathbb{Z}/\ell^n$-sheaves (plus some conditions) whereas a definition was a priori already available. What is an example of a $\mathbb{Z}_\ell$-constructible sheaf $\mathcal{F}=(\mathcal{F}_n)$ defined as a projective system which does not come from a "naive" constructible sheaf over the ring $A=\mathbb{Z}_\ell$? Why is the former the "right" definition?*

In the broad sense in which you state your question, and if you are looking for rigorous results, then the answer is surely no.

Keep in mind that the existence of the Lorenz attractor was only proved formally (using a computer-assisted proof) less than twenty years ago. This shows two things:

- In particular cases, it can be possible to show the existence of attractors by formal calculation, and probably a heuristic approach as you are describing would be a good start to develop an idea of where the attractors are
- It is highly unlikely that there is a general-purpose method that will work in all situations.

A few further comments: By the Newhouse phenomenon, in higher-dimensional systems, you will find places where topologically generic systems have infinitely many attractors, so you will never find "all" of them. Moreover, the structure of attractors will change radically when moving parameters. This shows that there is no general method of the type that you are talking about.

On the other hand, the Palis conjecture says that, at least for a dense set of systems, you will have finitely many attractors that you will detect from a positive-measure set of starting values. So you might not really expect to "see" Newhouse-type behaviour in actual applications.

135713192361995594135182884375730186410384861537121@skysurf3000: yes, but that's special to bicategories. It's special to $n \le 2$ that there aren't any interesting Whitehead brackets, and in fact strict $2$-groupoids can model all homotopy $2$-types. But that just stops being true for $n \ge 3$.I'm a master's student at the Turin University. My main interest is geometric, topological and combinatorial group theory and related topics such as the study of spaces with non-positive curvature. Unfortunately, I'm far from being an expert of the field. I'm also interested in computational geometry, numerical linear algebra, gpgpu, parallel programming and mathematical typography.129772@FrancescoPolizzi: See http://mathoverflow.net/questions/1878/extending-functions-on-closed-submanifolds-of-cn for instance.1138207I suspect that if you start with a model of determinacy and force to adjoin a nontrivial ultrafilter on $\omega$ (using infinite subsets of $\omega$ as forcing conditions), then the only ultrafilters in the resulting model should be the ones you get from the generic one and principal ones by iteratively taking limits (of previously produced ultrafilters along previously produced ultrafilters). I further suspect that this will produce altogether just continuum many ultrafilters. 1429224814385It's Theorem 5 in section VI.3 (page 181). It works on Lipschitz domains, which I believe includes your case.10994766967861977194@Kostya_I: Okay, but this application uses the fact "$\mathbb{R}^n$ is not a null set" which is surprisingly nontrivial, and which I would think of as one of the first significant results of measure theory.10777522140400195043816817821714843668806For two complex matrices $A,B \in \mathbb{C}^{n\times m}$ how to prove that: \begin{equation} \overline{\sigma}(B-A) \ge \underline{\sigma}(B) - \underline{\sigma}(A) \end{equation} where $\underline{\sigma}(A)$ and $\overline{\sigma}(A)$ denote the smallest and respectively the largest singular value of $A$.

4010231676327@Christian Remling. Maybe you are right, an there is no simple characterization. Unfortunately, I do not have a more specific conjecture :( Anyway, thanks for this comment!1247833128759016049892265286I'm not sure that I agree with your claim that combining Dirichlet's theorem with PNT adds only 10% to the length of the proof, but it is nice to see some texts where the two are treated simultaneously. Thanks especially for the link to your notes. 6Exactness of pure functors178100732046Cofibre of the $n$-fold transfer $\mathbb{R}P_+^{\wedge n}\to S^0$177275449640196885988557 +1000 !It may be a good approximation, but is certainly not an equation.17808331798897895642144831213467424@JoelDavidHamkins I guess that is just the MO policy/tradition for "big-list" questions. Now of course you, more than anybody else here, may feel free to use your privilege of "un-CW-ing" what some other mod made CW. But anyway, supposedly not many people will post an answer only because of getting some virtual reward for that. Sure I agree that such a reward is always nice for one's ego, but there are rewards which are far more precious in life. :) :)18468171767196In *Points fixes des applications compactes dans les espaces ULC* published in in the arXiv in 2010 Robert Cauty wrote

*il y a d’ailleurs une erreur dans la demonstration du
lemme 3 de [2], qu’il n’y a plus de raison de corriger, vu la superiorite de la
nouvelle approche*

(there is, by the way, an error in the proof of lemma 3 from [2] for which there is no need for correction in view of the superiority of the new approach)

It seems thus that Cauty still (or again) claims that the Schauder conjecture is settled.

Edit (August 2016). I was quite surprised that apparently there is no version of the mentioned article published in an international journal. While searching the web I learned that Robert Cauty died in 2013.

217478450601731437Can you say what you mean by "ordered group". A group with a total ordering that is both left and right invariant?827775@D_S: An unrelated question is why you've included the tag 'reductive-groups'? Probably 'group-actions' would be a better choice if you need another tag.1453462VWhy are $S$-arithmetic groups interesting?2193574PThis is also very instructive solution.As far as I understand, you can multiply your matrix from both sides by the matrix $\left(\matrix{E_p&0\cr 0&-E_r}\right)$ with a suitable sizes of unit matrices in order to obtain a matrix $B'$ with positive entries. Then Sinkhorn's theorem is applicable, as Felix mentioned.

(Surely, this theorem provides TWO diagonal matrices $D_1$, $D_2$ with positive numbers on the diagonal such that $D_1B'D_2$ is doubly stochastic. But, since these matrices are unique up to the scaling, they should coincide up to a scalar factor.)

What does the $p$-adic closure of an arithmetic lattice look like?1116271As Freyd and Scedrov show in "Categories, Allegories", you can think of a category as some kind of partial monoid. Such a monoid has a set of partial units that take over the role of objects in standard presentations of categories.

593017346082 In the book by Moriyoshi-Natsume, and in the MO question you reference, "operator algebras" are algebras of bounded operators on Hilbert spaces. That doesn't seem to be the same sense in which the term "operator algebra" is used in the book by Petkovšek, Wilf and Zeilberger, since there they are talking about operators on spaces of (possibly) unbounded sequences (e.g. the Fibonacci sequence). On the other hand, as you say, unbounded sequences do arise all the time as spectra of unbounded operators, so it is possible there is more to this than just a coincidence of terminology.related: http://mathoverflow.net/questions/19829/topologizing-free-abelian-groups5488106free Z-modules: Bases etc.22563552728322589761824843695553257748120103568222Does there exist finite dimensional irreducible representation of Euclidean or Poincare group in which translation and rotation both act nontrivially?jSubgroup property stronger than being characteristic94716612142051129286Sorry---I accidentally hit return and prematurely terminated my comment. I wanted to say that even if G is perfect, so G' = G, it may happen that not every element is a commutator. Also if H is any group other than an abelian two-generator group, then for all sufficiently large abelian groups A, the wreath product A wr H provides an example where not every element in the commutator subgroup is a commutator. This appears in Problem 4A.12 of my group theory book.105444524861720206622Although András' comment already answers the question, I think it is worthwile to give a few more details explicitely here, in order to point out that the analyticity is in fact a consequence of the *resolvent identity* only and has nothing to do with the operator whose resolvent we consider:

Let $X$ denote a complex Banach space, let $[X]$ denote the space of all bounded linear operators on $X$ and let $U \subseteq \mathbb{C}$ be non-empty and open.

**Definition.** A mapping $\mathcal{R}: U \to [X]$ is called a *pseudo-resolvent* if it fulfils the so-called *resolvent identity*
\begin{align*}
\mathcal{R}(\lambda) - \mathcal{R}(\mu) = (\mu - \lambda) \mathcal{R}(\lambda)\mathcal{R}(\mu)
\end{align*}
for all $\lambda, \mu \in U$.

Note that the resolvent of every closed linear operator on $X$ is a pseudo-resolvent, but there are also pseudo-resolvents which are not the resolvent of a closed linear operator (for instance the constant mapping $\lambda \mapsto 0$).

**Proposition.** Let $\mathcal{R}: U \to [X]$ be a pseudo-resolvent, let $\lambda,\lambda_0 \in U$ and assume that $|\lambda-\lambda_0| < \|\mathcal{R}(\lambda_0)\|^{-1}$ (where we define $0^{-1} := \infty$). Then
\begin{align*}
\mathcal{R}(\lambda) = \sum_{n=0}^\infty (\lambda_0 - \lambda)^n \mathcal{R}(\lambda_0)^{n+1},
\end{align*}
where the series converges absolutely in $[X]$ (which is endowed with the operator norm). In particular, the mapping $\mathcal{R}: U \to [X]$ is analytic.

**Proof.** It follows from the resolvent identity that
\begin{align*}
\mathcal{R}(\lambda) \big[\operatorname{id} - (\lambda_0 - \lambda) \mathcal{R}(\lambda_0)\big] = \mathcal{R}(\lambda_0).
\end{align*}
The operator norm of $(\lambda - \lambda_0) \mathcal{R}(\lambda_0)$ is strictly smaller than $1$ by assumption, so we conclude from the Neumann series that the operator in square brackets is invertible and that its inverse is given by $\sum_{n=0}^\infty (\lambda_0 - \lambda)^n \mathcal{R}(\lambda_0)^n$. This proves the assertion.

**A few references:**

For the case where the pseudo-resolvent is actually a resolvent of a closed operator:

"K.-J. Engel and R. Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)", Proposition IV.1.3 (as already pointed out by András in the comments)

"W. Arendt, C. Batty, M. Hieber, F. Neubrander: Vector-Valued Laplace Transforms and Cauchy Problems (2011)", Corollary B.3 in the appendix

For pseudo-resolvents:

- "M. Haase: The Functional Calculus for Sectorial Operators (2006)": in the appendix of this book, pseudo-resolvents are treated as resolvents of so-called
*multi-valued operators*.

*Personal remark concerning the Math.SE post mentioned in the question:*
To derive the analyticity of the resolvent as an "immediate" consequence of the resolvent identity without noting that this "proof" requires continuity to be shown first is, at least in my experience, an easily made mistake. I recall falling into this trap myself some time ago.

The paper The Tits alternative for $\operatorname{Out}(F_n)$ I by Bestvina, Feighn and Handel and the paper Automorphisms of free groups and Outer space by Vogtmann both state that $\operatorname{Aut}(F_n)$ and $\operatorname{Out}(F_n)$ are not linear groups for $n \geq 3$ respectively $n \geq 4$. They both cite The automorphism group of a free group is not linear by Formanek and Procesi. This text proves the claim about the automorphism group of a free group, but does not mention the claim about the outer automorphism group.

I was wondering if some properties about linear groups imply this result, but I couldn't figure out what property would. Any help would be appreciated.

**Once again, not really sure if this is the right place to post this question. If not, I'll take it down (or if someone knows how to relocate it to mathstack, that would be great too)**

Hy

First: take $\varphi_n(t) w(x)$ where $\varphi(t)\in C_0([0,T))$ and $w\in H^1_0(\Omega)$. If $\varphi_n$ is well chosen and converges to the characteristic function of the set $[0,t]$ you will understand the formulation.

Second: to be short, you have to mix the fact that $\int_0^t f_n(s,x)$ is a Cauchy sequence in $L^1(\Omega)$, that ($n$ fixed) $\int_0^t f_n(s,x)$ is continuous in $L^1(0,T; L^1(\Omega))$ and the De Giorgi regularity result. Since $\bar{v}_n$ is monotonically inscreasing in $n$ you have almost convergence to some $\bar{v}$ and then proving that $\bar{v}_n$ converges strongly to $v$ in $L^1(0,T;L^1(\Omega))$ is equivalent to prove that $\int_0^T\int_\Omega \bar{v}_n dxdt$ goes to $\int_0^T\int_\Omega \bar{v} dxdt$. From the previous result (the Cauchy sequence in $L^\infty(0,T;W^{1,q}_0(\Omega))$) you obtain the result.

90090720425531738758@Timothy: That's all right, in fact it is always good to give more detail! 970117187191972178431136212685012800092240017385654128385571309912577019783901Dear Elizabeth S. Q. Goodman, just a silly question, probably I have not correctly undestood, but if $df(0)=0$ then shouldn't be $\nabla f(0)=0$, independently of the metric $g$, and so unequal to $\Sum c_i\partial_i|_0$ independently by the diffeomorphism? Excuse me if I am not correct.This may turn out to be a bit embarassing, but here it is. It is probably not true that the ascending and descending central series (*) of a nilpotent group have the same terms. (Or at least one of MacLane-Birkhoff, Rotman and Jacobson would have mentioned it.) However, I have been unable to find an example where they are different. I thought I had a sketch of proof that they are always equal, but there is a gap, of the kind where you feel it is not patchable.

I've proved it for a few nilpotent groups (dihedral of the square, any group of order p^3, the Heisenberg groups of dimension 3 and 4 over any ring -- I think the argument extends to any dimension), and checked a few more exotic examples in the excellent Group Properties Wiki. So,

What is the simplest (preferably finite) nilpotent group such that its a.c.s. and d.c.s. are different?

and

Do the a.c.s. and d.c.s. coincide in some interesting general case?

(*) For completeness, the ascending central series of a group G is defined by Z_0 = 1, Z_{i+1} = the pullback of Z(G/Z_i(G)) along the projection, and the descending central series by G_0 = G, G_{i+1} = [G,G_i]. The group G is nilpotent iff ever Z_m = G or G_m = 1. It turns out that if such m exists it is the same for both.

2184274Smoothing the dual, we get a strongly convex primal. Then, is it true that the primal is smooth because the primal is strongly convex?669650306431687250705543145699623488581171807Are 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)Sorry, Ben. I'm familiar with non-abelian Hodge theory and Higgs bundles and whilst they are superficially related, they are really about something different. See the second half of Tony Pantev's answer.>http://www.ics.uci.edu/~prigor1I don't know any alternate geometric construction of the commutativity contraint, but there is a natural way to explain this, as a simple fact of homological algebra (and assuming very optimistic conjectures about mixed motives).

Let $DM_{gm}(k)$ be Voevodsky's triangulated category of geometric (=constructible) motives over $k$, with rationnal coefficients. For any (smooth) $k$-scheme, we have its homological motive $M(X)$ in $DM_{gm}(k)$ (this construction being functorial in $X$). Voevodsky proved that the full pseudo-abelian subcategory of $DM_{gm}(k)$, closed under arbitrary Tate twists, spanned by objects of shape $M(X)^\wedge=\underline{Hom}(M(X),\mathbf{Q})$ for $X$ smooth and projective is equivalent to the category $Chow(k)$ of Chow motives (the cohomological version, constructed from rational equivalence on cycles).

Now comes the conjectural part. Assume that there exists a motivic $t$-structure on $DM_{gm}(k)$; see this paper of Beilinson for what this means and for the link with the standard conjectures. Let us denote by $MM(k)$ the heart of this $t$-structure. The objects of the category $MM(k)$ are, by definition, *mixed motives*. This category is tannakian: for any prime number $\ell$, the $\ell$-adic realization functor defines a conservative $t$-exact and symmetric monoidal functor from $DM_{gm}(k)$ to the bounded derived category of finite dimensional $\mathbf{Q}_\ell$-vector spaces (this is part of what it means to be a motivic $t$-structure).

Call a mixed motive $M$ *pure of weight $n$* if $M[-n]$ is pure of weight zero in the sense of Bondarko (see this paper for the link between $t$-structures and weight structures): this simply means that $M[-n]$ belongs to $Chow(k)$, seen as a subcategory of $DM_{gm}(k)$. For $X$ smooth and projective over $k$, as $M(X)^\wedge$ is pure of weight zero, for any integer $n$, the cohomology object $H^n(X)=\tau^{\leq n}\tau^{\geq n}(M(X)^\wedge)[n]$ is pure of weight $n$. This implies that we have a (non canonical) isomorphism $M(X)^\wedge\simeq \oplus_n H^n(X)[-n]$ (the so called `Chow-Künneth decomposition'.

Denote by $M(k)$ the full subcategory of $MM(k)$ which consists of finite sums of objects $M$ of $MM(k)$ which are pure of weight $n$ for some integer $n$. We then have a natural functor
$$X\mapsto \oplus_n H^n(X)$$
from the category of smooth and projective $k$-schemes to $M(k)$. This functor factors (by construction) through the category $Chow(k)$:
$$Chow(k)\to M(k)$$
and factors through *numerical equivalence* (just because the motivic $t$-structure is compatible with the monoidal structure, using Poincaré duality in $DM_{gm}(k)$), inducing an equivalence of categories
$$Chow(k)/\text{numerical equivalence} \simeq M(k)$$
Now, the point is that $M(k)$ is a tannakian subcategory of $MM(k)$, and the modified commutativity constraint of the tensor product on the category of pure motives up to numerical equivalence is the one inherited from the natural monoidal structure of $M(k)$ through the equivalence above.

Finite groups are solvable if they have no nontrivial perfect subgroup. But I am sure that for infinite groups, the two notions diverge. Is there standard terminology for groups with no perfect subgroups?

25882018394981818226I always prefer to have error bounds for the CLT, so my favorite reference for your question is the paper "A Lyapunov type bound in $\mathbb{R}^d$" by Vidmantas Bentkus (Theory of Probability & Its Applications 49(2), 311--323, 2005).

From the abstract: Let $X_1, \dots, X_n$ be independent, mean-zero, $\mathbb{R}^d$-valued random variables. Let $S = X_1 + \cdots + X_n$ and let $C^2$ be the covariance matrix of $S$, assumed invertible. Let $Z$ be a $d$-dimensional Gaussian with mean zero and covariance $C^2$. Then for any convex subset $A \subseteq \mathbb{R}^d$,

$$|\Pr[S \in A] - \Pr[Z \in A]| \leq O(d^{1/4}) \cdot \beta,$$ where $\beta = \sum_{i} \mathbf{E}[|C^{-1}X_i|^3]$. This is a $d$-dimensional generalization of the Berry--Esseen Theorem.

11248550No. Take an extension of an elliptic curve by a torus which is of infinite order in the Ext group and take B the torus. (Reference: Serre, Groupes algébriques et corps de classes, Ch VII).

Xiao-Gang -- no worries; yes, yes and yes; if the action is trivial, i.e. via the trivial homomorphism $G\to H$, then $X$ is the quotient of $EH\times H$ by the trivial action of $G$. i.e., $H$ up to homotopy equivalence. I presume you are trying to understand a specific example, in which case it may be easier if you described it in your posting. Michael -- if $f:G\to H$ is a group homomorphism we can define an action of $G$ on $H$ by setting $h\cdot g=hf(g),h\in H,g\in G$.56167232983515904912516791829184176808710750922151576%As was pointed out by Deane Yang and Igor Khavkine in the comments, this feels like a fact that should be "looser" than Riemannian geometry. Indeed, as I will show below, your formula makes sense in the broader setting of smooth manifolds equipped with a volume form. I don't know a reference for this fact.

If $(M, \Omega)$ is a manifold equipped with volume form, ${\bf V}\in \operatorname{Vect}(M)$ is a vector field on $M$, and ${\bf X}_t:M\to M$ is the flow of ${\bf V}$, then:

1) One has $J_t\in C^\infty(M)$, (i.e., a smooth $J:\mathbb{R}\times M\to M$), with $J_0\equiv 1$, given by, $$J_t(x)=(\Omega_x)^{-1}\otimes \Omega_{{\bf X}_t(x)}\otimes \Lambda^n(d{\bf X}_t|_x),$$ well-defined since $\Lambda^n(d{\bf X}_t|_x):\Lambda^n( T_xM)\to \Lambda^n (T_{{\bf X}_t(x)}M)$ as Igor said.

2) One has a well-defined $\operatorname{div}({\bf V})$, namely (up to a sign, I forget whether $\pm$) $$\operatorname{div}({\bf V})=d(\iota_{\bf V}\Omega)/\Omega.$$

So the "hopefully-an-identity" $$\left.\frac{d}{dt}\right\rvert_{t=0}J_t=\operatorname{div}({\bf V})$$ (for simplicity I state just the $t=0$ formula) is well-formed. You can then prove it by choosing local co-ordinates in which $\Omega\equiv 1$. (There is surely an intrinsic proof, too, but I can't think of one now.)

**Afterthought, 30 Oct:** I imagine the intrinsic proof consists of identifying both sides with the Lie derivative $(\mathscr{L}_{\bf V}\Omega)/\Omega$.

**Update, 30 Oct, in response to questions of OP in comments:** If I calculate correctly, then yes, the formulas
\begin{align*}
\frac{dJ_t}{dt}(x)&=\operatorname{div}({\bf V})|_{{\bf X}_t(x)} J_t(x)\\
J_t(x)&=\exp\left(\int_0^t\operatorname{div}({\bf V})|_{{\bf X}_s(x)}ds\right)
\end{align*}
are also correct.

And yes, although the formula for "$\tfrac{d}{dt}\left[\det\left(d{\bf X}_t\right)\right]$" depends only on the volume form, I think one needs a Riemannian metric $g$ in order to make sense of "$\tfrac{d}{dt}\left[d{\bf X}_t\right]$" (really, the $g$-covariant derivative $D_t$ of the section $d{\bf X}_t$ of the bundle $T_x^*M\otimes TM$ along the curve ${\bf X}_t(x)$). Probably $$D_t\left[d{\bf X}_t\right]|_{t=0}=\nabla {\bf V},$$ where $\nabla$ is the Levi-Civita connection of $g$. (Please check this calculation before relying on it!)

79880324883558772187218Not sure he does your question, though, after reading those bits.11630921147439504681730631As mentioned in the comments, you can replace the $16\pi$ term by applying the Gauss-Bonnet theorem which gives you a Einstein-Hilbert functional term. Why not just use that for higher dimensions? Then you can test its convergence to the ADM mass (up to a constant multiple, of course). This is actually done by Miao, Tam and Xie, Quasi-local mass integrals and the total mass (arxiv).

43577I figured it out based on Keith Conrad's suggestion. Thanks everyone!82207150027141370818112436467053257873812962065153Tom, that's essentially a study of free algebras, isn't it? Determine the degree $n$ part of the free algebra on $n$ generators, for all $n\geq 0$, and check they don't assemble to an operad defining Lie algebras. This gives a definite answer, for sure.535741423891nDefinitions of Hochschild Cohomology $HH^{\bullet}(A)$The question says "where c[p] is a functional ...", so $c$ is A functional such that the first equation holds. The statement does not imply any functional $c$ would do. The statement does preclude the set of such $c$ being empty either. More directly, find $f$ such that there is some $c$ such that the first equation holds. Regarding your edited first comment, $c[\delta_z] = \exp(-kz)$ satisfies your condition $c[\delta_1]=c[\delta_{\frac{1}{2}}]$, or any similar condition. So it does not violate 1) and your conclusion does not hold.1026690Let $M$ be a smooth manifold endowed with $\nabla$ a flat torsion-free connection. Let $\operatorname{Diff}(M)$ be the group of smooth diffeomorphisms of $M.$ Obviously if $ \phi \in \operatorname{Diff}(M)$, then $\phi^* \nabla$ is torsion-free and flat, but equivalent (through $\phi$) to $\nabla.$

Is it known whether the set of nonequivalent affinely flat connections is infinite (if it is not empty)?. In the case of the $2$-dimensional torus it is known that the set of flat, torsion free connections quotient out by the group of diffeomorphisms is a $4$-dimensional manifold.

156593It has a limit if the argument of the function is expressed in degrees.@NoahS To answer your question about least Woodin, Magidor proved from Con(ZFC+supercompact) that there can be a supercompact with no strong compacts below. But supercompact is always a limit of Woodins.90754158164513212561675997156890416672552014501162479512886921148836 Welcome, Amri !1676725493786A simple upper bound is \begin{equation*} \text{tr}(AB^{-1}) \le \min\left(\lambda_{\max}(A)\text{tr}(B^{-1}), \text{tr}(A)\lambda_{\max}(B^{-1})\right). \end{equation*} Both these bounds are numerically "easy" to compute using Lanczos. For computing $\text{tr}(B^{-1})$ a randomized trace estimator can be used (following the more general idea outlined below).

Here is a simple approach, motivated by this nice book:

- First compute $\alpha=\|B\|$ approximately using Lanczos
- Now consider $B=\alpha I - C$, so that $B^{-1}=(I-\alpha^{-1}C)^{-1}/\alpha$
- After that, consider \begin{equation*} \text{tr}(AB^{-1}) = \frac1\alpha\text{tr}(A^{1/2}(I-\alpha^{-1}C)^{-1}A^{1/2}) \end{equation*}
- Now use the von Neumann series \begin{equation*} \text{tr}(A(I-\alpha^{-1}C)^{-1})=\sum_{k\ge0} (-1)^k\alpha^{-k}\text{tr}(A^{1/2}C^kA^{1/2}) \end{equation*}
- Let $u \sim \mathcal{N}(0,I)$ be a mean-zero spherical Gaussian rv. Then, we approximate the above quantity by taking $m$ samples, $u_1,\ldots,u_m$ and iteratively computing $u_i^TA^{1/2}C^kA^{1/2}u_i$ for $1\le i \le m$. Observe that the key subroutine that we have is to iteratively compute $z^TC^kz = z^TC(C^{k-1}z)$.
- In expectation this will be an estimator for the trace in question since $E[\text{tr}(u^TA^{1/2}C^kA^{1/2}u]=E[\text{tr}(A^{1/2}C^kA^{1/2}uu^T)]$ and $E[uu^T]=I$ by assumption.

If you do not have access to $A^{1/2}$ (or a Cholesky factorization of it) then an additional level of approximation arises by building a subroutine to compute $A^{1/2}u$. Such $f(A)b)$ family of subroutines are the subject of research interest in numerical linear algebra (see e.g., Nick Higham's webpage and his book on *Functions of Matrices* for further information).

Does anybody know of any finitely generated semigroups that are not residually finite and whose group of units (if there is an identity) is trivial? Basically, I'm looking for finitely generated semigroups that are not residually finite, but I don't just want to punt and look at groups (like Thompson's groups or something). Thanks!

2206569706941user147215998557701699638620115362792197943348230Can anyone point out some articles for the conjecture: the dual of an algebraic matroid is algebraic? Thank you!

It is known that some plane curves can be drawn with a tool. For instance, I heard at a web site that Archimedes created his spiral in the third century B.C. by fooling around with a compass and others.

Let’s however look at the spiral defined by the equation: $r'(\theta)^2+r(\theta)^2=\theta^2$, $r(\theta=0)=0$

I am looking for a method ( a tool) which could help to plot the spiral on paper ( I named it as Archimedean-Galileo spiral. For large $\theta$, the curve represents Archimedean spiral: $r=\theta$. When $\theta$ is small it transforms in Galileo spiral $r=\theta^2$) .

The spiral has a property that the junction point of the curve and the ray uniformly rotated in the origin coordinates when the junction point moves with uniform acceleration.

Do you think that there is a way to draw it without computer, but with other special curves (tools)?

I thought about the spiral of Theodorus, but I am not sure how the spiral of Theodurus is connected with the equation.

1133277Oddly enough, the wikipedia page on Grassmannian doesn't mention the dimension explicitly, as far as I can see.596962127165022360151629687XIs there a name for such operator anywhere?75956175730920005141800331728927134882917838742044699222239@QiaochuYuan The compatibility is the following: the 2-form must be closed for the Lie algebra cohomology of $\mathfrak{n}$. Like in [this paper](http://arxiv.org/pdf/1510.08212v1.pdf)18311macmania314I encounter the following problem of the type of degree theory in Hilbert spaces.

Consider the Hilbert space $L^2(\mathbb T)$ with the natural inner product.

Taking Fourier expansion with bases $e^{i2\pi k x}$, we have a direct sum decomposition into $E^-\oplus E^0\oplus E^+$, where $E^-,E^0,E^+$ correspond to subspaces with $k<0$, $k=0$ and $k>0$ respectively.

Next take the unit sphere in the subspace $E^-\oplus E^0$, denoted by $A$, and take the ball of radius one in $E^0\oplus E^+$ centered at $(1,0)\in E^0\oplus E^+$ and denote it by $B$.

It is clear that $A$ and $B$ intersect at one point $(0,1,0)\in E^-\oplus E^0\oplus E^+$.

Now we take the ball $B$, fix its boundary and consider any $B'$ homotopic to $B$ relative to its boundary. The question is:

Does $B'$ always intersect $A$?

Is there a topological degree theory explaining this, or a counter-example? The Leray-Schauder degree theory does not apply to this problem.

1382851919535 It really has to do with how intertwined gauge theory and algebra are. The most successful quantum field theories rely on the mathematical background of group theory because they're all gauge theories and the properties of groups are inherently important to gauge theory. If you imagine a world where calculus wasn't invented to solve physics problems, you might be able to ask the same question along the lines of "why is calculus so heavily utilized in mechanics?" It provides the best mathematical models of the physical phenomena. Why nature is this way, that's probably a philosophical question.477158AcuriousmindOr even http://stats.stackexchange.com, which is more statistics-focussed.1890021163316122493128622648Failure probability formula8576921142139898641Well, it's a fun device, but aside from the obvious objections given before, it also doesn't give any insight into why the theorem is true. Isn't that really what a proof is good for?810794@Noldorin: (1) I am not familiar enough with the subsystems of second-order arithmetic and how easy/difficult it is to construct semantics for the real numbers with them. (2) What about accessing objects larger than $\frak c$?1310198767590915282https://www.gravatar.com/avatar/b8810d9ef6520d9ece64673604464061?s=128&d=identicon&r=PG&f=15016801630574Let $a_n$ be a holonomic sequence. By definition, that means there exists a linear differential equation of finite order which annihilates $F(x)$, where $F(x):=\sum a_n x^n$. Similarly let $b_n$, $c_n$,$d_n$ be holonomic sequence.

Now we know few methods to get holonomic sequences out of already known holonomic sequences are as follows

1)$ \textit{Cauchy product}$

$\alpha_n := \sum_{k=1}^{n} a_{k}b_{n-k}$

2) $\textit{Hadamard product}$

$\beta_n = a_n b_n $

I am wondering about the following expression being holonomic.

$$ \sum_{k=1}^{n}a_{k}b_{n-k}\sum_{l=0}^{k-1} c_{l}d_{n-l} $$

It's true special cases of $a_n,b_n$.I wonder if I am missing some tricks.

In particular I want if the series $\gamma_n:=\sum_{k=1}a_k b_{n-k}s_{k,1^{n-k}}$ is holonomic. We obtain the series by taking $c_{l}=s_{(l)}$ and $c_{(d-l)}=s_{1^{d-l}}$. Where $s_{\lambda}$ denote the schur function.

Another standard example is Littlewood's theorem that the number of primes less than x is sometimes greater than Li(x). His proof used different arguments depending on whether the Riemann hypothesis is true or false. See http://en.wikipedia.org/wiki/Skewes%27_number

19671861195054118677621742 649512340133914502149630bDiscounted total reward vs. Average total reward191270018231646How to find a sub-forcing?74735821708588389383786739805249The existence of a specific partition of the edge set of $K_{2n}$9305841740115194968610673511075434$K$ is not contained in the boundary of $B^n$; it does not even touch the boundary. Any topos whose object of truth values has more than two truth values will contain freakish subsets of the singleton. Consider the Grothendieck topos $Sh(\mathbb{R})$ of sheaves over $\mathbb{R}$. The subsets of the singleton $\lbrace 0 \rbrace$ correspond to the open subsets of $\mathbb{R}$, but only two of those, $\emptyset$ and $\mathbb{R}$, are finite. There is also a computational reason why you should not expect subsets of a finite set to be finite. Consider the intersection of a c.e. set and a finite set. Why do you expect to be able to produce a finite list of its elements?@MaxHorn say $R=Z$. Yes, it is a polynomial ring in two variables.1787586943721Vit, I'm not familiar with the details, but it seems to me that the cohomology associated with any elliptic complex of differential operators, such as Dolbeault cohomology as well as operators on vector bundles, depend on more than the topology of the domain.776991744410225685210465752046947Note however that the wikipedia reference that you provide contains a big mistake. If $X$ is a singular variety, the collection $Sing(X), Sing(Sing(X)),...$ is not a Whitney stratification in general.157041314489301871280129456691294504On the contrary, the graphs can be disconnected, but they need to be serial, i.e. there are no "dead ends", otherwise we would have problems with infinite paths. @Pablo: The splitting field if often non-solvable. Consider $A=E^n$ for an elliptic curve $E$ over $k$ that is not geometrically CM. The Aut-scheme of $A$ is the constant $k$-group ${\rm{GL}}_n(\mathbf{Z})$, which is also the Aut-scheme of ${\rm{GL}}_1^n$, so the sets of isomorphism classes of $k$-forms of $A$ and of $n$-dimensional $k$-tori are in natural bijection. For a finite Galois extension $k'/k$, the Weil-restricted $k$-torus ${\rm{R}}_{k'/k}({\rm{GL}}_1)$ splits over a finite separable $F/k$ if and only if $k'$ embeds $F$ over $k$, so enough that there is such $k'/k$ non-solvable.@MarkGrant: I am not sure if the paper of Levine contains a similar result, he seems to be talking about the (meta)stable range (i.e. something like $4i=n$ or $4i=n-1$). The question, however, seems to be talking about $i$ much larger than $n$, i.e. much above the stable range.@HJRW and your approach can also show that the product of finitely many subgroups does not contain a coset of a finite index subgroup, right?1177998Why? Because I was being stupid and misread the question :D. I don't really see why that construction has any better chance than any other, but I concede it *might* work.2184439I agree with Yemon. I don't think that this question should be closed!L@JeffStrom I wondered the same thing.Scounged159588811120923326579There is a map from $S^3$ to the Poincare homology sphere $P$ ($S^3$ is its universal covering) and there is a map $P\rightarrow S^3$ ($P$ can be constructed usin a dodecahedron identifying some faces. Collapsing the whole boundary gives $S^3$ and the map). I guess it is better to consider the latter map (but it is just a feeling).1968924zA formal condition for a functor to preserve compact objects11533472294785880154@wccanard: Yes, your remarks are what I had in mind when I said that the decomposition laws for $\mathfrak{S}_3$-extensions can be recovered from Langlands. And yes, Satgé's theorem does seem to apply to a finite simple group $G$ (take $H$ to be trivial), but then the split primes are no longer characterised by quadratic forms but by forms of degree equal to the order of $G$.104182025481416950581450001646281277595@JeremyRickard It seems I have both the first and the second (in the first edition it's called Cor 5.9.1b), but in both it states that the results (a),(b),(c) of 5.9.1 hold for $M(n) = \widehat{\mathbb{Z}}[[t]]$ and $F^{\text{nilp}}$ in place of $\mathbb{Z}_p[[t]]$ and $F$ the free pro-$p$ group of rank $n$. In particular, it seems that the analogue of (c) in the case $n = 1$ should be $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]] = \widehat{\mathbb{Z}}[[t]]$...13197721783475260221953921122434If $X\subset\mathbb{P}^n$ is a smooth subvariety and the normal bundle splits then, by adjunction, $X$ is subcanonical i.e. $\omega_{X}\cong\mathcal{O}_X(k)$. This is important for the following result in codimension two.

In *"Bénédicte Basili and Christian Peskine, Décomposition du fibré normal des surfaces lisses de $\mathbb{P}^4$ et structures doubles sur les solides de $\mathbb{P}^5$, Duke Math. J.
Volume 69, Number 1 (1993), 1-245"*, you can find the following result:

Let $X\subset\mathbb{P}^N$, $n\geq 4$ be a smooth codimension two subvariety. If $N_X$ splits then $X$ is a complete intersection.

For curves in $\mathbb{P}^3$ is quite different and the following is still an open problem: "Let $C\subset\mathbb{P}^3$ be a smooth, connected curve. Is it true that if $N_C = \mathcal{O}_C(a)\oplus\mathcal{O}_C(b)$, then $C$ is a complete intersection?"

4Oops. Silly me. Sorry. :(12467931170145194297279116213964551034873Two corrections: first, there were several previous results that completely characterized a quantum complexity class in terms of a classical class (for example, QRG=EXP, NQP=coC_{=}P, PostBQP=PP, and BQP_CTC=PSPACE). Second, while PSPACE in QIP is nonrelativizing, the "new" direction (QIP in PSPACE) is relativizing.1825556606553117935911544849Indeed there was a bounty added by the person who wrote the other answer, who didn't specify its purpose. Anyway this additional answer is very welcome (and possibly the OP is still active under another nickname).Let $\lambda$ be an integer partition: $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq0)$. Further, let $h_u$ denote the hook-length of the cell $u$.

We call $\lambda$ a $t$-core partition if none of its hooks $h_u$ equals $t$. Define $c_t(n)$ to be the number of partitions of $n$ that are $t$-core partitions. It's well-known that $$\sum_{n\geq0}c_t(n)\,q^n=\prod_{k=1}^{\infty}\frac{(1-q^{tk})^t}{1-q^k}.$$ For example, $\sum_{n=0}^{\infty}c_2(n)\,q^n=\sum_{k=0}^{\infty}q^{\binom{k+1}2}$.

Now, consider only those partitions of $n$ with distinct parts and let $d_t(n)$ be the number of such partitions that are $t$-cores. Then it is easy to see $d_2(n)=c_2(n)$.

QUESTION.Is this true? $$\sum_{n\geq0}d_3(n)\,q^n=\sum_{k\geq0}q^{k^2} +\sum_{k\geq1}q^{2\binom{k+1}2}.$$

Note that I have simplified the generating function from $$\frac12\prod_{n\geq1}(1-q^{2n})(1+q^{2n-1})^2+\prod_{n\geq1}(1-q^{2n})(1+q^{2n})^2-\frac12.$$

Is there a good survey article listing all the theorems of Euclidean geometry that are equivalent to the parallel postulate?

90148414593554293691379254111090818966459008Looking for Spock here? Live long and prosper, my friends.

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1346326222218220209441530470 Can't help you.1087005I disagree with your last statement. Saying "a bit" is a bit of an understatement.770765>Automorphy Factors and BundlesAssume the Dirichlet form $\varepsilon$ adimits a Carre du champ $\Gamma$ and introduce the multilinear form $\Gamma_2$ $$ \Gamma_2 [f,g;\phi]:=\frac12 \int_X (\Gamma (f,g)L\phi -(\Gamma(f,Lg)+\Gamma(g,Lf))\phi) dm \quad (f,g,\phi)\in D(\Gamma_2) $$ where $D(\Gamma_2):=D_V(L)\times D_V(L)\times D_L^\infty(L)$,$V = \{ f:\varepsilon (f) < \infty \} $ and $$ D_V(L)=\{f\in D(L): Lf \in V\},\quad D_L^\infty(L):=\{\phi \in D(L)\cap L^\infty(X,m):L\phi \in L^\infty(X,m)\}. $$ We say that the strongly local Dirichlet form $\varepsilon$ satisfies the $BE(K,\infty)$($BE(K,N)$) condition, if it admits a Carre du Champ $\Gamma$ and $$ \Gamma_2 [f,f;\phi] \ge K\int_X \Gamma(f)\phi dm \quad \mbox{for every}\quad (f,\phi)\in D(\Gamma_2),\quad \phi \ge 0.\qquad (BE(K,\infty)) $$ $$ \Gamma_2 [f,f;\phi] \ge K\int_X (\Gamma(f) +\frac{1}{N} (Lf)^2) \phi dm \quad (f,\phi)\in D(\Gamma_2), \phi \ge 0.\qquad (BE(K,N)) $$

for Alexandrov spaces with curvature bounded below, do they satisfy $BE(K,N)$ and $BE(K,\infty)$?

Since Alexandrov spaces satisy $CD(K,N)$ which is equivalent to RCD(K,N). And $RCD(K,\infty)$ implies $BE(K,\infty)$, so Alexandrov spaces satisfy $BE(K,\infty)$.

But I don't know the relationship between $RCD(K,N)$ and $BE(K,N)$, so I don't know whether satisfy $BE(K,N)$.

2125729557709506601197607512870801439723120904521066151228037 cron6355061624590271125ts32922577533610924042281607301153374As far as I can tell you are correct. In fact, any map in the category of simplices coming out of a nondegenerate simplex must be injective. If $a:\Delta^n\to X$ is a nondegenerate simplex and $f:\Delta^n\to\Delta^m$ is a map to some other simplex $b:\Delta^m\to X$, factor $f=hg$ as a surjection followed by an injection. If $g$ is not the identity, then it makes $a$ a degeneracy of the simplex $bh$ in $X$. Thus $g=1$ and $f=h$ is injective.

165085710813881035918226492As a side comment, the sum resembles in a way Legendre's formula for counting primes http://mathworld.wolfram.com/LegendresFormula.html. You are counting integers up to $N^2$ that are divisible by $N$,...,$xN$, with multiplicities. Perhaps, like in Legendre's formula, there is an inclusion-exclusion approach to this. If that works though, the sum might be reduced to functions like $\pi(N^2)$, explaining the strange fluctuations in the error term but proving that it will be difficult to get an exact answer :)And I think at least in the case of a separable $E$, one can pass from a linear extensor for the scalar case to one for E-valued mappings, by the argument in Bill Johnson's answer.277794989838The answer is yes. I think it should be an exercise in any book on representation theory. Since $H$ has finite index in $G$, and $V$ is finite dimensional, so is the representation $W=Ind _H^G (V)$ induced to $G$. In characteristic zero, this is equivalent to proving that the Zariski closure of $G$ in $GL(W)$ is reductive.

The connected component ${\mathcal G}^0$ of the Zariski closure ${\mathcal G}$ of $G$ in $GL(W)$ is unchanged if we replace $G$ by any finite index subgroup $K$. We take $K$ to be the intersection $\cap gHg^{-1}$ as $g$ varies over $G$; this is a finite intersection since $G/H$ is finite.

Restricted to $K$, the representation is semi-simple since $W$ restricted to $K$ is the span of the restriction to $K$ of the semi-simple $gV$ as $g$ varies over $G/H$. Therefore, the unipotent radical of ${\mathcal G}^0$ acts trivially on $gV$ for every $g$ and hence on $W$. Thus, ${\mathcal G}^0$ has no unipotent radical.

[Edit] I should have said that I am assuming $Char (K)=0$.

114798422490221112440781488735814jA generalization of the Grauert direct image theoremMy character theory is rather weak, so excuse me if this is a triviality.

I have read on the encyclopedia of maths that for any group $G$, every block of $G$ contains an irreducible character of height zero. Is there a straightforward way to see this? Less importantly, is there a 'canonical' choice, like the trivial character for the principal block ('canonically' can be taken as vaguely as you like)?

Thanks for your time.

Ah, you patched (1) already :-). I just wanted to add that the Hahn-Banach theorem can be used to dualise (3), and it seems to say something like a (tempered distributional) solution to an inhomogenous heat equation with compactly supported forcing term and zero initial data cannot be compactly supported at any future time, which is a sort of "parabolic unique continuation" result.855726No - there are triangle-free, vertex transitive graphs of infinite chromatic number, see for instance this article.

Does $\Lambda$ by any chance assign to each tuple its stabiliser in $U$?Ok, I think there are examples where $\Omega(\log n)$ colors are needed.

Here’s an example, let $a_{ij} = \frac{1}{i}$ for $j < i$ and $a_{ij} = \frac{1}{j^2}$ for $j > i$. Then $\sum_{j} a_{ij} = \frac{i-1}{i} + \sum_{j > i} \frac{1}{j^2} = O(1)$. Of course, the bound is $O(1)$ instead of $1$, but that can be normalized and all that.

However, note that $\sum_j a_{j1} = \Omega(\log n)$ and if we only have $o(\log n)$ partitions, this sum cannot be "distributed" into small enough parts.

230893@David: Algebraic closure is not unique in the sense that $k \to k^{alg}$ cannot be extended to a functor such that $k \subseteq k^{alg}$ is a natural transformation. This shows that everytime you have to choose an algebraic closure. Two algebraic closures are isomorphic, but the isomorphism is not unique and not constructive. Thus the lnab-the does not fit here. Concerning your other question: I think that the canonical homomorphism $Gal(k^{alg} / k) \to Aut(k^{sep} / k)$ is an isomorphism. @Torsten: You may post this as an answer, I'll upvote it.DInadmissibility of Simpson's ruleBased on the comments, it looks like this is not a question specific to Pell's equation, and that you just want to evaluate a single binomial with big inputs as quickly as possible.

If you check the equation directly using fast multiplication algorithms (e.g., Schönhage-Strassen), you can expect the calculation to require about $(\log z )\cdot (\log \log z)$ operations, where $z = \max \{ x, y, D \}$.

If you want a way to quickly find a negative answer, you can check relative sizes and leading digits, then try reduction modulo small integers. If you start with a random negative answer, you can expect it to be eliminated after at most a few divisions (i.e., about $\log z$ operations).

To find a positive answer using modular arithmetic, you can use the Chinese remainder theorem. To prove that the identity holds, it suffices to check it modulo $n$, for $n$ ranging over a collection of positive integers whose least common denominator is larger than $x^2$ and $Dy^2$. It is common to check modulo a large collection of small primes, and this will require about $\log z$ primes and $(\log z)^2$ operations. Another natural choice with a binary computer is Fermat numbers, of the form $2^{2^n}+1$, since the division-with-remainder can be optimized - this ends up looking a lot like direct calculation.

In summary, the advantage of checking modulo small primes is that it lets you quickly eliminate negative answers, and the disadvantage is that (if I'm not mistaken) it is roughly quadratically slower than direct calculation when you have a positive answer. You can choose your method depending on exactly what sort of calculation you plan to do.

1978584907114988657I would like to study some family of sequences of real numbers. What axioms do you think I should adopt?3862359569820059472109467In particular, we may restrict $\mu$ to the set of tempered principal series representations, which a are parameterized by unitary characters of the maximal torus $T$ of $G$.

$\mathbf{Question:}$ Is there a good formula for the Mellin transform of this measure (this is a function on $T$)? I am mostly interested in the case $G=SL(2, F)$ for a local non-archimedian field $F$.

1115955Nathan Cook104619238675163057862023987156915382985211809581948995@AkivaWeinberger : Maybe. To be honest I don't immediately see how to turn the idea in the video into an actual proof. How do we know that the configurations arising from the construction in the video really cover all possibilities?@Igor: have you actually done that? I made the pictures at http://neil-strickland.staff.shef.ac.uk/talks/durham.pdf using Blender, and found it extremely painful to produce anything that looked smooth and natural.543017@BjørnKjos-Hanssen Please see the background, which I state above.19115912041153163355user3368556787525831I edited my answer, hopefully this is now complete. Your idea of using the closure of the fundamental Weyl chamber was the right call.I don't know what algorithm Maple is using, but a 100x100 SNF computation, if the matrix is complicated or if the algorithm isn't very smart, can be a disaster. A sparse 120x120 matrix in a primitive SNF algorithm can lead to disaster. You can very easily go past a 64-bit integer size, or start swapping if you are using arbitrary-precision integers. Hafner-McCurley and Havas-Holt-Rees have fairly smart SNF algorithms that can easily handle 400x400 matrices on contemporary computers.Also, btw, closing with "kind wishes" does not disturb me in the least!Thanks for the response! I don't exactly have much background here, but from what I can tell you're confirming and elaborating on what Pete mentioned in the comments. I guess I should mention that I would be perfectly happy with an embedding into a higher-dimensional projective space as long as its defining equations could be written explicitly in terms of a Hauptmodul. Does one exist for N = 1, 2?22322682257590577736 This is in a number of advanced math methods for physics, and graduate electromagnetism texts, but one offhand I know it's in is "Electrodynamics of Continuous Media" by Landau and Lifshitz, chapter 1 section 3, on methods of solving electrostatic problems. But there are other books out there with more involved and more sophisticated treatments, but offhand I don't know any titles. If you search amazon's "search inside this book" for "the method of conformal mapping" you can find part of the discussion. But the basics of it are elementary enough that that book should be sufficient.3907841381926287718925191247048329892080653The last sentence is just wrong. For instance, CH is $\Delta_2$, but not upwards absolute.Hmm, I don't think my $Z$ is a smooth submanifold of $\mathbb{R}^4$. At least, not in any way that I can think of.692074Vrational rotation vector and closed curvesvHow can I tell if y is a function of x in a random sample?<Penumbra? Cointerior?

@DmitriPavlov: I suspect you and Andy are using different criteria for "play a role". Likely Andy is setting the bar at "is used to prove a foundational result in the theory" or something like that. For example, Teichmuller theory and geometric flows are used to prove foundational results in 3-manifold theory. Saying that I *can* (if I choose) use subject X to study subject Y is a far lower bar. Do people in subject Y need that tool for something? Answering that would get you somewhere but it would still be significantly below Andy's bar.660018@JeremyRickard nice! Yes, I this looks exactly like what I was looking for. Could you post as an answer so that I can accept it?Ah, that's what "The Joy of Cats" (a category theory book) referred to.2251535This is an interesting, but hard, problem. To add some context, Schofield (Bull. London Math. Soc. 17 (1985), no. 4, 393-394) has shown that there is a function $f$ such that $g \leq f(\dim A)$. But I do not even know if $g \leq \dim A$ is true in general.14049841725933@marc: So you want the two vector fields to be orthonormal, not just orthogonal?14914692162771626187187427Without much certainty, if G is a profinite abelian group I think the infinity topos completion of the topos of continuous action of G should fit into such a spectrum object without being etale.The numbers $a_{n,m}$ are in fact the Fourier coefficients of the polynomial $$P_n(x)=\prod_{j=1}^{n-1} \Big( \frac{nx}{2} + \frac{n}{2}-j\Big) $$ with respect to the Chebyshev measure $d\sigma:=(1-x^2)^{-1/2}dx$ on $[-1,1]$, and its orthogonal bases of the Chebyshev polynomials of the first kind. Precisely, for $0\le m\le n$ $$a_{n,m}=\frac{1}{\pi} \int_{-1}^1 P_n(x)T_m(x)d\sigma \ .$$

Changing variable, we have a trigonometric version: $$a_{n,m}=\frac{1}{\pi}\int_0^\pi P_n(\cos \theta)\cos (m\theta)d\theta \ .$$

Note that the polynomials $P_n$ and $T_m$ are odd resp. even, according to the parity of $n-1$, respectively $m$, so the integrand $P_n(x)T_m(x)$ has the same parity of $n+m-1$. On the other hand, the Chebyshev measure is symmetric, which explains the vanishing property $a_{n,m}=0$ whenever $n+m$ is even. Moreover, for odd $n+m$ the integrand is positive and concentrated about $\pm 1$; this should hopefully yield to the desired estimate $a_{n,m}>0$. I'll try some computation and in case add details later.

To compute the integral we may also use the Chebyshev-Gauss Quadrature formula on $N$ nodes, which is exact on polynomials of degree less than $2N$. Therefore, for $2N\ge n+m$ we have $$a_{n,m}=\frac{1}{N}\sum_{k=1}^{N} P_n\Big(\cos\big( \frac{2k-1}{2N}\pi \big) \Big)\ \cos\big( m\frac{2k-1}{2N}\pi\big) \ .$$

605681576892What is missing from your description is a requirement that all the vertices of the polygon must be among the given points. If the polygon can have vertices not in the given set, then one can easily make a star-shaped polygon with area $\epsilon$ for any $\epsilon > 0$. (I assume this is what Noah meant by his comment.)1779139---I see mathematics as a messy bag of tricks to solve all kinds of messy problems, not as "a set of definitions and statements.--- Here we totally disagree. I see mathematics as a well-organized toolbox to solve all kinds of messy problems. "Definitions and statements" are just labels on various compartments and if you screw your labeling, you'll have hard time finding the right tool.2244814132508Depends what the bounds in Lagarias-Odlyzko are and I don't have them in my head right now. You might look at a recent paper of Kowalski and Zywina which carries out a computation of this kind in real life.1817122114296232058922870414785763541151242903137673518887881402913848811528035user1618618489882813801084852Max MenziesIf it's dissipative on $C_0({\bf R}^d)$ then it should be dissipative on $C_b({\bf R}^d)$ too.

First check this in the case that $a$ has bounded support. The point is that if $a$ is supported on a ball of radius $r$ about the origin then $Af(x)$ only depends on values of $f(y)$ for $|x - y| < r$. So if $f \in C_b({\bf R}^d)$ and $x \in {\bf R}^d$ then for $R \gg r$ the function $$g(y) = \cases{\big(1 - \frac{|x - y|}{R}\big)f(y)&if $|x - y| < R$\cr0&if $|x - y| \geq R$\cr}$$ will satisfy $Ag(y) \approx (1 - \frac{|x-y|}{R})Af(y)$ for $|x - y| < R$. Thus $(\lambda I - A)g(y) \approx (1 - \frac{|x-y|}{R})(\lambda I - A)f(y)$ and the dissipation inequality for $g$ will imply approximately the same inequality for $f$ at $x$.

For general $a$, approximate by $a$ with bounded support. This part is easy.

Also it's not surprising that the error term for $\check{m}(x)$ is often much smaller than $O(x^{-1/2})$; it's the same reason that $|M(x)|/\sqrt{x}$ is usually smaller than $1$, namely that you have an expansion of the normalised error term of these summatory functions as (essentially) almost periodic functions of the form $\sum_{\gamma} c_{\gamma} x^{i\gamma}$, and "most" of the time, there will be a ton of cancellation in these sums. You can make this precise via the Rubinstein-Sarnak method of determining the limiting logarithmic distributions of these normalised error terms.@Bullmoose: Probabilists are used to expressing your "sequence of sequences" as a triangular array: a family of random variables $X_i^n$ indexed by $n \ge 1$ and $1 \le i \le n$, as Stephan Sturm suggested.1912358214974189963497520113599319777167136151368371Thanks. 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.2142731Haefliger proved that $S^k$ in $\mathbb{R}^n$ is unknotted if $2n > 3(k+1)$. Therefore, any 2-sphere embedded in $\mathbb{R}^n$ with $n\ge 5$ is unknotted. On the other hand, as Mark Grant points out in his comment, there are knotted 2-spheres in $\mathbb{R}^4$ (see this paper by Andrews and Curtis, for example).

354467JI like 2,3,10,10 allowing exponents.4Have you looked in ATLAS?163541356058313087261210963168635120778612719061851992136678219840982237059763566jWhat is correct name of the following construction?76861279229224905189575019075591572964'For this discussion, $G$ is a compact semisimple Lie Group.

For many of the common representations of compact groups, there is a realization of the representation as the automorphisms of some algebraic structure. For example, the adjoint representation of any group $G$ is realized as the automorphisms of $LG$, the standard representation of $SO(n)$ is as the stabilizer of the volume form on $\mathbb{R}^n$, the 7-dimensional representation of $G_2$ is realized as the automorphisms of the imaginary octonions or as the stabilizer of a certain 3-form on $\mathbb{R}^7$, and so forth.

My particular interest for this question is in the automorphisms of algebras over $\mathbb{R}$ (although for Question 1, I don't care about the ground field).

Question 1:Given a representation of $G$ (not necessarily irreducible) is there a general method of constructing an algebra (over the appropriate ground field) such that $G$ is the group of automorphisms of this algebra acting by the prescribed representation?

My gut feeling is that a solution to this question in general would be related to the construction of Cartan Products, but even given the irreducible summands of the representation I don't see how to make an algebra which is not a direct sum of algebras (the problem with a direct sum of algebras is that it can have larger automorphism group than $G$).

Since this is such a general question, I understand it may have no easy answer. However, I do have a specific example in mind; some of my recent experimentation has suggested that if a representation has certain properties, any algebra on which this representation acts as the full set of automorphisms should also have certain nice properties. One specific case I would like to know about is a 16-dimensional representation of $G_2$ which contains two trivial summands and one copy of the 14-dimensional adjoint representation.

Question 2:Does anyone have an example of a 16-dimensional $\mathbb{R}$-algebra whose automorphism group is $G_2$ with the decomposition of the automorphic action given by $1\oplus1\oplus 14$. For this it would be enough to describe the bilinear multiplication.

Note that I do not care if the algebra is a unital ring, is commutative, associative, or any other particular property; any 16-dimensional algebra structure whose automorphism group is provably $G_2$ with the given representation will suffice for me to test my ideas.

145460914993962174911Sam, for that matter, can there be a map of finite-type algebras over a field that is surjective but not an isomorphism? I'm guessing not. Maybe still not even if you replace surjection" by "epimorphism"?621654@CarloBeenakker - yes I have taken a look, but the intuitive explanation is rather vague, and the examples rather technical. I'm really looking for something in between!17616806064215384432290323Btw, being rectifiable is the same as having density $1$ a.e., by a theorem of Preiss.mas19:Submodule of a Kisin moduleLet $\Omega$ be a bounded smooth convex domain in $\mathbb{R}^3$, then consider the following minimization problem:

$$\inf_{c \in \overline{\Omega}} f(c), \quad f(c) := \int_{\partial \Omega} |x-c| d \sigma_x,$$ where $\sigma$ is the surface measure.

If $\Omega$ is a ball, then it is easy to prove that the infimum is attained only at the center of the ball. I guess for a generic fixed convex domain $\Omega$, the infimum should be attained at the "center" of $\Omega$, but I've no idea how to prove or disprove it, and also I don't know if the "center" is the standard meaning, that is, $c=\frac{\int_{\Omega} x dx}{|\Omega|} $.

Any comment or hint would be really appreciated.

47493615364781188221To a connected graph $G$, quantum field theory attaches the integral $$ I_G=\int_{\sigma} \frac{\Omega_G}{\Psi_G^2} $$ where $N_G$ is the number of edges of the graph, $\sigma$ is the simplex of points $(x_1: \ldots : x_{N_G}) \in \mathbb{P}^{N_G-1}(\mathbb{R})$ such that $x_i \geq 0$ for all $i$, $\Omega_G$ is the volume form $$ \Omega_G=\sum (-1)^i x_i dx_1 \wedge \cdots \widehat{dx_i} \cdots \wedge x_{N_G} $$ and $\Psi_G$ is the first Symanzik polynomial, defined by $$ \Psi_G=\sum_{T \subseteq G} \prod_{e \notin T} x_e, $$ where the sum runs over spanning trees.

The integral is in general divergent, but converges for a class of graphs called *primitive log-divergent*. These are those for which $N_G=2b_1(G)$ (first Betti number) and $N_\gamma>2b_1(\gamma)$ for any proper subgraph $\gamma$. I have found this statement in several places but no proof.

Does anybody know how to prove the convergence of $I_G$ in the primitive log-divergent case?

Is it an if and only if?

Any intuition about what the condition means?

2924051285991255087716377Amin Hosseini253157273114112322831372663133881213096546099774This is how I would approach the problem:

Let $\theta_{ij}$ be the probability that i beats j and let $w_{ij}$ be the number of times i wins over j in $n_{ij}$ trials. Then, the likelihood function is:

$P(w_{ij}|n_{ij}, \theta_{ij}) = B(n_{ij},w_{ij})$ $\theta_{ij}^{w_{ij}}$ $(1-\theta_{ij})^{n_{ij} -w_{ij}}$

Choose a beta prior for $\theta_{ij}$ and your posterior for $\theta_{ij}$ would be a beta distribution. Hope that helps.

@Sasha you are right. I assume that $E$ is the pullback of a sheaf on $X$.2339972072222099647RW: Yes, I know this; I was talking about what type of _questions_ one asks. I also should have added "ergodic theory" to "dynamics". Of course, the border between geometry and dynamics is fuzzy. Also, I suspect that people in noncommutative geometry have their own questions in this setting, but to me, most of noncommutative geometry has very little to do with geometry. 840481In answering this MO question, the issue was raised of characterizing when a given endofunctor $R:C\to C$ has the form $U\circ F$ where $F:C\to D$ is left adjoint to $U:D\to C$, i.e. which admit a monad structure. Is there an algebraic or purely categorical characterization of such $R$?

We know $F$ has to preserve colimits and $U$ has to preserve limits. I'd be happy with an answer saying $R$ is of the form $U\circ F$ iff $R$ preserves $\langle$ fill in the blank $\rangle$. This seems like it should be known classically (e.g. in Categories for the Working Mathematician), but I never learned it and feel like it would be useful to know.

From this MO question we know that in order to be a composition of adjoints $R$ must be a homotopy equivalence on the nerve $N(C)$. According to this MO question, the converse fails, so this is not a characterization.

19304976769041400925Thank-you for the answer, Francesco. I will have a look at your preprint. However, the case of a complete intersection in $\mathbb{P}^n$ won't actually help me directly.2105599Vel, I've found the title misleading. Finding a homotopy equivalent (or just homology equivalent) simplicial set is *not* triangulating.18030041613493Sorry for the delay. I have the asymptotics for the bounded step discrete case neat and clean now (including "explicit" formulae for the related parameters and checking against numerics), but the continuous case still gives me a headache, especially when the steps are unbounded. Unfortunately, you cannot just pass to the limit from the discrete bounded step version if you want the accuracy I declared and I'm reluctant to settle for less.142376416014725635375315092132342Criterion for existence of solution to nonlinear second order ODE20630283850891769987859941Knowing Sandor, I will heartily second what he said. On second mention, I will say that Mumford's red book on algebraic geometry begins with 5-10 pages (depending on your edition) called "some algebra". This consists of the following subset of Sandor's list: noether's normalization lemma, and cohen - seidenberg's going up lemma, plus corollaries such as the "weak" nullstellensatz. Hence this seems to be a minimum of algebra to know.

I remark that this is also in sync with the post by Dmitry that not that much is necessary.

@GuillaumeAubrun It would be interesting to understand what happens for the critical exponent $p=1$. Suddenly Hamming balls stop becoming optimizer, and some kind of transition happens.2261198In the examples I gave, by choosing a different decomposition of $X$ into tori the action of the open torus does extend to all of $X$. Maybe a relevant question for you is whether any irreducible torified variety $X$ is a toric variety (after possibly changing the decomposition of $X$ as a union of tori). Oh, I've used "nonzero" whenever I talk about slim Kurepa trees.Here's an elementary observation, in the definition of non-splitting you can restrict your attention to those H's generated by a two (not necessarily distinct) elements of c.

In particular, if no powers of elements of c are in c (for example, if c consists of involutions) and any two distinct elements of c generate all of G then it follows that (G,c) is nonsplitting.

As FC points out two involutions always generate a dihedral group. As JSE points out (G,c) for c a class of involutions is nonsplitting iff the resulting dihedral groups are all of 2*odd order. In particular, (G,c) is nonsplitting for c an involution iff the product of any two elements of c has odd order.

21068262041234@DerekHolt I would gladly accept an answer of yours explaining these things. I could not find a full version of your book available online, and I am not sure where exactly in the book can I find the result. Also, the 1911 paper of Hartley is out of my reach. Thanks a lot!509613xDatabase of integer edge lengths that can form tetrahedrons@GHfromMO if so, maybe explicit bounds combined with known values allow to find the answer to OP's question of max $\theta(x)/x$?16640311870990The right hand side of the equation describes the derived category of coherent sheaves on the stack of ^{L}G torsors with connection. In the traditional Langlands correspondence, one side is described by certain homomorphisms of a Galois group into the complex reductive group that is Langlands dual to G. In the function field setting, the (unramified) Galois group is the etale fundamental group of a proper algebraic curve, and conjugacy classes of homomorphisms to ^{L} G are in natural bijection with ^{L}G torsors with connection. These torsors are in turn in bijection with certain skyscraper sheaves on the corresponding moduli stack. Passing to coherent sheaves is a natural expansion of the category (but I imagine there are better motivations that escape me at the moment).

There is a more symmetric quantized version of this conjecture, due to Feigin, E. Frenkel, and Stoyanovsky. Coherent sheaves on the right hand side are replaced by twisted D-modules on the stack of ^{L}G bundles. There is a discussion of it in the introduction to Gatsgory's paper on the twisted Whittaker model (on the arXiv).

does anyone know if there is a model structure on stacks where the cofibrations are the monomorphisms ?

As far as I know usually to get a model structure on stacks one localizes a model structure on presheaves of groupoids or sheaves of groupoids or categories fibered in groupoids and this process doesn't change the cofibrations so my question may be reformulated as: is there a model structure on those categories above where the cofibrations are the monomorphisms ?
I would be grateful for a reference if any.

Best

1693322128775568245020114449954698Question about Conic Bundle208235not sure if helpful, but I would look into the literature on "small ball probability" eg. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.146.3738&rep=rep1&type=pdfI am interested if one can design an efficient (polynomial) algorithm telling whether the first player has a winning strategy for a game described below.

The board is a string consisting of only pluses and minuses, for example: `---+--+`

. There are two players. The players make moves alternately. In each move a player replaces two neighboring minuses with pluses, for example: `--+---`

-> `--+-++`

. If a player can't do any move then he wins.

Are there any papers on this problem?

1043117162217012658051191749178875933641438668373125https://www.gravatar.com/avatar/18a164a5092a90beb5c7d0cbf3c0ab7e?s=128&d=identicon&r=PG&f=11971772Looks like I am about to do some reading. Thanks a lot for the detailed answer. 594493Diederichsen's treatment of $\mathbb Z/4$ is actually incorrect, see http://www.ams.org/mathscinet-getitem?mr=1244181155846138304719504883179581808833For an index set $A$ consider the locally convex direct sum $X_A := \bigoplus_{\alpha \in A} \mathbb{R}_\alpha$ of $|A|$-many lines $\mathbb{R}_\alpha = \mathbb{R}$. Then $X_A$ is complete. It is known that if $A$ is countable then $X_A$ is $B$-complete (and even a Krein-Smulian space) and if $|A|^{\aleph_0} = |A|$ then $X_A$ is not $B$-complete (and not even $B_r$-complete).

The standard proof for the last fact is as follows. Let $X$ be an inf.-dim. vector space. If $A$ is a Hamelbasis of $X$ then $X = X_A$ as vector spaces. It is known that $X$ can be equipped with a complete norm if and only if $|A|^{\aleph_0} = |A|$ [Arthur H. Kruse, "Badly incomplete normed linear spaces", Mathematische Zeitschrift, 1964]. So in this case, take such a complete norm ($X$ is a Banach space) and denote the norm topology by $u$. Denote by $\tau^{lf}$ the finest locally convex topology on $X$ in which case $(X, \tau^{lf}) = X_A$ as locally convex spaces. Then the identity $id : (X, \tau^{lf}) \to (X, u)$ is continuous and almost open but not open since $\tau^{lf}$ is not normable and thus $\tau^{lf} \supsetneq u$. This implies that $X_A$ is not $B_r$-complete.

Question: Is it provable in $\textrm{ZFC}$ that $X_A$ is not $B$-complete for any uncountable set $A$?

54952121791271230422470161fseek another proof of a result in Fourier analysisLet $p$ be an odd prime number, $n>1$ be an integer, and $\mathrm{tr}$ be the trace map of the field extension $\mathrm{GF}(p^{2n})/\mathrm{GF}(p)$. For which pair $(p,n)$ does there exists $x\in\mathrm{GF}(p^{2n})$ such that $x^{2(1+p^n)}=1$ and $\mathrm{tr}(x^{1+p^j})=0$ for $j=0,1,\dots,n-1$?

After computing the gcd of the corresponding $n+1$ polynomials over $\mathrm{GF}(p)$ for small $p$ and $n$, I believe that the answer should be ``if and only if $p$ divides $n$" and moreover, when $p$ divides $n$, all such $x$ would form a subgroup of $\mathrm{GF}(p^{2n})^\times$ of index $2(1+p^{n/p})$.

**Update.** If $(p,n)$ is such a pair, then denoting by $X$ the $(2n)\times(2n)$ matrix with $(i,j)$th entry $x^{p^{i+j-2}}$ we deduce from the conditions that $$
X^2=
\begin{pmatrix}
0&2naI_n\\
2naI_n&0
\end{pmatrix},
$$
where $a=x^{1+p^n}=\pm1$. This yields two necessary conditions:

(1). If $n$ is not divisible by $p$ then $n$ is odd.

*Proof.* Suppose on the contrary that $n$ is even. Then
$$
|X|^2=
\begin{vmatrix}
0&2naI_n\\
2naI_n&0
\end{vmatrix}
=(-1)^n(2na)^{2n}=(2na)^{2n}
$$
is a square in $\mathrm{GF}(p)$, which implies that $|X|\in\mathrm{GF}(p)$. However, changing columns shows that $-|X|=(-1)^{2n-1}|X|$ is the determinant of the matrix with $(i,j)$th entry $x^{p^{i+j-1}}$. Hence $-|X|=|X|^p$ and so $|X|\notin\mathrm{GF}(p)$, a contradiction.

(2). If $n$ is not divisible by $p$ then $x,x^p,\dots,x^{p^{2n-1}}$ is a normal basis of $\mathrm{GF}(p^{2n})/\mathrm{GF}(p)$.

*Proof.* $|X|^2=(-1)^n(2na)^{2n}\neq0$ implies that $|X|\neq0$ and so $x,x^p,\dots,x^{p^{2n-1}}$ are $\mathrm{GF}(p)$-linearly independent.

This question is a repost from stack.exchange. It didn't get a lot of attention there. Perhaps it is badly written (or silly?). If so, I'd be happy to get comments/suggestions about that.

Let $X$ be a (nice) connected topological space. Let $LX=Map(S^1,X)$ be the free loop space and $\Omega X = Map_*(S^1,X)$ the subspace of based loops (with some choice of base point for X). Now, there is a fiber sequence $$ \Omega X \to LX\to X $$

where the map $LX\to X$ is evaluation at the base point. I am interested in the following question:

Under what general conditions this sequence splits and we have a (weak) homotopy equivalence $LX \simeq X\times \Omega X$ ?

This is of course not true in general. In the case of a classifying space $BG$ of a discrete group $G$, we have $LBG$ equivalent to a disjoint union of $BC(x)$ where $x$ runs over conjugacy classes of $G$ and $C(x)$ is its centralizer subgroup (as explained nicely in the answer to this MO question). Hence, the splitting holds if and only if $G$ is abelian.

I have managed to prove this in the case where $X$ itself is of the homotopy type of a based loop space. i.e. $X\simeq \Omega Y$ for some $Y$ and a little bit more generally, when $X\simeq Map_*(W,Y)$ for some (nice) connected pointed space $W$ and some pointed space $Y$. There are several ways to see this, but the slickest proof I got is just a sequence of standard adjunctions. Denote by $X_+$ the based space obtained from $X$ by adding a disjoint base point. On the one hand,

$$ LX = Map(S^1,Map_*(W,Y))=Map_*(S^1_+,Map_*(W,Y))=Map_*(S^1_+\wedge W,Y) $$ and on the other, $$ \Omega X\times X=Map_*(S^1 \vee S^0,Map_*(W,Y))=Map_*((S^1 \vee S^0)\wedge W,Y) $$

The only difference between $S^1_+$ and $S^1\vee S^0$ is in the selection of the base point, but since $W$ is *connected*, after smashing with it we get a connected space so the choice of the base point does not matter.

Generalizing the loop space case, it is natural (I think) to ask whether this is true for a general H-space. I also have a feeling that this has something to do with vanishing of whitehead products. So I'll add a more specific version of the question:

Wow... suddenly I feel proud I found this same solution on my own! (didn't even seem that hard) Anyway, my question was about *uniqueness* of this solution. Thank you for the context, though, and I like the fractal solution!1875475@matthew: I didn't see that part of the question. The answer is not always. Take $X = \mathbb G_m$, $Y=\mathbb G_m$, with the map the squaring map. Then the fiber product is the disjoint union of two copies of the base, so any base change will be irreducible.8042915096176079501544451596798153441216702931915607Well, the easy proof is as follows: assuming contrary, consider all $n$ remainders with remainder $r$ modulo $n$, sum up. We get that sum of all elements of $A_1$ and $A_2$ equals $r+(r+n)+\dots+r+(n^2-n)$. But this does depend on $r$. Contradiction! 350212I'm from the Eastern Townships of Quebec, live and teach in Quebec City and am an adjunct at Bishop's University in Sherbrooke. I got my PhD in Differential Geometry/Geometric Analysis at Stony Brook.Thanks! So basically we can't get information about $\chi(G)$ from neither of the two types of local data. 5084Manchester19035152288478@urpzilmöräqÜ You don’t really need to talk about formulas here. If a bijection fails to be an isomorphism, it is because it fails to preserve some relation or function symbols from the signature, by the definition of isomorphism. So, fix one such symbol for each bijection, and voila, you have a finite sublanguage $L'$ such that the reducts of $\mathcal A$ and $\mathcal B$ to $L'$ are still nonisomorphic.13896498709761471057VFormal n-buds from BU(n) rather than SU(n)177515227342718191378Sorry, I meant $C/G_1G_2$. 13229248388654565421258713user3343719128142129018Sorry for the confusion, but in 2), I wanted to say $\mathbb{R}^d$, a d-dimensional random variable, not $\mathbb{R}^n$.46326185274065587813210554890562603101801621511215867121127425221723641503914FIntuitionistic algebraic topology?I doubt it, but I haven't thought about this. If you desperately need to know about 2-sites, you should contact Mike Shulman.@Qfwfq I am following almost they way that you have said... "learn everything from the perspective of Lie groupoids"... I am following Orbifolds as stacks by Eugene Lerman and Differential stacks and gerbes https://arxiv.org/abs/math/0605694 for this.... " then ask myself when a Lie groupoid /stack is étale/Deligne-Mumford i.e. an orbifold," when a stack is geometric then it is called an orbifold.... I am yet to find a reference that says when a morphism of stakcs makes source as gerbe over target.... can you suggest something....50569710664451159070557916@Anton: The problem here is to represent a *covariant* functor Sch -> Set, right? 4The limits of parallelismIn other words, the question is an open problem that looks well known among the people that know this sort of problems. Notice there is a minisection in the FAQ http://mathoverflow.net/faq about this.2252981119057078000 I tried to reproduce the rank calculation of the 10-points example using the Sage software. If I'm not mistaken, it can be done this way. sage: R.Is it true that if all whitehead products of $X$ vanish, then the above sequence splits?

I don't know whether this satisfies all your conditions, but $$\pmatrix{1&1&1&1\cr1&-1&-1&-1\cr1&\alpha&\beta&\gamma\cr1&-\alpha&-\beta&-\gamma\cr}$$ is singular, and no two rows or columns are identical.

ychemama103223864720230326210717712694521286284Forget the ramification indices: they're nearly always 1. Let's focus on the residue field degrees. Let $E$ be the Galois closure of $L$ over $K$. For a prime $P$ in $K$, write $f_P(E/K)$ for the common residue field degree of all primes in $E$ over $P$. Since $E$ is determined (up to isom. as an extension of $K$) by $L$, one can imagine there might be a *formula* for $f_P(E/K)$ in terms of the $f(Q_i|P)$'s as $Q_i$ runs over the primes in $K$ that lie over $P$. There is such a formula if $P$ is unramified in $E$: $f_P(E/K) = \text{lcm } f(Q_i|P)$. (This is basically because a composite of finite fields over a particular finite field has degree equal to the lcm of the degrees of the extensions.)

Now imagine that for all $P$ the numbers $f(Q_i|P)$ are equal. Then their lcm is that common value, so when $P$ is unramified in $E$ we have $f_P(E/K) = f(Q_i|P)$ for all $Q_i$ over $P$ in $L$. Then $f_{Q_i}(E/L) = 1$ for all $Q_i$, so all but finitely many primes in $L$ are split completely in $E$. By Chebotarev's density theorem that implies $E = L$, so $L/K$ is Galois. (Strictly speaking we are using a result that is logically much weaker than Chebotarev: that the primes which split completely determine a Galois extension is due to Bauer from 1916 or so, before Chebotarev proved his general theorem.)

Since Chebotarev works with a density 0 set of primes taken out, instead of assuming for all $P$ that the numbers $f(Q_i|P)$ (when $Q_i|P$) are equal you only need to assume that for all but finitely many $P$ or even for all but a density 0 set of $P$.

215220712045842Yes, it is a smooth manifold. No, it is not orientable.

For the first, just think geometrically, i.e., without bases: Fix a $3$-dimensional vector space $V$ and consider the homogeneous quadratic polynomials on it, which is a $6$-dimensional vector space isomorphic to $\mathsf{S}^2(V^\ast)$. Let $\mathbb{P}V$ be the projectivization of $V$, i.e., the space of $1$-dimensional subspaces of $V$. Given $L\in\mathbb{P}V$, the set of elements of $\mathsf{S}^2(V^\ast)$ that represent conics singular at $L$ is simply the $3$-dimensional space $\mathsf{S}^2(L^\perp)\subset \mathsf{S}^2(V^\ast)$, where $L^\perp\subset V^\ast$ is the space of linear functions on $V$ that vanish on $L$. In particular, the set $E\subset \mathbb{P}V\times \mathsf{S}^2(V^\ast)$ that consists of the pairs $(L,f)$ such that $f\in \mathsf{S}^2(L^\perp)$ is a smooth subbundle of rank $3$ over $\mathbb{P}V$ of the trivial bundle $\mathbb{P}V\times \mathsf{S}^2(V^\ast)$. Your space $X$ is simply $\mathbb{P}E$, the projectivization of this smooth, rank $3$ bundle. In particular, it is a smooth submanifold of $\mathbb{P}V\times \mathbb{P}\bigl(\mathsf{S}^2(V^\ast)\bigr)$ of dimension $4 = 2 + (3{-}1)$, and the projection $\mathbb{P}E\to\mathbb{P}V$ is a submersion that is a locally trivial fiber bundle.

Second, $X=\mathbb{P}E$ cannot be orientable because no projectivization of a locally trivial, rank $3$ bundle $E$ over a smooth manifold $M$ is orientable. You don't need spectral sequences to see this, you just need to produce one closed loop around which the orientation bundle is nontrivial, but these are easy to find: Just take a loop in a single fiber $\mathbb{P}E_x$ that generates that fiber's fundamental group (which is isomorphic to $\mathbb{Z}_2$). This is clearly an orientation-reversing loop in $\mathbb{P}E$.

**Response to Question 3:** The OP wondered why the bundle $\pi:E\to \mathbb{P}V$ is locally trivial. Here is one way to see this: Let $L_0\in \mathbb{P}V$ be a $1$-dimensional subspace and let $X,Y,Z\in V^\ast$ be a basis of the linear functions on $V$ such that $X$ and $Y$ vanish identically on $L_0$ (and hence are a basis of $(L_0)^\perp$) while $z$ does not identically vanish on $L_0$ (and hence only vanishes on $L_0$ at $0\in L_0$). Let $U\subset\mathbb{P}V$ be the open set consisting of those $L\in \mathbb{P}V$ such that $Z$ is not the zero linear functional when restricted to $L$. On $U$, there are two well-defined, smooth functions $x,y:U\to\mathbb{R}$ such that $x(L) = X(v)/Z(v)$ and $y(L) = Y(v)/Z(v)$ for some (and, hence, any) nonzero $v\in L$. Then, for each $L\in U$, the linear functions $X-x(L)\,Z$ and $Y - y(L)\,Z$ are a basis of $(L)^\perp\subset V^\ast$, and hence the quadratic forms
$$
A(L) = \bigl(X-x(L)\, Z\bigr)^2,\quad
B(L) = \bigl(X-x(L)\, Z\bigr)\bigl(Y-y(L)\, Z\bigr),\quad
C(L) = \bigl(Y-y(L)\, Z\bigr)^2
$$
are a basis of $E_L = \mathsf{S}^2(L^\perp)$ for each $L\in U$. Thus, they define a smooth trivialization of $\pi:E\to\mathbb{P}V$ over $U$. Obviously, $\mathbb{P}V$ can be covered by such open sets $U$, and it is easy to see that the transitions on overlaps are smooth. Thus $E$ is a smooth bundle with local smooth trivializations.

Consider a model category $\mathcal{C}$ and a sequence of cofibrations $0 \to X_0 \to X_1 \to X_2 \to \dots$ lying in $\mathcal{C}$. Let $X$ be the colimit of this sequence. Suppose furthermore that we have two maps $f,g : X \to Y$, where $Y$ is fibrant, such that the restrictions $f_n : X_n \to Y $ and $g_n : X_n \to Y$ are homotopic for every $n$.

Does it follow that $f$ is homotopic to $g$?

A simpler version: if $h : X \to Y$ is a morphism such that the restrictions $h_n$ are equivalences, then $h$ is an equivalence, since $X$ is the homotopy colimit of the $X_i$.

Context: I saw this result when studying rational homotopy theory, in the context of Sullivan algebras in the category of CDGAs. The result was stated in terms of a filtration of *minimal* Sullivan algebras, however. Perhaps some additional condition is required above, then.

Let $E$ be a vector bundle on projective space ${\bf P}^n$ whose Hilbert polynomial is the same as $\mathcal{O}^{{\rm rank}(E)}$.

Does there exist a vector bundle over ${\bf P}^n \times {\rm Spec}(R)$ where $R$ is a DVR so that the general fiber is $\mathcal{O}^{{\rm rank}(E)}$ and the special fiber is $E$?

The intuition is that the cohomology of twists of $E$ must be at least as large as the cohomology of the corresponding twists of the trivial bundle, and I'd like to see that realized as a flat degeneration.

The real goal is to use this to prove that $h^i(E \otimes F) \ge {\rm rank}(E) \cdot h^i(F)$ for all $i$ and all vector bundles $F$.

19646311746279218141761092I think $\theta$ actually converges to 1 quickly as $x$ grows large. For just an asymptotic formula for $\Gamma(x)$ (as in the common version of Stirling's formula), there would be no reason to mention $\mu(x)$ or $\theta$ at all, since the $e^{-\mu(x)}$ term converges to 1. It sounds like the author is stating on page 24 an exact formula involving a constant $\theta$, when it's actually not a constant. I suspect it's a mistake of the translator originating from a misunderstanding on page 22.20685234Translation seems correctCan you give a specific page reference (and especially explain a little more about the parenthetic comment)? Presentations using elementary matrices are the most natural, translated into the language of BN-pairs and Bruhat decomposition in general, but relations over rings and fields then lead to complications. I'm not sure what is the most elegant answer for finite general linear groups over arbitrary fields.Let $E$ be a vector space over the real (the complex case is interesting too). We consider functions $f:E\rightarrow\mathbb R$ which satisfy the homogeneity property $$f(\lambda x)=|\lambda|\,f(x).$$ In particular, we have $f(0)=0$.

I am interested in the generalized Hlawka inequality (GH$n$) at order $n$. Given $n$ vectors $x_1,\ldots,x_n$, and $I\subset[\![1,n]\!]$, define $x_I=\sum_{i\in I}x_i$. Then (GH$n$) says that $$\sum_{I\subset[\![1,n]\!]}(-1)^{{\rm card}\, I}f(x_I)\le0,\qquad\forall x_1,\ldots,x_n\in E.$$ For instance, (GH1) and (GH2) are respectively $0\le f(x)$ and $f(x+y)\le f(x)+f(y)$; thus, a function $f$ satisfying (GH1) and (GH2) is a norm over $E$. (GH3) is Hlawka inequality, which is satisfied by Euclidian norms and some others, though not by all norms.

Let me point out that (GH$n$) is sharp, in the sense that the equality holds true when either one vector is $0$, or all the vectors are positively collinear. If $n$ is odd, it is also an equality when $\sum_1^nx_i=0$, while if $n$ is even, this choice in (GH$n$) yields (GH$(n-1)$). One may also derive (GH2) from (GH3) by choosing $x_3=x_1$.

First question :

Do Euclidian norms satisfy (GH$n$) for all $n$ ?

Second question :

Conversely, if a norm satisfies (GH$n$) for all $n$, is it Euclidian ?

Third question :

Can we derive (GH$(n-1)$) from (GH$n$) when $n$ is odd ? (True if $n=3$).

**Edit.** Guillaume, I liked your suggestion. It does give a (new ?) proof of the classical Hlawka inequality ($n=3$). However it fails as soon as $n=4$. A counter-example is $(1,1,1,-2)$, where the sum is $2$. Therefore the answer to the very first question above is **No**, even in one space-dimension.

Given a minimal parabloic subgroup we know that conjugation by the longest element in the weyl group takes it to the opposite parabolic.

Can we do the same thing if we choose a standard parabolic subgroup? Can we always find an element in the weyl group such that conjugation by this element takes it to the opposite parabolic? I am guessing this should be possible by modifying the longest element in the weyl group. Is this true? Any help is appreciated.

15215251982943458094522933435631497007Jon SchallerTHe $E_\pi$ is a space of endomorphisms. The integral $\int$ is a Hilbert space integral -- the space of $L^2$-sections of a Hilbert space bundle over a measure space. Try looking around for "direct integral of Hilbert spaces" or "spectral measure" to understand better.15497274@j.c. Yes, you are right.2157184It depends on what you mean by "application": I could argue that theory of semisimple Lie group and symmetric spaces is application of affine Coxeter groups since these are equivalent to root systems. FYep, thanks for pointing this out!Nathan HurstConsider the identity

$$(x+1)^5+(y+1)^5 = (2x + 6y) ( -x +y )^4 + f(x,y)$$ where $f(x,y)=-x^5 + 2*x^4*y + 12*x^3*y^2 - 28*x^2*y^3 + 22*x*y^4 - 5*y^5 + 5*x^4 + 5*y^4 + 10*x^3 + 10*y^3 + 10*x^2 + 10*y^2 + 5*x + 5*y + 2$.

The curve $C : f(x,y)=0$ is genus 0 and has infinitely many rational points. I don't know if it has infinitely many integral points: suppose not. For integral points on $C$ the identity is:

$$(x+1)^5+(y+1)^5 = (2x + 6y) ( -x +y )^4. \qquad (1)$$

Infinitely many solutions on $C$ with $\gcd(x+1,y+1)=1$ contradict $abc$, so $abc$ implies either finitely many integral solutions, or sufficiently large $\gcd(x+1,y+1)$ (clearing a small gcd will still give abc triples of sufficiently good quality).

The integral points might be growing exponentially, so abc for polynomials doesn't appear to apply.

Q1 Is there a finite extension of $\mathbb{Z}$ where $C$ has infinitely many integral points? How do I find integral points there?

Q2 Is there a similar identity where $f$ is divisible by a quadratic with infinitely many integral points?

I tried to solve Q2 by equating coefficients, but couldn't solve the system.

Other similar identities exist.

If $f$ is divisible by $g= x^2+xy -y^2+1$, integral points on $g$ are consecutive fibonacci numbers $(F_{2n},F_{2n+1})$. If the lhs is still a sum of powers of linear polynomials, it appears unlikely to me that the common factor will always be large, so probably such an identity doesn't exist.

14545711853901101373If $\text{dim}(X \times X) = 2\text{dim}(X)$, does $\text{dim}(X^n) = n\text{dim}(X)$?11770492214660477420155143@MikeShulman So, it is an open problem which projective lines are `integrable` or `realizable` (i.e. come from projective planes). And as I suggested earlier, I believe this is a very difficult problem, given that any really "useful" solution probably would also settle the question of whether there are projective planes of non-prime-power order or not. With "useful solution" I mean that one would want an intrinsic characterization of integrable projective lines which is powerful enough to decide for a concrete projective line whether it is integrable or not.@Turbo No, it is over the rationals. In general the number of solutions is finite (possibly only unless the third equation is identically zero). From the finite solutions find the natural ones.8793161317030Requiring the fundamental unit not to have lattice points other than its vertices excludes two-dimensional counter-examples (since the only lattice polygons with that property are a unit parallelogram and half of it) and it also excludes counter-examples constructed using a Minkowski sum of segments as a fundamental unit. The latter was convenient for counter-examples since Minkowski sums have few different edge vectors. But I would expect the statement to still fail in this restrict version.1791900https://www.gravatar.com/avatar/23b1514e70b5cc6a07b91d48c49cbf62?s=128&d=identicon&r=PG&f=1105916470830115861672274893Hmmm. Sounds like you're hoping for some kind of Fourier information. I don't have a good feeling for these things, but my guess is that you would need to store *vast* amounts of information and do major calculations to get anything out of this (similar to what was suggested in the previous comments).FWIW the algorithm I suggested (with the numbers that you offered in the question) would be expected to terminate in 256 lookups, each taking 24 steps. 118600248621 This really is necessary for doing certain things. I'm not very knowledgeable about Sage, but MAGMA definitely has some gaps. For example, MAGMA knows about affine Weyl groups, but many basic functions regarding affine Weyl groups are unavailable. Coxeter3 can do such things. I also have several functions that fill other gaps in MAGMA, like the near-total inability of MAGMA to handle extended affine Weyl groups--MAGMA knows about weights and it knows about Coxeter groups, but there is no code linking the two (example: find a reduced expression for the affine Weyl component of a translation).1561988226237512259332946410198103615868045207992275360

This is part of the question I asked on 5191. On my observation primes categorized themselves into 3 types, a)primes that doesn't divide the form of 2^x + c, b)primes that divides but a smaller had divided it already and c)primes that divides a new x usually larger primes. Once a prime divides a form in this case 2^x + 3 it will divide it infinitely periodically, so I believe even the form of p^k|2^x + c will be true. All primes that has a period of p-1 (totient = p-1) will always divide any forms. On the case of wiefirich prime is a special form, all integer divides a mersenne, so no problem with any p^k dividing a mersenne, but tying the order of mersenne 2^k - 1 to prime k, it could just be one of the small numbers properties that the 2 wiefirich primes what they are.

155047316598191714142josh_whitneyThanks for clarifying, but I think you mean "Cartesian square". I don't think I've ever seen just "square" used to mean that.1399108589051 gLre413984805874123951In a comment on Tom Goodwillie's question about relating the Alexander polynomial and the Iwasawa polynomial, Minhyong Kim makes the cryptic but tantalizing statement:

In brief, the current view is that the Iwasawa polynomial=p-adic L-function should be viewed as a path in K-theory space.

As somebody who has been trying, for a long time, to understand the Alexander polynomial, and who believes there is something "more basic" underlying its appearances in mathematics in various guises (my particular interest has to do with why it turns up as the wheels part of the Kontsevich invariant, as per Melvin-Morton-Rozansky), I'd really like to understand this comment.

1477202I'm writing a paper in which I show that a variation of the number above is an l^2-Betti number of something (cf. http://arxiv.org/abs/1004.2030)@abx: Sorry, forgot to mention $X$ is connected. If $X$ has two connected components with each mapping surjectively to $Y$ then the fiber over $y$ will not be irreducible, as mentioned in the question.1879457@individ It may help you to read http://zakuski.math.utsa.edu/~kap/Timofeev_1963_Uspekhi.pdf The best explanation of this new material is this entire document http://zakuski.math.utsa.edu/~kap/Jagy_Encyclopedia.pdf especially pages 36-42.21915718Volume in tropical geometryQuestion: What is "K-theory space" supposed to be, and why should a path in it be a natural object to be examining? And why should we expect such a thing to give equivalent data to the Iwasawa polynomial/ Alexander polynomial?

Actually, I suppose that the answer is technically "yes," since computing the permanent is #P-complete, but that's not very satisfying. So here's what I mean:

Kirchhoff's theorem says that if you take the Laplacian matrix of a graph, and chop off a row and a column, then the determinant of the resulting matrix is equal to the number of spanning trees of the graph. It would be nice to have some analogue of this for other points of the Tutte polynomial, but this is in general too much to ask: the determinant can be computed in polynomial time, but problems such as counting 3-colorings are #P-hard.

However, if we use the permanent instead of the determinant, we don't run into these complexity-theoretic issues, at least. So, given a graph G on n vertices, can we construct a matrix of size around nxn whose permanent is the number of 3-colorings of G?

(The secret underlying motivation here is a vague personal hope that we can extend the analogy between the Laplacian matrix and the Laplacian operator [no, the naming isn't a total coincidence] to analogies between other matrices and general elliptic operators, and then prove some sort of "index theorem," which could [even more speculatively, here] help us understand why graph isomorphism is hard, prove or construct a counterexample to the reconstruction conjecture, prove the Riemann hypothesis, and achieve world peace forever.)

But a nice application of Liouville's Theorem is the description of the graded algebra of meromorphic functions in ${\mathbb C}$ with two given periods $1$ and $\omega$.6145281412199129456842756281977976@JasonStarr What representation of $PGL_2(\mathbb R)$ will you take? Surely not the adjoint representation...zWhat if Current Foundations of Mathematics are Inconsistent?Actually, some permutation have all zero entries in the diagonal. In fact, J-I has such a regular sequence where all but the 1 by 1 matrix has a zero diagonal. For any regular matrix, determinant computation by Laplace expansion will give you a sequence. In fact, it will give you at least n! many of them, if you count by original index sets.The module of all sections of $V$ that vanish to infinite order at a given point of the manifold will not be finitely generated (unless the bundle has rank zero or the manifold has dimension zero).

1631935452175170377You are looking at the solution of the **Abel functional equation** for your function
$$ f(z) = \eta^z = e^{\frac{z}{e}}.$$
(Note that this function is conformally conjugate to the map $\zeta \mapsto e^\zeta-1$ via $z = e\cdot (\zeta +1)$, which may be more convenient as its fixed point is at $0$ instead of $e$.)

This function has a fixed point at $z=e$ with derivative $1$. The solution of the Abel functional equation for such functions is well-understood, and referred to as the **Fatou coordinate**. In particular, you can extend the map by analytic extension along horizontal lines to all of
$$\newcommand{\C}{\mathbb{C}}\C\setminus (-\infty,-2]$$
(where I am using your normalisation, where the function maps $-1$ to the omitted value at $0$, although I am not sure why you would use this normalisation).

All you need to know to justify this is that the map $f$ is a covering map $f\colon \C\to \C\setminus \{0\}$ (so that the lift / analytic continuation is defined), and that your domain is simply-connected so that you can use the monodromy theorem.

As you note, you cannot extend the function continuously to any point of $(-\infty,-2]$. However, the \emph{inverse} $\phi$ of your function $F$ can easily be extended, via the functional relation, to the entire **basin of attraction**, i.e. all points converging to the fixed point $e$ under iteration of $f$.

This gives an alternative view on your result: The interval $(-\infty,-2]$ contains all the **singular values** of this function $\phi$ (i.e., values that don't have a neighbourhood over which $\phi$ is a covering map). So what you are doing is taking a branch cut through all the singular values of $\phi$, and taking one of the (infinitely many) branches of $\phi^{-1}$ defined on this set.

This addresses the second question "What is known about finite subgroups of $SU(n)$". A special case of the Margulis lemma implies that for each $n$, there is an $m(n)$ such that any finite subgroup of $O(n)$ has an abelian subgroup of index $m(n)$ (see Corollary 4.2.4 of Thurston's book). Thus, there is a normal abelian subgroup of index at most $m!$. So one may make a statement: there are finitely many finite groups so that any finite subgroup of $SU(n)$ is an abelian extension (of rank at most $n-1$) of one of these finitely many groups. It would be quite interesting to obtain an estimate of the function $m(n)$, which should be possible by giving an effective proof of Margulis' theorem. I did a literature search once to see if anyone had attempted this, but I didn't find anything, and I would be curious if someone knows something.

Addendum: Working backwards from Weisfeiler's paper referenced in Keivan's comment, I found a result of Collins implies that a finite linear subgroup of $GL(n,C)$ has an abelian normal subgroup of index at most $(n+1)!$ when $n\geq 71$ (and gives the bound for all $n$). Since finite subgroups of $GL(n,C)$ are conjugate into $U(n)$, this bound works for $SU(n)$. See also Collins paper on primitive representations, which has some historical discussion of this problem.

190284416641671165100810688891874452When I teach our "Introduction to Mathematical Reasoning" course for undergraduates, I start out by describing a collection of mathematical "facts" that everybody "knew" to be true, but which, with increasing standards of rigor, were eventually proved false. Here they are:

**Non-Euclidean geometry:**The geometry described by Euclid is the only possible "true" geometry of the real world.**Zeno's paradox:**It is impossible to add together infinitely many positive numbers and get a finite answer.**Cardinality vs. dimension:**There are more points in the unit square than there are in the unit interval.**Space-filling curves:**A continuous parametrized curve in the unit square must miss "most" points.**Nowhere-differentiable functions:**A continuous real-valued function on the unit interval must be differentiable at "most" points.**The Cantor Function:**A function that is continuous and satisfies f'(x)=0 almost everywhere must be constant.**The Banach-Tarski paradox:**If a bounded solid in R^3 is decomposed into finitely many disjoint pieces, and those pieces are rearranged by rigid motions to form a new solid, then the new solid will have the same volume as the original one.

Suppose, we have an ODE $$ \frac{dy}{dt}= f(t,y;p',a)$$ or alternatively $$ \frac{dy}{dt}= f(t,y;p)$$ where the set of all parameters $p = (p',a)$. We only need to estimate part of parameter set $p'$ from a collection of $\sum_{i=1}^{M} {a}_{i}$ data points: $$({t}_{1},{y}_{1,1}),({t}_{2},{y}_{1,2}),...,({t}_{{N}_{1}},{y}_{1,{N}_{1}}) $$ $$({t}_{1},{y}_{2,1}),({t}_{2},{y}_{2,2}),...,({t}_{{N}_{2}},{y}_{2,{N}_{2}})$$ $$...$$ $$({t}_{1},{y}_{M,1}),({t}_{2},{y}_{M,2}),...,({t}_{{N}_{M}},{y}_{2,{N}_{M}})$$

where $\left\{ ({t}_{j},{y}_{i,j})\right\},j=1,...,N_j$ denotes data points for $i$-th value of $a$. One could now determine optimal values for the parameter $p'$ by minimizing error $$E(p')=\sum_{i=1}^{M}\sum_{j=1}^{{N}_{i}}\left| y({t}_{j};p',{a}_{i}) -{y}_{i,j}\right|^2$$

The problem is if M is very large say M = 20, it is very cumbersome to write 20 slightly different function for each a. I am just wondering what tips are needed to call the solver fminsearch in Matlab in case of partly estimated parameters problems.

1698966I'm a graduate student in Tehran university.

My specific research topics are machine learning and optimization.

'I can give you a negative answer and a kind of a positive answer to your question. First, similar to what Henry Cohn says, the conventional view of your construction is that it is a restatement of the factoring problem rather than a step to an algorithm. A function with a lot of oscillation is at first glance a function with high information content. So the fact that your setup is numerical rather than discrete doesn't necessarily help anything. In fact, standard numerical integration methods won't look all that different from trial division. Of course it is impossible to rule out some excellent special-purpose numerical integration method, especially because if you apply your transformation backwards, any factoring algorithm is such a special method. And in my opinion subexponential factoring algorithms are a little bit miraculous. You suggest setting up factoring as a 1-dimensional integration problem. But I know of very few numerical analysis algorithms that give a more-than-polynomial gain in speed over obvious algorithms in low dimensions. The main example that comes to mind is spectral methods for solving differential equations or integrals to high accuracy --- but that already starts to look like your construction in reverse.

On the positive side, Shor's quantum algorithm for factoring does look a little bit like your oscillation picture. The algorithm does not look for an oscillation frequency that matches a factor of the number $n$. Instead, it computes the exponents of elements in $(\mathbb{Z}/n)^\times$, which is enough to factor $n$ when $n$ is odd. Using $a^x$ as a periodic function of $x$, it makes a vector $\psi$ in $L^2(\mathbb{Z})$ which is approximately periodic on a large scale with period the exponent of $a$. It then takes an approximate Fourier transform of this vector, and then measures a Fourier mode $k$ weighted according to the square amplitudes of the transform of $\psi$. So this is extracting information from oscillatory integrals! However, crucially, the integral is "computed" with quantum probability, in only the weak sense that a simulated random walk "computes" return probabilities. Quantum amplitudes are NOT stored numbers, they are probability-like state. So their approximate integration in quantum computation is very restricted. However, you have the exotic advantage of handling exponentially many amplitudes, just as randomized algorithms have probabilistic access to exponentially many possibilities.

8350022907971831732I have a question about the effect of applying a linear transformation $M$ in $\mathbb{R}^{n \times n}$ to a vector $v \in \mathbb{R^n}$.

I know that if $M$ has p-norm $\|M\|_p = \lambda$, then by definition I can guarantee that for every vector $v \in \mathbb{R^n}$ $$\|Mv\|_p \leq \lambda \|v\|_p.$$

Is there a corresponding lower bound? In general the answer is no, (for example, choose $v \in Nullspace(M)$ if the matrix is not full rank. )But I am interested in a corresponding lower bound where

$v$ is chosen ``generically'', so that the probability of it being in a (fixed in advance) subspace is 0.

$\|v\|_p$ is sufficiently large.

In this case, is it true that $$\|Mv\|_p \geq C \|v\|_p$$ where $C$ is some positive constant greater than zero?

1342261Let $R$ be a local singularity (for example $R=\mathbb{C}[[x_1, \ldots , x_n]]/I$) ring over $\mathbb{C}$. Let $\mathbb{L}_{R}$ be a cotangent complex of $R$, then one can define tangent (Andre-Quillen) cohomology $$ T^i_R(M)=Ext^i(\mathbb{L}_R, M), T^i_R=T^i_R(R). $$ I want to use tangent cohomology to study infinitesimal deformations of $R$.

If $A$ is an artinian ring over $\mathbb{C}$ then there are obstruction for lifting of deformations over $A$ to deformations over $\tilde{A}$, where $\tilde{A} \in Ex_{\mathbb{C}}(A, \mathbb{C})\cong T^1_A(\mathbb{C})$. In such settings, how obstruction map $$ o : Ex_{\mathbb{C}}(A, \mathbb{C}) \to T^2_R, $$ could be defined? I know construction for such map in case of explicit description of $T^2_R$, but it should be possible to define it just using properties of tangent cohomology.

57788010011291447623I think I understand now why it is not path-connected: correct me if I'm wrong, but it's because there can't for example be a path from $(0,0)$ to $(1/2,0)$ because it will have a lift to a path in $\mathbb{R}^2$ from $(0,0)$ to $(1/2,0)$ which lies in $(x,\pi x) + (x+1/2,\pi x) + \mathbb{Z}^2$, which is a contradiction, right? I'm still not sure about the connected part, though.1569963167112164092612470535301112418936254309415844Hello! My name is... well... I think I will stick with the number 23823. After all it is not that hard to remember... or is it?

I am on the following sites:

**Mathematics**: I am most interested in branches of mathematics that have direct application in physics, especially topology, differential geometry and Lie groups.

**Physics**: This is my major, but I am still trying to wrap my head around quantum field theory...

**Astronomy**: Celestial mechanics gets me excited.

**Mathematica**: Fascinated by functional programming, I use *Wolfram Mathematica* to do much of my work.

**LaTeX**: Do not actually like it but have to use it anyway...

**Japanese Language**: I probably know more Kanji than the average Japanese. (This is a joke.)

**Latin Language**: Not as good as my Japanese but still learning.

**Stack Exchange**: Python is my main programming language. I also have an interest in web app development and cryptography.

**Anime**: Do not watch anime often, but a big fan of *Yuru Yuri*.

I would like to share my experience and knowledge with everyone here.

>Crepant Morphisms of Varieties323694One comment on the "excision" property for Euler characteristics of complex algebraic varieties: $\chi(X) = \chi(Z) + \chi(X\backslash Z)$. There is a short proof of this in Fulton's *Introduction to Toric Varieties*, p. 142, which avoids the sticky triangulation issue and doesn't seem to be so well-known. In fact, it's equivalent to the statement that $\chi(X) = \chi_c(X)$ for any algebraic variety (the latter being the Euler characteristic for compactly supported cohomology).

This doesn't exactly help with the first question (about an algebraic definition), but it does show why the Euler characteristic is well-defined for any variety, per Arend's comment.

1547164221981057002718505162940441197722210624Ursa Major551512What is “as quickly as possible”? I don’t see how this is a well-defined problem.89481912727711045766117092915937726315161025515Oliver Knitter289957747803167496718022872286687zadded an explicit expression for the probability in question1558420FUnits can be seen as monomials... 709118https://www.gravatar.com/avatar/45651555e8d48a9ae47c692ee0f3d3b6?s=128&d=identicon&r=PG&f=111749291243171I think that does something slightly different, but it does helpfully rephrase the question: Taking the levelwise nerve puts us in bisimplicial sets, and takes levelwise equivalences to levelwise homotopy equivalences, i.e. to weak equivalences (w.e.) in the Bousfield-Kan or Reedy structures. Taking the diagonal of a bisimplicial set preserves w.e. in the Moerdijk structure (by definition), but not necessarily in the BK or Reedy structures. So, I'm looking for a functor from bisimplicial sets to simplicial sets which preserves BK or Reedy w.e. 872228247411840427173714914125632025645@JimHumphreys: My statement can also be expressed as the fact that a choice of Coxeter element is the same thing as a choice of acyclic orientation of the Coxeter diagram. This shows up quite a bit in quiver representations. A non-quivery source is https://arxiv.org/pdf/0710.3188.pdf -- see the first paragraph of section 1.67616931817827909918jGrothendieck duality for resolution of singularities14467561756410470703354551133394516416245160443<@Georges: nice catch. Thanks!Try checking out Palm measure. I think this is exactly what you want.10922911345577186554922640322019678Let $\mathcal{F}_t$ be a continuous filtration on a probability space, and let $B$ be a standard $\mathbb{C}$-valued $\mathcal{F}_t$-Brownian motion. Let's call a complex-valued process $X$, possibly defined only up to a stopping time, *analytic*, if it is a continuous $\mathcal{F}_t$-local martingale with stochastic differential $dX_t = X^{(1)}_t dB_t$, such that $X^{(1)}$ is again a continuous local martingale, $dX^{(1)}_t = X^{(2)}_t dB_t$, and so on. Equivalently, for every $n$ $X_t$ is a polynomial in $B_t$ of degree $n$ up to an $(n+1)$-fold stochastic integral.

The question is: can we describe such processes explicitly?

More precisely, the obvious examples are $f(B_t)$ for complex analytic $f$, and the analytic continuations along $B_t$ of complex analytic functions, defined up to the time when their radius of convergence shrinks to $0$. Note that these may be called *strongly analytic*, in the sense that they are analytic with respect to the filtration induced by $B_t$. Are these the only ones? Do there exist analytic processes which are not strongly analytic?

An obvious remark: if all $X^{(n)}$ are, say, $L^p,p>1$ integrable martingales then $X_t$ is an entire function of $B_t$ (since we can reconstruct the chaos decomposition, which happens to be a power series in $B_t$), so I'm interested in cases when they're not.

@BobbyGrizzard You're welcome. I would have accepted Joe's answer too!440128118840516602971976126@KConrad Oh sorry for my misunderstanding( which surprised me), thanks for your reference. Can I cite your comments in my course paper with this hyperlink? Thanks for your patience!22929462778241625840130704By solving it, I mean expressing $F(x)$ in a closed form; without the $\sum$ symbol (I then want to derive and further change it). Thank you for very insightful wiki-link! It's good to know when to stop pushing calculations in some direction.303936The Copeland–Erdős constant is formed by concatenating all the primes base 10, and is known to be normal.

The short answer to your specific question is no, the resolution of FLT via modularity of elliptic curves does not seem to be helpful in dealing with rational points on higher dimensional varieties. The first two equations you list, $a^5+b^5=c^5+d^5$ and $a^6+b^6=c^6+d^6$, are surfaces of general type in $\mathbb{P}^3$, so conjectures of Bombieri, Lang, Vojta, would say that their solutions, and solutions of similar equations in which the terms might have non-unit coefficients, lie on finitely many curves. The second two equations are 4-folds in $\mathbb{P}^5$. For $x^5+y^5+z^5=u^5+v^5+w^5$, the canonical bundle is ample, and I'm going to guess that the 4-fold is rational(?), so there will be lots of solutions. For $x^6+y^6+z^6=u^6+v^6+w^6$, the canonical bundle is trivial, so you've got a Calabi-Yau. At least conjecturally, there should be a number field $K$ such that the $K$-rational points are Zariski dense. Maybe for this 4-fold, one can take $K=\mathbb{Q}$?

17242031969193Aha - thanks! So you mean that if $f=\sum_{n=1}^\infty c(n)q^n$, I don't take the form \[ g = \sum_{n=1}^\infty \chi(n)c(n)q^n \] but I write $g$ as $g=f_\chi + g'$, where $f_\chi$ is a newform? This would make perfect sense, of course. (But it is a bit confusing that people write in several places things like $f$ has CM if and only if it is equal to a twist of itself.)There might be a more abstract/direct way to prove it, but I would try to prove the statement by using:

- Almost everywhere differentiability of Lipschitz mappings (giving that almost everywhere the boundary is like a hyperplane on small scales).
- The fact that we have a Lipschitz domain so that for the parts where the boundary is not sufficiently close to a hyperplane we have bounded estimates between the measure of the boundary and the corresponding measure of its $\epsilon$-neighbourhood.

This is a common first step. But the problem is that you can't assure that such method will converge. Because of this, you have to find a way to justify that your method gives "good" approximations.

643137If the map $e_X$ is required to be surjective, then any compact Hausdorff space with a unique non-isolated point is non-homogenizable.3078101187656138127814278031779791Let $X$ be a random variable with infinitely divisible and symmetric distribution $F$ distributed on $\mathbb{R}$.

It is well known that the characteristic function of $X$ has a canonical representation of the form: \begin{align} \phi(t)=e^{ -\frac{\sigma^2}{2}t^2-\int_{-\infty}^\infty (1-cos(tx) ) dV(x)} \end{align} where $V$ is a non-negative measure such that $V(\{0\})=0$ and $\int_{-\infty}^\infty \min(1,x^2) dV(x)<\infty$.

The measure $V$ is called Levy measure and here we are interested in its properties.

My questions are:

- Under what condition on $F$ is $V$ an absolutely continuous measure? For example, is absolute continuity of $F$ enough?
- Can we say anything about $V$ based on the tail behavior of $F$?
- Can we say anything about $V$ based on the tail behavior of $\phi(t)$?
I know it is generally difficult to determine $V$ but what can generally be said about $V$ from basic properties of $F$ and $\phi(t)$?

Also, any good reference would be appreciated.

Paul Howard, Kyriakos Keremedis, Jean E. Rubin, Adrienne Stanley,

*Paracompactness of metric spaces and the axiom of multiple choice*[MLQ 46 (2000), 219–232; MR1755811]Omar De la Cruz, Eric Hall, Paul Howard, Kyriakos Keremedis, Jean E. Rubin,

*Metric spaces and the axiom of choice*[MLQ 49 (2003), 455–466; MR1998077]

They show that the Nagata–Smirnov and Bing metrization theorems are equivalent to Stone's Theorem (paracompactness of metric spaces) though the backward implications are provable in ZF.

1712456227083Oh, I misattributed that. It's Alexander and Kolmogoroff that defined the cup product, in 1935. Alexander's definition was problematic and fixed by Cech and Whitney in 1936. Whitney is the one that gave cup product its name and notation. DDear Haim, thanks for references.17995241367461165027583525917240610569291130138It's easy to compute presentations, but the problem is that they won't usually be nice. For example, $J_1$ has a presentation $\langle a,b \mid a^2,b^3,(ab)^7, [a,b]^{10}, uvuuvuuvvuvuuvvuvvuvuvuuvvuvvuvuv \rangle$, where $u=ab$ and $v=ab^{-1}$.340701583272Let $n$ be a positive integer.

Let $S \subseteq \mathbb{R}^n$. Is the Hausdorff dimension of the boundary of $S$ always smaller than the Hausdorff dimension of $S$?

I have not found anything concerning those questions in some looked up books, I was not able to prove one of the statements, and I failed finding a counterexample. Does anybody know something about that?

174404zMaximal subgroups of infinite index and profinite completion7414865986131360269107146015732368508801585533Made my way from the Olympus of Complexity Theory, Probabilistic Combinatorics and Property Testing to the down-to-earth domain of Heterogeneous and GPU Comuting, and now I'm hoping to bring the gospel of GPU and massive-regularized parallelism to DBMS architectures. I'm post-doc'ing at the DB architecture group in CWI Amsterdam to do (some of) that.

I subscribe to most of Michael Richter's critique of StackOverflow; you might want to take the time to read it.

If you listen closely you can hear me muttering "Why am I not socratic again already?"

661815lcomparison of de Rham cohomology and etale cohomology766149This is in general as hard as computing the volume of an Euclidean polytope, but there are reductions for even dimensional polytopes to volumes of lower-dimensional things (of which there may, of course, be an exponential number). See http://www.math.ru.nl/~heckman/Heck_7.pdf (he mostly talks about the hyperbolic case, but the spherical case is identical).

Regarding a), I certainly agree that one of the major goals is to learn to do* mathematics, but regarding b) & c), I am led to remember Prof Tom Korner's essay "In Praise of Lectures". Roughly speaking, he argues that the real point of a lecture is for students to actually *see mathematics be done* . So a good lecturer actually does mathematics, rather than just lists what needs to be learnt. Thus e.g. the basketball example must be replaced by e.g reviewing videos of a basketball game, which can be instructive to someone who wants to play the sport well. (1/2)2+1, very concise answer.1818572ShahjalalBThere are some words " implicitly solving a transcendental equation?" in the question of Terry Tao. I want give a weak edition " explicitly solving a transcendental equation?" The method shown below suits 'Riemannian surfaces with an explicit distance function?' too. I don't know if it's a useful idea.

If we denote multivariate function composition $f (x_1, \ldots, x_{i-1}, g(x_1, x_2, \ldots, x_n), x_{i+1}, \ldots, x_n)$ like (f.g) for unary function composition as following three forms,

1 $(fC_{i}g)(x_{1},\cdots,x_{n})$ like an operation

2 $[C_{i}(f,g)](x_{1},\cdots,x_{n})$ like a function

3 $[C_{i}\frac{f}{g}](x_{1},\cdots,x_{n})$ like a fraction

For example,equation $x+x^{a}=b$,the left of it is $x+x^{a}$. It can be obtained by substituting $x_{1}$ in $x+x_{1}$ by $x^{a}$,so

$x+x^{a}=[f_{a}C_{2}f_{p}](x,a)$

$x+x^{a}=[C_{2}(f_{a},f_{p})](x,a)$

$x+x^{a}=[C_{2}\frac{f_{a}}{f_{p}}](x,a)$

In which $f_{a}(x_{1},x_{2})=x_{1}+x_{2}$ and $f_{p}(x_{1},x_{2})=x_{1}^{x_{2}}$

$C_{2}(f_{a},f_{p})$ is a binary function. If we define inverse binary function like define it to power operation $f_{p}$,

$z=f_{p}(x,y)=x^{y}$, $x=[I_{1}(f_{p})](z,y)=f_{r}(z,y)=\sqrt[y]{z}$ and $y=[I_{2}(f_{p})](x,z)=f_{l}(x,z)=\log_{x}z$,

In which $f_{r}(z,y)=\sqrt[y]{z}$ and $f_{l}(x,z)=\log_{x}z$. We can extend $I_{i}$ to multivariate functions.

$C_{2}(f_{a},f_{p})(x,a)=b$ then $x=\{I_{1}[C_{2}(f_{a},f_{p})]\}(b,a)$. Is this " explicitly solving a transcendental equation"?

If you answer 'yes',let us solve $x^{a}+x^{b}+x^{c}=d,(a,b,c,d\geq0)$

$f_{a2}{\{}f_{a1}[f_{p1}(x,a),f_{p2}(x,b)],f_{p3}(x,c){\}}=d,$

There are more than one additions or powers so we distinguish them by their subscript.

First,there are four parameters,x,a,b,c. So we obtain:

$f_{p1}(x,a)=P^4_{1,2}(f_{p})(x,a,b,c)$,

$f_{p2}(x,b)=P^4_{1,3}(f_{p})(x,a,b,c)$,

$f_{p3}(x,c)=P^4_{1,4}(f_{p})(x,a,b,c)$,

$f_{a1}(x_{1},x_{3})=P^4_{1,3}(f_{a})(x_{1},x_{2},x_{3},x_{4})=x_{1}+x_{3}$,where $ x_{1}$ or $ x_{3}$ is transitional variable.

$f_{a2}(x_{3},x_{4})=P^4_{3,4}(f_{a})(x_{1},x_{2},x_{3},x_{4})=x_{3}+x_{4}$, where $ x_{3}$ or $ x_{4}$ is transitional variable.

$P^n_{i,j}$ is called function promotion, take $P^3_{1,3}$ as an example, $[P^3_{1,3}(f_{a})](x_{1},x_{2},x_{3})=x_{1}+x_{3}+O(x_{2})=f_{a}(x_{1},x_{3})+O(x_{2})$,where $O(x)\equiv0$.That is say $P^n_{i,j}$ change a binary function f to a especial function of n variables and take two variables of f as the i-th and j-th variable of $P^n_{i,j}(f)$ respectively.

Substituting $P^4_{1,2}(f_{p})$ to $x_{1}$ and $P^4_{1,3}(f_{p})$ to $x_{3}$ of $P^4_{1,3}(f_{a})(x_{1},x_{2},x_{3},x_{4})=x_{1}+x_{3}$ respectively,

$C_{1}[P^4_{1,3}(f_{a}),P^4_{1,2}(f_p)]$.

$C_{3}{\{}C_{1}[P^4_{1,3}(f_{a}),P^4_{1,2}(f_p)],P^4_{1,3}(f_p){\}}$.

Substituting $C_{3}{\{}C_{1}[P^4_{1,3}(f_{a}),P^4_{1,2}(f_p)],P^4_{1,3}(f_p){\}}$ to $x_{3}$ and $P^4_{1,4}(f_{p})$ to $x_{4}$ of $P^4_{3,4}(f_{a})(x_{1},x_{2},x_{3},x_{4})=x_{3}+x_{4}$ respectively,

$C_{3}\frac{P^4_{3,4}(f_{a})}{C_{3}{\{}C_{1}[P^4_{1,3}(f_{a}),P^4_{1,2}(f_p)],P^4_{1,3}(f_p){\}}}$.

$C_{4}[C_{3}\frac{P^4_{3,4}(f_{a})}{C_{3}{\{}C_{1}[P^4_{1,3}(f_{a}),P^4_{1,2}(f_p)],P^4_{1,3}(f_p){\}}},P^4_{1,4}(f_{p})]$.

This is the structure of the left of the equation $x^{a}+x^{b}+x^{c}=d$ described by multivariate function composition .The equation will be:

${\{}C_{4}[C_{3}\frac{P^4_{3,4}(f_{a})}{C_{3}{\{}C_{1}[P^4_{1,3}(f_{a}),P^4_{1,2}(f_p)],P^4_{1,3}(f_p){\}}},P^4_{1,4}(f_{p})]{\}}(x,a,b,c)=d$

The expression for the solution of the equation is:

$x=I_{1}{\{}C_{4}[C_{3}\frac{P^4_{3,4}(f_{a})}{C_{3}{\{}C_{1}[P^4_{1,3}(f_{a}),P^4_{1,2}(f_p)],P^4_{1,3}(f_p){\}}},P^4_{1,4}(f_{p})]{\}}(d,a,b,c)$

For such an expression $I_{3}{\{}C_{4}[C_{3}\frac{P^4_{3,4}(f_{a})}{C_{3}{\{}C_{1}[P^4_{1,3}(f_{a}),P^4_{1,2}(f_p)],P^4_{1,3}(f_p){\}}},P^4_{1,4}(f_{p})]{\}}$,we never mind how complex it is. We consider it as a multivariate function being composition results of two other multivariate functions being composition results and/or promotion results.

25982170796581907Lower bound construction for the extremal number of $C_{2k}$-free bipartite graph2204798128525613434341784962This seems quite difficult, for an example of results (and an indication of the difficulty) see Ganesan's 2011 paper.

75141914264401470663961276139761676611546354201483202643910547442215137Yes Will, that is exactly it. I realized that if I knew how to do that I could solve this, but I haven't had any luck yet. Also, I unfortunately do not have any resident mathematicians available to me. =)because of some reasons in integrability of multivariable functions1065439107625236078614160952164541I realize I forgot to address the part of the question about the $\sigma$-algebra on $P(X)$. You can get some of the distance with that by using the theory of the Giry monad.Maximally nontrivial theories must exist for any finite $d$. For a nontrivial $\psi$-epistemic theory with given $\phi,\psi$ you only need to verify that $p=\inf_M\max_i min(\vert\nu_i^*\phi\vert,\vert\nu_i^*\psi\vert) > 0$. This holds by compactness+continuity so that the extrema are obtained, and must also hold in some neighborhood of $\phi,\psi$. You can boost this up by taking convex combinations of such theories and using compactess to cover all possible $\phi,\psi$.1082517Thanks! I didn't know about Gröbner bases. Somehow this seems like it can't possibly be that complicated, but maybe it is.(I assumed that $i$ in front of $y_j$ denotes the imaginary unit)2391431862020The manifold $X$ is spin iff the loop space $LX$ is orientable. But I don't think this will help. If you look at $spin^c$, almost complexe manifold $M$ is $spin^c$. All of the 4 dimention manifold is $spin^c$. So if more over, $det^(1/2) TM$ exist, $M$ is spin. 2909321930550102439131392625495312275207795618Thanks! I'm now reading your two volumes of system of conservation laws, I found it clear and interesting. Thank you for your reply. I will read your recommended book.355657Doesn't proceedings of steklov institute allow long papers? You could also write a book.369408Given a simple closed curve in the plane, there is a homeomorphism from the unit open disk to the interior of the curve. The homeomorphism can be taken conformal, this is the uniformisation theorem.

Is there a version of this result for closed curves that are not simple?

For example, given a $C^1$ closed curve $\gamma: S^1 \rightarrow {\bf R}^2$ with finitely many self-intersections, all of them transverse, is there a continuous map (maybe even conformal) from the unit open disk to the plane, such that the number of preimages of any point in ${\bf R}^2\backslash \gamma(S^1)$ is equal to the absolute value of the number of turns the curve makes around the point?

1307475What is $\mu(2)$? Nilpotent matrices there are easy to describe. I calculate that the exceptional solution has both sides equal -7. Gerhard "Is Quite Fond Of Sevens" Paseman, 2018.05.27.B@Carl - correct for both counts.The solution of this problem for detecting polynomials $f$ which are squares or more generally a power $g^p$ is in my article "On Hilbert covariants" with Chipalkatti in Canadian J. Math. 66 (2014), no 1, 3--30. The idea in Greg's answer works in general and is in fact due to Hilbert. In our article we also give an alternate formula for the relevant covariant in determinantal form (eq. 12 on page 9 in the arXiv version) which may be more useful for practical computations.

Thanks for the reply. It was not exactly what I was asking but I think it still answers my question in that I infer that you did *not* (significantly) change your mind regarding what should or should not be on MO.1852042812884951015448635543914935681590071380134Let $M$ $ $ be a differential manifold, and $f$ a diffeomorphism on $M$ which is isotopic to $id$. Assuming that $x\in M$ is a fixed point of $f$ and the orbit of $x$ under the isotopy is a trivial loop(trivial in $\pi_1(M)$). How to prove that there is an isotopy from $id$ to $f$ relative to $x$? (i.e. there is an isotopy $f_t$ satisfied that $f_0=id$, $f_1=f$ and $f_t(x)=x$.

1595284123824555597115499577176varsop418426226019210776094817052214731833565668807his every point of a Berkovich space a Shilov point?614491448003vSerre intersection formula and derived algebraic geometry?1685346In a finite $p$-group maximal subgroups are normal and $H$ is contained in a maximal subgroup. So probably the result will differ between nilpotent groups and other groups.2212699085736548599904571523162180693"properly discontinuously" is a single word (also called, more simply "properly"). It's not "properly and discontinuously". The property of acting properly on a CAT(0) cube complex is known as "Property PW".2189696465480Maybe this is relevant: [Ovseevič, A. I. Abelian extensions of fields of CM type. Funkcional. Anal. i Priložen. 8 (1974), no. 1, 16–24.](https://link.springer.com/article/10.1007%2FBF02028302)2167734I am wondering if the family of degenerate abelian surfaces constructed by K. Hulek, C. Kahn and S.H. Weintraub in "Moduli spaces of Abelian Surfaces: Compactification, Degenarations, and Theta Functions" is a toroidal degeneration, i.e. if we let $\mathcal{X}_2 \to \mathcal{A}_2$ be the universal family of principally polarized abelian surfaces and $\tilde{\mathcal{X}} \to \tilde{\mathcal{A}}$ its degenerate famiy (here $\tilde{\mathcal{A}}$ stands for the Igusa compactification) then is it true that $(\mathcal{X}, \mathcal{A})$ is a toroidal compactification?

47804222605711175540For larger $n$ we can still make a counterexample, can't we?

Let $S$ be a set of $n+1$ points in general position, and choose them to have rational coordinates. Any linear map which fixes these points must be the identity (on the projective space), and any semilinear map which fixes them must therefore be induced by a field automorphism, and must therefore fix every rational point. Now let $K$ consists of $S$ and two more rational points, and let $\phi:K\to K$ be the transposition interchanging these two points.

1596876567314408993585142dg.differential-geometry7465682198196@JDH: Are there any papers that have results regarding nontrivial elementary embeddings j:$V$$\rightarrow$$V$ in $ZF^{-f}$+$BAFA$? Is it possible to prove an analogue to your Thm.1 in this system? Is the consistency strength of $BAFA$ no greater than $ZF$ as well?1255804 This isn't really an answer (so I'm writing it as a comment) but experimentally, it appears that in the particular case of the Hilbert scheme of points on an ADE resolution being able to do the computations you're asking about tells you the codimensions of supports of modules for the corresponding symplectic reflection algebra. I have the impression that something similar should be true for Springer fibers and $W$-algebras (though I'm less sure about this). But this is a topic of ongoing research! So in this sense maybe your question is really an ``open question''? Personally, 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...14457@MichaelRenardy You can look at the link on MSE, it's a problem but achille hui was thinking that perhaps this problem is false, but as I said for $n=1$ is true because we can prove that $6(A^3+B^3+C^3)+I_n\ge 5(A^2+B^2+C^2)$ with proviso that $A+B+C=I_n$.I believe most of what you want is in http://front.math.ucdavis.edu/0308.5101 , especially the polynomiality you're looking for. Note that that was first proven in [H. Derksen, J. Weyman] "On the Littlewood-Richardson polynomials," http://www.math.lsa.umich.edu/~hderksen/preprints/lrpoly.dvi .

Judging from the comments it seems to me that the only reason why this doesn't appear to be well covered in the literature is that you're looking in the wrong places (or insisting on keywords that aren't used in most intro algtop books). In books like Hatcher's, they use the word "bundle", not "locally constant sheaf". Instead of "local system" words like "bundle of groups" are used. Moreover it looks like you prefer not to think about bundles of groups, but the induced vector bundles from the construction Arapura describes below. 814585LI added the word "many" to the title.9608685161390720263081124720if my answer is not right one, Under what conditions (hopefully regularity requirements) is my answer correct?140073313310401933124229754887458564661196107231116381809480403829@landweber exact functor theorem1049215@unknown: That's a fair comment. Still, if the goal is to find conjectures that are accessible to the general math-loving public that they may not have heard of before, I think the decimal expansion problem counts. Perhaps David Feldman can clarify whether he really means that 90% of non-number theorists haven't heard of the conjecture of which this happens to be a corollary, or whether he means something weaker than that.173608112420581088018You can represent rationals in the algorithm exactly as fractions (pairs of integers). Making it fit in two-byte integers may be a bigger problem, especially since your input numbers do not seem to fit there in the first place (32768).2122630http://en.wikipedia.org/wiki/Vector_measure#Lyapunov.27s_theorem74970838832272108325153372Please accept apology if this question is vague. (Would you please comment rather then downvote, I may be stopped to ask more questions. I will delete my question if required.)

It is related to the link which describes white noise theory to deal with stochastic differential equation. https://www.duo.uio.no/bitstream/handle/10852/10633/pm02-03.pdf?sequence=1&isAllowed=y On the other hand another theory relates to the same area as described in Wikipedia. https://en.wikipedia.org/wiki/Rough_path

Both of them shows that iterated integrals (different measures) has uniqueness. Rough path theory is very popular as it has has been used in machine learning however, white noise theory is not that popular and it seems it has not been used as extensively as the other. Is it possible to have comparison between the two? Any comments would be highly appreciated. Any reference would also be very helpful. I am looking for some insight to compare this two approach.

80877Actually covering space theory is rigid so it will do in this case, see my answer.I have to admit, this is one common misgiving I've never been able to really understand. That is, among the many non-constructive isomorphisms we encounter in mathematics, why is the one between $C$ and $\bar{Q_p}$ so odious? Isn't it supposed to be just a reflection of a sort of 'uniformity' of algebraically closed fields, as with vector spaces? I suppose the wildly different topologies give us pause, but why shouldn't a rather 'bare' structure like a field have several distinct topologies? @DavidSpeyer Being redundant here but this is beautiful! Thank you!8940461605324809668350352I 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?18298414014976035996194753244811018748Is it cheating to remark that A_5 is isomorphic to PSL_2(F_5) and therefore this can't invalidate my answer?15697Previous comment, continued: If so, I can see no substantial difference between your suggestion and my example with prime and natural numbers. Of course, that example has no relevance in number theory, and I still see no relevance of your suggestion to probability theory (say). I still see the essence of your suggestion as hardly more than the suggestion to call push-forward measures conditional expectations: "pushforward a finite measure [...]. This is the conditional expectation", in your words.TIrreducible Polynomials from a Reccurence608614David, thanks for pointing out my error. It should be correct now.Have you looked at Exposés IV and V in SGA 6? The whole theory is made in the general framework of a ringed topos. This should give you at least all the formal part, though it probably doesn't say much about computing concrete examples.15368901083541%Let $\mathbb{S}$ denote Sacks forcing. Let $\mathbb{S}_{\omega_2}$ denote the $\omega_2$-length countable support iteration of Sacks forcing.

Let $G_{\omega_2} \subseteq \mathbb{S}_{\omega_2}$ be generic over $L$ for $\mathbb{S}_{\omega_2}$. By the usual results around proper forcing, you can show that $L[G_{\omega_2}] \models 2^{\aleph_0} = \aleph_2$.

You can show that no Cohen reals over $L$ are added by showing that every real of $L[G_{\omega_2}]$ is contained in a ground model coded closed meager set. Note that Cohen forcing is forcing equivalent to $\mathbb{P}_{I_\text{meager}}$, the forcing of nonmeager Borel sets. Hence a Cohen real is not contained in any ground model meager set.

For simplicity, let first consider one Sacks forcing $\mathbb{S}$. Let $\tau \in V^{\mathbb{S}}$ and $\tau$ be a name for a real not in the ground model. Then do a fusion argument to produce a condition $p$ so that at every split node $s$ of $p$, $p_{s0}$ and $p_{s1}$ determines a certain finite amount of $\tau$ and what $p_{s0}$ determines about $\tau$ and what $p_{s1}$ determines about $\tau$ differs in at least two places. If you let $T$ be the set of all finite strings $t$ so that $t$ is an initial segment of what $p_s$ can determined about $\tau$ for some $s \in {}^{<\omega}2$, then you get a tree so that at each split node the two sides differ two times before splitting again. By the definability of forcing, $T \in L$. You can show that body $[T]$ is meager closed set and $p \Vdash \tau \in [\check T]$.

If you understand how to do this for 1-Sacks forcing, now do an iterated Sacks forcing version of this. This tends to be quite heavy in notation. See Geschke and Qickert $\textit{On Sacks Forcing and the Sacks property}$ for more information. Essentially the argument above is a modified version of their proof of the 2-localization property. Also see their paper on how to do the iterated Sacks version of the 2-localization property.

It seems that I understood the question as is it ever possible to force the continuum to be $\aleph_2$ without adding Cohen generic over $L$. It seems the far more interesting question is whether for any universe $V$ (perhaps not equal to $L$) which does not have Cohen generic over $L$, is there a forcing extension of $V$ which does not add Cohen generics over $L$.

228721287654By "converse inclusion" do you mean is the RHS included in the LHS? It doesn't look like it. For example what if, with probability 1, $Y_s=0$ for all $s$? So $X_s=X_0$ for all $s$. Then the LHS is trivial but (unless $X_0$ is also constant) the RHS is not. But actually I don't even believe the original inclusion as you state it. Don't you need some extra assumption such as right-continuity of $Y$ at $t$? Otherwise how can the path of $X$ on $[t,\infty)$ tell you the value of $Y_t$?7071857180472217618Andreas had posted an answer to the effect that there must be $\mathbb{Z}$ chains consisting of composites, but this doesn't by itself answer Jaap's question, and so he deleted it.Jon Bannon22562805559289965084181101881497By definition of freeness, $Hom_R(A,X)$ is naturally (with respect to $X$) isomorphic to the direct product of $n$ copies of $X$. Apply this in both the domain and codomain of $\Phi$, and use the fact that $-\otimes B$ distributes over finite products (because they're the same as finite sums in abelian categories). Is that "another" description? The content is the same but the viewpoint seems a bit different.

173473714875831590091I don't see why you would necessarily want to throw in the annihilation operators; couldn't you work in non-self-adjoint algebras rather than C* algebras? In either case, to have a Banach algebra norm on the quotient you will need to close your tensor algebra and ideal, and I do not see how this will relate directly to the orginal construction.1811572Despite the editing, the question still asks about the "same homopoy groups" 7872721836079208246322097316Why is (1) "easy to see"?74880815803142188377If I made no mistake the first six generating functions are irreducible polynomials over Q.6885612239901345411K.Amine108190713234669630071You wrote: the tower

⋯→An-spaces→An−1-spaces→⋯→A2-spaces,

I'm not sure how to interpret that A_n-spaces means the cat of A_n-spaces?? OR A_n-space means X together with the structure maps OR A_n-space X<--> XP(n), the projective space but then XP(n)\subset XP(n+1) ??? but for none of these do I see Ωn−2F(X[n],X)???

what am I missing?

and one notes that the n-th layer is given by Ωn−2F(X[n],X), which is the (n−2)-fold loop space of the function space of based maps from the n-fold smash product of X to X. Then the k-invariants (aka the maps inducing the d1-differential in the homotopy spectral sequence) Ωn−2F(X[n−1],X)→Ωn−2F(X[n],X),n≥2 can be computed explicitly and the formula for these is reminiscent of the Hochschild cohomology differential.

2442415998212160021Consider two (distinct) octahedral diagrams i.e. diagrams mentioned in the octahedron axioms of triangulated categories (with four 'commutative triangular faces' and four 'distinguished triangular faces'). Is it true than one can extend to a morphism of such diagrams: 1. any morphism of one of the 'commutative faces' of the octahedron 2 any morphism of the pair of morphisms whose target is the upper vertex of the octahedron (i.e. a morphism of commutative triangles not lying on the faces of the octahedrons)?

Is there any text where I could look for various facts of this sort?

P.S. It seems that the answer is 'no' in general. Having a morphism of 'commutative faces', one can extend it to a morphism of three neighbouring 'triangulated faces'. Thus one obtains morphisms of each of six vertices. Yet (all possible) compositions of edges of the 'first' commutative triangles and the neigbouring distinguished faces do not yield all edges of the octahedron; two of the edges (in the 'lower hat') are missing.

Yet it would be very interesting to know which additional conditions are needed in order for the morphism of the octahedrons desired to exist. I would be deeply grateful for any comments!!

383990A question on haussdorf case: Assume that $X$ is haussdorf and satisfies the condition of my question. Is $X$ necessarilly locally compact?dSurface dissection of regular tetrahedron to cube2401091956176Thanks for finding the relevant literature. I believe you have misread my answer, and it is in fact accurate. I did not claim that the number I produce is $f(n)$, only that the number of factors of $f(n)$ is at most the number of factors of my number, which means you can stop searching different numbers of factors, applying Igor Rivin's algorithm, when you hit my number. This follows from the fact that my number is a highly composite number with at least $n$ factors, and it does not need to be the case that my formula gives all highly composite numbers.PAnother close call is $\pi_{9+n}(S^n)$.2271594837470408035521331Does it somehow follow easily from Theorem 4a in that paper? I don't quite see it.rLength of $\mathbb{Z}_p$-modules for quaternion algebras:Recursively dependent types?1247772279477The expected constant in the Bateman-Horn conjecture is $$\frac1d \prod_p\frac{1-\frac{n_p}{p}}{1-\frac1p},$$ where $n_p$ is the number of roots of $f(x)$ modulo $p,$ and $d$ is the degree of $f(x).$ For the particular polynomial in question, this converges quite rapidly, and when the product is taken over the first 10000 primes, the constant is approximately $1.6235,$ which does not disagree with the experimental result.

For posterity, here is the *Mathematica* program:

```
f[x_] := 29160 x^3 + 30132 x^2 + 8046 x + 643
nn[p_] := Length[Solve[ff[x] == 0, x, Modulus -> p]]
rat[p_] := (1 - nn[p]/p)/(1 - 1/p)
bh[n_] := Product[rat[Prime[k]], {k,1,n}]/3
```

1511910A nice property of $\mathbb P^n$ is:

Property 1: Two subvarieties $U,V$ such that $\operatorname{dim} U +\operatorname{dim} V \geq n$ always intersect.

(for example, any 2 curves in $\mathbb P^2$ intersect)

There are other smooth varieties $X$ when Properties 1 holds. For example, a sufficient condition is that the ranks of $\text{CH}^i_{num}(X)$ are $1$ for $i\leq n/2$. Here $n = \operatorname{dim} X$ and $\text{CH}^i_{num}(X)$ is the Chow group of codimension $i$ modulo numerical equivalences.

My question is whether some converse is true:

Question: Let $X$ be a smooth projective variety satisfying Property 1. Does that impose some upper bounds on the ranks of $\text{CH}^i_{num}(X)$ for $i\leq n/2$?

Let's assume we are over $\mathbb C$, but I am also interested in results over any ground fields. One can ask the same questions for the ranks of $\text{CH}^i_{hom}(X)$ (I think they are conjectured to be the same). The baby case is $i=1$, where the question asks if Property 1 tells us something about the rank of the Neron-Severi group of $X$.

I am aware that the question is a little vague (upper bound as function of what?), but that was because of my ignorance, so comments to improve the question are welcome.

22478711983541In this case I think the answer should be no (see my answer below). Maybe assuming the stronger condition $H^k(\pi(A))=0$ *for any projector $\pi$ with finite dimensional range* allows a positive answer?841682149765821215235072361658855830429759031&$8=2^2$? $9=2^3$??542041717716I'm trying to find an approximation for the optimal path for a material point, minimizing the integral associated with the total energy.

I managed to write the exact formula for the energy along a path covering the points $(x,y(x))$, which should be:

$$ E(y) = \int_{x_0}^{x_f} \sqrt{a-by(x)}\sqrt{1+(y^\prime(x))^2} \ \ \mathrm{d}x $$

Where $a$ and $b$ are constants related to mass, initial position and speed, and gravitational constant.

If I knew the path, numerically computing the integral would be trivial, but I don't know any methods to estimate $$ y = \arg\min_{z} J(z) $$

I suppose that $y$ should be at least a $\mathcal{C}^1$ function (possibly $\mathcal{C}^\infty$?), but the only way to approach the problem in a general way that I can think about is to use a (linear) spline function, replacing differentiation with incremental ratios.

I could start with a straight line path, split it in $n$ parts and find a greedy optimization for each segment of the spline, moving the node $(x_1,y_1)$ upwards or downwards along the $y$-axis until an approximate minimum is found, then repeating the same with the node $(x_2,y_2)$, but this is obviously not going to work: .

I could instead try to move all the nodes at once (except the starting and ending ones, fixed at $(x_0,y_0)$ and $(x_f,y_f)$), but this seems too computationally intensive to pursue when using a finer grid.

Is there an efficient method to find an optimal path?

2179745Limbo20558371130641147995944733$E \times_H \mathbb{R}^n$ is isomorphic to the total space of the tautological bundle $\gamma^n$ over $G_n(\mathbb{R}^{n+k})$?The formula I've shown, but it is still not satisfied. It is not clear why? For quadratic forms $ax^2+bxy+cy^2=j$ It is necessary that the ratios between them were connected through the square. Or so $t^2=ja$ Or so $t^2=j(a+bk+ck^2)$ . You can pick up a different configuration, but the task still comes down to this.206013714756991212641**Edit: this does not answer the OP question, as it considers $1$-dimensional projections instead of $2$-dimensional projections.**

For every vector $(x_1,x_2,x_3) \in \mathbf{R}^3$, we have $$ |x_1| + |x_2| + |x_3| \geq \sqrt{x_1^2+ x_2^2+x_3^2} $$ with inequality only for multiples of basis vectors.

Summing this inequality over all edges of $P$ yields that $|P_1|+|P_2|+|P_3| \geq |P|$, with inequality for polytopes all whose edges are parallel to basis vectors, which are essentially cubes (up to dilatation in each of the three directions).

27722440593623922132911328878013199491948132Gabriel actually gave a short explanation himself in [Gabriel, Peter. Unzerlegbare Darstellungen. I. (German) Manuscripta Math. 6 (1972), 71--103]:

Für einen solchen 4-Tupel schlagen wir die Bezeichnung

Köchervor, und nicht etwa Graph, weil letzerem Wort schon zu viele verwandte Begriffe anhaften.

Attempt at translation: For such a 4-tuple we suggest the name *quiver*, rather than graph, since the latter word already has too many related concepts connected to it.

(This is community wiki, so anyone can add a proper English translation.)

499939Your conjecture is true. See the proof of Theorem 2.1.2 in http://www.math.uni-sb.de/ag/gekeler/PERSONEN/Lengler/Dissertation_Lengler.pdf for a combinatorial proof. I have an idea for a conceptual proof (continued...)1029952I guess a weaker version of Ascoli's theorem is needed (sorry this is meant as a comment but I don't have the privilege!).2229091160600518895991112819Let me point out why $K_0$ and $K^0$ is great notation: the functorial properties of these groups are analogous to those of homology and cohomology, respectively. $K^0$ is a ring, and $K_0$ is a $K^0$-module. If we have a map $\mathrm{Spec}\ R \to \mathrm{Spec}\ S$, then we get induced maps $K_0(R) \to K_0(S)$ and $K^0(S) \to K^0(R)$.20975881692453One often looks at $d^2$ instead of $d$. This at least fixes the smoothness issues along the diagonal mentioned by Ryan. Still, on any comapct manifold, $d^2$ will necessarily fail to be smooth at some points.323072119921837886:I have edited it now... yes.The answer is surely no.

If $D$ itself does not have a minimal log smooth resolution, then certainly $(V,D)$ couldn't have such a log resolution you need. On the other hand, there are bunch of isolated surface singularities whose minimal resolution is not log smooth, e.g., the log canonical surface singularity whose minimal log resolution has its exceptional locus an nodal rational curve.

1716513123968217929211567891512266NigelB129995739771407698When is the tangent bundle of a manifold naturally a complex manifold?9250007299201763691729092@Bingo, the point is that the canonical bundle of the blow up is the pullback of the canonical bundle of the surface that we blow up. (this is because we blow up double points, i.e. we have what is called "a crepant resolution"). On the other hand, that canonical bundle on the singular surface is just $O(d-4)$, where $d$ is the degree of the surface. This is clearly ample. It follows that the canonical bundle on the blow up is both semiample and big (as a pull back of an ample line bundle).11314492254151That is intriguing John. Any suggestions for how I may prove this?sho5547152014528 @tomasz Oh yes, you're right. How does this sound: for input $n$, the machine halts if the Collatz (3n+1) iteration eventually reaches 1. The halting set is conjectured to include every $n$, but it *might* be uncomputable (some generalizations of the Collatz problem are known to be $\Pi^0_2$-complete). Each $n$ gives rise to a $\Pi^0_1$ proposition (the sequence never converges after any number of steps) that might be undecidable. So the distances may also be uncomputable. I feel like it's possible to concoct a definitely uncomputable example along these lines, but don't see one offhand.3973241761146862596johannsbj8320009761978455@Zhaoting Wei even if you allow fixed points you won't get much. it's easy to see that the only order 2 diffeomorphisms of $\mathbb CP^1$ that commute with the action of $SL(2,\mathbb R)$ are the identity and the complex conjugation.7690591067818if this tropical variety (in fact, hypersurface) is a union of two tropical varieties then the tropical polynomial is a tropical product of two tropical polynomials547223:(Of course we *can* do that)424205NOkay, this seems fairly clear. Thanks.$(2^{\aleph_0})^+$ is the smallest cardinal larger than $2^{\aleph_0}$.1651161Rescaling time, $\tau = \theta t$, the equation becomes $\partial L/\partial \tau = \max\{\partial L/\partial x_1,\partial L/\partial x_2\}$; so you can assume that $\theta=1$.697735660744When cross-posting, it's usually a good idea to include a link to the other question, so that we don't end up having two independent discussions on the topic. In this case the other question has no answers, so we're safe, but here's the link anyway: http://cstheory.stackexchange.com/questions/1066/integer-relation-detection-for-subset-sum-or-npp @Matthew: Sorry, I'm not right. I delete my answer. – Boris Novikov 34 mins ago13924358Mathematical statistician

Below a list of some questions I have responded to, where I consider my response most interesting:

http://stats.stackexchange.com/questions/177102/what-is-the-intuition-behind-svd/179042#179042

http://stats.stackexchange.com/questions/123367/estimating-parameters-for-a-binomial/123748#123748

http://stats.stackexchange.com/questions/72479/general-sum-of-gamma-distributions/137318#137318

http://stats.stackexchange.com/questions/35956/taleb-and-the-black-swan/36048#36048

1800775520610174541@Wojowu I'm asking about "Known" short axiomatizations of $NF$, this means something that had already been there "before" I've presented the above axiomatization, not something that would come after.8002481218654This is not an answer but a shadow tangent Mohammad Ghomi has a wonderful paper concerning a converse question: what are neccessary and sufficient conditions on the shadows which insure the surface to be convex. His answer: for all projection directions, the shadow must be connected and simply connected. See: Solution to the shadow problem in 3-space, in Minimal Surfaces, Geometric Analysis and Symplectic Geometry, Adv. Stud. Pure Math, 34 (2002) 129-142. Which you can find on his web page.

I'm interested in data visualization, image processing, ODE/PDE, natural language processing, etc.

Email addr: `aquilaabz`

AT `gmail.com`

; or:

`1:eJztk8ENwzAIRUkHyA5ZqdfeMkC6/605Vl8Gg8EykUD6imMb8/hxjvP7vl5EtN/63NpKpVIpqSgBg4XzKbwzvP6PJ/Baa2BvxMxzea1caV2al3zQnId7pHqS1yPsEkPPa8/Yk4c8MzmwjuSNtydp3DsHg/PMwov7o74/ntni4f6nSAbuXcPK8YzyYqy+16u8tnrmYYnq07LOee3pKcJrzX3XrHPnjdaz/if47NVqsWl4tBrJmalsPJF9ZegNYzVPKZ9+c/YGIg==`

I *think* this works. Suppose $B(X)$ with WOT is second countable, so there is a countable base to the topology $(U_n)$. This means that if $(T_i)$ is a net in $B(X)$, then $T_i\rightarrow 0$ WOT if and only if, for each $n$ with $0\in U_n$, there is $i_n$ so that $i\geq i_n \implies T_i\in U_n$.

Each WOT open set is of the form
$$ \mathcal U(T_0,(x_j),(x_j^*)) := \{ T\in B(X) : |\langle (T-T_0)(x_j), x_j^* \rangle|\leq 1 \ (j=1,\cdots,n)\} $$
where $T_0\in B(X), (x_j)_{j=1}^n\subseteq X, (x_j^*)_{j=1}^n\subseteq X^*$.

Let $Q\subseteq X^*$ be the collection of all $x_j^*$ which occur for some $U_n$. Thus $Q$ is countable, and so the rational linear combinations of $Q$ is also a countable set.

Towards a contradiction, suppose that $X^*$ is not separable. Hence the closed linear span on $Q$ is not all of $X^*$. Thus there is $f_0\in X^*$ and a (bounded) net $(x_i)$ in $X$ such that $f_0(x_i)=1$ for each $i$ but $x^*(x_i)\rightarrow 0$ for each $x^*\in Q$.

Choose some $f_1\in X^*$ and let $T_i$ be the rank-one operator $x\mapsto f_1(x) x_i$. Let $U_n$ have the form $\mathcal U(T_0,(x_j),(x_j^*))$. As each $x_j^*\in Q$ we have that $$ \langle T_i(x_j), x_j^* \rangle = f_1(x_j) x_j^*(x_i) \rightarrow 0 $$ and so as $0\in U_n$, if $i$ is sufficiently large also $T_i\in U_n$.

This is a contradiction, as if $f_1(x_1)=1$ say, then $\langle T_i(x_1), f_0 \rangle = f_0(x_i) = 1$ for all $i$, and so $T_i\not\rightarrow 0$ WOT.

Thus $X^*$ is separable.

1056712dCovering properties of strongly compact embedding29432831203002The obstruction to lifting is given by a central extension $E$ of $T$ by $\mathbb{G}_m$. Such an extension must be commutative because the commutator induces a bi-homomorphism $T\times T\to\mathbb{G}_m$, and every such map is trivial. So $E$ is a torus, and the extension is dual to an extension of the constant group scheme $\underline{\mathbb{Z}}$ by $\underline{\mathbb{Z}}^r$, hence trivial. The argument works over any base scheme $S$ satisfying $H^1(S,\underline{\mathbb{Z}})=0$.

7467536580912556811928615195805080512%In A more general abc conjecture, p. 7 Paul Vojta conjectures:

If $x_0,\ldots x_{n-1}$ are nonzero coprime integers satisfing $x_0 + \cdots x_{n-1}=0$

$$ \max\{|x_0|,\ldots |x_{n-1}|\} \le C \prod_{p\mid x_0 \cdots x_{n-1}}p^{1+\epsilon}\qquad (1) $$

for all $x_0 , \ldots, x_{n-1}$ as above **outside a proper Zariski-closed subset**.

Similar claim here p.5.

For $F_n$ the fibonacci numbers and $L_n$ lucas numbers, $y,x=F_{2n+1}^2,F_{2n}^2$ satisfy $$ -x^2 + 3xy - y^2 + 2x - 2y - 1=0$$ and $x,y=L_{2n}^2,L_{2n+1}^2$ sastisfy $$ x^2 - 3xy + y^2 - 10x + 10y + 25=0$$

Take $x_i$ to be the monomials of any of the aboves.

In both cases $x,y$ are squares. $\max |x_i|=y^2$ and the radical is $O(y)$ because of the squares.

This gives abc quality of about $2$ and both identities are in Vojta's exceptional set.

In identities like this, one of $x,y$ being perfect power gives sufficiently large abc quality.

I conjecture that in general parametrizations of the forms $a_0 n^2,a_1n^2+a_2n+a_3$ give explicit genus $0$ curves of degree at most $2$ which give tuples violating Vojta's conjecture.

E.g. the parametrization $n^2,19n^2+n+1$ satisfies

$$ -y^2 + 38 y x - 361 x^2 + 2y - 37x - 1=0 $$

and the abc quality is about $1.5$.

Is the above true?

If this is true, there are infinitely many genus $0$ curves of degree $2$ where $x$ is perfect power and all of them violate the conjecture. Found some more binary recurrences for which $a_{2n}^2,a_{2n+1}^2$ are on degree $2$ genus $0$ curves (e.g. $a_n=2a_{n-1}+a_{n-2}$).

The requirement for perfect power can be strengthened. Let $f(x,y)=0$ be genus $0$ curve of degree $2$ with infinitely many integral points and $x,y$ are coprime infinitely often Assume $\log|x| \sim \log|y|$.

Fix $\epsilon > 0$

If $rad(x) < |x|^{1-\epsilon}$ or $rad(y) < |y|^{1-\epsilon}$ infinitely often the curves are in the exceptional set.

According the conjecture all them must be on some other varieties.

3944824759511728285567844904031938432@WillSawin Maybe these correspond to the surfaces of constant curvature? sphere, plane, hyperbolic disc1099448815259@Richard: I like this "walking complex" viewpoint! Thanks for the suggestion.6965624225114585741498622135303520034601689269I have non-constant rational parametrization to $x^3+y^3+z^3=3$. Must x,y,z be positive (not sure my parametrization is)? Is it worth posting the parametrization on MO?1863563vreference help needed on a fact about poles of L-functions885670A surprising and nice example was an answer to a question in ring theory - construct rings which are not isomorphic to their opposites: http://mathoverflow.net/questions/64370/simplest-examples-of-rings-that-are-not-isomorphic-to-their-opposites/64384#64384Theta functions with half integral weight are best thought of as vector-valued forms which transform under the Weil representation of the metaplectic group which is a double cover of $SL(2, \mathbb{Z})$.8Geodesics on a Grassmannian6835627659761080856It's a reasonable assumption, but certainly not the only reasonable one. From the viewpoint of computability theory, the most natural way to define an "infinite chessboard" is as a function that takes a location and returns the contents of that location. That way also generalizes to the case of infinitely many pieces, while the canonical index method does not. But the deeper issue, as usual, is that the format of the input can make a difference. I will also comment below. 11265831317871580304994771If yes, what is the proper Zariski-closed subset in the above examples?

Your operator is self adjoint on $ L^2 (M , e^{-f}) $: If your operator is $L$. Then we have for all fonctions u,v $\int uLv e^{-f}=\int (<du, dv>+huv) e^{-f}$

Hence , the equation $ Lu=0$ has a positive solution if and only if 0 is the lowest eigenvalue of $L $.

Let's me elaborate on this

This is a consequence of the so called Barta's lemma : If $u>0$ then a integration by parts formula implies that for every v: $$\int_M [|d(uv)|^2+h(uv)^2]e^{-f}= \int_M (Lu)uv^2+|dv|^2 u^2e^{-f}$$ Hence if there is a positive function $u$ such that $Lu\ge 0$, we get that the spectrum of $L$ is non negative. (This is the content of Barta's lemma: J. Barta, Sur la vibration fondamentale d'une membrane, {\em C. R. Acad. Sci. Paris} \textbf{204} (1937), 472--473.)

This Barta's lemma implies that if the equation $Lu=0$ has a positive solution, then $0$ is the lowest eigenvalue of $L$.

The other implication follows from the quadratic form : $$Q(v)=\int_M [|dv|^2+hv^2]e^{-f}= \int_M (Lv)v e^{-f}$$ The lowest eigenvalue , called it $\lambda_0$, of $L$ has a variationnal formulation : $$\lambda_0 =\inf\{ Q(v), v\in W^{1,2}, \int_M v^2 e^{-f}=1\}$$

And the eigenspace associated to $\lambda_0$ is precisely $$\{v\in C^\infty(M), Q(v)=\lambda_0\int_M v^2 e^{-f}\}$$ (the fact that these eigenfunctions are smooth is a consequence of Elliptic regularity).

But if $v\in W^{1,2}$ then $|v|\in W^{1,2}$ and $Q(|v|)\le Q(v)$ , $\int_M |v|^2 e^{-f}=\int_M v^2 e^{-f}$, hence if $u$ is an eigenvalue of $L$ associated to $\lambda_0$ then $|u|$ is also an eigenvalue and by linearity $u+|u|$ and $|u|-u$ also. But the unique continuation properties implies that if $|u|\not =u$ , then the function $|u|+u$ vanishes on the open subset where $u$ is non positive hence we must have $u+|u|=0$, hence $u$ is non negative. So that there is always a non negative eigenfunction associated to $\lambda_0$. The fact that such a non negative eigenfunction is positive is a consequence of the Harnack inequality (see Gilbarg-Trudinger subsection 8.8).

There is another approach that is more abstract (in the context of Dirichlet form see : The Allegretto-Piepenbrink Theorem for Strongly Local Dirichlet Forms, by D. Lenz, P. Stollmann, I.Veselic publish in {\em Documenta Mathematica} \textbf{14} (2009) 167--189).

For complete non compact manifold, the existence of a positive solution of $Lu=0$ is equivalent to the fact that the spectrum of $L$ is non negative. This is called the Allegretto-Piepenbrink principle and has been adapted to manifold by Fischer-Colbrie and Schoen (see W.F. Moss and J. Piepenbrink. Positive Solutions of Elliptic Equations. {\em Pacific J. Math.}, \textbf{75}:219--226, 1978.

D. Fischer-Colbrie and R. Schoen. The Structure of Complete Stable Minimal Surfaces in 3-manifolds of Non-negative Scalar Curvature.{\em Comm. Pure Appl. Math.}, \textbf{XXXIII}:199--211, 1980.

and the lemma 3.10 in the beautiful book : S. Pigola, M. Rigoli, and A. Setti. {\em Vanishing and Finiteness Results in Geometric Analysis.} Birkhäuser, 2008.

2958866514091Just for completeness, since the Wikipedia article linked above leaves a lot to be desired, here is a more detailed source on the Acyclic Models Theorem and on Categories with Models: http://amathew.wordpress.com/2010/09/11/the-method-of-acyclic-models/1091759One of the canonical references for questions like this is Serre's "Quelques applications du theoreme de densite de Chebotarev", Publ. Math. IHES 54. He proves, for example, that the number of primes $0\leq p \leq X$ with $\tau(p)=0$ is $\ll X (\log{X})^{-3/2}$ unconditionally, and is $\ll X^{\frac{3}{4}}$ under GRH.

https://www.gravatar.com/avatar/a42a4ecbc1afad613840a68e0a18c6ad?s=128&d=identicon&r=PG&f=11935275This is a version of the coupon collector's problem, so you'd probably find the answer in the related literature.1132761935806I am competing in a programming contest where the submission phase can be stated abstractly as follows : There is a known universe set, $U$, and a hidden target $T \subset U$. I submit $S \subset U$, and for feedback I am given $|S \cap T|$.

Two questions :

1) What is a good/optimal strategy for finding $T$ in as few submissions as possible given no assumptions on the distribution of $T$?

2) What if for each $u \in U$, there is a known probability, $P(u \in T)$, and the elements of $T$ are chosen independently according to those probabilities? Is it a good/optimal strategy to pick the set $S$ that gives the largest expected information gain given the feedback you have received so far? Is there an algorithm that would compute this set in a reasonable amount of time if $|U|$ is in the tens of millions and $|T|$ has a known size of around 5000?

1757630666553T@ benblumsmith No, it is not commutative.1018622226333915622371948478207589311940822193158393432932315226871211847511328221OK, Thank you. Then I guess I will have to wait until it will appear in my library.Welcome to Math Overflow, Roman! Note that I have a feeling that on this site, links are given only to papers that are freely accessible. Therefore, while the link to Losev's paper is perfectly OK, the link to Brion's paper seems problematic (even though there is no other link!).@Relating percentiles to moments1711367I guess you mean weakly 1-convergent rather than weakly 1-summable. The answer is no; consider the summing basis for $c_0$.Smoothness is an etale local property, and you can check that symmetric power sends etale maps to etale maps. So checking on $\mathbb{A}^n$ is fine, since all smooth varieties over an algebraically closed field is etale locally isomorphic to it.

Of course, this is secretly the same as VA's answer, since stuff which makes sense etale locally is exactly the stuff remembered by the completion of the local ring, but I like the sound of it better.

21235511573706170551622735901631818503681Although it may also not fully count as an example discovered via a proof assistant involving fully formalized mathematics, the classification of maximal subgroups of all finite classical groups of dimension $\le 12$ [*Bray, John N.; Holt, Derek F.; Roney-Dougal, Colva M.*, **The maximal subgroups of the low-dimensional finite classical groups.**, London Mathematical Society Lecture Note Series 407. Cambridge: Cambridge University Press (ISBN 978-0-521-13860-4/pbk; 978-1-139-19257-6/ebook). xiv, 438 p. (2013). ZBL1303.20053] involved large pieces of Magma code by which it was able to correct/fill gaps in the earlier theoretical results of [*Kleidman, Peter; Liebeck, Martin*, The subgroup structure of the finite classical groups, London Mathematical Society Lecture Note Series, 129. Cambridge etc.: Cambridge University Press. x (1990). ZBL0697.20004].

It's clear that you might start by looking inside a maximal torus of the given compact (say connected) Lie group. But given the long history of such problems, naive methods are unlikely to get very far with this type of question. Two useful sources are (1) a Bourbaki seminar report by Serre here and (2) the detailed discussion with references given earlier on MO here.

1178752@That is correct. Thanks Andreas125440467140611783156277921599838I have proved following functors are continuous: 1) $B\otimes_{max}(-)$ for any $C^*$-algebra $B$

2) $B\otimes_{min}(-)$ for exact $C^*$-algebra $B$

3) unitization of $C^*$- algebras.

399497Avshalom, your definition of $\delta$-strongly compact cardinals has the property that if $\kappa$ is $\delta$-strongly compact, then every larger cardinal $\kappa'>\kappa$ is also $\delta$-strongly compact. Is this what you intend? (I am used to considering $\theta$-strongly compact cardinals only when $\kappa\leq\theta$, but you have it the other way around.)915962265029C.A. V.N.2107008I am trying to compute the weigth of an ellipse with the density: $\exp(-r^2/2)$ or:

$M_2=\int_{U} \frac{1}{(2\pi)}\exp\Big(-\frac{(x^2+y^2)}{2}\Big)\; dx dy \quad\text{et}\quad U=\Big\{ \frac{x^2}{a^2}+\frac{y^2}{b^2}<1\Big\}$ and more generally:

$M_n=\int_{U}\frac{1}{{(2\pi)}^{n/2}}\exp\Big(-\frac{1}{2}\Big(x_1^2+\cdots+x_n^2\Big)\Big)\; dx_1\cdots dx_n\quad\text{et}\quad U=\{ \frac{x_1^2}{a_1^2}+\cdots+\frac{x_n^2}{a_n^2}<1\}$

Simple variable substitutions (polar, rational) don't work. I believe that one needs use improper convergent integrals and use the residue theorem. This seems to be a delicat matter.

Thanks for your help.

JP

184354875785(This is not a complete answer, but the following result might useful. Basically it says that if the poles of $f(x)$ are negative integers, then the time to compute $\sum_{k=0}^{p-1}f(k)$ is bounded by a constant times the time to compute binomial coefficients ${kp\choose p}$ modulo powers of $p$.

**Claim**: Suppose $f(x)\in\mathbb{Q}(x)$ has poles contained in the negative integers, and let $n\in\mathbb{Z}$ be given. Then there is a polynomial in $p$, $p^{-1}$, and binomial coefficients ${kp\choose p}$, whose coefficients are rational numbers independent of $p$, which is congruent to $\sum_{m=0}^{p-1}f(m)$ modulo $p^n$ for all sufficiently large $p$.

**Proof sketch**: Using a partial fraction decomposition, it suffices to prove the claim for $f(x)=x^b$, $b\geq 0$, and for $f(x)=(x+a)^{-b}$, $a,b\geq 1$. For $f(x)=x^b$, we can use the formula for power sums, and the sum is a polynomial in $p$. For $f(x)=(x+a)^{-b}$, the sum agrees, up to a rational function of $p$ (which can be expanded in a Laurent series), with
$$
\sum_{j=1}^{p-1}\frac{1}{j^b}.
$$

Write $e_{i,p}$ for the $i$-th elementary symmetric function evaluated at $1^{-1},2^{-1},\ldots,(p-1)^{-1}$. For every integer $j$, we have $$ \begin{align*} {jp\choose p}&=j\sum_{i=0}^p (j-1)^i p^i e_{i,p}\\ &\equiv j\sum_{i=0}^{n+b-1}(j-1)^i p^i e_{i,p} \mod p^{n+b}. \end{align*} $$ This is a linear system whose coefficient matrix is Vandermonde, hence we can solve for the terms $p^i e_{i,p}$ modulo $p^{n+b}$ as a linear combination of binomial coefficients ${jp\choose p}$, and the coefficients are independent of $p$. For $i\leq b$, we get an expression for $e_{i,p}$ modulo $p^n$ as a linear combination of ${jp\choose p}$. Finally, we can use Newton's formula to expression $\sum_{j=1}^{p-1}\frac{1}{j^b}$ as a polynomial (with coefficients independent of $p$) in the $e_{i,p}$, $1\leq i\leq b$.

I wrote a Sage script to compute these expressions. A random example: for all $p$ sufficiently large, $$ \sum_{n=0}^{p-1}\frac{n}{(n+1)^2(n+2)}= -2 \, p^{2} - \frac{1}{36} \, {\left(\frac{{3 p \choose p}}{p} - \frac{6 \, {2 p \choose p}}{p} + \frac{9}{p}\right)}^{2} + 2 \, p + \frac{{3 p \choose p}}{3 \, p^{2}} - \frac{{2 p \choose p}}{p^{2}}+O(p^3). $$

I'll mention that such sums can also be written in terms of $p$-adic zeta values. My software does more optimization here, so the expressions tend to be simpler. For example: $$ \sum_{n=0}^{p-1}\frac{n}{(n+1)^2(n+2)}=-p^{-2}+2p-2p\zeta_p(3)-2p^{2}+2p^{3}-4p^{3}\zeta_p(5)-2p^4 + O(p^{5}). $$

49306229874931442338Let M be a symmetric non-negative definite $n\times n$ matrix. Let $K_n$ denote the complete graph on $n$ vertices. Under what conditions is it possible to assign edge weights to $K_n$ in such a way that $M$ is the corresponding graph Laplacian? Obviously the nullspace of M must contain the constant vector $(1,1,...)$, but are there any other obstructions?

I would be interested in answers for either the normalized or unnormalized Laplacian.

189851151684221113591820506I agree with Carl that an important part of Martin's original proof of Borel Determinacy looks like a priority argument, but (if I remember correctly) Martin himself once told me that he didn't think of it that way. 1219950I'll try to make something out of this also. Thanks for the tip.17840707920912213345Ilya, thanks, I corrected the question, one has to require condition 1) only for the interior of $\mathbb B^k$. In such a case condition 1) alone is not sufficient for existence of a critical point.math110@Suvrit. Plz, have you a reference for the OP's formula. Thanks.420220254242Right, and I was saying that that answer couldn't possibly right for all ribbon knots. I think (based on Noah's answer) that's it's right for 0-framed ribbon knots, but that means there are some other funny scalars for other framings.1837987175395013534515679671749215525881274931330986hConstructing spherical embeddings from colored fansThe problem of classifying finite groupoids is essentially of the same order of difficulty as classifying finite groups, which as far as I am aware we are very, very far away from doing. (There is a classification theorem for finite *simple* groups. I don't know what is meant by classifying "finite subgroups".)

The basic idea is that groupoids are disjoint unions of connected groupoids, and connected groupoids are equivalent (in the technical sense of categorical equivalence) to groups as 1-object categories. Specifically, if you have a connected groupoid $G$ and choose an object $x$, then $G$ is equivalent to the group of automorphisms $\hom_G(x, x)$ (which I will abbreviate to $G(x, x)$.

So for example, I claim that a finite connected groupoid $G$ is classified by the cardinality of its object set $G_0$ together with the isomorphism type of a typical automorphism group $G(x, x)$. In other words, if $G$, $H$ are finite connected groupoids $G$, $H$ and there exists a bijection $F_0: G_0 \to H_0$ between their objects sets and a group isomorphism $\phi: G(x, x) \to H(y, y)$ between typical automorphism groups (supposing WLOG that $y = F_0(x)$), then $G$ and $H$ are isomorphic as groupoids.

The proof is easy. Let $x_0 = x$, $x_1, \ldots, x_n$ be the objects of $G$. For each $j > 0$, choose at random a morphism $g_j: x_0 \to x_j$, and let $g_0 = 1_{x}$; similarly choose at random a morphism $h_j: F_0(x_0) \to F_0(x_j)$ (but again with $h_0 = 1_{F_0(x)}$. Define a functor $F: G \to H$ to be $F_0$ at the object level. To define $F$ at the morphism level, notice that any morphism $f: x_i \to x_j$ is of the form $g_j \circ g \circ g_{i}^{-1}$ for some unique $g \in G(x, x)$. Then define $F(f)$ to be $h_j \circ \phi(g) \circ h_{i}^{-1}$. Then check that this defines a functor and indeed an isomorphism between $G$ and $H$; the details are straightforward.

1185045619184I got $b_{12} - b_{23}=k$. Anyway, Mark's example works out. Very nice!@JacquesCarette, by the way I think its cool that you're "doing research in "mechanized mathematics", which aims to join symbolic computation and theorem proving into a single system." That sounds awesome.1641064220756614651938560314181005214153620450184Real closed fields in HOD9950091395429@John: Natural transformations are not necessarily invertible, so the relation of "the existence of a natural transformation" is not symmetric. Also, it's true that if you invert the functors which induce weak homotopy equivalences of nerves, then you get a homotopy category equivalent to that of spaces, but I don't think that's the same thing as quotienting by the equivalence relation generated by natural transformations; not all objects in the Thomason model structure are fibrant and cofibrant.71221812153449Sorry for my terminology. I didn't take care of it hoping that if someone is familiar with this then he could help, without digging in details.229762697704404522+1 for the first line...195677818451322021925157173184011970698413141672943400287940@IgorRivin The link with $PGL_2(O_3)$ is almost clear : the two first basis elements furnish an hyperbolic factor. It's orthogonal is even and negative definite of determinant 3 and thus isometric to -Norm on $O_3$. This gives an isometry between your form and det on the module of $2\times 2$ herm. matrices over $O_3$...575142Thank you Mateusz! I now realize this is something I should know, for example as a step in proving Sudakov's minoration.+1. Why don't all books approach the JCF theorem the way suggested in this post?767452759400417121determinant of a coherent sheaf, locally free on a big open setThere is a general statement on the Jacobson radical of the path algebra of a quiver (without relations), see [this mathoverflow question](http://mathoverflow.net/q/200712/15887).71231515830420364341081189720735348007iOS / iOT Developer / Game Developer

Main Dev Languages: Swift, Obj-C, C# Main Dev IDE: XCode, VS/Unity3D

616169149886533603313466275503682085211510670No. The $S$ in the lemma is $T+\neg\phi$. $k$ satisfies universal consequences of that $S$, hence it extends to a model of $S$. Note BTW that the way you apply the lemma is in fact incorrect (or rather, the statement of the lemma is incorrect, it should read ”model“ instead of “field”), it is *not* true in general that a model $M$ satisfying universal consequences of $\neg\phi$ can be extended to a model of $\neg\phi$ together with universal sentences valid in $M$. You’d need the theory of fields to be *existentially* axiomatized for that to work.Let $\pi_m$, $m \geq 0$, be the unitary irreps of $\mathrm{SU}(2)$. The Clebsch–Gordan decomposition then gives that $$ \pi_m \otimes \pi_n = \bigoplus_{k=0}^{\min(m,n)}\pi_{m+n-2k}.$$ But suppose I want to think of this decomposition as matrices. Evaluating at a point $x \in \mathrm{SU}(2)$, on the left I have $$ (\pi_m(x))_{ij} (\pi_n(x))_{pq}.$$ How do the indices $i$, $j$ and $p$, $q$ correspond to the indices on the big matrix on the right?

16741302018559Elyse Yeager and I have constructed examples for an upper bound. Basically, if you understand the dimension of subposets of $P$, then you understand the dimension of the subposets of the lexicographic power $P^k$. Starting with an appropriate standard example then gets you a sublinear upper bound on $F_d(n)$; this bound depends on $d$, and in particular we have $F_2(n)\leq n^{0.8295}$.

Our note is on the arXiv: http://arxiv.org/abs/1404.0021

314029602540103675216391151242646144039218924184393707799524855241165104198270921702746Haar Measure on a Quotient87182916846451438682 41071534439Let $[0]_q:=0$ and $[n]_q:=\frac{1-q^n}{1-q}=1+q+\cdots+q^{n-1}$, for $n\geq1$.

Question.Is there a closed formula (with proof) for the determinant of the matrix of $(i,j)$-entries $$[i+j\bmod n]_q, \qquad i,j=1,2,\dots,n.$$

**Remark.** To bring in some context to the problem, this determinant is the specialization
$$x_{i+j\bmod n}\rightarrow [i+j\bmod n]_q$$
in the group determinant of Frobenius for the (finite) additive group $\mathbb{Z}_n$.

**EDIT.** Sorry, I was meant to write $[0]_q=0$ not $[0]_q=1$.

Let's assume that we have the following collection of structures:

- Some
*space*$P$. - Monoids $(M_{i+1},\circ_{i+1})$, and
- Actions $\bullet_{i+1}:M_{i+1}\times M_i\to M_i$, for $i\ge 0$
- And $\bullet_{0}:M_0\times P\to P$.

satisfying

- ($\bullet$ is a monoid action): $(m\circ_{i+1}m')\bullet_{i+1} n = m\bullet_{i+1}(m'\bullet_{i+1} n)$ and
- ($m\bullet-$ is a homomorphism): $m\bullet_{i+1}(n\circ_{i}n')=(m\bullet_{i+1}n)\circ_{i} (m\bullet_{i+1} n')$.

In my application, $P$ corresponds to computer programs. $M_0$ are modifications to elements of $P$. If you wish, you can think of $M_0$ as some kind of structured patch. Then each $M_{i+1}$ are *higher-order modifications* of the modifications in $M_i$.

The hierarchy isn't necessarily infinite.

I'm curious to know what kind of structure I'm looking at. I originally felt that I was defining some kind of $n$-category with one object at each level, namely the endomorphism, but one reader commented that my structures were too floppy, meaning that there were not enough equations.

It seems that the structure I'm interested in is related to the automorphism tower for groups, except that I'm interest in monoids, and rather than automorphism, I'm only concerned with endomorphism, and I am working indirectly through monoid actions, rather than having the endomorphism apply to the morphisms at the level below.

Have I defined a known structure?

What natural equations would one expect to link the various levels with each other?

What additional properties does it satisfy? What reasonable properties should it satisfy?

Are there conditions under which it becomes degenerate?

Any pointers would be appreciated.

Bugs, nice to read again from you on MO! (you (and Greg Kuperberg) answered my first question here). Nice argument. 161562This problem is given a [purported - TT] proof. http://arxiv.org/pdf/1507.01270v6.pdf8This is a fantastic answer!94754673244412794582**EDIT.** I found papers in which the problem of decay and integrability of $\widehat{1_A}$ has been studied and largely answered, at least for domains with some regularity.

This recent paper by Lebedev proves the following:

If $A\subset \mathbb{R}^n (n\geq 2$) has $C^2$-boundary (and finite measure), then the exponent $p=\frac{2n}{n+1}$ is ''critical'' for the integrability of the Fourier transform $\widehat{1_A}$, that is, $\widehat{1_A}\in L^p$ if and only if $p>\frac{2n}{n+1}$. Actually, even the case where $A$ is a ball shows that this cannot be improved. The same conclusion was proved by Herz if $A$ is convex but no regualrity is assumed.

A bit surprisingly, if $\partial A$ has less smoothness, namely it is just $C^1$, then for $n=2$ it is possible that $\widehat{1_A}\in L^p$ for $p>1$.

Also the decay of $\widehat{1_A}$ has been studied; for example in This paper of Svensson and This paper of Brandolini.

According to the first paper, for bounded sets $A$ that are closures of open sets with $C^{\infty}$-boundary, we have $\widehat{1_A}(\xi)=O(|\xi|^{-\frac{n+2}{2}})$ if and only if the Gaussian curvature of $A$ is nonzero everywhere.

Despite these results, it might be that for general measurable sets we cannot say anything; I have not found references yet.

I obtained some partial results for my question.

- By the Hausdorff-Young inequality, $\|\widehat{1_A}\|_q\leq \|1_A\|_p=m(A)^{\frac{1}{p}}$ for every $q\geq 2$, where $p=\frac{q}{q-1}$.
- The Fourier transform $\widehat{1_A}$ does not generally extend to an entire function. Indeed, suppose that $\widehat{1_A}$ is always analytic in $\Omega\subset \mathbb{C}^n$. Then also $\hat{s}$ is analytic in $\Omega$ for every simple function $s$ which is nonzero only in a set of finite measure. Simple functions with bounded support are dense in $L^1(\mathbb{R}^n)$ (for every $k$, choose a function $s_k$ that differs from $f1_{[-k,k]^n}1_{\{|f|\leq k\}}$ by at most $2^{-k}$), so for any $f\in L^1(\mathbb{R}^n)$ we find simple functions $s_k$ with bounded supports such that $\|f-s_k\|_1\to 0$. But then also $\|\hat{f}-\hat{s_k}\|_{\infty}\to 0$, and by Weierstrass' theorem, $\hat{f}$ is analytic in $\Omega$ as a uniform limit of analytic functions. To see the contradiction, it is now enough to take $f(x)=\frac{\sin^2 x}{x^2}$ (or a similar function in higher dimensions) whose Fourier transform is a triangular wave.

However, the first result does not tell whether $\widehat{1_A}$ is in $L^p$ for $1<p<2$ (this would also imply that the set would be closed in $L^p$), and the second one concerns only analyticity instead of weaker forms of smoothness.

16301101353228I'm wondering why the differential in homological Spectral Sequences $(E^*_{p,q},d^r)$ is defined as; $d^r_{p,q}: E^{r}_{p,q} \rightarrow E^{r}_{p-r,q+r-1}$ .

From Weibel's homological algebra(p122), or in any other texts, I don't see how/why Koszul or Leray defined the differential in that way. Are there any deep geometric meaning/mathematical techniques for that matter?

1232119158873764609hgroup theoretical transfer map and its consequences279656143598181825121763467765010469Have your upvote back, and another for your troubles!! Oh, P.S. covexity was there to stop us just thickening an arbitrary k sphere and calling it a neighbourhood (or else the conjecture would hold trivially)- suppose contractible would be easier though 1262574Elementary proof of a special case of Chebotarev's density theorem87068151443528911381197228@flatness and derived completion1214871Are "reproducing kernels" the same as Hilbert-Schmidt operators?Using homological perturbation theory, one can repair this defect. More precisely, on the space of harmonic forms, there is an $A_\infty$ structure with no differential whose 2-ary operation (multiplication) is constructed by wedging two harmonic forms then projecting the result back to the space of harmonic forms. See "Strong homotopy algebras of Kahler manifolds" by S.A. Merkulov (Int. Math. Res. Lett. no. 3 153--164) for details of the construction.

EDIT: Also, if the manifold is compact then the natural inclusion of harmonic forms into arbitrary forms becomes an equivalence of $A_\infty$ algebras, where the space of all forms has its usual dg-algebra structure.

1595768219299575731414849284811981122992397007320212481971538Khalid, are there only countably many isomorphism types of fg nilpotent pro-p groups? There are only countably many isomorphism classes of pro-p completions of fg nilpotent groups since fg nilpotent groups are fp.2071232Most books by Krantz are very good but this is one I don't have in my personal library. I'll certainly look into it.1061370@John: The key is that Morley rank (and all good ordinal rankings) is absolute for models that have the same ordinals. This is also the key idea behind Shoenfield absoluteness. IIRC, another way to see it is that *scattered* compact Hausdorff spaces are also absolute.Nice answer, Martin! Here's a link to supplement your last line: http://planetmath.org/encyclopedia/ExampleOfRightNoetherianRingThatIsNotLeftNoetherian.html2258137R tensor product with projective topology2051917BReplaced "radius" by "diameter".For (c): Suppose that two points, $x_1$ and $x_2$, of $X$ map to the same point of $y \in X_i$. Let $f^{-1}(x_j)=z_j$; we must show that $z_1=z_2$. If not, then $f^{-1}(z_1)$ and $f^{-1}(z_2)$ are disjoint. They are each covered by sets of the form $p_i^{-1}(y')$, for various $y'$ in $X_i$. The only $y'$ for which $p_i^{-1}(y')$ contains $x_1$ is $y$. So $p_i^{-1}(y) \subseteq f^{-1}(z_1)$. But, similarly, $p_i^{-1}(y) \subseteq f^{-1}(z_2)$. This contradicts that $f^{-1}(z_1)$ and $f^{-1}(z_2)$ are disjoint. One is because I clicked wrongly and I don't know how to undo it.2215433the example which shows that exp(zw) is not equal to exp (exp(z)^ w) the carrot sign means raised to the power another one a continuous function of a complex variable need not have primitive in a region.the example is f(z) = square ( | z| ).

2275396949188ALet $\mathcal C$ be a monoidal category. Recall that the *(Drinfel'd) center* of $\mathcal C$ is the braided monoidal category $Z(\mathcal C)$ with:

- Objects: pairs $M \in \mathcal C$ and $\mu: M\otimes(-) \overset\sim\to (-)\otimes M$ a natural iso, which is required to satisfy: $\mu: M \otimes \mathbb 1 \to \mathbb 1 \otimes X$ is the canonical isomorphism $X\otimes 1 \to M \to 1\otimes M$, and $\mu_{A\otimes B} = (\operatorname{id}_A \otimes \mu_B)\circ (\mu_A\otimes \operatorname{id}_B)$ as maps $M\otimes A \otimes B \to A\otimes B \otimes M$. (I will drop all associators.) So $\mu$ is tryin to be a "braiding" — the second axiom is some sort of "hexagon" axiom.
- Morphisms: a map $(M,\mu) \to (N,\nu)$ is a morphism $f: M \to N$ so that $\nu \circ (f\otimes \operatorname{id}) = (\operatorname{id}\otimes f) \circ \mu$ as natural transformations $M\otimes (-) \to (-) \otimes N$. (Here $f\otimes \operatorname{id}$ is the natural transformation $M\otimes \to N\otimes$, etc.)
- The monoidal structure is $(M,\mu)\otimes (N,\nu) = (M\otimes N, (\mu \otimes \operatorname{id}_N) \circ (\operatorname{id}_M \otimes \nu))$.
- The braiding $(M,\mu) \otimes (N,\nu) \to (N,\nu)\otimes (M,\mu)$ is given my $\mu_N : M\otimes N \to N\otimes M$. It is a fun calculation to check that this is a (bi-natural) isomorphism in the category, and satisfies all requirements to be a braiding.

An important example: when $\mathcal C$ is the representation theory of a finite abelian group $G$, and writing $G^\vee$ for the Pontrjagin dual to $G$ (it is isomorphic for finite abelian groups), then as a monoidal category, *but not as a braided monoidal category*, $Z(\mathcal C)$ is (equivalent to) the representation theory of $G\times G^\vee$.

When $\mathcal C$ is braided (with braiding $\beta_{M,N} : M\otimes N \to N\otimes M$), there is a full, faithful injection $\mathcal C \hookrightarrow Z(\mathcal C)$, which is usually not an essential surjection, given on objects by $M \mapsto (M,\beta_{M,-})$. (For the finite abelian group case, this map is dual to the projection $G\times G^\vee \to G$.)

In addition to representations of monoids, another important source of categories are as categories of modules of commutative rings. I think I can prove that if $R$ is a commutative ring, then the injection $\operatorname{Mod}(R) \to Z(\operatorname{Mod}(R))$ is an equivalents of categories (the monoidal structure on $\operatorname{Mod}(R)$ is $\otimes_R$). (Although I've read that the center of $\operatorname{Mod}(R)$ is supposed to be more like sheaves on the loop space of $\operatorname{Spec}(R)$, so I might have made an error.)

My question is about the converse.

The weak statement I do not expect to be true:

Question (weak assumptions):Suppose that $(\mathcal C,\otimes)$ is a monoidal category (abelian, complete, cocomplete, etc. ... maybe also some statement along the "tensor products distribute over filtered limits" variety) and that there exists a monoidal equivalence $\mathcal C \to Z(\mathcal C)$. Must $(\mathcal C,\otimes)$ be equivalent to $(\operatorname{Mod}(R),\otimes_R)$ for some commutative ring $R$?

A more reasonable question is:

Question (stronger assumptions):Suppose that $(\mathcal C,\otimes,\beta)$ is abraidedmonoidal category (with more adjective?) such that the canonical map $\mathcal C \to Z(\mathcal C)$ is an equivalence of braided monoidal categories. Must $(\mathcal C,\otimes,\beta)$ be equivalent as a braided category to $(\operatorname{Mod}(R),\otimes_R,\operatorname{flip})$ for some commutative ring $R$?

Probably (super)commutative rings in SuperVectorSpaces will provide counterexamples. And so the real question is:

646607

Real question:What is / is there a nice characterization of those monoidal categories that are equivalent to their centers?

Treating both sides as coefficients of $x^k$ in a power series with variable $x$, the left-hand side turns into a power-series product, one factor representing $\sqrt{1-x}$. We end up with a special case of the known hypergeometric identity $$(1-x)^{a+b-c} {}_2F_1(a,b;c;x) = {}_2F_1(c-a,c-b;c;x)$$ with $a = m+1$, $b = -m$, $c = 1/2$. Cf. (34) of this MathWorld entry.

1985371272812159701Consider two unitary irreducible representations on vector spaces $V_1$ and $V_2$ of a Lie group $G$. For $G$ is compact and $V_1$ and $V_2$ finite dimensional there is a unique decomposition of $V_1 \otimes V_2$ into a set of irreducible representations $V_k$, $k \in K$. The Clebsch-Gordan coefficients can then be viewed as *intertwiners*, so $G$-equivariant maps from $V_1 \otimes V_2 \to V_k$. Conversely, I think that nontrivial intertwiners $V_1 \otimes V_2 \to V_m$ can only be nonzero if $m \in K$ and then they are essentially the Clebsch-Gordan coefficients.

Now if $G$ is non-compact (but $V_1$ and $V_2$ still unitary) then it seems that I can define intertwiners from $V_1 \otimes V_2$ to another $V_m$ which does not appear in the decomposition of $V_1 \otimes V_2$. For example, if $V_1$ and $V_2$ are in the principal series representation of $SL(2,\mathbb R)$ then I can obtain such intertwiners by analytic continuation of the Clebsch-Gordan coefficients.

I was first of all wondering if the above story is true. If so, is there a way to understand the domain of these intertwiners? Do we understand for which triplet of representations they exist? Is there a general theory of their properties? If not, maybe just for $SL(2,\mathbb R)$?

A small update: My answer mainly addresses how to transform a general dissimilarity function into a metric. The original question was more related to an even more basic step: Turning a similarity function into a dissimilarity function. One way, as used in the thesis above, is $d(u,v) = s(u,u) + s(v,v) - 2s(u,v)$, for example. Or, assuming that similarity decays exponentially with distance (common assumption in psychology), you'd have the relationship $s(u,v) = e^{-c\cdot d(u,v)}$, for some constant $c$. This can, of course, be combined with the symmetry fixing mentioned earlier.16639761473551:What is the character basis?1567697130324717420461922055292418663365zIs the model category of Complete Segal Spaces right proper?330846 9702$F_{\ell+1}$ is just the number of ways to tile an interval of length $\ell$ by tiles of lengths up to $k$. Now take the interval of length $2m-3$ and look at what happens at the mark $m-2$. You may have it as a boundary point, which gives you $F_{m-1}F_m$ configurations or you may have it splitting one of the intervals into two subintervals with the left one of length $a\in\{1,\dots,k-1\}$ , which gives you at most $F_{m-1-a}F_{m}$ configurations. Thus, the total number of configurations is at most $\left[\sum_{a=0}^{k-1}F_{m-1-a}\right]F_m=F_m^2$, so, indeed, $F_{2m-2}\le F_m^2$.

1436508Yes. Moreover, $A$ is closed. Indeed, assume that $X$ is given by equations $f_1,\dots,f_n$. First, consider the set $B$ of lines through $p$ contained in $X$. Note that the set of all lines through $p$ is $P^{N-1}$ and $B$ is a closed subset of $P^{N-1}$. Indeed, assume for example that $p = (1,0,\dots,0)$. Then a point $(y_1,\dots,y_N)$ is in $B$ if for any $t$ the point $(1,ty_1,\dots,ty_N)$ is in $X$. But $$ f_i(1,ty_1,\dots,ty_N) $$ is a polynomial in $t$ with coefficients being polynomials in $y_1,\dots,y_N$. The condition that for all $t$ this is zero is equivalent to all coefficients of this polynomial in $t$ being zero, that is a finite number of polynomial equations in $y_1,\dots,y_N$.

Thus $B$ is closed. Finally, $A$ is the cone over $B$ (with vertex in $p$) hence is also closed.

By the way, note that irreducibility of $X$ is not required.

6idempotents of a character2296398153128514791105198892295501@DavidWhite thanks, I thought it would be better to continue to ask here because the questions were strictly connected. I'll consider your recommendation for future. Thanks again.2057155291636You don't need a fancy reference here, this is very easy to do by hand (and definitely not suitable for a MO question). The only slightly non-trivial thing is that the equidistribution theorem holds for Riemann integrable functions as well, but that follows from density + continuity. Do you know how to deduce quantitative equidistribution results for nicer functions? Say your favorite trigonometrical polynomial?1453355477156ChenChaoThe point is that the coefficients are wrong. De Rham cohomology is isomorphic with the singular cohomology with $\mathbb{R}$ coefficients. This does not recover the cohomology with coefficients in $\mathbb{Z}$.2026099Well what I was looking for is:

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

the Link helped, thx.

247241165033128214782176219313481887580693290:I suspected as much, thanks.But the quotient of $\mathbb C$ by an integral lattice of rank 2 is simply connected really? You said "A projective homogeneous G-variety is equivariantly isomorphic to a partial flag variety $G/P$" can you give a reference for it?13042122901521702160275658Tuples of 2x2-matrices simultaneously conjugate to matrices with integer entriesThere is a very nice category of mathematical results (that are also relevant to culture) : the negative results.

For example there is no solution problem X (Say Fermat last Theorem) is negative and usually the result is not very interesting or motivated for laymen. But think of the following : There no program checking that a program has no bugs ( by classical diagonal argument: how would this program test itself). In this case we prove something is impossible and so we save a hell of a lot of time: no need to search any more. Negative results are extremely useful : in a negative way you avoid loosing money and in fact it is a good justification for pure research.

Yet beware that tough simple negative statements are not always understood some people say : "Oh Fermat equation has no solution , that is because they did not try hard enough , I will do it". This is akin to the trisectors and other poor souls looking for perpetual motion machines.

1585093I imagine that the difficulty would be in what you must do for limit ordinals.Is there any elementary argument showing that there exist uncountably many distinct quasi-isometry classes of elementary amenable groups? How about solvable groups?

For amenable groups it follows from the result of Grigorchuk (proved in the 80's) stating that there are uncountably many groups of intermediate growth with pairwise incomparable growth functions.

22397181706977510902Incidentally, the smallest solution to $a^6+b^6+c^6 = d^6+e^6+f^6$, namely, $$(-23)^6+ 10^6+ 15^6 = 3^6+ 19^6+ (-22)^6$$ belongs to another family that is also true as $$3a+b+c = d+e+3f$$10176291389208ccarminat7974476483391571909Perhaps, the most interesting things around this are finitely generated simple and bisimple semigroups -- you may find quite a lot of beautiful papers by Byleen about this19819171025901188745521289408516811682330Greg G16802218487262214844171266013623767146031359795514969Eden, your last post is false! Let the $(n_i=m_i^2)$ be given non-negative numbers and $T=\sum_k c_kS^k$. We search the $(c_i)$ (assumed to be real numbers) s.t. the eigenvalues of $TT^T$ are the $(n_i)$. If $n=2$, then $CharPoly(TT^T)=X^2-(2c_0^2+c_1^2)X+c_0^4$ that has $2$ non-negative roots. More generally, we must solve a system of $n$ algebraic equations with $n$ unknowns. The calculation of $c_0\geq 0$ is staightforward. Then it "remains" to solve $n-1$ equations in $n-1$ unknowns. We have a look at the case $n=4$: we use the theory of Grobner basis. $c_1$ is some root of a polynomial of degree $48$ (recall that $c_0$ is known). After, it is easy: $c_2$ and $c_3$ are solutions of polynomials of degree $1$ (recall that $c_0,c_1$ are known). Some numerical calculations seem to "show" that the required result is true.

8820678875251395192053627374131520735654199446928bCan a "weak" topological space be a Moore space?263269And I am sure Gödel knew how to prove the fundamental theorem of arithmetic. Perhaps I should have been more precise: when Gödel does mathematics is allowed to use whatever mathematics is considered "standard" by his peers. My pet peeve is this: why do "ordinary" mathematicians think that "logicians" must secure the ground on which they stand, when those same mathematicians stand on the same ground together with logicians?A discrete subgroup $G$ of translations is actually a discrete subgroup of $V$ thought of as an abelian group. Therefore it is free and hence is a lattice, thought not necessarily of maximal rank. Let $W$ denote the complex span of $G$. Choose a splitting $V = W \oplus U$. Then $V/G = W/G \times U$, where $U \cong \mathbb{C}^p$ and $W/G$ is a complex torus (i.e., an abelian variety).199988JFinite-index subgroups of the ideles<+1 because I'm so shallow. :D2208173651733220114113315429545707 61831697959178160394315818117411846687143637@index theory, I am working on a [C/Z index theorem](https://arxiv.org/abs/1707.03453). Composing the C/Q Chern character with [the e-invariant](http://ac.els-cdn.com/0040938394E0023D/1-s2.0-0040938394E0023D-main.pdf?_tid=b6cb4866-711a-11e7-92fb-00000aacb35e&acdnat=1500974701_06f273e80eefe3878a870d7b4d3265b4), one gets a C/Q cohomology class. Its real part might be related to the real part of the Cheeger-Chern-Simons classes. Feel free to email me, and maybe we can discuss more.188026720190111344098Sadly, I think I was confused in my previous comment. You are right, unitarizability implies that $B(G) = \cup_{c} B_c(G)$ but does not immediately give that $B(G) = M_{cb}A(G)$.502767582959@Multivariablecalculus: any complex manifold is also a real manifold. I think it would be much better if you read/study Chapter 10 of Besse in its entirety. Stating one theorem is not going to help in your case.29615702206278Ice 9 is a popular account of such a local min not being the global min177277104395714756417325632049560@Bischof, that 's not quite true. You cannot turn $k[x]/(x^2)$ into a Hopf algebra if the characteristic of $k$ is not two.It would be useful to see this comment promoted to an answer I think@Jan-ChristophSchlage-Puchta the numbers you use are not pairwise realtively prime229528519857648@ToddTrimble Answer edited.157959812728220899691391675295784416150201130022@IgorRivin No, why boatloads? A generic spherical quadrilateral is not circular, so umbilic vertices of degree four are rare. But on the other hand, a generic octahedron will have no umbilics. So, at the moment I don't see anything promising.642833I've accepted the answer, but if anyone can nail the exceptional groups, I'll be interested.2226312In the above comment I mean, "if the modularity law holds for ideals generated by squarefree monomials on a regular system of parameters".1499786@wccanard. It's funny, I think I know you in real life, though I can't be sure :-) I like your Euler's example: but did he know the prime where $x^3-2$ splits completely with a proof ? That would be surprising, since the hadn't a proof of the simpler quadratic reciprocity. Then, who proved it? regarding your second comment, I completely agree, this question is much older than Langlands, in a sense it dates back to Euler, and in the nineteen's century I think it was a well-known problem and motivation in some form.1934215Thanks, that was a mistake. The number ob bases is $2^{2^{\aleph_{0}}}$@TomekKania On the other hand, if their argument is drastically different, then it still may make some sense for me to post mine.179901818842595016841713425xExistence en regularity of elliptic PDE with mixed boundaryI agree with Andres that this is a very ambitious question. Let me throw in a tiny little bit of information:

As far as I know, Jensen's construction of a model of CH without Souslin trees was one of the first uses of both countable support iteration and master conditions (conditions generic over a countable elementary submodel of a sufficiently large initial part of the universe).

One of the first publications using proper forcing seems to be Shelah 100, with the hilarious title "Independence results" (JSL 45 (1980), 563-573).

This paper not only introduces proper forcing (without proofs of the the iteration theorem and so on), but also oracle-ccc forcing and oracle-proper forcing.

In response to your first question, yes; a monoidal category structure on Rep(H) for an algebra H is equivalent to having a quasi-bialgebra structure on H. And braiding gives you what's called a QTQBA. If you also ask for duals, you get a quasi-Hopf algebra and QTQHA respectively. You can recover the axioms for the associator Φ and the R matrix as endomorphisms of the functor to vector spaces, similar to Tannakian type constructions. Also the antipode S which satisfies rather strange looking identities. [edit: Earlier I misread the parenthetical part assuming it's already Rep(A) of some algebra, and then said almost the opposite =).]

Before talking about braiding, we'll want to ask the category have a tensor product structure on its 2 category of module categories, which will of course be extra structure. I guess this will be equivalent to asking for a functor from \Delta: C to C x C satisfying various categorical versions of the bialgebra axioms. There should be a co-associativity constraint, which should be coherent in some straightforward way, and there should be an isomorphism of functors categorifying the assertion that the coproduct of a bialgebra is an algebra morphism from A to A ⊗ A. All of this together should endow your 2-category of modules with the structure of a monoidal 2-category.

You will then want a natural transformation of functors from \Delta\to \Delta^{op}, also satisfying various naturality properties.

How does that sound (as a start)?

The way PW construed it is another standard problem: for instance, a variation on it it appears in Halmos' "automathography". It was honestly not clear to me what the OP intended. I think both answers are good. 1016201298237I believe that the question you are asking is still open. It has only (relatively) recently been shown that $s(n)=s(n+1)=A$ has infinitely many solutions for $A\ge 3$. This was shown by Schlage-Puchta in 2003. This article by Goldston, Graham, Pintz, and Yildirim discusses this and related questions:

http://arxiv.org/abs/0803.2636

Remark: your arithmetic function $s(\cdot)$ is usually denoted $\omega(\cdot)$ nowadays, but was denoted $\nu(\cdot)$ by Ramanujan.

52598In evolutionary game theory the Fisher information metric is called the Shahshahani metric. See "Evolutionary Game Theory" by Hofbauer and Sigmund. The information divergence can be used to give a Lyapunov function for the replicator equation. I believe this goes back to Akin; others use an exponentiated form (e.g. Hofbauer and Sigmund).748304560042248926240303737974562234372392An $\infty$-categorical version of this statement is now proved as Theorem 3.8 here: http://arxiv.org/pdf/1510.03150v1.pdf.

The fact that this restricts to relative 1-categories as expected follows from Remark 3.2, and the fact that this applies in particular to marked quasicategories (which can be thought of as "presentations" of relative $\infty$-categories) follows from Remark 3.3.

@NoahStein Thanks for the correction. I will edit my answer and remove everything other than the SDP formulation. I will also ask the OP that your answer be accepted, rather than mine.1382092David, I've posted a solution, so 1) the question is not all that difficult, or 2) my solution is wrong, or 3) I've settled the Riemann Hypothesis. RH, I think, is equivalent to sharp estimates of Farey discrepancy, but less-than-sharp estimates are not so hard to come by, and can still be useful. ... However, I have yet to see a sensible formulation of "not being able to iterate" in this and similar contexts. 14595671653379621403@few_reps: I assumed on the right, and there still something wrong with your thing (one of the angles between the faces is ArcCos[1/(4 Sqrt[3])], which is not a submultiple of pi, as far as I can tell.12634448513449403424Very nice argument, Ilya!187397118365948423871045093119693147958411412692660019DThat's my want! Thanks very much!@M.Lin : I wrote: "It appears that the following stronger inequality holds for $n=2$". That was based on some, rather extensive, numerical experiments. However, I don't have a proof of that. I am sorry if my comment was not clear enough.21648474497412234436@Francois: Since you have not responded to my comment (and perhaps with good reason, too, if it is something you think I should already know...), I will offer up a possible line of reasoning (at the risk of this line of reasoning being a 'straw man', so to speak...) that would make productivity in SO a proper class. Here goes: the *-recursive sets form the smallest inner model of SO (the model being the 'universe' of sets of ordinals). Since the productive 'object' ( I call it an object because it is to be determined whether it is to be a proper class or set,77076310855491493641880849So very true! It's a weird term, and its everyday usage doesn't really correspond to its one in logic – so all the stranger that authors rarely take the time to explain it.1421925Incidentally, I wasn't aware of the distinction between MathOverflow and Math.StackExchange, so I apologize if this is the wrong place to ask. Even so, I do think it's an interesting pure maths question, even if you take my particular use case out of it.2190423Here is one possible answer. An essential feature of any board game, in the way I am thinking about it, is that there are only finitely many game states that are realizable on the board. This fact, combined with a rule about whether repeated states are allowed or cause a draw or whatever, means that in any such board game, there will be only finitely many isomorphism types realized as lower cones in the game tree.

So one possible answer to the question is that a given game tree is the game tree of a board game just in case it has only finitely many isomorphism types of its lower cones.

For the one direction, as I've mentioned, any board game has this property.

Conversely, if a game tree does have this property, we can realize it as a board game by making a "board" out of the finitely many isomorphism types, and a piece that moves from one type to another, according to the rule that such a type is realizable from another for a particular player if and only if the game tree allows it.

A3. A Local ring has one maximal ideal, but it's not artinian in general (for example, $\mathbb{Z}_{(2)}$).

822102542221480062jClassifying length $4$ modules over $\mathbb C[x,y]$1662215<17*43 == 731 is NOT a prime!!1160763@Greg: there was some error/bug with the LaTeX formatting. I've cleaned it up.11967095199216832301107339Yes, except that X can be any topological space (not necessarily a manifold) and G any topological group (not necessarily a Lie group) acting continuously on X. Note that EG is usually infinite-dimensional (as is the Borel space).21339771718631Is it possible to make a more definite statement, perhaps conjecture, about the programme outline here, not for all motives, but in the restricted case of motives that arise from Calabi-Yau varieties? 8205720367771491250Update 2017.09.26: As I've just learnt from Tamas Kiraly, this notion is well-studied and has several names, such as k-way cut, k-terminal cut, multiway cut. For at least three colors, the problem becomes $NP$-complete for general graphs, see E. Dahlhaus, D. S. Johnson, C. H. Papadimitriou, P. D. Seymour, and M. Yannakakis: The complexity of multiterminal cuts.

Earlier stuff: For two colors this can be solved in polynomial time for any graph, as it can be converted to a MinCut problem. Just connect all red vertices into a single vertex $s$, and all blue vertices into another single vertex $t$, obtaining a multigraph (or if you prefer, a graph with edge weights). The Minimum cut in this new graph between $s$ and $t$ gives the partition that gives the coloring maximizing the number of the monochromatic edges.

@cameroncounts: does the link provided by Maxime tell you what you wanted to know?56714517482632264782837940744413Lucky timing, I just worked out a couple of examples for the class I'm teaching this semester. Let A be a commutative local ring, m the maximal ideal, and k=A/m the residue field. Let's consider two rings: M = M_{2}(A) (2-by-2 matrices with entries in A), and I = the Iwahori order consisting of 2-by-2 matrices with entries in A whose lower-left entry is in m, i.e. matrices of the form

[ A A ]

[ m A ]

We'll compute rad(M) and rad(I) using the fact that rad = intersection of annihilators of simple left modules = { x : 1 - xy is a unit for all y }.

First let's compute rad(M). M acts naturally on k^2 by left multiplication (via M_{2}(A) ->> M_{2}(k)), and this is a simple M-module. The annihilator of k^2 is M_{2}(m) = 2-by-2 matrices with entries in m, so rad(M) is contained in M_{2}(m). On the other hand every element x in M_{2}(m) has the property that 1-xy is a unit in M_{2}(A) for all y in M_{2}(A) (since xy is in M_{2}(m), the determinant of 1-xy is a unit); therefore rad(M) contains M_{2}(m), and we've shown rad(M) = M_{2}(m).

Next let's compute rad(I). Now k is a simple left I-module in two ways: first by multiplication by the upper-left entry (mod m), and second by multiplication by the lower-right entry (mod m). (Check that if x,y are in I then the upper-left entry of xy is congruent mod m to the product of the upper-left entries of x and y, and similar for the lower right.) The annihilator of the first I-module is therefore matrices of the form

[ m A ]

[ m A ]

and the second is matrices of the form

[ A A ]

[ m m ]

and the intersection of these annihilators is the ring of matrices of the form

[ m A ]

[ m m ]

which therefore contains the Jacobson radical. On the other hand every matrix x of that form has the property that det(1-xy) is a unit in A for y in I, and therefore the above collection of matrices is equal to the Jacobson radical.

For $n,m \geq 3$, define $ P_n = \{ p : p$ is a prime such that $ p\leq n$ and $ p \nmid n \}$ .

For example : $P_3= \{ 2 \}$ $P_4= \{ 3 \}$ $P_5= \{ 2, 3 \}$, $P_6= \{ 5 \}$ and so on.

Claim: $P_n \neq P_m$ for $m\neq n$.

While working on prime numbers I formulated this problem and it has eluded me for a while so I decided to post it here. I am not sure if this is an open problem or solved one. I couldn't find anything that looks like it. My attempts haven't come to fruition though I have been trying to prove it for a while. If $m$ and $n$ are different primes then it's clear. If $m \geq 2n$, I think we can find a prime in between so that case is also taken care of. My opinion is that it eventually boils down to proving this statement for integers that share the same prime factors. My coding is kind of rusty so would appreciate anybody checking if there is a counterexample to this claim. Any ideas if this might be true or false? Thanks.

PS: I asked this question on mathstackexchage and somebody recommended I post it here as well. Here is the link to the original post

https://math.stackexchange.com/questions/2536176/a-conjecture-regarding-prime-numbers

21242011319651004987The space $S^\infty$ is actually homeomorphic to $\mathbb{R}^\infty$. To see this, put \begin{align*} B_n &= \{x\in\mathbb{R}^\infty\::\: \|x\|\leq n, x_k = 0\text{ for } k \geq n\} \\ C_n &= \{x\in S^\infty\::\: x_n \leq 1/2,\; , x_k = 0\text{ for } k > n\}. \end{align*} Then $\mathbb{R}^\infty$ is the colimit of the spaces $B_n$, and $S^\infty$ is the colimit of the spaces $C_n$. Moreover, $B_n$ and $C_n$ are both homeomorphic to the $n$-ball, and they are contained in the interiors of $B_{n+1}$ and $C_{n+1}$ respectively. One can cook up compatible homeomorphisms $f_n\:B_n\to C_n$, and pass to the colimit to obtain a homeomorphism $\mathbb{R}^\infty\to S^\infty$.

One can prove along similar lines that various other standard contractible spaces are also homeomorphic to $\mathbb{R}^\infty$, for example $\mathbb{R}^\infty\setminus A$ (for any finite subset $A$), or $F(\mathbb{R}^\infty\setminus A,n)$, or the Stiefel manifold $V_n(\mathbb{R}^\infty)$. However, the linear isometry space $L(\mathbb{R}^\infty,\mathbb{R}^\infty)$ is homeomorphic to $\prod_{i=0}^\infty\mathbb{R}^\infty$, which is different.

261494The Milnor-Wolf theorem says that any solvable group has either polynomial or exponential growth. I wonder about the existence of alternative proofs of this fact. I have an impression that the original proof is more brutal and less ideological. If you have a citation or a suggestion of an alternative proof, please post.

205807200192798827187128935005210995982617541682505Is there any global constant $C$ such that $$C<\frac{\sum_{i=1}^{n}x_{i}\log x_{i}-x_{i}+(1-x_{i})\log(1-x_{i})}{\sum_{i=1}^{n}x_{i}}+\log(\sum_{i=1}^{n}x_{i})-\log(\sum_{i=1}^{n}x_{i}^{2})$$ for all vectors $x\in\mathbb{R}^n$ such that $0<x_i<1$ for all $i$? The main sticking point seems to be that the last term is not convex, so i can't just fix $\sum_i x_i $ and apply symmetry.

4082532287869Contemporary work in biology has lots of examples. With the wealth of sequencing data coming out of the machines at ever lower costs there is a huge need for new methods and models to analyze and understand the data.

Google is another good example. Better (or even acceptable) internet search was desired and it took a mathematical advance to move the technology forward.

42610610180593193772298071My sister had this saying involving a rolling donut. Or doughnut.294223868560919608156Fractional Leibniz formula22049241940871What is the problem with taking $z_n=2\pi n i$ and $p(x,y) = y-1$?16647361044039@Karl, There is a small issue with your argument I think. Why should a minimal associated prime containing $f$ be in both the maximal ideals? For example, let the maximal ideals be $(x,y-1)$ and $(x-1,y)$ in $k[x,y]$. Let $f=xy$. 1614171It is well known that the saturated ideal of the 10-dimensional spinor variety $S^{10}\subset\mathbb{P}^{15}$ is generated by 10 quadrics. Is it possible to get $S^{10}$ as scheme-theoretic intersection of less than 10 quadrics of $\mathbb{P}^{15}$ ?

Thanks.

926125264834@RazvanD: when fedja says "dimension" he means 3 dimensions as in 3D movies. He doesn't mean the actual side lengths of your box.16804351957476371103864218233847677416If you aren't bounding the diameter of $C$, then the space isn't compact, and there isn't a worst, since thinner and thinner rectangles do arbitrarily badly.I 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!1694646How bad can a tacnode be for a polynomially parametrized curve?There are natural situations where non-unit preserving homomorphisms are needed. For instance if $R$ is a unital ring and $e$ an idempotent, then $eRe$ is a unital ring with identity $e$ and the inclusion $eRe\to R$ is not unital. Nonetheless it plays a role in Morita theory and in Chapter 6 of Green's lecture notes on polynomial representations of $Gl_n$. It also plays a role in the construction of irreducible representations of finite monoids.

1159249($P(-x) = x^nP(1/x)$A big plus for W|A is that I can access it through my cell phone.DWhat is a subchain of a powerset?22890162193656142173@AlexandreEremenko For the sake of giving a definition (and only for it), we say that a new theory *has been useful* if it has been used to solve an existing problem that has been posed before the new theory and that recordedly has withstood at least one previous attempt not using the new theory. Please don’t ask me again to do something like this.1747948Wait, did you just tell Vaughan Jones about his own paper? MO has jumped the shark.8...also to Watts, I mean. 1338360625585515083%You have to be kind of careful. For the complex numbers, there is no problem as BCnrd wrote above. For a general field $K$, you are asking a basic question from the field of geometric algebra (there's an AMS GTM volume by Grove on the topic, and Dieudonne's books on classical groups).

If you take a nondegenerate quadratic form $q$, then we know:

- If $q$ is isotropic over $K$, then $Spin(q)(K)$ is "projectively simple" (i.e., the quotient by its finite center is simple)
- If $q$ is anisotropic, then $Spin(q)(K)$ can be far from simple. You can construct examples using valuations, as you suggest.

But you asked about $SO(q)(K)$. Then you use Galois cohomology (where I assume that $K$ has characteristic different from 2 for simplicity):

$1 \to \mu_2(K) \to Spin(q)(K) \to SO(q)(K) \to K^{\times}/K^{\times 2}$

where the last map is the spinor norm. You should think of the spinor norm as usually having a big image, so $SO(q)(K)$ will be pretty far from simple.

You asked specifically about $SO(q)$ where $q$ is a sum of squares. Then the spinor norm map has image products of sums of squares (typically not actually squares themselves), so you should expect the image of $Spin(q)(K)$ to be a normal subgroup in $SO(q)(K)$ of large index. (That is all very imprecise, but a precise answer depends on the arithmetic of $K$ and the dimension of $q$.) But now you can ask: *Is $Spin(q)(K)$ modulo its center a simple group?*

Here you have a strong advantage. If $q$ is isotropic over $K$ (i.e., you can write $0$ as a sum of a small enough number of nonzero squares), then you know from classical results that the answer is "yes".

If $q$ is not isotropic over $K$, then $Spin(q)$ is anisotropic (as an algebraic group) but it is split by the quadratic extension $K(\sqrt{-1})$. For such groups, under some hypotheses on $K$ (maybe as weak as characteristic zero), you know that the answer is again "yes" by Chernousov. See Theorem 9.7 on page 514 of the book "Algebra & Number Theory" by Platonov and Rapinchuk. You will have to inspect the proof (which starts on p.546) to see the precise hypotheses they need on $K$, or you can consult some of the original papers, where the proof is slightly different.

14228303295581765963$-\Delta u=1$ on the unit disk excluding a solid sector (i.e. a round cake with piece taken) with homogeneous Dirichlet boundary condition. There is an explicit characterization of regularity loss on polygonal domains in e.g. Grisvard's book. Maybe you can work out the case of e.g. the plane with a quadrant missing.

1908891762531178913236823610632541659497183262511672485837712523041707562430302931922Also, here is a feeling towards your problem. One way to have your conjecture be true is to have M' be much smaller than M, so that homomorphisms to M' do not have as much room to play. So consider an algebra M with small subalgebra lattice. M might be an n-valued Post algebra or something in an arithmetic variety in which M has no subalgebras. Then homomorphisms from powers to M might be essentially n-ary, but there may be no way to find M' to reduce the arity since there are no smaller candidates. Gerhard "Ask Me About System Design" Paseman, 2010.08.18I don't understand something. I thought, $y$ must be an element of ${\mathcal B}(L_2(F))$?2f1 hypergeometric function??? I wish I had the slightest idea of how to use it for anything I need. :-). Anyway, any time you see double summation versus single summation, the chances that it is just a telescopic sum in disguise are about 20%, so you can always try to bet a decent amount of your time on that. This case is no exception:

$$ \sum_{i+k=q}(-1)^i{n-p-1\choose i}{p\choose k}-\sum_{i+k=q-1}(-1)^i{n-p-1\choose i}{p\choose k} \\ \sum_{i+k=q}(-1)^i\left[{n-p-1\choose i}{p\choose k}+{n-p-1\choose i-1}{p\choose k}\right] \\ =\sum_{i+k=q}(-1)^i{n-p\choose i}{p\choose k} $$ (I just paired $(i,k)$ with $(i-1,k)$ with the usual convention that ${m\choose -1}=0$). Now add from $q=0$ to $q=\frac{n-1}2$ to get the identity you want with Sum I in your list on the right hand side.

4380791097701527556XVaught conjecture for uncountable languagesResistance across opposite vertices of n-dimensional cube with each edge at one ohm resistance is $$R_n=\sum_{k=0}^{n-1}\frac1{(n-k){n\choose k}}=\frac1n\sum_{k=0}^{n-1}\frac1{{n-1\choose k}}.$$ The proof is simple: we can identify nodes with equal voltages, and there are ${n\choose k}$ nodes with a distance $k$ to a given point.

From another hand $$R_n=\frac1{2^n}\sum_{k=1}^{n}\frac{2^k}{k},$$ because both sums satisfy the same recurrence relation $$R_n=\frac1n+\frac12R_{n-1}.$$

We can consider the sum $\sum_{k=1}^{n}\frac{2^k}{k}$ as a partial sum of $p$-adic logarithm $$\log_p(1+x)=-\sum_{k=1}^{\infty}\frac{(-x)^k}{k}$$ at the point $x=2$ (it is well defined for $p=2$). Using two formulas for $R_n$ we can get a simple application. We can find the value of $2$-adic logarithm at the point $-2$: $$-\log_2(-2)=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2^k}{k}= \lim_{n\to\infty}\frac{2^n}n\sum_{k=0}^{n-1}\frac1{{n-1\choose k}}=0.$$ It is not a surprise (see § 4.4.11 from Cohen (2007), Number theory, Volume I: Tools and Diophantine equations).

**First question:** why does $2$-adic logarithm arise in combinatorial problem?

**Second question:** do you know any more connections between combinatorial and $p$-adic objects?

The cycle has this property. For instance, the distance matrix for a 6-cycle is:

$A=\begin{bmatrix} 0 & 1 & 2 & 3 & 2 & 1 \\\\ 1 & 0 & 1 & 2 & 3 & 2 \\\\ 2 & 1 & 0 & 1 & 2 & 3 \\\\ 3 & 2 & 1 & 0 & 1 & 2 \\\\ 2 & 3 & 2 & 1 & 0 & 1 \\\\ 1 & 2 & 3 & 2 & 1 & 0 \\\\ \end{bmatrix}$

The question is: is the cycle the only graph with this property?

1379665830375And I think this conjecture being called "the second Hardy-Littlewood conjecture" - which means that if the first Hardy-Littlewood conjecture is true, the second one is false :-)18165274436431956071Maybe one way to explain this answer is to start by a simpler lower bound, $\pi/2$, which can be obtained by simply projecting centrally (=gnomonically) the unit cube to the (inscribed) sphere of radius $1/2$: this map is $1$-Lipschitz so we get a lower bound of $\pi/2$ on the cube from the sphere. The surprising fact that Anton shows is that by essentially patching together $2d$ azimuthal equidistant projections, one for each side of the cube, we can get a bound of $2$.185799219263349675977946277506@A potential resolution of $R/r$Has not been mentioned yet: James Milne's course notes http://www.jmilne.org/math/CourseNotes/index.html and his books http://www.jmilne.org/math/Books/index.html, especially the one on Arithmetic Duality Theorems.

@DerekHolt Please consider adding your Magma(?) construction of the group as an answer (and, if you want, explain your thought process a little bit). I'm sure the op would appreciate seeing this as well.4741846591318888411908872169490537783141657This is a follow-up to normal form for some finite groups, extending the small groups library.

Not being familiar with groups, I wonder whether it is possible to check efficiently whether a group (given as a permutation group) is isomorphic to a generalized symmetric group.

Initial computer experiments indicate that the parameter $m$ in $\mathbb Z_m\wr\mathfrak S_n$ might be twice the index of the derived subgroup in the group.

From a practical point of view, I am trying to do this with GAP.

02011-03-28T15:23:59.9835671519122602216507432694325326 77971764649The result that you mention in the first sentence is due to Riemann, and the map is not unique, unless normalized correctly.60640093352901737178@Braverman: I don't think this is obvious at all... since now you are asking that $A$ is not Cohen-Macaulay. I know some families of non-normal varieties are that regular in codimension 1, but don't have a good way to get their equations / Hilbert series (precisely because they are not normal!). They have explicit descriptions, and if it would be of any help to have more such examples (to see if they do not apply to your situation or whatever), you can see my paper http://arxiv.org/abs/1101.4604 (Lemma 9.4; definition of varieties in Section 8).You seem to assume that $H^1(X,\mathbb{Z})$ is torsion-free: is that so?1676704367202905699Let's call an Artin stack $X$ *concentrated* iff it is quasi-compact and quasi-separated (the latter usually being included in the definition of an Artin stack). The category of quasi-coherent sheaves $\mathrm{Qcoh}(X)$ is a locally presentable abelian Grothendieck category.

**Question.** Is $\mathrm{Qcoh}(X)$ locally finitely presentable?
In other words, is every quasi-coherent sheaf a directed colimit of quasi-coherent sheaves of finite presentation?

This is known if $X$ is a concentrated scheme (EGA I, 6.9.12). It is also known when $X$ is noetherian (Lurie's Tannaka duality, Lemma 3.9). Is anything known about the general case?

Added: In his work about noetherian approximation (Theorem A + Prop. 2.7), David Rydh has shown that this is true for concentrated Deligne-Mumford stacks.

16596362472297814906489991945857For the time being, the “fair amount of detail” I am referring to may be found in Proposition 3.18 and in Section 4.2 of [arXiv:1308.0796](http://arxiv.org/abs/1308.0796).15230936Existence of limit measure"The argument can be recollected from the old literature, but I do not know a coherent reference": Maybe you should give this to a masters student. The masters thesis (assuming it's well written) could then become the missing "coherent reference".To Hirzebruch surfaces, you can add $\mathrm{CP}^2$, its $k$-folds blow-ups, $1\leq k\leq 8$, and some irrational ruled surfaces.

Related to this question is the determination of the symplectic cone. This is now understood for rational $4$-manifolds, ruled $4$-manifolds and their blow-ups, and also for some elliptic fibrations.

There is a nice survey by Tian-Jun Li of the relations between symplectic and Kahler cones for $4$-manifolds (and complex surfaces). See arXiv:0805.2931.

2141393BVery insgightful answwe! thanks!Thank you. But what will happend if both $n$ and $p$ goes to $\infty$, thinking that $n$ goes to $\infty$ faster then $p$ does?103603915981911317505http://planetmath.org/encyclopedia/ScatteredSet.html (wiki seems to be missing it) @Qiaochu Yuan Thank you for the answer. What is $H^4(B^2 \pi_2, \pi_3)$ when $\pi_3=U(1)$ and $\pi_2=Z_2$?629303I'm going out on a limb and say that the paper does not contain an actual proof. I don't know whether the author plans to publish a follow up with more details or not, but for now I still consider the problem open.2262821569277328355I think this is just the Davenport-Hasse Theorem, which Wikipedia calls the Hasse-Davenport lifting relation.

13270891293450109904311465201888083Please check http://tea.mathoverflow.net/discussion/183/comments-that-vanish/ if you're concerned about the recent disappearing comments here.59822713722462249111315410933@Lucia I guess f(s) may have certain speciality that makes the answer so simple. However, g(s) looks very similar, yet its residues are complicated, which also involves Euler–Mascheroni constant and also higher derivatives of digamma function.10216941878472@DouglasZare: Excellent question! I have little intuition here...1901751Paul B419157577345@DenisNardin: The filtration is by the kernels of $K^0 BG \to K^0 (BG)_n$ for $n$-skeleta $(BG)_n$; these kernels contain $\widehat I(G)^n$ and induce the same topology on $\widehat R(G)$. My inclination is that this means the associated graded shouldn't have torsion, but I don't have anything stronger than that.Yes but the complete toric varieties that I refereed to are not smooth.One can get an asymptotic formula for problems like this by using the saddle point method. This is classical, but one reference may be these course notes: http://math.berkeley.edu/~moorxu/oldsite/notes/155/155main.pdf , see page 44 and following. That deals with the partition problem with generating function $\Gamma(s)\zeta(s)\zeta(s+1)$ (see page 47); you need to make similar calculations with $\Gamma(s)\zeta(s)^2\zeta(s+1)$. There won't be the modular forms miracle as with partitions, but one doesn't need that for an asymptotic.5057002410252142098 @ Liviu Nicolaescu: I have to be sincere, the Alexander polynomial is the link invariant that I know less. I looked for some information and I read that it is an invariant of oriented links (see https://books.google.it/books?id=keElBQAAQBAJ&pg=PA230&lpg=PA230&dq=alexander+polynomial+change+of+orientation&source=bl&ots=2hl6PoutxA&sig=la95RcNCRORg1QZVSyE20iWSZ-c&hl=it&sa=X&ved=0CDwQ6AEwBWoVChMI4Y-C67ziyAIVB2kUCh3xeQnM#v=onepage&q=alexander%20polynomial%20change%20of%20orientation&f=false ,Murasugi, The knot theory and its applications, p. 230). I hope I understood everything well.xCompelling evidence that two basepoints are better than one215399211576702160853171661117341518728681756543I'm voting to close this question as off-topic because it should be on https://tex.stackexchange.com/ and not here2027100725539How do you sheafify presheaves with values in $A$ if $A$ doesn't have small filtered colimits? (A small abelian category that has small filtered colimits is necessarily trivial.)Yes, I checked for all primes less than 500 and connectedness becomes much easier for larger primes since all coefficients are smaller than $\sqrt p$. If the graph is not connected, then there exists a partition of $\lbrace 1,\dots,\lfloor \sqrt{p}\rfloor\rbrace$ into two disjoint subsets $A\cup B$ with one connected component having only coefficients in $A$ and all other components only coefficients in $B$. Since the number of vertices is much larger than $\sqrt{p}$ this is very unlikely. 1156061I wonder what you mean for the ODEs, because we know we can't stay in the algebraic category; but then there *is* all that differential Galois theory, that you get by expanding the language and repeating the thm => def'n ploy.You may enjoy googling around for "synthetic differential geometry".4Do you mean any or every?2154816@ManfredWeis: Thanks, but I don't see how to actually compute the optimal arrangement in 3D for arbitrary $n$, using the stereographic projection viewpoint. Perhaps it is like packing points on a sphere, with no clear theory except for special values of $n$.616429657697I am working on the chamber homology for $SL(2,F)$, and stuck at some basic stuff on D.S. reps of $SL(2,F)$.

Let $ I=\left( \begin{array}{cc} \mathcal{O}_{F} & \mathcal{O}_{F} \\ \varpi_{\mathbb{F}}\mathcal{O}_{F} & \mathcal{O}_{F}\\ \end{array} \right)\cap SL(2, F)$. Now, let $ w_{0}= \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right)$ and $ w_{1}= \left( \begin{array}{cc} 0 & -\varpi^{-1}_{F} \\ \varpi_{F} & 0 \\ \end{array} \right)$, then $J_{0}=I \cup Iw_{0}I$ and $J_{1}= I \cup Iw_{1}I$ are the two maximal compact subgroups of $SL(2,F)$ where $\varpi_{\mathbb{F}}$ is the uniformizer.

Just wondering if anybody knows how can I induce a cuspidal reps(D.S.) from a charachter belong to $J_{0}$ or/and $J_{1}$?

8920167197263669991093477I am trying to understand the Dedekind and Cauchy real objects in topoi concretely by looking at presheaf categories over small (tiny?) categories. For example, consider the topos which is the category of spans of sets. That is, this is the presheaf category over $\nearrow \nwarrow$. How can we construct the Dedekind and Cauchy objects concretely in this category?

6967854249853061410078391717547pWhat distributions fit this strange experimental model?21991704707055846372076186199605820102251391371572314174723319006567859That is true, and you can prove it in a very explicit fashion by taking a locally free presentation of $E_{\overline{k}}$ by locally free sheaves of the form $\bigoplus \mathcal{O}(-d_i)$, resp. $\bigoplus \mathcal{O}(-e_j)$. The "matrix" of this presentation is a finite sequence of maps $\mathcal{O}(-e_j)\to \mathcal{O}(-d_i)$. Define $L$ to be the subextension of $\overline{k}/k$ generated by all coefficients of those maps with respect to the standard monomial basis. A more thorough answer follows from EGA $IV_3$, Sections 8, 11.2, and 11.3.3495111244153Thank you! Did you try with $3\times 3$ matrices? How did you find this example?6A New Analytic Inequality 510394)Perhaps you are looking for an exposition of von Staudt's theory of complex elements in synthetic projective geometry.

A bit of history: Poncelet, in his work of the early 19th century, which launched the explosion of projective geometry, reasoned synthetically, but somewhat mysteriously, with imaginary points. At mid-century, von Staudt attempted to create a purely synthetic (and entirely non-metrical) foundation of projective geometry, which would include the imaginary objects that Poncelet and others found convenient to reason with. The new synthetic theory would, for example, make precise the sense in which a line and a conic are either tangent or meet in two distinct points, which may be a pair of conjugate imaginary points. The theory of von Staudt was successful as far as it went, but synthetic methods in general fell out of fashion as the analytic techniques of invariant theory and the study of higher-order curves and surfaces became more popular. One can, however, find discussions of von Staudt's theory in old texts written in the synthetic spirit.

For an overview of the subject in English, one might look at Coolidge's *A History of Geometrical Methods*. The part of that long book addressing this chapter in the history of projective geometry is essentially the same as Coolidge's lecture published as "The Rise and Fall of Projective Geometry" in *The American Mathematical Monthly*, Vol. 41, No. 4 (Apr., 1934), pp. 217-228. (That lecture contains a few comments on Kötter specifically as well, which you may find amusing, if you are interested in his work.) A search in Google books will yield many of the old textbooks discussing von Staudt's approach. One might read, for example, Chapter XX of *Projective Geometry* by G. B. Mathews. If you read German, you can look at von Staudt's original treatment in his *Beiträge zur Geometrie der Lage*, various editions of which are freely available online. All of these treatments are missing a satisfying treatment of completeness and order ("sense," which the old authors tried to use to distinguish the two complex points corresponding to an elliptic involution) for real projective geometry. To fill in the missing details, one can see volume 2 of Veblen and Young's comprehensive *Projective Geometry*, which, I believe, was written by Veblen alone. (Some hints on the theory of order--based the four-term separation predicate rather than the three-term betweenness predicate of formalized Euclidean geometry--can also be found in more recent texts such as Coxeter's *Non-Euclidean Geometry*.)

The following example comes from Neeman's ``Some new axioms\dots" (J. Algebra, 1991). A triangle is contractible if it is a direct sum of (translations of) triangles of the form $0\rightarrow X\rightarrow X \rightarrow 0$. Contractible triangles are exact, but exact triangles are seldom contractible. Assume that $X\rightarrow Y\rightarrow Z\rightarrow\Sigma X$ is not a contractible exact triangle. Then $$\begin{array}{ccccccc} X&\stackrel{f}\rightarrow&Y&\stackrel{i}\rightarrow&Z&\stackrel{q}\rightarrow&\Sigma X\\ {\scriptstyle 0}\downarrow&&{\scriptstyle 0}\downarrow&&{\scriptstyle q}\downarrow&&{\scriptstyle 0}\downarrow\\ Y&\stackrel{i}\rightarrow&Z&\stackrel{q}\rightarrow&\Sigma X&\stackrel{-\Sigma f}\rightarrow &\Sigma Y \end{array}$$ is a morphism of triangles and $q$ cannot be obtained in the way you described. Otherwise, $h$ would be simply a morphism $h\colon \Sigma X\rightarrow Z$ and, up to sign, $qhq=q$ necessarily. This leads to contractibility of the initial exact triangle.

This argument is very general, so we rather see it `in action' in $D(\mathbb Z)$. We can take the exact triangle associated to the short exact sequence $\mathbb Z/2\hookrightarrow \mathbb Z/4\twoheadrightarrow \mathbb Z/2$

$$\begin{array}{ccccccc} \mathbb Z/2&\stackrel{f}\rightarrow&\mathbb Z/4&\stackrel{i}\rightarrow&\mathbb Z/2&\stackrel{q}\rightarrow&\Sigma \mathbb Z/2\\ {\scriptstyle 0}\downarrow&&{\scriptstyle 0}\downarrow&&{\scriptstyle q}\downarrow&&{\scriptstyle 0}\downarrow\\ \mathbb Z/4&\stackrel{i}\rightarrow&\mathbb Z/2&\stackrel{q}\rightarrow&\Sigma \mathbb Z/2&\stackrel{-\Sigma f}\rightarrow &\Sigma \mathbb Z/4 \end{array}$$ Here $q\in \operatorname{Ext}_{\mathbb Z}^1(\mathbb Z/2,\mathbb Z/2)\cong \mathbb Z/2$ is the generator, but there are not degree $-1$ maps $\mathbb Z/2\rightarrow \mathbb Z/2$, i.e. the only map $\Sigma\mathbb Z/2\rightarrow\mathbb Z/2$ is the trivial map. Therefore, the standard completion of the first commutative square would be the trivial one, not the former.

The last explicit counterexample turned out to be more obvious than I remembered.

17501892525075 869620475405117166From a more topological viewpoint, I think it is worth reading the famous paper of Dennis Sullivan, *Infinitesimal computations in topology* (Publications Mathématiques de l'IHES, 1977), in which he builds algebraic models of manifolds from their differential forms. He explains how to construct recursively such models and gives several examples. He obtains from this construction deep results about homotopy types and diffeomorphism types of manifolds.

I haven't seen the following proof mentioned, which I learned from Hai Dang at Mississippi State.

Suppose the reals are countable, and let $a_1, a_2, a_3, \dots$ be an enumeration. For each $j$, let $I_j$ be an interval centered at $a_j$ and having length $1 / 2^j$.

Since the sequence $\{a_j\}$ enumerates the reals, it follows that $\bigcup_{j=1}^\infty I_j = \mathbb{R}$. But since the sum of the lengths of the $I_j$ is the geometric series $$ \sum_{j=1}^\infty \frac{1}{2^j} = 1,$$ this is nonsense.

(Pretty much the same proof yields that $[0,2]$ is uncountable.)

There's probably a relation with the outer measure proof previously posted, but this one seems more concrete.

*Update:* Two extended comments:

1) As mentioned below in my comment below, proving that [0,2] can't be covered by open intervals of length summing to 1 is easily done with compactness. I didn't quickly see how to prove it without compactness. And of course compactness of a closed interval uses the Nested Interval Theorem, as the original poster was trying to avoid.

2) I presented a proof along these lines in an Analysis I class. I liked how it came out a great deal, because it gave me a good reason to show students a typical application of compactness. Compactness is abstract and difficult for beginning students to grasp, and I usually find it difficult to find good applications. Students (or at least some of them) seemed to like this one a fair bit.

83421870318117690395In my paper, *A variant of Mathias forcing that preserves $\mathsf{ACA}_0$* [Archive for Mathematical Logic 51 (2012), 751–780; arXiv:1110.6559, doi:10.1007/s00153-012-0297-4], I show that $F_\sigma$-Mathias forcing preserves $\mathsf{ACA}_0$. When restricted to $\omega$-models the main preservation theorem gives:

**Theorem.** If $\mathfrak{X}$ is an arithmetically closed Turing ideal and $G$ is $F_\sigma$-Mathias generic over $\mathfrak{X}$, then the Turing ideal generated by $\mathfrak{X}\cup\{G\}$ is also arithmetically closed.

The meaning of "$F_\sigma$-Mathias generic over $\mathfrak{X}$" is that the forcing consists $F_\sigma$-Mathias conditions coded in $\mathfrak{X}$ and that the generic $G$ should meet all of the dense sets that are definable over $\mathfrak{X}$ in the language of second-order arithmetic. (As usual, there is some flexibility in the second requirement.)

To derive the above from Theorem 4.3 of my paper, observe that the names I used are a way of formalizing Turing functionals with built-in oracles from $\mathfrak{X}$. Therefore, the evaluation of a $G$-total name is a total computable function with respect to some oracle $G \oplus X$ where $X \in \mathfrak{X}$, and all such functions can be represented by a $G$-total name in this way. It follows that the evaluation of all $G$-total names, which is how I define the generic extension, is simply the Turing ideal generated by $\mathfrak{X}\cup\{G\}$.

The result you want then follows from the following:

**Lemma.** For every partial name $F$ coded in $\mathfrak{X}$ the collection of $\mathfrak{X}$-coded conditions $(a,A,\mu)$ such that, for some $x$, $(a,A,\mu)$ forces that $F(x) \neq C(x)$ is dense.

Indeed, we may assume that $(a,A,\mu)$ forces that $F$ is total, otherwise some extension forces that $F(x)$ is undefined for some $x$. Then, if $(a,A,\mu)$ does not decide all values of $F$, then some extension will decide a value $F(x)$ to be different from $C(x)$. Finally, if $(a,A,\mu)$ decides all values of $F$ then the resulting evaluation of $F$ is computable from $F$ and $(a,A,\mu)$. Therefore, $F \neq C$ since $C \notin \mathfrak{X}$.

Note that the above is really a meta-lemma since $F(x) \neq C(x)$ is expressible in $\mathfrak{X}$ for fixed input $x$ but not for all inputs at once. Nevertheless, the lemma gives a family of open dense sets for $F_\sigma$-Mathias forcing over $\mathfrak{X}$ and any generic $G$ that meets all of these dense sets (and all those that are definable from $\mathfrak{X}$) will be as required by the theorem: the arithmetic closure of $\mathfrak{X}\cup\{G\}$ will not contain $C$.

Note that the same result cannot be achieved with ordinary Mathias forcing since Blass has shown that every Mathias generic set computes all hyperarithmetic reals.

An earlier version of this answer outlined a tweaking of the cone avoiding result at the end of my paper to achieve the same goal. While that solution did work, it introduced some unnecessary complexity since the point of that result is that one can still arrange that the generic $G$ does not compute $C$ even if $C \in \mathfrak{X}$, provided that $C$ is not computable.

2171555Does "defined over $Q$" means "definable over Q"? when you say algebraic subgroups, do you mean that $G$ is an $Q$-subgroup? an $R$-subgroup? do you mean that $H$ is an $R$-subgroup?1143239I as a Chemist by education find it extremely enlightening to read Mathematicians takes on this subject. Thank you very much!22410973093169598022803221614300@Craig Westerland : Of course, the real interest is in infinite fundamental groups. And things can definitely be more complicated; for instance, the groups need not be finitely generated (eg $\pi_2(S^1 \vee S^2)$).@ Anton Petrunin: Thanks a lot. Indeed i encountered this problem for sublevel sets of a convex function, say $f$. If a sublevel $C$ corresponds to a nonminimal value $a$ i see that $K_pC$ is a sublevel of the differential $df_p$. But this is wrong if $a$ is minimal. Any hint what function to consider here? ps. Anyhow the general question might be of interest21863921634233I doubt that there is any site where this question would be appropriate. 1612143An entertaining topological party trick that I have seen performed is to turn your pants inside-out while having your feet tied together by a piece of string. For a demonstration, check out this video.

I have heard some testimonial evidence that it is also possible to turn your pants *backwards*, again with the constraint of having your feet tied together. This second claim seems pretty dubious to me.

**Question.** Is it indeed possible to turn your pants backwards, while having your feet tied together by a piece of string? A set of instructions or a video demonstration would suffice for a yes answer. A precise mathematical formulation of the problem together with a proof of impossibility would suffice for a no answer.

Let $w \in S^1$, and let $\pi^{-1}(w)=\{z_1, z_2 \}$.

Let $U$ be a sufficiently small open set containing $w$, so that $\pi^{-1}(U)$ is the *disjoint* union of two open sets $V_1$ and $V_2$, with $V_i$ containing $Z_i$.

Then, by definition, the space of sections of $\pi_* \mathbb{Z}$ over $U$ is the direct sum of the spaces of sections of $\mathbb{Z}$ over $V_1$ and $V_2$. So we are reduced to compute the monodromy action of the fundamental group of $S^1$ on the fibre $\pi^{-1}(w)$.

Let us fix the point $1$ as a base point for $\pi_1(S^1) \cong \mathbb{Z}$. When we consider the action of the generator $1 \in \pi_1(S^1)$ on $\pi^{-1}(w)$, one easily checks that $z_1 \to z_2$ and $z_2 \to z_1$.

Hence the monodromy matrix is $$\left(\begin{matrix}0 & 1 \cr 1 & 0 \end{matrix}\right).$$

119931922195192162202For n=3, one turn takes you 2 of one color and 1 of another. To get to a single color you need to pick one of the 2 (prob=2/3), then pick the odd one (prob=1/2). So the expected number of turns is 1+3=4

2237664A generalization of Deligne's construction may be found here: http://arxiv.org/abs/0911.4979. To recoved Deligne's construction one simply takes the perspective that any abelian category is a module category over $Vec$. Here we define $V\otimes N$, for any finite dimensional vector space $V$ and object $N$ to be the unique object representing the functor $N\mapsto Hom(V, Hom(N, N))$ (really internal hom).

In general let $\mathcal{M}, \mathcal{N}$ be right, left $\mathcal{C}$-module categories for $\mathcal{C}$ any tensor category. Then the $relative$ $tensor$ $product $ $\mathcal{M}\boxtimes_{\mathcal{C}}\mathcal{N}$ is defined as the unique (up to a unique equivalence) universal object for right exact in each variable $\mathcal{C}$-balanced bifunctors from the cartesian product $\mathcal{M}\times\mathcal{N}$. As such it follows that $\mathcal{M}\boxtimes_{Vec}\mathcal{N}=\mathcal{M}\boxtimes\mathcal{N}$ where $\mathcal{M}\boxtimes\mathcal{N}$ denotes the product of abelian categories defined in ``Categories Tannakiennes".

1094377leopragi200662913908173185601820115z@Henry and Mark: it is the same case as the one I mentioned.491185:@KConrad: This is an answer.221493022693631149712What about when $G$ is Lie group and we take $G/B$ as flag variety?83721649165119614322230261Good point. $Ded(\kappa)The answer to the second question is also yes, because the Dirichlet class number formula implies that $\zeta_{F}(s)$ also has a simple pole at $s = 1$.

1781653829091317704I wrote a note some time ago extending Chebyshev's method to number fields which gives an elementary and simple proof that the number of split primes is at least $x^{1/d}/\log x$. http://jtnb.cedram.org/jtnb-bin/fitem?id=JTNB_2000__12_1_81_01473268203606392573By the way, maybe your idea could shed a new light on a question of mine, namely http://mathoverflow.net/questions/154373/differential-galois-number-theoryhIt follows from GRH. Not known on its own.

22234472245842 Pedagogically speaking, I have had some success teaching this kind of Zariski topology -- at least on affine n-space and its subsets -- to non-algebraic geometers. It is often the case that simply having "Zariski closure" in your vocabulary makes apparently nontrivial theorems transparent. For instance, the fact that for $n \times n$ matrices $A$ and $B$ over a field, $AB$ and $BA$ have the same char. poly. In this regard, finite fields *are* a bit of a pain -- no point in taking the closure in a discrete top.! -- but this can often be circumvented by base change to the algebraic closure.Ialready knew there are no solutions for n=2, but thanks anyway.LA characterization of Hilbert spaces?@Qiaochu: I agree it is not too surprising from the topology side, but I do not know how to see it from the analysis side or make it rigorous enough so that I can compute more complicated spaces/operators.17821155605873800001184200Let $A$ be a $n \times n$ matrix all of whose entries has modulus 1.

Suppose the matrix $A$ is singular.

We will assume without loss of generality that all the entries in the first row and the first column of the matrix are 1.

Observe when $n=2$ the matrix $A$ can be then singular if and only if $a_{2,2}=1$ as well.

A slightly less trivial observation is that the same thing happens when $n=3$, that is the matrix $A$ is singular if and only if two of the rows or columns are identical.

\begin{equation} \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \alpha_{2,2} & \alpha_{2,3} \\ 1 & \alpha_{3,2} & \alpha_{3,3} \\ \end{array}\right| = 0 \end{equation}

So the matrix $A$ is singular iff $(\alpha_{2,2}-1)(\alpha_{3,3}-1)=(\alpha_{2,3}-1)(\alpha_{3,2}-1)$.

Let us assume without loss of generality that $\alpha_{2,2} \neq 1$ and $\alpha_{3,2} \neq 1$.

Consider the circle $C_1(t)= (\alpha_{2,2}-1) (e^{2 \pi i t}-1) $ and $C_2(t)=(\alpha_{2,3}-1) (e^{2 \pi i t}-1), t\in [0,1]$.

Since, the two circles either are identical and in that case $\alpha_{i,2}=\alpha_{i,3}$ that is the second and third columns are identical, or else as two distinct circles can intersect in at most two points we get similarly two of the rows or columns are identical.

Now, probably it is too much to expect the same result for all $n$.

But my requirement is only for $n=4$, is it true that a similar result holds for $n=4$ ?

Edit: I forgot to mention that I am interested in the case when the matrix is singular > > and none of its sub matrices are singular. (thanks @ Gerry Myerson for pointing it out)

Thankyou,

1326259110436222000441963621774999Perhaps one can construct $H$ as a stack as follows. Define a presheaf of groupoids $H'$ by letting $H'(U)$, for a scheme $U$ over $K$, to consist of tuples $(Y,\pi, f)$ where $\pi:Y \to X$ is a finite etale morphism and $f:U \to Y$ is any morphism. Morphisms from $(Y,\pi,f)$ to $(Y', \pi',f')$ are given by isomorphisms $g:Y \to Y'$ so that $gf = f'$. The main point is that we ignore $\pi$, $\pi'$ in defining morphisms. Let $H$ be the stackification of $H'$. There is a tautological map $X \to H$ which should presumably give an atlas.222121175465870716110497531138258I've read (somewhere) that types in typed first-order logic is just 'syntactic sugar'.

But, surely there is more to than that?

For example: the upper-bound property of the reals can't be expressed in first order logic, but it can be in second. Now with Henkin semantics this is the same as typed first-order logic, so although the property can't be expressed in first-order logic, it can be if it is typed.

Are there useful ideas/theorems that make essential use of types in typed first-order logic?

By putting $y=w\exp(\int\cos(2\pi t)/2dt)$ you kill the first derivative term in the second order linear equation, and obtain the equation of the form $w^{\prime\prime}+Qw=0$, where $Q$ is a trigonometric polynomial. This is called Hill's equation and your problem is an eigenvalue problem for it. Such eigenvalue problems were studied a lot, and I recommend the book of Whittaker-Watson. There is also a book on Hill's equation, by Magnus and Winkler.

8597582224211Let $[n]=\{1,...,n\}$ and $[\hat n]=\{\hat 1,...,\hat n\}$. Realize the hyperoctahedral group $H_n$ as the centralizer of the permutation $(1\hat 1)\cdots (n \hat n)$. It has $2^n n!$ elements.

Let $\pi$ be the special permutation $\pi=(12\cdots n)(\hat 1)\cdots(\hat n)$, i.e. the elements of $[n]$ are arranged in a cycle while the elements of $[\hat n]$ are all fixed points.

The question I would like to answer is this: How many elements $h\in H_n$ are there such that the product $h\pi$ has cycle type $\mu\vdash 2n$?

For example, if $\mu=1^{2n}$ then $h$ would have to be $\pi^{-1}$; since $\pi\notin H_n$, the solution is zero.

Numerics suggest that for $\mu=(2n)$ the solution is $2^{n-1}(n-1)!$

8336321358167Take the conjugacy class of your (not necessarily semisimple) matrix inside $GL(n, \mathbb{C})$, and take its Zariski closure (or its closure in the usual topology). This contains a unique semisimple conjugacy class. 1483043104882919042888Ways to prove an inequality873210It is consistent without choice (in Solovay's model) that this is the case, although that's not very helpful here I suppose. :-)348225You can turn that series into a theta function with little effort: http://en.wikipedia.org/wiki/Theta_function217015517853151569238 @Y Macdisi: Let $V$ be the $2$-dimensional rep of $\mathrm{SU}(2)$ and $W$ be the $3$-dimensional rep of $\mathrm{SU}(3)$. Then $U = V\otimes W$ has dimension $6$ and is irreducible as a rep of the product $G = \mathrm{SU}(2)\times \mathrm{SU}(3)$. Since $G$ is compact and since the top exterior power of $U = V\otimes W$ is clearly trivial as a $G$-rep, there is some $G$-invariant, definite Hermitian form on $U$. Since $G$ lies in the stabilizer of this Hermitian form and complex volume form, it follows that $G$ lies in a copy of $\mathrm{SU}(6)$ acting on $U$, and this action must be irred.11119034878562130719162864110148481448131323422533671233637Software Engineer by Profession. Involved in development of Java EE web based systems for various commercial applications.

Let $\sigma :\mathbb{N}\rightarrow\mathbb{R}$ an injective sequence of real numbers.

There exists an infinite set $A=$ { ${a_{1},a_2,\ldots ,a_n,\}\ldots$ } $ \subset{N}$ such that

i) $\sigma_{|A}$ is monotone

ii) $a_n=O(n^2)$ ?

3613620669692119219843328It's possible to converge to a multiplicity 2 plane which is perpendicular to P though.6@marc: no offence intended1248674329301287162Bourbaki's corollary even says that $B$ is a projective $A$-module. Precisely, if $A \subseteq B$, $A$ is regular, $B$ is a noetherian integral domain which is finitely generated as an $A$-module, then: $B$ is Macaulay iff $B$ is a projective $A$-module.13305941826427Brilliant. Googling didn't give me a generalized equation. Thank you MO and thank you Kovalev, Budney and DeVito.2172744209812318773531514556293564103613138399775521912427095414012161851139425@FedericoPoloni That's well said. In view of the answers and the comments a migration, this thread would probably appear a bit odd at academia.sx. I retracted my close vote…5205802270967655388281383654357$SU(2,\mathbf{R})$ is the $\mathbf{R}$-points of the special unitary group associated to $\mathbf{C}/\mathbf{R}$ (since the latter is an algebraic group over $\mathbf{R}$). This group is the standard group of two-by-two unitary matrices with complex entries and determinant 1 that you might see in a first course on linear algebra. A paper you can try looking at would be Arthur's 2002 paper "A Note on the Automorphic Langlands Group" where he gives a conjectural construction of a *global* Langlands group which could help you understand why the local group is what it is.What is known about unstable bundles on curves ? What is "maximally unstable bundle" and what means it corresponds to Schwarzian differential equation ?

Consider the P^1 with pairs of points glued p1=q1, p2=q2, ..., what can be said for such curves ?

Let me mention that for such curves bundles can be described just by the set of matrices M1, M2, .... which correspond to "gluing the fibers". This can be made precise algebraically as follows: functions on "glued curve" are function on P^1 which satisfy conditions f(p1)=f(q1), f(p2)=f(q2),... then modules can be described as vector functions on P1, such that s(p1)=M1 s(q1) , s(p2)=M2 s(q2) ,... clearly this is a module over the algebra above. It has been further studied in Serre's "Algebraic groups and class fields". Let me also mention our papers: http://arxiv.org/abs/hep-th/0303069 Hitchin system on singular curves I, http://arxiv.org/abs/hep-th/0309059 Hitchin systems on singular curves II. Gluing subschemes

PS Questions are inspired by discussion with Misha Kapovich after his answer on the following MO question What is the number field analogue of the Narasimhan-Seshadri theorem ?

15302722146232495102169417872671922740578262712296151For $G$ 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.pdfIs there a number k such that every natural number can be written as $\sum_{i=1}^k \binom{a_i}{3}$ for some natural numbers $a_i$'s?

BoppreH97181BMaybe add the "k3-surfaces" tag?Dear darij: The link with subspaces of a vector space seems not so obvious (though intuitively appealing) since in contrast with group-index there are examples (even finitely generated in char. 0) of fields $K$ and (necessarily proper) subfields $k$ and $k'$ such that $[K:k]$ and $[K:k']$ are finite but $[K:k \cap k']$ is infinite (hint: there are infinite groups generated by a pair of finite subgroups). I haven't given the matter any real thought, but anyway it may require an idea (perhaps tiny) to make the argument work. 617528https://lh5.googleusercontent.com/-5bM8iyaEtp8/AAAAAAAAAAI/AAAAAAAAAA4/cLW1HR9o8xQ/photo.jpg1815473528387154442Can we force with Fraisse filters to solve Vaught's conjecture?15046224444991805245Hi Willie, thanks for your reply and I agree with your idea on heuristic approach. I converted the function by applying a smoothing function on it. Then, the curve of the function becomes more smooth such that in most case, it is convex. Even in the case that it has multiple local minima, they are quite close to the global minimum, which is good enough to my problem.The Riemannian manifolds in the answer are tacitly assumed to be connected.@Sándor: Thanks. Hope the edited answer is more clear on the isomorphism $O_Y\simeq \pi_\star O_X$ (under rather weak hypothesis) and the equality $h^0=1$ which holds under OP's hypothesis. Of course $h^1(O_F)=0$ is somehow a strong condition. On the other hands, I belieave that if $X,Y$ are both smooth, then $h^0(O_F)=1$ holds (over $\mathbb C$). Note that by Bertini, one can reduce to the case $\dim Y=1$. Then this is true in relative dimension 1 case. 140243817323916It's certainly $O(x^2)$ :)1335234640922379114The last paragraph of this answer might be relevant: http://mathoverflow.net/a/12480/100. Are there supposed to be equivalences in $\mathbb{KK}$ that are not isomorphisms?hWhen is/isn't the monoidal unit compact projective?9034451525708166817612026161822780Is every convex subset of a Borel-linearly ordered space measurable?190104218807881029279599597296014#Can anybody advise mathematical journals that publish research announcements? (I mean little papers without proofs.)

It sometimes happens that a proof is so long that it takes years to review and few journals are willing to accept it. I think in this case it would be useful (apart from posting your paper in arXiv) to announce the result in a short communication (this is how this problem is resolved in Russia, but I am asking about the rest of the world).

I am working in Functional analysis and in Geometry.

EDIT. More generally, I am curious,

how do people solve the problem of long texts?

Suppose you write a long text, where the main results are just several propositions, say, 5 theorems, and the rest are various technical lemmas, many of them, say, 100. Your text is devoted to the explanation of one idea, those 5 theorems appear only in the end, and it is impossible to separate them so that some of them could be proved in the middle of the text. As a corollary, it is impossible to divide your text into several papers, on 40-50 pages, so that they could be sent to usual journals.

You have to send this long paper to a journal, that publishes long texts, they will spend some time on finding a reviwer, he will check everything in your text, after that they will put your paper into the queue, and this is a long process, you understand... It can take 2-3-5 years before your paper will be published.

Of course, it will be difficult to explain to your employer, what you were doing the prevoius 5 years, when you were writing this paper.

So how do people resolve this problem? In Russia there is a possibility to write several little papers, without proofs, but with the formulations of the main results (those 5 theorems), then send them to some journals (together with that long text with accurate proofs), then the reviewer checks everything, agrees, they publish these little papers (this is much quicker, since these papers are little, there is no need to put them into a long queue), and everything is OK. As far as I understand, in France the situation is more or less similar. What about the rest of the world?

525823Yes, thank you for the comment. For $A$ f.g we know how $R$ looks like.For your second point, you can ask Mathematica, which tells me that the integral is $$\log (u) \left(\frac{u \log (u)}{u+1}-2 \log (u+1)\right)-2 \text{Li}_2(-u).$$ The integral in your second displayed equation is, according to it, $$k x (-(1/(1 + x^k)) + {}\_2F\_1(1/k, 1, 1 + 1/k; -x^k)).$$

1786230847829A nice review is [Kelly and Shmitt](https://arxiv.org/abs/math/0501383), see Prop 3.6. They refer back to [Kelly's book](http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html) Thm 5.35, but in the unenriched case I'm not sure what an original reference would be. For particular classes of colimits, like all $\kappa$-small colimits or all $\kappa$-filtered colimits, this might be in SGA 4...If there is a problem with the "continuity argument" in the first line, an alternate is to combine items with the same weight into a single item with the sum of the weights. This just reduces the dimension of the problem so it seems harmless.4750381814658913621267059942971414237- Missoula .NET Users Group Founder
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A degree $k$ map $S^n\to S^n$ induces multiplication by $k$ on all the homotopy groups $\pi_m(S^n)$.

(Not sure if this is a common error, but I believed it implicitly for a while and it confused me about some things. If you unravel what degree $k$ means and what multiplication by $k$ in $\pi_m$ means, there's no reason at all to expect this to be true, and indeed it is false in general. It *is* true in the stable range, since $S^n$ looks like $\Omega S^{n+1}$ in the stable range, "degree k" can be defined in terms of the H-space structure on $\Omega S^{n+1}$, and an Eckmann-Hilton argument applies.)

You might also check out the GNU Scientific Library (GSL) at http://www.gnu.org/software/gsl/, for which there are bindings and wrappers for other languages. Or you could forgo OOP and write your programs in C, which does at least have the advantage of being extremely fast and efficient. And I believe you can compile GSL so it is optimized for your system. I believe there is (or was) a Haskell wrapper for GSL, but I can't find it.

Most high level systems (Matlab, Octave, Sage etc) allow you to write routines in C and link them in, thus getting the best of both worlds: the speed of C and the ease of use of a higher level computing environment. I guess it all comes down to what sort of algorithms you are using: big slow ones (such as Monte Carlo routines with 10^20 random numbers) or something quick and straightforward, such as quadrature.

Over the years I've used Maple, Matlab, MuPAD, Sage, Octave, C, Java, and others, and I've written (or at least blogged) about them all. I can't prescribe one over the other - you have to decide for yourself which best suits your usage, expertise, and requirements.

481702583659838031424818106509717344961354489If you have a new question, please ask it by clicking the [Ask Question](http://mathoverflow.net/questions/ask) button. Include a link to this question if it helps provide context.1449217This is closely related to the [homotopy hypothesis](http://ncatlab.org/nlab/show/homotopy+hypothesis), and some definitions of higher groupoids make it essentially a tautology. Of course, this is not much use for computations unless you have a definition of higher groupoid from which you can actually compute the homotopy groups of a colimit.1669087531507139307910981804568581633583885729472472170666%Suppose that $V, W$ are vector spaces and $f, g : V \to W$ are two parallel morphisms such that $f \le g$, for some preorder $\le$ satisfying your conditions.

Claim:$f$ is a scalar multiple of $g$.

*Proof.* The first condition implies that if $v : 1 \to V$ is any vector in $V$ (here $1$ denotes the $1$-dimensional vector space) then $f \circ v \le g \circ v$. The second condition implies that there is a monomorphism from the image of $f \circ v$ to the image of $g \circ v$ (**Edit:** here I interpret the condition to mean that the monomorphism must be compatible with the inclusion into $W$, but as Eric Wofsey's answer shows, this extra assumption is unnecessary). This is equivalent to the condition that for all $v \in V$ there is a scalar $\lambda(v)$, possibly zero, such that

$$f(v) = \lambda(v) g(v).$$

Observe that

$$f(v + v') = \lambda(v + v') (g(v) + g(v')) = f(v) + f(v') = \lambda(v) g(v) + \lambda(v') g(v').$$

Hence if $g(v)$ and $g(v')$ are linearly independent it follows that $\lambda(v) = \lambda(v') = \lambda(v + v')$. Now we split into cases.

**Case:** $\text{im}(g)$ has dimension at least $2$. Then the equivalence relation on vectors in $\text{im}(g)$ generated by "linearly independent" identifies every vector, and it follows that $\lambda(v)$ is a constant $\lambda$ and $f = \lambda g$.

**Case:** $\text{im}(g)$ has dimension $1$. Then $\text{im}(f)$ is contained in $\text{im}(g)$, and hence has dimension at most $1$. Picking $v_0 \in V$ such that $g(v_0) \neq 0$, we then have $f(v_0) = \lambda(v_0) g(v_0)$, and hence it again follows that $f = \lambda g$ (where $\lambda = \lambda(v_0)$).

**Case:** $\text{im}(g)$ has dimension $0$. Then $f = g = 0$.

This concludes the proof. $\Box$

So possibilities are limited. Over any field, we can take $f \le g$ iff $f = \lambda g$ for some $\lambda$. This is a preorder, but it's pretty boring: linear maps which are nonzero scalar multiples of each other are isomorphic, and the only order relation which isn't an isomorphism is that $0 \le f$ for any nonzero $f$. Over $\mathbb{R}$ we can take $f \le g$ iff $f = \lambda g$ for some $0 \le \lambda \le 1$. This is even a partial order, but it's still not very interesting.

2030072skupersI believe, SuperMario, that Fedor's comment was meant to draw your attention to that particular polynomial as an example that might help you to answer your question. You might learn from it something about your kind of polynomial, and its roots, or lack thereof, on the unit circle.Suppose $\Phi$ maps $m+d$ dimensional Hermitian matrices to $n$ dimensional Hermitian matrices. We have the following SDP constrain on $m$ dimensional Hermitian matrix $X$ and $d$ dimensional Hermitian matrix $Y$, $$\Phi(X\oplus Y)\leq A$$ where $X\oplus Y$ denotes an $m+d$ dimensional block diagonal Hermitian matrix with $X$ being its upper $m$ diagonal submatrix and $Y$ being its lower $n$ diagonal submatrix.

Now we only want the constrain of $X$ itself, is it still an SDP constain as $$\Psi(X)\leq B$$ for some Hermitian preserving linear map $\Psi$ and Hermitian $B$. If so, how to compute them?

XIsogenies in multidimensional formal groupsBy the way, there are exact formulas for all of the quantities you ask about here in the case of Brownian motion. Look up "Levy's arcsine law" or read any textbook treatment of the zeroes of Brownian motion.Thanks, I suppose I was too quick to assume that because these conditions look subtly different, they must actually be different!723543124555776104252381487711165866163340463091$Generalize your definition slightly - Let $M_{e,k,q,r}$ be the max over $W$ and $S \subseteq \mathbb F_q^\times$ of cardinality $q-1-r$ of the number of squarefree monic polynomials over $\mathbb F_q$ lying in $W$ whose roots all lie in $S$.

Your lower bound clearly becomes

$$M_{e,k,q,r} \geq {q-1-e-r+k\choose k} $$

This is sharp when $e=k$ by the same argument you gave.

I claim this is sharp (for $k\leq e$ and $e+r \leq q-1$ so this is well-defined). This follows inductively from the following inequality:

$$M_{e,k,q,r} \leq \max \left( M_{e-1,k,q,r+1} , \frac{q-1-r}{e}M_{e-1,k-1,q,r+1} \right)$$

The argument is induction on $e$, with case $e=0$ trivial, and induction step

$$M_{e,k,q,r} \leq \max \left( {q-1-e-r+k\choose k} , \frac{q-1-r}{e}{q-1-e-r+k-\choose k-1} \right)= \max\left( 1,\frac{q-1-r}{e} \frac{k}{q-1-e-r+k}\right){q-1-e-r+k\choose k} $$

If we let $a=k$, $b=e-k$, $c=q-1-e-r$ then $a,b,c \geq 0$ so $(a+b+c)a \leq (a+b) (a+c)$ so $\frac{a+b+c}{a+b} \frac{a}{c+a} \leq 1$ so we get $M_{e,k,q,r} = {q-1-e-r+k\choose k}$ as desired.

It suffices to verify that inequality. There is either an element $\alpha$ in $S$ such that all elements of $W$ vanish on $S$ or not.

If there is, then we divide each element of $W$ by $(x-\alpha)$, and all elements that previously had roots lying in $S$ now have roots lying in $S - \{\alpha\}$. Thus we reduce the degree by $1$, leave the dimension of the subspace fixed, and increase $r$ by one, without reduceing the number of solutions.

If there is not, then we bound the number of pairs of an element of $W$ monic, squarefree, with all roots in $S$ and a root $\alpha$ of $W$. The number of these pairs is exactly $e$ times the number of $W$ monic, squarefree, with all roots in $S$, so we will divide by $e$ afterwards. For each element $\alpha \in S$, a codimension $1$ subspace of $W$ vanishes on it. Then, as in the previous case, we can divide all elements of that subspace by $x-\alpha$, producing a new space of polynomials one degree lower, and one dimension lower, where we are counting squarefree monic polynomials whose roots all lie in $S- \{\alpha\}$. Multiplying by $|S|$, we obtain the number of ordered pairs, and dividing by $e$, we get the upper bound.

2203980

Questions:

Have there been attempts to either prove or disprove, that every general matching problem can be transformed into a bipartite matching problem, from whose solution the solution of the original problem can determined efficiently?

How would a proof of the existence or non-existence of such a transformation affect matching theory?

The reason for asking is that apparently the time complexity of finding minimum weight perfect bipartite matchings equals that of minimum weight perfect general matching, which seems strange in view of the different techniques that are used for solving the problems and, while implementations for weighted bipartite matching abound, implementations for general weighted matching are very rare.

2102991H$a+1/a\ge 2$. Add and divide by 2. >Do you see how to improve QRH?121815786942158486288475Thanks again. Which sections are these examples discussed in? I had a quick look at both references; the A_2 case is mentioned on pg 27 of the J. Alg. paper, and in Section 10.5 of the monograph (though I'm not sure if the projective covers are discussed in either example). Also, the monograph deals primarily with algebraic groups - is it possible to transfer information about projective covers to the Lie algebra setting (or equivalently, to that of the Frobenius kernel $G^{(1)}$)?1291438157053146791414481314I don't think the power-set structure helps you estimate entropy better. The following extreme cases all give the same entropy for $p_B$, but entropy of the power set observation ranges from 0 to $N$ (the full range).

- Extreme case 1, if you only observe $B$, your entropy $H(X)$ is zero, while entropy of $p_B$ is maximum, $\log_2(N)$.
- Extreme case 2, if you observe $\{1\}, \ldots, \{N\}$ uniformly, entropy is $H(X)=\log_2(N)$, and entropy of $p_B$ is also $\log_2(N)$.
- Extreme case 3, if you observe all subsets equally likely, entropy $H(X)$ is $N$, while the entropy of $p_B$ is $\log_2(N)$

If entropy of $p_B$ is very small, that might restrict the maximum possible entropy for the power set observation, but I do not see an inequality. Even if there's an inequality, I don't think it can be tight.

As @guest suggested, using generic entropy estimators that perform well in the undersampled regime might be the best bet. @guest suggested my paper (thanks!), here are some more choices.

18999291217152Perhaps you could try using Groebner bases. The two examples that I computed using Macaulay2 (displayed below) suggest that there is a Groebner basis for $(f_n, f_{n+1})$ consisting of polynomials with leading terms of degree $n$. (These are $f_n, yf_{n-1}, y^2f_{n-2}, \cdots, y^n$, up to signs.) The examples also suggest that when we start reducing x^{2n-2} with respect to this Groebner basis, $y^{n-1}$, which cannot be reduced to zero, shows up at some stage.

```
Macaulay2, version 1.5
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases,
PrimaryDecomposition, ReesAlgebra, TangentCone
i1 : R = kk[x,y];
i2 : f = n -> sum apply(floor(n/2)+1, i -> (-1)^(n-i)*binomial (n-i,i)*x^(n-2*i)*y^i)
o2 = f
o2 : FunctionClosure
i3 : gens gb ideal (f 5, f 6)
o3 = | y5 xy4 x2y3-y4 x3y2-2xy3 x4y-3x2y2+y3 x5-4x3y+3xy2 |
1 6
o3 : Matrix R <--- R
i4 : x^8 % ideal (f 5, f 6)
4
o4 = 14y
o4 : R
i5 : x^8 + x^3*(f 5)
6 4 2
o5 = 4x y - 3x y
o5 : R
i6 : x^8 + (x^3+4*x*y)*(f 5)
4 2 2 3
o6 = 13x y - 12x y
o6 : R
i7 : x^8 + (x^3+4*x*y)*(f 5) - 4*y^2*(f 4)
4 2 4
o7 = 9x y - 4y
o7 : R
i8 : x^8 + (x^3+4*x*y)*(f 5) - 13*y^2*(f 4)
2 3 4
o8 = 27x y - 13y
o8 : R
i9 : x^8 + (x^3+4*x*y)*(f 5) - 13*y^2*(f 4) - 27*y^3*(f 2)
4
o9 = 14y
o9 : R
i10 : gens gb ideal (f 6, f 7)
o10 = | y6 xy5 x2y4-y5 x3y3-2xy4 x4y2-3x2y3+y4 x5y-4x3y2+3xy3 x6-5x4y+6x2y2-y3 |
1 7
o10 : Matrix R <--- R
i11 : x^10 % ideal (f 6, f 7)
5
o11 = 42y
o11 : R
```

Perhaps there is a pattern above which can be exploited to prove (if it indeed is true!) that $x^{2n-2}$ reduces to $y^{n-1}$.

For a locally invertible holomorphic function on the unitary disk $f:\Delta\rightarrow \mathbb{C}$ we have a result of Becker (1972): if $(1-|z|^2)\left|\frac{zf''(z)}{f'(z)}\right|\leq 1$, then $f$ is invertible. We have also two criteria of Nehari involving the Schwarzian derivative $S(f)$ of $f$: if $|S(f)(z)|\leq\frac{2}{(1-|z|^2)^2}$ or $|S(f)(z)|\leq \frac{\pi^2}{2}$, then $f$ is invertible. A good reference is the book "geometric function theory in one and higher dimensions" by Graham-Kohr.

5456341673005530109105841614251244853149807810892251633587621792(Expanding on Yuan's point about the irrelevance of "machine language" and JT's about "the wrong question to ask":)

In a sense, one oughtn't even be able to ask whether two topological spaces (i.e., objects of Top) are equal, only whether they are isomorphic; if one can't ask about equality, then one certainly can't speak of cardinality (with respect to such equality), and so there is no notion of the cardinality of the collection of objects isomorphic to a given one.

However, if you construe every topological space as implicitly carrying extra non-topological information via which such a non-topological notion of equality is defined (e.g., if you take the points of topological spaces to furthermore be elements of the cumulative hierarchy of well-founded sets of sets of sets..., allowing one to ask whether points in distinct spaces are equal by appeal to this extra structure, and accordingly whether spaces themselves are equal by virtue of an isomorphism sending points to equal points), then, of course, the question can be answered (in the given example, as noted above, the answer will be that the isomorphism classes form proper classes). But this is not really a question about the category of topological spaces, as such; this is a question about the particular manner in which one may choose to realize the intuitive theory of topological spaces within the ontology of another (meta)theory/implement the structure of the category of topological spaces within a context imposing further structure as well.

If one avoids selecting such "implementation details", then the question is meaningless, in precisely the same way as questions such as "Is the integer 9 an element of the rational -3/5?" are meaningless in the abstract.

42077529862644107871690824@Rahman.M, Yes. A topological property is a property that preserved by homeomorphisms.76480612551131502683:link to a paper by Ramanujan3314211691521789531996314204363As maybe you know, the most good decoding algorithm for LDPC codes, for example iterative decoding, has not provable efficiency, except in special cases. But, it is believed that variable nodes with degree $2$, behave weaker in error protection against the higher degree variable nodes. Also, the important part of efficiency of the decoding algorithms depends to the first round executing. So, degree $2$ variable nodes shows their effects in the first stage for the failing of the algorithm.

For more details you can see the book:

"LDPC Coded Modulations" by "Michele Franceschini, Gianluigi Ferrari, Riccardo Raheli"

Also, there are some good references in the book which help you for finding good answer for your question.

Note that, the degree $2$ variable nodes effect on trapping sets and stopping sets of the LDPC codes which are good parameters for the efficiency of the decoding algorithm.

5403131886149No, there is no such gadget.

Suppose to the contrary that you want to allow flow from vertex x to either vertex y or vertex z, but that you want it to remain unsplit. If there exist two flows F1 and F2, both with m units into vertex x but with those units all going to vertex y in flow F1 and all going to vertex z in flow F2, then for any 0 ≤ p ≤ m there exists a flow F3 with p units from x to y and m – p units from x to z: simply let F3 = (p/m)F1 + ((m–p)/m)F2. It's easy to verify that, if F1 and F2 obey the flow constraints at each vertex and edge, then so does F3.

In order to force the flow to be unsplit, you can't remain within the formulation of a maximum flow; you'd have to extend the problem to include additional side constraints, and by doing so most likely make it NP-hard.

There's one possible exception: if every node of the graph has the same value m and you want to quantize flow in units of m rather than in integers. Then you can just divide everything by m and use an integer flow.

p(Cyclic) edge-connectivity for lifts of voltage graphs?766147@Feldmann Denis Thanks for your comment! My assumption that Apéry's séries is divergent is based on the following remark in his paper: "(Nous écrivons $\equiv $ au lieu de $=$ pour tenir compte des répugnances des mathématiciens qui considérent avec Abel, Cauchy et d' Alembert les séries *divergentes* comme une invention du diable; en fait, nous n' utilisons jamais qu' une somme finie de termes, mais le nombres de termes croit avec $x$)." Standard $2$-sphere = geometrical $2$-sphere (not topological), *i.e.* Euclidean sphere of $\mathbb R^3$. I know a colleague of mine worked on (and solved) a similar problem in a bounded region of the plane with some boundary condition, I'll ask him.rExtension of some feature of SDE Ornstein-Uhlenbeck typeFor that, as abx wrote, just take a smooth Calabi-Yau complete intersection of even dimension. For instance if you take a degree $4$ smooth surface in $\mathbb{P}^3$ (which is a $K3$ surface) you get an instance of my example.18109941902075516172122319I've found a solution using toric geometry and Stanley-Reisner theory, which should be similar to your description.179613611019991552017While I'm confident the embedding of $V$ into $H$ here is continuous, I'm doubting now that it's compact. This may disqualify this answer, as the question is now worded.796841280655352688I'm not sure how "pseudo-reflection" would be defined in case $V$ is infinite dimensional. On the other hand, there may be some possibilities when the action of $G$ is assumed to be locally finite dimensional. I'm not sure what the constraints should be on $G$ and $V$ to get a meaningful generalization.1162182600683446902316477602196647Thanks for the links, they did give me some ideas for searching the literature.1694883@Tom (and anyone else who's bothered about this): There are some threads over on meta where the issues of "when to close" are being discussed. Examples are: http://tea.mathoverflow.net/discussion/84 and http://tea.mathoverflow.net/discussion/75 1893484@quid: it is one thing for the journal to acknowledge receipt, and a whole other thing, based on the two factors I enumerated, for the editor to have a reliable time-estimate (hence the "shortly after"). 559728The most obvious candidate would seem to be to try the case $t = p.$156871416008329518381178452Caveat to the statement about this working on more general spacetimes: I'm making implicit use of the fact that Minkowski spacetime is globally hyperbolic, so that I can work on a spacelike slice.258425Nonnegative integers represented by $\prod_{i=1}^m \sum_{j=1}^n a_{i,j} x_j $, where the $a_{i,j}$ are positive integers@Francesco Polizzi Thank you! Do you know any references about this?173822116033821331069148955631513244234@Surfaces singular along a curve213148120590541404095https://www.gravatar.com/avatar/917157dcb83cd31f886c78235c939f74?s=128&d=identicon&r=PG&f=1474948I think all but the last three ones are either important or natural to formulate, so it is interesting per se to ask oneself what kind of relationship do they bear.@AsafKaragila If $0^\sharp$ exists then it is in HOD, so I would assume that $\mathrm{HOD}^{L[0^\sharp]}=L[0^\sharp]$.There's a very famous group, the largest sporadic simple finite group, sometimes called a **monster** whose size is quoted below. What's the explanation that the primes appearing in it,

`#`

{Monster} = 2^{46} * 3^{20} * 5^9 * 7^6 * 11^2 * 13^3 * 19 * 23 * 29 * 31 * 41 * 47 * 59 * 71

are exactly the **supersingular primes** (and here's a separate question about those)?

My notes contain some mystic reference that it's "related to famous modular function $j(\tau) = q^{-1} + 744 + 196884q + \cdots$ by some compactification of bosonic strings on a Leech lattice". But perhaps there could be a more purely number-theoretic direction?

Also, here's a Wikipedia article with some references.

4499051684229On the contrary, Wlod, I'm sad to think that you have been provoked, even sadder to think I made the provocation. But I'll get over it.For $1\leq p <+\infty$, $0<s<1$ and $\Omega\subset R^n$ domain, the fractional Sobolev space $W^{s,p}$ is defined as

$$W^{s,p}(\Omega):=\big\{f \in L^p(\Omega)\colon \int_{\Omega} \int_{\Omega} \frac{|f(x)-f(y)|^p}{|x-y|^{s p + n}}dx dy<+\infty .\big\}$$

I wonder if this definition makes sense for $s=0$ for bounded $\Omega$, in particular, can one describe functions $f$ such that

$$\int_{K} \int_{K} \frac{|f(x)-f(y)|^p}{|x-y|^{n}}dx dy<+\infty$$

for any compact $K\subset R^n$?

@Torsten: that's not right. The 1x1 matrix (p) has char poly in Z_p.LSurfaces that can be rolled by a ball614305googling "higher codimension restriction estimates" bring up [this paper](http://arxiv.org/pdf/math/0104095.pdf). Its corresponding MathSciNet entry is [here](http://www.ams.org/mathscinet-getitem?mr=1650964). Maybe you can get some answers/references from it.6915322102799The vertex normal information yo desire is contained in the image you generated. If `pic` is the result of your `ParametricPlot` command, then you can extract the vertex normals via `First[Cases[pic,HoldPattern[VertexNormals -> vn_] -> vn, Infinity]]`. I could type up a more complete response, if this is information that you are still interested in.8466321306360Let $H$ be a complex biquadratic Galois extension of $\mathbb{Q}$ such that the galois group of $H$ is isomorphic to the Klein Group. Let $H_{\infty}$ be an anticyclotomic $\mathbb{Z}_p$-extension of $H$ and $L_{\infty}$ be the maximal abelian unramified $p$-extension of $H_{\infty}$. Assume that $p$ splits in $H_{+}$, where $H_{+}$ is the maximal totally real subfield of $H$ and $p$ doesn't totally split in $H$.

Let $X$ be the galois group of the extension $L_{\infty}/H_{\infty}$ and $X^{-}$ be the part of $X$ on which $\sigma \ne 1 \in \mathrm{Gal}(H_{+}/\mathbb{Q})$ acts by $-1$.

Can we relate the characteristic ideal of the Iwasawa module $X$ (resp. $X^{-}$) to some L $p$-adic functions ? Can we say anything about the $\mathbb{Z}_p$-rank of $X^{-}$ or $X$ ?

1797439criticalthI completely agree and I only have to add to this that a lot of the power graduate programs in mathematics are beginning to reorganize thier graduate analysis courses in precisely this way-see Columbia. It may be that most programs will follow suite. As for giving a rigorous theory of the integral at the undergraduate level,I think the generalized Riemann (HK integral) is a terrific option that isn't considered enough.Indeed-you can have your cake and eat it too since the Generalized Riemann integral can only be simply defined in Euclidean spaces.(continued)322181730949137099410571086139788Can you explain that a bit?@RyanBudney The Riemannian metric from your second suggestion is explicitly described in this 1967 paper http://www.pnas.org/content/57/3/589.full.pdf+html which also describes geodesics and other interesting stuff.Inspired by a recent question on the multiplicative group of fields. Necessary conditions include that there are at most $n$ solutions to $x^n = 1$ in such a group and that any finite subgroup is cyclic. Is this sufficient? (Edit: Well, no, it's not, since the only such groups which are finite are the cyclic groups of order one less than a prime power. Hmm.)

33965610846468Representation of Groupoids52423521196871846125290929218100172305020180795506285I posted the same question on Math SE since this one got put on hold.

Link to Math SE question:Polygonal Mersenne numbers

Polygonal numbers are of the form $\cfrac {n^2(s-2)-n(s-4)}{2}$, where $s$ is the number of sides e.g. when $s=5$ we get pentagonal numbers, and $n$ is which one in order it is i.e. the $n^{th}$ $s$-gonal number.

Mersenne numbers are of the form $2^p-1$. We usually speak of Mersenne primes, but in my case I am looking at all Mersenne numbers.

So to find numbers that are both polygonal and Mersenne, we get
$$\cfrac {n^2(s-2)-n(s-4)}{2}=2^p-1$$
where $n,p,s \in \mathbb Z^+$. At $n=1$, we simply get $1$, that is to say the first $s$-gonal number is $1$, and at $n=2$ we get $s=2^p-1$, that is to say the second $s$-gonal number is $s$, so to avoid trivial solutions we say $n \ge 3$. Also $s \ge 3$, since we are talking about **poly**gons.

To start of I transform the equation into $$\frac{2(2^{p} + n^{2} - 1 - 2n)}{n(n-1)} = s$$ Then through a bit of modular artihmetic we can get three cases of $n$ 1 $$n \equiv 2 \pmod 4$$ $$n \equiv 3 \pmod 4$$ $$n \equiv 5 \pmod 8$$

It is also possible to transform the equation into a Mordell type equation ($x^2=y^3+k$) [2]

If $3+p=3a$ $$((s-2)(2(s-2)n-s+4))^2+(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^a)^3$$ If $3+p=3b+1$ $$(2(s-2)(2(s-2)n-s+4))^2+4(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{b+1})^3$$ If $3+p=3c+2$ $$(4(s-2)(2(s-2)n-s+4))^2+16(s-2)^2(-s^2+16s-32)=((s-2)\cdot2^{c+2})^3$$ So this way we can find all possible solutions $n,p$ for a given $s$, but calculating solutions to Mordells equation for a large $k$ is quite the task, so this is not what we are ultimately looking for.

One can also prove that the equation has a finite amount of solutions $n,p$ for each $s$ through transforming it into $$x^2+D=AB^y$$ where $D,A,B$ are given and $x,y$ are the variables $$(2(s-2)n-s+4)^2+(-s^2+16s-32)=(s-2)\cdot2^{p+3}$$ and since it has been proven that for every case of $x^2+D=AB^y$ there are finite solutions $x,y$, there are finite solutions $n,p$ for a given $s$

It is also quite easy to prove that $2^p-1$ cannot be prime

If $n=2a$ $$a(2as-4a-s-4)=2^p-1$$ $a=1,n=2$ but we do not count $n=2$, as stated earlier

If $n=2b+1$ $$(bs-2b+1)(2b+1)=2^p-1$$ either $b=0,n=1$ or $s=2$, but we do not count those, because of the reasons stated earlier.

My main goal is to find a way to easily calculate all solutions $n,p$ for any given $s$. An idea that I have is to find $n_{max},p_{max}$ in terms of $s$, so that when I give an $s$, I can easily just make the range of $n,p$ $$n_{max} \geq n \geq 3, p_{max} \geq p \geq 0$$ but this assumes that $n_{max},p_{max}$ will be small.

So my question is: Does anyone have any ideas on how to find $n_{max},p_{max}$, and do they even exist. Any other ideas on how to solve this problem (finding every possible $n,p$ for a given $s$).

Any insight, tips, or just generally help would be greatly appreciated. Also please ask if you require further details.

(Is it still too vague what I am asking for?)

```
[1] Greg Martin
[2] user236182
```

-redelectrons

845912the boundary constraint is $A(1,1)$, i.e. in the first iteration the size is 1. 3495461465323898@BrettParker and Chris Gerig, I think you are implicitly assuming that because one cannot symplectically displace a ball through a thin tube, one cannot displace a large part of its volume through it.70601227722312101841177616679071714664121512396The Riemannian distance has the following property: it is the so called length space. I recall (one of) the definitions of the length space. Having a distance function $d$, we can define the length of any curve $c(t)$, $t\in [0,1]$ , as the supremum of the sum $\sum_{i}d(c(t_i), c(t_{i+1}))$ over all $0=t_0\le t_1\le ... \le t_k=1$. Having this definition of the length of the curve, you can construct a new distance $\hat d$ by putting $\hat d(x,y)= $infimum of the lengths of all curves connecting $x$ and $y$. Then, the length spaces are characterized by the property $d=\hat d$.

Riemannian distance is evidently a length space. One could easily construct a distance such that it satisfies your condition but is not a length space and therefore not a Riemannian metric: put $d(x,y)= min(d_{riem}(x,y), 1)$. It still satisfies your property since we changed nothing in a small neightborhood, and is not a length space and hence not a Riemannian metric anymore.

4866931360353You've shown that there are $2^c$ models $B$ but only $c$ corresponding functions $f$, so indeed many $B$'s must yield the same $f$. Since you say you don't see how it's possible for even two $B$'s to yield the same $f$, here's an example. Consider the countably many sentences $S$ such that neither $S$ nor its negation is a tautology. For each such $S$, choose one classical model $Y(S)$ in which $S$ is true and one classical model $N(S)$ in which $S$ is false. Let $B$ be the set of all these $Y(S)$'s and $N(S)$'s. The corresponding $f$ maps each of these $S$'s to $-1$ (and it maps tautologies to 1 and the rest of the sentences, those whose negations are tautologies, to 0). Note that $B$ is countable (as it has just two members $Y(S)$ and $N(S)$ for each of countably many sentences $S$), so there are plenty of classical models not in $B$. Now let $B'$ be the union of $B$ with any nonempty subcollection of these other classical models. Then the $f$ associated to $B'$ is the same as that associated to $B$.

2270842227481343599634768580697138648098716518899697544480388663727113@urpzilmöräqÜ As far as I can see, Joel is using the same meaning of "atomic formula" as the wikipedia article that you linked to. Why do you think these formulas don't suffice as reasons for non-isomorphism? In fact, one could limit further to atomic formulas that are of one of the forms $R(x_1,\dots,x_n)$ for some predicate symbol $R$ and $F(x_1,\dots,x_n)=y$ for some function symbol $F$.1473937@RichardStanley Oh, I see. Thanks. But you have not come across the statement from my question, right? Do you at least believe it to be true?lTannakian categories equivalent as abelian categories1042337201484230421774929878902109457I accepted the answer, but something remains unclear to me. The degree of an element of $ M $ can be any integer, so how do you prove that $ F\otimes G $ belongs in $ M $ whenever $ F $ and $ G $ do ?20320111791117652879<@DylanWilson Yes, that works!2218974Finiteness follows from the following version of the Hilbert syzygy theorem: If A is a regular local ring of dimension n, M a finitely generated A module, and F_{n-1} --> ... --> F_1 --> F_0 --> M --> 0 a resolution of M by free A modules, then the kernel of F_{n-1} --> F_{n-2} is free. Use Hartshorne III.6.5.1 to build the first n vector bundles of your complex, and then use this theorem to conclude that the kernel of the last map is another vector bundle. 209707 Thank you very much! I've finally managed to locate your thesis, so I'll sit down this morning and try and work through the calculations. (Hopefully I've fixed in the typo's in the first question now. I guess my problem in part 3 is - how does one do the coproduct $\psi(\xi_1^2 \overline{\xi}_2)$?) 53842718313311959000xHave you seen this sort of an anti-involution on a lattice?14185762048427We are reading John Roe's book *Lectures on Coarse Geometry*. We come across a question in P27 line 9:
Suppose $X$ is a paracompact and locally compact Hausdorff space, $\bar{X}$ is a compactification of $X$, how to use Urysohn's Lemma to choose continuous functions $f,g:\bar{X}\times\bar{X}\to \mathbb{R}^+$ such that $f$ vanishes **only** on the diagonal of $\bar{X}\times\bar{X}$, and $g$ vanishes **only** at infinity (that is, on $\bar{X}\times\bar{X}\backslash X\times X$)?

If $\bar{X}\times\bar{X}$ is perfectly normal, or the diagonal and the infinity are both $G_\delta$ sets, we can have that according to P213 in Munkres's book *Topology*. But we fail to verify the condition(we are not familiar with paracompact Hausdorff space).

Or any other way?

6873381984725303937265234188310521984171653949837669PWell, an infinite number of exceptions.@Subprojectivity of $L_{p}(p>2)$305598As far as I can tell, hypercalc just uses standard floating point representations of iterated logs. This suggests that it cannot handle the sort of precision necessary here.2247476228722Thanks to you all for the discussion, answers, and references. It is really helpful. 2167951`(I have edited the question appropriately now.)2Time to recheck my work.14074082296779 There was a fairly recent example of someone randomly generating a math paper which was accepted for publication. The stunt was to show that pay-to-publish journals were terrible and incentivized to get anything published, rather than good things. But I think it fits into the wider spectrum of the growing inadequacies and flaws with the current journal system. In my own research I have gotten papers published before a paper it references has even gotten a response from a referee. And my field can take a long time to get things published. They're overwhelmed on all counts.2650322063469The eigenvalues of a square matrix $A$ are the roots of the characteristic polynomial, and are analytic except where their multiplicities change. Thus if (in a certain open region of parameter space) there is one eigenvalue, of algebraic multiplicity $m$, inside a positively oriented closed contour $\Gamma$ in $\mathbb C$, and no eigenvalues on $\Gamma$, that eigenvalue is

$$ \frac{1}{2\pi im}\oint_\Gamma \frac{z P'(z)}{P(z)}\; dz$$ where $P(z) = \det(A - z I)$, and this is analytic in the entries of $A$. The corresponding eigenvectors (with an appropriate normalization) can be found by solving a linear system depending linearly on the parameters and the eigenvalue, and these are also analytic.

20109181815549116533220794151073429In this paper with Iosif Pinelis, we give the exact constants for the lower bound (whose dependence on $\epsilon$ is $\frac1{\epsilon^2}$): https://arxiv.org/abs/1606.08920

Actually, our result is stronger since the bound holds for *any* estimator -- not just the natural one you get by counting the outcomes and normalizing by the number of flips.

This might get fixed in the future, but at the time of this writing, *Wolfram Alpha* gets apparently sometimes confused by logarithms of complex numbers:

Wolfram Alpha -- $\log(1+ \frac{1}{2}i) - \log(1 - \frac{1}{2} i)$

For reference, should the problem get fixed: it claims that $2i = 2i\cot^{-1}(2) \approx 0.9272$.

Curiously, the numerical approximation is correct, but the symbolic form seems to be wrong.

573980202842218071461493551469284How to derive explicit bound for the solution of following equation?@Daniel Marahrens and Leitz: There is a little typo I think, $k_1$ should start at 0 and not at 1, but this comes from a typo in the question itself, where the definition of $\bar{s}_k$ should include $s_0$. Best regards100348187422618487731947010@Friedrich Knop: Thanks for your comment, this is quite useful to know.5939024351561456844OK, but then it's hard to get a good lower bound on the problem's difficulty.1250048Perhaps somebody else can say more about the relation to Pólya's Positivstellensatz and on whether the conjecture is true.306804(This is in response to Gerhard Paseman's answer.)

```
.0001 .001
maxC/M 2.528183 2.528183
calculated s 4.459300 8.917000
total error 2.703720 2.718460
```

The above represents two runs - one incrementing by .0001 and the other .001 (from 0 to 20 in both cases). One trend is that the difference in total error between .001 and .0001 is absolutely consistent with the above, regardless of the data. The total error is in this case divided between 6 elements, each element averaging about 200 in size. If maxC/M were used as s, the total error would have been something like 7 or higher.)

But as to your observation regarding maxC/M being a good starting point for finding s, you can see above what s turned out to be. When using .001 as the increment, s was often much higher than with .0001, (with a consistent reduction in the size of the values for t). So maybe with increasing precision in the search, s converges towards the vicinity of maxC/m, but the example above has thus far seemed to be fairly typical. Haven't completely digested your remarks above yet. Thanks for letting a non-mathematician like me crash the party here.

21671931313128@RodrigoA.Pérez : It's news to me that Young diagrams and Ferrers diagrams have different orientations. I've seen different orientations for both. The difference, in my experience, is whether you draw a polyomino or a bunch of dots.69826618009121622462I am inclined to think that this is not a mere coincidence. Rather, that this is some sort of reflection of the fact that mathematicians tend to think in predicable ways.Iwasawa theory studies abelian extensions of fields $K$ where $K$ is a $\mathbb{Z}_p$-extension of $\mathbb{Q}$, that is the Galois group of $K/\mathbb{Q}$ is $\mathbb{Z}_p$. The corresponding Galois groups (of extensions of $K$) and class groups (of really subfields) of $K$, suitably interpreted, become $\mathbb{Z}_p$-modules and there are interesting relations between these modules and $p$-adic $L$-functions. It is a vast subject. Washington's book, Introduction to Cyclotomic Fields, is a good entry point.

547390507513Are there standard algorithms for enumerating flats in matroids?18528573226683111591588732726144831046I am teaching Calc I, for the first time, and I haven't seriously revisited the subject in quite some time. An interesting pedagogy question came up: How misleading is it to regard $\frac{dy}{dx}$ as a fraction?

There is one strong argument against this: We tell students that $dy$ and $dx$ mean "a really small change in $y$" and "a really small change in $x$", respectively, but these notions aren't at all rigorous, and until you start talking about nonstandard analysis or cotangent bundles, the symbols $dy$ and $dx$ don't actually mean anything.

But it gives the right intuition! For example, the Chain Rule says $\frac{dy}{du} \cdot \frac{du}{dx}$ (under appropriate conditions), and it looks like you just "cancel the $du$". You can't literally do this, but it is this intuition that one turns into a proof, and indeed if one assumes that $\frac{du}{dx} \neq 0$ this intuition gets you pretty close.

The debate about how rigorous to be when teaching calculus is old, and I want to steer clear of it. But this leaves an honest mathematical question: Is treating $\frac{dy}{dx}$ as a fraction the road to perdition, for reasons beyond the above, and which have not occurred to me?For example, what (if any) false statements and wrong formulas will it lead to?

(Note: Please don't worry, I have no intention of telling students that $\frac{dy}{dx}$ *is* a fraction; only, perhaps, that it can usually be treated as one.)

In Huang & Lepowsky's series of papers *A theory of tensor products for module categories for a vertex operator algebra*, they defined for a rational vertex algebra $V$ the $P(z)$ tensor product of two modules $W_1$ and $W_2$ to be
$$
W_1\boxtimes_{P(z)} W_2=\coprod_k{(\mathcal{M}[P(z)]^{M_k}_{W_1~W_2})}^*\otimes M_k
$$
where $\mathcal{M}[P(z)]^{M_k}_{W_1~W_2}$ is the (finite dimensional) vector space of conformal blocks for modules $M'_k$, $W_1$ and $W_2$ at points $\infty$, $z$ and $0$ on the compact Riemann surface $\mathbb{P}^1$. Here $M'_k$ is the module contragedient to $M_k$. They showed that under certain conditions this tensor product gives a vertex tensor category. Now if we define the tensor product by using conformal blocks on an arbitary compact Riemann surface, can we get a vertex tensor category (and hence a braided tensor category) under certain reasonable conditions? If this is true, can we show furthermore the rigidity and the modularity for this category?

For a very mathematical version of statistics, my favorite is on line lecture notes from two MIT courses. The instructor is named Panchenko and the course is called 'Statistics for Applications'. There are course notes that read like a book for the course in 2003 and 2006. I have enjoyed browsing through both of them. Here is a link: http://ocw.mit.edu/courses/mathematics/.

766180I'm almost certain that you need some kind of Noetherian property. E.g. take $Y = Spec(R)$ and $B = B' = R$, then this asks whether Ext commutes with base extension from a field in full generality (take $R$ to be an infinite product $\prod k$).BNice pun! Was it intentional? 168211811527941483259@Simon: Oops... But won't you attend the conference at CIRM, two weeks from now?8607688060062155376Given a natural number k, are there only finitely many finite simple groups with the property that all elements have order at most k?

This holds if I only look at the finite simple groups I understand (e.g. alternating groups and SL(k,finite field)), but it's not clear to me whether this holds for all finite simple groups, even using their classification.

There was a big investigation on projections in operator algebras made by D. Blecher and his students. This might help http://arxiv.org/pdf/1109.5347.pdf, http://arxiv.org/pdf/1109.5171.pdf2256799@David: Indeed. But it seems to me the point of view of GKZ is too `toric' for what I have in mind. I would like a Feynman diagram type expansion in terms of tensor contractions where the natural group of coordinate transformations is the full linear group.1545980By taking K = a simplex and L = its boundary you can show that |A| -> |X| is an isomorphism on all homotopy groups (do surjectivity and injectivity separately). Then apply Whitehead's theorem.

3813197786011@Leonid, I think you're right and you should post your observation as an answer -- it could be accepted then.180732819479102251941Analytic sections of a GIT quotient lying in the Kempf-Ness set172093749568220892642126958199486710771832353391802158509049Because the field of fractions of $B \otimes_A Q(A)$ is the same as the field of fractions of $B$ (localization is transitive).21912924868471829129Your operator $A$ is continuous from $C_0^\infty(M)$ into $C^\infty(M)$, so the adjoint $A^*$ is continuous from $\mathcal E'(M)$ into $\mathscr D'(M)$. Now, the operator $A^*$ is also a pseudodifferential operator of the same order than $A$, whose principal symbol is the complex conjugate of the symbol of $A$: this fact can be established by looking at the chart expression of the pseudodifferential operator.

More precisely in $\mathbb R^n$, if $a$ is the symbol of $A$, the symbol of $A^*$ is $$ a^*=\exp(2iπ D_\xi\cdot D_x)\bar a, $$ so that starting with the continuity property of $op(a^*)$ (the operator with symbol $a^*$), you obtain the sought property of $op(a^{**})$ which is $op(a)$.

2047596266012107614922526661265056120412617858455540751381671What kind of underlying topological space should the borel measures have? A polish space? A locally compact space? Something else?>Paul, thanks for your comment!129540041606512181754373775Wasn't the 1965 notice just an abstract? That's what I had understood, and by a closer look at the paper cited in the OP (I was addressed to it by some colleagues who explained the result was first proved there, which is the "second-hand information" I was referring to in the above) I've just realized there is a footnote on p. 397 (actually, the very 1st page of the article), where the author explains that "The main results of this paper were proved in the spring of 1966."7746531660264I think this expository article: Algebraic Dynamics, Canonical Heights and Arakelov Geometry of Xinyi Yuan is useful.

896395137718344307514098222522165448362173128525483159260321714391168099$|L|$ may consist of a single surface that is **not** what you want!20289245361965947877898061433229thank you very much. What is the antipode of $A[\hbar]/\hbar^2$?Poincare defined the fundamental group and the homology groups and proved that $H_1$ was $\pi _1$ abelianized. So the question came up whether there were other groups $\pi_n$ whose abelianization would give the $H_n$. Cech defined the higher $\pi_n$ as a proposed answer and submitted a paper on this. But Alexandroff and Hopf got the paper, proved that the higher $\pi_n$ were abelian and thus not the solution, and they persuaded Cech to withdraw the paper. Nevertheless a short note appeared and the $\pi_n$ started to be studied anyway...

Taken from http://www.intlpress.com/hha/v1/n1/a1/ ,page 17

1525926Do you need a nondegenerate example? There is one with all the sinks coinciding. 15547822Internal Day convolution362793@JoergSixt If $\text{char} K=p$, then $KG $ has trivial units iff either 1) $K=F_2$ and $G=C_2 $ or $C_3$ or 2) $K=F_3$ and $G=C_2$. See [sehgal](https://books.google.co.in/books/about/An_Introduction_to_Group_Rings.html?id=7m9P9hM4pCQC&hl=en) page 234@ChristianRemling: Isn't $Q_n$ forced to be a projection? If $v \in V_n$ then we have $Q_n v - v \in V_n$ and 1 says that $Q_n v - v \in V_n^\perp$. Hence $Q_n v = v$.7097577414831330465191846913082534841161006575It looks like $\widetilde{K_0}(R)$ is maybe the ideal class group is this context? In which case note that the direct sum of two cancellative monoids is cancellative.I realize this is a very old question, but you are correct in saying the actual probability of vertices u and v being connected is $$\frac{deg(u) * deg(v)}{2m - 1}$$ For large m, however, $2m - 1 \approx 2m$. Such reasoning can be found here.

826860773040Such a statement is true Zariski locally. See EGA IV, Proposition 18.1.1.BConstruct embedding given metricFor 241, the discriminant is 2^240 11^478 241^238 -- clearly a square.181878912337221878947 Another solution, this one which works even if the other person does not have a strategy, but assumes you can see a finite distance away. From where you start, make a bunch of circles by slicing planes through the circle perpendicular to the radius pointing at you. Make the distance between the circles small enough that you can see from one circle to the next. Walk each circle twice, leaving the message "do not cross this circle" along each path. You should eventually trap the other player between two circles, and you will find them on the second pass around the circle.32585741713713195429014231708444139961410121892089440hIs every ring the direct limit of Noetherian rings?21799888018732300022771349)Regarding the specific example, the construction of $C'$ can be tightened up as follows. The first functor, from $C$ to $C'$ is not just hom-preserving (closed) but bijective on objects. The second functor, from $C'$ to $D$, has the property that it is fully faithful on maps out of the monoidal unit.

(This is enough to determine $C'$ uniquely. As is implicit in Scott's comment, you need more than just the fact that the first leg is closed, otherwise you could take the first map to be the identity.)

So you could consider the (2-)category of symmetric monoidal closed categories, and (strong) symmetric monoidal functors. Any such morphism $F:C\to D$ comes with a canonical comparison $F[c,d]\to[Fc,Fd]$, and we can call it (strongly) closed if these comparisons are invertible. There's a class E of morphisms consisting of those which are bijective on objects and closed, and a class M of morphisms consisting of those for which $F$ induces a bijection between maps $i\to c$ and maps $Fi\to Fc$, for all $c$, where $i$ is the monoidal unit. (You might call these maps pre-fully-faithful, or something like that.) These classes E and M are closed under composition. I haven't checked in detail, but it looks like they have a reasonable chance of being orthogonal to each other, and so you would have a factorization system.

I don't see a way of doing something similar with mere functors between monoidal categories. But perhaps this is too much to expect. In the other example, although the internal hom is not preserved, there is a canonical comparison. So perhaps rather than looking at plain functors you should look at the lax monoidal ones (often just called monoidal, with the word strong being added to mean preservation up to isomorphism). Now it is true that every lax monoidal functor $C\to D$ factorizes as a lax monoidal $C\to C'$ followed by a strict monoidal $C'\to D$, moreover in a universal (initial) way. For any monoidal category $C$ there is a lax monoidal $p:C\to C'$ with the property that composition with $p$ induces a bijection between lax monoidal $C\to D$ and strict monoidal $C'\to D$, for any $D$. (Notice that the order is opposite to that in your example: the strict map comes second.)

This situation is quite common, and holds for many different types of structure. It is much less common, but does sometimes happen that there's a universal factorization $C\to D'\to D$ with the map $C\to D'$ strict and the map $D'\to D$ non-strict.

Your example involves an extra ingredient involving the bijective-on-objects and pre-fully-faithful conditions.

15495162951281361466221161310014272998291844967@GregoryArone If you still have the draft of that paper you could copy-paste that proof here as an answer and finally get it out in the open. (A reference to a MO post is better than nothing...)20639176656313695251133064816684351026226:Basic questions about stacks720736998870440687998237Andrew, your use of the word 'presentation' is not standard, you should either use standard terminology or explain clearly what you mean.192932Thanks for the info. But what I wanted to point out is that for the "lower" classical complexity classes (PSPACE and below) this is the best, is that correct? Although the NQP=coC_{=}P result seems to be at a really low level.255374Actually my first guess isn't quite right: 1089109891089 is OK. So instead of XXX..X it should be a palindrome X1X2X3...X3X2X1 with either all the X_i of the form 10999...89 or all of the form 21999..9978.10214888413501417967419756xNearest neighbor for planar Poisson is normally distributed801591482962196489399875https://www.gravatar.com/avatar/eaa78a7dde5462700fb6c131d7bd9a5b?s=128&d=identicon&r=PG&f=156466213956271418063Let's fix a scheme $S$ and consider some Grothendieck topology $\tau\in\{Zariski,\acute{e}tale,fppf\}$ on the category of schemes over $S$. Define a $\tau$ fiber bundle with fiber type $F$ to be a map $f:X\to Y$ of schemes over $S$ that $\tau$-locally on $Y$ looks like a product with $F$.

Given $\tau,\tau'\in\{Zariski,\acute{e}tale,fppf\}$ with $\tau'$ coarser than $\tau$, are there conditions on $F$ that imply that every $\tau$ fiber bundle with fiber type $F$ is also a $\tau'$ fiber bundle?

I've seen examples of etale fiber bundles that aren't Zariski bundles, but it's my understanding that for $F=\mathbb{P}^1$ this can't happen (I'm not positive about this), and I'm wondering about other $F$'s. I don't know if there are fppf fiber bundles that aren't etale bundles, and if there are I'd love to see an example.

2912424748224Nonlinear matrix equation67643815935114346361792795612238336279188994840751880438I appreciate your interest, but the answer you gave would have been better left as a comment. Many of us are tempted to be the first to answer a question to achieve self-actualization and earn points on MathOverflow (myself included), but for open problems such as this one, slow and steady wins the race.also, where does the expression for $P_{m}(x,z)$ come from? IS this some determinant? 4141791294200Will: The only requirements for the existence of maximal ideals are that the ring is nonzero and that we assume AC (Zorn's lemma, more directly). The union of a chain of ideals is an ideal, and so every chain has an upper bound. By ZL, there exists a maximal ideal. This can be modified slightly to guarantee that every non-unit lies in a maximal ideal. And Pace: Thanks a lot, non-commutative algebra is something that I view as impractically mysterious, and I plan to do away with that one day.418932@Justin: This implies the answer to my question because ${\mathbb Z}_2^\infty$ as a metric space embeds into ${\mathbb Z}^\infty$. Now we can refer to your proof also in our paper with A. Dranishnikov. Also see the related question: http://mathoverflow.net/questions/37529/covers-of-zk@DelioMugnolo Are there examples of domains which are isospectral to a disk (for Dirichlet problem with zero Cauchy data)?196339150443613982315863822129862740078"In 1957, Erik Sparre Andersen proposed using a renewal risk model to describe the behavior of the insurers surplus

$$U(t)=u+ct-\sum\limits_{i=1}^{\Theta(t)}Z_i, \quad t \geqslant 0$$

where:

$u \geqslant 0$ denotes the initial insurer's surplus;

$c > 0$ denotes the premium rate per unit of time;

the cost of claims $Z_1, Z_2,\ldots$ are independent copies of a nonnegative random variable $Z$;

the inter-occurrence times of claims $\theta_1, \theta_2, \ldots$ are another sequence of independent copies of a nonnegative random variable $\theta$ which is not degenerate at zero;

the sequences $\{Z_1, Z_2, \ldots\}$ and $\{\theta_1, \theta_2, \ldots\}$ are mutually independent;

$\Theta(t) = \#\{n \geqslant 1: T_n \in[0,t]\}$ is the renewal process generated by random variable $\theta$, where $T_n = \theta_1 + \theta_2 + \cdots + \theta_n$.

It is known that, for $\mathbb{E}Z-c\mathbb{E}\theta<0$ and $u=0$, the ruin probability

$$\psi(u)=\mathbb{P}\left(\sup_{n\geqslant1}\sum_{i=1}^{n}(Z_i-c\theta_i)>u\right)$$ has the following expression $$ \psi(0)=1-\exp\left\{-\sum_{n=1}^{\infty}\frac{1}{n}\mathbb{P}\left(\sum_{i=1}^{n}(Z_i-c\theta_i)>0\right) \right\}. $$

It is also known that $\psi(0)=1$ for $\mathbb{E}Z -c\mathbb{E}\theta\geqslant0$.

I want to see how the series

$$\sum_{n=1}^{\infty}\frac{1}{n}\mathbb{P}\left(\sum_{i=1}^{n}(Z_i-c\theta_i)>0\right)$$

diverges as $\mathbb{E}Z-c\mathbb{E}\theta$ approaches zero.

Let $X$ denote a r. v. such that $\mathbb{E}X=-\varepsilon$, where $\varepsilon>0$ and $\lim_{\varepsilon\to0}\mathbb{E}X^2>0$.

Can you show that

- $\mathbb{P}(X>0)>0$?

Or in general

- $\mathbb{P}\left(\sum_{i=1}^{n}X_i>0\right)>0$ for all $n\in\mathbb{N}$, where $X_i$ are independent copies of $X$.

It is natural to expect that both statements may hold only as $\varepsilon$ approximates zero.

2178467@TimothyChow In fact, there is a new volume in 2016 on algebraic topology. See [here](http://www.springer.com/fr/book/9783662493601) or [here](http://www.bourbaki.ens.fr/ta14-tdm.pdf) for more details.608490What is the principal value of the integral $$\int \limits _0^\infty \left( \frac {1}{x^2}-\frac{\cot(x)}{x} \right) dx ?$$ Maple finds $PV \int_0^\infty \tan(x)/x dx = \pi/2.$ Such integrals arise in physics. I unsuccessfully asked it in SE.

1584546The roots of your denominator $x^6 - x^4 - x^2 - 1$ are $\pm \sqrt{r_i}$ where $r_1, \ldots, r_3$ are the roots of $z^3 - z^2 - z - 1$, namely $$ \eqalign{r_1 &= \dfrac{1}{3} + \dfrac{1}{3} (19 + 3 \sqrt{33})^{1/3} + \dfrac{4}{3} (19+3 \sqrt{33})^{-1/3}\cr r_2, r_3 &= \dfrac{1}{3} - \dfrac{1}{6} (19 + 3 \sqrt{33})^{1/3} - \dfrac{2}{3} (19 + 3 \sqrt{33})^{-1/3}) \cr &\pm \frac{i \sqrt{3}}{6} \left((19 + 3 \sqrt{33})^{1/3} - 4 (19 + 3 \sqrt{33})^{-1/3}\right) } $$

L@Suvrit. See my partial answer below.2168384153917222836591788378Very Interesting and i finally learned the correct pronounciation of Erdos through this documentary, thanks!1011994136529Let X be a complex hermitian manifold with hermitian form $\omega$. How can you prove that if $\omega$ has negative holomorphic sectional curvature, then its scalar curvature is negative, too?

42494061994218Suppose $Y$ is a pair of pants with a hyperbolic structure and $\gamma_i; i = 1, 2, 3$ are the geodesic boundaries of length $l_i; i=1, 2, 3$ respectively. Now consider a essential simple arc $\sigma$ in $Y$ with end points on a same boundary component of $Y$ and $l$ denote the length of unique geodesic in the homotopy class of $\sigma$. Then, my question is weather the inequality, $l$ $\geq$ $\frac{1}{2}$ min{$l_i | i= 1, 2, 3$} hold or not for every pair of pants $Y$.

39186052269531147092718180152040560531771632544927830For your second comment, the quantum dimension being 0 or not, it's always true that an endormophism could *potentially* carry more information than a scalar, but it's not obvious to me that it's indeed the case in this context. Do you have an example ?3944891582787@Nik Weaver yes.This is how "discrete-analytic" functions behave.10987088466:extension of smooth function340298If it is the way my eyes see, it is still not difficult to compute, I've got$$ \left(2^7\times3\right)^{2^{n-3}}(n+1)^2(n+2)^2n^6\left(\prod_{k=5}^{n−1}k^{2^{n−k}}\right)^5$$@Jesko -- As I believe others point out below, van der Kallen's argument is completely correct and certainly is not circular. If you take an algebraic group like $\mathbf{G}_m\times \mathbf{G}_m$, a product of two copies of the multiplicative group, then the group of characters has rank $>1$, i.e., it is not the case that all characters are related to, i.e., commensurate with, a single "determinant" character. Thus if you modify your linear representation as van der Kallen suggests by these characters, the associated determinants will also be non-commensurate.1734102Yes, thank you, not sure what happened to my division sign there.I have described my search in 1965-74 for higher order Seifert-van Kampen theorems as: "An idea for a proof in search of a theorem." It took 9 years to find the gadgets to get a theorem. I think it is quite good to look at the work we do as a search for theorems. There is an apochryphal dedication of a PhD Thesis: "I am deeply grateful to Professor X, whose wrong conjectures and fallacious proofs, led me to the theorems he had overlooked." Sounds like excellent supervision!! It also leads to the question: What is a theorem? 11625782124260482834Here I'll show how the MAG-numbers of n=4 are computed. I set up a system of linear equations and solve.

I've simply extended this to n=5 but either I must have a bug in my data or there is a more severe problem which I might have overlooked - the rank of the system of linear equations is defective by 2 ranks. I'm still bug-tracking but I'm confident I'll get it working.

** Background** For some base $t$, with $\small 1 \lt t \lt e$ for the iterable function $\small f(x)=t^x-1$ let's write its logarithm $\small u=\log(t)$ .

*(for the following see the more detailed article APT-decomposition so far; I'll include the essentials here)*

The Carleman-matrix **F** for $\small f(x)$ is lower triangular with powers of $u$ in the diagonal and has coefficients from the Stirling numbers 2nd kind, similarity scaled by the factorials, so that in the second column the entries gived the coefficients for the powerseries of $\small f(x)$ Here is the top left of that matrix
$$ F=\begin{bmatrix} \Tiny \begin{array}{}
1 & . & . & . & . & . \\
0 & u & . & . & . & . \\
0 & 1/2 u^2 & u^2 & . & . & . \\
0 & 1/6 u^3 & u^3 & u^3 & . & . \\
0 & 1/24 u^4 & 7/12 u^4 & 3/2 u^4 & u^4 & . \\
0 & 1/120 u^5 & 1/4 u^5 & 5/4 u^5 & 2 u^5 & u^5 & \cdots \\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots & \ddots \end{array}
\end{bmatrix} $$

For *fractional* iterations we need *fractional* powers of that matrix. We can do this by diagonalization (which can even be done keeping $u$ symbolical)

The diagonalization gives three matrices such that $\small F = M \cdot D \cdot W$ where also $\small W=M^{-1}$.
The columns of $M$ (being the matrix of eigenvectors) can arbitrarily be scaled; I usually norm that such that the diagonal has all units. Then $M$ is also the Carlemanmatrix for the Schröder-function for $f(x)$ - that means, it has the coefficients of its formal power series in its second column. That coefficients can be decoded into polynomials in $u$ in the numerator and factorials and other polynomials in $u$ in the denominator and display precisely your formula involving the MAG-numbers.

$$ M[,1]=\small \begin{bmatrix} \Tiny \begin{array} {rll} 0 & & \\ 1 & \cdot u/1! & /u \\ 1 & \cdot u^2/2! & /u(1-u) \\ 2u+1 & \cdot u^3/3! & /u(1-u)(1-u^2) \\ 6u^3+5u^2+6u+1 & \cdot u^4/4! & /u(1-u)(1-u^2)(1-u^3) \\ 24u^6+26u^5+46u^4+45u^3+24u^2+14u+1 & \cdot u^5/5! & /u(1-u)(1-u^2)(1-u^3)(1-u^4)\\ \vdots \end{array}\end{bmatrix}$$

The inverse $W = M^{-1}$ , which is the Carlemanmatrix for the inverse Schröder-function, has the numbers which you are after:

$$ W[,1]=\small \begin{bmatrix} \Tiny \begin{array} {rll} 0 & & \\ 1 & \cdot u/1! & /u \\ -1 & \cdot u^2/2! & /u(1-u) \\ u+2 & \cdot u^3/3! & /u(1-u)(1-u^2) \\ -1u^3-5u^2-6u-6 & \cdot u^4/4! & /u(1-u)(1-u^2)(1-u^3) \\ 1u^6+9u^5+24u^4+40u^3+46u^2+36u+24 & \cdot u^5/5! & /u(1-u)(1-u^2)(1-u^3)(1-u^4)\\ \vdots \end{array}\end{bmatrix}$$

*(I hope I did not make sign errors here, the reproduction from the Pari/GP-output is a bit messy)*

The problem is, that the diagonalization is a recursive procedure, so the coefficients are defined by recursion which is what you did not want.

In my linked essay on the full symbolic decomposition I proceeded to express from $F^h = M \cdot D^h \cdot W$ (which is the Carlemanmatrix for the h'th iterate $f^{\circ h}(x)$) the symbolic representation of that fractional iterate.

**Key-topic**: I observed, that the Taylor-coefficients of $f^{\circ h}(x)$ are expressible as bivariate polynomials in $k$ (the index of the coefficient) and in $u$ and (separately!) $u^h$ as arguments and I displayed the coefficients of that polynomials $a_k(u,u^h)$ as matrices $A_k$ such that $$a_k(u,u^h) = V(u)\cdot A_k \cdot V^\tau(u^h) $$

$ \qquad \qquad $ (where $V(u)$ means $\text{rowvector}(1,u,u^2,u^3,...)$ up to the appropriate dimension).

For instance the matrix $A_4$

and the matrix $A_5$

$ \qquad \qquad $ $ \qquad \qquad $ *The related MAG-coefficients for $n=4$ and $n=5$ are in the last columns*

The denominators are the same as in the coefficents for the Schröder-function, shown in the picture of $M[,1]$ and for the inverse Schröderfunction in the picture of $W[,1]$.

The key for the possibility to compute the MAG-coefficients directly is now, that for each integer $h$ the columns are down- or upshifted and the full evaluation of the polynomial at $u^h$ must be a multiple of the denominator.

From this a system of linear equations can be set, determined by such integer multiplies of numerator and denominator-polynomials.

As starting values we find, that in the first and last rows are Stirlingnumbers of first and second kind (the latter scaled by factorials) and the rest of numbers are unknown and shall be found by solving that system of linear equations. Apriori- knowledge is (well, hypothese) that for $h=0$ the complete expression vanishes and for $h=1$ the numerator must exactly be equal to the denominator. This seems to make the problem solvable.

*end background*

Here is the solution for $k=n=4$.

As stated above, the matrix $A_4$ provides the coefficients for the polynomial in the numerator of the term at $x^4$ *(in the Taylorseries for $f^{\circ h}(x)$)* . Actually written as a polynomials in $u$ and $u^h$ the first few such polynomials look (with their matrix-arrangement kept):

The red coefficients are unknown when we want to avoid the recursive procedure. To make it more explicit this is the polynomial with the symbolic names for the coefficients where we know that of the first and last row:

The denominator $D_4(u) $ of the 4'th term is the product

Now my hypothese, from which I construct a system of linear equations is, that for the integer iterates the numerator is a polynomial multiple of the denominator - which, btw., explains then that the limit for $u \to 1$ exists for integer iteration heights, but that is doesn't exist for fractional heights (this is completely equivalent to the q-factorials which exist too if $q$ goes to 1 by the limit consideration).

For $h=0, u^h=1$ we have $A_4(u,1)= 0$ and from the hypothese we must have

$$\begin{array} {rrl} & A_4(u,1) &= z \cdot D_4(u) \\
& 0 &= z \cdot D_4(u) \\ \to & z&=0 \\
\end{array} $$
and also follow some more equations by the observation that this means that the rowsums in the matrix must equal zero.

For $h=1, u^h=u$ we have $A_4(u,u)= ?? $ with the highest exponent equalling the highest exponent in $D_4(u)$ and from the hypothese we must have

$$\begin{array} {rrl} & A_4(u,u) &= y \cdot D_4(u) \\
&&\text{also we "know", that }a_1=-1,d_4=1 \\
\to & y&=1 \\
\end{array} $$
From this follow again some more equations for the unknowns.

For $h>2$ (and higher $h$) the polynomial $A_h(u,u^h)$ has higher order, so -if the hypothesis holds- it must be a polynomial multiple of $D_4(u)$ and actually we have $$ A_4(u,u^2) = (x_1+x_2\cdot u + x_3 \cdot u^2 + x_4 \cdot u^3)\cdot D_4(u)$$ This gives 4 new unknowns (where however the first and last are "trivial") and again a new subset of linear equations.

I constructed the following system of linear equations:

*(where I marked the known coefficients with green color and put the 2x4 systematically known coefficients to the rhs of the system to solve it for the unknowns $a_2,...,e_3$ and $x_1...x_4, w_1...w_7$)*

Gaussian elimination provided the sought coefficients $d_1,d_2,d_3,d_4$ of the inverse Schröder-function which are the "MAG$\,_4$" numbers (and interestingly $a_1,a_2,a_3,a_4$ that of the non-inverse Schröderfunction).

So this solves your problem for the fourth coefficients $n=4$ in your post.

This all looks very suggestive to simply proceed to higher $n$, but currently, with $n=5$ (meaning $A_5(u.u^h)$ ) I get the system with defective rank ( 2 missing determinations) and I've a catch that if the $y$ to $u$ informations are used for the Gaussian algorithm then a set of $8 $ coefficients of $A_5()$ form a (minimal) collinear subset.

I do not yet know how to resolve this and am studying the reasons for it. If I cannot resolve the problem, this method becomes possibly useless (which were a pity...)

21321105484559190071674919707081824537How many operad structures are there on the symmetric sequence of simplices / finitely-supported probability measures?2096878dVanishing sum of roots of unity with fixed weight403079

Is there a theorem which classifies irreducible representations of semi-direct product of **finite** groups $G \rtimes A$, where $A$ is a finite abelian group and hence write down the character table for $G \rtimes A$?
In particular, I want to write down the character table for $M_{12} \rtimes \mathbb{Z}_2$ from the character table of $M_{12}$.

Serre's book on representation theory has a theorem on groups of the form $A \rtimes G$ for $A$ abelian, but I am studying groups of the form $G \rtimes A$.

Let $f:X\to S$ be a universal homeomorphism of schemes. Assume $X(S')\neq\emptyset$ for some étale surjective $S'\to S$. Does $f$ have a section?

The answer is yes if $S$ is reduced, by descent. Indeed, note that if $S_1$ is a reduced $S$-scheme then $X(S_1)$ has at most one element. Apply this to $S_1=S'\times_S S'$.

Interesting special case: if $S$ has prime characteristic $p$, let $G$ be a finite locally free $S$-group scheme with connected (i.e. "infinitesimal") fibers, such as $\alpha_p$ or $\mu_p$. Is $H^1_{\mathrm{et}}(S,G)$ trivial?

^"Complete for level $\omega$" looks plausible.279568Also, the $p^{p-3}$ part comes from looking at the Newton polygon.That looks really tempting. There seems to be an off-by one mismatch, (I'm really interested in the coefficients $a_{nk^2 + 1}$), but maybe a bit more care in writing my coefficients down may clear that up.894949If you replace the colouring number with the chromatic number, the question is equivalent to the Erdos-Faber-Lovasz conjecture (http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Faber%E2%80%93Lov%C3%A1sz_conjecture) -- see http://arxiv.org/abs/math/0305073Francois (forgive my limited typeset), I had a visualization difficulty too, until I cut out the rectilinear cube and saw that I just had a union of thin prisms adjacent to one vertex left to cover. It makes me wonder if I can use a "volume-greedy" cover which involves 2 rectilinear cubes and (n-2) other cubes to cover a large enough neighborhood which surrounds the uncovered ring of dimension (n-2) "edges". Gerhard "Ask Me About System Design" Paseman, 2010.12.181694667737788Yes, I got it now. Perfect.713150@KonstantinosGaitanas no for every prime $p$ we can find some $F_{t,k,\ell,\dots,r}$.6http://www.adriansilva.orgI am looking for an appropriate method or hint to solve the following constrained nonlinear least square problem:

$\operatorname{argmin}_X \sum_{i\in I} \|\mathbf{X}_i - \mathbf{X}_{i+1}\|_2^2 + \sum_{i\in I} \|\mathbf{X}_i - \mathbf{Z}_i \|_2^2$ $\text{s.t.}\quad \forall i,j \quad (\mathbf{X}_{i,0} - \mathbf{X}_{i,1})^2 = (\mathbf{X}_{j,0} - \mathbf{X}_{j,1})^2, (\mathbf{X}_{i,3} - \mathbf{X}_{i,4})^2 = (\mathbf{X}_{j,3} - \mathbf{X}_{j,4})^2$

where $\mathbf{X}_i, \mathbf{Z}_i \in\mathbb{R}^d$, and $I$ is a set of samples. The number of variables in $\mathbf{X}$ is about 1 million ($d\approx 100$). $\mathbf{Z}$ is a constant, full-rank matrix.

Thanks for suggestions!

I agree completely that the case of field coefficients and finite type is very important and is used all the time. My comment about unnatural and unuseful was directed just at the general version where the Tor term comes into play, which I took to be the version under discussion. Of course the "unnatural" appellation is subjective so opinions on that may differ.4950581609852That makes a lot of sense! What is the intuition for cuspidal representations?1030150149279819370431607554@JohnBaez is at the same time active on Mathoverflow and contributor of the "[Albert algebra](http://en.wikipedia.org/wiki/Albert_algebra)" page on Wikipedia (Richard Borcherds also).21655651558079This is a second answer (think of it as part two of the other, already long, answer), in response to the clarification of the question by Keshav Srinivasan in the comments on that question.

First, note that there's no such thing as a $\Delta^1_1$ formula; there are pairs of formulas $\phi(X,x)$ and $\psi(Y,x)$ where $\forall X\phi(X,x)$ is $\Pi^1_1$, $\exists Y\psi(Y,x)$ is $\Sigma^1_1$, and $\forall X\phi(X,x)$ is provably equivalent to $\exists Y\psi(Y,x)$. This is important to understanding why you don't get the behavior you might expect: we have a computable list of things that *might* define $\Delta^1_1$ sets, but we don't have a computable way to tell which ones actually do.

The theory which states that every $\Delta^1_1$-definable set exists---that is, the axiom $$(\forall x[\forall X\phi(x,X)\leftrightarrow\exists Y\psi(x,Y))\rightarrow(\exists Z\forall x(\forall X\phi(x,X)\leftrightarrow x\in Z)$$ is called $\Delta_1^1-CA_0$ and has proof-theoretic ordinal $\Gamma_0$; it is actually slightly weaker than $ATR_0$. The reason you don't get more is because there are nonstandard models which get the $\Delta^1_1$ sets wrong---models in which $\forall X\phi$ fails to be equivalent to $\exists Y\psi$ even though in the standard model they would be equivalent, and therefore a set that "ought" to exist fails to.

21613482085759Yes, square (2) is OK for me. What I really want to do is the following: I have 3 circles as input, and I have to find three points p1, p2 and p3 in these circles (a point in each circle) s.t. the angles of the triangle (p1, p2, p3) satisfy the following: the angle at p1 is equal to a given a1. ... p2 ...a2. ... p3 ...q3. Thank you.15092261248301Chris Melekian7488092395784608901130432After quite a bit of tinkering, I decided that the example and a more fully realized generalization merited separate answers, not least because my initial answer entered community wiki due to the number of edits I made.

Let $N$ be a null space matrix for $A$, i.e., the columns of $N$ are annihilated by $A$. We want vectors $w,w',\dots,w^{(m-n-2)}$ s.t. $(Nw)^{\ell+1} = Nw^{(\ell)}$. Define $z^{(\ell)}$ by $z^{(\ell)}_{i(\alpha)} := w^\alpha$, where $i(\alpha)$ is the grlex index of

$\alpha \in$ $ X_{m-n,\ell+1} \equiv$ {$\beta \in \mathbb{Z}^{m-n}: \sum_k \beta_k = \ell + 1$}.

Now let $P^{(\ell)}$ be the $m \times |X_{m-n,\ell+1}|$ matrix with entries given by

$P^{(\ell)}_{j,i(\alpha)} :=$ coefficient of $w^\alpha$ in $(\sum_k N_{jk}w_k)^{\ell+1}$.

To obtain this coefficient explicitly, note that

$(\sum_k N_{jk}w_k)^{\ell+1} = > \sum_{\alpha \in X} > \binom{\ell+1}{\alpha}N_{j,\cdot}^\alpha > w^\alpha$

whence $P^{(\ell)}_{j,i(\alpha)}$ equals

$\binom{\ell+1}{\alpha} > N_{j,\cdot}^\alpha$.

For example, with $N_{j,\cdot} = > (2,3,5,7)$, $\ell = 2$, and $\alpha = > (0,1,1,1)$, so that $i(\alpha) = 6$, we have that $P^{(\ell)}_{j,i(\alpha)} > =$

$\binom{3}{1,1,1} > N_{j,\cdot}^{(0,1,1,1)} = 3! \cdot 3 > \cdot 5 \cdot 7 = 630$.

Extending this example,

$P^{(2)}_{j,\cdot} = > (343,343,735,525,125,441,630,225,189,135,27,294,420,150,252,180,54,84,60,36,8)$.

Then the existence (ignoring distinctness of entries) of $w^{(\ell)}$ s.t. $(Nw)^{\ell+1} = Nw^{(\ell)}$ is equivalent to the existence of a solution to

$(Nw)^{\ell+1} = P^{(\ell)}z^{(\ell)}$.

Note that for all $\ell$ this is really an equation in the components of $w$, viz.

$\left(\sum_k N_{jk} w_k \right)^{\ell+1} = \sum_{\alpha \in X} \binom{\ell+1}{\alpha} N_{j,\cdot}^\alpha w^\alpha$

and it should (at least) be amenable to solution in a computer algebra routine.

53023Commutativity of the Yoneda algebra cannot tell a non-Hopf bialgebra (or even a quasi-Hopf algebra) from a Hopf algebra, though.799778723436188073418647111627101I think you're confusing two different theorems, Whitehead theorem (Wikpiedia: "Whitehead theorem states that if a continuous mapping f between topological spaces X and Y induces isomorphisms on all homotopy groups, then f is a homotopy equivalence provided X and Y are connected CW complexes. ") and Hurewitz theorem ("if X is (n−1)-connected, the Hurewicz map is an isomorphism for all k ≤ n." ). Note the Whitehead theorem doesn't require simple-connectedness.294090784845714446521827617179643929449361170904r$p$-adic Hodge Theory for rigid spaces, after P. ScholzeSee *Graffiti* by Siemion Fajtlowicz. But the great majority of these problems (mainly on graph theory) were not by Fajtlowicz directly but by *Graffiti* itself (only some were jointly obtained by Graffiti and Fajtlowicz), while *Graffiti* is a computer program created by Siemion. (Paul Erdos liked the Graffiti conjectures, so you may too). :-)

Is there any way (general procedure, i mean) to determine if a Euclid Number (En = pn# + 1) is prime or composite? Any research papers exploring this theme are also welcome. Thanks!

1243545295112313069401595071019293hWhy are matrices ubiquitous but hypermatrices rare?1380167 yyzz525991383807amritasu18402551243340oh, good. And why complete intersection is generally irreducible?1987241Let $R$ be an infinite, characteristic zero, commutative ring. I can furthermore suppose it is reduced and indecomposable (no nontrivial nilpotents or idempotents).

My question is whether there is a nonzero polynomial $f\in R[x]$ which is identically zero on $R$.

Note: it is easy to show that there are polynomials with infinitely many roots: let $R=\mathbb Z[s]/(2s)$ and consider $f\in R[x]$ given by $f(x)=\overline{s}x^2+\overline{s}x$. All integers are roots of $f$.

But my question is whether we can have $f$ vanishing on all of $R$, not just on an infinite subset.

On the other hand, if we further mod out by $s^2$, turning $f$ (I believe) identically vanishing, we create a nilpotent element.

A technique that I tried is trying and produce a Vandermonde matrix $V$ associated to the elements $a_1,...,a_k$ of $R$ that be a nonzero element, so that $V$ multiplied by the matrix of coefficients of the canonical basis $e_i$ of polynomials of degree up to $k$, with its $i$-th element replaced by the coefficients of $f$, $f_i$'s, would have two proportional columns and yield $\det(V)\,f_i=0$ and therefore, if I can manage to make $\det(V)$ regular, I will get $f_i=0$.

But I believe we may have reduced rings where all elements are zerodivisors, so I am currently trying to modify this Vandermonde argument, by using the very coefficients of $f$ as $a_i$'s and create a nice Vandermonde lattice.

19653941622085Although the original poster has not said so, I am interpreting connected locally finite to mean essentially that (there exists something which is) the smallest connected component of the poset containing the finitely many chosen elements (and) is itself finite as well as convex within the given poset. Even under this restriction, one does not need large antichains but only infinitely many of them to construct examples with uncountable automorphism group. Gerhard "Ask Me About System Design" Paseman, 2011.09.183856582212839I performed some computations in wolfram alpha looking at the behavior of the values of $|\zeta(e^{ni})|$ trying to predict a lower bound. I have got the following result:

For $n > 19 :|\zeta(e^{ni})|\leq \log(n)$

**My question here is:**
Is the above result true and if it is, how can I show it?

**Note** :$i$ is the unit imaginary part of complex number and $n$ is a positive integer

Thank you for any help

Sounds cool. Can you add a reference so we can read more about it?7844282196978559635This seems like it's exactly what you need:

http://link.springer.com/article/10.1023%2FA%3A1007978107063#

If I understand correctly, your data consists of a set of pixels together with an associated percent value. This is best viewed as a grayscale image, since any metric comparison in 3-space would attribute the units of the (x,y) values to the z values, which is not what you want since these values are percents. The authors introduce a topological metric that quantifies similarity between gray-scale images in a very robust fashion. Namely, the metric you're looking for is the one they call $\Delta_g$. Their analysis concludes the metric is more robust than the aforementioned Sobolev norm.

839232199675420929212044107Dear Franz. If you have access to one of those book, it would be very informative to give one example. Do the strategies introduced have a title? Is the format something like this: title: draw a picture; examples...I assume this is over $\mathbb C$. If they don't commute, about all you can say is that the determinant (which is the product of the eigenvalues, counted by algebraic multiplicity) of $AB$ is the product of the determinants of $A$ and $B$.

1257897233849358158It's still possible to have an example where the parametric definition is a smooth (as in $ C^{\infty} $) function of time but you get a corner, right? Focusing on your specific situation, here's what's happening: Given a pointed map $K \rightarrow X$ and a path $\gamma$ in $X$, we can try to imagine `shifting' this map along the path (think monodromy) and ending up with a new map where the basepoint of $K$ is sent to the endpoint of the path. Any map is basepoint-presrving for *some* basepoint of $X$ (trivially). Choosing a path from that one to $x$ gives a way of taking an arbitrary map and making it basepoint preserving. But there's ambiguity because there are different homotopy classes of paths between these two points.I 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.22497501269630It is completely analytic because it is a limit of polynomial sequence.LBrownian motion on Homogeneous spacesQuestion: a) Is the affine plane curve $y^a=f(x)$ irreducible for all $a\geq 2$ and $d\geq 3$?

b) Same question as in a) but with $k$ of positive characteristic.

1116208625577336861making sure: the forcings you describe kill Suslin trees only at one $\kappa$ at a time, not all the $\kappa$-Suslin trees between $\delta$ and the weakly compact above it?Thanks. I looked at the link, but it seems Macaulay2 is more efficient for me than Singular. If only I can get this working through Sage, where I have a bunch of other code. You don't by chance know how to load M2 packages into Sage? Ideally I'd like to define a function in Sage which takes as input a Sage matrix and a partition and returns the Schur functor applied to that matrix as a Sage matrix. But I couldn't find in the documentation how to load M2 packages in Sage.5616672030778Suppose the attainment of supremum occurs at $h_{0}\in\mathcal{H}$,$Pr\left[sup_{h\in\mathcal{H}}\left|\widehat{E}(h,S_{1})-\widehat{E}(h,S_{2})\right|>\frac{\epsilon}{2}\right]=Pr\left[\left|\widehat{E}(h_{0},S_{1})-\widehat{E}(h_{0},S_{2})\right|>\frac{\epsilon}{2}\right]\leq_{1-2\delta}Pr\left[\left|\widehat{E}(h_0',S'_{1})-\widehat{E}(h_0',S'_{2})\right|>\rho(h_{0})\right]\leq Pr\left[sup_{h'\in\mathcal{H}}\left|\widehat{E}(h',S'_{1})-\widehat{E}(h',S'_{2})\right|>\rho(h_{0})\right]$ is what they argued.11713031737987Your conclusion is correct, but you don't openess of $f$ : if $y\in Y$ and $x\in f^{-1}(y)$, then $O_{Y,y}\to O_{X,x}$ is faithfully flat. 28263Let $X$ be a smooth projective geometrically connected curve over $\mathbf{Q}$ of genus at least two. Fix an algebraic closure $\overline{\mathbf{Q}}$ of $\mathbf{Q}$ and let $G_{\mathbf{Q}}$ be the absolute Galois group of $\mathbf{Q}$. Moreover, fix a rational base point $x$ in $X(\mathbf{Q})$.

Let $\sigma:G_{\mathbf{Q}}\to \pi_1(X)$ be a section of the exact sequence of groups $$ 1\to \pi_1(X_{\overline{\mathbf{Q}}})\to \pi_1(X) \to G_{\mathbf{Q}}\to 1.$$

Let $K\subset \overline{\mathbf{Q}}$ be a finite field extension of $\mathbf{Q}$. (I don't want to assume $K$ to be Galois, but please do if this helps.)

Let $G_K$ be the absolute Galois group of $K$. Then $G_K$ is an open subgroup of $G_{\mathbf{Q}}$. Note that $\pi_1(X_K)$ (with the same base point $x$) injects into $\pi_1(X)$.

**Question.** Does there exist a section $\sigma^\prime:G_{\mathbf{Q}}\to \pi_1(X)$ which is a $\pi_1(X_{\overline{\mathbf{Q}}})$-conjugate of $\sigma$ such that the image of $\sigma^\prime|_{G_K}$ lies in $\pi_1(X_K)$?

*Motivation.* If $a\in X(\mathbf{Q})$, then $a\in X(K)$. Thus, if $\sigma$ is a section associated to $a$, then the answer to the above question is positive. My question is really about sections that *a priori* do not come from a rational point.

*Note.* I always use the base point $x$ to define the fundamental group and we can replace $\mathbf{Q}$ by any number field.

Let me try again. Let $R_n$ be the random variable denoting the longest contiguous run of heads for $n$ independent $p$-biased coin tosses.

It is well-known 1 that $E R_n\sim\log_{1/p}((1-p)n)$ plus small correction terms (the variance is $O(1)$).

This means that $S_n=\sum_{i=1}^n X_i$ grows at least as $\log n$ for any fixed $p$, and proves almost-sure escape to $\infty$.

17819901249970@Donu: Note that the extension is not required to be a *finite* étale map.I'm going to assume $L$ is base point free. I think it is clear how to change what I have written in the case where there are base points (numbers goes down by the degree of the base locus). In general, if $L$ is a line bundle of degree $d$ and $h^0(C,L)= r+1$, then the Clifford index of $L$, written Cliff(L) $= d-2r$. Cliffords theorem is Cliff(L) >= 0. Your line bundle satisfies Cliff(L) = 2. Two ways to achieve that if for the curve $C$ to have a $g^1_4$ or be a plane sextic. The Clifford index of a curve is the min{cliff(L)| $h^0(L)$ & $h^1(L)$ >=2}. For a general curve $C$, Cliff(C) is the floor of (g-1)/2. So for a general curve this can't be done. It is absolutely true that A-C-G-H will have more details and I believe a careful perusal will give the classification of all curves with Clifford Index 2. I think it is only the cases I mention.

101911515724401691542432272745451491592Suppose $X$ is a smooth proper curve over $\mathbb{F}_p$ for some prime number $p$. Let $l\neq p$ be a prime, and suppose $L$ is a rank 2 local system over $X$ with coefficients in $\mathbb{Z}_l$ such that $L$ "looks like" it comes fro an elliptic curve. In paricular:

1) $L$ is of weight 1. I.e. for any $\mathbb{F}_{p^m}$ point of $X$, the Frobenius acting on $L$ has eigenvalues that are Weil numbers of absolute value $\sqrt{p^m}$, and

2) $L$ has coefficient field $\mathbb{Q}$. That is, the corresponding $\pi_1(X)$ representation has traces in $\mathbb{Q}$.

$\textbf{Question}$: Does $L$ come from the Tate module of an elliptic curve over $X$?

$\textbf{Things I know:}$ I know that by work of L.Lafforgue $L$ is motivic, but I don't know how to extract the above statement.

Also, It sees like Drinfelds work on elliptic modules might allow you to find $L$ in the jacobian of some modular curve, which would be enough. This would require $L$ to be Steinberg at one place - though I'm not entirely sure what this means. In any case, I don't quite understand the details well enough.

331829The classical simplicial dual cell approach to PD, as taken in G-H, and modeled on Lefschetz, is done well in Seifert and Threlfall, according to Clint McCrory. He also said the intersection theory attempted by Lefschetz, is worked out in Pontryagin and Glezerman, but not as far as Poincare duality.1994692I think this question doesn't qualify as research level though, so you might want to take it to math.stackexchange.com1152716Out of curiousity, could you add some background about where this problem arose?8The Jantzen-Schaper theorem2084385Is there a bound on the length of the longest Morse trajectory?Let me add an alternative proof I just found.

Using the relation $(-x)_k=(-1)^k(x-k+1)_k$ and $(\frac12)_{n-k}(\frac12-n)_k=(-1)^k(\frac12)_n$, rewrite
$$\sum_{i=0}^{j-1}\frac{(\frac12)_i^2}{i!^2}\frac1{j-i}
=\sum_{k=0}^{j-1}\frac{(\frac12)_{j-k-1}^2}{(j-k-1)!^2}\frac1{k+1}
=\frac{(\frac12)_j^2}{j!^2}\sum_{k=0}^{j-1}\frac{(-j)_{k+1}^2}{(\frac12-j)_{k+1}^2}\frac1{k+1}.$$
The assertion amounts to $\sum_{k=0}^{j-1}\frac{(-j)_{k+1}^2}{(\frac12-j)_{k+1}^2}\frac1{k+1}=\sum_{k=0}^{j-1}\frac4{2k+1}$. Apply $(x)_{k+1}=x(x+1)_k$ so that
$$\sum_{k=0}^{j-1}\frac{(-j)_{k+1}^2}{(\frac12-j)_{k+1}^2}\frac1{k+1}
=\frac{j^2}{(\frac12-j)^2}
\sum_{k=0}^{j-1}\frac{(1-j)_k^2}{(\frac32-j)_k^2}\frac1{k+1}
=\frac{j^2}{(\frac12-j)^2}
\sum_{k=0}^{j-1}\frac{(1-j)_k^2(1)_k^2}{(\frac32-j)_k^2(2)_kk!}.$$
The RHS is a balanced ${}_4F_3$ series and can be transformed by Bailey tract [*Generalized Hypergeometric series*, Stechert-Hafner, New York, 1964, p.56] or any classical reference:
\begin{align}
4F_3[x,y,z,-m;u,v,w]&=\frac{(v-z)_m(w-z)_m}{(v)_m(w)_m} \times \\
&{}_4F_3[u-x,u-y,z-m;1-v+z-m,1-w+z-m,u].
\end{align}
Let $y=z=1, x=1-j, m=j-1, u=v=\frac32-j$ and $w=2$. Then,
\begin{align}
\sum_{k=0}^{j-1}\frac{(1-j)_k^2(1)_k^2}{(\frac32-j)_k^2(2)_kk!}
&={}_4F_3[1,1,1-j,1-j;\frac32-j,\frac32-j,2] \\
&=\frac{(\frac12-j)_{j-1}(1)_{j-1}}{(\frac32-j)_{j-1}(2)_{j-1}}
{}_4F_3[1,\frac12,-j+\frac12,1-j;-j+\frac32,\frac32,1-j] \\
&=\frac{2j-1}j\sum_{k=0}^{j-1}\frac{(\frac12)_k(-j+\frac12)_k}{(\frac32)_k(-j+\frac32)_k} \\
&=\frac{(2j-1)^2}j\sum_{k=0}^{j-1}\frac1{(2k+1)(2j-1-2k)} \\
&=\frac{(2j-1)^2}{2j^2}\sum_{k=0}^{j-1}\left[\frac1{2k+1}+\frac1{2j-1-2k}
\right] \\
&=\frac{(2j-1)^2}{j^2}\sum_{k=0}^{j-1}\frac1{2k+1}.
\end{align}
This completes the proof.

It is GI-complete. Take two arbitrary connected graphs with degrees at least 2 (obviously a GI-complete class). For each vertex $v$, add a new vertex $v'$ and join it only to $v$. The pairs $\{v,v'\}$ have the property you describe.

![Incorrect answer, perhaps mendable, perhaps not, sorry, see my comments below]

The answer is yes, the statement is in fact equivalent to AC.

AC is equivalent to "every set of pairwise disjoint nonempty sets has a choice set". So to see that the statement implies AC we can let ${\cal A}\subseteq{\cal P}(X)$ consist of pairwise disjoint sets.

Let $M\subseteq X$ be a *minimal cover* for ${\cal A}$ , that is: for all $m\in M$ the set $M\setminus \{m\}$ is not a cover of ${\cal A}$. Then $M$ is a choice set for ${\cal A}$.
$$ $$
To see that AC implies the statement, we use Zorn's lemma.

For given ${\cal A}\subseteq{\cal P}(X)$ with the property that $A\neq B \in {\cal A}$ implies $|A\cap B|\leq 1$, let $H\subset {\cal P}({\cal A})\times{\cal P}(X)$ be defined by: $$ $$ $$H := \ \{(K,L)\in {\cal P}({\cal A})\times{\cal P}(X)\ |\ L\ \mbox{is a minimal cover for}\ K\ \}$$

$$ $$ We define a partial order on $H$ using set-inclusion: $$(K,L)\leq(K',L'):= K\subseteq K' \wedge L\subseteq L'$$ $$ $$ For $J=(K,L)\in H$ let $J_0=K, J_1=L$ be the coordinate projections.

We verify that every ascending chain in $(H,<)$ has an upper bound. Let $G\subseteq H$ such that $<$ is a total order on $G$. Then the upper bound $T$ for $G$ is given by:

$$T = (\bigcup_{J\in G} J_0, \bigcup_{J\in G} J_1)$$

It is straightforward to see that $\bigcup_{J\in G} J_1$ is again a minimal cover for $\bigcup_{J\in G} J_0$, so indeed $T$ is in $H$.

It is easy to see that there are non-empty ascending chains. By Zorn's lemma, $(H,<)$ contains a maximal element $U=(V,W)$. We claim that $V={\cal A}$, or in other words: $W$ is the desired minimal cover for ${\cal A}$.

To see that $V={\cal A}$, suppose there is $X\in{\cal A}, X\not\in V$.

case 1) $X\cap W \not= \emptyset$. Then $(V\cup\{X\},W)$ is larger than $(V,W)$, contradiction.

case 2) $X\cap W = \emptyset$. Pick $x\in X$. Then $(V\cup\{X\},W\cup\{x\})$ is larger than $(V,W)$, again contradiction.

9059065080082029156173160111690341648472MrAestheticFor some specific examples the result holds, but I wanted to know if the continuous and dense embedding implied the density of $C(B,g)$, @Michael. I imagine that if $C(B,g)$ contains only 0, the same would happen to $C(H,g)$. Thanks.169874113211I also put this question on MSE here

Let $\Gamma\subset \Omega\subset \mathbb R^N$ be such that $\mathcal H^{N-1}(\Gamma)<+\infty$ (this also implise that $\Gamma$ is Hausdorff measurable).

Let $\mathcal S^{N-1}$ be the unit sphere in $\mathbb R^N$ and let $\nu\in \mathcal S^{N-1}$ be a fixed direction. We set \begin{align}\label{slicing_notation} \begin{cases} \pi_\nu: = \{x\in\mathbb R^N:\,\left<x,\nu\right>=0\};\\ \Omega_{x,\nu}:=\{t\in\mathbb R:\, x+t\nu\in\Omega\}\,\,\text{ for }x\in\pi_\nu;\\ \Omega_\nu: = \{x\in\pi_\nu:\,\Omega_{x,\nu}\neq \varnothing\}. \end{cases} \end{align} That is, the sets $\Omega_{x,\nu}$ are the one-dimensional slices of $\Omega$ indexed by $x\in\pi_\nu$.

We assume that $\Gamma$ has property such that $$ \mathcal H^0(\Omega_{x,\nu}\cap\Gamma)\geq 2 $$ for each $x\in\Omega_\nu$

My question is: Would it be possible to extract a subset $\Gamma'$ from $\Gamma$ such that $\Gamma'$ is $\mathcal H^{N-1}$ measurable and satisfies $$ \mathcal H^0(\Omega_{x,\nu}\cap\Gamma')= 1 \tag 1 $$ for each $x\in\Omega_\nu$?

To satisfy $(1)$ is easy, we just need to choose one point from each set $\Omega_{x,\nu}\cap\Gamma$ for each $x\in\Omega_\nu$ to form $\Gamma'$. However, I found it is hard to make a good choice so that $\Gamma'$ is $\mathcal H^{N-1}$ measurable...

Please advise!

728662@Miha: Of you're absolutely right. I guess that what happens when it's 21:00 and you still haven't had breakfast... :)12752881948654688505`Detecting a hidden convex body with line probes178572521916701892072The following papers prove this:

MR0682456 Reviewed Joris, Henri Une C∞-application non-immersive qui possède la propriété universelle des immersions. (French) [A nonimmersive C∞ mapping having the universal property of immersions] Arch. Math. (Basel) 39 (1982), no. 3, 269–277.

MR0833407 Reviewed Duncan, John; Krantz, Steven G.; Parks, Harold R. Nonlinear conditions for differentiability of functions. J. Analyse Math. 45 (1985), 46–68. (Reviewer: Wiesław Pleśniak) 26E10 (58C25)

MR2179865 Reviewed Myers, Robert An elementary proof of Joris's theorem. Amer. Math. Monthly 112 (2005), no. 9, 829–831. (Reviewer: Clifford E. Weil) 26A24

1271480127293715177207908020978761914690@AliEnayat The new edition of the Shelah and Horowitz paper [Madness and regularity properties](http://de.arxiv.org/pdf/1704.08327.pdf) addresses the question. See theorems 19 and 20 of their paper21545237863747150292270470417087598592ultrafilters' succession79842928022223676613076802295425fHow can I calculate $ \sum_{i=1}^{n} (n \bmod i) $605297It might help to remind people how "tilting object" is defined in this setting. Also, I don't see the point of "linearly" here, since you look only at tori. I guess a similar argument could be mafde for any nontrivial reductive group, at least in characteristic 0. Having infinitely many isomorphism classes of simple modules in a semisimple category seems to be the main ingredient ruling out a tilting object in the derived category.1766941The fact that it's implemented doesn't mean it is solved! CASes have all sorts of routines which 'solve' undecidable problems... because all undecidable problems have (often large) sub-classes which are semi-decidable. It turns out that, for this problem, there is a complete algorithm which is guaranteed to terminate and find all roots. As far as I know, none of the CASes actually implement that (it's much too slow), instead they all implement algorithms which might fail (but with extremely low probability). @Xiao-GangWen: j.c.'s reference is probably as good as any. I'm not sure when it was originally proved. But the basic argument is fairly simple. The action of the mapping class group on $H_i(S^1 \times S^2)$ gives an epi-morphism to $\mathbb Z_2^2$. A little cut-and-paste topology shows you that diffeomorphisms in the kernel must preserve fibres of the bundle $S^1 \times S^2 \to S^1$ (projection onto the first factor), moreover, they must preserve orientation of $S^2$. So by Smale's theorem on $Diff(S^2)$, such an element is in $\pi_1 Diff^+(S^2) = \pi_1 SO_2 = \mathbb Z_2$.H@Andreas: great! thanks for postingAn elliptic curve $C$ over a field $k$ is a smooth, genus 1 curve defined over $k$ with an associated $k$-rational point. If char$(k) \ne 2$, we can show that $C$ has a model of the form $y^2 = f(x)$ where deg$(f) = 3$. The method I have seen in Silverman and Hartshorne finds this model from a change of coordinates of the general Weierstrass equation which you find via Riemann-Roch. My question involves whether you can see this transformation through a different method.

One can show that $C$ admits a degree 2 map to $\mathbb{P}_k^1$. So, as char($k) \ne 2$, the corresponding extension of function fields is a Kummer extension and hence generated by a square root. Thus, we have an affine curve $y^2 = f(x)$ which is birational to $C$. By Riemann-Hurwitz, there should be 4 ramification points which occur either as roots of $f$ or occur at infinity. Thus, $f$ has degree either 3 or 4.

Is there some way to see that the polynomial $f$ can be chosen to have degree 3? That is, can one see **without explicitly writing our the transformation** that one of the ramification points can be chosen to be at infinity over a **non-algebraically closed** field?

The motivation for this question comes from the case of genus 2 where a hyperelliptic curve of genus 2 has a model of the form $y^2 = f(x)$ where the degree of $f$ is either 5 or 6. One can choose a model where $f$ is of degree 5 only when $f$ has a root over the given field $k$.

Essentially my question boils down to, is there some way to view from the function field side that you can choose a model $y^2 = f(x)$ where deg($f) = 3$? Also, from the function field side, can you see that there should be a difference in the case of genus 2?

6986388381631181343802179117551265087Theorem 2.2. in "PIF" says that Namba is semiproper (equivalently: there is some semiproper forcing changing $cf(\omega_2^V)$ to $\omega$) iff player II has a winning strategy in this game: In step $n$, Player I chooses a function $F_n:\omega_2\to \omega_1$, and Player II replies with a value $i_n<\omega_1$. In the end, let $i_\infty:=\sup\{i_n:n \in \omega\}$. Player II wins iff the set $\{\, t\in \omega_2: \sup\{ F_n(t):n\in\omega\}\le i_\infty\,\}$ is unbounded.366091349453Good evening,

In *Riemann surfaces* by Otto Forster there is the following theorem : Let $X$ be a Riemann surface and $P(T)=T^n+c_1T^{n_1}+\ldots + c_n\in\mathcal{M}(X)[T]$ an irreducible polynomial of degree $n,$ where $\mathcal{M}(X)$ is the set of all meromorphic functions on $X.$ Then there exist a Riemann surface $Y,$ a branched holomorphic n-sheeted covering $\pi : Y\to X$ and a meromorphic function $F$ on $Y$ such that $(\pi^{\ast}P)(F) = 0$.

We call $Y$ the algebraic function defined by the polynomial $P(T).$ (I don't restate the uniqueness of this Riemann surface).

**My question :** If $X$ is a compact Riemann surface, can we consider it as an algebraic function defined by some irreducible polynomial $P(T)\in\mathcal{M}(\mathbb{P}^1)[T]$?

I'm thinking of meromorphic functions on $X,$ which we can consider them as holomorphic mappings $X\to\mathbb{P}^1,$ having the smallest positive degree,i.e the inverse image of each point of $\mathbb{P}^1$ contains the smallest number of points. But I'm not sure.

2062576Let $X$ be a smooth projective curve over field $k$. Let $a\in A[n], \mathcal{L}\in A^\vee[n]$.

Recall the definition of Weil pairing (Mumford's Abelian varieties, IV.20): since $\mathcal{L}\in\mathrm{Pic}^0(A)$, we know $n^*\mathcal{L}=\mathcal{L}^{\otimes n}=\mathcal{O}_A$, so after pulling back by $n\colon A\to A$,the line bundle $\mathcal{L}$ becomes trivial, thus corresponds to a factor of automorphy $\chi\colon A[n]\to\mu_n$, the Weil pairing is defined by $e_n(a,\mathcal{L})=\chi(a)$.

To show the Weil pairing coincides with cup product, using SGA41/2 (6.2.2.3)(Duality, Proposition 3.4), it suffices to show the composition $$A^\vee[n]\to H^1(A[n],\mu_n)\to A^\vee[n]$$ is identity. The first map is taking factor of automorphy of the $n$-torsion line bundle (or equivalently, the $\mu_n$-torsor), the second map is collapsing the $A[n]$-torsor $$0\to A[n]\to A\overset{n}{\to} A\to 0$$ to a $\mu_n$-torsor. When we collapse such a torsor, the $n^{2g}$ fibers correspond to $A[n]$ are identified, the scaling is given by a homormophism $A[n]\to\mu_n$. Think this through one see the composition is identity.

1564247178280316255492Impressive calculations!21887441448602215305961831002750998Oh, wait: Will provides examples where the image of $\sigma^2$ has order $p$, so the image of $\sigma$ can have order $2p$. The order $2 \cdot p^9$ for the group suggests it comes from the symplectic form, indeed.What is wrong with this counterexample to primality test assuming GRH?2255869Here is an attempt to prove that the answer is **yes**.

**Claim 1**: Median graphs are bipartite.

This surely appears in the literature and is easy to verify. (Consider for a contradiction the shortest odd cycle and a median of 3 vertices on it: a pair of adjacent ones and a third one "opposite" of this pair.)

**Claim 2**: If $z \neq m(x,y,z)$ then there exists a vertex $z'$ adjacent to $z$ such that $d(x,z')=d(x,z)-1$ and $d(y,z')=d(y,z)-1$. Further, for each such vertex $z'$ we have $m(x,y,z')=m(x,y,z)$.

Let $m=m(x,y,z)$ and let $P(z,m)$ be as in Tony's comment. Then the neighbor of $z$ on $P(z,m)$ satisfies the claim. The second part of the claim holds as one can extend to $z$ the shortest paths between $z'$ and $x$ and $y$.

**Main argument**: By induction on $d(x,y)+d(x,z)+d(y,z)$. By Claim 1 $d(x,y)=d(x',y)\pm 1$ and $d(x,z)=d(x',z)\pm 1$. If the signs in both of these identities are the same then $m(x,y,z) = m(x',y,z)$ by Claim 2. Thus, wlog, $d(x,y)=d(x',y)+1$ and $d(x,z)=d(x',z)-1$.

If $z \neq m(x,y,z)$ then let $z'$ be as in Claim 2. We have $m(x,y,z')=m(x,y,z)$. As $d(x',z') \leq d(x,z')+1 = d(x',z)-1$, by the second part of the claim we have $m(x',y,z')=m(x',y,z)$. We can now replace $z$ by $z'$ and apply induction hypothesis.

We assume therefore that $z = m(x,y,z)$. Symmetrically, $y=m(x',y,z)$. We have

$(d(x,z) + d(z,y)) + (d(x',y)+d(y,z))=d(x,y)+d(x',z) \leq (d(x',y)+1)+(d(x,z)+1)$.

Thus $d(y,z) \leq 1$, as desired.

17779471119794193252618162631784204One can construct a certain non-degenerate form $l$ when $k$ divides $n$. See http://arxiv.org/pdf/1310.1193v1.pdf for example.I post this as an auto-answer mainly not to leave the question open.

After googling a bit better, I discovered two recent works by M. Lapidus and L. Hung (both available on Lapidus' webpage )

“Nonarchimedean Cantor Set and String”, Journal of Fixed Point Theory and Applications 3 (2008), pp. 181-190, (Special issue dedicated to Vladimir Arnold on the occasion of his Jubilee. Vol. I.)

“Self-Similar $p$-Adic Fractal Strings and Their Complex Dimensions”, $p$-Adic Numbers, Ultrametric Analysis and Applications (Russian Academy of Sciences, Moscow, and Springer-Verlag), No. 2, 1 (2009), pp. 167-180.

which seem to pose a "good" definition for a $p$-adic fractal. The definition follows the usual self-similarity one, attaching to each family $\{\Phi_1,\dots,\Phi_n\}$ of similarity contractions $$ \Phi_j:\mathbb{Z}_p\longrightarrow\mathbb{Z}_p $$ the unique non-empy, compact, fixed subset $\mathcal{S}\subseteq \mathbb{Z}_p$ such that $\mathcal{S}=\Phi_j(\mathcal{S})$ for each $1\leq j\leq n$. They develop some theory for such objects and define its Minkowsy dimension, mainly following the box-counting dimension definition . In a closing remark of the second paper, they also say that "it would be interesting to generalize this theory from subspaces of $\mathbb{Q}_p$ to Berkovich spaces" but I was unable to find anything more on the subject.

For $G=\mathbb{R}^d$ I know that a stationary point process $X$ either has 0 or infinitely many points, a.s. Daley and Vere-Jones refer to this as the 0-Infinity dichotomy. They hint that this fact is known in a more general setting. What is the most general setting for which the answer is known? Does it hold for all (locally compact second countable hausdorff topological) groups $G$? References appreciated.

Edit: I'm really interested in when the dichotomy DOES hold. If you have some example where it doesn't hold, please post as a comment unless you feel that your example really does answer my question.

Edit: Obviously $G$ must be non-compact for the dichotomy to hold, else a Poisson process will give a counterexample.

Is $\mathscr{M}_{1,1,\mathbb{Z}}$ isomorphic to a quotient stack by a finite group?Sure, for example, $f$ could behave like $e^{-\log^2 x}$ at infinity.15336841029712 @Gene: [I think it's "projective", not productive?] Regarding your question on first-order formulas, do you allow existential quantifiers over other variables, and do you allow "not equals to" in your formula? If existential quantifiers are allowed, I can think of counterexamples. Specifically, you can use first-order formulas to determine whether an element of the group is a commutator. For instance, we have an inductive family $A_n \hookrightarrow S_n \hookrightarrow A_{n+2} \hookrightarrow S_{n+2} \dots$. In $A_n$, all elements are commutators, in $S_n$, only half are. So, it oscillates.Certainly Cantor diagonalization? You have Russell's paradox, which is perfectly understandable to the lay person, going to the uncountability of reals, going to Godel's incompleteness. I think each incarnation passes Andrew Stacey's tests...

21011542095149405935=I think that the relation of contact structures and CR structures to classical G-structures is a bit more complicated than suggested by Ben McKay's reply, although I certainly agree with his comments on the method of equivalence. Starting with contact structures, the corresponding G-structure corresponds to the stabilizer of a hyperplane in $\mathbb R^{2n+1}$. So such a structure is equivalent to a co-rank $1$ subbundle $H$ in the tangent bundle of a manifold $M$ of dimension $2n+1$. The structure function is the tensorial map $H\times H\to TM/H$ induced by the Lie bracket of vector fields. Contact structures then exactly correspond to the case that this structure function is non-degenerate in each point. For such structures, there is a further reduction of structure group of the associated graded vector bundle $H\oplus (TM/H)$ of the tangent bundle to the conformal symplectic group $CSp(2n,\mathbb R)$ (which is much smaller than the stabilizer mentioned above). The conceptual way to view this in my opinion is as a special case of the concept of a filtered manifold.

In the CR case, the story is a bit more complex. The underlying G-structure is given by a complex subbundle $H\subset TM$ of complex rank $n$ on a manifold of dimension $2n+1$. Note that the homogeneous model of this is $\mathbb C^n\times\mathbb R$, so this has infinite dimensional automorphism group. The structure function again is the tensorial map $H\times H\to TM/H$ induce by the Lie bracket of vector fields. Now as before, you can require this to be non-degenerate, but then there arises a compatibility condition with the complex structure on $H$. Namely you have to require that the above tensor is of type $(1,1)$ with respect to the complex structure, which in CR terms is sometimes called "partial integrability". If this condition is satisfied then you can view the partially integrable almost CR structure as the filtered analog of a G-structure: You have a filtered manifold (a contact manifold as described above) and a further reduction of the structure group of the associated graded vector bundle of $TM$ to the conformal unitary group $CSU(p,q)\subset CSp(2n,\mathbb R)$. Here $(p,q)$ is the signature of the CR structure, with $(n,0)$ corresponding to the strictly pseudoconvex case. For this, the homogeneous model is the unit sphere $S^{2n+1}\subset\mathbb C^{n+1}$, which has finite dimensional automorphism group (isomorphic to $PSU(n+1,1)$). This filtered G-structure now has an intrinsic torsion, namely the Nijenhuis tensor of the partially integrable almost CR structures, whose vanishing is equivalent to integrability of the CR structure.

**Edit** (in view of the comment by @FrancoisZiegler): I am not completely sure what is meant by an almost contact structure. The main point I wanted to make is that while contact forms, contact distributions, and almost CR structures do have an underlying G-structure, one has to impose an additional non-degeneracy condition on the intrinsic torsion (which I referred to as "structure function" in my reply). For contact distributions, this can be equivalently phrased as a filtered manifold structure modelled on a Heisenberg algebra (i.e. as a frame bundle for the associated graded vector bundle to the tangent bundle). In the almost CR case, there are further invariants related to integrability of the almost CR structure. These cannot be defined uniformly for all G-structures of the same type. However, if you impose the non-degeneracy condition as well as partial integrability, then you can view a partially integrable almost CR structure as a filtered analog of a G-structure. Then you can recover the Nijenhuis tensor, which is the natural obstruction to integrability of the CR-structure (and by the way does not show up in the 3-dimensionl case treated in the reference in the answer by @BenMcKay ) as a filtered analog of an intrinsic torsion.

As a consequence of Euler's refection and Gauss's multiplication formulas (mentioned in the Elkies answer), all values $\Gamma(a)$ with $24a\in\mathbb{Z}$ can be expressed algebraically in terms of these values:
$$\Gamma\!\left(\frac12\right)=\sqrt{\pi},\quad \Gamma\!\left(\frac13\right), \quad \Gamma\!\left(\frac14\right), \quad \Gamma\!\left(\frac18\right), \quad \Gamma\!\left(\frac1{24}\right).$$ For example,
$$\Gamma\!\left(\frac5{24}\right)=\frac{\sqrt{\pi}\,\sqrt{\sqrt2-1}\,\sqrt{\sqrt3-1}}{2^{1/6}\,\sqrt{3}}\,\Gamma\left(\frac13\right)^{\!-1}\,\Gamma\!\left(\frac1{24}\right),\\
\Gamma\!\left(\frac7{24}\right)=\frac{\sqrt{\pi}\,\sqrt{\sqrt3-1}\,\sqrt{\sqrt3-\sqrt2}}{2^{1/4}\,3^{3/8}}\,\Gamma\left(\frac14\right)^{\!-1}\,\Gamma\!\left(\frac1{24}\right).$$ These expressions can be found in this paper, along with basic expressions of $\Gamma(a)$ with $60a\in\mathbb{Z}$ that additionally involve

$$\Gamma\!\left(\frac15\right),\quad \Gamma\!\left(\frac25\right), \quad \Gamma\!\left(\frac1{15}\right), \quad \Gamma\!\left(\frac1{20}\right), \quad \Gamma\!\left(\frac1{60}\right), \quad \Gamma\!\left(\frac7{60}\right).$$ Those formulas give $$\frac{\Gamma(\frac{1}{24})\,\Gamma(\frac{11}{24})}{\Gamma(\frac{5}{24})\,\Gamma(\frac{7}{24})} = \frac{\sqrt{6}}{\sqrt3-1}=\sqrt{\frac32}\,(\sqrt{3}+1).$$ This is a simpler answer! Consequently, $$\sqrt{2+\sqrt3}=\frac{\sqrt3+1}{\sqrt2}.$$

I am teaching a topics course on Riemann Surfaces/Algebraic Curves next term. The course is aimed at 1st and 2nd year US graduate students who have have taken basic coursework in algebra and manifold theory, but may not have had much expose to algebraic geometry. I will loosely follow the book Introduction to Algebraic Curves by Griffiths. In particular, I hope to spend a minimum amount of time developing basic machinery (e.g. sheaf theory) and to start doing concrete geometry (e.g. canonical models of curves of genus up to 4) as soon as possible.

My question is: what are some good concrete, accessible geometric topics in Riemann Surface/Curve theory that aren't in the standard textbooks?

Let's say that the standard textbooks are the book I mentioned and those discussed: here.

700488815224598827you might find something useful on [https://books.google.nl/books?id=4FAyDwAAQBAJ&pg=PA68&lpg=PA68&dq=constructive+caratheodory+measure&source=bl&ots=_8kGNWcr7c&sig=P_sMVgDLyUivyBZWz8lSotL0cns&hl=nl&sa=X&ved=2ahUKEwjao5nWgtneAhXS_qQKHQ-WAJgQ6AEwBXoECAcQAQ#v=onepage&q=constructive%20caratheodory%20measure&f=false](pp. 64-75) of Martin Väth's book _Integration theory- a second course_ , this seems to give some possibly criterion for a constructive extension. But I'm not sure and have no time to check...20619762516691493448@user74230: also, could you elaborate on how to lift the polarization (to get Zarhin going)? I realize that in my previous comment I overlooked that a polarization over the Artinian base is necessary.Let $R$ be a commutative ring and $A$ and $B$ two $R$-module. Suppose that $A$ is free of rank $n$ with basis $a_1,\dots,a_n$. Then there is an isomorphism $\Phi: Hom_R(A,B) \to Hom_R(A,R)\otimes_R B$ defined by $\Phi(\sigma)=\sum_{i=1}^n \psi_i\otimes \sigma_i$, where $\sigma_i=\sigma(a_i)$ and $\psi_i$ is the map such that $\psi_i(a_j)=\delta_{ij}$. Is there another way to describe $\Phi$?

148145012874111006451vTannakian Formalism for the Quaternions and Dihedral Group18967921580729460872172371411097831537754223230>@HJRW: Yes, I have fixed that.8434791616557%I have a couple questions regarding the proof of Proposition $3$ (see page $10-11$ of arxiv.org/abs/math/0102039) in Bezrukavnikov's paper "Quasi-exceptional sets and equivariant coherent sheaves on the nilpotent cone" . For simplicity, assume $G$ is simply connected, simple algebraic group (so that $G$ is its own universal cover). Also, $\alpha$ is a simple root, let $P_{\alpha}$ be the associated minimal parabolic and $u_a$ its unipotent radical; $G/B, G/P_{\alpha}$ are the flag variety and a partial flag variety with projection map $p_{\alpha}: G/B \rightarrow G/P_{\alpha}$; $\lambda, \lambda' \in \Lambda$ lie in the weight lattice and satisfy $s_{\alpha}(\lambda) = \lambda - n \alpha, s_{\alpha}(\lambda') = \lambda' - (n-1) \alpha$; and $O_{G/B}(\lambda')$ for instance is the line bundle associated to $\lambda'$. Set $\tilde{\mathcal{N}} = G \times_B \mathfrak{u}$ to be the Springer resolution ($u$ is unipotent radical, as usual).

Question $1$: Set $V_{\lambda'} = p_{\alpha}^* p_{\alpha *} O_{G/B}(\lambda')$.

On the first paragraph of pg $11$ it is claimed that there is a $G$-invariant filtration on $V_{\lambda'}$ with the quotients being $\mathcal{O}_{G/B}(\lambda'-i \alpha)$ for $0 \leq i \leq n-1$; why is this?

Question $2$: Set $\tilde{\mathcal{N}_{\alpha}} = G \times_B u_a$ (I am under the impression this is the same as what Roman defines on pg 11; if not, please correct me). How is the following exact sequence obtained (equation $23$ in the paper)? $0 \rightarrow O_{N'}(\alpha) \rightarrow O_{N'} \rightarrow O_{N'_{\alpha}} \rightarrow 0$

[In the above equation, I have used a ' to replace the tilde, since I cannot get Latex to work.]

Question $3$: Consider the natural projection $\pi': \tilde{\mathcal{N}_{\alpha}} \rightarrow T^{*}(G/P_\alpha)$.

The fibers are clearly projective lines. If $p': \tilde{\mathcal{N}_{\alpha}} \rightarrow G/B$ is the projection, in the last line of the proof (just before Prop $4$), it seems to be claimed that $p'^*(V_{\lambda'}(\lambda - \lambda'-\alpha))$ is isomorphic to a direct sum of copies of $\mathcal{O}_{\mathbb{P}^1}(-1)$ when restricted to any fibre of $\pi'$. Why is this?

EDIT: In all of the above, I had to simplify many of the Latex formulas to make them work (e.g. replace tilde by '). I think everything should be readable now, at the cost of bad notation.

2028761849587https://www.gravatar.com/avatar/70b2072de3d0d43e1af48fd2d416157f?s=128&d=identicon&r=PG&f=11510697While the basic facts were established much earlier, there is a short direct proof by Bryant-Kovacs, Tensor products of representations of finite groups. *Bull. London Math. Soc.* 4 (1972), 133–135For an $n$-tuple of vectors, what does "both" parenthesizations refer to?The Lojasiewicz theorem says that every semi-algebraic subset of $\mathbf{R}^n$ can be triangulated. Moreover, there is a similar statement for pairs of the form (a semi-algebraic set, a closed subset). See e.g. Hironaka, Triangulations of algebraic sets, Arcata proceedings 1974 and references therein (including the original paper by Lojasiewicz).

The case of an arbitrary (not necessarily quasi-projective) complex algebraic variety follows from Nagata's theorem (every variety can be completed) and Chow's lemma (every complete variety can be blown up to a projective one).

148959118999841140905Dave Anderson1446411264652257734279436735808Just operators on functions? The apparatus of BGG gives a description of the invariant operators on sections of invariant vector bundles; see the book of Baston and Eastwood.18093052940917BTrue; text modified accordingly.8Perturbation in Besov space1748384*I suggest using Barton-Sudbery method for $\mathbb K \otimes \mathbb L$ with one adjustment. In space $Tri (\mathbb K) +Tri (\mathbb L)+\mathbb K\otimes \mathbb L +\mathbb K\otimes \mathbb L+\mathbb K\otimes \mathbb L$ change sign in result of bracket of two spaces $\mathbb K \otimes \mathbb L$. I tried this method in 2008 when I created GAP script for generation of all algebras in magic square. However I did not finish my work, so I do not give 100% guarantee.

Here is example with description whan sign should be changed. I show the difference on $so(8,1)$ and $so_9$ Lie algebras example. Let $e_{i,j}$ be matrix having $-1$ on $i,j$ coordinate and $1$ on $j,i$ and zeros otherwise. Let $f_{i,j}$ be matrix having $1$ on $i,j$ and $j,i$ coordinates and zeros otherwise. Let $V_8$ be space generated by $e_{i,9}$ or $f_{i,9}$ for $i=1..8$. One can easily verify following results.

- $[e_{i,j}, e_{k,j}]=e_{i,k}$
- $[f_{i,j}, f_{k,j}]=-e_{i,k}$
- $e_{i,j}$ for $1 \leqslant i < j \leqslant 9$ generate $so_9$
- $e_{i,j}$ for $1 \leqslant i < j \leqslant 8$ and $f_{i,9}$, $i=1..8$ generate $so_{8,1}$

As one can see the $so_{8,1}$ can be obtained from $so_9$ by changing sign on results of $[V_8,V_8]$ brackets.

The same method can be applied to Barton-Sudbery Lie algebras by changing sign on $[A,A]$, $[B,B]$ brackets where $A, B, C$ are three components of $\mathbb K \otimes \mathbb L$.

In my opinion there is no good definition of exceptional Lie groups. One would expect Lie group as being symmetry group of something. For example $F_4$ is defined as isometry group of $\mathbb O P^2$ projective plane over octonions. How to define $\mathbb O P^2$ ? We can use Jordan algebra do define it in algebraic way. How to define it in geometric way ? Next question is how to define "bioctonionic projective plane" $\mathbb C \otimes \mathbb OP^2$ ? Baez in his "The Octonions" defined it as quotient of $E_6$ by the subgroup. Lie group $E_6$ is defined from it's Lie algebra. Lie algebra $e_6$ is defined in abstract way as bracket on some 78-dimensional vector space.

Similar for $E_7$ and $E_8$ there is missing nice definition we expect. Rosenfeld seemed to had some ideas but as you say - it is difficult to read and understand. Freudenthal worked on "octonionic projective geometry" but for me is also difficult to read and understand.

*All things are difficult before they are easy* (Thomas Fuller). Exceptional Lie groups looks difficult, so it means they are not well understood yet.

Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x_n$ to be the sequence given by $x_i=0$ for $1 \leq i \leq n$, $x_i=1$ for $n+1 \leq i \leq 2n$, and $x_{2n+i}=x_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x_n)]= (x_{n+1})$. We have $T^{2n}x_n=x_n$ for every $n \geq 1$, so the measure $\mu_n$ defined by $\mu_n:=(2n)^{-1}\sum_{j=0}^{2n-1}\delta_{T^jx_n}$ is an ergodic invariant Borel probability measure for $T$. Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim_{n \to \infty} \mu_n = \frac{1}{2}(\delta_{\overline{0}}+\delta_{\overline{1}})$, and this limit is not ergodic (since the set containing only the point $\{\overline{0}\}$ has measure 1/2 but is invariant).

There is a nice paper by Parthasarathy - called, I think, "On the category of ergodic measures" - which shows that for this particular dynamical system and some of its generalisations, the set of all ergodic measures and the set of all non-ergodic measures are both weak-* dense in the set of all invariant measures, so this phenomenon can actually happen quite a lot.

(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason. Anyway, it's supposed to be the set of all one-sided infinite sequences of zeroes and ones.)

258050251315487181bHow to choose contour for rational approximation1708736Let $\psi(x):=\sum_{n\leq x}\Lambda(n)$ denotes the 2nd Chebyshev function, where $\Lambda$ stands for the von Mangoldt function. Are there any known (and 'nice') estimates for the change rates $\psi(x+h)-\psi(x)$ for general or special $x$ and $h$?

Thanks in advance,

efq

93634563541536384123894Washington's proof appears in Ribenboim's book 'The New Book of Prime Number Records'. The citation was: "1980, Washington, L. C. The infinitude of primes via commutative algebra. Unpublished manuscript."160549664044393657821467791533731703137123458019067271972156210323617774223266462146891690142367756766943This answer and mine are in much the same spirit. See also Section 2 of my (old, but unfinished) manuscript: http://math.uga.edu/~pete/biconic.pdf. I heartily agree that these examples should be more prominent in introductory courses. @NoamD.Elkies in fact, it seems that Vinberg's algorithm is at least formulated only for forms looking like $-k x_0^2 + \sum_{i=1}^n x_i.$ I don't know how crucial that is...171577315413621722709593301@HassanJolany, this again talks about a NC? Can we say something about reduced pullbacks of irreducible components?5085941711338719727In a recent paper Alexei Pirkovskiy defines a class of Frechet algebras such that the subclass of commutative algebras corresponds to Stein complex manifolds (under a technical assumption of of finite embedding dimension). This may be viewed as noncommutative complex analysis. Link: http://arxiv.org/abs/1204.4936

138903014788176Have you tried asking GAP?What is the difference between

connected

strongly-connected and

complete?

My understanding is:

**connected**: you can get to every vertex from every other vertex.

**strongly connected**: every vertex has an edge connecting it to every other vertex.

**complete**: same as strongly connected.

Is this correct?

5559702048107157064821630872944826Richard, I agree with you. Particularly if the student has actively tried to solve the problem first, and run into a problem or an insurmountable level of frustration after whatever amount of effort was reasonable to them. Even more so if reading and copying the solution really does teach them the "trick", or how that difficult step could have been conceived of if they had tried for a little longer or tried when they weren't so tired..300438996332877221To make explicit what Tim was hinting at, note that a matrix defines a bounded linear operator on $\ell_1$ iff the columns form a bounded sequence in $\ell_1$, in which case the norm of the operator is the supremum of the $\ell_1$ norms of the columns. 261750215601in the same spirit, by the same author: http://arxiv.org/abs/1305.15598649132091811556786Looks like what you have is a product not of eta functions but of Klein forms, which is almost as good. It's $q\prod_{n=1}^\infty (1-q^n)^{c_n}$ where $c_n = 2,0,1,-2$ according as $n \equiv 0, \pm 1, \pm 2, \pm 3 \bmod 7$. cf. the formulas on p.84 of my chapter "The Klein Quartic in Number Theory" = http://library.msri.org/books/Book35/files/elkies.pdf in *The Eightfold Way: The Beauty of Klein's Quartic Curve* = http://library.msri.org/books/Book35/contents.html . 774318... and conversely, any such function will have a FT in $C_0^{\infty}$.CI'm assuming that you have a copy of Dominic Joyce's 1992 JDG article, "Compact hypercomplex and quaternionic manifolds" handy. Write $\frak{g} = \frak{su}(3)$ as a direct sum $$ \frak{su}(3) = \frak{b}\ \oplus\ \frak{d}\ \oplus\ \frak{f}\ , $$ where $$ {\frak{b}} = \left\{\begin{pmatrix}ia&0&0\\0&ia&0\\0&0&-2ia\end{pmatrix}\ \Biggl|\ a\in\mathbb{R}\ \right\}\simeq \mathbb{R} $$ $$ {\frak{d}} = \left\{\begin{pmatrix}ip&q+ir&0\\-q+ir&-ip&0\\0&0&0\end{pmatrix}\ \Biggl|\ p,q,r\in\mathbb{R}\ \right\}\simeq{\frak{su}}(2) $$ $$ {\frak{f}} = \left\{\begin{pmatrix}0&0&-\overline z\\0&0&-\overline w\\z&w&0\end{pmatrix}\ \Biggl|\ z,w\in\mathbb{C}\ \right\}\simeq \mathbb{C}^2\simeq\mathbb{H}. $$ One can easily check that this decomposition satisfies all of the conditions of Lemma 4.1 on page 751 (where, since $n=1$ in this case, I'm omitting the subscripts and summation signs). Now, in Theorem 4.2, we have $k=m=0$, and we can define the complex structures $I_1$, $I_2$, $I_3$ on ${\frak{su}}(3)$ satisfying $I_1I_2=I_3$ as on pages 753–754. (This depends on choosing a basis of $\frak{b}$, so there is a $1$-parameter family of such choices.) By Lemma 4.3, these extend by left-invariance to integrable almost complex structures on $G=\mathrm{SU}(3)$, and so they define a left-invariant hypercomplex structure on $\mathrm{SU}(3)$.

**Addendum:** The OP's request was for a 'description of (B) using the language of (A)', and the above answer doesn't quite do that. Moreover, while thinking about the best way to honor the original request, I took another look at Joyce's paper and realized that there is a sign mistake in one of his formulae that is liable to confuse whoever tries to carry out that description (maybe the OP?) and lead to some frustration.

First, the sign error: When Joyce goes to construct the three complex structures $I_1$, $I_2$, and $I_3$ on ${\frak{f}} \subset {\frak{su}}(3)$ at the top of page 754, he gives the formula $I_a(v) = \bigl[v,\phi_j(i_a)\bigr]$, but this would lead to $I_1I_2=-I_3$ on $\frak{f}$ instead of the desired $I_1I_2=I_3$, so his definition should instead be $I_a(v) = \bigl[\phi_j(i_a),v\bigr]$, and then everything works out correctly.

The upshot is that, if we then introduce a basis $\theta_i$ for the left-invariant $1$-forms on $\mathrm{SU}(3)$ in such a way that the canonical left-invariant form is $$ \gamma = g^{-1}\,\mathrm{d}g = \begin{pmatrix} i(\theta_1+\theta_2) & i\theta_3-\theta_4 & \theta_5 + i\theta_6\\ i\theta_3+\theta_4 & i(\theta_1-\theta_2) & i\theta_7+\theta_8\\ -\theta_5+i\theta_6 & i\theta_7 - \theta_8 & -2i\theta_1 \end{pmatrix}, $$ then the $\mathbb{H}^2$-valued left-invariant $1$-form $$ \theta = \bigl(\theta_1+i\theta_2 + j\theta_3 +k\theta_4\,,\ \theta_5+i\theta_6 + j\theta_7 +k\theta_8 \bigr) $$ satisfies $\theta(I_1v) = i\theta(v)$ and $\theta(I_2v) = j\theta(v)$ for all $v\in {\frak{su}}(3)$. As a result, the $\mathbb{C}^4$-valued $1$-forms $$ \alpha = \bigl(\ \theta_1{+}i\theta_2\ \ \theta_3{+}i\theta_4\ \ \theta_5{+}i\theta_6\ \ \theta_7{+}i\theta_8\ \bigr) =(\alpha_1\ \alpha_2\ \alpha_3\ \alpha_4) $$ and $$ \beta = \bigl(\ \theta_1{+}i\theta_3\ \ \theta_2{-}i\theta_4\ \ \theta_5{+}i\theta_7\ \ \theta_6{-}i\theta_8\ \bigr) =(\beta_1\ \beta_2\ \beta_3\ \beta_4) $$ satisfy $\alpha(I_1v) = i\alpha(v)$ and $\beta(I_2v) = i\beta(v)$ for all tangent vectors $v\in T\mathrm{SU}(3)$. Consequently the $\alpha_i$ are a basis for the $(1,0)$-forms on $\mathrm{SU}(3)$ with respect to the almost-complex structure $I_1$ while the $\beta_i$ are a basis for the $(1,0)$-forms on $\mathrm{SU}(3)$ with respect to the almost-complex structure $I_2$.

Finally, using the identity $\mathrm{d}\gamma = -\gamma\wedge\gamma$ to get the formulae for $\mathrm{d}\theta_i$, we find that $$ \mathrm{d}\alpha_i\equiv 0\ \mod\ \alpha_1,\alpha_2,\alpha_3,\alpha_4 $$ while $$ \mathrm{d}\beta_i\equiv 0\ \mod\ \beta_1,\beta_2,\beta_3,\beta_4\,. $$

Thus, $I_1$ and $I_2$ (and hence $I_3 = I_1I_2=-I_2I_1$) are integrable complex structures on $\mathrm{SU}(3)$, and thus the triple $(I_1,I_2,I_3)$ defines a hypercomplex structure on $\mathrm{SU}(3)$.

13520131125029If $m\neq0$ then pick $0\neq x\in m$. The map $r\mapsto xr$ is a nonzero right module homomorphism from $R$ to $m$. So $R/m$ is a subquotient (as right module) of $m$ and so $|m|\geq |R/m|$.

But $|R|=|m||R/m|$, so if $R$ is infinite then $|m|=|R|$.

1448291690306See the OP's update http://mathoverflow.net/questions/250266/average-distance-between-objects-in-a-cube#comment615365_2502661310462115510918154131587577As Margaux has already observed, there exist groups $K$ which consist only of semi-simple elements and yet need not be contained in the normalizer of a maximal torus.1413625That sum looks a little bit like the McShane identity. http://en.wikipedia.org/wiki/McShane%27s_identityWikipedia isn't too bad... http://en.wikipedia.org/wiki/Frölicher–Nijenhuis_bracket The basic properties are probably pretty easy to prove.253370:This is very useful- thanks!`@horaceT updated another source and hope helps!1293256FGreat, this is much more concreet!For a smooth, real surface $\Sigma$, its bundle of symmetric, bi-linear forms $S^2T\Sigma$ reduced to a $PGL(2,\mathbb{R})$ structure. A similar reduction(with different structure group) can be done for other tensor bundles.

Where can I find a good reference on this type of reductions?

$PGL(2,\mathbb{R})$ acts by conjugation on the space of square matrices $M(3,\mathbb{R}).$ Can one find any invariant polynomials associated to this action?

See

17034481072332389748true but the Op restricted the range to $m\ge 2$ so I didn't mention the case $m=1$.7472951|Does existence of a proper class model imply the consistency?1288226732443I'm still thinking over your email (thanks! it's very interesting). I hope the answer wasn't overly harsh or unfair. There are probably lots of different frameworks that could be considered.187756350252516748664719358716292959315992738271021888869899047842661I see, I wasn't aware that there are spaces with finite homological dimension, but infinite *co*homological dimension! I wonder if one would still find counter-examples if both dimensions are assumed to be finite....3198985430141591354221061If $E$ has no rational $3$-torsion point, will $\rho$ still have the same representation $?$427140Perhaps if a topology containing more information were found, like how the etale topology gives us crucial information that the Zariski topology misses, it might be worth it to weaken our assumptions to this weaker topology, but going from hausdorff to general topologies is a step in the opposite direction.161460520268022094692It’s not known to me at any rate. It actually looks plausible to me that the recursive complex numbers are an elementary submodel, but I’m not really an expert on these matters.7969016036049910853992693333489746357Consider the subgroup $N:=\langle k\rangle\subset \mathbb{Z}^n$. There exists a basis $f_1,\dots,f_n$ of $\mathbb{Z}^n$ such that $uf_1$ is a basis of $N$, where $u\in \mathbb{Z}$, $u>0$, see Vinberg, A Course in Algebra, Thm. 9.1.5, or Lang, Algebra, 3d ed., Thm. III.7.8. Changing, if necessary, $f_1$ to $-f_1$, we may think that $k=uf_1$, $u>0$. If you assume that your vector $k$ is not divisible by any positive integer different from 1, you obtain that $k=f_1$. Let $e_1,\dots,e_n$ be the standard basis of $\mathbb{Z}^n$. If follows easily that $S:=k^\perp=f_1^\perp$ is isomorphic to $e_1^\perp=\mathbb{T}^{n-1}$.

fBounding near the boundary for a Sobolev function.665415@Greg Graviton--No, Hausdorff was NOT a supporter of Nazis (please, use pronouns responsibly). It seems that Gentzen was; Gentzen was a member of Nazi parties and organizations. I hope that Gentzen was only trapped by History, that he was not actively any monster. Except for Genzen's Nazi memberships (as repulsive as they were) I didn't read anything damaging about Gentzen.(It works in characteristic 0 because we have the exponential sequence.)120045user32033@a result of soul theorem,right?15387476703621980850770764676027817520932775042052137See, e.g., http://math.stackexchange.com/questions/36798/what-is-the-math-behind-the-game-spot-it and http://stackoverflow.com/questions/6240113/what-are-the-mathematical-computational-principles-behind-this-game.2083538110548420246151534740It's a great book! Biggs, Lloyd & Wilson each write great papers on the history of mathematics.635509248616146058725295311028015I would like to see a good answer to this question! What I write below is a collection ideas that I think are relevant.

Cluster algebras provide one way to generate non-trivial instances of the Laurent phenomenon, yet there seem to be many different kinds of recurrence relations which exhibit such magic, some of them highly nonlinear, such as $$x_{n+3}x_n^3x_{n-1}=x_{n+2}^3x_{n-1}^3-x_{n+2}^2x_{n+1}^3x_{n-2}+a(x_{n+1}x_n)^6.$$ Much of what I'm saying here comes from an article by A. Hone, "Laurent Polynomials and Superintegrable Maps". One can view a recurrence relation $$x_{n+k}=F(x_n,\dots,x_{n+k-1}) \mathrel{\mathop :}= F(\mathbf{x}_n),$$ as an iteration of the map $$\varphi:(x_0,\dots,x_{k-1})\to (x_1,\dots,x_{k-1},F(\mathbf{x}_0)),$$ and therefore as a discrete dynamical system, say over $\mathbb R^k$ or $\mathbb C^k$. It turns out that a lot of the combinatorial properties of the recurrent sequence are in agreement with the behavior of $\varphi$ as a discrete dynamical system.

I interpret the method that you sketch in your question about "linearising" using joint recurrences as a sort of analog of "separation of variables". Being able to use separation of variables is one of the characterizing properties of what people call integrable systems. Therefore it makes sense to look for an answer among the recurrences which give rise to *discrete integrable systems* (I understand there is a large literature on these).

From this perspective, it becomes evident that linearising using joint recurrences should have something to do with having "conserved quantities", i.e. expressions in the terms of the sequence that remain constant as the index varies.

With this in mind, let us look at the example of the Somos-4 sequence $$x_{n+4}x_{n}=\alpha x_{n+3}x_{n+1}+\beta x_{n+2}^2.$$ I believe the reference here is an earlier paper, "Integrality and the Laurent phenomenon for Somos 4 sequences", by C. Swart and A. Hone. Where they use the fact that the corresponding discrete dynamical system is integrable to conclude the Laurent phenomenon.

The expression $$T=\frac{x_{n-1}x_{n+2}}{x_nx_{n+1}}+\frac{\alpha x_n^2}{x_{n-1}x_{n+1}}+\frac{x_{n-2}x_{n+1}}{x_{n-1}x_n},$$ turns out to be independent of $n$. Denoting $\mathcal{I}=\alpha^2+\beta T$, the authors prove that, in fact, we have $x_n\in \mathbb Z[\alpha, \beta, \mathcal{I}, x_1^{\pm}, x_2,x_3,x_4]$.

This is done by introducing the sequence $w_n$ satisfying $w_1=1, w_2=-\sqrt{\alpha},w_3=-\beta,w_4=\mathcal{I}\sqrt{\alpha}$, as well as $$w_{2m+1}=w_m^3w_{m+2}-w_{m+1}^3w_{m-1} \quad, \quad w_{2m+2}=\frac{w_{m+2}^2w_{m+1}w_{m-1}-w_{m}^2w_{m+1}w_{m+3}}{\sqrt{\alpha}}.$$ Now the desired property follows from examining the recurrences $$x _{2m+1}=\frac{w _m ^2x_mx _{m+2}-w _{m+1}w _{m-1}x _{m+1} ^2}{x _1}$$ and $$x _{2m+2}=\frac{w _{m+2}w _{m-1}x _{m+1}x _{m+2}-w _mw _{m+1}x _m x _{m+3}}{\sqrt{\alpha}x_1}.$$

This kind of auxiliary recurrences might have not been what you had in mind, but I thought it might be relevant, and perhaps attract some expert's opinion. It would be great if the connection between discrete integrable systems and the Laurent phenomenon was better understood, and we could treat such results systematically.

180853118591971837702user797312940251310967Note that the group of the square and the quaternion group have identical character tables but very different subgroup structures.1553193you're right, I forgot the ^3 - it's corrected now. I'll look at the references!226633There is a free open-source implementation at http://sourceforge.net/projects/fastgausslegendrequadrature/.

49398168526811218331425134Sorry. I just now realized n=(2k+1)2^t was a parameter. t+2 then.728916212613617903331459085@GerryMyerson Ah okay, so then the answer would be "no" there isn't a simple algorithm because it's at least NP-hard. Should I delete my question then? Or do you want to add that as an answer?4941787Doug Zare's answer says $d=c$ is impossible for large enough $n$.18073051935487170700515187328Very enlightening. Thanks!51022521189112030931The choice of $\Phi$ is equivalent to the choice of a splitting, since $\Omega_F$ is profinite and $\widehat{\mathbf{Z}}$ is the ``free profinite group on one element." In my answer I'm making the resulting splitting homomorphism (semi-)explicit.240367470167@Allen: these conventions of index theory confused me as well at first but actually they seem to make sense after a while. Analytical index is related to kernel of some differential operator, which involves solving differential equations, an obviously analytical task. On the other hand, topological index only involves algebraic topology. Of course, differential forms can appear if you work within de Rham setting but this is not necessary -- one can prefer to work with K-theory instead, a purely topological object.1961581zContext for "Coronidis Loco" from Weil's Basic Number Theory596555I guess you want S to start off being the set of row vectors of F.'I'm hoping that the following are true. In fact, they are probably easy, but I'm not seeing the answers immediately.

Let $M$ be a smooth $m$-dimensional manifold with chosen positive smooth density $\mu$, i.e. a chosen (adjectives) volume form. (A *density* on $M$ is a section of a certain trivial line bundle. In local coordinates, the line bundle is given by the transition maps $\tilde\mu = \left| \det \frac{\partial \tilde x}{\partial x} \right| \mu$. When $M$ is oriented, this bundle can be identified with the top exterior power of the cotangent bundle.) **Hope 1: Near each point in $M$ there exist local coordinates $x: U \to \mathbb R^m$ so that $\mu$ pushes forward to the canonical volume form $dx$ on $\mathbb R^m$.**

Hope 1 is certainly true for volume forms that arise as top powers of symplectic forms, for example, by always working in Darboux coordinates. If Hope 1 is true, then $M$ has an atlas in which all transition maps are volume-preserving. My second Hope tries to describe these coordinate-changes more carefully.

Let $U$ be a domain in $\mathbb R^m$. Recall that a change-of-coordinates $\tilde x(x): U \to \mathbb R^m$ is *oriented-volume-preserving* iff $\frac{\partial \tilde x}{\partial x}$ is a section of a trivial ${\rm SL}(n)$ bundle on $U$. An *infinitesimal change-of-coordinates* is a vector field $v$ on $U$, thought of as the map $x \mapsto x + \epsilon v(x)$. An infinitesimal change-of-coordinates is necessarily orientation-preserving; it is volume preserving iff $\frac{\partial v}{\partial x}(x)$ is a section of a trivial $\mathfrak{sl}(n)$ bundle on $U$. **Hope 2: The space of oriented-volume-preserving changes-of-coordinates is generated by the infinitesimal volume-preserving changes-of-coordinates, analogous to the way a finite-dimensional connected Lie group is generated by its Lie algebra.**

Hope 2 is not particularly well-written, so Hope 2.1 is that someone will clarify the statement. Presumably the most precise statement uses infinite-dimensional Lie groupoids. The point is to show that a certain *a priori* coordinate-dependent construction in fact depends only on the volume form by showing that the infinitesimal changes of coordinates preserve the construction.

** Edit:** I have preciseified Hope 2 as this question.

I'm looking for a reference on linear, bounded, self-adjoint operators defined on the product space, $T:E\times F\to E\times F$ such that $$Tx = \begin{pmatrix}A & B \\ C & D \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$$

In particular, what is the relation between the spectrum of $T$ and the spectrum of $A,B$, are there any inequalities with norms? Currently, I can only find some scattered references, like http://www.sciencedirect.com/science/article/pii/S0024379598102197 which says that $$\sigma(T)\subset\sigma(A)\cup\left\{\lambda\in \rho(A):\|(A - \lambda)^{-1}\|^{-1}\leq \|C\|\right\} \cup\sigma(B) \cup \left\{\lambda\in \rho(D):\|(D - \lambda)^{-1}\|^{-1}\leq \|B\|\right\}$$ Is there a systematic treatment of such things?

1285473For students, probably the most elementary thing to do would be to restrict yourself to regions in the plane that can be triangulated with finitely many triangles (with straight sides). Accepting the area of a triangle as known, you then define the area of a region by adding up the areas of the triangles in a finite triangulation. The one thing you have to check is that this is well-defined. For this, I would first prove that any two finite triangulations of a region in the plane have a common subdivision (if you're reasonably clever about it, this can be done very quickly; certainly in 3-5 pages), and then prove that your notion of area is invariant under subdivisions.

The nice thing about this is that all the main properties you want (eg that areas behave correctly under linear maps and translations) come for free from the analogous properties of triangles, which are easy.

Unfortunately that is not true. Let $Z$ be a smooth, projective, genus $0$ curve. Let $f:Y\to Z$ be a Hirzebruch surface fibered over $Z$ whose "directrix" $H$ has negative self-intersection, i.e., any Hirzebruch surface other than $\mathbb{P}^1\times \mathbb{P}^1$. The divisor class of $H$ is $f$-relatively ample, yet the only global sections of $\mathcal{O}_Y(m\underline{H})$ are scalar multiples of the section whose zero scheme is $m\underline{H}$.19883551284737My question is about overlapping a random Voronoi cell and a circle.
Suppose there are some Poisson Voronoi cells generated by a homogeneous Poisson Point Process with density λ and Voronoi partitioning. For an arbitrary cell region, consider a circle with fixed radius r centred at the seed (or site) of the cell, what's the average perimeter of the intersection of the circle and Voronoi cell region?

I can see that the overlapped area might have a mixture of Voronoi cell & circular boundary or purely Voronoi or purely circular depending on the realization of the Voronoi cell. Another thing I found is that the average boundary of a Poisson Voronoi cell is $\dfrac{4}{\sqrt{\lambda}}$, which should give us an upper bound of the perimeter.
Any hint on how to approach this problem? Much appreciated.

"Sufficient unto the day is the rigor thereof."-E.H.Moore.

There's a lot of discussion over not only the role of rigor in mathematics,but whether or not this is a function of time and point in history.Clearly,what was a rigorous argument to Euler would not pass muster today in a number of cases.

Passing from generalities to specific cases,I think the prototype of statements which were almost universally accepted as true without proof was the early-19th century notion that globally continuous real valued functions had to have at most a finite number of nondifferentiable points.Intuitively,it's easy to see why in a world ruled by handwaving and graph drawing,this would be seen as true. Which is why the counterexamples by Bolzano and Weirstrauss were so conceptually devastating to this way of approaching mathematics.

Edit: I see Jack Lee already mentioned this example "below" in his excellent list of such cases.
But to be honest,I don't think his first example is really about rigor so much as a related but more profound change in our understanding how mathematical systems are created. The main reason no one thought non-Euclidean geometries made any sense was because most scientists believed Euclidean geometry was **an empirical statement** about the fundamental geometry of the universe.Studies of mechanics supported this until the early 20th century;as long as one stays in the "common sense" realm of the world our 5 senses perceive and we do not approach relativistic velocities or distances,this is more or less true. Eddington's eclipse experiments finally vindicated not only Einstein's conceptions,but indirectly,non-Euclidean geometry-which until that point,was greeted with skepticism outside of pure mathematics.

I think there is a typo in the references to "Supersymmetry and Morse theory", [21] should be replaced by [22]="Constraints on supersymmetry breaking". The quantization of non-linear sigma models and its relation with the de Rham complex is discussed in Section 10 of this paper.

In addition to the big "Mirror symmetry" book already mentionned in the comments, a useful reference is in the book "Quantum Fields and Strings: A Course for Mathematicians" (http://bookstore.ams.org/qft-1-2-s/ ), Part 1, Supersolutions, Chapter 3 (p261-276), where one can find a detailed derivation of the supersymmetry of the Lagrangian, and Chapter 4 (p279-289), where a more conceptual point of view on this Lagrangian is given via some superspace formalism (in particular, in superspace formalism, the expression of the Lagrangian looks as simple as a single quadratic term, and one can recovers all the complicated terms such as the Riemann curvature-4-fermions terms by a simple systematic expansion in components).

Here is a brief sketch of the relation with the de Rham complex. Fields of the theory are $\phi^i$, $\psi^i$, $\overline{\psi}^i$. Upon canonical quantization, the Hilbert space of the quantum theory is the $L^2$-space of differential forms on $M$, quantum supercharges are $d$ and $d^{*}$ operators and so the quantum Hamiltonian is the Laplacian $\Delta=\{d,d^{*}\})$.

The relation between the explicit form of the Lagrangian and the Laplacian=quantum Hamiltonian is quite subtle because of the general phenomenon of ordering ambiguities in quantization. It is a good exercise to write explicitely the classical Hamiltonian from $L$ but it will not tell you immediately what the quantum Hamiltonian is the Laplacian because there are ordering ambiguities. The point of the supersymmetric story is that in this case the Hamiltonian is constructed from the simplest objects which are the supercharges, and that there is no ordering ambiguity to construct the quantum supercharges.

bVolume of normal cone of a simplex (at a vertex)8682514http://www.nomad-labs.comSpecifically, let $F_n$ be the based self homotopy equivalences of $S^n$, and $G_n$ the unbased self homotopy equivalences of $S^{n-1}$. There is a map $G_n \to F_n$ given by unreduced suspension. The relation between the Madsen-Milgram definition and the classical one is given by the composite $O_n \to G_n \to F_n$, since the homotopy groups of $F_n$ are the homotopy groups of spheres (as $F_{n}$ is the degree $\pm 1$ components of $\Omega^n S^n$ and the latter is an associative $H$-space). 89850660847I understand your point of view. The last line of this posting's description "not just so I can write more honest research proposals, or in case it comes up in an argument, but so I have an answer for myself" lead me to believe that my post was relevant. Sometimes things become clear when you attempt to answer them from a more general point of view.167304614139681333850361099imsky473677373015It's entirely possible that I flubbed a constant factor somewhere, but the amplitude seems right: for $n=101$ I get $f(0) = 7.9988...$ and $(2n/\pi)^{1/2} = 8.0186...$ .1349673For searching, note that "linegraph" is often written "line graph" or "line-graph". Of course, in mathematics (as in any other aspect of life, really) one can't always get the sort of answers one wants.223822147004221628021231481430827428986225308143299660599410150861834849731338I do not understand your answer! if the answer is no, you need to prove the following: for a fixed fibrant object $a\in M$ there is no fibrant object $b\in N$ such that $a\rightarrow U(b)$ is an equivalence.Thanks to all three! I more or less saw that `$\mathbb{Z}_p^t/(p)$` is local with nilpotent maximal ideal, but I couldn't go further and I guess that one can't say much more than that. If the comments are turned into an answer, I'll accept it.LWhich of these are are cohomological?ThomasMBerndt2144997203955223167842663Indeed, I need it to solve another problem in astronomy. It seems that the system in nonlinear but I am not sure about it, maybe if we do some substitutions then easily can be rewrite it in a matrix form and solve it; I do not know other possible approaches. I have tried stack exchange but have not get any answer yet. Yes the square root cover x+y in both equations1121542946840(Sorry, I can't make the symbols work.) Isn't this just $G*_A A\rtimes B*_B H$? @ZeraouliRafik: Barry Cipra's answer in the stackexchange question answers your first question here. https://math.stackexchange.com/a/2826712/32285 As for your second question, the sequence $a_n$ might not converge, but one can show (using Barry Cipra's argument) that $a_{2n}$ is decreasing, $a_{2n+1}$ is increasing, such that $a_{2n+1} < a_{2m}$, and hence each sequence has a limit. It appears that numerically the two subsequences have different limits, and hence it seems unlikely that the sequence $a_n$ has a limit. So I'm guessing that both of your questions are unfounded.I'm wondering what the statement is that one has to prove for the Millenium Problem "Quantum Yang-Mills Theory".

According to the official article, it is required to show that for every simple Lie group G there exists a YM quantum field theory for G with a mass gap. Finding a quantum field theory amounts to finding a Hilbert space H, a representation of the restricted Lorentz group by unitary transformations of H and operator valued tempered distributions $\varphi_1,...,\varphi_m$ which satisfy density conditions, transformation properties under the Poincare-group, (anti-)commutativity of field operators for test functions of space-like separated supports, an asymptotic completeness property and existence of a unique vacuum state, the field operators acting on this vacuum state span (a dense subspace of) H.

A quantum field theory has a mass gap if the spectrum of the energy operator is contained in $\{0\} \cup [a,\infty)$ for $a>0$.

Now these are straightforward definitions, but what turns a quantum field theory into a YM theory for a group? I know of classical YM theory which gives a Lagrangian and thus equations of motion for the curvature components (fields). Does that mean that $\varphi_1,...,\varphi_n$ have to satisfy these equations? In what way? I could not find any reference for this. Is this millenium problem a mathematical statement at all?

Well, thanks for writing this. An advantage of C having finite colimits and finite limits is that Ind(C) is lfp and Pro(C) is co-lfp i.e. they can be approximated in some sense. It also holds for the examples I have in mind, which I will now add.21234981760202Can an algebraic variety over a field $k$ be the union of proper closed subsets $(S_i)_{i\in I}$ with $I < k$1046105@FanZheng: Thanks for the comment. But, we need concrete proofs if you don't mind my saying.1849789763557@AntonKlyachko I've added a followup question for our friend $F_2$ http://mathoverflow.net/questions/138672/sets-from-f-2n-which-are-not-fixed-by-any-non-identity-isomorphism1856392737403^Product of random diagonals on the unit circle2177802 Ryan -- yes, I meant to say the spectral sequences degenerate and one can compute each $E^{pq}_\infty$; these will be finite dimensional and one can in principle construct explicit cohomology classes and evaluate them on each knot. In general (that is if one considers embeddings whose source is not necessarily the circle) codimension 3 does not suffice: one needs the discriminant to be of real codimension $>1$ in order for the spectral sequence to degenerate, so by counting parameters if $n$ is the dimension of the source, the dimension of the target must be at least $2n+2$.32487678785129538461633815112775816786868@Praphulla Random thought!2952862Ivan V.1392751536667416342Its limit set should be homeomorphic to a circle. Do you want an algebraic or geometric criterion? Do you have an application in mind?89645648187114322782124281916476Let \begin{align} \Omega=\begin{bmatrix} L_1 & \cdots & L_{n-1} \\ M_1 & \cdots & M_{n-1} \end{bmatrix}\end{align} be a matrix of linear forms on $\mathbb{P}^n$, i.e. homogeneous polynomials of degree $1$, such that for any $[\lambda, \mu] \in \mathbb{P}$, the linear forms $\lambda L_1 + \mu M_1,\dots,\lambda L_{n-1}+\mu M_{n-1}$ are linearly independent. For each $[\lambda, \mu]$, the equations $\lambda L_1 + \mu M_1=\dots=\lambda L_{n-1}+\mu M_{n-1}=0$ define a line $\ell_{\lambda,\mu}$ of $\mathbb{P}^n$. The union of all the lines as $[\lambda,\mu]$ varies is precisely the rank-1 locus variety of $\Omega$.

**Question:** Does there exist an open set $\mathcal{U}$ of hyperplanes of $\mathbb{P}^n$, such that every hyperplane $H \in \mathcal{U}$ does not contain any $\ell_{\lambda,\mu}$? (Here we parametrize the hyperplanes of $\mathbb{P}^n$ by elements of $\mathbb{P}^n$: each such element defines the normal to a hyperplane.) In other words, can we say that a general hyperplane $H$ of $\mathbb{P}^n$ does not contain any of the lines $\ell_{\lambda,\mu}$?

In Joel David Hamkins's "Well-founded Boolean Ultrapowers as Large Cardinal Embeddings", it is mentioned that if $U \in \mathbf{V}$ is an ultrafilter of a complete Boolean algebra $\mathbb{B}$ and $U$ is non-generic over $\mathbf{V}$, then either $U$ misses a countable maximal antichain of $\mathbb{B}$ in $\mathbf{V}$, or it misses a maximal antichain of size a measurable cardinal.

What I am confused about is, if we force in $\mathbf{V}$ with an atomless forcing notion that does not add any countable sequence of elements of $\mathbf{V}$ (eg. any atomless countably-closed forcing), we get to a generic extension $\mathbf{V}[G]$ with $G$ non-generic over $\mathbf{V}[G]$, and all countable maximal antichains of $\mathbb{B}$ in $\mathbf{V}[G]$ are already in $\mathbf{V}$. So $G$ meets all countable maximal antichain in $\mathbf{V}[G]$, and since $\mathbb{B}$ remains atomless, $G$ is not $\mathbf{V}[G]$-generic. Does this imply the existence of a measurable cardinal in $\mathbf{V}[G]$?

1221712203717@Turbo I still don't get it you mean for every $p$ we can find a $F_{t,k,l,...,r}=p$?702526For n=2 the answer is yes.https://www.gravatar.com/avatar/83c77e18eacdab7b1414b7c9ae5ae01a?s=128&d=identicon&r=PG&f=1If I understand your question correctly, an answer should appear in the paper "On lexicographically shellable posets" of Anders Bj\"orner and Michelle Wachs, in Transactions of the AMS 277, pp. 323-341.

372920mathymathy417241Let $(a_{n})_{n \ge 1}$ be a sequence of integers such that for all $n \ge 2$:

$0\le a_{n-1}+\frac{1-\sqrt{5}}{2}a_{n}+a_{n+1} <1$.

Prove that the sequence $(a_{n})$ is periodic.

This question was asked at the Miklos Schweitzer Competition 2005, problem 2 (in Hungarian).

Since $(a_{n})$ is integer it follows that $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic.

Some discussion regarding this question can be found at Mathlinks, and apparently we can choose $a_{1}$ and $a_{2}$ to make this period as large as we want.

Any help would be appreciated, thanks.

I would like to clarify some points about this problem:

1) There is a ceiling function ($\lceil x \rceil$) at the recursive sequence which makes it considerably harder. The period is not 5 as claimed by some answers.

2) Miklos Schweitzer is not a conventional competition. This competition for undergraduate students is unique. The contest lasts 10 days with 10-12 problems, which are quite challenging and basically of research level. Moreover any literature can be used.

3) An example of Miklos Schweitzer problem can be found here at MO. It was indeed a very nice question with an even nicer solution. I'm not sure Art of Problem Solving would be better.

I am sorry for any inconvenience caused and I hope this question do not get closed.

145887785197349281678727426651Suppose $\lambda$ is an non-orientable lamination on a closed orientable surface. How to construct an oriented double cover of $\lambda$?

14440651368250Your assumption means that the metric g is locally flat. Now you can solve your own problem. In any case, this is more suitable for the MSE, since it is not at research level.213463122121551368723Minimal expression of 0 as a sum of kth powers in a finite field49165419566243253251344388322812226744More generally, the Tits alternative states that any linear group either contains a solvable subgroup of finite index or a non-abelian free subgroup. Incidentally, you build free subgroups via a famous argument called *Ping-Pong* (usually attributed to Tits or even Klein). That would be one thing to try to do here.http://mathoverflow.net/questions/42709/question-about-hodge-number2020353460074477154~How to find which subset of bitfields xor to another bitfield?And what do you mean by embedded? If I say that a ring $A$ is embedded in another ring $B$, I mean that there is an injective ring homomorphism from $A$ to $B$. What kind of morphism do you want from your associative ring to your nonassociative "ring"? 29538941532981697185127650911666221861100Is the category of categories (finitely) (co)complete, as required by the definition of a model category? It appears that a problem might already arise at this level.l$f_{\epsilon_0}$ and provably total functions in $PA$766148100881502011-03-28T14:41:02.217Thanks for these! I've been hoping to see more Spivak for a long time.J@ Joachim- oops . Remove the "-1" .Your claim that we can reconstruct a homotopy type from its rationalization and completions only applies to nilpotent spaces. The kind of spaces seen by $\ell^1$-homology are very far from being nilpotent.6454241804294212068015444347741519357571237693329828:Relations in symmetric group$[X,Y]$ is not a commutator here, isn't it? I am sure that it is not but these notations are REALLY confusing. I will give a counterexample because I can't resist these notations. Of course the triangle inequality doesn't hold for the ""commutator" volumes. We can take $X,Y,Z$ -- the generators of Heisenberg group. $[X,Z]=[Y,Z]=0$ but $[X,Y]=Z \neq 0$ so the left-hand side ofthe inequality is positive and the right-hand side is zero. Consider it more like a facetious comment.5234321457317682445117407319579881764822104470941093516292902072418MHMertens5635910161588794517Sorry, my mistake. It's $B\otimes_kQ(A)$, and not $B\otimes_AQ(A)$, where $A$ and $B$ were $k$-affine domains to start with.105233319234712032003$g_l(x)$ and $g_r(x)$ are both elements of $SU(2)$ and usually chosen (atleast in the literature I am familiar with) such that $g_l(x)g_r(x)^{-1} = x$ $(A,B)$ is an element of $SU(2)\times SU(2)$ and $\sigma(x)$ is also an element of $SU(2)\times SU(2)$. So the $.$ is just group multiplication in $SU(2)\times SU(2)$. More explicitly if $\sigma(x)=(g_l(x),g_r(x))$ then $(A,B).\sigma (x) = (A.g_l(x),B.g_r(x))$. 11534455516998Proof without distributions140215219668791939686118851011792811753287:Thomas, read what you wrote.6048802056653This is actually fantastic. (I think the devil in algebraic geometry really is in the details, and things like this make a big difference in assuaging confusion.) It is indeed confusing that the subscript $x$ is used for both stalks and for fibers. Am I understanding you correctly that you would propose that if $\mathcal{F}$ is a quasicoherent sheaf on $X$, that $\mathcal{F}[x]$ be used for the fiber over a point $x \in X$? This would prevent a lot of confusion. 2154204657161smallest square containing k non-overlapping equal rectangles at any orientation221521941324926546It might be mentioned that "generically" this polynomial will have degree $6$, resulting in $6$ solutions (usually not in integers, of course). In special cases you may have more solutions or less.Except the original motivation, is it that important to restrict the question to plane 2-dimensional figures ? (Actually the only sets for which I can answer the question are balls !)17020622235938664517457431672231229912212649171716339Gerhard, was this the poset to which you had alluded to in the other post? 17082026846546743021536895450452115785806928217763519726774449476Thank you for the reference. While it looks very interesting, one needs to assume that the symplectic manifold admits a metaplectic structure in order to study symplectic spinors. This is not exactly in the spirit of my question because assuming an extra structure besides the symplectic form can of course lead the question ad absurdum: For example, assume that M is Kaehler. Then it is both symplectic and Riemannian and one could argue that there is indeed a spectral geometry of symplectic manifolds, as long as they are Kaehler :-)74379davidkobilnykI agree with you, as my preference is detailed pseudocode, in-line in the paper if possible, or in the appendix, with an URL pointing to detailed or complete source code available on your website or ArXiv. Most readers may not be computer knowledgeable, but those who wish to implement (to test or use) your algorithm will probably also want to use it on larger matrices and will want it to run quickly. Having access to the real source code will make that possible.1816631454250**Digital Video & Film Lab, LLC**

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21469501483600125081416478821280262If $A$ is a Gaussian random matrix as you describe, then the ensemble of matrices given by $A^TA$ is known as the Wishart ensemble, or the Laguerre ensemble. It has been extensively studied, and you can find information in standard books about random matrix theory.

The average of the largest eigenvalue of $A^TA$ is $4n$. The distribution around the average is given by the Tracy-Widom function. Distribution of large deviations from the mean are also known.

You can start collecting references about largest eigenvalues by looking here:

*Large Deviations of the Maximum Eigenvalue in Wishart Random Matrices*, by Pierpaolo Vivo, Satya N. Majumdar, Oriol Bohigas, https://arxiv.org/abs/cond-mat/0701371

Dualizable is equivalent to $BG$ being finitely dominated (a retract up to homotopy of a finite cell complex). In Wall's "Finiteness conditions on CW complexes, I" Wall shows that finite domination is equivalent to $\pi_1$ being finitely presented and also, the chain complex $C_*(EG)$ being chain finitely dominated over $\Bbb Z[G]$. This means there is a finite resolution of $\Bbb Z$ by finitely generated projective $\Bbb Z[G]$-modules. In the group cohomology literature, this last condition is called "FP"

581200NHodge-Tate weights of etale cohomology:So simple and clear, thanks!1305329(Reference Request) Tensor product of chain complexes in terms of strict $\infty$-categories4510201680849@Scott: not quite. An unbounded operator by definition is a linear transformation $A$ defined on some subspace $D$ of $H$. $D=H$ is allowed, in which case $A$ is "everywhere defined". But we mostly care about closed operators, and the closed graph theorem says a closed everywhere defined operator is bounded. So as far as useful examples, you are right. Moreover, unbounded everywhere defined operators (which are not closed) typically (necessarily?) require the axiom of choice to define. For instance, it's easy to do if you have a Hamel basis for $H$. 557915This question proposed in SEM but no answer and it's interesting to know connection

between Fourier analysis and number theory .

Mathematica knows that the logarithm of $n$ is:

$$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)$$

The von Mangoldt function should then be:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}.$$

Setting the first term of the von Mangoldt function $\Lambda(1)$ equal to the harmonic number $H_{\operatorname{scale}}$ where scale is equal to the scale of the Fourier transform matrix, one can calculate the Fourier transform of the von Mangoldt function with the Mathematica program at the link below:

In the program I studied the function within the limit for the von Mangoldt function, and made some small changes to the function itself:

$f(t)=\sum\limits_{n=1}^{n=k} \frac{1}{\log(k)} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n} \frac{\mu(d)}{d^{(1/2+i \cdot t-1)}}$ as $k$ goes to infinity.

(Edit 20.9.2013: The function $f(t)$ had "-1" in the argument for the zeta function.)

The plot of this function looks like this:

While the plot of the Fourier transform of the von Mangoldt function with the program looks like this:

There are some similarities but the Fourier transform converges faster towards smaller oscillations in between the spikes at zeta zeros and the scale factor is wrong.

Will the function $f(t)$ above eventually converge to the Fourier transform of the von Mangoldt function, or is it only yet another meaningless plot?

Now when I look at it I think the spikes at zeros comes from the zeta function itself and the spectrum like feature comes from the Möbius function which inverts the zeta function.

```
scale = 50; (*scale = 5000 gives the plot below*)
Print["Counting to 60"]
Monitor[g1 =
ListLinePlot[
Table[Re[
Zeta[1/2 + I*k]*
Total[Table[
Total[MoebiusMu[Divisors[n]]/Divisors[n]^(1/2 + I*k - 1)]/(n*
k), {n, 1, scale}]]], {k, 0 + 1/1000, 60, N[1/6]}],
DataRange -> {0, 60}, PlotRange -> {-0.15, 1.5}], Floor[k]]
```

Dirichlet series:

```
Clear[f]
scale = 100000;
f = ConstantArray[0, scale];
f[[1]] = N@HarmonicNumber[scale];
Monitor[Do[
f[[i]] = N@MangoldtLambda[i] + f[[i - 1]], {i, 2, scale}], i]
xres = .002;
xlist = Exp[Range[0, Log[scale], xres]];
tmax = 60;
tres = .015;
Monitor[errList =
Table[(xlist^(1/2 + I k - 1).(f[[Floor[xlist]]] - xlist)), {k,
Range[0, 60, tres]}];, k]
ListLinePlot[Im[errList]/Length[xlist], DataRange -> {0, 60},
PlotRange -> {-.01, .15}]
```

Fourier transform:

Matrix inverse:

```
Clear[n, k, t, A, nn]
nn = 50;
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1, nn}], {n, 1,
nn}];
MatrixForm[A];
ListLinePlot[
Table[Total[
1/Table[n*t, {n, 1, nn}]*
Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000, 60,
N[1/6]}], DataRange -> {0, 60}, PlotRange -> {-0.15, 1.5}]
```

```
Clear[n, k, t, A, nn]
nnn = 12;
Show[Flatten[{Table[
ListLinePlot[
Table[Re[
Total[1/Table[n*t, {n, 1, nn}]*
Total[Transpose[
Inverse[
Table[Table[
If[Mod[n, k] == 0, N[1/(n/k)^(1/2 + I*t - 1)], 0], {k,
1, nn}], {n, 1, nn}]]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
60, N[1/10]}], DataRange -> {0, 60},
PlotRange -> {-0.15, 1.5}], {nn, 1, nnn}],
Table[ListLinePlot[
Table[Re[
Total[1/Table[n*t, {n, 1, nn}]*
Total[Transpose[
Inverse[
Table[Table[
If[Mod[n, k] == 0, N[1/(n/k)^(1/2 + I*t - 1)], 0], {k,
1, nn}], {n, 1, nn}]]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
60, N[1/10]}], DataRange -> {0, 60},
PlotRange -> {-0.15, 1.5}, PlotStyle -> Red], {nn, nnn, nnn}]}]]
```

12 first curves together or partial sums:

```
Clear[n, k, t, A, nn, dd]
dd = 220;
Print["Counting to ", dd]
nn = 20;
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}];
Monitor[g1 =
ListLinePlot[
Table[Total[
1/Table[n*t, {n, 1, nn}]*
Total[Transpose[
Re[Inverse[
IdentityMatrix[nn] + (Inverse[A] - IdentityMatrix[nn])*
Zeta[1/2 + I*t]]]]]], {t, 1/1000, dd, N[1/100]}],
DataRange -> {0, dd}, PlotRange -> {-7, 7}];, Floor[t]]
mm = N[2*Pi/Log[2], 20]
g2 = Graphics[
Table[Style[Text[n, {mm*n, 1}], FontFamily -> "Times New Roman",
FontSize -> 14], {n, 1, 32}]];
Show[g1, g2, ImageSize -> Large]
```

Matrix Inverse of matrix inverse times zeta function (on critical line):

```
Clear[n, k, t, A, nn, h]
nn = 60;
h = 2; (*h=2 gives log 2 operator, h=3 gives log 3 operator and so on*)
A = Table[
Table[If[Mod[n, k] == 0,
If[Mod[n/k, h] == 0, 1 - h, 1]/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}];
MatrixForm[A];
g1 = ListLinePlot[
Table[Total[
1/Table[n*t, {n, 1, nn}]*
Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
nn, N[1/6]}], DataRange -> {0, nn}, PlotRange -> {-3, 7}];
mm = N[2*Pi/Log[h], 12]
g2 = Graphics[
Table[Style[Text[n*2*Pi/Log[h], {mm*n, 1}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 32}]];
Show[g1, g2, ImageSize -> Large]
```

Matrix inverse of Riemann zeta times log 2 operator:

```
Show[Graphics[
RasterArray[
Table[Hue[
Mod[3 Pi/2 + Arg[Zeta[sigma + I t]], 2 Pi]/(2 Pi)], {t, -30,
30, .1}, {sigma, -30, 30, .1}]]], AspectRatio -> Automatic]
```

Normal or usual zeta:

```
Show[Graphics[
RasterArray[
Table[Hue[
Mod[3 Pi/2 +
Arg[Sum[Zeta[sigma + I t]*
Total[1/Divisors[n]^(sigma + I t - 1)*
MoebiusMu[Divisors[n]]]/n, {n, 1, 30}]],
2 Pi]/(2 Pi)], {t, -30, 30, .1}, {sigma, -30, 30, .1}]]],
AspectRatio -> Automatic]
```

Spectral zeta (30-th partial sum):

```
Clear[n, k, t, A, nn, B]
nn = 60
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}]; MatrixForm[A];
B = FourierDCT[
Table[Total[
1/Table[n, {n, 1, nn}]*
Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000,
600, N[1/6]}]];
g1 = ListLinePlot[B[[1 ;; 700]]*Table[Sqrt[n], {n, 1, 700}],
DataRange -> {0, 60}, PlotRange -> {-60, 600}];
mm = 11.35/Log[2];
g2 = Graphics[
Table[Style[Text[n, {mm*Log[n], 100 + 20*(-1)^n}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 16}]];
Show[g1, g2, ImageSize -> Large]
```

Mobius function -> Dirichlet series -> Spectral Riemann zeta -> Fourier transform -> von Mangoldt function:

Larger von Mangoldt function plot still wrong amplitude: http://i.stack.imgur.com/02A1p.jpg

```
Clear[n, k, t, A, nn, B, g1, g2]
nn = 32
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}];
MatrixForm[A];
B = FourierDCT[
Table[Total[
1/Table[n, {n, 1, nn}]*
Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 0, 2000,
N[1/6]}]];
g1 = ListLinePlot[B[[1 ;; 2000]], DataRange -> {0, 60},
PlotRange -> {-5, 50}];
2*N[Length[B]/1500, 12]
mm = 13.25/Log[2];
g2 = Graphics[
Table[Style[Text[n, {mm*Log[n], 7 + (-1)^n}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 40}]];
Show[g1, g2, ImageSize -> Full]
```

Plot from program above: http://i.stack.imgur.com/r6mTJ.jpg

Partial sums of zeta function, use this one:

```
Clear[n, k, t, A, nn, B]
nn = 80
mm = 11.35/Log[2];
A = Table[
Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1,
nn}], {n, 1, nn}];
MatrixForm[A];
B = Re[FourierDCT[
Monitor[Table[
Total[1/Table[
n, {n, 1, nn}]*(Total[
Transpose[Inverse[A]*Sum[1/j^(1/2 + I*t), {j, 1, nn}]]] -
1)], {t, 1/1000, 600, N[1/6]}], Floor[t]]]];
g1 = ListLinePlot[B[[1 ;; 700]], DataRange -> {0, 60/mm},
PlotRange -> {-30, 30}];
g2 = Graphics[
Table[Style[Text[n, {Log[n], 5 - (-1)^n}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 32}]];
Show[g1, g2, ImageSize -> Full]
```

Edit 17.1.2015:

```
Clear[g1, g2, scale, xres, x, a, c, d, datapointsdisplayed]
scale = 1000000;
xres = .00001;
x = Exp[Range[0, Log[scale], xres]];
a = -FourierDCT[
Log[x]*FourierDST[
MangoldtLambda[Floor[x]]*(SawtoothWave[x] - 1)*(x)^(-1/2)]];
c = 62.357
d = N[Im[ZetaZero[1]]]
datapointsdisplayed = 500000;
ymin = -1.5;
ymax = 3;
p = 0.013;
g1 = ListLinePlot[a[[1 ;; datapointsdisplayed]],
PlotRange -> {ymin, ymax},
DataRange -> {0, N[Im[ZetaZero[1]]]/c*datapointsdisplayed}];
Show[g1, Graphics[
Table[Style[Text[n, {22800*Log[n], -1/4*(-1)^n}],
FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 12}]],
ImageSize -> Large]
```

```
Show[Graphics[
RasterArray[
Table[Hue[
Mod[3 Pi/2 +
Arg[Sum[Zeta[sigma - I t]*
Total[1/Divisors[n]^(sigma + I t)*MoebiusMu[Divisors[n]]]/
n, {n, 1, 30}]], 2 Pi]/(2 Pi)], {t, -30,
30, .1}, {sigma, -30, 30, .1}]]], AspectRatio -> Automatic]
```

@FriederLadisch : I thinkk a comment is sufficient. Feel free to expand as an answer of your own if you wish to.1144420It seems to me that your definition is equivalent to the usual definition of finite set in set theory (a set is finite if it has $n$ elements for some natural number $n$).

It is clear that any finite set satisfies your definition, since we can easily build cycles on an $n$-element set.

Conversely, if we have such a cyclic path on a set $S$ as you describe, then fix any element $b\in S$, to be used as a base point. We can define the point that is distance $n$ from $b$ in each direction. If there are points at every distance, then we get a $\mathbb{Z}$-chain and we can violate the cycle property as you did in the comments. So the points must all be at bounded finite distance, and in this case we can find a bijection with a natural number.

48344Well, the identity component $H^0$ of $H$ would have to be abelian, since, otherwise, it would have a compact simple component, and the induced map on $\pi_3$ would then be nontrivial.

If $H^0$ has positive dimension and is abelian, then what you are asking is whether $\pi_1(H^0)\to\pi_1(\mathrm{SO}(n))\simeq \mathbb{Z}_2$ is trvial or not. This does happen, of course. For example, if $H^0=S^1$ and it represents the zero element in $\pi_1(SO(n))\simeq \mathbb{Z}_2$, then the induced map on homotopy is trivial in $\pi_1$ and hence in all higher cases as well. This first happens for $n=4$, as there is such an $S^1$ sitting in $\mathrm{SO}(4)$ (in fact, a countably distinct family of them). It also happens for all $n\ge 4$.

As far as classification goes: Given $n$, there will be a countable number of codimension $1$ subtori $T^{r-1}$ in a maximal torus $T^r\subset\mathrm{SO}(n)$ (where $r$, the rank is the greatest integer in $n/2$) such that the inclusion $T^{r-1}\hookrightarrow T^r$ induces an inclusion $\pi_1(T^{r-1})\subset \pi_1(T^r)$ so that all the elements of $\pi_1(T^{r-1})$ are zero in $\pi_1(\mathrm{SO}(n))$. Then the abelian groups that you want are the ones that are conjugate to a subgroup of such a sub-maximal torus.

568880219996014767086306923A more rigorous version of Scott's answer: If a topology on a group $G$ is translation-invariant, then it also defines a *uniformity* on $G$, by definition a distinguished set of neighborhoods of the diagonal $G \times G$ that is analogous to a metric. Actually, in the present example with $G = \mathbb{Z}$, the uniformity comes from a metric. Like the metric spaces that they generalize, uniform spaces have completions. The completion of $\mathbb{Z}$ with respect to the uniformity cited by Furstenberg is exactly the adelic profinite completion of $\mathbb{Z}$. Or if $G$ is any group, there is a similar topology generated by finite-index subgroups, and a uniformity, and the completion is the profinite completion.

Let $\Phi_n(X,Y)$ be the modular polynomial that are the canonical equations for the modular curve $X_0(n)$. They parameterise pairs of elliptic curves related by a cyclic isogeny of degree $n$.

I am looking for proofs for the following two results:

- If $m=p^a$, where $p$ is prime and $a>1$, then $$\Phi_m(X,Y)=\begin{cases} \dfrac{\prod_{i=1}^{\Psi(p^{a-1})}\Phi_p(X,\xi_i)}{\Phi_{p^{a-2}}(X,Y)^p}\,, & a>2\,,\\[0.5cm] \dfrac{\prod_{i=1}^{p+1}\Phi_p(X,\xi_i)}{(X-Y)^{p+1}}\,, & a=2\,, \end{cases}$$ where $\xi_i$ are the roots of $\Phi_{p^{a-1}}(X,Y)=0$, and $\Psi(m)=m\prod_{p\mid m}(1+1/p)$.
- If $p^e$ is the $p$-primary part of $n$ and if we write $n=n'p^e$, then $\Phi_n(X,Y)$ splits modulo $p$ as follows: $$\Phi_n(X,Y) = \Phi_{n'}\left( X^{p^{e-1}}, Y \right) \Phi_{n'}\left( X, Y^{p^{e-1}} \right)\prod_{i=1}^{e-1} \Phi_{n'}\left( X^{p^{e-i-1}}, Y^{p^{i-1}} \right)^{p-1}\pmod{p}\,.$$

I found the first result as (13.14) in David Cox's *Primes of the form $x^2+ny^2$* but he refers the reader to Weber's *Lehrbuch der Algebra*.

The next result is found in Igusa's *Kroneckerian Model of Fields of Elliptic Modular Functions* and the reader was asked to refer to *Die typen der multiplikatorenringe elliptischer funktionenkörper* by Max Deuring.

I will be very happy if someone could provide a reference to proofs of these in english or direct me to english readings that would allow me to develop these proofs. THANKS!

@FrodeBjørdal I think you should un-accept my answer until I (or someone) has addressed your question around special paths, since that's an important part of your question, and I believe you were already familiar with all the information in my answer.18642379481031348052@Olivier: By coincidence I had been corresponding with Colliot-Thélène earlier today, and a few hours ago I sent him an email which drew his attention to this question. So you needn't ask him if you haven't already. Can you please give me a reference for the six blocks difference fact?145327419472141060517As long as ZFC is the arbiter of such matters, there's nothing more to say than CH is independent of our axioms. So if you want to make a definite case one way or the other, you're asking people to change what is currently a pretty well entrenched convention. You'd need to offer some good reasons, of some sort or another. I recommend looking up "intrinsic" and "extrinsic" justification of axioms in relation to Godel, and perhaps some of the writings of Peter Koellner if you're interested in such matters.2) Can $X$ be calculated by using poly($n$) memory (if such $X$ exists)for npc manifolds, https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem ---7146061499230>Yes I am agree with Reimundo 113884359861788713254031428887013637311766249Hey Ayman! I'm fine, thanks. I'll swing by your office this June. In the meanwhile, let me try to make sense of this comment by adding a link to a "related" question: https://mathoverflow.net/questions/268407/.1667464Do you know the behaviour of these sums mod 3 or mod 7? Gerhard "Prefers To Add Smaller Sums" Paseman, 2018.07.0418122403Suppose we have ring $R[M]$ over monoid $M$ in real number $R$.The number of generators for the monoid, is finite. Now suppose that every ring element $r$ has decomposition in finite linear basis of n-elements. So ring is subspace of n-dimensional vector space.

**How many rings, depending on n has this property?** Could You give me some references ( books, preprints, articles) when such theorem is stated or proved?

**Motivation:** Here monoid ring and some structure within it - how is it called? in a comment to the question, Scott Carnahan wrote:

The existence of the decomposition of elements of R[M] implies the ring is a subspace of a 4-dimensional real vector space. There are many such rings, but only finitely many monoid rings of dimension less than 5.

**Clarification of the problem:**

I have finitely presented noncommutative monoid with unity and two generators $g_1,g_2$: M = F/Rel where $Rel = \{g_1^2 = e , g_2^2 = g_2 \}$, $e$ is unit element and $F$ is free monoid over two generators. Because of relations $Rel$ every elemet in monoid has form for example $g = g_1g_2g_1g_2...g_1g_2$ ( alternating finite sequence with subscripts 1212... or 2121...). Different monoid elements contains different number of multiplications. It is very simple although infinite multiplicative structure.

Then I consider monoid ring over reals $R[M]$. Every element in $R[M]$ has form:

(1) $t = r_1g_1 + r_2g_2 r_3g_1g_2+r_4g_2g_1+ r_5g_1g_2g_1 + ...+r_p g_ig_kg_i...g_s+ ...$ and so on. $r_i \in R$ and $g_i \in M$.

Note that in general monomial element $g_ig_k...g_s$ every subscript has value in $\{1,2\}$ and no two following each other subscripts are the same ( they alternate like in sequence like $1212..$ or $2121..$. Of course this is standard ring definition.

In structure, I would like to describe You here, I have strange additional property: there is element $g_3$ in ring $R[M]$ ( **but it is not monoid element!**) which allows following decomposition:

For every $r \in R[M]$ we have

$r= r_0 e + r_1 g_1 +r_2 g_2 +r_3 g_3$

Look: there are **only four terms in decomposition, even if You decompose general ring element in the form of (1)**. However after such decomposition I may **only multiply** such elements and **not add them**. In some way it looks like vector space. From the other side such decomposition may be treated as another monoid. So in fact decomposition as above, I trying to treat as some kind of "parametrization" of ring elements. Indeed it is element of some vector space, but there are further requirements on coefficients $r_i$ ( which I do not need to describe here, they in fact are part of monoid definition).

As far as I know this is not standard ring property - maybe I am wrong. If I think about for example polynomial ring (that in simple case is real ring over multiplicative monoid generated by one generator $x$) such decomposition is not possible.

1415622\De Rham isomorphism for noncompact manifolds?The sequence $a(n)$ of minimal products (starting, for convenience, at $n=0$), which Will Jagy found to begin

$$1,2,4,3,6,12,4,8,16,12,5,10,\ldots$$

can be defined using the recursion

$$a(n) = \min\left( 2a(n-1),3a(n-3),4a(n-6),5a(n-10)\ldots\right) $$

for $n>0$ (with the understanding that $a(n)$ is undefined for $n<0$). For example,

$$a(11) = \min(2a(10),3a(8),4a(5),5a(1))= \min(10,48,48,10)=10.$$

This should allow a reasonably rapid tabulation up to around $n=10000$.

Thank you, I might look there if I can't come up with some more elementary way of seeing this.227082If you were to capitalize Cartesian, would you also write coCartesian?Brad Hein1423337What precisely is a "known transcendental constant"? It seems to me that we don't know a whole load more about $\Gamma(1/4)$ or $\gamma$ than we do about $\sum_n 1/2^{2^n}$, so in what sense can you say that the first two are "known" while the third is not?2187716869946The question is inspired by this previous MO question. There it was shown that it's possible to label the edges of a cube by the numbers $\{1,2,\ldots,6,8,9, \ldots, 13\}$ in such a way that:

- Three edges meeting at a vertex sum to 21.
- Four edges constituting a face sum to 28.

The solution is unique up to isometry (numbers in red correspond to invisible edges) :

Similary, is it possible the label the edges of a regular icosahedron (or dodecahedron) with distinct integers in such a way that the two conditions above are satisfied (with different sums of course). If yes, is the solution unique up to isometry?

Note that for the regular tetrahedron, the answer is no: a computation shows that two opposite edges must be equal.

Moreover, what would be natural generalizations of this problem for polytopes in $n$-dimensional space?

214458262136680124448170011992356569221542592120144510739881788638204793211272488690301467689227602118168351150990520788Friedlander and Suslin distinguish between polynomial functors, defined in terms of cross effects, and strict polynomial functors, defined in terms of polynomial maps hom(U,V)→hom(F(U),F(V)).17228151797076435154Actually, I am interested in showing that the tube domain in ${\Bbb C}^2$ with base given by $y>x^3$ is Kobayashi-hyperbolic. If the question that I have asked had a positive answer, it would mean that the domain is Brody-hyperbolic, which is a somewhat weaker statement.2031579@NoahSnyder: I've afraid I wouldn't know. Some pairs have the same convex hull, but the pair $U_{32}$ and $U_{72}$ do not. The former has a _truncated icosahedron_, while the latter has a _truncated dodecahedron_.5188995Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/73679/discussion-between-peter-heinig-and-lancejpollard).7834834A quotient stack question2811816279123391181008704196067@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.113059311075L+1 for the f.g. but non-free example.893233In fact, for a smooth projective curve $X$ over $\mathbb{Q}$, it seems that the trace of Frobenius $a_p$ should be the coefficients modular form if and only if $X$ has genus $1$.47221518752735163997253871527524Let $\pi(x)$ denote the number of primes $\leq x$. What is the asymptotic form for

$$\sum_{r=1}^{\pi(x)-1} \Bigg(\frac{(\pi(x))!}{r!(\pi(x)-r)!}-\frac{(\pi(x))!}{(r-1)!(\pi(x)-r+1)!} \Bigg) $$ ?

The idea of applying the Prime Number Theorem that $\pi(x) \sim x/\log x$ seems to encounter serious troubles in considering the asymptotics of the summands, hence the bound that can be obtained via this approach seems to be quite naive.

16171651909558676902We have Johnson's homomorphism $\tau_{2}\colon \mathcal{I}_{g,1} \to \bigwedge^{3}H$, where $\mathcal{I}_{g,1}$ - is a Torelli group, i.e. it consists of all elements of a mapping class group that acts trivially on homology of a surface $\Sigma_{g,1}$, that is an oriented surface of genus g without a disk. And $\tau_2$ is Johnson's homomorphism that was introduced by Johnson in "A survey of the Torelli group".

I'm interested in studying a preimage of certain subgroups of $\bigwedge^{3}H$. For example, let's consider a torus $T^2$ and fix two disks $D_{-}, D_{+}$, then we may take $g$ copies of a torus $T_{i}$ and glue $T_{i}$, $T_{i+1}$ pasting $D_{+}$ on $T_{i}$ to $D_{-}$ on $T_{i+1}$.So we have $V_1\#V_2\#...\#V_g$. Now we want to cut off some disk let's take $D_{-}$ on $V_{g}$, and that would be our surface with one boundary component. $x_i$ is a homology class of a longitude of $V_i$, and $y_{i}$ is a class of a meridian of $V_i$.

Now it's interesting to understand what is the preimage of a subgroup $W_{x}$ of $\bigwedge^{3}H$ that doesn't contain $x_{i}\wedge x_{j} \wedge x_{k}$ for all $i,j,k$. It's easy to see that we can add some BP-maps for certain curves to Johnoson's kernel, so we would get some subgroup $K$ (for example we may add just one BP map that's a preimage of $x_{1}\wedge y_{1}\wedge y_{2}$). I can't understand whether that would be enough to generate a preimage of $W_{x,y}$ (that's almost the same as $W_{x}$ but it also doesn't contain $y_{i}\wedge y_{j} \wedge y_{k}$ for all $i,j,k$).

Are there any useful tools for such problems?

73454422821651260064@@Jeremy: Very interesting idea!#I'm not a mathematician but I working on a problem that feels like it an example of a more general kind of problem and I'm hoping that someone might be able to point me in the right direction.

The problem is trying to find a convex combination of different ways of ranking n items that satisfy certain constraints. Let me be more concrete: Suppose we have $n$ individuals with endowments $b_1 > b_2 > b_3 \ldots b_n$. There are positions $1 \dots n$, each of which gives a value of $v_1 > v_2 > v_3 \ldots v_n \ge 0$ that the individuals can be assigned to.

I want to create a non-deterministic algorithm for assigning individuals to positions that is in some sense "equitable." so that for each individual, their expected proportional value from their positions is equal to their proportional endowment. E.g., suppose we have some method that assigns individual $i$ to position with 1 with probability $p_{i1}$, to position 2 with probability $p_{i2}$ and so on, I want it so that the algorithm generates an allocation such that:

$\forall i, \frac{\sum_{j=1}^n p_{ij} v_j}{\sum_{j=1}^n v_j} = \frac{b_i}{\sum_{j=1}^n b_j} $

For this to work even in the n=2 case, we need to constrain $b_1$ to that it isn't proportionally larger than $v_1$ (otherwise even always placing individual 1 at position 1 wouldn't be enough).

I originally thought this could be framed as a linear programming problem, where the goal is to find weights for each of the $n!$ possible orderings. Maybe this would work, but it would be computationally infeasible.

A particularly nice approach might be one that sequentially assigns positions by having each remaining individual "buy" probability shares with their budget and then draw a winner. Unfortunately, this doesn't have the equitable property above, but I was thinking that perhaps if we thought of the allocation as happening repeatedly, we could give the "losers" (payoff from position too small) a bigger endowment, taken from the "winners" in such a way that expected payoff converges to the equitable outcome.

Anyway, thanks for reading this far and I appreciate any comments, suggestions, answers etc.

15342371092724119183922721341135452914729837066620078550629Andrew: Oh, a question. Where am I taking limits/colimits? I never really took the time to learn infinite-dimensional manifolds properly --- I only really ever think about function spaces. I tend to believe the "topology" is secondary to the smooth structure: I should tell you all the smooth maps from R into my space, this collection should satisfy some axiom, and the rest of the data follows from there. Anyway, there are other trivializations of P. For example, we know the smooth structure on Hom([0,1],R^n) is, and P is isomorphic to Hom([0,1],R^n) \times R.918062The image is contained, at least, in the annihilator of the annihilator of $V$. 2242145838694577743Let $A$ be a semi-abelian variety defined over (a subfield of) $\mathbb C_p$. Consider its $p$-adic topology with some (non-canonical) metric. Can we bound the distance of torsion points to $0$ with respect to this metric ? In other words (to state this topologically), can we find a sequence of torsion points which leaves any $p$-adically open proper subset of $A$ ? This is true in abelian varieties and tori, because one may take an explicit integral model, but appear to be more complex for non-trivial extensions.

This is somehow related to the Tate-Voloch conjecture, since it states that torsion points cannot be too close to the subvariety at infinity in some compactification of $A$.

@Jean: why metrics on $R^2$? I would think "inner products on $R^3$" instead. 275213122624712224371179432214543344388189621978145dDoes the fundamental equality control finiteness?Sorry about $p=1297$. I mis-typed something. And I verified that $p=3889=3\cdot6^4+1$ does have $n=6$. I figured there had to be a $n=6$. :)I dont really understand your question. First of all, there are perhaps too many question marks, isn't it? Second, space open in which other space? The Picard variety of degree d?Very clever, Will! I see now that the discrepancy lies in my ambiguous second condition, "$d(L_1,L_2)$ increases with the degree of skewness between the lines." Your metric has the property that lines at, say, $\pi/2$ can be closer than lines at $\pi/4$, depending on how they sit w.r.t. the origin. But if you fix $L_1$ and rotate $L_2$, the distance is monotonic in $\theta$. So it satisfies the 2nd condition in one interpretation and fails in another.1314932@user44191 and Igor no you not only misread Breuillard's assumption but modify it in your comment. No Breuillard's notes are not confusing, but you seem to be confused anyway and this comment in your post is confusing. It's not "*generated by neighborhoods of 1*" (which is unclear and could be interpreted as "*generated by any neighborhood of 1*" but "**generated by a compact neighborhood of 1**", which is also known as "compactly generated", meaning that some compact neighborhood of 1 generates $G$, and is trivially fulfilled if $G$ is any compact group.8081410 P. Faisal Islam19250161176496757335932171x@François G. Dorais, thanks for the interesting reference!191315514437402043452I would like to ask the following two:

- For the integral: \begin{equation} \int_{0}^{t}\left( 1-s^p \right)^{\frac{1-p}{p}}ds \end{equation} I know that it is reduced to the following product involving the hypergeometric function: \begin{equation} \int_{0}^{t}\left( 1-s^p \right)^{\frac{1-p}{p}}ds={}_{2}F_{1}\left(1-\frac{1}{p}, \frac{1}{p}; 1+\frac{1}{p};t^p \right)t \end{equation} I am not yet able to prove that, but my thoughts are to expand $\left( 1-s^p \right)^{\frac{1-p}{p}}$ using the binomial theorem, then use that to approximate the series for the hypergeometric function mentioned.
- Also, for the following: \begin{equation} x={}_{2}F_{1}\left(1-\frac{1}{p}, \frac{1}{p}; 1+\frac{1}{p};t^p \right)t \end{equation} I would like to ask whether it is possible to solve explicitly for $t$. ${}_{2}F_{1}(a,b;c;z)$ is the Gauss ordinary hypergeometric function and $p>1$. Since I am not familiar with this kind of manipulations, I would like to ask if there is any transformation for this kind of functions, which would help in this situation. Thanks!

Denote the dominating number of a graph $G$ by $\gamma(G)$. I have found a number of upper bounds on $\gamma(G)$. For example, in Theorem 1.2.2 of Alon & Spencer's book, named "The Probabilistic Method", for a graph $G$ with minimal degree $d$, we have: $\gamma(G)\leq n\cdot\frac{1+ln(d+1)}{d+1}$. Unfortunately, this bound is not suitable for my work. I want to know if a better upper bound exists.

20541901274122791517227112817317This was cross-posted at TeX-SX: http://tex.stackexchange.com/q/55326/86 where it was linked to this question: http://tex.stackexchange.com/q/17031/86 which led to a TikZ/PGF package for drawing cobordism diagrams.10190045939941443554I've CWed.The book to read is **Opera de Cribro**, by Friedlander and Iwaniec.10201671074283The answer is NO. Consider the two-dimensional example $F=(f,g)$ with
$$
f(x,y) = \sqrt{2}e^{x/2}\cos(ye^{-x})
$$
$$
g(x,y) = \sqrt{2}e^{x/2}\sin(ye^{-x})
$$
The determinant of the Jacobian matrix is 1 at all points $(x,y)$
in the plane. However, $f(0, 2\pi)=f(0,0)$ and $g(0,2\pi)=g(0,0)$,
hence the map $F=(f,g)$ is not injective. It is still an open question
(the Keller Jacobian Conjecture, 1939) if a *polynomial* map $F:C^n\rightarrow C^n$
whose Jacobian determinant is identically 1 is injective.

There is a geometric (space level) way of realizing Eric's answer. In fact, this is the subject of section 1 in a paper by Ravenel and Wilson (MathSciNet). They used the iterated simplcial bar construction B^nG as a model for K(G,n) so that, when G is a ring, the ring multiplication

G\times G -> G

induces

G\times BG -> BG

by fixing the first factor and then

BG\times BG -> B(BG)=K(G,2)

by fixing the second factor, and so on. An explicit description is found in page 700 in their paper, where they prove the resu