GrassmannCalculus`
SimpleQ  |
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Details and Options
Examples
(1)
Basic Examples
(1)
In[1]:=
<<GrassmannCalculus`
In a 3 - space all m-elements are simple.
In[2]:=
★A;{a+b,x+y,x⋀y+y⋀z+z⋀x,x⋀y⋀z}
[%]
 | SimpleQ |
Out[2]=
{a+b,x+y,x⋀y+y⋀z+z⋀x,x⋀y⋀z}
Out[2]=
{True,True,True,True}
Elements which simplify to zero in the declared space are also considered to be simple.
In[3]:=
0,x⋀x,x⋀y+y⋀x,,+
[%]
α
4
α
4
α
6
| SimpleQ |
Out[3]=
0,x⋀x,x⋀y+y⋀x,,+
α
4
α
4
α
6
Out[3]=
{True,True,True,True,True}
In a space of any dimension n, elements of grade, 0, 1, n-1, and n are always simple. (Since elements of grades greater than n are zero, they are by definition also considered to be simple.)
In[4]:=
 | ★ℬ |
10
α
9
β
10
γ
20
| SimpleQ |
Out[4]=
5,x,,,
α
9
β
10
γ
20
Out[4]=
{True,True,True,True,True}
Elements of grades 2 through n-2 may, or may not, be simple. In the example below we take a general 2-element in 4-space and test for simplicity. We first compose the element using .
ComposeBivector
In[5]:=
| ★ℬ |
4
 | ComposeBivector |
Out[5]=
a
1
e
1
e
2
a
2
e
1
e
3
a
3
e
1
e
4
a
4
e
2
e
3
a
5
e
2
e
4
a
6
e
3
e
4
Applying gives a :
SimpleQ
ConditionalExpression
In[6]:=
X1=
[X]
| SimpleQ |
Out[6]=
ConditionalExpression[True,-+0]
a
3
a
4
a
2
a
5
a
1
a
6
We see that the element is simple contingent on a constraint between its coefficients. If this constraint is satisfied, the element is simple. We can express this several ways.
In[7]:=
Refine[X1,-+0]Refine[X1,X1〚2〛]X1/.-
a
3
a
4
a
2
a
5
a
1
a
6
a
6
a
2
a
5
a
3
a
4
a
1
Out[7]=
True
Out[7]=
True
Out[7]=
True
Suppose now a 2-element Y in which this constraint is satisfied.
In[8]:=
Y=X/.-
a
6
a
2
a
5
a
3
a
4
a
1
Out[8]=
a
1
e
1
e
2
a
2
e
1
e
3
a
3
e
1
e
4
a
4
e
2
e
3
a
5
e
2
e
4
(-+)⋀
a
3
a
4
a
2
a
5
e
3
e
4
a
1
SimpleQ
Y
In[9]:=
| SimpleQ |
Out[9]=
True
This can be confirmed by .
ExteriorFactorize
In[10]:=
| ExteriorFactorize |
Out[10]=
a
1
e
1
a
4
e
3
a
1
a
5
e
4
a
1
e
2
a
2
e
3
a
1
a
3
e
4
a
1
Here is an example verifying that the regressive product of simple elements is simple. First we declare a 4-space, and then compose the regressive product of two 3-elements. Since all (n-1)-elements are simple, and the regressive product of simple elements is simple we expect the resulting 2-element to be simple. (In general, a 2-element in a 4-space is not simple).
In[11]:=
★A;
;Z=
⋁
| ★ℬ |
4
| ComposeBasisForm |
x
3
y
3
Out[11]=
(⋀⋀+⋀⋀+⋀⋀+⋀⋀)⋁(⋀⋀+⋀⋀+⋀⋀+⋀⋀)
x
3,1
e
1
e
2
e
3
x
3,2
e
1
e
2
e
4
x
3,3
e
1
e
3
e
4
x
3,4
e
2
e
3
e
4
y
3,1
e
1
e
2
e
3
y
3,2
e
1
e
2
e
4
y
3,3
e
1
e
3
e
4
y
3,4
e
2
e
3
e
4
In[12]:=
X=
[Z]
[X]
| ToCommonFactor |
| SimpleQ |
Out[12]=
★c((-+)⋀+(-+)⋀+(-+)⋀+(-+)⋀+(-+)⋀+(-+)⋀)
x
3,2
y
3,1
x
3,1
y
3,2
e
1
e
2
x
3,3
y
3,1
x
3,1
y
3,3
e
1
e
3
x
3,3
y
3,2
x
3,2
y
3,3
e
1
e
4
x
3,4
y
3,1
x
3,1
y
3,4
e
2
e
3
x
3,4
y
3,2
x
3,2
y
3,4
e
2
e
4
x
3,4
y
3,3
x
3,3
y
3,4
e
3
e
4
Out[12]=
True
As a comparison, suppose we change just one of the coefficients in X, say the first element of the first coefficient. We no longer expect the result to be simple.
In[13]:=
X1=
((-a+)⋀+(-+)⋀+(-+)⋀+(-+)⋀+(-+)⋀+(-+)⋀);Q=
[X1]
| ★c |
y
3,1
x
3,1
y
3,2
e
1
e
2
x
3,3
y
3,1
x
3,1
y
3,3
e
1
e
3
x
3,3
y
3,2
x
3,2
y
3,3
e
1
e
4
x
3,4
y
3,1
x
3,1
y
3,4
e
2
e
3
x
3,4
y
3,2
x
3,2
y
3,4
e
2
e
4
x
3,4
y
3,3
x
3,3
y
3,4
e
3
e
4
| SimpleQ |
Out[13]=
ConditionalExpression[True,-(a-)(-)0]
x
3,2
y
3,1
x
3,4
y
3,3
x
3,3
y
3,4
In[14]:=
Clear[X,X1,Y,Z]
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""