GrassmannCalculus`
ExpandAndSimplifyVectorSpaceComplements  |
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Details and Options
Examples
(1)
Basic Examples
(1)
In[1]:=
<<GrassmannCalculus`
Set a 3-dimensional point space.
In[2]:=
★A;
;
 | ★ |
3
ExpandVectorSpaceComplements
SimplifyVectorSpaceComplements
ExpandAndSimplifyVectorSpaceComplements
ConvertComplements
In[3]:=
X=%//
%//
X//
%//
a+b⋀+c⋀⋀
e
1
e
1
e
3
e
1
e
2
e
3
 | ExpandVectorSpaceComplements |
| SimplifyVectorSpaceComplements |
| ExpandAndSimplifyVectorSpaceComplements |
| ConvertComplements |
Out[3]=
a+b⋀+c⋀⋀
e
1
e
1
e
3
e
1
e
2
e
3
Out[3]=
a
e
1
b⋀
e
1
e
3
c⋀⋀
e
1
e
2
e
3
Out[3]=
a+b⋀+c⋀⋀
e
1
e
1
e
3
e
1
e
2
e
3
Out[3]=
a+b⋀+c⋀⋀
e
1
e
1
e
3
e
1
e
2
e
3
Out[3]=
c-b+a⋀
e
2
e
2
e
3
ExpandVectorSpaceComplements
In[4]:=
X=(a+b)⊖(z+w)+X1=
[X]
1+bx+p⋀q⊖(ay)
 | ExpandVectorSpaceComplements |
Out[4]=
(a+b)⊖(w+z)+
1+bx+p⋀q⊖ay
Out[4]=
(a+b)⊖(w+z)+++
1
bx
p⋀q⊖ay
Applying to factors the scalars out of the Grassmann complements.
SimplifyVectorSpaceComplements
X1
In[5]:=
| SimplifyVectorSpaceComplements |
Out[5]=
(a+b)⊖(w+z)++b+a
1
x
p⋀q⊖y
However, applying directly to the expression gives the same result. evaluates the unit n-basis. converts to the normal Grassmann complements and further evaluates the remaining complements. Then a rule that implements the complement axion (equation 5.3) reduces down to complements of the symbolic vectors. reduces to scalar weighted basis vectors.
ExpandAndSimplifyVectorSpaceComplements
ConvertComplements
ToGrassmannComplements
ToScalarProducts
ToCommonFactor
In[6]:=
| ExpandAndSimplifyVectorSpaceComplements |
| ToGrassmannComplements |
| ToScalarProducts |
f1_⋀f2_
f1
f2
| ConvertComplements |
| ToCommonFactor |
| ★c |
Out[6]=
(a+b)⊖(w+z)++b+a
1
x
p⋀q⊖y
Further evaluation:
Out[6]=
(a+b)⊖(w+z)++b+a
★
★⋀x
★⋀(p⋀q⊖y)
Out[6]=
★
★⋀p
★⋀q
★⋀x
Out[6]=
a(q⊖y)⋁⋀⋀-a(p⊖y)⋁⋀⋀-b⋁⋀⋀+⋀⋀
p
e
1
e
2
e
3
q
e
1
e
2
e
3
x
e
1
e
2
e
3
e
1
e
2
e
3
Out[6]=
(b〈⋀〉+a〈⋀〉(p⊖y)-a〈⋀〉(q⊖y))⋀+(-b〈⋀〉-a〈⋀〉(p⊖y)+a〈⋀〉(q⊖y))⋀+(b〈⋀〉+a〈⋀〉(p⊖y)-a〈⋀〉(q⊖y))⋀+⋀⋀
e
3
x
e
3
q
e
3
p
e
1
e
2
e
2
x
e
2
q
e
2
p
e
1
e
3
e
1
x
e
1
q
e
1
p
e
2
e
3
e
1
e
2
e
3
Note that is neither expanded, nor simplified to zero, since it is a Grassmann complement rather than a vector space complement.
(a+b)⊖(z+w)
You can also use new symbols as long as you assert their grades, or you can override the grades of currently declared symbols. Here we assert that is of grade 2, making zero and non-zero. (Note also that although has been asserted to be a 5-element, it has not been simplified to zero in this 4-space, since it is not a vector space complement).
x
y⊖x
x⋀x
A
In[7]:=
| ★ℬ |
4
1+x⋀x+y⊖x
| ExpandAndSimplifyVectorSpaceComplements |
★Λ
5
★Λ
2
Out[7]=
A+
1+y⊖x+x⋀x
Out[7]=
A++
1
x⋀x
In[8]:=
Clear[X,X1]
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""