GrassmannCalculus`
ExpandAndSimplifyGrassmannComplements  |
|
| | ||||
|
| | ||||
Details and Options
Examples
(1)
Basic Examples
(1)
In[1]:=
<<GrassmannCalculus`
Set the book default to establish scalars and vectors.
In[2]:=
★A;
ExpandGrassmannComplements
SimplifyGrassmannComplements
ExpandAndSimplifyGrassmannComplements
ConvertComplements
In[3]:=
X=%//
%//
X//
%//
a+b⋀+c⋀⋀
e
1
e
1
e
3
e
1
e
2
e
3
 | ExpandGrassmannComplements |
| SimplifyGrassmannComplements |
| ExpandAndSimplifyGrassmannComplements |
| 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
As a further symbolic example, expands only the Grassmann complements in a Grassmann expression.
ExpandGrassmannComplements
In[4]:=
X=(a+b)⊖(z+w)+X1=
[X]
1+bx+p⋀q⊖(ay)
 | ExpandGrassmannComplements |
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.
SimplifyGrassmannComplements
X1
In[5]:=
| SimplifyGrassmannComplements |
Out[5]=
(a+b)⊖(w+z)++b+a
1
x
p⋀q⊖y
However, applying directly to the expression gives the same result.
ExpandAndSimplifyGrassmannComplements
In[6]:=
 | ExpandAndSimplifyGrassmannComplements |
Out[6]=
(a+b)⊖(w+z)++b+a
1
x
p⋀q⊖y
Note that is neither expanded, nor simplified to zero, since it is not a Grassmann 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 Grassmann complement).
x
y⊖x
x⋀x
A
In[7]:=
| ★ℬ |
4
1+x⋀x+y⊖x
| ExpandAndSimplifyGrassmannComplements |
★Λ
5
★Λ
2
Out[7]=
A+
1+y⊖x+x⋀x
Out[7]=
A++
1
x⋀x
In[8]:=
Clear[X,X1]
 |
|
 |
|
 |
|
""