Grassmann Algebra

ExploringHypercomplexAlgebra.nb 17
� m = 1
� m = 2
� m = 3
� m = 4
Α1
4
����Α2
4
Α1
2
����Α2
2
Α1 ���� Α2
3 3
�Α2
4
����Α1
4
Α1 ���� Α2 �Α2 ���� Α1 �� �2�Α1 ���� Α2
�Α2
2
����Α1
2
2,0,2
�� 2�� Σ �Α1 � Α2 �Α1
2 2 2
�Α2����Α1
�� 2�� Σ
3 3
3,1,3 �Α1
3
����� 1 �Α2
3
4,0,4
�� 2�� Σ �Α1 � Α2 � Σ
4 4
The body of a symmetrized sum
4,2,4 �Α1
�Α1
3
4
����� 2 �Α2
4
���� Α2�
2
���� Α2�
3
�Α1
4
���� Α2�
4
We now break this sum up into its body (scalar component) and soul (non-scalar
components).
The body of Α1 ���� Α2
m m
m,m,m
m, that is: 2 Σ ��Α1
�Α2����Α1
m m
m
���� Α2
m
� .
11.34
11.35
11.36
11.37
is the inner product term in which Λ becomes equal to
m,m,m
And since constraint 11.28 requires that Σ ��1, the body of Αm ���� Α becomes:
m
Body�Α ����Α � �Α1����
Α1
m m m m
���Α1
m
���� Α1
m
� 2�Α1
m
���� Α2
m
� Body�Α1 ���� Α2
m m
�Α2
m
�Α2
m
���� Α2�
��Α���� Α
m m m
����Α1 � �Α2����Α2
m m m
Hence, even if a component element like Α is not simple, the body of its hypercomplex
m
square is given by the same expression as if it were simple, that is, its inner product
square.
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