Grassmann Algebra

ExpTheGeneralizedProduct.nb 32
����Γ
�p
�
� Α�������� �
m�
p �
�
���à p
�
� ��� �
k�
�
�
By interchanging Γ, Α, and Β
p m k
����Γ
�p
���à p
�
� Α � Β��� ���� Γ
m k�
p
we also have two simple alternate expressions
�
� Α�������� �
m�
p �
�
���à p
����Α
�
� �������� �
�m
p�
p �
�
���à p
����Α
�
� �������� �
�m
p�
p �
�
��� k
�
� ��� �
k�
�
�
���à p
�
� Α � Β��� ���� Γ
m k�
p
�
� ��� �
k�
����Α
�
� Γ � Β��� ���� Γ
�m
p k�
p
�
� ��� �
p�
����Α
�
� Β � Γ��� ���� Γ
�m
k p�
p
Again we can demonstrate this identity by converting to scalar products.
Flatten�
Table�A � ToScalarProducts� ����Γ
�
� Α�������� �
�p
m�
p ����Γ
�
� ��� �
�p
k�
����Γ
�
� Α � Β��� ���� Γ�; �p
m k�
p
Print���m, k, p�, A��; A,�m, 0, 3�, �k, 0, 3�, �p, 0, 3���
10.33
�0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0�
10.13 The Generalized Product of Orthogonal
Elements
� The generalized product of totally orthogonal elements
It is simple to see from the definition of the generalized product that if Α m and Β k
orthogonal, and Λ is not zero, then their generalized product is zero.
We can verify this easily:
2001 4 26
Α m ����� Λ �Β k
� 0 Αi ���� Βj � 0 Λ�0
Flatten�Table�ToScalarProducts�Α����� �Β��. m Λ k
OrthogonalSimplificationRules���Α, Β���, m k
�m, 0, 4�, �k, 0, m�, �Λ, 1,Min�m, k����
are totally
�0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0�
10.34