Grassmann Algebra

ExploringCliffordAlgebra.nb 23
But ��1�
1
Λ �m�p�Λ� � ���� 2 Λ �Λ�1��m Λ Λ �p�Λ� �
� ��1� 1
����
����Α
�
� Γ��� �Β
�m
p�
k
� Testing the formulae
�Α� m ����Γ
�p
2 Λ �Λ�1� , hence we can write:
�
�Β��� Αi ���� Βj � 0
k�
12.27
To test any of these formulae we may always tabulate specific cases. Here we convert the
difference between the sides of equation 12.27 to scalar products and then put to zero any
products whose factors are orthogonal. To do this we use the GrassmannAlgebra function
OrthogonalSimplificationRules. We verify the formula for the first 50 cases.
? OrthogonalSimplificationRules
OrthogonalSimplificationRules���X1,Y1�,�X2,Y2�,��� develops a list of
rules which put to zero all the scalar products of a 1�element from
Xi and a 1�element from Yi. Xi and Yi may be either variables,
basis elements, exterior products of these, or graded variables.
Flatten�Table�ToScalarProducts� ���Α
�
� Γ�� �Β�Α� �m
p�
k m ���Γ
�
���� �.
� p k�
OrthogonalSimplificationRules���Α, Β���, m k
�m, 0, 3�, �k, 0, m�, �p, 0, m � 2���
�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�
12.9 The Clifford Product of Intersecting
Orthogonal Elements
Orthogonal union
We have shown in Section 12.7 above that the Clifford product of arbitrary intersecting
elements may be expressed by:
���Α
� Γ
†���
�
� m p �
�
�
��à p
���Α
� Γ
†���
�
� m p �
�
�
��à p
�
� �� � ��1�
k�
mp � � ��
�
���Α
�
� �� �
�Β�� ���� Γ
�m
p�
k�
p
�
� �� � ��1�
k�
mp � � ��Α �
� m �
�
��à p
�
� �� k�
���
���� Γ
� p
But we have also shown in Section 12.8 that if Α m and Β k
2001 4 26
are totally orthogonal, then: