Grassmann Algebra

TheRegressiveProduct.nb 21
� Example: Applying the Common Factor Theorem
In this section we show how the Common Factor Theorem generally leads to a more efficient
computation of results than repeated application of the Common Factor Axiom, particularly
when done manually, since there are fewer terms in the Common Factor Theorem expansion,
and many are evidently zero by inspection.
Again, we take the problem of finding the 1-element common to two 2-elements in a 3-space.
This time, however, we take a numerical example and suppose that we know the factors of at
least one of the 2-elements. Let:
Ξ1 � 3e1 � e2 � 2e1 � e3 � 3e2 � e3
Ξ2 � �5�e2 � 7�e3��e1
We keep Ξ1 fixed initially and apply the Common Factor Theorem to rewrite the regressive
product as a sum of the two products, each due to one of the essentially different rearrangements
of Ξ2 :
Ξ1 � Ξ2 � �Ξ1 � e1���5�e2 � 7�e3� � �Ξ1 ��5�e2 � 7�e3�� � e1
The next step is to expand out the exterior products with Ξ1 . Often this step can be done by
inspection, since all products with a repeated basis element factor will be zero.
Ξ1 � Ξ2 � �3�e2 � e3 � e1���5�e2 � 7�e3� �
�10�e1 � e3 � e2 � 21�e1 � e2 � e3��e1
Factorizing out the 3-element e1 � e2 � e3 then gives:
Ξ1 � Ξ2 � �e1 � e2 � e3����11 e1 � 15 e2 � 21 e3�
� 1
����� ���11 e1 � 15 e2 � 21 e3� ��11 e1 � 15 e2 � 21 e3
�
Thus all factors common to both Ξ1 and Ξ2 are congruent to �11 e1 � 15 e2 � 21 e3 .
� � Check using GrassmannSimplify
We can check that this result is correct by taking its exterior product with each of Ξ1 and Ξ2 .
We should get zero in both cases, indicating that the factor determined is indeed common to
both original 2-elements. Here we use GrassmannSimplify in its alias form �.
2001 4 5
���3e1 � e2 � 2e1 � e3 � 3e2 � e3����11 e1 � 15 e2 � 21 e3��
0
���5�e2 � 7�e3��e1 ���11 e1 � 15 e2 � 21 e3��
0