Grassmann Algebra

TheRegressiveProduct.nb 33
���� e1 � 2e4 � 6e5���e2 � e4 � e5����3 e3 � 4e4 � 12 e5��
3e1 � e2 � e3 � 4e1 � e2 � e4 � 12 e1 � e2 � e5 �
3e1 � e3 � e4 � 3e1 � e3 � e5 � 8e1 � e4 � e5 �
6e2 � e3 � e4 � 18 e2 � e3 � e5 � 12 e3 � e4 � e5
¥ Comparing this product to the original 3-element Α 3 verifies a final factorization as:
Α 3 � � e1 � 2e4 � 6e5���e2 � e4 � e5����3 e3 � 4e4 � 12 e5�
This factorization is, of course, not unique. For example, a slightly simpler factorization could
be obtained by subtracting twice the first factor from the third factor to obtain:
Α 3 � � e1 � 2e4 � 6e5���e2 � e4 � e5����3 e3 � 2e1�
Factorization of (nÐ1)-elements
The foregoing method may be used to prove constructively that any (nÐ1)-element is simple by
obtaining a factorization and subsequently verifying its validity. Let the general (nÐ1)-element
be:
Α
n�1
n
� �
i�1
ai�ei ����� , a1�0 where the ei ����� are the cobasis elements of the ei .
Choose Β 1
� e1 and Βj � ej and apply the Common Factor Theorem to obtain (for j ¹1):
Α
n�1 ��e1 � ej� � � Α � ej��e1 � � Α � e1��ej
n�1 n�1
n
� ������
�
�
������
ai�ei ����� � ej
i�1
�
� e1 � ������
�
�
������
ai�ei ����� � e1 � ej
i�1
�
� ��1� n�1 ��aj�e n � e1 � a1�e n � ej�
� ��1� n�1 �e n ��aj�e1 � a1�ej�
n
Factors of Α are therefore of the form aj�e1 � a1�ej , j ¹ 1, so that Α is congruent to:
n�1 n�1
Α
n�1 � �a2�e1 � a1�e2���a3�e1 � a1�e3��� ��an�e1 � a1�en�
The result may be summarized as follows:
2001 4 5
a1�e1 ����� � a2�e2 ����� � � � an�en �����
� �a2�e1 � a1�e2���a3�e1 � a1�e3��� ��an�e1 � a1�en�,
a1 � 0
3.39