Grassmann Algebra

TheInteriorProduct.nb 11
We can also see that Α2 lies in the 2-element e1 � e2 (that is, Α2 is a linear combination of e1
and e2 ) by taking their exterior product and expanding it.
e1 ���e1 � Α2� ���� e1��Α2 � e1 ���e1 ���� e1� Α2 � �e1 ���� Α2� e1��Α2 � 0
We will develop the formula used for this expansion in Section 6.4: The Interior Common
Factor Theorem. For the meantime however, all we need to observe is that e2 must be a linear
combination of e1 and Α2 , because it is expressed in no other terms.
We create the rest of the orthogonal elements in a similar manner.
e3 � �e1 � e2 � Α3� ���� �e1 � e2�
e4 � �e1 � e2 � e3 � Α4� ���� �e1 � e2 � e3�
�
In general then, the (i+1)th element ei�1 of the orthogonal set e1 , e2 , e3 , � is obtained from
the previous i elements and the (i+1)th element Αi�1 of the original set.
ei�1 � �e1 � e2 � � � ei � Αi�1� ���� �e1 � e2 � � � ei�
Following a similar procedure to the one used for the second element, we can easily show that
ei�1 is orthogonal to e1 � e2 � � � ei and hence to each of the e1, e2, �, ei , and that
Αi lies in e1 � e2 � � � ei and hence is a linear combination of e1, e2, �, ei .
This procedure is equivalent to the Gram-Schmidt orthogonalization process.
6.4 The Interior Common Factor Theorem
The Interior Common Factor Formula
The Common Factor Axiom introduced in Chapter 3 is:
�
�
���Α m � Γ p
����
�
�
����Β
�
� ��� �
�k
p�
�
�
���Α m � Β k
�
� Γ��� � Γ
p�
p
Here Γ is simple, the expression is a p-element and m+k+p = n.
p
The dual of this expression is also a p-element, but in this case m+k+p = 2n.
�
�
���Α m � Γ p
����
�
�
����Β
�
� ��� �
�k
p�
�
�
���Α m � Β k
�
� Γ��� � Γ
p�
p
We can write this dual formula with the alternative ordering:
����Γ
�p
�
� Α��� �
m�
����Γ
�
� ��� �
�p
k�
����Γ
�p
Then, by replacing Α m and Β k
2001 4 5
� Α m � Β k
����
� Γ
� p
by their complements we get:
6.26