Grassmann Algebra

TheRegressiveProduct.nb 14
Here we see that the regressive product of two intersecting elements is (apart from a possible
scalar factor) equal to the regressive product of their union e1 � e2 � e3 � e4 and their
intersection e2 � e3 .
It will turn out (as might be expected), that a value of a4 equal to 1 gives us the neatest results
in the subsequent development of the Grassmann algebra.
Furthermore since � 4 has only the one basis element, the unit n-element 1 n can be written
1 � � e1 � e2 � e3 � e4 , where � is some scalar yet to be determined. Applying axiom �8
n
then gives us:
�e1 � e2 � e3���e2 � e3 � e4� � 1
�����
� e2 � e3
This is the formula for �e1 � e2 � e3���e2 � e3 � e4� that we were looking for. But it only
determines the 2-element to within an undetermined scalar multiple.
The Common Factor Axiom
Let Α, Β, and Γ
m k p
be simple elements with m+k+p = n, where n is the dimension of the space.
Then the Common Factor Axiom states that:
�
�
���Α m � Γ p
����
�
�
����Β
�
� ��� �
�k
p�
�
�
���Α m � Β k
Thus, the regressive product of two elements Α m � Γ p
�
� Γ��� � Γ
p�
p
to the regressive product of the 'union' of the elements Α m � Β k
(their 'intersection').
m � k � p � n
3.15
and Β � Γ with a common factor Γ is equal
k p
p
� Γ p
with the common factor Γ p
If Α and Β still contain some simple elements in common, then the product Α � Β
m k
m k
hence, by the definition above �
�
Α � Β � Γ is not zero.
m k p
Since the union Α m � Β k
���Α m � Γ p
unit n-element: Α � Β � Γ
m k p
regressive product of two elements Α � Γ
m p
factor.
2001 4 5
� Γ is zero,
p
����
�
�
����Β
�
� ��� is also zero. In what follows, we suppose that
�k
p�
� Γ is an n-element, we can write it as some scalar factor Κ, say, of the
p
�
�
�Κ1 n . Hence by axiom �8 we derive immediately that the
���Α m � Γ p
����
�
�
����Β
�
� Γ��� �Γ
�k
p�
p
and Β � Γ with a common factor Γ is congruent to that
k p
p
m � k � p � n
3.16