Grassmann Algebra

TheRegressiveProduct.nb 23
Or, we could have used ToCongruenceForm on the original regressive product to give the
same result.
ToCongruenceForm�X � Y � Z�
x3 ��y2 z1 � y1 z2� � x2 �y3 z1 � y1 z3� � x1 ��y3 z2 � y2 z3�
�������������������������������� �������������������������������� �������������������������������� �������������������������������� ����������
�2 We see that this scalar is in fact proportional to the determinant of the coefficients of the
bivectors. Thus, if the bivectors are not linearly independent, their regressive product will be
zero.
3.7 The Regressive Product of Simple Elements
The regressive product of simple elements
Consider the regressive product Α � Β, where Α and Β are simple, m+k ³ n. The 1-element
m k m k
factors of Α and Β must then have a common subspace of dimension m+k-n = p. Let Γ be a
m k
p
simple p-element which spans this common subspace. We can then write:
Α m � Β k
� ����
Α
�
m�p � Γ p
����
�
�
����
�
 � ��� �
�k�p
p�
����
�
Α � Β � Γ��� � Γ
�m�p
k�p p�
p
The Common Factor Axiom then assures us that since Γ p
product of simple elements Α � Β. m k
In sum: The regressive product of simple elements is simple.
� The regressive product of (nÐ1)-elements
�Γ p
is simple, then so is the original
Since we have shown in Chapter 2: The Exterior Product that all (nÐ1)-elements are simple, and
in the previous subsection that the regressive product of simple elements is simple, it follows
immediately that the regressive product of any number of (nÐ1)-elements is simple.
Example: The regressive product of two 3-elements in a 4-space is simple
As an example of the foregoing result we calculate the regressive product of two 3-elements in a
4-space. We begin by declaring a 4-space and creating two general 3-elements.
2001 4 5
DeclareBasis�4�
�e1, e2, e3, e4�
X � CreateBasisForm�3, x�
x1 e1 � e2 � e3 � x2 e1 � e2 � e4 � x3 e1 � e3 � e4 � x4 e2 � e3 � e4