Grassmann Algebra

TheRegressiveProduct.nb 13
3.5 The Common Factor Axiom
Motivation
Although the axiom sets for the progressive and regressive products have been developed, it still
remains to propose an axiom explicitly relating the two types of products. This axiom will have
wide-ranging applications in the Grassmann algebra. We will find it particularly useful in the
derivation of theoretical results involving the interior product, the factorization of simple
elements, and in the calculation of geometric intersections.
The axiom effectively endows the dual exterior and regressive product structures with the
properties we would like associated with unions and intersections of elements and spaces.
The regressive product of two intersecting elements is equal to the regressive product of
their union and their intersection.
The formula we need is called the Common Factor Axiom. In this section we motivate its basic
form up to the determination of a (possible) scalar factor with a plausible argument. (If we had
been able to present a rigorous argument, then the axiom would instead have attained the status
of a theorem!)
Specifically, since axiom �6 states that the regressive product of an m-element and a k-element
is an (m+kÐn)-element, the formula should enable determination of that element as an exterior
product.
For concreteness, suppose we are in a 4-dimensional space with basis �e1, e2, e3, e4�.
Consider the regressive product �e1 � e2 � e3���e2 � e3 � e4� with the common factor
e2 � e3 . By axiom �6, �e1 � e2 � e3���e2 � e3 � e4� is a 2-element. We will show that
this 2-element is a scalar multiple of the common factor e2 � e3 .
The basis of � is �e1 � e2, e1� e3, e1� e4, e2� e3, e2� e4, e3� e4�. The most
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general 2-element expressed in terms of this basis and which maintains the homogeneity of the
original expression (that is, displays the elements in products the same number of times) is:
a1��e3 � e2 � e3 � e4���e1 � e2�
�a2��e2 � e2 � e3 � e4���e1 � e3�
�a3��e2 � e3 � e2 � e3���e1 � e4�
�a4��e1 � e2 � e3 � e4���e2 � e3�
�a5��e1 � e2 � e3 � e3���e2 � e4�
�a6��e1 � e2 � e3 � e2���e3 � e4�
Notice however that all but one of the terms in this sum are zero due to repeated factors in the 4element.
Thus:
2001 4 5
�e1 � e2 � e3���e2 � e3 � e4� � a4��e1 � e2 � e3 � e4���e2 � e3�