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Grassmann Algebra

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TheRegressiveProduct.nb 3<br />

3.2 Duality<br />

The notion of duality<br />

In order to ensure that the regressive product is defined as an operation correctly dual to the<br />

exterior product, we give the defining axiom set for the regressive product the same formal<br />

symbolic structure as the axiom set for the exterior product. This may be accomplished by<br />

replacing � by �, and replacing the grades of elements and spaces by their complementary<br />

grades. The complementary grade of a grade m is defined in a linear space of n dimensions to be<br />

nÐm.<br />

� ��� Α m � Α<br />

n�m<br />

� m � �<br />

n�m<br />

Note that here we are undertaking the construction of a mathematical structure and thus there is<br />

no specific mapping implied between individual elements at this stage. In Chapter 5: The<br />

Complement we will introduce a mapping between the elements of � and � which will lead to<br />

m n�m<br />

the definition of complement and interior product.<br />

For concreteness, we take some examples.<br />

Examples: Obtaining the dual of an axiom<br />

The dual of axiom �6<br />

We begin with the exterior product axiom �6:<br />

The exterior product of an m-element and a k-element is an (m+k)-element.<br />

�Α m �� m , Β k<br />

�� k � � Α m � Β k<br />

� �<br />

m�k<br />

To form the dual of this axiom, replace � with �, and the grades of elements and spaces by their<br />

complementary grades.<br />

� Α<br />

n�m � �<br />

n�m , Β<br />

n�k<br />

� �<br />

n�k � � Α<br />

n�m � Β<br />

n�k<br />

� �<br />

n��m�k�<br />

Although this is indeed the dual of axiom �6, it is not necessary to display the grades of what<br />

are arbitrary elements in the more complex form specifically involving the dimension n of the<br />

space. It will be more convenient to display them as grades denoted by simple symbols like m<br />

and k as were the grades of the elements of the original axiom. To effect this transformation<br />

most expeditiously we first let m' = nÐm, k' = nÐk to get<br />

2001 4 5<br />

� Α m' �� m' , Β k'<br />

�� k' � � Α m' � Β k'<br />

� �<br />

�m'�k'��n<br />

3.1

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