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Grassmann Algebra

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Exploring<strong>Grassmann</strong><strong>Algebra</strong>.nb 20<br />

� The inverse of a <strong>Grassmann</strong> number<br />

Let X be a general <strong>Grassmann</strong> number in 3-space.<br />

�3; X� Create<strong>Grassmann</strong>Number�Ξ�<br />

Ξ0 � e1 Ξ1 � e2 Ξ2 � e3 Ξ3 �Ξ4 e1 � e2 �<br />

Ξ5 e1 � e3 �Ξ6 e2 � e3 �Ξ7 e1 � e2 � e3<br />

Finding an inverse of X is equivalent to finding the <strong>Grassmann</strong> number A such that X�A�1 or<br />

A�X�1. In what follows we shall show that A commutes with X and therefore is a unique<br />

inverse which we may call the inverse of X. To find A we need to solve the equation X�A-1�0.<br />

First we create a general <strong>Grassmann</strong> number for A.<br />

A � Create<strong>Grassmann</strong>Number�Α�<br />

Α0 � e1 Α1 � e2 Α2 � e3 Α3 �Α4 e1 � e2 �<br />

Α5 e1 � e3 �Α6 e2 � e3 �Α7 e1 � e2 � e3<br />

The calculate and simplify the expression X�A-1.<br />

��X � A � 1�<br />

�1 �Α0 Ξ0 � e1 �Α1 Ξ0 �Α0 Ξ1� � e2 �Α2 Ξ0 �Α0 Ξ2� �<br />

e3 �Α3 Ξ0 �Α0 Ξ3� � �Α4 Ξ0 �Α2 Ξ1 �Α1 Ξ2 �Α0 Ξ4� e1 � e2 �<br />

�Α5 Ξ0 �Α3 Ξ1 �Α1 Ξ3 �Α0 Ξ5� e1 � e3 �<br />

�Α6 Ξ0 �Α3 Ξ2 �Α2 Ξ3 �Α0 Ξ6� e2 � e3 �<br />

�Α7 Ξ0 �Α6 Ξ1 �Α5 Ξ2 �Α4 Ξ3 �Α3 Ξ4 �Α2 Ξ5 �Α1 Ξ6 �Α0 Ξ7� e1 � e2 � e3<br />

To make this expression zero we need to solve for the Αi in terms of the Ξi which make the<br />

coefficients zero. This is most easily done with the <strong>Grassmann</strong>Solve function to be<br />

introduced later. Here, to see more clearly the process, we do it directly with Mathematica's<br />

Solve function.<br />

S � Solve���1 �Α0 Ξ0 � 0, �Α1 Ξ0 �Α0 Ξ1� � 0, �Α2 Ξ0 �Α0 Ξ2� � 0,<br />

�Α3 Ξ0 �Α0 Ξ3� � 0, �Α4 Ξ0 �Α2 Ξ1 �Α1 Ξ2 �Α0 Ξ4� � 0,<br />

�Α5 Ξ0 �Α3 Ξ1 �Α1 Ξ3 �Α0 Ξ5� � 0,<br />

�Α6 Ξ0 �Α3 Ξ2 �Α2 Ξ3 �Α0 Ξ6� � 0,<br />

�Α7 Ξ0 �Α6 Ξ1 �Α5 Ξ2 �Α4 Ξ3 �Α3 Ξ4 �Α2 Ξ5 �Α1 Ξ6 �Α0 Ξ7� �<br />

0�� �� Flatten<br />

�Α7 � 2 Ξ3 Ξ4 � 2 Ξ2 Ξ5 � 2 Ξ1 Ξ6 �Ξ0 Ξ7<br />

�������������������������������� �������������������������������� �����������<br />

Α6 �� Ξ6<br />

�������<br />

Ξ 0 3<br />

, Α4 �� Ξ4<br />

�������<br />

Ξ2 0 , Α1 �� Ξ1<br />

�������<br />

Ξ2 0 , Α2 �� Ξ2<br />

�������<br />

Ξ2 0 , Α3 �� Ξ3<br />

������� 2 Ξ0 ,<br />

Ξ 0 2 , Α5 �� Ξ5<br />

�������<br />

Ξ2 0 , Α0 � 1<br />

������� �<br />

We denote the inverse of X by Xr . We obtain an explicit expression for Xr by substituting the<br />

values obtained above in the formula for A.<br />

2001 4 5<br />

Ξ0

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