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Grassmann Algebra

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ExploringClifford<strong>Algebra</strong>.nb 55<br />

numbers, quaternions, or complex quaternions. To show this, we begin by writing a general<br />

element in terms of the basis elements of the <strong>Grassmann</strong> algebra:<br />

X � Create<strong>Grassmann</strong>Number�a�<br />

a 0 � e1 a 1 � e2 a 2 � e3 a 3 � e4 a 4 � a 5 e1 � e2 � a 6 e1 � e3 �<br />

a 7 e1 � e4 � a 8 e2 � e3 � a 9 e2 � e4 � a 10 e3 � e4 � a 11 e1 � e2 � e3 �<br />

a 12 e1 � e2 � e4 � a 13 e1 � e3 � e4 � a 14 e2 � e3 � e4 � a 15 e1 � e2 � e3 � e4<br />

We can then factor this expression to display it as a nested sum of numbers of the form<br />

Ck � a i � a j e1 . (Of course any basis element will do upon which to base the factorization.)<br />

X � �a 0 � a 1 e1 � � �a 2 � a 5 e1 ��e2 �<br />

��a 3 � a 6 e1 � � �a 8 � a 11 e1��e2��e3 � ��a 4 � a 7 e1� �<br />

�a 9 � a 12 e 1 ��e 2 � ��a 8 � a 13 e 1� � �a 14 � a 15 e 1 ��e 2 ��e 3 ��e 4<br />

Which can be rewritten in terms of the Ci as<br />

X � C1 � C2 � e2 � �C3 � C4 � e2 ��e3 � �C5 � C6 � e2 � �C7 � C8 � e2 ��e3 ��e4<br />

Again, we can rewrite each of these elements as Qk � Ci � Cj � e2 to get<br />

X � Q1 � Q2 � e3 � �Q3 � Q4 � e3 ��e4<br />

Finally, we write QQ k � Qi � Qj � e3 to get<br />

X � QQ 1 � QQ 2 � e4<br />

Since the basis 1-elements are orthogonal, we can replace the exterior product by the Clifford<br />

product to give<br />

X � �a 0 � a 1 e1 � � �a 2 � a 5 e1 � � e2 �<br />

��a 3 � a 6 e1 � � �a 8 � a 11 e1� � e2� � e3 � ��a 4 � a 7 e1� �<br />

�a 9 � a 12 e1 � � e2 � ��a 8 � a 13 e1� � �a 14 � a 15 e1 � � e2 � � e3 � � e4<br />

� C1 � C2 � e2 � �C3 � C4 � e2 � � e3 � �C5 � C6 � e2 � �C7 � C8 � e2� � e3� � e4<br />

� Q1 � Q2 � e3 � �Q3 � Q4 � e3 � � e4<br />

� QQ 1 � QQ 2 � e4<br />

12.17 Rotations<br />

To be completed<br />

2001 4 26

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