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Grassmann Algebra

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ExploringClifford<strong>Algebra</strong>.nb 16<br />

The Clifford product expressed by decomposition of both<br />

factors<br />

We can similarly decompose the Clifford product in terms of both Α and Β. The sign<br />

m k<br />

��1� 1<br />

���� 2 Λ �Λ�1� i<br />

can be absorbed by taking the reverse of either Α<br />

Λ<br />

Min�m,k�<br />

�<br />

Λ�0<br />

� m<br />

Λ �<br />

� m<br />

Λ �<br />

� ���<br />

i�1<br />

���<br />

i�1<br />

� k<br />

Λ �<br />

j�1<br />

Α m �Α 1<br />

Λ<br />

� k<br />

Λ �<br />

j�1<br />

Α m �Β k<br />

�<br />

��1� Λ��m�Λ� ��Α i<br />

Λ<br />

Α m �Β k<br />

Min�m,k�<br />

� �<br />

��1� Λ��m�Λ� ��Α i<br />

Λ<br />

Λ�0<br />

† j<br />

���� Β<br />

Λ<br />

�<br />

or Βj .<br />

Λ<br />

� Α i<br />

� Βj<br />

m�Λ k�Λ<br />

† i<br />

���� Βj � Α � Βj<br />

Λ m�Λ k�Λ<br />

� Α1<br />

m�Λ �Α2�<br />

Α2<br />

Λ m�Λ � � Β k<br />

Β 1<br />

Λ<br />

� Β 1<br />

k�Λ<br />

� Β 2<br />

Λ<br />

� Β 2<br />

� �<br />

k�Λ<br />

�<br />

12.16<br />

12.6 Writing Down a General Clifford Product<br />

� The form of a Clifford product expansion<br />

We take as example the Clifford product Α �Β and expand it in terms of the factors of its first<br />

4 k<br />

factor in both the D form and the C form. We see that all terms except the interior and exterior<br />

products at the ends of the expression appear to be different. There are of course equal numbers<br />

of terms in either form. Note that if the parentheses were not there, the expressions would be<br />

identical except for (possibly) the signs of the terms. Although these central terms differ<br />

between the two forms, we can show by reducing them to scalar products that their sums are the<br />

same.<br />

2001 4 26

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