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Grassmann Algebra

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ExploringClifford<strong>Algebra</strong>.nb 45<br />

C5 � C4 �. Thread�First�C4 � � Basis���� �.<br />

Thread��First�C4� ��Basis����<br />

��1, e1 ,e2 ,e3 ,e1 � e2 ,e1 � e3 ,e2 � e3, e1 � e2 � e3 �,<br />

�e1 ,1,e1 � e2 ,e1 � e3 ,e2, e3 ,e1 � e2 � e3 ,e2 � e3 �,<br />

�e 2 , ��e1 � e2 �, 1,e2 � e3 , �e1 , ��e1 � e2 � e3 �, e3, ��e1 � e3��,<br />

�e 3 , ��e 1 � e 3 �, ��e 2 � e 3 �, 1,e 1 � e 2 � e 3 , �e 1 , �e 2 ,e 1 � e 2 �,<br />

�e1 � e2, �e2 ,e1 ,e1 � e2 � e3 , �1, ��e2 � e3�, e1 � e3 , �e3 �,<br />

�e 1 � e3, �e3 , ��e1 � e2 � e3�, e1 ,e2 � e3 , �1, ��e1 � e2 �, e2 �,<br />

�e2 � e3, e1 � e2 � e3 , �e3 ,e2 , ��e1 � e3�, e1 � e2 , �1, �e1 �,<br />

�e 1 � e2 � e3 ,e2 � e3 , ��e1 � e3 �, e1 � e2 , �e3, e2 , �e1 , �1��<br />

Step 4: Verification<br />

Verify that this is the table with which we began.<br />

C 5 � C 2<br />

True<br />

� ��3 � : The Quaternions<br />

Multiplication of two even elements always generates an even element, hence the even elements<br />

form a subalgebra. In this case the basis for the subalgebra is composed of the unit 1 and the<br />

bivectors ei � ej .<br />

C4 � EvenCliffordProductTable�� ��<br />

CliffordToOrthogonalScalarProducts ��<br />

ToMetricForm; PaletteForm�C4 �<br />

1 e1 � e2 e1 � e3 e2 � e3<br />

e1 � e2 �1 ��e2 � e3 � e1 � e3<br />

e1 � e3 e2 � e3 �1 ��e1 � e2 �<br />

e2 � e3 ��e1 � e3 � e1 � e2 �1<br />

From this multiplication table we can see that the even subalgebra of the Clifford algebra of 3space<br />

is isomorphic to the quaternions. To see the isomorphism more clearly, replace the<br />

bivectors by �, �, and �.<br />

2001 4 26<br />

C5 � C4 �. �e2 � e3 � �, e1 � e3 � �, e1 � e2 � ��; PaletteForm�C5 �<br />

1 � � �<br />

� �1 �� �<br />

� � �1 ��<br />

� �� � �1

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