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Grassmann Algebra

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TheComplement.nb 29<br />

ComplementPalette<br />

� BASIS COMPLEMENT<br />

� 0<br />

� l<br />

� l<br />

1<br />

e1<br />

e2<br />

� 2 e1 � e2<br />

e2 g11 ���������������������������<br />

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

2<br />

�g11 g22 �g 12<br />

e2 g12 ���������������������������<br />

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

2<br />

�g11 g22<br />

�g 12<br />

e1 �e2 ���������������������������<br />

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

�g2 12 �g11 g22 �<br />

�<br />

e1 g12 ���������������������������<br />

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

2<br />

�g11 g22 �g 12<br />

e1 g22 ���������������������������<br />

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

2<br />

�g11 g22<br />

�g 12<br />

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

2 �g12 � g11 g22<br />

5.9 Geometric Interpretations<br />

The Euclidean complement in a vector 2-space<br />

Consider a vector x in a 2-dimensional vector space expressed in terms of basis vectors e1 and<br />

e2 .<br />

x � ae1 � be2<br />

Since this is a Euclidean vector space, we can depict the basis vectors at right angles to each<br />

other. But note that since it is a vector space, we do not depict an origin.<br />

Graphic of x and two orthogonal basis vectors at right angles to each other.<br />

The complement of x is given by:<br />

�����<br />

x � ae1<br />

����� � be2<br />

����� � ae2 � be1<br />

Remember, the Euclidean complement of a basis element is its cobasis element, and a basis<br />

element and its cobasis element are defined by their exterior product being the basis n-element,<br />

in this case e1 � e2 .<br />

It is clear from simple geometry that x and x<br />

����� are at right angles to each other, thus verifying<br />

our geometric interpretation of the algebraic notion of orthogonality: a simple element and its<br />

complement are orthogonal.<br />

Taking the complement of x<br />

����� gives -x:<br />

����� �����<br />

x � ae2<br />

����� � be1<br />

����� ��ae1 � be2 ��x<br />

Or, we could have used the complement of a complement axiom:<br />

����� ����� 1��2�1�<br />

x � ��1� �x ��x<br />

Continuing to take complements we find that we eventually return to the original element.<br />

����� �����<br />

x<br />

�<br />

��x<br />

�����<br />

2001 4 5

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