Grassmann Algebra

TheComplement.nb 16
����� � ��1� n�1 �� 2 �
�����
ei
n
j�1
n
gij�� gjk ������ ek
k�1
In Chapter 2 that the sum over j of the terms gij�gkj was shown to be equal to the determinant
������
�gij� whenever i equals k and zero otherwise. That is:
n
�
j�1
gij�gkj � �gij��∆ik
������
Note that in this sum, the order of the subscripts is reversed in the cofactor term compared to
that in the expression for ei
����� �����
.
Thus we conclude that if and only if the array gij is symmetric, that is, gij � gji , can we
express the complement of a complement of a basis 1-element in terms only of itself and no
other basis element.
����� ����� n�1 2
ei � ��1� �� �gij��∆ik ek �� ��1� n�1 �� 2 �gij� ei
Furthermore, since we have already shown in Section 5.3 that � 2 �gij� � 1, we also have
that:
����� ����� n�1
ei � ��1� ei
In sum: In order to satisfy the complement of a complement axiom for m-elements it is
necessary that gij � gji . For 1-elements it is also sufficient.
Below we shall show that the symmetry of the gij is also sufficient for m-elements.
The complement of the complement of a basis m-element
Consider the basis m-element e1 � e2 � � � em . By taking the complement of the complement
of this element and applying the complement axiom twice, we obtain:
����������������������������
����������������������������
e1 � e2 � � � em � e1
����� � e2
����� � � � em
�����
����������������������������
� e1
����� �����
� e2
����� �����
� � � em
����� �����
But since ei
����� ����� n�1
� ��1� �ei we obtain immediately that:
����������������������������
���������������������������� m��n�1�
e1 � e2 � � � em � ��1� �e1 � e2 � � � em
But of course the form of this result is valid for any basis element ei . Writing ��1�
m
m��n�1� in
the equivalent form ��1� m��n�m� we obtain that for any basis element of any grade:
2001 4 5
����� ����� m��n�m�
ei � ��1� �ei
m
m
5.29