Grassmann Algebra

TheComplement.nb 8
����� � �
ei
n
j�1
aij�ej
�����
For reasons which the reader will perhaps discern from the choice of notation, we extract the
scalar factor � as an explicit factor from the coefficients of the complement mapping.
����� � � �
ei
n
j�1
gij�ej
�����
This then is the form in which we will define the complement of basis 1-elements.
5.13
The scalar � and the scalars gij are at this point entirely arbitrary. However, we must still
ensure that our definition satisfies the complement of a complement axiom 4
����� . We will see that
in order to satisfy 4
����� , some constraints will need to be imposed on � and gij . Therefore, our
task now in the ensuing sections is to determine these constraints. We begin by determining �
in terms of the gij .
Determining the value of �
A consideration of axiom 5
����� in the form 1 � 1
���� ����
enables us to determine a relationship
between the scalar � and the gij .
1 � 1
���� ���� ���������������������������������
� � �e1 � e2 � � � en
����������������������������
� � �e1 � e2 � � � en
� � e1
����� � e2
����� � � � en
�����
� � ��n��gij� e1 ����� � e2 ����� � � � en �����
� � ��n 1
��gij� ������������
�n�1 �� e1 � e2 � � � en
���������������������������� �
� �2��gij� 1
Here, we have used (in order) axioms 5
����� , 2
����� , 3
����� , the definition of the complement [5.13], and
formula 3.35. The symbol �gij� denotes the determinant of the gij .
Thus we have shown that the scalar � is related to the coefficients gij by:
� ��
1
��������������������
���������������
�gij�
However, it is not the sign of � that is important, since we can always redefine the ordering of
the basis elements to ensure that � is positive (or in a 1-space change the sign of the only basis
element). For simplicity then we will retain only the positive sign of �.
When the determinant �gij� is negative, � will be imaginary. This is an important case, which
can be developed in either of two ways. Either we allow the implications of an imaginary � to
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