Grassmann Algebra

TheRegressiveProduct.nb 29
remaining basis 1-element ei in the product. This effectively cancells out (up to congruence)
the product e2 � e3 from the original term.
�e2 � e3 � ei��e2 � e3 ������������� � ��e2 � e3���e2 � e3 ������������� �� � ei � ei
Thus we can further reduce Α � e2 � e3
3 �������������
to give:
Α1 �Α�e2 � e3
3 ������������� � a1 e1 � a7 e4 � a8 e5
Similarly we can determine other factors:
Α2 �Α��e1 � e3
3 ������������� � a1 e2 � a4 e4 � a5 e5
Α3 �Α�e1 � e2
3 ������������� � a1 e3 � a2 e4 � a3 e5
It is clear from inspecting the product of the first terms in each 1-element that the product
requires a scalar divisor of a 1 2 . The final result is then:
Α 3 � 1
���������
a1 2 �Α1 � Α2 � Α3
�� 1
���������
a1 2 ��a1 e1 � a7 e4 � a8 e5��
�a1 e2 � a4 e4 � a5 e5���a1 e3 � a2 e4 � a3 e5�
Verification and derivation of conditions for simplicity
We verify the factorization by multiplying out the factors and comparing the result with the
original expression. When we do this we obtain a result which still requires some conditions to
be met: those ensuring the original element is simple.
First we declare a 5-dimensional basis and create a basis form for Α. Using
3
CreateBasisForm automatically declares all the coefficients to be scalar.
�5; Α 3 � CreateBasisForm�3, a�
a1 e1 � e2 � e3 � a2 e1 � e2 � e4 � a3 e1 � e2 � e5 �
a4 e1 � e3 � e4 � a5 e1 � e3 � e5 � a6 e1 � e4 � e5 �
a7 e2 � e3 � e4 � a8 e2 � e3 � e5 � a9 e2 � e4 � e5 � a10 e3 � e4 � e5
To effect the multiplication of the factored form we use GrassmannSimplify (in its
shorthand form).
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