Grassmann Algebra

GeometricInterpretations.nb 23
� Lines in a 3-plane
Lines in a 3-plane �3 have the same form when expressed in coordinate-free notation as they
do in a plane �2 . Remember that a 3-plane is a bound vector 3-space whose basis may be
chosen as 3 independent vectors and a point, or equivalently as 4 independent points. For
example, we can still express a line in a 3-plane in any of the following equivalent forms.
L � �� � x���� � y�
L � �� � x���y � x�
L � �� � y���y � x�
L � � ��y � x� � x � y
Here, x and y are independent vectors in the 3-plane.
The coordinate form however will appear somewhat different to that in the 2-plane case. To
explore this, we redeclare the basis as �3 .
�3
��, e1, e2, e3�
�X � CreateVector�x�, Y� CreateVector�y��
�e1 x1 � e2 x2 � e3 x3, e1 y1 � e2 y2 � e3 y3�
L � �� � X���� � Y�
L � �� � e1 x1 � e2 x2 � e3 x3���� � e1 y1 � e2 y2 � e3 y3�
Multiplying out this expression gives:
L � ���� � e1 x1 � e2 x2 � e3 x3���� � e1 y1 � e2 y2 � e3 y3��
L � � ��e1 ��x1 � y1� � e2 ��x2 � y2� � e3 ��x3 � y3�� �
��x2 y1 � x1 y2� e1 � e2 �
��x3 y1 � x1 y3� e1 � e3 � ��x3 y2 � x2 y3� e2 � e3
The scalar coefficients in this expression are sometimes called the PlŸcker coordinates of the
line.
Alternatively, we can express L in terms of basis elements, but without specific reference to
points or vectors in it. For example:
L � � ��ae1 � be2 � ce3� � de1 � e2 � ee2 � e3 � fe1 � e3
The first term � ��ae1 � be2 � ce3� is a vector bound through the origin, and hence
defines a line through the origin. The second term d e1 � e2 � ee2 � e3 � fe1 � e3 is a
bivector whose addition represents a shift in the line parallel to itself, away from the origin. In
order to effect this shift, however, it is necessary that the bivector contain the vector
�a e1 � be2 � ce3�. Hence there will be some constraint on the coefficients d, e, and f. To
determine this we only need to determine the condition that the exterior product of the vector
and the bivector is zero.
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