I discovered Clifford Algebra recently and I am deeply impressed with its explanatory power and geometrical intuitiveness. I've been playing with the GAViewer
(highly recommended!) and I'm trying to understand Clifford multiplication of vectors intuitively. If the two vectors are parallel (e.g. they are both multiples of e1) then multiplication scales their lengths, just as scalar multiplication scales their magnitudes. This provides a beautiful geometrical analogy to multiplication, revealing it to be an analog scaling function. If the vectors are not colinear, then the product becomes a wedge product, an oriented surface with a twist that rotates in the direction from a to b in the case of c = a*b.
You can see the effect of this product by multiplying the bivector with another vector, for example d = bc, or d = b(a*b), and you will see that the bivector scales b by the ratio of a to b, and rotates b by the angle from a to b. That is a beautiful and powerful notion that accounts for many of the extraordinary invariances in Clifford algebra especially in the projective and conformal models.
Here's my problem: Multiplication is an expansive operation, creating a multiplicity of its multiplicands, this is demonstrated quite literally by the bivector c = a*b which can be pictured as sweeping vector a through b, or vice-versa, and that results in a grade increase. The product of b and the bivector c however results in a grade reduction; vector * bivector = vector. What gives?
I understand this is only true because a, b, & c all lie in the same plane, whereas multiplying by a vector out of the plane would produce a volumetric trivector. But it is still a case of multiplication producing a grade reduction. I can't picture the "sweep" to explain this operation. Can anyone help my intuition? Is the second multiplication effectively a division, as is scalar multiplication by a fractional value? Or is this an effect of the "squares-to-one" property of Clifford algebra whereby a*a = 1?
I can picture the sweep of, for example, of ab/b as vector a swept along vector b, then swept back again collapsing the surface like a venetian blind and resulting back in a. I can picture ab/a as the same sweep-and-collapse, but this time the rotational component is reversed (ab = -ba) so the collapse leaves the vector rotated by the angle a*b. But how are we to imagine a vector times a bivector within the plane?
((I happen to believe that "seeing the picture" is the most important part of mathematical understanding, and Clifford algebra is particularly revealing of the geometrical spatial principles hidden within algebra.))