Clifford Algebra Multiplication Intuition

Question

I discovered Clifford Algebra recently and I am deeply impressed with its explanatory power and geometrical intuitiveness. I've been playing with the GAViewer

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(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.))

Answer 1

When you say $c = ab$ is a bivector, you're taking $a \perp b$ implicitly. The geometric product will in general have both terms: $ab = a \cdot b + a \wedge b$, remember? A scalar and a bivector.

It is, admittedly, pretty hard to visualize what that is geometrically. Adding vectors to vectors or bivectors to bivectors is sensible; adding vectors and bivectors strains thinking (and I know; I tried to explain this at a talk and people repeatedly insisted it was nonsense, but how is adding real and imaginary numbers any more sensible?).

Beyond that, it is possible to think about the generalized wedge and dot products separately and to imagine a geometric interpretation. The wedge product of a vector and a bivector produces the trivector with magnitude of the parallelepiped, as you know. The dot product produces the vector in the plane that is orthogonal to the first vector.

You can think of the geometric product, then, as describing a decomposition: given a bivector $B$ and a vector $v$, $v \wedge B = v_\perp B$ extracts the part of the vector orthogonal to the surface and builds a volume from that. $v \cdot B = v_\parallel B$ extracts the part of the vector that lies in the plane and gives the orthogonal vector in the plane that could be used to reconstruct $B$ with.

You talk about sweeping vectors over each other to form the wedge product, and that's a sensible way of looking at things, but what image do we need to imagine the dot product of a vector and a bivector? I can picture taking the bivector like a flat disc of clay, putting my hands on it, and squeezing it along the direction of $v_\parallel$. That's the best I can conceive of it. That's what the dot product is, after all--a reduction operation, one that forms vectors from planes and planes from volumes.

Thinking of the dot product as "division" does seem like a sensible interpretation.