What is a vector in Euclidean space? Some might say it's an entity characterized by possessing a magnitude and a direction. But scholars of the geometric algebra (such as Eric Chisolm and Dorst et al.) tell us that it's better to decompose the idea of direction into the two ideas of subspace attitude (our vector's quality of living in a particular line) and orientation (its quality of pointing in a particular direction in that line, and not the other). On this view, a vector is an attitudinal oriented length element. But having done this, it becomes inevitable that we should want to talk about attitudinal oriented area (volume, 4-hypervolume, &c.) elements. To this end we introduce the outer or wedge product ∧ on vectors. It is bilinear, it is anticommutative (swapping the order of arguments swaps the sign, so a∧b = –b∧a), and that's all you need to know.
Suppose we have two vectors a and b in Euclidean space and also a basis for the subspace that the vectors live in, e1 and e2, so that we can write a := a1e1 + a2e2 and b := b1e1 + b2e2. Then the claim is that the outer product a∧b can be said to represent a generalized vector (call it a 2-blade—and in general, when we wedge k vectors together, it's a k-blade) with a subspace attitude of the plane that our vectors live in and a magnitude equal to the area of the parallelogram spanned by them. Following Dorst et al., let's see what happens when we expand a∧b in terms of our basis—
a∧b = (a1e1 + a2e2)∧(b1e1 + b2e2)
= a1e1∧(b1e1 + b2e2) + a2e2∧(b1e1 + b2e2)
= a1e1∧b1e1 + a1e1∧b2e2 + a2e2∧b1e1 + a2e2∧b2e2
But the anticommutativity property implies that the outer product of a vector with itself is zero, because e∧e = –e∧e. So we have
(a1b2 – a2b1)e1∧e2
It's a determinant! And since determinants tell us about the oriented volumes of parallelepipeds, we can see why these blades defined by this outer product are a sensible generalization of the vector idea. And none can doubt that they shall play but ever such an essential role in our vaunted geometric algebra!
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