There's a dot product and cross product on ℝ³. But there's also a dot product and cross product on ℝ⁷, obeying a lot of the same identities. There's nothing really like this in other dimensions.
This stuff is pretty well-known: the group of linear transformations of ℝ³ preserving the dot product and cross product is called the rotation group SO(3). We say SO(3) has an 'irreducible representation' on ℝ³ because there's no linear subspace of ℝ³ that's mapped to itself by every transformation in SO(3).
Much to my surprise, it seems that SO(3) also has an irreducible representation on ℝ⁷ where every transformation preserves the dot product and cross product on ℝ⁷.
It's not news that SO(3) has an irreducible representation on ℝ⁷. In physics we say ℝ³ is the spin-1 representation of SO(3), or at least a real version thereof, while ℝ⁷ is the spin-3 representation. It's also not news that this representation of SO(3) on ℝ⁷ preserves the dot product. But I didn't know it also preserved the cross product on ℝ⁷, which is a more exotic thing!
In fact I still don't know it for sure, but @pschwahn asked me a question that led me to guess it's true:
https://mathstodon.xyz/@pschwahn/112435119959135052
and I think I almost see a proof, which I outlined after a long conversation on other things.
@johncarlosbaez A very basic question: What does "the cross product on R^7" mean? I'm familiar with defining an antisymmetric product in n dimensions in the form of the exterior product, but I'm only familiar with how to map the result to an individual vector in 3 dimensions (via the Hodge duality). @pschwahn
@johncarlosbaez What I meant was that, while for other operations like the dot product the definition is independent of dimension, the only definition I know for a cross product is specific to 3 dimensions, so I wasn't really sure what the claim "it's the 7-dimensional cross product" is supposed to imply, exactly.
Based on your mention of it "obeying the usual identities," I'm guessing on an arbitrary vector space it's defined to be a mapping V^2 -> V that's bilinear, antisymmetric, associative, and then obeys the same identities with the dot product as in R^3.
@internic - The cross product is not associative! But you got the idea. Let me spell it out in painful detail. A "vector cross product algebra" is a vector space with an inner product I'll call the dot product and denote by ⋅: V² → ℝ and also a bilinear antisymmetric operation called the cross product × : V² → V obeying
u ⋅ (u × v) = 0
and
(u ⋅ u) (v ⋅ v) = (u × v)(u × v) + (u ⋅ v)(u ⋅ v)
That's what I meant by "the usual identities". Another important identity holds in 3d but not in 7d, namely the cyclic identity
u ⋅ (v × w) = v ⋅ (w × u)
so we don't include that in the definition of "vector cross product algebra".
There only exist four vector product algebras! The only interesting ones are the 3d and 7d ones, since in the 0d and 1d examples the cross product is zero.
@johncarlosbaez Oops, yes, not associative. Thanks for clarifying on the other identities. I was wondering, for example, about the cyclic identity.
@ppscrv @johncarlosbaez One might argue that spelling things out in painful detail is basically the definition of pure mathematics. ;-P
@johncarlosbaez Actually, on that point, I was curious: Are you typing that all out using unicode characters? That's what it looks like to me. @ppscrv
@internic - what do you mean by "mean"? Only in 3 and 7 dimensions can we define a dot product and cross product obeying the usual identities. In 3 dimensions we can define the cross product using the exterior product and Hodge duality as you say, but in 7 dimensions we cannot: the only way I know uses octonions.