There's a dot product and cross product of vectors in 3 dimensions. But there's also a dot product and cross product in 7 dimensions obeying a lot of the same identities! There's nothing really like this in other dimensions.
We can get the dot and cross product in 3 dimensions by taking the imaginary quaternions and defining
v⋅w= -½(vw + wv), v×w = ½(vw - wv)
We can get the dot and cross product in 7 dimensions using the same formulas, but starting with the imaginary octonions.
The following stuff is pretty well-known: the group of linear transformations of ℝ³ preserving the dot and cross product is called the 3d 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 in 7 dimensions!
It's not news that SO(3) has an irreducible representation on ℝ⁷. In physics we call ℝ³ the spin-1 representation of SO(3), or at least a real form thereof, while ℝ⁷ is called the spin-3 representation. It's also not news that the spin-3 representation of SO(3) on ℝ⁷ preserves the dot product. But I didn't know it also preserves the cross product on ℝ⁷, which is a much 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:
@johncarlosbaez Here's an idea: Let's encode the cross product as a 3-form, and understand the irreducible representation of SO(3) on ℝ⁷ as the space of harmonic cubic homogeneous polynomials on ℝ³, i.e. ℝ⁷≅Sym³₀ℝ³. (Here Sym means the symmetrized tensor power, and the subscript 0 means we take only those elements which are trace-free, corresponding to harmonic polynomials).
So what we are looking for is an invariant 3-form on this space, i.e. an SO(3)-invariant element of Λ³(Sym³₀ℝ³)*.
My hunch is that this may be written as the restriction of an SL(3,ℝ)-invariant element of Λ³(Sym³ℝ³)* (dropping the 0 because taking a trace is not an SL(3,ℝ)-invariant operation). Indeed, a calculation in the software LiE confirms that both representations have a one-dimensional trivial subspace.
Since Sym³ℝ³ is spanned by the cubes X³ for X∈ℝ³, and the alternating powers are spanned by wedge products, my naive guess for an invariant 3-form on this space would be to map
X³∧Y³∧Z³ ↦ det(X,Y,Z)³.
One would of course have to check that this extends to a well-defined 3-form on the whole of Sym³ℝ³, which I am currently not sure how to do.
But supposing we have found this 3-form, we can then restrict it to Sym³₀ℝ³ and then re-interpret it as a cross product (raising/lowering indices is still an SO(3)-invariant operation).
@pschwahn - In our previous conversation I conjectured a representation of SO(3) on the imaginary octonions, and by now I've checked it really is a representation. Once we can know it's irreducible, we know it's the spin-3 irrep.
To get this rep we start by picking a basic triple of imaginary octonions, say i, j, ℓ. They span a 3d space. Any rotation of this 3d space rotates our basic triple to some other basic triple. This gives an automorphism of the octonions and thus a transformation of the space of imaginary octonions. So we get a rep of SO(3) on the 7d space of imaginary octonions! That's all there is to it.
We could either prove this is irreducible, or check by computation that it's the spin-3 rep. I can explain more explicitly how rotating our basic triple defines an automorphism of the octonions. Showing this gives the spin-3 rep may be easier at the Lie algebra level.
@johncarlosbaez Yes, but it this the right representation? Perhaps you missed my earlier reply, but I think this representation is not irreducible. Aren't span{i,j,ℓ}, span{k,iℓ,jℓ} and span{kℓ} invariant subspaces?
@pschwahn - Oh my god, how dumb I am! span{i,j,ℓ} is definitely invariant because I'm rotating those vectors into linear combinations of themselves. So... back to the drawing board.
By the way, it was not completely trivial to check that if we rotate i, j, ℓ we get a new basic triple, because the concept of basic triple is defined using octonion multiplication. So when I succeeded, I thought I was being clever. I seem to be getting a bunch of SO(3) subgroups of G₂ (each basic triple gives me one), but none that act irreducibly on the imaginary octonions.
I suspect that a 'generic' SO(3) subgroup of G₂ will act irreducibly. Sometimes 'generic' things are the ones for which there's no simple formula. But I still hope there's something nice going on here. I want to connect 3d geometry to 7d or 8d geometry in a nice way.
There's a nice way to build the octonions from ℝ³ where you take the exterior algebra Λ(ℝ³), which is 8-dimensional, and give it a multiplication. SO(3) acts on Λ(ℝ³) in the usual way, and this preserves the octonion multiplication, but unfortunately it acts reducibly since it preserves each grade Λⁱ(ℝ³).
@johncarlosbaez It seems this maximal SO(3) is more mysterious than anticipated!
Looking at Dynkin's paper (Table 16), there are four conjugacy classes of 3-dimensional subalgebras of 𝔤₂. Two of those are the 𝔰𝔲(2) factors of the 𝔰𝔬(4) subalgebra, one belongs to the SO(3)⊂SO(4) acting reducibly on Im 𝕆, and one belongs to the maximal subgroup SO(3)ᵢᵣᵣ.
Each 𝔰𝔬(3) subalgebra contains what Dynkin calls a "defining vector", that is, an element in a fixed maximal torus of 𝔤₂. Not just any element of the torus can be a defining vector: Dynkin gives the possible coordinates in the basis of simple roots.
From this point of view all 𝔰𝔬(3) subalgebras are on equal footing, so I'm not sure whether one can speak of "generic" objects here. At a glance it looks like they all have the same degrees of freedom, namely a choice of maximal torus in 𝔤₂ plus a choice of simple roots.
@pschwahn - There is a space of all 𝔰𝔬(3) subalgebras of 𝔤₂, and I was guessing that those subalgebras that act irreducibly on Im(𝕆) are 𝑑𝑒𝑛𝑠𝑒 in this space. That's what "generic" means in this context.
Another thing I guess is that the conjugacy class of 𝔰𝔬(3) subalgebras acting irreducibly on Im(𝕆) has higher dimension than the other 3 conjugacy classes. This should be a lot easier to check.
@johncarlosbaez@pschwahn
Naively, it seems to me that a proof follows from the usual generation of a series of algebras of space 2^n with the Caley-Dickson construction, then noting that the cross product is defined only on spaces of dimension 2^n - 1,
and then by Hurwitz's theorem that normed division algebras are only possible in 1,2,4,8 dimensions, and "The cross product is formed from the product of the normed division algebra by restricting it to the 0, 1, 3, or 7 imaginary dimensions of the algebra, giving nonzero products in only three and seven dimensions.".
So there isn't one associated with 2^4 Sedonion cross products in 2^4 - 1 spaces.
(Although no doubt one can get something in 2^4 - 1 that generalizes cross product without actually being cross product, by dropping some property of cross product and then looking at the properties of the resulting operator.)
But I may be misunderstanding; I don't know about SO(3)ᵢᵣᵣ etc. I have much to learn on all these topics.
@dougmerritt - you've got the right idea. There's an equivalence between "normed division algebras" (of which there are only four: ℝ, ℂ, ℍ and 𝕆) and "vector product algebras" (which are vector spaces equipped with a "dot product" and "cross product" obeying some familiar axioms).
To get from a normed division algebra to a vector product algebra you take its vector space of "imaginary" elements - something you can define systematically - and define a dot product and cross product on that space of imaginary elements using the formulas I gave:
v⋅w= -½(vw + wv), v×w = ½(vw - wv)
Conversely there's a way to go back from a vector product algebra to a normed division algebra.
So, just as there are only four normed division algebras, there are only four vector product algebras. The normed division algebras have dimensions 1, 2, 4 and 8. The vector product algebras have dimensions 0, 1, 3, and 7. Only the last two are interesting!
@johncarlosbaez
Thank you, that helps. I also note that, while I understand the general idea of building spaces and looking at properties that do or don't exist, I don't have the gauge for what's interesting that you have and were looking for in this thread.
@dougmerritt - the vector cross product in 7 dimensions is a very mysterious thing, but luckily it's just a spinoff of the octonions: to the extent we understand the octonions we understand it. I've been thinking about it for a long time. I thought I understood it pretty well.
The thing that's interesting me now - the thing I learned just yesterday! - is that just as 3d rotations act on 3d vectors in a way that preserves their dot product and cross product, 3d rotations also act on 7d vectors in a way that preserves their dot product and cross product. That's bizarre.
The "irreducibility" business is a way of saying that we're not getting this to happen using a cheap trick. If we dropped the irreducibility condition, we could think of 3d rotations as 7d rotations that just happen to only mess around with 3 of the coordinates. We are not doing that here!
So this is weird. By the way, in general, 7d rotations DON'T act on 7d vectors in a way that preserves their dot product and cross product. Only certain special ones do.
(It takes 28 numbers to specify a general rotation in 7 dimensions, and they all preserved the dot product of 7d vectors. Far fewer preserve the cross product too: those can be specified using only 14 numbers.)
@dmm - don't be shy about asking questions, and don't be scared to ask 'elementary' questions. I really like explaining math. Whatever might help you, I'm willing to give it a try.
@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
@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.
@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 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 ⋅ (v × w) = v ⋅ (w × u)
and
(u × v) × u = (u ⋅ u) v - (u ⋅ v) u
These are what I meant by "the usual identities". They imply a bunch more identities.
There exist only 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.
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