The n-Category Café

The first and second axiom for vector cross products together amount to the trilinear form

being a 3-form, i.e. completely antisymmetric. In our case this trilinear form is given (perhaps up to some scalar factor in the choice of inner product on Sym$^3$V) by the restriction of

to $W$ - and here we see easily the antisymmetry in all 3 arguments.

About the third axiom, I’m also not sure how to prove it directly, and I’m guessing for my operation it holds only up to some yet to be determined scalar factor. Since it is equivalent to the restriction

being orthogonal, it should be possible to find this scalar factor by plugging in two orthogonal vectors in $W$ and comparing lengths.

I’m currently still trying to weed out the errors in my calculations, but will report back when I’m done!

Could you explain, what “a kond of cube” is? ;P

Cool post!

As I noted in the other post, the cross product on $W$ can also be viewed as coming from the cross product on $V\otimes V\otimes V$, where we view $$W \subset Sym^3V \subset V\otimes V\otimes V.$$ Similarly, the dot product on $W$ is the dot product on $V\otimes V\otimes V$. I think the normalization is correct.

So it might be easier to check that the third axiom holds on $V\otimes V\otimes V$ rather than on $W$ or on $Sym^3V$.

In particular, it becomes somewhat simple if you just pick an orthonormal basis on $V\otimes V\otimes V$ induced by an orthonormal basis on $V$, and just take $a$ and $b$ to be basis vectors. The desired axiom holds on each factor, and you multiply together and remember that for basis vectors of $V$, $(c\times d)\times c$ is $0$ if $c = d$ and is $d$ otherwise.

Not quite sure yet about a basis-free way of doing this.

I could imagine choosing the standard basis of $\mathrm{Im}(\mathbb{O})$ in two steps. Step 1 is the hard part: choose a subgroup of the automorphism group isomorphic to $(\mathbb{Z}/2\mathbb{Z})^3$, acting with multiplicity one by all seven nontrivial irreps. Step 2 is just to fix some scale factors.

Inside of $G_2$, the interesting subgroup $(\mathbb{Z}/2\mathbb{Z})^3$ is interesting because it is maximal abelian. It can be chosen as follows. There are not very many conjugacy classes in $G_2$ of order $2$. One of them, call it “class 2A”, has centralizer $(\mathrm{SU}(2) \times \mathrm{SU}(2)) / (\mathbb{Z}/2\mathbb{Z}) \cong \mathrm{SO}(4)$, where the quotient is by the diagonal copy of the centre. All seven nontrivial elements in this interesting $(\mathbb{Z}/2\mathbb{Z})^3$ are in class 2A. This tells you how to build this interesting $(\mathbb{Z}/2\mathbb{Z})^3$: pick one element of class 2A, and then pick a $(\mathbb{Z}/2\mathbb{Z})^2$-subgroup of its centralizing $\mathrm{SO}(4)$ that does not meet the centre (so that together with the centre you get a $(\mathbb{Z}/2\mathbb{Z})^3$-subgroup). Here’s the best way to do the latter: note that there is a standard $\mathrm{SO}(3)$-subroup of $\mathrm{SO}(4)$, which in terms of the description as $(\mathrm{SU}(2) \times \mathrm{SU}(2)) / (\mathbb{Z}/2\mathbb{Z})$ is the diagonal copy of $\mathrm{SU}(2) / (\mathbb{Z}/2\mathbb{Z})$; note also that $\mathrm{SO}(3)$ contains an interesting $(\mathbb{Z}/2\mathbb{Z})^2$-subgroup which is not simultaneously diagonalizable, because it lifts to a quaternion subgroup of $\mathrm{SU}(2)$.

As an aside, you can also build $\mathbb{O}$ from $(\mathbb{Z}/2\mathbb{Z})^3$. Namely, observe that $(-1)^{\det} : ((\mathbb{Z}/2\mathbb{Z})^3)^3 \to \mathbb{R}$ is a 3-cocycle — any alternating multilinear form is a cocycle — and that this cocycle represents the trivial class in cohomology. So choose a primitive $\beta \in C^2((\mathbb{Z}/2\mathbb{Z})^3; \mathbb{R}^\times)$ solving $\mathrm{d}\beta = (-1)^{\det}$. I cannot right now remember if you have to be clever in this choice at all; perhaps so, but not very clever. (What you really want is a primitive invariant under $\mathrm{SL}_3(\mathbb{Z}/2\mathbb{Z})$, as that’s the normalizer in $G_2$ of this interesting $(\mathbb{Z}/2\mathbb{Z})^3$-subgroup.) In any case, for a suitable choice of $\beta$, the twisted group algebra $\mathbb{R}^\beta[(\mathbb{Z}/2\mathbb{Z})^3]$ is isomorphic to $\mathbb{O}$. What I’m really telling you is that, if you work in the standard basis, then the associator for $\mathbb{O}$ is precisely $(-1)^\det$.

Back to the question at hand. Unfortunately, I cannot immediately eyeball whether this class 2A meets the $\mathrm{SO}(3)_{\mathrm{irr}}$ or not. If they do meet, then they will meet along (some union of) conjugacy class(es) in $\mathrm{SO}(3)_{\mathrm{irr}}$. You could imagine identifying this class, and thereby get part of the way towards building your standard $\mathrm{O}$-basis, and maybe from there you can get the rest of the basis in some reasonably by-hand way.

This is a vague, probably superficial comment but what you are talking about reminds me a bit of the twistor transform. The twistor transform is way of realizing $\mathbb{R}^4$ (with its $\SO(4)$ action) as the space of real lines over $\mathbb{CP}^1$. Only here you are trying to realize $\mathbb{R}^7$ (with its $G_2$ action) as a space of harmonic cubic polynomials over $\mathbb{R}^3$.

I would like to do a bunch of computations and post them here one bit at a time, as comments. I’ll start out very slow, to make it easy to follow.

Fix a 3-dimensional real inner product space $V$ with an orientation. Give it an arbitrary orthonormal basis which I’ll call $x,y,z$. (Part of me wants to use $i,j,k$, but when I start manipulating polynomials, expressions like $x^2 y$ make my brain feel happier than $i^2 j$.)

I’ll be proving results about things like $x^3$ and $x^2 y$ and $x y z$, so it’s good to remember that $x,y,z$ here could be any orthonormal basis.

What’s the right inner product on the space $S^3(V)$ of degree-3 homogeneous polynomials on $V$? As Layra recalled here, $V^{\otimes 3}$ has an inner product defined by

$$\left\langle a_1\otimes a_2 \otimes a_3, b_1 \otimes b_2 \otimes b_3\right\rangle = \left\langle a_1, b_1\right\rangle\left\langle a_2, b_2\right\rangle\left\langle a_3, b_3\right\rangle$$

for all $a_1, a_2, a_3, b_1, b_2, b_3 \in V$.

We can think of the $S^3(V)$ as the subspace of $V^{\otimes 3}$ consisting of symmetric tensors. (We can also think of it as a quotient space, which is arguably morally better, but never mind.) Then we think of the product $a_1 a_2 a_3 \in S^3(V)$ as the symmetrized tensor product:

$$a_1 a_2 a_3 = \frac{1}{3!} \sum_{\sigma \in S_3} a_{\sigma(1)} \otimes a_{\sigma(2)} \otimes a_{\sigma(3)}$$

This lets us compute the inner product on the 10-dimensional space $S^3(V)$ very explicitly. This basis of monomials is orthogonal:

$$x^3, y^3, z^3$$ $$x^2 y, x^2 z, y^2 x, y^2 z, z^2 x, z^2 y$$ $$x y z$$

but it’s not orthonormal, since the three kinds of monomials (listed in the three rows above) have three different lengths. Let’s compute them in enough examples so we know the general formula. First:

$$x^3 = \frac{1}{3!} \sum_{\sigma \in S_3} x \otimes x \otimes x = x \otimes x \otimes x$$

so

$$\langle x^3 , x^3 \rangle = \langle x \otimes x \otimes x, x \otimes x \otimes x \rangle = 1$$

Second:

$$x^2 y = \frac{2}{3!} (x \otimes x \otimes y + x \otimes y \otimes x + y \otimes x \otimes x)$$ $$= \frac{1}{3} (x \otimes x \otimes y + x \otimes y \otimes x + y \otimes x \otimes x)$$

so

$$\langle x^2 y, x^2 y \rangle = \frac{1}{9} \langle x \otimes x \otimes y + x \otimes y \otimes x + y \otimes x \otimes x , x \otimes x \otimes y + x \otimes y \otimes x + y \otimes x \otimes x \rangle$$

and thus

$$\langle x^2 y, x^2 y \rangle = \frac{1}{3}$$

Third:

$$x y z = \frac{1}{3!} ( x \otimes y \otimes z + \text{all permutations})$$

and since you’re getting quicker at these, you’ll quickly see

$$\langle x y z , x y z \rangle = \frac{1}{6}$$

Next I’d like to study the orthogonal projection

$$p: S^3(V) \to W$$

from $S^3(V)$ to the subspace $W$ consisting of harmonic homogeneous degree-3 polynomials. For this, let’s first make sure we understand $W$ and its orthogonal complement $W^\perp$.

$S^3(V)$ is 10-dimensional and a convenient basis is

$$x^3, y^3, z^3$$ $$x^2 y, x^2 z, y^2 x, y^2 z, z^2 x, z^2 y$$ $$x y z$$

$W \subset S^3(V)$ is 7-dimensional and one basis is

$$x^3 - 3x y^2, y^3 - 3y x^2$$ $$y^3 - 3y z^2 , z^3 - 3z y^2$$ $$z^3 - 3z x^2, x^3 - 3x z^2$$ $$x y z$$

You can just see that the Laplacian of each of these polynomials is zero. The orthogonal complement $W^\perp$ of $W$ in $S^3(V)$ is thus 3-dimensional, and a convenient basis is

$$x(x^2 + y^2 + z^2), \; y(x^2 + y^2 + z^2), \; z(x^2 + y^2 + z^2)$$

Now I want to apply the projection $p: S^3(V) \to W$ to the cube of some element of $V$, because Paul Schwahn’s proposed definition of the cross product on $W$ is based on the rule

$$a^3 \times b^3 = p((a \times b)^3)$$

where $a \times b \in V$ is the usual cross product of the vectors $a, b \in V$, and $(a \times b)^3$ is its cube, as an element of $S^3(V)$.

To compute $p$ on the cube of an arbitrary element of $V$, it’s enough to do it for an element of norm $1$ and then rescale. So let’s do it for an element of norm $1$. So we might as well do it for $x$, which is one element of our arbitrary orthonormal basis $x,y,z \in V$.

I claim

$$p(x^3) = x^3 - \alpha x (x^2 + y^2 + z^2)$$

for some number $\alpha$, which we will determine.

By definition, $p(x^3)$ is the unique element of $S^3(V)$ such that $p(x^3)$ is harmonic and $x^3 - p(x^3) \in W^\perp$. Clearly

$$x^3 - p(x^3) = \alpha x (x^2 + y^2 + z^2) \in W^\perp$$

since it’s a constant times one of basis vectors for $W^\perp$ listed above. So we just need to find $\alpha$ that makes the proposed $p(x^3)$ harmonic. For this we rewrite it as

$$p(x^3) = (1 - \alpha) x^3 - \alpha x y^2 - \alpha x z^2$$

Then we take its Laplacian and get

$$\nabla^2 (p(x^3)) = 6(1-\alpha) x - 2 \alpha x - 2 \alpha x = (6 - 10\alpha) x$$

This is zero iff $6 - 10 \alpha = 0$, so we need $\alpha = \frac{3}{5}$. Thus,

$$p(x^3) = x^3 - \frac{3}{5} x (x^2 + y^2 + z^2)$$

Now, following the same method we used to show

$$p(x^3) = x^3 - \frac{3}{5} x (x^2 + y^2 + z^2)$$

let’s compute $p(x^2 y)$ and $p(x y z)$. Thanks to the symmetry between $x,y$ and $z$ this means we’ll know $p$ on all 10 basis vectors of $S^3(V)$.

First, let’s show

$$p(x^2 y) = x^2 y - \beta y (x^2 + y^2 + z^2)$$

for some constant $\beta$. We use the same method as before. The thing we’re subtracting off here lies in $W^\perp$, because its a constant times one of our basis vectors for $W^\perp$. Thus, we just need to show that for some choice of $\beta$ our proposed $p(x^2 y)$ is harmonic. So, we calculate the Laplacian of our proposed $p(x^2 y)$ and see what choice of $\beta$ makes it vanish.

Since

$$p(x^2 y) = (1 - \beta) x^2 y - \beta y^3 - \beta y z^2$$

we have

$$\nabla^2 p(x^2 y) = 2 (1 - \beta) y - 6 \beta y - 2 \beta y = (2 - 10 \beta) y$$

which vanishes iff $\beta = \frac{1}{5}$. Thus,

$$p(x^2 y) = x^2 y - \frac{1}{5} y (x^2 + y^2 + z^2)$$

Next, let’s show

$$p(x y z) = x y z$$

For this we just note that $x y z$ is already harmonic, so its projection onto the space of harmonic homogeneous degree-3 polynomials is itself.

So we’re done:

$$\begin{array}{ccl} p(x^3) &=& x^3 - \frac{3}{5} x (x^2 + y^2 + z^2) \\ p(x^2 y) &=& x^2 y - \frac{1}{5} y (x^2 + y^2 + z^2) \\ p(x y z) &=& x y z \end{array}$$

Next, I’ll try to use these to study the cross product on $W$.

Our goal is to study the cross product on $W$ and show it’s isomorphic to the usual cross product of imaginary octonions. Given our earlier work, it’s enough to prove

for all $a,b \in W$. The only way I know to prove this is to compute both sides. Maybe someone will have a better idea, but it can’t hurt to become able to compute with the cross product on $W$.

Remember, Paul Schwahn’s plan is to define a cross product on a larger space

$$\times \colon S^3(V) \times S^3(V) \to W$$

using the formula

$$v^3 \times w^3 = p((v \times w)^3)$$

and then restrict this down to $W \times W$. Here we rely on the fact that while not every element of $S^3(V)$ is the cube of a vector in $V$, they’re all linear combinations of cubes—and we’re demanding that the cross product be bilinear.

Schwahn’s formula is a bit inconvenient for proving (1), so it’s hard to know where to start if we want to avoid a brutal computation. So let’s just try tto do those computations, and see where it first gets hard.

First let’s pick a right-handed orthonormal basis $x, y, z$ of $V$ and try to compute $x^3 \times y^3$. This is actually not hard, given what we’ve done:

$$x^3 \times y^3 = p((x \times y)^3) = p(z^3)$$

and we’ve shown

$$p(z^3) = z^3 - \frac{3}{5} z (x^2 + y^2 + z^2)$$

so that’s it:

$$x^3 \times y^3 = z^3 - \frac{3}{5} z (x^2 + y^2 + z^2)$$

and similarly for cyclic permutations of $x, y, z$. By antisymmetry we also have

$$y^3 \times x^3 = -z^3 + \frac{3}{5} z (x^2 + y^2 + z^2)$$

and

$$x^3 \times x^3 = 0$$

and similarly for cyclic permutations of $x, y, z$. So great: we know how to compute cross products of cubes.

But now suppose we want to compute cross product of guys in $W \subset S^3(V)$. These are never cubes! So one question is: how do we take guys in $W$ and write them as linear combinations of cubes?

We know a basis for $W$:

$$x^3 - 3x y^2, y^3 - 3y x^2$$ $$y^3 - 3y z^2 , z^3 - 3z y^2$$ $$z^3 - 3z x^2, x^3 - 3x z^2$$ $$x y z$$

Let’s at least see how to write the first six as linear combinations of cubes. They’re all alike, so let’s just do $x^3 - 3x y^2$. It’s natural to write it as a linear combination of these cubes:

$$(x+y)^3 = x^3 + 3x^2 y + 3 x y^2 + y^3$$ $$(x-y)^3 = x^3 - 3x^2 y + 3 x y^2 - y^3$$

Since

$$(x+y)^3 + (x-y)^3 = 2x^3 + 6 x y^2$$

we have

$$x^3 + 3 x y^2 = \frac{1}{2} \left( (x+y)^3 + (x-y)^3 \right)$$

This and similar formulas should allow us to compute a bunch of cross products of elements of $W$.

One thing I’m looking forward to as a followup to this discussion, and that I hadn’t even considered before:

So far, we’ve seen that the space of harmonic, cubic polynomials in 3 variables admits an action of SO(3) and an SO(3) invariant product, expressible in terms of an invariant quadratic differential operator, quadratic polynomial, and antisymmetric product, all on $\mathbb{R}^3$. Furthermore, this product yields a vector product algebra!

$\mathbb{R}^7$ has a $G_2$ invariant quadratic differential operator, quadratic polynomial and antisymmetric product. So should there be a similar vector product algebra on the 29-dimensional space of harmonic, cubic polynomials in 7 variables, with product invariant under $G_2$?

Hurwitz’s theorem says no, so what goes wrong?

The two possibilities I can think of is either that the third axiom fails due to some combinatorial requirement that works out for 3 but not for 7 (or works for 7 but not for 29) or it fails because the 3-D cross product obeys the Jacobi identity but the 7-D product doesn’t (these reasons may be equivalent).

One thing I’m looking forward to as a followup to this discussion, and that I hadn’t even considered before:

So far, we’ve seen that the space of harmonic, cubic polynomials in 3 variables admits an action of SO(3) and an SO(3) invariant product, expressible in terms of an invariant quadratic differential operator, quadratic polynomial, and antisymmetric product, all on $\mathbb{R}^3$. Furthermore, this product yields a vector product algebra!

$\mathbb{R}^7$ has a $G_2$ invariant quadratic differential operator, quadratic polynomial and antisymmetric product. So should there be a similar vector product algebra on the 29-dimensional space of harmonic, cubic polynomials in 7 variables, with product invariant under $G_2$?

Hurwitz’s theorem says no, so what goes wrong?

The two possibilities I can think of is either that the third axiom fails due to some combinatorial requirement that works out for 3 but not for 7 (or works for 7 but not for 29) or it fails because the 3-D cross product obeys the Jacobi identity but the 7-D product doesn’t (these reasons may be equivalent).