## June 4, 2024

### 3d Rotations and the 7d Cross Product (Part 2)

#### Posted by John Baez

On Mathstodon, Paul Schwahn raised a fascinating question connected to the octonions. Can we explicitly describe an irreducible representation of $SO(3)$ on 7d space that preserves the 7d cross product?

I explained this question here:

This led to an intense conversation involving Layra Idarani, Greg Egan, and Paul Schwahn himself. The result was a shocking new formula for the 7d cross product in terms of the 3d cross product.

Let me summarize.

There are two equivalent ways to say what’s been done:

Theorem 1. We can explicitly describe an $SO(3)$ subgroup of $\mathrm{G}_2$ such that the 7d irreducible representation of $\mathrm{G}_2$ remains irreducible when restricted to this subgroup.

Theorem 2. We can explicitly describe an irreducible representation of $SO(3)$ on the imaginary octonions that preserves their dot product and cross product.

These are equivalent thanks to several well-known facts. The group of automorphisms of the octonions is $\mathrm{G}_2$. Its action on the space of imaginary octonions, $Im(\mathbb{O})$, is the unique 7d irreducible representation of $\mathrm{G}_2$. This action preserves the usual dot product and cross product of imaginary octonions, given by

$v \cdot w= - \frac{1}{2}(v w + w v), \qquad v\times w = \frac{1}{2}(v w - w v)$

Conversely, any linear transformation of $Im(\mathbb{O})$ that preserves the cross product also preserves the dot product, and all such transformations come from the action of $\mathrm{G}_2$.

Either way we state the theorem, the only novelty — if any — is that we now have an explicit description. The existence seems to go back to old work by Dynkin:

• E. B. Dynkin, Semisimple subalgebras of semisimple Lie algebras, American Mathematical Society Translations, Series 2, Volume 6, 1957.

In fact we seem to have two explicit descriptions. Unfortunately it takes some serious calculation to prove that either of them actually works. I think a truly conceptual proof still awaits us, though the second description points a way forward.

Our first description will start by building the unique 7-dimensional irreducible representation of $SO(3)$ in a familiar way. Then we will equip it with an isomorphism to $Im(\mathbb{O})$, and proof that the resulting action of $SO(3)$ on $Im(\mathbb{O})$ preserves the dot product and cross product. Layra Idarani proved this using fairly brutal calculations, which Greg Egan checked using Mathematica.

We start with the 3-dimensional inner product space $V$ with orthonormal basis $x, y, z$. Then we let $W$ be the space of harmonic homogeneous degree-3 polynomials in $x, y,$ and $z$. This is 7-dimensional, since it has a basis

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

Since $V$ has an inner product we get an isomorphism $V \cong V^\ast$, so we can also think of these polynomials as functions on $V$. $SO(3)$ acts on functions on $V$, preserving the conditions of being harmonic and homogeneous of degree 3, so it acts on $W$. This is well-known to give the unique 7-dimensional irreducible representation of $SO(3)$.

The next step is to choose a vector space isomorphism between $W$ and $Im(\mathbb{O})$. For this, we use a well-known orthonormal basis of the imaginary octonions:

This pictures shows the Fano plane, with 7 points and 7 lines (one of which is drawn as a circle). Each line contains 3 points, and the arrows indicate a cyclic ordering of these 3 points. Each point corresponds to a basis element $e_i$ of the imaginary octonions. The cross product obeys

$e_i \times e_j = e_k$

whenever $i, j, k$ are a cyclically ordered triple of points on a line. For example, $e_1 \times e_2 = e_4$ and $e_5 \times e_6 = e_1$, but $e_6 \times e_5 = -e_1$ because the cross product is anticommutative.

Layra Idarani chose this isomorphism between $\mathrm{Im}(\mathbb{O})$ and $W$:

$e_1 \mapsto \sqrt{\frac{3}{5}}(2x^3 - 3x y^2 - 3x z^2), \; e_2 \mapsto \sqrt{\frac{3}{5}}(2z^3 - 3x^2 z - 3y^2z), \; e_4 \mapsto \sqrt{\frac{3}{5}}(2y^3 - 3y z^2 - 3x^2 y)$

$e_3 \mapsto 3 x z^2 - 3x y^2 , \; e_5 \mapsto 3y x^2 - 3y z^2, \; e_6 \mapsto 3z y^2 - 3z x^2$

$e_7 \mapsto 6x y z$

Using rather lengthy calculations, Layra and Greg checked that if we use this isomorphism to transfer the cross product on $\mathrm{Im}(\mathbb{O})$ to $W$, we get a cross product on $W$ that is $SO(3)$-invariant.

Besides the calculations required, the main downside to this argument is that it relies on cleverly choosing an isomorphism $\mathrm{Im}(\mathbb{O}) \cong W$. Luckily, Paul Schwahn came up with a second approach that elegantly defines a cross product on $W$ without choosing an isomorphism $\mathrm{Im}(\mathbb{O}) \cong W$. Alas, it still requires a hard calculation to show this cross product is isomorphic to the usual cross product on $\mathrm{Im}(\mathbb{O})$, but Layra says he has done that calculation.

Here’s the cool part: this second approach defines the 7d cross product as a kind of ‘cube’ of the 3d cross product! That came as a big surprise to me.

Here’s how it works.

Let $S^3 V$ be the space of homogeneous degree-3 polynomials on our 3d inner product space $V$. This gets an inner product from $V$, so let

$p : S^3 V \to W$

be the orthogonal projection onto the subspace of harmonic homogeneous degree-3 polynomials. If we pick an orientation on $V$, we can define the usual 3d cross product

$\times : V \times V \to V$

using the right-hand rule.

Then Schwahn defines a bilinear operation

$\bullet : S^3 V \times S^3 V \to S^3 V$

in a cunning way. First, note that we can cube any element $v \in V$ and get an element $v^3 \in S^3 V$. Then, let

$u^3 \bullet v^3 = (u \times v)^3$

for all $u, v \in V$. It would take work to show that there exists a unique bilinear operation $\bullet$ obeying this formula. I haven’t done all this work. But for uniqueness, it’s enough to note that any degree-3 polynomial on $V$ is a linear combination of cubes, which follows from the ‘polarization identity’ for cubic maps.

Next Schwahn defines a cross product on $W \subseteq S^3 V$ by

$a \times b = p(a \bullet b)$

for all $a,b \in W$.

With this definition, it’s obvious that this cross product

$\times : W \times W \to W$

is $SO(3)$-invariant. The work comes when we try to choose an isomorphism $W \cong \mathrm{Im}(\mathbb{O})$ that carries this cross product to the usual cross product of imaginary octonions! And currently this seems to require a hard computation.

An alternative approach would be to check that this cross product on $W$, together with the inner product on $W$, obeys the axioms of a vector product algebra:

$a \times b = - b \times a$

$a \cdot (b \times c) = b \cdot (c \times a)$

$(a \times b) \times a = (a \cdot a) b - (a \cdot b) a$

The first one is obvious because $\bullet$ is already antisymmetric… but I haven’t figured out how to show the next two!

Posted at June 4, 2024 10:23 AM UTC

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### Re: 3d Rotations and the 7d Cross Product (Part 2)

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

(1)$W^3\to W:\qquad (a,b,c)\mapsto\langle a\times b,c\rangle$

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

(2)$(u^3,v^3,w^3)\mapsto\det(u,v,w)^3$

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

(3)$a\times -:\qquad a^\perp\to a^\perp$

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!

Posted by: Paul Schwahn on June 4, 2024 6:26 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

Sorry, I meant to say that $a\times -$ should be orthogonal for any unit vector $a\in W$, not just any vector.

Posted by: Paul Schwahn on June 4, 2024 6:27 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

I like how you disposed of the second axiom so efficiently by phrasing the first two axioms in terms of the trilinear form $a \cdot (b \times c)$ instead of the cross product!

Let me try to show that given the first two axioms, the third follows from $a \times -$ being an orthogonal transformation of $a^\perp$ when $a$ is a unit vector. My approach is a bit roundabout.

I’ll use an alternative set of axioms for a vector product algebra, adapted from Wikipedia. We start with an inner product space $W$ and demand that the map $\times : W \times W \to W$ obey

(1)$(a \times b) \cdot a = 0$
(2)$a \times b = - b \times a$

and

(3)$a \cdot b = 0 \implies \|a \times b\| = \|a\| \, \|b\|$

These are equivalent to the three I listed in my article.

Now suppose $a \cdot (b \times c)$ is totally antisymmetric and $a \times -$ is an orthogonal transformation of $a^\perp$ when $a$ is a unit vector. Then the latter condition implies 1) and 3) while the former implies 2). So yes, these conditions imply we have a vector product algebra!

Posted by: John Baez on June 4, 2024 10:57 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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

Cool post!

Posted by: Stéphane Desarzens on June 4, 2024 7:58 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

Fixed.

Posted by: John Baez on June 4, 2024 10:13 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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.

Posted by: Layra Idarani on June 4, 2024 9:59 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

Thanks, Layra. I am thinking about this, trying to figure out what it means to say the third axiom holds on $V \otimes V \otimes V$. Are you putting a cross product on $V \otimes V \otimes V$ where you take two elements of $V \otimes V \otimes V$, use the map $\times \otimes \times \otimes \times$ to get a new element of $V \otimes V \otimes V$, and then project that down to $W \subset V \otimes V \otimes V$?

Posted by: John Baez on June 5, 2024 12:21 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

I’m just sticking to $V\otimes V\otimes V$ for now, with the cross product $\times \otimes \times \otimes \times$ and the dot product $\cdot \otimes \cdot \otimes \cdot$.

Then getting the third axiom applied to $W$ would just be a restriction and a projection.

But I figure that working it out for $V\otimes V\otimes V$ rather than $W$ might be easier because the connection back to $V$ is clearer and it’s a little easier to write down the cross product explicitly.

Note: The other axiom you gave, in terms of norms, definitely doesn’t work on $V\otimes V\otimes V$ because of all the $0$ divisors.

Posted by: Layra Idarani on June 5, 2024 8:50 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

Whoops; forgot that non-linear things can’t just be checked on basis vectors. It turns out that the third axiom is actually false on $V\otimes V\otimes V$ in general. It holds if $a$ and $b$ are pure tensors of basis vectors, but not for mixed tensors.

I should probably try to write down a basis-independent description of the projection to $W$.

Posted by: Layra Idarani on June 6, 2024 12:09 AM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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.

Posted by: Theo Johnson-Freyd on June 5, 2024 11:09 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

In fact there is only one conjugacy class of order 2 in G_ 2.

The conjugacy classes in a simple simply connected compact group are in correspondence with the Weyl alcove, here a 30-60-90 triangle. If we double the Weyl alcove (with which to study squares of elements), we get a bigger triangle made of 2^2 copies of the smaller one, reflected around using the affine Weyl group. The important thing to notice is that the identity corner (the 30 degree angle) only reflects to two places – itself, and to the 90 degree corner. So those are the two conjugacy classes of order dividing 2, one of which is the identity-element conjugacy class.

To check one’s understanding – there are two conjugacy classes of order 3, one with centralizer SU(3) the other with centralizer U(2).

Posted by: Allen Knutson on June 9, 2024 4:08 AM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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 $\mathbb{R}^3$.

Posted by: anon on June 6, 2024 5:23 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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

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

Posted by: John Baez on June 7, 2024 12:26 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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,

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

Posted by: John Baez on June 7, 2024 2:24 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 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 $W$.

Posted by: John Baez on June 7, 2024 5:33 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

We can see from John’s computations that the projection operator from degree-3 polynomials to harmonic degree-3 polynomials is $I - \frac{1}{10}\Delta_2\nabla^2$ where I’m using $\Delta_2$ as a shorthand for $x^2+y^2+z^2$ to indicate that it’s an invariant, and also to make the operator look symmetrical. The projection operator on $V\otimes V \otimes V$ is very similar, using inner products instead of the Laplacian, which makes the constant slightly different.

Posted by: Layra Idarani on June 8, 2024 11:10 AM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

Nice! It may be good to have a manifestly basis-independent formula for $p \colon S^3(V) \to W$, and this one is easy to verify starting from my basis-dependent formulas. But you’re right, we may even want such a formula for the orthogonal projection from $V \otimes V \otimes V$ down to $W$. We’ll see!

Posted by: John Baez on June 10, 2024 6:51 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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

(1)$(a \times b) \times a = (a \cdot a)b - (a \cdot b) a$

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 $W$.

Posted by: John Baez on June 10, 2024 9:38 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

Your computation is not quite finished. We get $3x y^2 = \frac{1}{2}\big((x+y)^3+(x-y)^3\big)-x^3$ and so $x^3-3x y^3 = 2x^3-\frac{1}{2}\big((x+y)^3+(x-y)^3\big).$

Posted by: Andrew Hubery on June 10, 2024 10:10 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

Whoops! Thanks.

Posted by: John Baez on June 10, 2024 10:41 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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).

Posted by: Layra Idarani on June 12, 2024 5:36 AM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

In the light of day, I realized that it definitely can’t just be that the octonion product isn’t Jacobi, because if Jacobi were enough then the adjoint rep of any simple Lie algebra would work, having a bilinear form and an antisymmetric product for which the first two axioms hold and the antisymmetric product obeys the Jacobi identity.

Posted by: Layra Idarani on June 12, 2024 10:50 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 2)

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).

Posted by: Layra Idarani on June 12, 2024 5:37 AM | Permalink | Reply to this

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