## May 15, 2024

### 3d Rotations and the 7d Cross Product (Part 1)

#### Posted by John Baez

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.

The following stuff is well-known: the group of linear transformations of $\mathbb{R}^n$ preserving the dot and cross product is called $SO(3)$. It consists of rotations. We say $SO(3)$ has an ‘irreducible representation’ on $\mathbb{R}^3$ because there’s no linear subspace of $\mathbb{R}^3$ that’s mapped to itself by every transformation in $SO(3)$, except for $\{0\}$ and the whole space.

Ho hum. But here’s something more surprising: it seems that $SO(3)$ also has an irreducible representation on $\mathbb{R}^7$ where every transformation preserves the dot product and cross product in 7 dimensions!

That’s right—no typo there. There is not an irreducible representation of $SO(7)$ on $\mathbb{R}^7$ that preserves the dot product and cross product. Preserving the dot product is easy. But the cross product in 7 dimensions is a strange thing that breaks rotation symmetry.

There is, apparently, an irreducible representation of the much smaller group $SO(3)$ on $\mathbb{R}^7$ that preserves the dot and cross product. But I only know this because people say Dynkin proved it! More technically, it seems Dynkin said there’s an $SO(3)$ subgroup of $G_2$ for which the irreducible representation of $\mathrm{G}_2$ on $\mathbb{R}^7$ remains irreducible when restricted to this subgroup. I want to see one explicitly.

We can get the dot and cross product in 3 dimensions by taking the space of imaginary quaternions, which is 3 dimensional, and defining

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

The multiplication on the right-hand side of these formulas is the usual quaternion product.

We can get the dot and cross product in 7 dimensions using formulas that look just the same! But we start with the space of imaginary octonions, which is 7 dimensional, and we use the octonion product.

In both cases we get a ‘vector product algebra’. A vector product algebra is a finite-dimensional real vector space with an inner product I’ll call the dot product and denote by

$\cdot \colon V^2 \to \mathbb{R}$

together with a bilinear operation I’ll call the cross product

$\times \colon V^2 \to V$

obeying three identities:

$u \times v = - v \times u$

$u \cdot (v \times w) = v \cdot (w \times u)$

$(u \times v) \times u = (u \cdot u) v - (u \cdot v) u$

These imply a bunch more identities.

You can get a vector product algebra from a normed division algebra by taking the subspace of imaginary elements, namely those orthogonal to $1$, and defining a dot and cross product using the formulas above. You can also reverse this process. Since there are only four normed division algebras, $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$, there are only four vector product algebras! But you can also run this argument backwards, which is nice because there’s a great string diagram proof that there are only four vector product algebras:

The four vector product algebras have dimensions 0, 1, 3, and 7. But only the last two are interesting, since in the first two the cross product is zero.

In fact the category of normed division algebras and algebra homomorphisms (which automatically preserve the inner product these algebras have) is equivalent to the category of vector product algebras. Thus the group of automorphisms of the 7-dimensional vector product algebra is isomorphic to the group of automorphisms of $\mathbb{O}$. This group is called $\mathrm{G}_2$.

Recently on Mastodon Paul Schwahn wrote:

The compact Lie group $\mathrm{G}_2$, usually defined as automorphism group of the octonion algebra $\mathbb{O}$, has (up to conjugacy) three maximal connected subgroups:

• the subgroup preserving the algebra of quaternions $\mathbb{H} \subset \mathbb{O}$ which is isomorphic to $SO(4)$,
• the subgroup preserving some imaginary element like $i$, which is isomorphic to $SU(3),$
• the subgroup $SO(3)_{\text{irr}}$ given by the image of the irreducible, faithful 7-dimensional real representation of $SO(3)$. This representation may be realized as the space of harmonic cubic homogeneous polynomials on $\mathbb{R}^3,$ or if you are a chemist, the space of $f$-orbital wavefunctions.

Now I wonder whether $SO(3)_{\text{irr}}$ also has some interpretation in terms of the octonions. What irreducible action of $SO(3)$ on the imaginary octonions is there?

@johncarlosbaez, do you perhaps have an idea?

Alas, I’m a bit stuck!

I know the compact Lie group $\mathrm{G}_2$ acts irreducibly on the 7-dimensional space $Im(\mathbb{O})$ of imaginary octonions. In fact it’s precisely the group of linear transformations that preserves cross product on $Im(\mathbb{O})$, and all these automatically preserve the dot product (by the argument here).

But I don’t know a concrete example of an $SO(3)$ subgroup of $\mathrm{G}_2$ that acts irreducibly on $Im(\mathbb{O})$. Apparently they’re all conjugate, so we can pick any one and call it $SO(3)_{\text{irr}}$. Can anyone here describe one concretely?

I do know how to get $SO(3)$ to act irreducibly on a 7-dimensional space. It’s called the spin-3 representation of $SO(3)$, or more precisely it’s the real form of that. Up to isomorphism this is the only irreducible representation of $SO(3)$ on a 7d real vector space, so the representation of $SO(3)_{\text{irr}} \subset \mathrm{G}_2$ on $Im(\mathbb{O})$ must be this.

There may be an explicit description of $SO(3)_{\text{irr}}$ in here:

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

But I haven’t dug into this text yet. I’ll try. But if you happen to know such an explicit description, please tell me!

Posted at May 15, 2024 4:36 PM UTC

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### Re: 3d Rotations and the Cross Product in 7 Dimensions

We need Robert Bryant to turn up and answer your question about the $SO(3)$ in $G_2$

Posted by: David Roberts on May 17, 2024 1:21 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

I may pester him by email. Or Linus Kramer, who mentioned this fact in a book.

Posted by: John Baez on May 17, 2024 9:23 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

This subgroup preserves a certain three-dimensional $X \subset V \otimes V$, and I believe is precisely the subgroup of $G_2$ preserving $X$ (this is just from the decomposition of the tensor square of the seven-dimensional irrep of SO(3)). However, this description is in some sense circular - the inclusion $X \into V \otimes V$ is adjoint to a map $X \otimes V \to V$, which is just the action of $\mathfrak{so}(3)$ on $V$. Still, perhaps one could come up with some nice characterization of which three-dimensional subspaces of $V\otimes V$ are invariant under some irreducible SO(3) action.

To get a more hands-on view of this subgroup, then, it feels easier to me to go in the other direction: starting from $W$, the seven-dimensional irrep of SO(3), there is a one-dimensional space of invariant tensors in $W \otimes W \otimes W$, and this lies in the exterior cube. Therefore, there is an essentially unique invariant 3-form on $W$, which should be isomorphic to the adjoint of the 7-dimensional cross product. It’s possible to write out this invariant 3-form explicitly and, up to chasing down some signs and factors of 2 that I still need to do, confirm that it agrees with the Fano plane description of octonion multiplication. One even gets formulae for how $\mathfrak{so}(3)$ acts on $V$, although not particularly nice ones.

Posted by: William Ballinger on May 17, 2024 11:35 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Nice! I see why there’s a 1-dimensional space of $SO(3)$-invariant tensors in $W \otimes W \otimes W$ when $W$ is the 7d irrep of $SO(3)$. Here’s one way to see that using just stuff physicists all know.

There’s a unique $2j+1$-dimensional irrep of $SU(2)$ for each spin $j = 0,\frac{1}{2}, 1, \frac{3}{2}, \dots$. Let’s call it the spin-j irrep or simply $[j]$. When $j$ is an integer this factors through $SO(3)$, and we get all finite-dimensional irreps of $SO(3)$ this way. The rules for tensoring these irreps are called the Clebsch–Gordan rules:

$[j] \otimes [k] \cong [|j-k|] \oplus [|j-k|+1] \oplus \cdots \oplus [j+k]$

In this notation the 7d irrep of $SO(3)$ is called $[3]$, and we get

$[3] \otimes [3] \cong [0] \oplus [1] \oplus [2] \oplus [3] \oplus [4] \oplus [5] \oplus [6]$

and thus

$[3] \otimes \big([3] \otimes [3]\big) \cong [3] \otimes \big([0] \oplus [1] \oplus [2] \oplus [3] \oplus [4] \oplus [5] \oplus [6]\big)$

This gives a big pile of crud if we expand it using the Clebsch–Gordan rules, but in this big pile of crud $[0]$ shows up just once, since $[3] \otimes [j]$ contains $[0]$ as a summand if and only if $j = 3$.

More later!

Posted by: John Baez on May 19, 2024 11:40 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

This was a fun puzzle! Mostly fun because it turned out not to require too much tedious linear algebra.

Solution sketch:

The general strategy has been outlined:

1: Take the 7-dimensional rep $W$ of $SO(3)$

2: Find an $SO(3)$-invariant element of $\wedge^3 W$

3: Interpret invariant element as multiplication.

Of course, the devil of the details is in the eating:

1: Finding the $7$-dimensional rep: John already mentioned one way: harmonic degree-3 polynomials in 3 variables (a,b,c). But we should pick a basis! Which one? Well, the more symmetrical the better, usually. We have three variables, so our basis should be symmetrical in the three variables. Under the symmetric group $S_3$, we get one basis element that is invariant, one set of 3 basis elements that is invariant, and one set of 3 basis elements that transforms as the sign representation. We might as well choose our elements to also have integer coefficients, to make our lives easier.

2: Okay, next up: $SO(3)$-invariant elements of $\wedge^3 W$. As usual, when trying to figure out the action of a Lie group, it’s easier to figure out the action of the corresponding Lie algebra. In particular, it’s easier to figure out the action of the elements of a nice basis of the corresponding Lie algebra. To keep on the idea of symmetry, we might as well take the basis of $so(3)$ that sends one of $(a,b,c)$ to $0$ and swaps the other two with one picking up a minus sign, e.g. $a\mapsto 0, b \mapsto -c, c\mapsto b$.

Once we have the action of $so(3)$ on each of our basis elements of $W$, we find the action of $so(3)$ on basis elements of $\wedge^3 W$. $\wedge^3W$ has a somewhat natural basis, given in terms of triples of basis elements of $W$, but because it’s the exterior power, you have to be careful about signs. So part of the fun is figuring out a sign scheme that 1: makes it easy to see the 3-cyclic symmetries still inherited from (a,b,c), and 2: doesn’t cause too many headaches to implement and track.

Now we get to figuring out what linear combination of these basis elements of $\wedge^3 W$ is actually invariant under $so(3)$ without doing a bunch of linear algebra, and here’s where all that symmetry preservation comes in handy: If one triple is in the linear combination with some coefficient, then applying an element of $S_3$ gives another triple that has the same coefficient, because our action of $so(3)$ is $S_3$-equivariant.

This cuts down on the number of things whose image under the $so(3)$ map you actually have to check, and also cuts down on how many basis elements of $so(3)$ you actually have to compute the action of.

And because math likes us, there are some things that clearly can’t be in the combination at all, and then the invariant pops out pretty quickly.

3: Finally, interpreting the invariant as a multiplication. The invariant we get happens to have 7 terms in it. Each basis element from $W$ shows up in three of those terms. The signs are compatible with the Fano plane. So it looks like that, up to scaling, our basis for $W$ are exactly the standard basis for the imaginary octonions, and so we can read an action of $SO(3)$ on the imaginary octonions from the action of $SO(3)$ on our basis of $W$. It’s not the nicest action, nor does it appear terribly enlightening, but it is writable by a human being, and even readable by a human being!

But about that scaling: not all of the terms have the same coefficient. Oops! Have we ended up with a slightly squashed octonions? Not to worry! The interpretation of this triple as a multiplication will save us! In particular, to go from an element of $\wedge^3 W$ to an element of $Hom(\wedge^2 W, W)$, we use the inner product on $W$. What is the inner product on $W$? It’s the one inherited from $[1]\otimes[1]\otimes[1]$. In particular, not all degree-3 monomials in (a,b,c) have the same length in this inner product! If we wanted our basis elements for $W$ to all have the same length, we couldn’t use all integer coefficients. Oh well.

But, doing this out carefully, we get that while the invariant element of $\wedge^3 W$ we’ve gotten, written in the basis we’ve picked, doesn’t have the same coefficient for all terms, the corresponding element of $Hom(\wedge^2W,W)$ does, because the discrepancy in the inner product exactly fixes the discrepancy in the coefficients. Huzzah!

So to unsquash our octonions, we just rescale so that all of our basis elements are the same length.

But seriously, that discrepancy makes the description a little uglier than it could be.

Posted by: Layra Idarani on May 20, 2024 11:22 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

This is really cool, Layra! Let me see if I can just start redoing your calculations. I won’t get very far today, but I thought I’d record my progress and where I get stuck.

We start with the 3-dimensional inner product space $V = \mathbb{R}^3$ with standard basis $x, y, z$. (You called them $a, b, c$.) Your approach makes heavy use of the symmetric group $S_3$ that acts on this basis, giving an inclusion

$S_3 \subset \mathrm{O}(3)$

which restricts to an inclusion

$A_3 \subset \mathrm{SO}(3).$

Then we let $W$ be the space of harmonic homogeneous degree-3 polynomials in $x, y,$ and $z$. We can 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 be an irreducible representation of $SO(3)$.

The space $W$ is 7-dimensional, since it has a basis

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

Our goal is to think of $W$ as the imaginary octonions. We want to give it an inner product and cross product isomorphic to those on the imaginary octonions — and invariant under the action of $SO(3)$. The inner product is fixed up to scale by $SO(3)$ invariance because the action of $SO(3)$ on $W$ is irreducible, so the really hard part is defining the cross product.

To accomplish this we want to pick a basis for $W$ that corresponds to the 7 points of the Fano plane:

Decreeing this basis to be orthonormal will fix the inner product on $W$. The cross product will be defined by the arrows in the picture, e.g. $e_1 \times e_2 = e_4$ because we’re going along with the arrows and $e_5 \times e_1 = - e_6, e_6 \times e_5 = -e_1$ because we’re going against them.

The Fano plane has an obvious $S_3$ symmetry, but only its subgroup $A_3$ acts in a way that respects the cross product. $A_3$ also acts on $W$ by permuting the variables $x, y, z$ in our harmonic homogeneous degree-3 polynomials. We want to make these actions agree.

I may be off here, but following your hints I’ll try this. The point in the middle of the Fano plane is invariant under the $S_3$ action so we choose

$e_7 = x y z$

The three points on the circle in the Fano plane form another $S_3$ orbit so we choose

$e_1 = (x^3 - 3x y^2) + (y^3 - 3y x^2)$

$e_2 = (y^3 - 3y z^2) + (z^3 - 3z y^2)$

$e_4 = (z^3 - 3z x^2) + (x^3 - 3x z^2)$

Note that the three expressions at right simply get permuted when we permute the variables $x, y, z$.

The three corners of the Fano plane form a third $S_3$ orbit, and we need to complete our basis of harmonic homogeneous degree-3 polynomials, so we use

$e_3 = (x^3 - 3x y^2) - (y^3 - 3y x^2)$

$e_5 = (y^3 - 3y z^2) - (z^3 - 3z y^2)$

$e_6 = (z^3 - 3z x^2) - (x^3 - 3x z^2)$

or something like that. When we permute the variables $x, y, z$ these expressions get permuted — but they also pick up a minus sign when the permutation is odd. Luckily, for permutations in $A_3$, i.e. cyclic permutations of $x, y, z$, there’s no minus sign.

But I haven’t tried yet to get the ‘right’ correspondence between the polynomials at right and the basis vectors $e_3, e_5, e_6$. When we do things correctly, the $A_3$ action on the Fano plane points $e_1, \dots, e_7$ should extend to an $SO(3)$ action on the vector space they span, $W$, and this action should preserve the cross product!

At least that’s what I’m hoping. Besides fiddling with who is $e_3$, who is $e_5$ and who is $e_6$, we may also need to fiddle with normalizations a bit.

Posted by: John Baez on May 22, 2024 9:09 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Also, as Layra noted, to check $SO(3)$-invariance of the 7d cross product it’s enough to check it infinitesimally. This is nice because $\mathfrak{so}(3)$ acts on $\mathbb{R}^3$ in a very simple way: in fact $\mathfrak{so}(3) \cong \mathbb{R}^3$ and its action

$\alpha \colon \mathfrak{so}(3) \times \mathbb{R}^3 \to \mathbb{R}^3$

is just the 3d cross product! We then extend this action to an action of $\mathfrak{so}(3)$ on polynomials in $x, y, z \in \mathbb{R}^3$ by demanding that $\mathfrak{so}(3)$ act as derivations.

Let’s do an example. We can think of $x,y,z \in \mathbb{R}^3$ as a basis of $\mathfrak{so}(3)$. The basis vector $x \in \mathfrak{so}(3)$ then acts on $\mathbb{R}^3$ by

$\alpha(x) x = x \times x = 0$ $\alpha(x) y = x \times y = z$ $\alpha(x) z = x \times z = -y$

It then acts on the polynomial $x y z$ using the product rule:

$\alpha(x)(x y z) \; = \; (\alpha(x)x)y z + x (\alpha(x)y) z + x y (\alpha(x)z) \; = 0 + x z^2 - y z^2$

Hmm, this suggests a basis of $W$ different than the one I was guessing before: we could use one containing $x y z$ and the six polynomials coming from $(x-y)z^2$ by permuting variables.

Anyway, to check the $SO(3)$ invariance of a candidate cross product on $W$, it suffices to show

$\alpha(v)(w \times w') = \alpha(v)(w) \times w' + w \times \alpha(v)(w')$

for all $v \in \mathbb{R}^3$, $w,w' \in W$. And of course it’s enough to check this for basis vectors.

Posted by: John Baez on May 22, 2024 12:56 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

I believe this works, giving an action of $A_3$ commuting with the cross product, modulo a couple of minor typos.

You left out a power of 2 in the second terms in your first 6 homogeneous polynomials. These need to be $x^3-3 x y^2$, etc.

And you swapped the order of $e_5$ and $e_6$ in the second cycle of imaginary octonions. The expressions on the right hand side that you have assigned are in the correct order for the cycle $e_3, e_6, e_5$ rather than $e_3, e_5, e_6$.

When I make the identifications with that correction, I get a matrix of order 3 acting on the 7 imaginary octonions that commutes with the cross product:

$\left( \begin{array}{ccccccc} 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right)$

Posted by: Greg Egan on May 22, 2024 2:22 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Thanks, Greg! I’m really glad that works.

Someone on Mathstodon caught my error where I wrote things like $x^3 - 3x y$ instead of $x^3 - 3x^2 y$, so I fixed that above before seeing your post, but apparently after you posted it — in case anyone is wondering why you seem to be correcting a nonexistent error.

Posted by: John Baez on May 22, 2024 3:10 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

To be clear, I only checked that the $A_3$ action is compatible with the cross product. I don’t think the so(3) action works out yet.

Posted by: Greg Egan on May 22, 2024 3:20 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

I may just wait for Layra to give us a hint.

Posted by: John Baez on May 22, 2024 3:25 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

So I ended up with a slightly different basis for W than than you did.

I started with $P = 2x^3 - 3x y^2 - 3x z^2$ $Q = 2y^3 - 3y z^2 - 3y x^2$ $R = 2z^3 - 3z x^2 - 3z y^2$ $S = 3x z^2 - 3x y^2$ $T = 3y x^2 - 3y z^2$ $U = 3z y^2 - 3z x^2$ $V = 6x y z$

which has the nice property that everything is clearly orthogonal since monomials are orthogonal and monomials in an $S_3$ orbit are the same length. But now $P, Q$ and $R$ look quite different from $S, T$ and $U$.

$\alpha(x)$ sends $x$ to $0$ and almost just swaps $y$ and $z$, but the sign change in the swap means that $\alpha(x)Q$ involves $U$ and $\alpha(x)R$ involves $T$, etc. The relation between $Q$ and $T$ is the same as that between $R$ and $U$ (again with possibly some sign issues), so that cuts things down a little bit. Another misfortune: at least in the above basis, the action of $\alpha(x)$ gives half-integer coefficients rather than integer coefficients, so it turns out that $\alpha(2x)$ is better.

Then we get the issue of computing $\alpha(2x)(P\wedge Q \wedge R)$ etc. This is also not particularly nice.

But! We get to win a little bit: we know that the invariant is built out of $A_3$-invariant pieces. So that cuts down the amount of things to calculate by almost 1/3, but there are 35 things to calculate and the cutdown only reduces that to 13. We get some additional cutdown by noticing that some of the $A_3$ orbits are almost $S_3$ orbits, so we only end up with like 10 possible coefficients. But there are so many terms that figuring out those 10 coefficients isn’t too bad.

The end result is that $3Q P R + 5 P T U + 5 Q U S + 5R S T + 5P S V + 5Q T V + 5 R U V \in \wedge^3W$ is $SO(3)$-invariant. This is pretty promising! All our basis elements are orthogonal and, ignoring the coefficients, this is the usual multiplication on unit octonions!

That leading $3$ coefficient should cause brief concern until one realizes that $P, Q$ and $R$ have square length $10$ while $S, T, U$ and $V$ have square-length $6$.

So we rescale $P, Q$ and $R$ by $\sqrt{3/5}$, i.e. our eventual basis is

$P = \sqrt{\frac{3}{5}}(2x^3 - 3x y^2 - 3x z^2)$ $Q = \sqrt{\frac{3}{5}}(2y^3 - 3y z^2 - 3x^2 y)$ $R = \sqrt{\frac{3}{5}}(2z^3 - 3x^2 z - 3y^2z)$ $S = 3 x z^2 - 3x y^2$ $T = 3y x^2 - 3y z^2$ $U = 3z y^2 - 3z x^2$ $V = 6x y z$

in which case $\alpha(2x)$ acts as $x \mapsto 0, y \mapsto -2z, z \mapsto 2y$ $P \mapsto 0$ $Q \mapsto +3R - \sqrt{15}U$ $R \mapsto -3Q - \sqrt{15}T$ $S \mapsto +4V$ $T \mapsto +\sqrt{15}R - U$ $U \mapsto +\sqrt{15}Q + T$ $V \mapsto -4S$ which I don’t find terribly enlightening, although it is at least antisymmetric now that we’ve rescaled, and the invariant is $Q P R + P T U + Q U S + R S T + P S V + Q T V + R U V$

Note that I could have done the rescaling earlier, since I knew that $P, Q$ and $R$ were the “wrong” length. But that would have given me the action of $\alpha(x)$ above, and thus made the rest of the calculation a little irritating, having to carry around those $\sqrt{15}$s.

Posted by: Layra Idarani on May 22, 2024 11:31 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Hi everyone! Awesome that you’re all putting in the effort trying to answer my question!

I think I know how to give an SO(3)-invariant element of $\Lambda^3W$ in a basis-free fashion (check out my other Mastodon post) - even better, it directly relates to the cross product on $\mathbb{R}^3$.

Let $\mathrm{Sym}^3\mathbb{R}^3$ be the SO(3)-representation of cubic homogeneous polynomials. Now we know this splits (orthogonally) into

(1)$\mathrm{Sym}^3\mathbb{R}^3= W\oplus(\mathbb{R}^3\cdot g),$

where $W$ is the subspace of harmonic polynomials, and the second summand are multiples of the square-norm $g=x^2+y^2+z^2$.

Define an operation $\bullet$ on $\mathrm{Sym}^3\mathbb{R}^3$ in the following way: on cubes of vectors/linear polynomials $X,Y\in\mathbb{R}^3$, it is given by

(2)$X^3\bullet Y^3\; :=(X\times Y)^3.$

If we let $P: \mathrm{Sym}^3\mathbb{R}^3\to W$ be ortho-projection, then I think the cross product on W can be defined by

(3)$\times:\quad W\times W\hookrightarrow \mathrm{Sym}^3\mathbb{R}^3\times \mathrm{Sym}^3\mathbb{R}^3 \stackrel{\bullet}{\longrightarrow} \mathrm{Sym}^3\mathbb{R}^3 \stackrel{P}{\longrightarrow}W.$

With a bit of elbow grease one can work out a formula for $P$(any monomial), set up a multiplication table for $\times$, and find a suitable identification with the imaginary octonions. I am currently trying to do this.

It would be nice to see whether we arrive at the same results this way!

Posted by: Paul Schwahn on May 23, 2024 3:17 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Wow, this sounds like a very beautiful approach if it works! Let us know how it goes.

Posted by: John Baez on May 24, 2024 10:16 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Ooh, this is very nice.

A note, at least for myself: The $\bullet$ operation is the induced operation from the cross-product on $(\mathbb{R}^3)^{\otimes 3}$. So clearly $SO(3)$-invariant, and also gives a tip on how to compute it on things that aren’t cubes.

I’ve checked that it matches my results (up to some scalar factor), although I’m less clear on how to pick a good identification without getting lucky on the basis of $W$. Not all orthonormal bases of $W$ will work, but just shoving $W$ into $Sym^3\mathbb{R}^3$ or projecting from $Sym^3\mathbb{R}^3$ to $W$ doesn’t pull out any obviously good bases either, as far as I can tell.

Posted by: Layra Idarani on May 26, 2024 9:23 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

So the way that I got the lengths was to think of the polynomials as symmetrized elements of $(x,y,z)^{\otimes 3}$. $(x,y,z)^{\otimes 3}$ has a natural inner product on it, i.e. $\left\langle a\otimes b \otimes c, e\otimes f \otimes g\right\rangle = \left\langle a, e\right\rangle\left\langle b, f\right\rangle\left\langle c, f\right\rangle.$

Then we assume that $x,y,z$ are all orthonormal and this induces a inner product on homogeneous degree-3 polynomials.

The thing to note is that, for instance, the monomial $x^2y$ is not $x\otimes x \otimes y$, but rather $\frac{1}{3}(x\otimes x \otimes y + x \otimes y \otimes x + y \otimes x \otimes x)$ and hence has square length $1/3$. Compare to $x^3$, which is actually just $x\otimes x\otimes x$ and thus has square length $1$, and $xyz$ which is $\frac{1}{6}(x\otimes y\otimes z + x\otimes z \otimes y + y\otimes x \otimes z + y\otimes z \otimes x + z\otimes x\otimes y + z\otimes y\otimes x)$ and hence has square length $1/6$.

Posted by: Layra Idarani on May 26, 2024 1:12 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

I stand in awe. It would have taken me about a month of hard work to do that.

Posted by: John Baez on May 22, 2024 11:54 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

I agree, Layra’s calculations are awesome!

FWIW, I’ve managed to confirm the results, with help from Mathematica.

One identification with the imaginary octonions, numbered in the usual way, is:

$\{e_1,...,e_7\} = \{P,R,S,Q,T,U,V\}$

With this identification, each line in the Fano plane [as John showed it a few replies upthread] corresponds to one term in Layra’s invariant sum in the 3-fold exterior product.

I got a bit confused thinking that the actions on the imaginary octonions of the basis elements of so(3) ought to commute with the 7D cross product, but of course they don’t; this only works for the SO(3) group action. But once you exponentiate the matrices representing the Lie algebra elements, the resulting matrices do commute with the cross product.

Posted by: Greg Egan on May 23, 2024 7:32 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Layra wrote:

That leading 3 coefficient should cause brief concern until one realizes that $P,Q$ and $R$ have square-length 10 while $S,T,U$ and $V$ have square-length 6.

Can you say how you’re getting this, Layra? I like to use the inner product on polynomials such that monomials are orthogonal and $x^\ell y^m z^n$ has square-length $\ell! m! n!/(\ell + m + n)!$. This is natural for various reasons and physicists use it in ‘Fock space’.

But with this inner product the square-length of $x y z$ is $1$ so that of $V = 6 x y z$ is $6$. That looks good.

With this inner product the square-length of $x y^2$ or $x z^2$ is $1/3$, and they’re orthogonal, so the square-length of $S = 3 x z^2 - 3 x y^2$ is $2$, not $6$ as you say. You seem to be wanting the square-length of $x y^2$ and $x z^2$ to be $1$.

Even with the latter convention, I don’t see how you’re getting the square-length of $P=2x^3−3x y^2−3x z^2$ to be $10$.

I could be making mistakes.

Posted by: John Baez on May 23, 2024 1:18 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

John, you’re forgetting that for squared length, the coefficient gets squared. So $3x^2y$ has squared length $3^2 \cdot |x^2y|^2 = 3^2\cdot 1/3 = 3$ etc.

Posted by: Layra Idarani on May 26, 2024 1:19 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

If you integrate these functions on the unit sphere with the standard measure then normalise so that |1|^2 = 1, then the original |P|^2 = |Q|^2 = |R|^2 = 4/7, while the other four functions have squared length 12/35, so the ratio is 3 : 5 or 6 : 10.

Posted by: Greg Egan on May 23, 2024 1:44 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Thanks, Greg! I find integration over the unit sphere to be a bit distracting when trying to think about the inner product on homogeneous degree-$n$ polynomials. There’s a nice purely algebraic way to define an inner product. But the two inner products should be proportional, and I don’t seem to be getting even that.

Can you tell me the squared length of the polynomials $x^3, x^2 y$ and $x y z$ using integration? (In theory I could figure it out given what you said.)

Posted by: John Baez on May 23, 2024 3:08 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

$\|x^3\|^2 = \frac{1}{4\pi} \int _0^{\pi }\int _0^{2 \pi } \sin ^7(\theta ) \cos ^6(\phi ) d\phi d\theta = 1/7$ $\|x^2 y\|^2 = \frac{1}{4\pi} \int _0^{\pi }\int _0^{2 \pi } \sin ^7(\theta ) \sin ^2(\phi ) \cos ^4(\phi ) d\phi d\theta = 1/35$ $\|x y z\|^2 = \frac{1}{4\pi} \int _0^{\pi }\int _0^{2 \pi } \sin ^5(\theta ) \cos ^2(\theta ) \sin ^2(\phi ) \cos ^2(\phi ) d\phi d\theta = 1/105$

Posted by: Greg Egan on May 23, 2024 11:43 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

The 3D cross product is the Hodge dual of the wedge/outer/exterior product of two vectors in 3D, which readily generalises to all finite dimensions greater than or equal to 2. Is there something similar for the 7D cross product to generalise it to other dimensions?

### Re: 3d Rotations and the Cross Product in 7 Dimensions

While the 3d cross product

$\times \colon \mathbb{R}^3 \otimes \mathbb{R}^3 \to \mathbb{R}^3$

is a classical structure (i.e. its symmetry group is the classical group $SO(3)$, and it admits a natural generalization to any dimension, namely the exterior product $\wedge \colon V \otimes V \to \Lambda^2(V)$), the 7d cross product

$\times \colon \mathbb{R}^7 \otimes \mathbb{R}^7 \to \mathbb{R}^7$

is an exceptional algebraic structure (i.e. its symmetry group is the exceptional Lie group $G_2$). So, while we can reformulate the 7d cross product in other ways, I don’t think it generalizes to other dimensions. This is why it’s interesting.

The special nature of the 0d, 1d, 3d and 7d cross products are related to the fact that only the 0-sphere, 1-sphere, 3-sphere and 7-sphere have a trivializable tangent bundle:

• The 0-sphere and 1-sphere are the abelian Lie groups $\mathrm{O}(1) \cong \mathbb{Z}/2$ and $\mathrm{U}(1) \cong SO(2)$, so their Lie algebras endow $\mathbb{R}^0$ and $\mathbb{R}^1$ with a vanishing cross product.

• The 3-sphere is the nonabelian simple Lie group $Sp(1) \cong SU(2)$, and its Lie algebra endows $\mathbb{R}^3$ with its usual cross product.

• The 7-sphere is the Lie loop of unit octonions. This is a nonassociative generalization of a Lie group, so its tangent space at the identity is not a Lie algebra: it has a bracket, which is the 7d cross product, but this bracket doesn’t obey the Jacobi identity.

You would probably enjoy this:

This discusses generalizations of the cross product that take $n$ arguments instead of just $2$. But I think when you learn about these generalizations you’ll be even more convinced that the 7d cross product is an exceptional algebraic structure.

Posted by: John Baez on May 21, 2024 9:48 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

The Wikipedia article referenced above explains that the seven dimensional cross product can be written as the contraction of the exterior product with a certain trivector. Wikipedia writes the trivector as $v = e_{124} + e_{235} + e_{346} + e_{457} + e_{561} + e_{672} + e_{713}$. I believe this is the same thing as the invariant that Layra has been discussing ($QPR + \dots$). This seems like a good starting point for seeing the SO(3) invariance.

Posted by: Bob on May 28, 2024 7:11 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

That trivector

$v = e_{124} + e_{235} + e_{346} + e_{457} + e_{561} + e_{672} + e_{713}$

is visible in the Fano plane here:

It has one term for each of the 7 lines, and the numbers in each term are listed in a cyclic order matching the arrows on the picture. The Fano plane also lets us read off the cross product of imaginary octonions immediately, e.g.

$e_1 \times e_2 = e_4$ and cyclic permutations

arise from the line (shaped like a circle) containing the points 1, 2, and 4, while

$e_2 \times e_3 = e_5$ and cyclic permutations

arise from the line containing the points 2, 3, and 5.

So, all this is just a way of thinking about how the Fano plane determines the cross product of imaginary octonions. To get ahold of an $SO(3)$ subgroup acting irreducibly on the imaginary octonions and preserving the cross product takes further work, which is what Layra and Paul are doing.

Posted by: John Baez on May 28, 2024 10:18 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

According to a theorem I learned in Serre’s book “Lie Groups and Lie Algebras”, a reductive group G acting faithfully on a rep V is characterized by its tensor invariants.

In the case at hand let V be the 7-rep of G_2, and F the mysterious SO(3). Then as discussed in other answers, there is only 1-dim space of F-invariants in V@V and V@V@V, each of which are also G-invariant.

So we move on to 4-forms, invariants in V@V@V@V. Since V@V is multiplicity-free and self-dual (for either F or G) the dimension of the space of invariants in V@V@V@V is equal to the number of summands in V@V. For F this is 7 (see John’s discussion of Clebsch-Gordan), whereas for G it is only 4 (a calculation I did using the LieTypes package in Macaulay 2).

If we pick any element of this 7-d space of F-invariants that isn’t in the 4-d space of G-invariants, its stabilizer will include the SO(3) but be smaller than G2. If we’re to believe the claim that the SO(3) is maximal, then apparently we’ve found that SO(3).

We should be able to nail this down further using the S4 action on V@V@V@V. By Schur-Weyl, the decomposition into S4 x GL(7) irreps is a sum over the five partitions of 4. Then we’d need to pick out the F- and G-invariants in those GL(7)-reps.

I managed to do that for F. The dimensions are 4: 2 3+1: 0 2+2: 2 (which shows up twice, as the S_4-rep is 2-dim) 2+1+1: 0 1+1+1+1: 1 for the expected total of 7. I don’t see how to easily calculate the branching from GL(7) to G2 and thereby figure out how that 7 cuts down to 4. The Macaulay2 package will do it if I tell it how G2’s coroots expand in GL(7)’s coroots.

Posted by: Allen Knutson on May 21, 2024 7:33 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

I don’t know if this is compatible with the usual basis of the usual 7-dimensional representation of $\mathsf G_2$, but, in Bourbaki’s numbering, the coroots $a^\vee$ of $\mathsf G_2$ are related to the coroots $b^\vee$ of $\mathsf A_6$ by $a_1^\vee = b_1^\vee$ and $a_2^\vee = -\frac2 3 b_1^\vee - \frac1 3 b_2^\vee$.

Posted by: L Spice on May 22, 2024 10:53 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Oops, sorry, as anyone except apparently me could have noticed, that’s just some embedding of the $\mathsf G_2$ coroots in $\mathbb{R}^3$, not anything to do with $\mathsf A_6$.

Posted by: L Spice on May 22, 2024 11:02 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Looking for tensorial invariants in order to nail down this subgroup is a good idea!

The branching rules from G$_2$ to the maximal SO(3) are available for example in the software package LiE.

The decomposition of $V\otimes V$ into irreps of G$_2$ is well-known:

(1)$\Lambda^2V=\Lambda^2_{14}\oplus\Lambda^2_7,\qquad\mathrm{Sym}^2V=\mathrm{Sym}^2_0V\oplus\mathbb{R}g,$

where $\Lambda^2_{14}\cong\mathfrak{g}_2$, $\Lambda^2_7\cong V$, $\mathrm{Sym}^2_0V$ is the trace-free part, and $g$ is the invariant inner product on $V$.

Using John’s notation for the irreps of SO(3), the G$_2$-irreducible parts of $V\otimes V$ branch as follows:

(2)$\Lambda^2_{14}\cong[1]\oplus[5],$
(3)$\Lambda^2_7\cong[3],$
(4)$\mathrm{Sym}^2_0V\cong[2]\oplus[4]\oplus[6].$

Note that $[1]\cong\mathfrak{so}(3)$, thus [5] can be understood as the isotropy representation of the homogeneous space $\mathrm{G}_2/\mathrm{SO}(3)_{\mathrm{irr}}$.

As you noted, the fourth-order tensorial invariants correspond to the irreducible parts of $V\otimes V$. Now there is one that I, as a Riemannian geometer, am particularly interested in: the SO(3)-invariant element in

(5)$\mathrm{Sym}^2[1]\subset\mathrm{Sym}^2\Lambda^2V.$

An example of such an element would be the curvature tensor of a connection with holonomy $\mathrm{SO}(3)_{\mathrm{irr}}\subset\mathrm{G}_2$ on a 7-dimensional manifold.

An indeed, such a connection exists: the canonical homogeneous connection on the so-called Berger space $\mathrm{SO}(5)/\mathrm{SO}(3)_{\mathrm{irr}}$. The last $\mathrm{SO}(3)_{\mathrm{irr}}$ refers to the maximal subgroup of SO(5) that is embedded via the 5-dimensional representation [2]. The isotropy representation (i.e. tangent space) of this homogeneous space is again precisely $V\cong[3]$!

The aforementioned connection on the Berger space is metric (for homogeneous metrics), but has torsion. (To be more precise, it is an example of a geometry with parallel skew-symmetric torsion.) The Berger space has a series of very interesting geometric properties; for example, it is one of the few examples of homogenous nearly parallel G$_2$-manifolds – but that might lead too far…

Posted by: Paul Schwahn on May 23, 2024 2:49 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Does LiE tell you how to decompose the GL(7) reps corresponding to partitions with 4 boxes, under G_2?

It’d be enough to know how to decompose Sym^4(C^7) and Alt^4(C^7).

Posted by: Allen Knutson on May 30, 2024 4:58 AM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

Indeed it does. Here is the console output:

 > setdefault(G2) > t=1X[1,0] > dim(t) 7 > plethysm([4],t) 1X[0,0] +1X[2,0] +1X[4,0] > plethysm([3,1],t) 1X[0,1] +1X[1,0] +1X[1,1] +1X[2,0] +1X[2,1] +1X[3,0] > plethysm([2,2],t) 1X[0,0] +1X[0,2] +1X[1,1] +2X[2,0] > plethysm([2,1,1],t) 2X[0,1] +2X[1,0] +1X[1,1] +1X[2,0] +1X[3,0] > plethysm([1,1,1,1],t) 1X[0,0] +1X[1,0] +1X[2,0] 

These go from Sym$^4$ to $\Lambda^4$. Representations are denoted by giving their highest weight in the basis of fundamental weights. So [1,0] is $\mathbb{C}^7$, [0,1] is $\mathfrak{g}_2^{\mathbb{C}}$, [2,0] is $\mathrm{Sym}^2_0\mathbb{C}^7$, and so on.

Posted by: Paul Schwahn on May 31, 2024 7:59 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

OK good, so while the SO(3) invariants in Sym^4, Schur_ {2,2}, Alt^4 have dimensions 1,2,1 respectively, the G_2 invariants have dimensions 1,1,1 respectively.

This leads to a nearly-unique answer as to how to characterize the SO(3) as the stabilizer of a 4-form. Give V = C^7 the unique (up to scale) compact-G_ 2 invariant Hermitian form. Then we get a 2-d space W of SO(3)-invariants inside Schur_ {2,2}(V), containing a 1-d space W’ of G_ 2-invariants. There’s a unique vector (up to scale) w in W that’s perpendicular to W’.

This is stated a little weirdly, though, because it presupposes knowing the SO(3), which was kinda the point of the exercise. Really the idea is to start with Schur_ {2,2}(V), consider the 1-d space W’ of G_ 2-invariants, and pick a vector w not in there. Then hope that Stab_ {G_ 2}(w) is some SO(3).

Obviously the dimension of the space of such SO(3)s is dim(G_ 2/N(SO(3))) = 11. Whereas the dimension of Schur_ {2,2}(V) = 196 I think. So one needs an additional idea to know how to pick a good 4-form inside Schur_ {2,2}(V), whose stabilizer will be an SO(3).

Probably the place to start is to think about how SO(3)’s Cartan H should sit inside G_2’s, and only pick 4-forms with the right H-weight.

Posted by: Allen Knutson on June 8, 2024 2:30 PM | Permalink | Reply to this

### Re: 3d Rotations and the Cross Product in 7 Dimensions

The basis of Schur_ {2,2}(V) is indexed by 2x2 SSYT with entries from 1..7. The H-weights of the basis 1..7 are -3,-2,-1,0,1,2,3. So to get the right H-weight (zero), we want to consider such SSYT with average entry 4. There are 20 of these.

To sum up:

We’re looking for an SO(3) lying inside G_ 2 and containing the circle H. We know that any such SO(3) has a 2-dimensional invariant space inside Schur_ {2,2}(V), where G_ 2 only has a 1-dimensional invariant space. Any SO(3)-invariant is of course H-invariant, and the space of H-invariants is 20-dimensional.

Taking alpha_ 1 = the long simple root of G_ 2 and alpha_ 2 = the short,

(1) H = the kernel of omega1 + omega2 (2) the raising operator of the SO(3) lives in g_ {2 alpha1 + 3 alpha2} + g_ { - alpha1 - 2 alpha2}, roots spaces of G_ 2.

So the next computer calculation is to compute the 2-dim kernel of such a raising operator acting on that 20-dim space. 1-dim of it will be G_ 2-invariants.

Posted by: Allen Knutson on June 8, 2024 3:36 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 1)

For some possible applications of this mathematics to chemistry, again pointed out by Paul Schwahn, see another blog article of mine:

This article mentions how the 14-dimensional representation $[3] \otimes [1/2]$ of $SO(3)$ describes the orbital and spin angular momentum of electrons in the f shell. Given that the 7-dimensional representation $[3]$ of $SO(3)_{irr} \subset \mathrm{G}_2$ extends to a representation of $\mathrm{G}_2$, I wonder if it’s too crazy to hope that this 14-dimensional representation also extends to a representation of $\mathrm{G}_2$. There’s an obvious candidate, the adjoint representation! But maybe this is too good to be true.

Posted by: John Baez on May 28, 2024 12:28 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 1)

Ah, now if you bring spin into the mix, that breaks things - because $[3]\otimes[1/2]\cong[5/2]\oplus[7/2]$ is not a representation of the group SO(3)!

Also the 14-dimensional irrep of $\mathrm{G}_2$ splits as $[1]\oplus[5]$ under its restriction to $\mathrm{SO}(3)_{\mathrm{irr}}$.

Posted by: Paul Schwahn on May 31, 2024 9:55 PM | Permalink | Reply to this

### Re: 3d Rotations and the 7d Cross Product (Part 1)

Aargh, what a dumb idea of mine that was. I was momentarily bewitched by the fact that there are 14 lanthanides, Racah somehow used $\mathrm{G}_2$ for understanding lanthanides, and $\mathrm{G}_2$ is 14-dimensional.

Posted by: John Baez on June 1, 2024 3:54 PM | Permalink | Reply to this
Read the post 3d Rotations and the 7d Cross Product (Part 2)
Weblog: The n-Category Café
Excerpt: Two ways to get SO(3) to act irreducibly on the imaginary octonions while preserving their cross product.
Tracked: June 4, 2024 10:11 PM

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