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October 6, 2025

A Complex Qutrit Inside an Octonionic One

Posted by John Baez

Dubois-Violette and Todorov noticed that the Standard Model gauge group is the intersection of two maximal subgroups of F 4\mathrm{F}_4. I’m trying to understand these subgroups better.

Very roughly speaking, F 4\mathrm{F}_4 is the symmetry group of an octonionic qutrit. Of the two subgroups I’m talking about, one preserves a chosen octonionic qubit, while the other preserves a chosen complex qutrit.

A precise statement is here:

Over on Mathstodon I’m working with Paul Schwahn to improve this statement. He made a lot of progress on characterizing the first subgroup. F 4\mathrm{F}_4 is really the group of automorphisms of the Jordan algebra of 3×33 \times 3 self-adjoint octonion matrices, 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}). He showed the first subgroup, the one I said “preserves a chosen octonionic qubit”, is really the subgroup that preserves any chosen Jordan subalgebra isomorphic to 𝔥 2(𝕆)\mathfrak{h}_2(\mathbb{O}).

Now we want to show the second subgroup, the one I said “preserves a chosen complex qutrit”, is really the subgroup that preserves any chosen Jordan subalgebra isomorphic to 𝔥 3()\mathfrak{h}_3(\mathbb{C}).

I want to sketch out a proof strategy. So, I’ll often say “I hope” for a step that needs to be filled in.

Choose an inclusion of algebras 𝕆\mathbb{C} \hookrightarrow \mathbb{O}. All such choices are related by an automorphism of the octonions, so it won’t matter which one we choose.

There is then an obvious copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}) sitting inside 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}). I’ll call this the standard copy. To prove the desired result, it’s enough to show:

  1. The subgroup of F 4\mathrm{F}_4 preserving the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}) in 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) is a maximal subgroup of F 4\mathrm{F}_4, namely (SU(3)×SU(3))/ 3(\text{SU}(3) \times \text{SU}(3))/\mathbb{Z}_3.

  2. All Jordan subalgebras of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) isomorphic to 𝔥 3()\mathfrak{h}_3(\mathbb{C}) are related to the standard copy by an F 4\mathrm{F}_4 transformation.

Part 1. should be the easier one to show, but I don’t even know if this one is true! (SU(3)×SU(3))/ 3(\text{SU}(3) \times \text{SU}(3))/\mathbb{Z}_3 is a maximal subgroup of F 4\mathrm{F}_4, and Yokota shows it preserves the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}) in 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}). But he shows it also preserves more, seemingly: it preserves a complex structure on the orthogonal complement of that standard copy. Is this really ‘more’ or does it hold automatically for any element of F 4\mathrm{F}_4 that preserves the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C})? I don’t know.

But I want to focus on part 2). Here’s what we’re trying to show: any Jordan subalgebra of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) isomorphic to 𝔥 3()\mathfrak{h}_3(\mathbb{C}) can be obtained from the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}) by applying some element of F 4\mathrm{F}_4.

So, pick a Jordan subalgebra A of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) isomorphic to 𝔥 3()\mathfrak{h}_3(\mathbb{C}). Pick an isomorphism A ≅ 𝔥 3()\mathfrak{h}_3(\mathbb{C}).

Consider the idempotents

e 1 = diag(1,0,0) e 2 = diag(0,1,0) e 3 = diag(0,0,1) \begin{array}{ccl} e_1 &=& diag(1,0,0) \\ e_2 &=& diag(0,1,0) \\ e_3 &=& diag(0,0,1) \end{array}

in 𝔥 3()\mathfrak{h}_3(\mathbb{C}). Using our isomorphism A𝔥 3()A \cong \mathfrak{h}_3(\mathbb{C}) they give idempotents in AA, which I’ll call f 1,f 2,f 3f_1, f_2, f_3. Since A𝔥 3(𝕆)A \subset \mathfrak{h}_3(\mathbb{O}) these are also idempotents in 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}).

Hope 1: I hope there is an element gg of F 4\mathrm{F}_4 mapping f 1,f 2,f 3𝔥 3(𝕆)f_1, f_2, f_3 \mathfrak{h}_3(\mathbb{O}) to e 1,e 2,e 3𝔥 3()e_1, e_2, e_3 \in \mathfrak{h}_3(\mathbb{C})𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}).

Hope 2: Then I hope there is an element hh of F 4\mathrm{F}_4 that fixes e 1,e 2,e 3e_1, e_2, e_3 and maps the subalgebra gAg A to the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}) in 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}).

If so, we’re done: hgh g maps AA to the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}).

Hope 1 seems to be known. The idempotents e 1,e 2,e 3e_1, e_2, e_3 form a so-called ‘Jordan frame’ for 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}), and so do f 1,f 2,f 3f_1, f_2, f_3. Faraut and Korányi say that “in the irreducible case, the group KK acts transitively on the set of all Jordan frames”, and I think that implies Hope 1.

As for Hope 2, I know the subgroup of F 4\mathrm{F}_4 that fixes e 1,e 2,e 3e_1, e_2, e_3 contains Spin(8)\text{Spin}(8). I bet it’s exactly Spin(8)\text{Spin}(8). But to prove Hope 2 it may be enough to use Spin(8)\text{Spin}(8).

Let me say a bit more about how we might realize Hope 2. It suffices to consider a Jordan subalgebra BB of 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) that is isomorphic to 𝔥 3()\mathfrak{h}_3(\mathbb{C}) and contains

e 1 = diag(1,0,0) e 2 = diag(0,1,0) e 3 = diag(0,0,1) \begin{array}{ccl} e_1 &=& diag(1,0,0) \\ e_2 &=& diag(0,1,0) \\ e_3 &=& diag(0,0,1) \end{array}

and prove that there is an element hh of F 4\mathrm{F}_4 that fixes e 1,e 2,e 3e_1, e_2, e_3 and maps the subalgebra BB to the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}) in 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}). (In case you’re wondering, this BB is what I was calling gAg A.)

Hope 3: I hope that we can show BB consists of matrices

(a 1 z * y * z a 2 x y x * a 3) \left( \begin{array}{ccc} a_1 & z^\ast & y^\ast \\ z & a_2 & x \\ y & x^\ast & a_3 \end{array} \right)

where a 1,a 2,a 3a_1, a_2, a_3 are arbitrary real numbers and x,y,zx, y, z range over 2-dimensional subspaces V 1,V 2,V 3V_1, V_2, V_3 of 𝕆\mathbb{O}. This would already make it look fairly similar to the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}), where the subspaces V 1,V 2,V 3V_1, V_2, V_3 are all our chosen copy of \mathbb{C} in 𝕆\mathbb{O}.

If Hope 3 is true, the subspaces V 1,V 2,V 3V_1, V_2, V_3 don’t need to be the same, but I believe they do need to obey V 1V 2V 3V_1 V_2 \subseteq V_3 and cyclic permutations thereof, simply because BB is closed under the Jordan product.

So, we naturally want to know if such a triple of 2d subspaces of 𝕆\mathbb{O} must be related to the ‘standard’ one (where they are all \mathbb{C}) by an element of Spin(8)\text{Spin}(8), where Spin(8)\text{Spin}(8) acts on the three copies of 𝕆\mathbb{O} by the vector, left-handed spinor, and right-handed spinor representations, respectively — since this is how Spin(8)\text{Spin}(8) naturally acts on 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O}) while fixing all the diagonal matrices.

This is a nice algebra question for those who have thought about triality, and more general ‘trialities’.

So, that’s where I am now: a bunch of hopes which might add up to a clarification of what I mean by “the subgroup of symmetries of an octonionic qutrit that preserve a complex qutrit”.

Posted at October 6, 2025 5:24 PM UTC

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Re: A Complex Qutrit Inside an Octonionic One

There is a result due to Kac that reads off the classification of (conjugacy classes of) torsion elements of adjoint forms of Lie groups from their Dynkin diagrams; see e.g. Theorem 8.3.3.1 of https://categorified.net/LieQuantumGroups.pdf and references therein. In the case of F 4F_4, there are exactly three conjugacy classes of order-33 elements. I think that the centralizers of these conjugacy classes are: (SU(3)×SU(3))/ 3(SU(3) \times SU(3)) / \mathbb{Z}_3, U(1)×Spin(7)U(1) \times Spin(7), and U(1)×Sp(3)U(1) \times Sp(3) (where Sp(n)Sp(n) is the automorphisms of n\mathbb{H}^n that preserve the real norm and the \mathbb{H}-action, and I’m confident of the Lie algebras but I might have made a mistake about the specific group forms).

If you get stuck on a “good” proof of Part 2, you can probably at least supply a “bad” proof by showing that the automorphisms of an 𝔥 3()𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{C}) \subset \mathfrak{h}_3(\mathbb{O}) has a central 3\mathbb{Z}_3 and cannot be either of the latter two groups. This should be straightforward. For example, you already win if you can show that the map to Aut(𝔥 3())=PSU(3)Aut(\mathfrak{h}_3(\mathbb{C})) = PSU(3) has image larger than U(1)U(1).

Doesn’t Part 1 follow from the statements “Yokota shows [(SU(3)×SU(3))/ 3(SU(3) \times SU(3)) / \mathbb{Z}_3] preserves the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}) in 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O})” together with “(SU(3)×SU(3))/ 3(SU(3) \times SU(3)) / \mathbb{Z}_3 is a maximal subgroup of F 4F_4”?

Posted by: Theo Johnson-Freyd on October 6, 2025 8:18 PM | Permalink | Reply to this

Re: A Complex Qutrit Inside an Octonionic One

Theo wrote:

Doesn’t Part 1 follow from the statements “Yokota shows [(SU(3)×SU(3))/ 3(SU(3) \times SU(3)) / \mathbb{Z}_3] preserves the standard copy of 𝔥 3()\mathfrak{h}_3(\mathbb{C}) in 𝔥 3(𝕆)\mathfrak{h}_3(\mathbb{O})” together with “(SU(3)×SU(3))/ 3(SU(3) \times SU(3)) / \mathbb{Z}_3 is a maximal subgroup of F 4F_4”?

Wow, I think it does! It’s hard for me to believe, because to me it’s a completely insight-free proof, relying completely on the maximality of that subgroup. Ideally I’d like to see how to functorially construct the extra structure that gets preserved from the structure I know is preserved. But I’ll take what I can get!

I’m not following the logic of your “bad” proof of Part 2, but I’m probably just being thick, and maybe I’ll see it in a while. Thanks!

Posted by: John Baez on October 7, 2025 4:45 PM | Permalink | Reply to this

Re: A Complex Qutrit Inside an Octonionic One

Okay, now I’m following the logic. Yes, that would be another brilliant insight-free proof. (At least for me, since I’d be using results I don’t understand.)

Thanks! Right now my proposed strategy seems like it might be working, and the lemmas involved are giving me more understanding of the P 2\mathbb{C}\text{P}^2’s in an 𝕆P 2\mathbb{O}\text{P}^2. But the 3\mathbb{Z}_3 business is also very important. The “good” 3\mathbb{Z}_3’s in F 4\text{F}_4, those whose centralizer is a copy of (SU(3)×SU(3))/ 3(\text{SU}(3) \times \text{SU}(3))\!/\!\mathbb{Z}_3, are discussed in Yokota’s Section 2.12. They’re a spinoff of the nontrivial inner automorphisms of the octonions, which all have order 3, and each determine an inclusion 𝕆\mathbb{C} \hookrightarrow \mathbb{O}.

There’s a beautiful story here, which someday I hope to tell nicely.

Posted by: John Baez on October 7, 2025 5:41 PM | Permalink | Reply to this

Re: A Complex Qutrit Inside an Octonionic One

I wrote:

I know the subgroup of F 4\mathrm{F}_4 that fixes e 1,e 2,e 3e_1, e_2, e_3 contains Spin(8)\text{Spin}(8). I bet it’s exactly Spin(8)\text{Spin}(8).

Yes it is! Paul Schwahn showed it is:

I also think the subgroup fixing each of e₁, e₂, e₃ is Spin(8). Here’s a sketch of the argument:

The stabilizer of e₁ is Spin(9), and this also preserves the 𝔥₂(𝕆)-subalgebra containing e₂+e₃. I think you have shown somewhere else that this Spin(9) acts on this 𝔥₂(𝕆) by automorphisms, via the double covering Spin(9) → SO(9) = Aut 𝔥₂(𝕆).

Now we are one level lower: the set of points (trace 1 idempotents) in 𝔥₂(𝕆) is 𝕆ℙ¹ ≅ SO(9)/SO(8) ≅ Spin(9)/Spin(8), so the subgroup of Spin(9) fixing in addition the point e₂ (and therefore its complement e₃) must be Spin(8).

Posted by: John Baez on October 7, 2025 4:54 PM | Permalink | Reply to this

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