The n-Category Café

Very roughly speaking, $\mathrm{F}_4$ 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:

• Can we understand the Standard model using octonions?

Over on Mathstodon I’m working with Paul Schwahn to improve this statement. He made a lot of progress on characterizing the first subgroup. $\mathrm{F}_4$ is really the group of automorphisms of the Jordan algebra of $3 \times 3$ self-adjoint octonion matrices, $\mathfrak{h}_3(\mathbb{O})$. 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 $\mathfrak{h}_2(\mathbb{O})$.

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 $\mathfrak{h}_3(\mathbb{C})$.

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 $\mathbb{C} \hookrightarrow \mathbb{O}$ 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 $\mathfrak{h}_3(\mathbb{C})$ sitting inside $\mathfrak{h}_3(\mathbb{O})$. I’ll call this the standard copy. To prove the desired result, it’s enough to show:

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

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

Part 1. should be the easier one to show, but I don’t even know if this one is true! $(\text{SU}(3) \times \text{SU}(3))/\mathbb{Z}_3$ is a maximal subgroup of $\mathrm{F}_4$, and Yokota shows it preserves the standard copy of $\mathfrak{h}_3(\mathbb{C})$ in $\mathfrak{h}_3(\mathbb{O})$. 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 $\mathrm{F}_4$ that preserves the standard copy of $\mathfrak{h}_3(\mathbb{C})$? I don’t know.

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

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

Consider the idempotents

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

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

Hope 1: I hope there is an element $g$ of $\mathrm{F}_4$ mapping $f_1, f_2, f_3 \mathfrak{h}_3(\mathbb{O})$ to $e_1, e_2, e_3 \in \mathfrak{h}_3(\mathbb{C})$ ⊂ $\mathfrak{h}_3(\mathbb{O})$.

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

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

Hope 1 seems to be known. The idempotents $e_1, e_2, e_3$ form a so-called ‘Jordan frame’ for $\mathfrak{h}_3(\mathbb{O})$, and so do $f_1, f_2, f_3$. Faraut and Korányi say that “in the irreducible case, the group $K$ 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 $\mathrm{F}_4$ that fixes $e_1, e_2, e_3$ contains $\text{Spin}(8)$. I bet it’s exactly $\text{Spin}(8)$. But to prove Hope 2 it may be enough to use $\text{Spin}(8)$/

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

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

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

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

$$\left( \begin{array}{ccc} a_1 & z^\ast & y^\ast \\ z & a_2 & x \\ y & x^\ast & a_3 \end{array} \right)$$

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

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

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