The copy of $(SU(3) \times SU(3)) / \mathbb{Z}_3 \subset F_4$ is one of my favourite groups. There are many reasons why it is fun.

A very basic reason is that it is the centralizer of an order-$3$ element in $F_4$. Up to conjugacy, I count three order-$3$ elements of $F_4$. Two of them have nonsemisimple centralizers, i.e. centralizers whose Lie algebras contain a $U(1)$ factor. This one has a semisimple centralizer.

A second reason is that the two $SU(3)$s are very different inside $F_4$, and in particular are not conjugate to each other. One way to see this is that the roots of the two copies have different lengths. (Said another way, the Killing form for $F_4$ restricts differently to the two $\mathfrak{su}(3)$s — off by a factor of $2$, I think.) This is presumably related to your comment about the strong force versus a different one.

A third reason is that $SU(3)$ contains a copy of the finite Heisenberg group $3^{1+2}_+$ (the unique noncommutative group of order $27$ in which all nontrivial elements have order $3$) as what a quantum computer programmer would call the “clock” and “shift” matrices, but if you take the diagonal copy $diag(3^{1+2}_+) \subset SU(3) \times SU(3)$ [or maybe the antidiagonal copy, depending on what coordinates you choose for the two $SU(3)$s], then its image in $(SU(3) \times SU(3)) / \mathbb{Z}_3$ is just a $\mathbb{Z}_3^2$. Together with the residual centre of $(SU(3) \times SU(3)) / \bZ_3$, you get a copy of $\mathbb{Z}_3^3 \subset F_4$ which turns out to be maximal abelian. In particular, it does not fit inside of any torus. Serre (I believe) explained that having a finite abelian $p$-subgroup of a Lie group which does not fit inside any torus exactly corresponds to having nontrivial $p$-local cohomology.

The normalizer of this $\mathbb{Z}_3^3 \subset F_4$ is a nonsplit extension $\mathbb{Z}_3^3 \cdot SL_3(\mathbb{F}_3)$. This explains why the number $26$ shows up so much in the representation theory of $F_4$: any representation must diagonalize over $\mathbb{Z}_3^3$, but the 26 nontrivial eigenspaces must have the same dimension because they are permuted by the $SL_3(\mathbb{F}_3)$. Going the other way, Griess explained how to build the exceptional Jordan algebra, and hence $F_4$, just from this group.

This $\mathbb{Z}_3^3 \subset F_4$ is part of a trilogy. There is are analogous subgroups $\mathbb{Z}_2^3 \cdot SL_3(\mathbb{F}_2) \subset G_2$ and $\mathbb{Z}_5^3 \cdot SL_3(\mathbb{F}_5) \subset E_8$. The former can be used to provide a construction of $\mathbb{O}$ directly from $\mathbb{Z}_2^3 \cdot SL_3(\mathbb{F}_2)$. It is an easier version of Griess’s construction of the exceptional Jordan algebra and hence of $G_2$.

There seems to be no such construction of $E_8$ starting with $\mathbb{Z}_5^3 \cdot SL_2(\mathbb{F}_5)$. The problem seems to be the following. The “Weyl group” $SL_3(\mathbb{F}_p)$ suggests that you look at the determinant form $\det$ on $\mathbb{Z}_p^3$. This is an alternating trilinear form, and so, after including $\mathbb{F}_p \subset U(1)$, you get a 3-cocycle for ordinary $U(1)$-valued cohomology of $\mathbb{Z}_p^3$. It gives you as a class in $H^4(B\mathbb{Z}_p^3; \mathbb{Z})$, which I believe that it is the restriction along $\mathbb{Z}_p^3 \subset G$ of the generator of $H^4(BG; \mathbb{Z}) \cong \mathbb{Z}$, where $G = G_2,F_4,E_8$ for $p=2,3,5$. Anyway, the problem is that for $p \geq 5$, this $U(1)$-valued cocycle is not a coboundary: the class is nontrivial. For $p \leq 3$, it is a coboundary. The reason is that in three dimensions $\det$ is a sum of $3! = 6$ terms, and these terms are themselves cocycles, all of which are cohomologous in $U(1)$-cohomology. So as a cohomology class $\det$ is “divisible by 6”. Indeed, $H^3(\mathbb{Z}_p^3; U(1))$ has a copy of $\mathbb{Z}_p$ inside of it (living as the “essential cohomology”, i.e. the kernel of all restrictions to proper subgroups), and this copy is generated by “$\frac16 \det$”. The reason this is a problem for constructing $E_8$ is because the constructions of $\mathbb{O}$ and the exceptional Jordan algebra both begin by choosing an antiderivative of this specific cocycle.

## Re: Can We Understand the Standard Model Using Octonions?

Typo in the second slide deck: On slide 15, “10” should be “10d” twice.