The n-Category Café

$(SU(3) \times SU(3))/(diag(\mathbb{Z}/3))$ is not isomorphic to $SU(3) \times PSU(3)$, so your “I think it works like this” isn’t quite right. It is true that the standard embedding $SU(2) \subset SU(3)$ has centralizer a $U(1)$ living as $diag(\alpha,\alpha,\alpha^{-2})$, and the combination is an $(SU(2) \times U(1)) / (\mathbb{Z}/2) \subset SU(3)$.

There aren’t that many ways to form a group of shape $(SU(3) \times SU(2) \times U(1))/(\mathbb{Z}/6)$. You need to map $\mathbb{Z}/6 = \mathbb{Z}/3 \times \mathbb{Z}/2$ into the center of $SU(3) \times SU(2) \times U(1)$, which is a $\mathbb{Z}/3 \times \mathbb{Z}/2 \times U(1)$. Assuming you want your final answer not to be a product, your map must be nonzero on all three terms. There are two non-zero maps $\mathbb{Z}/6 \to \mathbb{Z}/3$, but they are related by an outer automorphism of $SU(3)$; there is one non-zero map $\mathbb{Z}/6 \to \mathbb{Z}/2$; and when I said “nonzero” I meant prime-by-prime, so I need the unique (up to outer automorphism) injection $\mathbb{Z}/6 \to U(1)$. So there really is no ambiguity in the expression “$(SU(3) \times SU(2) \times U(1))/(\mathbb{Z}/6)$”.

This discussion of $(SU(3) \times SU(3))/(diag(\mathbb{Z}/3))$ reminds me of a puzzle that I learned from David Treumann. The puzzle involves the three most exceptional groups: $G_2 = Aut(\mathbb{O})$, $F_4 = Aut(\mathfrak{h}_3(\mathbb{O}))$, and $E_8$. Among the ways that they are “most exceptional” is that they are the only groups whose simply-connected forms have no center.

There are many constructions of $\mathbb{O}$, and hence many constructions of $G_2$. In particular, it is easy to construct $G_2$ in such a way as to make manifest a maximal abelian subgroup $2^3 \subset G_2$. (By the group “$p^n$”, I always mean the elementary abelian group of that order: $(\mathbb{Z}/p)^n$.) See, Cartan subgroups are maximal abelian, and are the unique connected maximal abelian subgroups of compact Lie groups, but they are not the unique maximal abelian subgroups. A finite group is maximal abelian iff the corresponding grading of the Lie algebra has nothing in degree zero. Let me investigate this $2^3$ a bit more. Its “Weyl group”, i.e. its normalizer mod centralizer, is an $SL_3(\mathbb{F}_2)$ acting in the obvious way. Choose any $2 \subset 2^3$. Then its centralizer in $G_2$ has shape $(SU(2) \times SU(2)) / diag(2)$. (Note that this group still has center of order $2$.) If you take a complementary $2^2 \subset (SU(2) \times SU(2)) / diag(2)$, it does not lift to $SU(2) \times SU(2)$, but rather its preimage is a quaternion group $Q_8 = 2^{1+2}$. A $2^3 \subset G_2$ grades the Lie algebra, and so you get a system of subalgebras corresponding to a flag in $G_2$, which goes $\mathfrak{c} \subset \mathfrak{su}(2) \times \mathfrak{su}(2) \subset \mathfrak{g}_2$, where $\mathfrak{c}$ is the Cartan.

It turns out that there are similar pictures of $F_4$ and $E_8$. Namely, you’ve already mentioned that $F_4$ has a subgroup $(SU(3) \times SU(3)) / diag(3)$. Similarly, $E_8$ contains a $(SU(5) \times SU(5))/diag(5)$. (Warning: there are two central maps $5 \to SU(5)$ up to outer automorphism; you need to use both of them.) These subgroups can be found by centralziing regular elements of orders $3$ and $5$ respectively. There are interesting copies of the extraspecial groups $3^{1+2}$ and $5^{1+2}$ inside $SU(3)$ and $SU(5)$ respectively. If you take a certain “antidiagonal” copy inside the product, you find a copy of $3^2 \subset (SU(3) \times SU(3)) / diag(3)$ or $5^2 \subset (SU(5) \times SU(5))/diag(5)$, and the results give subgroups $3^3 \subset F_4$ and $5^3 \subset E_8$ that turn out to be maximal abelian. The Weyl groups are $SL_3(\mathbb{F}_3)$ and $SL_3(\mathbb{F}_5)$. The gradings on the Lie algebras give “flags” $\mathfrak{c} \subset \mathfrak{su}(3) \times \mathfrak{su}(3) \subset \mathfrak{f}_4$ and $\mathfrak{c} \subset \mathfrak{su}(5) \times \mathfrak{su}(5) \subset \mathfrak{e}_8$.

Greiss found a construction of $F_4$ that makes its maximal abelian $3^3$ manifest in A Moufang loop, the exceptional Jordan algebra, and a cubic form in 27 variables. The puzzle is to do the same for $E_8$.

I gather this is distinct from Furey’s ideas, which I also haven’t had time to evaluate?

I haven’t read the papers you linked. It is clear that they have found a copy of the Standard Model Gauge Group inside of $F_4$ as the centralizer of an element of order $6$. (Whence the flavor symmetry?)

If we were talking about 3d rather than 4d gauge theory, I would tell you the following story. A major ingredient in 3d gauge theory is the “level”, which for $G$ simply connected lives in $H^4(BG;\mathbb{Z})$. (When $G$ is not simply connected, the correct thing seems to be to use not ordinary cohomology, but rather generalized cohomology with coefficients in the Anderson dual to the sphere spectrum.) Any map of groups $G' \to G''$ induces a map $H^4(BG'') \to H^4(BG')$ of levels. Recall that if $G$ is simple simply-connected, then $H^4(BG) \cong \mathbb{Z}$; the generator is called “level 1”, and $c_2(adjoint rep)$ is twice the dual coxeter number (or maybe minus that).

I mentioned above an embedding $(SU(2) \times SU(2))/2 \subset G_2$. If I’m calculating correctly, the induced map $H^4(BG_2) \to H^4(BSU(2)) \times H^4(BSU(2))$ takes the generator to the pair $(2,6) \in \mathbb{Z}^2$. I haven’t done the exact $F_4$ computation, but it should’t be too hard. I do know that $H^4(BF_4) \to H^4(BSU(3))^2$ takes the generator to some multiple of $(1,2)$. In the $E_8$ case in my puzzle, the induced map $H^4(BE_8) \to H^4(BSU(5))^2$ takes the generator to $(1,1)$. The ratios $6/2$ and $2/1$ are the ratios of long to short roots in $G_2$ and $F_4$ respectively, and the ratio $1/1$ is the fact that $E_8$ is simply-laced.

If you had handed me $(SU(2) \times SU(2))/2$, I may or may not have handed you back $G_2$. (I mean, I probably would have, personally, because I’m stuck on my puzzle.) For instance, I might have instead said “this is $SO(4)$, what more do you want?”. But if you had handed it to me not just as a group, but as a group equipped with the level $(2,6)$, then I probably would have given you $G_2$ back, because the most natural levels for $SO(4)$ are the multiples of $(1,1)$. Specifically, you could hand me $(SU(2) \times SU(2))/2$ with the level $(2,6)$ and ask me to find the “simplest” overgroup $G \supset (SU(2) \times SU(2))/2$ such that $(2,6)$ is the restriction of a level of $G$. The unique answer to this is $G_2$, if I am not mistaken. Here’s how you do it: build $(SU(2) \times SU(2))/2$ gauge theory with level $(2,6)$, take its low-energy effective TFT, look for an irreducible chiral conformal boundary condition, and work out the symmetry group of the boundary condition.

Do “levels” show up also in the Standard Model? Something about the choice of matter fields, for example? In particular, is there a sense in which the levels of $(SU(3) \times SU(2) \times U(1))/6$ are some prescribed things? The map $SU(2) \times U(1) \to SU(3)$ takes the generator of $H^4(BSU(3))$ to $(1,3) \in H^4(BSU(2)) \times H^4(BU(1))$ if I am not mistaken. And the two $SU(3)$s in $F_4$, of course, have levels differing by a factor of $2$. So if you are handed $(SU(3) \times SU(2) \times U(1))/6$ with levels in the ratio $(1:2:6)$ or $(2:1:3)$, then you should be motivated to build $F_4$. But if you are handed some other levels, then you should reject $F_4$.

Urs Schreiber points out that the subgroup

$$\frac{\mathrm{SU}(3) \times \mathrm{SU}(3)}{\mathbb{Z}/3} \subset \mathrm{Aut}(\mathfrak{h}_3(\mathbb{O})) \; \;\colon\!\!= \mathrm{F}_4$$

is characterized in Section 2.12.2 of

• Ichiro Yokota, Exceptional Lie groups.

In Theorem 1.9.4, Yokota looks at a $3 \times 3$ diagonal matrix in $\mathrm{SU}(3)$ that’s a cube root of unity:

$$w = \mathrm{diag}(\omega,\omega,\omega) \in \mathrm{SU}(3) \subset \mathrm{Aut}(\mathbb{O}) \;\; \colon\!\! = \mathrm{G}_2$$

where

$$\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \in \mathbb{O},$$

$i$ being our chosen unit imaginary octonion. He shows that the subgroup of $\mathrm{Aut}(\mathbb{O}) = \mathrm{G}_2$ consisting of elements that commute with $w$ is the same as the subgroup that commute with multiplication by $i \in \mathbb{O}$, namely $\mathrm{SU}(3)$.

This should come as no surprise, since an element of $\mathbb{O}$ commutes with $i$ iff it commutes with $\omega$. But still, there’s a bit to check.

In Section 2.12.2 he treats $w$ as an element of $\mathrm{F}_4$ using the standard embedding

$$\mathrm{G}_2 \subset \mathrm{F}_4.$$

He shows that the subgroup of $\mathrm{F}_4$ consisting of elements that commute with $w$ is

$$\frac{\mathrm{SU}(3) \times \mathrm{SU}(3)}{\mathbb{Z}/3}.$$

So, there’s just a small argument required to get from here to the claim I called 1) in my post: one has to check that the elements of $\mathrm{F}_4$ commuting with $w$ is the same as the subgroup that preserves the splitting

$$\mathfrak{h}_3(\mathbb{O}) = \mathfrak{h}_3(\mathbb{C}) \oplus \mathfrak{h}_3(V)$$

arising from the splitting $\mathbb{O} = \mathbb{C} \oplus \mathrm{V}$$. But this should be fairly easy, since$\mathbb{C} \subseteq \mathbb{O}$is precisely the subalgebra of elements that commute with$\omega$.

Taking the coincident brane perspective, where $n$ branes acquire enhanced $U(n)$ symmetry on the worldvolume, one expects 3+2+1=6 branes behind the SM gauge symmetry. (Picture three worldvolumes, with three, two and one coincident branes, respectively.) From noncommutative geometry, there are three algebras, say $A_1\oplus A_2\oplus A_3$ of “noncommutative functions” over the 3+2+1=6 points. The $\mathbb{Z}_6$ symmetry geometrically implies one can freely permute the branes, between the three worldvolumes.

As the points are usually assigned to pure states, it is sensible to assign primitive idempotents of (formally real) Jordan algebras to them.