### A Group Theory Problem

#### Posted by John Baez

Preparing a talk on octonions and the Standard Model, I’m struggling with a calculation in this paper:

- Michel Dubois-Violette and Ivan Todorov, Exceptional quantum geometry and particle physics II.

and I’d like your help. The essence of the problem is nothing about octonions, it’s about Lie groups — and pretty simple Lie groups too, like $\mathrm{SU}(2)$ and $\mathrm{SU}(3)$. So, there’s a good chance you can help me out. I’ll explain it.

The essence of the problem is this. The **true gauge group** of the Standard Model is a Lie group

$G_{SM} = \frac{\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)}{\mathbb{Z}_6}$

but this notation, with no further explanation, is ambiguous — there are exactly 12 normal subgroups of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ that are isomorphic to $\mathbb{Z}_6$, and we need to say which one we’re talking about!

In Section 4 of their paper, using octonions, Dubois-Violette and Todorov get their hands on a Lie group of the form

$\frac{\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)}{\mathbb{Z}_6}$

They argue that it’s isomorphic to $\mathrm{SO}(3) \times \mathrm{U}(3)$, and they also seem to claim (in equation 4.1) that it’s isomorphic to $G_{SM}$. But they don’t give an argument that it’s isomorphic to $G_{SM}$.

**Problem.** Is $\mathrm{SO}(3) \times \mathrm{U}(3) \cong G_{SM}$?

Right now I think not, but I don’t have a proof yet.

So let me tell you what $G_{SM}$ is, more precisely. By definition

$G_{SM} = \frac{\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)}{N}$

where $N$ is this normal subgroup isomorphic to $\mathbb{Z}_6$:

$N = \{ (\zeta^n, (-1)^n, \omega^n ) : \; n \in \mathbb{Z} \} \subset \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$

where

$\zeta = e^{2 \pi i / 6} , \quad \omega = e^{2 \pi i / 3}$

Here and in what follows I’ll freely turning numbers into matrices by multiplying them by an identity matrix.

It may help you a bit to note that

$\mathrm{SO}(3) \times \mathrm{U}(3) \cong \frac{\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)}{M}$

where $M$ is a *different* subgroup isomorphic to $\mathbb{Z}_6$:

$M = \{ (\omega^n, (-1)^m, \omega^n) : \; n, m \in \mathbb{Z} \} \subset \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$

Why is this true? Well, there’s an onto homomorphism

$\begin{array}{ccc} \mathrm{U}(1) \times \mathrm{SU}(3) &\to& \mathrm{U}(3) \\ (\alpha, h) &\mapsto & \alpha^{-1} h \end{array}$

whose kernel is

$\{ (\omega^n , \omega^n ) \; : n \in \mathbb{Z} \} \subset \mathrm{U}(1) \times \mathrm{SU}(3)$

We can take the product of this with the identity on $\mathrm{SU}(2)$ and get an onto homomorphism

$\begin{array}{ccc} f \colon \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) &\to& \mathrm{SU}(2) \times \mathrm{U}(3) \\ (\alpha, g, h) &\mapsto & (g, \alpha^{-1} h) \end{array}$

whose kernel is

$\{ (\omega^n , 1, \omega^n ) \; : n \in \mathbb{Z} \} \subset \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$

We can then compose $f$ with this map built from the double cover $p \colon \mathrm{SU}(2) \to \mathrm{SO}(3)$:

$\begin{array}{ccc} g \colon \mathrm{SU}(2) \times \mathrm{U}(3) &\to& \mathrm{SO}(3) \times \mathrm{U}(3) \\ (g, h) &\mapsto & (p(g), h) \end{array}$

The composite

$\begin{array}{ccc} g \circ f \colon \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) &\to& \mathrm{SO}(3) \times \mathrm{U}(3) \\ (\alpha, g, h) &\mapsto & (p(g), \alpha^{-1} h) \end{array}$

has kernel

$M = \{ (\omega^n, (-1)^m, \omega^n) : \; n, m \in \mathbb{Z} \} \subset \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$

So, we get

$\mathrm{SO}(3) \times \mathrm{U}(3) \cong \frac{\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)}{M}$

Of course, the mere fact that $N \ne M$ does not imply

$\frac{\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)}{N} \ncong \frac{\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)}{M}$

## Re: A Group Theory Problem

I have two thoughts, one merely cosmetic. Put $G = \operatorname{U}(1) \times \operatorname{SU}(2) \times \operatorname{SU}(3)$.

By applying conjugation to the third factor in $G$, I may replace your $M = \langle(\omega, -1, \omega)\rangle$ by $\langle(\omega, -1, \omega^{-1})\rangle$. The advantage of this is that $N' = \langle(\omega, 1, \omega^{-1})\rangle$ is now an order-$2$ subgroup of both $M$ and $N$, so we may work in $G/N'$ and ask whether its quotients by $\langle(\zeta, -1, \omega)N'\rangle = \langle(\zeta\omega, 1, 1)N'\rangle = \langle(-1, 1, 1)N'\rangle$ and $\langle(1, -1, 1)N'\rangle$ are isomorphic. Whether or not this is easier I don’t know, but it looks prettier to me.

Since these are compact, connected Lie groups, to show that they are isomorphic, it suffices to show that they have the same Lie algebra and the same fundamental group. Since they are quotients of the same group by a discrete, normal (necessarily central) subgroup, they have the same Lie algebra, so it may be most tractable to try to analyse their fundamental groups instead of the entire Lie group.