## March 18, 2021

### 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:

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.

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

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

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

Posted at March 18, 2021 11:52 PM UTC

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### Re: A Group Theory Problem

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

1. 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.

2. 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.

Posted by: L Spice on March 19, 2021 3:52 AM | Permalink | Reply to this

### Re: A Group Theory Problem

In fact, doesn’t that do it? In the notation above, we have the isomorphism $G/N' \cong \operatorname{SU}(2) \times \operatorname{U}(3)$ given by $(z, g_2, g_3) \mapsto (g_2, z g_3)$. When we take the quotient by $(-1, 1) \in \operatorname{SU}(2) \times \operatorname{U}(3)$, we get $\operatorname{SO}(3) \times \operatorname{U}(3)$, as you say, with fundamental group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$. On the other hand, when we take the quotient by $(1, -1) \in \operatorname{SU}(2) \times \operatorname{U}(3)$, we get a group with the same fundamental group as $\operatorname{U}(3)/\langle{-1}\rangle$.

Now the fundamental group of a compact Lie group is the cocharacter lattice $P^\vee$ of a maximal torus modulo the coroot lattice $Q^\vee$. If we identify the cocharacter lattice of $\operatorname{U}(3)$ with $\mathbb{Z}^3$ in the obvious way, then $Q^\vee$ consists of those $(a, b, c) \in \mathbb{Z}^3$ such that $a + b + c = 0$, and $P^\vee$ is $\mathbb{Z}^3 + \frac1 2(1, 1, -1)\mathbb{Z}$ (where $\frac1 2(1, 1, -1)$ is $t \mapsto \operatorname{diag}(\sqrt t, \sqrt t, \sqrt t^{-1})$). Then $\pi_1(G/\langle(1, -1)N'\rangle) = P^\vee/Q^\vee$ is infinite cyclic, with generator $\frac1 2(1, 1, -1)$. In particular, it’s not $\pi_1(G/(-1, 1)N'\rangle)$.

Posted by: L Spice on March 19, 2021 5:22 AM | Permalink | Reply to this

### Re: A Group Theory Problem

Hurrah! Your answer agrees nicely with Marc Nardmann’s below. He also showed the $\pi_1$’s don’t match, but he took a very different approach. They’re both illuminating.

Posted by: John Baez on March 19, 2021 6:13 AM | Permalink | Reply to this

### Re: A Group Theory Problem

Thanks again!

1. Yes, that sounds right. At first when I read “by applying conjugation” I thought you meant conjugation by a group element, but now I think you mean complex conjugation. I’m not sure there’s an inner automorphism of $SU(3)$ mapping $\omega$ to $\omega^{-1}$ — in fact I think there’s not — but I know there’s an automorphism mapping the first to the second, namely complex conjugation applied to each matrix entry, and that’s good enough.

2. I agree that working with fundamental groups may help. But I don’t think that compact connected Lie groups with isomorphic Lie algebras and isomorphic fundamental groups are necessarily isomorphic. For example, $SU(2) \times SU(4)$ is simply connected, and $(-1,1)$ and $(1,-1)$ are two central elements of order 2 in this group. So

$\frac{ SU(2) \times SU(4)}{\langle (-1,1) \rangle} \cong SO(3) \times SU(4)$

and

$\frac{ SU(2) \times SU(4)}{\langle (1,-1) \rangle} \cong SU(2) \times \frac{SU(4)}{\{ \pm 1 \}}$

are two compact connected Lie groups with the same Lie algebra, both with fundamental group $\mathbb{Z}_2$. But they’re not isomorphic: you can check all the isomorphisms between their Lie algebras and show none integrates to give an isomorphism of Lie groups.

For a more touchy example, which confuses physicists sometime, consider

$\frac{ SU(2) \times SU(2)}{\langle (-1,-1) \rangle} \cong SO(4)$

and

$\frac{ SU(2) \times SU(2)}{\langle (-1,1) \rangle} \cong SO(3) \times SU(2)$

One can show the second of these has an outer automorphism of order 2, while the first does not.

I think one true fact similar to what you said is that a homomorphism of connected Lie groups that induces an isomorphism of Lie algebras and also an isomorphism of fundamental groups must be an isomorphism of Lie groups.

Posted by: John Baez on March 19, 2021 5:50 AM | Permalink | Reply to this

### Re: A Group Theory Problem

This raises another problem: The compact connected Lie groups $G = S(U(2)\times U(3))$ and $\hat{G} = U(1)\times SU(2)\times SU(3)$ have isomorphic Lie algebras and isomorphic fundamental groups. Prove that they are not isomorphic.

Posted by: Marc Nardmann on March 19, 2021 7:32 AM | Permalink | Reply to this

### Re: A Group Theory Problem

This is a good puzzle, Marc! A systematic approach to this general kind of puzzle would be nice. To fully describe a compact Lie group we need its Lie algebra, its fundamental group, but also some extra data. For a compact semisimple Lie groups, I believe there’s a standard approach to describing this extra data. The groups you’re talking about are not semisimple, only reductive, but I think the approach generalizes: the big difference is that the fundamental group can be infinite.

L Spice’s answer to my original puzzle seems to use this generalization. I think maybe Frank Adams’ book Lectures on Lie Groups talks about this generalization. He develops the theory of weights, etc. for arbitrary compact Lie groups.

However, for your particular puzzle I feel there should be an answer physicists will enjoy more. The category of representations of $G$ should be inequivalent to the category of representations of $\hat{G}$ in some way that’s fairly easy to describe.

Indeed, I believe there should be ‘more’ isomorphism classes of 2-dimensional irreps of $\hat{G}$ than of $G$, in some precise sense. There’s a countable infinity of them in each case, so what I really mean is that there are representations of $\hat{G}$ that have properties that no representations of $G$ have. In physics, this is the reason switching from $\hat{G}$ to $G$ is so interesting: it places more restrictions on the allowed particles!

Posted by: John Baez on March 20, 2021 2:57 AM | Permalink | Reply to this

### Re: A Group Theory Problem

I’ve recently been thinking about modal fracture for higher groups (slides from my recent talk here), and it strikes me that there is a story about how to recover a group from its fundamental group, its Lie algebra, and some extra data coming from the modal fracture square.

Modal fracture gives us, among other things, that a Lie group $G$ is the pullback of its fundamental groupoid $\Pi_1 G$ and the classifier $\Lambda^1_{\text{cl}}(\mathfrak{g})$ for closed 1-forms valued in its Lie algebra, taken over the delooping $B\flat \tilde{G}$ of its universal cover considered as a discrete group. The two maps into the delooping of the discrete universal cover are given by $\Pi_1 G \to B\flat \tilde{G}$ delooping the inclusion of the fundamental group and $\Lambda^1_{\text{cl}}(\mathfrak{g}) \to B \flat \tilde{G}$ interpreting a closed 1-form as a flat connection on the trivial $\tilde{G}$-principal bundle. The third bit of data required is therefore an identification between these two maps as coming from $G$, namely the identification of the flat connection on the trivial $\tilde{G}$-principal bundle on $G$ one gets from the Mauer-Cartan form with the one that comes from differentiating the monodromy action of the fundamental group (considered as a parallel transport).

This doesn’t, strictly speaking, recover the actual group structure, but it is enough to tell us if a homomorphism is an isomorphism.

Posted by: David Jaz Myers on March 22, 2021 8:59 PM | Permalink | Reply to this

### Re: A Group Theory Problem

I wrote:

Indeed, I believe there should be ‘more’ isomorphism classes of 2-dimensional irreps of $\hat{G}$ than of $G$, in some precise sense. There’s a countable infinity of them in each case, so what I really mean is that there are representations of $\hat{G}$ that have properties that no representations of $G$ have. In physics, this is the reason switching from $\hat{G}$ to $G$ is so interesting: it places more restrictions on the allowed particles!

Okay, this turns out to be true in a very simple way. The center of

$\hat{G} = \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$

is

$\mathrm{U}(1) \times \mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathrm{U}(1) \times \mathbb{Z}_6$

By Schur’s lemma every unitary irrep of $\hat{G}$ assigns a phase to each element of the center of $\hat{G}$, giving rise to a character

$\chi : \mathbb{Z}_6 \to \mathrm{U}(1)$

Since $G$ is the quotient of $\hat{G}$ by this central $\mathbb{Z}_6$, the irreps of $G$ correspond precisely to the irreps of $\hat{G}$ for which this character $\chi$ is trivial.

We can make this sound a bit more fun as follows.

Since characters of $\mathbb{Z}_6$ correspond to elements of the Pontryagin dual $\mathbb{Z}_6$, the symmetric monoidal category $Rep(\hat{G})$ of unitary reps of $\hat{G}$ is $\mathbb{Z}_6$-graded. That is, every rep is a direct sum of irreps, each of which has a well-defined ‘grade’ in $\mathbb{Z}_6$, given by the character $\chi$. The grade of a tensor product of two reps is the sum of their grades.

The category $Rep(G)$ is the grade-zero part of $Rep(\hat{G})$.

Posted by: John Baez on March 23, 2021 6:44 AM | Permalink | Reply to this

### Re: A Group Theory Problem

While we’re discussing categories of representations, I have a related question: Is the restriction functor

(1)$Res: Rep(U(2)\times U(3)) \to Rep(S(U(2)\times U(3)))$

invertible?

Posted by: Marc Nardmann on March 23, 2021 7:58 AM | Permalink | Reply to this

### Re: A Group Theory Problem

Forget the question. Everything about the relation between $\Rep(U(2)\times U(3))$ and $\Rep(S(U(2)\times U(3)))$ follows already from my point B.) here. In particular, yes, every rep of $S(U(2)\times U(3))$ on some vector space $V$ extends to a rep of $U(2)\times U(3)$ on $V$; but, no, not in a unique way.

Posted by: Marc Nardmann on March 23, 2021 11:28 AM | Permalink | Reply to this

### Rep(Ĝ)

Which is why, from low-energy physics alone, you can’t conclude that the gauge group is $G$ rather than $\hat{G}$. The simplest modification of the Standard Model is to introduce massive vector-like quarks in a rep $R$ with nonzero grade in $R(\hat{G})$. (“Massive” means $R$ is real: $R\simeq \overline{R}$.)

In the low-energy theory (where the massive quarks are integrated out), the avatar of their presence is a $\mathbb{Z}_6$ (or $\mathbb{Z}_3$, depending on the choice of $R$ above) 1-form global symmetry.

Posted by: Jacques Distler on March 23, 2021 1:44 PM | Permalink | PGP Sig | Reply to this

### Re: A Group Theory Problem

This restriction functor has a bunch of nice properties: for example, it’s a symmetric monoidal $\mathbb{C}$-linear functor between symmetric monoidal $\mathbb{C}$-linear categories. If we tack on enough adjectives like this, we can reconstruct a compact group from its category of representations, and also reconstruct a group homomorphism $f \colon G \to H$ from the functor $f^\ast : Rep(H) \to Rep(G)$ that it gives rise to. One avenue to making this precise is the Doplicher–Roberts theorem.

So, if we have an inclusion of groups

$i: G \hookrightarrow H$

and we demand that the restriction functor

$i^\ast : Rep(H) \to Rep(G)$

have an inverse (up to isomorphism) with enough nice properties, that forces $i$ to be an isomorphism, and thus $G = H$.

But if we don’t demand a bunch of nice properties, the subject gets a lot more tricky!

For example, there are two 64-element groups that are not isomorphic, whose categories of representations are equivalent as monoidal $\mathbb{C}$-linear categories:

$i: S(\mathrm{U}(2) \times \mathrm{U}(3)) \hookrightarrow \mathrm{U}(2) \times \mathrm{U}(3)$

it’s pretty easy to show that the monoidal functor

$i^\ast : Rep(\mathrm{U}(2) \times \mathrm{U}(3)) \to Rep(S(\mathrm{U}(2) \times \mathrm{U}(3)))$

doesn’t have an inverse (up to isomorphism) that’s a monoidal functor.

The reason is that $Rep(\mathrm{U}(2) \times \mathrm{U}(3))$ has a bunch of objects that are invertible with respect to $\otimes$: these are the 1-dimensional representation. You can see that the group of isomorphism classes of these invertible objects is $\mathbb{Z} \times \mathbb{Z}$. On the other hand, the group of isomorphism classes of invertible objects in $Rep(S(\mathrm{U}(2) \times \mathrm{U}(3)))$ is just $\mathbb{Z}$.

I could easily spend the whole thinking about this stuff, but I can’t afford to — not today.

Posted by: John Baez on March 23, 2021 8:10 PM | Permalink | Reply to this

### Re: A Group Theory Problem

As I think John is alluding to, one way to answer puzzles like this is to use the full root datum.

In this case, $G_1 = \operatorname{S}(\operatorname{U}(2) \times \operatorname{U}(3))$ has character lattice $X_1 = (\mathbb{Z}e_1 \oplus \mathbb{Z}e_2 \oplus \mathbb{Z}f_1 \oplus \mathbb{Z}f_2 \oplus \mathbb{Z}f_3)/\mathbb{Z}(e_1 + e_2 + f_1 + f_2 + f_3)$, root system $\Phi_1 = \{\pm(e_1 - e_2), \pm(f_1 - f_2), \pm(f_2 - f_3), \pm(f_1 - f_3)\}$, cocharacter lattice $Y_1 = \operatorname{ker}(\mathbb{Z}e_1^\vee \oplus \mathbb{Z}e_2^\vee \oplus \mathbb{Z}f_1^\vee \oplus \mathbb{Z}f_2^\vee \oplus \mathbb{Z}f_3^\vee \to \mathbb{Z})$, and coroot system $\Phi_1^\vee = \{\pm(e_1^\vee - e_2^\vee), \pm(f_1^\vee - f_2^\vee), \pm(f_2^\vee - f_3^\vee), \pm(f_1^\vee - f_3^\vee)\}$.

On the other hand, $G_2 = \operatorname{U}(1) \times \operatorname{SU}(2) \times \operatorname{SU}(3)$ has character lattice $X_2 = \mathbb{Z}e_1 \oplus (\mathbb{Z}f_1 \oplus \mathbb{Z}f_2)/\mathbb{Z}(f_1 + f_2) \oplus (\mathbb{Z}g_1 \oplus \mathbb{Z}g_2 \oplus \mathbb{Z}g_3)/\mathbb{Z}(g_1 + g_2 + g_3)$, root system $\Phi_2 = \{\pm(f_1 - f_2), \pm(g_1 - g_2), \pm(g_2 - g_3), \pm(g_1 - g_3)\}$, cocharacter lattice $Y_2 = \mathbb{Z}e_1^\vee \oplus \operatorname{ker}(\mathbb{Z}f_1^\vee \oplus \mathbb{Z}f_2^\vee \to \mathbb{Z}) \oplus \operatorname{ker}(\mathbb{Z}g_1^\vee \oplus \mathbb{Z}g_2^\vee \oplus \mathbb{Z}g_3^\vee \to \mathbb{Z})$, and coroot system $\Phi_2^\vee = \{\pm(f_1^\vee - f_2^\vee), \pm(g_1^\vee - g_2^\vee), \pm(g_2^\vee - g_3^\vee), \pm(g_1^\vee - g_3^\vee)\}$.

Any isomorphism $G_1 \to G_2$ must induce a Weyl-invariant isomorphism $X_1 \to X_2$. However, the Weyl-invariant lattices $X_1^{W_1} = \{0\}$ and $X_2^{W_2} = \mathbb{Z}e_1$ cannot be identified.

Posted by: L Spice on March 20, 2021 2:21 PM | Permalink | Reply to this

### Re: A Group Theory Problem

Excuse me, $X_1^{W_1}$ is $(\mathbb{Z}(e_1 + e_2) \oplus \mathbb{Z}(f_1 + f_2 + f_3))/\mathbb{Z}(e_1 + e_2 + f_1 + f_2 + f_3)$, which is isomorphic to $\mathbb{Z}e_1$. We have to look closer at the putative isomorphism $X_1 \to X_2$ to see a contradiction. I’ll try to fix up my mistake later, but for now I have to teach a class!

Posted by: L Spice on March 20, 2021 2:54 PM | Permalink | Reply to this

### Re: A Group Theory Problem

Sorry about that; let’s have another go. (I was going to wait until after class, but these kinds of questions are catnip to me.)

Our isomorphism $X_1 \to X_2$ must carry $\Phi_1$ onto $\Phi_2$, and so preserve irreducible components of the root systems. As such, it carries $e_1 - e_2 \in X_1$ to $\pm(f_1 - f_2) \in X_2$. Upon post-composing with a Weyl conjugation in the $\operatorname{SU}(2)$ factor if necessary, we may, and do, assume that $e_1 - e_2 \in X_1$ is sent to $f_1 - f_2 \in X_2$.

Now $e_1 + e_2$, which generates the rank-$1$ lattice $X_1^{W_1}$, must be carried to one of the two generators $\pm e_1$ of the rank-$1$ lattice $X_2^{W_2} = \mathbb{Z}e_1$. Upon post-composing with inversion in the $\operatorname{U}(1)$ factor if necessary, we may, and do, assume that $e_1 + e_2 \in X_1$ is sent to $e_1 \in X_2$.

Now the problem is that there’s nowhere for $e_1 \in X_1$ to go: the image of $2e_1 = (e_1 - e_2) + (e_1 + e_2) \in X_1$ is $(f_1 - f_2) + e_1 \in X_2$, but that doesn’t lie in $2X_2$.

Posted by: L Spice on March 20, 2021 3:13 PM | Permalink | Reply to this

### Re: A Group Theory Problem

I think one true fact similar to what you said is that a homomorphism of connected Lie groups that induces an isomorphism of Lie algebras and also an isomorphism of fundamental groups must be an isomorphism of Lie groups.

Yes, maybe that’s what I was thinking. Or maybe—it’s true for simple Lie groups, isn’t it, for the simple (har) reason that it’s so difficult to cook up simple groups with isomorphic fundamental groups, and since we get lucky and the order-2 subgroups of $\operatorname{Z}(\operatorname{Spin}(2n))$ are all conjugate by outer automorphisms? Anyway, I agree it doesn’t work here, so it’s a lucky thing I wound up claiming a disproof rather than a proof of isomorphism.

At first when I read “by applying conjugation” I thought you meant conjugation by a group element, but now I think you mean complex conjugation.

That ambiguity somehow didn’t occur to me! Yes, I meant componentwise complex conjugation, though it occurred to me afterwards that it may have been easier to refer instead to inversion on the $\operatorname{U}(1)$ factor rather than componentwise complex conjugation on the $\operatorname{SU}(3)$ factor.

Posted by: L Spice on March 19, 2021 10:01 AM | Permalink | Reply to this

### Re: A Group Theory Problem

Or maybe—it’s true for simple Lie groups, isn’t it, for the simple (har) reason that it’s so difficult to cook up simple groups with isomorphic fundamental groups, and since we get lucky and the order-2 subgroups of $\operatorname{Z}(\operatorname{Spin}(2n))$ are all conjugate by outer automorphisms?

Let me quit while I’m behind—not only is this false, I explained why it was false while claiming it was true! The conjugacy I claimed only holds when $n = 2$ or $n$ is odd. For even $n \mathrel{>} 2$, we have non-conjugate order-2 central (as I forgot to say originally) subgroups that give rise to, respectively, $\operatorname{SO}(2n)$ and the semispin group.

(By the way, what’s the proper way to type ‘greater-than’ in math mode? The parser chokes on a raw ‘greater-than’ character, and doesn’t space an escaped entity correctly. I fell back on using mathrel manually, but surely that’s not the intended mode!)

Posted by: L Spice on March 19, 2021 2:02 PM | Permalink | Reply to this

### Re: A Problem With < and >

The problem with $\lt$ and $\gt$ is that these characters play an important role in HTML, and thin veneer of LaTeX here rests on a deeper layer of HTML. Luckily the solution is simple enough: use \lt and \gt in math mode.

Posted by: John Baez on March 19, 2021 5:56 PM | Permalink | Reply to this

### Re: A Group Theory Problem

The fundamental groups (and hence the groups themselves) are not isomorphic. The fundamental group of $SO(3)\times U(3)$ is of course $\mathbb{Z}_2\times\mathbb{Z}$. The fundamental group of $G=G_{SM}\cong S(U(2)\times U(3))$ is $\mathbb{Z}$. I’ll just show that $\pi_1(G)$ is torsion-free: We have the normal-subgroup inclusion $G\to E=U(2)\times U(3)$ and the map $E\to U(1)$ given by $(A,B)\mapsto det(A)det(B)$, which yields a Lie group isomorphism $E/G \cong U(1)$. In the homotopy sequence of the fibration $G\to E\to U(1)$, the group $\pi_2(U(1))$ is trivial, and $\pi_1(E)$ is $\mathbb{Z}\times\mathbb{Z}$. Hence the group morphism $\pi_1(G)\to\pi_1(E)$ is injective into a torsion-free group. Thus $\pi_1(G)$ is torsion-free.

Posted by: Marc Nardmann on March 19, 2021 5:28 AM | Permalink | Reply to this

### Re: A Group Theory Problem

Hurrah! Thanks!

Does the fact that $\pi_1(S(U(2) \times U(3))) \cong \mathbb{Z}$ show up somewhere in Standard Model physics? $\pi_3(G)$ is relevant to solitons in 3+1-dimensional field theory, and $\pi_4(G)$ is relevant to instantons in 3+1-dimensional field theory, but what about $\pi_2$ and $\pi_1$?

Posted by: John Baez on March 19, 2021 6:11 AM | Permalink | Reply to this

### Re: A Group Theory Problem

Well, $\pi_2$ of a Lie group is not very interesting. As for $\pi_1$, I don’t know. Probably it’s at least interesting over spacetimes with nontrivial topology?

Posted by: Marc Nardmann on March 19, 2021 6:27 AM | Permalink | Reply to this

### Nikolay

Aaronov-Bohm?

Posted by: Nikolay Yakovlev on March 19, 2021 6:43 AM | Permalink | Reply to this

### Strings

$\pi_1(G/H)$ labels the strings produced in the phase transition where $G$ is broken to $H$. In the case at hand, $H=U(3)$, with a particular embedding in $G$ that I’ll let John explain.

Since $\pi_1(H)=\mathbb{Z}$ and $\pi_1(G)=\mathbb{Z}$, $\pi_1(G/H)$ is finite. I believe, in fact, that it vanishes. But it’s been more than 30 years since I’ve given that question more than a passing thought.

Ah, to relive the 1980s …

Posted by: Jacques Distler on March 19, 2021 8:28 AM | Permalink | PGP Sig | Reply to this

### Re: A Group Theory Problem

Two remarks.

A.) The manifolds

• $U(1)\times SU(2)\times SU(3)$
• $S(U(2)\times U(3))$
• $U(2)\times SU(3)$
• $SU(2)\times U(3)$

are diffeomorphic. (This is a much stronger statement than $\pi_1(S(U(2)\times U(3))) \cong \mathbb{Z}$.)

PROOF: Let’s only prove that $G = S(U(2)\times U(3))$ and $\hat{G} = U(1)\times SU(2)\times SU(3)$ are diffeomorphic; the other diffeomorphisms are constructed analogously. The maps $f: G\to\hat{G}$ and $h: \hat{G}\to G$ given by

(1)\begin{aligned} f(A,B) &= \left(\det(A), \begin{pmatrix}\det(A)^{-1}&\\ &1\end{pmatrix}A, \begin{pmatrix}\det(A)&&\\ &1&\\ &&1\end{pmatrix}B\right) ,\\ h(z,C,D) &= \left(\begin{pmatrix}z&\\ &1\end{pmatrix}C, \begin{pmatrix}z^{-1}&&\\ &1&\\ &&1\end{pmatrix}D\right) \end{aligned}

are well-defined, smooth, and inverse to each other. QED.

B.) The Lie groups

• $U(1)\times S(U(2)\times U(3))$
• $U(2)\times U(3)$

are isomorphic. (This improves what I used before: that $U(2)\times U(3)$ is the total space of a fibre bundle over $U(1)$ with typical fibre $S(U(2)\times U(3))$.) More precisely, let

• $p: U(2)\times U(3)\to U(1)$ be the map $(A,B) \mapsto \det(A)\det(B)$,
• $\pi: U(1)\times S(U(2)\times U(3)) \to U(1)$ be the projection to the first component,
• $\iota\colon S(U(2)\times U(3)) \to U(2)\times U(3)$ be the inclusion,
• $i\colon S(U(2)\times U(3)) \to U(1)\times S(U(2)\times U(3))$ be the map $g\mapsto (1,g)$.

There is a Lie group isomorphism $F: U(2)\times U(3) \to U(1)\times S(U(2)\times U(3))$ such that $F\circ\iota = i$ and $\pi\circ F = p$.

PROOF: Define

(2)\begin{aligned} F &: U(2)\times U(3) \to U(1)\times S(U(2)\times U(3)) ,\\ H &: U(1)\times S(U(2)\times U(3)) \to U(2)\times U(3) \end{aligned}

by

(3)\begin{aligned} F(A,B) &= \Big(\det(A)\det(B), \big(\det(A)\det(B)\big)^{-2}A, \det(A)\det(B)B\Big) ,\\ H(z,C,D) &= \Big(z^2C, z^{-1}D\Big) . \end{aligned}

$F$ is $U(1)\times S(U(2)\times U(3))$-valued because

(4)$\det\big(\det(A)^{-2}\det(B)^{-2}A\big) \det\big(\det(A)\det(B)B\big) = \det(A)^{-4+1+3}\det(B)^{-4+3+1} = 1 .$

Hence $F,H$ are well-defined Lie group morphisms. Obviously $\pi\circ F = p$ and $F\circ\iota = i$.

(5)\begin{aligned} &F(H(z,C,D))\\ &= \Big(\det\big(z^2C\big) \det\big(z^{-1}D\big), \det\big(z^2C\big)^{-2} \det\big(z^{-1}D\big)^{-2} z^2C, \det\big(z^2C\big) \det\big(z^{-1}D\big) z^{-1}D\Big)\\ &= \Big( z^{4-3}\det(C)\det(D), z^{-8+6+2}\big(\det(C)\det(D)\big)^{-2}C, z^{4-3-1}\det(C)\det(D)D\Big)\\ &= (z,C,D) . \end{aligned}
(6)\begin{aligned} H(F(A,B)) &= \Big(\big(\det(A)\det(B)\big)^2 \big(\det(A)\det(B)\big)^{-2}A, \big(\det(A)\det(B)\big)^{-1} \det(A)\det(B)B\Big)\\ &= (A,B) . \end{aligned}

QED.

Posted by: Marc Nardmann on March 22, 2021 6:26 AM | Permalink | Reply to this

### Re: A Group Theory Problem

Marc wrote:

The manifolds

• $\mathrm{U}(1)\times SU(2)\times SU(3)$

• $\mathrm{S}(\mathrm{U}(2)\times \mathrm{U}(3))$

• $\mathrm{U}(2)\times SU(3)$

• $\mathrm{SU}(2) \times \mathrm{U}(3)$

are diffeomorphic.

Cool! I seem to recall that similar tricks can be used to show

$SO(3) \times SU(2) \cong (SU(2) \times SU(2))/\langle (-1,1) \rangle$

and

$SO(4) \cong (SU(2) \times SU(2))/\langle (-1,-1) \rangle$

are diffeomorphic, though not isomorphic. There’s a diffeomorphism sending $(\pm g,h)$ to $(\pm g, \pm g h)$, where $g,h \in SU(2)$.

And by the way — just a sanity check for me — what’s the center of $\mathrm{S}(\mathrm{U}(2) \times \mathrm{U}(3))$, and what’s the center of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$?

Posted by: John Baez on March 23, 2021 2:30 AM | Permalink | Reply to this

### Re: A Group Theory Problem

The centre of $U(1)\times SU(2)\times SU(3)$ is indeed isomorphic to $U(1)\times\mathbb{Z}_2\times\mathbb{Z}_3$, in an obvious way.

The centre of $S(U(2)\times U(3))$ is obviously isomorphic to $C = \big\{ (a,b)\in U(1)\times U(1) \big| a^2b^3=1 \big\}$, which is in turn isomorphic to $U(1)$: The group morphisms $f: U(1)\to C$ and $h: C\to U(1)$ given by

(1)\begin{aligned} f(z) &= (z^3,z^{-2}) \,,\\ h(a,b) &= a b \end{aligned}

are inverse to each other:

(2)\begin{aligned} f(h(a,b)) &= (a^3b^3,a^{-2}b^{-2}) = (a,b) ,\\ h(f(z)) &= z^3z^{-2} = z . \end{aligned}

(One can also draw a nice picture of $C$ in $\mathbb{R}^2/\mathbb{Z}^2$ to see that it’s connected.)

Posted by: Marc Nardmann on March 23, 2021 5:54 AM | Permalink | Reply to this
Read the post Cosmic Strings in the Standard Model
Weblog: Musings
Excerpt: Prompted by some posts by John Baez, a little calculation with an unsurprising result.
Tracked: March 21, 2021 6:28 AM

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