## Can We Understand the Standard Model?

I’m giving a talk in Latham Boyle and Kirill Krasnov’s Perimeter Institute workshop Octonions and the Standard Model on Monday April 5th at noon Eastern Time.

This talk will be a review of some facts about the Standard Model. Later I’ll give one that says more about the octonions.

Can we understand the Standard Model?

Abstract. 40 years trying to go beyond the Standard Model hasn’t yet led to any clear success. As an alternative, we could try to understand why the Standard Model is the way it is. In this talk we review some lessons from grand unified theories and also from recent work using the octonions. The gauge group of the Standard Model and its representation on one generation of fermions arises naturally from a process that involves splitting 10d Euclidean space into 4+6 dimensions, but also from a process that involves splitting 10d Minkowski spacetime into 4d Minkowski space and 6 spacelike dimensions. We explain both these approaches, and how to reconcile them.

You can see the slides here, and later a video of my talk will appear. You can register to attend the talk at the workshop’s website.

Here’s a puzzle, just for fun. As I’ll recall in my talk, there’s a normal subgroup of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ that acts trivially on all known particles, and this fact is very important. The ‘true’ gauge group of the Standard Model is the quotient of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ by this normal subgroup.

This normal subgroup is isomorphic to $\mathbb{Z}_6$ and it consists of all the elements

$(\zeta^n, (-1)^n, \omega^n ) \in \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$

where

$\zeta = e^{2 \pi i / 6}$

is my favorite primitive 6th root of unity, $-1$ is my favorite primitive square root of unity, and

$\omega = e^{2 \pi i / 3}$

is my favorite primitive cube root of unity. (I’m a primitive kind of guy, in touch with my roots.)

Here I’m turning the numbers $(-1)^n$ into elements of $\mathrm{SU}(2)$ by multiplying them by the $2 \times 2$ identity matrix, and turning the numbers $\omega^n$ into elements of $\mathrm{SU}(3)$ by multiplying them by the $3 \times 3$ identity matrix.

But in fact there are a bunch of normal subgroups of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ isomorphic to $\mathbb{Z}_6.$ By my count there are 12 of them! So you have to be careful that you’ve got the right one, when you’re playing with some math and trying to make it match the Standard Model.

Puzzle 1. Are there really exactly 12 normal subgroups of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ that are isomorphic to $\mathbb{Z}_6$?

Puzzle 2. Which ones give quotients isomorphic to the true gauge group of the Standard Model, which is $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ modulo the group of elements $(\zeta^n, (-1)^n, \omega^n)$?

To help you out, it helps to know that every normal subgroup of $\mathrm{SU}(2)$ is a subgroup of its center, which consists of the matrices $\pm 1.$ Similarly, every normal subgroup of $\mathrm{SU}(3)$ is a subgroup of its center, which consists of the matrices $1, \omega$ and $\omega^2.$ So, the center of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ is $\mathrm{U}(1) \times \mathbb{Z}_2 \times \mathbb{Z}_3.$

Here, I believe, are the 12 normal subgroups of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ isomorphic to $\mathbb{Z}_6.$ I could easily have missed some, or gotten something else wrong!

1. The group consisting of all elements $(1, (-1)^n, \omega^n).$
2. The group consisting of all elements $((-1)^n, 1, \omega^n).$
3. The group consisting of all elements $((-1)^n, (-1)^n, \omega^n).$
4. The group consisting of all elements $(\omega^n, (-1)^n, 1).$
5. The group consisting of all elements $(\omega^n, (-1)^n, \omega^n).$
6. The group consisting of all elements $(\omega^n, (-1)^n, \omega^{-n}).$
7. The group consisting of all elements $(\zeta^n , 1, 1).$
8. The group consisting of all elements $(\zeta^n , (-1)^n, 1).$
9. The group consisting of all elements $(\zeta^n , 1, \omega^n).$
10. The group consisting of all elements $(\zeta^n , 1, \omega^{-n}).$
11. The group consisting of all elements $(\zeta^n , (-1)^n, \omega^n).$
12. The group consisting of all elements $(\zeta^n , (-1)^n, \omega^{-n}).$

### 17 Responses to Can We Understand the Standard Model?

2. mitchellporter says:

Pardon me for expressing a thought on octonions in advance of the talk… But if I think skeptically rather than optimistically about the idea of obtaining the standard model from octonions, I would ask whether the properties that distinguish the octonions – especially nonassociativity – are still there, playing a role in the physics, when the derivation is complete. If not, wouldn’t that suggest that the role of octonions per se is somewhat formal, and that such a derivation could proceed without them, starting with a simpler structure?

• John Baez says:

The octonions are perfectly simple, they’re just nonassociative. Anyway:

1) This talk won’t say much about the octonions; the next one will, so it might make sense to delay this discussion until then.

2) Nobody knows a “derivation” of the Standard Model from the octonions yet, in any good sense of the term, so it’s premature to worry about whether what happens when the derivation is complete. What we know so far are some intriguing connections between the octonions and the Standard Model gauge group and its representation on fermions.

3) It’s unlikely that the ideas involving octonions will succeed, but at this moment it’s unlikely that any one attempt to make progress in figuring out the fundamental laws of physics will succeed. It’s an incredibly difficult business.

3. Hendrik Boom says:

In your equation defining your favorite primitive cube root of unity, the (I presume) omega is missing, at least in my browser, so the equation starts with the equal sign. The other roots appear to be defined properly.

4. Gabriel Verret says:

It can’t be true that all normal subgroups are contained in the center. For example, there is the whole group, but also $\mathrm{U}(1)\times \mathrm{SU}(2)$, etc..

Perhaps it’s true for all finite normal subgroups?

Here’s a quick way to see the answer to Puzzle 1 is yes. Our group of order $6$ must be contained in $\mathrm{U}(1)\times \mathbb{Z}_2\times \mathbb{Z}_3$. (Normality will be automatic since this is the center.) It must be contained in the unique subgroup of order $6$ of $U(1)$, so we might as well assume we are working in $\mathbb{Z}_6\times \mathbb{Z}_2\times \mathbb{Z}_3\cong\mathbb{Z}_6\times \mathbb{Z}_6$.

So we are just asking for the number of subgroups of order $6$ in $\mathbb{Z}_6\times \mathbb{Z}_6$. Such a group must be cyclic, so is uniquely determined by a generator. Each group has two different generators, so it is the number of elements of order $6$ in $\mathbb{Z}_6\times \mathbb{Z}_6$ divided by $2$.

An element in $\mathbb{Z}_6\times \mathbb{Z}_6$ has order $6$ if one of its coordinate does, or if the coordinates have order $2$ and $3$. There are $(36-4\cdot 4)=20$ of the first type, and $2\cdot2=4$ of the second type, for $24$ elements of order $6$ in total (there are other ways to do this count), and $12$ groups of order $6$.

• John Baez says:

Gabriel:

It can’t be true that all normal subgroups are contained in the center. For example, there is the whole group, but also $\mathrm{U}(1)\times \mathrm{SU}(2)$, etc..

Whoops! All normal subgroups of a so-called ‘simple’ Lie group like $\mathrm{SU}(n)$ are contained in the center, but this is not true for $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ or even $\mathrm{SU}(2) \times \mathrm{SU}(3)$. (Simple Lie groups are defined not to be Lie groups that are simple as groups, but Lie groups that are connected and have simple Lie algebras. Equivalently, a Lie group is simple if it’s connected and every normal subgroup is either discrete or the whole group.)

I feel that every finite normal subgroup of $\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)$ is a subgroup of the center, but I’m not going to try to prove this right now.

I like your solution to Puzzle 1 very much!

• Gabriel Verret says:

Ah, right. Well if it’s true for the factors, it must be true for the whole group: a finite normal subgroup has to project to a finite normal subgroup in each component and so, by your argument, be contained in the center in each component.

• John Baez says:

Nice! That does it, then.

5. allenknutson says:

Back when I was a symplectic geometer I observed that if you take a coadjoint orbit of compact G_2 (so, 12-dim) and reduce by SU(3) at a generic value (multiply by a coadjoint orbit 6-dim; then subtract twice dim SU(3)) the result is a CP^1, which could be given an action of U(2). In some sense this should be about taking a G_2-rep and extending its SU(3)-action to an (SU(3) x U(2))-action, though I don’t see the details there. When I tried to ask Sternberg about it he asked if I could compute the mass of the neutrino, so that went nowhere fast. I don’t suppose the relevant SU(3) reps have any hope of combining to a G_2 rep?

6. John Baez says:

Allen wrote:

When I tried to ask Sternberg about it he asked if I could compute the mass of the neutrino [….]

I don’t suppose the relevant SU(3) reps have any hope of combining to a G_2 rep?

If you can’t even compute the mass of the neutrino I’m not sure I want to talk to you. But anyway, the answer to your question is yes.

I think this is how interest in octonions and the Standard Model first started. In 1973, Murat Günaydin and Feza Gürsey published a paper “Quark structure and octonions” where they noticed that SU(3) is the subgroup of G2 fixing a unit imaginary octonion (this was surely known earlier), and for each quark there is a lepton such that together their representations of SU(3) combine to give the resulting representation of SU(3) on the complexified octonions. So, this rep extends to a rep of G2.

• allenknutson says:

That’s most of what I want; I also want the $U(2)$ action on the $SU(3)$ pieces to be derivable from the overall $G_2$ action. (Obviously $SU(3)\times U(2)$ is nothing like a subgroup of $G_2$ — a rank 4 group doesn’t sit inside a rank 2 — so the derivation would have to be in a subtler sense I haven’t quite hit on.)

7. Keith Harbaugh says:

The following seemed interesting. I wonder if you might have time to make a quick comment on it. (Or maybe you already did?)

https://www.quantamagazine.org/a-new-map-of-the-standard-model-of-particle-physics-20201022/

{Dated, as indicated, 2020-10-22.)

• John Baez says:

It’s nice—much better than the usual pictorial depictions of the Standard Model. But for me, at least, the animated picture is less useful than the discussion in words that explains it. For example, if you don’t know that the W can change left-handed quarks of one flavor to left-handed quarks of another flavor, but not the same for right-handed quarks, it’s not clear that playing with the animation will teach you that! But maybe other people learn differently than I do.

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