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March 17, 2021

Can We Understand The Standard Model?

Posted by John Baez

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 explain, there’s a normal subgroup of U(1)×SU(2)×SU(3)\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 U(1)×SU(2)×SU(3)\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) by this normal subgroup.

This normal subgroup is isomorphic to 6\mathbb{Z}_6 and it consists of all the elements

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

where

ζ=e 2πi/6 \zeta = e^{2 \pi i / 6}

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

ω=e 2πi/3\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(-1)^n into elements of SU(2)\mathrm{SU}(2) by multiplying them by the 2×22 \times 2 identity matrix, and turning the numbers ω n\omega^n into elements of SU(3)\mathrm{SU}(3) by multiplying them by the 3×33 \times 3 identity matrix.

But in fact there are a bunch of normal subgroups of U(1)×SU(2)×SU(3)\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) isomorphic to 6\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 U(1)×SU(2)×SU(3)\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) that are isomorphic to 6\mathbb{Z}_6?

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

To help you out, it helps to know that every normal subgroup of SU(2)\mathrm{SU}(2) is a subgroup of its center, which consists of the matrices ±1\pm 1. Similarly, every normal subgroup of SU(3)\mathrm{SU}(3) is a subgroup of its center, which consists of the matrices 1,ω1, \omega and ω 2\omega^2. So, the center of U(1)×SU(2)×SU(3)\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) is U(1)× 2× 3\mathrm{U}(1) \times \mathbb{Z}_2 \times \mathbb{Z}_3.

Here, I believe, are the 12 normal subgroups of U(1)×SU(2)×SU(3)\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) isomorphic to 6\mathbb{Z}_6. I could easily have missed some, or gotten something else wrong!

  1. The group consisting of all elements (1,(1) n,ω n)(1, (-1)^n, \omega^n).

  2. The group consisting of all elements ((1) n,1,ω n)((-1)^n, 1, \omega^n).

  3. The group consisting of all elements ((1) n,(1) n,ω n)((-1)^n, (-1)^n, \omega^n).

  4. The group consisting of all elements (ω n,(1) n,1)(\omega^n, (-1)^n, 1).

  5. The group consisting of all elements (ω n,(1) n,ω n)(\omega^n, (-1)^n, \omega^n).

  6. The group consisting of all elements (ω n,(1) n,ω n)(\omega^n, (-1)^n, \omega^{-n}).

  7. The group consisting of all elements (ζ n,1,1)(\zeta^n , 1, 1).

  8. The group consisting of all elements (ζ n,(1) n,1)(\zeta^n , (-1)^n, 1).

  9. The group consisting of all elements (ζ n,1,ω n)(\zeta^n , 1, \omega^n).

  10. The group consisting of all elements (ζ n,1,ω n)(\zeta^n , 1, \omega^{-n}).

  11. The group consisting of all elements (ζ n,(1) n,ω n)(\zeta^n , (-1)^n, \omega^n).

  12. The group consisting of all elements (ζ n,(1) n,ω n)(\zeta^n , (-1)^n, \omega^{-n}).

Posted at March 17, 2021 2:45 AM UTC

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18 Comments & 2 Trackbacks

Re: Can We Understand The Standard Model?

These are very interesting relations between the Standard Model group and the 10d Euclidean/Minkowski spacetimes, but beside the efort to understand the algebro-geometric origin of the SM group, it is also important to understand the representation of the SM group which gives the known elementary particles, as well as the different generations. Because it is not only the group/algebra which is important, but also a representation as well, one needs some category theory. For one such attempt, please see Standard Model and 4-groups, A. Mikovic and M. Vojinovic.

Posted by: Aleksandar Mikovic on March 17, 2021 8:39 AM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

I’ve read that paper! It’s interesting!

Posted by: John Baez on March 18, 2021 12:57 AM | Permalink | Reply to this

Spin(9) GUT?

Since S(U(2)×U(3))Spin(9)S(U(2) \times U(3)) \subset Spin(9), I presume someone has tried to build a GUT with Spin(9)Spin(9) gauge group, but I feel like I never saw it. Maybe I just missed it? Or does something fail? Something about the representation on fermions? That can’t be it, because I could just restrict the Spin(10)Spin(10)-representation…

Posted by: Theo Johnson-Freyd on March 17, 2021 5:04 PM | Permalink | Reply to this

Re: Spin(9) GUT?

I’ve never seen a Spin(9) GUT worked out. As you can tell from my talk, my interest in Spin(9) was piqued by this paper:

but he doesn’t try a SO(9) GUT.

There are some quite elaborate studies of the lattice of subgroups of Spin(10) containing S(U(2) × U(3)), and the gauge theories with all these gauge groups, worked out by physicists trying to understand symmetry breaking in the Spin(10) theory. Spin(9) should be hiding somewhere in this work. Here’s one paper:

It has some really nice diagrams showing a bit of the lattice of subgroups of Spin(10) containing S(U(2) × U(3)). But I don’t see Spin(9) — they would call it SO(9) — in here.

I want to learn a bit more about this stuff, though it’s kind of a rabbit hole. One important thing is that Spin(10) has SU(5) × U(1) or maybe (SU(5) × U(1))/ 5\mathbb{Z}_5 as a maximal subgroup. The generator of the U(1) here is called the X charge, and it’s a linear combination of hypercharge (called Y) and baryon number minus lepton number (called B-L).

Posted by: John Baez on March 17, 2021 6:58 PM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

On slide 19, “the true gauge of the Standard Model” should be “the true gauge group of the Standard Model”.

Posted by: Blake Stacey on March 17, 2021 8:25 PM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

Whoops — thanks! I fixed it.

I don’t have such strong opinions about gauge-fixing.

Posted by: John Baez on March 18, 2021 12:59 AM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

Gabriel Verret solved Puzzle 1 assuming that every normal subgroup of order 6 in U(1)×SU(2)×SU(3)\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) is actually contained in the center, U(1)× 2× 3\mathrm{U}(1)\times \mathbb{Z}_2\times \mathbb{Z}_3. I actually think this is true for every finite normal subgroup.

His argument then goes as follows:

Our group of order 66 must be contained in U(1)× 2× 3\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 66 of U(1)\mathrm{U}(1), so we might as well assume we are working in 6× 2× 3 6× 6\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 66 in 6× 6\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 66 in 6× 6\mathbb{Z}_6\times \mathbb{Z}_6 divided by 22.

An element in 6× 6\mathbb{Z}_6\times \mathbb{Z}_6 has order 66 if one of its coordinate does, or if the coordinates have order 22 and 33. There are (3644)=20(36-4\cdot 4)=20 of the first type, and 22=42\cdot2=4 of the second type, for 2424 elements of order 66 in total (there are other ways to do this count), and 1212 groups of order 66.

Posted by: John Baez on March 18, 2021 3:07 AM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

A finite normal subgroup NN of a connected group GG is indeed central. Indeed, take the orbit of a point of NN under the conjugation action of GG. Since GG is connected, the orbit is connected. Since NN is finite, it must be a singleton.

Posted by: L Spice on March 19, 2021 3:11 AM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

Nice! I’d heard this result for simple Lie groups, but didn’t realize it was more general. What a nice simple proof!

Posted by: John Baez on March 19, 2021 5:20 AM | Permalink | Reply to this
Read the post A Group Theory Problem
Weblog: The n-Category Café
Excerpt: I've got a problem with Lie groups.
Tracked: March 19, 2021 7:57 PM

Re: Can We Understand The Standard Model?

So, Problem 1 has been nicely settled by Gabriel Verret and L Spice: there are exactly 12 normal subgroups HU(1)×SU(2)×SU(3)H \subset \mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3) that are isomorphic to 6\mathbb{Z}_6. These should be the ones listed in my post, but I’m not sure anyone but me has checked that list.

Nobody has taken a stab at Problem 2, so I’ll give a hint: at least two of the 12 normal subgroups HH on the list have

U(1)×SU(2)×SU(3)HG SM \frac{\mathrm{U}(1) \times \mathrm{SU}(2) \times \mathrm{SU}(3)}{H} \cong G_{SM}

where

G SM=S(U(2)×U(3)) G_{SM} = \mathrm{S}(\mathrm{U}(2) \times \mathrm{U}(3))

is the true gauge group of the Standard Model. One of them is number 11 on my list: that’s what I said in my post.

I have not proved that there are only two.

Posted by: John Baez on March 19, 2021 8:23 PM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

Regarding your slides, in the page of the 2x2 matrices, would quaternions fit the role of the 6d matrices. It would be neat to have all normed division algebras corresponding to the SM.

Posted by: Daniel de Franša on March 20, 2021 7:42 PM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

Of course the quaternions are 4-dimensional, not 6-dimensional. But in the talk series Octonions and the Standard Model there’s been some talk of approaches to Standard Model that use all 4 normed division algebras, e.g. Cohl Furey and Mia Hughes’ talk ‘Division algebraic symmetry breaking’, which uses 𝕆\mathbb{R} \otimes \mathbb{C} \otimes \mathbb{H} \otimes \mathbb{O} in a manner reminiscent of Geoffrey Dixon’s work. (The \mathbb{R} is here just for decorative purposes, since it can be left out without changing the algebra.)

It would probably be good to put more thought into the chain of inclusions

𝔥 2()𝔥 2()𝔥 2(𝕆)\mathfrak{h}_2(\mathbb{C}) \subset \mathfrak{h}_2(\mathbb{H}) \subset \mathfrak{h}_2(\mathbb{O})

Posted by: John Baez on March 20, 2021 9:48 PM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

Oh, I am sorry. What I meant is that in the page where you find “4d/10d Minkowski spacetime can be seen as the space of 2×2hermitian complex/octonionic matrices”, the quaternions would yield 6d 2x2 hermitian quaternionic matrices.

Posted by: Daniel de Franša on March 21, 2021 4:50 AM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

Yes, 𝔥 2()\mathfrak{h}_2(\mathbb{H}) is 6d Minkowski spacetime. I haven’t found any use for that fact in thinking about the Standard Model, but I also haven’t tried, so maybe I should! I’ve written about it elsewhere, e.g. here:

Posted by: John Baez on March 21, 2021 6:34 PM | Permalink | Reply to this

Re: Can We Understand The Standard Model?

I’ve looked at that paper and I since it is concerned about supersymmetry, I remembered that there is a very fertile field of studies in real 6d, or 3d complex, that is of Topological Sting theory. It also has the signature of AdS5.

Posted by: Daniel de Franša on March 21, 2021 7:19 PM | Permalink | Reply to this

Mitchell

Six dimensions is also the maximum dimension for superconformal field theory (Nahm’s theorem). Many such field theories occur as worldvolume theories of stacks of M5-branes.

The list at the bottom of page 6 in Baez & Huerta reminds me that Maldacena’s original (1997) conjectures for holography were for stacks of M2-branes, D3-branes, and M5-branes.

Posted by: Mitchell Porter on March 25, 2021 6:49 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 6:27 AM

Mozibur

I don’t really understand Connes non-commutative geometry. However, what I’ve gathered that he’s able to derive the classical Standard Model from his spectral action, which is, as far as I understand it, the non-commutative geometry version of the Einstein-Hilbert action. Thats one version of the Kaluza-Klein idea - expressing everything in terms of gravity.

To me, that sounds significant.

Especially after seeing the full Standard model written out in Veltman’s Diagrammatica - I think there was something like a hundred terms!

Posted by: Mozibur Ullah on April 13, 2021 8:42 AM | Permalink | Reply to this

Re: Mozibur

Connes gets the Standard Model from noncommutative geometry by putting it in. He can get lots of different quantum field theories from noncommutative geometry by making different choices. I give a link to some of his work at the start of my second talk.

He’s trying to figure out what makes the Standard Model special, and he’s figured out a lot of interesting stuff. But it’s certainly not true that only the Standard Model can pop out of his approach. So the question of understanding the Standard Model remains open. (If he’d already solved it I wouldn’t be thinking about!)

By the way, he basically uses a noncommutative generalization of the Einstein–Hilbert action to get the part of the action that only describes bosons and their interactions — that is, the part involving just gauge fields and the Higgs. But he needs another idea to describe how the fermions interact with the gauge fields and Higgs.

Posted by: John Baez on April 13, 2021 5:43 PM | Permalink | Reply to this

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