The n-Category Café

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