17
$\begingroup$

In this paper John Baez says the actual gauge group of the standard model is $SU(3) \times SU(2) \times U(1) /N$. Can someone explain the logic behind this line of thought?

  • Firstly, does this group $N$ has some name?

  • How can we see that $N$ acts "trivially on all the particles in the Standard Model" ?

  • Does this factoring change anything or why isn't this "real" gauge group mentioned anywhere else?

$\endgroup$
  • $\begingroup$ The logic behind this line of thought will, obviously, be contained in the paper. I think you should make your question (a lot) more specific: What exactly do you not understand? $\endgroup$ – Danu Apr 16 '15 at 18:26
  • $\begingroup$ Now that I've taken a look at the paper, it is even more unclear to me what your issue is. The reasoning is simple, almost trivial. There is a "redundant subgroup", so we quotient it out. That's it. What's your issue? $\endgroup$ – Danu Apr 16 '15 at 18:32
  • 2
    $\begingroup$ @Danu Could you elaborate why this is a "redundant subgroup"? Why redundant? Subgroup of which group? I can't see why $N$ "acts trivially on all particles"... And lastly, why isn't this mentioned anywhere else if it is that obvious? $\endgroup$ – Tim Apr 16 '15 at 19:48
  • 1
    $\begingroup$ Honestly, I don't really know why Baez starts by saying the things he does---I just think that the reasoning he presents is quite simple. Now that you've updated the question, I think it's much better and I have upvoted accordingly. $\endgroup$ – Danu Apr 16 '15 at 20:12
5
$\begingroup$

Baez actually has another paper (with Huerta) that goes into more detail about this. In particular, Sec. 3.1 is where it's explained, along with some nice examples. The upshot is that the hypercharges of known particles work out just right so that the action of that generator is trivial. Specifically, we have

Left-handed quark     Y = even integer + 1/3
Left-handed lepton    Y = odd integer
Right-handed quark    Y = odd integer + 1/3
Right-handed lepton   Y = even integer

Because these are the only values for known fermions in the Standard Model, that generator does nothing. So basically, you can just take the full group modulo the subgroup generated by $(\alpha, \alpha^{-3}, \alpha^2)$ -- where $\alpha$ is a sixth root of unity.

There's also this paper by Saller, which goes into greater detail about the "central correlations" of the Standard Model's gauge group, but in a more technical presentation. Saller also goes into some detail in chapter 6.5.3 of his book.

$\endgroup$
  • $\begingroup$ Thanks for the great reading recommendation. I found the explanation in books.google.de/… very illuminating, too $\endgroup$ – Tim Apr 17 '15 at 8:41
6
$\begingroup$

How can we see that the group $N$ generated by $$ g = (e^{2\pi i/3} I, -I, e^{i\pi /3}) \in SU(3)\times SU(2)\times U(1) $$ acts trivially on all fields in the Standard Model?

First of all, note that $g$ is in the center of $SU(3)\times SU(2)\times U(1)$. Therefore its representative in the adjoint representation is the identity. Since gauge bosons transform in the adjoint representation, $N$ acts trivially on them.

The left-handed lepton fields are in the trivial representation of $SU(3)$ and are an $SU(2)\times U(1)$ doublet, $$\Psi = \begin{pmatrix} \nu_L \\ \psi_L \end{pmatrix}.$$ Baez is using a normalization of charge such that these fields have $U(1)$ charge $-3$., so they also transform trivially under $N$. The right-handed lepton $\psi_R$ has $U(1)$ charge $-6$ in this system, so it too is trivial.

The left (right) handed quarks have $U(1)$ charge $1$ ($4$ or $-2$) in this system and transform under $SU(3)$. It is simple to see that they also transform trivially under $N$.

(Note that various sources put the $1/3$ ratio between quark and lepton charges in various places, so be careful comparing.)

$\endgroup$
5
$\begingroup$

The main point is that if one has a consistent gauge theory containing matter with gauge group $$G:=SU(3)\times SU(2)\times U(1),$$ and if one divides $G$ with a normal subgroup $N$, then the matter representations of the matter fields could potentially become multi-valued. However, it is possible to choose $N=\mathbb{Z}_6$ in such a way that the standard model matter fields are invariant. And $N=\mathbb{Z}_6$ happens to be the maximal possible in this case.

See also this & this Phys.SE posts and links therein for a similar discussion for the electroweak sector.

See also this related Phys.SE post.

References:

  1. J.C. Baez, Calabi-Yau Manifolds and the Standard Model, arXiv:hep-th/0511086.

  2. D. Tong, JHEP 07 (2017) 104, arXiv:1705.01853. (Hat tip: knzhou.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.