# Breaking a gauge group through a monopole$.$

I am trying to understand the paper arXiv:1712.08639 and, in particular, the discussion in §5.

In this section, the authors take a gauge theory with group $\mathrm{SO}(N)$, and they add a "unit of magnetic flux" in the $N,N-1$ direction in the gauge group. This breaks the group from $\mathrm{SO}(N)$ to $(\mathrm O(2)\times\mathrm O(N-2))/\mathbb Z_2$. Moreover, a fermion in the rank-2 symmetric representation of the gauge group breaks into

• a Dirac fermion of charge $2$ under $\mathrm O(2)$ and uncharged under $\mathrm O(N-2)$. This field has two zero-modes in the spin $j=1/2$ representation of the Lorentz group.

• A Dirac fermion of charge $1$ under $\mathrm O(2)$ and in the vector representation of $\mathrm O(N-2)$. This field has $N-2$ zero modes in the $j=0$ representation of the Lorentz group.

• $\frac12(N^2-3N+2)$ fermions that are neutral under $\mathrm O(2)$ and that have no zero-modes (and therefore play no role).

Finally, the authors claim that the monopole is "effectively abelian".

All this is mentioned very casually, which makes me think it should all be obvious. But after thinking about it for a week or so, I still cannot understand where this all comes from.

1. By a "unit of magnetic flux", they mean a GNO monopole, right? And the GNO charge is just the generator of rotations in the $N,N-1$ plane, right?

2. How can I understand the breaking of the symmetric field into its components? What if the field were e.g. anti-symmetric instead of symmetric?

3. Why is the monopole "effectively abelian"? is it because $\mathrm O(2)$ is abelian?

• This might be helpful. web2.ph.utexas.edu/~gsudama/pub_bak/1984_005.pdf – user93146 Jul 8 '18 at 16:24
• If you figure out the paper I cited, I'd be curious to see what you come up with. That was one of the papers I read in my grad work, and I spent a lot of time being confused, thinking I understood and then realizing I really didn't, etc. I'm pretty sure topology exists, but I'm really very bad at it. – user93146 Jul 8 '18 at 22:35

First, have a look at the papers https://arxiv.org/abs/1602.04251 and https://arxiv.org/abs/1605.02391 where the authors (Seiberg and Witten) have a more careful analysis of the properties of the monopole operators, which is extremely useful in the paper you have cited.

Now go back to your question.

1&3. Short answer, yes. But let me clarify a little bit (just skip it if you know it already). In 3+1 $D$ $U(1)$ Abelian gauge theory we normally have the Bianchi identity $$\partial_\mu \tilde{F}^{\mu\nu} = 0.$$ But if we allow a Dirac singularity in the space (say at $\textbf{x}=0$ the origin), we can have the following $$\partial_\mu \tilde{F}^{\mu\nu} = \frac{2\pi}{e}\delta^3(\textbf{x}),$$ the quantization of which comes from single-valuedness of electron wave-function (or in more mathematical terminology, consistency of the gauge bundle). In order to generalize the above results to non-abelian gauge theory, let us examine the non-abelian Bianchi identity $$D_\mu \tilde{F}^{\mu\nu} = 0$$ which replaces ordinary derivative $\partial_\mu$ with covariant derivative $D_\mu$ and contains a piece like $[A,F]$. In order to mimic the above construction, the magnetic charge has to commute with each other so that the contribution of $[A,F]$ vanishes, and the charges can be chosen to lie in the weight lattice (as shown in the original paper of GNO). Therefore, GNO monopole is essentially Abelian.

After inserting some monopole operator into the theory, only group element that commutes with the monopole charge survives, which in this case explains why the remaining gauge group is $O(2)\times O(N-2)/\mathbb{Z}_2$. Here the $\mathbb{Z}_2$ comes from the restriction that the determinant of the new gauge group, i.e. determinant of $O(2)$ part multiplied by the determinant of $O(N-2)$ part, should still be equal to 1.

2.This is simply a representation theory argument. The fermions are in the symmetric representation of $SO(N)$. Now the gauge group is broken down to $O(2)\times O(N-2)/\mathbb{Z}_2$. So we need to analyze how the fermions transform in the new representation. Besides the formal representation theory argument, a heuristic argument is the following.

Think of the fermions as elements in some $N\times N$ symmetric traceless matrix $A$, which has $\frac{1}{2}(N^2+N) - 1$ elements. The action of $SO(N)$ elements $U$ on the matrix is $A\rightarrow U^\text{T}AU$. Think of the upper-left $(N-2)\times(N-2)$ block of $U$ as the subgroup $O(N-2)$ and the lower-right $2\times 2$ block as the subgroup $O(2)$. It is easy to see that the corresponding upper-left $(N-2)\times(N-2)$ block of $A$ transform in the symmetric (but not traceless) representation of $O(N-2)$ and does not transform under $O(2)$. These $\frac{1}{2}(N^2-3N+2)$ fermions are neutral. Similarly, The lower-right $2\times 2$ block of $A$ (except the trace part, which we include in the upper-left block and do not transform under $O(2)$) transform in the symmetric traceless representation of $O(2)$ (i.e. spin 2 representation) but do not transform under $O(N-2)$. Finally, the rest should transform in the vector representation of both $O(N-2)$ and $O(2)$. This kind of analysis can be easily generalized to antisymmetric representation and so does formal representation theory argument, which can be found in any standard group theory textbook.

• If possible, I also want to discuss with you about the details of the paper further. There is so much that I do not understand about the paper as well. – Weicheng Ye Jul 14 '18 at 15:23
• This is a great answer, thank you very much! I won't award the bounty for now, so that the question stays in the featured list and you get a few more upvotes. Cheers! – AccidentalFourierTransform Jul 14 '18 at 15:39