How common is quantum symmetry?

Question

It is well-known that, if I have a two-dimensional quantum field theory, $\mathcal{Q}$, with a finite, abelian, non-anomalous global symmetry $\Gamma$, and I gauge it, the resulting theory (which I denote $\mathcal{Q}/\!\!/\Gamma$) has the Pontryagin dual group, $\widehat{\Gamma}:=\mathrm{Hom}(\Gamma,\mathrm{U}(1))$, as a global symmetry. This is known as quantum symmetry or Ponryagin dual symmetry.

Relatively recently it was understood that the story is quite a bit more general. In $d$ dimensions with a $p$-form symmetry, the dual symmetry is a $(d-p-2)$-form symmetry, while if $\Gamma$ is finite but non-abelian, the Pontryagin dual symmetry is a non-invertible symmetry, usually denoted as $\mathrm{Rep}(\Gamma)$ or variations thereof depending on the dimensionality and/or other data that specify the symmetry category.

My question is, how common is it, actually, in a generic gauge theory?

For that, let's imagine the following scenario. I have a pure $G$-gauge theory where $G$ is any group (I could specialise to a subcategory of groups, such as connected or simply connected, if technically necessary, but I think it's fine as is). I want to think about it as starting off with the trivial theory with $G$-symmetry and gauging $G$, so I'll denote it $\bullet/\!\!/G$. In this type of gauge theories, there isn't necessarily Pontryagin dual symmetry for all of $G$ because the gauging involves summing over non-flat bundles (for the continuous pieces of $G$) and hence it is not an invertible process -- which Pontryagin duality (or more generally Tannaka-Krein duality and generalisations) would imply. However what I could do, starting with the trivial $G$-symmetric theory, $\bullet$, is choose any finite (probably need normal) subgroup $\Gamma\subset G$ and gauge $G/\Gamma$ to get $\bullet/\!\!/(G/\Gamma)$, which still has global symmetry $\Gamma$, since I didn't gauge it. Then gauge $\Gamma$, to get $\left[\bullet/\!\!/(G/\Gamma)\right]/\!\!/\Gamma \cong \bullet/\!\!/G$. Since in the last step I only gauged a finite group, I expect to get back a Pontryagin dual global symmetry, $\widehat{\Gamma}=\mathrm{Rep}(\Gamma)$.

The above argument implies the following:

Any pure $G$-gauge theory has a $(d-2)$-form Pontryagin dual global symmetry $\widehat{\Gamma}_\text{max}=\mathrm{Rep}(\Gamma_\text{max})$, where $\Gamma_\text{max}$ is the maximal (normal?) finite subgroup of $G$.

Is this true? It seems a bit too strong to be true. If not, what is wrong with the argument?

Moreover, in a simple example, this is not how it seems to work. Let's take $G=\mathrm{SU}(2)$ and $\Gamma= Z(\mathrm{SU}(2))\cong\mathbb{Z}_2$. Then $G/\Gamma\cong \mathrm{SO}(3)$ and $\bullet/\!\!/(G/\Gamma)$ is the pure $\mathrm{SO}(3)$-gauge theory. To go to the $\mathrm{SU}(2)$-gauge theory one gauges the centre $\mathbb{Z}_2$ as a one-form symmetry! Which would result in a $\widehat{\mathbb{Z}}_2$ dual symmetry that is a $(d-3)$-form symmetry, instead of a $(d-2)$-form symmetry as the above argument would suggest. Or could one gauge $\mathbb{Z}_2$ either as a zero-form symmetry (which acts on nothing in $\bullet/\!\!/\mathrm{SO}(3)$), or as a one-form symmetry and obtain different quantum symmetries in either case?

Answer 1

In the following I assume that the $G$ symmetry is faithful.

I think the caveat is the step of gauging $G/\Gamma$, which is not necessarily normal.

Start from a theory with global symmetry $G$, and suppose $\Gamma$ is a non-anomalous normal subgroup. Just to make things simple, let's assume $G$ is 0-form. Then what we know for sure is that we can definitely gauge $\Gamma$ and the resulting theory has $G/\Gamma$ symmetry. You can further gauge $G/\Gamma$ to gauge the full symmetry $G$.

Now, suppose you want to gauge $G/\Gamma$ first instead. Let us again assume that there is no anomaly. The problem now is that $G/\Gamma$ is generally not normal. But we do not need this conclusion.) There are two things that can happen:

1. Gauging $G/\Gamma$ breaks $\Gamma$. A simple example is $G=S_3=Z_3\rtimes Z_2$. Take $\Gamma=Z_3$ (which is normal). Gauging $G/\Gamma=Z_2$ breaks $\Gamma$.

2. You can not gauge $G/\Gamma$ without also gauging (full or part of) $\Gamma$. This happens for example when $G$ is a nontrivial central extension of $G/\Gamma$ by $\Gamma$, and is exactly the case of your $G=SU(2), \Gamma=Z_2$ example.

Edit: the comment pointed out that the case 1 might be more subtle according to footnote 1 in https://arxiv.org/abs/1712.09542. In 2d, the resulting theory has a non-invertible symmetry.

Edit: In case 2, here is a heuristic argument for the SU(2) example: since the symmetry group is $SU(2)$, SU(2) must act faithfully, so there exist operators that transform under all irreps of SU(2). In that case, when SO(3) is gauged, the $Z_2$ part, which is $2\pi$ rotation in SO(3), must be gauged as well for consistency. In fact, we do not need the faithfullness assumption here. The faithfullness just guarantees that the $Z_2$ acts nontrivially, but we do not need it for the argument.

Your argument works when $G=\Gamma \times G/\Gamma$. It seems that one of the above must happen if it is not a direct product.