On-site symmetries can be gauged, but is a gaugeable symmetry necessarily on-site?

I've always liked lattice QFT because it's mathematically unambiguous and non-perturbative, but it does have two drawbacks: (1) the lattice is artificial, and (2) some features are messy. One of those messy features is chiral symmetry. As an example, the lattice model of a free Dirac fermion in even-dimensional spacetime is easy to construct, but if we construct it in such a way that the non-chiral symmetry $$\psi\to\exp(i\theta)\psi$$ is on-site, as we usually do, then the Nielsen-Ninomiya theorem tells us that the chiral symmetry $$\psi\to\exp(i\theta\Gamma)\psi$$ cannot be on-site. (I'm using $$\Gamma$$ for the product of all the Dirac matrices.)

Instead of thinking of that messiness as a drawback, we can also think of it in a more positive way: it almost helps us understand why the chiral anomaly must exist. On-site symmetries can always be gauged, but we have no such guarantee for non-on-site symmetries. If that last part could be strengthened to "non-on-site symmetries cannot be gauged," then this really would help us understand why the chiral anomaly must exist.

Question: It is true that non-on-site symmetries cannot be gauged? In other words, is a gaugeable symmetry necessarily on-site in every lattice version of the theory?

I'm guessing that this is not true. I don't know why we'd ever want to use a lattice version in which a gaugeable symmetry group is realized as a non-on-site symmetry, because that would be unnecessarily messy,$$^\dagger$$ but is it actually impossible?

My question is similar to another unanswered question, but that one asks about ensuring the absence of anomalies, while I'm asking about ensuring the presence of an anomaly.

$$^\dagger$$ I say "unnecessarily" because I'm assuming that anomaly-free symmetries can always be on-site in some lattice version. I don't know if that's true, either. That probably-more-difficult question has already been asked on Physics SE.

Also related: When can a global symmetry be gauged?

• Thank you for bringing me back to my senses! Now I see what I was missing: we can make either $U(1)$ on-site (but then not the other one), and we can gauge either one, but after gauging one of them, the other one is no longer a symmetry at all. On-site symmetries can be gauged, but non-symmetries cannot be gauged (of course). Still not obvious to me exactly how to gauge the non-on-site $U(1)$, but I should be able to work it out in this simple example. Thank you (+1)! Commented Aug 4, 2020 at 12:32
• @ChiralAnomaly I'd be quite interested if you figure out an answer to the puzzle of how to gauge the hidden U(1)! I thought about it a little bit before but it seems like it's very nonlocal. If we write the charge density in k-space as $\int dk \rho(k) a^\dagger(k) a(k)$, then $\rho(k)$ is a step function, leading to a $1/x$ non-locality of $\rho(x)$... Commented Aug 4, 2020 at 16:47
• After thinking about this more, I'm not sure I follow the argument. Either one of the two $U(1)$ factors can be on-site (and the other one off-site) in a lattice formulation, and either one of the two $U(1)$ factors can be gauged, but those are two separate statements, so this doesn't seem to imply that the non-on-site factor can be gauged in any given lattice formulation. As far as I can see, it only implies that we can gauge either $U(1)$ factor by choosing a lattice formulation in which that factor is on-site. Am I missing a step in the argument? Commented Sep 19, 2020 at 14:26
• Ok, I was just over-thinking that first sentence. It sounds like we're in sync: in any given lattice QFT, gauging a non-on-site symmetry may be difficult, but it's not known to be always impossible. I spent some more time playing with the $1+1$ dimensional free Dirac field, using a simple lattice version with an exact chiral $U(1)$ that is not on-site. I don't see how to gauge it, but I also don't see why it would be impossible, so I'm in limbo... Anyway, thank you for your patient replies! Commented Sep 19, 2020 at 21:45