With massless fermions, does a finite chiral rotation only affect the coefficient of the theta term? In quantum electrodynamics (QED), the chiral current is not conserved:
$$
 \partial_\mu(\overline\psi\gamma^\mu\gamma_5\psi)\sim F\wedge F
  + \text{mass term}.
\tag{1}
$$
It would be conserved if the chiral transformation
$$
 \psi\to \exp(i\alpha\gamma_5)\psi
\tag{2}
$$
were a symmetry, so we infer that (2) is not a symmetry, not even when the fermions are massless, at least not when the chiral rotation angle $\alpha$ is close to zero. This conclusion holds for any spacetime topology, because (1) is a local relationship.
In other contexts with massless fermions, many sources (refs 1,2,3,4) seem to say that for finite $\alpha$, the only effect of the chiral rotation (2) is to change the coefficient of the so-called theta term $\int F\wedge F$ in the action. But this term is zero for an abelian gauge field in topologically trivial spacetime, so if that really were the only effect, it would imply that (2) is a symmetry of QED — when spacetime is topologically trivial and the fermions are massless.
The first argument says that (2) is not a symmetry, but the second argument says it is. What's the resolution of this apparent contradiction? I must be making a mistake somewhere, but where?
(In case this isn't clear: The question is about the case where $\alpha$ is independent of the coordinates and the fermions are massless. Otherwise, (2) wouldn't even be a symmetry of the action, much less of the quantum field theory.)

References:

*

*Page 457 in Weinberg's The Quantum Theory of Fields Volume II


*"The Electroweak Vacuum Angle" (arXiv:1402.6340)


*Chapter 3 in Tong's Lectures on Gauge Theory (http://www.damtp.cam.ac.uk/user/tong/gaugetheory.html)


*This answer to the question Why is there no theta-angle (topological term) for the weak interactions?
Related: Is it the chiral anomaly which is solely responsible for having instanton effects (and therefore, the $\theta-$term) in the QCD action?
 A: I found at least two mistakes in my reasoning. First, I wasn't paying close enough attention to the difference between compact QED, whose gauge field is based on the compact group $U(1)$, and noncompact QED, whose gauge field is based on the noncompact group $\mathbb{R}$. Most texts gloss over this because they look the same in perturbation theory, but nonperturbatively it does make a difference. According to section 2.5 in ref 1,

*

*The global chiral rotation is not a symmetry of compact QED, not even when spacetime is topologically $\mathbb{R}^4$.


*The global chiral rotation is a symmetry of noncompact QED when spacetime is topologically $\mathbb{R}^4$.
Second, I wasn't paying close enough attention to the difference between a dynamic gauge field, which is what QED has, and a background gauge field, which is often used as a tool for studying chiral anomalies. According to section 2.5 in ref 1 again,

*

*The global chiral rotation is a symmetry when the gauge field is just a background field (if spacetime is $\mathbb{R}^4$), even when the gauge field is based on the compact group $U(1)$.

The details behind these assertions are spelled out in the cited paper. The basic idea is that in compact QED, the would-be symmetry is obstructed by the presence of additional observables that are constructed from the compact-$U(1)$ gauge field itself. Those observables are absent when the gauge field is merely a background field, or when it's a dynamic field based on the noncompact group.
So I was making at least two different mistakes, namely conflating the compact and noncompact cases and conflating the dynamic and background gauge field cases. This answers my question, in the sense that it identifies mistakes in my reasoning. (Thanks to users mike stone and Prof. Legolasov for comments/chat that helped direct my attention toward these issues.)
My understanding is still incomplete, though. If a chiral rotation is a good symmetry in the compact $U(1)$ case when the gauge field is only a background field, then a nonperturbative regulator that preserves that symmetry ought to exist. When the background gauge field is absent, we can use a lattice formulation witih the overlap Dirac operator (reviewed in ref 2), which preserves an exact chiral-rotation symmetry, without doublers. When the gauge field is nonzero, we can generalize that transformation to one that is still an exact symmetry of the lattice action, but now that transformation depends explicitly on the gauge field, so its effect on the path-integral measure must be considered carefully. That's where the chiral anomaly is hiding. This lattice formulation of compact QED is actually what I had in mind when writing the question, even though I didn't say that out loud. I still don't fully understand how it all plays out in the lattice formulation (I still have a lot to learn about the subject that inspired my username!), but that's beyond the scope of the question I asked here. One step at a time.


*

*Harlow and Ooguri, Symmetries in Quantum Field Theory and Quantum Gravity (https://arxiv.org/abs/1810.05338)


*Creutz, Confinement, chiral symmetry, and the lattice (https://arxiv.org/abs/1103.3304)
A: It is only the global rotation effects that is zero for topologically trivial spacetimes. There are still local effects that are captured by your eq 1.
