# How does the global $G^2G'$ anomaly make all the $\theta$-vacua associated to the gauge group $G$ physically equivalent?

Consider a gauge group $$G$$ and suppose that there is a $$\theta$$-term associated to it. According to this answer, the existence of a global anomalous symmetry $$G'$$ which rotates the $$\theta$$-term, ensures that the different $$\theta$$-vacua are physically indistinguishable. In that regard, it is enough to check for whether the $$G^2G'$$ anomaly cancels. If it does not, we are saved from $$\theta$$-worries.

I am trying to understand the mechanism.

1. How does a global anomalous symmetry ensure the physical equivalence of all the $$\theta$$-vacua? My naive interpretation would be the opposite. Since the group is anomalous, the symmetry is not respected by the quantum degrees of freedom and hence all the $$\theta$$-vacua obtained by the application of this "not-respected" group are physically distinguishable (otherwise, it is a respected symmetry).

2. Why is it sufficient to check for the $$G^2G'$$ anomaly? What's so special about the triangle diagram? There must be some physical reason for this choice. What is that?

It is perhaps better to think of it as though the anomaly is what connects the symmetry (which is still present in a sense) to the topological features of the gauge group that allow us to talk about a vacuum angle. The chiral symmetry naively just acts on fermions in Lagrangian. It can be present even if there is no gauge field. The vacuum angle is due to the topology of gauge field configurations, and is present even if there are no fermions. When it is possible to define a vacuum angle, instantons wind around the field configuration space and come back to the same vacuum which is a pure gauge field (i.e. $$F_{\mu\nu}=0$$). If we imagine the discrete vacua connected by topologically not-trivial paths are in fact distinct (i.e. we talk about a covering space), then we can define a $$\theta$$ angle which is a lot like momentum on lattice. The $$\theta$$ vacuum is a superposition of all of these discrete vacua, just like a definite momentum state is a superposition of all position eigenvectors.
Now when the anomaly is present, the fermions are connected to these topological features of the gauge field. The triangle diagram is just a convenient litmus test for the anomaly being present. In all the cases I am familiar with, it connects the divergence of the global current to a topological current of the gauge field. What this means at the level of states (at least in the Schwinger model) is that those discrete vacua in the covering space have different chiral charge. So the $$\theta$$ vacuum is a superposition over all possible states of definite chiral charge, just like a momentum eigenvector is a superpostion over position states. A chiral transformation, which still exists at the quantum, level shifts the $$\theta$$ vacuum in just the same way as the position operator shifts the momentum eigenstates.
What is key is that the chiral current is conserved in the Lagrangian. This is what ensures that all these $$\theta$$ vacua are equivalent. The anomaly connects the $$\theta$$ vacua to chiral charge but it does not ensure that they are equivalent. If we add a mass term to the Schwinger model, chiral symmetry is explicitly broken (not just through the anomaly). The anomaly is still present connecting vacua to chiral charge, but now a particular $$\theta$$ value is singled out as the unique lowest energy state. Changing the phase of the mass term, we can make this particular $$\theta$$ value be whatever we like.
In terms of the path integral, this is all seen because the $$\theta$$ angle is multiplying a term counting the instanton number. If we redefine the fields in accordance with the global symmetry (which we are always free to do), we pick up another term in the Lagrangian proportional to the instanton number coming from the path integral measure. This is the Fujikawa interpretation of the anomaly. So we can shift the $$\theta$$ angle to be whatever we like by doing a global transformation. But we can only say the physics is independent of the $$\theta$$ angle if the rest of the Lagrangian does not change, which is only the case if it was classically chirally invariant.