If a regularization procedure respects a symmetry, is this symmetry unbroken in perturbation theory?

Question

I read in this paper the statement that a proof that SUSY is preserved in perturbation theory would be the existence of a regularization procedure which respects SUSY (for a particular theory).

Is there a clear proof that the existence of a regularization procedure which respects a symmetry implies the absence of its anomaly? This makes sense at some level; step by step in my calculation, my formula for $\partial_\mu j^\mu$ respects this symmetry.

The contrapositive to this statement is that if there is an anomaly, there is no regularization procedure which respects the symmetry. This can be made plausible through examples, of course, but I'd like to be sure of it.

Answer 1

Here is a sketch of the philosophy.

A regularisation is a deformation of the theory that renders it finite. Let us assume for simplicity that there is a single deformation parameter, say $x\in\mathbb R$, and let us choose the origin (the undeformed theory) at $x=0$. Some typical examples for $x$ are

• $x=\mu/M$, where $\mu$ is some fixed mass scale, and $M$ a Pauli-Villars mass.
• $x=d-n$, where $d$ is the dim-reg dimension, and $n$ the physical dimension (e.g., $n=4$).
• $x=\mu\epsilon$, where $\mu$ is some fixed mass scale, and $\epsilon$ is a small distance used in point-splitting.

As the undeformed theory typically contain divergences, observables are usually $\mathcal O(1/x^a)$ for some $a\ge0$. In particular, this is so for the Noether current, which is a composite operator (and thus ambiguous, i.e., divergent).

The regulator may spoil some symmetry, but the same is formally recovered in the $x\to0$ limit, and so the divergence of the Noether current is formally $\mathcal O(x^b)$, for some $b\ge0$.

Combining these remarks, we obtain that the divergence of the regulated current is actually $\mathcal O(x^{b-a})$, for some parameters $a,b$. If $b>a$, the current is conserved in the physical limit, and the symmetry is non-anomalous. If $a=b$, the regulator leaves a finite piece in the physical limit, and the symmetry is anomalous. See e.g. this PSE post for an explicit example of this approach to anomalies.

Now comes the key point. If the regulator respects the symmetry, the divergence of the current is identically zero, even for finite $x$. Therefore, $dj=0$ for all $x$, and $dj=0$ in the $x\to0$ limit. No finite piece may survive the physical limit, because the current is exactly conserved in the deformed theory.

Note that this philosophy also holds for non-perturbative anomalies. If there is some regulator (defined non-perturbatively, e.g. lattice) that respects the symmetry, the latter is necessarily non-anomalous. This is why anomalies are typically induced by massless fermions only: if a mass term is allowed by the symmetry, then a Pauli-Villars regulator is allowed too, and the corresponding fermion cannot contribute to the anomaly. See e.g. 1909.08775 for more details.