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I am trying to understand an argument for why anomalies are one-loop exact, given by Bilal in Lectures on Anomalies. The relevant paragraph is reproduced here:

Let us first explain why the anomaly we have just computed corresponds to a one-loop effect. One can introduce a formal loop-counting parameter by rescaling the action as $S \rightarrow \frac{1}{\lambda}S$. Then in computing Feynman diagrams, the propagators get an extra factor $λ$ while the vertices get an extra $\frac{1}{\lambda}$. Thus every Feynman diagram comes with a factor $λ^{ I−V}$ , where $I$ is the number of internal lines and $V$ the number of vertices. By a well-known relation, one has $I − V = L − 1$ with $ L$ being the number of loops, and one sees that $\lambda$ is a loop counting parameter. Since the classical action comes with a $\frac{1}{\lambda}$ it is the tree-level contribution ($L = 0$) to the effective action, while the anomaly, like any determinant, has no factor of $\lambda$ and corresponds to a one-loop contribution ($L = 1$).

I understand everything up until the section in italics. What does it mean that any determinant has no factor of $\lambda$? Am I meant to somehow be inferring that the anomaly is invariant under such a rescaling and so it is a determinant, or is it the other way around (i.e. determinant implies invariance)?

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  • $\begingroup$ Both are consistent. The Jacobean is invariant, and hence the anomaly, which he also computes perturbatively. $\endgroup$ Jan 12 '20 at 19:43
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  1. From the Boltzmann factor perspective $e^{\frac{i}{\hbar}S}$, if we exponentiate the path integral measure $\rho= e^{\frac{i}{\hbar}\frac{\hbar}{i}\ln\rho}$ in the path integral $Z=\int\!{\cal D}\phi~\rho~e^{\frac{i}{\hbar}S}$, the anomaly in the measure is formally a 1-loop effect, cf. the $\hbar$/loop-expansion. This seems to be essentially Bilal's argument in a nutshell, with $\lambda$ playing the role of $\hbar$.

  2. A more rigorous argument can be given by the fact that boson- and massive fermion-lines have a parity-conserving gauge-invariant UV regularization. Also recall that anomalies only appear when the regulator fails to conserve the pertinent symmetries. Therefore any loop diagram (without divergent subdiagrams) that contains a loop-participating boson- and massive fermion-line cannot be anomalous. In other words, the anomalous loops must be made purely from massless fermions. It turns out that a higher-loop diagram (without divergent subdiagrams) must contain a boson loop-line, and is hence anomaly-free. We should stress that in the presence of a one-loop anomaly, it will create anomalous higher-loop diagrams with divergent subdiagrams. On the other hand, 0-loop/tree diagrams are always finite. So in that sense, anomalies are a 1-loop effect.

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