# Loop counting for determinants and anomalies

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)?

• Both are consistent. The Jacobean is invariant, and hence the anomaly, which he also computes perturbatively. Jan 12, 2020 at 19:43

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 $$\hbar$$.