# Does the vanishing of the one-loop beta-function imply no running to all orders?

This question sounds ridiculous, but bear with me. I am having a hard time reconciling the following two facts:

• Classical global symmetries can become anomalous upon quantization, and the anomalous non-conservation of the associated current is a one-loop effect (since the effect is based on the path integral measure failing to be invariant, and the transformations in the path integral measure contain no powers of $$\hbar$$, which is a loop-counting parameter).
• The non-vanishing of the beta function in a classically scale-invariant theory constitutes anomalous dilitation symmetry violation.

These two facts seem to imply that if the beta-function of a classically scale-invariant theory vanishes at one-loop, then it vanishes at all orders.

This seems obviously not true, otherwise people wouldn't have spent so long proving that $$\mathcal{N}=4$$ super-Yang-Mills is conformally invariant to all orders, since the 1-loop calculation is relatively easy to check. So, my question is "Where do the above statements fail to apply?"

The path integral measure is not one-loop exact in general. In the example of the anomaly for the chiral symmetry, the path integral measure for the fermions are one-loop exact because the integral is gaussian on the fermions $$\psi$$ and $$\bar\psi$$ (with the gauge field $$A_{\mu}$$ fixed, as well as any scalar in the Yukawa coupling). The anomaly comes from the fact that there is a conflict in trying to preserve both the gauge symmetry and the chiral symmetry in the process of regularization and renormalization of the measure. If you regulate the measure in order to preserve the gauge symmetry, then instead of a smooth cut-off like $$f(\displaystyle{\not}\partial)$$ you are going to use $$f(\displaystyle{\not}\partial+i\displaystyle{\not}A)$$ regulator, that breaks the chiral symmetry. After the shift of $$k$$ you get terms involving $$\displaystyle{\not}A$$ multiplying $$\gamma_5$$ what makes the trace in the space of fields being nonzero. For a general regulator, not necessarily a gauge invariant one, the finite counterterms that you are going to add in order to recover the gauge symmetry at the limit will also break the chiral symmetry. You can show that there is no available set of finite counterterms that preserve both the gauge symmetry and the chiral symmetry.