# Beta function of Seiberg-Witten

In Seiberg and Witten's seminal paper, a key role is played by the monodromy of $$\tau$$ around infinity. This monodromy can be computed in the weakly coupled regime, and it is given by the one-loop beta function. It is important, as stressed by the authors, that higher-loop corrections vanish (and the full answer is only modified non-perturbatively, by instantons).

Now, the fact that the higher-order loop contributions to the beta function vanish seems to be a consequence of supersymmetry. Indeed, Seiberg's holomorphy argument shows that physics must be holomorphic in $$\tau\sim\theta+i/e$$. But perturbation theory can only see $$e$$, so it cannot generate holomorphic corrections. Thus, these corrections must vanish.

We are often taught that having a beta function that vanishes beyond one-loop is a rather trivial statement. Indeed, 't Hooft showed that these higher orders are scheme-dependent, and can be made to vanish by a convenient choice of renormalization scheme. Thus, the fact that the beta function in $$\mathcal N=2$$ is one-loop exact does not seem like a very deep statement: one can always choose this to be true, for any QFT, regardless of supersymmetry.

So, on the one hand, Seiberg-Witten insist on the importance of having only a one-loop monodromy; and, on the other hand, we have 't Hooft argument that this is a trivial fact. Needless to say, both claims are definitely true, and I am just missing a piece of the puzzle. What is special about the beta function of $$\mathcal N=2$$? How is it different from that of regular QFTs?

't Hooft's argument, if I recall correctly, only shows that the higher order parts of the beta function can be made to vanish at one chosen value of the renormalization scale $$\mu$$. Seiberg's argument shows that these values vanish at all values of $$\mu$$, but only for the $$\mathcal{N}=2$$ QFTs.
• Thanks! I'm sure this is correct but I'm particularly slow today. If I can choose these orders to vanish at some particular $\mu$, what prevents me from making the same choice continuously as a function of $\mu$? i.e., I use $\mu$-dependent counterterms that ensure they vanish for all $\mu$. What am I missing? Apr 3, 2021 at 19:42