4
$\begingroup$

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?

$\endgroup$

1 Answer 1

4
$\begingroup$

'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.

$\endgroup$
1
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Apr 3, 2021 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.