I have some questions regarding the $D=4 $ ${\cal N}=4$ super-Yang-Mills theory (the one with a really long action which can be acquired by compactifying the 10-dimensional ${\cal N}=1$ theory).

I have heard that the theory has a vanishing beta-function, so the superconformal symmetry is not anomalous. Because there are beta-functions involved, I am obligated to ask whether this is a perturbative result. I am almost sure that it isn't, but still.

  1. If it is exact, how can I show that? I am not asking for the absolutely rigorous proof, but I need something better then "Feynman diagrams vanish", or at least why can we treat this non-perturbative problem perturbatively.

  2. Another (related) question: is this theory exactly solvable? If so, how does one recover the correlations? (My assumption would be that some kind of a bootstrap method can be used, since the superconformal symmetry is exact).

  3. And finally, does it mean that all anomalous dimensions (of fields and couplings) are zero (meaning that the classical RG flow completely determines how the theory is influenced by rescalings)? Is it correct to say that, unlike the usual Yang-Mills theory, N=4 SYM shows no fractal behaviour?

P.S. I am adding a string-theory tag despite the fact that this question has little to do with string theory, because I would expect string theorists to know a lot about SYM because of the AdS/CFT. It is also the reason I am studying it.

  • $\begingroup$ One can show that N=4 SUSY implies that all Feynman diagrams come in pairs with a different sign. As far as I know the fact that the beta function vanishes relies on this pertubative approach but it is exact. $\endgroup$
    – quan
    Jan 30, 2015 at 0:23
  • $\begingroup$ @quan I added an UPD to my question. Can you, please, extend your answer a little? If something relies on the perturbative approach, why do we say that it is exact? $\endgroup$ Jan 30, 2015 at 0:27
  • 1
    $\begingroup$ A more hopefully helpful explanation would be: Already in N=2 SUSY we know that the beta function consists only of a one-loop contribution and a non-pertubative part. This is related to the holomorphic structure of SUSY theories. By explicit computation the one-loop contribution vanishes. The non-perturbative corrections come from instating. However, again by explicit computation one realises that the instatons in N=4 SUSY do not correct the beta function. Instead they add higher derivative terms to the action. $\endgroup$
    – quan
    Jan 30, 2015 at 0:35
  • $\begingroup$ @quan They add higher derivative terms to the action - does it mean that the answer to my third question is no (the theory does actually depend on scale)? $\endgroup$ Jan 30, 2015 at 0:38
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    $\begingroup$ There is an answer at physicsoverflow.org/26677 $\endgroup$ Mar 8, 2015 at 13:32

1 Answer 1

  1. There is a non-perturbative proof that the theory is a CFT. See this paper. The idea of that paper is to use the fact that N=2 supersymmetry restrict the low energy effective action to be analytic in the N=2 prepotential and that the prepotential have good properties under $U(1)_{R}$ symmetry.
  2. About your second question I can only refer to you this review.
  3. The action does not receive anomalous corrections since the theory is a CFT. The theory has only one coupling, and this coupling sits in front of the action so it does not receive anomalous corrections to the dimension as well. The fields in general will receive anomalous corrections to the dimension, only some special family of operators that does not receive corrections such as the BPS operators that are protected by supersymmetry. The Lagrangian of the theory is one of these BPS operators (this is another way to see why the theory is a CFT).

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