# How to prove quantum N=4 Super-Yang-Mills is superconformal?

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I'm especially interested in elegant illuminating proofs which don't involve a lot of straightforward technical computations

Also, does a non-perturbative proof exist?

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retagged Apr 19, 2014

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In any supersymmetric theory you can choose the gauge coupling to be the coefficient of W_\alpha^2 in the superpotential. This gauge coupling runs only at one-loop, which is a fundamental consequence of the non-renormalization theorems. The other possible running coefficients are the kinetic terms, Z(\mu)QQ^\dagger. These generally get renormalized to all orders in perturbation theory.

In N=4 the one-loop coefficient is zero. This is trivial (just counting the fields). Hence, the gauge coupling (as defined above) does not run. But N=4 relates the gauge particles with the chiral superfields (all the matter particles sit in one big representation of N=4) and so the latter cannot get renormalized either.

This is a slick and intuitive argument... Similar logic operates in many N=2 theories as well.

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answered Nov 2, 2011 by (650 points)
Nice answer. Can you provide a ref pls? Do I understand correctly this argument is perturbative only? If so, is there a way to extend it to a nonperturbative one?

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It is non-perturbative, because the gauge coupling does not run non-perturbatively. I have not seen it explicitly written anywhere, but I am sure I am (by far) not the first one who had this thought :)

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Your argument is definitely nicer :)

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Yuji: I think you may need to resort to this argument anyway, even if you do Leigh-Strassler. This is because the latter only implies a one-dimensional line of CFTs, but it does not prove that this one-dimensional line coincides with the line on which N=4 sits. So at some point you have to invoke the higher symmetry.

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Ok, so can you provide a ref to the nonrenormalization theorem you are using?

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Page 96 in http://www.physics.uc.edu/~argyres/661/susy1996.pdf reviews the argument. In general you can prove that it is one-loop+non-perturbative contributions. In N=4 since there is no one-loop there is no \Lambda and so also the non-perturbative contributions are absent.

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@Zohar I was just wondering if in reference to the original question one might refer to Seiberg's paper http://www.sciencedirect.com/science/article/pii/0370269388912658 Isn't this paper the first non-perturbative argument for the superconformality of N=4 SYM ? And also is the argument for superconformality in N=4 SYM conceptually different from the one needed in this one of my earlier questions - http://physics.stackexchange.com/questions/11438/argument-for-quantum-theoretic-conformality-of-caln-2-super-chern-simons-t

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Yes that's right. The flavor of the argument there seems very similar to what I said. I assume a similar approach can be taken to theories in 3d, or more generally, any theory where the notion of holomorphy makes sense.

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Regard it as an N=1 super Yang-Mills theory with three adjoint chiral superfields, and apply the non-perturbative analysis of Leigh-Strassler.

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answered Oct 27, 2011 by (1,390 points)
It would be nice if you can add a short summary of this method

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