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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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    $\begingroup$ Regard it as an N=1 super Yang-Mills theory with three adjoint chiral superfields, and apply the non-perturbative analysis of Leigh-Strassler. $\endgroup$ – Yuji Oct 27 '11 at 14:39
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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 $\mathcal{N}=4$ the one-loop coefficient is zero. This is trivial (just counting the fields). Hence, the gauge coupling (as defined above) does not run. But $\mathcal{N}=4$ relates the gauge particles with the chiral superfields (all the matter particles sit in one big representation of $\mathcal{N}=4$) and so the latter cannot get renormalized either.

This is a slick and intuitive argument... Similar logic operates in many $\mathcal{N}=2$ theories as well.

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  • $\begingroup$ 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? $\endgroup$ – Squark Nov 2 '11 at 20:28
  • $\begingroup$ 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 :) $\endgroup$ – Zohar Ko Nov 2 '11 at 20:53
  • $\begingroup$ Your argument is definitely nicer :) $\endgroup$ – Yuji Nov 3 '11 at 14:20
  • $\begingroup$ 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. $\endgroup$ – Zohar Ko Nov 3 '11 at 15:18
  • $\begingroup$ Ok, so can you provide a ref to the nonrenormalization theorem you are using? $\endgroup$ – Squark Nov 3 '11 at 20:26

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