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

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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Regard it as an N=1 super Yang-Mills theory with three adjoint chiral superfields, and apply the non-perturbative analysis of Leigh-Strassler. – Yuji Oct 27 '11 at 14:39

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.