What evidence do we have for S-duality in N=4 Super-Yang-Mills?

Do we have anything resembling a proof*? Or is it just a collection of "coincidences"?

Also, do we have evidence from lattice gauge theory computations?

*Of course I'm not talking about a proof in the mathematical sense, only a compelling albeit heuristic argument e.g. an argument involving formal manipulations with path integrals.

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Just a quick feedback: I doubt there is something resembling a proof in a mathematical sense of anything at all for a fully fledged QFT (which has infinitely many states and divergences -- leading to non-trivial critical exponents), so I am not sure this is productive. Asking for evidence seems interesting though a bit broad, maybe there is a place to add some motivation. I am also not sure by what you mean by “numerical” and why you single out such evidence. – user566 Oct 26 '11 at 20:12
Thanks, much better. – user566 Oct 26 '11 at 20:59

1 Answer

There is a "proof" of S-duality in free abelian gauge theory. See, for example, Witten's On S-Duality in Abelian Gauge Theory or his IAS lecture. Morally, you should think of this as a Fourier transform. The non-Abelian version, of course, is much harder and requires N=4 SUSY. But this isn't a surprise as the non-Abelian Fourier transform is a tricky thing.

Now, at least from the math side, you should be thinking about Geometric Langlands which is pretty closely related and has the same distinction between the Abelian case (which I understand to reduce to some class field theory stuff) and the more difficult non-Abelian version. Long before Kapustin-Witten and all that has come after, this connection was first shown in Harvey, Moore and Strominger and Bershadsky, Johansen, Sadov and Vafa.

There is, of course, a long, long list of "coincidences", too.

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Geometric Langlands is very close to being proved nowadays. However, it only has to do with the topological part of the gauge theory - I don't think there is a mathematical approach to non-supersymmetric observables. – Alexander Braverman Oct 27 '11 at 3:30
Definitely (and other topological twists also evince S-duality, i.e., Vafa and Witten's hep-th/9408074). The point was to draw the analogy between the Abelian and non-Abelian Fourier transform as the desired heuristic. – Aaron Oct 27 '11 at 11:42
Welcome, Aaron! – user566 Oct 28 '11 at 17:38
I must be thoroughly confused. Isn't the Abelian case just ordinary electric-magnetic duality? If so, it just acts as the Hodge star on the field strength tensor. How does Fourier transform enter the picture? – Squark Oct 28 '11 at 18:11
Geometric Langlands is, in large parts, proved. Langlands for number fields is much harder. – Aaron Nov 10 '11 at 17:52