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The AdS/CFT correspondence states that string theory in an asymptotically anti-De Sitter spacetime can be exactly described as a CFT on the boundary of this spacetime.

Is the converse true? Does any CFT in a suitable number of spacetime dimensions have an AdS/CFT dual? If no, can we characterize the CFTs which have such a dual?

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2 Answers 2

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The answer is not known, but many believe it is: "Yes, every CFT has an AdS dual." However, whether the AdS dual is weakly-coupled and has low curvature -- in other words whether it's easy to do calculations with it -- is a different question entirely. We expect, based on well-understood examples (like $\mathcal N=4$ SYM dual to Type IIB strings on $\mathrm{AdS}_5 \times S^5$), that the following is true:

  • For the AdS dual to be weakly-coupled, the CFT must have a large gauge group.
  • For the AdS curvature scale to be small (so that effective field theory is a good approximation), the CFT must be strongly-coupled. In well-understood examples, the CFT has an exactly marginal coupling which when taken to infinity decouples stringy states from the bulk spectrum. By contrast, at weak CFT coupling, the AdS dual description would involve an infinite number of fields and standard EFT methods would not apply. (This doesn't necessarily mean calculations are impossible: we would just need to better understand string theories in AdS -- something which is actively being worked on.)

As far as I know, appropriate conditions for CFTs without exactly marginal couplings to have good AdS EFTs are not known. Also, well-understood AdS/CFT dual pairs where the CFT violates one or both of the above conditions are scarce.

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Welcome David. This is a good answer, but perhaps the criteria could be phrased in CFT language, not assuming auxiliary structures like gauge invariance. The first criterion is the CFT having a large central charge, the second is the existence of a gap in the spectrum of conformal dimensions, so there are few operators with dimensions of order one, and most operators have large dimension of order $\sqrt{\lambda}$. – user566 Nov 6 '11 at 0:23
You're right -- those are better ways to state the above conditions. – davidsd Nov 6 '11 at 17:20
Regarding weak coupling, it seems like what one really wants is that the CFT is "almost factorized" -- that is, there is a notion of single-trace and multi-trace operators, and a small parameter that controls deviations from mean field theory. A large central charge reflects this, but does it imply it? – davidsd Nov 6 '11 at 17:31
It depends if you are shooting for a weakly coupled string dual, or just a gravity dual. In my mind the specifics of large N counting has to do with the former, where you can think about large N as the classical limit, and large $\lambda$ as the low curvature limit. For an M-theory dual you have just have one parameter (usually some flux) controlling the size of the geometry. It must be that large flux gives factorization, or clustering (maybe by virtue of having a gap in the spectrum of conformal dimensions) but is unrelated to deviations from mean field theory (which are never small). – user566 Nov 6 '11 at 17:48

A recent work on this:

I hope davidsd or Moshe can clarify on what they meant by 'mean field theory' (and deviations from it) in large-N CFT.

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This is an old question, but since someone edited it, I'll take the opportunity to comment. 'Mean field theory' in CFT means that all operators factorize through Wick's theorem. (In other words, the theory is "almost free" or "almost Gaussian".) We know that this is true for free fields (i.e of dimension 1 in 4D), but you can define a CFT by choosing some field dimension $[\varphi] = \Delta$ and letting all correlation functions be exactly those predicted by Wick's theorem. – Vibert Jan 16 '13 at 11:57

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