An important aspect of the AdS/CFT correspondence is the recipe to compute correlation functions of a boundary operator $\mathcal{O} $ in terms of the supergravity fields in the interior of the $AdS_{n+1}$ (as we approach the boundary). Namely, $$\big< \exp \int_{\mathbb{S}^n} \mathcal{O} \phi_{0}\big > = Z_{s} \left[\phi \big|_{\partial(AdS)} = \phi_0\right],$$ where $Z_s$ is the supergravity partition function.

The review papers I have found (and Witten's original paper as well) explain how to use the above formula but fail to provide a satisfactory explanation why the formula ought to work, or even how it came about.

Can anyone explain if there is a logical (and/or insightful) path that would lead to the above correspondence between the generating function of the $n$-points correlators and supergravity/string theory?


1 Answer 1


First, you need to reflect what, in fact, is a CFT. The abstract answer is that

It's a set of correlation functions $\langle O_i(x) O_j(y) \cdots O_k(z) \rangle$ which satisfy certain axioms, like the conformal covariance or the short distance behavior when $x\to y$.

These multi-point functions can be encoded in the generating function, so the same set of axioms can be phrased as

It's a functional $\Gamma[\phi_i]= \langle \exp\int \sum_i O_i(x)\phi_i(x) d\,^nx\rangle$ which satisfies certain set of properties.

Now, consider a gravity theory in an asymptotically AdS spacetime, and consider its partition function given the boundary values of $\phi_i$. It gives a functional


This functional automatically satisfies the properties which a CFT generating function satisfies. Conformal covariance comes from the isometry of the AdS, for example. Therefore, abstractly, it is a CFT. (A duck is what quacks like a duck, as a saying goes.)

Now this line of argument does not say why Type IIB on AdS$_5\times$ S$^5$ gives $\mathcal{N}=4$ SYM. For that you need string theory. But everything above this paragraph is just about axiomatics.

So, when there is a consistent theory of gravity on AdS$_{d+1}$ other than string/M-theory, you still get CFT$_d$.

  • $\begingroup$ Thanks for this answer! is there a nice reference for the statement about N=4 SYM? $\endgroup$
    – Reimundo Heluani
    Oct 16, 2011 at 15:32
  • 1
    $\begingroup$ Nothing can beat arxiv.org/abs/hep-th/9905111 . $\endgroup$
    – Yuji
    Oct 17, 2011 at 4:17

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