Boundary stress-energy tensor form ADS/CFT

Question

In this article (http://arxiv.org/abs/1308.3716) authors derive linearised Einstein equations from ADS/CFT. At page 6 they use expression (1) $t_{\mu{\nu}}(x)=\frac{d}{16{\pi}G_{N}}H_{\mu\nu}$, where $t_{\mu{\nu}}(x)$ is stress-energy tensor on the boundary and $H_{\mu\nu}$ comes from perturbed ADS metric given by $ds^2=\frac{1}{z^{2}}(dz^{2}+dx_{\mu}dx^{\mu}+z^{d}H_{\mu\nu}(x,z)dx^{\mu}dx^{\nu})$
In article http://arxiv.org/pdf/hep-th/9903203.pdf the procedure to obtain (1) is suggested(page 1). One needs to follow $ADS_{d+1}/CFT_{d}$ prescription: $e^{-I_{Ads}[\phi]}=<\int{dx^{d}\phi_{0}(x)O(x)}>_{CFT}$
Yet in this case instead of O(x) we consider $t^{\mu\nu}$ and instead of $\phi_{0}(x)$ will be $h_{\mu\nu}$ - graviton field (metric perturbation).
At this point I have a problem. From what I know, Ads action in this prescription should be evaluated on shell. In case of graviton field we will have $h_{\mu\nu}$ satisfying linearised Einstein Equations in the bulk with some boundary conditions. But we can't use these equations because we are trying to prove them in the article. How to resolve this contradiction?
If there is another way to obtain (1) I will be glad to see it.

This post imported from StackExchange Physics at 2015-11-06 15:45 (UTC), posted by SE-user Yaroslav Shustrov

Answer 1

Your expression (1) can be taken as the statement of the correspondence, this is called the "extrapolate" dictionary. Namely if I have some bulk field $\phi$, then the CFT operator dual to $\phi$ can be obtained by pulling $\phi$ to the boundary of AdS and stripping off suitable factors of the radial coordinate. In your case the bulk field is the metric and the dual operator is the stress tensor. To see that the dual operator is actually the stress tensor, note that the mass of the graviton is protected by gauge invariance so the dual operator with spin 2 should have dimension exactly equal to d. Generically the only such operator is the stress tensor.

See Harlow and Stanford for the proof of the equivalence to the other form of the dictionary, http://arxiv.org/abs/1104.2621.