# AdS/CFT and geometrization of the RG flow

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It is considered that AdS/CFT correspondence is a geometrization of the RG flow.

In a RG flow, knowing the physics at a length scale $z_0$ means that one knows the physics at all length scales $z \geq z_0$

So,  is it possible that there exist a kind of "inside-AdS" correspondence that could be written :

$AdS_{| { z \geq z_0}}$ /  $(AdS \, slice)_{z=z_0}$, so that, in some sense, one recovers the AdS/CFT correspondence in the limit  $z_0 \to 0$.

If it makes sense, does this mean that there are possible relations between the AdS on-shell fields at $z \geq z_0$, and  the AdS  off-shell fields at $z = z_0$?

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This is part of the ordinary AdS/CFT story where we impose a UV cut-off in the CFT. This is important, for example in Ryu and Takayanagi's calculation of the entanglement entropy, which is a UV-divergent quantity.

UPDATE: There is a very interesting new paper that has just appeared about this : http://arxiv.org/pdf/1503.03542v1.pdf . The authors use ideas from AdS/MERA to propose a map from spacelike (spatial) codimension 1 hypersurfaces to states in the boundary conformal field theory. I think the AdS/MERA setting greatly clarifies the role of the RG flow in the correspondence.

answered Sep 6, 2014 by (1,925 points)
edited Mar 17, 2015
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It is confusing on what the $z$ coordinate you use is. Is it simply $z=1/r$ in the Poincare patch? If yes then this is what people do - imposing a regularization $z_0$ and performing holographic renormalization to compensate the infinity of the $AdS$ volume. Now, if you somehow mean on whether there is an "extension" of $AdS$/CFT on the bulk far inside compared to the $AdS$ radius infinity the answer is given to you by the mathematical statement of the correspondence which is exact (though not proven)

$$Z_{QFT} = \int [D\Phi] e^{iS_{ \text{string}}[\Phi] }$$

Note that we usually work on the deepest descend approximation (near $z=0$ or $r\to \infty$) where the SUGRA approximation gives us exact results in the dual theory. This by no means mean that in principle you cannot work in the strongly coupled regime inside the bulk where string theory is ruling, if you know how.

answered Mar 12, 2015 by (3,625 points)
edited Mar 13, 2015

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