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Bulk-boundary cutoffs in AdS/CFT

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I'm studying the holographic entanglement entropy (HEE) in this paper (Ryu-Takayanagi, 2006). In section 6.3 they compute the HEE for a segment in a 2D CFT. To do so, they obtain the corresponding geodesic in the bulk (in the Poincaré patch) and compute its length.

I understand all that process, but I'm having some trouble when they introduce the cutoff. The metric diverges when $z\to0$ so we introduce a cutoff $\epsilon>0$, I understand that. But then they say

Since $e^\rho\sim x^i/z$ near the boundary, we find $z\sim a$

Here, $\rho$ is the hiperbolic radial coordinate un the global coordinates for AdS,

$$ ds^2 = R^2(-\cosh^2\rho\ d\tau^2 + d\rho^2 + \sinh^2\rho\ d\Omega^2) $$

$x^i$ and $z$ are coordinates in the Poincaré patch,

$$ ds^2 = \frac{R^2}{z^2}(dz^2-dt^2+\sum_i(dx^i)^2) $$

And $a$ is the inverse of the UV cutoff of the CFT in the boundary, that is, the spacing between sites.

I have two problems:

1) First, I don't see why near the boundary $e^\rho\sim x^i/z$. I made up the relations between both coordinate systems and I find more complicated relations than that (even setting $z\sim0$).

2) Even assuming the previous point, I don't understand why we obtain that relation between the CFT and the $z$ cutoff.

This post imported from StackExchange Physics at 2014-09-02 07:57 (UCT), posted by SE-user David Pravos
asked Sep 1, 2014 in Theoretical Physics by David Pravos (35 points) [ no revision ]

1 Answer

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  1. To see why the relation has to hold, you have to acknowledge that it needs to transform the line elements you have written down into each other. One can see that this is true by taking the logarithm on both sides, which yields $$\rho=\log{x^i}-\log{z}.$$ Taking the derivative of this expression and squaring it gives $\mathrm{d}\rho^2=\mathrm{d}z^2/z^2$, which clearly transforms between the terms in the metric covering the holographic direction.
  2. As is explained in section 6.1, at the cutoff position $\rho_0$, which is close to the boundary, we have the relation $$\exp{\rho_0}\sim \frac{L}{a},$$ which implies $z\sim a$.
This post imported from StackExchange Physics at 2014-09-02 07:57 (UCT), posted by SE-user Frederic Brünner
answered Sep 1, 2014 by Frederic Brünner (1,060 points) [ no revision ]

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