A naive question about holographic dual of a massive QFT: The Ryu-Takanayagi formula for the entanglement entropy (see their paper) seem to suggest that the holographic dual of a massive QFT (e.g. a CFT, which has a good holographic dual, perturbed by a relevant perturbation) is a "capped off" AdS, so I'd like to know what the precise meaning of "capped off" AdS is? For example, a concrete metric?

Any comments or references are welcome, thank you.


At large energies, the relevant operator does not affect the field theory, so the UV region of the dual geometry of the perturbed CFT (i.e. close to the boundary) is simply AdS. However, when following the RG flow, at some scale the effects of the perturbation set in, this is where the geometry is deformed.

For a mass perturbed CFT below the mass scale, the mass gap prevents pair creation and the field theory "freezes out". This can be (roughly) modelled via AdS/CFT by capping of the holographic coordinate (and thus the spacetime) and restrict it to some range $0<z<z_0$, as everything that happens beyond $z_0$ cannot affect the mass deformed theory.

The entanglement entropy is of course altered by the deformation as pair produced particles are ultimately responsible for this entropy. The Ryu-Takayanagi formula $$ S=\frac{A}{4 G_N} $$ can capture this reduction of entanglement by cutting off the bulk homologous surfaces (ending on the entangling region on on the boundary) and thus reducing the surface area.

Strictly speaking, capping off is not well defined. There are however related (and well-defined) concepts like soft-wall models.

  • $\begingroup$ Thanks, one more question related: if we assume there is a space termination at $z_0$, in order to satisfy the Einstein equation, do we need a boundary condition, say, a massive membrane at $z=z_0$? How to understand that in holography? What's the dual object in QFT? $\endgroup$ – Yingfei Gu Jun 25 '15 at 16:34
  • $\begingroup$ I don't think people usually put boundary conditions such that the Einstein equations are satisfied. One simply takes a solution of Einstein's equations without cap off and caps off afterwards resulting in a spacetime, that obviously does not satisfy Einstein's equations. This is why I said the process is not well-defined. One only sets boundary conditions for the non-gravitational field, namely other fields have to vanish at $z_0$. $\endgroup$ – physicus Jun 25 '15 at 16:44

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