On page 131 of these notes, a precise formulation of the AdS/CFT correspondence is given by the GKPW dictionary

$$Z_{\text{grav}}[\phi_{0}^{i};\partial M] = \langle \exp \left( - \frac{1}{\hbar} \sum_{i} \int d^{d}x\ \phi_{0}^{i}(x)O^{i}(x) \right) \rangle_{\text{CFT on } \partial M}.$$

The index $i$ is said to run over all the light fields in the bulk effective field theory, and correspondingly over all the low-dimension local operators in CFT.

Equation (14.4) of the notes goes on to mention that the mass $m$ of the bulk scalar is related to the scaling dimension $\Delta$ of the CFT operator by

$$m^{2} = \Delta (d-\Delta) \qquad \qquad \Delta = \frac{d}{2} + \sqrt{\frac{d^{2}}{4}+m^{2}\ell^{2}}.$$

Primary question:

If you have a precise relation between the mass of the bulk scalar and the scaling dimension of the CFT operator, why would you then want to restrict yourself to light bulk fields and low-dimension CFT operators in the GKPW dictionary of the AdS/CFT correspondence?

Secondary question (skip this if you wish):

Does the consideration of light bulk scalar fields necessarily imply that we have reduced the quantum theory of gravity to a low-energy effective field theory?


1 Answer 1


Primary question:

You are in an $AdS_{d+1}$ space. Take $d$ to be 4 as an example.

If you raise the conformal/scaling dimension of the CFT operator too much you will get -eventually- a negative $m^2$. Now, there is the so-called Breitenlohner-Freedman bound which states that even if $m^2 < 0$ the states can be stable and thus the theory is free of tachyons. If you violate this limit, then you have tachyons in your theory, which is something you do not want.

Practically, the precise relation is to set bounds, calculate, and check what kind of states there are in the theory.

However, the way I have seen the relation is actually the other way around and this is a bit confusing... In any case, the same arguments regarding the BF-bound hold.

This is what I know.

$$m^2 = \Delta(\Delta - d)$$

where I set the $AdS$ radius to one.

$m^2 \geq 0$ only for $\Delta \geq 4$. There are certainly theories where $\Delta < 4$; the unitary bound is $∆ \geq 1$. Operators with $\Delta < 4$ correspond to fields with negative mass in $AdS_5$. However they are not tachyons, as explained above, since the energy is positive as long as the Breitenlohner-Freedman bound $m^2 \geq −4$ is satisfied. The curvature gives a positive contribution to the energy of a scalar field propagating in $AdS$. The minimal value for such a thing is $\Delta = 2$.

The notes for further study: http://laces.web.cern.ch/Laces/LACES09/notes/dbranes/lezioniLosanna.pdf



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