We know that in a CFT the spectrum of gauge invariant operators must contain an Identity operator (for the operator algebra to close). For those CFTs that admit a holographic dual what does the Identity operator correspond to in the bulk?
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1$\begingroup$ The Ads/CFT dictionary says that for scalar particles $\Delta_{\pm}=\frac12 (d\pm \sqrt{d^2+4m})$ so for the identity operator we would need $\Delta=0$ which is only possible for $\Delta_{-}$ when the mass is zero. $\endgroup$– PrasttNov 29, 2015 at 11:22
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$\begingroup$ $\Delta_-$ is not an allowed quantisation for a scalar of zero mass in AdS. So this doesn't help! $\endgroup$– OrbifoldMar 10, 2016 at 19:26
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The answer is that it is a mystery as far as I know.
I am giving two references in the end; in the first link you can find lecture notes -practically the same argument given there is pasted here in case one does not want to go through the notes- and in the second link it is the Klebanov-Witten paper on the matter.
Here is what is known and what is not known.
First of all, let's write down the $AdS_5$ mass, setting the $AdS$ radius to one
$$m^2 = \Delta(\Delta - 4)$$
If $\Delta \geq 4$ then $m^2 \geq 0$.
If $\Delta < 4$ then the $AdS$ mass squared (henceforth mass) can be negative, but the scalars are not tachyons as long as they do not violate the Breitenlohner-Freedman bound, $m^2 \geq - 4$.
As you mentioned, unitarity bound requires $\Delta \geq 1$. Using masses greater than $−4$ we can obtain all operators with $\Delta \geq 2$.
Case $1 \leq \Delta <2$: $\Delta$ is the largest solution to the equation of the $AdS$ mass, as you pointed out. This because typically only the largest solution is greater than the unitary bound. However, precisely for $−4 \leq m^2 \leq −3$, the equation above admits two different solutions both satisfying the unitary bound; one with $1 \leq ∆ \leq 2$ and another with $2 \leq ∆ \leq 3$. In turn, one then has two different choices for imposing boundary conditions: they amount to choice $\phi_0$ or $\phi_1$ as boundary value of the bulk field. These two different choices lead to correlation functions for two different operators one with $1 \leq ∆ \leq 2$ and another with $2 \leq ∆ \leq 3$.
Sources for further reading:
AdS/CFT Correspondence and Symmetry Breaking
Cheers!!!
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$\begingroup$ Why isn't the answer the obvious "the AdS vacuum"? $\endgroup$ Feb 3, 2018 at 0:08
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$\begingroup$ How can you make that conclusion? Is it something trivial that I am missing? $\endgroup$– user172341Feb 3, 2018 at 0:11
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2$\begingroup$ Well, by operator-state correspondence the identity operator creates the CFT vacuum, which by AdS/CFT is the same as AdS vacuum. If you don't use operator-state correspondence, the question becomes rather empty, since then you would try to interpret it in terms of correlation functions, but of course the identity operator insertions do not affect the correlation functions, so you could say that it is dual to nothing. $\endgroup$ Feb 3, 2018 at 9:57
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$\begingroup$ For a 2-dimensional CFT I agree with you, but I am not sure if anything changes in higher dimensions. $\endgroup$– user172341Feb 3, 2018 at 18:49
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