# Dual of the Identity operator (AdS/CFT)

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?

• 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. Nov 29, 2015 at 11:22
• $\Delta_-$ is not an allowed quantisation for a scalar of zero mass in AdS. So this doesn't help! Mar 10, 2016 at 19:26

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 $2 \leq ∆ \leq 3$.

Online notes.