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I saw a claim in this paper that holomorphic boundary CFT$_2$ primary operators correspond to massless states in the AdS$_3$ bulk. Specifically,

As always, we simplify the situation by assuming the absence of holomorphic primary operators. (These would have a little group different from that of a massive particle in the bulk of AdS; therefore for small $\Lambda = −L^{-2}$ they can only correspond to massless states, which do not have a rest frame, or else to states which do not propagate into the bulk of AdS at all.)

My question is: how did he arrive at this conclusion/where can I find an explanation? I can't figure it out, and nowhere near the claim does he give any relevant sources. It's certainly conceivable: holomorphic primary operators will include gauge fields, for example.

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Unitarity of the CFT imposes a lower bound on the conformal dimension $\Delta $ of any operator as is stated in page 32 : $$ \Delta \geq \frac{d-2} 2 $$ My guess is that for this case since $ d=2 $ we have $\Delta_{equality}=0 $ which corresponds to the primary operator and assuming the field is scalar this corresponds to $ m=0 $ in the AdS bulk

Does this help in any way? I'm no expert here.

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The little group is the subgroup of the Lorentz group that leaves an arbitrary four-momentum vector invariant, i.e. for an element of the group $g$ and momentum $V$ we have $gV=V$. This group is in general different for massive and massless particles.

If you now find that the little group of your holomorphic primaries corresponds to that of massive states, you know that you can only have such modes propagating in spacetime.

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