# Why are holomorphic boundary CFT2 primary operators massless in the AdS3 bulk?

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.

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
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.