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

+ 4 like - 0 dislike
51 views

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.

This post imported from StackExchange Physics at 2014-06-13 12:28 (UCT), posted by SE-user wittensdog
asked Apr 3, 2014 in Theoretical Physics by wittensdog (40 points) [ no revision ]
Minor comment to the post (v2): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0902.2790

This post imported from StackExchange Physics at 2014-06-13 12:28 (UCT), posted by SE-user Qmechanic

1 Answer

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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.

This post imported from StackExchange Physics at 2014-06-13 12:28 (UCT), posted by SE-user atabler
answered May 7, 2014 by atabler (0 points) [ no revision ]

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