Using the (extrapolate) dictionary, one can map a bulk field to a boundary CFT operator. The mapped operator is always a CFT operator? How is it guaranteed?

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    $\begingroup$ What's the extrapolate dictionary? $\endgroup$ – Ryan Thorngren Apr 24 at 12:47
  • $\begingroup$ @RyanThorngren Suppose r is a radial direction in AdS and r to infinity is a boundary. r^\Delta \phi equals to O is what the extrapolate dictionary states, where \phi is a bulk field and O is a boundary CFT operator. \Delta is the scaling dimension of O. $\endgroup$ – Amplituhedron Apr 24 at 12:52
  • $\begingroup$ I am talking about the classical limit of AdS/CFT. $\endgroup$ – Amplituhedron Apr 24 at 12:53
  • $\begingroup$ I think I still don't understand your question. Isn't every field on the boundary a CFT operator? Possibly there are some bulk fields (eg. massive ones) with faster than algebraic decay towards the boundary which map to the zero operator. $\endgroup$ – Ryan Thorngren Apr 24 at 13:13
  • $\begingroup$ @RyanThorngren To be honest, I’m not sure with the concrete definition of CFT operators. All QFT operators seem to be CFT operators (primaries and descendants) in the derivative expansion. But then, even the bulk fields become CFT operators and that is weird, isn’t it? $\endgroup$ – Amplituhedron Apr 24 at 13:43

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