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The asymptotic symmetry group of $AdS_{d+1}$ is $SO(d,2)$, which just so happens to be the conformal group of $d$-dimensional Minkowski spacetime. Therefore the boundary dual, if it exists, has conformal symmetry. This is the usual "first hint" towards AdS/CFT that one sees at the start of an introduction to the subject.

Does this conformal symmetry of the dual boundary theory get "enhanced" to the Weyl symmetry? It seems that in the literature, the boundary theory is assumed to have not only conformal symmetry but Weyl symmetry, at least up to an anomaly. But I don't see how to justify this based on symmetry considerations alone, since the Weyl group is much larger than $SO(d,2)$. Is there some subtle way to show that the boundary theory has Weyl symmetry?

nodumbquestions
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  • I tried to explain this in my answer to your previous post. I can elaborate if it didn't clarify your doubt. – KP99 Nov 26 '22 at 17:49
  • I don't understand how that answer helps, to be honest. Note that the title of this question should have a yes or no answer. Is the answer yes or no? – nodumbquestions Nov 26 '22 at 18:04
  • No. It seems your doubt has really nothing to do with AdS/CFT correspondence. $SO(d,2)$ is the connected component of isometry group of $AdS_{d+1}$, its just a subset of the asymptotic symmetry (AS) group. Generators of Weyl symmetry forms infinite dimensional abelian lie algebra, while its non-abelian for AS transformation – KP99 Nov 26 '22 at 18:28
  • I understand what you're saying about the bulk side, I think. But I'm surprised by the fact that the boundary CFT doesn't have Weyl symmetry. Weyl symmetry seems to always be assumed in the literature, up to an anomaly. I guess my worry would go away if a statement of the form "conformal => Weyl up to an anomaly" were true. – nodumbquestions Nov 26 '22 at 19:00
  • Since you mentioned asymptotic symmetry group, which involves point-wise transformation, it can't be Weyl transformation by definition. What you are instead referring to is "conformal mapping" , which is a point-wise transformation inducing a local rescaling of metric. I'm not yet familiar with conformal anomaly, so I can't tell if you will need to be worry about this. – KP99 Nov 26 '22 at 19:31
  • Regarding the ambiguity b/w weyl and conformal transformation, this should help – KP99 Nov 26 '22 at 19:42
  • Ah I just realised my question was poorly phrased: the second paragraph was meant to refer to the CFT (not the bulk). I will rephrase it - apologies. Note that my revised question is not answered by what you've written here or elsewhere. – nodumbquestions Nov 26 '22 at 19:54
  • Prahar's comment in https://physics.stackexchange.com/questions/738764/conformal-invariance-and-tracelessness-of-the-energy-momentum-tensor-contradict answered this question. To refute it you had to say that Weyl rescaling not being a symmetry of a CFT wa a matter of semantics which is not true. – Connor Behan Dec 01 '22 at 15:25
  • A CFT is an object where the metric is non-dynamical in the strong sense. Not only does it lack a kinetic term... it is not even included in the path integral so no transformation can act on it. The diff+Weyl invariant object which arises in AdS / CFT is not a CFT yet, just a stepping stone for getting one. – Connor Behan Dec 01 '22 at 15:27
  • Ok my bad, I shouldn't have called Weyl invariance a "symmetry". But it still makes perfect sense to ask whether correlators are invariant under rescaling $g\to\Omega^2 g$. Call this property W. Likewise, let C be the property that when the theory is put on Minkowski space, correlators are invariant under conformal maps. IHopefully we agree that W implies C but not the other way round. Clearly C applies to AdS/CFT; my question is whether W applies, and why. – nodumbquestions Dec 02 '22 at 17:06
  • Maybe I should delete this question and rephrase in terms of W and C to avoid confusion. – nodumbquestions Dec 02 '22 at 17:08

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