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I've read the the answer to this question where it is claimed that the BTZ black hole has a deficit angle. How can we see that in the BTZ case and in general?

VladimirA
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  • The linked answer makes no such claims. BTZ black hole is asymptotically AdS, deficit angle does not makes sense for it. – A.V.S. Feb 01 '22 at 19:09
  • @A.V.S. While I agree that the linked answer does not mention BTZ black holes (let alone whether they have a deficit angle), the singularity in a BTZ black hole is a conical singularity, and one can ascribe a deficit angle to it. – TimRias Feb 02 '22 at 08:23
  • @mmeent: The BHTZ paper explicitly states that their “singularity” at “$r = 0$ is not a conical singularity” (outside of unphysical case of negative mass, which is not a black hole). – A.V.S. Feb 02 '22 at 09:15
  • @A.V.S. That is because $r=0$ is not a singularity at all. The BTZ spacetime can be continued to negative $r$ (where it has a true conical singularity). – TimRias Feb 02 '22 at 09:50
  • @mmeent negative $r$ is the region of CTCs (which is why the authors excise it and call $r=0$ singularity of causal structure) Do you mean then “conical” in the Lorentzian sense (as opposed to 2D Riemmanian)? – A.V.S. Feb 02 '22 at 10:42
  • @A.V.S. Yes, in the Lorentzian sense (or I guess in this case in the locally AdS sense.) – TimRias Feb 02 '22 at 10:48
  • @mmeent But how are you defining the deficit angle on a Lorentzian cone? – A.V.S. Feb 02 '22 at 15:22

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