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In this paper https://homes.psd.uchicago.edu/~ejmartin/course/JournalClub/Basic_AdS-CFT_JournalClub.pdf, page 2, the authors state "The boundary of the conformal compactified $AdS_{d+1}$ is identical to the conformal compactification of the $d$-D Minkowski space, i.e., $R \times S_{d-1}$. This provides another motivation for $AdS_{d+1}/CFT_d$ correspondence".

  1. Does that mean that flat Minkowski spacetime is also asympotically AdS? Since they have the same conformal compactification?

  2. How does this provide motivation for the $AdS_{d+1}/CFT_d$ correspondence? How is this related at all?

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    $\begingroup$ Since they have the same conformal compactification? You misunderstood. It is the boundary of c.c. of $AdS_{d+1}$ that is identical to c.c. of $M_d$. $\endgroup$
    – A.V.S.
    Commented Jun 7 at 18:42
  • $\begingroup$ Uh so only the boundary after the conformal compactification is identical. But still, even if the boundary is identical, shouldn't both be asymptotically AdS? Which makes no sense to me. $\endgroup$
    – Βασίλης Γερμανίδης
    Commented Jun 7 at 19:24
  • $\begingroup$ @ΒασίληςΓερμανίδης - What do you think asymptotically AdS means? $\endgroup$
    – Prahar
    Commented Jun 7 at 19:54
  • $\begingroup$ In the paper I was reading, it states "In general, we call a spacetime Asympototically AdS, if it can be conformally compactified into a region which has the same boundary structure as one-half of the Einstein static universe". But I don't quite get that. Isn't normal AdS asymptotically AdS as well? And since the boundary of the conformally compactified AdS is identical to Minkowski, shouldn't Minkowski be asymptotically AdS too? I know it isn't, my logic is obviously flawed, but I don't know in what way. $\endgroup$
    – Βασίλης Γερμανίδης
    Commented Jun 7 at 20:40

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