# Does CFT in AdS/CFT live in flat spacetime?

As the title says, does CFT in AdS/CFT live in flat spacetime, or is it only approximately flat?

• The "boundary" of AdS in $d+1$ dimensions can be taken to be any conformally flat $d$-dimensional manifold. See physics.stackexchange.com/q/81131 – Ultima Jan 19 '18 at 18:57
• Conformally flat $\neq$ flat, in case you were wondering. – probably_someone Jan 19 '18 at 18:58
• The conformal boundary of $AdS_n$ is conformally flat. This means it can be mapped to a flat spacetime by a conformal transformation. It does not mean it is flat. – Lawrence B. Crowell Jan 19 '18 at 19:01

## 1 Answer

The CFT in AdS/CFT lives on the asymptotic boundary of the AdS space. The asymptotic boundary is a conformal manifold associated to any "visibility manifold". See this paper of Eberlein and O'Neil (1973), section 1. They define points on the asymptotic boundary as equivalence classes of geodesic rays where geodesic rays are considered equivalent if their distance is asymptotically bounded from above.

One can thus think about the tangent space of the asymptotic boundary to be the normal directions to a geodesic along which nearby geodesics can diverge. This lets one define a conformal structure. There is clearly no natural Riemannian structure on the boundary (and reparametrizations of the geodesics act like Weyl transformations) so if the boundary theory is to be defined in terms of the bulk, it has to be conformally invariant.

However this classical picture breaks slightly when quantizing, and in the end we do have to choose a metric to regulate our theory (this is the case with almost all CFTs). Then we get some mild dependence on the chosen metric through the Weyl anomaly.

• Hm. Yoy can't turn the generic conformally flat metric into the flat one by the conformal transformations - those are diffeomorfisms and won't change tensor quantities like $R$. What you need is the Weyl transformation. When we talk about 2d CFT the Weyl anomaly is proportional to the central charge $\langle T\rangle\sim c R$. So to not distinguish between various conformal metrices we need $c=0$. But when we look at AdS3/CFT2 the SUGRA limit in AdS corresponds to very large $c$ so the Weyl invariance in CFT is actually broken very strongly. Or do I misunderstand something? – OON May 13 at 15:02
• @OON Thanks, it's a good point. I would say the trace anomaly is quite mild compared to true symmetry breaking. One still has Ward identities for a CFT with $c \neq 0$. You can find a description in this physics paper arxiv.org/abs/hep-th/9806087 around eqn (3) "the metric Gˆ µν does not induce a unique metric g(0) on the boundary". Boundary metric comes about by choosing a finite-radius cutoff in AdS. – Ryan Thorngren May 13 at 15:14