# 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

• 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 '19 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 '19 at 15:14