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 Jul 2 awarded Popular Question May 20 awarded Popular Question May 5 awarded Yearling Apr 18 awarded Nice Question Jan 31 accepted Surface gravity of Kerr black hole Dec 15 accepted 'Easy way' of finding out the Killing vector fields? Oct 15 awarded Autobiographer Oct 15 accepted How would one experimentally prove AdS/CFT correspondence? Jul 30 awarded Popular Question Jul 2 awarded Inquisitive Jul 2 awarded Curious May 5 awarded Yearling Apr 27 comment Have we ever seen an atom? If not, how do we know they exists? Google STM (scanning tunneling microscope) or AFM (atomic force microscopy). And how do we know they exist... well the answer is physics... Feb 16 comment How would one experimentally prove AdS/CFT correspondence? Yeah, but that's only for string theory. AdS/CFT doesn't necessary need string theory in some cases (Kerr/CFT and AdS${}_3$/CFT${}_2$ where you work within the framework of asymptotic symmetries). Feb 16 asked How would one experimentally prove AdS/CFT correspondence? Feb 15 comment Wald General Relativity, Chap 7.1 Well, I think you need to just write it out. (7.1.19) is the metric in ($t,\phi,\rho,z$) basis, and V,X and W are defined on a page before. So you need to calculate (you can always use Mathematica and GRTensor for checking your result) the first equation, and then the second equation, and they should be the same. Feb 7 comment Warped AdS${}_3$ and symmetry breaking Oh, cool :D I was worried for a bit that I did something wrong :D Feb 7 comment Warped AdS${}_3$ and symmetry breaking Really? Because when I compare it to the near horizon extreme Kerr, with the $\Lambda(\theta)=1$ factor, I get the AdS${}_3$, where by just comparing I have $r\to\sinh\omega$ and $\varphi\to\sigma$ :S Plus I solved the Killing equation and got that for warped metric only $2\partial_\sigma$ is a Killing vector, and of all three of $SL(2,\mathbb{R})_R$ are too :S Feb 7 comment Warped AdS${}_3$ and symmetry breaking Thanks for the explanation :) Feb 6 comment Warped AdS${}_3$ and symmetry breaking So it's just putting them in Killing equation and seeing that they're no longer Killing vectors. I was thinking that there is some more profound way than that. How did they know that the warping will break the symmetry of $SL(2,\mathbb{R})$ to $U(1)$? Is it because the $\sigma$ part represents the rotation, and the wrapping factor is in front of that part of the metric which contains it?