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  How to show BTZ is a quotient of AdS3?

+ 0 like - 0 dislike

Maybe I'm missing something obvious, but how do you show explicitly that the BTZ metric

\(ds^2 = - (r^2 - 8 M) dt^2 + \frac{dt^2}{r^2-8M} + r^2 d \phi^2\)

is a quotient of the Poincare patch of \(AdS_3\)

\(ds^2 = -r^2 dt^2 + \frac{dr^2}{r^2} + r^2 d \phi^2\)

To me, it just looks like BTZ is asymptotically \(AdS_3\). Is the radial coordinate in BTZ not identified in \(AdS_3\)?

More generally, if that is the solution, why is this called quotienting and not compactifying?

asked Jul 6, 2018 in Q&A by Sam Makhoul (25 points) [ no revision ]

1 Answer

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In 2+1 dimensions, there is no propagating degree of freedom. The only way to construct different solutions is by taking quotient. i.e. considering the quotient 

where the tilde means the universal covering space, and $\Gamma$ is a discrete subgroup of the isometry group of $\tilde{AdS}_{3}$. 

For example, in 2 dimensions, you can have $\mathbb{C}$, and quotient spaces like cylinder $\mathbb{C}/\mathbb{Z}$ and torus $\mathbb{C}/\mathbb{Z}\oplus\mathbb{Z}$. 

For $AdS_{3}$, its more complicated. You can take quotient and construct the $BTZ$ black hole, and even worm holes that may or may not be asymptotic $AdS$. The $AdS_{3}$ has constant negative curvature, and can be viewed as the group manifold $SO(2,2)/SO(1,2)$, which is isomorphic to $SL(2,\mathbb{R})$. This manifold has no topological boundary since it's not compact. But you can consider its conformal compactification and visualize it as a solid torus with infinite radius. Its universal covering space is then viewed as a solid cylinder. Its isometry group is $SO(2,2)$. 

Now, the $BTZ$ black hole is then constructed by finding some discrete subgroup of $SO(2,2)$, say $\Gamma$, and take the quotient $\tilde{AdS}_{3}/\Gamma$. It was shown by Dieter R. Brill, Stefan Aminneborg, Ingemar Bengtsson, Soren Holst, Peter Peldan and Kirill Krasnov in gr-qc/9912079v1, gr-qc/9904083v2, gr-qc/9607026v1, gr-qc/9707036v1, hep-th/0005106v2 that this discrete subgroup of isometry is the integer number group $\mathbb{Z}$. i.e.


The idea is roughly as follows. The universal covering of $AdS_{3}$ is really a product space of a two dimensional hyperbolic manifold times the real line. i.e.


where $\mathbb{H}^{2}$ is the Poincare disk, and $\mathbb{R}$ is time. The quotient is then taken by considering the cyclic Fuchsian group of the Mobius transformation acting on the Poincare disk. This is well-explained in the Master thesis by Balt van Rees "Wormholes in 2+1 Dimensions".

What you consider in the $BTZ$ case is the so-called hyperbolic type of Mobius transformation. More specifically, such a Mobius transformation is an $PSL(2,\mathbb{R})$ element whose absolute value of trace is larger than 2. If we consider the cyclic group generated by such a transformation, then it can be shown that this group is isomorphic to $\mathbb{Z}$. The spatial slice of a $BTZ$ black hole is then the fundamental domain of the $\mathbb{Z}$ action on the Poincare disk. It has two asymptotic boundaries at infinity. However, such quotient disk will be singular at $t=-\pi/2$ and at $t=\pi/2$ in global $AdS$ coordinates. These two singularities are the $BTZ$ singularities where space and time terminate. You can even draw a vivid picture of the $BTZ$ black hole living inside the $AdS_{3}$. It looks like a diamond. 

answered Jul 6, 2018 by Libertarian Feudalist Bot (270 points) [ revision history ]
edited Jul 6, 2018 by Libertarian Feudalist Bot

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