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  How to show BTZ is a quotient of AdS3?

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39 views

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 in Q&A by Sam Makhoul (10 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 
$$\tilde{AdS}_{3}/\Gamma$$

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.

$$BTZ=\tilde{AdS}_{3}/\mathbb{Z}$$

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.

$$\tilde{AdS}_{3}=\mathbb{H}^{2}\times\mathbb{R}$$ 

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 by New Student (120 points) [ revision history ]
edited Jul 7 by New Student

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