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I have been reading Thomas Hartman's lecture notes on Quantum Gravity and Black Holes.


In page 97, he derives (9.4), which is the metric of AdS$_{3}$ in global coordinates:

$$ds^{2} = \ell^{2}(-\cosh^{2}\rho\ dt^{2} + d\rho^{2} + \sinh^{2}\rho\ d\phi^{2}).$$


In page 100, he states that, expanding the metric at large $r$ under the coordinate change \begin{align} t^{\pm} = t \pm \phi, \qquad \rho = \log(2r), \end{align} we can show that, to leading order, the induced metric on the hyperboloid AdS$_{3}$ becomes \begin{align} ds^{2} = \ell^{2} \left(\frac{dr^{2}}{r^{2}}-r^{2}dt^{+}dt^{-}\right). \end{align}


I find, under the coordinate change, that \begin{align} ds^{2} = \ell^{2} \left(\frac{dr^{2}}{r^{2}} - \frac{1}{4}dt^{+2} - \left(r^{2} + \frac{1}{16r^{2}} \right) dt^{+}dt^{-} - \frac{1}{4}dt^{-2} \right). \end{align}

Of course, the term in $1/r^{2}$ drops off at large $r$, but I am not able to get rid of the components in $dt^{+2}$ and $dt^{-2}$. Am I missing something here?


He then goes on to mention that these are Poincare coordinates, but does not make contact with the usual way in which the metric of AdS$_{3}$ is written in Poincare coordinates:

$$ds^{2} = \frac{\ell^{2}}{z^{2}}(dz^{2}-dt^{2}+d\vec{x}^{2}).$$

What am I missing here?

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    $\begingroup$ You could try appendix B of insti.physics.sunysb.edu/~cpherzog/stringtheory/AdSCFT.pdf. In answer to your first question, probably he is saying that $dt_\pm^2$ is small compared to $r^2 dt_+ dt_-$. In answer to the second, there should be relations $r \sim 1/z$ and $t_\pm = t \pm x$. $\endgroup$ – user2309840 Apr 26 '17 at 2:44
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It is likely that he cut the terms lower than $O(r)$ (except the $dr^2$ term), making your $dt^\pm$ terms, $O(1)$, vanish. I have derived a slightly different result but with a common result. I used the transformation \begin{equation} dt^2=(dt^+-d\phi)(dt^--\phi)=dt^+dt^-+d\phi^2 \end{equation} which brings me to the metric \begin{equation} ds^2=l^2\left(\frac{dr^2}{r^2}-\left(r^2+\frac{1}{16r^2}+\frac{1}{2}\right)dt^+dt^--d\phi^2\right). \end{equation} Cutting the small terms below $O(r)$ will lead you to the right answer. This is indeed in the form of Poincaré coordinate: \begin{equation} ds^2=\alpha^2\left(\frac{du^2}{u^2}+u^2(dx^\mu dx_\mu)\right). \end{equation}

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