# Does asymptotically AdS mean as $z \to 0$ or as $z \to \infty$ in Poincare metric of AdS?

The Poincare metric of AdS_3 is given by $ds^2 = \frac{R^2}{z^2}(dz^2 - dx_0^2 + dx_1^2)$. Using the coordinate transformation $\rho = \log(z)$, we can write this as, $ds^2 = R^2 (d\rho^2 + e^{-2 \rho} (-dx_0^2 + dx_1^2))$.

Now if I want to show that some other space time is asymptotically AdS, I should expect the metric to agree in the limit $\rho \to -\infty$ which is the limit $z \to 0$. Is that right?

(Nb: I am trying to show that BTZ metric is asymptotically AdS).

## 1 Answer

Now if I want to show that some other space time is asymptotically AdS, I should expect the metric to agree in the limit $\rho \to -\infty$ which is the limit $z\to 0$. Is that right?

Yes, this is correct. In this limit the leadin order of the metric components should agree with the pure AdS metric. There is however a more mathematically sound way of approaching the problem. One basically has to require that the asymptotic symmetry group is still the conformal group $O(2,d-1)$, which is the isometry group of AdS$_d$. The procedure is laid out in this paper by Henneaux and Teitelbom, where the focus is on AdS$_4$.

(Nb: I am trying to show that BTZ metric is asymptotically AdS).

Maybe, you want to have a look at the following physics.SE post.