Flat space limit of the $AdS$ metric: Very basic question

Question

Suppose I am given the following global coordinates in empty $AdS_n$:

$$ds^2 = \alpha^2\left(-\cosh^2\rho \, d\tau^2 + \, d\rho^2 + \sinh^2\rho \, d\Omega_{n-2}^2\right)$$

where the length scale $\alpha$ is related to the cosmological constant as $\Lambda = \frac{-(n - 1)(n - 2)}{2\alpha^2}$. I had two very basic questions:

1. In the limit $\alpha \to \infty$, the cosmological constant tends to 0. Is it correct to conclude that we thus obtain a flat space limit of $AdS_n$?
2. In that case, how do we obtain the standard Minkowski metric from the above metric using $\alpha \to \infty$?

Answer 1

Far far away from the boundary, AdS looks flat. We can see this by noting that AdS can be thought of as a spacetime with a "background potential" $V(\rho) \sim \rho^2/\alpha^2$ so the region in the neighborhood of $\rho=0$ is flat. As we move farther away from this point, the potential of AdS becomes stronger and the effects of curvature become more and more important.

OK, so in order to take the flat space limit, we must take a limit which allows us to zoom into this region. We do this by setting $r=\rho \alpha$, $t=\tau\alpha$ and then take $\alpha \to \infty$ keeping $r$ and $t$ fixed. In this limit we are simultaneously taking $\rho \to 0$ and $\tau \to 0$ which is the region we wish to zoom into. Note that $\tau$ has to be similarly rescaled so that we are sitting far away from ALL boundaries of AdS, not just the timelike boundary.

Doing this, we find \begin{align} ds^2 &= \alpha^2 \left[ - \cosh^2 \rho d\tau^2 + d\rho^2 + \sinh^2 \rho d\Omega_{n-2}^2 \right] \\ &= \alpha^2 \left[ - \alpha^{-2} \cosh^2 (r/\alpha) dt^2 + \alpha^{-2} dr^2 + \sinh^2 (r/\alpha) d\Omega_{n-2}^2 \right] \\ &\stackrel{\alpha\to\infty}{\to} - dt^2 + dr^2 + r^2 d\Omega_{n-2}^2. \end{align}