# AdS boundary global vs Poincare'

Is the global boundary of AdS the same of the boundary written in Poincare' coordinates?

• It would appear not – John Rennie May 10 '16 at 10:49
• Not sure what you're asking here. For example they are not even topologically equivalent, one is compact the other is not. One could be said to be a compactification of another, the infinity point on the boundary of half plane hyperbolic space is a single point on the boundary of poincare disk, which is a limit point of the rest of the image of the plane-to-disk map. – zzz May 17 '16 at 5:44

Here's what is written on wikipedia (it may be helpful):

Global coordinate:

$AdS _{n}$ is parametrized in global coordinates by the parameters $(\tau ,\rho ,\theta ,\varphi _{1},\cdots ,\varphi _{n-3}$) as:

\begin{cases}X_{1}=\alpha \cosh \rho \cos \tau \\X_{2}=\alpha \cosh \rho \sin \tau \\X_{i}=\alpha \sinh \rho \,{\hat {x}}_{i}\qquad \sum _{i}{\hat {x}}_{i}^{2}=1\end{cases} where $\hat {x}_{i}$ parametrize a $S^{n-2}$ sphere. i.e. we have $\hat {x}_{1}=\sin \theta \sin \varphi _{1}\dots \sin \varphi _{n-3}$, $\hat {x}_{2}=\sin \theta \sin \varphi _{1}\dots \cos \varphi _{n-3}$ etc. The ${AdS} _{n}$ metric in these coordinates is:

$ds^{2}=\alpha ^{2}(-\cosh ^{2}\rho \,d\tau ^{2}+\,d\rho ^{2}+\sinh ^{2}\rho \,d\Omega _{n-2}^{2})$ where $\tau \in [0,2\pi ]$ and $\rho \in \mathbb {R} ^{+}$. Considering the periodicity of time $\tau$ and in order to avoid closed timelike curves (CTC), one should take the universal cover $\tau \in \mathbb {R}$. In the limit $\rho \to \infty$ one can approach to the boundary of this spacetime usually called ${AdS} _{n}$ conformal boundary.

With the transformations $r\equiv \alpha \sinh \rho$ and $t\equiv \alpha \tau$ we can have the usual ${AdS} _{n}$ metric in global coordinates:

$ds^{2}=-f(r)\,dt^{2}+{\frac {1}{f(r)}}\,dr^{2}+r^{2}\,d\Omega _{n-2}^{2}$ where $f(r)=1+{\frac {r^{2}}{\alpha ^{2}}}$

Poincaré coordinates:

By the following parametrization:

\begin{cases}X_{1}={\frac {\alpha ^{2}}{2r}}(1+{\frac {r^{2}}{\alpha ^{4}}}(\alpha ^{2}+{\vec {x}}^{2}-t^{2}))\\X_{2}={\frac {r}{\alpha }}t\\X_{i}={\frac {r}{\alpha }}x_{i}\qquad i\in \{3,\cdots ,n\}\\X_{n+1}={\frac {\alpha ^{2}}{2r}}(1-{\frac {r^{2}}{\alpha ^{4}}}(\alpha ^{2}-{\vec {x}}^{2}+t^{2}))\end{cases} the ${AdS} _{n}$ metric in the Poincaré coordinates is:

$ds^{2}=-{\frac {r^{2}}{\alpha ^{2}}}\,dt^{2}+{\frac {\alpha ^{2}}{r^{2}}}\,dr^{2}+{\frac {r^{2}}{\alpha ^{2}}}\,d{\vec {x}}^{2}$ in which $0\leq r$. The codimension 2 surface $r=0$ is Poincaré Killing horizon and$r\to \infty$ approaches to the boundary of ${AdS} _{n}$ spacetime, so unlike the global coordinates, the Poincaré coordinates do not cover all ${AdS} _{n}$ manifold. Using $u\equiv {\frac {r}{\alpha ^{2}}}$ this metric can be written in the following way:

$ds^{2}=\alpha ^{2}\left({\frac {\,du^{2}}{u^{2}}}+u^{2}(\,dx_{\mu }\,dx^{\mu })\right)$ where $x^{\mu }=(t,{\vec {x}})$. By the transformation $z\equiv {\frac {1}{u}}$} also it can be written as:

$ds^{2}={\frac {\alpha ^{2}}{z^{2}}}(\,dz^{2}+\,dx_{\mu }\,dx^{\mu })$