Does the Topology of AdS4 Affect Global Hyperbolicity?

[ergospherical]

Is $\mathrm{AdS_4}$ globally hyperbolic?$$g = -\left(1+ \dfrac{r^2}{l^2} \right)dt^2 + \dfrac{dr^2}{1+ \dfrac{r^2}{l^2}}+ r^2d\Omega^2$$Letting $r = l \tan \chi$ then defining $\tilde{g} = g \cos^2 \chi$\begin{align*}
g &= \sec^2 \chi (-dt^2 + l^2 d\chi^2) + l^2 \tan^2 \chi d\Omega^2 \\ \\

\tilde{g} &= -dt^2 + l^2 (d\chi^2 + \sin^2 \chi d\Omega^2) \\
\tilde{g} &= -dt^2 + l^2 d\omega^2\end{align*}the topology is $\mathbb{R} \times S^3$. Does global hyperbolicity of $\tilde{g}$ $\iff$ global hyperbolicity of $g$?

 

[Twigg]

I don't know anything about string theory, but isn't that coordinate system a mapping to $\mathbb{R^2} \times S^2$ not $\mathbb{R^1} \times S^3$? It looks like only two angles and a radius to me (plus time). Am I missing something?

If I'm missing something obvious, ignore me. I just chimed in because the thread went unanswered.

 

[ergospherical]

My thinking was that $l^2(d\chi^2 + \sin^2 \chi d\Omega^2)$ is the round metric on a $3$-sphere of radius $l$ (or in fact since $\chi \in \bigg{[} 0, \dfrac{\pi}{2} \bigg{)}$ it'll only be half of the 3-sphere...)

 

[Twigg]

Just to make sure we're talking about the same thing, when you say $S^3$, you mean a surface embedded in $\mathbb{R}^4$ given by $x^2 + y^2 + z^2 + w^2 = 1$, right? As in, the sphere that is diffeomorphic to the special orthogonal group $\mathrm{SO}(3)$ and the special unitary group $\mathrm{SU}(2)$, right?

 

[ergospherical]

Yeah, and $(\chi, \theta, \varphi)$ would be the hyperspherical coordinates on the 3-sphere of radius $l$.

 

[Twigg]

I'm a dope and I only just caught that $d\Omega^2$ was a total solid angle over $S^2$. Now I'm on board with your claim about the topology. Sorry for derailing ya