I Does the Topology of AdS4 Affect Global Hyperbolicity?

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