Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 75 / 3 / a / Solution

Take the causally unwrapped universal cover of Anti-de Sitter spacetime, with curvature radius $L$. In global coordinates,The ordinary radial Penrose diagram is a half-strip: the left edge $\rho =0$ is a regular center and the right edge $\rho =\pi /2$ is a timelike conformal boundary. For displaying signed spatial motion, the figure doubles the radial diameter to $-\pi /2<\rho <\pi /2$. Its two ends represent antipodal spatial directions of the higher-dimensional boundary; they are not a claim that the full AdS boundary has two disconnected components.

The vertical edges are spatial infinity at every finite global time, forming the timelike boundary $\mathbb{R}\times S^{d-2}$. The arrows mark the future and past timelike ends $T\to \pm \infty$, conventionally $i^{+}$ and $i^{-}$. There are no separate null-infinity edges like Minkowski ${\mathcal{I}}^{\pm }$: radial null rays satisfy $dT=\pm d\rho$ and reach the timelike boundary in finite global time. Their physical affine parameter is nevertheless infinite there. The drawing is a finite excerpt of an infinitely tall strip, not an identification of its top and bottom. On the original hyperboloid global time is periodic and there are closed timelike curves; unwrapping it is essential to the usual physical maximally extended AdS spacetime.