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The local metric tensors do not determine the global coordinate identifications. The usual complete two-dimensional de Sitter spacetime is the hyperboloid $(X^0{)}^2-(X^1{)}^2-(X^2{)}^2=-1$ in ambient signature $(+--)$, withPulling back the ambient metric tensor gives $d{t}^2-{\cosh }^{2}td{\chi }^2$, withUnwrapping $\chi$ gives the universal cover of two-dimensional de Sitter spacetime instead. The usual embedded two-dimensional anti-de Sitter spacetime has $(X^0{)}^2+(X^1{)}^2-(X^2{)}^2=1$ in signature $(++-)$ andHere $r\in \mathbb{R}$ and $t$ is initially periodic modulo $2\pi$. The time circles are closed timelike curves. Its physically standard universal cover removes that identification:The two signs of $r$ describe two spatial ends; imposing $r\ge 0$ would retain only half of this complete model.

Both spacetimes have constant curvature and are geodesically complete. In the paper's convention their Ricci scalars are respectively $+2$ and $-2$. Their embeddings show that geodesics are intersections with planes through the ambient origin. They nevertheless have very different causal structures.

For the geodesics of two-dimensional de Sitter spacetime, use an affine parameter $\lambda$, normalization $\kappa ={\dot{t}}^2-{\cosh }^{2}t{\dot{\chi }}^2\in \{1,0,-1\}$ and conserved momentum $L={\cosh }^{2}t\dot{\chi }$. ThenTimelike geodesics extend to both infinite proper times; $\chi =\text{constant}$ is a simple example. Null geodesics obey $\sinh t=\pm |L|(\lambda -{\lambda }_0)$, so their affine parameter is also infinite at either temporal end. Spacelike geodesics have $|L|\ge 1$, obey $\sinh t=\sqrt{L^2-1}\sin (\lambda -{\lambda }_0)$ and are closed curves on the embedded cylinder; $t=0$ is its simplest spacelike geodesic circle. Unwrapping the spatial circle removes closure but not completeness.

For the geodesics of two-dimensional anti-de Sitter spacetime, the timelike Killing vector ${\partial }_t$ gives $E={\cosh }^{2}r\dot{t}$. The normalization is $\kappa ={\cosh }^{2}r{\dot{t}}^2-{\dot{r}}^2$, henceChoose a future-directed timelike geodesic, so $E\ge 1$ andFor $E>1$, it oscillates radially with proper period $2\pi$; $E=1$ gives the central geodesic $r=0$. In both cases, global time advances by $2\pi$ over a proper time interval $2\pi$. The curve is closed on the original hyperboloid, while its lift on the universal cover is nonclosed and future-directed. Null geodesics have $\sinh r=\pm E(\lambda -{\lambda }_0)$ and reach either spatial end only at infinite affine parameter. Spacelike geodesics satisfy $\sinh r=\sqrt{E^2+1}\sinh (\lambda -{\lambda }_0)$ and also have infinite proper length toward either end. Thus the finite coordinate time discussed below does not imply physical geodesic incompleteness.

The conformal cylinder of two-dimensional de Sitter spacetime follows by setting $\eta =\arctan (\sinh t)$:Its past and future conformal boundaries are spacelike circles. Null geodesics have $d\chi /d\eta =\pm 1$ and can travel only a finite angular distance over the entire infinite proper-time history. For the complete observer $\chi =0$, an event can send a signal to that observer precisely when its shortest angular distance $d(\chi ,0)$ is less than $\pi /2-\eta$. The observer horizons in two-dimensional de Sitter spacetime are therefore the null curves $d(\chi ,0)=\pi /2-\eta$; its past signal horizon has $d(\chi ,0)=\eta +\pi /2$. These are observer horizons, without curvature singularities. Their intersection bounds a static patch with metric tensorThe horizons at $\rho =\pm 1$ are regular Killing horizons beyond which this static chart fails, while the global coordinates remain smooth.

For the conformal strip of two-dimensional anti-de Sitter spacetime, set $\psi =\arctan (\sinh r)$. On its universal cover,The two conformal boundaries are timelike. A null ray from the center approaches a boundary after coordinate time $\pi /2$, though its affine parameter diverges. Every interior event can signal to the complete central static observer in finite global time, so that observer has no event horizon. A Poincaré horizon concerns a restricted coordinate patch, and accelerated observers can have different causal horizons.

Finally, the global de Sitter cylinder and its spatial cover are globally hyperbolic: constant-time slices are Cauchy hypersurfaces. The anti-de Sitter universal cover is not globally hyperbolic, because timelike infinity admits incoming signals within finite global time. A field's evolution therefore requires boundary conditions at its two conformal boundaries in addition to initial data. The opposite placement of the conformal boundaries explains why de Sitter observers have cosmological horizons despite global hyperbolicity, while the central anti-de Sitter observer has no event horizon despite the failure of global hyperbolicity.

Set $z=e^y$. The metric becomesso these are Poincare coordinates on anti-de Sitter spacetime with unit curvature radius. They cover one patch, not the whole global extension.

On the two-dimensional section with the three transverse $x_i$ suppressed, choose embedding coordinates $X_{-1}=\sec \rho \cos T$, $X_0=\sec \rho \sin T$, $X_d=\tan \rho$. The Poincaré transformation isChoose the connected component containing $T=\rho =0$. Its denominator is positive, givingIts two sloping edges are Poincaré horizons, null coordinate boundaries at $z\to \infty$. The right vertical edge is the $z\to 0$ timelike conformal boundary. Substituting the transformation into $z^{-2}(d{z}^2-d{t}^2)$ gives ${\sec }^2\rho (d{\rho }^2-d{T}^2)$, verifying the chart and its null edges.

The full coordinates in this part have $y\in \mathbb{R}$, hence the whole shaded $z>0$ patch. If one additionally retains only the Randall–Sundrum bulk $y\ge 0$ as in Question4, keep $z\ge 1$. The brane $y=0$, or $z=1$, has $\cos T=\cos \rho -\sin \rho$ and is drawn in orange; the retained side contains $T=0$, $\rho <0$. A patch horizon is not a curvature singularity or the timelike conformal boundary. A geodesic can cross it in finite proper time while the patch coordinate $t$ diverges.