= Solution
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 $(+--)$, with
$$
X^0=\sinh t,\qquad
X^1=\cosh t\cos\chi,\qquad X^2=\cosh t\sin\chi.
$$
Pulling back the ambient <metric tensor> gives $dt^2-\cosh^2t\,d\chi^2$, with
$$
\boxed{t\in\mathbb R,\qquad \chi\in\mathbb R/(2\pi\mathbb Z).}
$$
Unwrapping $\chi$ 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 $(++-)$ and
$$
X^0=\cosh r\cos t,\qquad
X^1=\cosh r\sin t,\qquad X^2=\sinh r.
$$
Here $r\in\mathbb R$ and $t$ is initially periodic modulo $2\pi$. The time circles are <closed timelike curves>. Its physically standard <universal cover> removes that identification:
$$
\boxed{r\in\mathbb R,\qquad t\in\mathbb R.}
$$
The two signs of $r$ describe two spatial ends; imposing $r\ge0$ 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 $\kappa=\dot t^2-\cosh^2t\,\dot\chi^2\in\{1,0,-1\}$ and conserved momentum $L=\cosh^2t\,\dot\chi$. Then
$$
\dot t^2=\kappa+\frac{L^2}{\cosh^2t}.
$$
<Timelike 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 $|L|\ge1$, 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 $E=\cosh^2r\,\dot t$. The normalization is $\kappa=\cosh^2r\,\dot t^2-\dot r^2$, hence
$$
\dot r^2=\frac{E^2}{\cosh^2r}-\kappa.
$$
Choose a future-directed <timelike geodesic>, so $E\ge1$ and
$$
\sinh r=\sqrt{E^2-1}\sin(\lambda-\lambda_0).
$$
For $E>1$, 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 $\eta=\arctan(\sinh t)$:
$$
ds^2=\sec^2\eta(d\eta^2-d\chi^2),\qquad
-\frac\pi2<\eta<\frac\pi2.
$$
Its past and future <conformal boundaries> are spacelike circles. <Null geodesics> have $d\chi/d\eta=\pm1$ 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 tensor>
$$
ds^2=(1-\rho^2)dT^2-\frac{d\rho^2}{1-\rho^2},\qquad |\rho|<1.
$$
The horizons at $\rho=\pm1$ 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,
$$
ds^2=\sec^2\psi(dt^2-d\psi^2),\qquad
-\frac\pi2<\psi<\frac\pi2,\qquad t\in\mathbb R.
$$
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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-68-conformal-diagrams.png]
{title=Conformal cylinder and observer horizons of de Sitter spacetime compared with the timelike boundaries of the anti-de Sitter universal cover}
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