= Solution
Use the original PDF's spatial Ricci component $R_{\chi\chi}=a\ddot a+2\dot a^2+2$; the converted TeX's $2\dot a\ddot a$ is a transcription defect. Continue with $c=1$ and signature $(-,+,+,+)$. Tracing the vacuum <Einstein field equations> with a <cosmological constant> gives $R=4\Lambda$, and hence $R_{ab}=\Lambda g_{ab}$. The time component therefore gives $\ddot a=(\Lambda/3)a$. Substituting this into the spatial component gives
$$
\boxed{\dot a^2+1=\frac\Lambda3a^2.}
$$
The left side is strictly positive, so \b[$\Lambda>0$] is necessary for a real positive closed <scale factor>. Define $\ell=\sqrt{3/\Lambda}$. The acceleration equation has solution $a=Ae^{t/\ell}+Be^{-t/\ell}$, and the first integral fixes $4AB=\ell^2$. Taking the positive scale-factor branch and shifting the origin to the minimum gives
$$
\boxed{a(t)=\ell\cosh(t/\ell).}
$$
The universe contracts for $t<0$, has its nonzero minimum $a=\ell$ at $t=0$, and expands for $t>0$. The minimum is a smooth bounce, not a curvature singularity.
Introduce the dimensionless time $\tau=t/\ell$. Then the <closed slicing of de Sitter spacetime> is
$$
ds^2=\ell^2[-d\tau^2+\cosh^2\tau\,d\Omega_3^2],\qquad
d\Omega_3^2=d\chi^2+\sin^2\chi(d\theta^2+\sin^2\theta\,d\phi^2).
$$
One can identify it explicitly: in five-dimensional Minkowski space, set $X^0=\ell\sinh\tau$ and $X^I=\ell\cosh\tau\,n^I$, where $\sum_{I=1}^4(n^I)^2=1$. These coordinates satisfy $-(X^0)^2+\sum_I(X^I)^2=\ell^2$, and their induced metric is the one above. Thus the solution is \b[global <de Sitter spacetime>], with spherical spatial topology and both contracting and expanding parts.
The time entering the printed compactification formula is this rescaled time. Write
$$
T=2\arctan(e^\tau)-\frac\pi2.
$$
Direct differentiation and elementary trigonometry give $dT/d\tau=\operatorname{sech}\tau$, $\sin T=\tanh\tau$ and $\cos T=\operatorname{sech}\tau$. Therefore
$$
\boxed{ds^2=\frac{\ell^2}{\cos^2T}\left[-dT^2+d\chi^2+\sin^2\chi\,d\Omega_2^2\right].}
$$
Multiplying by $(\cos T/\ell)^2$ gives the metric of a finite-time slab of the <Einstein static universe>, often called the Einstein cylinder. This is a <conformal transformation> of the metric and does not assert that the transformed matter content remains vacuum. The coordinate ranges are
$$
\boxed{-\frac\pi2<T<\frac\pi2,\quad0\leq\chi\leq\pi,\quad0\leq\theta\leq\pi,\quad\phi\in[0,2\pi).}
$$
The angular-coordinate degeneracies at the poles are the usual regular spherical ones.
Radial null lines satisfy $d\chi=\pm dT$, so the <Penrose diagram> is the square displayed below. Every interior point represents an angular two-sphere. Its top and bottom are the spacelike conformal infinities $\mathcal I^+$ and $\mathcal I^-$ at $T=\pi/2$ and $T=-\pi/2$; they are excluded from the physical spacetime and are not curvature singularities. The vertical sides $\chi=0$ and $\chi=\pi$ represent the regular north and south poles of the three-sphere, not spatial edges of the universe.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-54-de-sitter-penrose.png]
{title=Global de Sitter Penrose diagram with spacelike conformal infinities and a comoving observer's particle and event horizons}
A <particle horizon> at an observer's current time bounds the comoving sources whose light could have reached that observer over the available past conformal interval. Here the cosmological past extends to $t=-\infty$, but its conformal duration is finite. For a comoving observer at $\chi=0$,
$$
\boxed{\chi_p(t)=\int_{-\infty}^t\frac{dt'}{a(t')}=T+\frac\pi2.}
$$
Since $0<\chi_p<\pi$ at every finite time, a particle horizon exists. The corresponding null boundary in the diagram is $\chi=T+\pi/2$, emerging from the observer's past endpoint at $\mathcal I^-$. The antipodal point never has enough time to send a signal received at a finite proper time.
An <observer event horizon> bounds the events that can ever send a signal to the observer's future-inextendible worldline. The remaining conformal time fixes its distance:
$$
\boxed{\chi_e(t)=\int_t^\infty\frac{dt'}{a(t')}=\frac\pi2-T.}
$$
This too lies strictly between zero and $\pi$ at finite time, so the observer has a cosmological event horizon. Its null line ends at the observer's future endpoint on $\mathcal I^+$. Both horizon spheres have areal radius $a\sin\chi=\ell$, though their comoving angular positions differ except at $T=0$. These are the <cosmological horizons in global de Sitter spacetime>; changing the comoving observer relocates them by an isometry.
A <Cauchy horizon> is the boundary of the domain determined by data on a chosen partial <Cauchy hypersurface>; beyond it an inextendible causal curve can avoid that initial surface. Each complete constant-$T$ three-sphere here is a <Cauchy hypersurface>. To see this directly, a causal curve in the conformal metric has spatial speed at most one when parameterized by $T$. On any finite interior time interval its path therefore has a limit in the complete compact three-sphere. If an inextendible curve stopped at an interior time, that limit would allow its continuation, a contradiction. Its time range consequently extends between both conformal ends, and monotonicity makes it cross any complete constant-$T$ sphere exactly once. Its <domain of dependence> is the entire physical spacetime.
Thus \b[particle and cosmological event horizons exist for every comoving observer, while the complete global Cauchy slices have no <Cauchy horizons>.] Horizons of a restricted static patch or a deliberately partial initial surface are relative to that restriction; they do not alter the global Cauchy conclusion.
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