= Solution
The radial part of the <Minkowski metric> becomes $-du\,dv$, and $r=(v-u)/2$. With $u=\tan p$, $v=\tan q$ and the positive <conformal factor> $\Omega=2\cos p\cos q$, multiplication by $\Omega^2$ gives
$$
\Omega^2(-du\,dv)=-4\,dp\,dq=-dT^2+dR^2,
\qquad
\Omega^2r^2=(\cos p\cos q)^2(\tan q-\tan p)^2=\sin^2(q-p).
$$
Consequently the <Minkowski conformal compactification> is
$$
\boxed{ds^2=\Omega^{-2}\left(-dT^2+dR^2+\sin^2R\,d\Sigma^2\right),\qquad
\Omega=\cos T+\cos R.}
$$
The spatial metric $dR^2+\sin^2R\,d\Sigma^2$ is the round unit three-sphere. The extended unphysical metric is therefore that of the <Einstein static universe>, $\mathbb R\times S^3$ with unit spatial radius. Physical <Minkowski spacetime> occupies only the diamond-shaped causal patch of this cylinder, or a triangle after restricting to nonnegative spherical radius; it is not the whole cylinder.
The complete coordinate ranges and joint restrictions are
$$
\begin{gathered}
t\in\mathbb R,\quad r\in[0,\infty),\quad\theta\in[0,\pi],\quad\phi\in[0,2\pi),\\
u,v\in\mathbb R,\quad u\le v,\\
-\frac\pi2<p\le q<\frac\pi2,\\
-\pi<T<\pi,\quad 0\le R<\pi,\quad |T|+R<\pi.
\end{gathered}
$$
The polar angles have the usual coordinate degeneracies at their poles and at $r=0$. In the extended <Einstein static universe> one may let $T\in\mathbb R$ and $0\le R\le\pi$, but these are not the ranges of the original physical patch. In its interior $\Omega>0$. The line $R=0$ is the regular centre $r=0$, not a component of infinity.
The <conformal completion> has <future timelike infinity> $i^+=(\pi,0)$, <past timelike infinity> $i^-=(-\pi,0)$, and <spacelike infinity> $i^0=(0,\pi)$. The open sloping boundaries are
$$
\mathcal I^+:T+R=\pi,\quad0<R<\pi,
\qquad
\mathcal I^-:T-R=-\pi,\quad0<R<\pi,
$$
namely <future null infinity> and <past null infinity>. Each ordinary interior point of the radial diagram represents a two-sphere, and each point on either open null boundary represents a two-sphere of null directions.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-52-penrose.png]
{title=Minkowski Penrose triangle with timelike, spacelike and null infinities, the regular centre and representative causal and spacelike geodesics}
A <timelike geodesic> is an inertial worldline $\boldsymbol x=\boldsymbol b+\boldsymbol\beta t$ with $|\boldsymbol\beta|<1$. As $t\to+\infty$, both $t-r$ and $t+r$ tend to $+\infty$, giving $p,q\to\pi/2$ and endpoint $i^+$. At past infinity both tend to $-\infty$, giving $i^-$. These endpoints are reached at infinite physical <proper time>, although at finite conformal coordinate time $T$. A <null geodesic> has $r\sim|t|$. In its outgoing future, $u$ approaches a finite retarded time while $v\to+\infty$, giving an endpoint on $\mathcal I^+$; in its incoming past, $v$ has a finite limit and $u\to-\infty$, giving an endpoint on $\mathcal I^-$. A <spacelike geodesic> is a straight line with spatial tangent larger than its temporal tangent, so $u\to-\infty$, $v\to+\infty$ at either end and both endpoints are $i^0$ in the compactified radial description. All these physical geodesics are complete. Only the unparametrized <null geodesics> are necessarily preserved as geodesics of the unphysical metric; timelike and spacelike images need not be its geodesics.
For horizons, distinguish a property of the <spacetime> from a property of an observer or of an initial-data hypersurface. Complete inertial observers have \b[no particle horizon]: in the flat cosmological slicing $a(t)=1$, the past light-travel distance $\int_{-\infty}^{t_0}dt$ is infinite. There is no finite-age initial singularity truncating communication between comoving observers. This statement does not say that an event receives signals from spacelike-separated events at the same time.
There is also no black-hole <event horizon>, because every event can send a future-directed null ray to <future null infinity>. Nevertheless \b[accelerated timelike observers can have both future and past observer event horizons]. For an entire observer worldline $\gamma$, define them by $H^+_\gamma=\partial I^-(\gamma)$ and $H^-_\gamma=\partial I^+(\gamma)$, using its complete future and past respectively. An inertial observer's chronological past and future of the whole worldline are both all of <Minkowski spacetime>. By contrast, for the uniformly accelerated worldline in a Cartesian spatial direction,
$$
t=a^{-1}\sinh(a\tau),\qquad x=a^{-1}\cosh(a\tau),\qquad y=z=0,
$$
its null coordinates obey $t-x=-a^{-1}e^{-a\tau}<0$ and $t+x=a^{-1}e^{a\tau}>0$. An event with $t-x<0$ can signal to a sufficiently late point of this worldline, whereas an event with $t-x\ge0$ cannot; finite transverse displacements do not change that conclusion as $t+x\to\infty$. Similarly the worldline can signal to precisely the events with $t+x>0$. Thus its two <observer event horizons> are
$$
\boxed{H^+_\gamma:\ t-x=0,\qquad H^-_\gamma:\ t+x=0.}
$$
They are the <Rindler horizons>, and their accessible common region is $x>|t|$. They do not indicate a <curvature singularity> or a failure of global predictability.
Finally <Minkowski spacetime> is a <globally hyperbolic spacetime>: every inextendible <causal curve> crosses each complete spacelike plane $t=t_0$ exactly once. Indeed $t$ is strictly monotone along future-directed <causal curves>, and the speed bound $|d\boldsymbol x/dt|\le1$ prevents an inextendible curve from ending at finite $t$ by escaping to spatial infinity. Its time range is therefore all of $\mathbb R$. These planes are <Cauchy hypersurfaces> with full <domain of dependence>, so \b[there are no Cauchy horizons associated with complete Cauchy data].
A <Cauchy horizon> is, however, defined relative to a specified initial surface. Under a literal existence reading of “admits”, non-Cauchy initial surfaces can have such horizons even in flat <spacetime>. For example the edgeless spacelike hyperboloid $\Sigma_a:\ t=\sqrt{a^2+r^2}$ has $D(\Sigma_a)=\{t>r\}$. To see this, inside the future light cone $t-r$ and $t+r$ are positive and nondecreasing on future-directed <causal curves>. Their product $t^2-r^2$ increases past $a^2$ on every future-inextendible curve from below the hyperboloid; every past-inextendible curve from above it crosses down through $a^2$ before leaving the cone. Outside the cone there is an inextendible null curve avoiding the hyperboloid. The future light cone itself is its past <Cauchy horizon>. The time-reversed hyperboloid gives a future <Cauchy horizon>. This is the <dependence of a Cauchy horizon on the initial hypersurface>, not an intrinsic loss of predictability of full <Minkowski spacetime>. Thus a categorical claim that no choice of partial initial surface can ever have a Cauchy horizon would be too strong.
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