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Inverse stereographic projection identifies Euclidean four-space with a sphere minus one point. Multiplying its flat metric by $4/(1+|x{|}^2{)}^2$ gives the round metric, which extends smoothly over the added point. In the Klein quadric description, the compact space is the projectivized real null cone of signature $(1,5)$. It differs topologically from Minkowski conformal compactification.

This is the future endpoint of complete inertial timelike geodesics in the Minkowski conformal compactification, at $(T,R)=(\pi ,0)$. The endpoint has finite unphysical coordinate time but infinite physical proper time. Accelerated timelike observers need not have this endpoint; a Rindler horizon can accompany an observer asymptotic to a null direction instead.

Every event in complete Minkowski spacetime emits a future-directed null geodesic ending on future null infinity. Therefore its black-hole region $M\setminus J^{-}({\mathcal{I}}^{+})$ is empty and there is no intrinsic event horizon. The null edges of the Minkowski conformal compactification represent infinity, not hidden regions inside the physical spacetime. An accelerated observer can instead have an observer-dependent Rindler horizon.

In ordinary spherical coordinates, $t\in \mathbb{R}$, $r\in [0,\infty )$, $\theta \in [0,\pi ]$ and $\varphi \in [0,2\pi )$ with periodic identification. The regular coordinate chart excludes the origin and polar axes; the endpoint ranges include the usual spherical identifications. With null coordinates $u=t-r$ and $v=t+r$, both are real and their joint domain is $v\ge u$. The inverse relations giveThe arctangent coordinates obey $-\pi /2<p\le q<\pi /2$. Since $du={\sec }^{2}pdp$ and $dv={\sec }^{2}qdq$, the rescaled metric isNow $T=p+q$, $R=q-p$, so $4dpdq=d{T}^2-d{R}^2$. Thus the Minkowski conformal compactification hasThe full coordinate ranges are linked, not independent: $T\in (-\pi ,\pi )$ and $R\in [0,\pi )$ with $|T|+R<\pi$. Inverse coordinates are $p=(T-R)/2$, $q=(T+R)/2$, $t=[\tan q+\tan p]/2$, $r=[\tan q-\tan p]/2$. The conformal factor isThe unphysical metric extends as the unit-radius Einstein static universe on $\mathbb{R}\times S^3$, where $0\le R\le \pi$ with the endpoint two-spheres collapsed. Physical Minkowski spacetime occupies the triangular radial domain $R\ge 0$, $|T|+R<\pi$; the edge $R=0$ is the ordinary spatial center, not an infinity or physical boundary.

In the closure, the infinite-coordinate limits have locationsThese are future timelike infinity, past timelike infinity, spacelike infinity, future null infinity and past null infinity. For example ${\mathcal{I}}^{+}$ has $v\to +\infty$ with finite $u$, while ${\mathcal{I}}^{-}$ has $u\to -\infty$ with finite $v$. A diagram point with $0<R<\pi$ stands for a two-sphere, and each null boundary is topologically $\mathbb{R}\times S^2$; their three corners are points after the angular collapse. On the null boundaries $\Omega =0$ and $d\Omega \ne 0$, whereas the three corners have a degenerate zero gradient. Thus a compactifying metric can be smooth there even though those infinity points are not ordinary smooth null-boundary points.

The geodesic endpoints in Minkowski conformal compactification can be established directly. A physical inertial timelike geodesic has $t=t_0+E\tau$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{p}\tau$, with $E>0$ and $E^2-|\mathbf{p}{|}^2=1$. At future infinity $u,v\to +\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. At past infinity both tend to $-\infty$, giving $i^{-}$. Every complete inertial timelike path has these endpoints, regardless of velocity.

For a future-directed null geodesic, $t=t_0+E\lambda$, $\mathbf{x}={\mathbf{x}}_0+E\mathbf{n}\lambda$ with $|\mathbf{n}|=1$. As $\lambda \to +\infty$, $v\to +\infty$ but $u$ has a finite limit; the endpoint lies on ${\mathcal{I}}^{+}$. As $\lambda \to -\infty$, $u\to -\infty$ but $v$ has a finite limit, giving an endpoint on ${\mathcal{I}}^{-}$. A conformal transformation preserves these null paths as unparametrized geodesics. The null affine parameter under a conformal rescaling obeys $d\tilde{\lambda }/d\lambda ={\Omega }^2$; here ${\Omega }^2=O({\lambda }^{-2})$ at the null endpoints, so the unphysical affine parameter reaches them in finite time, although the physical affine parameter is infinite.

A physical spacelike geodesic has $t=t_0+a\lambda$, $\mathbf{x}={\mathbf{x}}_0+\mathbf{b}\lambda$, with $|\mathbf{b}|>|a|$. At either end $r>|t|$ asymptotically, $u\to -\infty$ and $v\to +\infty$, so both ends approach $i^0$. Timelike and spacelike physical geodesic paths are not generally geodesics of the rescaled metric. Their physical proper time or length is infinite, while $\Omega =O(|\lambda {|}^{-2})$ gives finite unphysical proper time or length near the endpoints. Infinity at finite diagram coordinates does not imply incompleteness of physical Minkowski geodesics.

The original Penrose diagram suppresses the angular spheres. The displayed physical paths illustrate their distinct endpoint types; the two sloping boundaries are null infinity, while the left edge is the regular center.

The original spherical coordinates have $t\in \mathbb{R}$, $0\le r<\infty$, $0\le \theta \le \pi$ and $0\le \varphi <2\pi$, with $\varphi$ periodic. The angular coordinates degenerate at the polar axis and all angles identify at $r=0$; a regular spherical chart uses $r>0$, $0<\theta <\pi$ and suitable overlapping angular patches.

The retarded time $u=t-r$ and advanced coordinate $v=t+r$ both range over $\mathbb{R}$, with the joint restriction $u\le v$. Their inverse relations are $t=(u+v)/2$ and $r=(v-u)/2$, soSet $p=\arctan u$, $q=\arctan v$. ThenThe conformal factor is $\Omega =2\cos p\cos q>0$. Since $(\tan q-\tan p)\cos p\cos q=\sin (q-p)$, multiplication by ${\Omega }^2$ givesFinally $T=p+q$ and $R=q-p$ give $4dpdq=d{T}^2-d{R}^2$. HenceThis is a region of the Einstein static universe, with positive time and negative spatial signature. The exact joint coordinate domain isor equivalently $R\ge 0$ and $|T|+R<\pi$. The regular radial chart has $R>0$; $R=0$ is the ordinary centre. The angular ranges are unchanged. It is important not to treat the stated individual ranges of $T,R$ as independent. The inverse map and conformal factor provide a useful check:

The resulting Minkowski conformal compactification has a triangular radial Penrose diagram. The two sloping boundary edges areThese are future null infinity and past null infinity. The three vertices are future timelike infinity $i^{+}=(\pi ,0)$, past timelike infinity $i^{-}=(-\pi ,0)$ and spacelike infinity $i^0=(0,\pi )$. They are ideal boundary points, not events of physical Minkowski spacetime.

For a complete inertial timelike geodesic, $\mathbf{x}=\mathbf{b}+\mathbf{v}t$ with $|\mathbf{v}|<1$. As $t\to +\infty$, both $t-r$ and $t+r$ tend to $+\infty$, so $p,q\to \pi /2$ and the curve ends at $i^{+}$. As $t\to -\infty$, both tend to $-\infty$, giving $i^{-}$. These endpoints occur at infinite physical proper time.

For a complete null geodesic, write $\mathbf{x}=\mathbf{b}+\mathbf{n}t$, $|\mathbf{n}|=1$. To the future, $r=t+\mathbf{n}\cdot \mathbf{b}+O(1/t)$, so $u\to -\mathbf{n}\cdot \mathbf{b}$ is finite and $v\to +\infty$. Thus $q\to \pi /2$ with $p$ finite: the endpoint lies on ${\mathcal{I}}^{+}$. To the past, $v\to -\mathbf{n}\cdot \mathbf{b}$ remains finite while $u\to -\infty$, giving ${\mathcal{I}}^{-}$. Physical affine parameter is infinite at both ends. Along a radial null path, $dT=\pm dR$, so the diagram represents its null directions at $45$ degrees.

For a complete spacelike geodesic, take $t=t_0+a\lambda$, $\mathbf{x}=\mathbf{b}+\mathbf{w}\lambda$ with $|\mathbf{w}|>|a|$. At either end $r>|t|$ asymptotically, so $u\to -\infty$, $v\to +\infty$ and $(T,R)\to (0,\pi )$. Both ends therefore reach $i^0$, again at infinite physical affine parameter. These are the geodesic endpoints in Minkowski conformal compactification.

Only conformal preservation of null geodesic paths is guaranteed. The pictured timelike and spacelike curves are images of physical geodesics, not generally geodesics of the unphysical metric. For a null path, an unphysical affine parameter satisfies $d\tilde{\lambda }/d\lambda \propto {\Omega }^2$. Since $\Omega =O(1/|\lambda |)$ near a generic null-infinity endpoint, the unphysical parameter can converge while the physical parameter diverges. At $i^{\pm }$ and $i^0$, the usual smooth-boundary condition $d\Omega \ne 0$ degenerates; these vertices should not be treated as regular points of null infinity.

Complete Minkowski spacetime has no black-hole event horizon. Every physical event can send an outgoing null signal to ${\mathcal{I}}^{+}$, so the black-hole region $\mathcal{M}\setminus J^{-}({\mathcal{I}}^{+})$ is empty. There is likewise no particle or future event horizon for an eternal inertial observer: any finite event can eventually signal that observer. Null infinity is a conformal boundary, not a physical horizon. Uniformly accelerated observers can have Rindler horizons, but that is an observer-dependent restriction and is not a horizon of the full inertial Minkowski spacetime.

The meaningful twistor-geometric interpretation of “colinear in $PN$” is that the entire projective line through two distinct projective null twistors lies in projective null twistor space. Mere membership of two points in a projective line would impose no condition, because every pair of distinct points in ${\mathbb{CP}}^3$ spans one.

Indeed, individual nullness givesThis vanishes for every $a,b\in \mathbb{C}$ precisely when $h(X,Y)=0$. Thus $W=\mathrm{span}(X,Y)$ is a maximal totally isotropic two-plane, a real point of Minkowski conformal compactification under the Lorentzian reality structure on twistor space. The two corresponding alpha-planes meet at this unique $W$. Hence their compact intersection is a real spacetime point, and their real null geodesics meet there.

There are two qualifications. The point can be at conformal infinity, so a finite affine meeting is not automatic. If $X,Y$ represent the same projective point instead, their alpha-planes coincide rather than having a unique intersection. If “colinear” is read only as spanning a line in ${\mathbb{CP}}^3$, individual nullness alone does not force a real intersection; the additional orthogonality condition above is indispensable.

For a concrete counterexample to the weaker reading, take $X=(0,0,1,0)$ and $Y=(1,0,0,1)$ in $({\omega }^0,{\omega }^1,{\pi }_{0^{\prime }},{\pi }_{1^{\prime }})$ coordinates. Each is a null twistor, but $h(X,Y)=1$. Their two incidence equations give $x=-i\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, which is not Hermitian. Thus their alpha-planes meet at a complex point that is not a real Minkowski point.

The radial part of the Minkowski metric becomes $-dudv$, 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 ${\Omega }^2$ givesConsequently the Minkowski conformal compactification isThe spatial metric $d{R}^2+{\sin }^{2}Rd{\Sigma }^2$ 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 areThe 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 arenamely 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.

A timelike geodesic is an inertial worldline $\mathbf{x}=\mathbf{b}+\mathbf{\beta }t$ with $|\mathbf{\beta }|<1$. 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 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 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,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 $t-x\ge 0$ 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 areThey 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 $|d\mathbf{x}/dt|\le 1$ 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 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.

Start with the four-dimensional Minkowski metric $d{s}^2=-d{t}^2+d{r}^2+r^{2}d{\Omega }^2$, where $r\ge 0$. Choose an arbitrary length $L>0$, and use the retarded and advanced null coordinates $u=t-r$, $v=t+r$. For the Minkowski conformal compactification, setBoth $p,q$ lie between $-\pi /2$ and $\pi /2$. Since $v\ge u$, we have $R\ge 0$; the remaining inequalities are $|T|+R<\pi$. Moreover,Multiply by the square of the conformal factor ${\Omega }_c=2\cos p\cos q/L$. The resulting metric is regular on the appropriate boundary pieces and preserves the null directions. Suppressing the angular two-spheres gives a triangular Penrose diagram with radial null geodesics at $45$ degrees.

The line $R=0$ is the ordinary timelike centre $r=0$. The upper sloping edge $T+R=\pi$ is future null infinity, reached with $v\to +\infty$ and finite $u$; the lower sloping edge $T-R=-\pi$ is past null infinity, reached with $u\to -\infty$ and finite $v$. The vertices $(T,R)=(\pi ,0)$ and $(-\pi ,0)$ are future and past timelike infinity, denoted $i^{+}$ and $i^{-}$. The vertex $(0,\pi )$ is spacelike infinity, $i^0$. These are limiting endpoints in the conformal completion, rather than ordinary physical events. In particular, finite diagram coordinates at null infinity do not imply finite physical affine parameter.

The four-dimensional radial diagram is the triangle $R\ge 0$, $|T|+R\le \pi$. If one instead draws two-dimensional Minkowski spacetime with a signed Cartesian spatial coordinate, the diagram is the full diamond. The centre is a boundary of the radial quotient, not a boundary of the physical four-dimensional Minkowski spacetime.

Past null infinity is the ideal boundary from which incoming null geodesics arrive in an asymptotically flat spacetime. In the Minkowski conformal compactification it is $T-R=-\pi$, with the retarded coordinate tending to $-\infty$ and the advanced coordinate finite.

This is the past endpoint of complete inertial timelike geodesics in the Minkowski conformal compactification, at $(T,R)=(-\pi ,0)$. It is the time reversal of future timelike infinity and is reached at infinite physical proper time to the past.

The five-real-dimensional space of projective null twistors. A projective line lies entirely in it precisely when its corresponding two-plane is totally isotropic for the twistor Hermitian form. Such lines represent real points of Minkowski conformal compactification. Two independent null twistors lie on one such line exactly when $h(X,Y)=0$.

Spacelike infinity is the ideal endpoint of spatial directions in the Minkowski conformal compactification, represented by $(T,R)=(0,\pi )$. Unlike null infinity, it is a vertex of the radial Penrose diagram.

Future and past timelike infinity are ideal endpoints of asymptotically escaping timelike trajectories in a conformal completion. For the Minkowski conformal compactification, they are the vertices $(T,R)=(\pi ,0)$ and $(-\pi ,0)$; these vertices are not physical spacetime events.