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For the superextreme Kerr metric,for all real $r$, so there are no coordinate singularities at horizon roots. The only stipulated curvature singularity is $\Sigma =0$, which forces $r=0$, $\theta =\pi /2$. In the supplied Cartesian coordinates this is the Kerr ring singularity $x^2+y^2=a^2$, $z=0$.

Write ${\rho }^2=x^2+y^2$. The coordinate transformation impliesand henceThe positive-radius Cartesian sheet usesIts zero set is the disk $z=0$, $\rho \le a$. At an interior disk point, $\rho <a$, one has $\cos \theta \ne 0$, so $\Sigma =a^2{\cos }^2\theta >0$. The disk interior is therefore not a singularity: in signed $(r,\theta ,\varphi )$ coordinates, the metric is analytic through $r=0$. In particular, $g_{rr}=\Sigma /\Delta$ is finite and the metric determinant $-{\Sigma }^2{\sin }^2\theta$ is nonzero away from the ordinary polar coordinate axis. The polar axes are regular using Cartesian angular charts and $2\pi$ periodicity.

The required two-sheeted extension of superextremal Kerr spacetime is constructed by taking a second Cartesian sheet with the negative root $r\le 0$, cutting both sheets along the disk, and gluing the upper bank of one to the lower bank of the other, and conversely. Signed $r$ then passes continuously through zero as the disk is crossed. The ring is omitted. A useful check near an interior disk point isso signed $r$ is a smooth transverse coordinate there. The apparent factors $z/r$ in the Cartesian metric represent $\cos \theta$ and have regular signed-coordinate limits; they do not justify deleting the whole disk. Identifying the two signs of $r$ on a single Cartesian copy would also identify two physically different metric values, since the Kerr-Schild coefficient changes sign.

Continue the two sheets out to $r\to \pm \infty$. Both ends are asymptotically flat; at the negative-radius end the mass term has the opposite sign, as $g_{tt}=-1+2M/r+O(r^{-2})$. There are no further finite-radius horizon boundaries requiring additional blocks. Removing only the curvature-singular ring, and removing coordinate artifacts with the charts described above, gives the standard maximal analytic extension. The ring cannot be filled by an analytic nondegenerate metric because its curvature diverges. This extension is not globally hyperbolic; analyticity of an extension should not be confused with unique evolution from Cauchy data.

Assume $M>0$ and use geometric units. The stationary Killing vector field $K={\partial }_t$ gives the conserved Killing energy $E=-g(K,U)=f\dot{t}$. For radial motion, normalization of the four-velocity gives $-f{\dot{t}}^2+f^{-1}{\dot{r}}^2=-1$. Combining these identities yieldsAt rest at infinity, $f\to 1$ and $\dot{r}\to 0$, so $E=1$. Choose the inward branch $\dot{r}=-\sqrt{2M/r}$. Its proper time from the Schwarzschild event horizon to the curvature singularity isIn ordinary units this is $4GM/(3c^3)$. The endpoint $r=0$ is not a regular centre of the spacetime. This is the Schwarzschild horizon-to-singularity proper time for fall from infinity, rather than a universal time for all infall energies.

For this geodesic congruence, set $w(r)=\sqrt{2M/r}$. Its four-velocity and metric dual areThe right side is an exact one-form because $w/f$ depends only on $r$. Thus $U_a=-{\partial }_a\tau$, with free-fall proper time in Schwarzschild spacetime determined byIndeed $U^a{\partial }_a\tau =-U^a{U}_a=1$, so this scalar advances by the actual proper time on every member of the infalling geodesic congruence, with a synchronized choice of additive offsets.

Substitution of $dt=d\tau -w{f}^{-1}dr$ in the Schwarzschild metric givesUsing $f=1-w^2$, the Painlevé–Gullstrand coordinates therefore haveAll coefficients are finite at $r=2M$; the determinant of the $\tau ,r$ block is $-f-w^2=-1$. Hence the horizon is a coordinate singularity of Schwarzschild time, not a degeneracy of the metric. The new spatial slices have $d{\ell }^2=d{r}^2+r^{2}d{\Sigma }^2$, the flat Euclidean metric, and $g^{ab}{\partial }_a\tau {\partial }_b\tau =-1$ makes their normals the freely falling four-velocities. By contrast, the curvature scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges at $r=0$; this genuine singularity is not removed.

For ingoing Eddington-Finkelstein coordinates, use $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$, for example $r_{\ast }=r+2M\log |r/(2M)-1|$. ThenBoth constructions cross the future horizon. Their time functions have different meanings: $v=\text{constant}$ is an ingoing null hypersurface, whereas $\tau =\text{constant}$ is a flat spacelike hypersurface orthogonal to the infallers. Thus the former is adapted to ingoing null geodesics, and the latter to a timelike free-fall clock congruence.

For the Kruskal diagram, choose Kruskal–Szekeres coordinates with $U_K=-e^{-(t-r_{\ast })/(4M)}$, $V_K=e^{(t+r_{\ast })/(4M)}$ in the right exterior and extend them across its horizon. They obeyThe four regions of Kruskal spacetime are I, the right exterior $(U_K<0,V_K>0)$; II, the future black hole $(U_K>0,V_K>0)$; III, the other exterior $(U_K>0,V_K<0)$; and IV, the past white hole $(U_K<0,V_K<0)$. The null axes are horizons, their intersection is the bifurcation two-sphere, and the spacelike boundaries $U_K{V}_K=1$ are the future and past curvature singularities. A point in this radial diagram represents a symmetry two-sphere.

The domain of the ingoing Painleve-Gullstrand chart is region I together with region II and their common future horizon. More precisely, it is $V_K>0$, $U_K{V}_K<1$, or $r>0$ with finite $\tau$. To verify the boundary rather than infer it from the locally regular metric, put $\rho =\sqrt{r/(2M)}$ and writeThis follows by subtracting $r_{\ast }$ from the preceding radial primitive, and $S$ is finite at $r=2M$. Thus $V_K=\exp [(\tau -S(r))/(4M)]>0$ throughout the chart. Conversely every point with $V_K>0$ and $r>0$ gives a finite $\tau$. The past horizon $V_K=0$, the bifurcation sphere, region III and region IV are absent. Reversing the sign of the radial term produces the outgoing chart through the white hole instead.

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.

Put $f(r)=1-2M/r$. An incoming radial null ray has $dt/dr=-1/f$. The tortoise coordinate $r_{\ast }=r+2M\log |r/(2M)-1|$ obeys $d{r}_{\ast }/dr=1/f$, so $v=t+r_{\ast }$ is constant along such a ray. For $r>2M$ this is exactly the printed logarithm. Substituting $dt=dv-dr/f$ gives the metric in ingoing Eddington-Finkelstein coordinatesThe radial null directions satisfy either $dv=0$ or $dr/dv=f/2$. Future-directed ingoing rays have decreasing $r$ on the first family. Outgoing rays increase $r$ outside the horizon, stay at $r=2M$ on its generators, and decrease $r$ inside. Thus inside the horizon every future timelike trajectory also moves toward smaller $r$.

The $(v,r)$ metric block has determinant minus one and is nonsingular at $r=2M$. The horizon is a null surface, not a curvature singularity. For example the Kretschmann scalar $48M^2/r^6$ is finite there and diverges at $r=0$.

Inside, set $A=2M/r-1>0$. For radial timelike motion, the Schwarzschild expression becomes $d{\tau }^2=A^{-1}d{r}^2-Ad{t}^2$, henceAngular motion would only reduce the available proper time further. Integrating from the horizon to zero, with $r=2M{\sin }^2\theta$, givesThe bound is sharp for the zero-energy radial geodesic with constant Schwarzschild $t$ in the interior. For a geodesic arriving from the exterior with strictly positive conserved energy, the proper time is strictly smaller; $\pi M$ is the supremum as that energy tends to zero. Thus the universally available horizon-to-singularity upper bound is $\pi M$, rather than a time attainable by every exterior infaller. Restoring units gives $\pi GM/c^3$.

The future domain of dependence $D^{+}(\Sigma )$ consists of points through which every past-inextendible causal curve meets the partial Cauchy hypersurface $\Sigma$. Data on $\Sigma$ determine evolution there for suitable hyperbolic equations. Its future boundary is the Cauchy horizon $H^{+}(\Sigma )$; one precise definition is $\overline{D^{+}(\Sigma )}\setminus I^{-}(D^{+}(\Sigma ))$.

For a nonextremal Reissner-Nordstrom spacetime, $0<|Q|<M$, the metric function is $F=1-2M/r+Q^2/r^2$, with $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The outer event horizon lies at $r_{+}$. The future trapped region leads toward the inner null surface $r_{-}$, a future Cauchy horizon for bridge data on the illustrated $\Sigma$. Beyond it some past causal curves arrive from other analytic blocks or timelike singularities without meeting $\Sigma$. The CP diagram repeats these analytic blocks; $r=0$ is a timelike curvature singularity.

Near the inner horizon, arbitrarily late exterior radiation is compressed into finite infaller proper time. Its frequency is exponentially blueshifted, with scale $e^{|{\kappa }_{-}|v}$, so even a decaying flux can have diverging measured stress. Generic ingoing and outgoing perturbations produce mass inflation, growing internal mass and curvature. The smooth RN Cauchy horizon is expected to be unstable to back-reaction. This is a generic instability argument, not a universal theorem from every specially tuned test trajectory; the approximation of a harmless test particle becomes inconsistent.

Use geometrized units $G=c=1$ and metric signature $(-,+,+,+)$. Put $f=1-2M/r$ in the Schwarzschild metric. The Schwarzschild tortoise coordinate satisfies $d{r}_{\ast }/dr=f^{-1}$, soSubstituting $dt=dv-f^{-1}dr$ gives the ingoing Eddington-Finkelstein coordinates:The radial metric tensor has determinant $-1$ and inverse components $g^{vv}=0$, $g^{vr}=1$, $g^{rr}=f$. Thus it is nondegenerate and analytic at $r=2M$. The same expression defines a Lorentzian metric for every $r>0$, extending the exterior across the future Schwarzschild event horizon into the black hole. It does not include the other exterior or the white hole of the full Kruskal spacetime. At $r=0$, the Kretschmann scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges, so this is a curvature singularity, not a removable coordinate singularity.

Use the subextremal four-dimensional Reissner-Nordstrom spacetime, with $0<|Q|<M$. Its static radial function isTake the time-symmetric two-ended bridge through the outer bifurcation surface. Its initial data have $K_{ij}=0$ andon each $r\ge r_{+}$ end. The bridge is smooth: in proper radial distance $s$, $r-r_{+}=f^{\prime }(r_{+})s^2/4+O(s^4)$. Each asymptotically flat end is an infinite Riemannian distance away. Hence the spatial Riemannian manifold is complete and admits no proper same-dimensional smooth isometric extension as a connected spatial manifold.

Its maximal Cauchy development includes the two exteriors and the adjacent future and past regions between $r_{+}$ and $r_{-}$. It ends at inner Cauchy horizons, not at a curvature singularity. Since the simple root at $r_{-}$ is removable in horizon-penetrating coordinates and all curvature invariants are finite there, the exact solution extends across these Cauchy horizons. The extension is no longer globally determined by the given initial data.

The shaded region is the maximal Cauchy development; dashed upper and lower null edges are inner Cauchy horizons. The displayed neighboring diamonds illustrate smooth continuation, rather than the entire infinite extension.

The strong cosmic censorship conjecture concerns generic admissible initial data, in a specified extension regularity. Exact charged spherical data are exceptional. Perturbations can produce mass inflation at the inner Cauchy horizon, obstructing suitably regular extensions; the precise conjecture depends on the matter model and whether extensions are required to be $C^0$, $C^2$, or another regularity. Thus this exact extendible example does not refute a generic strong cosmic censorship conjecture.

For an Einstein-Maxwell example the gravitational triple $(\Sigma ,h,K)$ must be accompanied by electromagnetic initial data. One may take zero magnetic field and the smooth radial electric flux $Q/r^2$ through the bridge. The charges at the two ends have opposite signs when measured with outward normals. This is an example in the electrovacuum theory, not vacuum initial data.

Use an affine parameter $\lambda$ along a generator of a null hypersurface. Its null twist vanishes by hypersurface orthogonality. The screen metric is positive definite, so ${\hat{\sigma }}^{ab}{\hat{\sigma }}_{ab}\ge 0$. The Einstein field equations and null energy condition imply $R_{ab}{U}^a{U}^b\ge 0$, giving from the Null Raychaudhuri equationStarting with ${\theta }_0<0$, the null expansion stays negative as long as the congruence remains regular. Set $z=-1/\theta >0$. Then $z^{\prime }={\theta }^{\prime }/{\theta }^2\le -1/2$, soA finite negative null expansion cannot persist to $\lambda -{\lambda }_0=2/|{\theta }_0|$. Thus the null focusing theorem givesThis is a focal breakdown of the transverse congruence, provided the generator and regular geometry extend that far; it is not by itself a curvature singularity or a statement that the individual null geodesic terminates there.

The Einstein field equations and the null energy condition imply the null convergence condition:The scalar-curvature and any cosmological constant terms vanish because $U^a{U}_a=0$. For generators of the null hypersurface, the supplied twist-free property gives $\hat{\omega }=0$. The null shear squared is nonnegative on the screen. Thus the Null Raychaudhuri equation impliesStarting from $\theta ({\lambda }_0)={\theta }_0<0$, the null expansion remains negative for as long as the regular congruence exists. HenceBefore the right-hand side reaches zero, inversion of the negative quantities givesSince the inverse of a finite negative null expansion cannot be nonnegative, the regular congruence cannot continue through the proposed upper limit. The null focusing theorem therefore givesProvided the geodesic itself extends this far, the transverse area collapses and $\theta \to -\infty$ at or before this bound. An earlier end of the affine geodesic would instead be geodesic incompleteness. A divergence of the null expansion marks a caustic or a conjugate point to a spacelike surface; it does not by itself establish a curvature singularity.

For the zero-angular-momentum curves, the fiber term in the induced metric tensor vanishes along the tangent, leavingfor a null geodesic. In the exterior, a future geodesic has positive conserved energy $E=(f/h)\dot{t}$. Its equations areDefine the tortoise coordinate, separately on intervals bounded by horizon radii, byFor an outgoing curve $\dot{r}=+E\sqrt{h}$,for an ingoing curve $\dot{r}=-E\sqrt{h}$, $d(t+r_{\ast })/d\lambda =0$. Thus $u=t-r_{\ast }$ and $v=t+r_{\ast }$ are the respective constant labels. The derivative $\sqrt{h}/f$, with its sign, is required for continuation; taking $\sqrt{h}/|f|$ would give the wrong labels between the horizons.

For a nonextremal horizon $r=r_H$, the tortoise coordinate has leading behaviorAt the double root of the extremal black hole it instead has a pole. These singularities are coordinate effects at a regular Killing horizon; they are not the curvature singularity at $r=0$. To cross the future horizon, the ingoing coordinatesgive the regular induced metric tensorThis also fixes the future branch used in the causal argument below.

Relative to a chosen asymptotically flat end, define the black hole region bywhere ${\mathcal{I}}^{+}$ is that end's future null infinity in a conformal completion. It consists of events unable to send a future causal signal to that infinity; its boundary is the corresponding future event horizon.

For $0<\alpha <r_0/2$, $h>0$ and $f<0$ on $r_{-}<r<r_{+}$. The gradient of $r$ has norm $g^{ab}{\nabla }_{a}r{\nabla }_{b}r=f$, so it is timelike there. Use the future extension regular in the ingoing coordinates above, reached by an exterior future ingoing null geodesic with $\dot{r}=-E\sqrt{h}<0$. This determines $\nabla r$ to be future timelike in the inter-horizon block. Every nonzero future causal tangent $V$ therefore obeysAt the future outer horizon the gradient becomes future null and gives the corresponding one-way inequality $dr/d\lambda \le 0$. No future signal can cross that horizon outward to the selected exterior. The future inter-horizon block is consequently within $\mathcal{B}$; a future-directed path can leave it only through the inner horizon, not return through the outer horizon to the selected infinity.

This statement requires a branch and an end. The maximal extension also contains time-reversed inter-horizon blocks, where future-directed curves have increasing $r$ and emerge into an exterior: they are white holes, not the selected black hole region. Thus the unqualified claim for every copy of $r_{-}<r<r_{+}$ in an arbitrary extension is false. Analytic continuation past a Cauchy horizon can also introduce other asymptotic ends; the displayed definition explicitly refers to the chosen component of infinity.

To sketch the causal structure, suppress the spacelike $\psi$ circle by its orbit-space projection. The induced three-dimensional metric tensor isThe causal projection along a spacelike circular fiber is exact: any causal curve projects to a $q$-causal curve, and every base curve has a lift $d\psi =\Omega dt$ with exactly its base norm. The diagrams therefore represent this two-dimensional orbit space, not a constant-$\psi$ slice of the three-dimensional submanifold.

There are three nonextremal block types. For $r>r_{+}$, $r_{\ast }$ spans the full real line from the outer horizon to infinity, producing an exterior diamond. For $r_{-}<r<r_{+}$, $r_{\ast }$ spans the full line with the time and space roles reversed; these diamonds contain the future black-hole or past white-hole branches. For $0<r<r_{-}$, $r_{\ast }$ is finite at zero and diverges at the inner horizon, producing a static half-diamond with a timelike singular edge. The supplied Kretschmann scalar diverges as $384{\alpha }^4{r}_0^4/r^{12}$ at $r=0$, and $f>0$ on the inner side, so that edge is a genuine timelike curvature singularity. In contrast, both horizon radii are regular in horizon-adapted coordinates.

The finite strip shows two exterior levels, a future trapped block, an inner static level with timelike singularities, and the next emerging block. Continuing through the inner Cauchy horizons repeats this pattern upward and downward in the maximal analytic extension. The inner boundary is not a spacelike Schwarzschild-type singularity. Such an ideal extension need not describe a physical collapse past its unstable inner horizon.

At $\alpha =r_0/2$, put $R=r_0/\sqrt{2}$. Then $r_{+}=r_{-}=R$, $f=(r^2-R^2{)}^2/r^4$, and the interval $r_{-}<r<r_{+}$ is empty. The lapse is positive on both sides, so there is no inter-horizon trapped diamond. The horizon is a degenerate Killing horizon, with $\kappa =0$. In the exterior $r_{\ast }\to -\infty$ as $r\downarrow R$, while inside $r_{\ast }\to +\infty$ as $r\uparrow R$; more precisely its leading pole is $-\sqrt{2}{R}^2/[4(r-R)]$. Static proper distance to the horizon is infinite. Future ingoing rays still cross it at finite affine parameter, as the regular ingoing metric tensor shows.

The extremal sketch gives the exterior diamond and singular interior half-diamond, with the complete gluing prescription: the exterior future horizon attaches to the interior past horizon; the interior future horizon attaches to the past horizon of another exterior. Repeating these attachments gives the maximal extension. The marked throat endpoints are conformal ideal endpoints, not bifurcation points of the spacetime. This block representation avoids incorrectly treating the extremal geometry as two transverse horizons with a collapsed trapped region.

Given an asymptotically flat spacetime with a conformal completion and future null infinity ${\mathcal{I}}^{+}$, its black hole region isThe event horizon is its boundary. For the magnetically charged dilaton black hole, first assume $0\le a=r_{-}<b=r_{+}$. The physical asymptotic component has $r>a$, and the ingoing chart extends regularly across $r=b$. Its future ingoing null direction is $-{\partial }_r$. For $a<r<b$,Hence $\nabla r$ is future timelike. Every nonzero future causal tangent $X$ satisfies $X(r)=g(\nabla r,X)<0$. A future causal curve from this interior can neither cross back to $r=b$ nor reach the asymptotic end $r\to \infty$, so it cannot reach ${\mathcal{I}}^{+}$. It lies in $\mathcal{B}$.

This proof uses the future black-hole extension. The radial interval alone does not distinguish it from the past white hole interior of the maximal extension; that interior has the same interval and can send signals to infinity. Thus the statement needs this future-component qualification if the maximal spacetime is intended.

The Kretschmann scalar confirms that $r=a$ is a genuine curvature singularity: for $0<a<b$, its numerator at $r=a$ is $3a^4(a-b{)}^2$, andFor $a=0$ it instead reduces to $12b^2/r^6$. Since $V<0$ on approaching the singularity from the interior, the singularity is spacelike. At $r=b>a$ the curvature is finite.

For the maximal extension, take $U=-e^{-u/(2b)}$, $W=e^{v/(2b)}$ in the right exterior and continue analytically across the horizons. ThenThe singularity is $UW=(1-a/b)e^{a/b}>0$ and the horizons are $U=0$ and $W=0$. Rescaling the null coordinates by the square root of this positive constant before arctangent compactification produces the usual qualitative Penrose diagram of Schwarzschild spacetime: two exteriors, a future black-hole region, a past white-hole region, and future and past spacelike singularities. The dashed lines below are the horizon branches; only the future interior is the black-hole region.

If $a>b$, the physical domain remains $r>a$, but now $V>0$ throughout it. The apparent zero $r=b$ is beyond the curvature singularity and is not an accessible horizon. The singularity is timelike, and outgoing radial null rays with $\dot{r}=E>0$ escape to infinity from arbitrarily near it. ThusThe excluded equality $r_{-}=r_{+}$ is a singular limiting case, not a regular horizon. These facts describe the causal structure of the magnetic dilaton black hole.

Curvature invariants and the electromagnetic invariant diverge at $r=0$, so it is a genuine curvature singularity, not a coordinate singularity. The causal structures are the higher-dimensional analogues of the familiar four-dimensional diagrams:

A boundary clock in Oppenheimer-Snyder collapse reads $\tau =a_{\max }(\eta +\sin \eta )/2$. Its event horizon crossing occurs at $R=2M$, giving the displayed angle, and the final curvature singularity occurs at $\eta =\pi$. The remaining proper time is $a_{\max }(\pi -{\eta }_H-\sin {\eta }_H)/2$. For $R_0\gg 2M$ this approaches $4M/3$, but it is not exactly that value for finite release radius.

For the Kruskal–Szekeres coordinates of a positive-mass Schwarzschild spacetime, $UV=(1-r/(2M))e^{r/(2M)}$. The displayed conformal compactification makes radial null geodesics lines of slope $\pm 1$. The Killing horizons are $T=\pm X$, and the spacelike curvature singularities $r=0$ lie at $T=\pm \pi /4$, with $|X|<\pi /4$. Two asymptotically flat exterior regions each have past null infinity, null infinity to the future, timelike infinity and spacelike infinity. The two exterior regions cannot communicate by any causal curve; future-directed curves that cross a future horizon instead end at the black-hole singularity.