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For a regular analytic stationary asymptotically flat vacuum black hole with a connected nondegenerate horizon, the stationary Killing vector field either generates the horizon or the spacetime admits an additional commuting axial Killing vector field. In the rotating case the horizon generator is $K+{\Omega }_H\Phi$, where $\Phi$ has closed rotational orbits. Together with the appropriate staticity and black-hole uniqueness theorem, this identifies an equilibrium uncharged collapse remnant with the Kerr metric. The theorem concerns an equilibrium geometry; it does not itself prove that arbitrary dynamical collapse settles to equilibrium.

On the equatorial plane of the Kerr metric, the displayed component is negative for sufficiently small negative $r$. A curve of constant $t,r,\theta =\pi /2$, with $\varphi$ increasing through $2\pi$, is therefore a regular closed timelike curve. It avoids the ring at $r=0$. Such circles occur on the negative-radius sheet of the exact analytic extension, not in the exterior region accessible from positive-radius infinity.

Ingoing Kerr coordinates cancel the explicit inverse-$\Delta$ singularities of the Boyer-Lindquist coordinates metric. They give a smooth, nondegenerate Kerr metric at the future outer horizon when $M>|a|$. Its stationary and axial Killing vector fields are ${\partial }_v$ and ${\partial }_{\chi }$. The advanced chart need not cover every horizon branch of the maximal extension.

The two-dimensional symmetry-axis restriction of the Kerr metric has simple horizons at $r_{\pm }$ for $0<|a|<M$, but is regular at $r=0$ because the ring is off-axis. Its static region below the inner horizon continues to an asymptotically flat end at negative infinity. The horizon blocks repeat in the maximal analytic extension. This ideal continuation is distinct from a claim of physical stability at the Cauchy horizon.

The Kerr black hole is the asymptotically flat stationary axisymmetric vacuum black hole with mass $M$ and angular momentum $J=Ma$.

A Kerr black hole is the Kerr metric with mass $M>0$ and spin parameter satisfying $|a|\le M$ in geometrized units; its outer horizon is $r_{+}=M+\sqrt{M^2-a^2}$.

The stationary axisymmetric Einstein–Maxwell family of asymptotically flat spacetimes extends the Kerr metric by including electric charge. Its horizon polynomial is the displayed $\Delta$; for $M^2>Q^2+a^2$ the roots are $r_{\pm }=M\pm \sqrt{M^2-Q^2-a^2}$. The outer Killing horizon has area $4\pi (r_{+}^2+a^2)$, angular velocity $a/(r_{+}^2+a^2)$ and electric potential $Q{r}_{+}/(r_{+}^2+a^2)$.

For nonzero spin, the curvature singularity of the Kerr metric is the ring $x^2+y^2=a^2$, $z=0$ in oblate Cartesian coordinates. The rest of the $r=0$ disk is regular and allows continuation to negative $r$. The entire disk must not be mistaken for a curvature singularity, and axial trajectories do not hit the ring.

An ergoregion of a stationary spacetime is where the stationary Killing vector field, normalized as time translation at infinity, is spacelike; a fixed-coordinate stationary observer cannot remain timelike there. For the Kerr metric, $k^2=-(1-2Mr/\Sigma )$, so the outer stationary-limit surface isFor a subextremal rotating hole $r_{\mathrm{E}}\ge r_{+}=M+\sqrt{M^2-a^2}$, with equality only at the poles. At the equator $r_{\mathrm{E}}=2M>r_{+}$. The ergosphere touches the horizon on the rotation axis and is outside it elsewhere. Its boundary is not the horizon, whose generator is $k+{\Omega }_{H}m$.

The drawing uses oblate Cartesian coordinates to display the intersection and is not an isometric embedding of the spatial geometry.

The inverse Kerr metric in ingoing Kerr coordinates gives the raised normal to a surface of constant $r$:On $r=r_{+}$, $\Delta =0$, so the normal is null and tangent to the horizon. It reduces toA constant linear combination of the two Killing vector fields is again a Killing vector field. Therefore $k+{\Omega }_{H}m$ is normal to this null hypersurface, establishing that it is a Killing horizon, with Kerr horizon angular velocityThe second equality uses $\Delta (r_{+})=0$. This normal calculation proves hypersurface orthogonality on the horizon, rather than merely proving that the proposed vector happens to have zero norm there.

Write $A=r^2+a^2$, $s=\sin \theta$, $\Sigma =A-a^2{s}^2$, and $\Delta =A-2Mr$. Use ingoing Kerr coordinates, so $dt=dv-Adr/\Delta$ and $d\varphi =d\chi -adr/\Delta$. To see the cancellations without expanding every term, write the Kerr metric in the equivalent formThe combinations becomeThe first square contributes $-\Sigma d{r}^2/\Delta$, cancelling the explicit radial term. Expanding the remaining terms givesThere is no denominator $\Delta$ in this Lorentzian metric. At the outer horizon $r_{+}=M+\sqrt{M^2-a^2}$, $\Sigma >0$ and the components are smooth. The determinant is $-{\Sigma }^2{\sin }^2\theta$, so away from the usual polar-coordinate degeneracy the metric is nondegenerate and extends across $r_{+}$. The axis can be covered by regular angular charts. Thus the Boyer-Lindquist coordinates are singular there, while the ingoing Kerr coordinates are regular at the future horizon.

For physical nonextremality the invariant parameter condition is $M>|a|$, $M>0$. The printed $M>a$ is sufficient when the rotation orientation has been chosen so that $a\ge 0$; without that convention it needs the absolute value.

Work in the Boyer-Lindquist coordinates exterior $r>r_{+}=M+\sqrt{M^2-a^2}$, away from the axes. There $\Delta >0$, $\Sigma >0$, andInverting the $t,\varphi$ block of the Kerr metric givesThus $W^a=-g^{ab}(dt{)}_b=-{\nabla }^{a}t$ has $g(W,W)=g^{tt}<0$, and the smooth nonvanishing timelike vector field $W$ defines a time orientation. Choose it future-directed, matching increasing $t$ near infinity. For every nonzero future causal vector $Z$, the Lorentzian inner product satisfies $g(W,Z)<0$, henceSo $t$ increases strictly along every regular future causal curve in this exterior. Notice that ${\partial }_t$ itself can be spacelike in the Kerr ergoregion; the timelike object used here is $-\nabla t$. Direct substitution also gives $D=\Delta {\sin }^2\theta$.

For $0<M<a$, $\Delta =(r-M{)}^2+a^2-M^2>0$, so no horizon blocks need adjoining. The oblate Cartesian coordinates of the Kerr metric satisfy the displayed relation, where ${\rho }^2=x^2+y^2$. Keeping only the positive root $r$ leaves a cut along the disk $z=0$, $\rho <a$. Continue $r$ through zero away from its boundary, the Kerr ring singularity, onto a negative-radius sheet. Equivalently cut two Cartesian sheets along the disk and join opposite banks, so $r$ changes sign continuously. The metric is analytic at the disk interior in signed oblate coordinates since $\Sigma =a^2{\cos }^2\theta \ne 0$ there. The extension has asymptotically flat ends at $r\to +\infty$ and $r\to -\infty$; only the singular ring remains inextendible by an analytic Lorentz metric.