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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 outer Killing horizon of a Kerr black hole is generated by $k+{\Omega }_{H}m$, with the stationary Killing vector field $k$ normalized at infinity and the axial generator $m$ having $2\pi$-periodic orbits. In ingoing Kerr coordinates its generators satisfy $d\chi /dv={\Omega }_H$. This is the horizon's angular velocity relative to the nonrotating asymptotic stationary frame.

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.

The change to ingoing Kerr coordinates adds functions of $r$ to $t$ and $\varphi$, and leaves $r,\theta$ unchanged. Differentiating at fixed $r,\theta ,\varphi$ therefore gives ${\partial }_{t}v=1$ and ${\partial }_t\chi =0$. Differentiating at fixed $r,\theta ,t$ gives ${\partial }_{\varphi }\chi =1$ and ${\partial }_{\varphi }v=0$. Hence the two Killing vector fields areThey remain the stationary and axial Killing vector fields; the coordinate change does not mix their generators with ${\partial }_r$.

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.

A generator of the Killing horizon is an orbit of $k+{\Omega }_{H}m$. In ingoing Kerr coordinates it has constant $r=r_{+}$ and $\theta$, with $d\chi /dv={\Omega }_H$. Since the coordinate shifts depend only on $r$, this is also $d\varphi /dt={\Omega }_H$ in the limiting Boyer-Lindquist coordinates description.

Thus ${\Omega }_H$ is the angular velocity of the horizon relative to the nonrotating stationary frame at infinity. The stationary Killing vector field $k$ is normalized to unit time translation there, while the axial Killing vector field has $2\pi$-periodic orbits. This normalization makes the Kerr horizon angular velocity physically definite. It describes the rotation of the null generators and the dragging of inertial frames, not a material solid surface rotating through space. In the Schwarzschild black hole limit $a=0$, ${\Omega }_H=0$.