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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.

A Killing horizon must be a null hypersurface whose normal is a Killing vector field. The outer stationary-limit surface instead has $s=r-r_e(\theta )=0$ and, in Boyer-Lindquist coordinates,On it, $r_e^2-2M{r}_e+a^2{\cos }^2\theta =0$, so $\Delta (r_e)=a^2{\sin }^2\theta >0$ away from the rotation axis. Its normal is spacelike and the surface timelike. Thus the Kerr stationary-limit surface is generically timelike and is not a Killing horizon, despite $K^2=0$ there. For example its stationary null vector is not normal, since $g(K,{\partial }_{\varphi })=-2Mar{\sin }^2\theta /\Sigma \ne 0$ away from the axis. The actual outer Killing horizon is $r=r_{+}=M+\sqrt{M^2-a^2}$, generated by $K+{\Omega }_H\Phi$, with Kerr horizon angular velocity ${\Omega }_H=a/(r_{+}^2+a^2)$. The two surfaces meet on the axis, which does not make the entire stationary-limit surface null.

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 $a=0$, ${\Omega }_H=0$.

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 $D=\Delta {\sin }^2\theta$.

On the north-pole surface, $\theta =0$ and $d\theta =d\varphi =0$, while $\Sigma =r^2+a^2$. Pulling back the Boyer-Lindquist coordinates metric therefore gives