= Solution
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>,
$$
g^{ab}\partial_as\,\partial_bs=\frac{\Delta(r_e)+[r_e'(\theta)]^2}{\Sigma}.
$$
On it, $r_e^2-2Mr_e+a^2\cos^2\theta=0$, 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 \b[is not a Killing horizon], despite $K^2=0$ there. For example its stationary null vector is not normal, since $g(K,\partial_\phi)=-2Mar\sin^2\theta/\Sigma\ne0$ 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.
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