= Solution
Axisymmetry preserves the horizon, so the axial <Killing vector field> $m$ is tangent to it, so its null normal $\xi=k+\Omega m$ obeys $\xi\cdot m=0$ and $\xi^2=0$. Therefore $k\cdot m=-\Omega m^2$ and
$$
0=k^2+2\Omega(k\cdot m)+\Omega^2m^2=k^2-\Omega^2m^2.
$$
\b[Thus $\boxed{k^2=\Omega^2m^2}$ there.] A generator has $dt/d\alpha=1$, $d\varphi/d\alpha=\Omega$, making $\Omega=d\varphi/dt$ its angular velocity relative to the nonrotating stationary frame at infinity. For Kerr it is the <Kerr horizon angular velocity> $a/(r_+^2+a^2)$; the argument itself is not Kerr-specific.
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