Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 58 / 2 / a / 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 $k\cdot m=-\Omega m^2$ andThus $\boxed{k^2={\Omega }^2{m}^2}$ there. A generator has $dt/d\alpha =1$, $d\phi /d\alpha =\Omega$, making $\Omega =d\phi /dt$ 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.