Past exam of the general relativity II course of the University of Oxford / 2017 / 3 / a / Solution

A regular hypersurface $f=0$, with $df\ne 0$, is a null hypersurface if its normal covector has zero squared norm:Its normal vector is then also tangent to the hypersurface; the induced metric tensor is degenerate.

For the Kerr metric the normal to a constant-$r$ surface has squared normFor $0<a<M$ the two roots are positive and distinct. To verify that they are genuine regular null hypersurfaces, rather than trusting singular Boyer-Lindquist coordinates, introduce ingoing Kerr coordinatesSubstitution givesThere are no ${\Delta }^{-1}$ terms. At either root $\Sigma >0$, and the determinant is $-{\Sigma }^2{\sin }^2\theta$ away from the angular coordinate poles. Thus the metric tensor extends smoothly across each root. The raised normal on either surface isThis is a nonzero null Killing vector field normal to the surface. Both surfaces are Killing horizons, with horizon angular velocities ${\Omega }_{\pm }$.

The event-horizon terminology in the paper needs qualification. In the usual asymptotically flat Kerr black hole extension, relative to a chosen exterior's future null infinity, the future branch of $r=r_{+}$ is the event horizon, the boundary $\partial J^{-}({\mathcal{I}}^{+})$. In the intervening black-hole region $\Delta <0$, the coordinate $r$ has timelike gradient and future-directed curves move toward smaller $r$; they cannot return across the outer boundary to that exterior. The inner surface $r=r_{-}$ is a Cauchy horizon, not a second event horizon for the same asymptotic end. Nullness is a local property; identifying an event horizon additionally requires this global causal structure.^[ref]

The allowed endpoint $a=0$ provides an especially concrete counterexample to the unrestricted description of both roots. The Kerr metric then becomes Schwarzschild spacetime, $r_{+}=2M$ is its regular event horizon, but $r_{-}=0$ is a curvature singularity:It is not a regular inner null hypersurface. Thus the two-horizon calculation above assumes $0<a<M$.