Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 / 1 / c / Solution

The normal to a surface of constant $r$ has squared normwhich vanishes at $r=r_{+}$. Thus this surface is a null hypersurface. The stationary Killing vector fieldhas $K^2=g_{tt}=-f$, so it becomes null there and generates a Killing horizon. Since $f$ has a simple zero, its surface gravity isA horizon cross-section has topology $S^3\times S^1$, where the circle is the periodic $z$ direction. Including a complete generator, the null hypersurface has topology