Exact Schwarzschild horizon data at the zero-rotation KN point

Substituting $M=m$, $a=q=0$ into [F1] gives $\mathrm{\Sigma }=r^2$, $f_0=f=1-2m\mathrm{/}r$, the displayed ingoing metric and $N={\mathrm{\partial }}_v-\frac{1+f^2}{2}{\mathrm{\partial }}_r$; the potential vanishes. The inverse of the $(v,r)$ metric block $\left(\begin{array}{cc}-f& 1\\ 1& 0\end{array}\right)$ is $\left(\begin{array}{cc}0& 1\\ 1& f\end{array}\right)$, so $g^{-1}(dr,dr)=f$. At $r_H=2m>0$, $f(r_H)=0$ and ${\chi }^{\mathrm{♭}}=g({\mathrm{\partial }}_v,\cdot )=-fdv+dr=dr$, with $dr(\chi )=0$. Thus $H=\{r=2m\}$ is a null hypersurface and $\chi$ is tangent, null and normal there. Since every coefficient is independent of $v$, ${\mathcal{L}}_{\chi }g=0$, so [F3] makes $\chi$ Killing; $\chi$ never vanishes, so the nonzero-generator part is all of $H$. Finally $g(\chi ,N)=dr(N)=-1\mathrm{/}2<0$ on $H$, so its generator is future-directed. This verifies the exact horizon conditions in [F2].

The Killing norm is $g(\chi ,\chi )=-f$, so $d(g(\chi ,\chi ))=-f^{\mathrm{\prime }}(r_H)dr$. From $f(r)=1-2m\mathrm{/}r$ and [F5], $f^{\mathrm{\prime }}(r)=2m\mathrm{/}{r}^2$, so at $r_H=2m$ this is $-\frac{1}{2m}dr$. Comparing with [F2] and ${\chi }^{\mathrm{♭}}=dr$ gives $-2{\kappa }_{H}dr=-\frac{1}{2m}dr$, hence ${\kappa }_H=1\mathrm{/}(4m)$. A section at fixed $v$ and $r=2m$ has induced metric $(2m{)}^2\gamma$. In spherical coordinates $\gamma =d{\theta }^2+{\sin }^2\theta d{\psi }^2$, so its area is $(2m{)}^2{\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\psi =16\pi m^2$. Substitution in [F4] gives the same area and surface-gravity functions and gives $j={\mathrm{\Omega }}_{\mathrm{K}\mathrm{N}}={\mathrm{\Phi }}_{\mathrm{K}\mathrm{N}}=0$.