 Solution

In the extremal case, the squared norm of the Killing vector field isFor $M=|Q|>0$, it is negative for every regular $r\ne M$ and vanishes precisely at $r=M$ in the extended spacetime. ThusThe static chart cannot itself test the geometry at its horizon. In the regular Eddington-Finkelstein coordinates, the normal to $h=r-M=0$ is $n=dr$, andAt $r=M$ this gives $n^{♯}={\partial }_v=K$, a nonzero null vector tangent to the surface. Therefore $r=M$ is a null hypersurface, even though $F$ has the same sign on its two sides.