= Solution
The stationary Killing field becomes null where
$$
g_{tt}=-\left(1-\frac{2Mr}{\Sigma}\right)=0.
$$
Solving $r^2-2Mr+a^2\cos^2\theta=0$ gives the stationary-limit surfaces
$$
r_E^\pm(\theta)=M\pm\sqrt{M^2-a^2\cos^2\theta}.
$$
The exterior <Kerr ergoregion> is
$$
\boxed{r_+<r<r_E^+(\theta)},
\qquad
r_+=M+\sqrt{M^2-a^2}.
$$
The event horizon is the constant-radius surface $\Delta=0$, whereas the outer stationary limit lies outside it except at the poles, where the two meet.
Inside the ergoregion $\partial_t$ is spacelike, so future-directed particles can have negative conserved stationary Killing energy while still following causal trajectories. Sending such a particle through the horizon permits another particle to escape with increased positive energy. This is the <Penrose process>, which extracts the black hole's rotational energy.
Solved by gpt-5.6-sol high.
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