I'm trying to find a statement and proof of the Hawking rigidity theorem for black holes, and I'm not having much success. The theorem basically says that given some assumptions, the event horizon of a stationary black hole is a Killing horizon. All the references I've found cite The large scale structure of space-time (1973), and some in particular refer to proposition 9.3.6:

Proposition 9.3.6

Let $(\mathcal{M}, \mathbf{g})$ be a stationary non-static regular predictable space, in which the ergosphere intersects $\dot{J}^- (\mathscr{I}^+, \bar{\mathcal{M}}) \cap J^+ (\mathscr{I}^-, \bar{\mathcal{M}})$. Then there is a one-parameter cyclic isometry group $\tilde{\theta}_\phi (0 \leq \phi \leq 2\pi)$ of $(\mathcal{M}, \mathbf{g})$ which commutes with $\theta_t$, and whose orbits are spacelike near $\mathscr{I}^+$ and $\mathscr{I}^-$.

However, like the text itself says, this theorem just seems to state that a non-static black hole is axisymmetric. Can this be related to the existence of a Killing vector normal to the horizon? Or does the rigidity theorem follow from somewhere else?

  • $\begingroup$ Intuitively, if the spacetime is stationary in the sense that it has a Killing vector $\xi$ that is timelike everywhere outside the horizon, then $\xi$ must be null on the horizon (because a horizon can't have a timelike tangent), which is almost the definition of a Killing horizon. (It would be, if $\xi$ is spacelike inside the horizon.) But apparently my intuition is too simplistic, because arxiv.org/abs/gr-qc/9610010 gives a much more elaborate proof of a "corrected version of the black hole rigidity theorem." I haven't studied it, but it looks like it might answer your question. $\endgroup$ Jun 3, 2020 at 2:55
  • $\begingroup$ Also, from pages 21-22 in web.physics.ucsb.edu/~pasi/Wald1.pdf, I get the impression that the rigidity theorem is an input to the proof of axisymmetry, instead of the other way around. Did you see references where it's the other way around? $\endgroup$ Jun 3, 2020 at 2:55
  • $\begingroup$ @ChiralAnomaly Your intuitive argument assumes that $\xi$ is tangent to the horizon, which to be honest I don't see why it should be the case. Still, thank you for that paper, it will surely be useful. Regarding the second PDF, it's not too clear to me whether the text below the theorem is a proof or a corollary! I do know I saw a paper (I don't remember which one now) that directly cited 9.3.6. $\endgroup$
    – Javier
    Jun 3, 2020 at 15:32
  • $\begingroup$ And if nothing else shows up for a few days, you could compile your comments into an answer, they have certainly helped. $\endgroup$
    – Javier
    Jun 3, 2020 at 15:33
  • 1
    $\begingroup$ @Mr.Feynman Unfortunately I didn't, and I sort of abandoned this subject later. $\endgroup$
    – Javier
    Feb 6 at 12:56


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