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It is conjectured that a rotating black hole has at its center a ring-shaped singularity.

Thus, at the center of the ring-shaped singularity the gravitational field must be zero (similar to gravitational field at center of dense object), and gravity must be minimized along the rotational axis. along the rotational axis, the gravitational field will be oriented towards the center from both sides, which will cause matter and eventually spacetime to flow towards the center.

Therefore, it seems plausible that the event horizon is a flattened ellipsoid that has upper and lower inward bulges that are rotationally symmetric around the rotational axis.

Also, for a very large black hole, the event horizon might open up around the rotational axis, such that the event horizon becomes toroidal (donut-shaped).

For such a rotating toroidal black hole, when matter/spacetime flows inward along the rotational axis (potentially from both sides), it seems plausible that some proportion instead of getting caught in the accretion disk will flow along the rotational axis through the center, staying outside the event horizon at all times.

Is this possible?

Most of the above was originally posted in this question: Can a rotating black hole have a donut-shaped event horizon? question, which was closed as too broad, and not reopened after editing to significantly narrow question.

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  • $\begingroup$ S. W. Hawking, Black holes in general relativity, Commun. Math. Phys. 25 (1972) 152–166. $\endgroup$ – MBN Sep 4 '17 at 21:02
  • $\begingroup$ projecteuclid.org/download/pdf_1/euclid.cmp/1103857884. Hawking does prove that Kerr solutions only admit a spherical topology for the event horizon. $\endgroup$ – Halfdan Faber Sep 5 '17 at 0:29

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