In his famous list of Problems in Low-Dimensional Topology, Kirby states the following as Problem 3.14 (B), which is attributed to Thurston:

Conjecture: Suppose $G$ (an arbitrary group I suppose) acts properly and discontinuously on a contractible 3-manifold with compact quotient. Suppose also that $G$ has no subgroup isomorphic to $Z\oplus Z$. Then $G$ is conjugate to a discrete group of isometries of hyperbolic 3-space.

Has this been settled, and if not would you still consider it interesting after the Geometrization theorem?

Going through Kirby's references I was not able to spot this question in one of Thurston's works; perhaps it is formulated differently there. A concrete reference would be appreciated.

  • 2
    $\begingroup$ Doesn't this follow from the geometrization conjecture? $\endgroup$
    – Marc Kegel
    May 4 at 11:51
  • $\begingroup$ @LeeMosher: Since the action is not assumed to be free, the quotient $O:=M/G$ is an orbifold, not a manifold. So you would need to apply the orbifold geometrization theorem. But the action is not assumed to be smooth either, so that that $O$ is a topological orbifold, not a smooth one. Do you still see how to put a $G$-invariant hyperbolic metric on $M$? $\endgroup$
    – Agelos
    May 4 at 14:27

1 Answer 1


This problem is answered in the literature, with a caveat. As indicated in the comments, it follows from the orbifold theorem + geometrization conjecture (to handle the case of orbifolds without fixed points).

The caveat is that the action needs to be assumed smooth (or PL). Otherwise, there exists wild involutions such as the Bing involution. One could incorporate this into an action on $\mathbb{R}^3$ by taking an orbifold quotient whose underlying space is $S^3$ and admits a reflection symmetry, then take a Bing involution preserving the orbifold locus by inserting into a part of the sphere being reflected. The resulting group action would have quotient which is not an orbifold. I think this was an underlying assumption of the problem as stated, but I thought I would clarify.

With the smoothness assumption, now this follows from the orbifold theorem (see Problem 3.46 of Kirby’s list posed by Geoff Mess). I won’t go through the history of the proofs of cases of this (see the link), but it follows for orientable orbifolds modulo previous results by the proof of the geometrization theorem by Perelman.

For the non-orientable case, take an index-two subgroup $G’<G$ which is orientation-preserving. By the orientable case (and no $Z+Z$ subgroup assumption), $\mathbb{R}^3/G’$ is an orientable hyperbolic 3-orbifold. By Selberg’s lemma, there is a finite-index subgroup $G’’ < G’$ for which $M’’=\mathbb{R}^3/G’’$ is a manifold. By passing to a further subgroup (the core), we may assume that $G’’\lhd G$, so $G/G’’$ acts as a finite group of transformations of $M’’$. Now one may apply a result of Dinkelbach-Leeb (Theorem H) to see that the quotient orbifold $M’’/(G/G’’)= \mathbb{R}^3/G$ is a hyperbolic orbifold.

  • 1
    $\begingroup$ This is a very helpful answer. So Kirby was obviously tacitly assuming smoothness of the action. Interestingly, Kirby's statement remains true without smoothness if one replaces "conjugate" by "isomorphic: such a group $G$ will admit a smooth action with the same properties (but not necessarily conjugate to the original one) by Pardon's theorem discussed here. $\endgroup$
    – Agelos
    May 6 at 15:25

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