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Can classical contractible manifolds such as Whitehead manifold admit a properly discontinuous cocompact group action?

Here "properly discontinuous" doesn't have to be fixed point free, but the stabilizer of each point must be finite. The question is inspired by an interesting question asked by Agelos.

Edit: Thanks to Igor Belegradek and Ian Agol's comments, the answer is negative due to the (orbiford) geometrization.

Myers ON MAPPING CLASS GROUPS OF CONTRACTIBLE OPEN 3-MANIFOLDS proved that the only finite groups which can act on the Whitehead manifold is cyclic group of order 2. For genus two example the only finite groups which can act are $\mathbb{Z}_2$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2$. Those actions are related to the involutions.

Is there any classification of finitely generated groups acting properly discontinuously and cocompactly on a simply connected open orientable 3-manifold?

Maillot Thm. 1.2 showed that finitely generated groups acting smoothly, properly discontinuously and cocompactly on an open orientable 3-manifold with infinite cyclic $\pi_1$ are virtually closed surface group. So what can one say if the manifold is simply connected?

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    $\begingroup$ Wouldn't the existence of a cocompact properly discontinuous cocompact group action on a Whitehead manifold contradict orbifold geometrization? $\endgroup$
    – Igor Belegradek
    Commented May 8, 2022 at 21:15
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    $\begingroup$ Yes, such an orbifold would be good and have a finite-sheeted manifold cover. It follows from geometrization that the universal cover is R^3. $\endgroup$
    – Ian Agol
    Commented May 9, 2022 at 4:17
  • $\begingroup$ Thank you both. Do you think there is any chance to find a counterexample for Agelos's question (asking if there exists a one-ended simply connected open 3-manifold admitting a properly discontinuous cocompact group action which is not R^3)? $\endgroup$
    – Shijie Gu
    Commented May 9, 2022 at 20:57
  • $\begingroup$ I don't know enough about orbifold geometrization, but here is a comment regarding the manifold case: any one-ended simply connected open 3-manifold that admits a free properly discontinuous cocompact action is aspherical. This goes back to 1949 paper of Specker as mentioned in theorem 15 in sites.ualberta.ca/~gepe/pdf/Peschke_TheoryOfEnds.pdf. In the orbifold case one can try to show the orbifold is good. $\endgroup$
    – Igor Belegradek
    Commented May 9, 2022 at 22:17
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    $\begingroup$ It would be good to clarify if the question is about smooth or topological actions. If the action is smooth then, as the comments point out, the quotient is an orbifold which can be geometrised, and the result follows. If the action is only topological but is free then we can appeal to Moise's theorem to get a smooth action. In the case of a general topological action one might need to appeal to John Pardon's answer here: mathoverflow.net/questions/417973/… . $\endgroup$
    – HJRW
    Commented May 12, 2022 at 11:28

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