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May 12, 2022 at 11:28 comment added HJRW 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/… .
May 11, 2022 at 18:06 comment added Shijie Gu @IgorBelegradek: Thanks. Is it possible to drop the “free”?
May 9, 2022 at 22:17 comment added Igor Belegradek 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.
May 9, 2022 at 21:13 history edited Shijie Gu CC BY-SA 4.0
Answered the first question based on comments
May 9, 2022 at 20:57 comment added Shijie Gu 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)?
May 9, 2022 at 4:17 comment added Ian Agol 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.
May 8, 2022 at 21:15 comment added Igor Belegradek Wouldn't the existence of a cocompact properly discontinuous cocompact group action on a Whitehead manifold contradict orbifold geometrization?
May 8, 2022 at 21:01 history asked Shijie Gu CC BY-SA 4.0