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Jun 1, 2021 at 19:33 history became hot network question
Jun 1, 2021 at 13:07 comment added Dmitri Panov What is funny is that the answer to this question is unknown in case you remove the assumption on $\Gamma$ to be finite (and act freely). This is a well known question asked by Gromov here: ihes.fr/~gromov/wp-content/uploads/2018/08/… page 12, second paragraph
Jun 1, 2021 at 12:52 history edited Daniele Tampieri CC BY-SA 4.0
Minor grammar improvement and Math Jaxing (formula hyperlinking)
Jun 1, 2021 at 12:44 comment added Sam Nead Oops, I did not see the finiteness assumption! My brain edited it out... ha!
Jun 1, 2021 at 12:35 review Close votes
Jun 10, 2021 at 3:06
Jun 1, 2021 at 12:26 comment added ABIM @abx I was about to ask about isometry groups acting freely and properly discontinuously..so these general answers preemptively helped there :)
Jun 1, 2021 at 12:20 comment added abx Note that the OP asks for finite isometry groups, which implies $\pi_1(X)$ finite... Any compact surface of genus $\geq 1$ is already a counter-example.
Jun 1, 2021 at 12:14 comment added Nick L Oh yeah, thanks!
Jun 1, 2021 at 12:07 comment added Sam Nead But avoiding the product of two-spheres... :)
Jun 1, 2021 at 12:04 comment added Nick L Take any closed, orientable, simply connected $4$-manifold which is not $S^4$.
Jun 1, 2021 at 12:02 vote accept ABIM
Jun 1, 2021 at 11:59 answer added Sam Nead timeline score: 8
Jun 1, 2021 at 11:57 comment added ABIM @user43326 Ah fair, which spaces have this property?
Jun 1, 2021 at 11:42 comment added user43326 This would imply that the universal cover is a product of spheres and euclidean spaces, which clearly is false.
Jun 1, 2021 at 11:30 history asked ABIM CC BY-SA 4.0