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Oct 28, 2018 at 13:05 comment added Liviu Nicolaescu In dimension 4 there are the so called Mazur manifolds that are contractible and not diffeomorphic to the ball. en.wikipedia.org/wiki/Mazur_manifold
Oct 27, 2018 at 18:55 comment added mme @Igor Thank you for the correction. I need to rely on something more precise than my weak memory!
Oct 27, 2018 at 18:53 comment added Igor Belegradek @MikeMiller: it is actually Stallings. The reference is in en.wikipedia.org/wiki/Simply_connected_at_infinity. Siebenman of course also worked on this topic starting from his thesis, see citeseerx.ist.psu.edu/viewdoc/….
Oct 27, 2018 at 18:37 comment added mme Siebenmann! Not Stallings. My mistake.
Oct 27, 2018 at 16:40 comment added mme In dimension 2 every simply connected manifold is homeomorphic to either the 2-sphere or the ball.
Oct 27, 2018 at 16:39 comment added mme There are contractible manifolds in every dimension at least 3 whose 'fundamental group at infinity' is not trivial, hence are not homeomorphic to a ball. It is a theorem of Stallings that any contractible manifold of dimension at least 5 with trivial fundamental group at infinity is homeomorphic to the ball, and a theorem of smoothing theory (unfortunately I don't know the names to cite here) that one may even say diffeomorphic there.
Oct 27, 2018 at 16:28 comment added MathBug Sure, stupid question. I was going to delete it.
Oct 27, 2018 at 16:21 comment added Najib Idrissi @RGC Yes, it does kill the last homology group. For a higher dimensional example, just take the product with $\mathbb{R}^n$.
Oct 27, 2018 at 16:20 review Close votes
Nov 1, 2018 at 3:05
Oct 27, 2018 at 16:19 comment added MathBug So is there a general result for higher dimensions? $dimM=4$ is still a low-dimensional case.
Oct 27, 2018 at 16:14 comment added MathBug You are right, thank you very much.
Oct 27, 2018 at 16:08 comment added Anubhav Mukherjee They are simply connected, so definitely oriented.
Oct 27, 2018 at 16:04 comment added MathBug Does that kill the last homology group, Najib?
Oct 27, 2018 at 16:01 comment added Najib Idrissi Or simply remove a point from the Poincaré sphere, no? The result has nontrivial $\pi_1$ so it's certainly not homeomorphic to an open ball.
Oct 27, 2018 at 16:01 comment added MathBug Are there examples that are oriented?
Oct 27, 2018 at 15:37 comment added Anubhav Mukherjee Freedman proved that every homology sphere bounfs a contractible 4 topological manifolds. Consider such a smooth one and take it's boundary out. Now the resultant manifold is not open ball.
Oct 27, 2018 at 15:20 review First posts
Oct 27, 2018 at 16:21
Oct 27, 2018 at 15:17 history asked MathBug CC BY-SA 4.0