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Jun 25, 2020 at 16:00 comment added Igor Belegradek @MichaelAlbanese: no worries. I was too lazy to post an answer, and it is good that you did that, since the answers get more visibility.
Jun 25, 2020 at 14:04 comment added Michael Albanese @IgorBelegradek: Sorry, I didn't see your comment when I posted my answer which is effectively your first sentence.
Jun 25, 2020 at 13:18 history edited Willie Wong
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Jun 25, 2020 at 11:44 answer added Michael Albanese timeline score: 23
Jun 25, 2020 at 11:26 comment added Igor Belegradek There are closed flat manifolds that are not homotopy equivalent but become diffeomorphic after product with $S^1$, see math.stackexchange.com/questions/396608/…. Also if $M$ is closed manifold of dimension $\ge 6$ (maybe $5$?) such that the Whitehead group of $\pi_1(M)$ is nonzero, then there is a $h$-cobordism $W$ from $M$ to a manifold $L$ not diffeomorphic to $M$, and $W\times S^1$ is trivial, so $M\times S^1$, $L\times S^1$ are diffeomorphic.
Jun 25, 2020 at 3:17 vote accept Mohammad Farajzadeh-Tehrani
Jun 25, 2020 at 3:16 answer added Anubhav Mukherjee timeline score: 29
Jun 25, 2020 at 3:06 comment added Mohammad Farajzadeh-Tehrani @ Mukherjee: Definitely means you know a result, please let me know.
Jun 25, 2020 at 3:04 comment added Mohammad Farajzadeh-Tehrani @ Chris: I am sure it is. I don't know where to find it. I googled but did not find anything. I asked our topologists; no response yet. The one with S^2 instead of S^1 is (negative) pretty famous and important.
Jun 25, 2020 at 3:04 comment added Anubhav Mukherjee Definitely not true if M is of dim 4.
Jun 25, 2020 at 3:02 comment added Chris Gerig This has to be in the literature somewhere... Internet suggests it's true for simply-connected n-manifolds with n > 4.
Jun 25, 2020 at 3:01 history edited Mohammad Farajzadeh-Tehrani
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Jun 25, 2020 at 2:43 history asked Mohammad Farajzadeh-Tehrani CC BY-SA 4.0