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May 22, 2019 at 0:54 comment added Anubhav Mukherjee I guess that is exactly the same argument mentioned by Ian Agol in the comment and a bit generalized result mentioned by Igor.
May 21, 2019 at 21:14 comment added Gro-Tsen How about this: if $M$ is an exotic $\mathbb{R}^4$ then $M\times\mathbb{R}$ is homeomorphic to $\mathbb{R}^5$, but since there is a single differentiable structure on $\mathbb{R}^5$, it is diffeomorphic to it, and restricting such a diffeomorphism to the submanifold $M\times\{0\}$ gives a submanifold of $\mathbb{R}^5$ diffeomorphic to $M$. Did I say something stupid? (I suspect I did, but I can't see it.)
May 21, 2019 at 12:08 vote accept Michael Albanese
May 21, 2019 at 2:03 answer added Igor Belegradek timeline score: 21
May 21, 2019 at 1:30 history migrated from math.stackexchange.com (revisions)
May 13, 2019 at 15:24 history asked Michael Albanese CC BY-SA 4.0