M.Freedman and R.Gompf's work show that there are at least 13 exotic structures in $S^3\times \mathbb{R}$, which is a open 4-manifold, so now I wonder whether there is an exotic structure in $S^3\times [0,\infty)$ such that its boundary $S^3\times \{0\}$ is a smooth 3-manifold?
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$\begingroup$ What about the structure induced by the inclusion as a submanifold with boundary of one of the exotic open manifolds you mention? $\endgroup$– B KCommented Apr 5, 2023 at 19:38
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3$\begingroup$ How did you get the number 13? $\endgroup$– Anubhav MukherjeeCommented Apr 5, 2023 at 21:12
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$\begingroup$ At least 13, as a result of connected sum of two some exotic R^4 $\endgroup$– Longteng ChenCommented Apr 6, 2023 at 9:14
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$\begingroup$ The boundary of a smooth $n$-manifold is always a smooth $(n-1)$-manifold (or is empty). So your question about the boundary $S^3 \times \{0\}$ does not make sense to me? $\endgroup$– Sam NeadCommented Apr 6, 2023 at 14:41
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2 Answers
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If you take an exotic $\mathbb{R}^4$ and remove an open ball lying in a chart, the result is an exotic smooth structure on $S^3 \times [0,\infty)$ (with smooth boundary). As Anubhav mentions, you can do this with all of the uncountably many exotic smooth structures on $\mathbb{R}^4$ constructed in Taubes' paper.
answered Apr 5, 2023 at 22:44
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1$\begingroup$ I am pretty sure that there is some exotic R^4 can't admit a smooth embedded S^3, for example the Freedman-Donaldson exotic R^4, so in this situation, your S^3*[0,\infty) is no longer a smooth manifold with boundary because the boundary is something fractal. $\endgroup$ Commented Apr 6, 2023 at 9:12
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1$\begingroup$ There are two different things going on here. You are correct that there is no $S^3$ `at infinity' in such an exotic $\mathbb{R}^4$. (In other words, there is a compact set K such that it is not enclosed by a smooth $S^3$; that's how you show it's exotic!) On the other hand, every smooth $n$-manifold contains a smoothly embedded $n$-ball (just work in a chart) and hence a smooth $n-1$ sphere. I was referring to the latter construction. $\endgroup$ Commented Apr 6, 2023 at 13:00
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One can use Gompf's end sum construction to produce uncountably many exotic structure on $S^3\times \mathbb R$. And I am pretty sure that the same proof holds for $S^3\times [0,\infty)$. One can generalize it for $Y\times \mathbb R$ as well, see https://arxiv.org/pdf/dg-ga/9604007.pdf [Theorem 4].
answered Apr 5, 2023 at 21:10