There do exist manifolds which do not admit any smooth structure at all. But the only examples I've heard of are all compact.
Are there any non-compact, non-smoothable manifolds?
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10$\begingroup$ Stupid example: take the union of a compact non-smoothable manifold with a noncompact manifold. $\endgroup$– Jim ConantCommented Mar 3, 2013 at 14:32
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1$\begingroup$ I think Kervaire's example embeds into Euclidean space... maybe you could take a small open neighborhood (that deformation retracts back down to the original)? The obstruction is homotopy invariant, so that should do it. $\endgroup$– Dylan WilsonCommented Mar 3, 2013 at 14:49
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15$\begingroup$ A small open neighborhood of anything in the Euclidean space does admit a smooth structure:)) $\endgroup$– Serge LvovskiCommented Mar 3, 2013 at 15:31
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$\begingroup$ eek! what has happened to me? :P $\endgroup$– Dylan WilsonCommented Mar 3, 2013 at 22:53
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1 Answer
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The Cairns-Hirsch theorem says that a PL manifold $M$ is smoothable if and only if $M\times \mathbb{R}$ is smoothable, so you can take $M$ to be any one of the known compact, PL examples such as Kervaire's manifold and then $M\times\mathbb{R}^n$ is non-smoothable for $n \geq 1$.
answered Mar 3, 2013 at 15:04
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4$\begingroup$ I wish to add for the benefit of the OP that any PL manifold $M$ is homotopy equivalent to a smooth manifold: properly embed $M$ into a a smooth $n$-manifold (e.g. Euclidean space) and take a regular neighborhood. I would be interested in any results on how (the smallest) $n$ depends on $M$. $\endgroup$ Commented Mar 3, 2013 at 16:03
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1$\begingroup$ @Igor: I think you have to be a little careful on how it's embedded, so that you can take a regular neighborhood which is homotopy equivalent. $\endgroup$– Ian AgolCommented Mar 3, 2013 at 23:54
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$\begingroup$ @Ian: the standard definition of a regular neighborhood of a subcomplex $K$ in a PL manifold is a closed neighborhood $R$ of $K$ that is a codimension zero PL submanifold, and that collapses to $K$ via finitely many elementary collapses. The last property implies that $R$ is simply homotopy equivalent to $K$. $\endgroup$ Commented Mar 4, 2013 at 0:15
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3$\begingroup$ @Ian: perhaps the concern is that the embedding should be PL? Yes, I thought this is understood since we start from a PL manifold. We do not want wild embedding here, and any PL manifold has a PL embedding into a Euclidean space. $\endgroup$ Commented Mar 4, 2013 at 0:49