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?

  • 9
    $\begingroup$ Stupid example: take the union of a compact non-smoothable manifold with a noncompact manifold. $\endgroup$
    – Jim Conant
    Mar 3 '13 at 14:32
  • 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. $\endgroup$ Mar 3 '13 at 14:49
  • 15
    $\begingroup$ A small open neighborhood of anything in the Euclidean space does admit a smooth structure:)) $\endgroup$ Mar 3 '13 at 15:31
  • $\begingroup$ eek! what has happened to me? :P $\endgroup$ Mar 3 '13 at 22:53

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 $n \geq 1$.

  • 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 $M$. $\endgroup$ Mar 3 '13 at 16:03
  • 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 Agol
    Mar 3 '13 at 23:54
  • $\begingroup$ @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 $K$. $\endgroup$ Mar 4 '13 at 0:15
  • 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. $\endgroup$ Mar 4 '13 at 0:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.