# Are there non-compact, non-smoothable manifolds?

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?

• Stupid example: take the union of a compact non-smoothable manifold with a noncompact manifold. – Jim Conant Mar 3 '13 at 14:32
• 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. – Dylan Wilson Mar 3 '13 at 14:49
• A small open neighborhood of anything in the Euclidean space does admit a smooth structure:)) – Serge Lvovski Mar 3 '13 at 15:31
• eek! what has happened to me? :P – Dylan Wilson 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 $n \geq 1$.
• 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 $M$. – Igor Belegradek Mar 3 '13 at 16:03
• @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 $K$. – Igor Belegradek Mar 4 '13 at 0:15