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If $M$ is a smooth manifold, then for any $p \in M$, $M\setminus\{p\}$ is an open subset of $M$ and therefore inherits a smooth structure from $M$. My question is about the converse.

Let $M$ be a topological manifold and $p \in M$. If $M\setminus\{p\}$ admits a smooth structure, does $M$?

Equivalently,

Is there a non-smoothable manifold that becomes smoothable after removing a point?

If such manifolds exist, are there compact examples?

The non-smoothable manifolds I know of are the simply connected, compact, four-dimensional manifolds with intersection form $mE_8\oplus nH$ where $m > 0$, $n \geq 0$ are integers satisfying $|m| \geq n$ - these are precisely the manifolds ruled out by Furuta's $\frac{10}{8}$ Theorem. But I have no idea whether any of these will provide an example.

Michael Albanese
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  • What about a cone and p as the tip of the cone? – lalala Apr 15 '17 at 18:08
  • @lalala: A double cone is not a manifold, whereas a single cone is homeomorphic to a ball so it admits a smooth structure. – Michael Albanese Apr 15 '17 at 18:37
  • There's a precise statement in Freedman and Quinn's book about constructing smooth structures on open 4-manifolds. It's not on hand, and I don't want to say it incorrectly offhand. I highly doubt there is anything stronger. – PVAL-inactive Apr 16 '17 at 06:05

1 Answers1

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Every open 4-dimensional manifold is smoothable; this is proven by Quinn. Therefore, any connected nonsmoothable 4-manifold will do.

F. Quinn, Ends of Maps III: Dimensions 4 and 5, Journal of Differential Geometry vol. 17 (1982).

Edit: Also, the 8-dimensional manifold $X$ in my answer here has the property that $X$ minus a point is smoothable, although $X$ itself is not. (You have to read Kuiper's paper to understand why.)

Michael Albanese
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Moishe Kohan
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