4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What about a cone and p as the tip of the cone? $\endgroup$ – lalala Apr 15 '17 at 18:08
  • $\begingroup$ @lalala: A double cone is not a manifold, whereas a single cone is homeomorphic to a ball so it admits a smooth structure. $\endgroup$ – Michael Albanese Apr 15 '17 at 18:37
  • $\begingroup$ 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. $\endgroup$ – PVAL-inactive Apr 16 '17 at 6:05
4
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ For those interested, the stated result is Corollary 2.2.3 of the aforementioned paper. $\endgroup$ – Michael Albanese Sep 9 '18 at 22:18

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.