Let $M$ be a smooth $n$-dimensional manifold with the property that any compact subset $K \subset M$ is contained in an $n$-dimensional smooth ball $K \subset B \subset M$.

If $M$ is open, does it follow that $M$ is diffeomorphic to $\mathbb{R}^n$?

Note that all of the homotopy groups of $M$ must vanish and therefore, by Whitehead's theorem, $M$ is contractible.

  • $\begingroup$ Do you consider manifolds with boundary? If yes, a smooth ball is a closed smooth ball? $\endgroup$ – Paul Frost Jan 8 '19 at 9:37
  • $\begingroup$ @PaulFrost - whoops sorry I fixed it - I am just interested in the open case $\endgroup$ – user101010 Jan 8 '19 at 16:14
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    $\begingroup$ What is an $n$-dimensional smooth ball? Is it simply a ball in $M$ that is diffeomorphic to $\mathbb{R}^n$? And finally isn't an exotic $\mathbb{R}^4$ a counterexample? $\endgroup$ – freakish Jan 8 '19 at 19:15
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    $\begingroup$ An $n$-dimension smooth ball in $B^n$ with the standard smooth structure. Is an exotic $\mathbb{R}^4$ a counterexample? I don't see why. If so, is it the only one? $\endgroup$ – user101010 Jan 8 '19 at 20:19
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    $\begingroup$ Can I assume each ball has compact closure diffeomorphic to a closed ball? $\endgroup$ – user98602 Jan 8 '19 at 20:58

Yes, this is true. First, orient your $n$-manifold $M$ (your hypotheses imply that $M$ is contractible, so this is possible).

First, by your hypothesis, you obtain an increasing exhaustion $M_k \subset M$ of compact sets, diffeomorphic to the $n$-ball, so that each $M_k$ is contained in the interior of $M_{k+1}$.

This is enough; here is the idea. Write $B(k)$ for the unit ball of radius $k$ in $\Bbb R^n$. We may construct, for each $k$, some oriented diffeomorphism $\phi_k: M_k \to B(k)$. If we had $\phi_k \big|_{M_{k-1}} = \phi_{k-1}$, then by taking the increasing union of these $\phi_k$, we define a bijection $M \to \Bbb R^n$ which is a diffeomorphism on the interior of any compact set, and hence is a global diffeomorphism.

In practice, each successive $\phi_k$ has nothing to do with the previous one. Here is how we will resolve this.

Consider the map $g_k: \phi_{k+1}\phi^{-1}_k: B(k) \to B(k+1)$. All we know about this map is that it is a smooth oriented embedding into the interior.

Lemma: There is only one oriented embedding of the $n$-disc into any oriented smooth open $n$-manifold $M$, up to isotopy.

This is a lemma of Cerf and Palais, independently; see here. The idea is to take any smooth embedding to the linear embedding in a chart given by the derivative at zero. In particular, we may find an ambient isotopy $f_t: B(k+1) \to B(k+1)$ which is the identity near the boundary, so that $f_0 = \text{Id}$ and $f_t g_k$ is a smooth isotopy between $f_0 g_k = g_k$ above and $f_1 g_k$ the canonical inclusion map.

Therefore, the map $f_1 \phi_{k+1}$ restricts to $\phi_k$ on $M_k$. So we choose this to be our given diffeomorphism $M_{k+1} \to B(k+1)$. Proceeding inductively, we have our desired result.

  • $\begingroup$ How do we know we can get an exhaustion? I see how to get an increasing union, but I don't see why it has to exhaust the space. $\endgroup$ – Cheerful Parsnip Jan 9 '19 at 19:05
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    $\begingroup$ @CheerfulParsnip Every smooth manifold has a compact exhaustion to begin with by finding a proper smooth function to $\Bbb R$, so we should choose discs using the OP's assumption that are larger than each of the compact sets in our chosen exhaustion. $\endgroup$ – user98602 Jan 9 '19 at 19:09
  • $\begingroup$ Ah okay. That makes sense. $\endgroup$ – Cheerful Parsnip Jan 9 '19 at 19:13
  • $\begingroup$ I think there must be an issue, at least smoothly. There are "small "exotic $\mathbb{R}^4}$'s which embed smoothly into $\mathbb{R}^4$. So these have the property that all compact things are in balls but are exotic. I think your proof works topologically though $\endgroup$ – user101010 Jan 20 '19 at 23:55
  • $\begingroup$ @user101010 Why do those have the property that all compact sets are contained in balls? (I do not think there is an issue.) $\endgroup$ – user98602 Jan 20 '19 at 23:56

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