# Characterization of $\mathbb{R}^n$?

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.

• Do you consider manifolds with boundary? If yes, a smooth ball is a closed smooth ball? – Paul Frost Jan 8 '19 at 9:37
• @PaulFrost - whoops sorry I fixed it - I am just interested in the open case – user101010 Jan 8 '19 at 16:14
• 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? – freakish Jan 8 '19 at 19:15
• 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? – user101010 Jan 8 '19 at 20:19
• Can I assume each ball has compact closure diffeomorphic to a closed ball? – 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.

• 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. – Cheerful Parsnip Jan 9 '19 at 19:05
• @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. – user98602 Jan 9 '19 at 19:09
• Ah okay. That makes sense. – Cheerful Parsnip Jan 9 '19 at 19:13
• 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 – user101010 Jan 20 '19 at 23:55
• @user101010 Why do those have the property that all compact sets are contained in balls? (I do not think there is an issue.) – user98602 Jan 20 '19 at 23:56