Question: Let $M$ be a contractible noncompact smooth manifold (without boundary) of dimension $n \geq 5$. Is $M$ diffeomorphic to an open subset of $\mathbb{R}^n$?
[I know that the $M$ is not necessarily diffeomorphic to $\mathbb{R}^n$ --- consider a Whitehead manifold. However, all the Whitehead manifolds I know of are diffeomorphic to an open subset of $\mathbb{R}^n$. EDIT: I assume $n \geq 5$ to rule out exotic $\mathbb{R}^4$.]
Shijie Gu in
Gu, Shijie, Contractible open manifolds which embed in no compact, locally connected and locally 1-connected metric space, Algebr. Geom. Topol. 21, No. 3, 1327-1350 (2021). ZBL1476.57027.
proves that for every $n\ge 3$ there exists an open contractible $n$-dimensional manifold $W$ which cannot embed (even topologically) in any compact $n$-dimensional manifold. (He proves much more than that.) The result was claimed earlier by Sternfeld in his dissertation, but his proof contained a mistake. In particular, the manifolds $W$ do not embed in $\mathbb R^n$. Gu does not state this, but his manifolds $W$ are smooth, by the construction. (Actually, all contractible topological manifolds of dimension $\ge 5$ are smoothable and smooth structure is unique up to a diffeomorphism.)
On the other hand, if $W$ is a tame $n$-dimensional manifold ($n\ge 5$) then $W$ embeds smoothly in $\mathbb R^n$. Here is a sketch of the proof. I will assume that $W$ is diffeomorphic to the interior of a smooth compact manifold with boundary $\bar{W}$. Let $M$ denote the double of $\bar{W}$ acros its boundary. One verifies that $M$ is a smooth $n$-dimensional homotopy sphere. Hence, after removing one point (any point) from $M$ we obtain a manifold diffeomorphic to $\mathbb R^n$. Hence, $W$ embeds smoothly in $\mathbb R^n$.
