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This question was prompted by my answer to this question.

An exotic $\mathbb{R}^4$ is a smooth manifold homeomorphic to $\mathbb{R}^4$ which is not diffeomorphic to $\mathbb{R}^4$ with its standard smooth structure. An exotic $\mathbb{R}^4$ is said to be small if it can be embedded in the standard $\mathbb{R}^4$ as an open subset. An exotic $\mathbb{R}^4$ which is not small is called large.

Does every large $\mathbb{R}^4$ embed in $\mathbb{R}^5$?

Freedman and Taylor showed there is a maximal exotic $\mathbb{R}^4$, into which all other $\mathbb{R}^4$'s can be smoothly embedded as open subsets. So it would be enough to show that this one embeds in $\mathbb{R}^5$.

By the Whitney Embedding Theorem, every large $\mathbb{R}^4$ embeds in $\mathbb{R}^8$, but I suspect one can do better.

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    $\begingroup$ How about this: if $M$ is an exotic $\mathbb{R}^4$ then $M\times\mathbb{R}$ is homeomorphic to $\mathbb{R}^5$, but since there is a single differentiable structure on $\mathbb{R}^5$, it is diffeomorphic to it, and restricting such a diffeomorphism to the submanifold $M\times\{0\}$ gives a submanifold of $\mathbb{R}^5$ diffeomorphic to $M$. Did I say something stupid? (I suspect I did, but I can't see it.) $\endgroup$ – Gro-Tsen May 21 at 21:14
  • $\begingroup$ I guess that is exactly the same argument mentioned by Ian Agol in the comment and a bit generalized result mentioned by Igor. $\endgroup$ – Anubhav Mukherjee May 22 at 0:54
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The product of any two smooth open contractible manifolds is diffeomorphic to a Euclidean space, see e.g. remark 5.3 in my paper; the result is of course not due to me but I don't know any other place where all references are collected.

In particular, this applies to the product of $\mathbb R$ and any exotic $\mathbb R^4$, so the latter embeds into $\mathbb R^5$. In this case the point is that the product is simply-connected at infinity, and hence by a result of Stallings it is PL homeomorphic to $\mathbb R^5$, but any PL structure on $\mathbb R^5$ is induced by a unique smooth structure (as proved by Munkres).

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  • $\begingroup$ Wow, that first result is amazing. Thanks! $\endgroup$ – Michael Albanese May 21 at 12:00
  • $\begingroup$ What can we say about the end of such embedding? $\endgroup$ – Anubhav Mukherjee May 21 at 15:04
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    $\begingroup$ @AnubhavMukherjee: I do not know what you mean an "end of an embedding" but I think the diffeomorphism of the product onto $\mathbb R^5$ is fairly explicit. Take a look at Stallings' paper maths.ed.ac.uk/~v1ranick/papers/stallings2.pdf. $\endgroup$ – Igor Belegradek May 21 at 15:21
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    $\begingroup$ This just follows from the fact that $\mathbb{R}^n$ has a unique smooth structure for any $n\neq 4$ from the Stallings paper you reference in your comment. In particular, there is a proper embedding en.wikipedia.org/wiki/Exotic_R4 $\endgroup$ – Ian Agol May 21 at 16:10

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