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

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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• 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.) – Gro-Tsen May 21 at 21:14
• I guess that is exactly the same argument mentioned by Ian Agol in the comment and a bit generalized result mentioned by Igor. – Anubhav Mukherjee May 22 at 0:54

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).
• @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. – Igor Belegradek May 21 at 15:21
• 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 – Ian Agol May 21 at 16:10