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Up to dimension 3, homeomorphic manifolds are diffeomorphic (in particular, homeomorphic open subsets of $\mathbb{R}^n$ ($n\leq 3$) are diffeomorphic). It is known that there are uncountably many mutually non-diffeomorphic open subsets of $\mathbb{R}^4$ that are homeomorphic to $\mathbb{R}^4$ (Demichelis & Freedman, 1992). For $n\neq 4$, $\mathbb{R}^n$ admits a unique smooth structure (up to diffeomorphism), thus the previously mentioned phenomenon is unique to dimension 4.

Questions: In dimension $n\geq 5$,

  1. is it possible for a manifold to be homeomorphic to an open subset of $\mathbb{R}^n$ but not diffeomorphic to it?

  2. is it possible that two homeomorphic open subsets of $\mathbb{R}^n$ may be non-diffeomorphic?

Are there examples of such open sets, in each dimension $n\geq 5$, having a simple description up to homeomorphism?

(the smooth structure on each open subset of $\mathbb{R}^n$ being the standard one).

Edit: If I understand it well, Misha Kapovich points out in his comment below that Robion Kirby told him that the answer to question (2) above is positive, in every dimension $n\geq 5$ (and thus in every dimension $n\geq 4$, with the difference that for $n\geq 5$ no such sets can be homeomorphic to $\mathbb{R}^n$). As (2) implies (1) this would settle the question.

It would be interesting to know if other mathematicians heard about these examples of Kirby. From the topological point of view (i.e. up to homeomorphism), what are these sets?

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    $\begingroup$ This question seems to be a duplicate of this one mathoverflow.net/questions/114528/… but I haven't voted to close since your specific question doesn't seem to be answered explicitly there. The comments of Tom Goodwillie and Igor Belegradek seem to state that the answer is no though. $\endgroup$ – j.c. Sep 20 '17 at 19:13
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    $\begingroup$ Question (2) is raised in mathoverflow.net/questions/114528/… but not answered there. The discussion seems to turn around the known existence of small exotic R4's on one hand, and contractible opens sets on the other. Apriori, the answer could be positive for (1) above and negative for (2). $\endgroup$ – Nautilus Sep 20 '17 at 19:36
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    $\begingroup$ About 10 years ago Rob Kirby told me about an example of open domains in $R^n$, $n\ge 5$, which are homeomorphic but homeomorphisms are not isotopic to diffeomorphisms. Sadly, I do not remember what the examples were. (I noted this in a list of open problems about boundaries of groups compiled at AIM in 2007.) One can ask Rob again, maybe he still remembers. $\endgroup$ – Misha Sep 25 '17 at 19:54
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    $\begingroup$ Misha: thank you very much for the useful hint. If i understand it well, what Kirby told you is that there are homeomorphic open subsets of $R^n$, $n\geq 5$, that are not diffeomorphic (each of them endowed with its standard smooth structure). Your expression "are homeomorphic but homeomorphisms are not isotopic to diffeomorphisms" is equivalent to what is written above, but somewhat less clear (a diffeomorphim is a homeomorphism and it is isotopic to itself, thus no diffeomorphism exists). $\endgroup$ – Nautilus Sep 26 '17 at 4:40
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I finally talked to Rob and did some literature search. Here are some examples of open subsets of Euclidean spaces which are homeomorphic but not diffeomorphic.

Let $\Sigma$ be an exotic $(n-1)$-dimensional sphere which can be realized as a Brieskorn variety (see here): $$ V(a):=\{z\in S_\epsilon^{2m+1} \, : \, z_0^{a_0} + \dots + z_m^{a_n} =0 \},$$ where $2m=n$ and $S_\epsilon^{2m+1}=\{z: ||z||=\epsilon\}$ for some sufficiently small $\epsilon>0$.

For instance, any 7-dimensional exotic sphere will do the job.

Next, if $M^{n-1}$ is an $(n-1)$-dimensional (smooth) Brieskorn variety $V(a)$, it has a natural embedding in $S^{n+1}$ and the image has trivial normal bundle (since it is the regular level set of a smooth map to ${\mathbb C}$; this smooth map comes from the defining equation of $V(a)$). Hence, our $\Sigma$ embeds in $R^{n+1}$ such that the tubular neighborhood $U_\Sigma$ of the image is diffeomorphic to $\Sigma\times R^2$. Let $S$ be the standard $n-1$-dimensional sphere; it also embeds in $R^{n+1}$ with trivial normal bundle, of course, and we get a diffeomorphism $U_S\to S\times R^2$ for the tubular neighborhood $U_S$ of $S$ in $R^{n+1}$. It is then clear that $U_\Sigma$ is homeomorphic to $U_S$.

Claim. The open subsets $U_\Sigma$ and $U_S$ of $R^{n+1}$ are not diffeomorphic; equivalently, the manifolds $\Sigma\times R^2$ and $S\times R^2$ are not diffeomorphic.

This is where the story gets a bit complicated. Rob just says that he knows this, but does not remember a reference. The claim about nonexistence of a diffeomorphism appears on page 150 (Theorem 1) in

S. Kwasik, R. Schultz, Multiplicative stabilization and transformation groups. Current trends in transformation groups, (2002) 147–165.

But instead of a proof they provide six references none of which contains this claim (at least in a recognizable form). I have no reason to doubt that the proof can be extracted from some combination of their references. Instead, I looked in their other paper:

S. Kwasik, R. Schultz, Toral and exponential stabilization for homotopy spherical spaceforms. Math. Proc. Cambridge Philos. Soc. 137 (2004), no. 3, 571–593.

In that paper they pretty much provide a proof of this claim in section 4. (Their argument is used for space-forms.)

Here is a sketch of their argument. Suppose that there is a diffeomorphism $$ f: \Sigma\times R^2\to S\times R^2. $$ Let $S_a^1\subset R^2, S_b^1\subset R^2$ be concentric circles of radii $a$ and $b$, where $a$ is much larger than $b$. Then the submanifolds $S\times S^1_b$ and $f(\Sigma\times S_a^1)$ cofound a smooth compact submanifold $W\subset S\times R^2$. Next, one verifies that $W$ is an h-cobordism; since $\pi_1(S\times S^1)\cong {\mathbb Z}$, this h-cobordism is actually an s-cobordism and, hence, by the s-cobordism theorem, is diffeomorphic to the product. In particular, $\Sigma\times S^1$ is diffeomorphic to $S\times S^1$. Now, one can either refer to Theorem 1.9 in

R. Schultz, Smooth Structures on $S^p\times S^q$, Annals of Mathematics, Second Series, Vol. 90 (1969), p. 187-198

or recycle the above argument, to conclude that $\Sigma$ is diffeomorphic to $S$. This proves the claim.

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Misha's answer provides examples (for certain $n$), of pairs of open subsets of $\mathbb{R}^n$ (both with the standard smooth structure), that are homeomorphic but not diffeomorphic to each other, thus giving a partial answer to the more general question 2). One of the open sets is built from an exotic $(n−2)$-dimensional sphere which can be realized as Brieskorn variety. Smoothly embedding in $\mathbb{R}^n$, it can be viewed as a "small exotic $\mathbb{S}^{n-2}\times\mathbb{R}^2$"

Meanwhile, I got the following partial answer from an expert (private communication; as above, dimension $4$ is excluded). It goes in the other sense, providing examples of large exotic structures on open subsets of $\mathbb{R}^n$ (again, not for all $n>4$):

"Strictly speaking, $\mathbb{R}^n\setminus 0$ answers your question for most values of $n$, since it is diffeomorphic to $\mathbb{S}^{n-1}\times\mathbb{R}$, and the first factor typically has exotic structures that remain nondiffeomorphic to the original after product with $\mathbb{R}$. However, the resulting exotic structures do not embed in $\mathbb{R}^n$. I am guessing that you want pairs of open subsets of $\mathbb{R}^n$ that are homeomorphic to each other but not diffeomorphic. I don't know if such pairs exist. Clearly, they would have to have trivial tangent bundles, but that does not preclude exoticness: there are exotic spheres $S$ that bound parallelizable manifolds, so $S\times\mathbb{R}$ has a trivial tangent bundle and is not diffeomorphic to a standard sphere $\times\mathbb{R}$. However, such an example cannot embed as an open subset of $\mathbb{R}^n$. (Otherwise, $S$ would bound a ball and hence be standard.)"

Two short comments:

1) this construction provides examples of manifolds homeomorphic to simple open subsets of $\mathbb{R}^n$, but which do not smoothly embed in $\mathbb{R}^n$ (summing up, an exotic structure in the $(n-1)$-sphere gives rise to an exotic structure on the punctured $\mathbb{R}^n$, which is "large" in the sense that it does not embed smoothly in $\mathbb{R}^n$).

2) It would be also interesting to know analogue (topologically) simple examples in every dimension $n>4$ for which the $(n-1)$-sphere exhibits no exotic structure (in the case $n=5$, for which no exotic structure is known).

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