# Are homeomorphic open subsets of $\mathbb{R}^n$ also diffeomorphic?

Let $U_1, U_2$ be open subsets of $\mathbb{R}^n$. Both are naturally differentiable submanifold, getting the differentiable structure from $\mathbb{R}^n$. Further, both are natural topological manifolds, as submanifolds of $\mathbb{R}^n$.

Question: If $U_1$ and $U_2$ are homeomorphic, are they also diffeomorphic?

Of course two general topological manifolds which are homoemorphic do not need to be diffeomorphic. But here the differentiable structure is a very special one.

The answer might depend on the dimension $n$. For $n=1,2,3$ it is yes, as there each topological manifold has a unique differentiable structure. For $n \geq 5$ and $U_1$ an open ball the answer is yes by the uniqueness of differentiable structures on $\mathbb{R}^n$ for $n \geq 5$.

Some special cases are:

1. What happens if $U_1$ (and hence $U_2$) is contractible?

2. What happens if $U_1$ is a ball and $n=4$? Is there an exotic $\mathbb{R}^4$ which can be realized as an open subset of the standard $\mathbb{R}^4$?

(The question came up because I encountered a sloppy definition of a manifold. One can view the above manifolds as being defined by only one chart. [That of course depends on your definition of chart, if you require it to start from a ball or not.] So the questions basically asks: How do manifolds with only one chart look like?)

In fact, there exist uncountably many small exotic smooth $\mathbb{R}^4$'s, i.e. smooth manifolds $X$ which are homeomorphic to $\mathbb{R}^4$ but not not diffeomorphic to it and which can be smoothly embedded as open subsets of $\mathbb{R}^4$. There are discussions of this in many places; I recommend first reading the appropriate part of Scorpan's book "The Wild World of 4-Manifolds" for a brief survey (it includes a nice bibliography of more detailed sources).

• (But this is something special about dimension $4$. For large enough $n$ the answer is yes.) – Tom Goodwillie Nov 26 '12 at 15:42
• Tom, for which question is the answer yes? That $\mathbb{R}^n$ has only one differentiable structure? Or my question 1. above? Or the general question, that homeomorphic subsets of $\mathbb{R}^n$ are diffeomorphic, for $n \geq 5$? – Mark Ullmann Nov 27 '12 at 14:25

Adding to Andy's answer: there are lots of contractible open subset of $\mathbb R^n$ that are not homeomorphic to $\mathbb R^n$. For example, any compact contractible manifold of dimension $n>4$ embeds into $\mathbb R^n$: the double of any compact contractible manifold is simply-connected and hence a homotopy sphere, which after removing a point becomes $\mathbb R^n$. For constructions of compact contractible manifolds, see Kervaire's paper Smooth homology spheres and their fundamental groups.

You mention a confusion about a "sloppy definition of a manifold" and ask which manifolds have one chart. By a chart you seem to mean any open subset of a manifold together with a homeomorphism onto an open subset of $\mathbb R^n$, which I think is a valid definition. Any atlas of charts whose transition functions are smooth defines a smooth structure on your manifold, which by definition is the set of all atlases compatible the given one. If there is only one chart, then the (only) transition function is the identity, which is smooth. However, this merely implies that your manifold with one chart it diffeomorphic to an open subset of $\mathbb R^n$.

• I am confused about the first part of your answer: I know that there are lots of open contractible subsets of $\mathbb{R}^n$ which are not homoemorphic to $\mathbb{R}^n$. I was told the construction of a <a href="en.wikipedia.org/wiki/Whitehead_manifold">Whitehead manifold</a> works in any dimension, which provides plenty of examples. However, how can you embed a compact $n$-manifold into $\mathbb{R}^n$? Isn't any $n$-dimensianal submanifold of $\mathbb{R}^n$ open? Do you assume your manifold has a boundary? (Otherwise, what is the double of that manifold?) – Mark Ullmann Nov 27 '12 at 14:06
• For the second part: Bröcker/Jänich in "Einführung in die Differentialtopologie"(p.4/7, German) state (my translation): "On an open subset $U$ of $\mathbb{R}^n$ there are a priori many different differential structures, but with an atlas with only one chart we do not get essentially different differentiable structures". They do not choose their atlas to be maximal, hence of course one has more than one choice for the atlas. But by Andy's answer, it is not true that all open subsets of $\mathbb{R}^4$ as topological manifolds have only one differentiable structure. – Mark Ullmann Nov 27 '12 at 14:19
• @Mark, of course compact contractible manifolds have boundary (no boundaryless compact manifold of positive dimension is contractible: just look at top-dimensional homology). Your wiki link is broken. I think you are generally confused about atlases and smoth structures, eg Andy's answer implies that an open subset of $\mathbb R^4$ can have more than one smooth structure. It has one given by its inclusion chart, and it has many others: highly nontrivial, but also given by a single chart, namely the homeomorphism to onto another exotic $\mathbb R^4$ and pulling back its smooth structure. – Igor Belegradek Nov 27 '12 at 15:15
• @IgorBelegradek: Hi Igor, just to clarify, are there open subsets of a numeric space $\mathbb{R}^n$ that are homeomorphic to $\mathbb{R}^n$ but not diffeomorphic to $\mathbb{R}^n$ ? Thank you! – orangeskid Jan 8 '15 at 1:07
• @orangeskid: the answer is yes for $n=4$, see en.wikipedia.org/wiki/Exotic_R4, for other $n$ the answer is no. – Igor Belegradek Jan 8 '15 at 3:15