What are the open subsets of $\mathbb{R}^n$ that are diffeomorphic to $\mathbb{R}^n$

Hello,

I would like to know if there is a known necessary and sufficient property on an open subset of $\mathbb{R}^n$ such that it is diffeomorphic to $\mathbb{R}^n$ :

For example :

1) Are all open star-shaped subsets of $\mathbb{R}^n$ diffeomorphic to $\mathbb{R}^n$ ?

2) Reciprocally, are all open subsets of $\mathbb{R}^n$ which are diffeomorphic to $\mathbb{R}^n$, star-shaped ?

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Huh, at first this looked suspiciously like a homework problem, but I see from the comments below that it's for real! –  Scott Morrison Nov 7 '09 at 6:37

Ad question 1): Yes, all open star-shaped subsets of $\mathbb{R}^n$ are diffeomorphic to $\mathbb{R}^n$.

This is surprisingly little-known and there is a proof due to Stefan Born. You can find this (fairly complicated) proof in Dirk Ferus's course notes

http://www.math.tu-berlin.de/~ferus/ANA/Ana3.pdf

page 154, Satz 237 [The notes are alas in German]

Added December 30, 2009: My excellent colleague Erwann Aubry informs me that this result is also proved more simply on page 60 of Gonnord & Tosel's book "Calcul Différentiel", ellipses,1998.

[This book is in French, and moreover published by "ellipses" a valiant little publisher, completely unknown outside of France because it caters to the idiosyncratic French academic system]

Kudos to any reference in honest English, rather than exotic foreign languages :)

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Why is this so complicated? –  Kevin H. Lin Jan 5 '10 at 10:17

You can certainly have a set diffeomorphic to R^n but not star-shaped. For example, for n=2, the Riemann mapping theorem implies that any simply connected open set is diffeomorphic to the plane. More concretely, you can take a ball and just deform it a little bit so it's very badly not convex (in particular, not star-convex) but still diffeormorphic to the ball. For example, you a thickened letter M in two dimensions.

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There are several characterizations of manifolds diffeomorphic to R^n when n>4, e.g. an open manifold that is simply-connected at infinity (Stallings), or the image of a degree one proper map from R^n (Siebenmann), but looks like this is not what you want. Surely tons of subsets of R^n that are diffeomorphic to R^n can be constructed by attaching "fingers" to a ball.

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No, not really. In dimension 4, for example, an open subset of R^4 can be homeomorphic to R^4 but not diffeomorphic, as there are exotic smooth R^4's that embed smoothly in R^4.

But in dimensions different from 4, R^n admits a unique smooth structure. So your neccessary and sufficient condition can be that the open subset is homeomorphic to R^n. That's probably not what you want to hear?

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Following up on Ryan Budney's response, there's some discussion of subsets of R^n which are homeomorphic to R^n here: math.niu.edu/~rusin/known-math/95/contractible . Contractibility is not enough, but I don't think any full necessary and sufficient conditions are given in that thread. –  j.c. Nov 7 '09 at 1:00
en.wikipedia.org/wiki/Simply_connected_at_infinity claims that contractibility and simple connectedness at infinity are equivalent to being homeomorphic to R^n. So I guess the results described in the page I linked to above go both ways after all. –  j.c. Nov 7 '09 at 1:11
Yeah, I think the argument goes like this: if its simply-connected at infinity, you apply Larry Siebenmann's dissertation to find a manifold compactification. Contractibility tells you this compactification is a topological n-ball. This argument requires a dimension restriction to n >= 6 though. –  Ryan Budney Nov 7 '09 at 1:19

The answer for 2) is no. Think of an annulus in R^2 with a radius removed.

1) seems much less trivial. It is true in 2 dimension, but the easiest way I can think of is to use the fact that star-shaped implies simply connected and use the Riemann mapping theorem. So complex analysis here yields a purely topological conclusion.

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So I guess that if $n\neq 4$, then the necessary and sufficient condition is precisely "contractible and simply connected at infinity". Here, there is only one possible differential structure.
Question : If $U$ is an open contractible simply connected at infinity subset of $\mathbb{R}^4$ on which we consider the standard differential structure. Then is $U$ diffeomorphic to $\mathbb{R}^4$ (with its standard differential structure) ?