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I would like to know if there is a known necessary and sufficient property on an open subset of $\mathbb{R}^n$ to be 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 ?

Thank you for your answers and proofs

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    $\begingroup$ Huh, at first this looked suspiciously like a homework problem, but I see from the comments below that it's for real! $\endgroup$ – Scott Morrison Nov 7 '09 at 6:37
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I sent an e-mail to Erwann Aubry (see Georges Elencwajg's answer) and asked for a scan of this proof from the book "Calcul Différentiel". He replied and send me the proof translated into English. Here is his original paper. And this is how the proof goes:

Theorem. Every open star-shaped set $\Omega$ in $\mathbb{R}^n$ is $C^\infty$-diffeomorphic to $\mathbb{R}^n.$

Proof. For convenience assume that $\Omega$ is star-shaped at $0.$

Let $F=\mathbb{R}^n\setminus\Omega$ and $\phi:\mathbb{R}^n\rightarrow\mathbb{R}_+$ (here $\mathbb{R}_+=[0,\infty)$) be a $C^\infty$-function such that $F=\phi^{-1}(\{0\}).$ (such $\phi$ exists due to Whitney extension theorem)

Now we set $f:\Omega\rightarrow\mathbb{R}^n$ by formula: $$f(x)=\overbrace{\left[1+\left(\int_0^1\frac{dv}{\phi(vx)}\right)^2||x||^2\right]}^{\lambda(x)}\cdot x=\left[1+\left(\int_0^{||x||}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot x.$$ Clearly $f$ is smooth on $\Omega.$

We set $A(x)=\sup\{t>0:t\frac{x}{||x||}\in\Omega\}.$ $f$ sends injectively the segment (or ray) $[0,A(x))\frac{x}{||x||}$ to the ray $\mathbb{R_+}\frac{x}{||x||}.$ Moreover $f(0\frac{x}{||X||})=0$ and $$\lim_{r\rightarrow A(x)}||f(r\frac{x}{||x||})||=\lim_{r\rightarrow A(x)}\left[1+\left(\int_0^{r}\frac{dt}{\phi\left(t\cdot\frac{rx}{||x||}\cdot||\frac{||x||}{rx}||\right)}\right)^2\right]\cdot r=\\ \left[1+\left(\int_0^{A(x)}\frac{dt}{\phi(t\frac{x}{||x||})}\right)^2\right]\cdot A(x)=+\infty.$$ Indeed, if $A(x)=+\infty,$ then it holds for obvious reason. If $A(x)<+\infty,$ then by definitions of $\phi$ and $A(x)$ we get that $\phi(A(x)\frac{x}{||x||})=0.$ Hence by Mean value theorem and the fact that $\phi$ is $C^1$ $$\phi\left(r\frac{x}{||x||}\right)\leqslant M(A(x)-r)$$ for some constant $M$ and every $r.$ As a result $$\int_0^{A(x)}\frac{dt}{\phi\left(t\frac{x}{||x||}\right)}$$ diverges. Hence we infer that $f([0,A(x))\frac{x}{||x||})=\mathbb{R_+}\frac{x}{||x||}$ and so $f(\Omega)=\mathbb{R}^n.$

To end the proof we need to show that $f$ has $C^\infty$-inverse. But as corollary from the Inverse function theorem we get that it is sufficient to show that $df$ vanish nowhere.

Suppose that $d_xf(h)=0$ for some $x\in\Omega$ and $h\neq 0.$ From definition of $f$ we get that $$d_xf(h)=\lambda(x)h+d_x\lambda(h)x.$$ Hence $h=\mu x$ for some $\mu\neq 0$ and from that $x\neq 0.$ As a result $\lambda(x)+d_x\lambda(x)=0.$ But we have that $\lambda(x)\geqslant 1$ and function $g(t):=\lambda(tx)$ is increasing, so $g'(1)=d_x\lambda(x)>0,$ which gives a contradiction.$\square$

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    $\begingroup$ One could make the definition of $f$ even more explicit by putting the construction of $\phi$ into the actual formula ;-) $\endgroup$ – David Roberts Jun 26 '16 at 1:49
  • $\begingroup$ @DavidRoberts If you provide me with construciton of $\phi,$ I will edit the answer. Btw. I am not huge fan of this Whitney extension theorem argument. $\endgroup$ – Fallen Apart Jun 26 '16 at 4:45
  • $\begingroup$ well, it wouldn't add much to the argument, but if I get a chance to look it up then I can edit it in. $\endgroup$ – David Roberts Jun 26 '16 at 6:55
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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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    $\begingroup$ Why is this so complicated? $\endgroup$ – Kevin H. Lin Jan 5 '10 at 10:17
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You can certainly have a set diffeomorphic to $\Bbb R^n$ 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 diffeomorphic to the ball. For example, a thickened letter M in two dimensions.

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

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

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    $\begingroup$ 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. $\endgroup$ – j.c. Nov 7 '09 at 1:00
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    $\begingroup$ 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. $\endgroup$ – j.c. Nov 7 '09 at 1:11
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    $\begingroup$ 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. $\endgroup$ – Ryan Budney Nov 7 '09 at 1:19
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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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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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  • $\begingroup$ Very nice proof :) $\endgroup$ – BLBA Dec 28 '20 at 15:39
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Thank you for your answers,

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.

In dimension 4, you have an infinity of possible differential structures. Is is true then that:

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) ?

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    $\begingroup$ Nope. As I mentioned earlier, there are exotic smooth R^4's that embed smoothly in R^4. So such open subsets of R^4 are contractible and simply connected at infinity (since they're homeomorphic to R^4), but not diffeomorphic to the standard R^4 as that's what "exotic smooth R^4" means. $\endgroup$ – Ryan Budney Nov 7 '09 at 3:57
  • $\begingroup$ Ok, but there is something I don't really understand : is the exotic smooth R^4 open subset of R^4 you consider provided with the differential structure induced by the standard R^4 ? Recall that I consider in my question only one differential structure (the standard one) on U and R^4. $\endgroup$ – Oliver Nov 7 '09 at 4:33
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    $\begingroup$ Yes. There is a subset U of R^4 which is an open submanifold and, with the induced topology, homeomorphic to R^4; however, the induced smooth structure on U coming from the standard smooth structure on R^4 makes U not diffeomorphic to the standard R^4. Such spaces are called "small" exotic R^4's, see e.g. en.wikipedia.org/wiki/Exotic_R4. $\endgroup$ – Tom Church Nov 7 '09 at 6:58
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Yet another proof in the case where $\Omega$ is star-shaped, in little details unless anyone is interested.

Assume for simplicity that $\Omega$ is star-shaped around $0$. Define the function $r$ mapping $\theta\in\mathbb S(\mathbb R^n)$ to the only $r(\theta)$ such that $\Omega\cap \mathbb R_+\theta = [0,r(\theta))\theta$ (possibly $r(\theta)=\infty$). Then $r:\mathbb S(\mathbb R^n)\to[0,+\infty]$ is lower semicontinuous and uniformly positive, so there exists a sequence $(r_k)_{k\geq0}$ of smooth uniformly positive functions such that $r=\sum_{k\geq0}r_k$ (e.g. use this post to write $r$ as an increasing limit of Lipschitz functions and reduce to the Lipschitz case, then iteratively fit a smooth function $r_{k+1}$ strictly between $(r-r_k)/2$ and $r-r_k$ by using a partition of unity). Without loss of generality, $r_0$ is actually a (positive) constant.

The idea is that we are looking for a diffeomorphism $\mathbb R^n\to\Omega$ such that the image of the sphere of radius $k$ is the surface $\lbrace |x|=r_k(x/|x|)\rbrace$. The final diffeomorphism will end up being a bit different, but I think it is a good image to keep in mind.

Find a collection of smooth functions $(\rho_k)_{k\geq0}$ from $\mathbb R_+$ to $[0,1]$ such that

  • for all $k>0$, $\rho_k$ starts at zero, then increases and finally stays constant equal to one, in such a way that $\rho_k'$ has support in $(k-1,k+2)$ and is (uniformly) positive over $[k,k+1]$;
  • $\rho_0$ starts at zero, then increases and finally stays constant equal to one, in such a way that $\rho_0'$ is one in a neighbourhood of zero, has support in $[0,2)$, and is (uniformly) positive over $[0,1]$.

Then $$\Phi:x=|x|\cdot\frac x{|x|}=r\cdot\theta\mapsto\sum_{k\geq0}\rho_k(r)r(\theta)\theta$$ is smooth at zero (it is a constant multiple of the identity) and everywhere else (it is locally a finite sum of smooth functions), it is injective (because it preserves rays and is strictly increasing on rays) and surjective with image $\Omega$ (because $r=\sum_{k\geq0}r_k$), and its differential is invertible (away from zero because it is strictly increasing with positive derivative along the rays and it preserves the angles). It means that $\Phi$ has an inverse $\Phi^{-1}$ that is locally smooth, so $\Phi$ is actually a diffeomorphism.

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