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So not so long ago, I asked for a simple proof that $\mathbf{R}$ has only one smooth structure. A proof that was communicated to me by Ryan Budney (link text) was the following:

So let me recall his argument: So let $X$ be a real line endowed with a "potentially" exotic smooth structure. We know that $X$ is Hausdorff and paracompact so for every open covering $\mathcal{U}$ of $X$ we have a partition of unity dominated by $\mathcal{U}$. Using this we can endow $\mathbf{R}$ with a Riemanian metric $ds^2$ (choose your favorite open covering which is locally finite!). Let $x_0$ be a point of $X$ so that $X-x_0=X^+\bigcup X^{-}$ is the disjoint union of the two components. Finally, note that one may integrate this metric against the Haar measure of $X$ using the velocity vectors $1$ and $-1$ in the fiber above $x_0$ to get two bijections

$f^+:X^+\rightarrow\mathbf{R}_{>0}$

and

$f^-:X^-\rightarrow\mathbf{R}_{<0}$.

Since the metric $ds^2$ is smooth we see that $f^+$ and $f^-$ are smooth and that they glue in a smooth way. So basically, the proof works because we can think of $\mathbf{R}$ as the union of two geodesics.

Q: Is there somekind of similar argument for $\mathbf{R}^2$ and $\mathbf{R}^3$ ?

Any simple proof along different lines is welcome!

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It would be strange (or interesting...) that a similar argument worked for $n=2$ and $n=3$, because nothing similar can work for $n=4$ :) –  Mariano Suárez-Alvarez Mar 14 '11 at 21:55
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A different way to phrase that argument is: we can pick a complete Riemannian metric on our $1$-dimensional smooth manifold $M$, see e.g. mathoverflow.net/questions/18844; the geodesic through one of its points is a surjective map $\mathbb R\to M$, according to Hopf-Rinow. The map is locally a diffeo by the inverse function theorem, and if it were not injective it would be periodic, by uniqueness of geodesics---but in that case $M$ would be compact. –  Mariano Suárez-Alvarez Mar 14 '11 at 22:06
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Ken, I think there's a confusion regarding the meaning of "smooth structure". One meaning is a maximal atlas of smooth charts. Another is a diffeomorphism class of such maximal atlases. Your example gives two distinct maximal atlases of smooth charts on $\mathbb{R}$. But these two maximal atlases give rise to diffeomorphic manifolds (note, though, that the diffeomorphism is not given by the identity map). I think this is explained in the first chapter of Spivak's book. –  Dan Ramras Mar 14 '11 at 23:22
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Hi Ken, the example you gave, say $X$, is equivalent to the usual smooth structure on $\mathbf{R}$. Indeed the map $\sqrt[3]:X\rightarrow\mathbf{R}$ is a diffeomorphism! –  Hugo Chapdelaine Mar 14 '11 at 23:35
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It seems to me that some of the modern proofs of the Uniformisation theorem have few to no topological prerequisites. I wouldn't be surprised if some of them did prove that any smooth structure on R2 can be refined to a complex structure equivalent to C or the unit disk, and is therefore diffeomorphic to R2. –  Maxime Bourrigan Mar 15 '11 at 1:02
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1 Answer 1

up vote 12 down vote accepted

Some comments alluded to the possibility to show this using the Riemann uniformization theorem (by paracompactness, any oriented $2$-manifold has an almost complex structure, which is integrable by Newlander-Nirenberg and by the uniformization theorem, it will be biholomorphically equivalent to the plane or the unit disc, hence diffeomorphic to $R^2$). This is not circular, but to claim that this is "simple" would be utterly absurd. The complete proof of the uniformization theorem is one of the hardest mathematical achievements of the early 20th century; the proof uses a lot of analysis and also a bit of algebraic topology.

Using Morse theory, you can argue as follows: let $U$ be a connected noncompact surface, pick a Morse function $f: U \to \mathbb{R}$. One can modify $f$ so that it has no critical points of index $2$ and precisely one critical point of index $0$, so let us assume that $f$ has this property. This is the most basic case of the handle-cancellation technique.

Now let $C_{\ast}(f)$ be the chain complex of the Morse function. $C_k (f)$ has the critical points of index $k$ as a basis. If $f$ is above, it follows that $C_0 (f)=Z$, $C_k (f)=0$ if $k \geq 2$. The differential $C_1 \to C_0$ will be zero and so $H_1 (U)= C_1 (f)$.

If $H_1 (U)=0$, we see that there is a Morse function $f:U \to \mathbb{R}$ with precisely one minimum. Use the flow lines of $f$ to cook up a diffeomorphism $f: U \to \mathbb{R}^2$.

I don't think you get this result much cheaper.

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Thanks Johannes, I think that your argument fulfils my requirement! And yes indeed, the proof of the uniformization theorem of simply connected Riemann surfaces is far from trivial and extremely deep! –  Hugo Chapdelaine Mar 15 '11 at 11:40
    
It is definitely far from simple and indeed was the source of great development in the XIXth and XXth century (if you allow me some advertisement pro domo, this book explains a part of this story: amazon.fr/…) but some modern proofs of the whole package aren't that long... –  Maxime Bourrigan Mar 15 '11 at 12:49
    
BTW, your proof confuses me, wouldn't the square of the norm be a Morse function with exactly one minimum on a small exotic R4 ? How do you "cook up" your diffeomorphism? –  Maxime Bourrigan Mar 15 '11 at 13:06
    
Assume that the minimum of $f$ is at $p$ and $f(p)=0$. For any $b>a >0$, the manifold $f^{-1}[a,b]$ is diffeomorphic to $[a,b] \times f^{-1}(a)$, see Milnor, Morse theory, Thm 3.1. Moreover, $f$ corresponds to the projection onto $R$ under this diffeomorphism (the phrase "use the flow lines" corresponds to this). Do this for many intervals; you get a diffeo $f^{-1}[a,\infty) \cong f^{-1}(a) \times [a,\infty)$. By the Morse lemma, $f^{-1}[0,a]$ is a disc when $a$ is small enough. These things can be glued together to give the asserted diffeo. –  Johannes Ebert Mar 15 '11 at 14:12
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If I remember correctly, the first step in the most standard proof of the uniformzation theorem is to show that a simply connected surface can be exhausted by relatively compact simply connected subsets (maybe with nice boundary). For that, a proper function and some algebraic topology is needed, in other words: quite a bit of what I sketched above. Only after this is done, the geometric analysis part of the proof begins. –  Johannes Ebert Mar 15 '11 at 19:11
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