Looking for a simple proof that R^2 has only one smooth structure

2011-03-14T21:45:14Z

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!

Answer by Johannes Ebert for Looking for a simple proof that R^2 has only one smooth structure

2011-03-15T09:42:43Z

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.

Looking for a simple proof that R^2 has only one smooth structure - MathOverflow

Looking for a simple proof that R^2 has only one smooth structure

Answer by Johannes Ebert for Looking for a simple proof that R^2 has only one smooth structure