# Uniqueness of smooth structure

1. Sep 6, 2010

### quasar987

How does one prove that the smooth structure on 2-manifolds is unique? Source?

Thx!

2. Sep 7, 2010

### lavinia

While googling I found an old post on PF that argues along the following lines - and may be in the right direction of a proof - not sure.

Every orientable smooth surface has an atlas of charts that define a conformal structure. With this atlas the manifold is a Riemann surface. There is a famous theorem that states that every simply connected Riemann surface is conformally equivalent (and therefore diffeomorphic) to either a bounded disk, the entire complex plane, or to the standard 2 sphere.

So for a simply connected Riemann surface there is only one differentiable structure.

I think that an arbitrary Riemann surface can be obtained from the simply connected surface by the action of a discrete group of holomorphic covering transformations. perhaps this can be used to get the general result.

Last edited: Sep 7, 2010
3. Sep 8, 2010

### lavinia

I am not sure this works.

Given two Riemann surfaces that are homeomorphic, choose fundamental domains in the unversal cover. These domains are diffeomorphic - by the theorem on conformal equivalence. Choose a diffeomorphism,D, between them.

The two respective groups of covering transformations are isomorphic. Choose an isomorphism,H.

Extend D to the entire covering space by the rule g(x) -> H(g)Dx.

Note that fg(x) -> H(fg)Dx -> H(f)(H(g)Dx).

This should define the diffeomorphism - I think.

I guess the idea is if the universal cover only has 1 differentiable structure then the manifold also only has 1.

Last edited: Sep 8, 2010
4. Sep 29, 2010

### Monocles

It should be noted that it depends on your definition of smooth structure. If you use the definition of smooth structure given in Lee's Introduction to Smooth Manifolds, for instance, there are actually an uncountable number of smooth structures on all manifolds that admit at least one smooth structure of dimension greater than 0.

I am actually still a bit confused on this point myself. I believe it is because, given an atlas on a smooth n-manifold where n>0, if you have a chart $$(U, \varphi)$$ and $$(V, \psi)$$ such that $$U \cap V \neq \emptyset$$, then $$\varphi$$ only needs to be homeomorphic on $$U \backslash V$$ but must be diffeomorphic on $$U \cap V$$ so that the transition function $$\psi \circ \varphi^{-1}$$ is smooth. So, given a subset of a chart that does not intersect with any other chart, the coordinate chart does not necessarily need to be a diffeomorphism on that subset, and so you can use this fact to construct an uncountable number of atlases that are not smoothly compatible.

5. Sep 29, 2010

### lavinia

I believe that unique smooth structure means unique up to diffeomorphism.