# Diffeomorphism vs. homeomorphism

1. Feb 20, 2009

### quasar_4

Is it fair to think of a diffeomorphism as being a "stronger" condition then a homeomorphism? I know this is probably a dumb question, but I'm trying to teach myself some topology, and still waiting for Munkres to come in the mail. :)

Last edited: Feb 20, 2009
2. Feb 20, 2009

### CompuChip

Basically, a diffeomorphism is a differentiable homeomorphism.
That is: a homeomorphism is a bijection which is continuous, with continuous inverse.
A diffeomorphism is a bijection which is differentiable with differentiable inverse.
A Cr-diffeomorphism is a bijection which is r times differentiable with r times differentiable inverse.

So, every diffeomorphism is a homeomorphism, but not vice versa.

3. Feb 20, 2009

### quasar987

An example to keep in mind it that a (full) square is homeomorphic to a disk, but not diffeomorphic to it.

This expresses the intuitively acceptable fact that one can continuously deform the square into the disk but not differentiably so because there are smoothness problems at the corners.

4. Feb 20, 2009

### wofsy

perhaps you could explain this more since the disk and any manifold homeomorphic to it has a unique differentiable structure.

It seems to me that the square is a non-differentiable embedding of the disk into the plane. Just because this embedding isn't differentiable doesn't meant that the manifold is not diffeomorphic to the disk. Take a simpler example. A line embedded in the plane with a kink in it. As a differentiable manifold it is just a line.

5. Feb 21, 2009

### WWGD

Just to add to what has been said, the square, as a subspace of R^2 , is
not a submanifold of R^2 (this can be made more rigorous by saying that not
every point can be given slice coordinates).
But you can show that if M is a smoot ( in Brooklyn,
or smooth; C^k anywhere else :) ) manifold , and N is just a topological space
that is homeomorph. to M, then N can be made into a manifold, by pulling back
the structure of M. ( this is a trick that happens very often).

I believe too, that , for n =/4 , there are no manifolds that are just C^0,
i.e., manifolds that are just topological manifolds, i.e., for n=/ 4, we can
always give a topological manifold a smooth structure. It is strange that
the smoothing that is done, e.g, for the square, cannot always be done
for n=4.

6. Feb 21, 2009

### wofsy

Your last remark about C^0 manifolds interest me and I would appreciate a reference or perhaps you could describe some examples. I know that in dimension 8 there are manifolds that have no differentiable structure. Apparently the first one discovered was 10 dimensional.

7. Feb 21, 2009

### WWGD

I'll look it up, Wofsy. All I remember is that it came up when dealing with a topic
related to this thread. I was suspicious of a (unqualified)claim that "the cone is an example
of a C^0 manifold that is not smooth, nor differentiable" . Again, it seems strange that
we cannot somehow homeomorphically smooth out any C^0 manifold into a smooth
manifold.
Anyway, I'll look it up.

8. Feb 21, 2009

### wofsy

thanks. I know that in the 8 dimensional case there is a combinatorial manifold i.e. a triangulated manifold that is not compatible with any differentiable structure. This is proved by showing that its combinatorial Pontriagin numbers are not integers. This does not mean though that the underlying topological manifold can not be smoothed (I don't think). But the 10 dimensional case is a topological manifold with no smooth structure. I don't know the proof.

9. Feb 24, 2009

### yyat

One of the simplest examples is the function f:R->R, f(x)=x^3 which is a homeomorphism and smooth, but not a diffeomorphism because the inverse is not differentiable at 0. This assumes that R is equipped with the usual smooth structure. (By changing the smooth structure on the domain or codomain, one can make any homeomorphism a diffeomorphism).

10. Mar 3, 2009

### wofsy

R has only one smooth structure

11. Mar 4, 2009

### yyat

R has only one smooth structure up to diffeomorphism. If I turn R (as a topological space) into a smooth manifold using the chart f(x)=x^3 instead of f(x)=x, then this will give a different smooth structure, i.e. a different set of smooth functions on R. But, as you said, these manifolds are of course diffeomorphic.

12. Mar 4, 2009

### WWGD

yyat: Re your comment: (sorry, I can't figure out the quoting thing here)
"R has only one smooth structure up to diffeomorphism. If I turn R (as a topological space) into a smooth manifold using the chart f(x)=x^3 instead of f(x)=x, then this will give a different smooth structure, i.e. a different set of smooth functions on R. But, as you said, these manifolds are of course diffeomorphic. "

Do you mean that the collection of functions from the manifold (R,x^3) into R, or other
manifolds will be different ? (i.e. if we have (M,Phi) , then Phi^-1 o f o x^3 , and this
would be smooth in certain cases). But, why classify a manifold based on the set of smooth
functions on it?

13. Mar 4, 2009

### yyat

Click on the "quote" button below the post.

Yes. Let M be R as a topological space but with smooth structure defined be the chart phi:R->M, phi(x)=x^3 (often charts are defined in the other direction, but to be consistent with what I wrote above...). Then the map f(x)=x^(1/3) is a smooth functions on M.

The smooth structure on a manifold M is in fact determined by which functions f:M->R are the smooth ones. This makes it possible to give an http://en.wikipedia.org/wiki/Smooth_manifold#Structure_sheaf" of a smooth manifold, not involving the chart transitions.

Last edited by a moderator: Apr 24, 2017
14. Mar 4, 2009