Intuitive meaning of Diffeomorphism

Question

Let $U\subset\mathbb{R}^n$, $V\subset\mathbb{R}^m$ and a bijection $f:U\to V$ is a diffeomorphism if $f$ and $f^{-1}$ are differentiable.

I would like to know the intuitive meaning of two open sets being diffeomorphic.

For example, if two spaces are homeomorphic, these spaces share the same topological properties. And we have a clear intuitive idea of two spaces being homeomorphs, like the classic relationship between a donut and a mug.

Is there a similar intuitive idea for diffeomorphism?

Edit: The proprieties that homeomorphism preserves is, for example, if one of them is compact, then the other is as well; if one of them is connected, then the other is as well; if one of them is Hausdorff, then the other is as well; their homotopy and homology groups will coincide.

So, what are the properties preserved by diffeomorphism?

I quoted homeomorphism, to indicate what I meant by an intuitive idea.

Answer 1

Diffeomorphisms are precisely the isomorphisms in the category of differentiable manifolds.

So it might be helpful to think about diffeomorphisms between differentiable manifolds in exactly the same way as you would think about homeomorphisms between topological spaces.

Since we consider spaces that are isomorphic to be "essentially the same" or "indistinguishable", this is what we have in mind in terms of differentiable manifolds whenever we talk about them being diffeomorphic.

Answer 2

It may help to illustrate in what ways two spaces can be homeomorphic but not diffeomorphic. Consider the half open square $X=(0,1)\times [0,1]\subset\mathbb{R}^2$ with usual differential structure. Topological properties of the space include that it is connected, simply connected, contractible.

What might a property of its differential structure be? One example is the fact that you can draw a smooth curve starting at one edge and ending at the other. In a way this is a bad example, because there is no "exotic" square, which is topologically the same as $X$, but where you cannot draw such a smooth curve.

However in dimension 4 this becomes more meaningful. There is a subset of $\mathbb{R}^4$ which is homeomorphic to $X \times X$, but where you cannot smoothly map a disk, so that the boundary of the disk goes once round the boundary of $X \times X$.

Answer 3

Regarding intuition for diffeomorphisms, it might be good to think of them as transformations that locally admit approximations by invertible linear maps. (More specifically this description correspond to $C^1$ diffeomorphisms; $C^r$ (for $r$ a positive integer), diffeomorphisms are similarly the transformations that locally admit approximations by invertible polynomial maps (think Taylor approximations). There are subtler classes of diffeomorphisms; these essentially almost always quantify the quality of the approximations.)

A consequence of this intuition is that diffeomorphisms preserve infinitesimal objects. Two specific instances of this phenomenon are:

• The change of variables formula for integration (think polar coordinates for instance). (For a (contravariant ($\dagger$)) change of variables formula homeomorphisms are not good enough, though it's also not the case that a diffeomorphism is necessary; there are broader formulas that work with transformations that are not quite diffeomorphisms (but still are more than homeomorphisms).) (($\dagger$) See Compute the density function of a pushforward measure)

• The adjoint operator on the space of continuous vector fields. If $f$ is a diffeomorphism and $X$ is a vector field, then under the adjoint operator of $f$ the vector field transforms into another vector field defined by

$$p\mapsto T_{f^{-1}(p)}f (X_{f^{-1}(p)}).$$

Again this operator would not be available if $f$ were merely a homeomorphism. (see What is the relationship between the vector fields of conjugate flows? for details.)