Intuitive meaning of Diffeomorphism

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.

• The diffeomorphisms are smooth so while a half cone is homeomorphic to the plane, it is not diffeomorphic.to it. Commented Jun 29, 2020 at 21:32
• @JohnDouma That example always bugs me. What smooth structure are you endowing the half cone with so as to make meaningful the statement that it's not diffeomorphic to the plane? It does not have a smooth structure naturally induced by its being a subset of $\mathbb{R}^3$, and in fact it does not have one satisfying your criteria at all, as pulling back such a smooth structure to the plane via our homeomorphism would imply the existence of an exotic $\mathbb{R}^2$. Commented Jun 29, 2020 at 21:50
• @GReyes I think you've overcomplicated things a bit-- there's no need to invoke connections or metrics to specify a smooth structure. In fact, it's the other way around. In order to induce a metric on the cone $C$ from the ambient $\mathbb{R}^3$ by restricting it from $T_p \mathbb{R}^3$ to $T_p C$, it is necessary (but not sufficient) to first choose a smooth structure on the cone, as $T_p C$ is a property of the smooth structure. This still cannot be done if the cone is to include its vertex $v$, however, as there is no way to identify $T_v C$ with a subspace of $T_v \mathbb{R}^3$. Commented Jun 29, 2020 at 22:31
• You are absolutely right. I misunderstood the question. It is about the existence of an isomorphism as differentiable manifolds... Commented Jun 29, 2020 at 22:38
• As well as the property I mentioned in my answer, another property that a differential $n$- manifold might have is that every compact subset may be enclosed by a smoothly embedded $n-1$ sphere. Clearly $\mathbb{R}^4$ has this property (compact subsets are bounded with respect to the usual metric, so you just need a sphere of large enough radius). However there exist manifolds homeomorphic to $\mathbb{R}^4$ that do not have this property. See planetmath.org/donaldsonfreedmanexoticr4 for a little detail.
– tkf
Commented Jun 30, 2020 at 2:13

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.

• What are the properties preserved by diffeomorphism? Commented Jun 30, 2020 at 0:10
• I think this is a useful related question, particularly Jack Lee's answer: math.stackexchange.com/questions/2373775/… Commented Jun 30, 2020 at 2:29

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.)

• Another instantiation of the idea that diffeomorphisms are closely related to linerizations is the fact that the diffeomorphism group of Euclidean space can be deformed to the general linear group; see math.stackexchange.com/q/4566967/169085 Commented Nov 8, 2022 at 6:09

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 $$X \times X$$.