# Why is the x->x^1/3 atlas on R diffeomorphic with the x->x atlas on R?

The fact that the atlas using $\phi: x \mapsto x^{1/3}$ on $\mathbb{R}$ is diffeomorphic to the trivial atlas using $\psi: x \mapsto x$ on $\mathbb{R}$ highlights my ignorance of diffeomorphisms and atlases. Apologies in advance for clustering several questions, but I'm not sure how to disentangle them.

First of all, which is the more relevant/correct statement here: that the atlases are diffeomorphic, that the manifolds are diffeomorphic ($M \mapsto N$), that the manifold-atlas pairs are diffeomorphic ($(M,\psi) \mapsto (N,\phi)$), or that the manifolds and their differential structures are diffeomorphic ($(M,\mathcal{A}) \mapsto (N,\mathcal{B})$)?

Is the relevant diffeomorphism patently obvious here given that the domain of $x \mapsto x^{1/3}$ is the same as that of $x \mapsto x$?

If we had two homeomorphic manifolds with atlases that were not diffeomorphic, would it be obvious?

• I suspect some of this might be terminology problems. "Atlases" are not objects that have a diffeomorphism relation. Smooth manifolds do. Topological manifolds can be enhanced to smooth manifolds by giving them a smooth atlas. All of these details are in the first few chapters of any textbook on abstract smooth manifolds (as opposed to submanifolds of euclidean space). – Ryan Budney Jul 29 '10 at 17:34
• Can you recommend a textbook? – Outis Jul 29 '10 at 18:17
• Conlon's "Differentiable Manifolds" would likely be suitable. – Ryan Budney Jul 29 '10 at 18:21

The fact that the atlas using $x \mapsto x^{1/3}$ on $\mathbb{R}$ is diffeomorphic to the trivial atlas using $x \mapsto x$ on $\mathbb{R}$ highlights my ignorance of diffeomorphisms and atlases.

Other differential topologists should weigh in on this to confirm or deny, but as a differential topologist I have never come across the notion of a diffeomorphism between two atlases, or even two smooth structures. Moreover, what you have here is not two atlases but two charts. These may seem like picky points, but if you find yourself getting confused then one good technique to learn is to sharpen your definitions. By that, I mean be a bit more careful about distinguishing between things that although often used synonymously are actually distinct.

So you have two charts, $x \mapsto x^{1/3}$ and $x \mapsto x$. As both of these have image $\mathbb{R}$, they each define an atlas: $\{x \mapsto x^{1/3}\}$ and $\{x \mapsto x\}$. Each of these atlases then defines a smooth structure on $\mathbb{R}$. Each of these smooth structures defines a smooth manifold with underlying topological space $\mathbb{R}$. Although each of these constructions follows in a unique way from the previous step, technically each is a different thing.

Back to the confusion about diffeomorphisms. We talk of two atlases being equivalent if they generate the same smooth structure, or if they define the same smooth manifold. In concrete terms, we can test this by looking to see if the identity map is smooth in both directions (using the atlases to test smoothness).

But I can have inequivalent atlases that nonetheless define diffeomorphic manifolds. This is because the condition of being equivalent is stronger than that of defining diffeomorphic manifolds. Equivalence rests on the smoothness of the identity map (in both directions), the manifolds being diffeomorphic rests on the smoothness of some map (and its inverse). So although the two atlases given are inequivalent, they define diffeomorphic manifolds because I'm free to take the map $x \mapsto x^{1/3}$ and its inverse to define the diffeomorphism.

It's a good exercise to help with sorting out the definitions to check that you really understand why these two manifold structures on $\mathbb{R}$ are diffeomorphic. That is, write down the map and write out its compositions with the transition maps and see that it works.

• Your first paragraph is makes pretty much the same point I made to the OP in mathoverflow.net/questions/33676/… I think, however, your third and fourth paragraphs are much better written than the equivalent ones in my post. – Willie Wong Jul 29 '10 at 19:31

In regards to your first question, the last three statements are essentially the same since to talk about diffeomorphisms between two manifolds you need to have a smooth structure on those manifolds, which is a choice of an (equivalence class of) atlas. One would usually simply say that the manifold $R$ with its regular smooth structure is diffeomorphic to the manifold $R$ with the smooth structure determined by the atlas $x \mapsto x^{1/3}$.

Here, the different smooth structure on $R$ is provided by an atlas that is a homeomorphism $F: R \to R$. This will always produce a manifold diffeomorphic to $R$ with the regular smooth structure since a diffeomorphism between them will be $F^{-1}: R \to R'$, where $R'$ is $R$ with the atlas $(R, F)$.

I do not know a lot about exotic structures, but I do not believe that there are examples where it is obvious that two smooth manifolds, that have the same underlying topological manifold, are not diffeomorphic. For example, I believe Milnor proved that some of his exotic 7-spheres were not diffeomorphic using Morse theory.

• Milnor used Morse theory to show his 7-manifolds were in fact homemorphic to $S^7$. He used other techniques (including the Hirzebruch signature theorem) to show that some of them were NOT diffeomorphic. – Jason DeVito Jul 29 '10 at 16:45
• «Obvious», of course, is a relative notion! :) – Mariano Suárez-Álvarez Jul 29 '10 at 16:45
• Thanks. Does an atlas uniquely determine the smooth structure of a manifold? – Outis Jul 29 '10 at 17:56
• See the sub-section on atlases and compatible atlases, here: en.wikipedia.org/wiki/Differentiable_manifold – Ryan Budney Jul 29 '10 at 17:59
• Ryan, thanks, I see the answer is yes. – Outis Jul 29 '10 at 18:16