A note on smooth maps between smooth structures

I’m still trying to catch up with the uni courses that I completely neglected during the quarantine days, I’m focusing on differential geometry right now which is a subject I always found hard to ‘get’, so many details and confusing notation!

Here is something very elementary that has puzzled me for some days coming from a very popular exercise. Suppose we have two charts defined on the whole $\mathbb R$ given by the identity and the cubing map $\varphi : x \mapsto x^3$ , these can be used to define two atlases ${\cal A}_1, {\cal A}_2$ which in turn define maximal atlases ${\cal A}_1^{\rm max}, {\cal A}_2^{\rm max}$ or smooth structures. The exercise usually asks to prove that the identity ${\rm Id} : (\mathbb R, {\cal A}_1^{\rm max}) \to (\mathbb R, {\cal A}_2^{\rm max})$ is not a diffeomorphism (a $\cal C^\infty$ map whose inverse is also $\cal C^\infty$ ) whereas the cubing map itself is.

Recall that a $\cal C^\infty$ map $F : M \to N$ between smooth manifolds needs, for every point $x \in M$ that there exist charts $(U, \alpha), (V, \beta)$ of $M, N$ respectively such that:

$F(U) \subseteq V$
The composition $\beta \circ F \circ \alpha^{-1}$ is infinitely differentiable (which makes sense since it’s a map between open sets of Euclidean space)

Now, while the spirit of this example is very clear to me, the usual solutions found online and in class were confusing me a bit: the idea is that taking the charts we started from $(\mathbb R, \varphi), (\mathbb R, {\rm Id})$ we get that ${\rm Id} \circ {\rm Id} \circ \varphi^{-1} = \varphi^{-1}$ should be a infinitely differentiable which is false, its derivative is discontinuous at $0$ !
However why are we allowed to use these particular charts? We have two whole maximal atlases filled with charts around $0$ , the above reasoning doesn’t seem to exclude the possibility of finding a pair that works.

Now I think I get it:

Proposition. Let $F : (M, {\cal A}) \to (N, {\cal B})$ be a map between differentiable manifolds with structure given by atlases ${\cal A} = \{(U_\alpha, \varphi_\alpha)\}$ and ${\cal B} = \{(V_\beta, \psi_\beta)\}$ . Then $F$ is $\cal C^\infty$ iff $\psi_\beta \circ F \circ \phi_\alpha^{-1}$ is infinitely differentiable for all $\alpha, \beta$ with $F(U_\alpha) \cap V_\beta \neq \emptyset$ .

Proof. If the last statement holds then $F$ satisfies the definition of $\cal C^\infty$ map since $\{U_\alpha\}$ is a covering of $M$ .
Conversely suppose that $F$ is a $\cal C^\infty$ map, then for each $x \in M$ we can find $(\tilde U, \tilde \varphi), x \in \tilde U$ and $(\tilde V, \tilde \psi), F(\tilde U) \subseteq \tilde V$ charts in the atlases $\cal A, B$ such that $\tilde \psi \circ F \circ \tilde \varphi^{-1}$ is infinitely differentiable. Now, the key point is that both of these charts must be compatible with all the other charts in their respective atlases, in particular there will be some $(U_\alpha, \varphi_\alpha)$ with $x \in U_\alpha$ and $(V_\beta, \psi_\beta)$ with $F(x) \in V_\beta$ such that all the maps $\tilde\varphi \circ \varphi_\alpha^{-1}, \varphi_\alpha \circ \tilde\varphi^{-1}, \tilde\psi \circ \psi_\beta^{-1}, \psi_\beta \circ \tilde\psi^{-1}$ are infinitely differentiable. Finally it’s just a matter of composing the right functions (recalling that the composition of infinitely differentiable maps is again infinitely differentiable):

$(\psi_\beta \circ \tilde\psi^{-1}) \circ (\tilde \psi \circ F \circ \tilde \varphi^{-1}) \circ (\tilde \varphi \circ \varphi_\alpha^{-1}) = \psi_\beta \circ F \circ \varphi_\alpha^{-1}$

and we are done. □

As I said, this is very elementary and was probably clear to everyone but me, but it feels good to write it out explicitly.

Partially Ordered Thoughts

A note on smooth maps between smooth structures

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A note on smooth maps between smooth structures

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