Definition of 'equivalent atlases' and 'smooth structure'

Question

I have a question to the following definition:

(I) Two atlases $A_1$ and $A_2$ of $M$ are equivalent, if $A_1\cup A_2$ is a smooth atlas.

(II) A smooth structure on $M$ is an equivalence class $\mathcal{A}$ of smooth atlases of $M$.

My question is: Does (I) define an equivalence relation? Because in (II) is spoken about equivalence classes, but what is the equivalence relation?

What is in (I) defined should not be an equivalence relation, if we say. It already fails in generel at the reflexivity.

$A_1\sim A_1$ if $A_1\cup A_1=A_1$ is a smooth atlas. But beeing smooth is not part of the definition, so in general this should not hold.

Symmetry however is trivial, but transitivity I do not see how you should proof it.

So either this definition is incomplete, or something else is meant.

So what is meant by 'equivalence class' in $\mathcal{A}$, when there is no equivalence relation?

Thanks in advance.

Answer 1

This is the first of many instances in differential geometry when people mean one thing, but write something else; a lot of times some things will just have to be inferred from context, and in this case, it's the use of the adjective "smooth". Anyway, here's a more precise statement:

Let $M$ be a (non-empty) set, and $r\geq 0$ an integer. Define a relation $\sim_r$ on the set of all $C^r$ atlases for $M$, by saying: \begin{align} A_1 \sim_r A_2 \end{align} if and only if $A_1 \cup A_2$ is a $C^r$ atlas for $M$.

Then, reflexivity and symmetry are obvious. Transitivity, while not immediately obvious, is something you should spend some time struggling with atleast once (I know I did when I first learnt this stuff) in order to write a complete proof. The basic idea though is pretty simple. Take $(U_1, \alpha_1) \in A_1$ and $(U_3, \alpha_3) \in A_3$; we have to show the map \begin{align} \alpha_3 \circ \alpha_1^{-1}: \underbrace{\alpha_1[U_1 \cap U_3]}_{\text{open in $\Bbb{R}^n$}} \to \underbrace{\alpha_3[U_1 \cap U_3]}_{\text{open in $\Bbb{R}^n$}} \end{align} is $C^r$ as a map between open subsets of $\Bbb{R}^n$. Well, to do this, we need to invoke the atlas $A_2$ somehow to bridge the gap. Since smoothness can be proven pointwise, my hint to you is to choose, for each $x \in \alpha_1[U_1 \cap U_3]$ (the domain) a suitable chart $(U_2, \alpha_2) \in A_2$ (so that all the various compositions I write down make sense), and then write \begin{align} \alpha_3 \circ \alpha_1 &= (\alpha_3 \circ \alpha_2^{-1}) \circ (\alpha_2 \circ \alpha_1^{-1}) \end{align} (equality being wherever the functions are defined). So, on the RHS, you're composing two $C^r$ functions, hence the LHS is also $C^r$ at $x$. But since $x \in \alpha_1[U_1 \cap U_3]$ was taken arbitrarily, it follows $\alpha_3 \circ \alpha_1$ is $C^r$ at each point of its domain, which is what we had to show.

I leave it to you to justify carefully why there always exist charts like $(U_2, \alpha_2)$ (and also where we used the hypothesis $A_1 \sim_r A_2$ and $A_2 \sim_r A_3$), and various other small details I may have left out (most intentionally, some probably not $\ddot{\smile}$).

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You can do what I did above for every $r \geq 0$, so of course you can also do it for $r = \infty$. Now that we have an equivalence relation, we can definitely talk about the equivalence classes. This is why the definition of a smooth structure as an equivalence class of $C^{\infty}$ (or more generally $C^r$) atlases $\mathcal{A} = [A]$ makes perfect sense.

Answer 2

The relation is only an equivalence on the set of smooth atlases. Nonsmooth atlases can essentially be excluded entirely, since no nonsmooth atlas is equivalent to a smooth one.

As a side note, it's also common to define smooth structures as maximal smooth atlases, i.e. atlases which contain every smoothly compatible chart. This is equivalent (the maximal atlas is the union of the equivalence class) and may or may be more intuitive.