# Manifold and maximal atlas

1) I didn't understand really what is a maximal atlas. Is it as set of compatible chart maximal in the sens that adding one more chart will yield the atlas not compatible ?

2) Let two atlas $\mathcal A$ and $\mathcal A'$. So if they are compatible, they are both in a maximal atlas $\hat{\mathcal A}$ ?

3) And if they are not compatible, there are two atlas $\hat{\mathcal A}$ and $\tilde{\mathcal A}$ such that $\mathcal A$ is for example in $\hat{\mathcal A}$ and $\mathcal A'\in\tilde{\mathcal A}$ ?

4) And if I understood well, $\hat{\mathcal A}$ gives smooth structure and $\tilde{\mathcal A}$ gives an other smooth structure ? But both are incompatible ?

I hope my question are clear enough.

(1) Yes, by the definition of maximal.

(2) Yes. Technical detains in Is Zorn's lemma required to prove the existence of a maximal atlas on a manifold? and Why maximal atlas.

(3) I understand that the "they" in "And if they are not compatible..." are two charts. Yes, each chart is in an atlas and the intersection of both atlas is empty.

(4) They give different structures. But can be diffeomorphic. Easy example: $\Bbb R$ and the two atlases $\{x\longmapsto x\}$ and $\{x\longmapsto x^3\}$.

• Perez Pinilla I think it might help if you add a little more to 4 i.e even though they have different structures they might still be diffeomorphic and in a sense they are not different – happymath Dec 2 '15 at 10:40
• @happymath, good idea. I've edited the answer. – Martín-Blas Pérez Pinilla Dec 2 '15 at 10:46
• Just a precision, do we have $\hat{\mathcal A}\cap\tilde{\mathcal A}=\emptyset$ ? To me it's this, but I just want to have a confirmation – MSE Dec 2 '15 at 12:20
• @MSE, yes, but the converse is false. – Martín-Blas Pérez Pinilla Dec 2 '15 at 15:09
• About 4th case, $\{x\longmapsto x^3\}$ and $\{x\longmapsto x\}$ are compatible, aren't they? I've been confused with another problem. – Fardad Pouran Sep 27 '17 at 12:46

(1) You are correct. $\mathcal{A}$ maximal atals is maximal in the sense that it contains all possible compatible charts.

(2) Yes. Every atlas $\mathcal{A}$ is contained in exactly one maximal atlas, and it is easy to desribe it: it is the set of all charts compatible with $\mathcal{A}$. Since $\mathcal{A}$ already covers $M$, it can be checked that any two such charts are compatible (i.e the corresponding transition maps are smooth) via going back and forth through charts in $\mathcal{A}$.

In particular, if ${\mathcal{A}}',{\mathcal{A}}$ are compatible, they are both contained in the same maximal atlas.

(3) Yes. same argument as (2).

(4) Yes.

It's worth noting why do we define smooth structure to be a maximal atlas:

We want each smooth structure (=maximal atlas) to define a unique sense of what does it mean for a function on the manifold (say from $M \to \mathbb{R}$) to be smooth.

We want a one-to-one corespondence between smooths structures and subsets of smooth functions.

Two compatible atlases are indistinguishable from this point of view, since they give rise to identical notions of smoothness of maps.

• Thanks you, it's clear now. – MSE Dec 2 '15 at 12:15