What is this $\mathfrak{D} =\mathfrak{D}(\mathfrak{U})$ in differentiable topology

Question

From the textbook: Introduction to Differential Topology by TH. Brocker and K. Janich.

Defintion 1.3: An atlas of a manifold is called differentiable if all its chart transformations are differentiable.

Then a few lines below:

If $\mathfrak{U}$ is a differentiable atlas on the manifold $M$, then the atlas $\mathfrak{D} =\mathfrak{D}(\mathfrak{U})$ contains precisely those charts for which every chart transformation with a chart from $\mathfrak{U}$ is differentiable.

I don't understand this at all. So a differentiable atlas is an atlas where every chart is differentiable. ok. However, I don't understand what "precisely those charts for which every chart transformation with a chart from $\mathfrak{U}$" means.

So what is $\mathfrak{D}(\mathfrak{U})$ and how is it different than $\mathfrak{U}$?

Answer 1

A chart on $M$ is any homeomorphism $\phi : U \to U'$ from an open subset $U \subset M$ to an open subset $U' \subset \mathbb R^n$. Let us denote by $\operatorname{Ch}(M)$ the set of all charts on $M$. An atlas on $M$ is a subset $\mathfrak A \subset \operatorname{Ch}(M)$ such that each $p \in M$ is contained in the domain $U$ of some chart $\phi : U \to U'$ in $\mathfrak A$.

Definition 1.3 introduces the concept of a differentiable atlas. This is a serious restriction; there exist atlases which are not differentiable.

For each differentiable atlas $\mathfrak A$ the authors define $$\mathfrak D(\mathfrak A) = \\ \{ \psi \in \operatorname{Ch}(M) \mid \forall \phi \in \mathfrak A : \text{ The chart transformations between } \psi \text{ and } \phi \text{ are differentiable}\} \\ = \{ \psi \in \operatorname{Ch}(M) \mid \mathfrak A \cup \{ \psi\} \text{ is a differentiable atlas} \} .$$

Clearly $\mathfrak A \subset \mathfrak D(\mathfrak A)$, but in general $\mathfrak D(\mathfrak A) $ is much bigger than $\mathfrak A$.

One can easily show that

1. $\mathfrak D(\mathfrak A)$ is a differentiable atlas.

2. Each differentable atlas $\mathfrak A'$ containing $\mathfrak A$ is contained in $\mathfrak D(\mathfrak A)$.

Certainly the authors have proved this.

In other words, there exists a unique maximal differentiable atlas containing $\mathfrak A$, and this atlas is $\mathfrak D(\mathfrak A)$.