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Consider the definition of differentiable structure (Lectures on Differential Geometry, S.S. Chern):

Suppose $M$ is an m-dimensional manifold. If a given set of coordinate charts ${\mathcal A} = \{(U,\phi_U),(V,\phi_V),(W,\phi_W),\cdots\}$ on $M$ satisfies the following conditions, then we call ${\mathcal A}$ a $C^r$-differentiable structure on $M$:

  1. $\{U,V,W,\cdots\}$ is an open covering of $M$;
  2. any two coordinate charts in ${\mathcal A}$ are $C^r$-compatible;
  3. ${\mathcal A}$ is maximal, i.e., if a coordinate chart $(\tilde{U},\phi_{\tilde{U}})$ is $C^r$-compatible with all coordinate charts in ${\mathcal A}$, then $(\tilde{U},\phi_{\tilde{U}})\in{\mathcal A}$.

An example for this definition in that book, is as following: For $M={\mathbb R}$, let $U=M$, and $\phi_U$ be the identity map. then $\{(U,\phi_U)\}$ is a coordinate covering of ${\mathbb R}$. This provides a smooth differentiable structure on ${\mathbb R}$, called the standard differentiable structure of ${\mathbb R}$.

I don't understand how $\{(U,\phi_U)\}$ provides a smooth differentiable structure. Is it maximal? Consider for example $\{(U,\phi_U),(V,\phi_V)\}$ where $V=(0,1)$ and $\phi_V$ the identity map. This puzzles me.

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You lack a theorem: to every coordinate covering there corresponds a unique differentiable structure. In your case, this differentiable structure is the collection $\{(U, \varphi)\}$ where $U$ is an open subset of the line and $\varphi: U \to V$ is a $C^r$-invertible mapping. Typically it is very difficult to exhibit explicitly all coordinate charts in a differentiable structure. – Giuseppe Negro Jun 10 '11 at 23:40
@dissonance: Hmm, good point. I think you are talking about the proposition that if a set $A'$ of coordinate chars satisfies 1) and 2) in the definition, then for any positive integer $s$, $0<s\leq r$, there exists a unique $C^s-$differentiable structure A such that $A'\subset A$. So to construct a differentiable manifold, we need only choose a covering by compatible charts instead of "exhibiting all coordinate charts". – Jack Jun 10 '11 at 23:59
Closely related: Why maximal atlas – t.b. Jun 11 '11 at 13:27
up vote 2 down vote accepted

You could of course talk about non-maximal differentiable structures -- just erase condition (3) from your definition. The point is that every one of these non-maximal differentiable structures sits in a unique maximal structure. To define the maximal differentiable structure from a non-maximal one, simply include all charts that are compatible with your non-maximal atlas.

So in your example, the differentiable structure compatible with the identity map on $\mathbb R$ is the collection of all diffeomorphisms between open subsets of $\mathbb R$.

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Evidently $\{(U, \phi_U)\}$ is not maximal, but you can always considerer the set of all charts compatible with $(U, \phi_U)$.

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