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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.

amWhy
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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". –  Jun 10 '11 at 23:59

4 Answers4

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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 $\mathbb R$.

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

AlbertH
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In Lee's Introduction to Smooth Manifolds 2.0, pg 14:

"If a topological manifold M can be covered by a single chart, the smooth compatibility condition is trivially satisfied, so any such chart automatically determines a smooth structure on M."

the definition of standard smooth structure is on page 17

"The Euclidean space ${\mathbb{R}}^n$ is a smooth n-manifold with the smooth structure determined by the atlas consisting of the single chart $({\mathbb{R}}^n,Id_{\mathbb{R}^n})$. We call this the standard smooth structure on ${\mathbb{R}}^n$ "

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I do not see any need for item "3" in the definition, although it is very customary.

IMO, usually a thing of type $T$ has at least two commonly used definitions, one is good for "checking whether or not something $T$" and the other one is good for "taking advantage of $T$'s properties".

For example, some define a $C^\infty$ differentiable $f: \mathbb{R}^p \rightarrow \mathbb{R}^q$ as a function where all partial derivatives can be differentiated any number of times. This is a very good definition if all you want is to ask your students to prove that a certain $f = (f_1, ..., f_q)$ is $C^\infty$. They will calculate all partial derivatives for all $f_j$ and that's it. Otherwise, this is a pretty useless definition.

It is much better in my opinion, if you define what it means to be differentiable (once!) at one point as a function that can be approximated by a linear transform [...]. This is a more advanced and more useful definition. But it has a steeper learning curve.

Another example is the definition of a measurable random variable. Some define $f: \Omega \rightarrow \mathbb{R}$ to be measurable when, for every $a \in \mathbb{R}$, $f^{-1}((\infty,a)) = \{\omega \in \Omega \,|\, f(\omega) < a\}$ is measurable. This, again is a pretty useless definition unless all you want is to check whether a certain "random variable" is or is not measurable. Check this for more details: Intuitively, how should I think of Measurable Functions?

Item "3" in the definition of a differentiable structure is there just make it easier to handle in many contexts. If you remove "3", you need to define what it means for two such structures to be equal or, maybe a better wording, equivalent. If the atlas $\mathcal{A}$ is maximal you define a chart as an element of $\mathcal{A}$. If the atlas is not maximal you might define a chart as a function that is compatible with $\mathcal{A}$. Or, maybe, you can define two different things: a chart (belongs to $\mathcal{A}$) and a compatible chart (compatible with everything in $\mathcal{A}$). This latter approach has the disadvantage that being a compatible chart does not imply being a chart! :-(

On the other hand, item "3" can get in the way, sometimes. I, personally, prefer the definition without it. When you define a $C^\omega$ or a $C^n$ differentiable structure and you demand maximal atlases, then a $C^\omega$ structure is not a $C^\infty$ structure, and a $C^\infty$ structure is not a $C^n$ structure. You would have to produce a maximal atlas, first.

So, I prefer to define an atlas as something that satisfies items "1" and "2". Then I define atlas compatibility and then maximal atlas. Note that maximal is a whole family of definitions as you have $C^\infty$-maximal, $C^n$-maximal, $C^\omega$-maximal and probably other kinds of maximal (I am not an expert in the subject).

The fact that there always is a maximal atlas is a theorem and in a sense, what it says is that atlases being compatible is an equivalence relation. And you can say two differentiable structures are ($C^\infty$, etc) equivalent when they have compatible atlases.

The differentiable structure is this abstract thing that is determined by the atlases (maximal or not). The same way, a topology is not just a family of open sets... it can be given by nets, filters, generators, closed sets, closure operator, neighbourhoods, neighbourhood basis, etc.