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I am currently studying a basic course on differentiable manifolds.I have read the following definition of differentiable atlas and manifolds:

Definition. Let $ \mathcal{A} = {(x_{\alpha},U_{\alpha})}_{\alpha \in A}$ be an atlas on a topological manifold $M$. Whenever the overlap $U_\alpha \cap U_\beta$ between two chart domains is nonempty we have the change of coordinates map $x_\beta \circ x_\alpha^{-1} : x_\alpha(U_\alpha \cap U_\beta) \to x_\beta(U_\alpha \cap U_\beta)$. If all such change of coordinates maps are $C^r$-diffeomorphisms, then $\mathcal {A}$ is called a $C^r$-atlas and a manifold endowed with maximal differentiable atlas is called differentiable manifold.

I really find it hard to understand the following:

Is there any intuitive idea from which the definition of diffferentiable atlas and manifolds emerge? How the local diffeomorphisms $x_\beta \circ x_\alpha^{-1} : x_\alpha(U_\alpha \cap U_\beta) \to x_\beta(U_\alpha \cap U_\beta)$ allow us to define a differential structure on manifold globally?

Moreover,why is it required for atlas to be maximal in order to define differentiable manifold

M. Winter
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abhishek
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    You can think about it the other way around. Instead of a topological space endowed with lots and lots of chart, think about small patches cut out of $\Bbb R^n$ and glued together to form e.g. a sphere or a torus. That we use a maximal atlas is because otherwise you can use different arrangements of patches to form the essentially same manifold. If two atlases are compatible, then they should describe the same manifold, and this is adressed by using a maximal atlas from the start. If then one asks if his atlas describes the same manifold, you just look if it is a subset of your maximal atlas! – M. Winter Jul 05 '17 at 10:35
  • A more intuitive definition can be given if you assume your manifold is a subset of $\mathbb R^n$. Some books take that approach (for example, Analysis on Manifolds by Munkres). – littleO Jul 05 '17 at 10:38

1 Answers1

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The user @M.Winter gives a very good explanation of why we have this maximal atlas. As far as the transition maps $x_{\beta} \circ x_{\alpha}^{-1}$, we require these to be diffeomorphisms because smoothness isn't invariant under homeomorphism. For instance, if $f: M \to \mathbb{R}$ is smooth, we mean that given any $p \in M$, there exists a local chart $(\phi,U)$ such that $f \circ \phi^{-1}: U \to \mathbb{R}$ is smooth.

Sometimes it is very useful to change your coordinates to make the problem simpler, but in order to do this without loss of generality, we need to know that if the problem says $f$ is smooth, then this is independent of the charts on $M$ that we use. This is what the transition map condition allows us to do. I hope this helps.

Faraad Armwood
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  • (+1) Your second paragraph cannot be emphasized enough: On a smooth manifold, smoothness of a function or mapping is dictated by the overlap maps (i.e. by smoothness of an arbitrary local representative), not by smoothness of a single local representative. Formulating smoothness of arbitrary representatives in terms of overlap maps allows us to check smooth compatibility some atlas once and for all instead of having to check smoothness of arbitrary representatives each time a new function or mapping comes along. – Andrew D. Hwang Jul 05 '17 at 12:46
  • @AndrewD.Hwang: Thank you. Also, thank you for the free pdf on general relativity. It wasn't until I read Lee (after reading from other books) that I understand this condition. For the OP, Lee's Introduction to Smooth Manifolds is the best. – Faraad Armwood Jul 05 '17 at 13:00
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    @faraad armwood:thanks a lot for your answer,but,I have a problem how do these transition maps which are diffeomorphism from subsets of $\Bbb R^n$ to subset of $\Bbb R^n$ define a differential structure on the manifold? – abhishek Jul 05 '17 at 16:00
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    Please read Lee's book. He addresses this as well. $C^o$ manifolds like the unit square with it's standard embedding in $\mathbb{R}^2$ i.e $[0,1] \times [0,1]$ doesn't have transition maps with this property on its $4$-corners. Hence there is something different about a square and the unit circle $S^1$. This is an example of the unit circle being a topological circle vs a smooth circle. The unit circle does have an atlas where the transition maps behave the way described i.e this must be some additional structure, so let's just call it a differentiable structure. – Faraad Armwood Jul 05 '17 at 17:37