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From my understanding of (smooth) manifolds, all you need is an atlas to describe a manifold. However, if you have some atlas 𝐴={($U_n$,$\phi_n$)} with $n$ charts, we still haven't defined our transition maps. My questions are:

  • Are the transition maps implied within the atlas (i.e. you can derive all transition maps from a given atlas) or do we have to store our transition maps along with our atlas to prove we have a smooth atlas?
  • If you have $n$ charts within an atlas, does that mean you are going to have something like $n!$ (maybe its a bit more complex than that) transition maps? For example, if $n=3$ and a chart $c\in A$, wouldn't you need a transition map from $c_1 -> c_2$, $c_1 -> c_3$, $c_2 -> c_3$ plus all of the inverses (that are implied)? When don't you need a transition map between two charts in the same atlas?
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  • $\begingroup$ You don't need a transition map if the charts have an empty intersection. If the intersection is non-empty then you can construct the transition map using the charts, which are part of the atlas. This is because the composition of homeomorphism and their inverses are also homeomorphism. $\endgroup$ Commented Nov 7, 2020 at 18:55

2 Answers 2

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You can simply define the transition maps, once the atlas is given.

There is a transition map which I shall denote $\psi_{m,n}$ for every pair of indices $m,n$ having the property that $U_m \cap U_n \ne \emptyset$.

The domain of $\psi_{m,n}$ is the set $\phi_m(U_m \cap U_n) \subset \mathbb R^k$ (I'm assuming implicitly that $k$ is the dimension of the manifold).

The range (or codomain) of $\psi_{m,n}$ is the set $\phi_n(U_m \cap U_n) \subset \mathbb R^k$.

And the formula for $\psi_{m,n} : \phi_m(U_m \cap U_n) \to \phi_n(U_m \cap U_n)$ is $$\psi_{m,n}(p) = \phi_n(\phi^{-1}_m(p)), \quad p \in \phi_m(U_m \cap U_n) $$

Also, once all of this is written down, one can use the definition of a manifold together with the Invariance of Domain Theorem to prove that the domain and range of $\phi_{m,n}$ are both open subsets of $\mathbb R^k$, and one can show that $\psi_{n,m}$ is an inverse map of $\psi_{m,n}$, hence each transition map is a homeomorphism from its domain to its range.

And once that is done, you can now ask yourself questions that are aimed at determining whether your manifold is a $C^\infty$ manifold, or a $C^2$ manifold, or a $C^1$ manifold or whatever smoothness property you want. Namely: Are the functions $\{\psi_{m,n}\}$ all $C^\infty$? or are they all $C^2$? or $C^1$?

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  • $\begingroup$ A transition map is smooth if the two charts that construct it are smooth as well, right? Or do you have to do some more extensive proofs? $\endgroup$
    – Tug Witt
    Commented Nov 7, 2020 at 21:26
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    $\begingroup$ That part you have backwards. If all the transition maps are smooth, then that defines a smooth structure on the manifold, after which it follows that every chart is smooth. See my answer here $\endgroup$
    – Lee Mosher
    Commented Nov 7, 2020 at 21:49
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Once you have the charts $\phi_n$, the transition maps are determined, as $\phi_m\circ\phi_n^{-1}$. (That uses my favorite convention for the direction of these maps; you might need to move the "inverse" if your convention is different.)

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