# Are Transition Maps Implied within an Atlas?

Mathematics Asked by Tug Witt on December 12, 2020

From my understanding of (smooth) manifolds, all you need is an atlas to describe a manifold. However, if you have some atlas ?={($$U_n$$,$$phi_n$$)} 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 $$cin A$$, 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?

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 $$C^1$$?

Correct answer by Lee Mosher on December 12, 2020

Once you have the charts $$phi_n$$, the transition maps are determined, as $$phi_mcircphi_n^{-1}$$. (That uses my favorite convention for the direction of these maps; you might need to move the "inverse" if your convention is different.)

Answered by Andreas Blass on December 12, 2020

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