# Are Transition Maps Implied within an Atlas?

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?
• 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. Commented Nov 7, 2020 at 18:55

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 $$C^1$$?

• 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? Commented Nov 7, 2020 at 21:26
• 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 Commented Nov 7, 2020 at 21:49

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.)