Why do we use only compatible charts in the Theory of Manifolds?

Question

I couldn't find a duplicate, although I think is a very common question.

Given two charts, ($U_{1},φ_{1}$), ($U_{2},φ_{2}$), on a n-dimensional topological manifold M, such that: $U_{1} \cap U_{2}\neq \emptyset$, we get transition maps:

$φ_{1}\circ φ_{2}^{-1} : φ_{2}(U_{1}\cap U_{2}) \rightarrow φ_{1}(U_{1}\cap U_{2})$, and

$φ_{2}\circ φ_{1}^{-1} : φ_{1}(U_{1}\cap U_{2}) \rightarrow φ_{2}(U_{1}\cap U_{2})$

Two charts, as above, are called compatible if the transition maps, as above, are homeomorphisms. If $U_{1} \cap U_{2} = \emptyset$, then they are compatible.

My question is, why do we need this behavior? In addition, if we want to define C$^{\infty}$-compatible charts, why do we need to take transition maps to be smooth, Euclidian mappings?

Answer 1

Imagine taking two nice pieces of tissue and gluing them together with the goal to obtain a larger piece of tissue which is still nice. Then you will put part of the second piece of tissue over part of the first one, make it so that there are no wrinkles and then glue. This basically what is happening with manifolds: you are gluing patches of euclidean space together in such a way that the resulting object is "nice", i.e. is again locally euclidean. If you want nicer objects, you will have to take better behaved gluings (you could accept that the resulting piece of tissue forms an angle somewhere, or you could ask for it to be smooth everywhere...)

Answer 2

Because we want use the charts to define differentiability. Differentiability depending of the choosing of some specific chart will be a nightmare.

Answer 3

Take an inclusion of sets $i:U\to M$, where $U$ is an open set in $\mathbb{R}^n$.

We'd like to be able to say that $i$ is a homeomorphism (continuous isomorphism) onto its image—or, for a smooth manifold, to say that $i$ is a smooth isomorphism. Or, for a piecewise-linear manifold, we'd like to say that $i$ is a piecewise linear isomorphism.

In any case, we'd like to check this behavior on charts, since we know how these concepts behave in $\mathbb{R}^n$. But this concept is not well-defined without transition maps with the right property. Otherwise, continuity/smoothness/piecewise linearity would be dependent on which charts you used to verify it, rather than a property of the manifold.

Answer 4

Yes, each individual chart $\phi \colon U \to \mathbb{R}^n$ is $C^\infty$, but remember that each chart in only defined on one open neighbourhood $U \subset M$. To allow us to consider behaviour on the whole of the manifold, it's important to be able to patch the $U$ neighbourhoods together nicely, going from one chart $\phi$ to another without having any non-differentiable issues. This is why we need transition maps to be smooth as well as the charts.