6
$\begingroup$

i have a question regarding the definition of differentiable functions on manifolds. Why is one atlas of the manifold not enough to define the differentiability of a function via local chart? Why is the maximal atlas, containing the original atlas + all compatible ones to it, necessary? Every introductory literature I have read on this subject does touch on this but I fail to understand why.

Kossinki:

For our purpose, which is to define differentiable functions, two different atlases may yield the same result. They certainly will if they are compatible, in the sense that their union is an atlas. This relation of compatibility is an equivalence relation; hence every atlas is contained in a maximal one: the union of all atlases compatible with it.

PennU Differential Geometry Script:

We must ensure that we have enough charts in order to carry out our program of generalizing calculus on $\mathbb{R}^n$ to manifolds. For this, we must be able to add new charts whenever necessary, provided that they are consistent with the previous charts in an existing atlas.

I'd be thankful if anyone was able to clear me up.

Kind regards.

$\endgroup$

1 Answer 1

6
$\begingroup$

One atlas is enough. You can define a manifold as a topological space together with one particular atlas.

But the same topological space with another atlas would then strictly speaking be another manifold, which would feel wrong if the two atlases are compatible. So either you relax your definition of manifold to be an equivalence class of what you previously called manifolds (the equivalence relation being compatibility of atlases), or else you require in the definition that the atlas is the maximal one.

In practice you just give one particular atlas anyway, since it uniquely defines the maximal atlas.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .