Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I pose the following question: Given a connected (maybe we need compactness) manifold or regular surface, can we find a single parametrization $\chi:B\to M$ from the open ball to the manifold, such that $\overline{\chi(B)}=M$ (the closure is to be understood in M)?

The sphere and torus have this kind of parametrization, so I wonder if this holds in general. I tried to answer the question using Zorn's lemma, that is, I defined the set of every parametrization from a set $U\cong B$ ($U$ isn't fix) to the manifold and ordered it with the relation: $\chi_{1}\le \chi_{2}$ iff $\textrm{dom}(\chi_{1})\le \textrm{dom}(\chi_{2})$ and they coincide in their domains. The problems have to do with the maximal element, how to prove that it works:

  1. How to control the size of the domains of the functions, so that the maximal element hasn't as domain all $R^n$ and $\overline{\chi(U)}\neq M$.
  2. Given a parametrization $\chi$ such that there exists $p\in M-\overline{\chi(U)}$, how to extend the parametrization to cover $p$.

This proof works for 1-dimension, because I overpassed both problems changing the parameter to arc-length, but in general, I cannot do that, because of the Egregium Theorem (It is impossible to use somehow like area-parameter).

I would like to know the answer, if it were true, we could classify surfaces up to homeomorphism working on quotient topologies in a ball. On the other hand, is easier to work with only one chart.

Thank you and I hope the question is not too obvious.

share|improve this question
3  
This answer shows it's always possible in the the compact (smooth, connected) case: math.stackexchange.com/questions/18083/…. I'm not entirely sure what happens in the noncompact case. –  Jason DeVito Feb 3 '13 at 2:44
    
Thank you for your comment, to know the answer for compact manifolds is important for me. –  user39490 Feb 3 '13 at 5:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.