Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Concerning charts on a manifold

  1. Jul 8, 2011 #1
    I'm new to manifolds, so please forgive me if this sounds ignorant. I was just wondering whether the charts of a smooth manifold (within some atlas) always "overlap". If I'm not mistaken they map to open subsets of R^n, and being homeomorphisms should have the inverse image as open. But I'm not entirely sure, so I'd appreciate it if somebody could either confirm my understanding or explain what I'm missing.
     
  2. jcsd
  3. Jul 9, 2011 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There can be charts that do not overlap any other. Take for instance R^n itself as a manifold. There is an atlas for this manifold consisting of the single chart id:R^n-->R^n. So trivially, this chart intersects no other since there are no other. More generally, if a manifold has a connected component C homeomorphic to an open subset U of R^n, then there is an atlas for this manifold with a chart mapping C to U and intersecting no other chart.
     
  4. Jul 9, 2011 #3
    Of course, you almost always deal with a maximal (possibly oriented) smooth atlas, which is defined to be an atlas which contains every possible chart that would be compatible with it (that's a loose paraphrase, but it conveys the idea). Then you can restrict a chart to a proper subset of its domain, and that will be a different chart also in the atlas, and of course they overlap.
     
  5. Jul 9, 2011 #4
    If your manifold is the union of open sets with pairwise-disjoint closure, then it is disconnected.
     
  6. Jul 9, 2011 #5
    Thanks you guys, that cleared a lot up. :smile:
     
  7. Jul 9, 2011 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm worried you guys are painting a misleading picture -- you're answering "can they overlap" but not "must they overlap".

    For any open interval contained in the real line, there is a coordinate chart that covers only that interval and nothing else.

    So, for example, a chart covering (0,1) and a chart covering (2,3) have empty overlap.


    (And for the record, the empty set is an open subset)
     
  8. Jul 10, 2011 #7
    Actually, using the fact that n is the Lebesgue Covering Dimension, an n-manifold
    has, in each component, a refinement of each cover in which each element is covered
    by a maximum of n+1 charts.

    To give a proof to the previous claim that a manifold M covered by a union
    of pairwise-disjoint open sets must be disconnected ( I think that disjoint closure
    is not necessary):

    Let U_(i in I) {V_i} cover M, with V_j/\V_k ={ }

    Then any V_k is open in M, but also closed in it, since its complement in M is the
    union of a collection of open sets V_k- \/{V_i}-V_k.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook