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Differential manifolds

  1. Feb 6, 2012 #1
    Hi everebody,
    I want to clear something.An n-dimentional differential manifoled is locally endowed by topologies defined by the metrices from the local parametrisations.I suppose that these topologies may all be different.Am i right?If i am mistaken ,then why?
    thank's
     
  2. jcsd
  3. Feb 6, 2012 #2

    mathwonk

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    i think you are mistaken. the local coordinate systems not only define the same topology on overlaps but also the same differentiable structure.
     
  4. Feb 7, 2012 #3

    Bacle2

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    I'm not sure I understand your question, but the manifold as a whole is a metric space, i.e., it has a global/general topology generated by ( with topological basis ) the open balls in the metric, where the distance is given by the length of the shortest path between points.
     
  5. Feb 7, 2012 #4
    It follows from definition on manifold. It must be locally homeomorphic to ℝn, so it inherits topology from there. A subset O of manifold is open iff for every chart (U; γ) of M a coordinate representation of intersection of O and U is open in ℝn.

    I might have not put it in the best way, but in fact its quite simple.

    You don't even have to define manifold as a topological space, but as a set with an atlas, and topology follows from that.

    Ordinary manifold shouldn't be a metric space I believe. You cannot measure lenghts on it.
     
  6. Feb 7, 2012 #5

    Bacle2

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    The standard definition/conditions for a topological space to be a manifold I know of, guarantees its metrizability: 2nd countable, paracompact, Hausdorff, etc. and then use some variants of Urysohn metrization theorem to guarantee its metrizable. I'll look up some refs. for a more detailed argument.
     
  7. Feb 7, 2012 #6

    Bacle2

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    I just checked Wiki and they state that 2nd countable manifolds are metrizable. Urysohn's metrization says that Hausdorff +2nd countable +regular => metrizable. In my use, these conditions are assumed..

    Still, I may be off in stating that in all manifolds the topology is determined by path length,(so that open balls B(p,r) are given by path length) only when we have a Riemannian metric, so that the manifold becomes a(n) intrinsic length space. Maybe someone can double-check on that. See the 'examples' section in:
    http://en.wikipedia.org/wiki/Length_space , for more, and a better explanation.

    Basically, a Riemannian structure on M allows us to define the length of paths
    between points. Then we define , for x,y in M , d_L(x,y) := inf {L(γ): γ:I→M; γ(0)=a,
    γ(1)=b} , where L is the length of the path. One can show this d_L is a metric, and the topology determined by this metric ( i.e., base elements are open balls B(p,r)) agrees with the intrinsic topology of the manifold.

    And I think we only need the manifold to be C^1 , for it to allow a Riemannian structure.
     
    Last edited: Feb 7, 2012
  8. Feb 7, 2012 #7
    I capitulate, since I don't understand this area, but I think you mistook metric and metrizable space.
     
  9. Feb 7, 2012 #8

    Bacle2

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    Well, I am making a distinction between the Riemannian metric --more accurately metric tensor-- and a metric as a distance function in the manifold, if that is what you meant. My point is that a Riemannian metric gives rise to a distance function metric on the manifold by the method I stated in my previous post.
     
  10. Feb 8, 2012 #9
    Thank's all for your constructive replies
     
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