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Manifold and metric

  1. Nov 20, 2007 #1
    Does a manifold necessarily have a metric?
    Does a manifold without metric exist? If it exists, what is its name?
  2. jcsd
  3. Nov 20, 2007 #2


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    Since a manifold is locally Euclidean, it must always have a local metric. However, it does not follow that there will be a "distance" between ANY two points and so there may not be a "global" metric.
  4. Nov 20, 2007 #3


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    the hierarchy is something like this. You take a set. Add a topology. It becomes a tpological space. Add an atlas, it becomes a topological manifold. Change for a differentiable atlas. It becomes a differential manifold. Add a Riemannian or Pseudo-Riemannian of Lorentzian or whatever metric and it becomes a Riemannian (resp. Pseudo-Riemannian, Lorentzian, whatever) manifold.
  5. Nov 20, 2007 #4
    Topological manifolds (sets with a topology locally homeomorphic to Rn) do not necessarily admit a metric. There are then many non-metrizable manifolds, such as the Prüfer manifold. Urysohn's metrization theorem will let you know if your manifold is metrizable.
    Last edited: Nov 20, 2007
  6. Nov 20, 2007 #5

    Chris Hillman

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    Ditto slider142 about topological manifolds. I have a hunch the OP will need to know that "metric space" is not the same thing as the notion of "metric tensor" from Riemannian geometry, although they are certainly related. But "metric tensor" from Lorentzian geometry is not much like "metric" from "metric space"!

    About smooth manifolds: you can give a smooth manifold additional structure, perhaps by defining a Riemannian or Lorentzian metric tensor. As Halls hinted, as per the fundamental local versus global distinction in manifold theory, even after defining a Riemannian or Lorentzian metric tensor, there will be multiple distinct notions of "distance in the large" which may or not correspond roughly to the notion of "metric" fro m "metric space". In particular, Lorentzian metrics get their topology from the (locally euclidean) topological manifold structure, not from the bundled indefinite bilinear form.

    (I'm being a bit more sloppy than usual due to PF sluggishness.)
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