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I Simple GR Question

  1. Jul 18, 2016 #1
    Is it fair to say, when talking about spacetime with a given metric, it would be redundant to state that the associated set has the metric topology placed on it. In other words, let ##M## be a set, ##O## the metric topology, ##\nabla## a connection, ##g## a metric, and ##T## be the direction of time, then
    ##(M,O,\nabla,g,T)=(M,\nabla,g,T)##

    Edit: I suppose what I'm asking is, does the metric ##g## induce the metric topology?
     
    Last edited: Jul 18, 2016
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  3. Jul 18, 2016 #2

    George Jones

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    What is "the metric topology" when ##g## is not positive-definite?
     
  4. Jul 18, 2016 #3

    robphy

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  5. Jul 18, 2016 #4
    You're right, I suppose I should make that precise. The metric topology I'm referring to has as open sets
    ##B_r(m_0)=\{m_0\in M : g(m_0,m)<r\ \forall m\in M \text{ where } r>0\}##
     
  6. Jul 18, 2016 #5

    George Jones

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    I am sorry, but I still do not understand. The "metric" ##g## (i.e., not a metric in the sense of metric spaces) takes as input two tangent vectors at the same point of the manifold, not two points of the manifold.
     
  7. Jul 18, 2016 #6
    I'm referring to ##g## in two different contexts which is what I think made my question unclear. So, let ##g## be the metric from GR and let ##d## be the metric from math class. Does defining ##g## on a differentiable manifold automatically induce ##d##?

    As an aside, this is interesting to me since I believe it to be key in the notion of a topological field theory. Of course I'm just starting on the subject so I could be wrong.
     
  8. Jul 18, 2016 #7

    Orodruin

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    The balls you refer to are not very good to use to build a topology when you have a pseudo Riemannian metric. You may want to use a topology based on the intersection of light cones instead.
     
  9. Jul 18, 2016 #8

    DrGreg

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    As I understand it, the topology of ##M## is independent of ##g##. It arises from the Euclidean topology of ##\mathbb{R}^4## mapped onto ##M## via the coordinate charts. You need the topology in order to define concepts such as "smooth" and "differentiable".

    The metric tensor ##g## doesn't define a "metric" in the "metric space" sense because it isn't positive definite.
     
  10. Jul 18, 2016 #9
    OK, I think I get it now. Thanks everyone!
     
  11. Jul 22, 2016 #10

    vanhees71

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    That's why it's better to talk about pseudo-Riemannian manifolds and pseudo-metric to make this clear from the very beginning. Analysis is based on a true metric not a pseudo-metric.
     
  12. Jul 26, 2016 #11
    As far as I understand it a given metric in general relativity could be valid for more than one topology.
     
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