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Manifold charts

  1. Jun 26, 2011 #1
    What is the least number of charts needed to specify a given smooth manifold, for simplicity of dimension 2? For instance the minimum number of charts to cover a torus, or a 2-sphere or a 2D 1-sheet hyperboloid?
    I would think it goes with the definition of manifold that in any case you need at least two charts are needed to cover the whole manifold, I think in the case of the 2-sphere is enough with two, but the torus needs 3, but another poster seems to think that in general it is enough with one chart of coordinates but I know that it is only possible trivially with the spaces like E^n in the case of Riemannian manifolds or M^n in case of pseudoriemannian, because by definition the Euclidean space R^n is a smooth manifold with a single chart, and the same in the case of a (-,+,+,+) signature with Minkowski space would need just one chart. But other than this I'd say you need more than one chart otherwise it defeats the manifold concept. Anyone agrees?
    Thanks in advance.
     
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  3. Jun 26, 2011 #2

    George Jones

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    What about Riemannian and semi-Riemannian differentiable manifolds diffeomorphic to R^n, but with different connections?
     
    Last edited: Jun 26, 2011
  4. Jun 26, 2011 #3
    What about them? Could you elaborate, or give some example?
     
  5. Jun 26, 2011 #4

    George Jones

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    Consider R^4 with the global chart (t,x,y,x). Take the subset of R^4 that has t>0, and use the metric

    ds^2 = -dt^2 + a(t)^2 (dx^2 + dy^2 + dz^2)

    with a(t) a smoothly increasing function of t.

    (t,x,y,z) with t>0 is a global chart on this semi-Rieimannian manfold, which has non-zero curvature tensor (from metric-compatible, torsion-free connection).
     
  6. Jun 26, 2011 #5
    But then that would not be locally diffeomorphic to R^4 but to a subset of R^4 with t>0, I'm not even sure that fulfills the definition for a 4-manifold.
    You are imposing an arbitrary condition on R^4 (that makes physical sense, but I was referring to the general, more abstract mathematical case) by considering only the t>0 subset of R^4.
     
  7. Jun 26, 2011 #6

    George Jones

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    This subset or R^4 is diffeomorphic to R^4.
     
  8. Jun 26, 2011 #7
    Sure, but then the manifold is not locally Minkowskian but locally a subset of Minkowski.
    Is this mathematically OK? Sounds like cheating ;)
     
  9. Jun 26, 2011 #8
    Are you maybe referring to manifolds with boundary? apparently this type can be defined with a correspondance to subsets of R^n instead of R^n.
     
  10. Jun 26, 2011 #9

    George Jones

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    I am not sure where we stand. I have exhibited a semi-Riemannian manifold that has both a global chart and non-zero intrinsic curvature.

    Do you agree?

    If not, why not?

    If so, does this have relevance to your original post?
     
  11. Jun 26, 2011 #10

    George Jones

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    No .
     
  12. Jun 26, 2011 #11
    The chart you showed is not the type I was referring to in the OP. I specified there that of course you can have global chart if your manifold is the Euclidean space or the Minkowskian space for a Semiriemannian case. Now your example is different in that it restricts R^n to a convenient subset of R^n that coincides with what the chart covers completely.

    A 2-sphere needs two charts to specify all its points in R^2. Imagine I have a chart that covers one half of a 2-sphere, and I decide that my manifold must only be locally diffeomorphic to the subset of R^2 it already covers. Now I have a global chart for that subset of R^2.
    In this way I can taylor make manifolds that are locally sub R^n instead of locally R^n and I will always be able to use just one global chart.
     
  13. Jun 26, 2011 #12

    George Jones

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    At first I though that you were excluding all Riemannian and semi-Riemannian isometric to (R^n , g), where there are global charts for which g is "constant" (and thus have zero intrinsic curvature). Now it seems that that you are excluding all Riemannian and semi-Riemannian diffeomorphic to R^n but not isometric to such (R^n , g).
     
  14. Jun 26, 2011 #13
    I'm just saying that if you define your manifold X so that every neighbourhood of a point of X is diffeomorphic to the subset of R^n that fits the coordinate chart that covers that subset of R^n, instead of just diffeomorphic to R^n, then yes, you have the recipe to having smooth manifolds with a global chart regardless their intrinsic curvature.
    But if you think of it, usually manifolds are thought of as the gluing of many different charts (atlas), so by restricting R^n (forinstance not including t= or< than 0) to get a global chart that covers the whole manifold, it seems that the whole motive that manifolds were invented for is wasted.
     
  15. Jun 26, 2011 #14

    George Jones

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    A manifold that is coverable by a single chart still has an atlas of compatible charts.
     
  16. Jun 26, 2011 #15
    Sure, so in your opinion why would one want to restrict the definition of manifold from imposing that every point in it has a neighbourhood diffeomorphic to R^n, to difeomorphic to a subset of R^n?
    For physical reasons as I suggested, like introducing a cosmic time that has only positive values?
     
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