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Integration of Ricci Scalar Over Surface

  1. Apr 12, 2015 #1
    Does this integration of Ricci scalar over surface apply in general or just for compact surfaces?

    ∫RdS = χ(g)

    where χ(g) is Euler characteristic.
    And could anybody give me some good references to prove the formula?
     
  2. jcsd
  3. Apr 13, 2015 #2

    Ben Niehoff

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    On a non-compact surface, the integral might fail to converge. Take hyperbolic space, for example.
     
  4. Apr 13, 2015 #3

    wabbit

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    This might still be true when the integral converges though ?
     
  5. Apr 13, 2015 #4

    Ben Niehoff

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    I'm not sure.

    Another way to look at this is an order-of-limits issue. Consider a modified integral for computing the Euler number of manifolds-with-boundary. This integral will contain a boundary term (sorry, I don't remember what it looks like). Then you can take any non-compact surface, and just cut a piece out of it, adding a boundary and making the integral finite. Perform the integral first, and then take the limit as the boundary goes to infinity.

    I think this will give you an Euler number of 1 for the infinite plane. It is homologous to the disk, and boundaries add 1 to the Euler number.
     
  6. Apr 13, 2015 #5

    wabbit

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    For a non compact surface, I was more thinking of compacification with a point/curve at infinity and hopefully making use of R=0 at infinity... But you're right the boundary term can make this whole business tricky.
     
  7. Apr 13, 2015 #6

    lavinia

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    In the non-compact case where the integral does converge, the total curvature can be less than 2πχ(S).

    The flat plane has zero total curvature 0 but has Euler characteristic 1.

    For a compact manifold with boundary one needs to add on the total geodesic curvature of the bounding curve.
     
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