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Torus and sphere with positive or negative curvature

  1. Nov 16, 2011 #1
    Hi I know that the sphere has positive curvature everywhere, and the torus has positive an negative curvature. Is there a space homeomorphic to the sphere such that this space has negative curvature everywhere?
    And, is there a space homeomorphic to the torus such that the curvature is always positive (negative)?
    Thanks
     
  2. jcsd
  3. Nov 16, 2011 #2

    lavinia

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    the answer is no in both cases.
     
  4. Nov 16, 2011 #3
    And how can i prove it?
     
  5. Nov 16, 2011 #4

    Ben Niehoff

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    You can prove it using the Gauss-Bonnet formula. The integral of the curvature over the whole manifold is proportional to the Euler characteristic. A sphere has Euler characteristic 2, and a torus has Euler characteristic 0.
     
  6. Nov 17, 2011 #5
    Here's some intuition for Gauss-Bonnet. Integrating curvature over a region gives you the holonomy around its boundary.

    Let me try to elaborate on that a bit. Holonomy is how much a vector gets rotated by when it is parallel-transported along a curve. On the sphere, you can see that, if you have a triangle, the sum of the angles is bigger than 180, by a factor proportional to its area--this is plausible by inspection and there's a simple geometric proof that I will skip to save time. The triangle is made out of geodesics. A geodesic, according to one definition, is a curve such its tangent vectors get parallel transported to tangent vectors of the same length. So, you can send a tangent vector to a geodesic on a little journey around the triangle by parallel transporting it. Start at a vertex, tangent to one of the sides. Parallel transport to the next vertex. Then, rotate the vector, so that it is tangent to the next side, but make it point outwards, so that you rotate by the angle at that vertex (vertical angles). Parallel transport to the next vertex and rotate again. Then repeat one more time. The vector just ends up 180 degrees from where it started. If you didn't do those rotations at each vertex, you would get the holonomy, which is the sum of the angle minus 180. So, on the sphere at least, if you understand this, then you should understand why holonomy is proportional to the integral of the curvature, which in this case (let's say unit sphere), is just the area.

    Next, you can understand curvature as essentially the holonomy around an infinitesimal loop. So, kind of like Stokes theorem (actually, this follows from a version of it, involving Lie algebra-valued differential forms), if you want the holonomy around a big loop, you can just consider the region it bounds subdivide it, add up the holonomies around all the little tiny subdivided loops. So, again, you are essentially integrating the curvature to get the holonomy.

    So, that's the local version of Gauss-Bonnet. Integrate curvature, get holonomy.

    What's relevant here, though, is the global version of Gauss-Bonnet. This requires the notion of an Euler characteristic of a surface.

    You can divide the surface up into geodesic triangles and see what happens when you add the integrals over each region together. I'm too lazy too carry the rest out, but the upshot is that the result has to be positive for a sphere, so it couldn't come from integrating a negative curvature over the sphere. It's easy to do the combinatorics for a simple example, but to prove that it doesn't depend on which example you choose requires the topological invariance of the Euler characteristic, which I would prove using homology theory.

    So, a space homeomorphic to the sphere can never have negative curvature everywhere.
     
  7. Nov 28, 2011 #6

    lavinia

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    It would be fun and maybe instructive to think of proofs other than Gauss Bonnet.

    An equivalent form of Gauss Bonnet shows the the integral of the Gauss curvature is the sum of the indices of a vector field with isolated zeros. In outline the proof is this:

    If v is a vector field with isolated zeros then v/||v|| is a unit length vector field with isolated singularities.

    It pulls back the exterior derivative of the connection 1 form and this pull back equals the Gauss curvature times the volume element of the surface. So this pull back just integrates to the total Gauss curvature of the surface.

    In the tangent circle bundle the one can use Stoke's Theorem to replace the surface integral by the integral of the connection one form on the boundary of the image of the vector field. Without being rigorous, this boundary is the fiber above the singularities multiplied by the indexes of the vector fields at the singularities. But the connection 1 form is just the volume element on each fiber so the integral is 2 pi times the sum of the indices of the vector field.

    A theorem of Hopf shows that the sum of the indices of a vector field is the same for every vector field. For the torus it is easy to find a vector field with no zeros so this integral must be zero. For the sphere it is easy to find a vector field with a single singularity of index 2 so the curvature can not be everywhere non-positive.
     
    Last edited: Nov 28, 2011
  8. Nov 29, 2011 #7

    lavinia

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    here is a special case. Suppose the surface has constant positive Gauss curvature.

    Then radial geodesics emanating from a point all converge to the same conjugate point. This means that there is a homeomorphic image of a sphere in the surface. But then the surface itself must be a sphere.
     
  9. Dec 15, 2011 #8

    lavinia

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    Here is another way to see that the integral of the Gauss curvature is the sum of the indices of a vector field with isolated zeros. It is really the same as the first way that I sketched but is done in coordinates so it is easier to see.

    In geodesic polar coordinates, (r,[itex]\theta[/itex]) the connection 1 form is

    [itex]\partial_{r}[/itex]G[itex]^{1/2}[/itex]d[itex]\theta[/itex] where G[itex]^{1/2}[/itex] is the length of [itex]\partial[/itex][itex]\theta[/itex].

    It is the pull back of the connection 1 form on the tangent unit circle bundle down to the surface under the radial vector field, [itex]\partial[/itex]r.

    The pull back under an arbitrary vector field is

    [itex]\partial_{r}[/itex]G[itex]^{1/2}[/itex]d[itex]\theta[/itex] -d[itex]\alpha[/itex]

    where [itex]\alpha[/itex]

    is the angular function between the vector field and [itex]\partial[/itex]r .

    Its integral over a polar circle is the total angular form plus the integral of

    [itex]\partial[/itex]rG[itex]^{1/2}[/itex]d[itex]\theta[/itex] over the circle.

    It is a standard theorem, basically just an application of L'Hopital's Rule, that the limit as r goes to zero of [itex]\partial_{r}[/itex]G[itex]^{1/2}[/itex] is 1. so in the limit the integral of the first term is just 2 pi and the limit of the entire integral is just the total rotation of the vector fields around the singularity.

    The computations that show these relationships are straight forward. the way I did it was to write the vector field as

    cos([itex]\alpha[/itex])[itex]\partial[/itex]r + sin([itex]\alpha[/itex])G[itex]^{-1/2}[/itex][itex]\partial[/itex][itex]\theta[/itex]

    then take the dual of its Lie bracket with its 90 degree rotation to calculate the pull back of the connection 1 form.

    So the integral of the pull back of the connection one form over all of the singularities is the sum of the indices of the vector field plus an error term that disappears as r goes to zero.

    Now apply Stokes theorem and integrate the exterior derivative of the connection 1 form over the surface. Again take the integral over the surface minus polar ice caps centered at the singularities and let the ice melt.
     
    Last edited: Dec 15, 2011
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