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Gaussian curvature

  1. Jan 15, 2012 #1
    are there two non isomorphic surfaces with the same gaussian curvature?
  2. jcsd
  3. Jan 15, 2012 #2


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    what do you mean by non isomorphic?
  4. Jan 15, 2012 #3
    i ment non isometric
  5. Jan 15, 2012 #4


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    One can have surfaces of zero Gauss curvature that are not even homeomorphic. There are,for instance,the flat torus, the flat Klein bottle, the flat Mobius band, flat Euclidean space ,and the flat cylinder. The only compact flat surfaces without boundary are the flat torus and the flat Klein bottle.

    This works in higher dimensions as well. In every dimension there are compact manifolds without boundary whose curvature tensor is identically zero that are not homeomorphic. In three dimensions there are compact flat Riemannian manifolds without boundary that are orientable, have the same holonomy groups, but do not have the same fundamental group. The number of non-homeomorphic compact flat Riemannian manifolds without boundary in any dimension is finite although it is not known in general how many.

    Additionally, any surface of genus greater that 1 admits a metric of constant negative Gauss curvature. But surfaces of different genus are not homeomorphic.

    One can also have surfaces that are homeomorphic and have constant Gauss curvature but are not isometric.A simple example would be two flat tori that have different areas.
    More generally, take two flat tori that are not conformally equivalent. Since an isometry must be a conformal equivalence, they can not be isometric -I don't think.

    The same things apply for surfaces of higher genus.

    For positive Gauss curvature the answer is also no. I just realized that the sphere and the projective plane can be given metrics of constant positive Gauss curvature.
    Last edited: Jan 15, 2012
  6. Jan 15, 2012 #5


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    I assumed the question meant: is a diffeomorphism that preserves gauss curvature also an isometry?
  7. Jan 15, 2012 #6


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    I didn't read it that way but any two flat tori are diffeomorphic. But you are right, your reading makes more sense. Mine maybe doesn't make any sense.
    Last edited: Jan 15, 2012
  8. Jan 16, 2012 #7


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    Mathwonk: I have confused myself and don't know what it means for a map to preserve Gauss curvature.

    I thought of two possibilities:

    1)There is a diffeomorphism so that the Gauss curvature at corresponding points is the same.

    2) the pull back of the exterior derivative of the connection 1 form is the exterior derivative of the connection 1 form.

    For possibility 1: The two spaces are both the open disc of radius pi/2 minus the origin.

    metric 1: ds^2 = dr^2 + cos^2(r)d[itex]\theta[/itex]^2

    metric2: ds^2 = dr^2 + sin^2(r)d[itex]\theta[/itex]^2

    The identity map is not an isometry but both metrics have constant curvature equal to 1.
    I think this works.

    For the case of curvature zero here are some more examples.

    consider any metric on the plane of the form ds^2 = exp(2h)(dx^2 + dy^2) where h is a harmonic function.

    The Gauss curvature is the Laplacian of h and so is always zero. Again the identity map is not an isometry for different h's.

    For possibility 2, not sure. There is a differential equation.
    Last edited: Jan 16, 2012
  9. Jan 19, 2012 #8
    Thank's a lot
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