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Hyperbolic Manifold With Geodesic Boundary?

  1. Nov 20, 2014 #1

    WWGD

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    Hi All,
    I am trying to figure out the details on giving a surface S a hyperbolic metric with geodesic boundary, i.e., a metric of constant sectional curvature -1 so that the (manifold) boundary components, i.e., a collection of disjoint simple-closed curves are geodesics under this metric. So far I know:

    1) There are genus constraints for the surface. Does this have to see with Gauss-Bonnet?

    2) Something; not sure exactly what, can be done by gluing pairs-of-pants http://en.wikipedia.org/wiki/Pair_of_pants_(mathematics [Broken])
    but not fully sure how this works.

    Not much more. Any ideas, refs., please?
     
    Last edited by a moderator: May 7, 2017
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  3. Nov 21, 2014 #2

    lavinia

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    Not sure what you mean by the genus of a surface with boundary but maybe this will help.

    Every compact hyperbolic surface without boundary that has constant Gauss curvature, -1, is the quotient of the Poincare disk by a group of isometries. Remove a small geodesic polygon from the disk and the quotient will be a surface of constant negative curvature with a geodesic boundary.

    For surfaces, the Gauss curvature times the volume element is cohomologous to the Euler class. One can see this from the Gauss Bonnet theorem or more simply by observing that the connection 1 form on the tangent circle bundle is a global angular form so its exterior derivative is the pullback (under the bundle projection map) of the Euler class.

    It follows that the Euler characteristic of a hyperbolic surface is always negative so this rules out the torus and the sphere. It is a simple exercise show that also must be even.
     
    Last edited: Nov 21, 2014
  4. Nov 22, 2014 #3

    lavinia

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    There are classical surfaces in 3 space that have constant negative Gauss curvature. Examples are the tractroid and Dini's surface.

    See if you can slice off part of one of these with a geodesic knife to get a surface with geodesic boundary.
     
  5. Nov 22, 2014 #4

    WWGD

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    Thanks Lavinia, for the sake of completeness, let me state the definition of manifold M with totally geodesic boundary is:

    A manifold M with non-empty boundary that admits an atlas {## \phi_{\alpha}: U_{\alpha} \rightarrow B_{\alpha} ##} to hyperbolic half-spaces bounded by geodesic hyperplanes ##H_{\alpha} \subset \mathbb H^n ## so that {## U_{\alpha} ##} covers M and every chart satisfies ##\phi_{\alpha}(U_{\alpha} \cap \partial M)=\phi_{\alpha}(U_{\alpha} \cap H_{\alpha}) ## , and overlap maps are restrictions of isometries.
     
    Last edited: Nov 22, 2014
  6. Nov 22, 2014 #5

    WWGD

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    Do you mean, if g is the geodesic polygon and P is the Poincare disk , that (P-g)/g has geodesic boundary, and so will this have totally geodesic boundary, i.e., will every boundary component (a simple-closed curve here) be a geodesic?
    If so, is this I guess the quotient metric?
    Thanks.
     
  7. Nov 23, 2014 #6

    lavinia

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    Yes.

    - I think also that with a tractroid you can cut off a neighborhood of the singular circle to get a surface in 3 space of constant negative curvature and geodesic boundary.


    - You could also just slice a hyperbolic manifold without boundary along one or more closed geodesic circles. For a surface of genus 2, three circles suitably chosen would give you your pair of pants.
     
    Last edited: Nov 23, 2014
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