Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is a realization of this surface in Euclidean space?

  1. Nov 27, 2011 #1

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I am interested to know how to realize this abstract surface as a subset of Euclidean space.

    The surface as a point set is the 2 dimenional Euclidean plane minus the origin.
    the metric is given by declaring the following 2 vector fields to be an orthonormal frame:

    e[itex]_{1}[/itex] = x[itex]\partial[/itex]x - y[itex]\partial[/itex]y

    e[itex]_{2}[/itex] = x[itex]^{2}[/itex][itex]\partial[/itex]x + e[itex]^{-xy}[/itex][itex]\partial[/itex]y

    I think that this surface has constant negative curvature equal to -1 but I do not think it is a Poincare disk minus a point because the origin is infinitely far away from any point on the x or y axes.

    BTW: Here is the curvature calculation but I am not 100% sure of it.

    the Lie bracket, [e[itex]_{1}[/itex],e[itex]_{2}[/itex]], equals e[itex]_{2}[/itex] so the covariant derivative can be defined using the following formulas:


    [itex]\nabla_{e_{1}}[/itex]e[itex]_{1}[/itex] = 0
    [itex]\nabla_{e_{1}}[/itex]e[itex]_{2}[/itex] = 0
    [itex]\nabla_{e_{2}}[/itex]e[itex]_{1}[/itex] = -e[itex]_{2}[/itex]
    [itex]\nabla_{e_{2}}[/itex]e[itex]_{2}[/itex] = e[itex]_{1}[/itex]

    Extend these formulas to arbitrary vector fields by linearity and the Leibniz rule.

    The connection 1 form is ω(e[itex]_{2}[/itex]) = -1, ω(e[itex]_{1}[/itex]) = 0.

    Since this connection is torsion free the exterior derivative of ω easily computes and is the volume element of the metric. Therefore the Gauss curvature is -1.

    It may be helpful to note that the rectangular hyperpolas y = k/x are geodesics as are the rays from the origin along the axes.
     
    Last edited: Nov 27, 2011
  2. jcsd
  3. Nov 27, 2011 #2
    Well I have never before tried to find the coordinates of a manifold from its tangent space but I will propose an idea. Let θ,λ be the coordinates with tangent vectors ∂θ and ∂λ. Then when pulled back to the euclidean spaca they are given as ∂x/∂θ∂x + ∂y/∂θ∂y
    and ∂x/∂λ∂x + ∂y/∂λ∂y . Looking at your equations this requires

    ∂x/∂θ = x(θ,λ), ∂y/∂θ= -y(θ,λ), and
    ∂x/∂λ = x2(θ,λ), ∂y/∂λ= e-xy(θ,λ).

    I think that if your tangent distribution is integrable, you should be able to solve these to get the manifold structure.
     
  4. Nov 27, 2011 #3

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    thanks Sina

    You are answering another problem.
    This problem is to find an embedding of this surface in euclidean space not to find coordinates.
     
  5. Nov 28, 2011 #4
    I see now what you mean by realization. However for instance I think if you repeat the same procedure for the circle, you get an embedding no? You would get (x,y) = (cos(θ),sin(θ))
     
  6. Nov 28, 2011 #5

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I don't think so. An embedding requires that the normal directions move around the surface in the right way, i.e. in a way that defines the surface's shape. this can not come from a coordinate system. further it is not obvious what dimension Euclidean space the surface can be realized in. Maybe 3 dimensions, maybe 4.
     
  7. Nov 28, 2011 #6
    Hmm you are right yes. It would only be a luck I suppose if finding the coordinates helped to find the embedding.

    I happen to recall that whitney's embedding theorem seemed constructive to me (though I am not sure) but I can not recall the details right now. However it did not make use of concepts such as curvature, connection etc that I am sure about. Maybe that would be also of help.
     
  8. Nov 28, 2011 #7

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    Right. The Whitney embedding theorem is for smooth embeddings not for isometric embeddings. That is the Nash embedding theorem but I do not know how it is proved. I have heard that is is monster hard.
     
  9. Nov 29, 2011 #8
    Hi Lavinia. This is of course just a guess, but why could it not be realized as isometric to a one-sheeted hyperbola?
     
  10. Nov 29, 2011 #9

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I don't know what a one sheeted hyperbola is.
     
  11. Nov 29, 2011 #10
    The Poincare disk model for the hyperbolic plane also has a realization as the upper sheet of z^2=x^2+y^2+1. This is of course simply connected, so one might try modifying the equation z^2=x^2+y^2-1. Think of fixing r=1/2 on the plane z=0 in R^3, and then blowing the boundary of the circle to z=+infinity, and at the same time blowing the origin out to z=-infinity, where the traces would be hyperbolas in two directions and circles in the other direction.

    Now, I can't see that this would work in your example because of the lack of symmetry, but it does give an example of constant negative curvature with the homotopy type of the circle. So my intuition would be that your surface has an isometric embedding in R^3, but the idea of solving the equations gives me indigestion :)
     
  12. Dec 1, 2011 #11

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I just realized that the two vector fields are dependent on the y axis so the metric can only be defined on one of the open half spaces x>0 or x<0
     
  13. Dec 4, 2011 #12
    You have a 2 dimensional surface which is simply connected and contractible with constant negative curvature. Surely that implies that it is isometric to the hyperbolic plane, no?
     
  14. Dec 4, 2011 #13

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    Yes Jamma. You are right.

    I originally thought that the metric extended to the plane minus the origin in which case the manifold would not have been contractible. But I realized that it does not extend to the y-axis so it is valid only on either open half plane.

    The problem was trying to see if a vector field of index negative 1 could be tangent to geodesics for some metric in a region of the plane. But this attempt fails. Any ideas?
     
  15. Dec 4, 2011 #14
    What do you mean by a vector field of index -1? Do you mean the index at a fixed point (I know about the index of a vector field at some fixed point-take the degree of the induced map into the sphere)?
     
  16. Dec 5, 2011 #15

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I mean exactly what you said. The vector field x[itex]\partial[/itex]x -y[itex]\partial[/itex]y has index -1 around the origin.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is a realization of this surface in Euclidean space?
  1. Non-Euclidean Space (Replies: 11)

Loading...