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Surface with prescribed metric

  1. Jan 17, 2007 #1
    Consider a Riemannian metric in R^2, e.g. consider g given at a point (x,y) by the matrix
    (1 x
    x 1+x^2 )

    Is there a surface S, embedded in R^3, which has the property that the metric on S which is induced by the Euclidean metric of R^3 coincides with the given metric g? If yes, what representation does S possess?
     
  2. jcsd
  3. Jan 17, 2007 #2

    Chris Hillman

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    This isn't HW, is it?

    Hi, KarstenKarsten,

    Sounds like you are asking for an embedded surface in E^3 which has the stated metric. A good keyword to look up in a good differential geometry textbook should be "Monge patch" or "embedding". In any case, the idea is simple: write down an undetermined basis (two vectors in E^3) for the tangent space to some point of S, form their E^3 inner product, and compare with the desired induced metric on S. See "Coordinate tutorial" at http://www.math.ucr.edu/home/baez/RelWWW/group.html for some radially symmetric examples.
     
  4. Jan 17, 2007 #3

    gvk

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    It looks like it is not embedded surface in E^3 (3D Euclidian space). If the surface is given to us in the form z = f(x, y), then the metric induced on the surface is

    [tex]g_{1,1} = 1 +f_{x}^{2}, g_{1,2}=g_{2,1} = f_{x} f_{y}, g_{2,2} = 1 + f_{y}^{2}[/tex].

    In Karsten case
    [tex]g_{1,1} = 1 , g_{1,2}=g_{2,1} = x, g_{2,2} = 1 + x^{2}[/tex].

    That means [tex]f_{x} =0[/tex] and should be [tex]g_{1,2}=g_{2,1} = 0[/tex].
    Hence, the surface is embedded in some noneucleadian 3D space.
     
  5. Jan 18, 2007 #4
    Thanks!

    I realize now that, at least, one can find a conformally equivalent metric
    g~_{ij} = c(x) g_{ij}
    with c(x) the solution of c(x)^2-c(x)(2+x^2) +1=0. And with g~ be metric of a in E^3 embedded surface. The components of g~ can then also be computed (and are slightly more complicated terms).
     
  6. Jan 19, 2007 #5

    Chris Hillman

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    Right, you don't need a noneuclidean embedding space.
     
  7. Jan 19, 2007 #6
    For me, it would somehow be counterintuitive that such an embedded surface should not exist.
    But can I obtain from the surface with the conformally equivalent metric a surface which actually has the very same metric I started with? How?
    (In the very beginning I started with the ansatz, as gvk proposed, of course...)
     
  8. Jan 23, 2007 #7

    gvk

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    The Riemannian tensor for your metric is [tex]R_{1,2,1,2}=-1[/tex]. That means your embedded 2D manifold has constant negative curvature. Of course, you always can orthogonize your metric to get conformal presentation in isothermal coords.
     
    Last edited: Jan 24, 2007
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