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Poincaré disk vs Riemann sphere

  1. Oct 25, 2012 #1
    Hello,

    it is well-known that with stereographic projection we can obtain a 1-1 correspondence between the points of the 2d Cartesian plane (plus the point at infinity), and the points on the Riemann sphere.

    What is the geometrical construction that corresponds to a 1-1 mapping between the Poincaré disk and the points on the 2d Cartesian plane?

    My attempt was to consider a point (x,y) on the Cartesian plane, then make a parallel projection on the upper sheet of the hyperboloid, thus obtaining (x,y,h), and finally apply a projective transformation with focal point (0,0,-1) and focal plane coincident with the Cartesian plane. This way all the points on the hyperboloid (upper sheet) are mapped onto the unit ball of the Cartesian plane, with the points on the unit circle representing the points at infininity.

    Is this the correct way of doing it?
     
  2. jcsd
  3. Oct 25, 2012 #2
  4. Oct 25, 2012 #3
    ok thanks.
    If that is true, then the symmetries on the Poincaré disk (Fuchsian groups) are basically the same as the symmetries described by the 17 wallpaper groups on the plane?
    We would just form a symmetric lattice on the 2d Cartesian plane, and project it as I described on the Poincaré disk.

    Am I right? Or am I missing something?
     
  5. Oct 25, 2012 #4
    That seems quite unlikely. That sort of reasoning would imply that if two spaces have the same topology then they have the same discrete isometry groups.

    I think what happens is that your mapping of the plane to hyperbolic space does not send isometries to isometries. So even though your wallpaper picture transforms into a nice wallpaper like picture of the disk, it does not work on the level of sending the group elements of one (isometries) into the group elements of the other.
     
  6. Oct 25, 2012 #5
    I see...then I got it wrong because I was essentially applying isometries on the 2d Cartesian plane, and then simply "deform" the 2d plane into the Poincaré disk.

    The question is then: in what space am I supposed to apply isometries?

    I make an analogy with the Riemann sphere, just for example.
    We can map the Cartesian 2d plane onto the Riemann sphere and define isometries (in 3d space) of the Riemann sphere, i.e. we can rotate it around any axis and translate it. By inverse stereographic projection we would map the Riemann sphere back onto the Cartesian 2d plane. "www.youtube.com/watch?v=JX3VmDgiFnY" [Broken]
    By considering the discrete isometry groups on the Riemann sphere we would then obtain some sort of "conformal symmetry" on the Cartesian plane.

    I bet an analogous procedure is what leads to the symmetry groups on the Poincaré disk. But how to do it?
     
    Last edited by a moderator: May 6, 2017
  7. Oct 30, 2012 #6

    lavinia

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    Stereographic projection maps the sphere minus a point conformally onto the complex plane. There is no conformal map of an open disk onto the whole complex plane.

    So this means that there is no right (conformal) way to map a disk onto the plane

    Still a disk is homeomorphic to the entire plane and is also diffeomorphic to it.

    What about trying stereographic projection from the center of the sphere rather than from the north pole (stand the sphere on the plane at the south pole). Follow its inverse by vertical projection onto the open unit disk.

    Notice also that the inverse of sterographic projection allows you to put a metric on Euclidean space that is bounded. Can you use this to show that the plane is homeomorphic to a disk?
    What about mapping a point in the tangent plane to the north pole to the point on the great circle whose distance to the north pole equals the Euclidean distance of the point to the origin in the tangent plane?
     
  8. Oct 30, 2012 #7
    Hi Lavinia!

    thanks for the answer. I did not think of constructing the Poincaré disk by stereographically projecting from the sphere center (0,0,0).

    However, once we map the 2d plane onto the lower hemisphere of the Riemann sphere, how do we define isometries that would describe the symmetry groups on the Poincaré disk?
     
  9. Oct 30, 2012 #8

    lavinia

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    Not sure what you are trying to do.

    The symmetries of the Poincare disk I think are the fractional linear transformations that preserve the unit circle. Not need for stereographic projection.
     
  10. Sep 4, 2013 #9
    Related to this topic. I have a cloud of points lying on the Riemann sphere.
    I understand I can use the stereographic projection to perform a mapping on a 2D plane.

    Besides of that, I would like to known if it makes sense trying to map my cloud of points from the sphere to the Poincaré disk. If yes, how could I do it?

    If it is not directly possible, my points can be interpreted as a four vectors of a Minkowski space, so can they be related to an hyperboiloid and consequently to the Poincaré disk?

    In a simple way, I have 4 dimension vectors (3 dimesions + norm) and I want to represent them on the Poincaré disk.

    Thank you
     
  11. Sep 4, 2013 #10
    Hi Aispin, and welcome to the forums.

    You have resumed an almost 1-year old thread.
    I think it makes sense to map points from the Riemann sphere to the Poincaré disk.
    In order to do that you could map from the Riemann sphere to the Cartesian plane, and then map from the Cartesian plane to the Poincaré disk. You have the formulas for both these mappings, so composing them should give you a mapping from the Riemann sphere to the Poincaré disk.

    The only problem is that this mapping won't be defined for the point at the pole of the Riemann sphere.
     
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