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Take, for example, the hyperboloid $x^2-y^2-z^2=1$ and the open disk (equipped with the Poincare metric) $x=1, \, y^2+z^2<1$. If we multiply a solution $(1,y,z)$ to the second set of equations by some scalar $s$ to get a solution to the first equation $(s, sy, sz)$, is the resulting map an isometry?

It's intuitive to me that if, given a two-sheeted hyperboloid, you center an open disk equipped with the Poincare metric at one of the hyperboloids vertices (is that what they're called?) such that the radius of the disk is the radius of the asymptotic cone at that point along its axis of symmetry, then performing the projection I described is an isometry of the hyperbolic plane. I haven't thought much about how to prove this, I feel as though I would need to know how to calculate distance on the hyperboloid, and trying to do so with an integral gives an equation that I'm pretty sure isn't solvable analytically.

Is my intuition correct? If so, How would you go about proving it? If not, there is an isometry between these spaces, because they're both models for the hyperbolic plane, right?

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2 Answers 2

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There are two relevant maps between disk and hyperboloid. The one you describe has the disk at one of the vertices, and the center of projection at the center of symmetry of the hyperboloid. The resulting disk is not the Poincaré disk model, but instead the Beltrami-Klein model:

Hyperboloid and Beltrami-Klein model

If you want to project to Poincaré disk model instead, you should place your disk at the plane of symmetry (i.e. at $z=0$) and your center of projection at one of the vertices (i.e. $(0,0,-1)$):

Hyperboloid and Poincaré disk

One way to reason about these things is the following: A hyperbolic line is uniquely defined by the two ideal points it's incident with, and a point uniquely defined by two lines intersecting with it. So for both of the above mappings, it is sufficient to show that the set of ideal points is projected in a metric-preserving fashion, and that the set of hyperbolic lines matches. In other words, that the set of hyperbolic geodesics on the hyperboloid maps to straight lines for Beltrami-Klein and to circles orthogonal with the unit circle for Poincaré.

Figures taken from my PhD. thesis Section 2.1.5 Figure 2.5 Page 20. Higher resolution available upon request.

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  • $\begingroup$ there is a version with all three matched up, hyperboloid, two disc versions, with an upper hemisphere as well. It might be in Hilbert and Cohn-Vossen, which I have here somewhere $\endgroup$
    – Will Jagy
    Jul 26, 2017 at 23:01
  • $\begingroup$ @WillJagy: I have a hemisphere in figure 2.3, but never saw all of them together. Would be interested to know details. $\endgroup$
    – MvG
    Jul 26, 2017 at 23:03
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    $\begingroup$ Alright, i will find something. It might be in Spivak five volumes. The hemisphere goes to the conformal disc model by stereographic projection out of the South Pole, the disc being in the plane of the equator. This is conformal: geodesics in the disc are mapped to semicircles in the hemisphere that meet the equator orthogonally. As a result, each such semicircle lies in a vertical plane, and a vertical projection sends each to a line segment in the plane $z=1,$ resulting in a Beltrami-Klein model. I guess I should add a 2-D slice of the thing as an answer. $\endgroup$
    – Will Jagy
    Jul 26, 2017 at 23:08
  • $\begingroup$ Nice diagrams, I wanted to include something like that in my question but I also didn't want to take the time to generate any. $\endgroup$ Jul 26, 2017 at 23:21
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    $\begingroup$ @WillJagy: Ah, so the center of projection used for the Beltrami-Klein disk is different from the others. Guess that can't be avoided, although at first I read your comment to imply that. Interesting to see that projecting from Beltrami to hyperboloid from origin apparently is the same as projecting Beltrami to hemisphere orthogonal to the disk, then projecting hemisphere from south pole. That would be the geometric theorem behind the “everything matches up” computation, I guess. If I find the time I'll create a povray with all of these, might even do a movie for it. $\endgroup$
    – MvG
    Jul 26, 2017 at 23:56
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picture of a single point in the hyperboloid model, projected to the Beltrami-Klein model and the Poincare disc model

enter image description here

If we think of the projection out of the origin as being an entire plane, we intersect with the hyperboloid in a geodesic. This is pulled back to a line segment in the Beltrami-Klein model (at $z=1$). This is projected vertically to a semicircle on the hemisphere $x^2 + y^2 + z^2 = 1, z \geq 0.$ Stereographic projection around the South Pole $(0,0,-1)$ takes that semicircle to a circular arc in the Poincare disc ($z=0$), which is orthogonal to the equator because the semicircle was orthogonal to the equator, while the equator is mapped to itself. Result in the xy plane depicted below, an entire geodesic arc in the Poincare disc.

enter image description here

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