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On http://en.wikipedia.org/wiki/Hypercycle_%28geometry%29 I found the statement.

The hypercycles through a given point that share a tangent through that point converge towards a horocycle as their distances go towards infinity.

But I don't understand it.

hypercycles are curves (equidistant to some line)

  • How do you get their tangent at a (given) point?

And what does the rest of the statement mean? can somebody give a proof and a picture?

Willemien
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1 Answers1

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Here is an illustration of what's going on.

Figure

Here I chose the common tangent to be orthogonal to the diameter through the given point. This ensures that the point lies on the symmetry axis of the crescent, so this is the most symmetric situation. For other tangents, the points where the hypercircle touches the boundary of the model would move at different speeds.

MvG
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  • Thanks that looks nice and does explains it, but with a scale going to 0 at the boundary line, I don't trust my eyes, is it just the way it looks on the Poincare Disk model or is it something real in the hyperbolic plane (and if so can it be proved?) – Willemien Jul 15 '14 at 10:39
  • @Willemien: This is something that's really happening in the hyperbolic plane. A hypercircle contains two ideal points, and your family of hypercircles can be arranged such that it converges to the case where these two ideal points coincide. This convergence can be readily seen in the model, but in this case the model is accurate. If I had to prove it, I'd indeed prove it using the model. – MvG Jul 15 '14 at 11:45
  • I think it depends on if you find the boundary circle a part of the hyperbolic plane, for the sake of argument, i would disagree with it. (it is no part of the finite real projective plane) but maybe this all becomes to philosophical – Willemien Jul 15 '14 at 16:34
  • Well, even if you don't count ideal points as parts of the hyperbolic plane, they are still inherent to the geometry and not an artifact of the model. For example, you could describe each ideal point by the family of limit parallel lines in that direction. Speaking of a real projective plane (while “finite” and “projective” don't fit well to one another, except when talking about projective planes with only a finite number of points), then that would suggest the Beltrami-Klein model, where not only ideal points but also hyper-infinite points or whatever are easily part of the game. – MvG Jul 15 '14 at 18:50
  • I see ideal and ultra ideal points (and verything else on or outside the unit circle ) just as (very) handy for constructions but that doesn't mean they are "real", but i guess it is just a philosophical standpoint, but where are the ideal points on the hyperboloid model ( http://en.wikipedia.org/wiki/Hyperboloid_model ) gone – Willemien Jul 15 '14 at 21:40
  • @W: That's probably one of the best reasons to avoid the hyerboloid model: it makes these seem even less real. But perhaps the proper view of the hyperboloid model is using homogeneous coordinates witha Minkowski inner product. Which means any vector in the inside of the cone of asymptotes can be used to represent a point, as you can simply project them onto the hyperboloid. In that setup, a vector outside that cone is usually considered as a representative of a line. An ideal point would be a vector on the cone. It is a limit between point and line. – MvG Jul 15 '14 at 21:47
  • @MvG: What software did you use to make this picture? – Lee Mosher Jul 16 '14 at 20:58
  • @LeeMosher: Sage and GIMP. Sage to write the animation and save it to a bunch of PNG files, and GIMP to do the color reduction and optimization for GIF. – MvG Jul 16 '14 at 21:00
  • @Willemien: You do not need to take a stance on this philosophical issue to appreciate the following mathematical facts about the Poincare disc model of the hyperbolic plane. Geodesics in the Poincare disc model are precisely the intersections (of the Poincare disc) with Euclidean circles that hit the unit circle at right angles. Hypercycles are precisely the intersections with Euclidean circles that hit the unit circle at two points. Horocycles are precisely the intersections with Euclidean circles that hit the unit circle at a single point. – Lee Mosher Jul 16 '14 at 21:05
  • @LeeMosher and MvG T realised I was wrong, I was to fixated on the points on the boundary circle, instead of the points near the given point. Am I correct that at the given point the curvature of the hypercycle approaches the curvature of the horocycle (and thus the points approuch each other) opens a whole other can of wurms, how to calculate the curvature of an hypercycle and the horocycle – Willemien Jul 17 '14 at 07:41