From: Douglas Zare
Subject: Re: Poincare disk model questions
Date: Fri, 24 Dec 1999 19:59:40 -0500
Newsgroups: sci.math
Keywords: constructing the geodesic joining two points on Poincare disk
The Blue Wizard wrote:
> Question #1:
> Suppose I have two distinct h-points sitting on a Poincare
> disk model of 2-D hyperbolic plane, how do I construct a
> h-line going through those two points using only compass
> and ruler? Of course this h-line would be a circle going
> through those two points and is orthogonal to the boundary
> edge of the Poincare disk. I know that the h-line becomes
> a line when both points and the center of the Poincare disk
> are collinear.
Given 3 points on a circle, you know how to construct the circle, right?
So you only need to construct a single extra point on the circle. Then
the following hint should help: The circle in the Euclidean plane
containing the hyperbolic line through the two points is fixed by a
circle inversion about the boundary of the disk. (See what circle
inversion does to the ideal points and the angles at the ideal points.)
> Question #2:
> Am I correct in guessing that the 3-D hyperbolic space
> can be modelled by a natural generalization of the Poincare
> disk model...to be more precise, a h-plane would be
> represented by a sphere orthogonal to the boundary edge
> of the "ball", an analogue of the Poincare disk, right? Of
> course when the h-plane passes through the center, the
> h-plane would be just a plane.
Yes.
Douglas Zare
==============================================================================
From: george.ivey@gallaudet.edu (G.E. Ivey)
Subject: RE: Poincare disk model questions
Date: 25 Dec 1999 23:20:50 -0500
Newsgroups: sci.math
The Blue Wizard wrote:
>Question #1:
>Suppose I have two distinct h-points sitting on a Poincare
>disk model of 2-D hyperbolic plane, how do I construct a
>h-line going through those two points using only compass
>and ruler? Of course this h-line would be a circle going
>through those two points and is orthogonal to the boundary
>edge of the Poincare disk. I know that the h-line becomes
>line when both points and the center of the Poincare disk
>are collinear.
Each point in the "Poincare" disk has an "inverse" outside the
bounding circle. It is the point on the same extended radius as the
original point such that the product of the distances from the center
of the circle is equal to the radius squared- in other words, the
radius is the "geometric mean" of the two distances. If you have read
about the "Poincare Disk Model", you have probably read that. What
you may not have read was a description of how to find the inverse of
a point in the circle.
A little work with proportions and similar triangles shows that the
two points lie on a right triangle. That is, the center of the circle
and the inverse point are ends of the hypotenuse of a right triangel
whose right angle lies on the circle and such that the perpendicular
to the hypotenuse crosses the hypotenuse at the original points (look
at the similar triangles and do the proportions). Of course the fact
that this is a right triangle means that the hypotenuse forms a
diameter of a semicircle passing through the original bounding circle
at the right angle.
THIS MEANS: to construct the inverse of a point inside the circle:
Draw the radius to the original point and construct a perpendicular at
the original point. Where this perpendicular crosses the circle,
construct the tangents to the circle (you really only need to
construct one of them). They will cross each other and the original
radius at the inverse point.
THIS MEANS: Given two points within the bounding circle, to
construct the "line" containing them, construct the inverse point to
either point. Now construct the circle passing through those three
points. This (the part inside the original bounding circle) is the
h-line through the two points. The only time this would not be
possible is if the orginal two points lie on the same radius so that
the inverse points are also on the same radius and we do not have
three NON-COLLINEAR points. Of course, in that case the h-line is the
"Euclidean" line passing through the two points.
>Question #2:
>Am I correct in guessing that the 3-D hyperbolic space
>can be modelled by a natural generalization of the Poincare
>disk model...to be more precise, a h-plane would be
>represented by a sphere orthogonal to the boundary edge
>of the "ball", an analogue of the Poincare disk, right? Of
>course when the h-plane passes through the center, the
>h-plane would be just a plane.
Yes.
By the way, the reason "inversion in the circle" is important is
that the easiest way to define "congruence" without first defining
distance is to use REFLECTION. Two figures (line segment, triangle,
etc.) are congruent if it is possible to map one into the other by a
series of reflections. Of course, we want reflection in a line to map
one side of the line into the other while preserving line
perpendicular to the reflecting line. In the Poincare disk model,
where "lines" are represented by circles, reflection in a line
corresponds to inversion in the corresponding circle.
Here are some other things to think about: Since, in the Poincare
disk model, circles othogonal to the bounding cirle represent lines,
what do circles NOT orthogonal to the bounding circle represent? What
do circles completely contained in the bounding circle represent?
What do circles internally tangent to the bounding circle represent?
(Hint: inversion in a circle maps some lines into circles and some
circles into lines but maps the class of "lines and circles" into
itself.)
==============================================================================
From: "The Blue Wizard"
Subject: Re: Poincare disk model questions
Date: Tue, 28 Dec 1999 03:25:52 GMT
Newsgroups: sci.math
G.E. Ivey wrote in article
...
> The Blue Wizard wrote:
> >Question #1:
> >Suppose I have two distinct h-points sitting on a Poincare
> >disk model of 2-D hyperbolic plane, how do I construct a
> >h-line going through those two points using only compass
> >and ruler? Of course this h-line would be a circle going
> >through those two points and is orthogonal to the boundary
> >edge of the Poincare disk. I know that the h-line becomes
> >line when both points and the center of the Poincare disk
> >are collinear.
>
> Each point in the "Poincare" disk has an "inverse" outside the
> bounding circle. It is the point on the same extended radius as the
> original point such that the product of the distances from the center
> of the circle is equal to the radius squared- in other words, the
> radius is the "geometric mean" of the two distances. If you have read
> about the "Poincare Disk Model", you have probably read that. What
> you may not have read was a description of how to find the inverse of
> a point in the circle.
[rest snipped to appease the Net gods]
No, I never have seen such discussion about the inversion of point in
the discussion about the Poincare disk model of hyperbolic plane. So
I searched the Web (ain't it sooo wonderful? ;) for "inversive geometry",
and found a good "webbooklet" on it:
http://www.maths.gla.ac.uk/~wws/cabripages/inversive/inversive0.html
[annotated quote of previous article deleted --djr]
The Blue Wizard