Here is a technical definition of how to compute distance.
Begin with any two points. If is the
-line on which they lie,
let
be the line on the back hemisphere that projects onto
.
Rotate the sphere so that one of the end points of
moves to the north
pole,
.
rotates into a new line
passing through
. The projection
of
is now a vertcal line,
. The points
and
have been
lifted to
rotated to
and then projected onto
. They are now
called
and
. We can take the ratio of the heights of
and
.
This is almost a distance. However, distance should be
symmetric. The ratio of the heights depends upon which point we name first.
Therefore, we take the absolute value of the natural log of the ratio of the heights to be
the distance between
and
.
Consider the two pairs of points
Let be the unit disc in the plane.
.
We saw earlier that
is the image of the lower half sphere under
stereographic projection. This is another model for the hyperbolic plane.
We will easily locate the h-lines once we see how this is related to
the upper half-plane.
Take the sphere. Rotate it so that the back hemisphere goes into the bottom hemisphere. Project the bottom hemisphere onto the unit disc. This procedure identifies the upper half plane (the image of the back hemisphere) with the unit disc (the image of the bottom hemisphere). An h-line in the upper half plane corresponds to a circle on the back hemisphere which is perpendicular to the prime meridian. Such a circle rotates into a circle on the bottom hemisphere that is perpendicular to the equator, and then projects to a circle in the plane that intersects the boundary of the unit disc at right angles. When we project onto the unit disc, we no longer have to worry about h-lines through infinity. Things look much more symmetric. However, we still have one weird type of h-line: a Euclidean straight line passing through the center of the disc. (See figure 24.)
Figure 25: Some hyperbolic cloth: A tiling of the hyperbolic plane by triangles with angles
Once we have a hyperbolic geometry, many new things are possible.
Enclosed is a picture of the tiling of the hyperbolic plane by triangles
whose angles are ,
and
. (See figure 25.)
The important thing to realize
about this picture is that ALL the trianglular tiles are congruent. That is,
even though the triangles near the boundary of
appear to be much
smaller than those in the center, their sides all have the same lengths.
To see this you just have to look through your hyperbolic glasses.