H2 last edited 31mar15

Lesson on Hyperbolic Distance
\textit{ $\C$  2010, 2015 Prof. George K. Francis, Mathematics Department,
University of Illinois}

\begin{document}
\maketitle

\section{Introduction}
In studying the Cartesian plane as a model for Euclidean geometry we discussed
Birkhoff's Ruler Axiom. Recall that it postulates the global concept of
\textit{distance} as a non-negative real number associated with any pair of
points, and the local concept of a \textit{ruler} on line which establishes
a bijection of the points on the line with the real numbers. And, there has
to be a correspondence between them in that the distance between two points
can be measured by any ruler on the line that joins them.

In the unit of Moebius transformations we saw that the crossratio is
invariant under Moebius transformations. In the previous lesson, we
identified the subgroup of hyperbolic MTs as the basis of \textit{congruence}.
These MTs, which map the unit circle to itself and the unit disk to itself,
are exactly the \textit{ rotated involutions}
$f(z)= e^{i\theta}\frac{z-\alpha}{\bar{\alpha}z - 1}, \, |\alpha|< 1.$

\section{Definition of Hyperbolic Distance}

The idea is to define the hyperbolic distance between two hyperbolic
points in terms of the crossratio of these two points, and together
with the two points where the
circline through them and perpendicular to the unit circle meets it.

\textbf{Definition of Hyperbolic Distance:}
In symbols, given points $\omega_0, z_0,z_1, \omega_1$, in that
order, on a circline perpendicular to the unit circle and crossing
it at $\omega_0, \omega_1$, define
$d_H (z_0,z_1) := |\log CR(\omega_0, z_0,z_1, \omega_1)| = |\log CR(z_0, \omega_0, z_1, \omega_1)|$

\subsection{Computing Distances}
In non-Euclidean geometry as in Euclidean geometry there are no distinguished
location, points, or lines. As in a perfect democracy, everybody is equal.
But in a model, this is not so. Recall that in Cartesian geometry the origin
and the axes are very special. And in the Poincare disk model, the center
and the diameters are special. We shall compute special distances in
increasing generality until we can replace the first definition of distance
above, with one that makes no reference to points other than the two we
are measuring. We next prove that

\textbf{Calculating the Hyperbolic Distance}
$d_H(z_0,z_1) = \log( \frac{|1-\bar{z_0}z_1| + |z_1 - z_0|} {|1-\bar{z_0}z_1| - |z_1 - z_0| })$

\subsection{Distance Along Diameters}
In this case, determining the $\omega$-points in terms of the hyperbolic
points is easy using vector calculus.  For notational purposes,
rename $p:= z_0$ and $q := z_1$.

Then the line $(pq)$ crosses the unit circle at
\begin{eqnarray*}
(\exists t > 1)( m &=& q + t(p-q) ) \\
(\exists s > 1)( n &=& p + s(q-p) ) \\
CR(p,m,q,n) &=& \frac{(t-1)(s-1)}{ts} > 0 \\
\end{eqnarray*}

Question 1.
Given four points $m,p,q,n$ on the same line in this order, so that
$m = q + t(p-q), n = p + s(q-p)$ for two reals numbers $t,s$, verify
theit crossratio is a real number
$CR(p,m,q,n) = \frac{(t-1)(s-1)}{ts} > 0$.

Unfortunately, this does not solve our problem of eliminating the need to
compute the omega points on an hline. Nor does it
generalize nicely to those hlines that are circular arcs perpendicualr to
the unit circle. We have to take a more judicious approach.

\subsection{Distance from the Origin}

We calculate the hyperbolic distance from $|\alpha| \lt 1$ to the origin
directly from the formula.

Question 2.
Show does the line $\ell_{0,z}$ through $0,z$, crosses the unit circle at
$\pm \frac{z}{|z|}$.
Thus
$d_H(\alpha,0) = |\log CR(\alpha, \frac{\alpha}{|\alpha|}, 0, -\frac{\alpha}{|\alpha|}) | = \log \frac{1+|\alpha|}{1-|\alpha|}$
\end{eqnarray*}

Question 3.
Perform the calculations, from the definition, that
$d_H(\alpha,0) = \log \frac {1+|\alpha|}{1-|\alpha|}.$
It may help to express $\alpha = re^{-i\theta}$.
What rule for logarithms did you applied in showing that
$- \log\frac{1-r}{1+r}= \log\frac{1+r}{1-r}$ ?

For the previous exercise we see that all parts of the definition are
really needed. The logarithm transforms the values from the positive
reals $(0,\infty)$ to all the reals $(-\infty, + \infty)$.
The crossratio goes to 1 when $\alpha=0$, whose logarithm is 0.

\subsection{Hyperbolic Distance along Circular Arcs}
We now justify using the crossratio in the first place, because
crossratios are invariant (do not change value) under a MT. In other words,
for any hyperbolic MT $f(z)$ we have that
$d_H(z_0,z_1) = d_H(f(z_0),f(z_1)).$
This finally justifies using the names \textit{hyperbolic isometry}
and \textit{hyperbolic congruence}.

So, given
two points $z_0,z_1$ in the unit disk, we first seek a hyperbolic
MT which takes $f(z_0) = 0$. Recall that rotating an involution
by $180^o$, namely
$V_{z_0, \pi (z)} = \frac{z-z_0}{1- \bar{z_0}z}$
does just that.  Applying this congruence to $z=z_1$ discovers that its
image $\alpha$ is as follows.

Question 4.
Calculate that $\alpha:=V_{z_0,\pi}{z_1} = \frac{z_1-z_0}{1- \bar{z_0}z_1} .$

Since we already know the hyperbolic distance from $\alpha$ to 0, you will obtain,
after some simplification, that

Question 5.
$d_H(z_0,z_1) = \log( \frac{|1-\bar{z_0}z_1| + |z_1 - z_0|} {|1-\bar{z_0}z_1| - |z_1 - z_0| }).$

\section{Summary}
TBA