H2 edited 23jul11, corrected 2aug11 with help from Tony Winkler

Lesson on Hyperbolic Distance

\textit{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{What you should study now.} This lesson replaces Theorem 8.11. The remainder of Hvidsten Section 8.2 is the same. \section{Motivation} In studying the Cartesian plane as a model for Euclidean geometry we discussed Birkhoff's Ruler Axiom. Since we did not study Hvidsten's Chapter 7 to which he refers in the proof of Theorem 8.11, we shall define a hyperbolic ruler and its associated hyperbolic distance in terms of cross ratios. Since the cross ratio is invariant under all Moebius transformations, it is invariant on the subgroup of Moebius transformations we have identified as the \textit{ hyperbolic group} of isometries in the Poincare disk model. To avoid confusion with Hvidsten's exposition, which is initially different than ours, the points in the figure here are labelled differently than in the textbook. Note we have drawn two cases of four points. Points $p,q$ are inside the unit disk, and $m,n$ are on on the unit circle. In one case the four points lie on the same (Euclidean) line, in the second, they lie on a Euclidean circle. So, in both cases they lie on a circline. Since the circle is drawn perpedicular to the unit circle, it is a hyperbolic Line in the Poincare model of the hyperbolic plane. Of course, only the circular arc minus its endpoints is the hyperbolic Line. Because the line on the left does not go through the origin, it is not interpreted as a hyperbolic Line. On the other hand, if we were discussing the Beltrami-Klein model, it would be the other way around. The chord of the circle would be a hyperbolic Line, and the circular arc would not. Therefore, it is best \textit{not} to make any interpretation yet, and just continue to consider both cases where the four points lie on the same \textit{circline}. Recall, by the \textit{ Symmetry Principle}, four points lie on the same circline if and only iff their cross ratio is a real number. Because we have abbreviated the exposition of this material to concentrate on the essentials, the Symmetry Principle is accepted here without proof. \section{Definition}. Given $p,q$ inside the disk, let $m,n$ be the points where the $\ell_{pq}$ crosses the unit circle. From vector geometry we recall that there are two real numbers $r,s$ such that \begin{eqnarray*} m &=& q + r(p-q) \ \mbox{with} \ r > 1 \\ n &=& p + s(q-p) \ \mbox{with} \ s > 1 \\ \end{eqnarray*} So we can calculate directly, without knowing additional properties of Moebius transformation, that \begin{eqnarray*} CR(p,m,q,n) &=& \frac{(r-1)(s-1)}{rs} > 0 \\ \end{eqnarray*}
Question 1.
Given four points $m,p,q,n$ on the same line in this order, so that $ m = q + r(p-q), n = p + s(q-p) $ for two reals numbers $r,s$, verify their crossratio is a real number $CR(p,m,q,n) = \frac{(r-1)(s-1)}{rs}$. But it is not so easy to verify that this fraction is positive directly from the definition of the cross ratio. Here is an alternative arguments \textit{using} geometric properties of MTs. Note that the Moebius transformation $z \mapsto CR(z;m,q,n)$ takes $m\mapsto 0, q\mapsto 1, n\mapsto \infty $. Therefore, it takes the point $p$ into the interval $0 < f(p) < 1 $. So the numerator of the fraction is the product of \textit{ two} negative real numbers. \subsection{The case of circles} Suppose now $m, p,q,n$ lie on a circle in that positive order. We have showed that the Moebius transformation $f(z) := CR(z;m,q,n) $ takes $f(m)=0, f(q) = 1, f(n)= \infty $, and it takes $circ(m,q,n)$ to the real line, preserving order. Thus the $arc(m,q,n)$ goes to the positive real line. The first assertion is easily recomputed from the defition of the cross-ratio. But the second assertion is another fact about MTs which we shall simply assume here as a lemma to be proved elsewhere, and at another time. \subsection{Logarithms} Now that we know the cross ratio discussed above is a positive real number, it is in the domain of the logarithm function. Indeed, as $p$ ranges over the entire portion of the circline between $m$ and $n$ (excluding the endpoints), it's logarithm ranges over the entire real line from $-\infty$ to $+\infty$ (excluding the "end points", of course.) So this fuction $ t_p = log_e(CR(p,m,q,n))$ is an excellent candidate for a non Euclidean ruler. \subsection{Distance} The absolute value of the logarithm we calculated above is a positive real number, which the distance between two points should be in any geometry. However, the concept of distance needs to have additional properties to be useful. \textbf{Axioms for Distance} \begin{eqnarray*} (\forall p,q \in \mathbb{C} ) \ d(p,q) & \ge & 0 \\ (\forall p,q \in \mathbb{C}) \ d(p,q) &=& d(q,p) \ \mbox{symmetry} \\ (\forall p,k,q \in \mathbb{C}) \ d(p,q) & \le & d(q,k) + d(k,p) \ \mbox{triangle inquality} \\ (\forall p,q \in \mathbb{C}) \ d(p,q)= 0 & \implies & p = q \\ \end{eqnarray*} Recall from our discussion of Birkhoff's Ruler Axiom that the rulers and the distance have to be related. Consider the ruler on the $\ell = circline(m,p,q,r)$. \begin{eqnarray*} t_\ell(z) & := & \ln CR(z,m,q,n) \\ t_\ell(m) &=& -\infty \\ t_\ell(q) &=& 0 \\ t_\ell(n) &=& +\infty \\ \end{eqnarray*} We have taken the (natural) logarithm to stretch the reals over the the positive and negatives. $t_\ell$ sets up a 1:1 and onto relation between the line-segment, or circular-arc $m,q,n$ and the entire real line, in both cases, the end points are not included (called $\pm \infty$ as limiting cases). Since the ruler is defined in terms of the cross ratio, it is invariant under Moebius transformations, in particular Moebius transformations that have additional properties. For example those in the hyperbolic group, which preserve the unit circle, and map the unit disk to itself. Thus we can now rightly call the MTs of the form \begin{eqnarray*} f(z) &=& e^{i\theta}\frac{z - \alpha}{\bar{\alpha}z -1} \\ \end{eqnarray*} the \textit{ isometries} in the Poincare model of non-Euclidean geometry. These then, are the correct interpretation of the primitive notion of \textit{ congruence} first formalized by Euclid. \subsubsection{Comment} Tony Winkler correctly pointed out that this section was abbreviated and so left the reader wondering what the point was to introduce the "axioms" for distance without further discussion. Missing is the verification that the functions we call hyperbolic distance does, in fact, satisfy these axioms. The first, second and fourth property are easily deduced directly from the definition and the properties of cross-rations and logarithms. The Triangle Inequality is more difficult to verify. Although it is one of the exercises in Hvidsten, it will be elaborated here in a future edition of these notes. \subsection{How about the other two models?} It is clear that part of the above approach, the definition of a rulers in terms of logarithms of cross ratios, will work in the other two models. \subsubsection{Beltrami-Klein model} Here the Lines are chords of the unit disk. But here, the interpretaton of "congruences" in terms of a group of transformations is more difficult and not treated in this course. \subsubsection{Upper Half Plane mdoel} Here the Lines are vertical rays in the UHP, and circular arcs perpendicular to the x-axis. Thus the Lines are again circlines. And the exact approach applies. It is not particularly difficult to compute the canonical form of the MTs which preserves this geometry. It turns out that we can use the same definitions for this model as well. However, more is true. There is a Moebius transformation which takes the Poincare disk model into the UHP models, namely the MT which takes $-i, 1, i$ to $0,1,\infty}. We shall discuss this issue further in a later lesson. \section{Calculation of Hyperbolic Distance} We saw why \[ d_H(z_0, z_1) = | log CR(z_0,\omega_0, z_1, \omega_1)| \] is the right choice for the distance function. We have changed the notation once again, to make it more mnemonic. Note that the subscripts match, so you can remember the order of the points on their circline, namely $\omega_0, z_0, z_1, \omega_1 $. Unfortunately, for the purpose of building the \textit{ Geometry Explorer}, this is computationally expensive. It requires finding the omega-points (on the unit circle) of the Line through the two points. So Hvidsten uses a simpler formula for computing the hyperbolic distance. In Exercise 8.2.13, Hvidsten tells you to compute a formula which depends only on the two given points, and does not require you to compute the omega-points on the unit circle. The lemma for this is Proposition 8.14, for which we give the following proof: \subsection{ Distance from the origin} Hyperbolic geometry has no preferred points, "every point is equal!" But in a particular model, "some points are more equal than others." In the Poincare disk model, the origin is a very special point (though a perfectly ordinary Point!) Similarly for diameters, especially the one on the real axis. Since hyperbolic distance is based on cross ratios, and cross ratios are invariant under Moebius transformations, we can measure distances between Points by moving their Lines into a special position using a hyperbolic isometry. So, one way to measure the hyperbolic distance $d_H(z_0,z_1)$ would be to first find an element in the group which takes $z_0 \mapsto 0$ and $z_1 \mapsto r \in \mathbb{R}$, and calculate $d_H(0,r)$. This is easier, because we know the end points on the real diameter, namely $-1,+1$, i.e. $d_H(0,r) = |log CR(0,-1,r,1)| $. Can you determine $r$? \subsection{The Motion Invariance Principle (MIP)} This idea is so important we will give it a fancy name for easier reference. Here it is in words: \textit{Any geometric property of a configuration of points which is invariant (remains the same) under a hyperbolic isometry, may be reliably investigated after the data has been moved into a convenient position in the model.} Note, for Euclidean geometry you have been following this principle throughout all of your school career. Afterall, when the teacher drew a particular figure and then proved a property of the figure, why is this a proof for every figure like it? She usually says something like, "without loss of generality", or perhaps, "since we made no particular assumptions about the points". Actually, it holds for all figures that are congruent to the given one, by the Euclidean MIP. In hyperbolic geometry, you have already met MIP in the discussion of a Lambert quadrilateral $ABCD$ with right angles at $A,B,C$. If we use a hyperbolic translation (discussed in the next section) moving the figure so that $B \mapsto 0$ and then rotate so that $C \mapsto c \in \mathbb{R}$, i.e. so that the new position of $C$ is on the real axis, then $A$ will have moved to the imaginary axis. Now the hyperblic perpendiculars at $A$ and $C$ are circular arcs that "bend away from each other." If they do meet at $D$, it will be at an acute angle there. Thus the angle at $D$ in the original figure must be the same, hence also acute. \subsection{Calculation from the definition} Since $0, z$ lie on a diameter, it's omega points are easily computed.
Question 2.
Show does the line $\ell_{0,z}$ through $0,z$, crosses the unit circle at $\pm \frac{z}{|z|}$. So $d_H(0,z) = log CR(0,-\frac{z}{|z|},z,\frac{z}{|z|})$ is easy to compute from the definition. \subsection{Using the MIP} First, rotate the entire disk by $z \mapsto e^{-i\theta}z$, where $\theta = \arg(z)$. This is an isometry, which takes $z$ to $r=|z|$. But now $CR(0, -1, r, 1) = \frac{1-r}{1+r} <1$ , so the distance is $d_H(0,r) = - \log\frac{1-r}{1+r}= \log\frac{1+r}{1-r}$ as claimed. Done.
Question 3.
What rule for logarithms were applied in showing that $ - \log\frac{1-r}{1+r}= \log\frac{1+r}{1-r}$ ? \subsection{General case} The trick is to reduce the general case (Exercise 8.2.13) to the particular case (Proposition 8.14). The hyperbolic translation $ f(z) = \frac{z-z_0}{1-\bar{z_0}z} $ takes $z_0 \mapsto 0$. Where must $f(z_1)$ be? Wherever it is, we can now measure its distance to the origin and clear some fractions to get the formula \begin{eqnarray*} 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| }) \\ \end{eqnarray*}
Question 4.
Calculate the hyperbolic distance $d_H(\frac{1}{2},\frac{1}{2}i)$.