## Lesson on the Hyperbolic Group

H1 30mar15 \\ \textit{ $\C$ 2010, 2015 Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{Introduction} From earlier experiments with geometrical construction software, such as Hvidsten's Geometry Explorer or Hohenwarter's Geogebra, recall that in Poincare Model of non-Euclidean geometry in the unit disk of the Euclidean (Cartesian) plane, we have the following interpretation. To distinguish between Euclidean points, lines, angles, distances and non-Euclidean Points, Lines, Angles, Distances, we shall use the capitalization convention for a while. \textbf{The Poincare Disk Model of the Hyperbolic Plane.} \begin{itemize} \item The Points are the points inside (but not on) the unit disk. \item The Lines are the circular arcs inside the unit disk which are perpendicular to the unit circle. In particular, the diameters of the unit circle are the only Lines which are also Euclidean line segments. \item Incidence is the same as in the Euclidean plane. \item Angles are the same as in the Euclidean plane. That is, the Euclidean angle between two arcs perpendicular to the unit disk is also the non-Euclidean Angle between two Incident (intersecting) Lines. In Birkhoff's terms, we can say that the hyperbolic Protractor is the same as the Euclidean protractor. \end{itemize} Note the absence of interpretations for Rulers, Distance and Congruence in this list. From your experience with GEX or GGB, you know there must be interpretations of all these other geometric entities as well. After all, there were tools for these on the palette of this software. From Euclidean geometry, you also know that distance and congruence are not arbitrary, they have to be related to each other: a congruence preserves distances between points and angles between rays. Moreover, from Birkhoff's Ruler Axiom, you also recall that distance along lines must be measurable along lines by means of the bijections with the real numbers, so so-called rulers. In this lesson we will discover these missing interpretations from properties of Moebius Transformations. \subsection{Klein's Erlangen Program} One of the truly revolutionary ideas in geometry was Felix Klein's suggestion (1872) that \textit{ all of geometry should based on the transformation groups.} Since he was professor at Erlangen at the time, it is called his \textit{ Erlangen Program}. Although the idea was quickly adopted by geometers and is now the foundation of geometry, over a century later it is still not universally taught in the schools. In fact, it is hardly ever taught in high school and rarely in college. Well meaning math education reforms, such as the New Math of the 1950s and the current Common Core State Standards pay lip service to the Erlangen Program. From all appearances, however, the authors of such reforms propose only some superficial and isolated concepts into their reforsm , possible because of the enormous resistance of the establishment to any significant educational innovations. In many of his proofs (for example SAS) Euclid himself appealed to an intuitive notion of moving and/or flipping one figure ontop of another to see whether they are congruent or not. This concept was so elusive that for two millenia, geometers avoided giving it a definition rigorous enough to be useful in proofs. They accepted it as one of the undefined terms (primitives) and left it at simply postulating its basic properties. With the Erlangen Program, however, we derive it analytically from our understandng of the real numbers and the Cartesian plane. \subsection{The Roadmap.} So, we shall follow Klein, and choose a group of Moebius transformations which preserve the unit circle, mapping the interior of the unit disk to itself. Thus they move Points to Points. In a previous lessson we saw that the set of such MTs form a group, which we call \textbf{the} \textit{ hyperbolic group}. Later, when we discuss other models, we have to be more careful to add which model is being considered. But for this lesson, we discuss only the Poincare Disk Model. Being MTs, the transformations in the hyperbolic group, also map circlines (circles and lines) to each other, preserving angles. In particular, the unit circle is mapped into itself, and hence it followd that hyperbolic Lines are mapped to Lines as well. At this point things get interesting, and we next define a Ruler along every Line, making sure that every MT in the hyperbolic group preserves the Rulers. This allows us to define Distance consistent with the previous choices. And we're done. Because it is a nuisance, we now drop the capitalization of hyperbolic primitives. From the context of the discussion you will know whether the points and lines refer to Euclidean or to hyperbolic points and lines. So you can, at any moment, put the capital letter back where it belongs. \subsection{Symmetry Properties of Moebius Transformations} Although we have investigated some geometric properties of MTs and alluded to others, there are more that are needed in the remainder of this lesson. We list them here. \begin{itemize} \item Circles and lines (circlines) are mapped to circlines. \item The orientation of a circline is preserved. \item Points on the left side of a circline (inside for a cyclic orientated circle) are mapped to points on the left side of the image. \item Angles between intersecting circlines are preserved. \item Pairs of symmetric points to a circline are mapped to symmetric pairs relative to the image circline. \end{itemize} Recall that relative to a circle $\bigodot{(c,r)}$ with center $c \in \mathbb{C}$ and radius $r \in \mathbb{R}$, the point $z, z^*$ are \mathit{symmetric} if \[ z^* = c + r^2 \frac{z - c}{|z-c|^2} \].Question 1.Show that $|z-c||z^*-c| = r^2 $. So a historical definition is that two points are symmetric to a circle if they are on the same radial ray and the radius is the geometric mean of their distance from the center. \section{First Canonical Form of a Hyperbolic Transformation} Some obvious examples of MTs that preserve the unit disk are the rotations about the center. These have the form $f(z)=\beta z, |\beta| =1 $. Note that $|\beta|=1$ is equivalent to $\beta = e^{i\theta}$ for some angle $\theta$, the angle of rotation.Question 2.Use Euler's formula and coordinates to show the formula takes $(x,y) \mapsto (x \cos \theta - y\sin \theta , x \sin \theta + y \cos \theta )$ for a rotation, as you learned it in calculus. Some MTs that decidedly do \texbf{not} preserve the unit disk are the translations, and dilations. And especially inversions are not in the hyperbolic group, whether we use the term in the sense of the anatomical primitive, $z \mapsto \frac{1}{z}$, or the other more geometrical meaning of inversion, where $z \mapsto 1/\bar{z}$. In either case the Point $0 \mapsto \infty$ and the point $\infty$ on the Riemann sphere is not a hyperbolic Point.Question 3.Which translations are in the hyperbolic group? Why? \subsection{The Involutions} A less obvious and much more interesting MT with this property depends on a Point in the model, i.e. a point $\alpha$ inside the unit disk. It is defined by \[ V_\alpha (z) := \frac{z-\alpha}{\bar{\alpha}z -1}.\] Substitution reveals that \begin{eqnarray*} V(0)&=& \alpha \\ V(1)&=& \frac{1-\alpha}{\bar{\alpha} -1} := \alpha_1 \\ V(\infty)&=& \frac{1}{\bar{\alpha}} := \alpha^* \\ \end{eqnarray*} By inspection, we see that $V(\alpha) = 0$ and that $V(\frac{1}{\bar{\alpha}})= \infty$. (Note that $\frac{1}{\bar{\alpha}} = \alpha^*$, the point symmetric to $\alpha$ with respect to the unit circle.) Also note the $:=$ symbol means that we define a shorter nickname for a more complicated expression. Thus $\alpha_1$ should remind you that this is where the MT $V$ takes the point 1.Question 4.Solve the equation $1=V_\alpha(z)$ and discover its solution to be $z = \alpha_1 = \frac{1-\alpha}{\bar{\alpha}-1}$. Do the calculation instead of just substuting the answer which is given here to let you know you succeeded. Thus $V_\alpha(z) = CR(z;\alpha,\alpha_1, \alpha^*) = V_\alpha^{-1}(z) $. In other words, every $V_\alpha$ is its own inverse transformation. Such things are called \textit{ involutions.}Question 5.Show that $V_\alpha(z)=\frac{z-\alpha}{\bar{\alpha}z -1} $ is its own inverse function. In other words, show that $V_\alpha (V_\alpha (z)) = z$ for all $z$. You can do this the hard way, calculating the algebra. Or you can be clever and invoke the Tripod Theorem. \subsection{The Involutions are in the Hyperbolic Group} We still have to verify that the $V_\alpha$ preserves the boundary and the interior of the unit disk. The best way to check this is in terms of the squared-distance $|w|^2 = w\bar{w}$ of a complex number from the origin. We calculate that, for $w = V_\alpha(z)$, \[ w\bar{w} = \frac{z-\alpha}{\bar{\alpha}z-1} \frac{\bar{z}-\bar{\alpha}}{\alpha\bar{z}- 1} = \frac{|z|^2 - 2\mathfrak{Re}(\alpha\bar{z}) +|\alpha|^2}{|\alpha|^2|z|^2 - 2\mathfrak{Re}(\alpha\bar{z}) +1}. \] where $\mathfrak{Re}$ means the "Real Part", i.e. $\mathfrak{Re}(w) = \frac{w+\bar{w}}{2}. $ (You'll have to check this on scratch paper before putting it into your Journal.) Note the the numerator and denominator are the same when $|z|^2=1$. Hence \[ (\forall z)( |z|=1 \Rightarrow |V_\alpha(z)|=1).\] When $|z|<1$, subtract the numerator of the fraction from its denominator to see that the numerator really is less than the denominator. And so \[ (\forall z)( |z|<1 \Rightarrow |V_\alpha (z)| < 1 ), \] and so $V_\alpha$ is in the hyperbolic group.Question 6.Calculate that $|\bar{\alpha}z - 1|^2 - |z -\alpha|^2 = (1-|\alpha|^2)(1-|z|^2).$ How does this clever identity show directly that $V_\alpha$ maps the inside, onside, and outside of the unit circle to the inside, onsine, and outside respectively?Question 7.Calculate directly that $\alpha_1$ lies on the unit circle by calculating directly that $ 1=|\frac{1-\alpha}{\bar{\alpha} - 1}|^2 $. \subsection{Generating the Hyperbolic Group} We combine the two discoveries and conclude that the composition of a rotation about the origin with an involution is in the Hyperbolic Group, $ z \mapsto e^{i\theta}V_\alpha(z) $. It is surprising that the converse also holds, namely that every hyperbolic transformation has such a canonical form. We prove this in detail now. \textbf{Theorem: } If a MT $f(z)$ perserves the unit disk, then there is an $\alpha$ inside the disk, and an angle $\theta$, so that \[f(z) = V_{\alpha,\theta}(z) := e^{i\theta} \frac{z - \alpha}{\bar{\alpha}z -1.} \] \textbf{Proof:} Note that we have stated the theorem in a slightly different manner, and we shall prove it in a slightly different manner too. That keeps life interesting. We have also extended out notation to make it easier to refer the transformation just by its "anchor" and "angle". We begin by giving the name $\alpha = f^{-1}(0)$ to the point which $f$ takes to the origin. We next calculate a $\theta$ as follows. By assumption, $f$ preserves the unit circle. So both $f(1)$ and $\alpha_1$ are points on the unit circle and we define $\theta = arg(f(1)) - arg(\alpha_1)$. Another way of expressing this, by Euler's Formula, is \[f(1) = e^{i\theta}\frac{1-\alpha}{\bar{\alpha} -1 }. \] Third, it is most convenient to use the \textit{Symmetry Principle} which states that any MT $f$ which preserves the unit circle, also preserves points symmetric with respect to the unit circle, namely that \[(\forall z) (w =f(z) \implies w^* = f(z^*) ),\] where $z*= \frac{1}{\bar{z}} $. In particular, $f(\alpha^*) = \infty $. \subsection{Applying the Tripod Theorem} We have shown that \begin{eqnarray*} f(\alpha) &=& 0 = V_{\alpha,\theta}(\alpha)\\ f(1) &=& e^{i\theta} \frac{1-\alpha}{\bar{\alpha}-1} = V_{\alpha,\theta}(1)\\ f(\alpha^*)&=&\infty = V_{\alpha,\theta}(\frac{1}{\bar{\alpha}})\\ \end{eqnarray*} Applying the Tripod Theorem, which says the value of an MT on three distinct points determines the transformation uniquely, we're done.Question 8.Show that $CR(z,\alpha, \alpha_1, \frac{1}{\bar{\alpha}}) = \beta \frac{z - \alpha}{\bar{\alpha}z -1} $ for some $\beta$. Find $\beta$. (Hint: This is not rocket science if you understood this lesson.) \subsection{Proof by Calculation } Thus we know the canonical form for the Hyperbolic Group. We already know it forms a group by an abstract argument. But, if we were to consider this set of transformations which have this form, without considering their geometrical meaning, showing they form a group would be more tedious. But if you're suspicious of all this abstraction, you could always calculate directly that the composition of two of them, $V_{\alpha, \theta} \circ V_{\omega,\psi}$ can again be put into the same form. This is a good exercise in complex numbers, and like all exercise, it is painful. And, unfortunately, that's all it is: Make work. You will learn nothing from the calculation you didn't know before. But, you may be more convinced of the truth of the statement. For the inverse, \begin{eqnarray*} V_{\alpha, \theta}^{-1} &=& V_{\alpha' , \theta' } \\ \mbox{ where } \alpha' &=& e^{i\theta}\alpha \\ \mbox{ and } \theta' &=& - \theta \\ \end{eqnarray*} however, it is worth doing the computations.Question 9.Show that $w=V_{\alpha,\theta}(z)$ implies that $z = e^{-i\theta} V_{e^{i\theta}\alpha}(w) =V_{e^{i\theta}\alpha, -\theta}(w)$ \section{Preview} In the next lesson we will discover the correct hyperbolic rulers.