Lesson on the Hyperbolic Group
H1 19april11 revised 17jul11, 2aug11 \\ \textit{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{Introduction} From your experiments with Hvidsten's Geometry Explorer you already know that in the Poincare model for non-Euclidean geometry has the following interpretation. \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. So, you can use any all of your knowledge about measuring the angle between two intersecting circles and/or lines. In terms of Birkhoff's work, we can say that the hyperbolic Protractor is the same as the Euclidean protractor. \end{itemize} And, from experience with GEX, you know there must be interpretations of all the other geometric entities, in particular Distance and Congruence. From Euclidean geometry, you also know that distance and congruence are not arbitrary, they have to be related. Moreover, from Birkhoff's Ruler Axiom, you also know that distance has to be measurable along lines using bijections with the real numbers. \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. 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. \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. One of the exercises was to show that these form a group, and we call this \textbf{the} \textit{ hyperbolic group}. Later, when we discuss other models, we'll have to be more careful to add which model we're talking about. These MTs, the ones that constitute the hyperbolic group, also move circlines (circles and lines) perpendicular to the unit circle to other circlines still perpendicular to the unit circle. So they do transform Lines to Lines. Since all MT are conformal (angle-preserving), also those in the hyperbolic group preserve angles. 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{Hyperbolic Geometry } There are more to prove about this model, namely that it is \textit{ categorical}, but we do not do it in this course. It means that all models of hyperbolic geometry are \textit{ isomorphic}, i.e. basically interchangeable. As you recall from the first lesson in the course, for a categorical model the Meta Theorem really does hold. In other words, we can discover and prove theorems about hyperbolic geometry by examining its models rather than deducing them logically from the axioms. Nobody today, except logician and some high school teachers, practice non-Euclidean geometry axiomatically. It is all done using the models using analytic methods. (specifically, complex numbers). \section{Cononical Form for the Hyperbolic Group} 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. Some MTs that decidedly do not preserve the unit disk are the translations, and dilations. And especially not inversion, whether we use the word in the sense of the anatomical primitive, $z \mapsto \frac{1}{z}$, or the other, geometrical meaning of inversion, where $z \mapsto \frac{1}{\bar{z}}$. In either case the Point $0 \mapsto \infty$ and the point $\infty$ on the Riemann sphere is not a Point. \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 origin. The star-operation, as in the symbol $\alpha^*$, is discussed in greater detail in the text and in the supplementary material. Here, it just means what we say it means.Question 1.Solve the equation $1=V_\alpha(z)$ and discover its solution to be $z = \alpha_1 = \frac{1-\alpha}{\bar{\alpha}-1}$. Don't just substitute in to see that the answer is correct. 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 2.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 the clever way by invoking 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 $w = V_\alpha(z)$, \[ \frac{z-\alpha}{\bar{\alpha}z-1} \frac{\bar{z}-\bar{\alpha}}{\alpha\bar{z}- 1} = \frac{|z|^2 - 2\mathfrak{Re}(\alpha\bar{z}) +|\alpha|^2}{|\alpha|^2|z|^2 - 2\mathfrak{Re}(\alpha\bar{z}) +1}. \] 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 \[ (\forall z)( |z|=1 \Rightarrow |V_\alpha(z)|=1).\] Next subtract the numerator of the fraction from its denominator to see that the numerator really is less than the denominator.Question 3.Obtain that $|\bar{\alpha}z - 1|^2 - |z -\alpha|^2 = (1-|\alpha|^2)(1-|z|^2).$ For $|\alpha| < 1 $, this is positive if and only if $|z| < 1 $. Hence $ |z|< 1 \Rightarrow |V_\alpha(z)| < 1. $ We leave it to you to show that $ |z|> 1 \Rightarrow |V_\alpha(z)| > 1. $Question 4.Calculate directly that $\alpha_1$ lies on the unit circle, i.e. that $ 1=|\frac{1-\alpha}{\bar{\alpha} - 1}|^2 $. \subsection{The Entire 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 transformation has this canonical form. [Hvidsten Theorem 8.10] \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 \[f(z) = V_{\alpha,\theta} := e^{i\theta} \frac{z - \alpha}{\bar{\alpha}z -1.} \] \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} [a special case of Hvidsten Theorem 8.9] 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 5.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, however, it is worth computing.Question 6.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.