## Exercises on Moebius Transformations

revised 21apr11, 13jul11\\ \textit{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{Introduction} This lesson is a commentary on Hvidsten's \textit{Exercises 8.2.1-8.2.15} on page 325. There is a great deal of geometry hidden in these problems, which is obscured by the brevity of their statement, or even their solution. The elaborations here is intended to help you use the exercises to understand `the facts and methods. A "solution" to a problem should be a mini essay on what the problem means, and not consist of just cranking out what looks like an answer to a direct question. In this spirit we proceed. \section{Discussion of the Exercises} \subsection{Transformations which form Groups} \begin{enumerate} \item \textbf{Ex. 8.2.1} "Show that the $w=e^{i\phi}+b$ form a group, the group of rigid Euclidean motions (a.k.a. Euclid's congruences) of the plane." This is special case of the group of \textit{similitudes} treated extensively in the lessons. \item Remember, to show that a subset of the group of all bijective functions of the extended plane is itself a group, all you need to do is to show that \begin{itemize} \item \textit{Closure: }The composition of two such function can be put into the same canonical form that defines the subset. \item \textit{Inverse: } The inverse of such a function can be put into the same canonical form. \end{itemize} Be sure you have this example in your Journal and are able to reproduce the definitions and proof of this theorem on a test. \item As Hvidsten says: \textbf{Ex 8.2.4 - 8.2.7} are steps in the proof that the set of Moebius transformations also form a group. Much of this is in the notes. These 4 problems help you to organize this material in a coherent manner in your Journal. As long as the actual calculations are short, a test question may ask you to check that a given set of MTs form a group. \begin{itemize} \item Ex 8.2.4 this is the closure property and is pure high school algebra. Substitute the result of the first transformation into the formula for the second and rewrite in the standard form of Moebius transformation. Use our definition, where $ad-bc=1$ is the condition on the coefficients. You'll be happier for it. \item Ex 8.2.5 As in our notes, the presence of the identity follows from the presence of the inverse (8.2.6) and closure (8.2.4). However, an immediate guess will solve this problem directly. But here there is an easier solution by inspection. \item Ex 8.2.6 you'll need to solve $w=\frac{az+b}{cz+d}$ for $ z =\frac{ez+f}{gz+h}$, by giving $e,f,g,h$ as functions of $a,b,c,d$. But Hvidsten suggests a simpler approach. For this to hold, you'll need to use his criterion $ad-bc\ne 0$. Recall why we can assume either condition. These calculations belong into your Journal. \item Ex 8.2.7 is done below. The answer is that transformations always are associative: $(k \circ h) \circ g = k \circ (h \circ g)$. All you need to remember is how to test that two functions are the same. \end{itemize} \item Solution to 8.2.7: An equation of functions holds provided the equation holds for every value in the domain of the functions. So applying LHS to an arbitrary complex number $z$ we see that $ (k\circ h)\circ g (z) = (k \circ h)(g(z)) = k(h(g(z)))$ which is what you get if you plug $z$ into the RHS. \item \textbf{Ex 8.2.8: } This exercise asks you prove that a particular subset of transformations which have a particular geometric property,forms a group. As we have seen, all you need to check is closure and inverses. It is unfortunate that Hvidsten uses the word "fix" instead of "preserve". He does not mean that each point on the unit circle remains fixed under the transformation. (That would mean it is the identity, right? since there are at least 3 points on the circle.)) By "$f$ fixes or preserves the unit circle" we mean it maps every point on the circle back to some (possibly different) point on the same circle. Algebraically, if $|z|=1$ then $|f(z)|=1$ as well. \end{enumerate} \subsection{CircLine Geometry} \begin{enumerate} \item \textbf{ Ex 8.2.2 } "Show that the equation of a line can be given as $\mathfrak{Im}(\alpha z + \beta) =0$." While you can prove this with just a little arithmetic applied to the definitions, it has a deeper meaning. Note that for any complex number $w$, $\mathfrak{Im}(w)=0$ is the same thing as saying $w \in \mathbb{R}.$ \item \textbf{Extension 8.2.2bis} Since a straight line is a special case of a circline, this problem is a special case of the general theorem that the equation of a circline through three points $z_0,z_1,z_\infty$is given by $\mathfrak{Im}CR(z,z_0,z_1,z_\infty)=0$. Equivalently that $CR(z,z_0,z_1,z_\infty) \in \mathbb{R}$. \item \textbf{ Ex 8.2.3} In my edition of the text has a misprint here. The correct equation of the unit circle is $|z|=1$. "Find a Moebius transformation that takes the unit circle to the x-axis." There are, of course, very many solutions. But there is a "canonical" solution in terms of cross-ratios, if you think of the unit circle as $\odot(1,i,-1)$. Use the theorem that an MT defined by $f(z)=CR(z,z_0,z_1,z_\infty)$ takes the circline $\odot(z_0,z_1,z_\infty)$ to the real-line. \item \textbf{Extension 8.2.3bis} Incidentally, if the three points are on any circle, and in positive (counter-clockwise) order, then this MT takes the interior of the circle to the upper halfplane, and the exterior to the lower halfplane. \end{enumerate} \subsection{The Hyperbolic Group} The set of MTs which "fix" the unit circle form the group of isometries in the hyperbolic plane, called the \textit{ Hyperbolic Group} for short. The way this is established is exploit . \begin{itemize} \item Calculate the canonical form of every such MT. \item Use one property of such an MT to define a hyperbolic line. \item Exploit a property of hyperbolic lines so defined in terms of a cross ratio. \item Define a ruler, in the sense of Birkhoff, for every line. \item Verify that Birkhoff's Ruler Axiom is indeed verified. \end{itemize} The Exercises in the textbook serve this program. \begin{enumerate} \item \textbf{ Ex 8.2.9:} This is the \textit{ transitivity } of the Hyperbolic Group. By \textif{isometry} here Hvidsten means a Moebius transformation which preserves the unit circle. In Prop.8.10 he proves that such a Moebius transformation must have the form \begin{eqnarray*} w = e^{i\theta}\frac{z-\alpha}{\bar{\alpha}z -1} & \mbox{where} & |\alpha|<1\\ \end{eqnarray*} So now the problem reduces to finding angle $\theta$ and point $\alpha$ which is moved to the origin under the isometry for given values for $z$ and $w$. Hvidsten he calls $z=P$ and $w=Q$. But Hvidsten gives you a hint, by asking the question for the case that $z$ remains arbitrary, but $w=0$. (Why is it enough to prove it just for this case? Don't forget to answer this in your solution.) Finally, for this special case the answer is now very easy, namely, to solve \begin{eqnarray*} 0 = e^{i\theta}\frac{P-\alpha}{\bar{\alpha}P -1} & \mbox{where} & |\alpha|<1\\ \end{eqnarray*} we can take $\theta = 0$, i.e. no rotation, and $\alpha = P$, which is legitimate, because we are given that $|P| < 1 $. The isometry is not unique, but it is unique up to a rotation. \end{enumerate} [The remaining exercises will be discussed here in the future.]