Z2. last edited 5mar15

Lesson on Moebius Transformations of the Plane

\textit{ $\C$ 2010, 2015 Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{Introduction} In this lesson we introduce the concept of a \texit{Moebius Transformation}, which is a function on the field of complex numbers whose values can be calculated in terms of four complex parameters, $a,b,c,d$ and the arithmetic of complex numbers, \[ f : \mathbb{C} \rightarrow \mathbb{C} \mbox{ such that } f(z) = \frac{az+b}{cz+d} \] Note that the numerator and the denominator have the same form, they both are complex \textit{monomials}, i.e. polynomials of degree 1. As a transformation, $w = az + b$ should remind you of the equation of a straight line, $y = mx +b$ you learned in highschool. We shall study these as a special kind of Moebius transformation, namely those for which $c=0$ and $d=1$. We also write the abbreviation MT for ``Moebius transformation". Moreover, we require that a the parameters of a Moebius transformation satisfy the condition \[ ad - bc \ne 0. \] This eliminates from the ranks of MTs the constant functions, for example, when $a=c=0$. But it only elimiminates the constant functions, as you can easily prove.
Question 0.
Show by pure algebra that if $ \frac{az_1 +b}{cz_1 +d}= \frac{az_2 +b}{cz_2 +d} $ but $z_1 \ne z_2$, then $ad -bc =0$. Hint: Of course you'll need to use the field property of Complex numbers, namely that $pq =0 \Rightarrow (p=0 \vee q=0)$ On the other hand, if we take $c=0$ but $d \ne 0$ and we are, in effect discussing functions of the form $f(z) = a_1 z + b_1 $ where $a_1=a/d, b_1 = b/d$. We drop the subscripts. While some authors call such Moebius transformations by the name of \textit{similitudes}, it's not a good term because it is easily confused with the much more general concept of similarity. An older alternative is the adjective \textit{linear}, which is even worse because not every linear transformation of the plane is a MT, and translations ($a = 1$) are not linear. Despite this, MTs were once called \textit{linear fractional}. This does not seem to be a good alternative either. So we'll call them \textit{similitudes} here . For the case $|a| =1$, the MT becomes $f(z) = e^{i\theta}z + b$. Thus we see that such a MT is a rotation of the plane followed by a translation. This is called a \textit{Euclidean Motion} or an \textit{isometry} of the plane, and constitute the mathematical concept Euclid had in mind as the concept of \textit{congruence}. The plan for the lesson is to study the transformation group of all MTs, and then certain subgroups in detail, in particular the \textit{ Euclidean} group for the Euclidean plane, and then the \textit{hyperbolic} group of the models of non-Euclidean geometry. \section{Algebraic-Geometric Properties of MTs} \subsection{Unitary Definition of a Moebius Transformation} Also, note that the four complex numbers defining a Moebius transformaton are by no means unique. Being a fraction, the values of the function do not change if we replace $a,b,c,d$ by multiples of the same non-zero complex factor. In particular, if we divide through by one of the two squareroots of the non-zero complex number, $\sqrt{ad-bc}$, we obtain a set of parameters for which $a_1 d_1 - b_1 c_1 = 1$. \subsection{Point Transformations of the Plane} Since each complex number is the name of a point in the plane and vice versa, we can interpret a bijection of the complex numbers as a \textit{point transformation of the plane.} That is, each point is transformed into a unique other point (which may or may not be itself.) Recall from MA347 (or the equivalent) that for a function to be \textit{bijective} it is both 1:1 and onto. This is a generalization from increasing (or decreasing) continuous functions on the reals which you studied in the calculus. Since complex numbers cannot be ordered, there is no easy way of checking that a function is bijective.
Question 1.
Show that $f(z)=z^2$ is onto (surjective) but no 1:1 (injective). Hint: For the former you must find at least one solution $z$ to the equation $w=z^2$ for \textit{every} $w$. For the latter it suffices to find two different solutions for \textit{one} particular value of $w$. \subsection{The Group of Similitudes of the Plane} Also recall from MA347 that having an inverse functions is the same as being a bijection. Thus similitudes are bijective, because for every $w$, the equation$w=az+b$ can be solved uniquely for $z$. Thus similitudes are closed under taking their inverses. Recall also that bijections on a set form a group under composition. The identity transformation is the compositional identity. So, to show that the similitudes form a group it remains only to demonstrate their closure under composition. A purely algebraic way of doing this is to Substitute the value of the first similitude $f_1(z) = a_1z +b_1$ into the argument of the second $f_2(z) = a_2z +b_2$ and then simplifying $ a_2(a_1z+b)+b_2$ to look like $f_3(z)= (f_2 \circ f_1)(z) = a_3z +b_3$. Thus similitudes are closed under composition, and together they form a transformation group of the plane.
Question 2.
Calculate the inverse of a similitude, and then show directly that the composition of a similitude with its inverse is the identity. \subsection{The Group of Moebius Transformations} An easy computation similarly shows that the Moebius transformations form a group (but not of the plane, as we shall see.) First, solve $w=f(z) = \frac{az+b}{cz+d}$ for $z$ in terms of $w$, as $z = \frac{dw-b}{-cw+a}$. Put your calculation into your Journal. Thus $f^{-1}(z) = \frac{dz-b}{-cz+a}$ and check that $da - (-b)(-c) =1$ still. Review the introduction to recall why we need to make this check.
Question 3.
For $f(z)=\frac{az+b}{cz+d}$ and $g(z)=\frac{Az+B}{Cz+D}$, calculate their composition $h(z)=\frac{\alpha z+\beta}{\gamma z+\delta}$ by calculating explicitly what the parameters $\alpha, \beta, \gamma, \delta $ are in terms of $a,b,c,d,A,B,C,D$. And then show that $\alpha \delta - \beta \gamma = 1$. \subsection{The Extended Complex Plane} The previous subsection illustrates how easy it is to fall into a gap in mathematics. Nowhere did we check that every MT is a bijection of the plane. It is not. For consider the MT with $a=b=0, b=c=0$, whose value is \textit{multiplicative inverse} of complex number, $f(z)=\frac{1}{z}$, more commonly called the \textit{reciprocal} of $z$. It is undefined at $z=0$. So we define it by adjoining one point, $\infty = \frac{1}{0}$, to $\mathbb{C}$ to obtain the \textit{Extended Complex Plane} $\hat{\mathbb{C}}$ and extend the arithmetic as follows. \begin{eqnarray*} \frac{1}{0} &=& \infty \\ \frac{1}{\infty} &=& 0 \\ \infty \pm b &=& \infty \\ \infty b &=& \infty \\ \frac{a \infty + b}{c \infty + d}&=& \frac{a}{c} \\ \end{eqnarray*} The way to understand this strange arithmetic is, in each case, to subsitute $\frac{1}{z}$ for each occurence $\infty$ and take the $ \lim_{z -> 0}$. It is instructive to illustrate this method for the last example when applying a MT to $\infty$. \begin{eqnarray*} \mbox{ To find: } \frac{a\infty +b }{b\infty +c} && \mbox{ substitute }\\ \frac{ a/z + b}{c/z + d} &=& \frac{a + zb}{c + zd} \\ \lim_{z->0} \frac{ a + zb}{c + zd} &=& \frac{a}{c}\\ \end{eqnarray*} \subsection{The Moebius Group and the Special Linear Group} Thus we can now confidently assert that the MTs form the \textit{Moebius Group} (MG). So far we have used only the original \textit{representation} of a MT as $f(z) =\frac{az+b}{cz+d}, \, ad-bc=1$. If you have had matrix algebra you will have recognized that both the inverse and composition of MTs you computed above (you did, didn't you?) has the same algebraic form as 2x2 matrix inverses and compositions. For example \[\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)^{-1}= \left( \begin{array}{cc} d & -b \\ -c & a \\ \end{array} \right) \] And composition becomes matrix multiplication. Thus we have a second representation of the MG as the group of complex 2x2 matrices with unit determinant, called the \textit{ Special Linear Group } $SU(2,\mathbb{C})$ and important in modern physics. \section{Geometric Properties of Moebius Transformations} \subsection{Anatomy of a MT} It is clear that to calculate the value of an MT all you need to know is how to add, subtract, multiply and divide complex numbers. From a transformational viewpoint, a MT is the composition of simpler MTs which go by special names. We have already met similitudes, reciprocals, and Euclidean transformations. For example, the anatomy of a similarity may be anlyzed as follows. Since we can write $a = re^{i\theta}$ we see that a similitude, $f(z)= re^{i\theta} z + b $, is just the composition of three familiar transformations, $f=f_3 \circ f_2 \circ f_1$, where: \begin{itemize} \item The \textit{rotation} $ f_1(z) = e^{i\theta}z $ \item is followed by the \textit{ dilation } $ f_2(z) = rz$ \item which is followed by the \textit{ translation } $ f_3(z)= z + b$. \end{itemize} The functional notation is often more conveniently replaced by the \textit{ maps-to} notation, as in \[ z \mapsto e^{i\theta}z \mapsto r(e^{i\theta}z) \mapsto (re^{i\theta}z) + b \] While you should have no difficulty doing a similar analysis of a MT (try it, and put it into your Journal), a more interesting exercise is to ask what the simplest such dissection might be. Of course ``simplest" is a matter of taste, but the most elegant is easier to determine.
Question 4.
Verify algebraically that \[\frac{az+b}{cz+d} = \frac{a}{c} - \frac{ad-bc}{c^2}\frac{1}{z+ d/c}\] and conclude that every MT is the composition of a translation, followed by an inversion, followed by a similarity. Indentify each of these three transformations. Surely this is an elegant solution of what the anatomy of a MT is. But it also has, as an application, a much easier, if longer, proof that the MTs form a group. Each separate component type MTs form very familiar groups themselves. \begin{itemize} \item The translations are the group of vectors in the the plane. \item The rotations (about the origin) are the group of reals modulo $2\pi$. \item The dilations (from the origin) are the multiplicative group of positive reals. \item The identity and inverse, $ \mathbb{Z}_1 = \{ \iota, \, \rho(z)=\frac{1}{z}\} $ is the group of order 2 under composition. \end{itemize} Since compositions of bijections of are bijections, and thus have an inverse, so do MT without the paint of calculating it explicitly. Even the composition theorem could be derived from this anatomical dissection. But it requires the lemma in each case, that the composition of a MT with each one of the four elementary transformations is again an MT. Can you prove this? Its a good training exercise to see if you understand the above. \subsection{The Riemann Sphere} Originally we extended the complex numbers to $\hat{\mathbb{C}}$ by closing it up into a sphere. You may remember from high school geography, that in addition to the familiar \textit{Mercator Projection} there is the \textit{stereographic map projection} of the earth to a plane, for example from the North Pole. This is a geometric demonstration of this algebraic trick, except that on the map there is no place for the North Pole itself. This is considered to be that point at $\infty$ extending the plane to the sphere. This is not an unusual mathematica process. The complex numbers were an algebraic extension of the reals which served, initially, the purpose of solving quadratic equations. Geometrically, the Cartesian plane and its vectors became a realistic picture of these numbers, just as the number line is a convenient picture of the reals for school children. Now, the Euclidean sphere itself has a natural group of transformations, namely the rotations about an axis, any axis. In a later lesson we shall study how to express rotations as MTs, and how not all MTs are rotations. In a subsequent lesson, we shall define the hyperbolic group as the subgroup of MT which map the unit circle into itself, the unit disk onto itself, and therefore the exterior of the unit disk onto itself. But, of course we want to know how to tell which MTs are hyperbolic. Finally, there is one transformation of complex numbers we have already met which is \textit{not} an MT, namely conjugation, $z \mapsto \bar{z}$. Since this is the reflection of the plane in the real axis, it reverses orientations, an angle is mapped into the negative of itself, a picture of a left hand into a picture of a right hand. And, as we shall see later, MT preserve orientation. Another transformation we have already seen is inversion in a circle. Recall that in vector form, $X \mapsto \bar{X}/|X|^2$ translates into the transformation $z \mapsto \frac{1}{\bar{z}} $. Being a reflection, we already knew circle inversions reverse orientation.