Z3 last edited 14mar15

$\C$ 2010, 2015 Prof. George K. Francis, Mathematics Department, University of Illinois.## Lesson on Crossratios and the Tripod Theorems

\begin{document} \maketitle \section{Introduction} In this lesson we introduce the \textit{ Crossratio}, which is a number obtained by taking a ratio of ratios of comparable geometric quantities. We shall define an alternative representation of the a Moebius Transformations (MT) in terms of the cross ratio. We shall that a TM is completely determined by its values on three points, any three points. Finally, we draw some useful geometric consquences from this. \subsection{Ratios} The Greeks based their geometry not on real numbers, as we do now, but on \textit{comparable geometric quantities}. Here is a familiar example. The circumference and the diameter of a circle have the ratio $\pi$. The Greeks knew that for avery circle this ratio was the same, but they did not consider it a number, like $49$ or $\frac{4}{9}$. The number $\pi$ also measures the ratio of the area of a disk to that of a square whose side is a radius of the disk. We might express this as a \textit{proportion}, in modern notation: \begin{eqnarray*} \frac{c}{d}&=&\frac{D}{S} \\ c&=& \mbox{ circumference} \\ d&=& \mbox{ diameter} \\ D&=& \mbox{ disk area} \\ S&=& \mbox{ squared radius } \\ \end{eqnarray*} But the Greeks would not have rearranged the proportion and compared the area of a disk to the length of its radius, and called it half of its circumference. That would have been a comparison of two different kinds of quantities, an area and a length in this case. Again, in the calculus you learned that two vectors are not comparable unless they happen to be parallel. Insofar as a vector expresses the difference of two points, and once we have identified points with complex numbers, the ratio of two vectors in the plane \textit{can} be made sense of as a complex number. Althrough the \textit{crossratio} was originally defined as a \textit{ratio of ratios} by Pappus, we prefer to write it as a product of two ratios of displacement vectors, thus \[ CR(p,q,r,s) = \frac{p-q}{p-s}\frac{r-s}{r-q} \] where $p,q,r,s$ are four distinct extended complex numbers (including $\infty$). The following exercise is important in algebra, but here, it serves only to help you remember the definition of the crossratio.Question 1.Show that, of the 24 permutations of the 4 terms in the crossratio, only six crossratios have different values. First show that \[ \lambda = CR(p,q,r,s) = CR(q,p,s,r) = CR(s,r,p,q)=CR(r,s,q,p) .\] The show that \[ \frac{1}{\lambda} ,\; \frac{1}{1-\lambda}, \; 1-\lambda, \; \frac{\lambda}{\lambda-1} ,\; \frac{\lambda-1}{\lambda} \] expresses the other five different values of the permuted crossratios. \subsection{Crossratio form of a Moebius Transformation} Recall from high school that the form $ax+by+c=0$ for the equation of a line was easy to handle algebraically, but the geometrical meaning of the parameters (or, more properly, their ratio $a:b:c$) was hard to remember. (Do you?) Rewriting (non-vertical) lines as $y=mx+b$ is more informative because $m$ is the slope and $b$ the y-intercept. So too, the form $w = \frac{az+b}{cz+d}$ for a Moebius tranformation is easy to handle algebraically, but the geometrical meaning of $a,b,c,d$ is not so easy to determine, even if we normalize them to $ad-bc=1$. The alternative studied here is to express a MT as function of one of the four points in a cross-ratio, keeping the other three points as parameters for the function. When convenient, we shall mnemonic names of the three parameters, as in $f(z) = CR(z;z_0,z_1,z_\infty)$. The next exercise explains:Question 2.Show that if $z_0 \ne z_1 \ne z_\infty \ne z_0$ then \[ CR(z;z_0,z_1,z_\infty) = \frac{az+b}{cz+d} \] has solution \[a = \frac{z_1 - z_\infty}{z_1-z_0} ,\; b = -az_0 ,\; c = 1 ,\; d = -az_\infty \] with $ad-bc \ne 0$ Conversely, you could solve for $z_0 \; z_1\; z_\infty$ in terms of $a \; b \; c \; d$ algebraically, but there is a less messy way by simply observing that $f(z)=\frac{az+b}{cz+d}$ sends \[ f(-\frac{b}{a}) =0 \; f(\frac{b-d}{c-a}) =1 \mbox{ and } f(-\frac{d}{c})=\infty . \] So does the MT $CR(z;z_0,z_1,z_\infty)$ if we set \[ -\frac{b}{a} =z_0 \; \frac{b-d}{c-a} =z_1 \mbox{ and } -\frac{d}{c} = z_\infty .\] To conclude from just these three cases that the two MTs are equal on every point requires the Tripod Theorem. \section{The Tripod Theorems} We will next prove that the value of a MT on three points completely determines the transformation. We do this in three steps, using the notion of \textit{fixed points} of an MT. By a fixed point of a MT we mean any solution of the equation $z = f(z)$. Since a translation $ w = z + b,\; b \ne 0$ moves every point in the plane, and an MT is a bijection, its only fixed point must be $\infty$. And indeed, $\infty = \infty + b$. This is also the case for every similarity $f(z) = az +b $ whose only fixed point must be $\frac{b}{1-a}$.Question 3.Show that $\infty = a\infty + b$ implies that $a = 1$. Hint: Replace $\infty$ by $\frac{1}{z},\; z \ne 0$ and take a limit $z \rightarrow 0$. But even if $c\ne 0$ then equation $z = \frac{az+b}{cz+d}$ is equivalent to a quadratic equation, which over the complex numbers always has a solution, but it has most two solutions ( unless it is the identity, of course.)Question 4.Solve $ z = \frac{az+b}{cz+d}$ when $ c \ne 0 $ by completing the square. Hint: Compare with the quadratic equation you remember from high school. It would be tempting to use the anatomy lesson to show that each elementary MT constituent of a MT has at most two fixed points. For examples $f(z)=\frac{1}{z}$ has the two fixed points $\pm 1$. But having a certain number of fixed points is not a property that is preserved under composition.Question 5.Find two MTs, neither of which has three or more fixed points, but their composition does. \subsection{The Tripod Theorem, Second Form} Since the MTs form a group, we can consider the MT $g^{-1}\circ f$ for two given MTs $f,g$. If $f$ and $g$ have the same values on the same three points, then composition $g^{-1}\circ f$ has 3 fixed points, and so is the identity. From $g^{-1}\circ f =\iota$ follows that $f=g$. \subsection{The Tripod Theorem, Third Form} Suppose we specify two sets of three distinct points $(z_0,z_1,z_2)$ and $(w_0,w_1,w_2)$ and want to find the MT $f(z)$ for which $f(z_i)=w_i, i=0,1,2$. Consider the MT $g(z) = CR(z; z_0,z_1,z_2)$ and $h(w) = CR(w; w_0,w_1,w_2)$. Since \[ g(z_0) = 0 = h(w_0),\; g(z_1) = 1 = h(w_1) \mbox{ and }g(z_2) = \infty = h(w_2) \] we see that $f= h^{-1}\circ g$ does what is required. Hence to actually find this MT we merely have to solve \[ \frac{z-z_0}{z-z_2}\frac{z_1-z_2}{z_1-z_2} = \frac{w-w_0}{w-w_2}\frac{w_1-w_2}{w_1-w_2}\] algebraically for $w=f(z)$. \section{Crossratios and the Moebius group} We close this lesson with the important property of crossratios, namely that its value does not change if all four of its arguments undergo the same Moebius transformation. \textbf{ Crossratio Invariance Theorem} If $f(z)$ is a MT then \[(\forall a,b,c,d)(CR(f(a),f(b),f(c),f(d))= CR(a,b,c,d))\] We give two proofs here, one which does use the decomposition of an MT into the composition of a translation, followed by an inversion, followed by a similarity. \textbf{ Proof:} Here we use the "anatomy" of a MT as discussed in the lesson on Moebius transformations. Because MTs form a group under composition, we need to prove this theorem only for similarities and for the reciprocal $f(z)=\frac{1}{z}$. \subsection{The case for the reciprocal.} Note that writing out the LHS we have \[ \frac{\frac{1}{a} - \frac{1}{b}}{\frac{1}{a} - \frac{1}{d}} \frac{\frac{1}{c} - \frac{1}{d}}{\frac{1}{c} - \frac{1}{b}} = \frac{b-a}{d-a}\frac{d-c}{b-c} = \frac{a-b}{a-d}\frac{c-d}{c-b} \] which is the RHS. \subsection{The case for a translation} Follow the same steps as before. But note, every numerator and every denominator is the difference of two complex numbers. But we know that translations preserve displacement vectors: \[(\forall z,w,m)( (z+m) - (w+m) = z - w )\] \subsection{The case of complex multiplication} The only case remaining is a dilation and a rotation about the origin, which is just a multiplication by a nonzero comples number. The distributive law of multiplication causes the multiplier to cancel out of the two fractions, leaving the cross ratio unchanged. \subsection{Immediate Proof by the Tripod Theorem} Suppose we want to prove that $CR(p,q,r,s) = CR(f(p),f(q),f(r),f(s))$. Consider the MT $g(z)=CR(z,q,r,s)$ and consider this composition $h=g\circ f^{-1}$. We see that \[ h(f(q))=0 , h(f(r))=1 \mbox{ and } h(f(s))= \infty .\] But the MT $CR(z,f(q), f(r), f(s))$ does exactly the same thing. Hence $h(z) = CR(z,f(q), f(r), f(s))$ for all $z$, in particular for $z=f(p)$. But, substituting and unpacking the definitions, \[ CR(f(p),f(q),f(r),f(s)) = h\circ f(p)= g(p) = CR(p,q,r,s) .\] So, whether you prefer the anatomical proof or the tripodal proof, it's a theorem. \subsection{Applications the Invariance Theorem.} This theorem has many applications, but we give a specific example. \textbf{Problem: } Solve $ CR(w,-1,0,1) = CR(z,-i,0,i) $ for $w=f(z)$ without using the definition of cross ratios. \textbf{Solution: } \begin{eqnarray*} CR(w,-1,0,1) &=& CR(z,-i,0,i)\\ &=& CR(i(z),i(-i),i(0),i(i))\\ &=& CR(iz,1,0,-1)\\ &=& CR(-iz,-1,-0,--1)\\ &=& CR(-iz,-1,0,1)\\ \end{eqnarray*} We work on the RHS only. The object is to get the three parameters of the two MTs to match exactly. Since ($z \mapsto iz$) is a MT, we apply it to all four terms in the second equation. In the fourth we note that we're off by a minus sign for the three paramaters. Since ($ z \mapsto -z$) is also a MT, we proceed to the last step. Both sides now express the value of the \textit{ same } MT (written in a cross ratio format} on two points $w$ and $-iz$. Since the two values are the same, the two points must be the same (MTs are bijective), so $w=-iz$. This is just an algebraic illustration. We can confirm it geometrically, because multiplying by $-i$ is just a rotation by $-90^o$. Since the LHS is an MT that maps the real axis back to itself, with $-1,0,1$ going to $0,1,\infty$, and the RHS maps the y-axis to the x-axis, the solution $w=f(z)$ must map the x-axis to the y-axis. And the rotation does just that. This is not necessarily any easier than the algebraic solution of such a problem but the method has other applications. So give the next problem a try.Question 6.Solve $CR(w, -1,0,1) = CR(z,-i, \infty, i) $ for $w=f(z)$ using both methods discussed above. Hint: Which MT takes $0 \mapsto \infty$ but leaves the unit circle fixed?