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?