## Remarks Concerning Certain Homework Problems

\textit{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle \section{Regarding Homework Problem 8.2.14} Unhappily, it appears that no-one in the class got this problem right. Almost everyone ignored one part or another of the hypothesis, and did whatever was convenient about those absolute value signs. Many of you ignored the class notes for W12 which begins with the explicit statement "This lesson replaces Theorem 8.11 in the book." Hvidsten's exposition is based on prior work which we did not do in this class. Moreover, here the difference between Hvidsten's cross ratio $(a,b,c,d)$ and our cross ratio $CR(a,b,c,d)$ really matters. \subsection{What should I remember most about this?} The formula $\frac{a-b}{a-d}\frac{c-d}{c-b}$ is entirely unambiguous. But there is a choice whether to write this as $CR(a,b,c,d)$ as we did, or as $(a,c,b,d)$ as Hvidsten, p. 317 chose to do it. Our choice is again explained after the solution. And a guess about why Hvidsten made his choise is given even further down. If your journal has a mixture of these, you should make the appropriate corrections as you use the journal to study for the final. At any event, on a test, it is wise to make it clear exactly what you mean by the abbreviation $CR(a,b,c,d)$. And watch the subscripts, if there are some. \section{Complete solution in utter detail.} I will first give a correct solution. And then discuss how your reviewing this section of the course can profit from the above errors. \subsection{Did you use the hypothesis?} The hypothesis, that $z_1$ lies \textit{between} $z_0, z_2$ cannot be ignored. Consider a counterexample: If $ z_0 = 0, z_1 = 2, z_2 = 1 $ then even for Euclidean distance $ 1 = d(0,1) \ne 3 = d(0,2) + d(2,1).$ \subsection{Normalize the problem.} In the notes on this section we repeatedly move the circline bearing the 3 points to the diameter by an isometry (so the distances do not change). So move $z_0 \mapsto 0$ by a hyperbolic translation. If if the diameter happens to be at a slant, a rotation (about the origin) will move it to the real diameter, namely so that $z_1 \mapsto r, z_2 \mapsto t$ and now the hypothesis $0 < r < t < 1$ comes into play. \subsection{Getting rid of the absolute values.} For $0 < r < 1$, we evaluated the crossratio and found that $d_H(0,r) = log \frac{1+r}{1-r} $. The absolute values bars disappear because $ 1 < \frac{1+r}{1-r} $. Since we calculated this one in the notes, let us calculate the second distance: \begin{eqnarray*} d_H(r,t)&=& | log CR(r,-1,t,+1)| \\ &=& | log \frac{r+1}{r-1}\frac{t-1}{t+1}| \\ &=& | log \frac{1+r}{1-r}\frac{1-t}{1+t}| \\ &=& - log \frac{1+r}{1-r}\frac{1-t}{1+t} \\ &=& + log \frac{1-r}{1+r}\frac{1+t}{1-t} \\ \end{eqnarray*} If you don't get that minus sign in line 4 right, you would have a reciprocal in the last line and the laws of logarithms wouldn't work as they need to thus: \begin{eqnarray*} d_H(0,r) + d_H(r,t) &=& log \frac{1+r}{1-r} + log \frac{1-r}{1+r}\frac{1+t}{1-t} \\ &=& log \frac{1+r}{1-r}\frac{1-r}{1+r}\frac{1+t}{1-t} \\ &=& log \frac{1+t}{1-t} \\ &=& d_H(0,t). \\ \end{eqnarray*} And that minus sign in the 4th line itself requires a simple proof. Namely, if $ 0 < r < t < 1 $ then which is bigger? You might guess (hope?) that \begin{eqnarray*} \frac{1+r}{1-r} & < ? & \frac{1+t}{1-t} \\ 1 + r - t - tr & < ? & 1 + t - r - rt \\ 2r & < ! & 2t \\ \end{eqnarray*} and so it is. Now this problem has been solved in the way we studied this material. \section{So, what about using the textbook?} I explained at some length that we used a different definition of the cross ratio symbol than Hvidsten does. For the final you'll need only understand and use the one we developed in class. Read your class notes! And you can safely ignore what follows here. However, in all fairness, I will go over the differences. There are no errors in either approach. I think Hvidsten's is more likely to lead to confusion than mine. But using both at once is guaranteed to get confusing. \subsection{The mnemonic} For present purposes, I will use $CR(a,b,c,d)$ and $HV(a,b,c,d)$. If $f(z) = CR(z,z_0,z_1,z_\infty) = \frac{z-z_0}{z-z_\infty}\frac{z_1 - z_\infty}{z_1-z_0}$ takes $ z_0 \mapsto 0, z_1 \mapsto 1, z_\infty \mapsto \infty$ then Hvidsten chose the order of his cross so that the same function would be written $ f(z) = HV(z,z_1,z_0,z_\infty) $. To avoid this transposition I chose the particular order we did. Moreover, in Hvidsten's distance formula, the one we did \textit{not} adopt, he uses a Ruler on the Poincare Line $\ell_{z_0, z_1}$ which orients it from $z_1$ to $z_0$, while our definition keeps the natural order from the first mentioned point to the second mentioned point. (I hope you noticed the capital R. This ruler is non-Euclidean, but it still has the additive property on Lines. That's what this problem is about.) The reason there is no contradiction is absolute values around the logarithms. If you compare the two cross ratios they differ by being reciprocals. But $| log \frac{n}{d} | = | - log \frac{d}{n} | = | log \frac{d}{n} |.$ Finally, you can't just ignore the absolute values when you want to take the "sum of two logarithms" to the "logarithm of the product". You have to make sure that you're working with positive logarithms i.e. that they are logarithms of real numbers \textit{bigger} than $1$. And that's were the order of the three points matters. This is stated in Hvidsten's proof of Theorem 8.11, but most students miss that hypothesis and get it wrong on an application. To summarize, there are two issues here. First, you don't take logarithms of complex numbers (though it can be defined, we don't take $log (1+i)$, for example.) So, we make sure that the cross ratios we take the logarithm of come out to be real. And that is, if and only if the four points lie on a circline. Second, when measuring a distance we want the logarithms to be positive so we can ignored those absolute value signs. For that the real number you take the logarithm of needs to be greater than 1. That's where the order of $z_0,z_1,z_2$ matters. Hvidsten reverses the ruler, I think, so that for $0 < r < 1$ we have $ d_H(0,r) = log \frac{1+r}{1-r} = log CR(0,+1,r,-1) = HV(0,r,1,-1) $ comes out directly, whereas we had to really use the absolute values or distance would come out negative. The moral of this story is that when there is a deviation from the approach taken in the textbook, try not to serve two masters, something won't come out right.