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

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?