Z3 revised 6jul11 and 16jul11
\section{What you should study now.}
This lesson immediately follows lesson on Moebius Transformations. It is
an elaboration and discussion of Hvidsten's exposition. Hvidsten uses a
style that labels some propositions as a "Lemma", "Theorem", "Corollary",
depending on whether the proposition is primarily a helper (lemma) for
a major fact (theorem) which has easy consequences (corollaries). We will
simply refer to them as Hvidsten's Proposition 8.1 through 8.15. Thus,
the final observation in the previous lesson, which we called the
"Anatomoy of a Moebius Transformation", is HP8.1.
\subsection{Notation}
Henceforth, we shall abbreviate "Moebius transformation" to just "MT".
Moreover, $CR(a,b,c,d)$ is an abbreviations for
$\frac{a-b}{a-d}\frac{c-d}{c-b}$, and the two notations are used
ingterchangeably.
\section{Introduction}
In this lesson we introduce the \textit{ Cross Ratio}, which is a number
obtained by taking a ratio of ratios of comparable geometric quantities.
By a \textit{comparable geometric quantities} we mean objects whose
comparison may be expressed as a number. For example, the diameter and
a the circumference of a circle have the ratio $\pi$, and both objects are
lengths. The number $\pi$ is also the ratio of the area of a disk to
a square whose side is the radius. The Greeks would not have compared
the area of the disk to the length of radius, as we do today. 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.
\section{Motivation}
Recall from high school that the form $ax+by=c$ for the equation of a line
was easy to handle algebraically, but the geometrical
meaning of the paramaters (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,
as in $f(z) = CR(z;b,c,d).
\section{The Tripod Theorems}
Hvidsten's Propositions 8.2, 8.3, 8.4 (a lemma, theorem and corollary)
say, in words, that
\textbf{Three points determine the Moebius Transformation}.
Hvidsten's exposition is very clear and there is no need to repeat it here.
But you should write it, preferably from memory, into your Journal. It
is based on solving the quadratic equation by "completing the square" as
you have learned in high school.
Lesson on Cross Ratios
\textit{ $\C$ 2010, Prof. George K. Francis, Mathematics Department, University of Illinois} \begin{document} \maketitle
\section{What you should study now.}
This lesson immediately follows lesson on Moebius Transformations. It is
an elaboration and discussion of Hvidsten's exposition. Hvidsten uses a
style that labels some propositions as a "Lemma", "Theorem", "Corollary",
depending on whether the proposition is primarily a helper (lemma) for
a major fact (theorem) which has easy consequences (corollaries). We will
simply refer to them as Hvidsten's Proposition 8.1 through 8.15. Thus,
the final observation in the previous lesson, which we called the
"Anatomoy of a Moebius Transformation", is HP8.1.
\subsection{Notation}
Henceforth, we shall abbreviate "Moebius transformation" to just "MT".
Moreover, $CR(a,b,c,d)$ is an abbreviations for
$\frac{a-b}{a-d}\frac{c-d}{c-b}$, and the two notations are used
ingterchangeably.
\section{Introduction}
In this lesson we introduce the \textit{ Cross Ratio}, which is a number
obtained by taking a ratio of ratios of comparable geometric quantities.
By a \textit{comparable geometric quantities} we mean objects whose
comparison may be expressed as a number. For example, the diameter and
a the circumference of a circle have the ratio $\pi$, and both objects are
lengths. The number $\pi$ is also the ratio of the area of a disk to
a square whose side is the radius. The Greeks would not have compared
the area of the disk to the length of radius, as we do today. 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.
\section{Motivation}
Recall from high school that the form $ax+by=c$ for the equation of a line
was easy to handle algebraically, but the geometrical
meaning of the paramaters (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,
as in $f(z) = CR(z;b,c,d).
\section{The Tripod Theorems}
Hvidsten's Propositions 8.2, 8.3, 8.4 (a lemma, theorem and corollary)
say, in words, that
\textbf{Three points determine the Moebius Transformation}.
Hvidsten's exposition is very clear and there is no need to repeat it here.
But you should write it, preferably from memory, into your Journal. It
is based on solving the quadratic equation by "completing the square" as
you have learned in high school.
Question 1.
Solve $az^2 + bz + c =0, a \ne 0 $ by completing the square. Hint:
Write the equation $a(z^2 + 2 \frac{b}{2c} + ? ) = a(z - ?)^2 = ? $.
He then concludes that two Moebius transformation which three (or more)
\textit{fixed points} must be the identity. Since we know that an MT is
the composition of similitudes, we'll check this fact for each separately,
although this is no substitute for Hvidsten's exposition.
\textbf{Theorem: }
If similitude $f(z)=az+b$ has 3 fixed points then $a=1, b=0$, i.e. $f=\iota$.
\textbf{proof} Suppose we examine two fixed points $p,q$.
\begin{eqnarray*}
ap + b &=& p \\
aq + b &=& q \\
a(p-q) &=& (p-q) \\
a &=& 1 \\
b &=& 0 \\
\end{eqnarray*}
It is in the 4th step that we use the assumption $p\ne q$. So, in fact, a
non-trivial similitude can have at most one solution to $f(z)=z, for
$z\in \mathbb{C}$.$ We also know that
rotations and dilations fix the origin and no other $z \in \mathbb{C}$..
And translation have no fixed points at all in the planee.
Actually, a similitude does have one extra fixed point on the
Riemann sphere $\hat\mathbb{C}$, namely $\infty$.
Question 2.
Show that $a\infty + b = \infty$ using the method described earlier.
The inversion also has two fixed points. It exchanges $\infty$ and $0$, and
$\frac{1}{z} = z $ has solutions $ z = \pm 1 $. So, we conclude that the
component transformations with 3 fixed points are the identity. But what
about composition? It is possible for the composition of two transformations to
have 3 fixed points without either of the components to have this property.
Question 3.
Find two MTs, neither of which has 3 fixed points, but their composition does.
Thus, even though checking a property for an MT by checking it seperately
for each of its anatomical parts is often useful, it is not so here. The reason
is that such a property must be known to be \textit{ invariant under composition}.
That means, if two MTs have the property, their composition must be have it
also.
\subsection{Summary}
A nontrivial MT ( $f \ne \iota$) can have at most 2 fixed points. Translations
all have a single fixed point at $\infty$. All other MTs have two.
The quadratic formula itself is based on \textit{completing the square},
which works for complex numbers as well as it does for real numbers.
\subsection{the theorem and its corollary}
Since the MTs form a group, we can consider $g^{-1}\circ g$ for two MTs.
If $f$ and $g$ have the same values on the the same three points, then
this composition has 3 fixed points, and so is the identity. Hence $f=g$.
But there is a much more interesting version of these theorem, for which
we need cross ratios.
\section{Cross Ratios}
A second, and even more important form to express a Moebius transformation is
as a \textit{cross ratio} defined thus:
\begin{eqnarray*}
CR(p;q,r,s) = \frac{(p-q)}{(p-s)}\frac{(r - s)}{(r - q)}
\end{eqnarray*}
Be aware that we have defined the cross ratio differently that Hvidsten does.
In all work for this course you must use this definition. The reason for this
variance is explained below.
We next show that we can write a Moebius transformation either in the
cross ratio form or in the fractional linear form.
\begin{eqnarray*}
\mbox{So } f(z) &=& \frac{z-z_0}{z-z_2}\frac{z_1 - z_2}{z_1 - z_0} \\
\mbox{Let } m &=& \frac{z_1 - z_2}{z_1 - z_0} \\
\mbox{Then} f(z) &=& \frac{mz - mz_0}{z - z_2} \\
\mbox{So for } a &=& m \\
\mbox{and } b &=& - mz_0 \\
\mbox{and } c &=& 1 \\
\mbox{and } d &=& - z_2 \\
\mbox{We have } f(z) &=& \frac{az+b}{cz + d} \\
\end{eqnarray*}
But for the converse we shall be more clever. Note first that
\begin{eqnarray*}
CR(z_0;z_0,z_1,z_2) &=& 0 \\
CR(z_1;z_0,z_1,z_2) &=& 1 \\
CR(z_2;z_0,z_1,z_2) &=& \infty \\
\end{eqnarray*}
Question 4.
Verify that $CR(a;a,b,c)=0, CR(b;a,b,c)=1, CR(c;a,b,c)=\infty$ by subtsitution.
So, given $f(z)=\frac{az+b}{cz+d}$, we solve
\begin{eqnarray*}
f(z_0)=0 & \mbox{ for } & z_0 = -b/a \\
f(z_1)=1 & \mbox{ for } & z_1 = (d-b)/(a-c) \\
f(z_2)= \infty \implies 1/f(z_2) = 0 & \mbox{ for }& z_2 = -d/c \\
\end{eqnarray*}
and so decide that $ \frac{az+b}{cz+d} = CR(z,z_0,z_1,z_2)$, which you
should check by calculation to keep up your algebraic skill.
\subsection{The Tripod Theorem}
Suppose we specify two sets of three distinct points,
$(z_1,z_2,z_3)$ and $(w_1,w_2,w_3)$ and we solve the equation
\[ CR(z; z_1,z_2,z_3) = CR(w; w_1,w_2,w_3)\] for the
MT $w =f(z)$. Then $w_i = f(z_i), i=1,2,3$. Conversely,
if $f(z)$ is given to be a MT and we know its value on
three distinct points, we know it value on every point, because
we can solve the cross ratio equation.
This is the high point of this lesson. You must understand the proof
of this. But it involves nothing more than a lot of algebra. There
is also a more elegant argument than brute force calculations. But you
won't appreciate the latter until you have gone through the former. When
you have figured out an efficient way of doing the calculations, you
should put a clean version of it into your Journal.
\subsection{Comments for the Variance from Hvidsten}
The author is grateful to Steve Dugan for pointing out some errors.
Moreover, note that our definition of the cross ratio differs from
Hvidstens, who sends $z_1 \mapsto \infty $ and $z_2 \mapsto 1 $. I prefer
the mnemonic which matches the subscripts, except, perhaps,
we should have written $z_\infty$ for $z_2$ in $CR(z;z_0,z_1,z_\infty)$.
\section{Crossratios and the Moebius group}
We close this lesson with a profound theorem [Hvidsten Theorem 8.5]
which says that Moebius transformations preserve cross ratios. That is:
\textbf{ Cross Ratio Invariance Theorem}
If $f(z)$ is a MT then
\[(\forall a,b,c,d)(CR(f(a),f(b),f(c),f(d))= CR(a,b,c,d))\]
\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 $f(z)=\frac{1}{z}$.
Note that Hivdsten give a different proof, using the Tripod Theorem. You
should understand both proofs and write them into your Journal. Supply the
albraic details if you don't see them immediately.
\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{Applications the theorem.}
As observed above, Hvidsten provides a proof using the Tripod Theorem.
This is useful, because there are many more theorems that are easily
proved by the is property of MTs. However, an application of the
"anatomy lesson" also reappears [Hvidsten, Lemma 8.8], so it's useful
to see in working in a simpler 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 $w=-iz$.
\subsection{Geometric observation}
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.
Note that we have used another deep geometrial property of MTs, which we
take up in the next lesson.
\section {Correlation with Hvidsten, updated.}
Note that we have now discussed the first five of Hvidsten's fifteen
propositions. The next four deal with geometric properties of MTs in general.
The theorem themselves and also Hvidsten's proof are independently
interesting. But since they are not essential to understand the final
five propositions [Hvisten 8.11 -- 8.15] we offer an alternative lesson
on the geometry of cross ratios in general, which may be skipped on
your way to the final lesson. This is where we give the interpretation of
Euclid's primitives: point, line, angle, distance and congruence, as in
the Poincare disk modle of hyperbolic geometry.