# The Cross Ratio Tool and the Koebe Principle

Let $j \subseteq t$ be intervals and let l, r be the components of $t \setminus j$. Then the Cross Ratio is defined as

$C(t,j) = (|t| \cdot |j|) / ( |l| \cdot |r|).$

Assume g is a $C^{3}$ monotone function on the interval t, and g(t)=T, g(j)=J, g(l)=L, g(r)=R. Then define

$B(g,t,j)=\frac{C(T,J)}{C(t,j)} = \frac{|T|\cdot |J|}{|L| \cdot |R|} \cdot \frac{|l|\cdot |r|}{|t|\cdot |j|}.$

Define the Schwarzian Derivative for $C^{3}$ function g,

$Sg(x)=\frac{D^{3}g(x)}{Dg(x)} -\frac{3}{2}(\frac{D^{2}g(x)}{Dg(x)})^{2}.$

Proposition 1. Assume f and g are $C^{3}$ functions, then

$S(f\circ g)(x)= Sf(g(x)) \cdot |Dg(x)|^{2}+ Sg(x).$

$S(f^{n})(x)= \sum_{i=0}^{n-1}(Sf(f^{i}(x)) \cdot |D(f^{i})(x)|^{2}.$

Proposition 2. If $f(x)=x^{\ell}+c$ for some $c\in \mathbb{R}$ and $\ell \geq 2$, then $Sf(x)<0$ for all $x \neq 0$.

Proposition 3. Minimum Principle.

Assume $I=[a,b]$, $f: I \rightarrow \mathbb{R}$ is a $C^{3}$ diffeomorphism with negative schwarzian derivative, then

$|Df(x)| > \min \{|Df(a), |Df(b)|\} \text{ for all } x \in (a,b).$

Theorem 1. Real Koebe Principle.

Let Sf<0. Then for any intervals $j \subseteq t$ and any n for which $f^{n}|t$ is a diffeomorphism one has the following. If $f^{n}(t)$ contains a $\tau-$scaled neighbourhood of $f^{n}(j)$, then

$(\frac{\tau}{1+\tau})^{2} \leq \frac{|Df^{n}(x)|}{|Df^{n}(y)|} \leq (\frac{1+\tau}{\tau})^{2} \text{ for all } x, y \in j.$

Moreover, there exists a universal function $K(\tau)>0$ which does not depend on f, n,  and t such that

$|l| / |j| \geq K(\tau),$

$|r| /|j| \geq K(\tau).$

Theorem 2. Complex Koebe Principle

Suppose that $D \subseteq \mathbb{C}$ contains a $\tau-$scaled neighbourhood of the disc $D_{1} \subseteq \mathbb{C}$. Then for any univalent function $f: D \rightarrow \mathbb{C}$ one has a universal function $K(\tau)>0$ which only depends on $\tau>0$ such that

$1/K(\tau) \leq \frac{|Df(x)|}{|Df(y)|} \leq K(\tau) \text{ for all } x, y \in D_{1}.$

Theorem 3. Schwarz Lemma (Original Form)

Assume $\mathbb{D}=\{ z: |z|<1\}$ is the unit disc on the complex plane $\mathbb{C}$, $f: \mathbb{D} \rightarrow \mathbb{D}$ is a holomorphic function with $f(0)=0$. Then $|f(z)|\leq |z|$ for all $z \in \mathbb{D}$ and $|f^{'}(0)| \leq 1.$ Moreover, if $|f(z_{0})|=|z_{0}|$ for some $z_{0}\neq 0$ or $|f^{'}(0)|=1,$ then $f(z)= e^{i\theta} z$ for some $\theta \in \mathbb{R}.$

Corollary 1.

Assume $\mathbb{D}$ is the unit disc on the complex plane $\mathbb{C}$, and $f: \mathbb{D} \rightarrow \mathbb{D}$ is a holomorphic function, then

$|\frac{f(z)-f(z_{0})}{1-\overline{f(z_{0})}f(z)}| \leq |\frac{z-z_{0}}{1-\overline{z}_{0}z}| \text{ for all } z, z_{0} \in \mathbb{D}.$

$\frac{|f^{'}(z)|}{1-|z|^{2}} \leq \frac{1}{1-|z|^{2}} \text{ for all } z \in \mathbb{D}.$

Corollary 2.

Assume $\mathbb{H}$ is the upper half plane of the complex plane $\mathbb{C}$, $f: \mathbb{H} \rightarrow \mathbb{H}$ is a holomorphic map. Then

$\frac{|f(z_{1})-f(z_{2})|}{|f(z_{1})-\overline{f(z_{2})}|} \leq \frac{|z_{1}-z_{2}|}{|z_{1}-\overline{z_{2}}|} \text{ for all } z_{1}, z_{2} \in \mathbb{H} .$

$\frac{|f^{'}(z)|}{\Im{f(z)}} \leq \frac{1}{\Im{z}} \text{ for all } z\in \mathbb{H} .$

Corollary 3. Pick Theorem

The hyperbolic metric on $\mathbb{D}$ is $\rho(z)= \frac{2}{1-|z|^{2}}dz$, assume $d(z_{1}, z_{2})$ denotes the hyperbolic distance between $z_{1}$ and $z_{2}$ on $\mathbb{D}$. Assume $f: \mathbb{D} \rightarrow \mathbb{D}$ is a holomorphic function, then

$d(f(z_{1}), f(z_{2}))\leq d(z_{1}, z_{2}) \text{ for all } z_{1}, z_{2} \in \mathbb{D}.$

Moreover, if $d(f(z_{1}), f(z_{2}))= d(z_{1}, z_{2})$ for some points $z_{1}, z_{2} \in \mathbb{D}$, then $f \in Aut(\mathbb{D})$, where

$Aut(\mathbb{D})=\{e^{i\theta}\frac{z-z_{0}}{1-\overline{z_{0}}z}: \theta \in \mathbb{R}, z_{0} \in \mathbb{D}\}.$

Background in hyperbolic geometry

Define

$\mathbb{C}_{J}=(\mathbb{C} \setminus \mathbb{R}) \cup J$

where $J \subseteq \mathbb{R}$ is an interval. It is easy to show that $\mathbb{C}_{J}$ is conformally equivalent to the upper half plane and define $D_{k}(J)$ as

$D_{k}(J)= \{ z: \text{the hyperbolic distance to J is at most k} \}.$

k is determined by the external angle $\alpha$ at which the discs intersect the real line. Moreover, $k=\ln \tan( \frac{\pi}{2}- \frac{\alpha}{4}) .$ Define

$D_{*}(J)=D(J,\frac{\pi}{2}) .$

Corollary 4. (NS) Schwarz Lemma

(1) Assume $G: \mathbb{C}_{I} \rightarrow \mathbb{C}_{J}$ is a holomorphic map, then $G((D_{*}{I})) \subseteq D_{*}(J).$

(2) Assume $F: \mathbb{C} \rightarrow \mathbb{C}$ is a real polynomial map, its critical points are on the real line. Assume $F: I \rightarrow J$ is a diffeomorphism, then there exists a set $D \subseteq D_{*}(I)$ such that $D\cap \mathbb{R} =I$ and

$F: D \rightarrow D_{*}(J)$ is a conformal map.

Corollary 5.

Assume $f: D \rightarrow \mathbb{C}$ is a univalent map and D contains $\tau-$scaled neighbourhood of $D_{1},$ and assume f maps the real line to the real line. For each $\alpha \in (\pi/2, \pi)$ there exists $\alpha^{'} \in (\alpha, \pi)$ such that if J is a real interval in $D_{1}$, then

$f(D(J,\alpha)) \supseteq D(f(J), \alpha^{'}).$

The Hyperbolic Metric On the Real Interval and Cross Ratio

As far as we know, the hyperbolic metric on the unit disc $\mathbb{D}=\{|z|<1\}$ is

$\rho_{D}(z)= \frac{2}{1-|z|^{2}}|dz| \text{ for all } z\in \mathbb{D}.$

Then the restriction to the real line is

$\rho_{I}(x)=\frac{2}{1-x^{2}} dx \text{ for all } x \in I=(-1,1).$

Moreover, from it, we can deduce the hyperbolic metric on the real interval $I=(a,b)$ is

$\rho_{I}(x)=\frac{b-a}{(x-a)(b-x)} dx \text{ for all } x \in I=(a,b).$

If $(c,d) \subseteq (a,b)$, then the hyperbolic length of the interval $(c,d)$ on the total interval $(a,b)$ is

$\ell_{(a,b)}((c,d))=\ell_{t}(j)=\ln(1+Cr(t,j)),$

where $l=(a,c), j=(c,d), r=(d,b), t=(a,b).$

Theorem 4. Assume $f: T \rightarrow f(T) \subseteq \mathbb{R}$ is a $C^{3}$ diffeomorphism with negative schwarzian derivative. Assume $J \subseteq T$, then

$\ell_{f(T)}(f(J)) \geq \ell_{T}(J).$

That means f expands the hyperbolic metric on the real interval.

Proof.  Since the schwarzian derivative of f is negative, $C(f(T),f(J)) \geq C(T,J).$

Therefore, $\ell_{f(T)}(f(J)) \geq \ell_{T}(J).$ That means f expands the hyperbolic metric on the real interval.

Remark. From Schwarz-Pick Theorem, for a holomorphic map $f: \mathbb{D} \rightarrow \mathbb{D}$, $f$ contracts the hyperbolic distance in the unit disc $\mathbb{D}$. Conversely, from above, for a $C^{3}$ diffeomorphism $f$ with negative schwarzian derivative, $f$ expands the hyperbolic distance in the real interval.

Exercise 1.  “Mathematical Tools for One Dimensional Dynamics” Exercise 6.5, Chapter 6

Let $f: I \rightarrow f(I) \subseteq \mathbb{R}$ be a $C^{3}$ diffeomorphism without fixed points ( $I$ being a closed interval on the real line). If $Sf(x)<0$ for all $x \in I$, then there exists a unique $x_{0} \in I$ such that $|f(x_{0})-x_{0}| \leq |f(x)-x|$ for all $x \in I$.

Proof.  If $f$ is a decreasing map, then the right boundary of the real interval I is the $x_{0}$. Therefore, assume that $f$ is an increasing map on the real interval I.

Since $f(x)$ has no fixed points on the real interval I, then $f(x)>x$ or $f(x) for all $x \in I$. Without lost of generality, assume $f(x)>x$ for all $x\in I$. Since $f(x)-x$ is a continuous function on the closed interval I, there exists $x_{0} \in I$ such that $|f(x_{0})-x_{0}| \leq |f(x)-x|$ for all $x\in I$.

By contradiction, there exist two distinct points $x_{0}, x_{1}$ $(x_{0} such that $|f(x_{0})-x_{0}| \leq |f(x)-x|$ and $|f(x_{1})-x_{1}| \leq |f(x)-x|$ for all $x\in I$. From here, we know that $|f(x_{0})-x_{0}|= |f(x_{1})-x_{1}|$.

From Langrange’s mean value theorem, there exists $\xi \in (x_{0}, x_{1})$ such that $(Df)(\xi)=1$. Since the schwarzian derivative of $f$ is negative, from the minimal principle, we get

$(Df)(\xi) > \min(Df(x_{0}), Df(x_{1})).$

i.e. $Df(x_{0})<1, Df(x_{1})<1$. However, from the definition of $x_{0}$ and $x_{1}$, we get

$Df(x_{0}) = \lim_{x\rightarrow x_{0}^{+}} \frac{f(x)-f(x_{0})}{x-x_{0}} \geq 1$

$Df(x_{1}) = \lim_{x\rightarrow x_{1}^{-}} \frac{f(x_{1})-f(x)}{x_{1}-x} \leq 1$

This is a contradiction. Therefore, the existence of $x_{0}$ is unique.

Assume $f: I \rightarrow f(I) \subseteq \mathbb{R}$ is a $C^{3}$ diffeomorphism, define the non-linearity of $f$ as

$f \mapsto Nf=\frac{D^{2}f}{Df} = D \ln Df \text{ whenever } Df \neq 0.$

Proposition 4.  $N(f \circ g)= (Nf \circ g) \cdot Dg+ Ng.$

Proposition 5. $Sf=\frac{D^{3}f}{Df} -\frac{3}{2}(\frac{D^{2}f}{Df})^{2}=D(Nf)-\frac{1}{2}(Nf)^{2}.$

Theorem 5. Koebe Non-linearity Principle.

Given $B, \tau>0$, there exists $K_{\tau,B}>0$ such that, if $f: [-\tau, 1+\tau] \rightarrow \mathbb{R}$ is a $C^{3}$ diffeomorphism into the reals and $Sf(t)\geq -B$ for all $t\in [-\tau,1+\tau],$ then we have

$|\frac{f^{''}(x)}{f^{'}(x)} | \leq K_{\tau,B}$

for all $0\leq x \leq 1.$ Show that $K_{\tau,B} \rightarrow 2/\tau$ as $B\rightarrow 0.$ (This recovers the classical Koebe non-linearity principle).