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


\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


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)<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}<x_{1}) 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).

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