Geometric conditions for matrix domination in two dimensions

TL;DR

This paper establishes geometric criteria for matrix domination in two dimensions, linking eigenvector and trace properties with hyperbolic geometry to enable explicit computation and construction.

Contribution

It introduces necessary and sufficient geometric conditions for domination in the special linear group, including an explicit algorithm for constructing dominated sets with given eigenvectors.

Findings

Derived explicit geometric conditions for matrix domination

Provided an algorithm for constructing dominated sets with prescribed eigenvectors

Connected matrix domination to two-dimensional hyperbolic geometry

Abstract

In this article we prove a necessary and a sufficient condition for a finite subset of the special linear group to be dominated. These conditions are purely geometric in nature, as they only involve the trace and the eigenvectors of the matrices, and can be computed explicitly. Our sufficient condition, in particular, provides a simple algorithm for constructing a dominated set with prescribed eigenvectors. The techniques involved in our proofs take advantage of the interaction between dominated sets and two-dimensional hyperbolic geometry.

Equations

$$g^n{f}^n(z)=z+2-\frac{1}{2^{n-1}}\to z+2,\text{as }n\to \infty .$$

$$C(f,g)=\frac{\alpha (f)-\alpha (g)}{\alpha (f)-\beta (g)}\frac{\beta (f)-\beta (g)}{\beta (f)-\alpha (g)}.$$

$$\tau (h)<\frac{1}{5}\min \left\{\frac{C(f,g)-1}{C(f,g)+3},1\right\},\text{for all }h\in \mathcal{F},$$

$$\tau (h)>4\max \{|\log |C(f,g)(C(f,g)-1)||\}+23,\text{for all }h\in \mathcal{F},$$

$$\tau (f_k)<\frac{1}{5}\min \left\{\frac{\frac{25}{4}-1}{\frac{25}{4}+3},1\right\}=\frac{21}{5\cdot 37}\approx 0.113,\text{ for }k=1,2\dots ,5,$$

$$\tau (f_k)>4\log 9(9-1)+13=4\log 72+13\approx 40.106,\text{ for }k=1,2\dots ,5,$$

$$C(f,g)=\{\begin{array}{ll}{\tanh }^2\frac{1}{2}d,& \text{if}0<C(f,g)<1,\\ {\coth }^2\frac{1}{2}d,& \text{if}C(f,g)>1.\end{array}$$

$$C(f,g)=-\frac{|\alpha (f)-\alpha (g){|}^2}{|\beta (g)-\alpha (f){|}^2}.$$

$$C(f,g)=\frac{(\lambda -\sigma {)}^2}{(\lambda +\sigma {)}^2}.$$

$$\cosh d=\frac{{\lambda }^2+{\sigma }^2}{2\lambda \sigma },$$

$$\tau (f),\tau (g)<\frac{1}{5}\frac{C(f,g)-1}{C(f,g)+3}.$$

$$\tau (f),\tau (g)>\log C(f,g)+\frac{3}{2}$$

$$\frac{1}{2}\left|\text{tr}(f\circ g)\right|=\left|\cosh d\sinh \frac{1}{2}\tau (f)\sinh \frac{1}{2}\tau (g)-\cosh \frac{1}{2}\tau (f)\cosh \frac{1}{2}\tau (g)\right|.$$

$$\tau (f),\tau (g)<\frac{1}{5}\frac{C(f,g)-1}{C(f,g)+3}.$$

$$\sinh b=\frac{1}{\sqrt{2}\sqrt{\cosh d-1}}=\frac{1}{2\sinh \frac{1}{2}d}.$$

$$\sinh a=\frac{\sqrt{2}}{3\sqrt{\cosh d-1}}=\frac{1}{3\sinh \frac{1}{2}d},$$

$$\cosh a=\frac{\sqrt{9\cosh d-7}}{3\sqrt{\cosh d-1}}$$

$$\frac{\partial h}{\partial x}(x,y)=\frac{\partial h}{\partial y}(y,x)=\cosh d\cosh x\sinh y-\sinh x\cosh y.$$

$$\frac{\partial h}{\partial x}(x,y)⩾\cosh d\frac{\sqrt{9\cosh d-7}}{3\sqrt{\cosh d-1}}\frac{\sqrt{2}}{3\sqrt{\cosh d-1}}-\frac{1}{\sqrt{2}\sqrt{\cosh d-1}}\frac{\sqrt{2}\sqrt{2\cosh d-1}}{2\sqrt{\cosh d-1}}.$$

$$\frac{\partial h}{\partial x}(x,y)⩾\sqrt{2}\frac{\cosh d\sqrt{9\cosh d-7}-\sqrt{4\cosh d-2}}{9(\cosh d-1)}.$$

$$\frac{1}{5}\frac{C(f,g)-1}{C(f,g)+3}<2(b-a).$$

$$\sinh (b-a)$$

$$>\frac{{\sinh }^2{b}^{\prime }{\cosh }^2{a}^{\prime }-{\cosh }^2{b}^{\prime }{\sinh }^2{a}^{\prime }}{2{\cosh }^2{b}^{\prime }}=\frac{5}{18}\frac{1}{4{\sinh }^2\frac{1}{2}d+1}$$

$$=\frac{5}{18}\frac{{\coth }^2\frac{1}{2}d-1}{{\coth }^2\frac{1}{2}d+3}>\frac{1}{4}\frac{{\coth }^2\frac{1}{2}d-1}{{\coth }^2\frac{1}{2}d+3}=\frac{1}{4}\frac{C(f,g)-1}{C(f,g)+3}.$$

$$2(b-a)>2\log \left(\frac{1}{4}\frac{C(f,g)-1}{C(f,g)+3}+\sqrt{(\frac{1}{4}\frac{C(f,g)-1}{C(f,g)+3}{)}^2+1}\right)>2\log \left(\frac{1}{4}\frac{C(f,g)-1}{C(f,g)+3}+1\right),$$

$$2(b-a)>2\frac{\frac{1}{4}\frac{C(f,g)-1}{C(f,g)+3}}{\frac{1}{4}\frac{C(f,g)-1}{C(f,g)+3}+1}=2\frac{C(f,g)-1}{5C(f,g)+11}>\frac{1}{5}\frac{C(f,g)-1}{C(f,g)+3}.$$

$$\tau (f),\tau (g)>\log C(f,g)+\frac{3}{2}.$$

$$\sinh b^{\prime }=\sqrt{\frac{2}{\cosh d-1}}=\frac{1}{\sinh \frac{1}{2}d}.$$

$$2b^{\prime }=2\log \left(\frac{1+\cosh \frac{1}{2}d}{\sinh \frac{1}{2}d}\right)<2\log (2\coth \frac{1}{2}d)=\log \left(4C(f,g)\right)<\log C(f,g)+\frac{3}{2}.\text{qed}$$

$$\tau (f),\tau (g)>\log C(f,g)+\frac{3}{2}.$$

$$\sinh \frac{1}{2}\tau (f),\sinh \frac{1}{2}\tau (g)>\frac{1}{\sinh \frac{1}{2}d},$$

Full text

1 Introduction

This paper studies semigroups of conformal isometries of the hyperbolic plane. For a semigroup generated by a finite collection of isometries, we calculate explicit geometric constraints on its generators which imply that elements of the semigroup do not accumulate at the identity transformation. Semigroups of Möbius transformations have been previously studied by Fried, Marotta and Stankiewitz [FMS], as a particular branch of the theory of semigroups of rational functions initiated by Hinkkanen and Martin [HiMa]. Our approach follows techniques similar to those used by Jacques and Short [JS], who further developed the material in [FMS], and incorporates well-known results from the theory of Fuchsian groups. Also, the work of Avila, Bochi and Yoccoz [ABY] on uniformly hyperbolic finitely-valued $\text{SL}(2,\mathbb{R})$-cocycles is closely related to the theory of semigroups and at the end of this section we will discuss how our main result can be used to infer the uniform hyperbolicity of a cocycle.

Let us start by setting up some notation. The group $\mathcal{M}$ of conformal automorphisms of the upper half-plane $\mathbb{H}$ consists of the Möbius transformations of the form $z\mapsto(az+b)/(cz+d)$, where $a,b,c,d\in\mathbb{R}$ and $ad-bc>0$. Note that $\mathcal{M}$ also acts on the extended real line $\overline{\mathbb{R}}=\mathbb{R}\cup\{\infty\}$, and in fact on the whole of the complex plane. If we endow $\mathbb{H}$ with the hyperbolic distance $\rho$ induced by the Riemannian metric $|dz|/\operatorname{Im}z$, then $\mathcal{M}$ is the group of conformal isometries of the complete metric space $(\mathbb{H},\rho)$. It is often convenient to transfer our arguments to the unit disc $\mathbb{D}$ using a Möbius transformation, and we will do so for all our figures.

The non-identity elements of $\mathcal{M}$ are classified as elliptic, parabolic or hyperbolic depending on whether they have one fixed point in $\mathbb{H}$, one fixed point in $\overline{\mathbb{R}}$, or two fixed points in $\overline{\mathbb{R}}$, respectively. For a hyperbolic transformation $f$, we shall denote by $\alpha(f)$ and $\beta(f)$ its attracting and repelling fixed points, respectively, and by $\text{Ax}(f)$ the unique hyperbolic geodesic joining $\alpha(f)$ and $\beta(f)$, called the axis of $f$. In addition, the translation length of $f$ is the distance $\rho(w,f(w))$, for any $w\in\text{Ax}(f)$, and will be denoted by $\tau(f)$. Observe that for each $k\in\mathbb{N}$, the $k$th iterate $f^{k}$ of a hyperbolic transformation $f$ is also hyperbolic, with translation length $k\,\tau(f)$.

By the term semigroup we mean a subset of $\mathcal{M}$ that is closed under composition. We say that a semigroup $S$ is generated by a set $\mathcal{F}\subset\mathcal{M}$, if every element of $S$ can be written as a composition of transformations of $\mathcal{F}$. A semigroup is called finitely-generated if the set of generators $\mathcal{F}$ is finite. If $f_{1},f_{2},\dots,f_{n}$ are transformations in $\mathcal{M}$, then $\langle f_{1},f_{2},\dots,f_{n}\rangle$ will denote the semigroup generated by $\{f_{1},f_{2},\dots,f_{n}\}$. A subset of $\mathcal{M}$ is called discrete if the topology it inherits from $\mathcal{M}$ is the discrete topology. For semigroups of Möbius transformations the authors of [JS] also make the following definition.

Definition**.**

A semigroup $S$ is called semidiscrete if the identity transformation is not an accumulation point of $S$. We say that $S$ is inverse-free if it contains no inverses of elements of $S$.

For subgroups of $\mathcal{M}$, the discrete and semidiscrete properties are equivalent. This is not the case for semigroups, however, as the following example illustrates (see [JS, Section 3]). Consider the transformations $f(z)=2z$ and $g(z)=\tfrac{1}{2}z+1$, and let $S$ be the semigroup generated by $f$ and $g$. It is easy to check that elements of $\langle f,g\rangle$, which are not iterates of $f$ or $g$, are of the form $2^{n+m}z+2^{n}+2^{n-1}c$, for some $n\in\mathbb{N}\cup\{0\},m\in\mathbb{Z}$ and $c\geqslant 0$. So, elements of $S$ cannot accumulate at the identity and $S$ is semidiscrete and inverse-free. It is not discrete, however, as

[TABLE]

Let $S$ be a semigroup generated by a finite collection of Möbius transformations $\mathcal{F}$. Suppose that there exists a union $C$ of $m$ open intervals in $\overline{\mathbb{R}}$, with disjoint closures, such that each element of $\mathcal{F}$ maps $C$ strictly inside itself. Also, suppose that $m$ is the smallest integer with this property, in the sense that if there exists another union $D$ of $n$ open intervals in $\overline{\mathbb{R}}$, with disjoint closures, such that each element of $\mathcal{F}$ maps $D$ strictly inside itself, then $m\leqslant n$. We then say that $S$ is a Schottky semigroup of rank $m$. The example presented above is a Schottky semigroup of rank one, as it maps the interval $(1,\infty)$ strictly inside itself. It is fairly straightforward to prove that every Schottky semigroup is semidiscrete and inverse-free (see [JS, Theorem 7.1]).

For two hyperbolic transformations $f,g$, we define the cross ratio $C(f,g)$ of $f$ and $g$ to be

[TABLE]

Since the fixed points of $f$ and $g$ are concyclic, we have that $C(f,g)\in\overline{\mathbb{R}}$. The value of $C(f,g)$ allows us to infer information about the geometric configuration of $\mathrm{Ax}(f)$ and $\mathrm{Ax}(g)$, as it is shown in Figure 1 (see, also, Lemma 2.1). It is easy to check that $C(f,g)=1$ if and only if either $f$ or $g$ is parabolic, and $C(f,g)=0$ if and only if either $\alpha(f)=\alpha(g)$ or $\beta(f)=\beta(g)$. Also, $C(f,g)=\infty$ if and only if $\alpha(f)=\beta(g)$ or $\beta(f)=\alpha(g)$.

We are now ready to present the main result of this paper.

Theorem 1.1**.**

Let $\mathcal{F}$ be a finite collection of hyperbolic transformations, such that $\alpha(f)\neq\beta(g)$ for all $f,g\in\mathcal{F}$ and $S=\langle\mathcal{F}\rangle$ is not a Schottky semigroup of rank one.

1. (i)

Suppose that

[TABLE]

where the minimum is taken over all pairs $f,g$ in $\mathcal{F}$ with $C(f,g)>1$. Then $S$ is not semidiscrete. 2. (ii)

Suppose that

[TABLE]

where the maximum is taken over all pairs $f,g$ in $\mathcal{F}$ with $C(f,g)\neq 0$. Then $S$ is semidiscrete and inverse-free.

Given a particular configuration of directed hyperbolic lines $\ell_{1},\ell_{2}\dots\ell_{n}$, and hyperbolic transformations $f_{1},f_{2}\dots f_{n}$, with $\mathrm{Ax}(f_{i})=\ell_{i}$ for $i=1,2,\dots,n$, Theorem 1.1 provides us with constraints on the translation length of each $f_{i}$ in order for the semigroup $\langle f_{1},f_{2},\dots,f_{n}\rangle$ to be semidiscrete and inverse-free. A strength of Theorem 1.1 is that these constraints are given by algebraically simple formulas which can be computing by considering the generators $f_{1},f_{2}\dots f_{n}$ in pairs. Note that because of our assumptions, $C(f,g)$ is never one or infinity.

Our main result also holds for some semigroups with pairs of generators whose attracting and repelling fixed points meet, but for the sake of simplicity we restrict ourselves to the class of semigroups described in Theorem 1.1.

We now present an example where we compute the constants in Theorem 1.1 explicitly. Suppose that the axes of the hyperbolic transformations $f_{1},f_{2},\dots,f_{5}$ are as shown in Figure 2 and consider the collection $\mathcal{F}=\{f_{1},f_{2},\dots,f_{5}\}$. The cross ratios of all the pairs are $C(f_{1},f_{5})=C(f_{3},f_{5})=-1,C(f_{1},f_{3})=C(f_{2},f_{4})=25/4,C(f_{2},f_{5})=1/9,C(f_{4},f_{5})=9$ and $C(f_{1},f_{2})=C(f_{1},f_{4})=C(f_{2},f_{3})=C(f_{3},f_{4})=0$. Therefore, if

[TABLE]

then the semigroup $\langle\mathcal{F}\rangle$ is not semidiscrete, whereas if

[TABLE]

then $\langle\mathcal{F}\rangle$ is semidiscrete and inverse-free, and in fact it is a Schottky semigroup of rank two.

Finally, let us use Theorem 1.1 to discuss uniformly hyperbolic $\text{SL}(2,\mathbb{R})$-cocycles. Avila, Bochi and Yoccoz [ABY] studied the parameter space $\mathcal{H}$ of all $N$-tuples $(F_{1},F_{2}\dots,F_{N})$ in $\text{SL}(2,\mathbb{R})^{N}$ for which there exists a finite union $X$ of open intervals in $\overline{\mathbb{R}}$, with disjoint closures, such that each $F_{i}$ maps $\overline{X}$ into $X$. It is easy to check that if $(F_{1},F_{2},\dots,F_{N})$ lies in $\mathcal{H}$, then $\{F_{1},F_{2},\dots,F_{N}\}$ generates a Schottky semigroup. Therefore, the parameter space of collections in $\mathcal{M}$ that generate Schottky semigroups, contains the parameter space of uniformly hyperbolic cocycles.

Suppose, now, that $F_{1},F_{2}\dots,F_{N}$ are hyperbolic transformations that satisfy the hypotheses of part (ii) of Theorem 1.1. Then our proof shows that each $F_{i}$ maps a finite union of open intervals compactly inside itself, and so $(F_{1},F_{2}\dots,F_{N})$ is uniformly hyperbolic. Hence, part (ii) of Theorem 1.1 acts as a quantitative test for an $N$-tuple in $\text{SL}(2,\mathbb{R})^{N}$ to lie in $\mathcal{H}$.

2 Two-generator semigroups

In this section we prove a version of Theorem 1.1 for two hyperbolic transformations $f$ and $g$. Semigroups generated by pairs of Möbius transformations were studied by Avila, Bochi and Yoccoz [ABY, Chapter 3] and Jacques and Short [JS, Section 12]. Our results will be obtained by following techniques similar to [JS, Theorem 1.4] and modifying well-known results for two-generator Fuchsian groups (see [BeaBook, Chapter 11]).

For a transformation $h(z)=(az+b)/(cz+d)$ with $ad-bc=1$, we define the trace of $h$ to be the number $\mathrm{tr}(h)=a+b$. Also, we say that and interval $I\subset\overline{\mathbb{R}}$ is symmetric with respect to $h$, if the hyperbolic geodesic in $\mathbb{H}$, with the same endpoints as $I$, is perpendicular to the axis of $h$. Finally, if the axes of two transformations $f$ and $g$ cross at a point $p\in\mathbb{H}$, we define the angle $\theta\in[0,\pi]$ between $\mathrm{Ax}(f)$ and $\mathrm{Ax}(g)$ to be the angle at $p$ of the hyperbolic triangle defined by $\alpha(f),\alpha(g)$ and $p$.

We start by establishing formulas that relate the cross ratio $C(f,g)$ of two hyperbolic transformations $f$ and $g$ with the geometric configuration of their axes. Throughout, two hyperbolic lines will be called disjoint if they do not cross in $\mathbb{H}$ and have distinct endpoints.

Lemma 2.1**.**

Suppose that $f$ and $g$ are hyperbolic transformations.

1. (i)

If the axes of $f$ and $g$ cross at an angle $\theta$, then $C(f,g)=-\tan^{2}\tfrac{1}{2}\theta$. 2. (ii)

If the axes of $f$ and $g$ are disjoint and a hyperbolic distance $d$ apart, then

[TABLE]

Proof.

Suppose that the axes of $f$ and $g$ cross at angle $\theta$. By conjugating $f$ and $g$ with a Möbius transformation so that $f(z)=az$ for some $a\neq 1$, we can see that $C(f,g)$ has to be negative. Now, conjugate $f$ and $g$ so that they act on the unit disc, and their axes meet at the origin (see Figure 3 on the left). Because $\lvert C(f,g)\rvert=-C(f,g)$, it is easy to see that

[TABLE]

Also, by the law of cosines, $\lvert\alpha(f)-\alpha(g)\rvert^{2}=2(1-\cos\theta)$ and $\lvert\beta(g)-\alpha(f)\rvert^{2}=2(1-\cos(\pi-\theta))$, which yield the desired equation.

Assume, now, that the axes of $f$ and $g$ are disjoint and a hyperbolic distance $d$ apart. Note that in this case we have $C(f^{-1},g)=1/C(f,g)$, and so it suffices to assume that $C(f,g)>1$. Conjugate $f$ and $g$ so that $f$ fixes $-\sigma,\sigma$ and $g$ fixes $-\lambda,\lambda$, for some $0<\sigma<\lambda$ (see Figure 3 on the right). Then,

[TABLE]

Observe that $d=\log\frac{\lambda}{\sigma}$, which implies that

[TABLE]

and using the half-angle formula for the hyperbolic cotangent completes our proof. ∎

We first consider the case of two hyperbolic transformations with disjoint axes. In order to prove Theorem 2.3, to follow, we will make use of the next theorem that comes as a direct application of Theorem 1.4, Theorem 1.5 and Lemma 12.9 from [JS].

Theorem 2.2** ([JS]).**

Suppose that $f$ and $g$ are hyperbolic transformations. The semigroup $\langle f,g\rangle$ satisfies one of the following possibilities: if $\langle f,g\rangle$ does not contain elliptic elements then it is a Schottky semigroup of rank two; otherwise, either $\langle f,g\rangle$ is itself a discrete group, or else it is not semidiscrete.

Theorem 2.3**.**

Let $f,g$ be two hyperbolic transformations with $C(f,g)>1$.

1. (i)

Suppose that

[TABLE]

Then $\langle f,g\rangle$ is not semidiscrete. 2. (ii)

Suppose that

[TABLE]

Then $\langle f,g\rangle$ is a Schottky semigroup of rank two.

Proof.

Let $d$ be the hyperbolic distance between the axes of $f$ and $g$. The proof revolves around evaluating the trace of the composition of $f$ and $g$, which is given by the following equation found in [BeaBook, Theorem 7.38.3]:

[TABLE]

Recall that $f\circ g$ is elliptic if $|\text{tr}(f\circ g)|<2$ and hyperbolic if $|\text{tr}(f\circ g)|>2$. Let $\mathbb{R}_{+}^{2}$ be the first quadrant of $\mathbb{R}^{2}$ and consider the function $h\colon\mathbb{R}_{+}^{2}\to\mathbb{R}$, with $h(x,y)=\cosh d\sinh x\sinh y-\cosh x\cosh y$. Suppose that

[TABLE]

We are going to prove that there exist positive integers $m,n$, such that $f^{m}\circ g^{n}$ is elliptic, and $\langle f,g\rangle$ is not a discrete group. Then, Theorem 2.2 would imply that $\langle f,g\rangle$ is not semidiscrete.

First, recall [BeaBook, Theorem 11.6.9], which states that if $h(\tfrac{1}{2}\tau(f),\tfrac{1}{2}\tau(g))<-\tfrac{1}{2}$, then the group generated by $f$ and $g$ is not discrete. Therefore, it suffices to prove that if $f$ and $g$ satisfy (2.1) then there exist positive integers $m,n$, such that $-1<h(\tfrac{m}{2}\tau(f),\tfrac{n}{2}\tau(g))<-\tfrac{1}{2}$.

Define $D=\{(x,y)\in\mathbb{R}_{+}^{2}\colon-1<h(x,y)<-\tfrac{1}{2}\}$, and let $b\geqslant 0$ be such that $h(b,b)=-\tfrac{1}{2}$. Solving this equation for $\sinh b$ yields

[TABLE]

Also, define $a\geqslant 0$ to be the unique solution of $h(a,a)=-\tfrac{7}{9}$. Then

[TABLE]

which implies that $a<b$. Let $C$ be the square with vertices $(a,b),(b,b),(b,a),(a,a)$ (see Figure 4). We will prove that the compact set $K$, bounded by the square $C$ lies in $\overline{D}$. Note that points $(x,y)$ in $K$ satisfy the inequalities $\sinh a\leqslant\sinh x,\sinh y\leqslant\sinh b$, or equivalently $\cosh a\leqslant\cosh x,\cosh y\leqslant\cosh b$, where

[TABLE]

We are going to show that $h$ is increasing on vertical and horizontal line segments in $K$, which will complete the proof of our claim. Because $h$ is symmetric with respect to the line $y=x$,

[TABLE]

Thus, for all $(x,y)\in K$ we have that $\frac{\partial h}{\partial x}(x,y)\geqslant\cosh d\cosh a\sinh a-\sinh b\cosh b$, which is

[TABLE]

So, we have that

[TABLE]

Since $\sqrt{9\cosh d-7}>\sqrt{4\cosh d-2}$, the function $h$ is increasing on horizontal and vertical line segments inside $K$.

Therefore, if $\tau(f),\tau(g)<2(b-a)$ then there exist positive integers $m,n$, such that the point $(\tfrac{m}{2}\tau(f),\tfrac{n}{2}\tau(g))$ lies in the interior of $K$. In order to finish the proof of the first part we are going to show that

[TABLE]

Observe that

[TABLE]

Taking the inverse hyperbolic sine yields

[TABLE]

and finally, since $\log(x+1)>\frac{x}{x+1}$ we have

[TABLE]

For the second part of the theorem, suppose that

[TABLE]

Because of Theorem 2.2, it suffices to prove that the semigroup $\langle f,g\rangle$ does not contain elliptic transformations. Let $b^{\prime}\geqslant 0$ be such that $h(b^{\prime},b^{\prime})=1$. Then,

[TABLE]

It is easy to check that $h(x,b^{\prime})$ and $h(x,x)$ are increasing functions of $x$. Hence, points $(x,y)$ in the domain bounded by the half-lines $\{(x,x)\in\mathbb{R}_{+}^{2}\colon x\geqslant b^{\prime}\}$ and $\{(x,b^{\prime})\in\mathbb{R}_{+}^{2}\colon x\geqslant b^{\prime}\}$ satisfy the inequality $h(x,y)\geqslant 1$.

By the symmetry of $h$, all points in the quarter-plane $Q=\{(x,y)\in\mathbb{R}_{+}^{2}\colon x\geqslant b\text{ and }y\geqslant b\}$ (see Figure 4) satisfy $h(x,y)\geqslant 1$ and therefore if $\tau(f),\tau(g)>2b^{\prime}$, then $h(\tfrac{m}{2}\tau(f),\tfrac{n}{2}\tau(g))>1$ for all positive integers $m,n$, which implies that the semigroup $\langle f,g\rangle$ does not contain elliptic transformations. The proof is complete upon observing that

[TABLE]

Corollary 2.4**.**

Let $f,g$ be two hyperbolic transformations with $C(f,g)>1$. Suppose that

[TABLE]

Then, there exist open intervals $A_{f},B_{f},A_{g},B_{g}$ in $\overline{\mathbb{R}}$, with pairwise disjoint closures, that satisfy the following properties: the intervals $A_{f},B_{f}$ are symmetric with respect to $f$ and $f(B_{f}^{c})\subset A_{f}$; the intervals $A_{g},B_{g}$ are symmetric with respect to $g$ and $g(B_{g}^{c})\subset A_{g}$.

Proof.

Suppose that $f$ and $g$ are hyperbolic transformations with $C(f,g)>1$, and let $d$ be the hyperbolic distance between their axes. Let $\ell$ be the unique hyperbolic line that is perpendicular to the axes of $f$ and $g$, and $\sigma$ the reflection in $\ell$. Also, define the reflections $\sigma_{f}=f\circ\sigma$ and $\sigma_{g}=\sigma\circ g$ and let $\ell_{f},\ell_{g}$ be their lines of reflection respectively (see Figure 5).

Assume that $\tau(f),\tau(g)>\log C(f,g)+\frac{3}{2}$. We are going to show that the desired intervals can be chosen as follows. Let $A_{f}$ be the open interval in $\overline{\mathbb{R}}$ with the same endpoints as $\ell_{f}$ and containing $\alpha_{f}$, and $B_{g}$ the open interval with the same endpoints as $\ell_{g}$ and containing $\beta_{g}$ . Also, let $B_{f}=\sigma(A_{f})$ and $A_{g}=\sigma(B_{g})$. It is obvious that $f(B_{f}^{c})\subset A_{f}$ and $g(B_{g}^{c})\subset A_{g}$, and so it suffices to prove that $A_{f}$ and $B_{g}$ have disjoint closures, as that would imply that the same holds for $B_{f}$ and $A_{g}$.

Since $\tau(f),\tau(g)>\log C(f,g)+\frac{3}{2}$, from the proof of part (ii) of Theorem 2.3 we see that $\tau(f)$ and $\tau(g)$ satisfy

[TABLE]

and

[TABLE]

If the line $\ell_{f}$ meets $\mathrm{Ax}(g)$ in $\overline{\mathbb{H}}$, then the lines $\ell_{f},\mathrm{Ax}(f),\ell$ and $\mathrm{Ax}(g)$ form a quadrilateral in $\mathbb{H}$, and so [BeaBook, Theorem 7.17.1] implies that

[TABLE]

which is a contradiction since $\sinh\tfrac{1}{2}d<\sinh d$. Therefore, $\ell_{f}$ does not cross $\mathrm{Ax}(g)$, and similarly $\ell_{g}$ does not cross $\mathrm{Ax}(f)$.

So, we can now see that if $\overline{A_{f}}\cap\overline{B_{g}}\neq\emptyset$, then the lines $\ell,\mathrm{Ax}(f),\ell_{f},\ell_{g}$ and $\mathrm{Ax}(g)$ define a pentagon in $\overline{\mathbb{H}}$, and thus [BeaBook, Theorem 7.18.1] yields

[TABLE]

This, however, contradicts (2.2) and therefore $\overline{A_{f}}\cap\overline{B_{g}}=\emptyset$. ∎

Before we move on to the case of hyperbolic transformations with crossing axes, we are going to need the next lemma about limit sets of semigroups. We define the forward limit set $\Lambda^{+}(S)$ of a semigroup $S$ to be the set accumulation points of $\{f(z_{0})\colon f\in S\}$ in $\overline{\mathbb{R}}$, where $z_{0}\in\mathbb{H}$, with respect to the chordal metric in $\overline{\mathbb{C}}$. The backward limit set $\Lambda^{-}(S)$ of $S$ is defined to be the forward limit set of $S^{-1}=\{f^{-1}\colon f\in S\}$. It is easy to check that the limit sets are independent of the choice of $z_{0}$, and $\Lambda^{+}(S)$ is forward invariant under transformations in $S$ whereas $\Lambda^{-}(S)$ is backward invariant. For more information on limit sets of semigroups we refer to [FMS].

Lemma 2.5**.**

Let $S$ be a semigroup with $|\Lambda^{-}(S)|\neq 1$ that is not a discrete group. Suppose that $\Lambda^{+}(S)^{o}\cap\Lambda^{-}(S)\neq\emptyset$, where $\Lambda^{+}(S)^{o}$ denotes the interior of the forward limit set. Then $S$ is not semidiscrete.

Proof.

Since $\Lambda^{-}(S)$ is the smallest closed set that contains all the repelling fixed points of hyperbolic elements of $S$, there exists a hyperbolic transformation $f$ in $S$ with repelling fixed point in $\Lambda^{+}(S)^{o}$. So, by the invariance of the limit sets under the semigroup, $f(\Lambda^{+}(S)^{o})=\overline{\mathbb{R}}\subset\Lambda^{+}(S)$ and therefore $\Lambda^{+}(S)=\overline{\mathbb{R}}$. Because $|\Lambda^{-}(S)|\neq 1$ and $\Lambda^{-}(S)\subset\overline{\mathbb{R}}=\Lambda^{+}(S)$, Theorem [JS, Theorem 1.9] tells us that if $S$ were semidiscrete, it would have to be a discrete group, which is a contradiction. ∎

Theorem 2.6**.**

Suppose that $f$ and $g$ are hyperbolic transformations whose axes cross at an angle $\theta\in(0,\pi)$ and suppose that $\beta(g)<\alpha(f)<\alpha(g)<\beta(g)$.

1. (i)

If $\tau(f),\tau(g)<\displaystyle\frac{1}{5}$, then $\Lambda^{+}\left(\langle f,g\rangle\right)=[\alpha(f),\alpha(g)]$. 2. (ii)

If $\tau(f),\tau(g)>\displaystyle\lvert\log\lvert C(f,g)\rvert\,\rvert+\frac{3}{2}$, then there exist open intervals $A_{f},B_{f},A_{g},B_{g}$ in $\overline{\mathbb{R}}$, with pairwise disjoint closures, that satisfy the following properties: the intervals $A_{f},B_{f}$ are symmetric with respect to $f$, and $f(B_{f}^{c})\subset A_{f}$; the intervals $A_{g},B_{g}$ are symmetric with respect to $g$, and $g(B_{g}^{c})\subset A_{g}$.

Proof.

For convenience, we conjugate $f$ and $g$ by a Möbius transformation so that they act on the unit disc $\mathbb{D}$, their axes cross at the origin and the Euclidean diameter landing at $i$ and $-i$ bisects $\theta$. As an abuse of notion, for $z,w\in\partial\mathbb{D}$ we denote by $[z,w]$ the closed arc of the unit circle that is oriented counter-clockwise and joins $z$ and $w$.

Let $\phi_{g}$ be the angle between the Euclidean radius landing at $g(\alpha(f))$ and the axis of $g$ (see Figure 6 on the left). Applying the hyperbolic sine and cosine laws on the triangle with vertices 0, $g(0)$ and $g(\alpha(f))$, we obtain

[TABLE]

and therefore

[TABLE]

Defining $\phi_{f}$ similarly and carrying out the same computations we can see that equation (2.3) holds if $g$ is replaced by $f$.

For convenience, define $I=[\alpha(f),\alpha(g)]$ and note that because $I$ is invariant under $\langle f,g\rangle$, the forward limit set of $\langle f,g\rangle$ is contained in $I$. Observe that if $\phi_{f},\phi_{g}>\tfrac{\theta}{2}$ then $g(I)\cup f(I)=I$. Thus, for every $x\in I$ there exists $f_{1}\in\{f,g\}$ so that $x\in f_{1}(I)$. We can thus recursively find a sequence $(f_{n})$ with $f_{n}\in\{f,g\}$, and such that $x\in f_{1}f_{2}\dots f_{n}(I)$, for all $n=1,2,\dots$. It is easy to check that the intervals $f_{1}f_{2}\dots f_{n}(I)$ are nested and their Euclidean length converges to 0 as $n\to\infty$. Hence, if $\phi_{f},\phi_{g}>\tfrac{\theta}{2}$, then every $x\in I$ is an accumulation point of either $\alpha(f)$ or $\alpha(g)$ under $\langle f,g\rangle$, which implies that $\Lambda^{+}(\langle f,g\rangle)=I$.

So, it suffices to prove that if $\tau(f),\tau(g)<\frac{1}{5}$, then $\phi_{f},\phi_{g}>\tfrac{\theta}{2}$. Note that these last inequalities for $\phi_{f}$ and $\phi_{g}$ are equivalent to $\cos\phi_{f},\cos\phi_{g}<\cos\tfrac{\theta}{2}$, which by substituting (2.3) (and the respective equation for $\cos\phi_{f}$) yield $\sinh\tau(f),\sinh\tau(g)<\varepsilon$, where

[TABLE]

Therefore, in order to complete the proof of the first part, it suffices to show that $\frac{1}{5}<\operatorname{arsinh}\varepsilon$, which follows from the

[TABLE]

For the second part of the theorem, recall that the diameters landing at $i,-i$ and $1,-1$, respectively, bisect the two complementary angles between the axes of $f$ and $g$. Let $\sigma_{g}$ be the angle between the axis of $g$ and the radius landing at $g(-1)$ (see Figure 6 on the right). We are going to prove that if $\tau(g),\tau(f)>\lvert\log\lvert C(f,g)\rvert\,\rvert+\frac{5}{2}$, then $g$ maps the complement of $[i,-1]$ in $\partial\mathbb{D}$ strictly inside $[-i,1]$, and $f$ maps the complement of $[1,i]$ strictly inside $[-1,-i]$. The proof will be carried out for $g$; the situation for $f$ is identical.

It suffices to prove that $g(-1),g(i)\in(-i,1)$. Considering the triangle with vertices $0,g(-1)$ and $g(0)$ and carrying out the same calculations we did for part (i), we obtain

[TABLE]

It is obvious that if $\sigma_{g}<\tfrac{\theta}{2}$ then $g(-1)\in(-i,1)$. So substituting the equation above to $\cos\sigma_{g}>\cos\tfrac{\theta}{2}$ and solving for $\sinh\tau_{g}$ results in the inequality $\sinh\tau(g)>M$, where

[TABLE]

One can easily check that working similarly in the triangle with vertices $0,g(0)$ and $g(i)$ yields the same estimate for the translation length of $g$.

So, in order to complete the proof, it suffices to prove that $\lvert\log\lvert C(f,g)\rvert\,\rvert+\frac{5}{2}>\operatorname{arsinh}M$. Observe that $M<4/\sin\theta$, and from Lemma 2.1 we have $\sin\theta=2\sqrt{-C(f,g)}/(1-C(f,g))$. Thus

[TABLE]

as required. ∎

Corollary 2.7**.**

Let $f$ and $g$ be hyperbolic transformations, whose axes cross at an angle $\theta\in(0,\pi)$ and suppose that $\beta(g)<\alpha(f)<\alpha(g)<\beta(g)$. Also, assume that $h$ is a hyperbolic transformation whose repelling fixed point lies in $(\alpha(f),\alpha(g))$. If $\tau(f),\tau(g)<\tfrac{1}{5}$, then the semigroup generated by $f$, $g$ and $h$ is not semidiscrete.

Proof.

Assume that $\tau(f),\tau(g)<\frac{1}{5}$. Then, from Theorem 2.6 we have $\Lambda^{+}\left(\langle f,g\rangle\right)=[\alpha(f),\alpha(g)]$, and we can apply Lemma 2.5 to the semigroup $\langle f,g,h\rangle$ in order to deduce that it is either a discrete group or else it is not semidiscrete. Our goal is to prove that if $\tau(f),\tau(g)<\tfrac{1}{5}$, then $\langle f,g,h\rangle$ is not a discrete group.

Recall [BeaBook, Theorem 11.6.8], which states that if the group generated by $f$ and $g$ is discrete, then

[TABLE]

Observe that if

[TABLE]

then $\sinh\tfrac{1}{2}\tau(f)\sinh\tfrac{1}{2}\tau(g)\sin\theta<\cos\frac{3\pi}{7}$ and thus $f,g$ do not generate a discrete group. Inequality (2.4) is equivalent to $\sinh\tau(f),\sinh\tau(g)<\lambda$, where

[TABLE]

Hence, it suffices to prove that $\frac{1}{5}<\operatorname{arsinh}\lambda$. To that end, observe that $\lambda>2\cos\frac{3\pi}{7}>\frac{2}{5}$ and thus

[TABLE]

We end this section by establishing the following lemma about collections of hyperbolic transformations that share an attracting fixed point.

Lemma 2.8**.**

Suppose that $f_{1},f_{2},\dots,f_{n}$ are hyperbolic transformations with $\alpha(f_{1})=\alpha(f_{2})=\dots=\alpha(f_{n})$. If $\tau(f_{i})>\log 5$, for $i=1,2,\dots,n$, then there exist open intervals $A,B$ in $\overline{\mathbb{R}}$, with disjoint closures, such that $f_{i}$ maps the complement of $B$ inside $A$, for all $i=1,2,\dots,n$.

Proof.

Without loss of generality, we can assume that all transformations $f_{1},f_{2},\dots,f_{n}$ fix infinity, and each $f_{i}$ can be written in the form $f_{i}=\lambda_{i}z+x_{i}(1-\lambda_{i})$, where $\lambda_{i}>1$ and $0\leqslant x_{i}\leqslant 1$.

Suppose that $\tau(f_{i})>\log 5$, for $i=1,2,\dots,n$. Since $\lambda_{i}=e^{\tau(f_{i})}$, this implies that $\lambda_{i}>5$, for $i=1,2,\dots,n$. Consider the intervals $A=[\infty,-\tfrac{3}{2})\cup(\tfrac{5}{2},\infty]$ and $B=(-\tfrac{1}{2},\tfrac{3}{2})$. Then, for $i=1,2,\dots,n$, we have

[TABLE]

Similarly, we have that $f_{i}(-\tfrac{1}{2})<-\frac{\lambda}{2}+x_{i}<-\tfrac{3}{2}$ for $i=1,2,\dots,n$, which concludes our proof. ∎

3 Proof of the main result

We are now ready to prove Theorem 1.1. Suppose that $f_{1},f_{2}\dots,f_{n}$ are hyperbolic Möbius transformations, such that $\alpha(f_{i})\neq\beta(f_{j})$ for all $i,j=1,2\dots,n$ and $S=\langle f_{1},f_{2}\dots,f_{n}\rangle$ is not a Schottky semigroup of rank one. These assumptions imply that we cannot partition $\overline{\mathbb{R}}$ into two intervals $I_{1},I_{2}$, such that $\alpha_{i}\in I_{1}$ and $\beta_{i}\in I_{2}$, for all $i=1,2\dots,n$. Hence, we can always find a pair of generators $f_{k}$ and $f_{m}$ such that, either the axes of $f_{k}$ and $f_{m}$ are disjoint and $C(f_{k},f_{m})>1$, or else their axes cross and there exists another generator $f_{l}$ with $\alpha(f_{k})<\beta(f_{l})<\alpha(f_{m})$. So, the first part of Theorem 1.1 comes as an application of the first part of Theorem 2.3 and Corollary 2.7.

Let us now focus on part (ii). For the rest of the proof we define $C_{i,j}=C(f_{i},f_{j})$. We are going to prove that if the translation lengths $\tau(f_{k})$ are big enough, for all $k=1,2,\dots,n$, then there exists a finite collection of open intervals in $\overline{\mathbb{R}}$, with disjoint closures, whose union is mapped strictly inside itself under each $f_{k}$.

Assume first that all the fixed points of the generators are distinct, which implies that $C_{i,j}\neq 0$ for all pairs $i,j$. Fix some $k\in\{1,2,\dots,n\}$. Applying Corollary 2.4 and the second part of Theorem 2.6 to all pairs $f_{i},f_{k}$ with either $C_{i,k}<0$ or $C_{i,k}>1$, yields constants $M_{i,k}=\lvert\log\lvert C_{i,k}\rvert\,\lvert+\frac{3}{2}$ with the following properties. For each pair $i,k$, the inequalities $\tau(f_{k})>M_{i,k}$ imply that there exist open intervals $A_{i,k}$ and $B_{i,k}$, symmetric with respect to $f_{k}$, such that $\alpha_{k}\in A_{i,k}$, $\beta_{k}\in B_{i,k}$ and $f_{k}$ maps the complement of $B_{i,k}$ inside $A_{i,k}$. From the symmetry of the intervals with respect to $f_{k}$, the collections $\{A_{i,k}\}_{i}$ and $\{B_{i,k}\}_{i}$ consist of nested intervals. Let $A^{k}\in\{A_{i,k}\}_{i}$ and $B^{k}\in\{B_{i,k}\}_{i}$ be the innermost interval in each collection. Note that $M_{i,j},A_{i,j},B_{i,j}$ are symmetric with respect to the indices $i,j$.

Because $A_{i,k}\cap B_{i,k}=\emptyset$ for $i=1,2,\dots,n$, the intervals $A^{k}$ and $B^{k}$ are disjoint. In principle, however, we may have that $A^{k}=A_{j,k}$ and $B^{k}=B_{k,l}$ for different pairs $j,k$ and $k,l$. Let $\ell_{A}$ be the hyperbolic line in $\mathbb{H}$ that has the same endpoints as $A^{k}$ and $\ell_{B}$ the hyperbolic line with the same endpoints as $B^{k}$. Also, denote the distance $\rho(\ell_{A},\ell_{B})$ between these two lines by $d_{k}$. Recall that the lines $\ell_{A}$ and $\ell_{B}$ have to be perpendicular to the axis of $f_{k}$. It is easy to see that if $\tau(f_{k})>d_{k}$ then $f_{k}$ maps the complement of $B^{k}$ strictly inside $A^{k}$. We claim that $M>d_{k}$, where

[TABLE]

and both maxima are taken over all pairs $f_{m},f_{n}$ (recall that $C_{m,n}\neq 0$ for all $m,n$).

If $d_{k}<M_{j,k}+M_{k,l}$, our claim is obvious because $M_{j,k}+M_{k,l}<2\max\{M_{m,n}\}$. Assume that $d_{k}\geqslant M_{j,k}+M_{k,l}$, and let $\gamma_{B}$ be the hyperbolic line with the same endpoints as $B_{j,k}$ and $\gamma_{A}$ the line with the same endpoints as $A_{k,l}$ (see Figure 7). In other words, $\gamma_{B}$ is the unique hyperbolic line, perpendicular to $\mathrm{Ax}(f_{k})$ that is a distance $M_{j,k}$ away from $\ell_{A}$ and closer to $\ell_{B}$. So, $d_{k}=M_{j,k}+M_{k,l}+\rho(\gamma_{B},\gamma_{A})$. Consider the hyperbolic half-plane $H(\gamma_{B})$ that is bounded by $\gamma_{B}$ and contains $\ell_{A}$, and define $H(\gamma_{A})$ similarly. Because $d_{k}\geqslant M_{j,k}+M_{k,l}$, these two half-planes are disjoint. Also, since $A^{k}$ and $B^{k}$ were chosen to be the innermost intervals in their respective collection, for the transformations $f_{j}$ and $f_{l}$ we have that $\mathrm{Ax}(f_{j})\subset H(\gamma_{B})$ and $\mathrm{Ax}(f_{l})\subset H(\gamma_{A})$. Hence, the axes of $f_{j}$ and $f_{l}$ are disjoint, which implies that $\rho(\gamma_{B},\gamma_{A})<\rho(\mathrm{Ax}(f_{j}),\mathrm{Ax}(f_{l}))$, proving our claim.

Since $k$ was chosen arbitrarily, if $\tau(f_{k})>M$ for all $k=1,\dots,n$, then for each $k$ there exist open intervals $A^{k}$, $B^{k}$, with disjoint closures, such that $f_{k}$ maps the complement of $B^{k}$ inside $A^{k}$. Also, as all the fixed points of the generators are distinct, the collection $\{A^{k},B^{k}\colon k=1,2,\dots,n\}$ consists of disjoint intervals, and thus each $f_{k}$ maps $\cup_{i}A^{i}$ strictly inside $A^{k}$, which implies that $\langle f_{1},f_{2},\dots,f_{n}\rangle$ is a Schottky semigroup. In order to complete the proof in this case it suffices to show that

[TABLE]

To that end, recall that if $\mathrm{Ax}(f_{m})$ and $\mathrm{Ax}(f_{n})$ are disjoint (i.e. $C_{m,n}>0$) then

[TABLE]

and so

[TABLE]

Therefore,

[TABLE]

and the maximum can be taken over all pairs $f_{m},f_{n}$ with $C_{m,n}\neq-1$ (recall that $C(f,g)=-1$ implies that the axes of $f$ and $g$ cross orthogonally).

If $\lvert C_{m,n}\rvert<1$, then

[TABLE]

On the other hand, for $1<\lvert C_{m,n}\rvert\leqslant 4$ we have

[TABLE]

Finally, assuming that $\lvert C_{m,n}\rvert>4$, we obtain

[TABLE]

where the last two steps are obtained by using the triangle inequality.

Combining these last inequalities with (3.2) yields

[TABLE]

where the maximum can be taken over all pairs $f_{m},f_{n}$.

The only case left to consider now is when a number of generators of $S$ share the same attracting fixed point (the case where they share the same repelling fixed point is identical). Suppose, without loss of generality, that $\alpha(f_{1})=\alpha(f_{2})=\dots=\alpha(f_{k})$ for some $k<n$, and all other fixed points of the generators are distinct. Also, assume that

[TABLE]

where the maximum is taken over all pairs $f_{m},f_{n}$ with $C_{m,n}\neq 0$. Then, as we saw earlier, for each $j=1,2,\dots,n$ there exist open intervals $A^{j},B^{j}$, with disjoint closures, such that $f_{j}$ maps the complement of $B^{j}$ inside $A^{j}$.

Observe that in this case $\cap_{i=1}^{k}A^{i}\neq\emptyset$. Also, as the constant $M$ does not take into account pairs of generators with zero cross ratio, it is possible that $A^{m}\cap B^{n}\neq\emptyset$, for some $m,n\leqslant k$. However, since we assumed that $\tau(f_{i})>23>\log 5$ for $i=1,2,\dots,k$, Lemma 2.8 implies that there exist intervals $A,B$, such that $f_{i}$ maps the complement of $B$ inside $A$, for all $i=1,2,\dots,k$. Hence, for $i=1,2,\dots,k$, each $f_{i}$ maps the complement of $B^{i}\cap B$ inside $\left(\cup_{i=1}^{k}A^{i}\right)\cap A$, and these intervals are disjoint for all $i=1,2,\dots,k$. In addition, $B^{i}\cap B$ and $\left(\cup_{i=1}^{k}A^{i}\right)\cap A$ are disjoint from all $A^{j}$ and $B^{j}$, for all $i=1,2,\dots,k$ and $j=k+1,k+2,\dots,n$. Therefore, the semigroup $S$ maps the union of $\left(\cup_{i=1}^{k}A^{i}\right)\cap A$ with $\cup_{j=k+1}^{n}A^{j}$ strictly inside itself, completing our proof.