Distortion in groups of Affine Interval Exchange transformations

TL;DR

This paper investigates distortion phenomena in the group of Affine Interval Exchange Transformations, revealing structural properties of distorted elements and implications for group actions and subgroup classifications.

Contribution

It characterizes distorted elements in AIET groups and shows that certain groups cannot act faithfully, providing new insights into the structure of AIET groups.

Findings

Distorted elements have iterates conjugate to products of infinite order rotations.

No faithful action of Baumslag-Solitar groups with |m| ≠ |n| on AIETs.

Finitely generated nilpotent subgroups of AIETs are virtually abelian.

Abstract

In this paper, we study distortion in the group $\mathcal A$ of Affine Interval Exchange Transformations (AIET). We prove that any distorted element $f$ of $\mathcal A$, has an iterate $f^ k$ that is conjugate by an element of $\mathcal A$ to a product of infinite order restricted rotations, with pairwise disjoint supports. As consequences we prove that no Baumslag-Solitar group, $BS(m,n)$ with $\vert m \vert \neq \vert n \vert$, acts faithfully by elements of $\mathcal A$, every finitely generated nilpotent group of $\mathcal A$ is virtually abelian and there is no distortion element in $\mathcal A_{\mathbb Q}$, the subgroup of $\mathcal A$ consisting of rational AIETs.

Equations

$${\sigma }_{f\circ g}(x)=\frac{D{f}_{+}(g_{+}(x))D{g}_{+}(x)}{D{f}_{-}(g_{-}(x))D{g}_{-}(x)}=\frac{D{f}_{+}(g_{+}(x))}{D{f}_{-}(g_{-}(x))}\times {\sigma }_g(x).$$

$$\frac{l_n\log (\mathrm{Inf}{D}_{+}{g}_i)}{n}\le \frac{\log (D_{+}{f}^n)(p)}{n}\le \frac{l_n\log (\mathrm{Sup}{D}_{+}{g}_i)}{n},$$

$$\#BP(f^n)\le \#BP(f)\times n\text{ and }\#B{P}_0(f^n)\le \#B{P}_0(f)\times n.$$

$$\{b=f^{-p}(a),f^{-p+1}(a),...,a,...,f^l(a)=c\},$$

$${\Delta }_{f^n}(f^{-k}(a))=f_{+}^n(f^{-k}(a))-f_{-}^n(f^{-k}(a))=f^{n-k}(a)-f_{-}^{n-k}(f_{-}^k(f^{-k}(a)).$$

$${\Delta }_{f^n}(f^{-k}(a))={\Delta }_{f^{n-k}}(a)\text{, for }0\le k\le n-1.$$

$$\#B{P}_0(f)\times n\ge \#B{P}_0(f^n)\ge n-N_a.$$

$$B{P}_0(f^n)\le \sum _{a\in A}2N_a+\sum _{a\in B}period(a),$$

$${\sigma }_{f^n}(f^{-k}(a))=\frac{D{f}_{+}(f_{+}^{n-1}(f^{-k}(a)))...D{f}_{+}(f_{+}^k(f^{-k}(a)))...D{f}_{+}(f^{-k}(a))}{D{f}_{-}(f_{-}^{n-1}(f^{-k}(a)))...D{f}_{-}(f_{-}^k(f^{-k}(a)))...D{f}_{-}(f^{-k}(a))}.$$

$$\frac{D{f}_{+}(f_{+}^{n-1-k}(a))...D{f}_{+}(f_{+}^{N+1}(a))}{D{f}_{-}(f_{-}^{n-1-k}(a))...D{f}_{-}(f_{-}^{N+1}(a))}\times \frac{D{f}_{+}(f_{+}^N(a))...D{f}_{+}(a)}{D{f}_{-}(f_{-}^N(a))...D{f}_{-}(a)}\times \frac{D{f}_{+}(f^{-1}(a))...D{f}_{+}(f^{-k}(a))}{D{f}_{-}(f^{-1}(a))...D{f}_{-}(f^{-k}(a))}.$$

$${\sigma }_{f^n}(f^{-k}(a))=\frac{D{f}_{+}(f_{+}^N(a))...D{f}_{+}(a)}{D{f}_{-}(f_{-}^N(a))...D{f}_{-}(a)}=:{\Pi }_a$$

$$\#BP(f)\times n\ge \#B{P}_1(f^n)\ge n-N_a.$$

$$\#B{P}_1(f^n)\le \sum _{a\in A}2N_a+\sum _{a\in B}period(a),$$

$$BP(f^n)\subset BP(g_{i_1})\cup g_{i_1}^{-1}BP(g_{i_2})....\cup (g_{i_{l_{n-1}}}...g_{i_1}{)}^{-1}BP(g_{i_{l_n}}),\text{ then }$$

$$\#BP(f^n)\le \#BP(g_{i_1})+\#BP(g_{i_2})+...+\#BP(g_{i_{l_n}})\le l_n\max \{\#BP(g_i),i=1,...s\}$$

$${\Delta }_{f^n}(f^{-k}(a))=0,\text{ for all }k\ge 0,n-k>N_a.$$

$$B{P}_0(EF{E}^{-1})\subset E{F}^{-1}(B{P}_0(E))\cup E(B{P}_0(F))\cup B{P}_0(E^{-1}).$$

$$B{P}_0(EF{E}^{-1})\subset E(B{P}_0(F)).$$

$$EF{E}^{-1}(E({\omega }_i))=(EF{E}^{-1}{)}_{-}(E({\omega }_i)).$$

$${\Pi }_{a_i}=\frac{D{f}_{+}(f_{+}^{N_i}(a_i))....D{f}_{+}(a_i)}{D{f}_{-}(f_{-}^{N_i}(a_i))....D{f}_{-}(a_i)}=1,\forall i\in \mathcal{I}.$$

$$\Pi (f):=\prod _{i\in \mathcal{I},0\le k\le N_i}{\sigma }_{f^{N+1}}(f^k(a_i))=1.$$

$$\displaystyle\sigma_{H\circ f\circ H^{-1}}\bigl{(}H(f^{k}(a_{i}))\bigr{)}$$

$$=\frac{{\sigma }_{f^{N+1}}(f^{k+1}(a_i))\times {\sigma }_f(f^k(a_i))}{{\sigma }_{f^{N+1}}(f^k(a_i))}=1,$$

$$\underset{n\to +\infty }{\lim }\frac{N_1(x,n)}{n}=\mu ([0,a])\text{ and }\underset{n\to +\infty }{\lim }\frac{N_2(x,n)}{n}=\mu ([a,1]).$$

$${\mathcal{N}}_j(D_{+}{B}^n(x))={\beta }_j.N_1(x,n)+{\delta }_j.N_2(x,n).$$

$$\frac{{\mathcal{N}}_j(D_{+}{B}^n(x))}{n}={\beta }_j.\frac{N_1(x,n)}{n}+{\delta }_j.\frac{N_2(x,n)}{n}\to {\beta }_j.(1-\rho )+{\delta }_j\rho =\rho ({\delta }_j-{\beta }_j)+{\beta }_j.$$

$$|{\mathcal{N}}_j(D_{+}{B}^n(x))|\le \sum _{m=1}^{l_n}|{\mathcal{N}}_j(D_{+}{g}_{i_m}(x_m))|\le l_{n}S,$$

$$e^{2i\pi \frac{{\alpha }_i}{l_i}{m}^s}=e^{\pm 2i\pi \frac{{\alpha }_i}{l_i}{n}^s}.$$

$$[u^p,v^q]=c^{pq}.$$

Full text

1. Introduction.

In the recent years, notions of distortion have attracted the interest of many people working on geometric group theory as well as rigidity theory (see [10] for a survey).

On one hand, some results established the existence of distorted elements in transformations groups. For instance, D. Calegari and M. Freedman, in [7], showed that all homeomorphisms of spheres are distorted. Moreover, in the case of the unit circle, they proved that every irrational Euclidean rotation is distorted inside the group of $C^{2-\varepsilon}$-diffeomorphisms for any $\varepsilon>0$. Requiring smoothness, Avila proved in [1] that irrational rotations are distorted in ${\textit{Diff }}^{\infty}(\mathbb{S}^{1})$. In higher dimensions, Militon (see [18], Theorem 1) showed that irrational translations of the $d$-dimensional torus are distorted in ${\textit{Diff }}^{\infty}(\mathbb{T}^{d})$.

On the other hand, a significant consequence of non existence of distortion is the proof of the Zimmer conjecture in dimension 2: ”any action of $SL(3,\mathbb{Z})$ by area preserving diffeomorphisms on a surface, has finite image”. For instance, Polterovich ([24]) and Franks-Handel([11]) proved that ${\textit{Diff }}^{1}_{\mu}(\Sigma^{2})$ does not contain distortion, where $\mu$ is a full support measure on a compact surface $\Sigma^{2}$.

Novak (in [23]) proved that there is no element of distortion in the group of Intervals Exchange Transformations: bijections of the unit interval that are piecewise increasing and isometric.

In this work, we deal with a closely related problem, namely the existence of distortion elements inside the group of AIETs: Affine Intervals Exchange Transformations, denoted by $\mathcal{A}$. Roughly speaking an AIET is a bijection of the unit interval that is increasing and affine on a finite number of intervals. If the endpoints of these intervals and their images are rational points, then the AIET is called a rational AIET. The set of rational AIETs is a subgroup of $\mathcal{A}$ and it is denoted by $\mathcal{A}_{\mathbb{Q}}$.

Finitely generated groups of AIETs have provided several algebraically interesting groups, as for instance the classical Thompson’s groups $F$, $T$ and $V$ [8] as well as some of their generalized versions [14], [2], the fundamental groups of orientable surfaces [12], the modular group [22], wreath products of the form $\mathbb{Z}\wr\mathbb{Z}\wr...\wr\mathbb{Z}$ ([3], [5], [20], [21]) etc.

Our main result proves that most elements of $\mathcal{A}$ are in fact undistorted.

Theorem 1**.**

For every distorted element $f$ of $\mathcal{A}$, there exists an integer $k>0$, such that $f^{k}$ is conjugate by an element of $\mathcal{A}$ to a product of infinite order restricted rotations, with pairwise disjoint supports.

This description of distorted elements of $\mathcal{A}$ enables us to prove the following statements.

Theorem 2**.**

The Baumslag-Solitar groups $BS(n,m)=<a,b\ |\ ba^{n}b^{-1}=a^{m}>$ with $m$, $n$ integers and $|m|\not=|n|$ do not act faithfully via elements of $\mathcal{A}$.

Theorem 3**.**

Every torsion free nilpotent subgroup of $\mathcal{A}$ is abelian and every finitely generated nilpotent subgroup of $\mathcal{A}$ is virtually abelian.

As corollary, the Heisenberg group and thereby $SL(3,\mathbb{Z})$ do not act faithfully via elements of $\mathcal{A}$.

Last two theorems were proved by Higman for Thompson’s group $V$ (see [14] and [25], chapter 2).

Theorem 4**.**

There is no distortion elements in $\mathcal{A}_{\mathbb{Q}}$.

This theorem extends to all groups of rational AIET, results of Burillo-Cleary-Röver (see [6]) and Hmili-Liousse (see [15]) on non existence of distortion in Thompson’s groups $V_{n}$. The main consequence of this theorem is that any group $G$ containing distortion elements has no faithful actions as rational affine interval exchange transformations. Moreover, if $G$ is almost-simple, such actions have finite image.

**This paper is organized as follows: **

• •

In Section 2, definitions and basic facts are given.

In sections 3 to 8, we establish propositions that play an essential role in the proof of Theorem 1, that will be given in section 9.

• •

In section 3, we prove that elements of $\mathcal{A}$ with semi-hyperbolic periodic orbits are undistorted.

• •

In section 4, it is shown that given $f\in\mathcal{A}$, the sequence whose general term is the number of break points of the iterate $f^{n}$ of $f$ (for simplicity, this sequence will be called “number of break points of $f^{n}$ ” and it will be denoted by $\#BP(f^{n})$) is either bounded or growths linearly. As a consequence, for any distortion element the number of break points of $f^{n}$ is bounded.

In following sections, we study $f\in\mathcal{A}$ without semi-hyperbolic periodic orbit and with bounded number of break points of $f^{n}$.

• •

In section 5, Theorem 5 (Extended ”Alternate Version of Li’s Theorem”), we establish that such an $f$ has an iterate that is conjugate by an element of $\mathcal{E}$ to a product of restricted PL-homeomorphisms $f_{i}$ such that numbers of break points of $f_{i}^{n}$ are bounded.

• •

For such a PL-homeomorphism, in section 6, we apply results of Minakawa [19] to prove that it is PL-conjugate to a PL-homeomorphism $B$ with at most two distinct slopes.

• •

In section 7, under the additional assumption that $f$ is distorted, we derive that $B$ is a rotation, by showing that the slopes are $1$.

• •

In section 8, Theorem 1 is proved.

• •

Section 9 is devoted to prove applications of Theorem 1: Theorems 2, 3 and 4.

Acknowledgments. We are deeply indebted to Andres Navas for bringing this problem to our attention, for allowing us to resume work on the unpublished manuscript111Liousse, I. & Navas, A. Distortion elements in $PL_{+}(S^{1})$ (2008). and for several stimulating discussions during this work.

We gratefully acknowledge several fruitful discussions with Ignacio Monteverde who also pointed out an error in a preliminary version of this work.

2. Preliminaries.

2.1. Affine Interval Exchange Transformations of $I=[0,1)$.

Definition 2.1.

— A bijection $f$ of $[0,1)$ is an Affine Interval Exchange Transformation (or AIET) of $[0,1)$ if there exists a finite subdivision $0=a_{0}<a_{1}<....<a_{p}=1$ of $[0,1)$ such that for all $i=0,...,p-1$, one has $f(x)=\lambda_{i}x+\beta_{i}$ for $x\in[a_{i},a_{i+1})$, where $\lambda_{i}\in\mathbb{R^{*}_{+}}$ and $\beta_{i}\in\mathbb{R}$.

— A break point is either the initial point [math] or a discontinuity of $f$ or a discontinuity of $Df$, the derivative of $f$.

— The set of break points of $f$ is denoted $\mathbf{BP}(f)$; it can be decomposed as the union of $\mathbf{BP_{0}}(f)$, the set consisting of [math] and the discontinuities of $f$ and $\mathbf{BP_{1}}(f)$, the set of [math] and the discontinuities of $Df$.

— We define $\Delta_{f}(x)=f_{+}(x)-f_{-}(x)$, if $x\in(0,1)$ and $\Delta_{f}(0)=f_{+}(0)-f_{-}(1)$, where $f_{+}(a)=\displaystyle\lim_{x\rightarrow a_{+}}f(x)=f(a)$ and $f_{-}(a)=\displaystyle\lim_{x\rightarrow a_{-}}f(x)$.

— The $\lambda_{i}$’s are the **slopes ** of $f$.

— The jump of $f$ at $x$ is defined by $\sigma_{f}(x)=\frac{D_{+}f(x)}{D_{-}f(x)}$, if $x\in(0,1)$ and $\sigma_{f}(0)=\frac{D_{+}f(0)}{D_{-}f(1)}$, where $Df_{+}(a)=\displaystyle\lim_{x\rightarrow a_{+}}\frac{f(x)-f_{+}(a)}{x-a}$ and $Df_{-}(a)=\displaystyle\lim_{x\rightarrow a_{-}}\frac{f(x)-f_{-}(a)}{x-a}$.

— The sets of slopes and jumps of $f$ are denoted respectively by $\Lambda(f)$ and $\sigma(f)$.

— An AIET $f$ of $[0,1)$ is called an **IET ** if $\Lambda(f)=\{1\}$.

Definition 2.2.

— We denote by $\mathbf{\mathcal{A}}$ the group consisting of all AIETs of $[0,1)$.

— We denote by $\mathbf{\mathcal{E}}$ the group consisting of all IETs of $[0,1)$.

Remark 1**.**

A homeomorphism $f$ of the circle $\mathbb{S}^{1}=[0,1]/(0=1)$ can be seen as the bijection of $[0,1)$ defined by $x\mapsto f(x)\ (mod\ 1)$.

Definition 2.3.

When this bijection is an AIET, $f$ is called a PL-homeomorphism of $[0,1)$, even if it may not be continuous at eventually one point of $(0,1)$.

In what follows, PL-homeomorphisms of $\mathbb{S}^{1}$ will be seen as AIETs of $[0,1)$. For example, the circle rotation by $\alpha$ is viewed as the element of $\mathcal{A}$, with break points [math] and $1-\alpha$, given by $f(x)=x+\alpha$ if $x\in[0,1-\alpha)$ and $f(x)=x+\alpha-1$ if $x\in[1-\alpha,1)$ which is called rotation of $[0,1)$.

Definition 2.4.

An IET $f$ is called a restricted rotation if there exists some interval $I=[a,b)\subset[0,1)$ such that the support of $f$ is $I$ and $f(x)=x+\delta$ if $x\in[a,b-\delta)$ and $f(x)=x+\delta-b+a$ if $x\in[b-\delta,b)$, where $\delta\in\mathbb{R}$, $0<\delta<b-a$.

An AIET, $f$ is called a restricted PL-homeomorphism if there exists some interval $I=[a,b)\subset[0,1)$ such that the support of $f$ is $I$ and $\#(BP_{0}(f)\cap(a,b))\leq 1$.

Here are some elementary properties of the sets of break points.

Property 1**.**

Let $f$ and $g$ be elements of $\mathcal{A}$.

– $BP(f^{-1})=f(BP(f)$,

– $BP(f\circ g)\subset BP(g)\cup g^{-1}(BP(f))$,

– $BP(f^{n})\subset BP(f)\cup f^{-1}(BP(f))\cup...\cup f^{-(n-1)}(BP(f))$, for all integer $n\geq 0$.

These still hold for $BP_{0}(f)$.

Unfortunately, such formulas do not hold for $BP_{1}(f)$, this is due to the following

Property 2**.**

Let $f$, $g$ in $\mathcal{A}$ and $x\in[0,1)$, one has

[TABLE]

If $x\notin BP_{0}(g)$ then ${\displaystyle\sigma}_{f\circ g}(x)={\displaystyle\sigma}_{f}(g(x))\times{\displaystyle\sigma}_{g}(x)$.

For $x\in BP_{0}(g)$, ${\displaystyle\sigma}_{f\circ g}(x)\not={\displaystyle\sigma}_{f}(g(x))\times{\displaystyle\sigma}_{g}(x)$, in general.

Property 3**.**

Let $f\in\mathcal{A}$ and $g$ is a PL-homeomorphism, then

$\forall x\in[0,1),{\displaystyle\sigma}_{f\circ g}(x)={\displaystyle\sigma}_{f}(g(x))\times{\displaystyle\sigma}_{g}(x)$* and thereby $BP_{1}({f\circ g})\subset g^{-1}(BP_{1}({f}))\cup BP_{1}(g)$.*

Indeed, if $g$ is a PL-homeomorphism and $g_{+}(x)\not=g_{-}(x)$, then $g_{+}(x)=0$ and $g_{-}(x)=1$ therefore $\frac{Df_{+}(g_{+}(x))}{Df_{-}(g_{-}(x))}={\displaystyle\sigma}_{f}(0)$.

2.2. Interesting subgroups of $\mathcal{A}$.

Numerous generalizations of Thompson’s groups have been defined and studied: we recall, for example, the works of Bieri-Strebel [2] and Higman [14].

Definition 2.5.

Let $\Lambda\subset\mathbb{R}^{+*}$ be a multiplicative subgroup and $A\subset\mathbb{R}$ be an additive subgroup, invariant by multiplication by elements of $\Lambda$ and such that $1\in A$.

We define $V_{\Lambda,A}$ as the subgroup of $\mathcal{A}$ consisting of elements with slopes in $\Lambda$, break points and their images in $A$, and $\mathbf{\mathcal{E}}_{A}$ as the subgroup of $\mathcal{E}$ consisting of elements with break points in $A$, in fact $\mathbf{\mathcal{E}}_{A}=\mathbf{\mathcal{E}}\cap V_{\Lambda,A}$.

The subgroup ${A}_{\mathbb{Q}}$ of rational AIETs is $V_{\mathbb{Q}_{>0},\mathbb{Q}}$.

Definition 2.6.

Let $n$ be a positive integer, the group $V_{<n>,\ \mathbb{Z}[{1}/{n}]}$ is denoted by $V_{n}$ and called a Higman-Thompson’s group.

Note that the classical Thompson’s group $V$ arises as $V_{2}$.

Among many things, Higman (see [14], [25]) proved that $V_{n}$ is finitely presented and satisfies the conclusions of Theorems 2 and 3.

Remark 2**.**

As established in [9], Thompson’s groups $F$, $T$ and $V$ are not subgroups of $\mathcal{E}$.

Remark 3**.**

Similarly, one can define AIETs of any interval $[a,b)\subset\mathbb{R}$. In anticipation of a more general use of results from [16], [23], [19] stated for $[0,1)$, we note that there exists a unique direct affine map sending $[a,b)$ onto $[0,1)$. Thus any AIET of $[a,b)$ is affinely conjugate to an AIET of $[0,1)$ with the same slopes set. However, arithmetic properties of break points might not be preserved.

Definition 2.7.

Let $I=[a,b)$, We will denote by $\mathcal{A}(I)$ [resp. $\mathcal{E}(I)$] the group consisting of AIETs [resp. IETs] of $I$.

2.3. Distortion.

Definition 2.8.

Let $\Gamma$ be a finitely generated group and $S=\{s_{1},...,s_{r}\}$ be a finite generating set of $\Gamma$.

The smallest integer $l$ such that $g=s_{i_{1}}^{\epsilon_{1}}...s_{i_{l}}^{\epsilon_{l}}$, with ${\epsilon_{j}}\in\{-1,1\}$ is called the length of $g$ relatively to $S$ and denoted by $l_{S}(g)$.

We set $l_{S}(e)=0$. The function $l_{S}:\Gamma\rightarrow\mathbb{N}$ is invariant by taking inverse and satisfies: $l_{S}(gh)\leq l_{S}(g)+l_{S}(h)$. In particular, for all $g$ in $\Gamma$, the sequence $l_{S}(g^{n})$ is sub-additive, thus the sequence $\frac{l_{S}(g^{n})}{n}$ converges. This leads to

Definition 2.9.

We say that $g$ is distorted (or of distortion) in $\Gamma=<S>$  if $g$ has infinite order and $\lim\limits_{n\rightarrow+\infty}\frac{l_{S}(g^{n})}{n}=0$.

Remark 4**.**

The property of being distorted does not depend on the generating set.

Definition 2.10.

More generally, if $G$ is not finitely generated, an element $g$ of $G$ is said to be distorted in $G$ if it is a distortion element in some finitely generated subgroup of $G$.

Properties 2.1**.**

• •

The following properties are equivalent

1. (1)

$g\in G$* is distorted in $G$,* 2. (2)

$\exists N\in\mathbb{Z}^{*}$* : $g^{N}$ is distorted in $G$,* 3. (3)

$\forall N\in\mathbb{Z}^{*}$* : $g^{N}$ is distorted in $G$.*

• •

If $\Phi:\Gamma\rightarrow G$ is a morphism and $g\in\Gamma$ is distorted in $\Gamma$ then its image $\Phi(g)\in G$ is either of finite order or distorted in $G$.

3. Semi-hyperbolicity prevents distortion.

Definition 3.1.

Let $f\in\mathcal{A}$, we say that $p$ is a semi-hyperbolic periodic point of period $l$, if either:

— $p$ is not a break point of $f^{l}$, $f^{l}(p)=p$ and $Df^{l}\not=1$ (hyperbolic) or

— $p$ is a break point of $f^{l}$, $f^{l}_{+}(p)=p$ and $Df_{+}^{l}(p)\not=1$ or $f^{l}_{-}(p)=p$ and $Df_{-}^{l}(p)\not=1$ (virtual).

Proposition 3.1**.**

If $f\in\mathcal{A}$ has a semi-hyperbolic periodic point then $f$ is undistorted in $\mathcal{A}$.

Proof. Let $p$ be a semi-hyperbolic periodic point of $f$. W.l.o.g, we can suppose that $f_{+}(p)=p$ and the right derivative of $f$ at $p$ : $Df_{+}(p)=\lambda\not=1$. For clarity, $Df_{+}$ will be denoted by $D_{+}f$.

By absurd, suppose that $f$ is distorted in a subgroup $G$ of $\mathcal{A}$ generated by $S=\{g_{1},...,g_{s}\}$. Then $f$ can be written as $f^{n}=g_{i_{l_{n}}}...g_{i_{1}}$ with $\displaystyle\lim_{n\rightarrow+\infty}\frac{l_{n}}{n}=0$.

We have: $\displaystyle D_{+}f^{n}(p)=D_{+}g_{i_{l_{n}}}(p_{l_{n}})\ ...\ D_{+}g_{i_{1}}(p_{1})$, where $p_{1}=p$ and $p_{j}=g_{i_{j-1}}\ ...\ g_{i_{1}}(p)$, for $j=2,...,l_{n}$.

Then $(\mathop{\rm Inf}D_{+}g_{i})^{l_{n}}\leq D_{+}f^{n}(p)\leq(\mathop{\rm Sup}D_{+}g_{i})^{l_{n}}$ and

[TABLE]

where $\displaystyle\mathop{\rm Inf}D_{+}g_{i}=\mathop{\rm inf}_{i,x}D_{+}g_{i}(x)$ and $\displaystyle\mathop{\rm Sup}D_{+}g_{i}=\mathop{\rm sup}_{i,x}D_{+}g_{i}(x)$.

As $f$ is distorted, one has $\displaystyle\lim_{n\rightarrow+\infty}\frac{\log(D_{+}f^{n})(p)}{n}=0$.

On the other hand, since $p$ is a fix point of $f$, $\displaystyle D_{+}f^{n}(p)=\lambda^{n}$ and then

$\displaystyle\lim_{n\rightarrow+\infty}\frac{\log(D_{+}f^{n})(p)}{n}=\log\lambda\not=0$, this is a contradiction. $\square$

4. Alternative for the growth of the number of break points.

Recall that $BP(f)$ the set of break points of $f$ is the union of the two following sets:

$BP_{1}(f)=\{a\in[0,1):\sigma_{f}(a)\not=1\}\cup\{0\}$ and $BP_{0}(f)=\{a\in[0,1):\Delta_{f}(a)\not=0\}\cup\{0\}$.

We denote by $\#BP_{*}(f)$ the cardinality of $BP_{*}(f)$.

According to Property 1, one has :

[TABLE]

Proposition 4.1**.**

If $f\in\mathcal{A}$, then either

• •

$\#BP(f^{n})$* has linear growth or*

• •

$\#BP(f^{n})$* is bounded.*

Proof. Let $a\in BP(f)$.

Property 4**.**

If $a$ is a $f$-periodic point, then for all integer $n$, the set $BP(f^{n})\cap\mathcal{O}_{f}(a)$ is finite and has cardinality less or equal than the period of $a$.

Property 5**.**

1. (1)

If $a$ is not a $f$-periodic point, then there exists a segment $S_{a}$ of the orbit of $a$:

[TABLE]

such that $b$ and $c$ belong to $BP(f)$ and for all $k\in\mathbb{N}^{*}$, $f^{-k}(b)\notin BP(f)$ and $f^{k}(c)\notin BP(f)$. Such a break point $b$ is called initial break point. Therefore: 2. (2)

If $a$ is an initial break of $f$ point then $S_{a}=\{\ a,\ f(a),\ ...\ ,\ f^{N_{a}}(a)\ \}$ and

$\mathcal{O}_{f}(a)\cap BP(f^{n})\subset\{f^{-(n-1)}(a),...,f^{N_{a}}(a)\}$, for all integer $n\geq 0$, in particular:

(a)- $f^{-k}(a)\notin BP(f^{m})$, for all integers $k\geq m\geq 0$,

(b)- $f^{\ k}\ (a)\notin BP(f^{m})$, for all integers $m\geq 0$, $k>N_{a}$ and

(c)- $f^{-k}(a)\notin BP(f^{-p})$, for all integers $p\geq 0$, $k\geq 0$.

Indeed, because $\#BP(f)$ is finite, if (1) does not hold then there would exist some $d\in BP(f)$, $m_{1}$ and $m_{2}$ distinct integers such that $d=f^{m_{1}}(a)=f^{m_{2}}(a)$, which contradicts the non periodicity of $a$. We derive the second item from Property 1.

Step 1: Alternative for $\#BP_{0}(f^{n})$.

Let $a\in BP_{0}(f)$ be a non periodic initial break point and $S_{a}=\{a,f(a),...,f^{N_{a}}(a)\}$; for simplicity of notation, we set $N=N_{a}$.

Claim 1**.**

• •

If $\Delta_{f^{N+1}}(a)=0$ then for all integer $l\geq 1$, $\Delta_{f^{N+l}}(a)=0$.

• •

If $\Delta_{f^{N+1}}(a)\not=0$ then for all integer $l\geq 1$, $\Delta_{f^{N+l}}(a)\not=0$.

Indeed, let $l\geq 1$, $f_{+}^{N+l}(a)=f^{N+l}(a)=f^{l-1}(f^{N+1}(a))$ and $f_{-}^{N+l}(a)=f_{-}^{l-1}(f_{-}^{N+1}(a))$.

– If $\Delta_{f^{N+1}}(a)=0$ then $\Delta_{f^{N+l}}(a)=f^{l-1}(f^{N+1}(a))-f_{-}^{l-1}(f^{N+1}(a))=0$, since $f^{l-1}$ is continuous at $f^{N+1}(a)$ by Property 5 (2b).

– If $\Delta_{f^{N+1}}(a)\not=0$, as $f^{l-1}$ is continuous at $f^{N+1}(a)$, the point $f^{l-1}(f^{N+1}(a))$ can not be equal to $f_{-}^{l-1}(c)$, for some $c\not=f^{N+1}(a)$. It follows that $f^{N+l}(a)=f^{l-1}(f^{N+1}(a))\not=f_{-}^{l-1}(f_{-}^{N+1}(a))=f_{-}^{N+l}(a)$, since $f^{N+1}(a)\not=f_{-}^{N+1}(a)$.

We turn now to the proof of Step 1, estimating $\#BP_{0}(f^{n})$ for a given positive integer $n$.

By Property 5 (2), $\mathcal{O}_{f}(a)\cap BP(f^{n})\subset\{\ f^{-(n-1)}(a),\ ...\ a,\ ...\ ,f^{N}(a)\ \}$.

Let us compute $\Delta_{f^{n}}(f^{-k}(a))$, for $0\leq k\leq n-1$, we get:

[TABLE]

Moreover, for all $k\geq 0$, one has $f^{k}_{-}(f^{-k}(a))=a$, according to Property 5 (2a).

Finally, $\Delta_{f^{n}}(f^{-k}(a))=f^{n-k}(a)-f^{n-k}_{-}(a)=\Delta_{f^{n-k}}(a)$.

Summarizing, we have:

Lemma 4.1**.**

Let $a$ be an initial break point and $n$ be a positive integer.

[TABLE]

Combining previous Lemma and Claim 1, we deduce that:

• •

If $\Delta_{f^{N+1}}(a)\not=0$ then $\Delta_{f^{n}}(f^{-k}(a))\not=0$, for all $n-k>N$.

Hence $\#(BP_{0}(f^{n})\cap\mathcal{O}_{f}(a))\geq n-N$.

• •

If $\Delta_{f^{N+1}}(a)=0$ then $\Delta_{f^{n}}(f^{-k}(a))=0$, for all $n-k>N$.

Hence $\#(BP_{0}(f^{n})\cap\mathcal{O}_{f}(a))\leq N+\left(n-(n-N)\right)=2N$.

Conclusion 1.

• •

If exists $a\in BP_{0}(f)$ non periodic initial break point such that $\Delta_{f^{N_{a}+1}}(a)\not=0$, then

[TABLE]

That is $\#BP_{0}(f^{n})$ has linear growth.

• •

If for all $a\in BP_{0}(f)$ non periodic initial break point, $\Delta_{f^{N_{a}+1}}(a)=0$, then

[TABLE]

where $A=\{a\in BP_{0}(f){\text{ non periodic initial}}\}$ and $B=\{a\in BP_{0}(f){\text{ periodic }}\}$.

That is $\#BP_{0}(f^{n})$ is bounded.

Step 2: Alternative for $\#BP(f^{n})$.

If $\#BP_{0}(f^{n})$ is not bounded then $\#BP(f^{n})$ has linear growth, by Step 1.

Now suppose that $\#BP_{0}(f^{n})$ is bounded.

Let $a\in BP(f)$ be a non periodic initial break point and $S_{a}=\{a,f(a),...,f^{N}(a)\}$ be the segment containing all the break points of $f$ in the orbit of $a$.

Let $n\geq N+1$, recall that $BP(f^{n})\cap\mathcal{O}_{f}(a)\subset\{f^{-(n-1)}(a),...,a,...,f^{N}(a)\}$.

Let us compute the jump of $f^{n}$ at the point $f^{-k}(a)$ for $k\geq 0$ and $n-1-k>N$, that is for $0\leq k<n-1-N$.

Iterating the composition formula given in Property 2, we get :

[TABLE]

According to Property 5 (2), $f^{-k}(a)=f_{-}^{-k}(a)$ and therefore $f_{-}^{l}(f^{-k}(a))=f_{-}^{l}(f_{-}^{-k}(a))=f_{-}^{l-k}(a)$, for any integer $l\geq 0$; in addition, if $l\leq k$ then $f_{-}^{l}(f^{-k}(a))=f^{l-k}(a)$.

Therefore, noting that $n-1-k>N$, we get: ${\displaystyle\sigma}_{f^{n}}(f^{-k}(a))=$

[TABLE]

As $BP_{1}(f)\cap\mathcal{O}_{f}(a)\subset S_{a}$, the third ratio is trivial.

Since $\#BP_{0}(f^{n})$ is bounded, by Conclusion 1 and Claim 1, for all $m\geq N+1$ the point $f^{m}(a)=f_{+}^{m}(a)=f_{-}^{m}(a)$ and does not belong to $BP_{1}(f)$ (by Property 5 (2)). Thus, the first fraction is also trivial. Finally,

[TABLE]

Note that this formula also holds for $n-1-k=N$, since the first fraction does not appear in ${\displaystyle\sigma}_{f^{n}}(f^{-k}(a))$.

Therefore, the following alternative holds.

• •

If $\Pi_{a}\not=1$ then for all $k$ integer such that $0\leq k\leq n-1-N$, $f^{-k}(a)\in BP_{1}(f^{n})$ and $\#(BP_{1}(f^{n})\cap\mathcal{O}_{f}(a))\geq n-N$.

• •

If $\Pi_{a}=1$ then for all $k$ integer such that $0\leq k\leq n-1-N$, $f^{-k}(a)\notin BP_{1}(f^{n})$ and $\#(BP_{1}(f^{n})\cap\mathcal{O}_{f}(a))\leq N+(n-(n-N))=2N$.

Conclusion 2.

• •

If exists $a\in BP_{1}(f)$ non periodic initial break point such that $\Pi_{a}\not=1$, then

[TABLE]

Hence $\#BP_{1}(f^{n})$ and $\#BP(f^{n})$ have linear growth.

• •

If for all $a\in BP_{1}(f)$ non periodic initial break point $\Pi_{a}=1$, then

[TABLE]

where $A=\{a\in BP_{1}(f){\text{ non periodic initial}}\}$ and $B=\{a\in BP_{1}(f){\text{ periodic}}\}$.

Hence $\#BP_{1}(f^{n})$ and $\#BP(f^{n})$ are bounded.

$\square$

Proposition 4.2**.**

If $f$ is distorted in AIET then $\#BP(f^{n})$ is bounded.

Proof.

By absurd, suppose that $\#BP(f^{n})$ is unbounded and $f$ is distorted in a subgroup $G$ of $\mathcal{A}$ generated by $S=\{g_{1},...,g_{s}\}$. This means that $f^{n}$ can be written as $f^{n}=g_{i_{l_{n}}}...g_{i_{1}}$ with $\displaystyle\lim_{n\rightarrow+\infty}\frac{l_{n}}{n}=0$. Therefore, by Property 1, we have:

[TABLE]

[TABLE]

and therefore $\displaystyle\frac{l_{n}}{n}\geq\frac{\#BP(f^{n})}{n}(\max\{\#BP(g_{i}),i=1,...s\})^{-1}$.

Thus $\displaystyle\lim_{n\rightarrow+\infty}\frac{l_{n}}{n}>0$, since ${\#BP(f^{n})}$ has linear growth, according to Proposition 4.1, this is a contradiction. $\square$

5. Extended ”Alternative Version of Li’s Theorem”.

The aim of this section is to prove an extended version of the ”Alternate Version of Li’s Theorem” of [23].

Theorem 5**.**

Let $f$ in $\mathcal{A}$ without periodic points and with $\#BP(f^{n})$ bounded then there exists an integer $q$, such that $f^{q}$ is conjugate in $\mathcal{E}$ to a product of restricted PL-homeomorphisms of disjoint support that are minimal when restricted to their respective supports.

Definition 5.1.

Let $f\in\mathcal{A}$. We say that $f$ satisfies pair property if

1. (1)

$f$ does not have periodic points, 2. (2)

$BP_{0}(f)=\{\beta_{1},....\beta_{s},\omega_{1},....,\omega_{s}\}$, any pair $(\beta_{i},\omega_{i})$ for $i=1,...s$ verifies $f(\beta_{i})=\omega_{i}$ and $\beta_{i}\notin BP_{0}(f^{2})$ and 3. (3)

the $f$-orbits of $\beta_{i}$ are disjoint.

Convention. Eventually re-indexing the $\omega_{i}$, we suppose that $0=\omega_{1}<\omega_{2}<...<\omega_{s}$.

Basic Properties

If $f$ has pair property, then any associated pair $(\beta_{i},\omega_{i})$, for $i=1,...s$, verifies

1. (1)

$\beta_{i}\in BP_{0}(f)\setminus BP_{0}(f^{2})$, 2. (2)

$\omega_{i}\in BP_{0}(f)\cap BP_{0}(f^{-1})$, 3. (3)

pair property is invariant by $C^{0}$-conjugation. 4. (4)

if $f$ has pair property with associated pairs $(\beta_{i},\omega_{i})$ for $i=1,...s$ then, for any $n\in\mathbb{N}$, $f^{n}$ has pair property with associated pairs $(f^{-n}(\omega_{i}),\omega_{i})$ for $i=1,...s$.

Using these properties, we get $f_{-}(\beta_{i})\in BP_{0}(f)\cap BP_{0}(f^{-1})\cup\{1\}$ so we can give

Definition 5.2.

Let $\pi$ be the permutation of $\{1,...s\}$ defined by:

either $j={\pi(i)}$ in the case that $f_{-}(\beta_{i})=\omega_{j}$ or ${\pi(i)}=1$ when $f_{-}(\beta_{i})=1$.

Hence, one has $f(\omega_{i})=f_{-}(\omega_{\pi(i)})$, for $i\not=\pi^{-1}(1)$, otherwise, for $i=\pi^{-1}(1)$, $f(\omega_{i})=f_{-}(1)$.

Definition 5.3.

Let $f\in\mathcal{A}$ with pair property, a pair $(\beta_{i},\omega_{i})$ is said removable if either:

(1) $\omega_{\pi(i)}<\omega_{i}$ or

(2) $\omega_{\pi(i)}>\omega_{i}$ and there exists $\omega_{j}\in(\omega_{i},\omega_{\pi(i)})$.

Lemma 5.1**.**

Let $f\in\mathcal{A}$ without periodic points and for which $\#BP_{0}(f^{n})$ is bounded, then there exists an iterate of $f$ that satisfies pair property.

Proof.

According to Section 4, $\displaystyle BP_{0}(f)\subset\bigcup_{a\scriptscriptstyle{\text{ initial break point}}}$ $\{a,....,f^{N_{a}}(a)\}$ , $\Delta_{f^{N_{a}+1}}(a)=0$ and

[TABLE]

Let $N$ be the maximum of $\{N_{a}\}$, let $F=f^{N+1}$, we have that $BP_{0}(F)\cap\mathcal{O}_{f}(a)\subset\{f^{-N}(a),...,a,...,f^{N_{a}}(a)\}$.

We note that $\Delta_{F}(f^{-(N+1)+N_{a}+l}(a))=0,$ for $l=1,...,N-N_{a}+1$ by formula (1). Hence $\displaystyle BP_{0}(F)\cap\mathcal{O}_{f}(a)\subset\bigcup_{l=1}^{N_{a}}\{f^{-(N+1)+l}(a),f^{l}(a)\}.$

We claim that any pair $(\beta,\omega)=(f^{-(N+1)+l}(a),f^{l}(a))$ for $l=1,...,N_{a}$ satisfies $F(\beta)=\omega$ and $\beta\notin BP_{0}(F^{2})$.

Indeed, obviously $F(\beta)=\omega$, we now compute $\displaystyle\Delta_{F^{2}}(\beta)=\Delta_{f^{2N+2}}(f^{-(N+1)+l}(a))=0$ by formula (1) with $n=2N+2,k=N+1-l$.

It follows that either $f^{-(N+1)+l}(a)\in BP_{0}(F)$ and $f^{l}(a)\in BP_{0}(F)$ or $f^{-(N+1)+l}(a)\notin BP_{0}(F)$ and $f^{l}(a)\notin BP_{0}(F)$. So $BP_{0}(F)$ is a finite union of pairs of the form $(f^{-(N+1)+l}(a),f^{l}(a))$.

Obviously, $F$ satisfies the conditions $(1)$ and $(3)$ of the pair property. ∎

Lemma 5.2**.**

Let $F\in\mathcal{A}$ with pair property, and $(\beta_{i},\omega_{i})$ a removable pair, then there exists $E\in\mathcal{E}$ such that $\#BP_{0}(EFE^{-1})\leq\#BP_{0}(F)-2$ and $EFE^{-1}$ has also pair property. Moreover, $BP_{0}(E)\subset BP_{0}(F)\cap BP_{0}(F^{-1})$.

Proof.

– We begin by considering the case where $\omega_{\pi(i)}<\omega_{i}$.

Let $E$ be in $\mathcal{E}$ with $BP_{0}(E)=\{0,\omega_{\pi(i)},\omega_{i}\}$ and permutation $(1,2,3)\mapsto(1,3,2)$.

According to second item of Basic Properties, $BP_{0}(E)\subset BP_{0}(F)\cap BP_{0}(F^{-1})$.

One has that

[TABLE]

As $F^{-1}(BP_{0}(E))\subset BP_{0}(F)$ and $BP_{0}(E^{-1})=E(BP_{0}(E))\subset E(BP_{0}(F))$, it holds that

[TABLE]

We prove that $E(\beta_{i})$ and $E(\omega_{i})$ do not belong to $BP_{0}(EFE^{-1})$, by computing the right and left values at these points.

• •

$EFE^{-1}(E(\beta_{i}))=EF(\beta_{i})=E(\omega_{i})=\omega_{\pi(i)}$ and since $\beta_{i}\notin BP_{0}(E)$,

• •

$(EFE^{-1})_{-}(E(\beta_{i}))=E_{-}F_{-}(\beta_{i})=E_{-}(\omega_{\pi(i)})=\omega_{\pi(i)}$.

This proves that $E(\beta_{i})\notin BP_{0}(EFE^{-1})$.

• •

$EFE^{-1}(E(\omega_{i}))=EF(\omega_{i})$ and

• •

$(EFE^{-1})_{-}(E(\omega_{i}))=(EFE^{-1})_{-}(\omega_{\pi(i)}))=(EF)_{-}(\omega_{\pi(i)})$.

Since, $F(\omega_{i})=F_{-}(\omega_{\pi(i)})$ and do not belong to $BP_{0}(E)$,

[TABLE]

This proves that $E(\omega_{i})\notin BP_{0}(EFE^{-1})$.

Finally, $\#BP_{0}(EFE^{-1})\leq\#BP_{0}(F)-2$.

We claim that $EFE^{-1}$ has pair property, with associated pairs of the form $(E(\beta_{j}),E(\omega_{j}))$.

Indeed, as $(F^{-2}(\omega_{i}),\omega_{i})$ is a pair for $F^{2}$, it holds that $BP_{0}(E)\subset BP_{0}(F^{2})\cap BP_{0}(F^{-2})$ and same arguments as in the beginning of this proof, show that $BP_{0}(EF^{2}E^{-1})\subset E(BP_{0}(F^{2})).$

Obviously $EFE^{-1}(E(\beta_{j}))=E(\omega_{j})$. Suppose, by absurd, that $E(\beta_{j})\in BP_{0}(EF^{2}E^{-1})$ then $\beta_{j}\in E^{-1}(BP_{0}(EF^{2}E^{-1}))\subset BP_{0}(F^{2})$, which is a contradiction.

It is clear that $EFE^{-1}$ also satisfies the conditions $(1)$ and $(3)$ of the pair property.

– Now, we consider the case $(\beta_{i},\omega_{i})$ removable and $\omega_{\pi(i)}>\omega_{i}$ and there exists $\omega_{j}\in(\omega_{i},\omega_{\pi(i)})$.

We identify [math] to $1$ to get a circle and then we cut this circle at the point $\omega_{j}$. We are in the previous case. This ends the proof of Lemma 5.2.

∎

Proof of Theorem 5.

Let $f\in\mathcal{A}$ without periodic points and with bounded $\#BP_{0}(f^{n})$.

By Lemma 5.1, there exists some integer $N$ such that $f^{N}$ has the pair property.

Applying Lemma 5.2 a finite number of times, we get that $G=E_{r}.....E_{1}f^{N}E_{1}^{-1}...E_{r}^{-1}$ has pair property and no removable pair.

Since associated pairs $(\beta_{i},\omega_{i})$ of $G$ are not removable, then $\omega_{\pi(i)}>\omega_{i}$, for any $i$ and intervals $(\omega_{i},\omega_{\pi(i)})$ are pairwise disjoint and disjoint from the last interval $(\omega_{s},1)$; note that $s=\pi^{-1}(1)$, since by absurd $\omega_{s}<\omega_{\pi(s)}<1$, a contradiction.

We claim that for any $1\leq i<s$, there exists a unique discontinuity point of $G$ in $(\omega_{i},\omega_{\pi(i)})$ and this still holds for $(\omega_{s},1)$.

Indeed as $G_{+}(\omega_{i)})=G_{-}(\omega_{\pi(i)})$, the interval $(\omega_{i},\omega_{\pi(i)})$ contains at least a point of $BP_{0}(G)$, similar argument shows that the same holds for $(\omega_{s},1)$.

Pairs are unremovable so this discontinuity point is a $\beta_{j}$.

Since the number of $\beta$’s is exactly the number of intervals of the form $(\omega_{i},\omega_{\pi(i)})$ or $(\omega_{s},1)$, then $\beta_{j}$ is unique.

Therefore $G([\omega_{i},\omega_{\pi(i)}))=[\omega_{j},\omega_{\pi(j)})$, since $G_{-}(\beta_{j})=\omega_{\pi(j)}$ and $G_{+}(\beta_{j})=\omega_{j}$, if $j\not=\pi^{-1}(1)=s$. If $j=s$, then $G([\omega_{i},\omega_{\pi(i)})=[\omega_{s},1)$.

This implies that $R=\mathop{\sqcup}\limits_{i=1}^{s-1}[\omega_{i},\omega_{\pi(i)})\mathop{\sqcup}[\omega_{s},1)$ is $G$-invariant. We claim that it is $[0,1[$.

Indeed, if not, the complementary of $R$ is a finite union of half open intervals that is $G$-invariant and $G$ is continuous on each interval (since such intervals do not contain $\beta$’s and $\omega$’s the discontinuity points of $G$). Thus, these intervals are periodic which contradicts that $G$ ($f$) does not have periodic points.

Moreover, there exists an iterate of $G$ such that $G^{l}([\omega_{i},\omega_{\pi(i)}))=[\omega_{i},\omega_{\pi(i)})$, $G^{l}([\omega_{s},1)=[\omega_{s},\omega_{1})$ and the restriction of $G^{l}$ to any $[\omega_{i},\omega_{\pi(i)})$,$i=1,...,s-1$ and to $[\omega_{s},1)$ has just one interior discontinuity point.

Finally, $G^{l}$ is a product of restricted PL-homeomorphisms $\Gamma_{i}$ with disjoint support and it is conjugated by $E=E_{r}.....E_{1}$ to $f^{lN}$.

As $f$ and then $G^{l}$ has no periodic points, by Denjoy’s Theorem for Class P circle homeomorphisms (see [13]), each $\Gamma_{i}$ is minimal when restricted to its support.

This ends the proof of Theorem 5. $\square$

Remark 5**.**

Note that endpoints of the supports of the restricted PL-homeomorphisms and discontinuities of the $E_{i}$’s are in the orbit of $BP_{0}(f)$.

In particular, if $f\in\mathcal{A}_{\mathbb{Q}}$ then endpoints of the supports of the restricted PL-homeomorphisms are rational and $E\in\mathcal{A}_{\mathbb{Q}}$.

6. PL conjugation.

Next proposition is due to Minakawa [19].

Proposition 6.1**.**

Let $f\in\mathcal{A}$ a PL-homeomorphism such that $\#BP(f^{n})$ is bounded. Then there exists a PL-homeomorphism $H\in\mathcal{A}$ such that $H\circ f\circ H^{-1}$ is an AIET $B=B_{\lambda_{1},\lambda_{2}}$ verifying $\Lambda(B)=\{\lambda_{1},\lambda_{2}\}$ and $BP(B)=\{0,B^{-1}(0)\}$. In particular, $DB(x)=\lambda_{1}$ on $[0,B^{-1}(0))$ and $DB(x)=\lambda_{2}$ on $[B^{-1}(0),1)$.

Remark 6**.**

The maps $B_{\lambda_{1},\lambda_{2}}$ are PL-homeomorphisms. They were studied in [4]. There it was proven that $B$ is $C^{0}$-conjugate to a rotation $R_{\rho}$, by a map of the form $\displaystyle x\mapsto\frac{(\omega^{x}-1)}{(\omega-1)}$ for some positive $\omega$ distinct from 1, when $\lambda_{1}\neq\lambda_{2}$ and if $\lambda_{1}=\lambda_{2}$ then $B$ is a rotation.

Remark 7**.**

We shall give a refinement of Minakawa’s proof which will enable us to preserve arithmetic properties of $f$, this will be explained in Remark 8. An alternative proof using a ”PL pair property” can be found in [17].

Proof.

As $\#BP(f^{n})$ is bounded, Conclusion 2 in the proof of Proposition 4.1 indicates that there exists a subset $\{a_{i},i\in\mathcal{I}\}$ of $BP_{1}(f)$ such that $BP_{1}(f)$ is contained in $\bigsqcup_{i\in\mathcal{I}}{\mathcal{S}}_{i}$ with ${\mathcal{S}}_{i}=\{f^{k}(a_{i}),\ k=0,...,N_{i}\}$ and

[TABLE]

Note that $\frac{Df_{+}(f_{+}^{k}(a_{i}))}{Df_{-}(f_{-}^{k}(a_{i}))}={\displaystyle\sigma}_{f}(f^{k}(a_{i}))$, since $f_{+}^{k}(a_{i})=f_{-}^{k}(a_{i})$ or $f_{+}^{k}(a_{i})=0$ and $f_{-}^{k}(a_{i})=1$.

Then $\displaystyle\Pi_{a_{i}}=\prod_{c\in{\mathcal{S}}_{i}}\sigma_{f}(c)=1$, for all $i\in\mathcal{I}$.

Consider a PL homeomorphism $H_{f}=H$ of $[0,1)$ such that:

– the break points of $H$ are the points $f(a_{i})$, …, $f^{N_{i}}(a_{i})$, for $i\in\mathcal{I}$,

– with associated jumps $\sigma_{H}({f^{k}(a_{i}))=\sigma_{f^{N+1}}(f^{k}(a_{i}))}$ for $k=1,\dots,N_{i}$, where $N=\max\{N_{i},i\in\mathcal{I}\}$.

Note that we also have $\sigma_{H}(a_{i})=\sigma_{f^{N+1}}(a_{i})=\prod_{n=0}^{N}\sigma_{f}(f^{n}(a_{i}))=1$.

At the end of this proof, we indicate a general lemma about existence of PL-homeomorphisms with prescribed break points and slopes. It implies that a necessary and sufficient condition for the existence of such a homeomorphism $H$ is that the product of the $H$-jumps is trivial, that is

[TABLE]

– If $\Pi(f)=1$, then we can define a map $H$ as above and normalize it by setting $H(0)=0$.

– If $\Pi(f)\not=1$, then we add a break point $c\notin\{a_{i},...,f^{N_{i}}(a_{i})\}$ and require that $\sigma_{H}(c)=(\Pi(f))^{-1}$; we normalize $H$ by setting $H(c)=0$.

Now, since $f$ and $H$ are PL-homeomorphisms, Property 3 implies that

• •

the set $BP_{1}(H\circ f\circ H^{-1})$ satisfies

$\begin{aligned} BP_{1}(H\circ f\circ H^{-1})&\subset BP_{1}(H^{-1})\cup H(BP_{1}(f))\cup H\circ f^{-1}(BP_{1}(H))\\ &\subset\{H(a_{i}),...,H(f^{N_{i}}(a_{i})),i\in\mathcal{I}\}\cup\{H(c),H(f^{-1}(c))\},\end{aligned}$

• •

for $i\in\mathcal{I}$, $0\leq k\leq N_{i}$, the jump of $H\circ f\circ H^{-1}$ at $H(f^{k}(a_{i}))$ is equal to

[TABLE]

• •

$\sigma_{H\circ f\circ H^{-1}}(H(c))=\Pi(f)$ and

• •

\sigma_{H\circ f\circ H^{-1}}\bigl{(}H(f^{-1}(c))\bigr{)}=\Pi(f)^{-1}.

Conclusion.

• •

If $\Pi(f)=1$, the AIET $B=H\circ f\circ H^{-1}$ has no break of slopes, it is a rotation.

• •

If $\Pi(f)\not=1$, the AIET $B=H\circ f\circ H^{-1}$ has exactly two break of slopes at $0=H(c)$ and $B^{-1}(0)=H(f^{-1}(c))$, that is $\Lambda(B)=\{\lambda_{1},\lambda_{2}\}$. $\square$

Lemma 6.1**.**

Given $0=c_{0}<c_{1}<...<c_{p}<1$ points in $[0,1)$ and $\sigma_{0},...,\sigma_{p}$ positive real numbers such that $\prod\limits_{i=0}^{p}\sigma_{i}=1$, there exists a PL-homemorphism $H$ such that :

– $BP_{0}(H)=\{c_{0},c_{1},...,c_{p}\}$ and

– $\sigma_{H}(c_{i})=\sigma_{i}$, for $i=0,...,p$.

Proof is left to readers, however we indicate some elements of the construction of $H$.

If we denote $\Lambda(H)=\{\lambda_{1},...,\lambda_{p+1}\}$, one has

– $\lambda_{i}=\sigma_{0}...\sigma_{i-1}\lambda_{1}$ and

– $\lambda_{1}=(|I_{1}|+\sigma_{1}|I_{2}|+...+\sigma_{1}...\sigma_{p}|I_{p+1}|)^{-1}$ (by computing the total length of $H([0,1))$).

In particular, if $c_{i}$ and $\sigma_{i}$ are rational numbers then $\lambda_{i}\in\mathbb{Q}$. Moreover, we can choose $H$ such that $H(c_{j})\in\mathbb{Q}$, for some $c_{j}$ and then $H\in\mathcal{A}_{\mathbb{Q}}$.

Note that if $\prod\limits_{i=0}^{p}\sigma_{i}\not=1$ such an $H$ does not exist since it should satisfy that $\lambda_{p+1}=\sigma_{0}...\sigma_{p}\lambda_{1}$ and $\sigma_{p+1}=\frac{\lambda_{p+1}}{\lambda_{1}}.$

Remark 8**.**

We have described explicitly the conjugating PL-homeomorphism $H$, we can deduce that if $f\in\mathcal{A}_{\mathbb{Q}}$ then the break points of $H$ and the jumps of $H$ belong to $\mathbb{Q}$, provided that the point $c$ is chosen in $\mathbb{Q}$. Therefore, if $f\in\mathcal{A}_{\mathbb{Q}}$ then conclusions of Proposition 6.1 hold with $H$ and $B$ belonging to $\mathcal{A}_{\mathbb{Q}}$.

7. The case of $\Lambda(B)=\{\lambda_{1},\lambda_{2}\}$.

Definition 7.1.

Let $\alpha_{1},...,\alpha_{s}$ generating a rank $s$ free abelian multiplicative subgroup $\Lambda$ of $\mathbb{R}^{+*}$. Therefore, given $\lambda\in\Lambda$, there exists a unique $(n_{1},...,n_{s})\in\mathbb{Z}$, such that $\lambda=\alpha_{1}^{n_{1}}\ ...\ \alpha_{s}^{n_{s}}$ and we define $\mathcal{N}_{j}(\lambda)=n_{j}$, for all $j\in\{1,...,s\}$.

Proposition 7.1**.**

Let $B=B_{\lambda_{1},\lambda_{2}}\in\mathcal{A}$, such that $\Lambda(B)=\{\lambda_{1},\lambda_{2}\}\subset\Lambda$, $BP_{1}(B)=\{0,a=B^{-1}(0)\}$ and $B$ is $C^{0}$-conjugate to an irrational rotation $R_{\rho}$.

If $(\lambda_{1},\lambda_{2})\not=(1,1)$ then exist $j\in\{1,...,s\}$ and $x\in[0,1)$ such that $\frac{\mathcal{N}_{j}(D_{+}B^{n}(x))}{n}\rightarrow\nu\not=0$.

Proof.

Noting that $B$ satisfies that $DB(x)=\lambda_{1}$ on $[0,a)$ and $DB(x)=\lambda_{2}$ on $[a,1)$, one has $D_{+}B^{n}(x)=\lambda_{1}^{N_{1}(x,n)}\lambda_{2}^{N_{2}(x,n)}$, where

$\displaystyle N_{1}(x,n)=\#\{x,f(x),...,f^{n-1}(x)\}\cap[0,a)=\sum_{k=0}^{n-1}\mathbb{I}_{[0,a)}(f^{k}(x))$ and

$\displaystyle N_{2}(x,n)=\#\{x,f(x),...,f^{n-1}(x)\}\cap[a,1)=\sum_{k=0}^{n-1}\mathbb{I}_{[a,1)}(f^{k}(x))$.

The map $B$ has a unique invariant probability measure $\mu$, since it is $C^{0}$-conjugate to an irrational rotation $R_{\rho}$. More precisely, consider $h$ such that $h\circ B\circ h^{-1}=R_{\rho}$, one has $\mu(A)=\lambda(h(A)$, for all measurable set $A$. In particular, $\mu([0,a])=\mu([0,B^{-1}(0)))=\lambda([h(0),h\circ B^{-1}(0)))=\lambda([h(0),R_{\rho}^{-1}(h(0))))=(1-\rho)$ and $\mu([a,1))=\rho$.

Birkhoff Ergodic Theorem implies that for $\mu$-almost every point $x\in[0,1)$, one has

[TABLE]

Now, let us write $\lambda_{1}$ and $\lambda_{2}$ in the basis $\alpha_{1},...,\alpha_{s}$ of $\Lambda$: $\lambda_{1}=\alpha_{1}^{\beta_{1}}...\alpha_{s}^{\beta_{s}}$ and $\lambda_{2}=\alpha_{1}^{\delta_{1}}...\alpha_{s}^{\delta_{s}}$ and compute the coordinates $\mathcal{N}_{j}(D_{+}B^{n}(x))$ of $D_{+}B^{n}(x)$ in this basis.

As $\displaystyle D_{+}B^{n}(x)=\lambda_{1}^{N_{1}(x,n)}\lambda_{2}^{N_{2}(x,n)}=\alpha_{1}^{\beta_{1}.N_{1}(x,n)+\delta_{1}.N_{2}(x,n)}\ ...\ \alpha_{s}^{\beta_{s}.N_{1}(x,n)+\delta_{s}.N_{2}(x,n)}$, one has

[TABLE]

It follows that

[TABLE]

Finally, suppose that $(\lambda_{1},\lambda_{2})\not=(1,1)$ then necessary $\lambda_{1}\not=\lambda_{2}$ and there exists $j$ such that $\delta_{j}\not=\beta_{j}$. Therefore, $\nu=\rho(\delta_{j}-\beta_{j})+\beta_{j}\not=0$, as $\rho\notin\mathbb{Q}$. $\square$

8. Proof of Theorem 1.

Let $f$ be distorted in $\mathcal{A}$, as $f$ has no a semi-hyperbolic periodic point, its periodic points are not isolated. Using in addition that $BP_{0}(f)$ is finite, we get that the set $Per(f)$ of $f$-periodic points is the union of a finite collection of half open intervals with endpoints in the orbits of $BP_{0}(f)$. Thereby, there exists some positive integer $p$ such that $Per(f)=Per(f^{p})=Fix(f^{p})$. It is easy to check that there exists $S\in\mathcal{E}$ whose discontinuities are endpoints of connected components of $Per(f)$ and such that $Fix(Sf^{p}S^{-1})$ is an interval $P=[0,a)$ and the restriction of $Sf^{p}S^{-1}$ to $M=[a,1)$ has no periodic points.

Applying Theorem 5 to the restriction of $Sf^{p}S^{-1}$ to $M$, there exist $q\in\mathbb{N}^{*}$ and $E\in\mathcal{E}$ such that $ES\circ f^{pq}\circ(ES)^{-1}=\displaystyle\prod\limits_{i=1}^{p}f_{i}$, where $f_{i}$ are restricted PL-homeomorphisms with pairwise disjoint support $I_{i}=[a_{i},b_{i})$, $f_{i}|_{I_{i}}$ is minimal and $\#BP(f_{i}^{n})$ is bounded (since $f$ is distorted).

Let $i\in\{1\ ...\ p\}$, by Proposition 6.1 to $f_{i}|_{I_{i}}$, it is conjugate by a PL-homeomorphism $H_{i}$ of $I_{i}$ to $B_{i}$ with $\Lambda(B_{i})=\{\lambda_{i,1},\lambda_{i,2}\}$ and $BP(B_{i})=\{a_{i},B_{i}^{-1}(a_{i})\}$. Since $f_{i}|_{I_{i}}$ is minimal, $B_{i}$ also is minimal and according to Remark 6, $B_{i}$ is $C^{0}$-conjugate to an infinite order rotation $R_{\rho}$ of $I_{i}$.

Let $H\in\mathcal{A}$ defined by $H(x)=H_{i}(x)$, if $x\in I_{i}$ and $H(x)=x$, if $x\notin\cup I_{i}$ and let $B=(HES)\circ f^{pq}\circ(HES)^{-1}$.

It is easy to check that $B|_{I_{i}}=B_{i}$ and $B$ is distorted in a subgroup $G=<g_{1},...,g_{q}>$ of $\mathcal{A}$, since $f$ is distorted in $\mathcal{A}$.

Let $\Lambda_{G}$ be the free abelian multiplicative subgroup $\Lambda$ of $\mathbb{R}^{+*}$ generated by $\{\ Dg_{k}(x),\ x\in[0,1),\ k\in\{1\ ...\ q\}\ \}$. It has finite rank $s$, we consider a basis $\alpha_{1},...,\alpha_{s}$ of it.

Let $i\in\{1\ ...\ p\}$, note that $\mathcal{N}_{j}(D_{+}B^{n}(y))=\mathcal{N}_{j}(D_{+}B_{i}^{n}(y))$, $\forall y\in I_{i}$. We suppose that $(\lambda_{i,1},\lambda_{i,2})\not=(1,1)$.

On one hand, by Proposition 7.1, there exist $j\in\{1,...,s\}$ and $x\in I_{i}$ such that $\frac{\mathcal{N}_{j}(D_{+}B^{n}(x))}{n}\rightarrow\nu\not=0$.

On the other hand, since $B$ is distorted in $G$, its iterates $B^{n}$ can be written $B^{n}=g_{i_{l_{n}}}\ ...\ g_{i_{1}}$ with $\displaystyle\lim_{n\rightarrow+\infty}\frac{l_{n}}{n}=0$.

Hence $D_{+}B^{n}(x)=D_{+}g_{i_{l_{n}}}(x_{l_{n}})\ ...\ D_{+}g_{i_{1}}(x_{1})$, where $x_{m}=g_{i_{m-1}}\ ...\ g_{i_{1}}(x)$. Then

[TABLE]

where $S=\max\{\ |\mathcal{N}_{j}(D_{+}g_{k}(y))|,\ y\in[0,1),\ 1\leq k\leq q\ \}.$

Finally, $\displaystyle\frac{l_{n}}{n}\geq\frac{|\mathcal{N}_{j}(D_{+}B^{n}(x))|}{nS}\rightarrow\frac{|\nu|}{S}>0$, this is a contradiction.

Consequently, for any $i\in\{1\ ...\ p\}$, $(\lambda_{i,1},\lambda_{i,2})=(1,1)$ and thereby $B_{i}$ is an infinite order rotation of $I_{i}$. Thus $B$ is a product of infinite order restricted rotations with pairwise disjoint supports.

In conclusion, we have proved that when restricted to $M$, there exists an iterate of $f$ that is conjugate in $\mathcal{A}$ to a product of infinite order restricted rotations with pairwise disjoint supports. We conclude by noting that $f_{|M^{c}}=Id_{|M^{c}}$. $\square$

9. Proof of Theorems 2, 3 and 4.

9.1. Proof of Theorem 2.

Let $a,b$ in $\mathcal{A}$ such that $ba^{m}b^{-1}=a^{n}$ with $m,n$ integers and $|m|\neq|n|$. We will prove that $a$ has finite order.

By absurd, since $a$ is distorted, eventually passing to a power of $a$ and conjugating $a$ and $b$ by an element of $\mathcal{A}$ we can suppose that $a$ is a product of infinite order restricted rotations $R_{\alpha_{i}}$ of disjoint supports $I_{i}$. We denote $\displaystyle a=\prod_{i=1}^{p}(R_{\alpha_{i}},I_{i})$.

The main tool of the proof is the following

Lemma 9.1**.**

Let $a,b\in\mathcal{A}$ with $a=\prod_{i=1}^{p}(R_{\alpha_{i}},I_{i})$ and $ba^{m}b^{-1}=a^{n}$ then there exists an integer $s$ such that $b^{s}$ maps $I_{i}$ to itself preserving the Lebesgue measure on $I_{i}$, $Leb|I_{i}$.

Let $X=\cup I_{i}$. In what follows we identify $a$ with its restriction to $X$.

By unique ergodicity of irrational rotations, one has that ergodic $a^{p}$-invariant probabilities on $X$ are $Leb|I_{i}$, for all $p\in\mathbb{Z}$.

As $Supp(ba^{m}b^{-1})=b(Supp(a^{m}))$ and $Supp(a^{n})=Supp(a^{m})$, then $b(X)=X$ and $b$ is identified to its restriction to $X$.

The image by $b$ of an ergodic $a^{m}$-invariant measure is an ergodic $a^{n}$-invariant measure.

Hence, for some permutation $\sigma$, $b_{\star}(Leb|I_{i})=Leb|I_{\sigma(i)}$. Thus there exists an integer $s$ such that $b^{s}_{\star}(Leb|I_{i})=Leb|I_{i}$. $\square$

Spectrum of irrational rotations viewed as IET are $Sp((R_{\alpha},I,Leb|I)=<e^{2i\pi\frac{\alpha}{l}}>$ where $l=|I|$.

As a consequence of $b^{s}a^{m^{s}}b^{-s}=a^{n^{s}}$, $b^{s}|I_{i}$ sends the generator of $Sp(a^{m^{s}},I_{i},Leb|I_{i})$ into a generator of $Sp(a^{n^{s}},I_{i},Leb|I_{i})$. Then

[TABLE]

Finally, $\frac{\alpha_{i}}{l_{i}}m^{s}=\pm\frac{\alpha_{i}}{l_{i}}n^{s}$ mod $\mathbb{Z}$. This is a contradiction since $\frac{\alpha_{i}}{l_{i}}\notin\mathbb{Q}$ and $|m|\neq|n|$.

Therefore $a$ has finite order, hence any action of $BS(m,n)$ with $|m|\neq|n|$ by elements of $\mathcal{A}$ is not faithful.

9.2. Proof of Theorem 3.

Let $G$ be a nilpotent subgroup of $\mathcal{A}$.

Suppose by absurd that $G$ is either non abelian torsion free or finitely generated and not virtually abelian.

Since $G$ is nilpotent there exist $u,v\in G$ such that $c=[u,v]$ commutes with $u$ and $v$. Furthermore, we can choose $c$ of infinite order because $G$ is either non abelian torsion free or finitely generated and not virtually abelian. This implies

Claim 9.1**.**

For any integers $p$ and $q$, it holds that

[TABLE]

In particular, $c^{n^{2}}=[u^{n},v^{n}],$ so $c$ is distorted.

Hence, eventually passing to a power of $c$ and conjugating by an element of $\mathcal{A}$ we can suppose that $c$ is a product of infinite order restricted rotations $R_{\alpha_{i}}$ of disjoint supports $I_{i}$. We denote $\displaystyle c=\prod_{i=1}^{p}(R_{\alpha_{i}},I_{i})$.

Applying Lemma 9.1 with $m=n$, $a=c$, and $b=u$ [resp. $b=v$], there exist $s_{u}$ [resp. $s_{v}$] such that $u^{s_{u}}_{\star}(Leb|I_{i})=Leb|I_{i}$ and $v^{s_{v}}_{\star}(Leb|I_{i})=Leb|I_{i}$.

Therefore $u^{s_{u}}|I_{i}$ and $v^{s_{v}}|I_{i}$ are IET which commute with the rotation $R_{\alpha_{i}}$. Finally, by Lemma 5.1 of [23], $u^{s_{u}}|I_{i}$ and $v^{s_{v}}|I_{i}$ are rotations so they commute.

According to claim 9.1, $[u^{s_{u}},v^{s_{v}}]=c^{s_{u}s_{v}}$, so $c^{s_{u}s_{v}}=Id$ on $I_{i}$ for any $i=1,...,p$. It follows that $c^{s_{u}s_{v}}=Id$. This contradicts that $c$ has infinite order.

9.3. Proof of Theorem 4.

In this section we prove that there is no distortion elements in $\mathcal{A}_{\mathbb{Q}}$.

Let $f\in\mathcal{A}_{\mathbb{Q}}$ distorted in $\mathcal{A}_{\mathbb{Q}}$, by Theorem 1, there exist a positive integer $s$ and $H\in\mathcal{A}$, $E\in\mathcal{E}$ and $S\in\mathcal{E}$ such that $(HES)f^{pq}(HES)^{-1}=\prod\limits_{i=1}^{p}(R_{i},I_{i})$ is a product of infinite order restricted rotations of disjoint supports $I_{i}$.

We first check that the conjugating maps $H$, $E$ and $S$ are in $\mathcal{A}_{\mathbb{Q}}$.

By definition of $S$, break points of $S$ are endpoints of connected components of $Per(f)$ so belong to the orbit of $BP_{0}(f)$, that is contained in $\mathbb{Q}$. Therefore $S\in\mathcal{A}_{\mathbb{Q}}$

According to Remark 5, $E\in\mathcal{A}_{\mathbb{Q}}$ and endpoints of the $I_{i}$’s are rational. Hence, by Remark 8, $H\in\mathcal{A}_{\mathbb{Q}}$.

Therefore $(HES)f^{pq}(HES)^{-1}\in\mathcal{A}_{\mathbb{Q}}$ and then $R_{i}\in\mathcal{A}_{\mathbb{Q}}$. This is a contradiction. $\square$