Product of two staircase rank one transformations that is not loosely Bernoulli

TL;DR

This paper constructs two specific staircase rank one transformations whose Cartesian product does not exhibit the loosely Bernoulli property, challenging assumptions about their typical behavior.

Contribution

The paper introduces a novel example of staircase rank one transformations with a non-loosely Bernoulli product, providing new insights into their dynamical properties.

Findings

Constructed two staircase rank one transformations

Their Cartesian product is not loosely Bernoulli

Challenges previous assumptions about typical behavior

Abstract

We construct two staircase rank one transformations whose Cartesian product is not loosely Bernoulli.

Equations

$$\begin{gathered} {\mathcal{C}}_{\gamma ,{\gamma }^{\prime }}:=\{T\in Rank(1):\text{ there exists }{n}_T^{\prime }\in \mathbb{N}\text{ such that for every }n⩾n_T^{\prime }, \\ {p}_n^T\in [(h_n^T{)}^{\gamma },(h_n^T{)}^{{\gamma }^{\prime }}),a_{n,i}^T>a_{n,i-1}^T\text{ for }i=2,\dots ,p_n^T,\text{ and }{a}_{n,p_n}^T⩽(h_n^T{)}^{{\gamma }^{\prime }}\}. \end{gathered}$$

$${\mathcal{L}}^{\gamma ,{\gamma }^{\prime },\delta }:=\left\{(T,S)\in {\mathcal{C}}_{\gamma ,{\gamma }^{\prime }}\times {\mathcal{C}}_{\gamma ,{\gamma }^{\prime }}:(h_n^T)\text{ and }(h_n^S)\text{ are }\delta -\text{alternating}\right\}.$$

$$h_n=p_{n-1}{h}_{n-1}+\sum _{i=1}^{p_{n-1}}{a}_{n-1,i}.$$

$$h_n⩾2^{n-1}.$$

$$\sum _{n=1}^{+\infty }\frac{1}{{\prod }_{k=1}^n{p}_i}\left(\sum _{i=1}^{p_n}{a}_{n,i}\right)<+\infty .$$

$$\prod _{i=1}^{n-1}{p}_i⩽h_n⩽K\prod _{i=1}^{n-1}{p}_i.$$

$$\sum _{i=1}^{p_{n-1}}{a}_{n-1,i}={ϵ}_{n-1}\prod _{i=1}^{n-1}{p}_i⩽{ϵ}_{n-1}{p}_{n-1}{h}_{n-1}$$

$$h_n⩽p_{n-1}{h}_{n-1}(1+{ϵ}_{n-1})$$

$$h_n⩽\left(\prod _{i=1}^{n-1}(1+{ϵ}_i)\right)\prod _{i=1}^{n-1}{p}_i⩽K\prod _{i=1}^{n-1}{p}_i,$$

$$d_H(x,y):=\frac{1}{h_{m(x,y)}},$$

$$\bar{f}(A,B)=1-\frac{r}{k},$$

$$\bar{f}({\mathcal{P}}_0^N(x),{\mathcal{P}}_0^N(y))<\epsilon .$$

$$I(M,w)=\{s\in \{1,\dots r\}:i_s\in [i_w,i_w+M]\}\subset [0,N]$$

$$J(M,w)=\{s\in \{1,\dots r\}:j_s\in [j_w,j_w+M]\}\subset [0,N].$$

$$I(M,w)\cap I(M,s)=\varnothing ,$$

$$\min \left(|I(K^{1+\xi },s)|,|J(K^{1+\xi },s)|\right)⩽2K,$$

$$|B_1|⩽2K.$$

$$|B_{\ell +1}|⩽2K.$$

$$B_1\cup \cdots \cup B_v=\{1,\dots ,r\},$$

$$|B_{\ell }|=\min (|I(K^{1+\xi },s_{\ell })|,|J(K^{1+\xi },s_{\ell })|)⩽2K\text{ for every }\ell \in \{1,\dots ,v\}.$$

$$B_{{\ell }_1}\cap B_{{\ell }_2}=\varnothing .$$

$$\frac{v}{2}⩽\frac{N}{K^{1+\xi }}.$$

$$r\le |B_1|+\cdots +|B_v|\le 2Kv\le \frac{4N}{K^{\xi }}.$$

$$(h_{n-1}^S{)}^{1+\gamma /4}<h_n^T\text{ and }(h_n^T{)}^{1+\gamma /4}<h_n^S.$$

$${\left(\prod _{i=1}^{n-2}{p}_i^S\right)}^{1+(\frac{1}{4}+\eta )\gamma }<\prod _{i=1}^{n-1}{p}_i^T\text{ and }{\left(K_T\prod _{i=1}^{n-1}{p}_i^T\right)}^{1+\gamma /4}<\prod _{i=1}^{n-1}{p}_i^S.$$

$$K_T\prod _{i=1}^n{p}_i^T⩾h_n^T⩾(h_{n-1}^T{)}^{1+\gamma }⩾(\prod _{i=1}^{n-1}{p}_i^T{)}^{1+\gamma }.$$

$${\left(\prod _{i=1}^{n_T^{\prime }+1}{p}_i^S\right)}^{1+(\frac{1}{4}+\eta )\gamma }=(2^{n_T^{\prime }+1}{)}^{1+(\frac{1}{4}+\eta )\gamma }⩽{\left(\frac{1}{K_T}{2}^{n_T^{\prime }+1}\right)}^{1+\gamma }⩽\prod _{i=1}^{n_T^{\prime }+2}{p}_i^T.$$

$$\left(\prod _{i=1}^w{p}_i^S\right)p_{w+1}^S\in \left[{\left(K_T\prod _{i=1}^{w+1}{p}_i^T\right)}^{1+\gamma /4},{\left(\prod _{i=1}^{w+2}{p}_i^T\right)}^{\frac{1}{1+(\frac{1}{4}+\eta )\gamma }}\right].$$

$$\begin{gathered} \left|{\left(\prod _{i=1}^{w+2}{p}_i^T\right)}^{\frac{1}{1+(\frac{1}{4}+\eta )\gamma /4}}-{\left(K_T\prod _{i=1}^{w+1}{p}_i^T\right)}^{1+\gamma /4}\right|⩾ \\ \left|{\left(\frac{1}{K_T}\prod _{i=1}^{w+1}{p}_i^T\right)}^{\frac{1+\gamma }{1+(\frac{1}{4}+\eta )\gamma }}-{\left(K_T\prod _{i=1}^{w+1}{p}_i^T\right)}^{1+\gamma /4}\right|⩾4\prod _{i=1}^{w+1}{p}_i^T, \end{gathered}$$

$$\frac{p_w^S}{{\prod }_{i=1}^{w-1}{p}_i^S}⩽\frac{{\prod }_{i=1}^{w+1}{p}_i^T}{{\left({\prod }_{i=1}^{w-1}{p}_i^S)\right)}^2}⩽\frac{{\prod }_{i=1}^{w+1}{p}_i^T}{{\left({\prod }_{i=1}^{w-1}{p}_i^T\right)}^2}=\frac{p_w^T{p}_{w+1}^T}{{\prod }_{i=1}^{w-1}{p}_i^T}.$$

Full text

1. 1 Introduction

2. 1.1 Statement of the main result

3. 2 Basic definitions

4. 2.1 Rank one maps

5. 2.2 Loosely Bernoulli transformations

6. 3 Preliminary lemmas

7. 3.1 A combinatorial lemma: lower bound on $\bar{f}$

8. 3.2 Proof of Lemma 1.2

9. 3.3 Distribution of points for maps from $\mathcal{C}_{\gamma,\gamma^{\prime}}$

10. 4 A proposition which implies Theorem 2

11. 5 Proof of Proposition 4.1

12. 5.1 Construction of $B$ and $C$ in Proposition 4.1

13. 5.2 Proof of $(P2)$ in Proposition 4.1

1 Introduction

In this note we study the loosely Bernoulli property for zero entropy, measure preserving transformations. Loose Bernoulliness was introduced by J. Feldman, [4], and A. Katok, [6]111In [6] loosely Bernoulli is called standard.. Recall that a zero entropy measure preserving transformation is loosely Bernoulli (LB for short) if it is isomorphic to a transformation induced from an irrational rotation of the circle. It follows from [3] that the class of loosely Bernoulli systems is broad: it is closed under taking factors, compact extensions and inverse limits. Moreover it contains all finite rank systems. First non-loosely Bernoulli examples were constructed by J. Feldman [4] by a cutting and stacking method. D. Ornstein, D. Rudolph and B. Weiss, [3], constructed a rank one transformation $T$ such that $T\times T$ is not loosely Bernoulli. The example in [3] is based on Ornstein’s construction in [7] of a class of almost surely mixing rank one transformation with random spacers. On the other hand, M. Gerber constructed in [5] a mixing rank one transformation whose Cartesian square is loosely Bernoulli. An algebraic example of a zero entropy non-loosely Bernoulli transformation comes from the work of M. Ratner, [9], where it was shown that the Cartesian square of the horocycle flow (on compact quotients) is not loosely Bernoulli, although horocycle flows are loosely Bernoulli, [8].

In this note we study the loosely Bernoulli property for a natural class of mixing rank one transformations: staircase rank one systems (mixing for this class was proved by Adams, [1]). Loose Bernoulliness for products of staircase rank one transformations is not known. Our main result implies in particular that there exit two staircase rank one transformations whose product is not loosely Bernoulli.

1.1 Statement of the main result

For a rank one system $T$, let $(p^{T}_{n})_{n\in\mathbb{N}}$, $(a_{n,i})_{i=1}^{p_{n}^{T}}$ and $(h_{n}^{T})_{n\in\mathbb{N}}$ denote the sequence of cuts, spacers and heights respectively (see Section 2.1). For $0<\gamma<\gamma^{\prime}<1$ define

[TABLE]

Notice that if $T$ is a staircase rank one transformation, i.e. $a^{T}_{n,i}=i$ for $n\in\mathbb{N}$ and $i\in\{1,\ldots,p^{T}_{n}\}$, and if $p^{T}_{n}\in[(h^{T}_{n})^{\gamma},(h^{T}_{n})^{\gamma^{\prime}}]$, then $T\in C_{\gamma,\gamma^{\prime}}$.

Definition 1.1** ($\delta$-alternating sequences).**

Fix $\delta>0$ and let $(a_{n})$ and $(b_{n})$ be two increasing sequences of positive integers. We say that $(b_{n})$ is $(a_{n})$-alternating with exponent $\delta$ if the following holds: there exists $n_{0}\in\mathbb{N}$ such that for every $n\geqslant n_{0}$, if $m(n)$ is unique such that $b_{m(n)}\leqslant a_{n}<b_{m(n)+1}$, then $b_{m(n)}^{1+\delta}<a_{n}$ and $a_{n}^{1+\delta}<b_{m(n)+1}$. Moreover, $(a_{n})$ and $(b_{n})$ are called $\delta$-alternating if $(a_{n})$ is $(b_{n})$-alternating with exponent $\delta$ and $(b_{n})$ is $(a_{n})$-alternating with exponent $\delta$.

For $0<\gamma<\gamma^{\prime}<1$ let

[TABLE]

Theorem 1**.**

If $T$ is weakly mixing and $(T,S)\in\mathcal{L}^{\gamma,\gamma^{\prime},\delta}$ with $\delta<\gamma$ then $T\times S$ is not loosely Bernoulli.

In Subection 3.2 we will also show the following:

Lemma 1.2**.**

For every $T\in\mathcal{C}_{\gamma,\gamma^{\prime}}$, $0<\gamma<\gamma^{\prime}<1/3$ there exists a staircase rank one $S\in\mathcal{C}_{\gamma/3,3\gamma^{\prime}}$ such that $(T,S)\in\mathcal{L}^{\gamma/3,3\gamma^{\prime},\gamma/4}$.

It follows from [2] that staircase rank one transformations are mixing. Therefore, Theorem 1 together with Lemma 1.2 has the following consequence:

Corollary 1.3**.**

There exists two staircase rank one transformations whose product is not loosely Bernoulli.

Acknowledgements: The authors would like to thank Jean-Paul Thouvenot for several discussions on the subject.

2 Basic definitions

2.1 Rank one maps

Recall that a rank one system is constructed by cutting and stacking: fix a sequence of *cuts * $(p_{n})_{n\in\mathbb{N}}$ and a sequence of spacers $(a_{n,i})_{i=1}^{p_{n}}$, $n\in\mathbb{N}$. We define $h_{1}=1$ and inductively

[TABLE]

The sequence $(h_{n})_{n\in\mathbb{N}}$ is the sequence of heights. The rank one system $T$ is constructed from the sequences $(p_{n})_{n\in\mathbb{N}}$ and $(a_{n,i})_{i=1}^{p_{n}}$, $n\in\mathbb{N}$ in the following way: we start with the interval $[0,1]$ which we cut into $p_{n}$ equal intervals $(I^{1}_{i})_{i=1}^{p_{n}}$. For every $i\in\{1,...,p_{n}\}$ we put $a_{1,i}$ spacers over $I^{1}_{i}$. Then we stack everything over $I^{1}_{1}$ and call this tower ${\mathcal{T}}_{2}$ with base $I^{1}_{1}$. The transformation $T$ on ${\mathcal{T}}_{2}$ just moves one level up except the last level, where it is not defined. Next, inductively at stage $n$, we cut the tower ${\mathcal{T}}_{n}$ with base $I^{n}_{1}$ into $p_{n}$ equal subtowers, over i’th subtower we put $a_{n,i}$ spacers and then we stack to get tower ${\mathcal{T}}_{n+1}$ with base $I^{n}_{1}$. Finally, the rank one map $T$ is defined almost everywhere, i.e. it is defined on ${\mathcal{T}}_{\infty}:=\bigcup_{n\in\mathbb{N}}{\mathcal{T}}_{n}$. Often, if there are more transformations involved, we will write a superscript $T$ in the sequences $(p_{n})$, $(a_{n,i})$, $(h_{n})$ and ${\mathcal{T}}_{n}$. In what follows we will always assume that $p_{n}>1$ for every $n$. This implies that

[TABLE]

We also assume below that the total measure of all the spacers is finite:

[TABLE]

Under this condition, $T$ preserves a probability measure $\mu_{T}$ given by the normalized Lebesgue measure on ${\mathcal{T}}_{\infty}$. It is a classical fact that rank one systems are uniquely ergodic. Hence $T$ acts on $({\mathcal{T}}_{\infty},\mathcal{B},\mu_{T})$. Moreover, (5) implies the following bounds on the sequence of heights: there exists $1\leq K<\infty$ such that for every $n\in\mathbb{N}$

[TABLE]

Indeed, the LHS is immediate from (3). The RHS follows by (5) and (3) since for $\epsilon_{n}=\frac{1}{\prod_{i=1}^{n}p_{i}}\left(\sum_{i=1}^{p_{n}}a_{n,i}\right)$, we have

[TABLE]

and so by (3)

[TABLE]

and recursively

[TABLE]

where $K:=\prod_{1}^{\infty}(1+\epsilon_{i})<\infty$ by (5). This shows (6).

For $n\in\mathbb{N}$ and $x,y$ in one level of ${\mathcal{T}}_{n}$ we define the horizontal distance of $x,y$, setting

[TABLE]

where $m=m(x,y)\geqslant n$ is the largest number such that $x,y$ are in one level of ${\mathcal{T}}_{m}$ (i.e. they are in different levels of ${\mathcal{T}}_{m+1}$). The set of rank one transformations (satisfying (5)) will be denoted $Rank(1)$.

2.2 Loosely Bernoulli transformations

We recall the definition of $\bar{f}$ metric introduced in [4]. For two finite words (over a finite alphabet) $A=a_{1}...a_{k}$ and $B=b_{1}...b_{k}$ a matching between $A$ and $B$ is

any pair of strictly increasing sequences $(i_{s},j_{s})_{s=1}^{r}$ such that $a_{i_{s}}=b_{j_{s}}$ for $s=1,...r$. The $\bar{f}$ distance between $A$ and $B$ is defined by

[TABLE]

where $r$ is the maximal cardinality over all matchings between $A$ and $B$.

Let $T:(X,\mathcal{B},\mu)\to(X,\mathcal{B},\mu)$ be a measure preserving automorphism. For a finite partition $\mathcal{P}=(P_{1},...P_{r})$ of $X$ and an integer $N\geqslant 1$ we denote $\mathcal{P}_{0}^{N}(x)=x_{0}...x_{N-1}$, where $x_{i}\in\{1,...,r\}$ is such that $T^{i}(x)\in P_{x_{i}}$ for $i=0,...,N-1$.

Definition 2.1**.**

The process $(T,\mathcal{P})$ is said to be loosely Bernoulli if for every $\varepsilon>0$ there exists $N_{\varepsilon}\in\mathbb{N}$ and a set $A_{\varepsilon}\in\mathcal{B}$, $\mu(A_{\varepsilon})>1-\varepsilon$ such that for every $x,y\in A_{\varepsilon}$ and every $N\geqslant N_{\varepsilon}$

[TABLE]

The automorphism $T$ is loosely Bernoulli (LB for short) if for every finite measurable partition $\mathcal{P}$, the process $(T,\mathcal{P})$ is loosely Bernoulli.

To prove Theorem 1, in view of Definition 2.1 it is enough to show that there exists a finite partition $\mathcal{P}$ such that the process $(T\times S,\mathcal{P})$ is not LB. Recall that we have an exhaustive sequence of towers $({\mathcal{T}}_{n}^{T})$ and $({\mathcal{T}}_{n}^{S})$ for $T$ and $S$ respectively. For $n\in\mathbb{N}$ let $\mathcal{Q}_{n}$ be the partition of ${\mathcal{T}}^{T}_{\infty}$ into levels of ${\mathcal{T}}_{n}^{T}$ and ${\mathcal{T}}^{T}_{\infty}\setminus{\mathcal{T}}_{n}^{T}$ (${\mathcal{T}}_{\infty}^{T}\setminus{\mathcal{T}}^{T}_{n}$ is the union of all spacers at stages $\geqslant n$). Let $\mathcal{R}_{n}$ be the analogous partition of ${\mathcal{T}}^{S}_{\infty}$ and define $\mathcal{P}_{n}:=\mathcal{Q}_{n}\times\mathcal{R}_{n}$. Then $\mathcal{P}_{n}$ is a partition of ${\mathcal{T}}^{T}_{\infty}\times{\mathcal{T}}^{S}_{\infty}$ (in fact, the sequence $(\mathcal{P}_{n})$ converges to partition into points). Theorem 1 follows by the following theorem:

Theorem 2**.**

Let $T$ be weakly mixing and $(T,S)\in\mathcal{L}^{\gamma,\gamma^{\prime},\delta}$ with $\delta<\gamma$. There exists $m_{0}$ such that the process $(T\times S,\mathcal{P}_{m_{0}})$ is not LB.

In the following subsections we will state some lemmas which will help us in proving Theorem 2.

3 Preliminary lemmas

3.1 A combinatorial lemma: lower bound on $\bar{f}$

In this section we state a general combinatorial lemma which allows to give a lower bound on the $\bar{f}$ distance. Let $(i_{s})_{s=1}^{r}$ and $(j_{s})_{s=1}^{r}$ be two increasing sequences of positive integers in $[0,N]$.

For $M\leqslant N$ and $1\leqslant w<r$ define

[TABLE]

and

[TABLE]

The following property of $I(M,w)$ and $J(M,w)$ is straightforward and will be useful in the proof of the lemma below: If $s>w$ and $s\notin I(M,w)$, then

[TABLE]

and analogously for $J(M,w)$. We have the following lemma:

Lemma 3.1**.**

Let $\xi\in(0,1)$. Let $(i_{s})_{s=1}^{r}\in[0,N]^{r}$ and $(j_{s})_{s=1}^{r}\in[0,N]^{r}$ be two increasing sequences of integers for which there exists a number $K\in\mathbb{N}$, $8K^{1+\xi}\leqslant N$ such that for every $s=1,\ldots,r$

[TABLE]

then $r<\frac{4N}{K^{\xi}}$.

Proof.

Set $s_{1}:=1$. If $|I(K^{1+\xi},s_{1})|\geqslant|J(K^{1+\xi},s_{1})|$, we set $B_{1}:=J(K^{1+\xi},s_{1})$ otherwise, we set $B_{1}:=I(K^{1+\xi},s_{1})$. By (10), we have

[TABLE]

We then proceed inductively: assume that for some $\ell\geq 1$ we have constructed $1=s_{1}<s_{2}<\cdots<s_{\ell}\leq r$ and subsets $B_{1},\ldots,B_{\ell}$ of $\{1,\dots,r\}$. If $B_{1}\cup\cdots\cup B_{\ell}=\{1,\dots,r\}$, we stop here and set $v:=\ell$. Otherwise, we define $s_{\ell+1}$ as the smallest element of $\{1,\dots,r\}\setminus B_{1}\cup\cdots\cup B_{\ell}$, and $B_{\ell+1}$ as either $I(K^{1+\xi},s_{\ell+1})$ or $J(K^{1+\xi},s_{\ell+1})$ (by choosing the set with the smallest cardinal). By (10), we always have

[TABLE]

We go on in this way until we obtain a finite sequence of set $\{B_{i}\}_{i=1}^{v}$, such that

[TABLE]

and such that

[TABLE]

Moreover, if $1\leq\ell_{1}<\ell_{2}\leq v$, then $s_{\ell_{2}}\notin B_{\ell_{1}}$. If we further assume that $B_{\ell_{h}}=I(K^{1+\xi},s_{\ell_{h}})$ for $h\in\{1,2\}$ (or $B_{\ell_{h}}=J(K^{1+\xi},s_{\ell_{h}})$ for $h\in\{1,2\}$), then by construction (see also (9))

[TABLE]

By pigeonhole principle, there exists a set $A\subset\{1,\ldots,v\}$, with $|A|\geqslant\frac{v}{2}$ and such that either for every $\ell\in A$, $B_{\ell}=I(K^{1+\xi},s_{\ell})$, or for every $\ell\in A$, $B_{\ell}=J(K^{1+\xi},s_{\ell})$. By (13), the subsets $B_{\ell}$, $\ell\in A$ correspond to disjoint subintervals of $[0,N]$, each of them of length $K^{1+\xi}$. We thus get

[TABLE]

Now, (11), (12) and (14) yield

[TABLE]

∎

3.2 Proof of Lemma 1.2

Proof of Lemma 1.2:.

Let $(h_{n}^{T})$ denote the sequence of heights for $T$. We will construct a staircase $S\in C_{\gamma/3,3\gamma^{\prime}}$ such that $(T,S)\in\mathcal{L}^{\gamma/3,3\gamma^{\prime},\gamma/4}$, i.e. the sequences $(h_{n}^{T})$ and $(h_{n}^{S})$ are $\gamma/4$- alternating. For this, we will show that there exists $n_{0}$ such that for every $n\geqslant n_{0}$, we have

[TABLE]

Fix $\eta<1/100$. Let $K_{T}$ denote the constant $K$ for $T$ in (6). We will first construct $(p_{i}^{S})_{i=1}^{+\infty}$ so that for $n\geqslant n_{0}$, we have

[TABLE]

Then we will show how to derive (15) from (16).

Since $T\in\mathcal{C}_{\gamma,\gamma^{\prime}}$ it follows that for $n>n^{\prime}_{T}$, by (6), we have

[TABLE]

Set $p^{S}_{n}=2$ for every $n\in\{1,...,n^{\prime}_{T}+1\}$ and $a_{n,i}^{S}=0$ for every $i\in\{1,2\}$ and $n\in\{1,...,n^{\prime}_{T}+1\}$. Then (by enlarging $n^{\prime}_{T}$ if neccesary), we have

[TABLE]

(The last inequality by (17) and since $p_{n}^{T}\geqslant 2$.) So the left inequality in (16) holds for $n=n^{\prime}_{T}+3$. We then proceed inductively: having defined $(p_{i}^{S})_{i=1}^{w}$ so that the left inequality in (16) holds for $n=w+2$, we choose $p_{w+1}^{S}\in\mathbb{N}$ so that

[TABLE]

We explain below why such a choice is always possible. Notice that by (17) for $n=w+2$, we have

[TABLE]

the last inequality by $\frac{1+\gamma}{1+(\frac{1}{4}+\eta)\gamma}\geqslant 1+\gamma/4$ (recall also that $K_{T}$ is a fixed constant and $w\geqslant n^{\prime}_{T}+3$ and $n^{\prime}_{T}$ can be made sufficiently large with respect to $K_{T}$.) Moreover, since we assume that the left inequality in (16) holds for $w+2$, it follows that $4\prod_{i=1}^{w+1}p_{i}^{T}\geqslant 4\prod_{i=1}^{w}p_{i}^{S}$. Therefore the length of the interval on the right of (18) is at least $4\prod_{i=1}^{w}p_{i}^{S}$, and so such $p_{w+1}^{S}\geqslant 2$ always exists. Recursively, it follows that (16) holds for $n\geq n_{0}:=n^{\prime}_{T}+3$.

To guarantee that $S$ is a staircase rank one system we set $a^{S}_{n,i}=i$ for $n\in\mathbb{N}$ and $i\in\{1,...,p_{n}\}$. It remains to show that $S\in\mathcal{C}_{\gamma/3,3\gamma^{\prime}}$ and, using (16), that (15) holds. Notice that by (16), for $w\geqslant n_{0}$, we have

[TABLE]

Since $T\in\mathcal{C}_{\gamma,\gamma^{\prime}}$ with $0<\gamma<\gamma^{\prime}<1/3$ and by (6), we get

[TABLE]

Notice that (19) and (20) together with the definition of spacers for $S$ and $\gamma^{\prime}<1/3$ imply that

[TABLE]

and hence (6) holds for $S$ (with constant $K_{S}$). Then, (16) and (6) (for $T$ and for $S$) ensure the validity of (15) for $n$ large enough. This proves that $(h_{n}^{T})$ and $(h_{n}^{S})$ are $\gamma/4$- alternating.

Now it remains to check that $S\in C_{\gamma/3,3\gamma^{\prime}}$. For this, we will use the following inequality: for $n$ large enough, using several times (6) and the fact that $T\in C_{\gamma,\gamma^{\prime}}$, we get

[TABLE]

By (15), it follows that for $n$ large enough, we have

[TABLE]

On the other hand, again by (15)

[TABLE]

Therefore and by (6), for $n$ large enough, we have

[TABLE]

Using (21) and the definition of spacers for $S$ ($S$ is a staircase transformation) we get that $S\in\mathcal{C}_{\gamma/3,3\gamma^{\prime}}$. This finishes the proof. ∎

3.3 Distribution of points for maps from

$\mathcal{C}_{\gamma,\gamma^{\prime}}$

In this section we will do quantitative estimates on recurrence of points. Fix $G\in\mathcal{C}(\gamma,\gamma^{\prime})$. Since $G$ will be fixed throughout this section, we drop the superscript $G$ and denote by ${\mathcal{T}}_{n}$, $B_{n}$, $h_{n}$ and $p_{n}$ the tower, base, height and the number of cuts at stage $n$ (recall that $p_{n}\in[h_{n}^{\gamma},h_{n}^{\gamma^{\prime}}]$ for $n\geqslant n^{\prime}_{G}$). For $i\in\{1,\ldots,p_{n}\}$, let ${\mathcal{T}}_{n,i}\subset{\mathcal{T}}_{n}$ denote the column over the $i$-th cut.

We will construct some subsets of ${\mathcal{T}}_{\infty}$, on which we control the dynamics well. First, we cut off the “boundaries” of ${\mathcal{T}}_{n}$: let

[TABLE]

Notice that since $p_{n}\geqslant h_{n}^{\gamma}$ for $n\geqslant n^{\prime}_{G}$ and every column has equal measure, we have $\mu_{G}(F_{n}^{G})\geqslant 1-\frac{4}{n^{2}}$ for $n$ large enough. Let

[TABLE]

where $n_{1}$ is such that $\mu(F^{G})\geqslant 1-10^{-5}$. In particular, if $x\in F^{G}$, then $x\in{\mathcal{T}}_{n_{1}}$. This will be used in the statement of the lemma below.

Lemma 3.2**.**

Fix $0<\xi\leq 100^{-1}\min(\gamma,1-\gamma^{\prime})$. There exists $m_{1}\in\mathbb{N}$ such that for every $n\geqslant m_{1}$ the following holds: for every $x,x^{\prime}\in F^{G}$ such that $d_{H}(x,x^{\prime})=\frac{1}{h_{n}}$ (see (7)), for every $r\in\{h_{n},h_{n}+1,\ldots,(h_{n})^{1+2\xi}\}$ for which $G^{r}x,G^{r}x^{\prime}\in F^{G}$, we have

[TABLE]

Moreover, for every $x,x^{\prime}\in F^{G}$ such that $d_{H}(x,x^{\prime})\leqslant\frac{1}{h_{n}}$ and for every $i,j\in\{0,...\frac{h_{n}}{n^{3}}\}$, $i\neq j$, we have

[TABLE]

Proof.

Let $n\geq n_{1}$ be large enough so that $h_{n}^{\gamma^{\prime}+2\xi}<h_{n}/n^{2}$, and let $x,x^{\prime}\in F^{G}$ be such that $d_{H}(x,x^{\prime})=\frac{1}{h_{n}}$. Fix $r\in\{h_{n},h_{n}+1,\ldots,h_{n}^{1+2\xi}\}$ for which $G^{r}x,G^{r}x^{\prime}\in F^{G}$. Let $0\leqslant\ell_{1},\ell_{2}\leqslant p_{n}$ be such that $x\in{\mathcal{T}}_{n,\ell_{1}}$ and $x^{\prime}\in{\mathcal{T}}_{n,\ell_{2}}$. Since $x,x^{\prime}\in F^{G}\subset F_{n}^{G}$ (see (22)), $\ell_{1},\ell_{2}\in[h_{n}^{\gamma/4},p_{n}-h_{n}^{\gamma/4}]$. Notice that $r\leqslant h_{n}^{1+2\xi}$ and the height of ${\mathcal{T}}_{n}$ is $h_{n}$. By the choice of $\xi$ it follows that

[TABLE]

Since $d_{H}(x,x^{\prime})=\frac{1}{h_{n}}$, we have $\ell_{1}\neq\ell_{2}$ (otherwise $d_{H}(x,x^{\prime})\leqslant\frac{1}{h_{n+1}}$). Assume WLOG that $\ell_{1}<\ell_{2}$. Let $\ell_{r}\geqslant\ell_{1}$ be such that $G^{r}x\in{\mathcal{T}}_{n,\ell_{r}}$. Notice that since $r\geqslant h_{n}$ and $G^{r}x\in F^{G}\subset{\mathcal{T}}_{n}$, it follows that in fact $\ell_{r}>\ell_{1}$. Define

[TABLE]

By $G\in C_{\gamma,\gamma^{\prime}}$, using (26) we get that $a_{n,\ell_{2}-\ell_{1}+i}-a_{n,i}>0$ (the spacers are monotonically placed). Since $\ell_{r}>\ell_{1}$ it follows that $w_{r}>0$. Moreover, $a_{n,i}\leqslant h_{n}^{\gamma^{\prime}}$ for $i\in\{1,\ldots,p_{n}\}$, hence

[TABLE]

Let $s$ be such that $G^{r}x\in G^{s}B_{n}$. Since $G^{r}x\in F^{G}\subset F_{n}^{G}$, we have $s\geq h_{n}/n^{2}$, and it follows by definition of $w_{r}$ and the fact that $w_{r}<s$ that $G^{r}x^{\prime}\in G^{s-w_{r}}B_{n}$. Hence $G^{r}x$ and $G^{r}x^{\prime}$ are not in the same level of ${\mathcal{T}}_{n}$ and (24) follows.

For (25), notice that $x,x^{\prime}\in F^{G}\subset F_{n}^{G}$ (see (22)) and by assumptions $x,x^{\prime}$ are in one level of ${\mathcal{T}}_{n}$, i.e. for some $s\in[\frac{h_{n}}{n^{2}},h_{n}-\frac{h_{n}}{n^{2}}]$, $x,x^{\prime}\in G^{s}(B_{n})$. But by the bounds on $s$ it follows that for $u\in\{i,j\}$, we have $G^{u}(x)\in G^{s+u}(B_{n})$ (and $0\leqslant s+u\leqslant h_{n}$). Since $i\neq j$, we have $s+i\neq s+j$ and hence $G^{i}(x)$ and $G^{j}(x^{\prime})$ are in different levels of ${\mathcal{T}}_{n}$. This gives (25). ∎

For $n\in\mathbb{N}$ and $x\in{\mathcal{T}}_{n}$ let $i_{n,x}\in\{0,\ldots,h_{n}-1\}$ be such that $x\in G^{i_{n,x}}(B_{n})$. Define

[TABLE]

Notice that, since $G\in C_{\gamma,\gamma^{\prime}}$, we have $\mu({\mathcal{T}}_{n})\geq 1-{\rm O}(n^{-2})$, hence $\mu(D_{G}^{x,n})\geqslant 1-{\rm O}(n^{-2})$. We define

[TABLE]

where $n_{3}$ is such that for all $x$, $\mu(D^{G}_{x})\geqslant 1-10^{-5}$.

Lemma 3.3**.**

There exists $m_{2}\in\mathbb{N}$ such that for every $x\in F^{G}$ (see (23)), every $x^{\prime}\in D^{G}_{x}$ every $n\geqslant m_{2}$ and every $i,j\in\left\{0,...,\left\lfloor\frac{h_{n+1}}{(n+1)^{2}}\right\rfloor\right\}$ for which $G^{i}x\in{\mathcal{T}}_{n}$, we have

[TABLE]

Proof.

Let $m_{2}:=2\max(n_{1},n_{3})$ and fix $n\geqslant m_{2}$. Let $x$, $x^{\prime}$ and $i,j$ be as in the assumptions of the lemma. Assume by contradiction that $d_{H}(G^{i}(x),G^{j}(x^{\prime}))<\frac{1}{h_{n}}$. By (7) this implies that $G^{i}(x)$ and $G^{j}(x^{\prime})$ are in the same level of ${\mathcal{T}}_{n+1}$. Since $x\in F^{G}\subset F_{n+1}^{G}$ (see (22)) and $0\leq i\leq\frac{h_{n+1}}{(n+1)^{2}}$ , this implies that $G^{i}x$ and $G^{j}(x^{\prime})$ are both located $i$ levels above $x$ in ${\mathcal{T}}_{n+1}$. Therefore $x$ and $G^{j-i}x^{\prime}$ are in the same level of ${\mathcal{T}}_{n+1}$. But $|j-i|\leqslant\frac{h_{n+1}}{(n+1)^{2}}$, which contradicts the fact that $x^{\prime}\in D^{G}_{x}\subset D_{G}^{x,n+1}$ (see (27)). The contradiction finishes the proof. ∎

4 A proposition which implies Theorem 2

For the rest of the paper $(T,S)\in\mathcal{L}^{\gamma,\gamma^{\prime},\delta}$ with $0<\delta<\gamma<\gamma^{\prime}<1$ are fixed. Since we deal with two transformations $T$ and $S$, we will denote the horizontal distance (see (7)) for $T$ and $S$ respectively by $d_{1}$ and $d_{2}$.

Recall from (23) the sets $F^{T}$ and $F^{S}$. Let $\mathcal{P}=\mathcal{P}_{n}$ for some $n\geq 3$. For $(x,y),(x^{\prime},y^{\prime})\in{\mathcal{T}}_{\infty}^{T}\times{\mathcal{T}}_{\infty}^{S}$, and $N\in\mathbb{N}$, consider a matching $\theta=(i_{s},j_{s})_{s=1}^{r}$ of $\mathcal{P}_{0}^{N}(x,y)$ and $\mathcal{P}_{0}^{N}(x^{\prime},y^{\prime})$. Then for each $k\in\mathbb{N}$ we define

[TABLE]

Notice that by definition of the partition $\mathcal{P}_{n}$ (every level of ${\mathcal{T}}_{n}^{T}\times{\mathcal{T}}_{n}^{S}$ is a different atom), the definition of the horizontal distance and (4), we have for each $s\in\{1,\ldots,r\}$

[TABLE]

Hence for $k<\frac{n}{2}$, we have

[TABLE]

The following proposition implies Theorem 2:

Proposition 4.1**.**

There exist $n_{0}\geqslant{100}$, $N_{0}\in\mathbb{N}$ and a set $B\times C\subset{\mathcal{T}}_{\infty}^{T}\times{\mathcal{T}}_{\infty}^{S}$, $(\mu_{T}\times\mu_{S})(B\times C)\geqslant 99/100$, for which the following holds: for every $(x,y)\in B\times C$ there exists a set $D_{x}\times D_{y}\in{\mathcal{T}}_{\infty}^{T}\times{\mathcal{T}}_{\infty}^{S}$, $(\mu_{T}\times\mu_{S})(D_{x}\times D_{y})\geqslant 99/100$ such that for every $(x,y)\in B\times C$, $(x^{\prime},y^{\prime})\in D_{x}\times D_{y}$ and every $N\geqslant N_{0}$

1. $(P1)$

for each $(z,w)\in\{(x,y),(x^{\prime},y^{\prime})\}$,

[TABLE] 2. $(P2)$

for every $N>N_{0}$, every $k\in\mathbb{N}$ and every matching $\theta=(i_{s},j_{s})_{s=1}^{r}$ of $(\mathcal{P}_{n_{0}})_{0}^{N}(x,y)$ and $(\mathcal{P}_{n_{0}})_{0}^{N}(x^{\prime},y^{\prime})$, we have

[TABLE]

The proof of Proposition 4.1 is the most technical part of the paper. We will devote a separate section to its proof. Let us first show how Proposition 4.1 implies Theorem 2.

Proof of Theorem 2.

To simplify the notation, set $\mathcal{P}:=\mathcal{P}_{n_{0}}$. We will prove the following:

Claim: For every $(x,y)\in B\times C$, $(x^{\prime},y^{\prime})\in D_{x}\times D_{y}$ and every $N\geqslant N_{0}$, we have

[TABLE]

Before we show the Claim, let us show how it implies the result. Assume by contradiction that the process $(T\times S,\mathcal{P})$ is LB. Let $\epsilon:=\frac{1}{200}$ and let $N_{\epsilon}\in\mathbb{N}$, $A_{\epsilon}\subset{\mathcal{T}}_{\infty}^{T}\times{\mathcal{T}}_{\infty}^{S}$, $(\mu_{T}\times\mu_{S})(A_{\epsilon})>1-\epsilon$ be from Definition 2.1. Take $N\geqslant\max(N_{0},N_{\epsilon})$. Notice that for every $(x,y)\in(B\times C)\cap A_{\epsilon}$, we have

[TABLE]

Indeed, if the set above is non-empty, then there exists $(x,y)\in(B\times C)\cap A_{\epsilon}$ and $(x^{\prime},y^{\prime})\in(D_{x}\times D_{y})\cap A_{\epsilon}$, but then by (32) and (8)

[TABLE]

and this is a contradiction. So (33) holds. This in turn contradicts $\mu_{T}\times\mu_{S}(B\times C),\mu_{T}\times\mu_{S}(D_{x}\times D_{y})\geqslant 99/100$ and $\mu_{T}\times\mu_{S}(A_{\varepsilon})\geqslant 1-\epsilon$. So the proof is finished up to proving the Claim.

To prove the Claim, we will show that for every matching $\theta=(i_{s},j_{s})_{s=1}^{r}$ of $\mathcal{P}_{0}^{N}(x,y)$ and $\mathcal{P}_{0}^{N}(x^{\prime},y^{\prime})$, we have $r\leqslant\frac{9N}{10}$. Fix any such matching $\theta=(i_{s},j_{s})_{s=1}^{r}$ and let

[TABLE]

By $(P1)$ we have $|\{1,\ldots,r\}\setminus H_{n}|\leqslant\frac{1}{10}N$, hence the Claim follows by showing that

[TABLE]

Notice that by definition of $A_{\theta}^{k}((x,y)(x^{\prime},y^{\prime}))$, we have

[TABLE]

Therefore, using (31) and remembering that $n_{0}\geq 100$, we get

[TABLE]

By $(P2)$, this implies that

[TABLE]

This shows (34), which concludes the proof of the Claim and the proof of Theorem 2. ∎

5 Proof of Proposition 4.1

This section will be devoted to the proof of Proposition 4.1. We will divide the proof into several steps.

5.1 Construction of $B$ and $C$ in Proposition 4.1

Recall that $(T,S)\in\mathcal{L}^{\gamma,\gamma^{\prime},\delta}$. Set $\xi:=\min(\gamma,1-\gamma^{\prime},\delta)/100$. To define the sets $B$ and $C$, we will use some notation from Section 3.3. First let $n_{0}$ be large enough so that

[TABLE]

Since the construction follows similar lines for $T$ and $S$ we will do it simultaneously for $G\in\{T,S\}$. First recall the definition of $F^{G}$ from (23). By the ergodic theorem, there exists a set $F_{erg}^{G}$, $\mu(F_{erg}^{G})>1-10^{-4}$ and a number $m_{3}\in\mathbb{N}$ such that for every $x\in F_{erg}^{G}$ and every $N\geqslant m_{3}$,

[TABLE]

We then define $B:=F^{T}\cap F_{erg}^{T}$ and $C:=F^{S}\cap F_{erg}^{S}$. We will write this in the product form:

[TABLE]

Notice in particular that $(\mu_{T}\times\mu_{S})(B\times C)\geqslant 99/100$ as required in Proposition 4.1.

Construction of $D_{x}$ and $D_{y}$.

We define

[TABLE]

where $D^{T}_{x}$ and $D^{S}_{y}$ come from (28) for $G=T$ and $G=S$ respectively (see also (27)). Notice that $(\mu_{T}\times\mu_{S})(D_{x}\times D_{y})\geqslant 99/100$ as in the statement of Proposition 4.1

Definition of $N_{0}$ and proof of $(P1)$ in Proposition 4.1

Let $N_{0}:=\max\{m_{3},m_{2},m_{1},n_{3},n_{1}\}$, where $m_{3}$ comes from (35), $m_{2}$ is from Lemma 3.3, $m_{1}$ is from Lemma 3.2, $n_{1}$ from (23) and $n_{3}$ from (28)222All the constants come in two copies: for $T$ and $S$, for instance we have $m_{3}^{T}$ and $m_{3}^{S}$ and we define $m_{3}:=\max(m_{3}^{T},m_{3}^{S})$. We define all other constants analogously.. Notice now that for $(x,y)\in B\times C$ and $(x^{\prime},y^{\prime})\in D_{x}\times D_{y}\subset B\times C$ (see (37)) and for $N\geqslant N_{0}\geqslant m_{2}$, $(P1)$ holds since by (35), for any $(z,w)\in B\times C$,

[TABLE]

5.2 Proof of $(P2)$ in Proposition 4.1

Proof.

We will use Lemma 3.2, Lemma 3.3 for $G=T$ and $G=S$ and then Lemma 3.1. Fix $N>N_{0}$, $(x,y)\in B\times C$ and $x^{\prime},y^{\prime}\in D_{x}\times D_{y}$. Consider any matching $\theta=(i_{s},j_{s})_{s=1}^{r}$ between $(\mathcal{P}_{n_{0}})_{0}^{N}(x,y)$ and $(\mathcal{P}_{n_{0}})_{0}^{N}(x^{\prime},y^{\prime})$, and let $k\in\mathbb{N}$. If $k<n_{0}/2$, then $A_{\theta}^{k}((x,y),(x^{\prime},y^{\prime}))=\emptyset$ (see (31)) and $(P2)$ follows trivially. We therefore assume that $k\geq n_{0}/2$.

Take any $s\in\{1,...,r\}\cap A_{\theta}^{k}((x,y),(x^{\prime},y^{\prime}))$. To simplify notation, denote $x_{s}=T^{i_{s}}x,x^{\prime}_{s}=T^{j_{s}}x^{\prime}$ and $y_{s}=S^{i_{s}}y,y^{\prime}_{s}=S^{j_{s}}y^{\prime}$. By definition of $A_{\theta}^{k}((x,y),(x^{\prime},y^{\prime}))$, we have

[TABLE]

Let $n,m\in\mathbb{N}$ be unique such that $(h_{n}^{T})^{-1}=d_{1}(x_{s},x^{\prime}_{s})$ and $(h_{m}^{S})^{-1}=d_{2}(y_{s},y^{\prime}_{s})$. We assume WLOG that $h_{n}^{T}<h_{m}^{S}$ (notice that, by the $\delta$-alternation, we cannot have $h_{n}^{T}=h_{m}^{S}$, and the case $h_{n}^{T}>h_{m}^{S}$ is analogous). Then by (38), we have

[TABLE]

Then, again by the $\delta$-alternation and (38) we know that

[TABLE]

We will first show that $2^{k(1+2\xi)+3}<N$. Indeed, otherwise we would have $N\leq 2^{k(1+\delta/50)+3}$ (recall that $\xi\leq\delta/100$). Then by (40) there would exist some $\eta=\eta(\delta)>0$ such that $h_{m}^{S}\geqslant N^{1+\eta}$; in particular, $\frac{h_{m}^{S}}{m^{2}}\geqslant N$ ($m$ is sufficiently large). By an application of Lemma 3.3 with $y\in F^{S}$, $y^{\prime}\in D_{y}^{S}$ and $S^{i_{s}}y\in F^{G}\subset{\mathcal{T}}_{m}$, and since $0\leqslant i_{s},j_{s}\leqslant N\leqslant\frac{h_{m}^{S}}{m^{2}}$, we would get

[TABLE]

a contradiction. Hence

[TABLE]

Let $v=v(n)$ be unique such that

[TABLE]

By the $\delta$-alternation, since $n$ is large enough and $\xi\leqslant\delta/100$, we know that

[TABLE]

Notice that by definition of $v$ and since $h_{n}^{T}<h_{m}^{S}$, we have $m\geqslant v+1$ and therefore $d_{2}(y_{s},y^{\prime}_{s})=(h_{m}^{S})^{-1}\leqslant(h_{v+1}^{S})^{-1}$. Moreover, $y_{s},y_{s}^{\prime}\in F^{S}$ (see (30)). Then, by (25) (for $n=v+1$, $y_{s}$, $y^{\prime}_{s}$ and $G=S$), for every $0\leqslant i,j\leqslant(h_{n}^{T})^{1+2\xi}<\frac{h_{v+1}^{S}}{(v+1)^{3}}$, $i\neq j$, we get

[TABLE]

(For the last inequality, we also used the fact that $k\geq n_{0}/2$ is large enough.) Hence, if for some $w>s$, we have

[TABLE]

then $i_{w}-i_{s}=j_{w}-j_{s}$. Indeed, if not then by (43) for $i=i_{w}-i_{s}$ and $j=j_{w}-j_{s}$

[TABLE]

This would contradict the definition of $A_{\theta}^{k}((x,y),(x^{\prime},y^{\prime}))$. So $i_{w}-i_{s}=j_{w}-j_{s}$. Let $w>s$ be such that $i_{w}-i_{s}=j_{w}-j_{s}\in[h_{n}^{T},(h_{n}^{T})^{1+2\xi}]$. By (24) (for $G=T$ and $x_{s},x_{s}^{\prime}\in F^{T}$) and since $x_{w},x^{\prime}_{w}\in F^{T}$ (see (30)) we get for $r_{w}:=i_{w}-i_{s}=j_{w}-j_{s}$,

[TABLE]

The last inequality holds because $T\in C_{\gamma,\gamma^{\prime}}$, hence $(h_{n-1}^{T})^{-1}\geq(h_{n}^{T})^{-\frac{1}{1+\gamma}}\geq(h_{n}^{T})^{-\frac{1}{1+2\xi}}$ and we remember the last inequality in (43). Therefore, if $w>s$ is such that

[TABLE]

then by (44), $w\notin A_{\theta}^{k}((x,y)(x^{\prime},y^{\prime}))$.

By (39) it follows that if

[TABLE]

is such that $w\in A_{\theta}^{k}((x,y),(x^{\prime},y^{\prime}))$, then

[TABLE]

Let $\{w_{s}\}_{s=1}^{r^{\prime}}:=A_{\theta}^{k}((x,y),(x^{\prime},y^{\prime}))\cap\{1,\ldots,r\}$ and consider the matching subsequence given by $\{i_{w_{s}},j_{w_{s}}\}_{s=1}^{r^{\prime}}$.

For $0\leq M\leq N$ and $w\in\{1,\ldots,r^{\prime}\}$, consider the sets $I(M,w)\subset\{1,\ldots r^{\prime}\}$ and $J(M,w)\subset\{1,\ldots r^{\prime}\}$ defined as in Section 3.1. Then by the above reasoning, it follows that for $K=2^{k}$, we have

[TABLE]

Since (41) holds, the assumptions of Lemma 3.1 are satisfied, and we get

[TABLE]

(The last inequality since $k\geq n_{0}/2$ is sufficiently large.) This finishes the proof of $(P2)$ and therefore also the proof of Proposition 4.1 ∎