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

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Preprint submitted on 9 May 2019

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is not loosely Bernoulli

Adam Kanigowski, Thierry de la Rue

To cite this version:

Adam Kanigowski, Thierry de la Rue. Product of two staircase rank one transformations that is not loosely Bernoulli. 2019. �hal-01958522v2�

not loosely Bernoulli

Adam Kanigowski and Thierry de la Rue 9th May 2019

Contents

1 Introduction 1

1.1 Statement of the main result . . . 2

2 Basic definitions 4

2.1 Rank one maps . . . 4 2.2 Loosely Bernoulli transformations . . . 6

3 Preliminary lemmas 7

3.1 A combinatorial lemma: lower bound onf¯. . . 7 3.2 Proof of Lemma 1.2 . . . 8 3.3 Distribution of points for maps fromC_γ,γ⁰ . . . 11

4 A proposition which implies Theorem 2 12

5 Proof of Proposition 4.1 14

5.1 Proof of(P2)in Proposition 4.1. . . 15 Abstract

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

1 Introduction

In this paper we study the loosely Bernoulli property for zero entropy, measure preserving transformations. Loose Bernoullicity was introduced by J. Feldman, [3], and A. Katok, [6]¹. 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 [8] that the class of loosely Bernoulli systems is broad: it is closed under taking factors, compact extensions and inverse limits. Moreover it contains allfinite rank systems. First non-loosely Bernoulli examples were constructed by J. Feldman [3] by a cutting and stacking method. D. Ornstein, D. Rudolph and B. Weiss [8] constructed arank

1In [6], zero entropy loosely Bernoulli is calledstandard.

1

onetransformation T such thatT×T is not loosely Bernoulli. The example in [8] 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 [4] 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, [10], 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 [9]. It was recently shown in [5] that there exist smooth flows onT²(Kochergin flows) whose Cartesian product is not loosely Bernoulli (provided that the exponents are different).

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 Bernoullicity for products of staircase rank one transformations is not known. Our main result implies in particular that there exist two staircase rank one transformations whose product is not loosely Bernoulli.

Our method uses some ideas from [5], but the main mechanism is different: systems considered in [5] have superlinear (in fact polynomial with degree close to2) growth, which is the main source of non-loose Bernoullicity. In our case, the orbit growth is linear and instead we take crucial advantage from the δ-alternation property (see Definition 1.1), which ensures that the heights of the towers of the two rank one transformations are never of the same order. Since our argument is essentially based on this assumption, it certainly does not apply to the case of the Cartesian square of a single rank-one transformation. In particular, new ideas will be required to answer a famous question by Jean-Paul Thouvenot whether the Cartesian square of Chacon’s transformation is loosely Bernoulli or not.

1.1 Statement of the main result

For a rank one systemT, let (p^T_n)n∈N, (a^T_n,i)^p_i=1^Tn and (h^T_n)n∈N denote the sequence of cuts, spacersand heightsrespectively (see Section2.1). For0< γ < γ⁰<1define

C_γ,γ⁰ :=n

T ∈Rank(1) : there exists n⁰_T ∈Nsuch that for every n>n⁰_T, p^T_n ∈h

(h^T_n)^γ,(h^T_n)^γ0 i

, a^T_n,i> a^T_n,i−1 fori= 2, . . . , p^T_n, and a^T_n,pn 6(h^T_n)^γ0 o

. (1) Notice that if T is a staircase rank one transformation, i.e. a^T_n,i = i for n ∈ N and i∈ {1, . . . , p^T_n}, and ifp^T_n ∈[(h^T_n)^γ,(h^T_n)^γ0], thenT ∈C_γ,γ⁰.

Definition 1.1(δ-alternating sequences). Fixδ >0and let(an)and(bn)be two increasing sequences of positive integers. We say that(b_n) is(a_n)-alternating with exponent δ if the following holds: there existsn₀ ∈Nsuch that for every n>n₀, if m(n) is unique such that b_m(n) 6a_n < b_m(n)+1, thenb^1+δ_m(n) < a_n and a^1+δ_n < b_m(n)+1. Moreover, (a_n) and(b_n) are called δ-alternating if (an) is(bn)-alternating with exponent δ and(bn) is(an)-alternating with exponent δ.

For0< γ < γ⁰ <1 let L^γ,γ0,δ :=

(T, S)∈ C_γ,γ⁰× C_γ,γ⁰ : (h^T_n)and (h^S_n) areδ-alternating . (2)

Theorem 1. If T is weakly mixing and (T, S) ∈ L^γ,γ0,δ with δ < γ then T ×S is not loosely Bernoulli.

In Subsection 3.2we will also show the following:

Lemma 1.2. For every T ∈ C_γ,γ⁰, 0 < γ < γ⁰ < 1/3 there exists a staircase rank one S∈ C_γ/3,3γ0 such that (T, S)∈ L^{γ/3,3γ0,γ/4}.

It follows from [2] that staircase rank one transformations are mixing. Therefore, Theorem 1together with Lemma1.2 has the following consequence:

Corollary 1.3. There exist two staircase rank one transformations whose product is not loosely Bernoulli.

Outline of the proof

A rank one transformationT is defined inductively (see Section2.1) by a sequence of nested towers (T_m)m∈N. Therefore, there is a natural notion of distance: two points x, y are at distance _h¹

m, if m ∈ N is the largest such that x, y are in the same level of T_m (see (7)).

To show that a system is not loosely Bernoulli, one needs to show that “typical” points are not close in thef¯metric (see Section2.2), i.e. that for typical points there is no good matching. For two rank one transformations T and S and typical points (x, y),(x⁰, y⁰) in the product space, consider some matching of pieces of orbits of length N. One can then split such a matching into sets A_k according to which interval of the form [2^−k−1,2^−k) the maximum over the distances (for T and S) of a given iterate of T ×S belongs (see (31)). The main goal is to show that the cardinality of A_k is bounded above by N k⁻² (see Proposition 4.1). Then the fact that the matching is of small cardinality follows by summing overk.

The main result towards this is Lemma3.2which describes how the horizontal distance behaves under iterates of T and S. First, due to the staircase nature of T and S, it is shown that if two points in the same level of T_n but not in the same level of T_n+1 reach the top of T_n (which happens before h_n), then their orbits will not be in the same level of T_n for time>h^1+2ξn (for someξ >0) (see (25)). This is a consequence of the fact that each time the orbits hit the top they split by a positive shear (due to the staircase nature), but the only way they could end up in the same level of T_n is if they split by>h_n, which does not happen before h^1+2ξn (the total amount of shear is to small). The second part of Lemma 3.2 (see (26)) shows that if x, x⁰ are in the same level of T_n (not too close to the top or bottom ofT_n), the only way that Tⁱx and T^jx⁰ lie in the same level of T_n for i, j≤ ^h_nⁿ₃ is when i=j.

We will use Lemma 3.2 for both T and S to show that Ak has cardinality 6 N k⁻², together with the fact that the sequences of heights are δ-alternating. Namely, if x, x⁰ are at distance _h¹_T

n and (y, y⁰) are at distance _h¹_S

m, (assume without loss of generality that

1 h^T_n < _h¹S

m), then by (25) (for S), the “vertical” matching (that is, matching(Tⁱx, Sⁱy)with (Tⁱx⁰, Sⁱy⁰)) does not work onA_k fori∈[h^S_m,(h^S_m)^1+2ξ](it only can work fori∈[0, h^S_m]).

On the other hand, using (h^S_m)^1+2ξ 6 ^h_n^Tn3 (by δ-alternation) and (26) forT, the only way to match(Tⁱx, Sⁱy) with(T^jx⁰, S^jy⁰) inAk for i, j∈[h^S_m,(h^S_m)^1+2ξ]is that i=j, i.e. the

matchinghas to bevertical. The inconsistency of the matching on[h^S_m,(h^S_m)^1+2ξ]gives the bound|A_k|6N k⁻².

Acknowledgements: The authors would like to thank Jean-Paul Thouvenot for sev- eral discussions on the subject, and the anonymous referee for the relevance of his advice and remarks.

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∈Nand a sequence ofspacers(a_n,i)^p_i=1ⁿ ,n∈N. We defineh₁ = 1and inductively

hn=pn−1hn−1+

pn−1

X

i=1

an−1,i, for n≥2. (3)

The sequence (h_n)n∈N is the sequence of heights. The rank one system T is constructed from the sequences (p_n)n∈N and (a_n,i)^p_i=1ⁿ , n∈ N in the following way: we start with the interval [0,1]which we cut intop₁ equal intervals(I_i¹)^p_i=1¹ . For every i∈ {1, ..., p₁}we put a_1,ispacers (extra intervals with the same length asI_i¹) overI_i¹, getting theith subcolumn.

Then westackeverything overI₁¹, putting the(i+ 1)th subcolumn over theith subcolumn for each i= 1, . . . , p₁−1, and call this new tower T₂ (with baseI₁¹). The transformation T on T₂ just moves one level up except the last level, where it is not yet defined. Next, inductively at stagen, we cut the towerT_nwith baseI₁ⁿ intopnequal subtowers, over the ith subtower we put an,i spacers and then we stack to get tower T_n+1 with base I₁ⁿ (see Figure 1). Finally, the rank one map T is defined almost everywhere, i.e. it is defined on T∞ := S

n∈NT_n. Often, if there are more transformations involved, we will write a superscript T in the sequences (pn), (an,i), (hn) and T_n. In what follows we will always assume thatp_n>1for every n. This implies that

hn>2ⁿ⁻¹. (4)

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

+∞

X

n=1

1 Qn

k=1pi pn

X

i=1

an,i

!

<+∞. (5)

Under this condition, T preserves a probability measure µ_T given by the normalized Le- besgue measure on T_∞. It is a classical fact, easily seen by a density point argument, that rank one systems are ergodic with respect to the Lebesgue measure. HenceT acts on (T_∞,B, µ_T). Moreover, (5) implies the following bound on the sequence of heights: there exists1≤K <∞such that for every n∈N

n−1

Y

i=1

pi 6hn6K

n−1

Y

i=1

pi. (6)

Figure 1: Stepnin the construction of a staircase rank-one transformation T. Each point not on the top of the figure is mapped to the point directly above him. The arrows show how the subcolumns are stacked together to form T_n+1. The figure also shows two points x,x⁰ in the same level ofT_n, but not in the same level ofT_n+1, and their images T^rxand T^rx⁰ for some h_n ≤ r ≤ (h_n)^1+2ξ as in Lemma 3.2. The points T^rx and T^rx⁰ are not any more in the same level of T_n because the corresponding orbits went through different numbers of spacers.

Indeed, the LHS is immediate from (3). The RHS follows by (5) and (3) since for n =

1 Qn

i=1pi(Ppn

i=1an,i), we have

pn−1

X

i=1

an−1,i=n−1 n−1

Y

i=1

p_i6n−1pn−1hn−1

and so by (3)

h_n6pn−1hn−1(1 +n−1) and recursively

hn6

n−1

Y

i=1

(1 +i)

!_n−1 Y

i=1

pi 6K

n−1

Y

i=1

pi,

whereK:=Q∞

1 (1 +i)<∞by (5). This shows (6).

The set of rank one transformations (satisfying (5)) will be denoted byRank(1).

For n ∈ N and x, y in the same level of T_n we define the horizontal distance of x, y, setting

dH(x, y) := 1

h_m(x,y), (7)

where m = m(x, y) > n is the largest number such that x, y are in one level of T_m (i.e.

they are in different levels of T_m+1). If x, y ∈ T∞ and there is non such that x, y are in the same level ofT_n, we set

d_H(x, y) := 1.

Observe that if T ∈C_γ,γ⁰, for nlarge enough we have µ(T_n+1^T )≤µ(T_n^T)

1 + (h^T_n)^γ

0−1

, and using (4), we get

µ(T_n^T)≥1−O(n⁻²). (8)

2.2 Loosely Bernoulli transformations

We recall the definition of f¯metric introduced in [3]. For two finite words (over a finite alphabet)A=a₁...a_k and B=b₁...b_k, amatching betweenA andB is any pair of strictly increasing sequences(i_s, j_s)^r_s=1 such that a_is =b_js for s= 1, ...r. The f¯distance between A andB is defined by

f¯(A, B) = 1− r k,

wherer is the maximal cardinality over all matchings between A andB.

Let T : (X,B, µ) → (X,B, µ) be a measure preserving automorphism. For a finite partition P = (P1, ...Pr) of X and an integerN >1we denote P₀^N(x) =x0...xN−1, where xi ∈ {1, ..., r}is such that Tⁱ(x)∈Pxi for i= 0, ..., N−1.

Definition 2.1. A zero entropy process (T,P) is said to be loosely Bernoulli if for every ε >0 there exists Nε∈N and a set Aε∈ B, µ(Aε)>1−ε, such that for every x, y∈Aε

and everyN >Nε

f¯(P₀^N(x),P₀^N(y))< ε. (9) A zero entropy automorphism T is loosely Bernoulli (LB for short) if for every finite measurable partitionP, the process (T,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 P such that the process (T ×S,P) is not LB. Recall that we have an exhaustive sequence of towers(T_n^T) and (T_n^S) for T and S respectively. For n∈Nlet Q_n be the partition ofT_∞^T into levels of T_n^T andT_∞^T \ T_n^T (T_∞^T \ T_n^T is the union of all spacers at stages>n). LetR_nbe the analogous partition ofT_∞^S and defineP_n:=Q_n× R_n. Then P_n is a partition of T_∞^T × T_∞^S (in fact, the sequence (P_n) converges to the partition into points). Theorem1 is a consequence of the following theorem:

Theorem 2. Let T be weakly mixing and (T, S) ∈ L^γ,γ0,δ with δ < γ. There exists m₀ such that the process (T ×S,P_m0) 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 f¯

In this section we state a general combinatorial lemma which allows to give a lower bound on thef¯distance. Let (is)^r_s=1 and(js)^r_s=1 be two increasing sequences of positive integers in[0, N].

ForM 6N and 16w < r define

I(M, w) ={s∈ {1, . . . r} : is∈[iw, iw+M]} ⊂[0, N] and

J(M, w) ={s∈ {1, . . . r} : j_s∈[j_w, j_w+M]} ⊂[0, N].

The following property of I(M, w) and J(M, w) is straightforward and will be useful in the proof of the lemma below: Ifs > w and s /∈I(M, w), then

I(M, w)∩I(M, s) =∅, (10)

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

Lemma 3.1. Let ξ ∈ (0,1). Let (is)^r_s=1 ∈ [0, N] and (js)^r_s=1 ∈ [0, N] be two increasing sequences of integers for which there exists a number K ∈ N, 8K^1+ξ 6 N such that for every s= 1, . . . , r

min

|I(K^1+ξ, s)|,|J(K^1+ξ, s)|

62K, (11)

then r < ^4N_Kξ.

Proof. Sets1 := 1. If |I(K^1+ξ, s1)|>|J(K^1+ξ, s1)|, we set B1 := J(K^1+ξ, s1) otherwise, we setB1:=I(K^1+ξ, s1). By (11), we have

|B₁|62K.

We then proceed inductively: assume that for some `≥ 1 we have constructed 1 = s1 <

s2 <· · · < s` ≤r and subsets B1, . . . , B` of {1, . . . , r}. If B1∪ · · · ∪B` = {1, . . . , r}, we stop here and setv:=`. Otherwise, we defines`+1 as the smallest element of{1, . . . , r} \ B1∪ · · · ∪B`, andB`+1 as eitherI(K<sup>1+ξ</sup>, s`+1)or J(K^1+ξ, s`+1)(by choosing the set with the smallest cardinal). By (11), we always have

|B_`+1|62K.

We go on in this way until we obtain a finite sequence of sets{B_i}^v_i=1, such that B1∪ · · · ∪Bv ={1, . . . , r}, (12) and such that

|B_{`</sub>|= min(|I(K<sup>1+ξ</sup>, s<sub>`})|,|J(K^1+ξ, s<sub>`</sub>)|)62K for every `∈ {1, . . . , v}. (13) Moreover, if 1 ≤ `1 < `2 ≤ v, then s_{`2</sub> ∈/ B<sub>`1}. If we further assume that B_{`h</sub> = I(K<sup>1+ξ</sup>, s<sub>`h}) for h ∈ {1,2} (or B_{`h</sub> = J(K<sup>1+ξ</sup>, s<sub>`h}) for h ∈ {1,2}), then by construction (see also (10))

B`1 ∩B`2 =∅. (14)

By pigeonhole principle, there exists a set A ⊂ {1, . . . , v}, with |A| > ^v₂ and such that either for every` ∈A, B<sub>`</sub> =I(K^1+ξ, s<sub>`</sub>), or for every `∈A,B_{`</sub> =J(K<sup>1+ξ</sup>, s<sub>`}). By (14), the subsetsB<sub>`</sub>,`∈A correspond to disjoint subintervals of [0, N], each of them of length K^1+ξ. We thus get

v

2 6 N

K^1+ξ. (15)

Now, (12), (13) and (15) yield

r ≤ |B₁|+· · ·+|B_v| ≤2Kv ≤ 4N K^ξ.

3.2 Proof of Lemma 1.2

Proof of Lemma 1.2: Let(h^T_n) denote the sequence of heights for T. We will construct a staircaseS ∈C_γ/3,3γ⁰ such that(T, S)∈ L^{γ/3,3γ0,γ/4},i.e. the sequences(h^T_n) and (h^S_n) are γ/4- alternating. For this, we will show that there exists n0 such that for every n >n0, we have

(h^S_n−1)^1+γ/4 < h^T_n and (h^T_n)^1+γ/4< h^S_n. (16) Fix η < 1/100. Let K_T denote the constant K for T in (6). We will first construct (p^S_i )^+∞_i=1 so that forn>n₀, we have

n−2

Y

i=1

p^S_i

!1+(¹₄+η)γ

<

n−1

Y

i=1

p^T_i and K_T

n−1

Y

i=1

p^T_i

!1+γ/4

<

n−1

Y

i=1

p^S_i. (17)

Then we will show how to derive (16) from (17).

Since T ∈ C_γ,γ⁰ it follows that for n > n⁰_T, by (6), we have K_T

n

Y

i=1

p^T_i >h^T_n+1>(h^T_n)^1+γ>

n−1

Y

i=1

p^T_i

!1+γ

. (18)

Setp^S_n = 2 for every n∈ {1, ..., n⁰_T + 1}. Then (by enlarging n⁰_T if necessary), we have





n⁰_T+1

Y

i=1

p^S_i





1+(¹₄+η)γ

= (2^n0T+1)^1+(14+η)γ6 1

KT

2^n0T+1 1+γ

6

n⁰_T+2

Y

i=1

p^T_i .

(The last inequality by (18) and since p^T_n > 2.) So the left inequality in (17) holds for n=n⁰_T+ 3. We then proceed inductively: having defined(p^S_i)^w_i=1so that the left inequality in (17) holds for n=w+ 2, we choosep^S_w+1≥2so that

w

Y

i=1

p^S_i

!

p^S_w+1 ∈



 KT w+1

Y

i=1

p^T_i

!1+γ/4

,

w+2

Y

i=1

p^T_i

! ¹

1+( 14+η)γ



. (19) We explain below why such a choice is always possible. Notice that by (18) forn=w+ 2, we have

w+2

Y

i=1

p^T_i

! ¹

1+( 14+η)γ

− KT w+1

Y

i=1

p^T_i

!1+γ

>

1 KT

w+1

Y

i=1

p^T_i

! ^1+γ

1+( 14+η)γ

− K_T

w+1

Y

i=1

p^T_i

!1+γ/4

>4

w+1

Y

i=1

p^T_i ,

the last inequality by ^1+γ

1+(¹₄+η)γ > 1 +γ/4 (recall also that K_T is a fixed constant and w>n⁰_T+ 3andn⁰_T can be made sufficiently large with respect toKT.) Moreover, since we assume that the left inequality in (17) holds forw+2, it follows that4Qw+1

i=1 p^T_i >4Qw i=1p^S_i. Therefore the length of the interval on the right of (19) is at least 4Qw

i=1p^S_i, and so such p^S_w+1 >2 always exists.

By (19), the first inequality in (17) is satisfied forn=w+ 3, and the second inequality in (17) holds for n=w+ 2. Recursively, it follows that (17) holds forn≥n₀ :=n⁰_T + 3.

To guarantee that S is a staircase rank one system we set a^S_n,i = i for n ∈ N and i ∈ {1, ..., p_n}. It remains to show that S ∈ C_γ/3,3γ0 and, using (17), that (16) holds.

Notice that by (17), forw>n₀, we have p^S_w

Qw−1 i=1 p^S_i 6

Qw+1 i=1 p^T_i Qw−1

i=1 p^S_i2 6

Qw+1 i=1 p^T_i Qw−1

i=1 p^T_i 2 = p^T_wp^T_w+1 Qw−1

i=1 p^T_i . (20)

SinceT ∈ C_γ,γ⁰ with0< γ < γ⁰<1/3 and by (6), we get p^T_wp^T_w+1

Qw−1

i=1 p^T_i 6 h^γw⁰h^γ_w+1⁰ Qw−1

i=1 p^T_i 6 K_T²

Qw−1 i=1 p^T_i 2γ⁰

(p^T_w)^γ0 Qw−1

i=1 p^T_i 6

K_T³ Qw−1

i=1 p^T_i 3γ⁰

Qw−1

i=1 p^T_i . (21)

Notice that (20) and (21) together with the definition of spacers forS andγ⁰<1/3imply that

+∞

X

n=1

1 Qn

i=1p^S_i

pn

X

i=1

a^S_n,i

! 6

+∞

X

n=1

(p^S_n)² Qn

i=1p^S_i 6

+∞

X

n=1

p^S_n Qn−1

i=1 p^S_i <+∞,

and hence (6) holds forS (with constantKS). Then, (17) and (6) (forT and forS) ensure the validity of (16) fornlarge enough. This proves that(h^T_n)and(h^S_n)areγ/4- alternating.

Now it remains to check thatS∈C_γ/3,3γ⁰. For this, we will use the following inequality:

for nlarge enough, using several times (6) and the fact that T ∈C_γ,γ⁰, we get h^T_n+2 ≤K_T

n+1

Y

i=1

p^T_i

=K_T p^T_n+1p^T_n

n−1

Y

i=1

p^T_i

≤K_T p^T_n+1p^T_nh^T_n

≤K_T (h^T_n+1)^γ0(h^T_n)^1+γ0

≤KT (KT p^T_nh^T_n)^γ0(h^T_n)^1+γ0

≤K_T² (h^T_n)^1+γ0γ⁰

(h^T_n)^1+γ0

=K_T² (h^T_n)^1+2γ0+γ02

≤(h^T_n)^1+3γ0. By (16), it follows that forn large enough, we have

KSh^S_n+1

h^S_n 6 KS(h^T_n+2)

1 1+γ/4

(h^T_n)^1+γ/4 6 KSh^T_n+2

h^T_n 6KS(h^T_n)^3γ0 6(h^S_n)^3γ0. On the other hand, again by (16)

h^S_n+1

h^S_n > (h^T_n+1)^1+γ/4

(h^T_n+1)^1+γ/41 >(h^T_n+1)

(1+γ/4)2−1

1+γ/4 >(h^S_n)^γ/2.

Therefore and by (6), for nlarge enough, we have p^S_n∈

"

h^S_n+1

h^S_n ,K_Sh^S_n+1 h^S_n

#

⊂h

(h^S_n)^γ/3,(h^S_n)^3γ0i

. (22)

Using (22) and the definition of spacers forS (S is a staircase transformation) we get that S∈ C_γ/3,3γ⁰. This finishes the proof.

3.3 Distribution of points for maps from Cγ,γ⁰

In this section we will do quantitative estimates on recurrence of points. FixG∈ C(γ, γ⁰).

Since G will be fixed throughout this section, we drop the superscript G and denote by T_n, B_n, h_n and p_n the tower, base, height and the number of cuts at stage n (recall that pn ∈[h^γn, h^γn⁰] for n >n⁰_G). For i∈ {1, . . . , p_n}, letT_n,i ⊂ T_n denote the column over the ith cut.

We will construct some subsets of T∞, on which we control the dynamics well. First, we cut off the “boundaries” ofT_n: let

F_n^G:=









hn−j

hn n2

k

[

i=j

hn n2

k

Gⁱ(Bn)









∩







pn−bh^γ/4_n c

[

i=bh^γ/4_n c

T_n,i





. (23) Notice that since pn > h^γn for n >n⁰_G and every column has equal measure, we have µG(F_n^G)>1−_n⁴₂ for n large enough. Let

F^G := \

i>n1

F_i^G, (24)

wheren₁ is such thatµ(F^G)>1−10⁻⁵. In particular, ifx∈F^G, then x∈ T_n1. This will be used in the statement of the lemma below.

Lemma 3.2. Fix 0 < ξ ≤100⁻¹min(γ,1−γ⁰). There exists m₁ ∈N such that for every n> m₁ the following holds: for every x, x⁰ ∈F^G such that d_H(x, x⁰) = _h¹

n (see (7)), for every r∈ {h_n, h_n+ 1, . . . ,(h_n)^1+2ξ} for which G^rx, G^rx⁰ ∈F^G, we have

d_H(G^r(x), G^r(x⁰))> 1 hn−1

. (25)

Moreover, for every x, x⁰ ∈ F^G such that dH(x, x⁰) 6 _h¹_n and for every i, j ∈ {0, ...^h_nⁿ₃}, i6=j, we have

d_H(Gⁱ(x), G^j(x⁰))> 1 hn−1

. (26)

Proof. Letn≥n1be large enough so thath^γn^0+2ξ< hn/n², and letx, x⁰ ∈F^G be such that dH(x, x⁰) = _h¹

n (see Figure1). Fix r ∈ {h_n, hn+ 1, . . . , h^1+2ξn } for whichG^rx, G^rx⁰ ∈F^G. Let 0 6 `<sub>1</sub>, `₂ 6 p_n be such that x ∈ T_{n,`1</sub> and x<sup>0</sup> ∈ T<sub>n,`2}. Since x, x⁰ ∈ F^G ⊂ F_n^G (see (23)), `1, `2 ∈[h^γ/4n , pn−h^γ/4n ]. Notice that r 6h^1+2ξn and the height of T_n is hn. By the definition of F_n^G it follows that

G^rx, G^rx⁰ ∈/

pn

[

i=pn−¹

2bh^γ/4_n c

T_n,i. (27)

Sinced_H(x, x⁰) = _h¹

n, we have `<sub>1</sub> 6=`₂ (otherwise d_H(x, x⁰)6 _hn+1¹ ). Assume without loss of generality that`<sub>1</sub> < `₂. Let`<sub>r</sub>>`₁ be such that G^rx∈ T<sub>n,`r</sub>. Notice that since r>h<sub>n</sub> andG<sup>r</sup>x∈F<sup>G</sup>⊂ T<sub>n</sub>, it follows that in fact`_r> `₁. Define

w_r :=

`r−1

X

i=`1

(a<sub>n,`2</sub>−`₁+i−a_n,i).

By G∈Cγ,γ⁰, using (27) we get thatan,`2−`1+i−an,i>0 (the spacers are monotonically placed). Since `<sub>r</sub> > `₁ it follows that w_r > 0. Moreover, a_n,i 6 h^γ_n⁰ for i ∈ {1, . . . , p_n}, hence

w_r6(`<sub>r</sub>−`₁)h^γ_n⁰ 6 h^1+2ξn

h_n h^γ_n⁰ =h^γ_n^0+2ξ< h_n/n².

Let s be such that G^rx ∈G^sB_n. Since G^rx ∈ F^G ⊂F_n^G, we have s≥ h_n/n², and it follows by definition ofw_r and the fact thatw_r< sthatG^rx⁰∈G^s−wrB_n. Hence G^rxand G^rx⁰ are not in the same level of T_n and (25) follows.

For (26), notice that x, x⁰ ∈ F^G ⊂F_n^G (see (23)) and by assumptions x, x⁰ are in one level of T_n, i.e. for some s ∈ [^h_nⁿ2, h_n− ^h_nⁿ₂], x, x⁰ ∈ G^s(B_n). But by the bounds on s it follows that for u∈ {i, j}, we have G^u(x)∈G^s+u(B_n) (and06s+u6h_n). Since i6=j, we haves+i6=s+j and henceGⁱ(x) andG^j(x⁰) are in different levels ofT_n. This gives (26).

Forn∈Nand x∈ T_n let i_n,x ∈ {0, . . . , h_n−1} be such thatx∈G^in,x(B_n). Define

D^x,n_G :=T_n\

jhn n2

k

[

k=−j

hn n2

k

G^k+in,x(Bn) (28)

Notice that, since G∈C_γ,γ⁰, by (8) we haveµ(D_G^x,n)>1−O(n⁻²). We define D_x^G:= \

n>n3

D^x,n_G , (29)

wheren₃ is such that for all x,µ(D^G_x)>1−10⁻⁵.

Lemma 3.3. There exists m₂ ∈N such that for every x∈F^G (see (24)), every x⁰ ∈D^G_x every n>m2 and everyi, j∈n

0, ..., j hn+1

(n+1)²

ko

, we have

d_H(Gⁱ(x), G^j(x⁰))> 1 hn

. (30)

Proof. Let m2 := 2 max(n1, n3) (where n1 and n2 are defined just after (24) and (29) respectively), and fix n >m2. Let x, x⁰ and i, j be as in the assumptions of the lemma.

Assume by contradiction thatdH(Gⁱ(x), G^j(x⁰))< _h¹

n. By (7) this implies thatGⁱ(x)and G^j(x⁰) are in the same level of T_n+1. Sincex∈F^G ⊂F_n+1^G (see (23)) and0≤i≤ _(n+1)^hn+12

, this implies thatGⁱx and G^j(x⁰) are both located i levels above x inT_n+1. Thereforex and G^j−ix⁰ are in the same level of T_n+1. But |j−i|6 _(n+1)^hn+12, which contradicts the fact thatx⁰ ∈D_x^G⊂D^x,n+1_G (see (28)). The contradiction finishes the proof.

4 A proposition which implies Theorem 2

For the rest of the paper(T, S)∈ L^γ,γ0,δ with0< δ < γ < γ⁰<1are fixed. Since we deal with two transformations T and S, we will denote the horizontal distance (see (7)) for T andS respectively byd₁ and d₂.

Recall from (24) the setsF^T andF^S. Letn0≥4, and setP :=P_n0. For(x, y),(x⁰, y⁰)∈ T_∞^T × T_∞^S, and N ∈N, consider a matchingθ= (is, js)^r_s=1 of P₀^N(x, y) and P₀^N(x⁰, y⁰)(see Section2.2). Then for eachk∈Nwe define

A^k_θ((x, y)(x⁰, y⁰)) :=

n

s∈ {1, . . . , r} :

(T^is×S^is)(x, y),(T^js ×S^js)(x⁰, y⁰)∈(F^T ∩ T_n^T

0)×(F^S∩ T_n^S

0) and 2^−k−1 6max d₁(T^isx, T^jsx⁰), d₂(S^isy, , S^jsy⁰)

<2^−ko . (31) (The dependence on n₀ is only implicit to avoid cumbersome notation.) Notice that by definition of the partitionP_n0 (every product of a level inT_n^T

0 with a level inT_n^S

0is a different atom), the definition of the horizontal distance and (4), we have for eachs∈ {1, . . . , r}

max d1(T^isx, T^jsx⁰), d2(S^isy, , S^jsy⁰)

≤max 1

h^S_n0, 1 h^T_n0

≤ 1 2ⁿ⁰⁻¹. Asn0≥4, for k < ⁿ₂⁰ we have 2^−k−1>2⁻ⁿ⁰⁺¹ and hence

A^k_θ((x, y)(x⁰, y⁰)) =∅. (32) The following proposition implies Theorem2:

Proposition 4.1. There exist n0 > 100, N0 ∈ N and a set B ×C ⊂ T_∞^T × T_∞^S, (µT × µS)(B×C)>99/100, for which the following holds: for every (x, y)∈B×C there exists a setDx×Dy ∈ T_∞^T × T_∞^S,(µT ×µS)(Dx×Dy)>99/100such that for every(x, y)∈B×C, (x⁰, y⁰)∈Dx×Dy and every N >N0

(P1) for each(z, w)∈ {(x, y),(x⁰, y⁰)},

{i∈[0, N] : (Tⁱz, Sⁱw)∈(T_n^T

0 ∩F^T)×(T_n^S

0∩F^S)}

> 9 10N;

(P2) for every N > N0, every k ∈ N and every matching θ = (is, js)^r_s=1 of (P_n0)^N₀ (x, y) and(P_n0)^N₀ (x⁰, y⁰), we have

A^k_θ((x, y)(x⁰, y⁰)) 6 N

k².

The proof of Proposition 4.1is the most technical part of the paper. We will devote a separate section to its proof. Let us first show how Proposition4.1implies Theorem2.

Proof of Theorem 2. We will prove the following:

Claim: For every(x, y)∈B×C,(x⁰, y⁰)∈Dx×Dy and every N >N0, we have f¯ P₀^N(x, y),P₀^N(x⁰, y⁰)

>1/100. (33)

(Here P is the partition P_n0, where n₀ is given by Proposition4.1.) Before we show the Claim, let us show how it implies the result. Assume by contradiction that the process (T ×S,P) is LB. Let := ₂₀₀¹ and let N ∈N,A ⊂ T_∞^T × T_∞^S, (µ_T ×µ_S)(A)>1−be

from Definition2.1. TakeN >max(N0, N). Notice that for every(x, y)∈(B×C)∩A, we have

(Dx×Dy)∩A =∅. (34)

Indeed, if the set above is non-empty, then there exists(x, y)∈(B×C)∩A and(x⁰, y⁰)∈ (Dx×Dy)∩A, but then by (33) and (9)

1/1006f¯ P₀^N(x, y),P₀^N(x⁰, y⁰)

6

and this is a contradiction. So (34) holds. This in turn contradictsµ_T ×µ_S(B×C), µ_T × µ_S(D_x×D_y) >99/100and µ_T ×µ_S(A_ε) >1−. So the proof is finished up to proving theClaim.

To prove the Claim, we will show that for every matching θ= (is, js)^r_s=1 of P₀^N(x, y) andP₀^N(x⁰, y⁰), we haver6 ^9N₁₀. Fix any such matchingθ= (is, js)^r_s=1 and let

H_N :=n

s∈ {1, . . . , r} : (T^isx, S^isy),(T^jsx⁰, S^jsy⁰)∈(T_n^T

0∩F^T)×(T_n^S

0)∩F^S)o . By(P1)we have |{1, . . . , r} \H_n|6 ₁₀¹N, hence theClaimfollows by showing that

|H_N|6 7

10N. (35)

Notice that by definition of A^k_θ((x, y)(x⁰, y⁰)), we have HN =

+∞

[

k=0

A^k_θ((x, y)(x⁰, y⁰)).

Therefore, using (32) and remembering that n₀≥100, we get

|H_N|6 X

k>n0/2

A^k_θ((x, y)(x⁰, y⁰)) ≤ X

k≥50

A^k_θ((x, y)(x⁰, y⁰)) . By(P2), this implies that

|H_N|6N X

k≥50

1 k² 6 7

10N.

This shows (35), which concludes the proof of theClaimand the proof of Theorem2.

5 Proof of Proposition 4.1

This section will be devoted to the proof of Proposition4.1. We will divide the proof into several steps.

Choice of n₀

Recall that (T, S) ∈ L^γ,γ0,δ. Set ξ := min(γ,1−γ⁰, δ)/100. Recall the definitions of m₁ from Lemma 3.2, m₂ from Lemma 3.3, n₁ from (24) and n₃ from (29)². We choose n₀≥max(m₁, m₂, n₁, n₂) large enough so that

µG(T_n^G₀)>1−2⁻¹⁰⁰ for G∈ {T, S}.

2All the constants come in two copies: forT andS, for instance we havem^T1 andm^S1 and we define m1:= max(m^T₁, m^S₁). We define all other constants analogously.

We also assumen0 large enough so that some inequalities depending only onξ and δ hold for eachk≥n0/2 (see (42), (44), (45), (46) and (47)).

Construction of B and C

Since the construction follows similar lines for T and S we will do it simultaneously for G∈ {T, S}. Recall the definition of F^G from (24). By the ergodic theorem, there exists a set F_erg^G , µ(F_erg^G ) >1−10⁻⁴, and a number m^G₃ ∈ N such that for every x ∈ F_erg^G and everyN >m^G₃,

j ∈[0, N −1] : G^jx∈F^G∩ T_n^G

0

>(1−10⁻⁴)N. (36) We then define B := F^T ∩F_erg^T and C := F^S∩F_erg^S . We will write this in the product form:

B×C= (F^T ∩F_erg^T )×(F^S∩F_erg^S ). (37) Notice in particular that(µT ×µS)(B×C)>99/100as required in Proposition4.1.

Construction of D_x and D_y. We define

D_x×D_y := (D_x^T ∩B)×(D_y^S∩C), (38) where D^T_x and D_y^S come from (29) for G = T and G = S respectively (see also (28)).

Notice that(µ_T ×µ_S)(D_x×D_y)>99/100as in the statement of Proposition 4.1.

Definition of N₀ and proof of (P1)in Proposition 4.1

LetN0 := max(m^S₃, m^T₃), wherem^G₃ comes from (36). Notice now that for (x, y)∈B×C and (x⁰, y⁰)∈D_x×D_y ⊂B×C (see (38)) and forN >N₀,(P1)holds since by (36), for any (z, w)∈B×C,

{i∈[0, N] : (Tⁱz, Sⁱw)∈(F^T ∩ T_n^T

0)×(F^S∩ T_n^S

0)}

>

1− 2

10⁴

N.

5.1 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 > N0, (x, y) ∈ B ×C and x⁰, y⁰ ∈ Dx ×Dy. Consider any matching θ = (is, js)^r_s=1 between (P_n0)^N₀ (x, y) and (P_n0)^N₀ (x⁰, y⁰), and let k ∈ N. If k < n0/2, then A^k_θ((x, y),(x⁰, y⁰)) = ∅ (see (32)) and (P2) follows trivially. We therefore assume that k≥n₀/2.

Take anys∈ {1, ..., r}∩A^k_θ((x, y),(x⁰, y⁰)). To simplify notation, denotex_s =T^isx, x⁰_s= T^jsx⁰ and y_s=S^isy, y⁰_s=S^jsy⁰. By definition ofA^k_θ((x, y),(x⁰, y⁰)), we have

2^−k−16max d1(xs, x⁰_s), d2(ys, y_s⁰)

<2^−k. (39)