Rank One Power Weakly Mixing Nonsingular

Transformations �

Terrence Adams y Nathaniel Friedman z Cesar E. Silva x

June 4, 1999; August 27, 2000

Abstract

We show that Chacon's nonsingular type III� transformation T�, 0 < � � 1,

is power weakly mixing, i.e., for all sequences of nonzero integers fk1; : : : ; krg,

Tk1�� : : :�T kr

�is ergodic. We then show that in in�nite measure, this condition

is not implied by in�nite ergodic index (having all �nite Cartesian products

ergodic), and that in�nite ergodic index does not imply 2-recurrence.

Key words and phrases: ergodic, weak mixing, Cartesian product, nonsingular mea-

sure, multiple recurrence

�2000 Mathematics Subject Classi�cation. Primary 37A40. Secondary 28DyOhio State University, Columbus, OH 43210zState University of New York at Albany, Albany, NY 12222xWilliams College, Williamstown, MA 01267

1

1 Introduction

In [KP], Kakutani and Parry de�ned an in�nite measure preserving transformation

T to have in�nite ergodic index if for all k > 0, the k-fold Cartesian products

T � : : :� T are ergodic. They also showed that it may happen that T � T is ergodic

but T�T�T is not ergodic, so T�T ergodic does not imply T of in�nite ergodic index.

Recently, in [DGMS], the authors de�ned a transformation T to be power weakly

mixing if for all �nite sequences of nonzero integers fk1; : : : ; krg, Tk1 � : : : � T kr

is ergodic; they also constructed examples of rank one in�nite measure preserving

transformations that are power weakly mixing. In this paper we show that in�nite

ergodic index does not imply power weak mixing, and that in�nite ergodic index does

not imply 2-recurrence. (In fact, in our example all �nite products T � : : : T are

ergodic but T � T 2 is not conservative, hence not ergodic.) We start by showing in

Section 2 that the nonsingular type III�, 0 < � � 1, Chacon transformations of [JS1]

are power weakly mixing. Our example of an in�nite ergodic index transformation

that is not power weakly mixing is presented in Section 3. However, Section 3 may

be read essentially independently of Section 2, it only uses a standard approximation

lemma frommeasure theory (Lemm2.6) and the notation established in the paragraph

previous to that lemma.

The nonsingular Chacon transformations we study here were shown to be prime

and to have trivial centralizer in [JS1] using nonsingular coding techniques; using

these same techniques the authors in [JS1] also had a proof (unpublished) that their

Cartesian square is ergodic. Transformations similar to these were shown to have

2-fold nonsingular minimal self-joings in [RS], a property which implies prime, trivial

centralizer and ergodic Cartesian square. However, it is not clear if 2-fold nonsingular

minimal self-joings as de�ned in [RS] implies power weak mixing. Furthermore, the

notion of n-fold nonsingular minimal self-joinings, for n > 2, is yet to be studied in

detail, as questions regarding the right extension of the notion and the properties

of a rational joining of more than two transformations remain open. (It was shown

in [RS] that, in the nonsingular category, to study a property such as minimal self-

joinings one must restrict the nonsingular joinings to a sub-class such as the rational

joinings.) The methods we use here to show power weak mixing are extensions of

[AFS], where other examples of rank one in�nite measure preserving and nonsingular

transformations are constructed.

In [JS2], all factors of 2-fold Cartesian products, and the centralizer of arbitrary

Cartesian products, of these transformations are classi�ed, where use is made of the

property of power weak mixing shown in this paper.

As remarked by the referee, in�nite measure preserving null recurrent Markov

shifts that are power weakly mixing have been available for some time. In fact, if

p = (p0; p1; : : : ) is a Kaluza sequence and Sp is its corresponding Markov shift (cf.

[R]) then it can be shown that Sp has in�nite ergodic index if and only if

1Xn=0

pdn =1

for all d � 1. In this case Sp is power weakly mixing. (Such a sequence pn can be

obtained by choosing, for example, pn = 1ln(n+e)

, n � 0.) However, these examples

are di�erent to our rank one constructions, as for Markov shifts of this type in�nite

ergodic index implies power weak mixing.

We would like to thank the referee for several remarks that impoved the exposition.

2 Type III� rank one power weakly mixing

We de�ne our transformations T = T� inductively by the process of \cutting and

stacking"[F], [JS1]. Let 0 < � < 1, C0 = f[0; 1)g, h�1 = 0 and h0 = 1. Assuming that

column Cn of height hn has been constructed, to obtain column Cn+1 cut each interval

(or level) of Cn in the proportions �1+2�

, 11+2�

, and �1+2�

to produce three subcolumns

of Cn, numbered from left to right as Cn;i; i = 0; 1; 2. Then place a spacer (a new

level) above the top of Cn;1, of the same length as the top level of Cn;1, and stack the

three sub-columns from left to right, i.e., the top level of Cn;0 is sent by an a�ne map

to the bottom level of Cn;1, the top level of Cn;1 is sent to the spacer, and the spacer

at the top of Cn;1 is sent by an a�ne map to the bottom of Cn;2. This de�nes a new

column Cn+1 of height hn+1 = 3hn + 1; levels in Cn+1 are numbered from 1 to hn+1,

bottom to top. On each level the transformation is de�ned by the (unique orientation

preserving) a�ne map that takes that level onto the one on top; the contraction or

expansion on the level is the value of the Radon-Nikodym derivative (or Jacobian) on

2

that level. For example, the Radon-Nikodym derivative on the �rst three levels of C1

has values 1=�, 1, and �. This de�nes a transformation T = T� on a �nite interval X

of IR; we then normalize the measure to be a probability measure �. It is clear that

the resulting transformation T is conservative ergodic and that the Radon-Nikodym

derivative are powers of �.

Given a column Cm, columns Cm;i, i = 0; 1; 2, are said to be copies of Cm in Cm+1.

In this way one de�nes, for any n > m, copies of Cm in Cn, and for any interval I in

Cm, copies of I in Cn.

De�nition 2.1. We say levels I and J in Cn are j`j apart if J is the i+ ` level in

Cn when I is the i level.

To show ergodicity of products of powers of T we show that for any two measurable

sets of positive measure, there is a forward iterate of one that intersects the other

in positive measure. We �rst obtain the necessary results for products of levels in

lemmas 2.2, 2.3, 2.4, and then use some approximation results to extend this to

arbitrary measurable sets in the product space.

Lemma 2.2. Suppose that k is a positive integer with ternary expansion

k =Xj2Rk

aj3j

where Rk is a subset of the nonnegative integers and aj 2 f1; 2g. Let tk = jRkj +Pj2Rk

ajhj�1. If n is a positive integer and I is a level in Cn such that T�tkI is in

Cn, then

�(T khnI \ T�tkI) � (�

1 + 2�)jRkj�(T�tkI);

�(T khnI \ T�tk+1I) � (�

1 + 2�)jRkj+1�(T�tk+1I):

Proof. Note that if p is a positive integer and I is a level in Cp such that T�1I

is in Cp, then both T hpI and T 2hpI contain the rightmost third of T�1I. Thus, if

q � p, then �(T hqI \ T�1I) � ( �

1+2�)�(T�1I), �(T 2hqI \ T�1I) � ( �

1+2�)�(T�1I),

and �(T hqI \ I) � ( �

1+2�)2�(I), �(T 2hqI \ I) � ( �

1+2�)2�(I), and furthermore both

T hqI \ T�1I \ Cq;2 and T 2hqI \ T�1I \ Cq;2 are each a union of levels in Cq+1. Now,

we use the ternary expansion of k to express khn as a sum of heights.

3

First, note that hj is formed by the relation

hj+1 = 3hj + 1:

From this relation, it follows that for m � 1 and j � 1,

3jhm = hm+j � hj�1:

Thus,

khn =Xj2Rk

aj3jhn (1)

=Xj2Rk

aj[hn+j � hj�1] (2)

=Xj2Rk

ajhn+j �Xj2Rk

ajhj�1: (3)

Let R =max(Rk) and

� =X

j2RknfRg

ajhn+j :

It may be checked, using induction, that T �I \ T�jRkj+1I contains a union of levels

in Cn+R whose measure is at least

(�

1 + 2�)jRkj�1�(T�jRkj+1I);

and T �I \ T�jRk j+1I is a union of levels in Cn+R. Hence,

�(T �+arhn+RI \ T�jRk jI) � (�

1 + 2�)jRkj�(T�jRkjI)

and

�(T �+arhn+RI \ T�jRkj+1I) � (�

1 + 2�)jRkj+1�(T�jRkj+1I):

Therefore, by translating �P

j2Rkajhj�1, we obtain Lemma 2.2.

Lemma 2.3. Suppose that k is a positive integer with ternary expansion

k =Xj2Rk

aj3j

4

where Rk is a subset of the nonnegative integers, R = max(Rk) and aj 2 f1; 2g. Let

tk = jRkj +P

j2Rkajhj�1. Let ` be a positive integer and suppose that n1 < n2 <

� � � < n` are positive integers such that ni+1 � ni � R + 1 for 1 � i � ` � 1. Let

H =P`

i=1 hni . If I is a level in Cn1 such that T�`tkI is in Cn1, then

�(T kHI \ T�`tk+iI) � (�

1 + 2�)`(jRkj+1)�(T�`tk+iI)

for 0 � i � `.

Proof. We obtain this lemma by invoking Lemma 2.2 ` times. First, by Lemma 2.2,

since n2 � n1 � R + 1 both T khn1 I \ T�tkI and T khn1 I \ T�tk+1I contain a union of

levels whose measure is at least ( �1+2�

)jRkj+1�(I).

Thus, by applying Lemma 2.2 on each of the levels from Cn2 contained in Tkhn1I\

T�tkI and T khn1I \ T�tk+1I, we get that all three of the sets T k(hn1+hn2 )I \ T�2tkI,

T k(hn1+hn2 )I \ T�2tk+1I and T k(hn1+hn2 )I \ T�2tk+2I have measure at least

(�

1 + 2�)2(jRkj+1)�(I):

Finally, if we apply Lemma 2.2 a total of ` times, we obtain Lemma 2.3.

Lemma 2.4. Let m0 and k1; : : : ; kr be positive integers. Let I0 and J0 be levels in

Cm0with I0 above J0. Let k = maxfk1; : : : ; krg and m = m0 + 2k. Let I be the

uppermost copy of I0 in Cm. Then there exists a copy J of J0 in Cm, an integer

s = s(m;k1; : : : ; kr) and Hm = H(m;k1; : : : ; kr) such that

�(T kiHI \ J) � (�

1 + 2�)s�(J)

for i = 1; : : : ; r. Moreover, if I 0 and J 0 are any sublevel of I and J in the same copy

of Cm in Cn then there exists a positive integer Hn = H(n; k1; : : : ; kr) such that for

i = 1; : : : ; r,

�(T kiHnI 0 \ J 0) � (�

1 + 2�)s�(J 0):

Proof. Let tki and Rki be as in Lemma 2.3; put t = maxftk1; : : : ; tkrg, and ` = hm0.

Then t � 2k. Then for any level L in the top Cm0-copy in Cm, T

�`tkiL 2 Cm for

i = 1; : : : ; r.

Put R = max(Sr

i=1Rki) and � = maxfjRkij : i = 1; : : : ; rg.

5

Let m1 = m and choose m2 < : : : < m` so that mj+1 � mj � R + 1 for j =

1; : : : ; `� 1. Let

Hm =

`Xj=1

hmj

as in Lemma 2.3.

Let I be the uppermost sublevel of I0 in Cm. Since `tki � hm0t � hm, for each

i = 1; : : : ; r, T�`tki+jI, as j = 0; : : : ; `, covers ` + 1 consecutive levels in Cm, and

so at least one of them must intersect a copy of J0 in Cm, call it J . Therefore, by

Lemma 2.3,

�(T kiHmI \ J) � (�

1 + 2�)`(jRki j+1)�(J)

for all i = 1; : : : ; r.

So let

s = `(� + 1):

Let n1 = n and choose n2 < : : : < n` so that nj+1�nj � R+1 for j = 1; : : : ; `�1.

Let

Hn =

`Xj=1

hnj

as in Lemma 2.3.

Let D be any copy of Cm in Cn and let I 0 be the sublevel of I in D. Since

`tki � hm0t � hm, for each i = 1; : : : ; r, T�`tk

i+jI 0, as j = 0; : : : ; `, covers ` + 1

consecutive levels in D, and so at least one of them must intersect a copy of J in D,

call it J 0. Therefore, by Lemma 2.3,

�(T kiHnI 0 \ J 0) � (�

1 + 2�)s�(J 0)

for all i = 1; : : : ; r.

Lemma 2.5. (1) For all n � 0,

�2 �d� � T�hn

d�(x) � ��2a:e:

6

(2) Let ` be a positive integer and let H =P`

i=1 hni be as in Lemma 2.4. Then for

any integers n � 0, k; k0 with 0 < jk0j � k,

�2k` �d� � T k0H

d�(x) � ��2k`a:e:

Proof. Part (1) is in Proposition 3.1 in [JS1]. Part (2) follows from the cocycle relation

and part (1) since H is a sum of column heights.

Lemma 2.6 below is the nonsingular version of a standard approximation lemma;

its proof may be obtained by standard measure theory arguments using a Martingale

Convergence theorem (a more elementary proof follows from the proof of a similar

lemma in [DGMS]). To apply this lemma to the nonsingular case we use Lemma 2.7

below (from [M-Z]), whose proof is immediate from the construction and is omitted.

Let r be a positive integer and � be product measure onQr

i=1X. A rectangle

from Cm is a set of the form I = I1 � : : :� Ir, where Ii are all levels of column Cm.

If

�(A \ I) > (1� �)�(A)

we say that I is (1 � �)�full of A. It is clear that there are 3n�m copies of Cm in

Cn. Let U(m;n) = f1; : : : ; 3n�mg index the copies of Cm in Cn, and let U(m;n)r =

U(m;n)� : : :� U(m;n) (r times).

Lemma 2.6. (Double Approximation) Let L = L1� : : :�Lr be a rectangle from Cm,

let � be product measure, and let 0 < � < 1. If A is any subset of L such that L

is (1 � �)-full of A, then for any �, 0 < � < 1 and any t, 0 < t < (1 � �), there

exists an integer N such that for all n � N , there is a set U 00 � U(m;n)r such that

for all u = (u1; : : : ; ur) 2 U 00, L(u) is (1 � �)-full of A and each L(u) is of the form

L(u) = Lu11 � : : :� Lur

r where Luii is a sublevel of Li in the ui copy of Cm in Cn, and

furthermore, if L00 = [u2U 00I(u) then

�(L004L) < (1� t)�(I):

Lemma 2.7. Let I and J be two levels in a column Cm. If I 0 and J 0 are any two

sublevels of I and J , respectively, in a later column Cn such that I 0 and J 0 are in the

same copy of Cm in Cn, then �(I) = �k�(J) implies �(I 0) = �k�(J 0).

7

Theorem 2.8. The type III� transformation T�, 0 < � < 1, is power weakly mixing.

Proof. Let k1; : : : ; kr be nonzero integers and let k = maxfjkij : 0 < i � kg. We will

show that S = T k1� � : : :� T kr

� is ergodic.

Let A and B be subsets of the product space �ri=1X with �(A) > 0 and �(B) > 0.

Choose I0 = I0;1 � : : :� I0;r and J0 = J0;1 � : : :� J0;r so that

�(A \ I0)

�(I0)> 1�

1

4

��2kr

�;�(B \ J0)

�(J0)> 1 �

1

4

��2kr

�;

and for all i = 1; : : : ; r, I0;i and J0;i are all in some column Cm, with m > k, I0;i is

above J0;i if ki is positive, and I0;i is below J0;i if ki is negative (this latter part can

be done by choosing a �ner approximation in an earlier column if necessary). Let

` = hm0, for each i = 1; : : : ; r and Rki be as in Lemma 2.3. For each i, let Ii and Ji

be the sublevels of I0;i and J0;i as in Lemma 2.4 (with the appropriate interpretation

when ki is negative), and s = s(m;k1; : : : ; kr). Let I = I1� : : :� Ir, J = J1� : : :�Jr

and A and B denote A\ I and B \J , respectively. By the choice of I0 and J0 above,

�(A\I)

�(I)> 3

4and

�(B\J)

�(J)> 3

4. Let �i be such that �(Ii) = �i�(Ji), � = �r

i=1�i and

�� = minf�i; 1g. Then �(I) = ��(J). Let

� = ��(�2k`)r:

Choose

0 < � <��2k`

�r� �

1 + 2�

�sr

(��)r:

By Lemma 2.5(2),

�(2k`)r �d� � SH

d�� �(�2k`)ra:e:

Apply Lemma 2.6 to I and J in Cm twice to obtain sets of indices U 00 and V 00

and rectangles I(u), and J(v) so that for all u 2 U 00, v 2 V 00, I(u) = Iu11 � : : : � Iurr

is (1 � �

3�)-full of A and J(v) = Jv1

1 � : : :� Jvrr is (1 � �

3)-full of B, Iu11 ; : : : ; Iurr and

Jv11 ; : : : ; Jvr

r are in the same column Cn, with n > m, and for each i, Iuii and Jvii are in

the ui and vi Cm-copy in Cn, respectively, and if I 00 = [u2U 00I(u) and J 00 = [v2V 00J(v)

then

�(I 004I) <

�1

2

��(I)

8

and

�(J 004J) <

�1

2

��(J):

Since the union of the intervals corresponding to the indices in U 00 covers more than

1=2 of I, Lemma 2.7 implies that

�

([u2U 00

J(u))4J

!<

�1

2

��(J):

Thus there must exist at least one index w in both U 00 and V 00; so we must have some

I(w) and J(w), which are more than 1 � �

3�and 1 � �

3full of A and B respectively.

Also, �(I(w)) = ��(J(w)).

Write I(w) = Iw1

1 � : : : � Iwrr and J(w) = Jw1

1 � : : : � Jwrr . As Iwii and Jwi

i are in

the same copy of Cm, by Lemma 2.4, if ki > 0,

�(T kiHIwii \ Jwii ) >

��

1 + 2�

�s

�(Jwii );

and if ki < 0,

�(T kiHIwii \ Jwii ) =

ZIwi

i\T�kiHJ

wi

i

d� � T kiH

d�d�

� �2k`�(Iwii \ T�kiHJwii )

���2k`

�� �

1 + 2�

�s

�(Iwii )

=��2k`

�� �

1 + 2�

�s

�i�(Jwii )

���2k`

�� �

1 + 2�

�s

���(Jwii ):

Thus

�[SHI(w) \ J(w)] ���2k`

�r� �

1 + 2�

�sr

(��)r�(J(w)) > ��(J(w)):

Now

�(SH(I(w) nA)) =

ZI(w)nA

d� � SH

d�d� � �(�2k`)r�(I(w) nA)

� �(�2k`)r�

3��(I(w)) =

�

3��(I(w)) =

�

3�(J(w)):

9

Therefore,

�(SHA \ B) � �(SHI(w) \ J(w))� �([SHI(w) \ J(w)] n [SHA \ B])

� �(SHI(w) \ J(w))� �[SH(I(w) nA)]� �(J(w) nB)

� ��(J(w))��

3�(J(w))�

�

3�(J(w)) > 0:

Remark. 1. By choosing log �1 and log �2 irrationally related in alternate times n in

the construction of T we obtain a type III1 transformation with the same properties.

Recently, after this research was completed, Hamachi and the third-named author

have constructed type III0 measures on Chacon's map for which they prove their

Cartesian square is ergodic; however, these measures do not satisfy the analogue of

Lemma 2.5 and their methods are di�erent.

2. The argument we have given applies to other families of transformations. For

example, the argument gives ergodicity for Cartesian products of powers of T�i for

di�erent values of �i, 0 < �i < 1.

3 In�nite Ergodic Index Not Power Weakly Mix-

ing

In this section we construct a transformation that has in�nite ergodic index but is

not power weakly mixing. We also show that our example is not 2-recurrent. We note

that the �rst-named author has studied other constructions where the ergodicity and

conservativity of two-fold products are analyzed [A].

We recall that a transformation T is said to be d-recurrent if for all sets of positive

measureA, there exists n > 0 such that �(A\T nA\: : :\T ndA) > 0. As is well known,

Furstenberg [Fu] has shown that every �nite measure preserving transformation is

multiply recurrent. In [EHH], Eigen, Hajian and Halverson showed that there exist

in�nite measure preseving transformations that are d recurrent but not d+1 recurrent,

for every d > 1; however their examples are in�nite measure preserving odometers

and hence do not have in�nite ergodic index. More recently, Aaronson and Nakada

[AN] have shown a theorem that in particular implies that an in�nite Markov shift

10

with in�nite ergodic index is d-recurrent for all d. They also construct an in�nite

odometer with some additional properties that is not 2-recurrent.

We now construct a family of transformations. Let an, bn, cn and dn be sequences

of nonnegative integers. Let Cn be a column of height Hn. Form Cn+1 by cutting Cn

into four subcolumns of equal width and placing an, bn, cn and dn spacers on the �rst,

second, third and fourth subcolumns, respectively; then stack the subcolumns from

left to right to form Cn+1, of height Hn+1. If we begin with C1 as an interval, then

the previous algorithm de�nes a Lebesgue measure preserving transformation on a

subset of IR. De�ne pn = Hn + an, `n = Hn + bn, qn = Hn + cn, mn = Hn + dn and

hn+1 = pn + `n + qn +Hn for n 2 IN. Denote the transformation by T = Tv where

v = f(pn; `n; qn;mn) : n 2 INg. Consider only transformations T = Tv where

qn � pn;

limn!1

pn

hn=1;

and `n and mn are chosen so that

`n > n(pn + qn + 2hn)

and

mn > n(pn + `n + qn +Hn):

Let us describe informally how `n and mn are chosen, n > 3. mn�1 is chosen large

enough so that once the bottom hn intervals of Cn have entered the top dn�1 levels of

Cn under T2, those levels remain in the dn�1-spacers until the bottom hn levels enter

the dn�1-spacers under T . `n is chosen large enough so that the bottom hn levels

of the third subcolumn of Cn have moved into the mn-spacers under T before the

bottom hn levels of the second subcolumn have exitted the `n-spacers under T2.

We �rst show that T has in�nite ergodic index; for this we assume cn = an+1 for

all n � 1.

The proof of the following lemma is as in Lemma 2.1 in [AFS].

Lemma 3.1. Let cn = an+ 1 for all n � 1. If I and J are any two levels in Cn that

are at most ` apart, with I above J , then there exists and integer H = H(n; `) such

that

�(THI \ J) �

�1

4

�`

�(J):

11

Theorem 3.2. Let cn = an + 1 for all n � 1. Then T has in�nite ergodic index.

Proof. We outline the main ideas since the proof is similar to that of Theorem 2.8

(and Theorem 2.2 of [AFS]). Let A and B be subsets of the product space of positive

measure. Choose I = I1 � : : : � Ir and J = J1 � : : : � Jr such that�(A\I)

�(I)> 3

4,

�(B\J)

�(J)> 3

4, and for all i = 1; : : : ; r, Ii and Ji are all in the same column Cm, and Ii

is above Ji. Let A and B denote A \ I and B \ J , respectively. Let ` > 0 be such

that Ii and Ji are at most ` apart, for all i. Choose

0 < � < (1

4)`r:

Let S = T � : : :�T (r times). Apply the double approximation lemma twice with

� = 1=4, t = 1=2, and �=3 above, to obtain sets of indices U 00 and V 00 and rectangles

Iu, and Jv so that for all u 2 U 00, v 2 V 00, Iu = I 01 � : : :� I 0r is (1 ��3)-full of A and

Jv = J 01� : : :�J 0r is (1��

3)-full of B, I 0i; J

0

i are in the same column Cn, and I 0i and J 0i

are in the same Cm-copy in Cn, for i = 1; : : : ; r. Since all levels in a column have the

same length we can conclude that U 00 and V 00 contain more than 1=2 of the indices in

U and V , respectively.

Thus there must exist at least one index w 2 U 00\ V 00 so so that Iw and Jw are

both more than (1� �

3)-full of A and B respectively.

By Lemma 3.1,

�[SHIw \ Jw] >

�1

4

�`r

�(Jw):

Then, as in Theorem 2.8, one obtains that

�(SHA \B) � ��(Jw)��

3�(Jw)�

�

3�(Jw) > 0:

Theorem 3.3. If the set

fn 2 IN : jqn � 2pnj � 3hng

is �nite, then T � T 2 is not conservative, hence T is not power weakly mixing. Fur-

thermore, T is not 2-recurrent.

12

Proof. Choose N such that N > 3 and jqn � 2pnj > 3hn for n � N . Let A be the

bottom level of CN+1. De�ne

� = fi > 0 : �� �((T � T 2)i(A�A) \ (A�A)) > 0g

and

�n = � \ (0;1

2Hn):

We will prove inductively on n, that �n = ;. This is true for 1 � n � N since

�(T iA \ A) = 0 for 0 < i < HN .

Suppose that our assertion holds for �n. We wish to verify that �n+1 = ;: Let D

denote the union of the bottom hn levels in Cn. The set A will be contained in D.

Let Dj be the part of D in the jth subcolumn of Cn, j = 1; : : : ; 4. Below we give a

table showing the integers i between 0 and 12Hn+1 such that T iDj \Dk 6= ;. The �rst

number in each rectangle is the left hand point of the interval and the second point

is the right hand point of the interval of intersection.

i D1 D2 D3 D4

T iD1 0 pn � hn pn + `n � hn pn + `n + qn � hn

hn pn + hn pn + `n + hn pn + `n + qn + hn

T iD2 0 `n � hn `n + qn � hn

hn `n + hn `n + qn + hn

T iD3 0 qn � hn

hn qn + hn

T iD4 0

hn

We want to show that there is no simultaneous intersection for i = j and i = 2j

among the 10 intervals listed. The intervals give the sets of intersection for T . To

obtain the sets of intersection for T 2 divide the endpoints of the intervals by 2.

By the choice of mn�1 we have that

mn�1

2>

mn�1

(n� 1)> hn:

Also, 12Hn > 1

2mn�1 > hn. By induction it follows that � \ (0; hn] = ; and so there

cannot be simultaneous intersection on this interval with any other interval in the

table.

13

Now `n was chosen so that

pn + qn + hn <`n � hn

n<

`n � hn

2:

Thus, `n was chosen large enough so that we only need to compare the intervals with

`n as a summand in the endpoints to other intervals with `n.

Still, `n and n are su�ciently large so that

pn + `n + qn + hn

2<

(n� 1)(pn + `n + qn + hn)

n

< pn + `n + qn + hn �1

n`n

< `n � hn:

Thus, T 2 will have pushed D1 beyond D4 before T pushes D2 into D3. This takes

care of the worst case of intersection between intervals with `n as a summand.

Finally, we consider the case of simultaneous intersection on the interval in the

�rst row, second column and on the interval in the third row, fourth column. The

assumption on N implies that

jqn

2� pnj >

hn

2+ hn:

Thus there is no possibility of intersection of D1 with D2 under T , and intersection of

D3 with D4 under T2 since the distance between the centers of each of the intervals

is greater than the sum of their radii. The remaining possibility cannot occur since

qn � hn >pn+hn

2as

qn > Hn > mn�1

> (n � 1)(pn�1 + qn�1 +Hn�1 + `n�1)

= (n � 1)hn

� 3hn(for n � 4),

by construction.

Finally, as � � �[(T � T 2)n(A�A) \ (A�A)] = 0 for all n 6= 0, it easily follows

that for all n 6= 0, �(A\T nA\T 2nA) = 0, hence T is not 2-recurrent. This completes

the proof of the theorem.

14

Finally we have the following corollary.

Corollary 3.4. In�nite ergodic index does not imply power weak mixing. Furthere-

more, in�nite ergodic index does not imply multiple recurrence.

Proof. Choose an = 3hn and cn = an + 1. Then qn = pn + 1 and 2pn � qn =

2Hn + 6hn �Hn � 3hn � 1 � 3hn. Now apply Theorems 3.2 and 3.3.

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15

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16