New York Journal of Mathematics
New York J. Math.
3A (1998) 125{134.
Finite Rank Zd Actions and the Loosely Bernoulli
Property
Aimee S. A. Johnson and Ayse A. Sahin
Zd actions and show that those nite
rank actions with a certain tower shape are loosely Bernoulli for d 1.
Abstract. We dene nite rank for
Contents
1. Introduction
2. Background
3. Finite Rank
4. The Matching Lemma
5. The Theorem for Finite Rank
References
1.
125
126
127
127
131
134
Introduction
In 4], the authors show that rank 1 Zd transformations are loosely Bernoulli,
extending the Z result from 7]. D. Ornstein, D. Rudolph, and B. Weiss also show
in 7] that all nite rank transformations are loosely Bernoulli. In this paper we
give the Zd generalization of this result.
Intuitively, a zero entropy loosely Bernoulli (LB) action has one name up to the
f metric on processes (cf 6, 2]). The proof in 4] rests on the fact that for a rank
one action most large enough names are well covered by towers of various sizes.
Given two large names we use this fact to identify towers in one name who have a
same size tower close by in the second name. We use these pairs of neighbours to
show that the names are f close.
In the rank r case with r > 1, it is still true that large names are well covered
by towers of many sizes, but now each size tower is of r dierent types. Hence,
Received February 20, 1998.
Mathematics Subject Classication. 28D15, 28D20.
Key words and phrases. ergodic theory, nite rank, Zd actions, loosely Bernoulli.
The research of the rst author was partially supported by the Swarthmore College Research
Fund.
The research of the second author was partially supported by the NSF under grant number
DMS-9501103.
125
c 1998 State University of New York
ISSN 1076-9803/98
Aimee S. A. Johnson and Ayse A. Sahin
126
while it is still possible to nd close by towers, the methods of 4] do not guarantee
that nearby towers are of the same type. In this paper we generalize the matching
argument of 4] to address this issue.
The next two sections contain the necessary denitions. There is then a section
containing a Generalized Matching Lemma. The nal section uses this lemma to
prove the result.
2.
Background
Let (X A ) be a Lebesgue probability space. Take T to be an ergodic Zd
action on (X A ). We can think of T as being generated by d commuting measure
preserving 1-dimensional transformations on X , fT~e1 ::: T~ed g, where f~e1 : : : ~edg is
the standard basis for Zd. Then T~v (x) = T~ev11 T~evdd (x), where ~v = (v1 ::: vd ).
We call (X A ) T a Zd-dynamical system. Often we will simply write (X T ).
Let P be a nite label set, or equivalently, a nite measurable partition P =
fp1 ::: ph g on X . (T P ) is then the usual process associated with T and the
partition P . Set jj~vjj = max fjvi j : 1 i dg, and for n 2 N ,
Bn = f~v = (v1 : : : vd ) 2 Zd : 0 vi ng:
For each x we can then dene its Pn -name to be Pn (x) : Bn ! P by Pn (x)(~v ) = i if
T~v (x) 2 pi . In order to dene a loosely Bernoulli process we start with : Bn ! Bn ,
a permutation of the indices in Bn , and dene a size for this permutation. This
idea is dened and extended in 1] and 5].
Denition 2.1. Let : Bn ! Bn be a permutation of the indices of Bn. We say
is of size , denoted by m() < , if there exists a subset S of Bn satisfying
(i) jS j > (1 ; )jBn j, where jS j is the cardinality of the set S ,
(ii) jj~u ; ~v ; (~u ; ~v )jj < jj~u ; ~v jj for every ~u~v 2 S .
Denition 2.2. Given two Pn -names and , we dene the f n-distance between
them to be
f n (
) = inf f > 0 : there exists a permutation of Bn such that
(i) m() < (ii) d(
) < g:
Here d(: :) denotes the Hamming metric which simply gives the proportion of locations of Bn on which the two names disagree.
Informally, we will think of as rearranging the name to make it d close to the
name and we will often refer to as acting on a name instead of the (technically
correct) set of indices. If , , and satisfy (ii) of the above denition we say matches a (1 ; )-proportion of and .
Intuitively, a zero entropy loosely Bernoulli process has one name up to f . Formally,
Denition 2.3. A zero entropy process (T P ) is loosely Bernoulli (LB) if for any
> 0 Wthere exists an integer N such that for any n N and -a.e. atoms ! and
! of ~v Bn T~v P ,
f n (! ! ) < :
0
2
0
Finite Rank Zd Actions and the Loosely Bernoulli Property 127
Denition 2.4. We say (X A ) T is LB if for every partition P of X , (T P ) is
LB.
3.
Finite Rank
Denition 3.1. Let r 2 N. We say
(X A ) T is a Zd rank r transformation
j
if there exists a sequence of sets Fi X , 1 j r, and Flner sequences Dij ,
1 j r, of subsets of Zd, such that for each i, fT~v Fij g are pairwise disjoint for
~v 2 Dij and 1 j r, and the partitions
Pi = fT~v Fij : ~v 2 Dij and 1 j r X ; rj=1 ~v D T~v Fij g
converge to A as i ! 1. We also assume that r is the smallest integer for which
2
j
i
the above sets can be found.
For each i, we will have r disjoint towers, ~v Dij T~v Fij , where 1 j r, which
will be denoted by ij . As in the one dimensional case, any transformation which
is rank r is ergodic and has zero entropy 3, 8].
In this paper we restrict our attention to rank r transformations with a special
tower shape.
Denition 3.2. (X A ) T is a Zd uniform square rank r transformation
if it is
j
rank r and there exists 1 such that for all i and j , the sets Di of Denition 3.1
satisfy:
1. Dij is a rectangle of dimensions fl1ji ::: ldji g.
2. If sji = k=1
mindflkji g and bji = kmax
fljig, and we set
=1 d k
2
si
= j=1
minrfs g
ji
bi
= jmax
fbji g
=1 r
then .
Let v = (si )d , the smallest possible volume of a tower at stage
i, and vib = (bi )d , the
vis
1
biggest possible volume of a tower at stage i. Note then vib , where = d 1.
s
i
si
bi
1
Denition 3.3. Given a rectangle R Zd of size l1 ld, the -interior of R
is the collection of indices in R which are at least a distance lk from the kth edge
of R. The -collar of R is the complement of the -interior and corresponds to the
set of indices within lk of the kth edge in the boundary of R.
Denote the volume of Dij by vij = l1ji ldji and notice that the volume of
the -interior of Dij is (1 ; 2)l1ji (1 ; 2)ldji = (1 ; 2)dvij . Notice also that
by Denition 3.2 we have that for 1 j k r
1 vis vij vib :
4.
The Matching Lemma
vib
vik
vis
The following result is a generalization of the Matching Lemma in 4]. We have
already discussed the dierence between the rank 1 case and the case with rank r
with r > 1. To deal with this dierence the proof of this result uses two applications
of the ergodic theorem. We pick a tower stage k such that there is some k-tower
128
Aimee S. A. Johnson and Ayse A. Sahin
which has measure approximately 1r . By the ergodic theorem we know that most
large enough names will visit this tower approximately r1 of the time. We match
even larger names which visit the previous large names frequently enough. This
second use of the ergodic theorem guarantees that there is \randomness" in the
location of the dierent types of towers.
Lemma 4.1 (The Generalized Matching Lemma). Let and c < r1 be xed. Let
a = 12 (1 ; 2)d(1 ; )dc2 .
There is an integer K ( c) > 0 such that for all k K ( c), if Pk is the partition
associated with the kth towers, then there exist integers N (k) and m(k) such that
for all n N (k), we can nd a set W with (W ) > 1 ; and for ! ! , (Pk )n -names
of two points x x 2 W , there exists a permutation : Bn ! Bn such that
d(! ! ) < 1 ; a
0
0
0
and the action of can be described as follows:
(a) translates all the indices of Bn by a vector ~v, with k~vk < m(k), except for
those indices i for which i + ~v 2= Bn . On these, (i) is dened to be one of
the indices vacated by the translation.
(b) For a subset of the kj 's occurring in !, moves the -interiors of these towers
by an additional amount which can vary for each tower but is always less in
magnitude than sk . The resulting location of the -interiors of these towers
matches perfectly with the corresponding interior of a kj in ! .
0
Proof. Let and c be given. Pick 1 satisfying
(1)
1 1 0 < 1 < min 4 3 ( r ; c) 2 satisfying
(2)
0 < 2 < 21 2 1 d(1 ; 31)
;
and choose K such that for all k K rj=1 kj > 1 ; 21 . Fix some k K .
For 1 j r set mj = kj , and notice that for some j we must have mj 1r ; 21 .
Say this is true for j = 1.
Choose n1 2 N such that
(3)
bk
1
n1 < 2d
and there is a set U with U > 1 ; 2 such that for all x 2 U ,
jf~v 2 Bn1 : T~v (x) 2 k1 gj > 1 ; :
(4)
nd1
r
1
Next, choose N > n1 large enough so that for every n > N
(5)
n1 < 1 n 2d
Finite Rank Zd Actions and the Loosely Bernoulli Property 129
and there is a set W with W > 1 ; so that for all x 2 W
jf~v 2 Bn : T~v (x) 2 U gj > 1 ; 2
(6)
2
nd
jf~v 2 Bn : T~v (x) 2 rj=1 kj gj > 1 ; 2
1
nd
1
jf~v 2 Bn : T~v (x) 2 k gj > 1 ; :
nd
r 1
(7)
(8)
Fix such an integer n and a set W . Thus points in W have n-names which are
all but 21 full of k-towers, are all but 22 full of U , and see k1 with frequency at
least 1r ; 1 .
Take two points in W and let ! and ! be their n-names. We will dene a permutation between them satisfying the statement of the lemma. Roughly speaking,
to do this we will rst identify the towers in ! which have a tower of the same type
occurring close by in ! . We will be able to match a subset of these towers.
Consider all the towers which occur in !. Notice that by conditions (5) and (7)
a proportion
0
0
> (1 ; 31 )
(9)
of ! lies in a complete tower which is a distance at least n1 away from the boundary
of !.
Enumerate these complete towers in !. Denote the position of the base point of
tower t by ~zt 2 Bn . Consider Bsk , the box of size sk sk , centered at ~zt in
! . We will say tower t is a good tower if the corresponding Bsk box in ! contains
the base point of an n1 -name from U .
Create an array whose rows correspond to complete towers in ! that lie at least
n1 away from the boundary of !, and whose columns are the elements of Bn1 listed
in some order. A row in this array corresponding to a good tower will be called a
good row.
Fix a good row of the array, say row t. So tower t in ! is a good tower, namely
there is an occurrence of a base point of a name from U in the Bsk box at position
~zt in ! . Now shift this box in ! by each ~v 2 Bn1 . In entry (t~v ) of the array, place
a if the box shifted by ~v contains the base point of a tower of the same type as
tower t in !.
Suppose the array is c% full of marks, for some c > 0. Then there will be a
column, say column ~v 2 Bn1 such that at least c% of the column will be full of marks.
The towers in the marked rows of column ~v will be the ones we will match.
We will dene the permutation from ! to ! as follows: will rst translate all
indices in Bn by the vector ~v, except for those in the n1 -collar of Bn which are
dislocated by the translation. These indices can be mapped into the indices in the
n1 -collar which are vacated by the translation. Note that the towers represented
in the array are still intact after this translation. Further, the towers whose entry
in column ~v has a are now within sk of a tower of the same type in ! . will
shift the -interiors of these towers by the amount needed to give an exact match
to that tower in ! .
0
0
0
0
0
0
0
130
Aimee S. A. Johnson and Ayse A. Sahin
This construction will give m(k) n1 , and the permutation will have matched
a proportion at least
(percentage of ! covered by complete towers within n1 of the boundary)
(the percentage of these towers moved close to a like tower in ! by ~v)
(the percentage of a tower inside its -interior):
By equation (9) and the denition of an -interior, this is at least
(10)
(1 ; 31 ) (percentage of array which is full of ) (1 ; 2)d:
Letting c be as above, we will nd a lower bound for c by computing the
(percentage of rows of the array which correspond to good towers of type 1)
(the proportion of such a row which is lled with * marks):
To nd the rst quantity, rst note that by (2), (6), and (9) less than
1
(11)
0
of the rows in the array are bad rows.
Next, note that by conditions (5) and (8) we have a proportion
1 ( 1 ; 2 )
r
1
of the rows of the array corresponding to towers of type 1 in !. Call these type 1
rows.
Hence the rows which are both good and of type 1 take up at least a proportion
1 ( 1 ; 2 ) ; 1 = 1 ( 1 ; 3 )
r
1
r
1
of the array. To nd a bound for c all we are missing is the proportion of a type 1
row which is guaranteed to be lled by marks.
For this computation we note that by conditions (3) and (4) we have that a name
from U sees at least
( 1r ; 21)nd1
vk1
base points of towers of type 1. Hence a good row of type 1 is a proportion at least
( r1 ; 21) d s 1 1
d
v1 vk ( r ; 21)
k
full of marks.
So the entire array is
(12)
> 12 ( 1r ; 31)( 1r ; 21)d > 12 ( 1r ; 31 )2 d
full of marks.
Finally, we plug (12) in for c in (10) to conclude that we have matched a proportion
> (1 ; 3 ) 1 ( 1 ; 3 )2 d(1 ; 2)d
1
2 r
1
Finite Rank Zd Actions and the Loosely Bernoulli Property 131
of !. By our choice of 1 in (1) this is
> (1 ; ) 1 c2 d(1 ; 2)d = a:
2
5.
The Theorem for Finite Rank
Theorem 5.1. Let r 2 N. If T is a uniform square rank r Zd action on a Lebesgue
probability space (X M ) then T is LB.
Proof. Let Q be an arbitrary partition on X and consider the process (T Q). Let
> 0 be xed.
We will show that there is an integer N such that for all n N we can nd a
set W with W > 1 ; such that for all x x 2 W if ! ! are the Qn-names of x
0
0
and x then
(13)
f (! ! ) < :
2
Let = 16d2 2 . Pick c < r1 and let a = 12 (1 ; 2)d(1 ; )d c2 .
For ease of exposition we assume (1 ; a)2 < 2 . The proof of the general case
can be constructed directly from our argument. The idea will be to apply the
Generalized Matching Lemma (GML) to obtain an integer K ( c) and to nd two
tower
Wt sizes K ( c) < k1 < k2 so that the partition Q is well approximated by
some t k2 . If W is the set from the GML associated to k2 , and ! !
i=1 Pi , for
W
t
are two ( i=1 Pi )n -names from W , then for large enough n the GML guarantees
the existence of a permutation 2 such that
d(! 2 ! ) < 1 ; a:
The trick is to choose k1 and k2 such that if k11 satises (k11 ) r1 , and Gc is the
set of unmatched indices of ! 2 , then for some n1 , Gc sees n1 -names which visit
k11 approximately r1 of the time. This is exactly the setup of the GML, and we can
construct a permutation 1 of Gc as in that lemma which will match a percent of
Gc .
We now choose our parameters and give the details of the argument.
Apply the GML with and c to obtain an integer K ( c) as in the statement of
that lemma. Pick 1 satisfying
1 1 (1 )
0 < 1 < min 4 3 ( r ; c)
and 2 satisfying
(2 )
0 < 2 < 21 2 1 d (1 ; 31 ):
0
0
0
0
0
0
Pick k1 > K such that
21 :
(
rj=1 kj1 ) > 1 ; 16
If mj = (kj1 ), then for some j we must have
21 > 1 ; 1 :
mj > 1r ; 16
r 4
Aimee S. A. Johnson and Ayse A. Sahin
132
Say this happens for j = 1.
Choose n1 2 N such that
1
<
n1 2d
2
and such that there is a set U with U > 1 ; 22 such that for all x 2 U
jf~v 2 Bn1 : T~v (x) 2 k11 gj > 1 ; 1 :
(4 )
r 2
nd1
Next choose k2 > k1 such that m(k2 ) > n1 and
n1 < 22 :
(5 )
sk2
16d
W
Choose t 2 N such that t > k2 and there is a partition Qt ti=1 Pi such that
d(Q Qt) < 8 . Then pick n > N (k2 ) large enough so that we can nd a set W
which not only satises the statement of the GML, but in addition, for all x 2 W
bk1
(3 )
0
0
0
we have
(14)
jf~v 2 Bn : T~v (x) 2 U gj > 1 ; 2
2
nd
j
jf~v 2 Bn : T~v (x) 2 rj=1 k2 gj > 1 ; 3 2
(15)
nd
21
2
m(k2 ) + n1 < 2
(16)
n
16d
and the (Qt )n -nameWof x and the (Qn ) name of x dier less than 4 of the time.
Consider ! ! two ( ti=1 Pi )n -names of points in W . We will dene a permutation
: Bn ! Bn such that
d(! ! ) < 2 and m() < :
We can then use this same on Q names, and by our choice of t, we will have
0
0
obtained (13).
Let ! ! be as above and apply the GML to ! and ! with tower k2 and to
obtain a permutation 2 : Bn ! Bn such that (a) and (b) of that lemma hold. Let
G be thec set of matched indices of ! 2 .
If Gnd < 2 then to complete the proof we need only show that m(2 ) < .
Suppose that
0
j
0
j
jG j :
(17)
nd 2
We have already chosen 1 and 2 appropriately (in (1 ) and (2 ) respectively)
and have chosen our parameters to guarantee (3 ),(4 ) and (5 ). So to construct
a permutation 1 as in the GML it remains to show that Gc is all but 22 full of
occurrences of U , is all but 21 full of occurrences of k1 towers and visits k11 all but
1
r ; 1 of the time. These estimates will be the analogs of (6),(7) and (8).
For all these estimates we will need to know what proportion of the indices in
Gc lie in an (m(k2 ) + 2n1 )-collar around the boundary of Gc . The boundary of Gc
consists of the boundary of Bn , and the disjoint union of boundaries of boxes of
dimensions at most (1 ; 2)bk2 (1 ; 2)bk2 . Using (3 ), (16), (17), and the
c
0
0
0
0
0
0
Finite Rank Zd Actions and the Loosely Bernoulli Property 133
fact that the largest number of k2 towers matched is < vnks , we see that the number
2
of indices within m(k2 ) + 2n1 of the boundary of Gc is at most
32 jGc j:
(18)
4
Using (2 ), (14), (16), (17), and (18) we have that for both ! 2 and !
jf~v 2 Gc : ~v is the base point of an n1 -name from U gj > 1 ; 2 :
(6 )
d
0
0
jGc j
0
2
Further, by (3), (15), (16), and (18),
jf~v 2 Gc : ~v is in a k1 tower gj > 1 ; 2 :
(7 )
1
jGc j
0
We also claim that a proportion
(9 )
> 1 ; 31
of ! 2 is covered by complete towers who are further than n1 to the boundary
of ! 2 . To see this note that the suspect indices are exactly those who are in a
2n1 collar of Bn . By (18) this is a set of indices of proportion less than 342 .
Finally, we claim that by (4 ), (6 ), and (18)
jf~v 2 Gc : ~v lies in a k11 towergj > 1 ; :
(8 )
0
0
0
jGc j
0
r
1
Matching up conditions (3 ),(4 ), (5 ),(6 ), (7 ),(8 ), and (9 ) with their analogs
in the proof of the GML, we see that the conditions for that argument are satised
and we proceed as in the proof of that lemma to obtain a permutation 1 which is
the identity on G, satises (a) and (b) of the GML on Gc with m(k1 ) = n1 , and
matches a% of Gc with the corresponding indices in ! .
Set = 1 2 : Bn ! Bn . It follows from our construction that d(! ! ) <
(1 ; a)2 < 2 . To nish the proof, it remains to show that m() < . We need, then,
to show that there is a set C Bn such that jC j > (1 ; )nd, and for all ~u~v 2 C ,
k~u ; ~v ; (~u ; ~v)k < k~u ; ~vk.
Recall that 2 matches the -interiors of some k2 -towers, and
p 1 matches the
-interiors of some k1 towers. Let C2 be the indices in the -interiors of the
boxes matched by 2 , and dene C1 similarly. As in the rank one argument we set
C = C2 C1 . The rest of the argument is identical to the rank one case we include
it below for completeness' sake.
To compute BCn we count the indices which were matched by , but are not in
p
C . The -collar we removed from the -interiors is a proportion
0
0
0
0
0
0
0
0
0
j
j
j
j
2
X
p
p
2d (the proportion of the n-name matched by i ) < 2d :
i=1
p
So by our choice of , BCn > 1 ; 2 ; 2d > 1 ; .
Now pick ~u~v 2 C . Suppose rst that ~u and ~v were both matched by 2 . Then,
if they are in the same k2 tower we have
k~u ; ~v ; (~u ; ~v )k = 0:
j
j
j
j
Aimee S. A. Johnson and Ayse A. Sahin
p
Otherwise, they lie in dierent k2 towers, so k~u ; ~vk > 2 sk2 . On the other hand,
k~u ; ~v ; (~u ; ~v)k < 2bk2
= k~u2;bk~v2 k k~u ; ~vk < 2pbk2 k~u ; ~vk
2 sk2
< k~u ; ~v k:
The argument for the case where ~u and ~v are both matched by 1 is similar.
p
Now suppose ~u is matched by 2 and ~v is matched by 1 . Then k~u ; ~vk > sk2 ,
134
and
k~u ; ~v ; (~u ; ~v)k < bk2 + m(k1 ) + bk1
bk2 + n1 + bk1 :
But bk1 < 21d n1 which by (3 ) is
0
So we have that
2
1 22 b :
2
< 21d 16d
sk2 <
2d 16d k2
k~u ; ~v ; (~u ; ~v)k < 2bk2
2bk2 k~u ; ~vk
= k~u2;bk~v2 k k~u ; ~vk < p
< k~u ; ~vk:
sk2
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Department of Mathematics and Statistics, Swarthmore College, Swarthmore,
PA 19081
aimee@cc.swarthmore.edu
Department of Mathematics, North Dakota State University, Fargo, ND 58105
sahin@plains.nodak.edu http://www.math.ndsu.nodak.edu/faculty/sahin
This paper is available via http://nyjm.albany.edu:8000/j/1998/3A-10.html.