The Pascal Adic Transformation is Loosely Bernoulli

pefav - Pascal Adic

Découvre YouScribe en t'inscrivant gratuitement

Je m'inscris

Obtenez un accès à la bibliothèque pour le consulter en ligne
En savoir plus

9 pages

English

Obtenez un accès à la bibliothèque pour le consulter en ligne
En savoir plus

A propos
Informations
Extrait

Description

lebesgue measure

called pascal adic

has zero

entropy measure preserving

pascal adic

transformation

Sujets

The Pascal Adic Transformation is Loosely Bernoulli
Elise Janvresse, Thierry de la Rue
CNRS - UMR 6085
Abstract
The Pascal adic transformation is one of the simplest examples of adic
transformations. We recall its construction by cutting and stacking and
prove that it is loosely Bernoulli.
Key-words : adic transformation, loosely Bernoulli systems.
Classi cation AMS 2000: 28D05
La transformation Pascal adique est lac^ hement Bernoulli.
Resume
La transformation Pascal adique est un des exemples les plus simples de
transformations adiques. Nous rappelons sa construction par decoupage
et empilement et montrons qu’elle est l^achement Bernoulli.
Mots-clefs : transformation adique, systemes l^achement Bernoulli.
1 Introduction
The notion of adic transformation has been introduced by Vershik (see e.g.
[5], [4]), as a model in which the transformation acts on in nite paths in some
graphs, called Bratteli diagrams. As shown by Vershik, every ergodic automor-
phism of the Lebesgue space is isomorphic to some adic transformation, with
a Bratteli diagram which may be quite complicated. Vershik also proposed
to study the ergodic properties of an adic transformation in a given simple
graph, such as the Pascal graph which gives rise to the so-called Pascal adic
transformation.
1.1 The Pascal adic transformation
Here we recall the construction and some basic properties of the Pascal adic
transformation with parameter p, following the cutting and stacking model
exposed in [2]. Our space X is the interval [0;1[, equipped with its Borel -
algebra A and the Lebesgue measure .
Let 0 < p < 1 be a xed parameter. We start by dividing X into two
def def def
subintervals P = [0;p[ and P = [p;1[. Let P = fP ;P g be the partition0 1 0 1
1obtained at this rst step. We also consider P and P as \degenerate" Rokhlin0 1
1 1towers of height 1, respectively denoted by and .0 1
On second step, P and P are divided in proportions (p;1 p). The trans-0 1
formation T is de ned on the right piece of P by sending it linearly onto the left0
piece of P ; note that both intervals have the same length p(1 p). This gives1
2 2 2a collection of 3 disjoint Rokhlin towers denoted by ; , , with respective0 1 2
heights 1, 2, 1 (see gure 1.1).
n nAfter step n, we get (n + 1) towers ;::: ; , with respective heights0 n
n n n n k k n;::: ; , the width of being p (1 p) . Denote bu F the basek k0 n
nof . At this step, the transformation T is de ned on the whole space except
k
the top of each stack. We then divide each stack in proportions (p;1 p), and
nde ne T on the right piece of the top of by sending it linearly onto the leftk
n n n k k+1piece of the base F of (both have the same length p (1 p) ).k+1 k+1
Repeting recursively this construction, T is nally de ned almost every-
where, and clearly preserves the measure .
P P
0 p 1 10
11
10
2
1
2 2
0 2
3 3
1 2
3 3
30
Figure 1: Cutting and stacking construction of the Pascal adic transformation
It is well-known (see e.g. the proofs given in [2]) that T is ergodic and has
zero entropy.
1.2 Loose Bernoullicity
In this section and in 2.1, we consider a general dynamical system (X;A ; ; T),
where T is an invertible measure-preserving transformation of the Lebesgue
probability space (X;A ;). The notion of loose Bernoullicity has been intro-
duced by Feldman in 1976 ([1]), then used by Ornstein, Rudolph and Weiss ([3])
to develop the study of Kakutani equivalence for measure preserving transfor-
mations. In the zero-entropy case, saying that a transformation T is loosely
Bernoulli is equivalent to say that T is isomorphic to a induced
2
tttttttttby an irrational rotation. The characterization of loose Bernoullicity given by
Feldman makes use of the so-called \P-name" of a point x.
Let P = fP ;::: ;P g be a nite measurable partition of (X;A ;). For0 k
def
x2 X, we set P(x) = j2f0;::: ;kg if x2 P . For m < n in , we de ne thej
P-name of x (from m to n) by
defnPj (x) = j j j ;m m+1 nm
def iwhere, for each m i n, j = P(T x). The entire P-name of x is thei
+1doubly-in nite sequence Pj (x). 1
To de ne the property of being loosely Bernoulli, Feldman introduced the
f distance between nite words. Let V = v v and w = w w be two1 l 1 l
words of length l on the same alphabet. The f distance between v and w is
de ned by
l sdef
f(v;w) = ;
l
where s is the greatest integer inf0;::: ;lg such that we can nd 1 i < i <1 2
< i l and 1 j < j < < j l with v = w (r = 1;::: ;s).s 1 2 s i jr r
De nition 1.1 Let T be a zero-entropy measure preserving transformation on
the probability space (X;A ;), and let P be a nite measurable partition of
X. The process (P;T) is said to be loosely Bernoulli (LB) if for all " > 0 and
for all su cien tly large l, we can nd A X with (A) > 1 " such that

l l8x;y2 A; f Pj (x);Pj (y) < ":0 0
The transformation T is said to be LB if for each nite partition P the
process (P;T) is LB.
Remark { In order to prove that a transformation T is LB, it is enough to
verify that (P;T) is LB for some generating partition P.
1.3 Main result
Theorem 1.2 The Pascal-adic transformation is loosely Bernoulli.
2 Proof of the loose-Bernoullicity
2.1 Equivalence of loy with seemingly weaker
properties
Lemma 2.1 Suppose that for all " > 0 and for all su cien tly large l, we can
nd B X X with
(B) > 1 " such that

l l8(x;y)2 B; f Pj (x);Pj (y) < ":0 0
then the process (P;T) is LB.
3Proof | Given " > 0, let B X X with
(B) > 1 " be such that

l l8(x;y)2 B; f Pj (x);Pj (y) < "=2:0 0
We can nd x2 X such that (B ) > 1 ", wherex
def
B = fy2 X j (x;y)2 Bg:x
0But, because of the triangular inequality for f, for all y and y in B we havex

l l 0f Pj (y);Pj (y ) < ":0 0
def
Thus, the de nition of LB is satis ed, with A = B . ox
Lemma 2.2 Suppose that for all " > 0 and for
almost every (x;y) 2
X X, we can nd an integer l(x;y) 1 such that

l(x;y) l(x;y)
f Pj (x);Pj (y) < ":0 0
then the process (P;T) is LB.
Proof | Let us x " > 0. For
-almost every (x;y)2 X X, we de ne
k kl(x;y) as the smallest integer k 1 such that f Pj (x);Pj (y) < "=3. Since0 0

(l(x;y) <1) = 1, there exists n2 such that
2
(l(x;y) n) < " =3:
For any l > 3n=", we consider
l 1X1def
M = :k kl fl(T x;T y)ngl
k=0
Using Markov’s inequality and the fact that T preserves the measure , one
can easily check that
2E(M ) " =3l
(M "=3) < = ":l
"=3 "=3
def
Therefore, the set B = fM < "=3g X X is such that
(B) > 1 ".l
l lLet us x (x;y)2 B. We want to show that f Pj (x);Pj (y) < ".0 0
k kWe say that k2f0; ;l 1g is bad if l(T x;T y) > n. Since (x;y)2 B, there
are less than l"=3 such k.
def
We de ne (j ) and (r ) recursively by j = r = inffr 0j r is not badg,i i0 i i0 0 0
and for i 1 such that j l n,i 1

j ji 1 i 1r = inf r 0j j + l(T x;T y) + r is not badi i 1
j ji 1 i 1j = j + l(T x;T y) + ri i 1 i
4
j j j j0 0 1 1
l (T x , T y) l (T x , T y)
j j j j l0 f1 2
r r r0 1 2 n
Figure 2: Covering off0; ;lg with good intervals and bad points.
We denote by f the greatest index i such that j is de ned: l j < n.i f
Recall the de nition of f.

l l(l + 1) f Pj (x);Pj (y)0 0
f 1 X
j ji+1 i+1 (j j ) f Pj (x);Pj (y) + (l j )i+1 i fj ji i
i=0
f 1 X j j j ji i i ij +l(T x;T y) j +l(T x;T y)j j i ii i l(T x;T y) f Pj (x);Pj (y)j ji i
i=0
f 1X
+ r + (l j )i f
i=0
f 1 X j j j ji i i il(T x;T y) l(T x;T y)j j j ji i i i= l(T x;T y) f Pj (T x);Pj (T y)0 0
i=0
f 1X
+ r + (l j )i f
i=0
f 1X" l"j ji i l(T x;T y) + + n < (l + 1)":
3 3
i=0
Therefore, we proved that for all su cien tly large l, we can nd B X X
l lwith
(B) > 1 " such that 8(x;y) 2 B, f Pj (x);Pj (y) < ". We0 0
conclude with Lemma 2.1. o
2.2 Some lemmas on the Pascal adic transformation
From now on, T is the Pascal adic transformation described in section 1.1, and
P is the partition fP ;P g given by the rst step of the cutting-and-stacking0 1
construction. For x2 X and n 1, we de ne k (x) as the element off0;::: ;ngn
ntelling in which tower of the level n x lies: for each n 1, x2 .
k (x)n
Lemma 2.3 P is a generating partition for the system (X;A ; ; T), i.e.
+1_
kT P = A :
k= 1
5Proof | As in [2], for each n 1, we de ne the basic blocks of level n Bn;k
(0 k n), which are words on the alphabetf0;1g, by the following induction
def def
: B = 0, B = 1, and for 1 k n 1,n;0 n;n
def
B = B B :n;k n 1;k 1 n 1;k

nIt is straightforward to verify that B is the P-name of length of anyn;k k
n npoint x lying in the base F of . We are now going to prove by induction
k k
non n that B characterizes the base of . More precisely, for any n 2 andn;k k
1 k n 1,
n 1( )k nif Pj (x) = B ; then x2 F : (1)n;k0 k
Indeed, (1) is clearly satis ed for n = 2. Next, suppose that (1) is satis ed for
n( ) 1
kn 1,