A class of nonstationary adic transformations

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www.elsevier.com/locate/anihpb

A class of nonstationary adic transformations

Xavier Méla

IML, 163, avenue de Luminy, case 907, 13288 Marseille cedex 09, France

Received 29 January 2004; received in revised form 7 January 2005; accepted 18 February 2005 Available online 23 September 2005

Abstract

We introduce a class of nonstationary adic transformations generalizing the Pascal adic transformation. We identify the invariant ergodic measures, show that there are no rational eigenvalues, and construct a topological model. We give upper and lower bounds for the complexity function and show that the systems are topologically weakly mixing. We also show that these adics are loosely Bernoulli.

Résumé

On introduit une classe de transformations adiques non-stationnaires généralisant la Pascal-adique. On détermine l’ensemble des mesures invariantes ergodiques et on montre qu’il n’existe pas de valeur propre rationnelle. Ces systèmes admettent des modèles en dynamique topologique pour lesquels nous donnons des bornes inférieures et supérieures pour la fonction de com- plexité et montrons qu’ils sont topologiquement faiblement mélangeants. On démontre aussi que ces transformations adiques sont lâchement Bernoulli.

MSC: primary 37A05, 37A35, 37A50; secondary 37B10, 37B40

Keywords: Adic transformations; Ergodicity; Weak mixing; Loosely Bernoulli; Complexity; Binomial coefficients

1. Introduction

In this paper we introduce a class of nonstationary adic transformationsT_ddefined on a Bratteli diagram identified to{0,1, . . . , d−1}^N, with the particularity that the number of finite paths from the root(0,0)to a vertex (n, k)is given in terms of certain generalized binomial coefficients. When the number of such paths is given by the ordinary binomial coefficients (the cased=2), the system is just the Pascal adic [20,15,16].

E-mail address: [email protected] (X. Méla).

doi:10.1016/j.anihpb.2005.02.002

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The ergodicity of the Bernoulli or Markov measures for these adics is related to the questions of exchangeability and triviality of tail sigma-algebras in probability theory. It is a well-known result (de Finetti, Hajian–Ito–Kakutani, Vershik, [7,21]) that the nonatomic invariant ergodic measures for the Pascal adic are the Bernoulli measures µ_α,1₋_α. Here we extend this result to the adicsT_d, establishing that the set of nonatomic invariant ergodic measures is formed by a one-parameter family of Bernoulli measures{µ_p(a)|a∈(0,1)}for a certain probability vector p(a)(Theorem 3.6). We show that these measures are totally ergodic, equivalently that there are no rational eigenvalues (Proposition 4.5). For both of the previous results we avoid direct calculations on the generalized binomial coefficients, using instead arguments of isotropy or measure-theoretic tricks. Alternate proofs could be obtained if we had a deeper understanding of the arithmetics of these coefficients.

In Section 3.2 we show that the adicT_d is isomorphic to the Pascal adic on a certain subshift, whose study was in fact the original problem. The ergodic Bernoulli measures forT_dare carried to a family of(d−1)-step Markov measures for the Pascal on the subshift (Theorem 3.11).

The adicsT_dare homeomorphisms but not on the whole space, and extensions lead to many discontinuities. Our topological model, which we present in Section 5, is “almost minimal” and metrically isomorphic to the original one. For this homeomorphism on a compact space, which can be viewed as a dynamical system defined by infinitely many substitutions (or infinitely many letters), we estimate the asymptotics of the complexity function and show that the system is topologically weakly mixing. The question of metric weak mixing (already difficult in the case d=2) seems however to require a more complete understanding of the divisibility properties of the generalized binomial coefficients.

Finally, in Section 6, using the result of de la Rue and Janvresse [3], we show that the adicsT_d are loosely Bernoulli.

2. Description of the systems 2.1. The graph construction

Letd∈Nbe fixed. Consider the infinite graph divided into levelsn=0,1, . . ., with a root vertex labelled(0,0) and with(d−1)n+1 vertices labelled(n, k)at each levelnfor k=0, . . . , (d−1)n. From each vertex leaved edges labelled 0 throughd−1 connecting to the next level in such a way that for all finite paths going from(0,0) to a given vertex(n, k)the sum of the labels of the edges is equal tok— see Fig. 1. Ford=2 this is the Pascal graph — see [15].

LetX_d be the space of infinite paths starting at the root and going down the graph, which we identify to the space{0, . . . , d−1}^Nvia the labelling of the edges. We denote by(n, k_n(x))the vertex crossed by the pathx∈X_d at leveln. We define a partial order onX_d, writingx < yforx, y∈X_dwheneverxandycoincide below a certain

Fig. 1. Bratteli diagram in the cased=4. The picture shows the pathx=102. . .and the label of its vertex at level 3.

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levelnandx_n< y_n. Equivalently,x < yif there is annsuch thatx_n< y_n,x_i =y_ifor alli > n, andk_n(x)=k_n(y).

LetX_d^maxandX^min_d be respectively the sets of maximal and minimal paths. The generalized Pascal adic is defined by

Td:Xd\X^max_d →Xd\X_d^min, T_d(x)=smallesty > x.

In order to determine the image ofx one has to look for the first appearance of a block of the type ij, where 1id−1 and 0j < d−1. Indeed, ifx=0^lij . . .andj =d−1, we can always find an edge on the right of j and leading to the same vertex asj (giving therefore room for a successor). If there is no such occurrence then we are in the case of a maximal path, and we see a string of 0’s followed byi(d−1)^∞, where 0id−1. The reverse situation happens for minimal paths; we therefore have:

X_d^max=

0^li(d−1)^∞|l∈N, i=0, . . . , d−1

∪

0^∞, (d−1)^∞ , X_d^min=

(d−1)^li0^∞|l∈N, i=0, . . . , d−1

∪

0^∞, (d−1)^∞ .

There is a way to extendT_das a bijection to the whole spaceX_dby sending maximal paths to minimal ones:

T_d0^li(d−1)^∞=(d−1)^li0^∞, T_d0^∞=0^∞,

T_d(d−1)^∞=(d−1)^∞.

The resulting extended transformation is not a homeomorphism onX_d, but the discontinuities only occur at the minimal and maximal paths. Furthermore, all but the two edge paths (0^∞and(d−1)^∞) have infinite orbits.

Remark 1. In [23,21] another generalized version of the Pascal adic is defined called then-dimensional Pascal adic. Its graph lies on the multidimensional latticeN^Nand is (a priori) not related to the one described here.

2.2. Generalized binomial coefficients

A simple combinatorial argument shows that the number of finite paths from the root(0,0)to a vertex(n, k)is equal to the coefficient ofx^k in the polynomial(1+x+x²+ · · · +x^d⁻¹)ⁿ:

(1+x+x²+ · · · +x^d⁻¹)ⁿ=

(d−1)n k=0

C_d(n, k) x^k. The coefficientsCd(n, k)can be obtained from the relations

C_d(n+1, k)=C_d(n, k)+C_d(n, k−1)+ · · · +C_d(n, k−d+1),

with the convention thatC_d(n, k)=0 if k > (d−1)nork <0. In the cased=2 we get the ordinary binomial coefficients satisfying the well-known relationC₂(n+1, k)=C₂(n, k)+C₂(n, k−1). The name “generalized binomial coefficients” has various different definitions in the literature, but here it will refer only to theCd(n, k).

The Pascal-type triangle formed by theC_d(n, k)is sometimes called a generalized Pascal triangle of orderd [2,1];

we will refer to it as the∆_d-Pascal triangle.

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Fig. 2. The picture showsT₃(01210∗)=20020∗.

Example 2.1. The∆₃-Pascal triangle starts as follows:

1

1 1 1

1 2 +3 +

= 2 1

1 3 6 7 6 3 1

1 4 10 16 19 16 10 4 1

. . . .

One can write C_d(n, k) in terms of regular binomial coefficients or as a sum of multinomial coefficients (see [1,2]):

Cd(n, k)=

[k/d] i=0

(−1)ⁱ n

i

n−1+k−di n−1

= n!

(n−k₁)!(k₁−k₂)!. . . (k_d₋₂−k_d₋₁)!k_d₋₁!, where the summation is over all(k1, . . . , kd−1)such thatk1+ · · · +kd−1=kandkiki−1.

3. A one-parameter family of ergodic measures

As one would suspect, the invariant ergodic measures for Td are Bernoulli measures. However, not every Bernoulli measure is invariant for the adic Td; for each d the set of invariant Bernoulli measures forms a one- parameter family. The fact that each invariant Bernoulli measure is ergodic is a consequence of the Hewitt–Savage 0,1 law. To show the converse, which is that every invariant ergodic measure is in fact a Bernoulli measure, we use an argument of “isotropy”, giving at the same time a new proof for the cased=2 of the Pascal adic.

Proposition 3.1. The set of Bernoulli measures invariant for the adicTdon{0, . . . , d−1}^Nforms a one-parameter family

µ_p(a)|a∈ [0,1] ,

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where the vectorp(a)∈ [0,1]^d⁻¹is eitherp(0)=(0, . . . ,0,1), or ifa=0 p(a)=

a, t_a,t_a²

a, t_a³

a², . . . ,t_a^d⁻¹ a^d⁻²

,

wheret_ais the unique solution in[0,1]to the equation

a^d⁻¹−a^d⁻²+a^d⁻³t+a^d⁻⁴t²+ · · · +at^d⁻²+t^d⁻¹=0.

Proof. Let’s start by examining the cased=3. Denote by[a₁a₂. . . a_r]the cylinder of all pathsx whose firstn edges are labelleda1a2. . . ar. Letp=(p0, p1, p2)withp0+p1+p2=1, and letµpbe a Bernoulli measure on {0,1,2}^N such thatµ_p[i] =p_i>0 fori=0,1,2. First note thatµ_p[20] =µ_p[11]is a necessary and sufficient condition forµto be invariant. Indeed, assume thatµ_p[20] =µ_p[11]holds and letC be a cylinder set. Since the set of maximal paths has measure zero (it is countable), it follows thatCis a disjoint union up to measure zero of (non-maximal) cylinders of the form

C₁= [0ⁱ11], C₂= [0ⁱ10], C₃= [0ⁱ2^j21], C₄= [0ⁱ2^j20], C₅= [0ⁱ12^j1], C₆= [0ⁱ12^j0], wherei0 andj >0. Then respectively

T C₁= [0ⁱ02], T C₂= [0ⁱ01], T C₃= [2^j10ⁱ2], T C₄= [2^j10ⁱ1], T C5= [2^j0ⁱ02], T C6= [2^j0ⁱ01].

Note thatµ_p(T C_l)=µ_p(C_l)is automatically satisfied forl=2,3,6, and that forl=1,4,5 the equality follows from our hypothesis thatµp[20] =µp[11]. Hence ifµp[20] =µp[11]thenµp(T C)=µp(C)for any cylinder setC. Thusµ_pis invariant under the action of the adicT₃if and only if the following relation holds:

p₀p₂=p₁².

Since in additionp₀+p₁+p₂=1, we can solve forp₁andp₂in terms ofp₀; lettingp₀=a, we get p1=

−a+

a(4−3a) /2, p₂=1−a+

a−

a(4−3a) /2.

For the general case, letp=(p₀, p₁, . . . , p_d₋₁)withp₀+p₁+ · · · +p_d₋₁=1, and let µ_p be a Bernoulli measure on{0,1, . . . , d−1}^N such thatµp[i] =pi >0 for i=0,1, . . . , d−1. Arguing as previously we see that µp is invariant if and only if pipj =µp[ij] =µp[kl] =pkpl for alli, j, k, l∈ {0,1, . . . , d−1}such that i+j =k+l. The previous condition impliesd(d−1)/2 nontrivial relations onp₀, p₁, . . . , p_d₋₁. Out of those d(d−1)/2 nontrivial relations onlyd−2 of them are independent; these arep₀p_j=p₁p_j₋₁forj =2, . . . , d−1 (the other ones follow from them). For simplicity we letp₀=aandp₁=t. For allj =2, . . . , d−1 we have

p_j=p₁ p0

p_j₋₁= t ap_j₋₁.

We can therefore express eachpj as a function ofaandt:

p₂= t

ap₁=t² a, p3= t

ap2= t³ a², ...

pd−2= t

apd−3=t^d⁻² a^d⁻³, p_d₋₁= t

ap_d₋₂=t^d⁻¹ a^d⁻².

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Now, using thatp₀+p₁+ · · · +p_d₋₁−1=0 we get a+t+t²

a + t³

a²+ · · · +t^d⁻¹

a^d⁻²−1=0.

Multiply bya^d⁻²on both sides to obtain

a^d⁻¹−a^d⁻²+a^d⁻²t+a^d⁻³t²+ · · · +at^d⁻²+t^d⁻¹=0.

To conclude thatp₁, p₂, . . . , p_d₋₁can all be expressed as a function ofp₀=ait remains to show that the previous polynomial has a unique root in[0,1]for eacha∈(0,1]. This can be seen by using the intermediate value theorem:

let f (t )=a^d⁻¹−a^d⁻²+a^d⁻²t+a^d⁻³t²+ · · · +at^d⁻²+t^d⁻¹. Then f (0)=a^d⁻¹−a^d⁻²0 and f (1)= a^d⁻¹+a^d⁻³+a^d⁻⁴+ · · · +a+1>0, showing thatf has a root in[0,1]. This root is unique sincef is strictly increasing on[0,1]:f(t )=(d−1)t^d⁻²+a(d−2)t^d⁻³+ · · · +2a^d⁻³t+a^d⁻²>0 for allt∈ [0,1].

Finally observe that ifp₀=a=0, the only “trivial” Bernoulli measure which is invariant isB(0, . . . ,0,1). 2 Proposition 3.2. Let µ_p be a Bernoulli measure on {0, . . . , d−1}^N, where p=(p₀, . . . , p_d₋₁). Ifµ_p is T_d- invariant, then it is ergodic.

Proof. Simply note that theσ-algebra of invariant sets for the adicTdis trivial by the Hewitt–Savage 0,1 Law. 2 Definition 3.3. Let(n, k)be a vertex in the graph. We defineΛ(n, k)to be the number of finite paths from(0,0)to (n, k). For every pathx∈XletΛ(xn)=Λ(n, kn(x)). For every cylinder setC⊂X and every pathx∈Xdefine Λ(C, xn)to be the number of paths inCwhich coincide withxafter thenth level.

Lemma 3.4. Letµbe an invariant nonatomic ergodic Borel probability measure for the adicTd. Then for every cylinder setC andµ-almost everyx∈X

µ(C)= lim

n→∞

Λ(C, xn)

Λ(xn) . (3.1)

Proof. Note that sinceµ is nonatomic, minimal and maximal paths form a set of measure zero. For any path x∈X_d\X_d^min∪X_d^max, leti_n(x)andj_n(x)be the integers such thatT⁻ⁱⁿ^(x)x is the smallest path agreeing withx after thenth level, andT^jⁿ^(x)xis the greatest path which agrees withxafter thenth level — see Fig. 3. Note that we have the relations

Λ(x_n)=i_n(x)+j_n(x) and

Λ(C, x_n)=

jn(x) i=−in(x)

1C

T_dⁱx .

The result follows from the Ergodic Theorem. 2

Remark 2. Ifµis invariant but not necessarily ergodic, then the limit in (3.1) is just the conditional expectation E(1C|I)(x), whereIis theσ-algebra ofT_d-invariant sets.

The following lemma shows in particular that the invariant ergodic measures for the adicT_dare invariant for the (one-sided) shiftσ on{0, . . . , d−1}^N.

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Fig. 3.

Fig. 4.

Lemma 3.5. Letσjx=j x₀x₁. . ., forj=0, . . . , d−1. If µis invariant and ergodic for the adic, then for any Borel setB

µ(B)=µ(σ₀B)+ · · · +µ(σ_d₋₁B)=µ(σ⁻¹B).

Proof. LetCbe a cylinder set. By Lemma 3.4, forµ-almost everyx we simultaneously have µ(C)= lim

n→∞

Λ(C, x_n) Λ(xn) , µ(σ_jC)= lim

n→∞

Λ(σ_jC, x_n)

Λ(xn) for allj=0, . . . , d−1.

For everyj=0, . . . , d−1 letx^j be the path whose vertexv^j_nat leveln−1 connects to(n, k_n(x))along the edge labelledj, i.e.v_n^j=(n−1, k_n(x)−j )— see Fig. 4. Every path inCwhich coincides withx after thenth level must go through one of the verticesv^j_n; therefore

Λ(C, x_n)=Λ C, x_n⁰

+ · · · +Λ

C, x_n^d⁻¹

. (3.2)

The following relation can be seen from the “isotropic” structure of the graph and is illustrated in Fig. 5:

Λ(σ_jC, x_n)=Λ C, x_n^j

, for allj=0, . . . , d−1. (3.3)

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Fig. 5. The number of paths inCwhich coincide withxafter thenth level is exactly the number of finite paths inside the dashed rectangle. The number of paths inσjCwhich coincide withx^jafter the(n+1)’th level is exactly the number of finite paths inside the dot-dashed rectangle.

Since these two rectangles have the same dimensions these numbers are equal.

Dividing (3.2) byΛ(x_n)and then using (3.3) we get Λ(C, x_n)

Λ(x_n) =Λ(C, x_n⁰)

Λ(x_n) + · · · +Λ(C, x_n^d⁻¹)

Λ(x_n) =Λ(σ₀C, x_n)

Λ(x_n) + · · · +Λ(σ_d₋₁C, x_n) Λ(x_n) .

The result follows for cylinder sets after taking the limit whenn→ ∞, and since cylinders generate the σ- algebraBdthe lemma is proved. 2

Remark 3. The previous result remains true if µ is not ergodic since the argument in the proof would yield E(χ_C|I)=E(χ_σ₀_C|I)+ · · · +E(χ_σ_d₋₁_C|I) µ-a.e. (see Remark 2), and integrating we would get the same result.

Theorem 3.6. The invariant ergodic measures for the adicT_dare the nonatomic Bernoulli measures described in Proposition 3.1, andB(1,0, . . . ,0)andB(0, . . . ,0,1), which are atomic.

Proof. Letµ be an invariant nonatomic ergodic probability measure. Ifµ[j] =0 for somej, thenµ[i] =0 for alli. Indeed, suppose that for somei=j µ[i]>0. Then, unlessµonly gives masses to maximal and minimal paths in[i], there has to exist ak∈Zsuch thatµ(T^k[i] ∩ [j]) >0, which contradictsµ[j] =0. In the following we assume thatµ[j]>0 for allj∈ {0, . . . , d−1}.

We first introduce some notation. For any cylinder setCandj, k∈ {0, . . . , d−1}, letC^kdenote the cylinderC extended by the edgekand letC^{j k}be the cylinderCextended by the edgesj k; more precisely, ifC= [a₁a₂. . . arr] thenC^k= [a1a2. . . ark]andC^{j k}= [a1a2. . . arj k]. To prove thatµis a Bernoulli measure it is enough to show that there is a numberp_k independent ofCsuch thatµ(C^k)/µ(C)=p_kfor allk∈ {0, . . . , d−1}. To show this we establish the equality

µ(C^k)

µ(C) =µ(C^{j k})

µ(C^j), j, k∈ {0, . . . , d−1}. (3.4)

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Then for any cylinder setC= [a₁a₂. . . a_r]we have µ(C^k)

µ(C) =µ[a₁a₂. . . a_r₋₁]^a^r^k

µ[a₁a₂. . . a_r₋₁]^a^r =µ[a₁a₂. . . a_r₋₁]^k

µ[a₁a₂. . . a_r₋₁] = · · · =µ[a₁]^k

µ[a₁]. (3.5)

On the other hand, the invariance ofµunderT_dimplies that for allj, k, l∈ {0, . . . , d−1} µ[j l] =µ[lj] and µ[j lk] =µ[lj k],

and again using (3.4) we get µ[j]^k

µ[j] =µ[j]^lk

µ[j]^l =µ[j lk]

µ[j l] =µ[lj k]

µ[lj] =µ[l]^{j k}

µ[l]^j =µ[l]^k

µ[l]. (3.6)

Together (3.5) and (3.6) show thatµ(C^k)/µ(C)is independent of the cylinderC.

We now give a proof of the identity (3.4). By Lemma 3.4 we can find a set of full measureEsuch that for all x∈Eand allj, k∈ {0, . . . , d−1},

µ(C^{j k}) µ(C^k) = lim

n→∞

Λ(C^{j k}, x_n)

Λ(C^k, x_n) and µ(C^k) µ(C) = lim

n→∞

Λ(C^k, x_n) Λ(C, x_n). Again using the “isotropic” aspect of the graph (see Fig. 6) we have that

Λ(C^k, x_n)=Λ

C^{j k}, (σ_jx)_n₊₁

and (3.7)

Λ(C, x_n)=Λ

C^j, (σ_jx)_n₊₁

. (3.8)

If we divide (3.7) by (3.8), we get Λ(C^k, x_n)

Λ(C, x_n) =Λ(C^{j k}, (σ_jx)_n₊₁) Λ(C^j, (σ_jx)_n₊₁).

Now choosingx ∈E such that σjx∈E and taking the limit asn→ ∞ yields (3.4). (SinceE is a set of full measure, ifµ(σ_jE)=0 for somej∈ {0, . . . , d−1}, then by Lemma 3.5

1=µ(E)=

k=j

µ(σ_kE)

k=j

µ[k]1.

This forcesµ[j] =0, which contradicts our assumption made in the beginning. Therefore µ(E∩σjE) >0 for allj. It follows thatE∩σ_jE= ∅and hence there is ay∈Esuch thaty=σ_jxfor somex∈E, proving that some x∈Eis such thatσ_jx∈E.) 2

Remark 4. In the previous proof we showed thatµ(C^k)/µ(C)was independent ofCavoiding calculations on the generalized binomial coefficients. On the other hand, letting(r, s)be the vertex at which the cylinderCends (where s=a₁+ · · · +a_r), we could have tried to show instead that lim_n_→∞C_d(n−r−1, k_n(x)−s−k)/C_d(n, k_n(x)) (=lim_n_→∞Λ(C^k, x_n)/Λ(C, x_n)=µ(C^k)/µ(C)) is independent ofC. But if such calculations are easy in the case of ordinary binomial coefficients they seem very messy in the general case. Furthermore our geometric argument implies that these calculations are true; in particular, forC= [0^r] we have thatC_d(n−r, k_n(x))/C_d(n, k_n(x)) converges top^r₀for almost everyx, which will be useful in Section 6.

3.1. The cutting and stacking equivalent

Start with the unit interval and cut it into d columns (subintervals in this case) I₁¹, I₂¹, . . . , I_d¹ of equal width (step 1). At each step n we cut every column I_iⁿ for i=1, . . . , (d−1)n+1 intod equal subcolumns

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Fig. 6. The number of paths inC^kwhich coincide withxafter thenth level is exactly the number of finite paths inside the dashed rectangle.

The number of paths inC^{j k}which coincide withσjxafter the(n+1)’th leveln+1 is exactly the number of finite paths inside the dot-dashed rectangle. Since these two rectangles have the same dimensions these numbers are equal.

Fig. 7. Cutting and stacking in the cased=3.

I_i,1ⁿ , I_i,2ⁿ , . . . , I_i,dⁿ of equal width, then stackI_i,jⁿ underI_iⁿ₊_1,j₋₁fori=1, . . . , dnandj=2, . . . , d. This produces (d−1)(n+1)+1 new columns I₁ⁿ⁺¹, . . . , I_(dⁿ⁺₋¹_1)(n₊₁₎₊₁. If we repeat indefinitely, the resulting mapT (which maps every open interval of each stack to the one above it) is defined everywhere except at thed-adic rationals (which correspond to the paths which are eventually diagonal in the graph construction). Denote bymLebesgue measure, and letB([0,1])be theσ-algebra of Borel sets in[0,1].([0,1],B([0,1]),T, m)is a measure-preserving system which we will refer to as the generalized binomial transformation. See Fig. 7 for an illustration of the cutting and stacking in the cased=3.

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We can show that the generalized Pascal adic transformation equipped with theσ-algebraBdof Borel sets and the Bernoulli measure with equal weights is isomorphic to the generalized binomial transformation:

Proposition 3.7. The systems([0,1],B([0,1]),T, m)and(Xd,Bd, Td, µp)wherep=(1/d, . . . ,1/d)are isomor- phic.

Proof. The isomorphism is given by ψ:[0,1] \ {d-adic rationals} →X\ {y: x∈X^min_d ∪X^max_d andy ∈O(x)}, which at everyxwithd-adic expansion_∞

i=1x_id⁻ⁱ associatesψ (x)=x₁x₂. . .. 2

Remark 5. If we cut and stack with proportionsp₀, . . . , p_d₋₁ where(p₀, . . . , p_d₋₁)is a probability vector as described in Proposition 3.1, which guaranties that the intervals that are being stacked one on top of the other have the same length, then Lebesgue measure carries to the Bernoulli measureB(p₀, . . . , p_d₋₁).

We now give a direct proof that the generalized binomial transformation is ergodic.

Proposition 3.8. The system([0,1],B([0,1]),T, m)is ergodic.

First we need a standard result from probability, which can be found in [19, p. 85]

Lemma 3.9. LetX_nandY_nbe two independent stationary processes on([0,1],B([0,1]), m)with values in{0,1} such that

X_ndm=

Y_ndm. Then for almost every (x, y) in X×X there are infinitely many n for which _n

i=0X_i(x)=_n

i=0Y_i(y).

LetI = [0,1] \ {d-adic rationals}. For anyx ∈I, letIn(x)be thed-adic interval of length 1/dⁿ to whichx belongs. Let(x, y)∈I×I haved-adic expansions

x=^∞

i=1

x_i

dⁱ and y=^∞

i=1

y_i dⁱ.

Then it is not difficult to prove by induction thatx andybelong to the samekth stack at stepnin the cutting and stacking iffk=_n

i=1x_i=_n

i=1y_i. Lemma 3.9 implies that for a.e.(x, y)∈I×I there are infinitely manynfor whichI_n(x)andI_n(y)belong to the same stack at leveln.

We will also need the following lemma which is a consequence of the Lebesgue density theorem and can be proved using a martingale argument:

Lemma 3.10. LetA⊆ [0,1]be a set of positive measure. Then for almost everyx∈A

nlim→∞

m(A∩I_n(x)) m(I_n(x)) =1.

Proof. Let Fn denote the σ-algebra generated by the d-adic intervals [k/dⁿ, (k+1)/dⁿ), for 0kn−1.

ThenFnB([0,1]), andX_n=E(IA|Fn)is an Fn-martingale. By the martingale convergence theorem (see for example [18]), X_n converges almost everywhere to m(A). Since for almost everyx ∈A E(IA|Fn)(x)= m(A∩In(x)),the result follows. 2

Proof of the proposition. LetAbe aT-invariant measurable set such that 0< m(A) <1. Then, by Lemma 3.10, for almost everyx∈Aand almost everyy∈A^c=X\A

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nlim→∞

m(A∩In(x))

m(In(x)) =1, (3.9)

nlim→∞

m(A^c∩I_n(y))

m(I_n(y)) =1. (3.10)

On the other hand, for a.e.(x, y)∈I ×I there is a sequencen_i ∞such thatI_n_i(x)andI_n_i(y)belong to the same stack at leveln_i. Therefore, sinceAisT-invariant andI_n_i(x)maps ontoI_n_i(y)(under some iterate ofT), we have

m

A∩I_n_i(x)

=m

A∩I_n_i(y) .

This contradicts (3.9) and (3.10), som(A)=0 or 1. 2 3.2. A relationship betweenT_d and the Pascal adicT₂

LetX_{₁₁_}be the subset ofX₂of all paths in the Pascal graph which do not go along the edge labelled 1 twice in a row.X_{₁₁_}can identified with the subshift of finite type of all infinite sequences not containing the block 11. Denote byT_{11} the Pascal adic on X_{11}, that is, the map defined forx∈X_{11} byT_{11}x =smallesty∈X_{11} greater thanx. We haveT_{₁₁_}0^r(10)^s100. . .=(10)^s0^r010. . .. In [21, Theorem 2.11] it was shown that the Pascal adic T_{₁₁_}on the subshift of finite typeX_{₁₁_}is isomorphic to the Pascal adicT =T₂on the space{0,1}^N. Furthermore, the preimage of a Bernoulli measureµ_pis the Markov measure with transition matrix

p 1−p

1 0

and initial distributionP (0)=pandP (1)=1−p. The Markov shift has the following graph representation:

0

p

1−p

1.

1

Let X_{₁d} be the subset ofX₂ of all paths in the Pascal graph which do not go along the edge labelled 1d times in a row.X_{₁d}can be identified with the subshift of finite type consisting of all sequences of 0’s and 1’s not containingd 1’s in a row. Denote byT_{₁d}the Pascal adic on the subshiftX_{₁d}, i.e. the map defined forx∈X_{₁d}

byT_{₁d}x=smallesty∈X_{₁d}greater thanx. We generalize the previous statement by relatingT_{₁d} andTd as follows:

Theorem 3.11. There is a continuous bijective mapψ from the generalized Pascal adicT_d onX_d to the Pascal adicT_{₁d}on the subshift of finite typeX_{₁d}. Furthermore,ψcommutes with the actions of the two adics, and the images of the Bernoulli measures are(d−1)-step Markov measures.

Proof. Define a mapψfromX_dtoX_{1d}by the following block codes:

0→0 1→10 2→110

... d−1→1^d⁻¹0.

It is easy to check thatψis continuous and bijective. To show thatψcommutes with the actions of the two adics, i.e.T_d◦ψ=ψ◦T_{₁d}, it is enough to check that for allx, y∈X_d ify > x thenψy > ψ x. Settingx=x₁x₂. . ., y=y₁y₂. . .and assumingy > x, there exists ak1 such that

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• y_k> x_k,

• x_j=y_j for allj > k,

• _k

i=1xi=_k

i=1yi. Then

ψ x=1^x¹01^x²0. . .01^x^k0. . .=1^x¹0. . . .0

x_k

1. . .1 0. . . ψy=1^y¹01^y²0. . .01^y^k0. . . =l1^y¹0. . .1 . . .1. . .1

y_k

0. . .

Therefore, lettingl=(x₁+1)+(x₂+1)+ · · · +(x_k₋₁+1)be the index of the 0 preceding the string ofx_k 1’s in ψ x, we have

• (ψy)_l=1>0=(ψ x)_l,

• (ψ x)j=(ψy)j for allj > l,

• _l

i=1(ψ x)_i=_k

i=1x_i+k=_k

i=1y_i+k=_l

i=1(ψy)_i, showing thatψy > ψ x.

The fact that the image underψ of a Bernoulli measure is a(d−1)-step Markov measure is straightforward, since in order to determine the probability that thenth symbol is 0 or 1 depends only on knowing thed−1 previous ones. 2

Example 3.12. In the cased=4,ψis defined by X₄−→^ψ X_{₁₁₁₁_}







0 →0,

10 →1,

110 →2, 1110 →3

and the subshift of finite type X_{₁₁₁₁_}, with Markov transition probabilities corresponding to B(p₀, p₁, p₂, p₃) onX₄, has graph representation:

001

p2+p3 p1+p2+p3

p1 p1+p2+p3

011

p1 p2+p3

p2 p2+p3

000

p0

1−p₀

101

p2+p3 p1+p2+p3

p1 p1+p2+p3

010

p0

1−p₀

111

1

100

p₀

1−p₀

110

p₀

1−p₀

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4. Existence of eigenvalues

In this section we show that the generalized Pascal adics do not have any “rational” eigenvalue, i.e. no complex numberλ=e^2πia/b, whereaandbare nonzero integers, can be an eigenvalue forT_d. The existence of “irrational”

eigenvalues (λ^2πiα,α /∈Q) is still an open question.

First note that if a pathxis such thatx_n₊₁x_n₊₂=10, thenxcomes back close to itself afterC_d(n, k_n(x))steps, more precisely:

Lemma 4.1. Letx∈X_d be path such thatx_n₊₁x_n₊₂=10. ThenT_d^C^d^(n,kⁿ^(x))x andxcoincide on the firstnedges.

Proof. Let r (resp.s) be the number of iterations ofT_d (resp.T_d⁻¹) necessary to bring the pathx to coincide with thenfirst coordinates of the maximal (resp. minimal) path going through(n, k_n(x)). Note thatr+s+1= C_d(n, k_n(x)). SinceT_d(T_d^rx)coincides with thenfirst coordinates of the minimal path going through(n, k_n(x)), it follows thatT_d^r⁺^s⁺¹x andxcoincide on the firstnedges. 2

This previous observation enables us to formulate a simple eigenvalue condition — cf. [16,8,13,22].

Proposition 4.2. Ifλis an eigenvalue for the generalized Pascal adicTd, then for almost every pathxinXd(with respect to a Bernoulli measureB(p0, . . . , pd−1))λ^C^d^(n,kⁿ^(x))converges to 1.

Proof. Letλbe an eigenvalue forTd andf be a corresponding measurable eigenfunction (f ◦Td=λf). Forn large, by Lusin’s Theorem we may assume thatf is continuous and almost constant on the cylinder sets determined by finite paths from (0,0) to the vertices(n, k) for k=0, . . . , (d−1)n. For every pathx the set of paths x very close tox such thatx_n₊₁x_n₊₂=10 has positive measure, i.e.µp([x₁. . . xn10]) >0. By Lemma 4.1, for all x∈ [x₁. . . x_n10]T_d^C^d^(n,kⁿ^(x))xandxcoincide on the firstnedges. Therefore for almost everyx∈ [x₁. . . x_n10] f (x)f (T_d^C^d^(n,kⁿ^(x))x)=λ^C^d^(n,kⁿ^(x))f (x), so thatλ^C^d^(n,kⁿ^(x))1. 2

In the cased=2 it is well known that the Pascal triangle modulo a prime number has a self-similar structure [2].

In particular, a consequence of Lucas’s Theorem [14] is that for any prime numberqand any natural numbersthe binomial coefficientC(q^s−1, k)is not divisible byqfor anyk=0, . . . , q^s−1. Hence, in the Pascal triangle there are infinitely many levels where all binomial coefficients are not divisible byq. Ifλ=e^2πia/b (gcd(a, b)=1) is an eigenvalue, thenλ^C(n,kⁿ^(x)) has to converge to 1 a.e., by Proposition 4.2. But ifq is a prime divisor ofb, this is possible only ifC(n, k_n(x))is divisible by q for allnlarge enough. The existence of entire rows of binomial coefficients not divisible byq, forming “blocking lines”, prevent this from happening.

In the general case things are a little trickier. One has an analogue of Lucas’s Theorem for generalized binomial coefficients, unfortunately it does not imply the existence of “blocking lines”. For example, in the case of the

∆₃-Pascal triangle modulo 2 it can be shown [1] that there are no such “blocking lines” — see Fig. 8. However, according to Proposition 4.2, in order to prove that a givenλ=e^2πia/b is not an eigenvalue, it is enough to show that the set of paths which hit infinitely many times vertices corresponding to nonzero coefficients moduloq(where q is a prime factor ofb) has positive measure. This is what we will do.

First recall the Lucas type theorem for the generalized binomial coefficientsCd(n, k):

Theorem 4.3 [1,2]. Letq be a prime number. Letn, k∈Nbe such that 0k(d−1)n. Then Cd(n, k)=

(r₁,r₂,...,rs)

s i=0

Cd(ni, ri)modq,

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Fig. 8. The∆₃-Pascal triangle modulo 2.

wheren=n₀+n₁q+ · · · +n_sq^s, with 0n_i < q, is theq-ary representation ofn, and the sum is taken over all (r₁, r₂, . . . , rs)such that

• r₁+r₂q+ · · · +r_sq^s=k,

• 0r_i(d−1)n_i

if such a representation exists; otherwiseC_d(n, k)=0 modq.

Note that ford=2 this is just Lucas’ Theorem [14]. It has the following immediate consequence:

Corollary 4.4 [2]. Letq be a prime number. Lets∈N. Then in the rown=q^s of the generalized Pascal triangle of orderd, we have

C_d(n, k)=

1 modq ifk=0, q^s,2q^s, . . . , (d−1)q^s, 0 modq otherwise.

This corollary tells us that at each leveln=q^s we see, moduloq, a row composed ofd 1’s each separated by n−1 0’s. The 1’s propagate from one level to the next forming diagonal lines (as shown in bold in Fig. 9) and we seed−1 similar adjacent triangles of 0’s. For example, in the cased=3, we see two adjacent such triangles below each leveln=q^s — see Fig. 9. Recall that in this case the paths have the freedom to go straight, left, or right down the graph. The 1’s in bold do not form a “hermetic barrier” playing a similar role as the “blocking lines” in the case d=2. But we will show that with positive probability (independently ofn) paths are forced to hit one of those 1’s.

This argument will be sufficient to show that there are no rational eigenvalues.

Proposition 4.5. Noλ=e^2πia/b, whereaandbare nonzero integers, can be an eigenvalue forT_d.