ERGODIC AND SPECTRAL PROPERTIES
S ´EBASTIEN FERENCZI, CHARLES HOLTON, AND LUCA Q. ZAMBONI
ABSTRACT. In this paper we present a detailed study of the spectral/ergodic properties of three- interval exchange transformations. Our approach is mostly combinatorial, and relies on the dio- phantine results in Part I and the combinatorial description in Part II. We define a recursive method of generating three sequences of nested Rokhlin stacks which describe the system from a measure- theoretic point of view and which in turn gives an explicit characterization of the eigenvalues. We obtain necessary and sufficient conditions for weak mixing which, in addition to unifying all previ- ously known examples, allow us to exhibit new examples of weakly mixing three-interval exchanges.
Finally we give affirmative answers to two questions posed by W.A. Veech on the existence of three- interval exchanges having irrational eigenvalues and discrete spectrum.
1. INTRODUCTION
An interval exchange transformation is given by a probability vector(α1, α2, . . . , αk)together with a permutationπof{1,2, . . . , k}.The unit interval[0,1)is partitioned intoksub-intervals of lengthsα1, α2, . . . , αkwhich are then rearranged according to the permutationπ.Katok and Stepin [20] used interval exchanges to exhibit a class of systems having simple continuous spectrum.
Interval exchange transformations have been extensively studied by several people including Keane [21] [22], Keynes and Newton [23], Veech [31] to [36], Rauzy [28], Masur [24], Del Junco [14], Boshernitzan [5, 6], Nogueira and Rudolph [27], Boshernitzan and Carroll [7], Berth´e, Chekhova, and Ferenczi [4], Chaves and Nogueira [10] and Boshernitzan and Nogueira [8].
While most articles in the area aim at establishing generic results for general interval exchange transformations, a few papers provide a detailed analysis of the dynamical behaviour/structure of specific families of interval exchanges. For instance, [14] describes the combinatorial structure of the symbolic sub-shifts associated to a restricted class of three-interval exchange transformations.
In [8], Boshernitzan and Nogueira further widen the class of weak mixing examples (see§5), while in [4] there are examples of three-interval exchanges which do not have discrete spectrum.
In this paper we give a detailed analysis of the spectral and ergodic properties of the three- interval exchange transformation T with probability vector (α, β,1−(α +β)), α, β > 0, and permutation(3,2,1)1defined by
(1) T x=
x+ 1−α ifx∈[0, α) x+ 1−2α−β ifx∈[α, α+β) x−α−β ifx∈[α+β,1).
Throughout the paper,T denotes the transformation onX = [0,1)defined in (1).
Date: October 6, 2003.
1991 Mathematics Subject Classification. Primary 37A25; Secondary 37B10.
Partially supported by NSF grant INT-9726708.
1All other nontrivial permutations on three letters reduce the transformation to an exchange of two intervals.
1
Our approach is mostly combinatorial and relies on the arithmetic results in [18] and the combi- natorial description in [19] of return words (with respect to the natural coding) to a special family of intervals. [18, 19] develop a theory for three-interval exchange transformations analogous to that developed by Morse-Hedlund [26], Coven-Hedlund [13], and Arnoux-Rauzy [3] which links together the diophantine properties of an irrational numberα,the ergodic/dynamical properties of a circle rotation by angleα,and the combinatorial/symbolic properties of a class of binary sequences known as the Sturmian infinite words: this is achieved through a vectorial algorithm of simultane- ous approximation, and a description of a class of sub-shifts of block complexityp(n) = 2n+ 1 which generalize Sturmian words. In the present paper we apply this description to prove spec- tral properties forT. In a forthcoming fourth part, we apply our theory to the study of joinings ofT. It is well known that each transformation T is induced by a rotation on the circle, and some properties ofT are readily traced back to the underlying rotation. For instance, under the assump- tion that T satisfies the infinite distinct orbit condition of Keane [21], the system is known to be both minimal and uniquely ergodic. Also, in the case of three intervals, the associated surface (obtained by suspending an interval exchange transformation via the so-called ‘zippered rectan- gles’ [33]) is nothing more than a torus devoid of singularities (see also [2]). We recall that, in the general case of interval exchanges, a celebrated result proved independently by Veech [33] and Masur [24] states that, if the permutationπis irreducible (π{1, . . . , l}={1, . . . , l}only ifl =k), for Lebesgue almost everyλ = (λ1, λ2, . . . , λk)in the setΛk ={λ ∈Rk, λi >0,1≤i ≤ k}the interval exchange transformation defined by the probability vector Σkλ
i=1λi and the permutationπis uniquely ergodic.
On the other hand, other properties of T appear not to be directly linked to the underlying ro- tation. These include for instance the existence and characterization of the eigenvalues of the associated unitary operator (in particular the weak mixing) and joinings (minimal self-joining and simplicity). In [20] Katok and Stepin prove that, for Lebesgue almost every (α, β) in the set {α >0, β >0, α+β <1},T is weakly mixing (see§5.1 below for the full result). This was later extended by Veech in [34]: ifπis irreducible and(1, . . . ,1)is not in the orthogonal complement inRkof the kernel of the operatorLdefined by the matrix((Lij))whereLij = 1 ifπ(i) > π(j), 0otherwise (this is true in particular for the permutation (k, k−1, k −2, . . . ,1)ifk is odd), for Lebesgue almost every λ ∈ Λk, the interval exchange transformation defined by the probability vector Σkλ
i=1λi and the permutationπis weakly mixing. In [27] Nogueira and Rudolph prove that if πis irreducible and not of rotation class (there is nolsuch thatπ(i) = i+l−1(modk+1) for alli), for Lebesgue almost everyλ∈Λk, the interval exchange transformation defined by the probability vector Σkλ
i=1λi and the permutation π has no non-constant somewhere continuous eigenfunctions and hence is topologically weakly mixing. More information on the behaviour of eigenfunctions and new proofs of weak mixing properties can be found in [10].
In this paper we obtain necessary and sufficient conditions onαandβforT to be weak mixing.
These conditions show that the weak mixing comes from the presence of either a spacer above a column of positive measure (like for Chacon’s map [9]), or of an isolated spacer above a column of small measure (like for del Junco-Rudolph’s map [15]). In addition, we exhibit new examples of weak mixingT. The conditions stem from a combinatorial recursive construction for generating three sequences of nested Rokhlin stacks which describe the system from a measure-theoretic point
of view, and which combined with a result of Choksi and Nadkarni [11] in the class of rank one systems, provide an explicit computation of the eigenvalues.
While it is known that, under the infinite distinct orbit condition, T is always topologically weakly mixing (see for instance [27]), in [30] Veech2proved the surprising existence of transfor- mations T with eigenvalue λ = −1.3 This was later extended by Stewart [29] who showed that for all rational numbers pq there exists a transformationT with eigenvaluee2πpiq . In this paper we give a simple combinatorial process for constructing the transformations of Veech and Stewart. In addition we exhibit examples ofT having ap-adic odometer as factor.
However the question concerning the existence of eigenvalues of the form e2πiγ, where γ is irrational, remained unsolved, in spite of some partial results due to Merrill [25] and Parreau (un- published), see the discussion in§6. In [34] Veech asks, forT satisfying the infinite distinct orbit condition:
Question 1.1 (Veech, [34], 1.9, (1984)). Do there existαandβ such thatT has an element of its point spectrum which is not a root of unity? Is it possible forT to have pure point spectrum?
In this paper we give affirmative answers to both questions:
Theorem 1.2. Let0 < γ < 1be an irrational number,[0;y1, y2. . .]its usual continued fraction expansion, andqk,k ≥1the denominators of its convergents, given byqk+1 =yk+1qk+qk−1. If
+∞
X
k=1
qk
yk+1 <+∞,
then there exists aT satisfying the i.d.o.c. condition, which is measure-theoretically isomorphic to the rotation of angleγ, and hence has discrete (pure point) spectrum.
We also show
Theorem 1.3. For every quadratic irrational number γ there exists a T satisfying the i.d.o.c.
condition, with eigenvaluee2πiγ.
Theorems 1.2 and 1.3 are extremes of one another in that in one case the partial quotients tend to infinity very quickly, while in the other they are eventually periodic. We do not know whether every complex number of modulus1is an eigenvalue of someT.
Theorems 1.2 and 1.3 suggest that not all properties ofT can be traced back to the underlying rotation: the irrational rotation by angleγ of Theorem 1.2 has no connection with the underlying rotation inducing T, and in the case of Theorem 1.3 a factor of T is a rotation with a quadratic angle, while the angle of the inducing rotation is a Liouville number.
ACKNOWLEDGEMENTS
The authors were supported in part by a Cooperative Research Travel Grant jointly sponsored by the N.S.F. and C.N.R.S.. The third author was also supported in part by a grant from the Texas Advanced Research Program. We are very grateful to E. Lesigne and A. Nogueira for numerous e-mail exchanges and many fruitful conversations.
2Although the result was actually established in [30], it was only first stated in the language of interval exchange transformations in [23], see also [34].
3To keep in line with the existing litterature, in this introduction we denote eigenvalues multiplicatively. However, in the core of the paper, we shall use an additive notation, see the beginning of§3 below.
2. DESCRIPTION OF THREE-INTERVAL EXCHANGE TRANSFORMATIONS
2.1. Preliminaries. T depends only on the two parameters 0 < α < 1and0 < β < 1−α. We note thatT is continuous except at the pointsαandα+β.
Set
A(α, β) = 1−α
1 +β and B(α, β) = 1 1 +β.
ThenT is induced by a rotation on the circle by angleA(α, β).More precisely,T is obtained from the 2-interval exchangeRon[0,1)given by
(2) Rx=
(x+A(α, β) ifx∈[0,1−A(α, β)) x+A(α, β)−1 ifx∈[1−A(α, β),1).
by inducing (according to the first return map) on the subinterval[0, B(α, β)),and then renormal- izing by scaling by1 +β.
We sayT satisfies the infinite distinct orbit condition (or i.d.o.c. for short) of Keane [21] if the two negative trajectories {T−n(α)}n≥0 and {T−n(α +β)}n≥0 of the discontinuities are infinite disjoint sets. Under this hypothesis,T is both minimal and uniquely ergodic; the unique invariant probability measure is the Lebesgue measureµon[0,1)(and hence(X, T, µ)is an ergodic system).
Because of the connection with the rotationR, T does not satisfy the i.d.o.c. condition if and only if one of the following holds:
• A(α, β)is rational, or equivalentlypα+qβ =p−q,
• B(α, β) = pA(α, β)−q, or equivalentlypα+qβ =p−q−1,
• B(α, β) = −pA(α, β) +q, or equivalentlypα+qβ =p−q+ 1
for some nonnegative integer p, and positive integer q. In the first case, T is not minimal; in the second and third cases, there is an immediate semi-topological conjugacy (hence a measure- theoretic isomorphism) betweenT and an irrational rotation.
2.2. Properties of the arithmetic algorithm. This subsection summarizes some results from [18].
LetIdenote the open interval(0,1), D0 ⊂R2the simplex bounded by the linesy= 0, x= 0,and x+y = 1,andDthe triangular region bounded by the linesx= 12, x+y = 1,and2x+y = 1.
We define two mappings onI×I, F(x, y) =
2x−1 x ,y
x
and G(x, y) = (1−x−y, y).
We check that if(α, β) ∈D0 is not inDand is not on any of the rational linespα+qβ =p−q, pα+qβ =p−q+ 1,pα+qβ =p−q−1then there exists a unique finite sequence of integersl0, l1, . . . , lksuch that(α, β)is inH−1DwhereH is a composition of the formGt◦Fl0 ◦G◦Fl1 ◦ G . . .◦G◦Flk◦Gs, s, t∈ {0,1}.
The functionH(α, β) is computed recursively as follows: we start with α(0) = α, β(0) = β.
Then, given(α(k), β(k)), we have three mutually exclusive possibilities: if(α(k), β(k))is inD, the algorithm stops; ifα(k) < 12, we applyG; if2α(k)+β(k) <1, we applyF.
Associated to each point(α, β) ∈ D0 there is a sequence (nk, mk, k+1)k≥1,wherenk andmk are positive integers, andk+1is±1.This sequence we call the three-interval expansion of(α, β), is a variation of the negative slope expansion defined in [18]; it is constructed as follows:
• For(α, β)inDwe put
x0 = 1−α−β
1−α and y0 = 1−2α 1−α and define fork ≥0
(xk+1, yk+1) =
n yk
(xk+yk)−1
o ,n
xk
(xk+yk)−1
o
ifxk+yk >1 n 1−yk
1−(xk+yk)
o ,
n 1−xk 1−(xk+yk)
o
ifxk+yk <1 (nk+1, mk+1) =
b(x yk
k+yk)−1c,b(x xk
k+yk)−1c
ifxk+yk >1
b1−(x1−yk
k+yk)c,b1−(x1−xk
k+yk)c
ifxk+yk <1
where{a}andbacdenote the fractional and integer part ofarespectively. Fork ≥0set k+1 = sgn(xk+yk−1).
We note that1 is always−1, hence we ignore it in the expansion.
• For(α, β)∈/ Dwe letH be the function above for which(α, β)∈H−1Dand put
¯ α,β¯
=H(α, β),
and define(nk, mk, k+1)as in the previous case, starting from α,¯ β¯
∈D.
The following proposition sums up what we need of [18]; when(α, β)is inD, it is a translation (taking into account the fact that the initial conditions are slightly different) of results in [18]; in the general case, it comes from these results and the definition of( ¯α,β).¯
Proposition 2.1. (1) If T satisfies the i.d.o.c. condition, then the three-interval expansion (nk, mk, k+1)of(α, β)is infinite.
(2) An infinite sequence(nk, mk, k+1)is the expansion of at least one pair(α, β)defining aT satisfying the i.d.o.c. condition, if and only ifnkandmkare positive integers,k+1 =±1, (nk, k+1)6= (1,+1)for infinitely manykand(mk, k+1)6= (1,+1)for infinitely manyk.
(3) Each infinite sequence(nk, mk, k+1)satisfying the conditions in (2) is the three-interval expansion of a countable family of couples(α, β), with exactly one couple in each of the disjoint triangles H−1D, where H has any of the possible forms defined earlier in this section, including the identity.
(4) A( ¯α,β) =¯ 1 2 + 1
m1+n1− 2
m2+n2− 3
m3+n3−. . .
(5) A(α, β) has bounded partial quotients (in the usual continued fraction expansion) if and only if in the three-interval expansion of (α, β)thenk+mk are bounded, as well as the lengths of strings of consecutive(1,1,+1).
(6) The following two properties are equivalent
– in the three-interval expansion of(α, β), there existsc > 0such that for everyM >0 there is aksuch thatmk+nk > M and
c <
mk−nk
mk+nk
<1−c.
– there existsc0 >0such that for every >0, there exist integersp,qsuch that
A(α, β)−p q
≤ q2 and for any integerr
B(α, β)− r q
> c0 q.
2.3. Combinatorial properties. This subsection summarizes some results from [19].
We define the natural partition
P1 = [0, α), P2 = [α, α+β), P3 = [α+β,1).
For every pointxin[0,1), we define an infinite sequence(xn)n∈Nby puttingxn = iifTnx ∈Pi, i= 1,2,3.This sequence, also denoted byx, is called the trajectory ofx. IfT satisfies the i.d.o.c.
condition, the minimality of the system implies that all trajectories contain the same finite words as factors.
LetI0be a set of the form∩n−1i=0T−iPki; we sayI0 has a name of lengthngiven byk0. . . , kn−1; note thatI0 is necessarily an interval andk0, . . . , kn−1 is the common beginning of trajectories of all points inI0.
For each intervalJ, it is known (see for example [12]) that the induced map of T on J is an exchange of three or four intervals. More precisely, there exists a partitionJi,1≤ i≤tofJ into subintervals (witht = 3ort = 4), andtintegershi, such thatThiJi ⊂J, and{TjJi},1≤i ≤t, 0 ≤ j ≤ hi − 1, is a partition of [0,1) into intervals: this is the partition into Rokhlin stacks associated toT with respect toJ. The intervalsJihave names of lengthhi, called return words to J.
In [19] we give an explicit construction of the trajectories of the points1−αand1−α−β. As a consequence we have the following structure theorem [19].
Theorem 2.2. LetT satisfy the i.d.o.c. condition, and let(nk, mk, k+1)k≥1, be the three-interval expansion of (α, β). Then there exists an infinite sequence of nested intervals Jk, k ≥ 1, which have namewk and exactly three return words,Ak,BkandCk, given recursively fork ≥ 1by the following formulas
Ak=Ank−1k−1Ck−1Bmk−1k−1Ak−1, Bk =Ank−1k−1Ck−1Bk−1mk , Ck =Ank−1k−1Ck−1Bk−1mk−1 ifk+1 = +1, and
Ak =Ank−1k−1Ck−1Bk−1mk , Bk=Ank−1k−1Ck−1Bk−1mk−1Ak−1,
Ck =Ank−1k−1Ck−1Bk−1mk Ak−1
ifk+1 =−1.
Fork = 0, the wordsA0,B0,C0 are defined as follows.
Jk1 Jk2 Jk3
T Jk1 T Jk2 T Jk3
Tak−1Jk1
Tbk−1Jk2
Tck−1Jk3
FIGURE 1. The three Rokhlin stacks
Proposition 2.3. When(α, β)is inD, we haveA0 = 13,B0 = 2,C0 = 12. When(α, β)is not in D, we defineHand( ¯α,β)¯ as in the previous subsection, and two substitutions,σF by
1 → 1 2 → 21 3 → 31 andσGby
1 → 3 2 → 2 3 → 1
and we define σH by replacing, in the expression of H, F by σF and G by σG. Then we have A0 =σH(13),B0 =σH(2),C0 =σH(12).
In all cases the lengths ofA0 andB0differ by±1.
Though we shall not use this fact in the sequel, if(α, β)is inDthenJkis an interval of continuity ofTnfor somensuch that1−α−βis inJkandαis inTnJk. Equivalently thewkare the bispecial factors (in the minimal subshift associated withT, see [19]) beginning in1and ending in2. When (α, β)is not inD, letJ¯kbe, for the three-interval exchange T0 defined by( ¯α,β), the intervals of¯ continuity ofT0nfor somen, such that1−α¯−β¯is inJ¯kandα¯is inT0nJ¯k; letw¯kbe the name of J¯k. ThenJk is the interval whose name isσH( ¯wk).
2.4. Rokhlin stacks. Theorem 2.2 gives an explicit construction of the Rokhlin stacks mentioned in the previous subsection.
Let wk be the name of Jk. Every trajectory under T is a concatenation of wordsAk, Bk, Ck, which we call thek-words. We say that ak-word occurs at its legalk-place in a trajectory if it is immediately followed by the word wk. A concatenation of k-words occurs at its legalk-place if each of itsk-words occur at their legalk-place.
We defineF Ak to be set ofx ∈ X such that (in the trajectory ofx) x0 is the first letter of the k-wordAk in its legalk-place, and similarlyF BkandF Ck. Note thatF Ak∪F Bk∪F Ck =Jk, and these are exactly the three intervals of continuity of the induced map ofT onJk.
Letak,bk,ckbe the lengths ofAk,BkandCk; it follows from Theorem 2.2 and Proposition 2.3 that|ak−bk|=|a0−b0|= 1andckis eitherak−ak−1 =bk−bk−1 orak+ak−1 =bk+bk−1. In particular, we haveck ≤ 2ak. ThenX = [0,1)is the disjoint union ofTjF Ak, 0 ≤ j ≤ ak−1, TjF Bk,0≤j ≤bk−1,TjF Ck,0≤j ≤ck−1; we denote byτ Akthe disjoint union ofTjF Ak, 0≤j ≤ak−1, and define similarlyτ Bkandτ Ck;τ Ak,τ Bkandτ Ckare called thek-stacks, and theTjF Ak,0≤j ≤ak−1are the levels of the stackτ Ak, and similarly forB andC. The levels are intervals of small diameter, as they have names of arbitrarily large length; hence any integrable functionf can be approximated (in L1 for example) by functions fk which are constant on each level of eachk-stack.
Hence, in the language of finite rank systems, see [16], the Theorem 2.2 implies that
Corollary 2.4. T is of rank at most three, generated by the stacks τ Ak, τ Bk, τ Ck; the recursion formulas in the Theorem 2.2 give an explicit construction by cutting and stacking of these stacks.
The rank at most three was known at least since [28] (while, by the above remark, the rank at most four was in the folklore and is proved in [12]), but the explicit construction of Theorem 2.2 is new, and from it comes all our knowledge of the system. The fact that Ak, Bk, Ck are return words of the same word is useful additional information, but it may be deduced from the explicit construction.
The finite rank structure is particularly relevant in the measure-theoretic study of the system (X, T, µ), and in this framework, the following lemma will be useful:
Lemma 2.5. Let T satisfy the i.d.o.c. condition. For any trajectory x, the Lebesgue measure µ(τ Ak)is the limit whenn goes to infinity of n1 times the total number of indices0 ≤ i ≤ n−1 such thatxi belongs to a wordAkin its legalk-place. Similarly forBkandCk.
Proof. This follows from the unique ergodicity ofT.
Remark An immediate consequence of the finite rank structure is thatT is completely known, up to measure-theoretic and topologic isomorphisms, from the associated three-interval expansion (nk, mk, k+1)k≥1, and the initial lengthsa0,b0,c0. Two three-interval exchange transformationsT andT0 with the same expansion (nk, mk, k+1)k≥1 need not be measure-theoretically isomorphic, see Proposition 6.2 below, but are related by induction on intervals: that is, there exist intervalsI00 andI0 such that the induced maps ofT onI00and ofT0 onI0 are the same. Also, it is noted in [19]
that, if we apply our recursion formulas, but starting from A0 = B0, the systems we get are just irrational rotations.
3. WEAK MIXING OFT
We recall two definitions of measure-theoretic ergodic theory: if(X, S, ν)is a measure-theoretic dynamical system, we say that a real number0≤γ <1is an eigenvalue ofS(denoted additively) if there exists a nonconstant f inL2(X,R/Z) such thatf ◦S = f +γ (inL2(X,R/Z))); f is then an eigenfunction for the eigenvalue γ. As constants are not eigenfunctions,γ = 0is not an eigenvalue ifS is ergodic. Sis weak mixing if it has no eigenvalue.
3.1. The fundamental lemma. In this subsection, we use the fact that there are two Rokhlin stacks of heights differing by one to kill eigenvalues: this is a trick inspired by Chacon’s map [9].
We first use it to give a short proof of a classical result of topological dynamics: the transformation T cannot have continuous eigenfunctions. Then we tackle the general case of measurable eigen- functions; the proofs of Proposition 3.1 and Lemma 3.2 below give some clues about the difference between continuous and non-continuous eigenfunctions: the presence of two nonempty stacks of heights differing by one is enough to prevent the existence of continuous eigenfunctions. However, to show that there are no measurable eigenfunctions, we must also know that the measure of these stacks is not too small.
In the sequel||x||denotes the distance ofxto the nearest integer.
Proposition 3.1. EveryT satisfying the i.d.o.c. condition is topologically weakly mixing: it has no continuous eigenfunction.
Proof. Letγ be an eigenvalue with a continuous eigenfunction f; then, for given , if k is large enough,|f(z)−f(y)|< (inR/Z) ifzandyare inJk. We takex∈F Ak⊂Jk; thenTakx∈Jk, hence,
||γak||=|f(Takx)−f(x)| ≤;
taking a pointx0 inF Bk, we get the same relation withbk=ak±1,henceγ = 0, which is not an
eigenvalue asT is ergodic.
Remark This result was known since [27]. The topology we use here is the one of the interval [0,1); but the same proof works if we look atT as the shift on the symbolic trajectories, equipped with the product topology on {1,2,3}N, as the bases of thek-stacks are cylinders associated to arbitrarily long words. Note that for a three-interval exchange transformation with a permutation such as(132), there are continuous eigenfunctions, asT is trivially conjugate to a rotation. In [1]
Arnoux exhibits an example of an interval-exchange transformation (on seven intervals) having a nontrivial continuous eigenfunction.
Lemma 3.2. Ifγ is an eigenvalue ofT, there exists a sequenceδk →0such that for allk
||γak||< δk µ(τ Ak),
||γbk||< δk µ(τ Bk),
||γck||< δk µ(τ Ck).
Proof. Supposef is an eigenfunction forT for the eigenvalue γ; then we haveR
|f −fk| < δk0, withδk0 →0andfkis constant on every level of the stackτ Ak,τ Bkorτ Ck.
To prove the first inequality, we define a new wordDk, and an integerk0(k)≤k−1such that
• k0(k)tends to infinity whenk tends to infinity,
• Dkis a concatenation ofk0-words,
• Dkin its legalk0-place is a prefix ofAk,
• the length ofDkisdk≥ a3k,
• in any trajectory x, each occurrence of a k-word in its legal k-place is followed by an occurrence ofDk in its legalk0-place.
The wordDkis defined as follows:
• if nk > 1 ormk > 1, we put Dk = Ank−1k−1Ck−1Bk−1mk−1, the common prefix of the three k-words; thusdk≥ a2k, and we setk0 =k−1;
• ifmk=nk= 1, letk0(k)be the largestl < ksuch that(nl, ml, l+1)6= (1,1,+1); then the recursion formulas give eitherAk =Ckk−k0 0Ak0,Bk =Ckk−k0 0Bk0,Ck =Ck0 (ifk+1 = +1), or Ak = Ckk−k0 0Bk0, Bk = Ckk−k0 0Ak0, Ck = Ckk−k0 0Bk0Ckk−k0 0−1Ak0 (if k+1 = −1); and ck0 > a2k0. Then we takeDk =Ckk−k0 0, from which it follows thatdk ≥ a3k.
Because (nk, mk, k+1) cannot be (1,1,+1) ultimately (Proposition 2.1), k0(k) → +∞ when k →+∞. The other properties ofDkare clear from its construction.
We callτk0 the union of the first (or lower)dklevels ofτ Ak; thus, forx∈ τk0,xandTakxare in the same level of the samek0-stack.
Now ifx∈τk0,fk0(Takx) =fk0(x)whilef(Takx) = γak+f(x); we have Z
τk0
|Takfk0−γak−fk0|= Z
τk0
|γak| ≥ ||γak||µ(τ Ak) 3
and Z
τk0
|Takfk0−γak−fk0| ≤ Z
τk0
|Takfk0 −Takf|+ Z
τk0
|fk0 −f|<2δk00
which gives||γak||<6 δ
0 k0
µ(τ Ak), hence the first inequality.
The second inequality is proved in the same way, as is the third, except that we replace Dk by
Ckandτk0 byτ Ck in the casemk =nk = 1.
Remark on the proof. In the proof of Lemma 3.2, we can alternatively usewkitself instead of buildingDk (which is a prefix ofwk), and make similar computations, replacingdkby the length d0kofwk, andτk0 by the union of the first (or lower)d0klevels ofτ Ak. In this case, the results follow from the fact thatTiJkis still an interval for0≤i≤d0k−1, and thatd0k ≥ak−2([19]). However, we prefer to prove Lemma 3.2 directly from the recursion formulas.
Corollary 3.3. If for a constantcand infinitely manykwe have bothµ(τ Ak)≥candµ(τ Bk)≥c, thenT is weakly mixing.
Proof. The first two inequalities of Lemma 3.2, together with the relationbk =ak±1give||γ||<
2cδk, henceγ = 0.
3.2. Study of possible non weakly mixing cases. Corollary 3.3 gives some sufficient conditions forT to be weakly mixing. We shall now focus on the cases when these conditions are not fulfilled, and show that this may happen only for very special sequences(nk, mk, k+1), namely those with a subsequence on which mk is much bigger (or smaller) than nk while outside this subsequence we must have (nk, k+1) = (1,+1) or (mk, k+1) = (1,+1) (or else (nk, k+1) = (1,−1) or (mk, k+1) = (1,−1), these last two cases being allowed only one step beforemk is much bigger or smaller thannk).
Lemma 3.4. If T is not weakly mixing, then for every integer M there exists an integer k0(M) such that for everyk > k0(M), one of the following two assertions is satisfied:
• there existsl ≥ k+ 1such thatml ≥ M nl, k+2 = . . .= l−1 = +1ifl > k + 2and, if l > k+ 1,nk+1 =. . .=nl−1 = 1ifl= +1,mk+1 =. . .=ml−1 = 1ifl =−1;
• there existsl ≥ k+ 1such thatnl ≥ M ml, k+2 = . . .= l−1 = +1ifl > k + 2and, if l > k+ 1,mk+1 =. . .=ml−1 = 1ifl = +1,nk+1 =. . .=nl−1 = 1ifl =−1.
Proof. By Corollary 3.3, ifT is not weakly mixing, then for everyδandksufficiently large, either µ(τ Ak)< δorµ(τ Bk)< δ.
We estimateµ(τ Ak), using Lemma 2.5 and Theorem 2.2. Ifnk+1 > 1, wordsAk in their legal k-place cover a proportion at least n nk+1−1
k+1+mk+1+1 of eachk+ 1-word (recall thatck <2akand that, kbeing large enough,bk=ak±1can be treated asak) and hence
µ(τ Ak)≥ nk+1−1 nk+1+mk+1+ 1.
Ifnk+1 = 1, wordsAk in their rightk-place do not appear in bothAk+1andBk+1, and we have µ(τ Ak)≥ 1
mk+1+ 2µ(τ Ak+1) ifk+2 = +1,
µ(τ Ak)≥ 1
mk+1+ 2µ(τ Bk+1) + 1
mk+1+ 3µ(τ Ck+1) ifk+2 =−1.
Hence, ifµ(τ Ak)is small, then
• eithermk+1is much bigger thannk+1,
• ornk+1 = 1,k+2 = +1, and m 1
k+1+2µ(τ Ak+1)is small,
• ornk+1 = 1,k+2 =−1and both m 1
k+1+2µ(τ Bk+1)and m 1
k+1+3µ(τ Ck+1)are small.
In the third case, we may havemk+2 = 1; but in that case, ask+2 =−1, we haveck+1 ≥ak+1, andµ(τ Ck+1)is at least m 1
k+2+2 because this is the minimal proportion of eachk+ 2-word which it covers. Hence, if we are in the third case and not in the first, we must havenk+2 much bigger thanmk+2.
Note that in the second case, however,µ(τ Ak)may be small even with theml andnlare bounded forl ≥k+ 1, provided there is a long string ofnl= 1withl+1 = +1ahead.
More precisely, the fact thatµ(τ Ak)is small implies that
• either from l = k + 1 to some k0 ≥ l, there is a (possibly empty), string of nl = 1, withl+1 = +1except maybe the last one, followed by annk0 much smaller thanmk0, (or followed by anmk0 much smaller thannk0 if the lastl+1 is a−1);
• or else that froml =k+ 1to somel0there is a very long string ofnl = 1, withl+1 = +1.
IfT is not weakly mixing, the hypotheses of Corollary 3.3 are not satisfied forklarge enough.
In view of the computations above, for every integer M there exists k0(M) such that, for every k > k0(M), at least one of the following four assertions is satisfied:
• the first assertion of this lemma;
• the second assertion of this lemma;
• nk+1 =. . .=nk+M = 1,k+2 =. . .=k+M+1 = +1;
• mk+1 =. . .=mk+M = 1,k+2 =. . .=k+M+1 = +1.
To prove the lemma, it remains to prove that the last two assertions alone are not sufficient to prevent weak mixing. The main argument here is that if the sequence(nk, mk, k+1)is made only of long strings (nk, k+1) = (1,+1) or(mk, k+1) = (1,+1), then transitions between strings of (nk, k+1) = (1,+1)and of(mk, k+1) = (1,+1)must occur infinitely often, and these transitions will always produce weak mixing. But, to prove this, we need some involved computations (using, as in Lemma 3.2, differences of 1between lengths of words) in the case where these transitions are made of strings of(nk, mk, k+1) = (1,1,+1).
Namely, we suppose that there are arbitrarily largek which do not satisfy the first and second assertion. Take such akand suppose for example thatnk+1 =. . .=nl = 1, k+2 =. . .=l+1 = +1, l −k > M; as we cannot have (nk, k+1) = (1,+1) ultimately, we can choosel such that either nl+1 > 1or l+2 = −1. Ifnl+1 > 1, we cannot have ml+1 > M nl+1 ask would satisfy the first assertion, hence eithernl+1 > M ml+1 orml+1 = 1, and this implies thatµ(τ Al)≥ 14. If nl+1 = 1, thenl+2 = −1and we still cannot have ml+1 > M nl+1; then eithernl+2 > M ml+2, but that is excluded askwould satisfy the second assertion, orml+1 = 1andml+2 > M nl+2, and this implies againµ(τ Al)≥ 14.
But also l−M + 3 must satisfy one of our four assertions; because of our assumptions onk and nl+1, this can only happen if ml+3−M = . . . = ml = 1, while still nk+1 = . . . = nl = 1, k+2 =. . .=l+1 = +1.
Suppose first that there exists somek+ 1 ≤ k1 ≤ l+ 2−M such thatmk1 > 1, and take the largest possible suchk1. Thennk1 = 1,mk1 =m >1,k1+1 = +1. Thenτ Bk1−1 has measure at least 14, and by Lemma 3.2 we have||γbk1−1||< δ, withδsmall ifkis large.
Now, Al = (Ck1)M0Ak1, with M0 ≥ M −4, andCk1 = Ck1−1Bkm−1
1−1, henceck1 > 12ak1 and µ(τ Ck1)≥ 12µ(τ Al)≥ 18. Hence by Lemma 3.2 it follows that||γck1||< δ.
We need to show also that||γM0ck1|| < 20δ; this is done by an argument similar to the proof of Lemma 3.2: we take M00 = [M4 ], and τ the set of those x for which x0 lies in a word Ck1 in its legal k1-place, followed by at least M00 words Ck1 in their legal k1-places, and estimate R
τ|TM00ck1fk1 − γM00ck1 − fk1|; as µ(τ) ≥ 12µ(τ Ck1) ≥ 161, we get the desired estimate for γM00ck1 and hence forγM0ck1 because of the last result.
Hence, as γal is close to 0 (modulo 1), we get that both γck1 and γak1 are close to 0. But Ck1 =Ck1−1(Bk1−1)m−1andAk1 =Ck1−1(Bk1−1)m−1Ak1−1, hence we get that||γak1−1||<100δ, and hence||γ||<101δ.
Suppose now that mk+1 = . . . = ml+2−M = 1. Then we cannot have nl+1 > M ml+1 or l+2 =−1,nl+1 = 1,ml+2 > M nl+2 as in both caseskwould satisfy the first or second assertion, hence nl+1 > 1, ml+1 = 1. And l cannot satisfy the first or second assertion as k would then satisfy it, nor the third one as nl+1 > 1. Hence we must have ml+1 = . . . = ml+M = 1, l+2 =. . .=l+M+1 = 0, and we can start the same reasoning withlreplacingk, and themandn exchanged; but we know thatnl+1 >0, hence we are in the case where this implies thatγ is close to0.
Hence this situation cannot occur infinitely many times ifT is not weakly mixing.
3.3. Sufficient conditions for weak mixing. Lemma 3.4 gives a necessary condition forT not to be weakly mixing; our aim now is to reduce the set of possibly non-weakly mixingT to an explicit set of rank-one (see below) transformations, ready to be studied by classical methods.
The main improvement over Lemma 3.4 is that, on the sequencek(j)wheremk is much smaller or bigger than nk, the ratios mmk(j)∧nk(j)
k(j)∨nk(j) must form a converging series, otherwise there is weak mixing by a Borel-Cantelli argument. Some technical conditions are also added to deal with the cases wheremk(j)∧nk(j) = 1.
Note that the condition depends only on the expansion(nk, mk, k+1)k≥1.
Theorem 3.5. LetT satisfy the i.d.o.c. condition , and let(nk, mk, k+1)k≥1, be the three-interval expansion of(α, β). Then, ifT is not weakly mixing, there exists a strictly increasing sequence of integersk(j),j ∈N, such that
• (Ma)
+∞
X
j=1
mk(j)∧nk(j)
mk(j)∨nk(j) <+∞;
• (Mb)mk(j)6=nk(j);
• (Mc) ifmk(j) > nk(j), then for every k(j −1)< k < k(j)−1, k+1 = +1; and for every k(j−1)< k < k(j),nk= 1ifk(j)= +1,mk= 1ifk(j) =−1;
• (Md) ifnk(j) = 1, thenmk(j+1) > nk(j+1) ifk(j)+1k(j+1) =−1or ifk(j + 1) =k(j) + 1 andk(j)+1 =−1,nk(j+1) > mk(j+1) otherwise;
• (Mc’) ifnk(j)> mk(j), then for everyk(j −1)< k < k(j)−1,k+1 = +1; and for every k(j−1)< k < k(j),mk= 1ifk(j) = +1,nk= 1ifk(j) =−1;
• (Md’) ifmk(j)= 1, thennk(j+1) > mk(j+1)ifk(j)+1k(j+1) =−1or ifk(j+ 1) =k(j) + 1 andk(j)+1 =−1,mk(j+1) > nk(j+1) otherwise.
Proof. AsT is not weakly mixing, by Lemma 3.4 we can find a sequencek(j)such that mk(j)∧nk(j)
mk(j)∨nk(j) →0
whenj tends to infinity, and such that Mc and Mc’ are satisfied. We can of course choose thek(j) so that alsomk(j) 6=nk(j).
We modify this sequence to get Md and Md’: if k(j) is such that nk(j) = 1 but Md is not satisfied, or thatmk(j)= 1and Md’ is not satisfied, then we look at the firstj0 > jsuch thatnk(j0)∧ mk(j0) > 1 ornk(j0) = 1 and Md is satisfied, or mk(j0) = 1 and Md’ is satisfied. Such aj0 does exist, as otherwise we would have either(nk, k+1) = (1,+1) ultimately, or(mk, k+1) = (1,+1) ultimately, or else we would have infinitely many situations similar to the following: nk = 1, mk =m, k+1 = 1,mk+1 = 1,nk+1 =n, so for exampleAk =Ck−1Bk−1m ,Ck =Ck−1Bmk−1Ak−1, and
Ak+1 = (Ck−1Bk−1m )n−1Ck−1Bk−1m Ak−1Ck−1Bk−1m .
Then, as in the proof of Lemma 3.4, we show (possibly by going to the nextmk0 >1) thatγbk−1, γak andγak+1 are all close to0(modulo1), and by substraction we get thatγak−1 is also close to 0, and this cannot happen ifkis large enough.
And, if we deletek(j), k(j+ 1), ...,k(j0 −1)from the sequencek(n), the modified sequence k(n) keep all its other properties: this is clear for Ma and Mb, but we have to check that the k(j −1) < k < k(j)0 −1do satisfy Mc and Mc’. The part of Mc and Mc’ concerning mk and nk follows from the properties ofk(j),k(j+ 1), ...,k(j0 −1), so we just have to check thek+1; now, ifk(j−1) + 1< k < k(j0)−1andk+1 =−1, we check that we get again a weakly mixing situation, and conclude that this cannot happen ifkis large enough.
We have now to prove condition Ma. We estimate the measures of τ Ak(j)−1 and τ Bk(j)−1. Suppose for examplemk(j) > nk(j); thenµ(τ Bk(j)−1) > 1−δifj is large enough. Ifnk(j) > 1, then
µ(τ Ak(j)−1)≥ nk(j)−1
nk(j)+mk(j)+ 1 ≥ nk(j)
2nk(j)+ 2mk(j)+ 2 ≥ nk(j) 2mk(j).
Suppose now nk(j) = 1, mk(j) = m, and suppose for example that k(j)+1 = k(j+1) = +1.
Because of Md and Md’, we must havenk(j+1) > mk(j+1). Letk2 be the firstk > k(j)such that nk2, mk2, k2+1 6= (1,1,+1); then eithernk2 is much bigger thanmk2, ornk2 >1andmk2 = 1, or
nk2 = mk2 = 1, k2 = −1which is excluded because of the hypothesis onk(j+1). In each case, we haveµ(τ Ak2−1)≥ 14. Ifk2 =k(j) + 1, we get
µ(τ Ak(j)−1)≥ 1
mk(j)+ 1µ(τ Ak(j))≥ 1
4mk(j)+ 4 ≥ nk(j)
5mk(j).
Ifk2 > k(j) + 1, ask(j)+1 = +1, we conclude like in the proof of Lemma 3.4 that bothγak(j)and γck(j) are close to0(modulo1), which together imply thatγak(j)−1 is close to 0, and that cannot happen if j is large enough. Similar reasonings take care of the other cases for nk(j+1), mk(j+1), k(j)+1, andk(j+1), except that, in the casek2 > k(j) + 1andk(j)+1 = −1, we use the fact that µ(τ Ck(j))≥ 12 and
µ(τ Ak(j)−1)≥ 1
mk(j)+ 3µ(τ Ck(j))≥ 1
2mk(j)+ 6 ≥ nk(j) 3mk(j)
. Suppose now that
+∞
X
j=1
mk(j)∧nk(j)
mk(j)∨nk(j) = +∞.
Then at least one of the four series
•
X
j;mk(j)>nk(j),ak(j)−1−bk(j)−1=1
nk(j) mk(j),
•
X
j;mk(j)>nk(j),ak(j)−1−bk(j)−1=−1
nk(j)
mk(j),
•
X
j;nk(j)>mk(j),ak(j)−1−bk(j)−1=1
mk(j)
nk(j),
•
X
j;nk(j)>mk(j),ak(j)−1−bk(j)−1=−1
mk(j) nk(j),
diverges. Suppose for example the first one diverges and calll(j)the (infinite) sequence of those k(j)for whichmk(j) > nk(j) andak(j)−1−bk(j)−1 = 1.
We fix some j0; let Uj = ∪j0≤i≤jτ Al(i)−1. We haveµ(Uj) → 1 when j → +∞ by Borel- Cantelli. Uj \Uj−1 is made of those x whose coordinate x0 lies in a word Al(j)−1 (in its legal l(j)−1-place) but not in a wordAl(i)−1(in its legall(i)−1-place) for anyj0 ≤i≤j−1; for these i, anl(i)−1-word (in its legall(i)−1-place) which is not anAl(i)−1is either aCl(i)−1, or aBl(i)−1 in a string(Bl(i)−1)ml(i)−1or(Bl(i)−1)ml(i) inside anl(i)-word, or a single(Bl(i)−1)at the beginning of anl(i)-word. LetZj be the set of thosexwhose coordinatex0 lies in a wordAl(j)−1(in its legal l(j)−1-place) but, for anyj0 ≤ i ≤ j −1, not in a word Al(i)−1 (in its legall(i)−1-place) nor in a wordCl(i)−1 (in its legall(i)−1-place), nor in a single (as defined above) word Bl(i)−1, nor in the first word Bl(i)−1 of a string (Bl(i)−1)ml(i)−1 or (Bl(i)−1)ml(i). Then, as the ml(i) are large, µ(Zj)> 12 ifj0is large enough.
Let Zj0 = Zj \ Zj−1. Let dl(j)−1 be defined as in the proof of Lemma 3.2 and Yj the set of the x in Zj0 whose coordinate x0 lies in the prefix of length dl(j)−1 of its word Al(j)−1. Then µ(Yj)≥ 13µ(Zj0), andµ(∪j≥j0Yj)≥ 16. TheYj are disjoint; and also all theTal(j)−1Yj are disjoint, as Tal(j)−1Yj0 is made of x whose coordinatex0 lies in the same l(i)-words than thex in Yj for
