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STRUCTURE OF THREE-INTERVAL EXCHANGE TRANSFORMATIONS III:ERGODIC AND SPECTRAL PROPERTIES

SEBASTIEN 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 methodof 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. Weobtain 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 transformationis given by a probability vector(α1, α2, . . . , αk) togetherwith a permutationπ of {1, 2, . . . , k}. The unit interval[0, 1) is partitioned intok sub-intervals oflengthsα1, α2, . . . , αk which 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], Berthe, 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 exchangetransformations, a few papers provide a detailed analysis of the dynamical behaviour/structure ofspecific families of interval exchanges. For instance, [14] describes the combinatorial structure ofthe 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), whilein [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 transformationT with probability vector(α, β, 1 − (α + β)), α, β > 0, andpermutation(3, 2, 1) 1 defined by

(1) Tx =

x + 1− α if x ∈ [0, α)

x + 1− 2α− β if x ∈ [α, α + β)

x− α− β if x ∈ [α + β, 1).

Throughout the paper,T denotes the transformation onX = [0, 1) defined in (1).

Date: October 6, 2003.1991Mathematics 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

2 S. FERENCZI, C. HOLTON, AND L.Q. ZAMBONI

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 familyof intervals. [18, 19] develop a theory for three-interval exchange transformations analogous tothat developed by Morse-Hedlund [26], Coven-Hedlund [13], and Arnoux-Rauzy [3] which linkstogether the diophantine properties of an irrational numberα, the ergodic/dynamical properties of acircle rotation by angleα, and the combinatorial/symbolic properties of a class of binary sequencesknown as theSturmian 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 + 1which 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 transformationT is induced by a rotation on the circle, and someproperties ofT are readily traced back to the underlying rotation. For instance, under the assump-tion thatT satisfies theinfinite distinct orbit conditionof Keane [21], the system is known to beboth 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, inthe general case of interval exchanges, a celebrated result proved independently by Veech [33] andMasur [24] states that, if the permutationπ is irreducible (π{1, . . . , l} = {1, . . . , l} only if l = k),for Lebesgue almost everyλ = (λ1, λ2, . . . , λk) in the setΛk = {λ ∈ Rk, λi > 0, 1 ≤ i ≤ k} theinterval exchange transformation defined by the probability vectorλ

Σki=1λi

and the permutationπ isuniquely ergodic.

On the other hand, other properties ofT appearnot to be directly linked to the underlying ro-tation. These include for instance the existence and characterization of the eigenvalues of theassociated unitary operator (in particular the weak mixing) and joinings (minimal self-joining andsimplicity). 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 laterextended by Veech in [34]: ifπ is irreducible and(1, . . . , 1) is not in the orthogonal complementin Rk of the kernel of the operatorL defined by the matrix((Lij)) whereLij = 1 if π(i) > π(j),0 otherwise (this is true in particular for the permutation(k, k − 1, k − 2, . . . , 1) if k is odd), forLebesgue almost everyλ ∈ Λk, the interval exchange transformation defined by the probabilityvector λ

Σki=1λi

and the permutationπ is weakly mixing. In [27] Nogueira and Rudolph prove that if

π is irreducible and not of rotation class (there is nol such thatπ(i) = i+l−1 (modk+1) for all i),for Lebesgue almost everyλ ∈ Λk, the interval exchange transformation defined by the probabilityvector λ

Σki=1λi

and the permutationπ has no non-constant somewhere continuous eigenfunctionsand hence istopologically weakly mixing. More information on the behaviour of eigenfunctionsand new proofs of weak mixing properties can be found in [10].

In this paper we obtain necessary and sufficient conditions onα andβ for T to be weak mixing.These conditions show that the weak mixing comes from the presence of either a spacer above acolumn of positive measure (like for Chacon’s map [9]), or of an isolated spacer above a column ofsmall measure (like for del Junco-Rudolph’s map [15]). In addition, we exhibit new examples ofweak mixingT . The conditions stem from a combinatorial recursive construction for generatingthree sequences of nested Rokhlin stacks which describe the system from a measure-theoretic point

STRUCTURE OF 3-IETS III 3

of view, and which combined with a result of Choksi and Nadkarni [11] in the class ofrank onesystems, provide an explicit computation of the eigenvalues.

While it is known that, under the infinite distinct orbit condition,T is always topologicallyweakly mixing (see for instance [27]), in [30] Veech2 proved the surprising existence of transfor-mationsT with eigenvalueλ = −1.3 This was later extended by Stewart [29] who showed that

for all rational numberspq

there exists a transformationT with eigenvaluee2πpi

q . In this paper wegive a simple combinatorial process for constructing the transformations of Veech and Stewart. Inaddition we exhibit examples ofT having ap-adic odometer as factor.

However the question concerning the existence of eigenvalues of the forme2πiγ, whereγ isirrational, 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 orbitcondition:

Question 1.1(Veech, [34], 1.9, (1984)). Do there existα andβ such thatT has an element of itspoint 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. Let 0 < γ < 1 be an irrational number,[0; y1, y2 . . .] its usual continued fractionexpansion, andqk, k ≥ 1 the denominators of its convergents, given byqk+1 = yk+1qk + qk−1. If

+∞∑k=1

qk

yk+1

< +∞,

then there exists aT satisfying the i.d.o.c. condition, which is measure-theoretically isomorphic tothe rotation of angleγ, and hence has discrete (pure point) spectrum.

We also show

Theorem 1.3. For every quadratic irrational numberγ there exists aT 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 tendto infinity very quickly, while in the other they are eventually periodic. We do not know whetherevery complex number of modulus1 is an eigenvalue of someT .

Theorems 1.2 and 1.3 suggest that not all properties ofT can be traced back to the underlyingrotation: the irrational rotation by angleγ of Theorem 1.2 has no connection with the underlyingrotation inducingT , and in the case of Theorem 1.3 a factor ofT is a rotation with a quadraticangle, 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 sponsoredby the N.S.F. and C.N.R.S.. The third author was also supported in part by a grant from the TexasAdvanced Research Program. We are very grateful to E. Lesigne and A. Nogueira for numerouse-mail exchanges and many fruitful conversations.

2Although the result was actually established in [30], it was only first stated in the language of interval exchangetransformations 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 parameters0 < α < 1 and0 < β < 1 − α. Wenote thatT is continuous except at the pointsα andα + β.

Set

A(α, β) =1− α

1 + βand B(α, β) =

1

1 + β.

ThenT is induced by a rotation on the circle by angleA(α, β). More precisely,T is obtained fromthe 2-interval exchangeR on [0, 1) given by

(2) Rx =

{x + A(α, β) if x ∈ [0, 1− A(α, β))

x + A(α, β)− 1 if x ∈ [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 theinfinite distinct orbit condition(or i.d.o.c. for short) of Keane [21] if thetwo negative trajectories{T−n(α)}n≥0 and{T−n(α + β)}n≥0 of the discontinuities are infinitedisjoint sets. Under this hypothesis,T is both minimal and uniquely ergodic; the unique invariantprobability measure is the Lebesgue measureµ on [0, 1) (and hence(X, T, µ) is an ergodic system).

Because of the connection with the rotationR, T doesnot satisfy the i.d.o.c. condition if andonly 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 integerp, and positive integerq. In the first case,T is not minimal; inthe 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].Let I denote the open interval(0, 1), D0 ⊂ R2 the simplex bounded by the linesy = 0, x = 0, andx + y = 1, andD the triangular region bounded by the linesx = 1

2, 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 inD and is not on any of the rational linespα + qβ = p − q,pα + qβ = p− q + 1, pα + qβ = p− q− 1 then there exists a unique finite sequence of integersl0,l1, . . . , lk such that(α, β) is in H−1D whereH is a composition of the formGt ◦ F l0 ◦ G ◦ F l1 ◦G . . . ◦G ◦ F lk ◦Gs, s, t ∈ {0, 1}.

The functionH(α, β) is computed recursively as follows: we start withα(0) = α, β(0) = β.Then, given(α(k), β(k)), we have three mutually exclusive possibilities: if(α(k), β(k)) is in D, thealgorithm stops; ifα(k) < 1

2, we applyG; if 2α(k) + β(k) < 1, we applyF .

Associated to each point(α, β) ∈ D0 there is a sequence(nk, mk, εk+1)k≥1, wherenk andmk

are positive integers, andεk+1 is±1. This sequence we call thethree-interval expansionof (α, β),is a variation of thenegative slope expansiondefined in [18]; it is constructed as follows:


• For (α, β) in D we put

x0 =1− α− β

1− αand y0 =

1− 2α

1− α

and define fork ≥ 0

(xk+1, yk+1) =

({

yk

(xk+yk)−1

},{

xk

(xk+yk)−1

})if xk + yk > 1({

1−yk

1−(xk+yk)

},{

1−xk

1−(xk+yk)

})if xk + yk < 1

(nk+1, mk+1) =

(b yk

(xk+yk)−1c, b xk

(xk+yk)−1c)

if xk + yk > 1(b 1−yk

1−(xk+yk)c, b 1−xk

1−(xk+yk)c)

if xk + yk < 1

where{a} andbac denote the fractional and integer part ofa respectively. Fork ≥ 0 set

εk+1 = sgn(xk + yk − 1).

We note thatε1 is always−1, hence we ignore it in the expansion.• For (α, β) /∈ D we letH be the function above for which(α, β) ∈ H−1D and put(

α, β)

= H(α, β),

and define(nk, mk, εk+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 in D, it is a translation(taking into account the fact that the initial conditions are slightly different) of results in [18]; inthe 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, εk+1) of (α, β) is infinite.

(2) An infinite sequence(nk, mk, εk+1) is the expansion of at least one pair(α, β) defining aTsatisfying the i.d.o.c. condition, if and only ifnk andmk are positive integers,εk+1 = ±1,(nk, εk+1) 6= (1, +1) for infinitely manyk and(mk, εk+1) 6= (1, +1) for infinitely manyk.

(3) Each infinite sequence(nk, mk, εk+1) satisfying the conditions in (2) is the three-intervalexpansion of a countable family of couples(α, β), with exactly one couple in each of thedisjoint trianglesH−1D, whereH has any of the possible forms defined earlier in thissection, including the identity.

(4) A(α, β) =1

2 +1

m1 + n1 −ε2

m2 + n2 −ε3

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 thelengths of strings of consecutive(1, 1, +1).

(6) The following two properties are equivalent– in the three-interval expansion of(α, β), there existsc > 0 such that for everyM > 0

there is ak such thatmk + nk > M and

c <

∣∣∣∣mk − nk

mk + nk

∣∣∣∣ < 1− c.


– there existsc′ > 0 such that for everyε > 0, there exist integersp, q such that∣∣∣∣A(α, β)− p

q

∣∣∣∣ ≤ ε

q2

and for any integerr ∣∣∣∣B(α, β)− r

q

∣∣∣∣ >c′

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 pointx in [0, 1), we define an infinite sequence(xn)n∈N by puttingxn = i if T nx ∈ Pi,i = 1, 2, 3. This sequence, also denoted byx, is called thetrajectoryof x. If T satisfies the i.d.o.c.condition, the minimality of the system implies that all trajectories contain the same finite wordsas factors.

Let I ′ be a set of the form∩n−1i=0 T−iPki

; we sayI ′ has anameof lengthn given byk0 . . . , kn−1;note thatI ′ is necessarily an interval andk0, . . . , kn−1 is the common beginning of trajectories ofall points inI ′.

For each intervalJ , it is known (see for example [12]) that the induced map ofT on J is anexchange of three or four intervals. More precisely, there exists a partitionJi, 1 ≤ i ≤ t of J intosubintervals (witht = 3 or t = 4), andt integershi, such thatT hiJi ⊂ J , and{T jJi}, 1 ≤ i ≤ t,0 ≤ j ≤ hi − 1, is a partition of[0, 1) into intervals: this is the partition intoRokhlin stacksassociated toT with respect toJ . The intervalsJi have names of lengthhi, calledreturn wordstoJ .

In [19] we give an explicit construction of the trajectories of the points1−α and1−α− β. Asa consequence we have the following structure theorem [19].

Theorem 2.2. LetT satisfy the i.d.o.c. condition, and let(nk, mk, εk+1)k≥1, be the three-intervalexpansion of(α, β). Then there exists an infinite sequence of nested intervalsJk, k ≥ 1, whichhave namewk and exactly three return words,Ak, Bk andCk, given recursively fork ≥ 1 by thefollowing formulas

Ak = Ank−1k−1 Ck−1B

mk−1k−1 Ak−1,

Bk = Ank−1k−1 Ck−1B

mkk−1,

Ck = Ank−1k−1 Ck−1B

mk−1k−1

if εk+1 = +1, andAk = Ank−1

k−1 Ck−1Bmkk−1,

Bk = Ank−1k−1 Ck−1B

mk−1k−1 Ak−1,

Ck = Ank−1k−1 Ck−1B

mkk−1Ak−1

if εk+1 = −1.

Fork = 0, the wordsA0, B0, C0 are defined as follows.


Jk1 Jk2 Jk3

TJk1 TJk2 TJk3

T ak−1Jk1

T bk−1Jk2

T ck−1Jk3

FIGURE 1. The three Rokhlin stacks

Proposition 2.3. When(α, β) is in D, we haveA0 = 13, B0 = 2, C0 = 12. When(α, β) is not inD, we defineH and(α, β) as in the previous subsection, and twosubstitutions, σF by

1 → 12 → 213 → 31

andσG by1 → 32 → 23 → 1

and we defineσH by replacing, in the expression ofH, F by σF and G by σG. Then we haveA0 = σH(13), B0 = σH(2), C0 = σH(12).In all cases the lengths ofA0 andB0 differ by±1.

Though we shall not use this fact in the sequel, if(α, β) is inD thenJk is an interval of continuityof T n for somen such that1−α−β is in Jk andα is in T nJk. Equivalently thewk are thebispecialfactors(in the minimal subshift associated withT , see [19]) beginning in1 and ending in2. When(α, β) is not inD, let Jk be, for the three-interval exchangeT ′ defined by(α, β), the intervals ofcontinuity ofT ′n for somen, such that1− α− β is in Jk andα is in T ′nJk; let wk be the name ofJk. ThenJk is the interval whose name isσH(wk).

2.4. Rokhlin stacks. Theorem 2.2 gives an explicit construction of the Rokhlin stacks mentionedin the previous subsection.

Let wk be the name ofJk. Every trajectory underT is a concatenation of wordsAk, Bk, Ck,which we call thek-words. We say that ak-word occurs at itslegal k-placein a trajectory if it isimmediately followed by the wordwk. A concatenation ofk-words occurs at its legalk-place ifeach of itsk-words occur at their legalk-place.


We defineFAk to be set ofx ∈ X such that (in the trajectory ofx) x0 is the first letter of thek-word Ak in its legalk-place, and similarlyFBk andFCk. Note thatFAk ∪ FBk ∪ FCk = Jk,and these are exactly the three intervals of continuity of the induced map ofT onJk.

Let ak, bk, ck be the lengths ofAk, Bk andCk; it follows from Theorem 2.2 and Proposition 2.3that|ak − bk| = |a0 − b0| = 1 andck is eitherak − ak−1 = bk − bk−1 or ak + ak−1 = bk + bk−1. Inparticular, we haveck ≤ 2ak. ThenX = [0, 1) is the disjoint union ofT jFAk, 0 ≤ j ≤ ak − 1,T jFBk, 0 ≤ j ≤ bk − 1, T jFCk, 0 ≤ j ≤ ck − 1; we denote byτAk the disjoint union ofT jFAk,0 ≤ j ≤ ak − 1, and define similarlyτBk andτCk; τAk, τBk andτCk are called thek-stacks, andtheT jFAk, 0 ≤ j ≤ ak − 1 are thelevelsof the stackτAk, and similarly forB andC. The levelsare intervals of small diameter, as they have names of arbitrarily large length; hence any integrablefunctionf can be approximated (inL1 for example) by functionsfk which are constant on eachlevel of eachk-stack.

Hence, in the language offinite rank systems, see [16], the Theorem 2.2 implies that

Corollary 2.4. T is of rank at most three, generated by the stacksτAk, τBk, τCk; the recursionformulas 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 atmost four was in the folklore and is proved in [12]), but the explicit construction of Theorem 2.2is new, and from it comes all our knowledge of the system. The fact thatAk, Bk, Ck are returnwords of the same word is useful additional information, but it may be deduced from the explicitconstruction.

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 trajectoryx, the Lebesgue measureµ(τAk) is the limit whenn goes to infinity of1

ntimes the total number of indices0 ≤ i ≤ n − 1

such thatxi belongs to a wordAk in its legalk-place. Similarly forBk andCk.

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, εk+1)k≥1, and the initial lengthsa0, b0, c0. Two three-interval exchange transformationsTandT ′ with the same expansion(nk, mk, εk+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 intervalsI ′′

andI ′ such that the induced maps ofT on I ′′ and ofT ′ on I ′ are the same. Also, it is noted in [19]that, if we apply our recursion formulas, but starting fromA0 = B0, the systems we get are justirrational rotations.

3. WEAK MIXING OF T

We recall two definitions of measure-theoretic ergodic theory: if(X, S, ν) is a measure-theoreticdynamical system, we say that a real number0 ≤ γ < 1 is aneigenvalueof S (denoted additively)if there exists a nonconstantf in L2(X, R/Z) such thatf ◦ S = f + γ (in L2(X, R/Z))); f isthen aneigenfunctionfor the eigenvalueγ. As constants are not eigenfunctions,γ = 0 is not aneigenvalue ifS is ergodic.S is weak mixingif it has no eigenvalue.


3.1. The fundamental lemma. In this subsection, we use the fact that there are two Rokhlinstacks 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 transformationT 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 differencebetween continuous and non-continuous eigenfunctions: the presence of two nonempty stacks ofheights 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 thesestacks is not too small.

In the sequel||x|| denotes the distance ofx to the nearest integer.

Proposition 3.1. EveryT satisfying the i.d.o.c. condition istopologically weakly mixing: it hasno continuous eigenfunction.

Proof. Let γ be an eigenvalue with a continuous eigenfunctionf ; then, for givenε, if k is largeenough,|f(z)− f(y)| < ε (in R/Z) if z andy are inJk. We takex ∈ FAk ⊂ Jk; thenT akx ∈ Jk,hence,

||γak|| = |f(T akx)− f(x)| ≤ ε;

taking a pointx′ in FBk, we get the same relation withbk = ak ± 1, henceγ = 0, which is not aneigenvalue 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, equippedwith the product topology on{1, 2, 3}N, as the bases of thek-stacks are cylinders associated toarbitrarily long words. Note that for a three-interval exchange transformation with a permutationsuch 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 anontrivial continuous eigenfunction.

Lemma 3.2. If γ is an eigenvalue ofT , there exists a sequenceδk → 0 such that for allk

||γak|| <δk

µ(τAk),

||γbk|| <δk

µ(τBk),

||γck|| <δk

µ(τCk).

Proof. Supposef is an eigenfunction forT for the eigenvalueγ; then we have∫|f − fk| < δ′k,

with δ′k → 0 andfk is constant on every level of the stackτAk, τBk or τCk.To prove the first inequality, we define a new wordDk, and an integerk′(k) ≤ k − 1 such that• k′(k) tends to infinity whenk tends to infinity,• Dk is a concatenation ofk′-words,• Dk in its legalk′-place is a prefix ofAk,• the length ofDk is dk ≥ ak

3,

• in any trajectoryx, each occurrence of ak-word in its legalk-place is followed by anoccurrence ofDk in its legalk′-place.

The wordDk is defined as follows:


• if nk > 1 or mk > 1, we putDk = Ank−1k−1 Ck−1B

mk−1k−1 , the common prefix of the three

k-words; thusdk ≥ ak

2, and we setk′ = k − 1;

• if mk = nk = 1, letk′(k) be the largestl < k such that(nl, ml, εl+1) 6= (1, 1, +1); then therecursion formulas give eitherAk = Ck−k′

k′ Ak′, Bk = Ck−k′

k′ Bk′, Ck = Ck′ (if εk+1 = +1),or Ak = Ck−k′

k′ Bk′, Bk = Ck−k′

k′ Ak′, Ck = Ck−k′

k′ Bk′Ck−k′−1k′ Ak′ (if εk+1 = −1); and

ck′ >ak′2

. Then we takeDk = Ck−k′

k′ , from which it follows thatdk ≥ ak

3.

Because(nk, mk, εk+1) cannot be(1, 1, +1) ultimately (Proposition 2.1),k′(k) → +∞ whenk → +∞. The other properties ofDk are clear from its construction.

We callτ ′k the union of the first (or lower)dk levels ofτAk; thus, forx ∈ τ ′k, x andT akx are inthe same level of the samek′-stack.

Now if x ∈ τ ′k, fk′(Takx) = fk′(x) while f(T akx) = γak + f(x); we have∫τ ′k

|T akfk′ − γak − fk′| =∫

τ ′k

|γak| ≥ ||γak||µ(τAk)

3

and ∫τ ′k

|T akfk′ − γak − fk′| ≤∫

τ ′k

|T akfk′ − T akf |+∫

τ ′k

|fk′ − f | < 2δ′k′

which gives||γak|| < 6δ′k′

µ(τAk), hence the first inequality.

The second inequality is proved in the same way, as is the third, except that we replaceDk byCk andτ ′k by τCk in the casemk = nk = 1. �

Remark on the proof. In the proof of Lemma 3.2, we can alternatively usewk itself instead ofbuilding Dk (which is a prefix ofwk), and make similar computations, replacingdk by the lengthd′k of wk, andτ ′k by the union of the first (or lower)d′k levels ofτAk. In this case, the results followfrom the fact thatT iJk is still an interval for0 ≤ i ≤ d′k−1, and thatd′k ≥ ak−2 ([19]). However,we prefer to prove Lemma 3.2 directly from the recursion formulas.

Corollary 3.3. If for a constantc and infinitely manyk we have bothµ(τAk) ≥ c andµ(τBk) ≥ c,thenT is weakly mixing.

Proof. The first two inequalities of Lemma 3.2, together with the relationbk = ak ± 1 give ||γ|| <2cδk, henceγ = 0. �

3.2. Study of possible non weakly mixing cases.Corollary 3.3 gives some sufficient conditionsfor T to be weakly mixing. We shall now focus on the cases when these conditions arenot fulfilled,and show that this may happen only for very special sequences(nk, mk, εk+1), namely those witha subsequence on whichmk is much bigger (or smaller) thannk while outside this subsequencewe must have(nk, εk+1) = (1, +1) or (mk, εk+1) = (1, +1) (or else(nk, εk+1) = (1,−1) or(mk, εk+1) = (1,−1), these last two cases being allowed only one step beforemk is much biggeror smaller thannk).

Lemma 3.4. If T is not weakly mixing, then for every integerM there exists an integerk0(M)such that for everyk > k0(M), one of the following two assertions is satisfied:

• there existsl ≥ k + 1 such thatml ≥ Mnl, εk+2 = . . . = εl−1 = +1 if l > k + 2 and, ifl > k + 1, nk+1 = . . . = nl−1 = 1 if εl = +1, mk+1 = . . . = ml−1 = 1 if εl = −1;

• there existsl ≥ k + 1 such thatnl ≥ Mml, εk+2 = . . . = εl−1 = +1 if l > k + 2 and, ifl > k + 1, mk+1 = . . . = ml−1 = 1 if εl = +1, nk+1 = . . . = nl−1 = 1 if εl = −1.


Proof. By Corollary 3.3, ifT is not weakly mixing, then for everyδ andk sufficiently large, eitherµ(τAk) < δ or µ(τBk) < δ.

We estimateµ(τAk), using Lemma 2.5 and Theorem 2.2. Ifnk+1 > 1, wordsAk in their legalk-place cover a proportion at least nk+1−1

nk+1+mk+1+1of eachk + 1-word (recall thatck < 2ak and that,

k being large enough,bk = ak ± 1 can be treated asak) and hence

µ(τAk) ≥nk+1 − 1

nk+1 + mk+1 + 1.

If nk+1 = 1, wordsAk in their rightk-place do not appear in bothAk+1 andBk+1, and we have

µ(τAk) ≥1

mk+1 + 2µ(τAk+1)

if εk+2 = +1,

µ(τAk) ≥1

mk+1 + 2µ(τBk+1) +

1

mk+1 + 3µ(τCk+1)

if εk+2 = −1.Hence, ifµ(τAk) is small, then

• eithermk+1 is much bigger thannk+1,• or nk+1 = 1, εk+2 = +1, and 1

mk+1+2µ(τAk+1) is small,

• or nk+1 = 1, εk+2 = −1 and both 1mk+1+2

µ(τBk+1) and 1mk+1+3

µ(τCk+1) are small.

In the third case, we may havemk+2 = 1; but in that case, asεk+2 = −1, we haveck+1 ≥ ak+1,andµ(τCk+1) is at least 1

mk+2+2because 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 biggerthanmk+2.Note that in the second case, however,µ(τAk) may be small even with theml andnl are boundedfor l ≥ k + 1, provided there is a long string ofnl = 1 with εl+1 = +1 ahead.

More precisely, the fact thatµ(τAk) is small implies that

• either from l = k + 1 to somek′ ≥ l, there is a (possibly empty), string ofnl = 1,with εl+1 = +1 except maybe the last one, followed by annk′ much smaller thanmk′, (orfollowed by anmk′ much smaller thannk′ if the lastεl+1 is a−1);

• or else that froml = k + 1 to somel′ there is a very long string ofnl = 1, with εl+1 = +1.

If T is not weakly mixing, the hypotheses of Corollary 3.3 are not satisfied fork large enough.In view of the computations above, for every integerM there existsk0(M) such that, for everyk > 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, εk+2 = . . . = εk+M+1 = +1;• mk+1 = . . . = mk+M = 1, εk+2 = . . . = εk+M+1 = +1.

To prove the lemma, it remains to prove that the last two assertions alone are not sufficient toprevent weak mixing. The main argument here is that if the sequence(nk, mk, εk+1) is made onlyof long strings(nk, εk+1) = (1, +1) or (mk, εk+1) = (1, +1), then transitions between strings of(nk, εk+1) = (1, +1) and of(mk, εk+1) = (1, +1) must occur infinitely often, and these transitionswill always produce weak mixing. But, to prove this, we need some involved computations (using,as in Lemma 3.2, differences of1 between lengths of words) in the case where these transitionsare made of strings of(nk, mk, εk+1) = (1, 1, +1).


Namely, we suppose that there are arbitrarily largek which do not satisfy the first and secondassertion. Take such ak and suppose for example thatnk+1 = . . . = nl = 1, εk+2 = . . . = εl+1 =+1, l − k > M ; as we cannot have(nk, εk+1) = (1, +1) ultimately, we can choosel such thateithernl+1 > 1 or εl+2 = −1. If nl+1 > 1, we cannot haveml+1 > Mnl+1 ask would satisfythe first assertion, hence eithernl+1 > Mml+1 or ml+1 = 1, and this implies thatµ(τAl) ≥ 1

4. If

nl+1 = 1, thenεl+2 = −1 and we still cannot haveml+1 > Mnl+1; then eithernl+2 > Mml+2,but that is excluded ask would satisfy the second assertion, orml+1 = 1 andml+2 > Mnl+2, andthis implies againµ(τAl) ≥ 1

4.

But alsol − M + 3 must satisfy one of our four assertions; because of our assumptions onkandnl+1, this can only happen ifml+3−M = . . . = ml = 1, while still nk+1 = . . . = nl = 1,εk+2 = . . . = εl+1 = +1.

Suppose first that there exists somek + 1 ≤ k1 ≤ l + 2 − M such thatmk1 > 1, and take thelargest possible suchk1. Thennk1 = 1, mk1 = m > 1, εk1+1 = +1. ThenτBk1−1 has measure atleast1

4, and by Lemma 3.2 we have||γbk1−1|| < δ, with δ small if k is large.

Now, Al = (Ck1)M ′

Ak1, with M ′ ≥ M − 4, andCk1 = Ck1−1Bm−1k1−1, henceck1 > 1

2ak1 and

µ(τCk1) ≥ 12µ(τAl) ≥ 1

8. Hence by Lemma 3.2 it follows that||γck1 || < δ.

We need to show also that||γM ′ck1 || < 20δ; this is done by an argument similar to the proofof Lemma 3.2: we takeM ′′ = [M

4], andτ the set of thosex for which x0 lies in a wordCk1

in its legal k1-place, followed by at leastM ′′ words Ck1 in their legalk1-places, and estimate∫τ|TM ′′ck1fk1 − γM ′′ck1 − fk1|; as µ(τ) ≥ 1

2µ(τCk1) ≥ 1

16, we get the desired estimate for

γM ′′ck1 and hence forγM ′ck1 because of the last result.Hence, asγal is close to0 (modulo 1), we get that bothγck1 andγak1 are close to0. But

Ck1 = Ck1−1(Bk1−1)m−1 andAk1 = Ck1−1(Bk1−1)

m−1Ak1−1, hence we get that||γak1−1|| < 100δ,and hence||γ|| < 101δ.

Suppose now thatmk+1 = . . . = ml+2−M = 1. Then we cannot havenl+1 > Mml+1 orεl+2 = −1, nl+1 = 1, ml+2 > Mnl+2 as in both casesk would satisfy the first or second assertion,hencenl+1 > 1, ml+1 = 1. And l cannot satisfy the first or second assertion ask would thensatisfy it, nor the third one asnl+1 > 1. Hence we must haveml+1 = . . . = ml+M = 1,εl+2 = . . . = εl+M+1 = 0, and we can start the same reasoning withl replacingk, and them andnexchanged; but we know thatnl+1 > 0, hence we are in the case where this implies thatγ is closeto 0.

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 tobe weakly mixing; our aim now is to reduce the set of possibly non-weakly mixingT to an explicitset ofrank-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 smalleror bigger thannk, the ratios

mk(j)∧nk(j)

mk(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 thecases wheremk(j) ∧ nk(j) = 1.Note that the condition depends only on the expansion(nk, mk, εk+1)k≥1.

Theorem 3.5. LetT satisfy the i.d.o.c. condition , and let(nk, mk, εk+1)k≥1, be the three-intervalexpansion of(α, β). Then, ifT is not weakly mixing, there exists a strictly increasing sequence ofintegersk(j), j ∈ N, such that


• (Ma)+∞∑j=1

mk(j) ∧ nk(j)

mk(j) ∨ nk(j)

< +∞;

• (Mb) mk(j) 6= nk(j);• (Mc) if mk(j) > nk(j), then for everyk(j − 1) < k < k(j) − 1, εk+1 = +1; and for every

k(j − 1) < k < k(j), nk = 1 if εk(j) = +1, mk = 1 if εk(j) = −1;• (Md) if nk(j) = 1, thenmk(j+1) > nk(j+1) if εk(j)+1εk(j+1) = −1 or if k(j + 1) = k(j) + 1

andεk(j)+1 = −1, nk(j+1) > mk(j+1) otherwise;• (Mc’) if nk(j) > mk(j), then for everyk(j − 1) < k < k(j)− 1, εk+1 = +1; and for every

k(j − 1) < k < k(j), mk = 1 if εk(j) = +1, nk = 1 if εk(j) = −1;• (Md’) if mk(j) = 1, thennk(j+1) > mk(j+1) if εk(j)+1εk(j+1) = −1 or if k(j + 1) = k(j) + 1

andεk(j)+1 = −1, mk(j+1) > nk(j+1) otherwise.

Proof. As T 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’: ifk(j) is such thatnk(j) = 1 but Md is notsatisfied, or thatmk(j) = 1 and Md’ is not satisfied, then we look at the firstj′ > j such thatnk(j′)∧mk(j′) > 1 or nk(j′) = 1 and Md is satisfied, ormk(j′) = 1 and Md’ is satisfied. Such aj′ doesexist, as otherwise we would have either(nk, εk+1) = (1, +1) ultimately, or(mk, εk+1) = (1, +1)ultimately, or else we would have infinitely many situations similar to the following:nk = 1,mk = m, εk+1 = 1, mk+1 = 1, nk+1 = n, so for exampleAk = Ck−1B

mk−1, Ck = Ck−1B

mk−1Ak−1,

and

Ak+1 = (Ck−1Bmk−1)

n−1Ck−1Bmk−1Ak−1Ck−1B

mk−1.

Then, as in the proof of Lemma 3.4, we show (possibly by going to the nextmk′ > 1) thatγbk−1,γak andγak+1 are all close to0 (modulo1), and by substraction we get thatγak−1 is also close to0, and this cannot happen ifk is large enough.

And, if we deletek(j), k(j + 1), ..., k(j′ − 1) from the sequencek(n), the modified sequencek(n) keep all its other properties: this is clear for Ma and Mb, but we have to check that thek(j − 1) < k < k(j)′ − 1 do satisfy Mc and Mc’. The part of Mc and Mc’ concerningmk andnk follows from the properties ofk(j), k(j + 1), ...,k(j′ − 1) , so we just have to check theεk+1;now, if k(j − 1) + 1 < k < k(j′)− 1 andεk+1 = −1, we check that we get again a weakly mixingsituation, and conclude that this cannot happen ifk is large enough.

We have now to prove condition Ma. We estimate the measures ofτAk(j)−1 and τBk(j)−1.Suppose for examplemk(j) > nk(j); thenµ(τBk(j)−1) > 1 − δ if j is large enough. Ifnk(j) > 1,then

µ(τAk(j)−1) ≥nk(j) − 1

nk(j) + mk(j) + 1≥

nk(j)

2nk(j) + 2mk(j) + 2≥

nk(j)

2mk(j)

.

Suppose nownk(j) = 1, mk(j) = m, and suppose for example thatεk(j)+1 = εk(j+1) = +1.Because of Md and Md’, we must havenk(j+1) > mk(j+1). Let k2 be the firstk > k(j) such thatnk2 , mk2 , εk2+1 6= (1, 1, +1) ; then eithernk2 is much bigger thanmk2 , or nk2 > 1 andmk2 = 1, or


nk2 = mk2 = 1, εk2 = −1 which is excluded because of the hypothesis onεk(j+1). In each case,we haveµ(τAk2−1) ≥ 1

4. If k2 = k(j) + 1, we get

µ(τAk(j)−1) ≥1

mk(j) + 1µ(τAk(j)) ≥

1

4mk(j) + 4≥

nk(j)

5mk(j)

.

If k2 > k(j)+ 1, asεk(j)+1 = +1, we conclude like in the proof of Lemma 3.4 that bothγak(j) andγck(j) are close to0 (modulo1), which together imply thatγak(j)−1 is close to0, and that cannothappen ifj is large enough. Similar reasonings take care of the other cases fornk(j+1), mk(j+1),εk(j)+1, andεk(j+1), except that, in the casek2 > k(j) + 1 andεk(j)+1 = −1, we use the fact thatµ(τCk(j)) ≥ 1

2and

µ(τAk(j)−1) ≥1

mk(j) + 3µ(τCk(j)) ≥

1

2mk(j) + 6≥

nk(j)

3mk(j)

.

Suppose now that+∞∑j=1

mk(j) ∧ nk(j)

mk(j) ∨ nk(j)

= +∞.

Then at least one of the four series• ∑

j;mk(j)>nk(j),ak(j)−1−bk(j)−1=1

nk(j)

mk(j)

,

• ∑j;mk(j)>nk(j),ak(j)−1−bk(j)−1=−1

nk(j)

mk(j)

,

• ∑j;nk(j)>mk(j),ak(j)−1−bk(j)−1=1

mk(j)

nk(j)

,

• ∑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 thosek(j) for whichmk(j) > nk(j) andak(j)−1 − bk(j)−1 = 1.

We fix somej0; let Uj = ∪j0≤i≤jτAl(i)−1. We haveµ(Uj) → 1 when j → +∞ by Borel-Cantelli. Uj \ Uj−1 is made of thosex whose coordinatex0 lies in a wordAl(j)−1 (in its legall(j)−1-place) but not in a wordAl(i)−1 (in its legall(i)−1-place) for anyj0 ≤ i ≤ j−1; for thesei, anl(i)−1-word (in its legall(i)−1 -place) which is not anAl(i)−1 is either aCl(i)−1, or aBl(i)−1

in a string(Bl(i)−1)ml(i)−1 or (Bl(i)−1)

ml(i) inside anl(i)-word, or a single(Bl(i)−1) at the beginningof anl(i)-word. LetZj be the set of thosex whose coordinatex0 lies in a wordAl(j)−1 (in its legall(j) − 1-place) but, for anyj0 ≤ i ≤ j − 1, not in a wordAl(i)−1 (in its legall(i) − 1-place) norin a wordCl(i)−1 (in its legall(i) − 1-place), nor in a single (as defined above) wordBl(i)−1, norin the first wordBl(i)−1 of a string(Bl(i)−1)

ml(i)−1 or (Bl(i)−1)ml(i) . Then, as theml(i) are large,

µ(Zj) > 12

if j0 is large enough.Let Z ′

j = Zj \ Zj−1. Let dl(j)−1 be defined as in the proof of Lemma 3.2 andYj the set ofthe x in Z ′

j whose coordinatex0 lies in the prefix of lengthdl(j)−1 of its word Al(j)−1. Thenµ(Yj) ≥ 1

3µ(Z ′

j), andµ(∪j≥j0Yj) ≥ 16. TheYj are disjoint; and also all theT al(j)−1Yj are disjoint,

asT al(j)−1Y ′j is made ofx whose coordinatex0 lies in the samel(i)-words than thex in Yj for


j0 ≤ i ≤ j − 1, but in anl(j)− 1-word just after anAl(j)−1, and this has to be either anAl(j)−1, oraCl(j)−1, or a singleBl(j)−1, or the firstBl(j)−1 of a string.

Let γ be an eigenvalue,f its eigenfunction,fk its approximation as in Lemma 3.2,k′j = l(j)−1.We fix δ, and choosej0 such that for allj ≥ j0 we have

∫|f − fk′j

| < δ and||γbl(j)−1|| ≤ δ. As inLemma 3.2, ifx ∈ Yj, fk′j

(T al(j)−1x) = x. Then

||γ|| ≤ 6

∫∪j≥j0

Yj

|γ| = 6∑j≥j0

∫Yj

|γ| < δ + 6∑j≥j0

∫Yj

|γ(bl(j)−1 + 1)| =

δ + 6∑j≥j0

∫Yj

|γal(j)−1|.

And ∑j≥j0

∫Yj

|γal(j)−1| =∑j≥j0

∫Yj

|T al(j)−1fk′j− γal(j)−1 − fk′j

| ≤

∑j≥j0

∫Yj

|T al(j)−1fk′j− T al(j)−1f |+

∑j≥j0

∫Yj

|fk′j− f | =

∑j≥j0

∫T

al(j)−1Yj

|fk′j− f |+

∑j≥j0

∫Yj

|fk′j− f | =

∫∪

j≥j0Tal(j)−1Yj

|fk′j− f |+

∫∪j≥j0Yj

|fk′j− f | ≤ 2δ.

Hence, if the series in condition Ma diverges,T is weakly mixing. �

4. CHARACTERIZATION OF THE EIGENVALUES

LetT satisfy the i.d.o.c. condition. We are now ready to give a necessary and sufficient conditionfor a number to be an eigenvalue ofT : eitherT has no eigenvalue, or, by Theorem 3.5,T has anexplicit construction as a rank one transformation, in which case we can apply a criterion due toChoksi and Nadkarni [11], which translates into a condition on the(nk, mk, εk+1) and(a0, b0, c0),to find the eigenvalues (which may still not exist).

Theorem 4.1. Let T satisfy the i.d.o.c. condition and conditions Ma to Md’ with the sequencek(j). Thenγ is an eigenvalue ofT if and only if both the following conditions are satisfied:

• (Eγ)+∞∑j=1

pk(j)||hjγ|| < +∞,

pk(j) = mk(j) ∨ nk(j), hj is the length ofAk(j)−1 if nk(j) > mk(j) and the length ofBk(j)−1

otherwise;• (Fγ)

+∞∑j=1

Qj(γ) < +∞,


where

Qj(γ) = 1−|Q′

k(j+1)−1|t′(k(j + 1)− 1)

with the following definitions:

• Q′k(j) = 1 andt′(k(j)) = 1 (henceQj = 0 wheneverk(j + 1) = k(j) + 1);

• if k(j) + 2 ≤ k(j + 1), andnk(j+1) > mk(j+1) with εk(j+1) = +1, or mk(j+1) > nk(j+1)

with εk(j+1) = −1

Q′k(j)+1 = 1 + e2πiεk(j)+1ak(j)−1γe−2πiak(j)γ

e−2πink(j)+1ak(j)γ − 1

e−2πiak(j)γ − 1,

t′(k(j) + 1) = nk(j)+1 + 1;

• if k(j) + 2 ≤ k(j + 1), andmk(j+1) > nk(j+1) with εk(j+1) = +1, or nk(j+1) > mk(j+1)

with εk(j+1) = −1

Q′k(j)+1 = e2πiεk(j)+1bk(j)−1γe−2πimk(j)+1bk(j)γ +

e−2πimk(j)+1bk(j)γ − 1

e−2πibk(j)γ − 1,

t′(k(j) + 1) = mk(j)+1 + 1;

• if k(j) + 1 < k ≤ k(j + 1) − 1 andnk(j+1) > mk(j+1) with εk(j+1) = +1, or mk(j+1) >nk(j+1) with εk(j+1) = −1,

Q′k = e−2πi(ak−1−ak−2)γ e−2πinkak−1γ − 1

e−2πiak−1γ − 1Q′

k−1 + Q′k−1 − e−2πi(ak−1−ak−2)γQ′

k−2,

t′(k) = (nk + 1)t′(k − 1)− t′(k − 2);

• if k(j) + 1 < k ≤ k(j + 1) − 1 andmk(j+1) > nk(j+1) with εk(j+1) = +1, or nk(j+1) >mk(j+1) with εk(j+1) = −1,

Q′k =

e−2πimkbk−1γ − 1

e−2πibk−1γ − 1Q′

k−1 + e−2πi(mkbk1−bk−2)γ(Q′

k−1 −Q′k−2),

t′(k) = (mk + 1)t′(k − 1)− t′(k − 2).

Proof. We suppose Ma to Md’ are satisfied. Ifnk(j) > mk(j), we putHj = Ak(j)−1, hj = ak(j)−1,H ′

j = Bk(j)−1. The total proportion of wordsH ′j in everyk(j)-word, and hence inHj+1, is at most

mk(j)

mk(j)+nk(j)−1≤ mk(j)

nk(j)and the total proportion of wordsCk(j)−1 in everyk(j)-word, and hence in

Hj+1, is at most 1mk(j)+nk(j)−2

≤ 2mk(j)

nk(j); the corresponding relations hold fornk(j) < mk(j), with

Hj = Bk(j)−1, hj = bk(j)−1, H ′j = Ak(j)−1, and them andn exchanged.

The system(X, T, µ) is then ofrank oneas the sequence of stacksτHj generate the wholespace, see for example [16] for precise definitions. The formulas in Theorem 2.2 allow us to writeHj under the form

Sj,0Hj−1Sj,1Hj−1 . . . Sj,t(j)−1Hj−1Sj,t(j)

whereSj,r is either empty or a concatenation of wordsH ′j−1 andCk(j−1)−1. Let sj,r be the length

of Sj,r. Because of the computations above, the total proportion of wordsSj,r in Hj is

1

hj

t(j)∑r=0

sj,r ≤ 3mk(j−1) ∧ nk(j−1)

mk(j−1) ∨ nk(j−1)


and is the general term of a convergent series.

We build a rank one system(X ′, T ′, µ′) by cutting and stacking in the following way, see [16]:we start from a setE with the required measure, which is cut intoH0 equal parts to make thefirst stack. To get thej-stack, we cut thej − 1-stack intot(j) columns, addsj,r spacers(that ispieces ofEc with the required measure) above ther-th column, and alsosj,0 spacers under the firstcolumn, and stack these columns by putting thet(j)-th above thet(j)− 1-th . . . above the first.T ′

is the transformation that sends each point in a stack, except those in the top level, to the point justabove. We build a measure-theoretic isomorphic between(X, T, µ) and(X ′, T ′, µ′), by sendingthej-th level ofτHk to thej-th level of thek-th stack forT ′: it is consistent by construction, and isdefined almost everywhere because of the condition on1

hj

∑t(j)r=0 sj,r. Henceforth we shall identify

(X, T, µ) and(X ′, T ′, µ′).

The eigenvalues of such a system are computed in [11]. Theorem 4 of [11] states that, providedsj,0 = 0 andsj,t(j) = 0 for all j, γ is an eigenvalue if and only if

+∞∑j=1

(1− 1

t(j)2|Pj(e

2πiγ)|2)

< +∞,

where

Pj(z) =

t(j)−1∑r=0

z−vj,r

and

vj,r = rhj−1 +r∑

r′=1

sj,t(j)−r′ .

We shall now translate this criterion in our setting. First, as indicated in [11], we can change thesystem so thatsj,t(j) = 0 for all j: in the construction, we replacesj,t(j) by 0, and, for1 ≤ r < t(j),sj,r by s′j,r = sj,r +

∑j−1j′=1 sj′,t(j′) (sj,0 is not changed). The new system is still, by the same trick as

above, measure-theoretically isomorphic toT , and hasj-stacks of heighthj = hj −∑j−1

j′=1 sj′,t(j′).

We check then that Choksi-Nadkarni’s criterion written with thes′j,r andhj is exactly the same asthe one written with thesj,r andhj, as thet(j) and thevj,r are not changed, and hence we can justsuppress the conditionsj,t(j) = 0. Similarly, we can suppress the conditionsj,0 = 0 by changingthes′j,r to s′j,r +

∑j−1j′=1 sj′,0 for 1 ≤ r ≤ t(j)− 1, and keep the same formula.

Suppose for exampleHj andHj+1 areA-words, andεk(j+1) = +1. To get our parameters, wewrite first the formulas

Ak(j) = Ank(j)−1

k(j)−1 Ck(j)−1Bmk(j)−1

k(j)−1 Ak(j)−1,

Ck(j) = Ank(j)−1

k(j)−1 Ck(j)−1Bmk(j)−1

k(j)−1 ,

or

Ak(j) = Ank(j)−1

k(j)−1 Ck(j)−1Bmk(j)

k(j)−1,

Ck(j) = Ank(j)−1

k(j)−1 Ck(j)−1Bmk(j)

k(j)−1Ak(j)−1,


and replace in them theBk(j)−1 andCk(j)−1 by strings of spacers (of the same length as the wordthey replace). Without changing the system up to measure-theoretic isomorphism (as the pro-portion we change is smaller than

mk(j)

nk(j), which is the general term of a convergent series), we

replace also the wordAk(j)−1 by a string of spacers, when this wordAk(j)−1 appears at the end ofa k(j)-word (this is just a cosmetic modification to make the formulas a little nicer). Then, eitherk(j) = k(j + 1)− 1, or, for everyk(j) < k ≤ k(j + 1)− 1, we haveAk = Ank−1

k−1 Ck−1Ak−1; Ck−1

is written above fork − 1 = k(j), while, fork − 1 > k(j) (if there are suchk), Ck−1 is justAk−1

deprived of its last wordAk−2; we use these formulas, taking into account the modifications wemade toAk(j andCk(j) and without further modifications. Applying the definitions, we get that

Pj+1(e2πiγ) =

e−2πi(nk(j)−1)hjγ − 1

e−2πihjγ − 1Q′

k(j+1)−1

andt(j + 1) = (nk(j) − 1)t′(k(j + 1)− 1),

with the claimed values of theQ′ andt′. A similar reasoning takes care of the other possibilitiesfor Hj, Hj+1 andεk(j+1) (whenHj is aB-word, we replace by spacers theAk(j)−1 andCk(j)−1,and also, for cosmetic reasons, one wordBk(j)−1 in each string(Bk(j)−1)

mk(j) appearing in ak(j)-word).

Let Qj be as claimed and

Rj = 1− 1

nk(j) − 1

∣∣∣∣e−2πi(nk(j)−1)hjγ − 1

e−2πihjγ − 1

∣∣∣∣ .

If γ is an eigenvalue, then1 − (1 − Qj)2(1 − Rj)

2 = zj where∑+∞

j=1 zj < +∞. As bothQj

andRj are real numbers between zero and one, this impliesQj andRj are close to zero, andzj isequivalent to2Rj+2Qj. SinceRj is close to zero, we deduce that bothe−2πihjγ ande−2πi(nk(j)−1)hjγ

are close to one (of course, this can also be proved directly as in Lemma 3.2 and Lemma 3.4) andthat

||(nk(j)−1)hjγ||(nk(j)−1)||hjγ|| is close to one. Hence, by developing the exponentials at order2 we get that

Rj is equivalent to12nj||hjγ||. Hence, ifγ is an eigenvalue, the two series stated in this theorem

converge. Conversely, ifnj||hjγ|| andQj are close to zero, we get thatRj is close to zero andequivalent to1

2(nk(j)−1)||hjγ||, and that1−(1−Qj)

2(1−Rj)2 is equivalent to2(Qj +Rj), hence,

if the two series in this theorem converge, the series in Choksi-Nadkarni’s criterion converges andγ is an eigenvalue. �

Corollary 4.2. T is weakly mixing if and only if, either Theorem 3.5 is not satisfied, or Theorem3.5 is satisfied with a sequencek(j) and, for that sequence and everyγ, Eγ or Fγ is not satisfied.

Proof. Immediate. �

The conditionFγ being a little cumbersome, we note that it implies a simpler necessary con-dition, which will be used to give new examples of weakly mixingT , or a simpler sufficientcondition, which will be used to give examples ofT with nontrivial eigenvalues.

Corollary 4.3. SupposeT satisfies conditions Ma to Md’ with the sequencek(j); let pk = mk∨nk;let gk be the length ofAk−1 if k = k(j) and nk(j) > mk(j), or if k(j) < k < k(j + 1) andnk(j+1) > mk(j+1) with εk(j+1) = +1, or if k(j) < k < k(j + 1) and mk(j+1) > nk(j+1 with


εk(j+1) = −1, and the length ofBk−1 otherwise.

If 0 < γ < 1 is an eigenvalue ofT , then Eγ is satisfied and

pk||gkγ|| → 0

whenk → +∞.

If 0 < γ < 1 and+∞∑k=1

pk||gkγ|| < +∞,

γ is an eigenvalue ofT .

Proof. The first assertion comes from the fact that, ifγ is an eigenvalue,pkγ is close to0 modulo1, as in the proof of Lemma 3.2 and Lemma 3.4. Conversely, if the second assertion is true, Eγ

holds, as does Fγ because, after equivalence arguments on quantities such ase−2πi(nk+2)ak−1γ−1

e−2πiak−1γ−1, we

can bound by recursion theQ′k, while Qj is bounded by 2

∑k(j+1)−2k=k(j)−1 pk||gkγ||. �

5. EXAMPLES OF WEAKLY MIXING T

In every example of weakly mixingT which we know, our criterion can be used to re-prove theweak mixing in a very quick way. And it allows us to find new examples.

5.1. Katok - Stepin’s examples.Katok and Stepin [20] showed that if there existsc > 0 suchthat for everyε > 0, there exist integersp, q with∣∣∣∣A(α, β)− p

q

∣∣∣∣ ≤ ε

q2

and for any integerr ∣∣∣∣B(α, β)− r

q

∣∣∣∣ >c

q,

thenT is weakly mixing. Now, by assertion 6 of Proposition 2.1, and by the computations at thebeginning of the proof of Lemma 3.4, the weak mixing in these cases follows from Corollary 3.3.

5.2. If A(α, β) has bounded partial quotients. In the unpublished [8], it is shown that when-everA(α, β) has bounded partial quotients,T is weakly mixing. This follows from assertion 5 ofProposition 2.1 and Lemma 3.4. This category includes the examples with nondiscrete spectrumin [4], which are in fact weakly mixing.

The examples in§5.1 and§5.2 can be said to be weakly mixinga la Chacon, i.e. the eigenvaluesare killed by the occurrence of a shift of one (equivalently, a spacer) above a column of positivemeasure, like in Chacon’s map [9]. Because of the above examples, we have that,for every irra-tional α′, there exists a setM(α′) of measure one such that, ifβ′ is in M(α′), the transformationT such thatA(α, β) = α′ andB(α, β) = β′ is weakly mixing. Consequently, for a set of(α, β) ofmeasure one, the associatedT is weakly mixing, as was known since [20].


5.3. Del Junco’s examples.Starting with an irrational0 < λ < 1, we putα = 13, β = 1

3− 2λ

3

if λ < 12, β = 1 − 2λ

3if λ > 1

2. Then it is shown in [14] that the three-interval exchange

transformationT is weakly mixing.This can also be shown by computing the three-interval expansion of(α, β): if λ > 1

2, (α, β)

is in the preferred triangleD, and we check thatx1 − y1 = 1, hencem1 = n1 + 1, x2 = y2, andmk = nk for everyk ≥ 2. The caseλ < 1

2splits into a countable number of subintervals according

to the value of(α, β), and we check that in each casex1 − y1 is an integer andmk = nk for everyk ≥ 2. Hence the weak mixing comes from Lemma 3.4.

Note that in all casesB(α, β) = 32A(α, β), hence we are never in Katok - Stepin’s situation.

5.4. New examples not satisfying Lemma 3.4.Consider an expansion made of unbounded stringsof (1, 1, +1) separated by isolated(1, 2, +1) and(2, 1, +1), each of them infinitely many times.

Then the weak mixing comes from the presence of an isolated spacer above a column of smallmeasure. This can be called weak mixinga la del Junco - Rudolph[15]; thus the examples of delJunco are, according to the continued fraction expansion ofλ, weakly mixing eithera la Chaconor a la del Junco - Rudolph.

5.5. New examples not satisfying conditions Ma to Md’.Consider an expansion withmk = 2,nk = k for everyk, εk+1 being arbitrary. Then the weak mixing comes from a Borel-Cantelliargument which seems completely new.

5.6. New examples not satisfying condition Eγ for any γ. Consider an expansion and(a0, b0, c0)such thatmk = ak−1, nk > k2mk, εk+1 = +1; then conditions Ma to Md’ are satisfied withk(j) = j, but asak+1 = (mk+1 + nk+1)ak − ak−1 − (mk+1 − 1)ek, whereek = ak − bk = 1 as alltheεk+1 are+1, Eγ implies thatγ = 0, andT is weakly mixing.

5.7. New examples not satisfying condition Eγ or Fγ for any γ. Consider an expansion suchthat a0 andc0 are even (for example, take(α, β) ∈ D), andm2k+1 = n2k+1 = 1, εk+1 = +1,m2k = 2, n2k = k2. Then this satisfies Ma to Md’ withk(j) = 2j, and we have

A2k = Ak2−12k−1C2k−1B2k−1A2k−1,

A2k+1 = Ak2−12k−1C2k−1B2k−1A

k2−12k−1C2k−1B2k−1A2k−1,

C2k+1 = Ak2−12k−1C2k−1B2k−1,

a2k+1 = (2k2 + 2)a2k−1 − 2a2k−3 − 2.

Hence Eγ implies that2γ is an integer, but asa2k is odd,γ = 12

is excluded by Fγ which impliesthat||a2kγ|| → 0, hence we have weak mixing.

6. EXAMPLES OF T WITH EIGENVALUES

The existence of non weakly mixingT (except in the trivial case when it does not satisfy thei.d.o.c. condition) follows from [30], see the discussion below. Here we re-prove in a few linesthe results of Veech, and give examples ofT (satisfying the i.d.o.c. condition) with irrationaleigenvalues. The idea is that, when conditions Ma to Md’ are satisfied, the system is of rank one,and generated by the stacks (for example)τAk, and the information on the eigenvalues is given bytheir heightsak. Using the latitude we have in choosing the(mk, nk, εk+1), while ensuring that Mato Md’ are satisfied, we can build examples with specific properties. In Proposition 6.1 theak areall multiples of an integerp; in Theorem 6.3 theak are the convergentsqk of some irrationalγ (in


a very particular class, as it has very large partial quotients), and this gives, by identification of thegenerating Rokhlin stacks, a measure-theoretic isomorphism betweenT and the rotation of angleγ; in Proposition 6.4 and Theorem 6.5 theak (or else, alternativelyak andbk) are a subsequenceof the convergentsqk of some irrationalγ (in wider classes, in particular the partial quotients maybe periodic), and this givesγ as an eigenvalue ofT .

6.1. Rational eigenvalues.

Proposition 6.1. For every rational number0 < γ < 1, there exists aT satisfying the i.d.o.c.condition, with eigenvalueγ.

For every integerp, there exists aT satisfying the i.d.o.c. condition, which has ap-adic odometer(see[16] for a definition) as a factor.

For every rational0 < γ < 1, for everyα′ with unbounded partial quotients, there exists acontinuumK(α′, γ) such that wheneverβ′ ∈ K, the transformationT such thatA(α, β) = α′ andB(α, β) = β′ satisfies the i.d.o.c. condition and hasγ as an eigenvalue.

Proof. It is enough to prove the first assertion forγ = 1p

with p an integer. We define a transforma-tion T such that for allk mk > 1 andnk > k2mk ; hence Theorem 3.5 is satisfied withk(j) = jfor all j. We choose for example(α, β) ∈ D, εk+1 = +1 for all k ≥ 2. We have then

ak+1 = (mk+1 + nk+1)ak − ak−1 −mk+1 + 1

with a0 = 2, a1 = 2n1 + m1 + 1; hence we may choosem1, n1, m2, n2 so thata1 anda2 aremultiples ofp, and thenmk+1 such thatmk+1 − 1 is a multiple ofp for everyk ≥ 3; henceak is amultiple ofp for everyk large enough.

Then we can apply Theorem 4.1, or re-prove it in this particular case by building an eigen-function for the eigenvalue1

p: we put φk(x) = j

pif x lies in therp + j-th level of τAk, for

0 ≤ j ≤ p − 1, 0 ≤ rp + j ≤ ak − 1. The hypotheses ensure thatT has rank one, with anexplicit way to buildτAk+1 by cutting and stacking fromτAk: we cut thek-stack intonk+1 equalcolumns, stack these columns by putting thenk+1 − 1-th above thenk+1 − 2-th . . . above the first,addmk+1ak−ak−1−mk+1+1 spacers above thenk+1−1-th column, and thenk+1-th column abovethe spacers. This and the fact that

∑+∞k=1

mk

nk< +∞ ensure that theφk converge inL2(X, R/Z) to

a functionφ, which satisfiesTφ = 1p

+ φ.

To get ap-adic odometer as a factor, we do the same construction, but choose themk andnk

such that, for eachr, ak is a multiple ofpr for k larger than somek(r).

We prove the third assertion for13≤ α′ ≤ 1

2; we expand it in the form

α′ =1

2 +1

r1 −ε2

r2 −ε3

. . .

with ri ≥ 2 andεi+1 = ±1; there are many ways to do this, as at each stage we have two choices,one withri = r andεi = −1 and one withri = r + 1 andεi = +1 (except ifr = 1, where the only


possible choice isri = 2, εi+1 = +1). We fix some integerszk such that+∞∑k=1

1

zk

< +∞

and expandA(α, β) with the following choices: from the beginning, we chooseεi+1 = +1 until,by this process, we get a string ofri = 2, εi+1 = +1 of length at leastz1; then (if this happens) wereplace the choice just before this string byεi+1 = −1, and this allows us to replace the string byanri ≥ z1, εi+1 = +1 (see [18]). Then we continue the choiceεi+1 = +1 until we get a string ofri = 2, εi+1 = +1 of length at leastz2, make the same modifications if this happens and continuein the same way. Now, any expansion ofα′, which has unbounded partial quotients, must haveeither unboundedri, or unbounded strings ofri = 2, εi = +1, by the same proof as assertion 5 ofProposition 2.1; it follows that in our expansion there is a sequencek(j) such thatrk(j) ≥ zj, andεk+1 is always+1 except maybe fork = k(j)− 1.

Let theεk+1 be as above; we choose now, for eachk(j), integersmk(j) ≥ 2 andnk(j) ≥ 2 suchthatrk(j) = mk(j)+nk(j) and either

nk(j)

mk(j)≥ 1

2zj or

mk(j)

nk(j)≥ 1

2zj, in such a way thatmk(j+1)−nk(j+1)

has the same sign asmk(j)−nk(j) if εk(j+1) = +1, the opposite sign otherwise ; fork(j)+1 ≤ k ≤k(j +1)−1 we put eithermk = 1, nk = rk−1, ormk = 1, nk = rk−1, in order to satisfy Mc andMc’. And we can choose the values ofmk(j) andnk(j) to ensure that bothk(j)-words used at thenext stage (eitherak(j) andck(j) or bk(j) andck(j)) are multiples ofp. Hence conditions Ma to Md’are satisfied, and the recursion formulas ensure thatgk is a multiple ofp for all k large enough;hence Corollary 4.3 ensures that1

pis an eigenvalue of the correspondingT .

Now, our expansion being given, we choose a corresponding(α, β) in the preferred triangleD,by assertion 3 of Proposition 2.1. Then, by construction and by assertion 4 of Proposition 2.1 itsatisfiesA(α, β) = α′. As we can choose from a continuum of sequences(nk, mk, εk+1), the resultis true for a continuum ofβ′ = B(α, β).

To get the result for other values ofα′, we choose one of the several possible trianglesH−1Dcompatible with the value ofα′, and work in this triangle: we choose a suitable(nk, mk, εk+1),knowing the values ofa0, b0, c0 imposed by the triangle, and all these will give ourβ′. �

Remarks. The same method allows us to build aT satisfying the i.d.o.c. condition, with acountable family of prescribed rational eigenvalues1

pn, or with a countable product ofpn-adic

odometers as a factor.The first and third assertions of this proposition were proved by Veech forp = 2; the result

is published in a different form [30]: Veech proved that for everyα′′ with unbounded partialquotients andβ′′ in a continuumK ′′(α′′), the map taking value1 on [0, β′′) and−1 on [β′′, 1) is acoboundary for the rotation of angleα′′, and this implies immediately that1

2is an eigenvalue for the

induced mapS of this rotation on[β′′, 1). The fact thatS is indeed a mapT satisfying the i.d.o.c.condition is mentioned first in [23], where the powerS2 provides an example of non-uniquelyergodic interval exchange transformation satisfying the i.d.o.c. condition; it is also mentioned in[34], 1.8. The results of Veech were generalized to the casep ≥ 2 by Stewart [29].

A by-product of the proof allows us to distinguish between two three-interval exchange trans-formations with the same expansion.

Proposition 6.2. There exist two three-interval exchange transformations with the same expansion(nk, mk, εk+1)k≥1, but with different eigenvalues, and hence they are not measure-theoreticallyisomorphic.


Proof. Choose an expansion such that for everyk ≥ 1 εk+1 = +1, nk > k2mk, mk andnk are odd.If we start froma0 = c0 = 2, b0 = 1, which happens if(α, β) is in D, we get that all theak are

even, and we apply the proof of the first assertion of Proposition 6.1 to get that12

is an eigenvalue.If we start froma0 = c0 = 3, b0 = 2, which happens if(α, β) is in F−1D, we get that all thea2k

are odd, and, becauseE 12

is not satisfied,12

is not an eigenvalue. �

6.2. Irrational eigenvalues.

Theorem 6.3. Let 0 < γ < 1 be an irrational number,[0; y1, y2 . . .] its usual continued fractionexpansion, andqk, k ≥ 1 the denominators of its convergents, given byqk+1 = yk+1qk + qk−1. If

+∞∑k=1

qk

yk+1

< +∞,

there exists aT , satisfying the i.d.o.c. condition, which is measure-theoretically isomorphic to therotation of angleγ, and hence has discrete (pure point) spectrum.

Proof. We define a transformationT such thata1 = q1, a2 = q2, ε2 = ε3 = +1, and, for allk ≥ 2,εk+2 = +1, mk+1 = qk − 2qk−1 + 1, nk+1 = yk+1 −mk+1 + 1. Then we have

ak+1 = (mk+1 + nk+1)ak − ak−1 −mk+1 + 1

andak = qk for everyk.We have

∑+∞k=1

mk

nk< +∞, henceT has rank one, and we do not change it up to measure-

theoretic isomorphism by replacing the last wordAk in Ak+1 by a string of spacers. Hence, to getthek + 1-stackτAk+1 we cut thek-stack intonk+1 − 1 equal columns, stack these columns byputting thenk+1 − 1-th above thenk+1 − 2-th . . . above the first, and add(mk+1 + 1)ak − ak−1 −mk+1 + 1 spacers above thenk+1 − 1-th column.

For the rotation of angleγ, the standard Sturmian trajectories (see [17] for example) are concate-nations of wordsA′

k andC ′k with C ′

k+1 = A′k andA′

k+1 = (A′k)

yk+1C ′k. As

∑+∞k=1

qk−1

yk+1qk< +∞,

we can make the following standard rank one construction: thek-stack has heightqk and, to getthek + 1-stack, we cut thek-stack intoyk+1 equal columns, stack these columns by putting theyk+1-th above theyk+1 − 1-th . . . above the first, and addqk−1 spacers above theyk+1-th column.

Hence if, in the construction of the rotation, we replace the lastyk+1 − nk+1 + 1 = qk −2qk−1 columns of thek-stack by strings of spacers of lengthqk, we get the construction ofT .As

∑+∞k=1

qk

yk+1< +∞, the two systems are measure-theoretically isomorphic, as in the proof of

Theorem 4.1. �

Remarks. Theorem 6.3 provides affirmative answers to both parts of Question 1.1, see theintroduction. An alternative method of answering the first of the two questions is to buildα′′

andβ′′ such that the map taking value1 on [0, β′′) ande2πiγ on [β′′, 1) is a coboundary for therotation of angleα′′. This approach was initiated by Merrill in [25] who builtquasi-coboundaries(rather than coboundaries) having the required properties, and is being extended by Parreau (workin progress), to give examples ofT with irrational eigenvalues.

In the last proof, we could also takemk+1 = rk+1qk−2qk−1 +1 andnk+1 = yk+1−mk+1 +rk+1

for any sequencerk such that∑+∞

k=1rk+1qk

yk+1< +∞, for example any bounded sequencerk. Now, if


we take(α, β) in D, a non-standard continued fraction expansion ofA(α, β) is

A(α, β) =1

2 +1

m1 + n1 −1

m2 + n2 −1

. . .

by assertion 4 of Proposition 2.1 andmk + nk = yk + rk. We see that for a givenα′ whose partialquotients grow fast enough, there is a continuum of numbers0 < γ < 1 such that aT induced bythe rotation of angleα′ is isomorphic to the rotation of angleγ. In particular, some of the possibleγare rationally independent ofα′. All the γ andα′ are Liouville numbers, and hence transcendental.

Proposition 6.4. Let0 < γ < 1 be an irrational number,[0; y1, y2 . . .] its usual continued fractionexpansion,qk, k ≥ 1 the denominators of its convergents. If there exists a sequence of numberszj

and a strictly increasing sequence of integersw(j) such that

•∑+∞

j=11zj

< +∞,

• yw(j)+1 ≥ z2j ,

• zjq2w(j) ≤ qw(j+1) ≤

yw(j)+1

zjq2w(j),

there exists aT satisfying the i.d.o.c. condition, which hasγ as an eigenvalue.

Proof. We define a transformationT satisfying the conditions in Theorem 3.5 withk(j) = j, andaj = qw(j). We choose all theεk+1 to be+1. We must findmj+1 andnj+1 such that

qw(j+1) = (mj+1 + nj+1)qw(j) − qw(j−1) −mj+1 + 1.

We writeqw(j+1) = Nqw(j) + l, 0 ≤ l ≤ qw(j) − 1, and takemj+1 = 2qw(j) − qw(j−1) − l + 1andnj+1 = N −mj+1 + 2. Thenmj+1 is betweenqw(j) − qw(j)−1 and2qw(j); hencemj+1 ≥ 1 andthe lower bound onqw(j+1) ensures thatnj+1 >

zj

2mj+1. The upper bound onqw(j+1) ensures that

nj+1 is at mostqw(j+1)

qw(j)≤ yw(j)+1

zjqw(j) and hence nj+1

qw(j)+1is at most 1

zj.

Hence we have+∞∑j=1

mj+1

nj+1

< +∞

and+∞∑j=1

nj+1

qw(j)+1

< +∞,

while

||qw(j)γ|| ≤1

qw(j)+1

.

We can then apply Theorem 4.1, or re-prove it in this particular case by building an eigenfunctionfor the eigenvalueγ: we putφk(x) = pγ if x lies in thep-th level ofτAk, for 0 ≤ p ≤ ak − 1. Thehypotheses ensure thatT has rank one, with an explicit way to buildτAk+1 by cutting and stackingfrom τAk as in the proof of Proposition 6.1, and this ensures that theφk converge inL2(X, R/Z)to a functionφ, which satisfiesTφ = γ + φ. �


Remarks. The same method also works if we putaj = pjqw(j), wherepj is an integer which issmall compared toqw(j). In this way, we can build aT having both an irrational eigenvalue (hencean irrational rotation as a factor) and an odometer (or a product of odometers) as a factor.

The possibleγ we get under the hypothesis of Proposition 6.4 have unbounded partial quo-tients (with a subsequence going to infinity quickly, but also possibly with bounded infinite subse-quences), although not everyγ with unbounded partial quotients satisfies this hypothesis.

Theorem 6.5. For every quadratic irrational number0 < γ < 1 there exists aT satisfying thei.d.o.c. condition, with eigenvalueγ.

Proof. We define a transformationT satisfying Theorem 3.5 withk(j) = j, andhj = qw(j), wherehj = aj andh′j = bj if mj+1 < nj+1, hj = bj andh′j = aj if mj+1 > nj+1. We choose all theεk+1

to be−1; then we have

hj+1 = (pj+1 + p′j+1)hj + hj−1 − p′j+1ej,

whereej = hj − h′j = ±1, pj+1 + p′j+1 = mj+1 + nj+1, and, depending on which words havelengthhj andhj+1, p′j+1 is eithermj+1 ∧ nj+1 or (mj+1 ∧ nj+1)− 1.

Let yk andqk be the partial quotients and the denominators of the convergents ofγ, andt be theperiod of the sequence(yk). For everyk large enough, we check that

qkt+t = Rqkt + (−1)t−1qkt−t,

R being a fixed positive integer. We have then, ift is even,

qkt+lt ≡ −qkt−lt mod qkt

for 0 ≤ l ≤ k − 1, and, ift is odd,

qkt+lt ≡ (−1)l+1qkt−lt mod qkt

for 0 ≤ l ≤ k − 1. Let r = 95; we have always

q[kr]t ≡ xkt mod qkt

where|xkt| ≤ q kt4

. We also have thatqk

cνk is close to one whenk is large enough, for some constantc and some real numberν > 1.

We start fromw(1) large enough and definew(j) by w(j) = w′(j)t, w′(j + 1) = [rw′(j)]. Wechoosemj+1 andnj+1 such thatej is always1 (this is done by alternatingm > n andn > m, sothathj will be alternativelya or b, while the sign ofaj− bj alternates because theεj+1 are−1). Wewrite qw(j+1) = Nqw(j) + l, with |l| ≤ |xw(j)| and takep′j+1 = qw(j−1) − l, pj+1 = N − p′j+1. Then

mj+1 ∧ nj+1 is betweenqw(j−1) − qw(j)4

andqw(j−1) + qw(j)4

, hence between1 and2q 3w(j)5

< ν2w(j)

3 ;

andmj+1 ∨ nj+1 is betweenν3w(j)

4 andν9w(j)

10 .Hence we have

+∞∑j=1

mj+1 ∧ nj+1

mj+1 ∨ nj+1

< +∞

and+∞∑j=1

mj+1 ∨ nj+1

qw(j)+1

< +∞,


while

||qw(j)γ|| ≤1

qw(j)+1

.

We conclude as in Proposition 6.4. �

Remark. In the examples of Theorem 6.5, if we take(α, β) in D, the (usual) continued fractionapproximation ofA(α, β) has partial quotients tending to infinity (they are ultimatelymk + nk

asεk+1 = −1), while γ has periodic partial quotients, and for example may be the golden rationumber.

Open questions

• Can any irrational number be an eigenvalue of a transformationT satisfying the i.d.o.c.condition?

• Can a transformationT satisfying the i.d.o.c. condition have two rationally independentirrational eigenvalues?

• Can a transformationT satisfying the i.d.o.c. condition be measure-theoretically isomor-phic to the Cartesian product of two irrational rotations?

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