Pinned Repetitions in Symbolic Flows: Preliminary Results
aa r X i v : . [ m a t h . D S ] A ug PINNED REPETITIONS IN SYMBOLIC FLOWS:PRELIMINARY RESULTS
MICHAEL BOSHERNITZAN AND DAVID DAMANIK
Abstract.
We consider symbolic flows over finite alphabets and study cer-tain kinds of repetitions in these sequences. Positive and negative results forthe existence of such repetitions are given for codings of interval exchangetransformations and codings of quadratic polynomials. Introduction
Fix some finite alphabet A and consider the compact space A Z of two-sidedsequences over this alphabet. Here we endow A with the discrete topology and A Z with the product topology.We want to study repetitions in sequences ω ∈ A Z that either begin at the originor are centered at the origin and hence are “pinned.” Moreover, we also want tostudy how many times arbitrarily long subwords are repeated in this way. Thus,for ω ∈ A Z , we define R n ( ω ) = 1 + 1 n sup { m : ω k = ω k + n for 1 ≤ k ≤ m } ,T n ( ω ) = 1 + 1 n sup { m : ω k = ω k + n and ω n +1 − k = ω − k for 1 ≤ k ≤ m } , and R ( ω ) = lim sup n →∞ R n ( ω ) ,T ( ω ) = lim sup n →∞ T n ( ω ) . For definiteness, we will declare sup ∅ = 0. Notice that R ( ω ) , T ( ω ) may be infiniteand that we always have T ( ω ) ≤ R ( ω ).Repetitions that begin at the origin have been studied, for example, by Berth´eet al. [1]. (In their terminology, R ( ω ) = ice( ω ), the initial critical exponent of ω ).Repetitions that are centered at the origin are of interest in the study ofSchr¨odinger operators; compare the survey articles [4, 5]. As explained there, if T ( ω ) is sufficiently large, one can prove results about the continuity of spectral mea-sures of an associated Schr¨odinger operator using the Cayley-Hamilton theorem.Denote the shift transformation on A Z by S , that is, ( Sω ) k = ω k +1 . A subshiftΩ is a closed, S -invariant subset of A Z . A subshift Ω is called minimal if the S -orbitof every ω ∈ Ω is dense in Ω. We denote by W ( ω ) the set of all finite words over A that occur somewhere in ω . If Ω is minimal, then there is a set W (Ω) such that W ( ω ) = W (Ω) for every ω ∈ Ω. Finally, we let W n (Ω) = W (Ω) ∩ A n . Date : October 31, 2018.D. D. was supported in part by NSF grant DMS–0653720.
Proposition 1.
Suppose that Ω is a minimal subshift. Then, R max (Ω) =max ω ∈ Ω R ( ω ) and T max (Ω) = max ω ∈ Ω T ( ω ) both exist, as elements of [1 , ∞ ] . More-over, the sets { ω ∈ Ω : R ( ω ) = R max (Ω) } and { ω ∈ Ω : T ( ω ) = T max (Ω) } areresidual.Proof. We prove the two statements for two-sided pinned repetitions. The one-sidedcase may be treated analogously.Note that it suffices to show that, for any ˆ ω ∈ Ω, the set M (ˆ ω ) = { ω ∈ Ω : T ( ω ) ≥ T (ˆ ω ) } is residual. Indeed, once this is shown, one may choose a sequencein { ˆ ω ( k ) } ⊂ Ω such that T (ˆ ω ( k ) ) → sup ω ∈ Ω T ( ω ) and then consider the residual set M = T k ≥ M (ˆ ω ( k ) ). By construction, T ( ω ) = sup ω ∈ Ω T ( ω ) for every ω ∈ M , andhence the sup is a max, and since the set { ω ∈ Ω : T ( ω ) = T max (Ω) } contains M ,it is residual.So let ˆ ω ∈ Ω be given. Choose a finite or countable strictly increasing sequence t , t , . . . with T (ˆ ω ) = sup m t m . Fix m . By minimality, for each length l , there is N l such that every word in W N l (Ω) contains all words from W l (Ω) as subwords.Thus, it is possible to find, for each w ∈ W n +1 (Ω) a word E ( w, m ) ∈ W (Ω) of oddlength that has w as its central subword of length 2 n + 1 and obeys (1) T ( E ( w, m )) ≥ t m − n . We may simply shift ˆ ω until the first occurrence of w is centered at the origin andthen take a sufficiently long finite piece that is centered at the origin as well. Bythe consequence of minimality mentioned above, it is clearly possible to ensure anestimate of the form (1) that is uniform in w ∈ W n +1 (Ω).Consider the open set [ w ∈W n +1 (Ω) [ E ( w, m )] , where, for a word x of odd length, [ x ] = { ω ∈ Ω : ω − | x |− . . . ω | x |− = x } . Noticethat T m = [ n ≥ m [ w ∈W n +1 (Ω) [ E ( w, m )] , is dense in Ω. Consequently, T = \ m ≥ T m is a dense G δ subset of Ω and for every ω ∈ T , we have by construction T ( ω ) ≥ T (ˆ ω ). (cid:3) Proposition 2.
Suppose that µ is an S -ergodic probability measure on Ω . Thenthere exist R µ and T µ such that R ( ω ) = R µ and T ( ω ) = T µ for µ -almost every ω ∈ Ω .Proof. Clearly, T ( · ) is invariant and hence it is µ -almost surely constant. While R ( · ) may not be globally invariant, we always have the inequality R ( Sω ) ≥ R ( ω )for every ω ∈ Ω. This implies that R ( Sω ) = R ( ω ) for µ -almost every ω and hence R ( ω ) is µ -almost surely constant. (cid:3) While T was defined above only for two-sided infinite words, a completely analogous definitioncan be given for finite words, where the central position plays the role of the origin. INNED REPETITIONS IN SYMBOLIC FLOWS 3
Remark.
There are some related results in [1]. They prove that, for minimalsubshifts, R ( · ) attains its maximum (without showing that it does so on a residualset). They also establish the almost sure constancy of R ( · ) with respect to anyergodic measure under the assumption of minimality and sublinear block complexity([1], Proposition 2.1). Our result (Proposition 2) holds in complete generality, andits proof is very short.In the remainder of the paper, we study pinned repetitions in symbolic flowsthat are generated by coding certain specific transformations of finite-dimensionaltori. In fact, given the space allotment, we will focus on two such classes – intervalexchange transformations and quadratic polynomials arising in the study of skew-shifts. We intend to continue our study of pinned repetitions in symbolic flows ina future work. We would also like to point out that the present study is related toour recent papers [2, 3] on the repetition property for dynamical systems on generalcompact metric spaces (which are not necessarily totally disconnected).2. Preliminaries
In this section, we present several results that will be useful later in our studyof pinned repetitions in certain specific models.2.1.
Irrational Rotations of the Circle.
For x ∈ R , we write h x i = dist ( x, Z ).Notice that d ( x, y ) = h x − y i is a metric on T = R / Z .Recall that the Farey sequence of order n is the sequence of reduced fractionsbetween 0 and 1 which have denominators less than or equal to n , arranged inorder of increasing size. Thus, for example, F = (cid:8) , (cid:9) , F = (cid:8) , , (cid:9) , F = (cid:8) , , , , (cid:9) , F = (cid:8) , , , , , , (cid:9) . If ab and cd are neighbors in a Fareysequence, then b + d > n and | ab − cd | = bd ; see, for example, [9]. Lemma 1.
Let x, y ∈ T and u > be such that for some n ≥ , the set { x + y, x + y, x + y, . . . , nx + y } does not intersect an arc J ⊆ T of length | J | = u . Then forsome positive integer q < min (cid:8) n, u (cid:9) , we have h qx i < n .Proof. Of course, we can assume without loss of generality that y = 0. Denote S = (cid:8) kx : 1 ≤ k ≤ (cid:4) u (cid:5)(cid:9) and consider the position of x relative to the Farey sequence oforder n − r s ≤ x ≤ r s . By the well-known properties of the Farey sequence oforder n − s j < n for j = 1 , s + s ≥ n .The point r + r s + s subdivides the interval [ r s , r s ]. Suppose x ∈ [ r s , r + r s + s ]. Then, (cid:12)(cid:12)(cid:12)(cid:12) x − r s (cid:12)(cid:12)(cid:12)(cid:12) ≤ r + r s + s − r s = 1 s ( s + s ) < s n . The case x ∈ [ r + r s + s , r s ] is analogous. Consequently, for pq equal to either r s or r s ,we have that(2) (cid:12)(cid:12)(cid:12)(cid:12) x − pq (cid:12)(cid:12)(cid:12)(cid:12) < qn . In particular, we must have q < n and h qx i < n . Moreover, since pq is a reducedfraction, the estimate (2) implies that the maximal gap (in T ) of { kx : 1 ≤ k ≤ q } is bounded from above by q . By assumption, we therefore must have u < q , thatis, q < u . (cid:3) M. BOSHERNITZAN AND D. DAMANIK
Continued Fraction Expansion.
Let us recall some basic results from thetheory of continued fractions; compare [9, 12]. Given an irrational number α ∈ T ,there are uniquely determined a n ∈ Z + , n ≥ α = 1 a + 1 a + 1 a + · · · . Truncation of this infinite continued fraction expansion after k steps yields the k -th convergent p k q k . We have the following two-sided estimate for the quality ofapproximation of α by the k -th convergent:1 q k ( q k + q k +1 ) < (cid:12)(cid:12)(cid:12)(cid:12) α − p k q k (cid:12)(cid:12)(cid:12)(cid:12) < q k q k +1 . The numerators and denominators of the convergents obey the following recursiverelations: p = 0 , p = 1 , p k = a k p k − + p k − for k ≥ ,q = 1 , q = a , q k = a k q k − + q k − for k ≥ . Discrepancy Estimates for Quadratic Polynomials.
In this subsectionwe discuss uniform distribution properties of quadratic polynomials. More precisely,given α, β, γ ∈ T , we consider the points(3) x n = αn + βn + γ ∈ T . The numbers D N = D N ( α, β, γ ) = sup intervals I ⊆ T (cid:12)(cid:12)(cid:12)(cid:12) N { n : 1 ≤ n ≤ N, x n ∈ I } − Leb( I ) (cid:12)(cid:12)(cid:12)(cid:12) measure the quality of uniform distribution of the given sequence and are called its discrepancy . Theorem 1.
Suppose α ∈ T is irrational and α = pq + θq , where p and q are relatively prime and | θ | ≤ . Then, for β, γ ∈ T and ε > arbitrary, we have D q ( α, β, γ ) < C ε q − − ε with some constant C ε that only depends on ε . In particular, the set { x , . . . , x q } intersects every interval I ⊆ T of length at least C ε q − − ε . We will use the following version of the Erd¨os-Tur´an Theorem, which holds infact for arbitrary real numbers x , . . . , x N ; compare [11, pp. 112–114]. Theorem 2 (Erd¨os-Tur´an 1948) . There is a universal constant ˜ C such that forevery m ∈ Z + , D N ≤ ˜ C m + m X h =1 h (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N N X n =1 e πihx n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)! . INNED REPETITIONS IN SYMBOLIC FLOWS 5
This estimate relates discrepancy bounds to bounds for exponential sums. Thusthe following lemma, which is closely related to a lemma given on p. 43 of [12], isof interest.
Lemma 2.
Suppose α ∈ T is irrational, β, γ ∈ T are arbitrary, and x n is given by (3) . Then, we have for N ∈ Z + , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 e πix n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ N + N X n =1 min (cid:18) N, h nα i (cid:19) . Proof.
This can be proved by a slight variation of the argument given on pp. 43–44of [12]. (cid:3)
Proof of Theorem 1.
By the Erd¨os-Tur´an Theorem, we have D q . m + m X h =1 h (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q q X n =1 e πihx n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) for every m ∈ Z + . Take m = ⌊ q δ ⌋ for some δ ∈ (0 , ε D q . q − δ + ⌊ q δ ⌋ X h =1 h (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q q X n =1 e πihx n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ q − δ + 1 q ⌊ q δ ⌋ X h =1 h q + q X n =1 min (cid:18) q, h hnα i (cid:19)! / . q − δ + 1 q ⌊ q δ ⌋ X h =1 h (cid:18) q + hε (cid:0) q δ − (cid:1) q ε (cid:19) / . q − δ + q δ − + ε . Here, we applied [10, Eqn. (153) on p. 75] in the third step. In the last line, theexponents coincide when δ = − ε . (cid:3) Given an irrational α , Theorem 1 gives a discrepancy estimate for those valuesof N that appear as denominators in the convergents associated with α . This willbe sufficient for our purpose. In some cases (e.g., α ’s of Roth type), it is possibleto use [12, Theorem 6 on p. 45] to modify the proof so as to cover all values of N . Moreover, if one is only interested in a metric result, the following improveddiscrepancy estimate is of relevance. Theorem 3.
Suppose α ∈ T is irrational. Then, for Lebesgue almost every β ∈ T and every γ ∈ T , we have D N ( α, β, γ ) < C ε N − (log N ) + ε . Since we will not use this result, we do not prove it. It may be derived from [7,Theorem 1.158 on p. 157]. We write a . b for positive a, b if there is a universal constant C such that a ≤ Cb . M. BOSHERNITZAN AND D. DAMANIK Pinned Repetitions in Codings of Interval ExchangeTransformations
In this section we study pinned repetitions occurring in codings of interval ex-change transformations. Interval exchange transformations are maps from an in-terval to itself that are obtained by partitioning the interval and then permutingthe subintervals.More explicitly, let m > m = { λ ∈ R m : λ j > , ≤ j ≤ m } and, for λ ∈ Λ m , β j ( λ ) = ( j = 0 P ji =1 λ i ≤ j ≤ m , I λj = [ β j − ( λ ) , β j ( λ )) , | λ | = m X i =1 λ i , I λ = [0 , | λ | ) . Denote by S m the group of permutations on { , . . . , m } , and set λ πj = λ π − ( j ) for λ ∈ Λ m and π ∈ S m . With these definitions, the ( λ, π )-interval exchange map T λ,π is given by T λ,π : I λ → I λ , x x − β j − ( λ ) + β π ( j ) − ( λ π ) for x ∈ I λj , ≤ j ≤ m. A permutation π ∈ S m is called irreducible if π ( { , . . . , k } ) = { , . . . , k } implies k = m . We denote the set of irreducible permutations by S m .Veech proved the following theorem in [13]. Theorem 4 (Veech 1984) . Let π ∈ S m . For Lebesgue almost every λ ∈ Λ m andevery ε > , there are N ≥ and an interval J ⊆ I λ such that (i) J ∩ T l J = ∅ , ≤ l < N , (ii) T is linear on T l J , ≤ l < N , (iii) | S N − l =0 T l J | > (1 − ε ) | λ | , (iv) | J ∩ T N J | > (1 − ε ) | J | . An interval exchange transformation T = T λ,π is said to satisfy property V if forevery ε >
0, there are N ≥ J ⊆ I λ such that the four conditionsof Theorem 4 are satisfied. This convention is motivated by the following result. Theorem 5.
Let T = T λ,π satisfy property V. Then the coding s of the T -orbitof Lebesgue almost every x ∈ I λ with respect to any finite partition of I λ obeys R ( s ) = T ( s ) = ∞ .Proof. Fix any finite partition I λ = J ⊔ · · · ⊔ J N .If lim sup ε → N ( ε ) < ∞ , it is not hard to see that T λ,π is a “rational rotation”and the assertion of the theorem holds trivially.Consider the other case and let ε n → N ( ε n ) → ∞ . By passing toa suitable subsequence { ε n m } , we can ensure that the Lebesgue measure of thosepoints x ∈ I λ that do not have m repetitions in both directions is bounded by2 − m . Thus, by Borel-Cantelli, almost every point has unbounded repetitions inboth directions, which implies the assertion. (cid:3) Corollary 1.
Let π ∈ S m . For Lebesgue almost every λ ∈ Λ m , the coding s of the T λ,π -orbit of Lebesgue almost every x ∈ I λ with respect to any finite partition of I λ obeys R ( s ) = T ( s ) = ∞ . INNED REPETITIONS IN SYMBOLIC FLOWS 7 Pinned Repetitions in Codings of Quadratic Polynomials
In this section we study sequences s of the following type. Suppose { J l : 1 ≤ l ≤ N } is a partition of T into finitely many intervals, ( α, β, γ ) ∈ T , and s denotesthe coding of αn + βn + γ with respect to the partition J . For example, assigningdistinct numbers λ l to the partition intervals, we may write s n = N X l =1 λ l χ J l ( αn + βn + γ ) . Sometimes we make the dependence of s on the parameters explicit and write s ( α, β, γ ) or s ( α, β, γ, J ). As explained in [8], there is close connection betweencodings of quadratic polynomials and codings of orbits of the skew-shift on T .We are interested in identifying the numbers R ( s ) and T ( s ) for such sequences s . We present a number of results regarding this problem. Roughly speaking, thesenumbers may take on the extreme values 1 and ∞ and they grow with the qualitywith which α can be approximated by rational numbers.4.1. Absence of Repetitions.
We first consider the case where α is not wellapproximated by rational numbers and show that there indeed are no repetitionsin the sense that R ( s ) = T ( s ) = 1. To make the argument more transparent,we begin by considering α ’s with bounded partial quotients. This means that thecoefficients { a n } in the continued fraction expansion are bounded; equivalently,inf q ∈ Z + q h qα i >
0. The set of such α ’s has zero Lebesgue measure. Theorem 6.
Suppose that α ∈ T has bounded partial quotients and each par-tition interval J l has length strictly less than / . Then, R ( s ( α, β, γ, J )) = T ( s ( α, β, γ, J )) = 1 for every ( β, γ ) ∈ T .Proof. Assume that R ( s ) > ν for some ν ∈ (0 , s has infinitely many(1 + ν )-repetitions starting at the origin. Let n be the length of such a prefix of s | Z + that is (1 + ν )-repeated.For 1 ≤ k ≤ νn , write y k = αk + βk + γz k = α ( n + k ) + β ( n + k ) + γd k = z k − y k = αn + βn + 2 αnk. Choose 0 < l < / l . Then, for 1 ≤ k ≤ νn , y k and z k must fall in the sameinterval of the partition. In particular, h d k i is bounded above by l for each such k .Consequently, d k avoids an arc J ⊆ T of length u = 1 − l > x = 2 αn and y = αn + βn , we find that there is apositive integer q < u such that h qx i < n . On the other hand, there is c = c ( u, α ) > h qx i = h αqn i > cn since α has bounded partial quotients. Combiningthe two estimates, we have that(4) cn < h qx i < n . Now assume in addition that n > νu and let m = ⌊ νnq ⌋ , so that 1 ≤ νnq ≤ m ≤ νn . Consider the points { d qk : 1 ≤ k ≤ m } . It follows from (4) that the diameterof this set belongs to the interval ( cνq , νq ). M. BOSHERNITZAN AND D. DAMANIK
For n sufficiently large, we obtain a contradiction because, as we saw above, thepoints y sk are well distributed on T and will come close to the partition points.Addition of the difference d qk for k with 1 ≤ k ≤ m sufficiently large will then go“across” such a partition point, which contradicts the assumption the y qk and z qk belong to the same interval of the partition. It follows that R ( s ) = 1, which alsoyields T ( s ) = 1. (cid:3) As pointed out above, this result covers only a set of α ’s that has zero Lebesguemeasure. Let us extend it to a larger set. Recall that α ∈ T is a Roth number iffor every ε >
0, there is a constant c ( ε ) such that h qα i > c ( ε ) q ε , for every q ∈ Z + .If we replace the qualitative well-distribution property with the quantitative dis-crepancy bound established above, virtually the same argument proves the followingresult, which covers a set of α ’s that has full Lebesgue measure. Theorem 7.
Suppose that α ∈ T is a Roth number and each partition interval J l has length strictly less than / . Then, R ( s ( α, β, γ, J )) = T ( s ( α, β, γ, J )) = 1 forevery ( β, γ ) ∈ T .Proof. The only change that needs to be made to the argument above is the fol-lowing. Assuming α to be Roth, instead of (4), we can prove(5) c ε n ε < h qx i < n for any ε >
0. Applying Theorem 1, we see that for every ˜ ε >
0, the points x , x , . . . , x n are C ˜ ε q − − ˜ εk -dense in T , where k is chosen such that q k ≤ n < q k +1 .Since the Roth condition also implies that the q k ’s associated with α obey q k +1 . q εk , it follows that x , x , . . . , x n is C ˜ ε n − (1+ ε )( +˜ ε ) -dense in T .Now we can conclude the proof as before by considering the points { d qk : 1 ≤ k ≤ m } . The addition of one of them to the corresponding y qk will take the pointacross a partition point by the estimates just obtained. (cid:3) Infinite Repetitions.
Let us now turn to the other extreme and start off bystudying the case of rational α . We will see that in this case, there are infinite repe-titions for almost every pair ( β, γ ). The next step will then be to identify situationsin which we have infinite repetitions for irrational α ’s that are well approximatedby rational numbers in a suitable sense.Denote by T w the subset of irrational numbers in T with unbounded partialquotients. It is well known that T w is a set of full Lebesgue measure in T .First we address the case α = 0. Lemma 3.
Suppose that β ∈ T w . Then, for Lebesgue almost every γ ∈ T , R ( s (0 , β, γ, J )) = T ( s (0 , β, γ, J )) = ∞ . Proof.
In this case s = s (0 , β, γ, J ) is a coding of βn + γ with respect to the fixedpartition J . The claim of the lemma follows from the fact that the 2-intervalexchange transformation T = T λ,π with λ = (1 − β, β ) , π = (21) is Veech if β ∈ T w (see Theorems 4 and 5 above).For an alternative, more direct argument, consult [6]. (cid:3) Theorem 8.
Suppose that α ∈ T is rational. Then, for every β ∈ T w , (6) R ( s ( α, β, γ, J )) = T ( s ( α, β, γ, J )) = ∞ INNED REPETITIONS IN SYMBOLIC FLOWS 9 for Lebesgue almost every γ ∈ T . In particular, (6) holds for Lebesgue almost every ( β, γ ) ∈ T .Proof. Write α = pq . Notice that αn is q -periodic. This serves as a motivationto begin with a study of repetitions along arithmetic progressions of step-length q .Put differently, we regard β ( qn ) as ( βq ) n and then add the constant α ( qn ) + γ .For every β ∈ T w , βq ∈ T w as well. By Lemma 3, there exists a sequence Q k → ∞ and a set ˜ G β ⊂ T of full measure such that for ˜ γ ∈ ˜ G β , the coding of( βq ) n + ˜ γ with respect to the given partition of T has k repetitions of length Q k toboth sides, for every k ≥ q , define G ′ β = T ∞ k =1 T Q k n =1 { γ ∈ T : α ( qn ) + β ( qn ) + γ ∈ ˜ G β } . As a countable intersection of sets of full measure, G ′ β has full measure. Wefind that for α = pq rational, β ∈ T w and γ ∈ G ′ β (and hence for almost every( β, γ ) ∈ T ), R ( s ) = T ( s ) = ∞ . (cid:3) Theorem 9.
There is a dense G δ set R ⊂ T such that for α ∈ R , we have R ( s ( α, β, γ, J )) = T ( s ( α, β, γ, J )) = ∞ for Lebesgue almost every ( β, γ ) ∈ T .Proof. Let r , r , r , . . . be a sequence of rational numbers that contains each fixed pq ∈ Q ∩ (0 ,
1) infinitely many times.Fix k and consider the coding of r k n + βn + γ with respect to the given partition,denoted by s k . By Theorem 8 we have that T ( s k ) = ∞ for almost every β, γ. For m ∈ Z + , we can therefore choose a set B m,k ⊂ T such that • B m,k is open, • Leb( B m,k ) > − m , • for ( β, γ ) ∈ B m,k , s k has m repetitions in both directions at least once, • for ( β, γ ) ∈ B m,k , the itinerary r k n + βn + γ does not contain any partitionpoint for n ’s from the finite interval on which we observe the m repetitionsin both directions.Next, choose K m,k ⊂ B m,k such that • K m,k is compact, • Leb( K m,k ) > − m .By compactness of K m,k , we have that for ( β, γ ) ∈ K m,k , the itinerary r k n + βn + γ for n ’s from the finite interval on which we observe the m repetitions inboth directions has a uniform positive distance from the partition points.Consequently, we can perturb r k slightly and not change the coding on the (large)finite interval that supports the two-sided repetition in question. In other words,there is an open set U k containing r k such that for α ∈ U k and ( β, γ ) ∈ K m,k , s has m repetitions in both directions at least once.Define R = T m ≥ S k ≥ m U k . Since r , r , r , . . . contains each fixed pq ∈ Q ∩ (0 , S k ≥ m U k is dense.Therefore, R is a dense G δ set.By construction, we have that for α ∈ R and Lebesgue almost every ( β, γ ) ∈ T , R ( s ) = T ( s ) = ∞ . Indeed, if α ∈ R , then α belongs to some U k . By Borel-Cantelli and the measure estimates for the sets K m,k , we have that for almost every( β, γ ) ∈ T , s has m repetitions to both sides for any m . (cid:3) The residual set obtained in Theorem 9 is not explicit. If we are willing to settlefor infinite repetitions for just one pair ( β, γ ) ∈ T , then the following result is ofinterest. It also has the advantage that the argument we give can treat generalpolynomials, not merely quadratic polynomials, and hence we state and prove theresult in this more general setting. For τ >
0, denote S τ = { α ∈ T : h qα i < q − τ for infinitely many odd positive integers q } . Clearly, S τ is a residual subset of T . Theorem 10.
Let r be a positive integer and ε > . Then, for every α ∈ S r + ε , thesequence s given by s n = χ [0 , / ( αn r + 1 / obeys R ( s ) = T ( s ) = ∞ .Proof. Fix an integer m ≥ q be an odd integer with h qα i < q − r − ε . Let p ∈ Z be such that h qα i = | qα − p | . Then, for | n | ≤ mq we have, on the one hand, h α ( n + q ) r − αn r i = * r − X j =0 (cid:18) rj (cid:19) αn j q r − j + ≤ r − X j =0 (cid:18) rj (cid:19) n j q r − − j h qα i≤ r − X j =0 (cid:18) rj (cid:19) m j q r − q r + ε = 1 q ε r − X j =0 (cid:18) rj (cid:19) m j and, on the other hand, h αn r ± / i ≥ (cid:28) pq n r ± (cid:29) − n r q r + ε ≥ q − m r q ε . Since for large enough q , we have that14 q − m r q ε > q ε r − X j =0 (cid:18) rj (cid:19) m j , and hence T ( s ) ≥ m . Since m was arbitrary, this shows T ( s ) = ∞ , which alsoimplies that R ( s ) = ∞ . (cid:3) References [1] V. Berth´e, C. Holton, and L. Zamboni, Initial powers of Sturmian sequences,
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Department of Mathematics, Rice University, Houston, TX 77005, USA
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