Hausdorff dimension of a class of three-interval exchange maps

In \cite{B} Bourgain proves that Sarnak's disjointness conjecture holds for a certain class of Three-interval exchange maps. In the present paper we slightly improve the Diophantine condition of Bourgain and estimate the constants in the proof. We further show, that the new parameter set has positive, but not full Hausdorff dimension. This, in particular, implies that the Lebesgue measure of this set is zero.

In where N ( , n) is the largest number of -separated points in Y using the metric d n : Y × Y → R + defined by d n (x, y) = max 0≤i≤n d(T i x, T i y).
A sequence f : Z → C is said to be deterministic if it is of the form for all n and some topological dynamical system (Y, T ) with zero topological entropy h(Y, T ) = 0, a base point x ∈ Y , and a continuous function F : Y → C.
con1 Conjecture 1 (Sarnak). Let f : N → C be a deterministic sequence. Then f0 f0 (1.1) S n (T (x), f ) = 1 n The conjecture, also known as the Möbius orthogonality or Möbius disjointness conjecture, is known to be true for several dynamical systems. Note, that in the simplest case, when f ≡ const, the conjecture is equivalent to the statement 1 N N n=1 µ(n) = o (1), which, in fact, is equivalent to the Prime Number Theorem. The orthogonality of the Möbius function to any sequence arising from a rotation dynamical system (X is the circle T and T (x) = x + α, α ∈ T) follows from the following inequality of Davenport([8]) for any A > 0. However this result predates Sarnak's conjecture and the methods used in the proof are number-theoretical. When (X, T ) is a translation on a compact nilmanifold it is proved in [17]. In [4] it is established also for the discrete horocycle flows. For orientation preserving circle-homeomorphisms and continuous interval maps of zero entropy the conjectures is proved in [20]. The conjecture has also been proved to hold in several other cases ( [1], [2], [3]). Another natural class of dynamical systems are the interval exchange maps. In [5] Bourgain, using the dynamical description of trajectories in ([10]-[12]) and the Hardy-Littlewood circle method, showed that under a certain diophantine condition Conjecture 1 holds for a certain class of three-interval exchange maps. In this paper we slightly improve the diophantine condition of Bourgain and estimate the Hausdorff dimension of the new parameter set (Theorem 7). We want to note, that using the criterion of Bourgain in [5] and the generalization of the self-dual induction defined in [15], for each primitive permutation, Ferenczi and Mauduit([14]) construct a large family of k-interval exchanges satisfying Sarnaks conjecture. In [6] Eskin and Chaika proved the Möbius orthogonality for three interval exchange maps satisfying a certain mild diophantine condition. Even though their result holds for almost all three interval exchange maps, the diophantine condition considered in their paper is essentially complementary to the one considered here. In [6] the continued fractions are required to have certain bound from above, while in Bourgains method they need to be uniformly large. We note that Eskin and Chaika, in fact, give two proofs of the fact that the Möbius orthogonality holds almost surely for three-interval exchange maps, however the second proof does not provide an explicit Diophintine condition. Their proof is based on the Katai [21] and Bourgain-Sarnak-Ziegler [4] criterion, while Bourgain uses a direct approach.
The present paper is a part of the authors Ph.D. thesis.
In order to present the main result of the paper we need to recall some facts and definitions from [10] and [5]. Set by inducing (according to the first return map) on the subinterval [0, B(α, β)] and then renormalizing by scaling by 1 + β. We say T satisfies the infinite distinct orbit condition (or i.d.o.c. for short) of Keane [23] if the two negative trajectories {T −n (α)} n≥0 and {T −n (α + β)} n≥0 of the discontinuities are infinite disjoint sets. Under this hypothesis, T is both minimal and uniquely ergodic; the unique invariant probability measure is the Lebesgue measure µ on [0, 1) (and hence (X, T, µ) is an ergodic system).
Clearly H is a countable set.
propos Proposition 2 ([10, Proposition 2.1, (4)]). For (α, β) ∈ D 0 , let (ᾱ,β) = H(α, β) as above, then We define the natural partition For every point x ∈ [0, 1), we define an infinite sequence (x n ) n∈N by putting x n = i if T n x ∈ P i , i = 1, 2, 3. The sequence (x 1 , x 2 , . . . ) is called the trajectory of x. If T satisfies the i.d.o.c. condition (see [23]), the minimality of the system implies that all trajectories contain the same finite words as factors. Let I be a set of the form ∩ n−1 i=0 T −i P k ; we say I has a name of length n given by k 0 , . . . , k n−1 ; note that I is necessarily an interval and k 0 , . . . , k n−1 is the common beginning of trajectories of all points in I .
For each interval J, there exists a partition J i , 1 ≤ i ≤ t, of J into subintervals (with t = 3 or t = 4), and t integers h i , such that is a partition of [0, 1) into intervals: this is the partition into Rokhlin stacks associated to T with respect to J. The intervals J i have names of length h i and are called return words to J.
We have the following theorem.
Ft Theorem 1 ([10, Theorem 2.2]). Let T satisfies the i.d.o.c. condition, and let be the three-interval expansion of (α, β). Then there exists an infinite sequence of nested intervals J k , k ≥ 1, which have exactly three return words, A k , B k and C k , given recursively for k ≥ 1 by the following formulas The initial words A 0 ,B 0 ,C 0 satisfy ||A 0 | − |B 0 || = 1 and they are simple combinations of the symbols 1, 2 and 3 (see Proposition 2.3,[10]).
Remark 2. In [6] Eskin and Chaika show, that for three interval exchange maps satisfying the conditions (A0)-(A9) (page 3), Sarnak's conjecture holds. We want to note, that their approach is also based on the fact that three interval exchange maps can be induced from a two interval exchange map. The numbers {a k } ∞ k=1 , in conditions (A0)-(A9), are the continued fractions of the rotation number of the two interval exchange map which induces the three interval exchange map with parameters (α, β). In our case this corresponds to the number A(α, β) (see (2.2)), hence the continued fractions of A(α, β) are the numbers {a k } ∞ k=1 and from Proposition 2 it is easy to see, that they are related to the numbers {(n k + m k )} ∞ k=1 . However in Bourgain's approach this numbers are required to be sufficiently large (see Theorem 4), while the conditions (A0)-(A9) essentially give upper bounds.
Using the dynamical descriptions of trajectories in [10], Bourgain [5], proves Sarnak's disjointness conjectures for a certain class of three interval exchange maps. Now we recall the statement of his theorem. Consider a symbolic system on the alphabet V with finitely many symbols and with order-n words W ∈ W n of the form where it is assumed that r remains uniformly bounded, r < C. It is also assumed the following property for the system {W n }. For W ∈ W n , which is expressed in words W ∈ W n−s , 0 < s ≤ n, by iteration of (2.9) we have, where f13 f13 (2.11) β(s) > C s 0 , for some s and sufficiently large constant C 0 .
thm Theorem 3 ([5, Theorem 2, page 126]). Let {W n ; n ≥ 1} be a symbolic system with properties (2.9)-(2.11) and σ be the shift on the system. Then, if W ∈ W n and |W | = N , one has To see how this implies Sarnak's conjecture, we recall the following inequality, which immediately follows from Parseval's identity which implies Sarnak's conjecture. For three interval exchange maps we have One can see, that if m k and n k are uniformly large, then the conditions (2.10)-(2.11) are satisfied and as a corollary from Theorem 3 one gets the following result: th1 Theorem 4 ([5, Theorem 3]). Assume T α,β is a three-interval exchange transformation satisfying the Keane condition and such that the associated three-interval expansion sequence for C 0 sufficiently large. Then T α,β satisfies Sarnak's disjointness conjecture.
rmk Remark 5. Note, that Bourgain's theorem is in fact more general than the form it is stated in Theorem 4. As it was mentioned above, in Theorem 4 it is assumed, that there is uniform expansion at each steps, i.e. (2.14), but as one can see from the conditions (2.10)-(2.11) it is sufficient to have this expansion after s many iterations, for some fixed s. More precisely, one can replace the condition (2.14) with for all large k and fixed s.
Next we prove a proposition, which will allow us to rewrite the Diophantine condition (2.14) above in even more general form. main-prop Proposition 3. In Theorem 4 the condition (2.14) can be replaced by Proof. As we have already mentioned, for three interval exchange maps we have W k = {A k , B k , C k }. From (2.10)-(2.11) it follows, that it suffices to show, that for any for sufficiently large k. One can check from the formulas (2.3)-(2.5), that (2.5) has the shortest length. Hence Assume W k−1 = C k−1 , then from (2.18) and (2.20) So we can assume, that We want to show, that for large enough k Since we have |a k − b k | = 1, then clearly We now assume that m k , n k = 1. If n k ≥ 3, then from (2.21) we will have In the same way, under the assumptions m k , n k = 1, we can show ( We now assume, that one of the numbers m k and n k is 1. First let m k = n k = 1. In this case, according to Proposition 1, k+1 can not be positive for infinitely many values of k, so we assume, that we have (m k , n k , k+1 ) = (1, 1, −1). Hence, W k is defined by the formulas (2.6)-(2.8), so the word C k has the largest length and in view of (2.24) we will have again for large values of k.
From this proposition and in view of Remark 5, we arrive at the following theorem: th2 Theorem 6. Assume T α,β is a three-interval exchange transformation satisfying the Keane condition and such that the associated three-interval expansion sequence of integers for all k ≥ k α,β and for some s ≥ 1 fulfills the conditions where C 0 is as in Theorem 4. Then T α,β satisfies Sarnak's disjointness conjecture.
We now turn to the estimation of the constant C 0 . In the proof of Theorem 3 Bourgain, first estimates the L 1 norm of the polynomials P W , namely the lemmas 3 and 4 in [5]. For the result it is also essential to slow down the growth of the L 1 norm of the polynomial P W , whenever |W k | → ∞ (see Lemma 4 in [5]). This condition is achieved by assuming that the lengths |W k | of the words in the symbolic representations (2.9) grow sufficiently fast, i.e. conditions (2.10) and (2.11). One of the key places, where this is used is Lemma 4. To estimate how big the constant C 0 has to be, we will follow Bourgains steps and give a more quantitative proof of this lemma.
clm Claim 1. The constant C 0 in Theorem 6 is at least required to satisfy C 0 > 24 12 .
Proof. First we note the following. Let W 1 → W 2 · · · → W n be a sequence of words with W k ∈ W k , where each W k participates in the symbolic representation of W k+1 . Then according to the assumptions (2.10) and (2.11) one has q1 q1 (2.26) In the same way We note that there can only be finitely many indices, where the above inequalities do not hold, but that will not affect the estimates that follow. Multiplying together the inequalities in (2.26) and (2.27) we will have The proof of (2.29), in its turn, is based on Lemma 3. We note, that in the proof of Lemma 3 in [5], Bourgain doesn't use any particular property of the polynomial P W and since the inequality (2.29) holds for any polynomial P W (θ), we can assume, that P W (θ) ≡ const. So we will end up with the following inequality where D k is the Dirichlet kernel for which one has (e.g., see [29]) So we conclude, that Dividing both sides by log k and tending k to infinity we get, that Next we estimate the ε in (2.15) Or dividing both sides by N c:11 c:11 (2.32) From the above inequality it follows, that if one of the numbers P, Q is sufficiently large, then the necessary estimate on the set V Q,K will be achieved. From here Bourgain concludes, that one can assume Q, K < N , for some . But we see from (2.32), that this argument will be possible only if the quantity N −τ /4 is small relative to N ε , or But from Lemma 6 in [5] we have that 0 < τ < 1/3. Hence ε < 1/12.
However we will neglect the coefficient 4 n in the above inequality (which amount to saying, that at each step of the iteration of (2.34) we have only 1 word, or W k+1 is a power of W k ). In other words, instead of (2.36) we will consider the following inequality which is clearly implied by (2.36). Since ε < 1/12, then the above inequality will also imply C log |W | 1 n + C log 3 Therefore, if (2.24) in [5] holds, then so does the inequality above. Now consider For the critical point of x c we have so we see, that if f (x) > 0 for some large x, then we must have x > x c = 24 12 . But we know from (2.28), that x = |W | 1 n > C 0 for large n. This finishes the proof of Claim 1.
As we see the constant C 0 in Theorem 6 must be very large. But in the present paper we will only assume that C 0 ≥ 20.

Estimates on Hausdorff dimension sec:2
We first recall the definition of Hausdorff dimension. Let X be a metric space. If S ⊂ X and d ∈ [0, ∞), the d-dimensional Hausdorff content of S is defined by there is a cover of S by balls with radii r i > 0 .
In other words, C d H (S) is the infimum of the set of numbers δ > 0 such that there is some (indexed) collection of balls {B(x i , r i ) : i ∈ I} covering S with r i > 0 for each i ∈ I that satisfies i∈I r d i < δ . Then the Hausdorff dimension of X is defined by dim H (S) := inf{d ≥ 0 : C d H (S) = 0}. We will need the following classical facts about this concept, (see, e.g. [27], Theorem 2).
Next we prove the following proposition. Proof. It is enough to show this for the maps F and G. By definition D 0 is the region bounded by the lines y = 0, x = 0, and x + y = 1. The inverse of F and G can be computed as . If x 1 + y 1 ≤ 1, then considering the two coordinates of F −1 we have To prove that they are Lipschitz it is sufficiently to show, that the partial derivatives are uniformly bounded in D 0 . ∂ ∂x 1 as x 1 , y 1 ≤ 1. Since for any H ∈ H, H −1 is a composition of Lipschitz functions, i.e.
then H −1 is also Lipschitz.
From the discussion at the beginning of Section 2 and the definition of the sequence Therefore, to estimate the Hausdorff dimension of P 0 , it is enough to estimate it for P, i.e. when (α, β) ∈ D and for this (α, β)'s one has the following relation . We now recall the definition of standard continued fractions. For any θ ∈ [0, 1] its continued fraction is an expression of the form θ = a 0 + 1 , and its n'th convergent is denoted by With the conventions p −1 = 1, q −1 = 0, p 0 = 0, q 0 = 1, we have p n+1 = a n+1 p n + p n−1 , q n+1 = a n+1 q n + q n−1 , (n ≥ 0), and c1 c1 (3.3) p n+1 q n − p n q n+1 = (−1) n , (n ≥ −1).
Let I n = I n a 1 ,a 2 ,..,an , where a 1 , a 2 , .., a n are positive integers, be the interval  that is (3.4) p n q n , p n + p n+1 q n + q n+1 .
Here, for a = b, we mean by (a, b) the closed interval with end-points a, b. This means, that we can also have b < a. From (3.3) it follows that the length |I n | of I n satisfies c2 c2 (3.5) |I n | = 1 q n (q n + q n−1 ) .
We will also work with more general kind of continued fractions, namely semi-regular continued fractions (SRCF). For θ ∈ [0, 1] its SRCF expansion looks like this f1 f1 (3.6) θ = 1 where k = ±1 and a k ≥ 2 for all k ≥ 1. For short we will write (3.6) in the following way θ = [ 1 /a 1 , 2 /a 2 , . . . , n /a n , . . . ]. Note, that if k = 1 for all k ≥ 1, then we get the standard continued fraction expansion of θ. The SRCF expansion is defined for a k ≥ 2, but we will also deal with the cases, when a k = 1 or a k = 0.
The following identity will be fundamental for us (see e.g. [18]). For a ∈ Z, b ∈ N + and x ∈ [0, 1) we have Using this identity we are going to find the standard continued fraction expansion from their SRCF expansion. According to Proposition 1, if (α, β) ∈ D, then To not carry the minus sign in (3.8) all the time in the computations we will simply replace the − k with k . So from now on (3.8) will look like this b:2 b:2 (3.9) 1 − α 1 + β = [1/2, 2 /(m 1 + n 1 ), . . . , k+1 /(m k + n k ), . . . ], where k = ±1. Now we will use the identity (3.7) to get rid of the negative k 's. If in (3.9) k+1 = −1 for some k, then from (3.7) b:4 b:4 (3.10) m k + n k + − 1 By definition m k , n k ≥ 1 and hence m k + n k ≥ 2. As we see from the equation above, any number (n k + m k ) can participate in at most two replacements, hence can be reduced by at most 2, i.e. become n k + m k − 2. This will be the case with m k+1 + n k+1 − 1 in (3.10) if k+2 = −1. If m k+1 + n k+1 − 2 > 0, then the replacement will be valid. The case m k+1 + n k+1 = 2 needs special considerations. From the left hand side in (3.10) b:3 b:3 (3.12) , and since m k+1 + n k+1 = 2, then Putting this back into (3.10) we get In a similar way, if we have m k+s + n k+s = 2, k+l+1 = −1 for all s = 1, . . . , l, and either m k+l+1 + n k+l+1 > 2 or k+s+2 = 1 and then one can show, that the equality above, can be rewritten as follows Observe, that according to (2.37), we should have l < k α,β . We see that as a result of this procedure we will get the continued fraction expansion of (1 − α)/(1 + β). The next proposition shows, that we have a (2.25) like property also for the standard continued fractions of (1 − α)/(1 + β): pro Proposition 5. Let (α, β) ∈ P and . . ] there is a number C α,β ∈ N so that for any large n there is a number s α,β ∈ N, with s α,β ≤ C α,β , such that (a n a n−1 · · · a n−s α,β ) Proof. We know from (2.25), that Let the number of 2's and 3's between the numbers {(m j +n j )} k j=k−s+1 be respectively equal to m 1 and m 2 . Therefore where l + m 1 + m 2 = s, and (n k j + m k j ) ≥ 4, for j = k 1 , . . . , k s . Therefore during the procedure described above have transformed into new r many digits, i.e corresponding to the standard continued fractions res res (3.16) {a i , a i−1 , . . . , a i−r+1 }.
In case we have digit 1's appearing on both sides of the continued fraction (n j + m j ), then in (3.16) we will include only the left digit 1.
In case j = −1 for all j = k − s + 1, . . . , k, we will have a new digit 1 appearing in between any two digits (n j + m j ) and (n j−1 + m j−1 ). Therefore the number of digits in (3.15) will at most double, i.e. est:m est:m (3.17) r ≤ 2s.
And since each digit may participate in at most two replacements, then clearly a i a i−1 · · · a i−r ≥ (n k 1 + m k 1 − 2) · · · (n k l + m k l − 2). Now note, that if m k j + n k j ≥ 4, then Therefore And since r ≤ 2s, then We can also see from (3.17) that r is uniformly bounded, since where the function A is defined in (2.1). Notice, that the set S S (3.20) S = {θ ∈ [0, 1] : ∃k θ , s θ ≥ 1, such that, if k ≥ k θ , then (a n a n−1 · · · a n+1−s θ ) log a k ≥ log Λ.
Therefore the Lebesgue measure of the set S is zero. From this one can see, that the two dimensional Lebesgue measure of (α, β) ∈ D 0 , for which A(α, β) ∈ S, is also zero. However this will also follow from Corollary 1.
In [9] the authors, alongside with other things, for each γ > 0, compute the Hausdorff dimension of the set of all x ∈ [0, 1], for which the following limits exists and equals lim n→∞ 1 n n j=1 log a j (x) = γ, or equivalently, if lim n→∞ (a 1 (x)a 2 (x) · · · a n (x)) 1 n = e γ .
In this paper we need to estimate the Hausdorff dimension of the set S 0 , where the continued fractions, in particular, satisfy the property (3.24). Therefore we need, in a sense, stronger result. We will show in the sequel, that the method used in [9] will allow to achieve this.
prop:3 Proposition 6. For C 0 ≥ 20, the Hausdorff dimension of the set S 0 satisfies the following bounds fin1 fin1 (3.25) where the function t(ζ) is defined in Theorem 10.
which is known to preserve the measure From Birkhoff's ergodic theorem we have, that For real numbers ζ, β ≥ 0 one considers the level sets of Khintchine exponents and Lyapunov exponents The following theorem holds.
Hence, in view of (3.31), (3.33), we will have s:5 s: Consider now the following two sets A 1 and A 2 : log a k (x) = ζ x , for some ζ x ≥ log Λ}.
But since ζ 1 was an arbitrary number between log Λ and ζ 0 , and t(ζ) is a continuous function, then it follows final final (3.40) dim H S ≤ t(log Λ).
Recall the definitions of the sets S 0 and S, (3.20), (3.19). Our goal is to estimate the Hausdorff dimension of the set S 0 . From (3.21) we have dim H S 0 ≤ dim H S ≤ t(log Λ).
We want to estimate the Hausdorff dimension of the set P 0 . From (2) we have, that dim H P 0 ≤ dim H P.
For this we refer to a standard fact from the theory of Hausdorff dimensions. If F : G ⊂ R 2 → R is F ∈ C 1 , F = 0 for all (α, β) ∈ G, then for any set E ⊂ R one has haus1 haus1 (4.1) dim H F −1 (E) = 1 + dim H E.