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1 ESTIMATING THE FRACTAL DIMENSION OF SETS DETERMINED BY NONERGODIC PARAMETERS A thesis submitted to the faculty of San Francisco State University In partial fulllment of The Requirements for The Degree Master of Arts In Mathematics by Joseph Paul Squillace San Francisco, California May 204

2 CERTIFICATION OF APPROVAL I certify that I have read Estimating the Fractal Dimension of Sets Determined by Nonergodic Parameters by Joseph Paul Squillace, and that in my opinion this work meets the criteria for approving a thesis submitted in partial fullment of the requirements for the degree: Master of Arts in Mathematics at San Francisco State University. Yitwah Cheung Associate Professor of Mathematics Eric Hayashi Professor of Mathematics Professor of Mathematics

3 ESTIMATING THE FRACTAL DIMENSION OF SETS DETERMINED BY NONERGODIC PARAMETERS Joseph Paul Squillace San Francisco, California 204 In 969, Veech introduced two subsets K (θ), K 0 (θ) of R/Z which are dened in terms of the continued fraction representation of θ R/Z. These subsets are known to give information about the dynamics of certain skew products of the unit circle. We show that the Hausdor dimension of the sets K i (θ) can achieve any value between 0 and, inclusive. I certify that the Abstract is a correct representation of the content of this thesis. Chair, Thesis Committee Date

4 TABLE OF CONTENTS List of Figures v Introduction Veech's Examples and a Billiard Interpretation of these Examples Upper Bound on Hausdor Dimension Reduction Specication of Self-Similar Covering A Self-similar covering of K i 0 0 (θ) Calculation A Nontrivial Example of θ Satisfying HdimK 0 (θ) = Example of θ for which HdimK (θ) = Falconer's Lower Bound Formula Lower Bound for HdimK (θ) HdimK i (θ) = δ Upper Bound Lower Bound Refefences iv

5 LIST OF FIGURES Table Page A Billiard table B α with a barrier of length α Unfolding of Billiard Table B α v

6 INTRODUCTION Let {x : 0 x < } be the compact group of real numbers modulo. Any irrational element θ in this set has a continued fraction representation a + a where (a i ) i= is a sequence of positive integers known as the terms or partial quotients ( ) m of the continued fraction. Take k n k to be the sequence of principal convergents k= of θ; that is, m k n k = [0; a, a 2,..., a k ] for each positive integer k. From the theory of continued fractions, we have n k+ = a k+ n k + n k () for k N, with n 0 := and n = a. Let (b k ) k N be a sequence of even integers satisfying b j a j+ for each j N. Given any integer m, ( ) m; b, b 2,... θ := mθ + b j n j θ mod denes a point of {x : 0 x < }. We call such a representation θ a θ expansion. From now on, we omit the mod notation. Veech [7] denes the sets j= K 0 (θ) := { m; b, b 2,... θ : b j is eventually even, lim j b j n j n j θ = 0} and

7 2 K (θ) := { m; b, b 2,... θ : b j is eventually even, } j= b j n j n j θ < where denotes the distance to the nearest integer. Clearly, K (θ) K 0 (θ). In Section 2 we nd an upper bound formula for the Hausdor dimension of K 0 (θ). In Section 4. we apply a result from [5] to obtain a lower bound formula for the Hausdor dimension of K (θ). Using these formulas, in Section 5 we show that for any δ [0, ] we can construct a number θ such that the Hausdor dimension of K i (θ) equals δ, where i = 0,.. Veech's Examples and a Billiard Interpretation of these Examples In [7], Veech constructs examples of minimal and not uniquely ergodic dynamical systems as follows (see [6]). Take two copies of the unit circle and mark o segment a J of length 2πα in the counterclockwise direction on each one with endpoint at 0. Now choose an irrational θ and consider the following dynamical system. Start with a point p in the rst circle. Rotate counterclockwise by 2πθ repeatedly until the the orbit lands in J; then switch to the corresponding point in the second circle, rotate by 2πθ until the rst time the point lands in J; switch back to the rst circle and so forth. Veech showed there exists irrational α for which this system is minimal and the Lebesgue measure is not uniquely ergodic. We may describe Veech's dynamical system by a ow on a surface arising from a billiard. Consider billiards in the table formed by a rectangle with a horizontal barrier of length α with one end touching 2 at the midpoint of a vertical side. We can identify the top half of the table as the positive side and the bottom half as the negative side.

8 3 Figure : A Billiard table B α with a barrier of length α. A standard unfolding of this billiard table is shown in Figure 2. We can view the new gure as having two (identied) barriers of length 2α. Figure 2: Unfolding of Billiard Table B α

9 4 (See [6]) If we consider the billiard ow in this rectangle with direction θ, there is an unfolding of this billiard such that the restriction of the ow to the map of rst return to the horizontal barrier is equivalent to the function described in the examples given by Veech. In [4], Cheung, Hubert, and Masur showed that if we x the parameter α then the dimension of the set of all θ for which this ow is not ergodic is either 0 or. Specically, Cheung, Hubert, and Masur show that the dimension is 2 2 if and only if k N log log q k+ q k < where {q k } k N is the sequence of denominators of the continued fraction representation of α. Our goal is to consider what happens in the case in which we x θ and allow α to vary. Moreover, the sets K (θ) and K 0 (θ) are known to give information (see [3]) about the dynamics of Veech's examples described above. 2 UPPER BOUND ON HAUSDORFF DIMENSION For A R n, we denote by HdimA the Hausdor dimension of A (see [4]). In Section 2. we give an upper bound for HdimK 0 (θ) by applying what Cheung [2] has dened as a self-similar covering to subsets which we call K0 i (θ) of K 0 (θ). In particular, the self-similar covering will give an upper bound for HdimK0 i (θ), and we show that this upper bound is also an upper bound for HdimK 0 (θ). The following denition is motivated by the fact that (i) Veech showed in [7] that if θ has bounded partial quotients (i.e., sup j N a j < ) then HdimK 0 (θ) = 0, and

10 5 (ii) Lothar Narins conjectured that HdimK (θ) = when a j. Denition. Let M N. An irrational θ with unbounded partial quotients is divergent relative to [, M] if the sequence of partial quotients formed by the terms that are greater than M diverges to. If θ is divergent relative to [, M], dene κ θ := {k i : a ki + > M} k 0 := min κ θ Theorem. Let θ be divergent relative to [, M], and let (k i ) i=0 enumerate the values of k satisfying a k+ > M. Then 2. Reduction HdimK 0 (θ) lim sup i log a ki + log n ki+ log n ki (2) We reduce computing an upper bound of HdimK 0 (θ) directly by introducing new sets K i 0 (θ) for nonnegative integers i. We will show that this suces for nding an upper bound of HdimK 0 (θ). Denition. Suppose θ is divergent relative to [, M]. K i 0 (θ) := { m; b, b 2,... θ K 0 (θ) : j > k i b j n j n j θ < } 4M

11 6 The sets K i 0 (θ) are nonempty for all suciently large i since, by denition of K 0 (θ), lim b j n j n j θ = 0 j Further, the next three lemmas along with the fact that we can nd an upper bound for HdimK i 0 (θ) that does not depend on i will furnish an upper bound for HdimK 0 (θ). Lemma. K 0 (θ) K0 i (θ) i=0 Proof. If y K 0 (θ), then there is a sequence m, b, b 2,... such that y = m; b, b 2,... θ, where b j n j n j θ 0. Hence, there is an i for which j > k i implies b j n j n j θ < 4M. y K i 0 (θ). Therefore Lemma 2. (Countable Stability) Hdim K0 i (θ) = sup { HdimK0 i (θ) : i 0 } i=0 Proof. For any set F R n and s 0, denote by H s (F ) the s-dimensional Hausdor measure of F (see [5]). HdimK i 0 (θ) Hdim i Ki 0 (θ) for each i, so sup i HdimK i 0 (θ)

12 7 Hdim i Ki 0 (θ). Conversely, let s = sup i HdimK i 0 (θ). It suces to show H s+ε i Ki 0 (θ) = 0 when ε > 0. Let δ > 0. For each i we can cover K i 0 (θ) by intervals A ij such that the sum of their radii by the power s + ε is less than 2 i δ. The union of all intervals A ij, over i and j, covers i Ki 0 (θ) and the sum of their radii raised by the power s + ε is less than δ. Therefore, H s+ε i Ki 0 (θ) = 0. Lemma 3. There exists an i 0 such that if j k i0, j κ θ and b j n j n j θ < 4M, then b j = 0. Proof. Take i 0 large enough such that k i0 min k θ. Since j κ θ, a j+ M. Hence 4M > b j n j n j θ > b j n j 2n j+ > b j 4a j+ Hence, a j+ > M b j. Since a j+ < M, b j = 0. A consequence of Lemma 3 is x K i 0 0 (θ) has a θ expansion m; b, b 2,... such that for all j > k i0 either a j+ > M or b j = 0.

13 8 2.2 Specication of Self-Similar Cover Given b i a i+ and an even b satisfying b a k+, dene by I (m; b, b 2,..., b k, b) the interval of length 8 n k centered at k mθ + b j n j θ + b n k θ j= Lemma 4. x I (m; b,... b k ) for any x K 0 (θ). Proof. Let x = m; b,... θ K 0 (θ). Since θ is between m k n k and m k+ n k+, it follows that m k θ m k n k m k+. n k+ Multiplying by m nk gives m k n k θ m k n k+. k n k+ n k Therefore, n k θ m k n k θ m m k+ k n k = n k+ n k+ We also note that since n k+2 > 2n k we have n k+ > 2 i n k+(2i+) and n k+2 > 2 i n k+(2i+2). Therefore, i=0 < i=0 2 n k+2. n k+(2i+) 2 i n k+, which is a geometric series that simplies to 2 n k+. Similarly, i= n k+(2i) <

14 9 Without loss of generality, suppose m = 0. Then k x b i n i θ = b j n j θ i= Therefore, x I (m; b,... b k ). = < j=k+ j=k+ b j n j θ b j n j+ j=k+ j=k+ a j+ n j+ n j j=k+ i=0 n k+(2i+) n k+ 4 n k+ n k+2 i= n k+2i Denition. (Section 3 of [2]) Let X be a metric space and J a countable set. Given σ J J and α J we let σ (α) denote the set of all α J such that (α, α ) σ. We say a sequence (α j ) j N of elements in J is σ-admissible if α j+ σ (α j ) for all j N; and we let J σ denote the set of all σ-admissible sequences in J. By a selfsimilar covering of X we mean a triple (B, J, σ) where B is a collection of bounded subsets of X, J a countable index set for B, and σ J J such that there is a map E : K 0 (θ) J σ that assigns to each x X a σ-admissible sequence ( ) αj x such j N that for all x X

15 0 (i) j= B ( α x j ) = {x}, and (ii) diam B ( α x j ) 0 as j, where B (α) denotes the element of B indexed by α. 2.3 A Self-similar covering of K i 0 0 (θ) We have access to a self-similar covering of K i 0 0 (θ). Dene J := { (m; b, b 2,..., b k ) : m Z, k κ θ, b j Z ( j k), b j a j+, if and b j = 0 if both k i0 < j < k, j κ θ } B := {I (β) : β J} and dene σ J J such that for each α ki = (m; b, b 2,..., b ki ) J we have σ (α ki ) = {( m; b, b 2,..., b ki+ 2, b ) : b a ki+ } dene Let J σ denote the set of all σ-admissible sequences in J. By Lemma 4 we can E : K 0 (θ) J σ x ( α x j ) j=0 where x I (α j ) = I (m; b,..., b j ) J 0 for each α j ( α x j ). Suppose x K i 0 0 (θ). Our triple satises (i) of the denition of a self-similar covering; apply Lemma 4 to As mentioned immediately after Theorem 3. of [2], we can take elements of B to be subsets of the ambient space X.

16 show x I (β j ) for all β j J 0. Since lim j diam I ( ) αj x = 0, j=0 I ( ) αj x = {x}. Thus, (ii) is also satised. Dene E i := I (m; b,..., b ki ) such that the union is over all nite sequences m, b,..., b ki, where k i κ θ and b j a j+ for each j, and dene E := i=0 E i 2.4 Calculation In this section we give an upper bound on HdimK i 0 0 (θ). The following lemma is a direct consequence of the denition of σ. Lemma 5. Let α J and k = α. #σ (α) a k+ (3) Proof. Let α be a sequence in J of length k i. Then α = (m; b,..., b ki ) where both j > k i0 and j κ θ imply b j = 0. Since α = k, α corresponds to an interval belonging to E i. Hence, the centers determined by the children of the intervals comprising E i are determined by all even integers b k satisfying b k a ki +.

17 2 For s 0, since the value of I (β) does not depend on the choice of β σ (α), a direct consequence of equation (3) is Further, for β σ (α) β σ(α) ( ) s ( ) s I (β) I (β) = #σ (α) (4) I (α) I (α) #σ (α) The critical value of s is s = ( ) s ( ) s I (β) nki #σ (α) I (α) n ki+ log #σ(α) log n ki+ log n ki. Let ε > 0, and let s = s + ε. Then β σ(α) ( ) s I (β) I (α) ( ) s nki n ki+ ( nki n ki+ ) ε #σ(α) (by (4)) Theorem 5.3 of [] implies HdimK 0 (θ) s. Since ε can be arbitrarily small, we have HdimK 0 (θ) s. Hence HdimK 0 (θ) lim sup i = lim sup i log #σ (α) log n ki + log n ki log a ki + (by (3)) log n ki+ log n ki

18 3 3 A NONTRIVIAL EXAMPLE OF θ SATISFYING HdimK 0 (θ) = 0 In [7], Veech showed that the Hausdor dimension of K i (θ) vanishes when θ has bounded partial quotients. Using our upper bound formula (2), we give an example for which sup j a j = and HdimK 0 (θ) = 0. Let 0 < δ be given and x M = 2 δ. We specify the continued fraction representation of θ recursively. We choose { } a, a 2,..., a k0 = M, where k 0 is the smallest index for which n δ k 0 > max 2M,. 2 δ Note that there exists an integer in between n δ k 0 and 2 δ n δ k 0 since n δ k 0 ( 2 δ ) >. Choose a k0 + Z such that n δ k 0 < a k0 + < 2n δ k 0. Recursively, given k i, dene δi log nki k i+ := k i + + log M (5) set a ki +2 = a ki +3 = = a ki+ = M, and choose a ki + such that n δ k i < a ki + < 2n δ k i. This completes the recursive denition of the sequence (a ki ) i 0. Moreover, a direct consequence of the denition of (5) is M k i+ k i n δi k i (6) Claim. Under the above construction, θ is divergent relative to [, M]. Proof. By construction of (a ki +) i 0, we have, for each i 0, a ki + > M and n δ k i < a ki + < 2n δ k i. Therefore a ki + diverges to. Theorem 2. Let θ be constructed as above. Then HdimK 0 (θ) = 0.

19 4 Proof. n ki+ > a ki+ n ki+ = Mn ki+ > Ma ki+ n ki+ 2 = M 2 n ki+ 2. = M k i+ k i n ki > n δi k i n ki (by (6)) = n δi+ k i Therefore, log n ki+ > (δi + ) log n ki, and this implies log n ki+ lim = (7) i log n ki Since θ is half-divergent, HdimK 0 (θ) lim sup i log a ki + log n ki+ log n ki (by (2)) log 2n δ k lim sup i i log n ki+ log n ki δ log n ki + log 2 lim sup i log n ki+ log n ki = δ lim sup log n i ki+ log n ki = 0 (by (7))

20 5 4 EXAMPLE OF θ FOR WHICH HdimK (θ) = 4. Falconer's Lower Bound Formula (See [4]) Let [0, ] = F 0 F F 2... be a decreasing sequence of sets, with each F k a union of a nite number of disjoint closed intervals (called kth level basic intervals), which each interval of F k containing at least two intervals of F k+, and the maximum length of kth level intervals tending to 0 as k. Then the set F = k=0 F k is a totally disconnected subset of [0, ]. The condition needed to apply Falconer's lower bound formula is ( ) Each (j )th level interval contains at least m j 2 jth level intervals (j =, 2,...) which are separated by gaps of at least γ j, where 0 < γ j+ < γ j for each j. Falconer's lower bound formula is HdimF lim inf k log (m 0 m m j ) log (m j+ γ j+ ) (8) 4.2 Lower Bound for HdimK (θ) Fix δ (0, ) and ε (0, δ). For the remainder of Section 4, let θ be an element for which there exists a k 0 suciently large so that n δ ε k 0 3, n ε k 0 9 and for all k k 0, n δ k < a k+ < 2n δ k, b k is even and b k < n δ ε k. Our strategy is to

21 6 construct an innite family of Cantor sets contained in K (θ), each of which satises the conditions needed to apply Falconer's lower bound formula, allowing us to give lower bounds of HdimK (θ) arbitrarily close to. Given b i a i+, dene by L (m; b, b 2,..., b k ) or L k the interval of length n k θ concentric with I (m; b,..., b k ). We require that for each k the length of the gaps between the intervals L k is constant. Hence, we have the following lemma. Lemma 6. The gaps between consecutive children intervals of L k are of length n k θ. Proof. The distance between the centers of adjacent intervals of L k+ is 2 n k θ. Between these centers is the gap between them as well as two half intervals of L k+. Hence, the size of the gap is n k θ. The following lemma gives a sucient condition for L k+ L k. Lemma 7. Suppose we are given m, b,..., b k, b. If b a k+ 8, then L (m; b,..., b k, b) L (m; b,..., b k ) Proof. We may suppose a k+ 9 since the claim is vacuously true otherwise. Let y L (m; b,..., b k, b), and let x be the center of the corresponding parent interval

22 7 Then L (m; b,..., b k ). Let us call x the center of the interval L (m; b,..., b k, b). x y x x + x y < b n k θ + n kθ 2 a k n k+ ( ak+ 4 2 = a k+ 4n k+ a k+ < 4a k+ n k = 4n k < n k θ 2 ) + 2 (since a k+ > 4) n k+ The following result was anticipated by Lothar Narins. Theorem 3. The number θ, as constructed in 4.2, satises the following: HdimK (θ) = Proof. Dene F j := L (m; b..., b k0 +j ) where the union is over all nite sequences such that m = b = = b k0 = 0. If k k 0, then

23 8 b k n δ ε k n δ ε k = nδ k n ε k < a k+ 8 so that, by Lemma 7, F j+ F j for j 0. Dene F (k 0, ε) := j=0 F j Lemma 8. F (k 0, ε) K (θ) Proof. If y F (k 0, ε), then we have a sequence of the form b, b 2,... such that y = m; b, b 2,..., b k0, b k0 +,... θ, where b = = b k0 = 0 and, by construction, each b k is even. We show that y satises the constraints imposed on elements of K (θ). b j n j n j θ j= j= j= n ε j= j n δ ε+ j n j+ n δ ε+ j n +δ j

24 9 Using the Ratio Test on the latter series, we have lim j n ε j n ε j+ = 0 ( ) ε nj lim j a j+ n j lim = 0 j a ε j+ Therefore, j= b j n j n j θ <. Hence y K (θ). For convenience, dene k j := k 0 + j. We show that F (k 0, ε) satises ( ). Denote by m j the number of children intervals L kj of F so that m j counts the number of even integers b satisfying b < n δ ε k j, so that m j = n δ ε k j Denote by γ j the length of the gaps between children intervals of L kj so that, by Lemma 6, γ j = nkj θ From basic continued fraction theory, < n k θ < (9) 2n k+ n k+ Since we have n kj +2 < 2n kj + for each j, we have 0 < γ k j + < γ kj for each j. By (9), lim j γ j = 0. It is the case that each interval L ( m; b,..., b kj ) contains at least 3 intervals of L kj and the gaps γ j decrease monotonically to 0 as j. Hence, the conditions for Falconer's lower bound formula (inequality 4.7 of [4]) are satised.

25 20 To simplify our calculation of the lower bound of HdimF (k 0, ε), we use the fact that k k 0 implies n k+ < (a k+ + ) n k < 3n +δ k Taking log of both sides of the expression n k+ < 3n +δ k gives log n k+ < ( + δ + log 3 ) log n k (0) log n k0 Further, m j+ = n δ ε k j+ > nδ-ε k j+ 2 and γ j+ = nkj+ θ > > > 2n kj+ + (by (9)) (since n k+ < 2a k+ n k ) 4a kj+ +n kj+ 8n +δ k j+ imply m j+ γ j+ > 6n +ε k j+ ()

26 2 Using Falconer's lower bound (8) gives HdimF (k 0, ε) lim inf j lim inf j lim inf j = lim inf j log (m 0 m j ) log (m j+ γ j+ ) (δ ε) ( ) log n k0 + log n k + + log n kj δ ε + ε ( δ ε + ε ( + ε) log n kj δ + log 3 log n k0 + δ + log 3 log n k0 ( + δ + log 3 ( +δ+ log 3 log n k0 +δ+ log 3 log n k0 ) ) j+ log n k0 ) j+ = δ ε ( ) ( + ε) δ + log 3 log n k0 HdimF (k 0, ε) as ε 0, k 0. Therefore, HdimK (θ) =.

27 22 5 HdimK i (θ) = δ Theorem 4. For any δ [0, ] there is a θ satisfying HdimK i (θ) = δ. The cases δ = 0 and δ = are handled by Theorem 2 and Theorem 3, respectively. Let us choose δ (0, ). Our strategy is to construct a half-divergent θ in terms of δ in a particular way which gives δ HdimK (θ) HdimK 0 (θ) δ Proof. Suppose M N 3. We construct a θ divergent relative to [, M]. In what follows we construct integers a k in terms of the indices k i by taking 3 a k+ M if k k i, (2) n δ k < a k+ < 2n δ k if k = k i Choose a, a 2,..., a k0 [3, M], where k 0 is large enough so that n δ k 0 > M. With this choice of k 0, a k+ > M when k = k i ; therefore, a k+ [3, M] if and only if k k i. Suppose we are given k i and that a, a 2,..., a ki, n, n 2,..., n ki + are dened according to (2). Choose k i+ to be the smallest k k i + such that n k < n 2 k i, and set a k+ [3, M] if k < k i+, n k n 2 k i, and set a k+ ( n δ k, ) 2nδ k if k = k i+

28 23 By this recursive denition, n ki+ n 2 k i and n ki+ < n 2 k i. If k i+ = k i +, then n ki+ = a ki +n ki + n ki < 2a ki +n ki < 4n +δ k i < (M + ) n 2 k i If k i+ > k i +, then k i+ is not of the form k j + for any j > i; for if j > i, then k j + > k j k i+. Therefore, in this case, we have a ki+ [3, M], so n ki+ = a ki+ n ki+ + n ki+ 2 < ( a ki+ + ) n ki+ < (M + ) n 2 k i Therefore, n 2 k i n ki+ < (M + ) n 2 k i (3) and n δ k i < a ki + < 2n δ k i Further, it will be used in the upper and lower bound calculations that (3) implies lim j log n kj log n kj+ = 2 (4)

29 24 5. Upper Bound As construction in Section 5, θ is divergent relative to [, M] since the subsequence (a ki +) i 0 of terms of (a k ) k which are larger than M also satisfy n δ k i < a ki + < 2n δ k i. Therefore, lim i a ki + =. Hence, HdimK 0 (θ) lim sup j lim sup j = lim sup j log a kj + log n kj+ log n kj (by (2)) log n δ k j + log 2 log n kj+ log n kj δ log n k j log n kj+ log n k j log n kj+ = δ (by (4)) Therefore, HdimK 0 (θ) δ. 5.2 Lower Bound Let ε (0, ). Choose k 0 k 0 suciently large such that k i k 0 implies a ε k i + < a ki+ 8. Dene F j := L ( ) m; b, b 2,... b k 0,..., b kj to be the union over all nite sequences of even terms b,... b kj such that if k = k i k 0 then b k is even and satises b k < a ε k+, otherwise bk = 0. WLOG, let m = 0. Dene

30 25 F ε = F (k 0, ε) := Lemma 7 implies F j+ F j for j N since if k i k 0, then b ki < a ε k i + j=0 F j a ε k i + < a k i+ 8 Lemma 9. F ε K (θ) Proof. Let y F ε. Then we are given a sequence of the form b, b 2,... such that y = m; b, b 2,..., b k 0, b k 0 +,... θ, b k is even and satises b k < a ε k+ if k = ki > k 0 and b k = 0 otherwise. Further, b j n j n j θ j= = a ε j+n j n j= j+ j= j= j= a ε j+n j a j+ n j a ε j+ n δ( ε) j

31 26 Proof. Using the Ratio Test on the latter series, we have n ( ε)δ j lim j n ( ε)δ j+ = 0 n ( ε)δ j lim j n ( ε)δ j+ lim j a ( ε)δ j+ Therefore, j= b j n j n j θ <. Hence y K (θ). Dene M j := a ε k j + so that M j is a lower bound on the number of intervals in F j+ contained in intervals of F j. Dene Γ j := nkj θ so that, by Lemma 6, Γ j is a lower bound on the gaps between the intervals of F j+. Since we have n kj+2 < 2n kj+ 2n kj + for each j, we have 0 < Γ j+ < Γ j for each j. Since < Γ j < n kj + for each j, lim j Γ j = 0. It is the case that each interval L ( m; b,..., b kj ) contains at least 3 intervals of Fkj+ and the gaps Γ j decrease monotonically to 0 as j. The conditions for Falconer's lower bound formula are satised.

32 27 To simplify our calculation of the lower bound of HdimF ε, we use the fact that and Γ j+ = nkj+ θ > > > 2n kj+ + (by (9)) 2 ( ) (n k+ < 2a k+ n k ) 2a kj+ +n kj+ 8n +δ k j+ M j+ = [ ] a ε k j+ + imply M j+ Γ j+ > Using Falconer's lower bound gives HdimF lim inf j = lim inf j log (M 0 M j ) log (M j+ Γ j+ ) aε k j+ + 2 > nδε k j+ 2 6n +δ δε k j+ (5) ε ( ) log a k0 + + log a k log a kj + (by (5)) ( + δ δε) log n kj+ ( ) log nk0 + log n k + + log n kj δε lim inf + δ δε j δε + δ δε δε + δ δε lim inf j log n kj+ ( ) 2 2 j+ (by (4))

33 28 HdimF (k 0, ε) δ as ε. Therefore, HdimK (θ) δ.

34 29 References [] Y. Cheung. Hausdor Dimension of the Set of Points on Divergent Trajectories of a Homogeneous Flow on a Product Space. Ergodic Theory and Dynamical Systems, 27 (2007), [2] Y. Cheung. Hausdor Dimension of the Set of Singular Pairs. Annals of Mathematics, 73 (20), [3] Y. Cheung, A. Eskin. Slow Divergence and Unique Ergodicity. Fields Institute Communications, 5 (2007), [4] Y. Cheung, P. Hubert, and H. Masur. Dichotomy for the Hausdor Dimension of the Set of Nonergodic Directions. Inventiones, 83 (200), [5] K. Falconer. Fractal Geometry: Mathematical Foundations and Applications. Chichester: John Wiley and Sons, 990. Print. [6] H. Masur and S. Tabachnikov. Rational Billiards and Flat Surfaces. Handbook of Dynamical Systems, Vol. A (2002), [7] W. Veech. Strict Ergodicity in Zero Dimensional Dynamical Systems and the Kronecker-Weyl Theorem Modulo 2. Transactions of the American Mathematical Society, 40 (969), -33.

UNIQUE ERGODICITY OF TRANSLATION FLOWS

UNIQUE ERGODICITY OF TRANSLATION FLOWS YITWAH CHEUNG AND ALEX ESKIN Abstract. This preliminary report contains a sketch of the proof of the following result: a slowly divergent Teichmüller geodesic satisfying