SVETLANA KATOK AND ILIE UGARCOVICI (Communicated by Jens Marklof)

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1 JOURNAL OF MODERN DYNAMICS VOLUME 4, NO. 4, 010, doi: /jmd STRUCTURE OF ATTRACTORS FOR (a, )-CONTINUED FRACTION TRANSFORMATIONS SVETLANA KATOK AND ILIE UGARCOVICI (Communicated y Jens Marklof) ABSTRACT. We study a two-parameter family of one-dimensional maps and related (a, )-continued fractions suggested for consideration y Don Zagier. We prove that the associated natural extension maps have attractors with finite rectangular structure for the entire parameter set except for a Cantor-like set of one-dimensional Leesgue zero measure that we completely descrie. We show that the structure of these attractors can e computed from the data (a,), and that for a dense open set of parameters the Reduction theory conjecture holds, i.e., every point is mapped to the attractor after finitely many iterations. We also show how this theory can e applied to the study of invariant measures and ergodic properties of the associated Gauss-like maps. CONTENTS 1. Introduction 637. Theory of (a, )-continued fractions Attractor set for F a, Cycle property Finiteness condition implies finite rectangular structure Finite rectangular structure of the attracting set Reduction theory conjecture Set of exceptions to the finiteness condition Invariant measures and ergodic properties 687 References INTRODUCTION The standard generators T (x) = x + 1, S(x) = 1/x of the modular group SL(, Z) were used classically to define piecewise continuous maps acting on Received March 9, 010; revised Septemer 6, Mathematics Suject Classification: 37E05, 11A55, 11K50. Key words and phrases: Continued fractions, attractor, natural extension, invariant measure. We are grateful to Don Zagier for helpful discussions and the Max Plank Institute for Mathematics in Bonn for its hospitality and support. IU: Partially supported y the NSF grant DMS IN THE PUBLIC DOMAIN AFTER AIMSCIENCES

2 638 SVETLANA KATOK AND ILIE UGARCOVICI the extended real line R = R { } that led to well-known continued fraction algorithms. In this paper we present a general method of constructing such maps suggested y Don Zagier, and study their dynamical properties and associated generalized continued fraction transformations. Let P e the two-dimensional parameter set P = { (a,) R a 0, a 1, a 1 }, and consider the map f a, : R R defined as x + 1 if x < a, (1.1) f a, (x) = 1/x if a x <, x 1 if x. Using the first-return map of f a, to the interval [a,), denoted y f ˆ a,, we introduce a two-dimensional family of continued fraction algorithms and study their properties. We mention here three classical examples: the case a = 1/, = 1/ gives the nearest-integer continued fractions considered first y Hurwitz in [7], the case a = 1, = 0 descried in [3, 9] gives the minus (ackward) continued fractions, while the situation a = 1, = 1 was presented in [1, 10] in connection with a method of symolically coding the geodesic flow on the modular surface following Artin s pioneering work [3]. We remark that in the case a = 1, the class of one-parameter maps f 1, with [0,1] is conceptually similar to the α-transformations introduced y Nakada in [18] and studied susequently in [4, 5, 15, 16, 17, 19, 0, ]. Also, in the case a =, the one parameter family f, with 1/ 1 is related to a class of maps (factors of α-transformations) studied in [4, 17]. The main oject of our study is a two-dimensional realization of the natural extension map of f a,, F a, : R R, = {(x, y) R x = y}, defined y (x + 1, y + 1) if y < a, (1.) F a, (x, y) = ( 1/x, 1/y) if a y <, (x 1, y 1) if y. Numerical experiments led Don Zagier to conjecture that such a map F a, has several interesting properties for all parameter pairs (a, ) P : REDUCTION THEORY CONJECTURE. (1) The map F a, possesses a gloal attractor set D a, = n=0 F n ( R ) on which F a, is essentially ijective. () The set D a, consists of two (or one, in degenerate cases) connected components each having finite rectangular structure, i.e., ounded y nondecreasing step-functions with a finite numer of steps. (3) Every point (x, y) of the plane (x y) is mapped to D a, after finitely many iterations of F a,. REMARK 1.1. If one identifies a pair (x, y) R with a geodesic on the upper half-plane from x to y, then F a, maps geodesics to geodesics. The attractor of

3 ATTRACTORS OF (a,)-continued FRACTIONS 639 F a, corresponds to reduced geodesics, hence the name of the conjecture and the reason why F a, is also called the reduction map. This approach also has applications to coding symolically geodesics on the modular surface [14] a FIGURE 1. Attracting domain for Zagier s example: a = 4 5, = 5 Figure 1 shows the computer picture of such a the set D a, with a = 4/5 and = /5. It is worth mentioning that the complexity of the domain D a, increases as (a,) approach the line segment a = 1 in P, a situation fully analyzed in what follows. The main result of this paper is the following: MAIN RESULT. There exists an explicit one-dimensional Leesgue measure zero, uncountale set E that lies on the diagonal oundary = a + 1 of P such that: (i) for all (a,) P E, the map F a, has an attractor D a, satisfying properties (1) and () aove; (ii) for an open and dense set in P E, property (3), and hence the Reduction theory conjecture, holds. For the rest of P E, property (3) holds for almost every point of R. We point out that this approach gives explicit conditions for the set D a, to have finite rectangular structure that are satisfied, in particular, for all pairs (a,) in the interior of the maximal parameter set P. At the same time, it provides an effective algorithm for finding D a,, independent of the complexity of its oundary (i.e., numer of horizontal segments). The simultaneous properties satisfied y D a,, attracting set and ijectivity domain for F a,, is an essential feature that has not een exploited in earlier works. This approach makes the notions of reduced geodesic and dual expansion natural and transparent, with a potential for generalization to other Fuchsian groups. We remark that for α-transformations [18, 16], explicit descriptions of the domain of the natural

4 640 SVETLANA KATOK AND ILIE UGARCOVICI extension maps have een otained only for a suset of the parameter interval [0, 1] (where the oundary has low complexity). The paper is organized as follows. In Section, we develop the theory of (a,)-continued fractions associated to the map f a,. In Section 3 we prove that the natural extension map F a, possesses a trapping region; it will e used in Section 6 to study the attractor set for F a,. In Section 4 we further study the map f a,. Although it is discontinuous at x = a,, one can look at two orits of each of the discontinuity points. For generic (a, ), these orits meet after finitely many steps, forming a cycle that can e strong or weak, depending on whether the composite transformations at the end of each side of the cycle are equal or not. The values appearing in these cycles play a crucial role in the theory. Theorems 4. and 4.5 give necessary and sufficient conditions for and a to have the cycle property. In Section 5 we introduce the finiteness condition using the notion of truncated orits and prove that under this condition the map F a, has a ijectivity domain A a, with a finite rectangular structure that can e computed from the data (a,) (Theorem 5.5). In Section 6 we define the attractor for the map F a, y iterating the trapping region, and identify it with the earlier constructed set A a, assuming the finiteness condition (Theorem 6.4). In Section 7 we prove that the Reduction theory conjecture holds under the assumption that oth a and have the strong cycle property, and that under the finiteness condition, property (3) holds for almost every point of R. In Section 8 we prove that the finiteness condition holds for all (a,) P except for an uncountale set of one-dimensional Leesgue measure zero that lies on the oundary = a +1 of P (Theorem 8.11), and we present a complete description of this exceptional set. We conclude the proof of the Main Result y showing that the set of (a,) P where a and have the strong cycle property is open and dense in P (Proposition 8.1). And, finally, in Section 9 we show how these results can e applied to the study of invariant measures and ergodic properties of the associated Gauss-like maps.. THEORY OF (a, )-CONTINUED FRACTIONS Consider (a,) P. The map f a, defines what we call (a,)-continued fractions using a generalized integral part function x a, : for any real x, let x a if x < a x a, = 0 if a x < x if x, where x denotes the integer part of x and x = x + 1. Let us remark that the first-return map of f a, to the interval [a,), fa, ˆ, is given y the function fˆ a, (x) = 1 x 1 = T 1/x a, S(x) if x 0, f (0) = 0. x a,

5 ATTRACTORS OF (a,)-continued FRACTIONS 641 We prove that any irrational numer x can e expressed in a unique way as an infinite (a, )-continued fraction x = n 0 n n 1... which we will denote y n 0,n 1,... a, for short. The digits n i, i 1, are nonzero integers determined recursively y (.1) n 0 = x a,, x 1 = 1 x n 0, and n i = x i a,, x i+1 = 1 x i n i. In what follows, the notation (α 0,α 1,...,α k ) is used to write formally a minus continued fraction expression, where α i are real numers. THEOREM.1. Let x e an irrational numer, {n i } the associated sequence of integers defined y (.1) and r k = (n 0,n 1,...,n k ). Then the sequence r k converges to x. Proof. We 1 start y proving that none of the pairs of type (p,1), ( p, 1), with p 1 are allowed to appear as consecutive entries of the sequence {n i }. Indeed, if n i+1 = 1, then x i+1 = 1 x i n i < + 1. Therefore, 1 x i n i < and n i < 0. If n i+1 = 1, then a 1 x i+1 = 1 x i n i < a, so 1 a 1 x i n i < 1 a. But a a 1, thus n i > 0. With these two restrictions, the argument follows the lines of the proof for the classical case of minus (ackward) continued fractions [9], where n i, for all i 1. We define inductively two sequences of integers {p k } and {q k } for k : (.) p = 0, p 1 = 1, p k = n k p k 1 p k for k 0; q = 1, q 1 = 0, q k = n k q k 1 q k for k 0. We have the following properties: (i) there exists l 1 such that q l < q l+1 < < q k < ; (ii) (n 0,n 1,...,n k,α) = (αp k p k 1 )/(αq k q k 1 ) for any real numer α; (iii) p k q k+1 p k+1 q k = 1. 1 The authors proved initially the convergence statement assuming 1 a 0 1, and two Penn State REU students, Tra Ho and Jesse Barour, worked on the proof for a and outside of this compact triangular region. The unified proof presented here uses some of their ideas.

6 64 SVETLANA KATOK AND ILIE UGARCOVICI Let us prove property (i) since the proof is more intricate than in the classical case. Oviously 1 = q 0 q 1 = n 1, q = n q 1 q 0 = n n 1 1. Notice that q > q 1 unless n 1 = 1, n = or n 1 = 1, n =. We analyze the situation n 1 = 1, n =. This implies that q 3 = n 3 (n n 1 1) n 1 = n 3 n 1, so q 3 > q, unless n 3 =. Notice that it is impossile to have n i = for all i, ecause x is irrational and the minus continued fraction expression consisting only of two s, (,,...), has numerical value 1. Therefore, there exists l 1 such that n l+1 1,. This implies that q l+1 > q l. We continue to proceed y induction. Assume that property (i) is satisfied up to k-th term, k > l. If n k+1, then q k+1 n k+1 q k q k 1 q k q k 1 > q k. If n k+1 = 1, then q k+1 = q k q k 1. Since q k = n k q k 1 q k with n k < 0, one gets q k 1 = q k + q k n k. We analyze the two possile situations If q k > 0 then q k < q k, so q k + q k > 0 and q k 1 < 0. This implies that q k+1 = q k q k 1 q k > 0. If q k < 0, then q k < q k, so q k + q k < 0 and q k 1 > 0. This implies that q k+1 = q k q k 1 < q k < 0. Thus q k < q k+1. A similar argument shows that the inequality remains true if n k+1 = 1. To finish the proof we use properties (i) (iii). First, we see that r k = p k /q k for k 0. Moreover, the sequence r k is a Cauchy sequence ecause r k+1 r k = 1 q k q k+1 1 (k l) for k > l. Hence r k is convergent. In order to prove that r k converges to x, we write x = (n 0,n 1,...,n k, x k+1 ), and look only at those terms (n 0,n 1,...,n k, x k+1 ) with x k+1 1. There are infinitely many such terms: indeed, if 1 a < 1, then x k+1 1 for all k 1; if a < 1 and x k+1 < 1, then x k+1 < 1, so x k+ = 1/(x k+1 1) 1; if > 1 and x k+1 < 1, then 1 < x k+1 < a, so x k+ = 1/(x k+1 + 1) 1. Therefore, the corresponding susequence r k = p k /q k satisfies p k x q = p k p k x k+1 p k 1 k q k q k x k+1 q = 1 k 1 q k (q k x k+1 q k 1 ) 1 q k ( q k x k+1 q k 1 ) 1 q k 0. We showed that the convergent sequence r k = p k /q k has a susequence convergent to x, therefore the whole sequence converges to x. REMARK.. One can construct (a, )-continued fraction expansions for rational numers, too. However, such expansions will terminate after finitely many steps if 0. If = 0, the expansions of rational numers will end with a tail of

7 ATTRACTORS OF (a,)-continued FRACTIONS 643 s, since 0 = (1,,,...). In this latter case, properties (i) (iii) listed aove still hold. REMARK.3. It is easy to see that if the (a,)-continued fraction expansion of a real numer is eventually periodic, then the numer is a quadratic irrationality. It is not our intention to present in this paper some of the typical numertheoretical results that can e derived for the class of (a, )-continued fractions. However, we state and prove a simple version aout (a, )-continued fractions with ounded digits. For the regular continued fractions, this is a classical result due to Borel and Bernstein (see [6, Theorem 196] for an elementary treatment). We are only concerned with (a, )-expansions that are written with two consecutive digits, a result explicitly needed in Sections 7 and 8. PROPOSITION.4. The set Γ (m) a, = { x = 0,n 1,n,... a, n k {m,m +1} } has zero Leesgue measure for every m 1. Proof. First, notice that if m = 1, then the set Γ (1) has oviously zero measure, a, since the pairs (,1) and (, 1) are not allowed in the (a,)-expansions. Assume m. Notice that Γ (m) a, Γ(m) 0, 1 since a formal continued fraction x = (0,n 1,n,...) with n k {m,m + 1} coincides with its minus (ackward) continued fraction expansion (a = 1, = 0), x = 0,n 1,n,... 1,0. The reason is that any sequence of digits n i gives a valid minus continued fraction expansion. In what follows, we study the set Γ (m) 0, 1. For practical reasons we will drop the suscript (0, 1). It is worth noticing that the result for Γ (m) 0, 1 does not follow automatically from the result aout regular continued fractions, since there are numers for which the (0, 1)-expansion has only digits and 3, while the regular continued fractions expansion has unounded digits. We follow the approach of [6, Theorem 196] and estimate the size of the set Γ (m) n 1,n,...,n k Γ (m) with the digits n 1,n,...,n k {m,m + 1} eing fixed. In this particular case, the recursive relation (.) implies that 1 = q 1 < q < < q k. If x Γ n (m) 1,n,...,n k, then (0,n 1,n,...,n k 1) x < (0,n 1,n,...,n k ). Using property (ii), the end-points of such an interval I n (m) 1,...,n k are given y (n k 1)p k 1 p k n k p k 1 p k and (n k 1)q k 1 q k n k q k 1 q k and the length of this interval is l(i n (m) 1 1,...,n k ) = (n k q k 1 q k )((n k 1)q k 1 q k ) = 1 q k (q k q k 1 ) y using that p k q k 1 p k 1 q k = 1 and q k = n k q k 1 q k. Denote y Γ (m) the set of numers in [ 1, 0) with ( 1, 0)-continued fraction k digits n 1,n,...,n k {m,m + 1}. The set Γ (m) is part of the set k = I n (m) 1,...,n k. I (m) k n 1,...,n k {m,m+1}

8 644 SVETLANA KATOK AND ILIE UGARCOVICI We have the following relation: I (m) k+1 = I n (m) 1,...,n k,m I (m) n 1,...,n k,m+1. n 1,...,n k {m,m+1} If x lies in I (m) n 1,...,n k,m I (m) n 1,...,n k,m+1, then (0,n 1,n,...,n k,m 1) x < (0,n 1,n,...,n k,m + 1). The length of this interval is l(i (m) n 1,...,n k,m I (m) n 1,...,n k,m+1 ) = ( (m + 1)qk q k 1 )( (m 1)qk q k 1 ). Now we estimate the ratio l(i (m) n 1,...,n k,m I (m) n 1,...,n k,m+1 ) l(i (m) n 1,n,...,n k ) q k (q k q k 1 ) = ((m + 1)q k q k 1 )((m 1)q k q k 1 ) q k (m + 1)q k q k 1 q k = 3q k q k 1 3 q k 1 /q k k k + 1 since q k 1 /q k (k 1)/k. Indeed, if n 1 = = n k =, then q k 1 /q k = (k 1)/k; if some n j >, then q k 1 /q k 1/ from (.). This proves that for every k 1 so I (m) k+1 k k + 1 I (m) k l(i (m) 4 (k ) (m) ) l(i k 1 ) 0 as k. 3 5 (k 1) Therefore, in all cases, l(i (m) ) 0 as k. Since Γ (m) I (m) for every k 1, k k the proposition follows. REMARK.5. By a similar argument, the set Γ ( m) = { x = 0,n a, 1,n,... a, n k { m, m 1} } has zero Leesgue measure for every m ATTRACTOR SET FOR F a, The reduction map F a, defined y (1.) has a trapping domain, i.e., a closed set Θ a, R with the following properties: (i) for every pair (x, y) R, there exists a positive integer N such that F N a, (x, y) Θ a,; (ii) F a, (Θ a, ) Θ a,.

9 ATTRACTORS OF (a,)-continued FRACTIONS 645 THEOREM 3.1. The region Θ a, consisting of two connected components (or one if a = 0 or = 0) defined as [, 1] [ 1, ] [ 1,0] [ 1/a, ] if 1, a 0, Θ u a, = if a = 0, [, 1] [ 1, ] [ 1,0] [min( 1, 1 a ), ] [0,1] [ 1 1, ] if 0 < < 1, and [0,1] [, 1/] [1, ] [, a + 1] if a 1, 0, Θ l a, = if = 0, [ 1,0] [, 1 a a+1 ] [0,1] [,max( a+1, 1 )] if a > 1, [1, ] [, a + 1] is the trapping region for the reduction map F a,. 4 4 a 0-0 a FIGURE. Typical trapping regions: case a < 1, 0 < < 1 (left); case 1 < a < 0 < < 1 (right) Proof. The fact that the region Θ a, is F a, -invariant is verified y a direct calculation. We focus our attention on the attracting property of Θ a,. Let (x, y) R, write y = n 0,n 1,... a,, and construct the following sequence of real pairs {(x k, y k )} (k 0) defined y x 0 = x, y 0 = y, and y k+1 = ST nk ST n 1 ST n 0 y, x k+1 = ST nk ST n 1 ST n 0 x. If y is rational and its (a,)-expansion terminates y = n 0,n 1,...,n l a,, then y l+1 = ±, so (x, y) lands in Θ a, after finitely many iterations. If y has an infinite (a,)-expansion, then y k+1 = n k+1,n k+,... a,, and y k+1 1/a or

10 646 SVETLANA KATOK AND ILIE UGARCOVICI y k+1 1/ for k 0. Also, Hence, y = T n 0 ST n 1 S T n k S(y k+1 ) = p k y k+1 p k 1 q k y k+1 q k 1 x = T n 0 ST n 1 S T n k S(x k+1 ) = p k x k+1 p k 1 q k x k+1 q k 1. x k+1 = q k 1x p k 1 q k x p k = q k q k q k (p k/q k x) = q k 1 + ε k. q k Since p k /q k y x, and q k, we otain that ε k < 1/q k for large enough k. But one of the properties of the sequence {q k } is that q k 1 < q k for large enough k (see property (i) in the proof of Theorem.1 and Remark..) Therefore, x k+1 q k for large enough k. q k q k We proved that there exists N > 0, such that F N a, (x, y) = ST nk ST n 1 ST n 0 (x, y) [ 1,1] ([ 1/a, ] [, 1/]). The point F N a, (x, y) =: ( x, ỹ) elongs to Θ a,, unless < 1 and ( x, ỹ) [0,1] [ 1/a, 1/( 1)] or a > 1 and ( x, ỹ) [ 1,0] [ 1/, 1/(a + 1)]. Let us study the next iterates of ( x, ỹ) [0,1] [ 1/a, 1/( 1)]. If ỹ + 1 then F a, ( x, ỹ) = ( x, ỹ ) [ 1,1] [ 1, ], so F a, ( x, ỹ) Θ a,. If it so happens that 1/a ỹ < + 1, then and F a, ( x, ỹ) = ( x 1, ỹ 1) [ 1,0] [0,] F a, ( x, ỹ) = ST 1 ( x, ỹ) [0, ] [ 1/, ] Θ a,. Similarly, if (x, y) [ 1,0] [ 1/, 1/(a + 1)], then F a, (x, y) Θ a,. Notice that if a = 0, then y k+1 1/ for all k 0 (so Θ u = ) and if = 0, a, then y k+1 1/a for all k 0 (so Θ l a, = ). Using the trapping region descried in Theorem 3.1 we define the associated attractor set (3.1) D a, = D n, where D n = n i=0 F i a, (Θ a,). REMARK 3.. In the particular cases when a = 0 and 1, or = 0 and a 1, or (a,) = ( 1,1), the trapping regions n=0 Θ 0, = [ 1,0] [, 1] [0,1] [,0] [1, ] [,1] Θ a,0 = [, 1] [ 1, ] [ 1,0] [0, ] [0,1] [1, ] Θ 1,1 = [, 1] [ 1, ] [ 1,0] [1, ] [0,1] [, 1] [1, ] [,0]

11 ATTRACTORS OF (a,)-continued FRACTIONS 647 are also ijectivity domains for the corresponding maps F a,. Therefore, in these cases the attractor D a, coincides with the trapping region Θ a,, so the properties mentioned in the introduction are oviously satisfied. In what follows, all our considerations will exclude these degenerate cases. 4. CYCLE PROPERTY In what follows, we simplify the notations for f a,, a,, fa, ˆ and F a, to f,, f ˆ and F, respectively, assuming implicitly their dependence on parameters a,. We will use the notation f n (or f ˆn ) for the n-times composition operation of f (or f ˆ ). Also, for a given point x (a,) the notation f ˆ(k) means the transformation of type T i S (i is an integer) such that fˆ k (x) = f ˆ(k) f ˆ(k 1) f ˆ() f ˆ (1) (x), where f ˆ(1) (x) = f ˆ (x). The map f is discontinuous at x = a,. However, we can associate to each a and two forward orits: to a we associate the upper orit O u (a) = {f n (Sa)}, and the lower orit O l (a) = {f n (Ta)}; to the lower orit O l () = {f n (S)} and the upper orit O u () = {f n (T 1 )}. We use the convention that if an orit hits one of the discontinuity points a or, then the next iterate is computed according to the lower or upper location. For example, if the lower orit of hits a, then the next iterate is Ta, if the upper orit of hits a then the next iterate is Sa. Now we explore the patterns in the aove orits. The following property plays an essential role in studying the map f. DEFINITION 4.1. We say that the point a has the cycle property if for some nonnegative integers m 1 and k 1, f m 1 (Sa) = f k 1 (Ta) = c a. We will refer to the set {Ta, f Ta,..., f k 1 1 Ta} as the lower side of the a-cycle, to the set {Sa, f Sa,..., f m 1 1 Sa} as the upper side of the a-cycle, and to c a as the end of the a-cycle. If the transformations f k 1 T and f m 1 S at the end of each side of the a-cycle are equal, we say that a has the strong cycle property. Otherwise, we say that a has the weak cycle property. Similarly, we say that has the cycle property if for some nonnegative integers m and k, f k (S) = f m (T 1 ) = c. We will refer to the set {S, f S,..., f k 1 S} as the lower side of the -cycle, to the set {T 1, f T 1,..., f m 1 T 1 } as the upper side of the -cycle, and to c as the end of the -cycle. If the transformations f k T and f m S at the end of each side of the -cycle are equal, we say that has the strong cycle property. Otherwise, we say that has the weak cycle property. A similar matching condition is used y the authors of [5, 0] to study the measuretheoretic entropy of α-transformations.

12 648 SVETLANA KATOK AND ILIE UGARCOVICI It turns out that the cycle property is the prevalent pattern. It can e analyzed and descried explicitly y partitioning the parameter set P ased on the first digits of S, STa, and Sa, ST 1, respectively. Figure 3 shows a part of the countale partitions, with B 1,B,... denoting the regions where S has the first digit 1,,..., and A 1, A,..., denoting the regions where Sa has the first digit 1,,... For most of the parameter region, the cycles are short: everywhere except for the narrow triangular regions shown in Figure 3 the cycles for oth a and end after the first return to [a,). However, there are Cantor-like recursive sets where the lengths of the cycles can e aritrarily long. Part of this more complex structure, studied in details in Section 8, can e seen as narrow triangular regions close to the oundary segment a = 1. A 1 A... 1 B 1 B a 0 FIGURE 3. The parameter set P and its partition By symmetry of the parameter set P with respect to the line = a, (a,) (, a), we may assume that a and concentrate our attention to this suset of P. The structure of the set where the cycle property holds for is descried next for the part of the parameter region with 0 < a < 1. We make use extensively of the first-return map f ˆ. THEOREM 4.. Let (a,) P, 0 < a < 1, and m 1 with a T m S < a + 1. (I) Suppose that there exists n 0 such that ( fˆ k T m ) S + 1, a + 1 for k < n, and [ f ˆ n T m S a, + 1 ].

13 ATTRACTORS OF (a,)-continued FRACTIONS 649 (i) If f ˆ n T m S (a, /( + 1)), then has the cycle property; the cycle property is strong if and only if f ˆ n T m S 0. (ii) If f ˆ n T m S = a, then has the cycle property if and only if a has the cycle property. (iii) If f ˆ n T m S = /( + 1), then does not have the cycle property, ut the orits of S and T 1 are periodic. (II) If f ˆ k T m S (/( + 1), a + 1) for all k 0, then does not have the cycle property. Proof. (I). In the case m = 1, and assuming a < T S < a + 1 we have (4.1) a < 1 1 < + 1, and the cycle relation for can e explicitly descried as T 1 1 S 1 1 T 1 (4.) S c = 1 S 1 T 1 In the particular situation that T S = a, the lower orit of hits a and continues to a + 1, while the upper orit hits /(1 ) = 1/a. This means that the iterates will follow the lower and upper orits of a, respectively, thus statement (ii) holds. Since the second inequality (4.1) is strict, the case (iii) cannot occur. For the case m = (and assuming T S a), we analyze the following situations: if < 1/, then 1/ < 0, and the cycle relation is T 1 1 S 1 1 ST T 1 (4.3) S S c = 1 1 T 1

14 650 SVETLANA KATOK AND ILIE UGARCOVICI If > 1/ we have 0 < 1/ /( +1), since we must also have 1/ < a +1, i.e., 1/(1 a), and the cycle relation is T 1 1 S 1 1 ST c = S 1 T 1 S 1 The aove cycles are strong. If = 1/ the cycle relation is T 1 1 S 1 1 T T S T c = 1 = 0 1 It is easy to check that this cycle is weak. In the particular situation when T S = a, the lower orit of hits a, and continues with a + 1, while the upper orit still hits /(1 ) = 1/a. This means that the iterates will follow the lower and upper orits of a, respectively, and statement (ii) holds. The relation 1/ = /( + 1) implies = 1+ 5 that does not have the cycle property and the orits of S and T 1 are periodic; this is the only possiility for (iii) to hold. The situation for m 3 is more intricate. First we will need the following lemmas. LEMMA 4.3. Suppose ST Sx = y. Then each of the following are true. (a) If T S x < a, then 1 y < a/(1 a). () If a x < /( + 1), then a/(1 a) y <. (c) If /( + 1) x < a + 1, then y < a/(1 a) + 1. (d) If x = 0, then y = 0. Proof. Applying ST S to the corresponding inequalities, we otain (a) 1 = ST ST S y < ST Sa = a 1 a () a = ST Sa y < ST ST ST = 1 a (c) = ST ST ST y < ST STa = T 1 Sa 1 1 a = a 1 a + 1,

15 ATTRACTORS OF (a,)-continued FRACTIONS 651 where the last inequality is valid for a 1 5, which is true in the considered region 1/( a). Relation (d) is ovious. LEMMA 4.4. Suppose that for all k < n (4.4) Then + 1 < ˆ f k T m S < a + 1. (i) for 0 k n, in the lower orit of, f ˆ (k) = T m S or T m+1 S; in the upper orit of, f ˆ (k) = T i S with i = or 3; (ii) there exists p > 1 such that (4.5) (ST S) ˆ f n T m S = (T S) ˆ f p T 1. Proof. (i). Applying T m S to the inequality (4.4), we otain a 1 T m 1 S = T m ST ST < T m S ˆ f k T m S T m STa T m S < a + 1. Therefore f ˆ(k+1) = T m S or T m+1 S. Since f ˆ(0) = T m S, we conclude that f ˆ(k) = T m S or T m+1 S for 0 k n. (ii). In order to determine the upper side of the -cycle, we will use the following relation in the group SL(, Z) otained y concatenation of the standard relations (from right to left) (4.6) (ST S)T i S = (T S) i 1 T 1 (i 1), and Lemma 4.3 repeatedly. The proof is y induction on n. For the ase case n = 1, we have + 1 < T m S < a + 1. Then for 1 i m 1, T i S satisfies Lemma 4.3(a). Hence 1 < (T S) i T 1 < a 1 a, which means that on the upper side of the -cycle f ˆ(1) = T 1 and f ˆ(i ) = T S for 1 < i m 1. Using (4.6) for i = m, we otain (ST S)T m S = (T S) m 1 T 1 = (T S) ˆ f m T 1. In other words, (4.5) holds with p = m. Now suppose the statement holds for n = n 0, and for all k < n we have + 1 < f ˆk T m S < a + 1. By the induction hypothesis, there exists p 0 > 1 such that (4.7) (ST S) ˆ f n 0 T m S = (T S) ˆ f p 0 T 1. But since + 1 < f ˆn 0 T m S < a + 1, the condition of Lemma 4.3(c) is satisfied. Hence, < (T S) ˆ f p 0 T 1 < a 1 a + 1,

16 65 SVETLANA KATOK AND ILIE UGARCOVICI which is equivalent to 1 < (T 3 S) f ˆp 0 T 1 < a 1 a, i.e., ˆ f (p 0 +1) = T 3 S. Using the relation (ST S)T S = T 1 (ST S), we can rewrite equation (4.7) as (4.8) (ST S)T S ˆ f n 0 T m S = (T 3 S) ˆ f p 0 T 1 = ˆ f p 0+1 T 1. Let ˆ f (n 0+1) = T q S. We have proved in (i) that q = m or m + 1, hence q 3. Let 0 = T S ˆ f n 0 T m S and c 0 = (T 3 S) ˆ f p 0 T 1. Then (ST S) 0 = c 0 y (4.8). Using the relation (ST S)T = T S(ST S), we otain (ST S)T i = (T S) i (ST S), and therefore, (4.9) (ST S)T i 0 = (T S) i (ST S) 0 = (T S) i c 0. Since, for 0 i < q, T i 0 satisfies condition (a) of Lemma 4.3, we conclude that 1 < (T S) i c 0 < a 1 a. Therefore f ˆ(i ) = T S and (4.9) for i = q give us the desired relation with p = p 0 + q. (ST S) ˆ f n 0+1 T m S = (T S) ˆ f p 0+q T 1 Now we complete the proof of part (I) of Theorem 4.. In what follows we introduce the notations ( ) ( a ) I l = a, and I u = a, and write I l, I u for the corresponding closed intervals. If f ˆn T m S I l, then condition () of Lemma 4.3 is satisfied, and (T S) f ˆp T 1 I u. It follows that f ˆ (p+1) = T S. Therefore (4.5) can e rewritten as (ST S) f ˆn T m S = f ˆp+1 T 1, which means that we reached the end of the cycle. More precisely, (i) If f ˆn T m S (0,/( + 1)), then f ˆp T 1 (0,), T S f ˆn T m S = S f ˆp T 1 = c, and 1 < f ˆj T 1 < a 1 a, for j < p. In this case, c < S. If f ˆn T m S (a,0), then f ˆp T 1 (a/(1 a), 0), S f ˆn T m S = T 1 S f ˆp T 1 = c, and 1 < f ˆj T 1 < a 1 a, for j < p. In this case, c > Sa. Since the cycle relation in oth cases is equivalent to the identity (4.5), the cycle property is strong, and (i) is proved. If f ˆn T m S = 0, then f ˆn T m S = f ˆp T 1 = 0 is the end of the cycle and 1 < f ˆj T 1 < a 1 a for j < p. In this case the cycle ends efore the identity (4.5) is complete, therefore the product over the cycle is not equal to identity, and the cycle is weak.

17 ATTRACTORS OF (a,)-continued FRACTIONS 653 (ii) If f ˆn T m S = a, then following the argument in (i) and using relation (4.5) we otain that the upper orit of hits T 1 S f ˆp T 1 = S f ˆn T m S = Sa = 1/a, while the lower orit hits the value a + 1, hence satisfies the cycle property if and only if a does. (iii) If f ˆn T m S = /( + 1), then following the argument in (i), we otain that (T S) f ˆp T 1 =. However, one needs to apply one more T 1 to follow the definition of the map f, hence f ˆ(p+1) = T 3 S, not T S, and the cycle will not close. One also oserves that in this case the (a,)-expansions of S and T 1 will e periodic, and therefore the cycle will never close. We prove now part (II). If f ˆk T m S I l for all k 0, y the argument in the part (I) of the proof, on the lower orit of each f ˆ(k) = T q S, where q = m or m + 1, and on the upper orit of each f ˆ(p) = T r S, where r = or 3, and for all p 1, f ˆ p T 1 I u. This means that for all images under the original map f on the lower orit of we have f k S ( 1 1 ) (, a ) + 1, a + 1 while for the images on the upper orit of ( f k T 1 a 1, 1 a ) (,1 1 ). a Since these ranges do not overlap, the cycle cannot close, and has no cycle property. A similar result holds for the a-cycles. First, if Sa has the first digit 1, i.e., Sa < + 1, then one can easily write the a-cycle, similarly to (4.1). For the rest of the parameter region we have the following. THEOREM 4.5. Let (a,) P, 0 < a < 1 with Sa +1 and m 1 such that a T m STa < a + 1. (I) Suppose that there exists n 0 such that ( fˆ k T m ) STa + 1, a + 1 for k < n, and [ f ˆ n T m STa a, + 1 (i) If f ˆ n T m STa (a, /( + 1)), then a has the cycle property; the cycle property is strong if and only if f ˆ n T m STa 0. (ii) If f ˆ n T m STa = /( + 1), then a has the cycle property if and only if has the cycle property. (iii) If f ˆ n T m STa = a, then a does not have the cycle property, ut the (a, )-expansions of Sa and Ta are eventually periodic. (II) If f ˆ k T m STa (/( +1), a +1) for all k 0, then a does not have the cycle property. Proof. The proof follows the proof of Theorem 4. with minimal modifications. In particular, the relation (4.5) should e replaced y relation (4.10) (ST S) ˆ f n T m ST = (T S) ˆ f p. ].

18 654 SVETLANA KATOK AND ILIE UGARCOVICI For (ii), since f ˆn T m STa = +1, on the lower side we have T S f n T m STa = S, and on the upper side, using (4.10), (T S) f ˆp =. As in the proof of Theorem 4., f ˆ p+1 = T 3 S, so (T 3 S) f ˆp = T 1. Therefore a has (strong or weak) cycle property if and only if does. Let us now descrie the situation when a 1. THEOREM 4.6. Let (a,) P with 0 < a and a 1. Then a and satisfy the cycle property. Proof. It is easy to see that a = 1 has the degenerate weak cycle: (4.11) a = 1 S 1 T T 1 0 while a < 1 satisfies the following strong cycle relation: S 1 a T 1 1 a 1 S a a + 1 T 1 (4.1) a T S c a = 1 a + 1 a + 1 In order to study the orits of, let m 0 such that a T m S < a + 1. If m = 0, then S = a (since S a), and the cycle of is identical to the one descried y (4.11). If m 1, then one can use relation (4.6) to construct the -cycle. More precisely, if a < T m S < a + 1, then we have: T 1 1 S 1 1 (ST ) m m T 1 S S c = 1 m 1 T m 1 m If T m S = a, then it happens again that the lower orit of hits a, and then Ta, while the upper orit hits Sa. Following now the cycle of a descried y (4.1), we conclude that satisfies the strong cycle property.

19 ATTRACTORS OF (a,)-continued FRACTIONS 655 If T m S = 0, i.e., = 1/m, then a minor modification of the aove -cycle gives us the following weak cycle relation: 1 S 1 1 T 1 (ST ) m 1 m + = 1 T 1 T 1 c = 0 S T 1 = m T m 1 1 The following corollaries are immediate from the proof of Theorems 4., 4.5, and 4.6. COROLLARY 4.7. If has the cycle property, then the upper side of the -cycle and the lower side of the -cycle do not have repeating values. {T 1, f T 1,..., f m 1 T 1 } {S, f S,..., f k 1 S} COROLLARY 4.8. If a has the cycle property, then the upper side of the a-cycle and the lower side of the a-cycle do not have repeating values. {Sa, f Sa,..., f m 1 1 Sa} {Ta, f Ta,..., f k 1 1 Ta} 5. FINITENESS CONDITION IMPLIES FINITE RECTANGULAR STRUCTURE In order to state the condition under which the natural extension map F a, has an attractor with finite rectangular structure mentioned in the Introduction, we follow the split orits of a and, and define the truncated orits L a and U a y O l (Ta) if a has no cycle property, L a = lower part of a-cycle if a has strong cycle property, lower part of a-cycle {0} if a has weak cycle property, O u (Sa) if a has no cycle property, U a = upper part of a-cycle if a has strong cycle property, lower part of a-cycle {0} if a has weak cycle property.

20 656 SVETLANA KATOK AND ILIE UGARCOVICI Similarly, L and U are defined y O l (S) L = lower part of -cycle lower part of -cycle {0} O u (T 1 ) U = upper part of -cycle lower part of -cycle {0} if has no cycle property, if has strong cycle property, if has weak cycle property, if has no cycle property, if has strong cycle property, if has weak cycle property. We find it useful to introduce the map ρ a, : R {T,S,T 1 } T if x < a ρ a, (x) = S if a x < T 1 if x in order to write f a, (x) = ρ a, (x)x and F a, (x, y) = (ρ(y)x,ρ(y)y). REMARK 5.1. It follows from the aove definitions that ρ(y) = S or T if y L a L, and ρ(y) = S or T 1 if y U a U. DEFINITION 5.. We say that the map f a, satisfies the finiteness condition if the sets of values in all four truncated orits L a, L, U a, and U are finite. PROPOSITION 5.3. The following statements are equivalent. (1) The set L is finite. () The set U is finite. (3) Either has the cycle property, or the upper and lower orits of are eventually periodic. Similar statements hold for the sets L a and U a. Proof. The two properties follow from Theorem 4. and its proof. If does not have the cycle property, ut its lower orit is eventually periodic, then one uses the argument in Lemma 4.4 to conclude that the upper orit of has to e eventually periodic. REMARK 5.4. If has the strong cycle property, then the set L coincides with the lower side of the -cycle and U coincides with the upper side of the - cycle. If does not have the cycle property, ut the lower and upper orits of are eventually periodic then L and U are identified with these orits accordingly, until the first repeat. THEOREM 5.5. Let (a,) P, a 0, 0, and assume that the map f a, satisfies the finiteness condition. Then there exists a set A a, R with the following properties. (i) The set A a, consists of two connected components each having finite rectangular structure, i.e., ounded y nondecreasing step-functions with a finite numer of steps.

21 ATTRACTORS OF (a,)-continued FRACTIONS 657 (ii) F a, : A a, A a, is a ijection except for some images of the oundary of A a,. Proof. We construct a set A a, whose upper connected component is ounded y a step-function with values in the set U a, = U a U that we refer to as upper levels, and whose lower connected component is ounded y a stepfunction with values in the set L a, = L a L that we refer to as lower levels. Notice that each level in U a and U appears exactly once, ut if the same level appears in oth sets, we have to count it twice in U a,. The same remark applies to the lower levels. Now let y l L a, e the closest y-level to S with y l S, and y u U a, e the closest y-level to Sa with y u Sa. Since each level in U a and in L appears only once, if y u = Sa, y u can only elong to U, and if y l = S, y l can only elong to L a. We consider the rays [, x ] {} and [x a, ] {a}, where x a and x are unknown, and transport them (using the special form of the natural extension map F a, ) along the sets L, U, L a, and U a, respectively, until we reach the levels y u and y l (see Figure 4). Now we set-up a system of two fractional linear equations y equating the right end of the segment at the level S with the left end of the segment at the level y l, and, similarly, the left end of the segment at the level Sa and the right end of the level y u. Sa ST 1 y u a x y l x a S STa FIGURE 4. Construction of the domain A a, In what follows, we present the proof assuming that 0 < a < 1. The situation a 1 is less complex due to the explicit cycle expressions descried in Theorem 4.6 and will e discussed at the end of this section.

22 658 SVETLANA KATOK AND ILIE UGARCOVICI LEMMA 5.6. The system of two equations at the consecutive levels y u and Sa, and y l and S, has a unique solution with x a 1 and x 1. Proof. Let m a,m e positive integers such that a T m a STa < a + 1 and a T m S < a + 1. For the general argument we assume that m a,m 3, the cases m a or m {1,} eing considered separately. The level y u may elong to U a or U, and the level y l may elong to L a or L, therefore we need to consider the following four possiilities. Case 1 : y u U a, y l L a. Then we have Sx a = T 1 S f ˆn 1 ( ) and Sx = T S f ˆn + T x a, where f ˆn 1 is a product of factors T i S (that appear on the upper orit of a) with i = or 3, and f ˆn + is a product of factors T i S (that appear on the lower orit of a) with i = m or m + 1. Using (4.10), we rewrite the first equation as x a = ST 1 S f ˆn 1 ( ) = ST 1 SST ST S f ˆk 1 + T m ST ( ) = T 1 f ˆk 1 + T m ST ( ). Since f ˆk 1 + is a product of factors T i S with i = m or m + 1, m 3, we conclude that T x a has a finite formal continued fraction expansion starting with m 3, i.e., T x a >, and x a > 1. Furthermore, it follows from the second equation that x = ST S f ˆn + T x a. Hence f ˆn + T x a has a finite formal continued fraction expansion starting with m 3, i.e., f ˆn + T x a >, and x <. Case : y u U a, y l L. Then Sx a = T 1 S f ˆn 1 ( ) and Sx = T S f ˆn + ( ). Like in Case 1 we see that x a > 1, and x = ST S f ˆn + ( ) <, since fˆ n + ( ) has a formal continued fraction expansion starting with m 3, and therefore is >. Case 3 : y u U, y l L a. Then Using (4.5), Sx a = T 1 S f ˆn 1 T 1 x and Sx = T S f ˆn + T x a. x a = ST 1 S ˆ f n 1 T 1 x = ST 1 SST ST S ˆ f k T m Sx, and using the second equation and simplifying, we otain T x a = f ˆk T m SST S f ˆn + (T x a) = f ˆk T m+1 S f ˆn + (T x a). Since all its factors are of the form T i S with i 3, the matrix f ˆk T m+1 S f ˆn + is hyperolic and its attracting fixed point T x a has periodic formal continued fraction expansion starting with m 3 (see Proposition 1.3 of [11]), hence x a > 1. Finally, as in Case 1, x = ST S f ˆn + T x a <, since f ˆn + T x a has formal continued fraction expansion starting with m 3, hence >. Case 4 : y u U, y l L. Then Sx a = T 1 S f ˆn 1 T 1 x and Sx = T S f ˆn + S( ).

23 ATTRACTORS OF (a,)-continued FRACTIONS 659 From the second equation we otain x = ST S f ˆn + S( ) < since fˆ n + S( ) has formal continued fraction expansion starting with m 3, hence >. Finally, x a = ST 1 S f ˆn 1 T 1 x = T 1 f ˆk T m+1 S f ˆn + S( ). Hence T x a = f ˆk T m+1 S f ˆn + S( ) > since it has formal continued fraction expansion starting with m 3, therefore x a > 1. Now we analyze the particular situations when m a or m {1,}, using the explicit cycle descriptions that exist for these situations as descried y Theorems 4. and 4.5. (i) If m a = m = 1, then relation (4.) for the -cycle and a similar one for the a-cycle shows that y l = and y u = 1 a 1, therefore x a = 1 and x = 1. (ii) If m a = 1, m =, following the explicit cycles given y (4.3) we otain y l = 1/ + 1, and y u = 1/( 1) 1, therefore x a =, x = 1. (iii) If m a = 1, m 3, using the cycle structure in Theorem 4. we otain y l = 1/ + 1 and y u = T 1 (ST ) m ST 1, therefore, x a = m, and x = 1. (iv) If m a =, m =, using the cycle structure in Theorems 4. and 4.5 we otain y l = 1 a+1 +1 and y u = 1 1 1, and a calculation in this particular case, like in Lemma 5.6, Case 3 implies that x a > 1 and x < 1. (v) If m a =, m >, an analysis of the four cases aove for this particular situation (with an explicit cycle relation for a) yields x a 1 and x 1. Indeed, in Case 1, we have y u = 1/a 1, hence x a = 1 and x =. In Case, we get x a = 1 and x <. Cases 3 and 4 are treated similarly. Now, since x a and x are uniquely determined, y transporting the rays [, x ] {} and [x a, ] {a} along the sets L, U, L a, and U a, we otain the x-coordinates of the right and left end of the segments on each level. DEFINITION 5.7. We say that two consecutive levels y 1 y of L a,, respectively, U a,, are connected y a vertical segment (we will refer to this as connected) if the x-coordinate of the right end-point of the horizontal segment on the level y 1 is equal to the the x-coordinate of the left end-point of the horizontal segment on the level y. In order to prove part (i) of Theorem 5.5, we show that all levels of L a, and all levels of U a, are connected. We first look at the levels in L a,. By Lemma 5.6 the levels y u and Sa, and the levels S and y l are connected. LEMMA 5.8. The levels S L and STa L a are two consecutive levels of L a, connected y a vertical segment at x = 0. The levels Sa U a and ST 1 U are two consecutive levels of U a, connected y a vertical segment at x = 0. Proof. Suppose there is y L a, such that STa y S. Then y L a or L. In either case, since y Lemmas 4.8 and 4.7 the truncated orits L a,l do not have repeated values, neither STa = y nor y = S is possile. Thus the only case we need to consider is STa < y < S. Then, either y = Sy for some y L a, (0 < y a + 1) or y = T y for some y L a,. These would imply

24 660 SVETLANA KATOK AND ILIE UGARCOVICI that either y > Ta, which is impossile, or T y < S, i.e., y < T 1 S, which is also impossile (if y < T 1 S then y = T y must e the end of the a-cycle, y Theorem 4.5). The x-coordinate of the right end-point of the segment at the level STa and of the left end-point of the segment at the level S is equal to 0. The second part of the proof is similar. The following proposition will e used later in the proof. PROPOSITION 5.9. Suppose that the set L a, is finite and y L a, with y > STa. (1) If y L a, then there exists n 0 > 0 such that ρ(f n y) = ρ(f n STa) for all 0 < n < n 0 and ρ(f n 0 y) ρ(f n 0 STa), or f n 0 y = 0. () If y L, then y > S and there exists n 0 > 0 such that ρ(f n y) = ρ(f n S) for all n < n 0 and ρ(f n 0 y) ρ(f n 0 S), or f n 0 y = 0. Proof. Suppose that y L a and a satisfies the cycle property. It follows that such an n 0 exists or f n 0 y is the end of the a-cycle. We will show that the latter is possile only if f n 0 y = 0, i.e., it is the end of a weak cycle. Suppose f n 0 y is the end of the a-cycle. Then if ρ(f n 0 1 y) = ρ(f n 0 1 STa) = S, we must have f n 0 1 y < 0 since otherwise the cycle would not stop at S, ut f n 0 1 (STa) > 0 since for STa we have not reached the end of the cycle. This contradicts the monotonicity of f n 0 1 and the original assumption y > STa, thus is impossile. The other possiility is ρ(f n 0 1 y) = ρ(f n 0 1 STa) = T. But this either implies that f n 0 1 y < T 1 S, and y monotonicity of f n 0 1, f n 0 1 (STa) < f n 0 1 y < T 1 S, which implies that we have reached the end of the cycle of STa as well, a contradiction, or, f n 0 y = 0, i.e., it is the end of a weak cycle. Now suppose y L. Then y Lemma 5.8 y S, ut since each level in L appears only once, we must have ut y > S. Now the argument that f n 0 y cannot e the end of the -cycle is exactly the same as for the a-cycle. In the periodic case, assume that no such n 0 exists. Then, in case (1), the (a,)-expansions of STa and y, which is the lower part of the former, are the same, i.e., (a,)-expansions of STa is invariant under a left shift. In case (), we have seen already that we must have y > S. Then the (a,)-expansions of S and y, which is the lower part of the former, are the same, i.e., (a, )-expansions of S is invariant under a left shift. The proof that this is impossile is ased on the following simple oservation: if σ = (a 1, a,..., a k, a k+1, a k+,..., a k+n ) is an eventually periodic symolic sequence with the minimal period n and invariant under a left shift y m, then σ is purely periodic and m is a multiple of n. By the uniqueness property of (a, )-expansions, this would imply that y = STa or y = S, a contradiction. Let y, y+ U a, e two consecutive levels with y < y+, and y a, y+ a L a, e two consecutive levels with y a < a y+ a.

25 ATTRACTORS OF (a,)-continued FRACTIONS 661 LEMMA There is always one level connected with level a + 1, and the levels y a and y+ a are connected y the vertical segment at x a. Proof. By Lemmas 5.6 and 5.8, we know that three consecutive levels STa S y l are connected. Moreover, their images remain connected under the same transformations in SL(,Z). Since each level in U a and in L a appears only once, at least one of the two inequalities must e strict, i.e., if STa = S, then STa = S < y l, and if S = y l, then STa < S = y l. First we prove that y l < T S. Suppose y l T S. Its preimage must e y l = T 1 y l since for any y, 0 < y < Ta, Sy < STa S < T S, and we would have S y l < y l that contradicts the assumption that y l is the next level aove S. Therefore, if the first digit in the (a,)-expansion of S is m, then the first digit of y l is (m 1) or m. In the first case, the three levels T m 1 S < a T m 1 y l are connected and satisfy T m 1 S = ya, T m 1 y l = y a +. T m S and a + 1 are connected. For the second case, we know that S y l and Therefore, the levels a T m S T m y l < a + 1. If S = y l, then y l L a, and STa < y l. If S < y l, then y l L, or y l L a and STa < y l. Let us assume that y l elongs to L a. Since STa < y l, y Proposition 5.9, there are two possiilities: (1) f n 0 y l is the end of a weak cycle. () There exists n 0 such that ρ(f n y l ) = ρ(f n STa) for all n < n 0, and ρ(f n 0 y l ) ρ(f n 0 STa). In the first case, we have f n 0 STa = y a and f n 0 S = y + a, or f n 0 S = y a and f n 0 y l = y + a. Therefore, either f n 0+1 STa or f n 0+1 S is connected with level a+1. In the second case, we notice that ρ(f n 0 1 y l ) = ρ(f n 0 1 STa) = T otherwise, ρ(f n 0 1 y l ) = ρ(f n 0 1 STa) = S would imply ρ(f n 0 y l ) = ρ(f n 0 STa) = T in contradiction with the choice of n 0. Further, there are two possiilities: (i) ρ(f n 0 STa) = S, ρ(f n 0 y l ) = T ; (ii) ρ(f n 0 STa) = T, ρ(f n 0 y l ) = S. In case (i) we otain f n 0 y l < a f n 0 STa which contradicts the monotonicity of f and the original assumption y l > STa. Thus the only possiility is f n 0 y l a > f n 0 STa.

26 66 SVETLANA KATOK AND ILIE UGARCOVICI By using the monotonicity of f n 0 we have f n 0 y l > f n 0 S > f n 0 STa and conclude that f n 0 STa = y a and f n 0 S = y + a, or f n 0 S = y a and f n 0 y l = y + a. Therefore, either f n 0+1 STa or f n 0+1 S is connected with level a + 1. The case when y l elongs to L is similar, and in this case f n 0 S = y a, f n 0 y l = y + a, and f n 0+1 S is connected with a + 1. By construction, in oth cases the common x-coordinate of the end-points is equal to x a. After an application of S the level connected with a + 1 will e connected with STa, and now, instead of 3 connected levels STa S y l (with at least one strict inequality) we have at least 4 connected levels y STa S y l (with no more than two equalities in a row). The process continues with a growing numer of connected levels, the highest eing a+1. Since on each step we cannot have more than two equalities in a row, the numer of distinct levels in this sequence will also increase. Therefore, we otain a sequence of connected levels (5.1) a + 1 y 1 y s > + 1 y s+1. It is evident from the construction that there are no unaccounted levels y L a,, a + 1 y y s+1. Now we prove a similar result for U a,. LEMMA There is always one level connected with level 1, and the levels y and y+ are connected y a vertical segment at x. Proof. By Lemmas 5.6 and 5.8, we know that the three consecutive levels y u Sa ST 1 are connected. It is easy to see that the first digit in (a,)-expansion of ST 1 is, and the first digit in (a,)-expansion of Sa is either 1 or. Therefore, the first digit in (a,)-expansion of y u is either 1 or. In the first case, either T 1 Sa < T 1 ST 1 or T 1 y u < T 1 Sa are the connected levels. Therefore either T 1 Sa = y and T 1 ST 1 = y +, or T 1 y u = y and T 1 Sa = y + are connected. So either T ST 1 or T Sa is connected with level 1. In the second case, we know that y u Sa and 1 T y u T Sa <. If y u = Sa, y u must elong to U, in which case y u < ST 1. If y u < Sa, then y u U a, or y u U and y u < ST 1. Let us assume that y u elongs to U. Since y u < ST 1, y Proposition 5.9 there are two possiilities: (1) f n 0 y u is the end of a weak cycle; () there exists n 0 such that we have ρ(f n y u ) = ρ(f n ST 1 ) for all n < n 0, and ρ(f n 0 y u ) ρ(f n 0 ST 1 ).