€¦ · journal of modern dynamics web site: volume 3, no. 3, 2009, 1–46 symbolic dynamics for...

JOURNAL OF MODERN DYNAMICS WEB SITE: http://www.math.psu.edu/jmdVOLUME 3, NO. 3, 2009, 1–46

SYMBOLIC DYNAMICS FOR THE GEODESIC FLOW ON HECKESURFACES

DIETER MAYER AND FREDRIK STROMBERG

ABSTRACT. In this paper we discuss a coding and the associated symbolic dy-namics for the geodesic flow on Hecke triangle surfaces. We construct an explicitcross section for which the first return map factors through a simple (explicit)map given in terms of the generating map of a particular continued fraction ex-pansion closely related to the Hecke triangle groups. We also obtain explicitexpressions for the associated first return times.

CONTENTS

1. Introduction 12. The geodesic flow on T 1M 33. λ-Continued fraction expansions 44. Construction of the cross-section 255. Construction of an invariant measure 386. Lemmas on continued fraction expansions and reduced geodesics 40

1. INTRODUCTION

Surfaces of negative curvature and their geodesics have been studied since the1898 work of Hadamard [15] (see in particular the remark at the end of §58). In-spired by the work of Hadamard and Birkhoff [6] Morse [31] introduced a coding ofgeodesics essentially corresponding to what is now known as ,,cutting sequences“and used this coding to show the existence of a certain type of recurrent geodesics[32].

Further ergodic properties of the geodesic flow on surfaces of constant negativecurvature given by Fuchsian groups were shown by e.g. Artin [5], Nielsen [36],Koebe [26], Lobell [29], Myrberg [33], Hedlund [16, 17, 18, 19], Morse and Hed-lund [30] and Hopf [20, 21]. In this sequence of papers one can see the subject ofsymbolic dynamics emerging. For a more up-to-date account of the ergodic prop-erties of the geodesic flow on a surface of constant negative curvature formulatedin a modern language see e.g. the introduction in Series [47].

2000 Mathematics Subject Classification: Primary: 37D40, Secondary: 37E05, 11A55, 11K50.This work has been supported by the German Research Foundation (DFG) under contract Ma

633/16-1. The authors appreciate the very helpful remarks of an anonymous referee.

1 c©2009 AIMSCIENCES

2 ENGLISHD. MAYER AND F. STROMBERG

Artin’s [5] approach was novel in that he used continued fractions to code geo-desics on the modular surface. After Artin, coding and symbolic dynamics on themodular surface have been studied by e.g. Adler and Flatto [1, 2, 3] and Series[48]. For a recent review of different aspects of coding of geodesics on the modularsurface see for example the expository papers by Katok and Ugarcovici [24, 25].

Other important references for the theory of symbolic dynamics and coding ofthe geodesic flow on hyperbolic surfaces are e.g. Adler-Flatto [4], Bowen and Se-ries [7] and Series [47].

In the present paper we study the geodesic flow on a family of hyperbolic sur-faces with one cusp and two marked points, the so-called Hecke triangle surfaces,generalizing the modular surface. Symbolic dynamics for a related billiard has alsobeen studied by Fried [12]. We now give a summary of the paper. Sections 1 and2 contain preliminary facts about hyperbolic geometry and geodesic flows. In Sec-tion 3 we develop the theory of λ-fractions connected to the coding of the geodesicflow on the Hecke triangle surfaces. The explicit discretization of the geodesic flowin terms of a Poincare section and Poincare map is developed in Section 4. As animmediate application we derive invariant measures for certain interval maps inSection 5. Some rather technical lemmas are confined to the end in Section 6.

1.1. Hyperbolic geometry and Hecke triangle surfaces. Recall that any hyper-bolic surface of constant negative curvature −1 is given as a quotient (orbifold)M = H/Γ. Here H = z = x + iy | y > 0, x ∈ R together with the metricds = |dz|

y is the hyperbolic upper half-plane and Γ ⊆ PSL2(R) ∼= SL2(R)/ ±Iis a Fuchsian group. Here SL2(R) is the group of real two-by-two matrices with de-terminant 1, I = ( 1 0

0 1 ) and PSL2(R) is the group of orientation preserving isome-tries of H. The boundary of H is ∂H = R∗ = R ∪ ∞, H∗ = H ∪ ∂H.If g =

(a bc d

) ∈ PSL2(R) then gz = az+bcz+d ∈ H for z ∈ H, gx ∈ ∂H for

x ∈ ∂H and we say that g is elliptic, hyperbolic or parabolic depending on whether|Tr g| = |a + d| < 2, > 2 or = 2. The same notation applies for fixed points of g.In the following we identify the elements g ∈ PSL2(R) with the map it defines onH∗. Note that the type of fixed point is preserved under conjugation g 7→ AgA−1

by A ∈ PSL2(R). A parabolic fixed point is a degenerate fixed point, belongs to∂H and is usually called a cusp. Elliptic points appear in pairs, z, z with z ∈ H andz belongs to the lower half-plane H− and Γz , the stabilizer subgroup of z in Γ, iscyclic of finite order. Hyperbolic fixed points appear also in pairs with x, x∗ ∈ ∂H,where x∗ is said to be the conjugate point of x. A geodesics γ on H is either ahalf-circle orthogonal to R or a line parallel to the imaginary axis and the end-points of γ are denoted by γ± ∈ ∂H. We identify the set of geodesics on H withG = (ξ, η) | ξ 6= η ∈ R∗ and use γ (ξ, η) to denote the oriented geodesic on Hwith γ+ = ξ and γ− = η. Unless otherwise stated all geodesics are assumed to beparametrized with hyperbolic arc length with γ (0) either at height 1 if γ is verti-cal or the highest point on the half-circle. The tangent of γ at γ (t) is denoted byγ (t). It is known that z ∈ H and θ ∈ [−π, π) ∼= S1 determine a unique geodesic(cf. Lemma 62) passing through z whose tangent at z makes an angle θ with thepositive real axis. This geodesic is denoted by γz,θ. It is also well known that a

JOURNAL OF MODERN DYNAMICS VOLUME 3, NO. 3 (2009), 1–46

ENGLISHSYMBOLIC DYNAMICS FOR HECKE SURFACES 3

geodesic γ (ξ, η) is closed if and only if ξ and η = ξ∗ are conjugate hyperbolicfixed points.

The unit tangent bundle ofH, T 1H =⊔

z∈H ~v ∈ TzH | |~v| = 1 is the collec-tion of all unit vectors in the tangent planes ofH with base points z ∈ H which wedenote by T 1

zH. By identifying ~v with its angle θ with respect to the positive realaxis we can view T 1H as the collection of all pairs (z, θ) ∈ H× S1. We may alsoview this as the set of geodesics γz,θ on H or equivalently as G ⊆ R∗2.

Let π : H → M be the natural projection map, i.e. π (z) = Γz and let π∗ :T 1H → T 1M be the extension of π to T 1H. Then γ∗ = πγ is a closed geodesicon M if and only if γ+ and γ− are fixed points of the same hyperbolic map gγ ∈Γ. For a more comprehensive introduction to hyperbolic geometry and Fuchsiangroups see e.g. [23, 27, 41].

DEFINITION 1. For an integer q ≥ 3 the Hecke triangle group Gq ⊆ PSL2(R)is the group generated by the maps S : z 7→ −1

z and T : z 7→ z + λ where

λ = λq = 2 cos(

πq

)∈ [1, 2). The corresponding orbifold (Riemann surface) is

Mq = Gq\H, which we sometimes identify with the standard fundamental domainof Gq

Fq = z ∈ H | |<z| ≤ λ/2, |z| ≥ 1with sides pairwise identified. Let ρ = ρ+ = e

πiq and ρ− = −ρ. We define the

following oriented boundary components of Fq: L0 is the circular arc from ρ− toρ+. L1 is the vertical line from ρ+ to i∞ and L−1 is the vertical line from i∞ toρ−. Thus ∂Fq = L−1 ∪ L0 ∪ L1 is the positively oriented boundary of Fq.

REMARK 2. Gq is a realization of the Schwarz triangle group(

π∞ , π

q , π2

)and it is

not hard to show (see e.g. [27, VII]) that Gq for q ≥ 3 is a co-finite Fuchsian groupwith fundamental domain Fq and the only relations

(1) S2 = (ST )q = Id – the identity in PSL2(R).

Hence Gq has one cusp, that is the equivalence class of parabolic points, and twoelliptic equivalence classes of orders 2 and q respectively. Note that G3 = PSL2(Z)–the modular group and G4, G6 are conjugate to congruence subgroups of themodular group. For q 6= 3, 4, 6 the group Gq is non-arithmetic (cf. [23, pp. 151-152]), but in the terminology of [9, 46] it is semi-arithmetic, meaning that it ispossible to embed Gq as a subgroup of a Hilbert modular group.

2. THE GEODESIC FLOW ON T 1MWe briefly recall the notion of the geodesic flow on a Riemann surface M =

Γ\H with Γ ⊂ PSL2(R) a Fuchsian group. To any (z, θ) ∈ T 1H ∼= H × S1 wecan associate a unique geodesic γ = γz,θ onH such that γ (0) = z and γ (0) = eiθ.The geodesic flow on T 1H can then be viewed as a map Φt : T 1H → T 1H withΦt (γz,θ) = Φt (z, θ) = (γz,θ (t) , γz,θ (t)) , t ∈ R satisfying Φt+s = Φt Φs. Thegeodesic flow Φ∗ on T 1M is then given by the projection Φ∗t = π∗ (Φt).



A more abstract and general description of the geodesic flow, which can be ex-tended to other homogeneous spaces, is obtained by the identification T 1H ∼=PSL2(R). Under this representation the geodesic flow corresponds to right mul-tiplication by the matrix a−1

t =(

et/2 00 e−t/2

)in PSL2(R) (cf. e.g. [11, Ch. 13]).

DEFINITION 3. Let Υ be a set of geodesics on H. A hypersurface Σ ⊆ T 1H issaid to be a Poincare section or cross section for the geodesic flow on T 1H for Υif any γ ∈ Υ intersects Σ

(P1): transversally i.e. non-tangentially, and(P2): infinitely often, i.e. Φtj (γ) ∈ Σ for an infinite sequence of tj → ±∞.

The corresponding first return map is the map T : Σ → Σ such that T (z, θ) =Φt0 (z, θ) ∈ Σ and Φt (z, θ) /∈ Σ for 0 < t < t0. Here t0 = t0 (z, θ) > 0 is calledthe first return time.

Poincare sections were first introduced by Poincare [40] to show the stability ofperiodic orbits. For examples of cross section maps in connection with the geodesicflow on hyperbolic surfaces see e.g. [4, 1].

The previous definition extend naturally to T 1M with Υ and Σ replaced byΥ∗ = π (Υ) and Σ∗ = π∗ (Σ). The first return map T is used to obtain a dis-cretization of the geodesic flow, e.g. we replace Φt (z, θ) by Φtk (z, θ) wheretk (z, θ) is a sequence of consecutive first returns. Incidentally this provides a re-duction of the dynamics from three to two dimensions and it turns out that in ourexample the first return map also has a factor map, which allows us to study thethree dimensional geodesic flow with the help of an interval map (see Sections 4.3and 5).

3. λ-CONTINUED FRACTION EXPANSIONS

3.1. Basic concepts. Continued fraction expansions connected to the groups Gq,the so-called λ-fractions, were first introduced by Rosen [42] and subsequentlystudied by Rosen and others, cf. e.g. [43, 44, 45]. For the purposes of naturalextensions (cf. Section 3.4) the results of Burton, Kraaikamp and Schmidt [8] areanalogous to ours and we occasionally refer to their results. Our definition of λ-fractions is equivalent to Rosen’s definition (cf. e.g. [42, §2]).

To a sequence of integers, a0 ∈ Z and aj ∈ Z∗ = Z\ 0 , j ≥ 1 (finite orinfinite) we associate a λ-fraction x = Ja0; a1, a2, . . .K. This λ-fraction is identifiedwith the point

x = a0λ− 1a1λ− 1

a2λ−. . .

= limn→∞T a0ST a1 · · · ST an 0

if the right hand side is convergent. When there is no risk of confusion, we some-times write x = x. For any m ≥ 1 we define the head x(m) and the tail x(m)ofx by x(m) = Ja0; a1, . . . , amK and x(m) = Jam+1, am+2, . . .K. Note that −x =J−a0;−a1,−a2, . . .K. If a0 = 0, we usually omit the leading J0; K. Repetitions ina sequence is denoted by a power, e.g. Ja, a, aK = Ja3K and an infinite repetition



is denoted by an overline, e.g. Ja1, . . . ak, a1, . . . , ak, . . .K = Ja1, . . . , akK. Sucha λ-fraction is said to be periodic with period k, an eventually periodic λ-fractionhas a periodic tail. Two λ-fractions x and y are said to be equivalent if they havethe same tail. In this case it is easy to see that, if the fractions are convergent, thenx = Ay for some A ∈ Gq.

The sole purpose for introducing λ-fractions is to code geodesics by identifyingthe λ-fractions of their endpoints with elements ofZN. For reasons that will be clearlater (Section 3.4), we have to consider also bi-infinite sequences ZZ and view ZNas embedded in ZZ with a zero-sequence to the left. On ZZ and ZN we always usethe metric h defined by h

(ai∞i=−∞ , bi∞i=−∞)

= 11+n where ai = bi for |i| < n

and an 6= bn or a−n 6= b−n. In this metric ZZ and ZN have the topological structureof a Cantor set and the left- and right shift maps σ± : ZZ → ZZ, σ± aj = aj±1are continuous. We also set σ+Ja1, a2, . . .K = Ja2, a3, . . .K.

3.2. Regular λ-fractions. In the set of all λ-fractions we choose a ,,good“ subset,in which almost all x ∈ R have unique λ-fractions and in which infinite λ-fractionsare convergent. The first step is to choose a ,,fundamental region“ Iq for the actionof T : R → R, namely Iq =

[−λ2 , λ

2

]. Then it is possible to express one property

of our ,,good“ subset as follows: If in the fraction x = Ja0; a1, a2, . . .K the firstentry a0 = 0, then x ∈ Iq. That means, we do not allow sequences with a0 = 0 tocorrespond to points outside Iq.

A shift-invariant extension of this property leads to the following definition ofregular λ-fractions:

DEFINITION 4. Let x = Ja0; a1, a2, . . .K be a finite or infinite convergent λ-fraction and let xj = σjx = J0; aj , aj+1, . . .K, j ≥ 1, be the j-th shift of x. Let xj

be the corresponding point. Then x is said to be a regular λ-fraction if and only if

(*) xj ∈ Iq, for all j ≥ 1.

A regular λ-fraction is denoted by[a0; a1, . . .

], the space of all regular λ-fractions

is denoted by Aq and the subspace of infinite regular λ-fractions with a0 = 0 isdenoted by A0,q.

For a finite fraction x = Ja0; a1, . . . , anK we get xj = J0; K and xj = 0 ∈ Iq forj > n.

We will see later that regular λ-fractions can be regarded as nearest λ-multiplecontinued fractions. In the case q = 3 or λ = 1, nearest integer continued fractionswere studied already by Hurwitz [22] in 1889. An account of Hurwitz reductiontheory can be found in Fried [13] (cf. also the H-expansions in [24, 25]). Forgeneral q this particular formulation of Rosen’s fractions was studied by Nakada[35].

For the remainder of the paper we let h = q−32 if q is odd and h = q−2

2 if q iseven. The following Lemma is an immediate consequence of [8, (4)].



LEMMA 5. The points ∓λ2 have finite regular λ-fractions given by

∓λ

2=

[(±1)h]

, for q even,[(±1)h ,±2, (±1)h]

, for q odd.

LEMMA 6. If q is odd, the point x = 1 has the finite regular λ-fraction

1 =[1; 1h

].

Proof. Since a =[1; 1h

]= T (ST )h 0, one has also −a = T−1

(ST−1

)h 0. Fromidentity (1) we get

Sa = (ST )h+1 0 =(T−1S

)h+2 0 = T−1(ST−1

)hST−1S0 = T−1

(ST−1

)h 0,

and hence −1/a = −a. Since a > 0, this implies that a = 1.

DEFINITION 7. Let bxc be the floor function defined by bxc = n ⇔ n < x ≤n + 1 for x > 0, respectively n ≤ x < n + 1 for x ≤ 0, and let 〈x〉λ =

⌊xλ + 1

2

⌋be the corresponding nearest λ-multiple function. Then define Fq : Iq → Iq by

Fqx =

− 1

x − 〈− 1x〉λλ, x ∈ Iq\ 0 ,

0, x = 0.

LEMMA 8. For x ∈ R the following algorithm gives a finite or infinite regularλ-fraction cq (x) =

[a0; a1, . . .

]corresponding to x:

(i) Set a0 := 〈x〉λ and x1 = x− a0λ.(ii) Set xj+1 := Fqxj = − 1

xj− ajλ, j ≥ 1, with aj = 〈−1

xj〉λ, j ≥ 1.

If xj = 0 for some j, the algorithm stops and gives a finite regular λ-fraction.

Proof. By definition we see that xj+1 = T−ajSxj , j ≥ 1, and it follows thatx = T a0ST a1 · · · ST anxn for any n ≥ 1. If x =

[a0; a1, . . . .

], then for j ≥ 1

xj = σjx =[0; aj , . . .

]corresponds to the point xj and condition (*) of Definition

4 is fulfilled, since Fq maps Iq to itself and x1 ∈ Iq.

REMARK 9. We say that Fq is a generating map for the regular λ-fractions. Itis also clear from Lemma 8 that Fq acts as a shift map on the space A0,q, i.e.cq (Fqx) = σcq (x).

An immediate consequence of Lemma 8 is the following corollary:

COROLLARY 10. If x has an infinite regular λ-fraction, then it is unique and equalto cq (x) as given by Lemma 8.

The above choice of floor function implies that Fq is an odd function and that〈±λ

2 〉λ = 0 in agreement with Lemma 5. The ambiguity connected to the choice offloor function at integers affects only the points x = 2

λ(1−2k) where −1xλ + 1

2 = k ∈ Zand Fqx = (k − bkc) λ − λ

2 = ±λ2 . By Lemma 5 we conclude, that any point,

which has more than one regular λ-fraction, is Fq-equivalent to ±λ2 and hence has

a finite λ-fraction.



We can produce in this way a regular λ-fraction cq (x) as a code for any x ∈ R.For the purpose of symbolic dynamics we prefer to have an intrinsic description ofthe members of the spaceA0,q formulated in terms of so-called forbidden blocks (orf.b. for short), i.e. certain subsequences which are not allowed. From Definition 4it is clear, which subsequences are forbidden and how to remove them by rewritingthe sequence, using the fraction with a leading a0 6= 0 for any point outside Iq

instead of a0 = 0.

LEMMA 11. Let q be even with q = 2h + 2. Then the blocks J(±1)h ,±mK form ≥ 1 are forbidden. The block Ja, (±1)h ,±m, bK with a 6= ±1 and b 6= ±1 ifm = 1, can be rewritten as

Ja, (±1)h ,±m, bK→

[a∓ 1, (∓1)h ,±m∓ 1, b

], if m ≥ 2,[

a∓ 1, (∓1)h−1 , b∓ 1], if m = 1 and h ≥ 2,[

a∓ 1, b∓ 1], if m = 1 and h = 1.

Proof. Since by Lemma 5∓λ2 =

[(±1)h]

, the blocks J(±1)h ,±mK for m ≥ 1 areforbidden. Using then the relation (ST )2h+2 = I we can restrict ourselves to thecase where the number of consecutive 1’s is smaller than or equal to h + 1. Thenthe rewriting rules follow immediately from the relation (ST )2h+2 = I .

REMARK 12. Obviously the rewritten block Ja ∓ 1, b ∓ 1K is forbidden itself ifh = 1, i.e. q = 4 and a = 2 and b ≥ 2 or a = −2 and b ≤ −2 in the case of minusand plus sign respectively. How to get allowed blocks in this case after repeatedrewriting will be discussed in Lemma 19.

LEMMA 13. Let q ≥ 5 be odd with q = 2h + 3. Then the blocks J(±1)h+1K andJ(±1)h ,±2, (±1)h ,±mK with m ≥ 1 are forbidden. The block Ja, (±1)h+1 , bKfor a 6= ±1, b 6= ±1 can be rewritten as

Ja, (±1)h+1 , bK→ [a∓ 1, (∓1)h , b∓ 1

],

the blocks Ja, (±1)h ,±2, (±1)h ,±m, bK with a 6= ±1 can be rewritten as

Ja, (±1)h ,±2, (±1)h ,±m, bK→[

a∓ 1, (∓1)h ,∓2, (∓1)h ,±m∓ 1, b], m ≥ 2,[

a∓ 1, (∓1)h ,∓2, (∓1)h−1 , b∓ 1], m = 1.

For q = 3 the blocks J±1K and J±2,±mK with m ≥ 1 are forbidden. The blockJa,±1, bK with a 6= ±1,±2 and b 6= ±1 can be rewritten as

Ja,±1, bK→ Ja∓ 1, b∓ 1K.The block Ja,±2,±m, bK with a 6= ±1,±2 can be rewritten as

Ja,±2,±m, bK→Ja∓ 1,∓2,±m∓ 1, bK, for m ≥ 2,

Ja∓ 1, b∓ 2, K, for m = 1.

Proof. Since (ST )2h+3 = I we can restrict ourselves again to blocks with no morethan h+1 consecutive 1’s. From Lemma 5 we know that∓λ

2 =[(±1)h ,±2, (±1)h]



and hence the forbidden blocks follow immediately. Using then the relation (ST )2h+3 =I gives the rewriting rules.

REMARK 14. In the first rewriting rule for q = 3 the rewritten block Ja∓1, b∓1Kis forbidden if a = 3 and b ≥ 2 or a = −3 and b ≤ −2 (in the case of minus andplus signs respectively) or if b = ±2. The rewritten block Ja∓ 1,∓2,±m∓ 1, bKis itself forbidden if m = 2 or m = 3 and b ≥ 1 respectively b ≤ −1 (in the case ofminus and plus signs respectively) or b = ±1. The rewritten block Ja∓ 1, b∓ 1K isforbidden in this case if a = 3 and b ≥ 3 or if a = −3 and b ≤ −3 (in the case ofminus and plus signs respectively). How to get allowed blocks in these cases afterrepeated rewriting will be discussed in the proofs of Lemmas 18 and 20.

REMARK 15. It is easy to see that Rosen’s λ-fractions [42] can be expressed aswords in the generators T, S and JS of the group G∗

q =< Gq, J >⊆ PGL2(R),where J : z 7→ −z is the reflection in the imaginary axis, i.e. JSx = 1

x for x ∈ R∗.Since J is an involution of Gq, e.g. JTJ = T−1 and JSJ = S, it is easy to see thatRosen’s and our notions of λ-fractions are equivalent: e.g. in the Gq-word identi-fied with our λ-fraction we replace any T−a by JT aJ , a ≥ 1. Algorithmically thismeans for a λ-fraction with entries aj that the corresponding Rosen fraction hasentries (εj , |aj |) where ε1 = −sign (a1) and εj = −sign(aj−1aj) for j ≥ 2.

From the definition of regular λ-fractions it is clear that Rosen’s reduced λ-fractions [42, Def. 1] correspond to a fundamental interval

[0, λ

2

]for the action of

the group 〈T, J〉 together with the choices made for finite fractions in [42, Def. 1(4)-(5)]. It is easy to verify, for example using the forbidden blocks, that a finitefraction not equivalent to ±λ

2 or an infinite regular λ-fraction correspond to a re-duced λ-fraction of Rosen. The main difference between our regular and Rosen’sreduced λ-fractions is that any λ-fraction equivalent to ±λ

2 has two valid regularλ-fractions. The root of this non-uniqueness is our choice of a closed interval Iq

which is in turn motivated by our Markov partitions in Section 3.5.It is then clear, that those results of [42] and [8] pertaining to infinite reduced

λ-fractions can be applied directly to our regular λ-fractions.

LEMMA 16. An infinite λ-fraction without forbidden blocks is convergent.

Proof. This follows from [42, Thm. 5] and Remark 15.

An immediate consequence of Definition 4 and Lemmas 11, 13 and 16 is thefollowing

COROLLARY 17. A λ-fraction is regular if and only if it does not contain anyforbidden block.

Rewriting a forbidden block may produce new f.b.’s and by rewriting a f.b. com-pletely we mean that we rewrite also all new f.b.’s that arises. For the reductionprocedure cf. Section 3.6 it is important that rewriting a f.b. completely can bedone in most cases without affecting the head of the λ-fraction (up to some point).

LEMMA 18. Suppose that the λ-fraction x = Ja0; a1, . . .K has the first forbiddenblock beginning at an, n ≥ 2. If q ≥ 5 or q = 4 and the forbidden block is not of



the type J12K, then the head of x up to n − 2, i.e. x(n−2) = Ja0; a1, . . . , an−2K isnot affected by rewriting this forbidden block completely.

Proof. For simplicity consider an initial f.b. containing +1’s, the blocks with −1’sare treated analogously. By using the relation (ST )q = 1 we may assume thatthere are no blocks of consecutive ±1’s of length greater than q

2 . The analogue ofLemmas 11 and 13 in this case is very simple: Ja, 1j , bK→ Ja−1, (−1)q−j−2 , b−1K for any j > q

2 .Suppose first that q is even and that the first f.b. begins with an, i.e.

x = Ja0; a1, . . . , an−2, an−1, 1h, an+h, an+h+1, . . .K, an−1 6= 1, an+h ≥ 1.

By applying Lemma 11 we rewrite x into either Ja0; a1, . . . , an−2, an−1−1, (−1)h , an+h−1, an+h+1, . . .K if an+h ≥ 2 or Ja0; a1, . . . , an−2, an−1 − 1, (−1)h−1 , an+h+1 −1, an+h+2, . . .K if an+h = 1 (here an+h+1 6= 1). If this rewriting did not produce anew f.b. we are done so suppose that a new f.b. was created. Note that unless q = 4and an = an+1 = 1 (this case will be treated in Lemma 19) we have an+h ≥ 2and h ≥ 1 or an+h = 1 and h ≥ 2. In this case there is a non-empty block of −1’sstarting at position n and any new f.b. has to either end before or begin after theblock of −1’s.

If the new f.b. appears to the left of position n it clearly has to end with an−1−1but sign (an−1) = sign (an−1 − 1) so any such f.b. had to be forbidden also beforethe rewriting, contradicting the assumption about the position of the first f.b. in x.Any new f.b. beginning directly after the −1’s has to begin with +1 so rewriting itwill only change the last −1 to −2 and there are at least two digits between an−2

and any new f.b.’s.Now suppose that q ≥ 5 is odd. There are two different types of f.b.’s but their

treatments are very similar. Assume first that we have

x = Ja0; a1, . . . , an−2, an−1, 1h+1, an+h+1, . . .K, an−1, an+h+1 6= 1,

then by Lemma 13 we rewrite x into x1 = Ja0; a1, . . . , an−2, an−1−1, (−1)h , an+h+1−1, . . .K. There are four possibilities to create a new f.b. in x1:

1. If the f.b. ends at an−1−1 then it was also forbidden in x, since sign (an−1) =sign (an−1 − 1), contradicting the assumption that the first f.b. begun withan.

2. If the f.b. ends with the (−1)h then x1 = Ja0; a1, . . . , (−1)h ,−2, (−1)h , an+h+1−1, . . .K with an+h+1 ≤ −2 and an−1 = · · · = an−2−h = −1 so the f.b.J(−1)h+1K beginning at an−2−h would have been present in x, also contra-dicting the assumption on the first f.b.

3. If the new f.b. begins with the (−1)h then an+2h+2 ≤ −2 ( an+2h+2 6= −1since otherwise x contains J(−1)h+2K and h+2 > q

2 ) and x1 = Ja0; a1, . . . , an−2, an−1−1, (−1)h ,−2, (−1)h , an+2h+2, . . .K. Rewriting gives

x2 = Ja0; a1, . . . , an−2, an−1, 1h, 2, 1h, an+2h+2 + 1, . . .K.JOURNAL OF MODERN DYNAMICS VOLUME 3, NO. 3 (2009), 1–46


Since an−1 6= 1 any new f.b. in x2 must begin with an+2h+2 + 1 = −1 inwhich case rewriting it only changes J. . . , an−1, 1h, 2, 1h, . . .K to J. . . , an−1, 1h, 2, 1h−1, 2, . . .Kand there are at least four digits between an−2 and any new f.b.

4. If the f.b. begins after the (−1)h then x1 = Ja0; a1, . . . , an−2, an−1 −1, (−1)h , 1h+1, . . .K or x1 = Ja0; a1, . . . , an−2, an−1−1, (−1)h , 1h, 2, 1h, an+3h+2 . . .Kwith an+3h+2 ≥ 1. Rewriting this f.b. thus changes the last −1 to −2 andthere are at least two digits between an−2 and any new f.b.

In the second case, assume that we have

x = Ja0; a1, . . . , an−2, an−1, 1h, 2, 1h, an+2h+1, . . .K, an−1 6= 1, an+2h+1 ≥ 1,

then by Lemma 13 we can rewrite x into either

x2 = Ja0; a1, . . . , an−2, an−1 − 1, (−1)h ,−2, (−1)h , an+2h+1 − 1, . . .K,if an+2h+1 ≥ 2, or

x3 = Ja0; a1, . . . , an−2, an−1 − 1, (−1)h ,−2, (−1)h−1 , an+2h+2 − 1, . . .K,if an+2h+1 = 1. If x2 or x3 contains a new forbidden block, a similar argument asabove tells us that it must begin with either with an+2h+1−1 = 1 or an+2h+2−1 =1 in which case rewriting it will only change the last−1 to a −2 if h ≥ 2 or the −2to a −3 if h = 1. In all cases there are at least 3 digits between an−2 and any newf.b.

We have shown that for any f.b. we can rewrite it without changing an−2 or anydigit to the left of it and after this rewriting any new f.b.’s are separated from an−2

by at least 2 digits. A recursive application of the above argument thus shows thatin any case any further rewriting will not change the head x(m−2).

LEMMA 19. Let q = 4 and suppose that

x = Ja0; a1, a2, . . . , an−l−1, (±2)l , (±1)2 , (±2)k , an+k+2, . . .Khas precisely one forbidden block, which begins with an = an+1 = ±1 andan−l−1, an+k+2 6= ±2. Then a complete rewriting of this forbidden block doesnot affect the head x(n−m) = Ja0; a1, . . . , an−mK where m = 0 if l = 0, m = 3 ifl ≥ 1 and k = 0 and m = min (k + 3, l + 3) otherwise. In all cases we assumethat n ≥ m.

Proof. Suppose without loss of generality that x = Ja0; a1, . . . , an−l−1, 2l, 12, an+2, . . .Kwith n ≥ l + 1 ≥ 1, an = an+1 = 1, an−l−1 6= 2 and an−1 6= 1.if l = 0 thenan−1 6= 1. The case of (−1)2 is analogous.

If l = 0 then one rewriting of x produces x0 = Ja0; a1, . . . , an−1 − 1, an+2 −1, . . .K which does not contain a new f.b. unless an+2 = 2. If an+2 = 2 andan+3 ≥ 2 the new f.b. is of the type covered in Lemma 18 and rewriting itcompletely does not change x(n−2). If an+2 = 2 and an+3 = 1 then an+4 ≤−1 and x0 = Ja0; a1, . . . , an−1 − 1, 12, an+4 . . .K which is rewritten into x1 =Ja0; a1, . . . , an−1 − 2, an+4 − 1, . . .K which contains no f.b., and hence x(n−2) isnot changed.



Suppose that l ≥ 1. If an+2 ≤ −1 one rewriting produces Ja0; a1, . . . , an−l−1, 2l−1, 1, an+2−1, . . .K which contains no f.b. and does not change x(n−2).

If an+2 ≥ 3 then two rewritings produce Ja0; a1, . . . , an−2 − 1,−1, an+2 −2, . . .K if l = 1 and x2 = Ja0; a1, . . . , an−l−1, 2l−2, 1,−1, an+2 − 2, . . .K if l ≥ 2.The first rewritten λ-fraction does not contain any f.b. but the second may containa new f.b. if an+2 = 3 and an+3 ≥ 1. If an+2 = 3 and an+3 ≥ 2 this is a f.b. of thetype J1,mK with m ≥ 2 and by Lemma 18 we can rewrite it completely withoutchanging x(n−3). If an+2 = 3 and an+3 = 1 then an+4 ≤ −1 and we rewrite x2

into Ja0; a1, . . . , an−l−1, 2l−2, 1,−2, an+4−1, . . .K which contains no f.b., withoutchanging x(n−3).

The remaining case is when an+2 = 2. Assume that x = Ja0; a1, . . . , an−l−1, 2l, 12, 2k, an+k+2, . . .Kwith an−l−1, an+k+2 6= 2 and k ≥ 1. It is easy to see that J2, 12, 2K is rewritten intoJ12K so if l ≥ k we rewrite x in k steps into x3 = Ja0; a1, . . . , an−l−1, 2l−k, 12, an+k+2, . . .Kwhich, as has been shown above, can be rewritten completely without changingx(n−k−3) (the last 2 before the 12 is an−(k+1)). If l < k we rewrite x in l

steps into Ja0; a1, . . . , an−l−1, 12, 2k−l, an+k+2, . . .K which is then rewritten intox4 = Ja0; a1, . . . , an−l−1 − 1, 1, 2k−l−1, an+k+2, . . .K. If either k ≥ l + 2 ork = l+1 and an+k+2 ≥ 3 then x4 contains a new f.b. of the type J1,mKwith m ≥ 2which, by Lemma 18, can be rewritten completely without changing x(n−l−2). Ifk = l + 1 and an+k+2 = 1 then an+k+3 ≤ −1 and x4 = Ja0; a1, . . . , an−l−1 −1, 12, an+k+3, . . .K which we have shown above can be rewritten completely with-out changing x(n−l−2).

We have shown that the f.b. in x = Ja0; a1, . . . , an−l−1, 2l, 12, 2k, an+k+2 . . .Kcan be rewritten completely without changing x(n−m) where m = 2 if l = 0 andm = 3 if l ≥ 1 and k = 0 respectively m = min (l + 3, k + 3) otherwise. (Ofcourse we can do better in certain cases but we are mainly interested in whether mis finite or not.)

LEMMA 20. Let q = 3 and suppose that x = Ja0; a1, . . .K has only one forbid-den block, which begins at an. Assume further that x contains at most one digit±1. Then a complete rewriting of x does not change the head x(n−m) where m =min (l + 3, k + 4) if x = Ja0; a1, . . . , an−l−1, (±3)l ,±2,±1, (±3)k , an+k+2, . . .K,l, k ≥ 0, with an−l−1 6= ±2,±3 and an+k+2 6= ±3 and m = 3 otherwise. We as-sume in all cases that n ≥ m.

Proof. Assume without loss of generality that an ≥ 1. The case of an ≤ −1 isanalogous. There are three different f.b.’s to consider. Suppose first that

x = Ja0; a1, . . . , an−2, an−1, 1, an+1, . . .K,where an−1 6= 1, 2, an+1 6= ±1 and if an+1 = ±2 then sign (an+2) = −sign (an+1).By Lemma 13 we rewrite x into x0 = Ja0; a1, . . . , an−2, an−1−1, an+1−1, an+2, . . .K.Since sign (an−1) = sign (an−1 − 1) there can be no new f.b. ending at an−1−1 sothere are only three possibilities to create a new f.b., either an−1 = 3 and an+1 ≥ 2or an+1 = 3 and an+2 ≥ 2 or an+1 = 2:



1. If an−1 = 3 then x0 = Ja0; a1, . . . , an−2, 2, an+1 − 1, an+2, . . .K and ifan+1 ≥ 3 we rewrite x0 into x1 = Ja0; a1, . . . , an−2 − 1,−2, an+1 −2, an+2, . . .K and if an+1 = 2 into x2 = Ja0; a1, . . . , an−2 − 1, an+2 −2, an+3, . . .K. Since an−2 6= −1 and an+2 ≤ −2 in x2 there can not beany new f.b.’s in x2. Since the tail Jan+1, an+2, . . .K does not contain anymore forbidden blocks we can only obtain a new f.b. in x1 if an+1 = 4 andan+2 ≥ 2 or an+1 = 3.(a) If an+1 = 4 and an+2 ≥ 2 then x1 = Ja0; a1, . . . , an−2−1,−2, 2, an+2, . . .K

which is rewritten into Ja0; a1, . . . , an−2− 1,−3,−2, an+2− 1, . . .K andthere is at least 3 digits between an−3 and any new f.b.

(b) If an+1 = 3 then x1 = Ja0; a1, . . . , an−2 − 1,−2, 1, an+2, . . .K which isrewritten into Ja0; a1, . . . , an−2 − 1,−3, an+2 − 1, an+3, . . .K and thereis at least 2 digits between an−3 and any new f.b.

2. If an−1 6= 3, an+1 = 3 and an+2 ≥ 2 then x0 = Ja0; a1, . . . , an−2, an−1 −1, 2, an+2, . . .K is rewritten into x3 = Ja0; a1, . . . , an−2, an−1−2,−2, an+2−1, . . .K and there is at least 3 digits between an−3 and any new f.b.

3. If an−1 6= 3, an+1 = 2 then x0 = Ja0; a1, . . . , an−2, an−1 − 1, 1, an+2, . . .Kis rewritten into x4 = Ja0; a1, . . . , an−2, an−1 − 2, an+2 − 1, an+3, . . .K andsince an+2 ≤ −2 there is at least 3 digits between an−3 and any new f.b.

We have shown that a single rewriting of a f.b. of the type ±1 does not change theλ-fraction more than two steps to the left of the beginning of the f.b..

The second type of forbidden block is:

x = Ja0; a1, . . . , an−1, 2, an+1, an+2, . . .K, an−1 6= ±1, 2,

where an+1 ≥ 2 and we rewrite x into x5 = Ja0; a1, . . . , an−1 − 1,−2, an+1 −1, an+2, . . .K. Then any new f.b. in x6 must begin with an+1 = 2 or an+1 = 3 andan+2 ≥ 2. For an+1 = 2 rewriting leads to x6 = Ja0; a1, . . . , an−1−1,−3, an+2−1, an+3, . . .K which does not contain any new f.b. since an+2 ≤ −2. For an+1 = 3rewriting leads to x7 = Ja0; a1, . . . , an−1− 1,−3,−2, an+2− 1, an+3, . . .K, henceany more rewriting of newly appearing f.b.’s does not change the head x(n−2).

We have shown that rewriting a single f.b. of the type J1K or J2, bK with b ≥ 2does not change the head x(n−3).

The third and most complicated type of f.b. is when an = 2 and an+1 = 1. Sup-pose that x = Ja0; a1, . . . , an−1−l, 3l, 2, 1, 3k, an+2+k, . . .K, l, k ≥ 0, an−1−l 6=2, 3, an+2+k 6= 3. By Lemma 13 it is clear that J3, 2, 1, 3K is rewritten into J2, 1Kand using this recursively we get different three cases depending on if l < k, = kor > k.

1. If l > k then x is rewritten first into Ja0; a1, . . . , an−1−l, 3l−k, 2, 1, an+2+k, . . .Kand then into y

0= Ja0; a1, . . . , an−1−l, 3l−k−1, 2, an+2+k − 2, . . .K which

may contain a new f.b. beginning with the 2 but an+2+k − 2 6= 1 and wehave shown above that an f.b. of the type J2,mK with m ≥ 2 can be rewrittenwithout changing any element in the sequence more than two steps to the leftof the f.b. Hence we can rewrite any new f.b. without changing the headx(n−k−4).



2. If l < k then x is first rewritten into Ja0; a1, . . . , an−1−l, 2, 1, 3k−l, an+2+k, . . .K,then into Ja0; a1, . . . , an−1−l − 1, 1, 3k−l−1, an+2+k, . . .K and next into y

1=

Ja0; a1, . . . , an−1−l−2, 2, 3k−l−2, an+2+k, . . .K if k ≥ l+2 or y2

= Ja0; a1, . . . , an−1−l−2, an+2+k − 1, an+3+k, . . .K if k = l + 1.(a) If k > l+2 we rewrite y

1into y

3= Ja0; a1, . . . , an−1−l−3,−2, 3k−l−3, an+2+k, . . .K

which can only contain a new f.b. if k = l + 3 and an+2+k ≤ −2 and wehave shown that rewriting such a f.b. does not affect the sequence morethan two steps to the left so any further rewriting can not change the headx(n−3−l).

(b) If k = l + 2 and an+2+k ≥ 2 we rewrite y1

= Ja0; a1, . . . , an−1−l −2, 2, an+2+k, . . .K into y

4= Ja0; a1, . . . , an−1−l−3,−2, an+2+k−1, . . .K

so any new f.b. in y4

is either beginning with an+2+k − 1 = 1 or withan−1−l − 3 = 1. If it is beginning with an−1−l − 3 = 1 then y

4is

rewritten into y5

= Ja0; a1, . . . , an−2−l − 1,−3, an+2+k − 1, . . .K whichcan only have a new f.b. beginning with an+2+k−1 − 1 = 1. In casean+2+k+1 − 1 = 1 we have shown that rewriting such an f.b. does notaffect the sequence more than two steps to the left and thus, in both cases,a complete rewriting will not change x(n−3−l).

3. If k = l then x is first rewritten into Ja0; a1, . . . , an−1−l, 2, 1, an+2+k, . . .K,then into z0 = Ja0; a1, . . . , an−1−l−1, an+2+k−2, . . .K and since an−1−l 6=2, 3 and an+2+k 6= 3 any new f.b. must either begin with an+2+k − 2 =0, 2. In case an+2+k = 2 we get z1 = Ja0; a1, . . . , an−1−l + an+3+k −1, an+4+k, . . .K and an+4+k ≤ −2 so any new f.b. has to be of the forman−1−l + an+3+k − 1 = ±1 or an−1−l + an+3+k − 1 = −2. In both caseswe have a f.b. which we have shown is possible to rewrite without changingthe sequence more than two steps to the left hence rewriting does not changethe head x(n−4−l).

We conclude that for all three types of f.b.’s, a complete rewriting of the initial f.b.leaves the head x(n−m) unchanged with m = 3 unless x = Ja0; a1, . . . , an−1−l, 3l, 2, 1, 3k, an+2+k, . . .Kin which case m = min (l + 3, k + 4).

3.3. Dual regular λ-fractions. To encode the orbits of the geodesic flow in termsof a discrete invertible dynamical system it turns out that we still need another kindof λ-fraction, the so-called dual regular λ-fraction. In the case q = 3 this wasalready introduced by Hurwitz [22], see also [25, p. 102].

Consider the set of λ-fractions y = J0; b1, . . .Kwhich do not contain any reversedforbidden block, i.e. a forbidden block given in Definitions 11 or 13 read in reversedorder.

Let R be the largest number in this set and define r = R−λ and IR = [−R, R].To give an explicit expression for R we need to investigate the connection betweenordering of points and their corresponding λ-fractions.

LEMMA 21. Let x, y ∈ Iq with cq (x) =[a1, a2, . . .

]and cq (y) =

[b1, b2, . . .

]and

suppose that x and y are not Gq-equivalent to ±λ2 and that a1 6= b1. Then x < y



if and only if b1 < 0 < a1 or b1a1 > 0 and a1 < b1. For one-digit λ-fractions thisordering is simply given by

[1]

<[2]

< · · · < 0 < · · · < [−2]

<[−1

].

Proof. Consider ϕn (x) = STn (x) = −1nλ+x then ϕn (Iq) =

[−2

λ(2n−1) ,−2

λ(2n+1)

]=

[ln, rn] and the intervals [ln, rn] only overlap at their endpoints which are imagesof ±λ

2 . It is also easy to verify that rn < rm if and only if m < 0 < n or mn > 0and n < m. Since x ∈ ϕa1 (Iq) and y ∈ ϕb1 (Iq) the lemma follows.

LEMMA 22. Let x = Ja1, . . . , an−1, aK, y = Ja1, . . . , an−1, bK and x, y the cor-responding points. Assume that

[a2, . . . , an−1, a

]and

[a2, . . . , an−1, b

]are both

regular. Then x < y if and only if b < 0 < a or ab > 0 and a < b.Furthermore, if y = Ja1, . . . , an−1K then x < y if and only if a > 0.

Proof. First consider ϕm (x) = STmx = −1mλ+x then ϕ′m (x) = (mλ + x)−2 > 0

for all x 6= −mλ so ϕm is increasing and positive in (−∞,−mλ) and increasingand negative in (−mλ,∞). Hence ϕm (x) = STmx < ϕm (y) = STmy ⇔ eithery < −mλ < x or x < y.

Define Aj = ST aj ST aj+1 · · · ST an−1 , xj = AjST a (0) , yj = AjST b (0),j = 1, . . . , n − 1. Then x1 = x, y1 = y and xj , yj ∈ Iq for all j = 2, . . . , n − 1.Hence, if m ∈ Z∗ then λ

2 < |mλ| so Lemma 21 implies that STmxj < STmyj ifand only if xj < yj , for all j = 2, . . . , n − 1 for all m ∈ Z∗. Using the fact thatxj = ST ajxj+1 and yj = ST ajyj+1 we see that ST a1x1 < ST a1y1 ⇔ x1 < y1

⇔ ST a2x2 < ST a2y2 ⇔ x2 < y2⇔ · · · ⇔ ST a (0) < ST b (0) ⇔ 1b < 1

a ⇔b < 0 < a or ab > 0 and a < b.

To prove the last equivalence, define zj = Aj (0) and proceed as above: A1ST a (0) <A1 (0) ⇔ ST a1x1 < ST a1z1 ⇔ · · · ⇔ ST an−1ST a (0) < ST an−1 (0) ⇔ST a (0) < 0⇔ a > 0.

Using Lemmas 21 and 22 it is easy to prove the following Lemma.

LEMMA 23. Let x, y ∈ R∞ with cq (x) =[a0; a1, a2, a3, . . .

], and cq (y) =[

b0; b1, b2, b3, . . .]. Then x < y if and only if either a0 < b0 or ai = bi for

i = 0, . . . , n − 1 and either bn < 0 < an or anbn > 0 and an < bn for somen ≥ 1.

Proof. Define xj = limm→∞ ST aj ST aj+1 · · · ST am and yj = limm→∞ ST bj ST bj+1 · · · ST bm

for j ≥ 1 and observe that we have xj , yj ∈ Iq for all j ≥ 1. It is clear that ifa0 < b0 then x < y and if a0 > b0 then x > y. Suppose that a0 = b0. If x 6= ythen there exists a smallest n ≥ 1 such that an 6= bn and ai = bi for 0 ≤ i ≤ n−1.Hence, just as in the previous proof of Lemma 22, x < y if and only if x1 < y1

⇔ · · · ⇔ xn < yn. Since xn, yn ∈ Iq have infinite λ-fractions we can applyLemma 21 and see that xn < yn ⇔ bn < 0 < an or anbn > 0 and an < bn.



LEMMA 24. The number R is given by the following regular λ-fraction

R =

[1; 1h−1, 2

], for q even,[

1; 1h, 2, 1h−1, 2], for q ≥ 5 odd,[

1; 3], for q = 3.

Proof. To obtain the largest number without reversed forbidden blocks we useLemma 22 recursively. I.e. the largest one-digit λ-fraction is J−1K and J(−1)2Kis the largest two-digit λ-fraction etc. For q 6= 3 J(−1)hK is the largest h-digit λ-fraction without reversed forbidden blocks but since J(−1)h+1K is forbidden andreversely forbidden the largest h + 1-digit λ-fraction without reversed f.b.’s isJ(−1)h ,−2K. Continuing like this inductively and observing that the λ-fractionwithout the first −1 is always regular we obtain the following expressions forR: R = J0; (−1)h ,−2, (−1)h−1K for even q, R = J0;−2,−3K for q = 3 and

R = J0; (−1)h ,−2, (−1)h ,−2, (−1)h−1K for odd q ≥ 5. The Lemma then fol-lows by rewriting these λ-fractions recursively into regular λ-fractions using Lem-mas 11 and 13.

LEMMA 25. For even q we have the identity R = 1.

Proof. Consider the action of S on R:

SR =[0; 1h, 2, 1h−1

]=

[−1; (−1)h−1 ,−2]

= −R.

Hence −1/R = −R and since R > 0 we must have R = 1.

LEMMA 26. For odd q we have λ2 < R < 1 and

a) −R = (TS)h+1 R,b) R2 + (2− λ) R− 1 = 0

Proof. From the explicit expansions of λ2 and 1 in Lemmas 5 and 6 together with

the the expansions of R in Lemma 24 it follows from Lemmas 22 and 23 thatλ2 < R < 1. By rewriting as in Lemma 13 for q ≥ 5 we get

SR = J0; 1h+1, 2, 1h−1, 2, 1hK =[−1; (−1)h−1 ,−2, (−1)h ,−2, (−1)h−1 ,−2

]

and deduce that R = ST−1(ST−1

)h−1ST−2T (−R) =

(ST−1

)h+1 (−R) andhence−R = (TS)h+1 R, which is identity a). A similar rewriting works for q = 3.Using the following explicit formula for the matrix (TS)n (cf. e.g. [8, p. 1279])

(2) (TS)n =1

sin2 πq

(Bn+1 −Bn

Bn −Bn−1

), where Bn = sin

nπ

q

and some elementary trigonometry gives (TS)h+1 R = −R+1−R+λ−1 = −R which

implies identity b). See also Lemma 3.3 in [8].

REMARK 27. Using the representation (2) one can also show that the map Ar

fixing r = R − 1 is given by Ar = (ST )h+1 T (ST )h T = 1(2−λ)2

(2−2λ λ−2λ2

λ 2+λ2

)

for odd q and Ar = (ST )h−1 ST 2 = 14 sin2 π

q

(2−λ2 7λ−3λ3

λ 2+λ2

)for even q.



DEFINITION 28. Let y = Jb0; b1, . . .K be a finite or infinite λ-fraction. Set y1

=J0; b1, . . .K, y

j= σj−1y

1= J0; bj , . . .K and yj , j ≥ 1, the corresponding point in

R. Then y is said to be a dual regular λ-fraction if and only if it has the followingproperties:

if b0 = 0 ⇒ y ∈ IR,(D1)

if b0 6= 0 ⇒ y1 ∈ sign (b0) [r,R] , and(D2)

yj+1 ∈ sign (−yj) [r,R] for all j ≥ 1.(D3)

A dual regular λ-fraction is denoted by[b0; b1, . . .

]∗, the space of all dual regularλ-fractions by A∗q and the subspace of all infinite sequences in A∗q with leading 0by A∗0,q.

Uniqueness of a subset of dual regular λ-fractions is again asserted using a gener-ating map.

DEFINITION 29. Let b·c be the floor function from Definition 7 and consider theshifted nearest λ-multiple function 〈y〉∗λ =

⌊ yλ + R

λ

⌋if y ≤ 0 and 〈y〉∗λ =

⌊ yλ − r

λ

⌋if y > 0. For IR = [−R, R] we define the map F ∗

q : IR → IR by

F ∗q y =

− 1

y − 〈− 1y 〉∗λλ, y ∈ IR\ 0 ,

0, y = 0.

LEMMA 30. For y ∈ R the following algorithm produces a finite or infinite dualregular λ-fraction c∗q (y) =

[b0; b1, . . .

]∗ corresponding to y:(i) Let b0 = 〈y〉∗λ and y1 = y − b0λ.

(ii) Set yj+1 = F ∗q yj = − 1

yj− bjλ, i.e. bj = 〈− 1

yj〉∗λ, j ≥ 1.

If yj = 0 for some j the algorithm stops and one obtains a finite dual regularλ-fraction.

Proof. It is easy to verify that 〈y〉∗λ = 0 ⇔ y ∈ IR and that in general x − 〈x〉∗λ ∈[r,R] for x ≥ R and x − 〈x〉∗λ ∈ [−R,−r] for x ≤ −R. It is thus clear that (D1)and (D2) are automatically fulfilled and it follows that F ∗

q maps [−R, 0] into [r,R]and [0, R] into [−R,−r]. Hence condition (D3) is also satisfied.

REMARK 31. We say that F ∗q is a generating map for the dual regular λ-fractions

and it is easily verified that F ∗q acts as a left shift map on A∗0,q.

It is easily seen that the points affected by the choice of floor function appearingin 〈·〉∗λ (cf. ±λ

2 in the regular case) are exactly those that are equivalent to ±r.Hence we obtain the following corollary.

COROLLARY 32. If y has an infinite dual regular λ-fraction expansions which isnot equivalent to the expansion of ±r then it is unique and is equal to c∗q (y).

LEMMA 33. A λ-fraction y = Jb0; b1, . . .K is dual regular if and only if the se-quence y

0does not contain any reversed forbidden blocks. Thereby y

0= y if

b0 = 0 and y0

= Sy = J0; b0, b1, . . .K if b0 6= 0.



Proof. For both even and odd q we use Lemma 21, 22 and 23 to compare pointsbased on their λ-fractions. We give the details for the case of forbidden blockscontaining +1’s. The case of −1’s is analogous.

Consider q even and a reversed forbidden block of the form Jm, 1hKwith m ≥ 1.If y

0contains such a forbidden block, we have y

j= Jbj , 1h, bh+j+1, . . .K < 0 for

some j ≥ 0 and bj ≥ 1. Hence yj+1

= J1h, bh+j+1, . . .K <[1h−1, 2

]= r, i.e.

yj+1 /∈ [r,R] so by Definition 28 (D2) (in case j = 0) or (D3) (in case j > 0) y isnot dual regular.

Consider q odd. If y0

contains a reversed f.b. of the form J1h+1K then y0

is notdual regular since J1h+1K <

[1h, 2, 1h, 2, 1h−1

]∗ = −R. If y0

contains a reversedf.b. of the form Jm, 1h, 2, 1hK for some m ≥ 1 then y

j= Jbj , 1h, 2, 1h, bj+2h+2, . . .K <

0 for some bj ≥ 1 and j ≥ 0. Hence yj+1

= J1h, 2, 1h, bj+2h+1, . . .K < r, i.e.yj+1 /∈ [r,R] so by Definition 28 (D2) (in case j = 0) or (D3) (in case j > 0) y isnot dual regular.

In the other direction, suppose that y = Jb0; b1, . . .K does not contain any re-versed forbidden blocks. It is clear that if b0 = 0 then y

0= y and y = y0 ∈

[−R, R] so y is dual regular. Suppose that b0 6= 0. Then y0

= Jb0, b1, b2, . . .K andy

j= Jbj , bj+1, . . .K. Suppose that y

0does not contain any reversed f.b. and that

bj > 0. If q is odd then yj+1

> J1h, 2, 1h−1, 2, . . .K ≥ r and if q is even then

yj+1

> J1h−1, 2, . . .K ≥ r. Hence yj+1 ∈ [r,R]. Similarly if bj < 0 we see thatyj+1 ∈ [−R,−r].

LEMMA 34. An infinite λ-fraction without reversed forbidden blocks converges.

Proof. This follows from the convergence of infinite regular λ-fractions using rewrit-ing. Note that the only case in which rewriting gives a non-convergent λ-fraction iswe produce an infinite sequence of new forbidden blocks all beginning at the sameposition. From the proofs of Lemmas 18 19 and 20 it is easy to see that this onlyoccurs for the λ-fraction x if either q = 4 and x has the tail J12, 2, 1K or q = 3 andx has the tail J1, 3K. These two tails both contain reversed forbidden blocks andhence can not occur.

REMARK 35. Just as the regular λ-fractions are equivalent to the reduced Rosen λ-fractions, one can show that the dual regular λ-fractions are essentially equivalentto a particular instance of so-called α-Rosen λ-fractions, see [10] and [34] (in thecase q = 3). Note that 〈y〉∗λ =

⌊ yλ + 1− R

λ

⌋for y > 0. Hence F ∗

q x = Tα (x) withα = R

λ for x < 0 where Tα is the generating map of the α-Rosen fractions of [10].

3.4. Symbolic dynamics and natural extensions. An introduction to symbolicdynamics and coding can be found in e.g. [28]. See also [4, 47] or [4, AppendixC]. Our underlying alphabet is infinite, N = Z∗ = Z\ 0. The dynamical sys-tem

(N Z+ , σ+)

is called the one-sided full N−shift. Since the forbidden blocks(cf. Definitions 11 and 13) imposing the restrictions on A0,q and A∗0,q all have fi-nite length it follows that (A0,q, σ

+) and (A∗0,q, σ+) are both one-sided subshifts of

finite type (cf. [4, Thm. C7]).



One can show that cq : Iq → Aq and c∗q : IR → A∗q as given by Lemmas 8and 30 are continuous (with respect to the metric h defined in Section 3.1) andwe call these the regular and dual regular coding map respectively. Let R∞ =x ∈ R | c∗q (x) infinite

= R\Gq (∞) be the set of ,,Gq-irrational points“ and set

I∞α = Iα ∩ R∞ for α = q, R. Since the set Gq (∞) of cusps of Gq is countableit is clear, that the Lebesgue measure of I∞α is equal to that of Iα, α = q,R. ByCorollaries 10 and 32 it follows that the restrictions cq : I∞q → A0,q and c∗q : I∞R →A∗0,q are homeomorphisms. Since σ+ = cq Fq c−1

q on A0,q and σ+ = c∗q F ∗q

c∗q −1 on A∗0,q it follows that the one-sided subshifts (A0,q, σ+) and (A∗0,q, σ

+) aretopologically conjugate to the abstract dynamical systems

(I∞q , Fq,

)and

(I∞R , F ∗

q

)respectively (see [4, p. 319]).

Consider the set of regular bi-infinite sequences Bq ⊂ A∗0,q×A0,q ⊂ ZZ consist-ing of precisely those

[. . . , b2, b1 ¦ a1, a2, . . .

]which do not contain any forbidden

block. Then (Bq, σ) is a two-sided subshift of finite type extending the one-sidedsubshift (A0,q, σ

+) , where σ = σ+ and σ−1 = σ−. If c∗q (y) =[b1, b2, . . .

]∗ ∈A∗0,q and cq (x) =

[a1, a2, . . .

] ∈ A0,q we define the coding map C : Iq×IR → ZZby C (x, y) = c∗q (y) ¦ cq (x) =

[. . . , b2, b1 ¦ a1, a2, . . .

]. For a given bi-infinite

sequence ζ =[. . . , b2, b1 ¦ a1, a2, . . .

]we let ζ

(+)=

[a1, a2, . . .

]and ζ

(−)=[

b1, b2, . . .]∗ denote the ,,future“ and ,,past“ respectively. In the next section we

will see that there exists a domain Ω ⊂ Iq × IR such that C|Ω∞ : Ω∞ → Bq isone-to-one and continuous (here Ω∞ = Ω∩ I∞q × I∞R , i.e. we neglect points (x, y)where either x or y has a finite λ-fraction). The natural extension, Fq, of Fq to Ω∞

is defined by the condition that (Bq, σ) is topologically conjugate to(Ω∞, Fq

),

i.e. by the relations σ+ = C Fq C−1 and σ− = C F−1q C−1, meaning that

Fq (x, y) =(Fqx, −1

y+a1λ

)with a1 = 〈−1

x 〉λ and F−1q (x, y) =

(−1

x+b1λ , F ∗q y

)with

b1 = 〈−1y 〉∗λ.

3.5. Markov partitions for the generating map Fq. To construct a Markov parti-tion of the interval Iq with respect to Fq we consider the orbits of the endpoints±λ

2 .For x ∈ Iq or y ∈ IR we define the Fq-orbit and F ∗

q -orbit of x and y respectivelyas

O (x) =F j

q x | j ∈ Z+

and O∗ (y) =

F ∗

qjy | j ∈ Z+

.

By Lemma 5 it is clear that O (±λ2

)is a finite set. Define κ = #

O (λ2

)− 1 =q−22 = h for even q and κ = q − 2 = 2h + 1 for odd q. Let −λ

2 = φ0 < φ1 <

· · · < φκ = 0 be an ordering of O (−λ2

), set Ij = [φj−1, φj) and I−j = −Ij

for 1 ≤ j ≤ κ. It is easy to verify that the closure of the intervals form a Markovpartition of Iq for Fq. I.e.

Ij

covers Iq, overlaps only at endpoints and Fq maps

endpoints to endpoints. Since the alphabet N is infinite there exist also anotherMarkov partition of the form Jn =

[−2

λ(2n−1) ,−2

λ(2n+1)

]∩ [−λ

2 , 0]

= −J−n with

n ∈ N and if q > 4 then n ≥ 1 and if q = 3 then n ≥ 2. The map − 1x − nλ

restricted to Jn is expanding and bijective and maps the interval Jn for n = 2 in



the case q = 3 respectively for n = 1 in case q > 3 not onto the entire interval Iq,that means the Markov partition is not proper.

From the explicit formula of Fq−1 it is clear that we also need to consider

the orbits of the endpoints of ± [r,R]. From Lemmas 25 and 26 we see that# O∗ (−R) = κ + 1. Set r0 = −R and let 0 > r1 > r2 > · · · > rκ = r >−R = r0 be an ordering of O∗ (−R) = rj. One can verify that rκ+1−j ∈ Ij ,1 ≤ j ≤ κ. Define the intervals Rj = [rj , R] = −R−j , 1 ≤ j ≤ κ, the rectanglesΩj = Ij × Rj , 1 ≤ |j| ≤ κ and finally the domain Ω = ∪|j|≤κΩj . We also setΩ∞= Ω ∩ I∞q ∩ I∞R .

REMARK 36. For even q we have φ0 = −λ2 =

[1h

], r =

[1h−1, 2

]and κ = h

where h = q−22 (see Lemma 5 and 24). It is then easy to verify that

φj = F jq (φ0) = −φ−j =

[1h−j

], 0 ≤ j ≤ h,

rj = F h−jq (r) =

[1j−1, 2, 1h−1

], 1 ≤ j ≤ h,

Ij = [φj−1, φj) =[[

1h+1−j],[1h−j

])= −I−j , 1 ≤ j ≤ h,

Rj = [rj , R] = −R−j , 1 ≤ j ≤ h.

REMARK 37. For odd q ≥ 5 we have φ0 = −λ2 =

[1h, 2, 1h

], r =

[1h, 2, 1h−1, 2

]

and κ = 2h+1 where h = q−32 (see Lemma 5 and 24). It is then easy to verify that

φ2j = F jq (φ0) =

[1h−j , 2, 1h

], 0 ≤ j ≤ h,

φ2j−1 = F h+jq (φ0) =

[1h+1−j

], 1 ≤ j ≤ h + 1,

r2j+1 =[1j , 2, 1h−1, 2, 1h, 2

], 0 ≤ j ≤ h and

r2j =[1j−1, 2, 1h, 2, 1h−1, 2

], 1 ≤ j ≤ h.

Hence

I2j+1 = [φ2j , φ2j+1) =[[

1h−j , 2, 1h],[1h−j

]), 0 ≤ j ≤ h,

I2j = [φ2j−1, φ2j) =[[

1h+1−j],[1h−j , 2, 1h

]), 1 ≤ j ≤ h,

Rk = [rk, R] = −R−k, 1 ≤ k ≤ 2h + 1.

For q = 3 we have κ = 1, φ0 = −12 =

[2], φ1 = 0 and r1 = r =

[3]. Hence

I1 =[−1

2 , 0)

= −I−1 and R1 = [r,R] = −R−1.

To establish the sought correspondence between the domain Ω∞ and Bq we firstneed a Lemma.

LEMMA 38. r is the smallest number y in IR such that C (x, y) ∈ Bq for allx ∈ Iκ.

Proof. Let q be even. We know from Lemma 24 and its proof that r =[1h−1, 2

]∗,−R =

[1, 1h−1, 2

]∗ and φκ−1 = φh−1 =[1]. Hence C (φh−1, r) ∈ Bq and for

−R ≤ y < r then c∗q (y) =[1h, bh+1, . . .

]∗ and C (φh−1, y) contains the forbiddenblock J1h+1K.JOURNAL OF MODERN DYNAMICS VOLUME 3, NO. 3 (2009), 1–46


FIGURE 1. Domains Ω and Ω∗ = Ω∗L ∪ Ω∗R for q = 7

ΩΩ∗

LΩ

∗

R

Let q ≥ 5 be odd. Then r =[1h, 2, 1h−1, 2

]∗, −R =[1, 1h−1, 2, 1h, 2

]∗ andφκ−1 = φ2h =

[2, 1h

]. Hence C (φ2h, r) ∈ Bq and if −R ≤ y < r then c∗q (y) =[

1h, 2, 1h, b2h+2, . . .]∗ and C (φ2h, y) contains the forbidden block J1h, 2, 1h, 2K.

We have shown that r is the smallest number such that C (φκ−1, r) does not containa forbidden block and it is easy to show that also C (x, r) ∈ Bq for any x ∈ Iκ =[φκ−1, 0). The same argument applies to q = 3.

LEMMA 39. (x, y) ∈ Ω∞⇔ C (x, y) ∈ Bq.

Proof. Just as in the proof of Lemma 38 it is not hard to verify that rj is the smallestnumber in IR with a dual regular expansion which can be prepended to the regularexpansion of φj and hence of all x ∈ [φj , 0).

DEFINITION 40. To determine the first return map we introduce a ,,conjugate“region Ω∗ = S (Ω∞) where S (x, y) = (Sx,−y), i.e. setting I∗j = S

(Ij ∩ I∞j

),

R∗j = −Rj ∩ I∞R and Ω∗j = I∗j ×R∗j we get Ω∗ = ∪Ω∗j .

Thus Ω with Ω∞ as a dense subset is the domain of the natural extension Fq ofFq. An example of Ω and Ω∗ is given in Figure 1. See [35] for another choice ofa ,,conjugate“ Ω∗ of Ω using the maps

(x, y−1

)and also [8] for the corresponding

domain for the reduced Rosen fractions. In the case q = 3 the domain Ω∗ wasconsidered also by Hurwitz in his reduction process for pairs of points under hiscontinued fraction expansion.

3.6. Reduction of λ-fractions. In a first step in our construction of a cross-sectionfor the geodesic flow we select a set of geodesics on H which contains at least onelift of each geodesic on Mq = H/Gq, i.e. a set of ,,representative“ or ,,reduced“geodesics modulo Gq. For an overview and a discussion of different reductionprocedures in the case of PSL2(Z) see [25, Sect. 3].

LEMMA 41. Let u 6= v ∈ R∞ both have infinite λ-fractions. Then there existsB ∈ Gq such that (Bu,Bv) ∈ Ω∗.

Proof. Set u′ = Su = −1u and v′ = Jv = −v. Assume without loss of gen-

erality that cq (u′) =[a0; a1, a2, . . .

]and c∗q (v′) =

[0; b1, b2, . . .

]∗. We nowextend the domain of definition of F−1

q from Iq × IR to R∞ × IR by setting



F−1q (x, y) =

(−1

x+bλ , F ∗q y

), b = 〈− 1

y 〉∗λ. Since x ∈ R∞ x + bλ 6= 0 so F−1q (x, y)

is well-defined. Then F−1q (u′, v′) =

(ST b1u′, T−b1Sv′

),

(F−1

q

)2(u′, v′) =

(ST b2ST b1u′, T−b2ST−b1Sv′

), etc. In fact, if n ≥ 1 then

(F−1

q

)n(u′, v′) =

(SAnSu′, JAnJv′) = S An (u, v) where An = T bnST bn−1S · · · T b1S. It iseasy to verify that JTmJ = T−m for any m ∈ Z.

If aj = −bj+1, j = 0, . . . , k − 1 and ak + bk+1 6= 0 and we take n > k it isclear that u′n = SAnSu′ has the formal λ-fraction u′n = J0; bn, . . . , bk+2, bk+1 +ak, ak+1, . . .K and v′n = JAnJv′ has the dual regular λ-fraction v′n =

[bn+1, bn+2, . . .

]∗.To simplify the notation we assume k = 0, i.e. a0 + b1 6= 0. Since cq (u′) ∈ Aq

and c∗q (v′) ∈ A∗q it is clear that any f.b. in u′n must include the term a0 + b1. Letn0 ≥ 1 be larger than the length of the largest f.b.. If q ≥ 5 then by Lemma 18 it isclear that we may choose n ≥ n0 large enough so that rewriting u′n into a regularλ-fraction does not change the beginning of u′l, up to bn−n0 .

If q = 4 and it is not the case that u′ =[±1− b1; (±2)

]and v′ =

[b1,±1, (±2)

]∗for some±b1 ≤ −1 (i.e. u′ and v′ have the same tail as±r), then by Lemma 19 weknow that we can choose n large enough so we can rewrite u′n completely withoutchanging the head u′(n−n0) for some n0 ≥ 0.

If q = 3 then aj 6= ±1 for j 6= 0 and bj 6= ±1 for all j. Hence there are onlythree kinds of f.b.’s possible in u′n for q = 3. Either the f.b. starts with b2 = ±2 and± (b1 + a0) ≥ 1 or it starts with b1 + a0 = ±1 or b1 + a0 = ±2 and ±a1 ≥ 2. Wecan thus apply Lemma 20 and in case v′ and u′ are not of the form v′ =

[b1, 2, 3

]∗and u′ =

[1− b1; 3

]for some b1 ≤ −1 we can choose n large enough that rewriting

of u′n does not change the head u′(n−n0) for some n0 ≥ 0.After a complete rewriting (if necessary) which does not change the head of u′n

up to bn−n0 we may assume that u′n =[bn, bn−1, . . . , bn−n0 , a1, a2, . . .

] ∈ Aq andv′n =

[bn+1, bn+2, . . .

]∗ ∈ A∗q . Hence C (u′n, v′n) =[. . . , bn+2, bn+1 ¦ bn, . . . , bn−n0 , a1, a2, . . .

]and since n0 is larger than the length of any f.b. it is clear that C (u′n, v′n) ∈Bq, i.e. (u′n, v′n) ∈ Ω or equivalently S (u′n, v′n) ∈ Ω∗ and hence S (u′n, v′n) =

S(S (Anu,Anv)

)= (Anu,Anv) ∈ Ω∗ with An = T bnS T bn−1S · · · T b1S ∈

Gq.We now have to treat the special cases which are left. Without loss of generality

we assume that we have f.b. containing a 1, i.e. we assume the plus sign in thestatement of the theorem. The case of minus sign is analogous.

Suppose that q = 3, v′ =[b1, 2, 3

]∗ and u′ =[1− b1; 3

]with b1 ≤ −2 (since

v′ ∈ A∗q). Then we can rewrite v′ into the regular λ-fraction[b1 − 1,−3

]. Hence

u = Su′ = −v′ = v = ST 1−b1 (r) but we assumed that u 6= v so this case can nothappen.

Suppose that q = 4, v′ =[b1, 1, 2

]∗ and u′ =[1− b1; 2

]with b1 ≤ −1

(since v′ ∈ A∗q). Then we can rewrite v′ into the following regular λ-fraction[b1 − 1,−2

]. Hence u = Su′ = −v′ = v = ST 1−b1 (r) but we assumed that

u 6= v so this case can not happen.



REMARK 42. In the previous lemma we actually showed that for any u ∈ R∞,which in case q = 3 or 4 is not Gq-equivalent to ±r, there exists a B ∈ Gq suchthat (Bu,Bu) ∈ Ω∗.

REMARK 43. For q = 3 the previous lemma and remark should be compared withHurwitz [22, §7].

In fact, one can do slightly better than in the previous Lemma by using theimportant property of the number r, namely that r and −r are Gq-equivalent butnot orbit-equivalent, i.e. O∗ (r) 6= O∗ (−r) (cf. Lemmas 25 and 26). Using theexplicit map identifying r and −r one can show that it is possible to reduce anygeodesic to one with endpoints in Ω∗ without the upper horizontal boundary.

LEMMA 44. If (x, y) ∈ Ω∗ and y has a dual regular expansion with the same tailas −r then there exists A ∈ Gq such that (Ax, Ay) ∈ Ω∗ and Ay has the same tailas r.

Proof. Using Fq we may assume that y = −r and Sx ∈ Iκ. Consider even q. LetA := T−1ST−1 with A (−r) = r (recall that −R = SR). Set y′ = Ay = r andx′ = Ax. By Remark 36 Iκ = Ih =

[− 1λ , 0

)hence if Sx ∈ Ih then x ≥ λ,

T−1x ≥ 0, ST−1x < 0, T−1ST−1x < −λ and finally ST−1ST−1x ∈ (0, 1

λ

) ⊆I−h. Hence (Sx′,−y′) ∈ Ω∞.

Consider odd q ≥ 5. There are three cases to consider if y = −r and Sx ∈ Iκ =I2h+1 = [φ2h, 0) =

[−1λ+1 , 0

)(cf. Remark 37): We have cq (Sx) =

[a1, a2, . . .

]

with either a1 ≥ 3 or cq (Sx) =[2, 1j , aj+2, . . .

]for some 0 ≤ j ≤ h − 1 for

aj+2 6= 1 or cq (Sx) =[2, 1h, ah+2, . . .

]for ah+2 ≤ −1. In the first case, set

A = T−1, y′ = T−1y = −R and x′ = T−1x. Then Sx ∈ [− 13λ , 0

), x ∈ [3λ,∞),

x′ = T−1x ∈ [2λ,∞) and Sx′ ∈ [− 12λ , 0

) ⊆ [−λ2 , 0

). Hence (Sx′, R) ∈ Ω∞.

In the second case, unless j = h − 1, ah+1 = 2, ah+2 = · · · = a2h+1 = 1 anda2h+2 ≥ 1 we also set A = T−1, y′ = −R and x′ = Ax. Thus cq (Sx′) =[1j+1, aj+2, . . .

] ∈ A0,q and it follows that Sx′ ∈ [−λ2 , 0

)and (Sx′, R) ∈ Ω∞.

In the remaining two cases, set A = T−1(ST−1

)hST−2, y′ = Ay = r and

x′ = Ax. Then Sx′ = SAx =(ST−1

)h+1ST−2T 2 (ST )j ST a2+j · · · (0) =(

ST−1)h+1−j

T a2+j ST a3+j · · · (0). If j = h then cq (Sx′) =[aj+2 − 1, . . .

] ∈A0,q and if j = h − 1, ah+1 = 2, ah+2 = · · · = a2h+1 = 1 and a2h+2 ≥ 1then Sx′ = J−1, 1h+1, a2h+1, . . .K which is rewritten into J−2, (−1)h , a2h+1 −1, . . .K which is either regular or contains a new f.b. beginning at a2h+1 − 1 =1. In any case, by Lemma 18 it is clear that after rewriting any f.b. in Sx′completely we either get cq (Sx′) =

[−2, (−1)h , a2h+1 − 1, . . .]

or cq (Sx′) =[−2, (−1)h−1 ,−2, . . .]

and in both cases C (Sx′,−y′) ∈ Bq .If q = 3 and y = −r then there are two possibilities, either Sx ∈ (−1

2 ,−13

)or Sx ∈ [−1

3 , 0). In the first case we let A = T−1ST−2 so that y′ = Ay =

r and Sx′ = ST−1ST−2x. Since x ∈ (2, 3) it is clear that ST−1ST−2x ∈ST−1S (0, 1) = ST−1 (−∞,−1) = S (−∞,−2) =

(0, 1

2

)hence (Sx′,−y′) ∈



Ω∞. In the second case, let A = T−1. Then y′ = −r − 1 = −R and Sx′ =ST−1x ∈ ST−1 (3,∞) = S (2,∞) =

(−12 , 0

)so (Sx′,−y′) ∈ Ω∞.

In all cases we have shown that (Sx′,−y′) ∈ Ω∞, i.e. (x′, y′) ∈ Ω∗ and y′ =−R or r.

3.7. Geodesics and geodesic arcs. If γ (ξ, η) is a geodesic in H oriented fromη to ξ (cf. Section 1.1) and A ∈ PSL2(R) we define the geodesic Aγ as Aγ =γ (Aξ, Aη) . If (ξ, η) ∈ Ω∗ we associate to (ξ, η) a bi-infinite sequence (code) to γ,C (γ) = C S (ξ, η) = cq (−η) .c∗q (Sξ) ∈ Bq.

DEFINITION 45. For an oriented geodesic arc c on H we let c denote the uniquegeodesic containing c and preserving the orientation, e.g. L1 = λ

2 + iy∣∣ y > 0

oriented upwards. Let c± denote the forward and backward end points of c and let−c denote the geodesic arc with endpoints−c±. Here−c should not to be confusedwith the geodesic c with reversed orientation, denoted by c−1.

For z, w ∈ H ∪ ∂H denote by [z, w] the geodesic arc oriented from z to wincluding the endpoints in H.

DEFINITION 46. Let

Bq =ζ ∈ Bq | (σ−nζ)(−) 6= c∗q (−r) , ∀n ≥ 0

be the set of bi-infinite sequences which does not have the same past as −r and let

Υ =

γ | C (γ) ∈ Bq

.

LEMMA 47. The coding map C : Υ → Bq is a homeomorphism.

Proof. Υ is identified with the set of endpoints in Ω∗, the map S : Ω∗ → Ω∞in Definition 40 is a continuous bijection and by Corollary 10 and 32 respectivelyLemmas 39 and 44 it is clear that C : Ω∞ → Bq is well-defined, continuous andthat each point (ξ, η) ∈ Ω∞ has a unique code unless η is equivalent to±r in whichcase it might have two codes. Since one of these is disregarded in the definition ofBq it is clear that C = C S : Υ → Bq is a homeomorphism. For the case ofGq = PSL2(Z) see also [25, p. 105].

The set Υ contains representatives of all geodesics on H/Gq. This property isan immediate corollary to the following Lemma which additionally also provides areduction algorithm.

LEMMA 48. Let γ be a geodesic on the hyperbolic upper half-plane with endpointsγ−, γ+ ∈ R∞, γ− 6= γ+ and γ− =

[b0; b1, . . .

]∗. Then there exists an integer

n ≥ 0 and A ∈ Gq such that

γ′ = A T−bnS · · · T−b1ST−b0γ

and C (γ′) ∈ Bq . Here A is one of the maps Id, T−1 and T−1ST−1 for even q,

respectively T−1(ST−1

)hST−2 for odd q.

Proof. This is an immediate consequence of Lemmas 41 and 44. Note that T−b0γ− =[b1, b2, . . .

]∗ and F ∗q

nT−b0γ− = T−bnS · · · T−b1ST−b0γ−.



COROLLARY 49. If γ∗ is a geodesic on H/Gq with all lifts having endpoints inR∞ then Υ contains an element of π−1 (γ∗).

LEMMA 50. If for ξ ∈ Iq cq (ξ) =[a1, . . . , an

] ∈ A0,q, then ξ is the attractivefixed point of the hyperbolic map A = ST a1 ST a2 · · · ST an with conjugate fixedpoint ξ∗ = η−1, where c∗q (η) =

[an, an−1, . . . , a1

]∗ ∈ A∗0,q. Conversely, if ξ is anhyperbolic fixed point of B ∈ Gq, then cq (ξ) is eventually periodic.

Proof. It is not hard to show, that Aξ = ξ and Aη−1 = η−1. Since η ∈ IR, withR ≤ 1 < 2

λ it is clear that ξ∗ 6= ξ and hence A is hyperbolic. The other statementin the Lemma is easy to verify by writing B in terms of generators, rewriting anyforbidden blocks and going through all cases of non-allowed sequences, e.g. if Bends with an S.

Since the geodesic γ (ξ, η) is closed if and only if ξ and η are conjugate hy-perbolic fixed points and since r and −r are Gq-equivalent we conclude fromLemma 50 that there is a one-to-one correspondence between closed geodesicson Mq = H/Gq and the set of equivalence classes, under the shift map, of purelyperiodic regular λ-fractions except for the one containing −r.

REMARK 51. Because Ω∞ only contains points with infinite λ-fractions, the setΥ does not contain lifts of geodesics which disappear out to infinity, i.e. with oneor both endpoints equivalent to ∞. The neglected set however corresponds to a setin T 1M of measure zero with respect to any probability measure invariant underthe geodesic flow. See e.g. the introduction of [24].

The subshift of finite type(Bq , σ

)is conjugate to the invertible dynamical sys-

tem(Υ, Fq

). Here Fq : Υ → Υ is the map naturally induced by Fq acting on Ω∞:

i.e. if γ = γ (ξ, η) , then F±1q (γ) = γ (ξ′, η′) where (ξ′, η′) = SF±1

q S−1 (ξ, η).Using the same notation for both maps should not lead to any confusion.

3.8. Reduced geodesics.

DEFINITION 52. A geodesic γ (ξ, η) with ξ, η ∈ R∞ is said to be reduced ifC (γ) ∈ Bq and |ξ| > 3λ

2 or ξη < 0. Denote by Υr the set of reduced geodesicsand by Ω∗r the corresponding set of (ξ, η) ∈ Ω∗. Then Ω∞r := S (Ω∗r) ⊆ Ω∞ andΩr = Ω∞r , the closure of Ω∞r inR2, i.e. Ω∞r = Ωr∩Ω∞, and Bq,r = C (Υr) ⊆ Bq .REMARK 53. We observe that for odd q, by Lemma 6, φκ−1 =

[2, 1h

]= −1

λ+1 andthus− 2

3λ < φκ−1. For even q on the other hand φκ−1 = −1λ and since −1

λ < −23λ we

have φκ−1 < − 23λ . Hence the shape of Ω∞r =

(u, v) ∈ Ω∞ | |u| ≤ 2

3λ or uv < 0

differs slightly between even and odd q. Set Λ1 :=(−λ

2 , 0) × [0, R] and Λ2 :=(

0, 23λ

) × [0,−r] for even q respectively Λ2 := (0, φκ−1) × [0,−r] and Λ3 :=(φκ−1,

23λ

)× [0, rκ−1] for odd q. Then we have:

Ωr =k⋃

j=1

Λj ∪ −Λj ,



FIGURE 2. Domain of reduced geodesics Ωr

with k = 2 for even and 3 for odd q. See also Figure 2 where Ωr is displayed forq = 5 and q = 6 as a subset of Ω. An even more convenient description of the setof reduced geodesics is in terms of the bi-infinite codes of their base points (ξ, η)

Bq,r =[

. . . b2, b1 ¦ a0, a1, . . .] ∈ Bq

∣∣ |a0| ≥ 2, or a0b1 > 0

.

LEMMA 54. If γ is a geodesic on H with C (γ) ∈ Bq then there exists an integerk ≥ 0 such that F k

q γ is reduced.

Proof. For k ≥ 0 let γk := F kq γ, cq

(γk

+

)=

[ak; ak+1, . . .

]and c∗q

(γk−

)=[

ak−1, . . . , a0, b1, . . .]∗. If |ak| ≥ 2 then

∣∣γk+

∣∣ > 3λ2 and if ak−1ak < 0 then

γk+γk− < 0. In both cases by definition F k

q γ ∈ Υr. Since an infinite sequence of1’s or −1’s is forbidden, it is clear that there exists a k ≥ 0 such that one of theseconditions apply.

Combining the above lemma with Lemma 41 we have shown the followinglemma.

LEMMA 55. Let γ (ξ, η) be a geodesic on the hyperbolic upper half-plane withξ 6= η ∈ R∞. Then γ is Gq-equivalent to a reduced geodesic.

4. CONSTRUCTION OF THE CROSS-SECTION

As a cross section for the geodesic flow on the unit tangent bundle T 1M of Mwhich can be identified with Fq × S1 modulo the obvious identification of pointson ∂Fq × S1, we will take a set of vectors with base points on the boundary ∂Fq

directed inwards with respect toFq. The precise definition will be given below. Fora different approach to a cross section related to a subgroup of Gq see [14]. For thesake of completeness we include the case q = 3 in our exposition but it is easy toverify that our results in terms of the cross-section, first return map and return timeagree with the statements in [24, 25].



FIGURE 3. Cross-Section (q = 5)

4.1. The cross section.

DEFINITION 56. Consider the fundamental domain Fq and its boundary arcs L0

and L±1 as in Definition 1. For the construction of the cross section we use theadditional arcs L±2 = ± [ρ, λ] = T±1SL±1 and L±3 = ± [ρ, ρ + λ] = T±1L0 .

DEFINITION 57. We define the following subsets of T 1M:

Γr =(z, θ) ∈ Lr × S1 | γz,θ (s) directed inwards at z

, r = 0,±1,

Σj =

(z, θ) ∈ Γj | γz,θ = γ (ξ, η) ∈ Υ, |ξ| > 3λ

2or ξη < 0

, −1 ≤ j ≤ 1,

Σ±2 =

(z, θ) ∈ Γ±1 | γz,θ /∈ Υ, γ = T±1Sγz,θ ∈ Υ,3λ

2< γ+ ≤ λ + 1

,

Σ±3 =(z, θ) ∈ Γ0 | γz,θ /∈ Υ, γ = T±1γz,θ ∈ Υ

.

If q is even let Σ := ∪2j=−2Σ

j and if q ≥ 5 is odd let Σ := ∪3j=−3Σ

j . If q = 3we drop the restriction on γ+ in the definition of Σ±2 and set Σ = ∪2

j=−2Σj .

We will show that there is a one-to-one correspondence between Σ and the setof reduced geodesics.

LEMMA 58. There exists a bijection P : Υr → Σ defined through P (γ) :=(z, θ) ∈ Σ with z = γ (s) ∈ ∂Fq for some s ∈ R and θ given by Argγ (s) = θ.

Proof. Consider Figure 3 and a geodesic γ1 (ξ, η) from η ∈ [−R,−r) to ξ > 3λ2 .

It is clear that either γ1 intersects L−1 ∪L0 inwards or L2 from the left to the right.Let z = γ (s) be this intersection and set θ = Argγ (s). In the first case we getPγ = (z, θ) ∈ Σ−1 ∪ Σ0. In the second case we either get Pγ = (z, θ) ∈ Σ2 ifξ ∈ (

3λ2 , λ + 1

)or, if q is odd and ξ > λ + 1, we get Pγ = (z′, θ′) ∈ Σ3, where

z′ = γ (s′) ∈ L3 and θ′ = Argγ (s′), since by Lemma 93 γ must intersect L3.Remember, that for q even, r = 1 − λ, so γ can not intersect L3 to the right of



FIGURE 4. Illustration of the first return map

λ2 . Consider next a geodesic γ (ξ, η) with η ∈ [−R, 0) and ξ ∈ (

2λ , 3λ

2

). Since the

geodesic SL−1 from 0 to 2λ intersects ρ, the intersection point of γ and L1 must lie

above ρ. Hence γ intersects either L−1 or L0 inwards, i.e. Pγ ∈ Σ−1 ∪ Σ0.The case ξ < 0 is analogous and the inverse map P−1 : Σ → Υr is clearly given

by P−1 (z, θ) = γz,θ if (z, θ) ∈ Σj , |j| ≤ 1, respectively P−1 (z, θ) = T±1Sγz,θ

if (z, θ) ∈ Σ±2 and P−1 (z, θ) = T±1γz,θ if (z, θ) ∈ Σ±3.

DEFINITION 59. For (ξ, η) ∈ Ω∗r we define P : Ω∗r → Σ by, P (ξ, η) :=Pγ (ξ, η).

A consequence of Lemma 54 is that for any reduced geodesic we can find aninfinite number of reduced geodesics in its forward and backward Fq-orbit (withinfinite repetitions if the geodesic is closed). Furthermore, since the base-arcs L±1

and L0 of Σ consist of geodesics, none of whose extensions are in Υr, it is clearthat any reduced geodesic intersects Σ transversally. The set Σ thus fulfills therequirements (P1) and (P2) of Definition 3 and is a Poincare (or cross-) sectionwith respect to Υr. Since any geodesic γ∗ on Mq which does not go into infinityhas a reduced lift we also have the following lemma.

LEMMA 60. π∗ (Σ) is a Poincare section for the part of the geodesic flow on T 1Mwhich does not disappear into infinity.

From the identification of Σ and Ω∗r via the map P we see that the natural ex-tension Fq of the continued fraction map Fq induces a return map for Σ, i.e. ifz = (z, θ) ∈ Σ then P Fn

q P−1z ∈ Σ for an infinite number of n 6= 0. We givea geometric description of the first return map for Σ and we will later see that thismap is in fact also induced by Fq.

DEFINITION 61. The first return map T : Σ → Σ is defined as follows (cf. Figure4): If z0 ∈ Σ and γ = P−1z0 ∈ Υr let wnn∈Z be the ordered sequence ofintersections in the direction from γ− to γ+ between γ and the Gq-translates of ∂Fwith w0 given by z0. Since γ+ and γ− have infinite λ-fractions they are not cusps



FIGURE 5.

of Gq and the sequence wn is bi-infinite. For each wn let An ∈ Gq be the uniquemap such that w′n = Anwn ∈ ∂F and γ′ = Anγ intersects ∂F at w′n in the inwardsdirection.

If γ′ ∈ Υr and Pγ′ = z′ we say that z′ ∈ Σ is a return of γ to Σ. If n0 > 0 isthe smallest integer such that wn0 gives a return to Σ we say that the correspondingpoint PAn0γ = z1 ∈ Σ is the first return and the first return map T : Σ → Σ isdefined by T z0 = z1 where z1 is the first return after z0. Sometimes T : Ω∗r → Ω∗rgiven by T = P−1 T P is also called the first return map.

After proving some useful geometric lemmas in the next section we will showin Section 4.3 that the first return map T is given explicitly by powers of Fq.

4.2. Geometric lemmas.

LEMMA 62. The map z = (z, θ) 7→ (γz,θ, s) ∼= (ξ, η, s) where γz,θ = γ (ξ, η) ,γz,θ (s) = z and Argγz,θ (s) = θ is a diffeomorphism for θ 6= ±π

2 .

Proof. Let z = x + iy, y > 0, and θ ∈ [−π, π) be given. First we want to showthat there exist ξ, η ∈ R∗ and s ∈ R such that for the geodesic γ = γ (ξ, η) onefinds γ (s) = z and Argγ (s) = θ. Without loss of generality we may assume thatθ ∈ (−π

2 , π2

)so that η < ξ. Set c = 1

2 (η + ξ) , r = 12 (ξ − η) and parametrize γ

as γ (t) = c + reit, 0 < t < π. It is easy to verify that if c = x + y tan θ, r = ycos θ

and t0 = θ + π2 then γ (t0) = z and Argγ (t0) = θ. See Figure 5. To find the

arc length parameter s we use the isometry A : z 7→ −z+(c−r)z−(c+r) mapping γ to iR+,

Aγ (t) = i(tan t

2

)−1 and A (c + ir) = i. It is then an easy computation to seethat s (θ) = d (z, c + ir) = d

(i/ tan t

2 , i)

= ln tan t2 = ln tan

(θ2 + π

4

). From the

above formulas one can easily deduce differentiability of the map (z, θ) → (ξ, η, s)as well as its inverse away from θ = ±π

2 .

COROLLARY 63. The map (x, y, θ) → (ξ, η, s) of Lemma 62 gives a change ofvariables on T 1H which is diffeomorphic away from θ = ±π

2 . Explicitly, ξ = x +y tan θ+ y

|cos θ| , η = x+y tan θ− y|cos θ| , s = ln tan

(θ2 + π

4

)and the corresponding

Jacobian is∣∣∣∂(x,y,θ)∂(ξ,η,s)

∣∣∣ = 12 cos2 θ.

Proof. This follows from the proof of Lemma 62 and a trivial computation.



DEFINITION 64. For z ∈ H and ξ ∈ R define

g (z, ξ) =|z − ξ|2=z

and for γ a geodesic with endpoints γ+ and γ− set g (z, γ) = g (z, γ+) .

LEMMA 65. If A ∈ PSL2(R) then g (Az, Aξ) = g (z, ξ)A′ (ξ) .

Proof. Let A =(

a bc d

), then =Az = =z

|cz+d|2 , |Az −Aξ|2 = |z−ξ|2|cz+d|2|cξ+d|2 and

since ξ ∈ Rg (Az, Ax) = g (z, ξ) (cξ + d)−2 = g (z, ξ) A′ (ξ) .

LEMMA 66. Let γ = γ (ξ, η) be a geodesic with ξ, η ∈ R, η < ξ and suppose thatzj , j = 1, 2, with η ≤ <z1 < <z2 ≤ ξ are two points on γ. Then

d (z1, z2) = ln g (z1, ξ)− ln g (z2, ξ) = ln

[=z2

=z1

∣∣∣∣z1 − ξ

z2 − ξ

∣∣∣∣2]

.

Proof. It is easy to verify, that the hyperbolic isometry B : w 7→ w−ηξ−w maps γ to

iR+ and if a < b then d (ia, ib) =∫ ba

dyy = ln

(ba

). Thus if =zj = yj we get

d (z1, z2) = d (Bz1, Bz2) = d (i=Bz1, i=Bz2) = ln

[y2

y1

∣∣∣∣z1 − ξ

z2 − ξ

∣∣∣∣2]

.

LEMMA 67. Let γ = γ (ξ, η) be a geodesic with ξ, η ∈ R, η < 0 < ξ and letz = γ ∩ iR be the intersection of γ with the imaginary axis. Then

z = i√−ξη.

Proof. With r = 12 (ξ − η) and c = 1

2 (ξ + η) any point on γ is given by γ (t) =c + reit for some 0 ≤ t ≤ π. Suppose z = γ (t0), then <γ (t0) = c + r cos t0 = 0and hence cos t0 = − c

r . But then sin2 t0 = 1 − c2

r2 and therefore z = ir sin t0 =i√

r2 − c2 = i√−ξη.

LEMMA 68. Let γ = γ (ξ, η) be a geodesic with ξ, η ∈ R, η < ξ. For ω a geodesicintersecting γ at w ∈ H let A =

(a bc d

) ∈ PSL2(R) be such that Aω = iR+. Then

w = w (ξ, η) =1

ad + bc + ac (ξ + η)

[acξη − bd + εi

√−lA (ξ) lA (η)

], and

g (w, γ) =|w − ξ|2=w

= (ξ − η)

√− lA (ξ)

lA (η)

where lA (ξ) = (aξ + b) (cξ + d) and ε = sign (ad + bc + ac (ξ + η)).



Proof. According to Lemma 65 g (w, γ) = g(A−1z, A−1γ′

)= g (z, γ′) A−1 ′ (Aξ)

where γ′ = Aγ and z = Aw ∈ iR. By Lemma 67 z = i√−ξ′η′ with ξ′ = Aξ and

η′ = Aη. We choose A, i.e. the orientation of γ′ such that η′ < 0 < ξ′ and hencesign (aξ + b) = sign (cξ + d) and sign (aη + b) = −sign (cη + d). Then

w = A−1(i√−AξAη

)=

(di

√∣∣∣∣aξ + b

cξ + d

aη + b

cη + d

∣∣∣∣− b

)(a− ci

√∣∣∣∣aξ + b

cξ + d

aη + b

cη + d

∣∣∣∣)−1

=

[i

√∣∣∣∣aξ + b

cξ + d

aη + b

cη + d

∣∣∣∣−(

ab + dc

∣∣∣∣aξ + b

cξ + d

aη + b

cη + d

∣∣∣∣)](

c2

∣∣∣∣aξ + b

cξ + d

aη + b

cη + d

∣∣∣∣ + a2

)−1

=− (ab |cξ + d| |cη + d|+ dc |aξ + b| |aη + b|) + i


c2 |(aξ + b) (aη + b)|+ a2 |cξ + d| |cη + d|

=ε (ab (cξ + d) (cη + d)− dc (aξ + b) (aη + b)) + i


c2ε (aξ + b) (aη + b)− a2ε (cξ + d) (cη + d)

=i√−lA (ξ) lA (η)− ε [ξηca− bd]−ε [(ξ + η) ca + (cb + ad)]

=acξη − bd− iε


ac (ξ + η) + (ad + bc)

where ε = sign ((aξ + b) (aη + b)) = −sign (ac (ξ + η) + ad + bc) since w ∈ H.For the function g we now have

g(z, γ′

)=|z − ξ′|2=z

=ξ′2 − ξ′η′√−ξ′η′

=(ξ′ − η′

)√

ξ′

−η′.

Since A−1 ′ (Aξ) = (−cAξ + a)−2 = (cξ + d)2 Lemma 65 implies that

g (w, γ) =(ξ′ − η′

)√

ξ′

−η′(cξ + d)2 =

(aξ + b

cξ + d− aη + b

cη + d

)√√√√aξ+bcξ+d

aη+bcη+d

(cξ + d)2

= ((aξ + b) (cη + d)− (aη + b) (cξ + d))

√(aξ + b) (cξ + d)(aη + b) (cη + d)

= (ξ − η)

√(aξ + b) (cξ + d)(aη + b) (cη + d)

= (ξ − η)

√− lA (ξ)

lA (η).

Application of the previous Lemma to vertical or circular geodesics yields thefollowing corollaries:

COROLLARY 69. Let γ (ξ, η) be a geodesic with η < ξ. Then



a) if η < a < ξ for some a ∈ R, then γ and the vertical geodesic ωv = a + iR+

intersect at

Zv (ξ, η) = a + i√

(ξ − a) (−η + a) ∈ H, and

gv (ξ, η) = g (Zv, γ) = (ξ − η)

√− ξ − a

η − a;

b) if η < c − ρ < ξ < c + ρ for some c ∈ R and ρ ∈ R+, then γ and the circulargeodesic ωc with center c and radius ρ intersect at

Zc (ξ, η) =ξη + ρ2 − c2

ξ + η − 2c+

i

|ξ + η − 2c|

√((ξ − c)2 − ρ2

)(ρ2 − (η − c)2

)∈ H, and

gc (ξ, η) = g (Zc, γ) = (ξ − η)

√− (ξ − c)2 − ρ2

(η − c)2 − ρ2.

The subscripts ,,v“ and ,,c“ above refer to intersections with vertical and circulargeodesics respectively.

COROLLARY 70. Let γ = γ (ξ, η) be an arbitrary geodesic on H with ξ, η ∈ R.For Zj (ξ, η) = γ ∩ Lj set gj (ξ, η) = g (Zj , γ), −2 ≤ j ≤ 2. If Zj (ξ, η) exists,the following formulas hold:

Z0 (ξ, η) =1

ξ + η

(1 + ξη + εi

√(ξ2 − 1) (1− η2)

), ε = sign (ξ + η) ,

Z±1 (ξ, η) = ∓λ

2+ i

√(ξ ∓ λ

2

)(−η ± λ

2

),

Z±2,±3 (ξ, η) =1

ξ + η ∓ 2c

(ξη + ρ2 − c2 + εi

√((ξ ∓ c)2 − ρ2

)(ρ2 − (η ∓ c)2

))

where ε = sign (x + y ∓ 2c).

Furthermore

g0 (ξ, η) = (ξ − η)

√ξ2 − 11− η2

,

g±1 (ξ, η) = (ξ − η)

√− ξ ∓ λ

2

η ∓ λ2

,

g±j (ξ, η) = (ξ − η)

√− (ξ ∓ c)2 − ρ2

(η ∓ c)2 − ρ2, j = 2, 3.

Here (ρ, c) =(λ− 1

λ , 1λ

)for j = 2 and (1, λ) for j = 3.



Proof. Taking A0 = 1√2

(1 −11 1

), A±1 =

(1 ∓λ

20 1

)and A±2 = 1√

2ρ

(−1 ±c−ρ1 ∓c−ρ

)it

is easy to verify that AjLj = iR+ preserving the orientation.

LEMMA 71. Let γ = γ (ξ, η) be a geodesic with ξ, η ∈ R. Then γ intersects thevertical arc L1 if and only if η < λ

2 < ξ and δ (ξ, η) < 0 where

δ (ξ, η) = η − ξλ− 22ξ − λ

.

For even q we have in particular δ (ξ, η) = η − (TS)h+1 ξ.

Proof. It is clear, that γ intersects L1 if and only if the intersection with L1 =λ2 + iR+ is at a height above sin π

q = =ρ. By Lemma 69 the point of intersection

is given by w (ξ, η) = λ2 + i

√(ξ − λ

2

) (−η + λ2

). We thus need to check the

inequality(ξ − λ

2

) (−η + λ2

)> sin2 π

q . With η < λ2 < ξ it is clear that =w

decreases as η increases for ξ fixed. Using λ2 = cos π

q we see, that(

ξ − λ

2

)(λ

2− η

)= sin2 π

q= 1− λ2

4⇔

λ− 2η =4− λ2

2ξ − λ⇔ η =

λξ − 22ξ − λ

= Aξ

where A = 14−λ2

(λ −22 −λ

) ∈ SL2(R). Hence =w > sin πq ⇔ η < Aξ ⇔ δ (ξ, η) <

0.Observe, that Aρ = ρ and A2 = Id, i.e. A is elliptic of order 2. The stabilizer

Gq,ρ of ρ in Gq is a cyclic group with q elements generated by TS. For evenq = 2h + 2 one can use the explicit formula (2) to verify that A = (TS)h+1 ∈ Gq.For odd q on the other hand there is no element of order 2 in Gq,ρ, so A /∈ Gq.

COROLLARY 72. Let γ = γ (ξ, η) be a geodesic with ξ, η ∈ R. Set δn (ξ, η) :=δ (ξ − nλ, η − nλ). Then γ intersects the line TnL1 if and only if η <

(n + 1

2

)λ <

ξ and δn (ξ, η) < 0.

4.3. The first return map. Our aim in this section is to obtain an explicit expres-sion for the first return map T : Σ → Σ. The notation is as in Definition 61, seealso Figure 4. The main idea is to use geometric arguments to identify possiblesequences of intersections wn and then use arguments involving regular and dualregular λ-fractions to determine whether a particular wn corresponds to a return toΣ or not.

LEMMA 73. If ξ =[1; 1j , aj+1, aj+2

]with cq (Sξ) =

[1j+1, aj+1, . . .

] ∈ A0,q

then ξ ∈((TS)j λ, (TS)j 3λ

2

)if aj+1 ≤ −1 respectively ξ ∈

((TS)j+1 3λ

2 , (TS)j λ)

if aj+1 ≥ 2.

Proof. Note that ξ = (TS)j ξ′ where cq (ξ′) =[1; aj+1, aj+1, . . .

]. If aj+1 ≤

−1 then ξ′ ∈ (λ, 3λ

2

)and since TSx = λ − 1

x is strictly increasing there ξ =



FIGURE 6. Geodesics through the point ρ for q = 7 and 8 (h = 2and 3)

(TS)j ξ′ ∈ (TS)j (λ, 3λ

2

)=

((TS)j λ, (TS)j 3λ

2

). If on the other hand aj+1 ≥ 2

then ξ′′ = ST−1ξ′ ∈ (3λ2 ,∞)

and ξ = (TS)j ξ′ = (TS)j+1 ξ′′ and therefore

ξ ∈ (TS)j+1 (3λ2 ,∞)

=((TS)j+1 3λ

2 , (TS)j λ)

.

DEFINITION 74. Define the geodesic arcs

χj := (TS)j L2 = (TS)j+1 TL−1, 0 ≤ j ≤ h,

ωj := (TS)j L3 = (TS)j+1 L0, 0 ≤ j ≤ h + 1

and set αj := (TS)j λ, βj := (TS)j 3λ2 and δj := (TS)j (λ + 1). Then χj =

[ρ, αj ] , ωj =[ρ, (TS)j (ρ + λ)

]⊂ [ρ, δj ] and αj < βj < δj , 0 ≤ j ≤ h + 1.

Note that αh = λ2 and χh = L′1 for even q while δh+1 = λ

2 and ωh+1 ⊆ L′1 for oddq (see Figure 6).

LEMMA 75. If γ = γ (ξ, η) ∈ Υ with cq (ξ) =[1; 1j , aj+1, . . .

]then γ has the fol-

lowing sequence of intersections with Gq∂F after passing L1: ω0,χ0,. . . , ωj−1,χj−1, ωj if ξ ∈ (αj , βj) (aj+1 ≤ −1) respectively ω0, χ0, . . . , ωj , χj if ξ ∈(βj+1, αj) (aj+1 ≥ 2).

Proof. See Figure 6. Since all arcs involved are hyperbolic geodesics it is clearthat γ does not intersect any other χi’s or ωi’s than those mentioned. Supposethat ξ ∈ (αj , βj) , then after χj−1 the geodesic γ may intersect either ωj or itsextension, i.e. [(TS)j (ρ + λ) , δj ]. If it intersects this extension it has to passfirst through the arc [(TS)j Tρ, αj−1]. But the completion of this arc is clearly(TS)j TL1 = [βj , αj−1] and hence γ can not intersect this arc and must passthrough ωj . The second case is analogous, except that we do not care about whetherthe next intersection is at ωj+1 or (TS)j+1 TL1.

LEMMA 76. If γ (ξ, η) ∈ Υr and cq (ξ) =[1; 1j , aj+1, . . .

]then T γ (ξ, η) =

F j+1q γ (ξ, η) .

Proof. Let (z, θ) = P (ξ, η) , then z ∈ L−1 ∪L0 and by Lemma 75 the subsequentintersections are w0, w1, w2, . . . , w2j+1, w2j+2 if aj+1 ≤ −1 and w0, w1, w2, . . . ,



w2j+2, w2j+3 if aj+1 ≥ 2. It is also easy to verify, that the corresponding maps areA2i+1 = T−1

(ST−1

)i and A2i =(ST−1

)i. Note that A2i+1 = T−1A2i and thatA2iγ = Fq

iγ, 0 ≤ i ≤ j + 1. It is thus clear, that A2iγ ∈ Υ and A2i+1γ /∈ Υ for1 ≤ i ≤ j + 1. If (ξi, ηi) = (Aiξ,Aiη) , then cq (ξ2i) =

[1; 1j−i, aj+1, . . .

]and

c∗q (η2i) =[(−1)i , b1, . . .

]∗ and hence A2iγ /∈ Υr for 1 ≤ i ≤ j but A2j+2γ ∈ Υr.

Thus in both cases the first return is given by w2j+2 and the return map is T =Fq

j+1.

DEFINITION 77. DefineK : R→ N and n : R→ Z as follows: if ξ =[a0; (ε)

k−1 , ak, . . .]

with k ≥ 1, and ε = sign(a0), then K (ξ) := k and

n (ξ) :=

ε · 3, k = h + 1, q odd,

ε · 2, k = h, ah ≥ 2, q even,

ε · 1, k = h + 1, q even,

0, else.

We also have to consider the return map for the second type of reduced geodesics.

LEMMA 78. For z ∈ Σ with P−1z = (ξ, η) ∈ Ω∗r and |ξ| > 3λ2 one has T z =

P F kq P−1z ∈ Σn where k = K (ξ) and n = n (ξ).

Proof. Consider z0 = Pγ ∈Σ with γ = γ (ξ, η) ∈ Υr and assume without loss ofgenerality that ξ > 0 with cq (ξ) =

[a0; 1j , aj+1, . . .

]for some j ≥ 0, aj+1 6= 1

and aj+1 6= ±1 if j = 0 (the case of −1’s is analogous). Recall the notationin Definition 61, in particular the sequence wnn∈Z and the corresponding mapsAn ∈ Gq. It is clear, that wn gives a return if and only if Anγ ∈ Υr.

There are two cases to consider: Either z0 ∈ Σ−1 ∪Σ0 respectively z0 ∈ Σ−1 ∪Σ0 ∪ Σ3 in the case of odd q or z0 ∈ Σ2. In Figure 7 these different possibilitiesare displayed, PγA ∈ Σ−1, PγB ∈ Σ0,PγC ∈ Σ2 and PγD ∈ Σ3. It is clear, thatif z0 ∈ Σ2 then w0 = TSz0 ∈ L2 and the sequence of wn is essentially differentfrom the case z0 6∈ Σ2 when w0 = z0.

Case 1: If z0 6∈ Σ2 (see geodesics γA, γB and γD in Figure 7), then wn ∈TnL−1 for 1 ≤ n ≤ k − 1 and k = a0 − 1, a0 or a0 + 1 depending on whetherz0 ∈ Σ0,1 or Σ3 and whether the next intersection is on T a0L0 or T a0−1L0. Theneither wk ∈ T a0L0 or wk ∈ T a0−1L0 (see geodesics γE and γF in Figure 8). SinceAn = T−n for wn ∈ TnL−1 and, as we will show in Lemma 88 T−nγ /∈ Υ noneof the wn ∈ TnL−1 for 1 ≤ n ≤ k − 1 gives a return to Σ. There are now twopossibilities:

(i) If wk ∈ T a0L0, then Ak = ST−a0 and γk = Akγ = Fqγ. If j = 0 it is clearthat γ′ ∈ Υr and T z0 = P Fqγ ∈ Σ0. If j ≥ 1, by Lemma 76 applied to γ′

we get T z0 = P F jq γ′ = P F j+1

q γ ∈ Σn(ξ).(ii) If wk ∈ T a0−1L0, then we will show in Lemma 90 and 94 that none of the

arcs emanating from T a0−1ρ gives a return (cf. Figures 8 and 6) except forthe next return at T a0−1

(ST−1

)h+1L1. Furthermore it follows, that T z0 =

P F kq P−1z0 ∈ Σn with k = K (ξ) (here h or h + 1) and n = n (ξ).



FIGURE 7. Geodesics leaving the Poincare section (q=7)

FIGURE 8. Geodesics returning to the Poincare section (q = 8)

Case 2: If z0 ∈ Σ2 then 3λ2 < ξ < λ + 1 and γ must intersect TL1 below Tρ.

By the same arguments as in Case 1 we conclude that the first return is given bywq−1 ∈ T

(ST−1

)h+1L1 and T z0 = P F k

q P−1z0 ∈ Σn where k = K (ξ) andn = n (ξ) as in Case 1 (ii). In all cases we see, that the first return map T : Σ → Σis given by T = P F k

q P−1 or alternatively by T = F kq where k = 1, h or h+1

depending on ξ.



By combining Lemma 76 and 78 it is easy to see, that the first return map P isdetermined completely in terms of the coordinate ξ:

PROPOSITION 79. If z ∈ Σ with P−1z = (ξ, η) ∈ Ω∗r then T z = P Fqk

P−1z ∈ Σn where k = K (ξ) and n = n (ξ).

Having derived explicit expressions for the first return map, in a next step wewant to get explicit formulas for the first return time, i.e. the hyperbolic lengthbetween the successive returns to Σ.

4.4. The first return time.

LEMMA 80. Let γ = γ (ξ, η) ∈ Υr with ξ =[a0; (ε)

k−1 , ak, . . .]

(ε = sign(a0))and Pγ = z0 = (z0, θ0) and let z1 = T z0. For w0 ∈ γ the point corresponding toz1, i.e. w0 ∈ Gqz1, one has

d (z0, w0) = ln g (z0, γ)− ln g(z1, T γ

)+ 2 lnF (γ)

where F (γ) =∏k

j=1 |ξj | with k = K (ξ) as in Definition 77 and ξj = SF jq Sξ.

Proof. Set γj := F jq γ = Bjγ and wj := Bjw0. By Lemma 65 g (w0, γ) =

g(B−1

1 w1, B−11 γ1

)= g (w1, γ1) ξ−2

1 . Applying the same formula to g (wj , γj) for

j = 1, . . . , k we get ln g (w0, γ) = ln g(z1, T γ

)− 2 ln

∏kj=1 |ξj |. The statement

then follows by Proposition 79 and Lemma 66.

LEMMA 81. If γ = γ (ξ, η) is a reduced closed geodesic with cq (Sξ) =[a1, . . . , an

]

of minimal period n and ξj = SF jq Sξ, then the hyperbolic length of γ is given by

(3) l (γ) = 2n∑

j=1

ln |ξj | = − lnn∏

j=1

∣∣[aj+1, . . . , an, a1, . . . , aj

]∣∣2 .

Proof. Denote by (zj , θj) ∈ Σ the successive returns of γ to Σ and let wj−1 ∈ γbe the point on γ corresponding to zj . If γ is closed, the set zjj≥0 is finite withN + 1 elements for some N + 1 ≤ n, i.e. zN+1 = z0. It is clear that the length ofγ is given by adding up the lengths of all pieces between the successive returns toΣ and a repeated application of Lemma 80 gives us

l (γ) =N∑

j=0

d (zj , wj) =N∑

j=0

(ln g

(zj , T jγ

)− ln g

(wj , T jγ

))

=N∑

j=0

(ln g

(zj , T jγ

)− ln g

(zj+1, T j+1γ

)+ 2 lnF

(T jγ

))

= 2N∑

j=0

lnF(T jγ

)= 2 ln

n∏

i=1

∣∣SF iqSξ

∣∣ .



REMARK 82. Formula (3) can also be obtained by relating the length of γ(ξ, η)to the axis of the hyperbolic matrix fixing ξ and observing that this matrix must begiven by the map Fn

q acting on ξ.In the case of PSL2(Z) and the Gauss (regular) continued fractions formula (3)

is well-known.

We are now in a position to discuss the first return time. By Lemma 66 it is clear

that we need to calculate the function gj (ξ, η) = |wj−ξ|2vj

, where wj = uj + ivj =Zj (ξ, η) for all the intersection points in Corollary 70.

DEFINITION 83. Let B ∈ PSL2(R) be given by Bz = 2−λzλ−2z . Set δn (ξ, η) :=

η − TnBT−nξ,

Ξ+ :=

(ξ, η) ∈ Ω∗r

∣∣∣∣ δ〈x〉λ−1 (ξ, η) ≥ 0

, and

Ξ− :=

(ξ, η) ∈ Ω∗r

∣∣∣∣ δ〈x〉λ−1 (ξ, η) < 0

.

PROPOSITION 84. The first return time r for the geodesic γ (ξ, η) is given by thefunction

r (ξ, η) = ln gA (ξ, η)− ln gn(ξ)

(FK(ξ)

q γ (ξ, η))

+ 2 lnFγ (ξ, η) for ξ > 0,

respectively

r (ξ, η) = r (−ξ,−η) , for ξ < 0.

Thereby A = A(ξ, η) is given by

A =

−1, −R ≤ η < −λ2 , ξ > −B (−η) ,

0, −R ≤ η < −λ2 , ξ < −B (−η) , or− λ

2 ≤ η < −r, ξ ≥ B (η) ,

2, 34λ− 1

λ < η < −r, 3λ2 < ξ < B (η) < λ + 1,

3, λ− 1 < η < −r, λ + 1 < ξ < B (η) ,

K (ξ) and n (ξ) are defined as in Definition 77, whereas the functions gj (ξ, η) =g (zj , γ) for zj ∈ Lj are given as in Corollary 70 and F (γ) is given as in Lemma80.

Proof. Consider γ = γ (ξ, η) ∈ Υr with ξ > 0 and suppose that Pγ = z0 ∈ Σ andT z0 = z1 ∈ Σ with w ∈ γ corresponding to z1. Since geodesics are parametrizedby arc length the first return time is simply the hyperbolic length between z0 andw, i.e.

r (ξ, η) = d (z0, w) = ln g (z0, γ)− ln g (w, γ)

= ln g (z0, γ)− ln g(z1, Fqγ

)+ 2 lnF (γ)

by Lemma 80. If z0 ∈ Lj , we set g (z0, γ) = gj (ξ, η) as given in Corollary 70. ByCorollary 72 it is easy to verify, that the sets in the definition of A (ξ, η) correspondexactly to the cases z0 ∈ L−1, L0, L2 and L3 respectively, where the last set is



empty for even q. It is also easy to see, that B(

34λ− 1

λ

)= 3λ

2 and B (λ− 1) =λ + 1. The statement of the Proposition now follows from the explicit formula forF (γ) in Lemma 80 and the domains in Proposition 79 for which T = F k

q . Thatr (−ξ,−η) = r (ξ, η) follows from the invariance of the cross-section with respectto reflection in the imaginary axis.

5. CONSTRUCTION OF AN INVARIANT MEASURE

By Liouvilles theorem we know that the geodesic flow on T 1H preserves themeasure induced by the hyperbolic metric. This measure, the Liouville measure,is given by dm = y−2dxdydθ in the coordinates (x + iy, θ) ∈ H × S1 on T 1H.Using the coordinates (ξ, η, s) ∈ R3 given by Corollary 63 we obtain the Liouvillemeasure in these coordinates

dm =dxdydθ

y2=

∣∣∣∣∂ (x, y, θ)∂ (ξ, η, s)

∣∣∣∣2

r2 cos2 θdξdηds =

2dξdηds

(η − ξ)2.

The time discretization of the geodesic flow on T 1M in terms of the cross-sectionΣ and the first return map T : Σ → Σ thus preserves the measure dm′ = 2dξdη

(η−ξ)2

on Ω∗r . We prefer to work with the finite domain Ω∞r ⊆ Ω∞. Hence the measuredµ (u, v) = dm′ (ξ, η) given by

dµ =2dudv

(1− uv)2

on Ω∞r is invariant under T (u, v) = FqK(u) (u, v) = (f1 (u) , f2 (v)) where

f1 (u) = FK(u)q u = Au and f2 (v) = A−1v. Because dµ is equivalent to Lebesgue

measure, we deduce that dµ is in fact an FK(u)q invariant measure on Ωr. If

πx (x, y) = x it is clear that πx T (u, v) = f1 (u) = f1 πx (u, v) so f1 is afactor map of T . An invariant measure µ of f1 : Iq → Iq can be obtained by inte-grating dµ in the v-direction. We get different alternatives depending on q being 3,even or odd greater than 3.

5.1. q = 3. In this case FKq = Fq and if we set U1 =

[−12 , 0

]= −U−1 and

V1 = [r,R] then dµ = χU1dµ1 + χU−1dµ−1 where

dµ1 (u) =∫ R

r

2dudv

(1− uv)2=

[1

u (1− uv)

]v=R

v=r

du

=1

u (1− uR)− 1

u (1− ur)du =

1− ur − 1 + uR

u (1− uR) (1− ur)du

=1

(1− uR) (1− ur)du

and dµ−1 (u) = −dµ1 (−u) = 1(1+uR)(1+ur)du. Here r =

√5−32 and R = r + 1 =

√5−12 .



5.2. Even q ≥ 4. Here U1 =[−λ

2 ,− 23λ

], U2 =

[− 23λ , 0

], V1 = [0, R] and

V2 = [r,R]. Hence

dµ1

du(u) =

∫ R

0

2dv

(1− uv)2=

1u (1− uR)

− 1u

=R

1− uR,

dµ2

du(u) =

∫ R

r

2dv

(1− uv)2=

1u (1− uR)

− 1u (1− ur)

=R− r

(1− uR) (1− ur)=

λ

(1− uR) (1− ur),

dµ−j

du(u) =

dµj

du(−u)

and the invariant measure µ of FKq for even q is given by

dµ (u) =3∑

j=−3

χUj (u) dµj (u)

where χIj is the characteristic function for the interval Uj . This measure is piece-wise differentiable and finite. The finiteness is clear since uR and ur 6= 1 foru ∈ Iq. If

∫Iq

dµ (u) = c then 1cdµ is a probability measure on Iq.

5.3. Odd q ≥ 5. Let U1 =[−λ

2 , −23λ

], U2 =

[− 23λ ,− 1

2λ

], U3 =

[− 12λ , 0

], V1 =

[0, R] , V2 = [rκ−1, R] and V3 = [r,R]. Then

dµ1

du(u) =

∫ R

0

2dv

(1− uv)2=

1u (1− uR)

− 1u

=R

1− uR,

dµ2

du(u) =

∫ R

rκ−1

2dv

(1− uv)2=

1u (1− uR)

− 1u (1− urκ−1)

=R− rκ−1

(1− uR) (1− urκ−1),

dµ3

du(u) =

R− r

(1− uR) (1− ur)=

λ

(1− uR) (1− ur),

dµ−j

du(u) =

dµj

du(−u)

and the invariant measure µ of FKq for odd q is given by

dµ (u) =3∑

j=−3

χUj (u) dµj (u)

where χIj is the characteristic function for the interval Uj . This measure is piece-wise differentiable and finite. The finiteness is clear since uR, ur and urκ−1 6= 1for u ∈ Iq. If

∫Iq

dµ (u) = c then 1cdµ is a probability measure on Iq.

REMARK 85. The geodesic flow on finite surfaces of constant negative curvaturehas been shown by Ornstein and Weiss [39] to be Bernoulli with respect to Liouville



measure. Since our cross section is smooth also the Poincare map T is Bernoulli[38] with respect to the induced measure dµ. But the factor map of a Bernoullisystem is again Bernoulli [37]. Hence also the map f1 : Iq → Iq is Bernoulli withrespect to the measure µ.

REMARK 86. For another approach leading to an infinite invariant measure seee.g. Haas and Grochenig [14].

5.4. Invariant measure for Fq. It is easy to verify that dm (ξ, η) = 2dξdη

(ξ−η)2is

invariant under Mobius transformations, i.e. if A ∈ PSL2(R) then dm (Aξ, Aη) =dm (ξ, η). By considering the action of Fq on Ω∗, i.e.

F Sq (ξ, η) = S Fq S (ξ, η) =

(SFqSξ,

1nλ− η

)=

(ST−nξ, ST−nη

)

it is clear that dm is invariant under F Sq : Ω∗ → Ω∗ and letting u = Sξ and v = −η

it is easy to verify that dµ (u, v) = 2dudv(1−uv)2

is invariant under Fq : Ω → Ω. We canthus obtain corresponding invariant measure dµ for Fq by projecting on the firstvariable. Let dµ (u) = dµj (u) for u ∈ Ij . Then

dµj

du(u) = 2

∫ R

rj

dv

(1− uv)2

= 2[

1u (1− uv)

]R

rj

=2u

[1

1−Ru− 1

1− rju

]=

2 (R− rj)(1−Ru) (1− rju)

and the invariant measure µ of Fq is given by

dµ (u) =κ∑

j=−κ

χIj (u) dµj (u)

where χIj is the characteristic function for the interval Ij . This measure is piece-wise differentiable and finite. If c =

∫Iq

dµ then 1cdµ is a probability measure on

Iq. It turns out that c = 14C where C−1 = ln

(1+cos π

q

sin πq

)for even q and C−1 =

ln (1 + R) for odd q (see Lemma 3.2 and 3.4 in [8]).

6. LEMMAS ON CONTINUED FRACTION EXPANSIONS AND REDUCEDGEODESICS

This section contains a collection of rather technical lemmas necessary to showthat the first return map on Σ in Lemma 78 is given by powers of Fq.

LEMMA 87. If γ ∈ Υ with 〈γ+〉λ = a0 intersects T a0−1L0 then

γ+ =

[a0;

(1h−1, 2

)l, 1h, a(l+1)h+1, . . .

]for some 0 ≤ l ≤ ∞,

a(l+1)h+1 ≤ −1, if q is even and[a0; 1h, ah+1, . . .

], with ah+1 ≥ 2, if q is odd.



Proof. Let x = γ+ − a0λ. By convexity γ does not intersect T a0−1L0 more thanonce. Hence x ∈ (−λ

2 , 1− λ]. If q is odd, then cq (1) =

[1; 1h

]and cq

(−λ2

)=[

1h, 2, 1h]

according to Lemmas 6 and 5 hence[1h, 2, 1h

]< cq (x) <

[1h

]and

by the lexicographic ordering (see proof of Lemma 24) it is clear, that cq (x) =[1h, ah+1, . . .

]with ah+1 ≥ 2. If q is even then cq

(−λ2

)=

[1h

]and 1 − λ = r

with cq (r) =[1h−1, 2

]so that

[1h

]< cq (x) ≤ [

1h−1, 2]. By the lexicographic

ordering it is clear, that cq (x) =[(

1h−1, 2)l

, 1h, a(l+1)h+1, . . .]

for some l ≥ 0(l = ∞ is allowed) and a(l+1)h+1 ≤ −1 if l < ∞.

LEMMA 88. Let γ ∈ Υ with 〈γ+〉λ = a0. Then T−sign(a0)nγ /∈ Υ for n ≥ 1.

Proof. Without loss of generality assume a0 ≥ 1 and let γn = T−nγ for n ≥ 1.Then γn

+ = γ+ + nλ > γ+ and γn− = γ− + nλ. Since γ ∈ Υ γ− ≥ −R ⇒γn− ≥ nλ − R = −r + (n− 1) λ ≥ −r. Hence γn− /∈ [−R,−r) so γn /∈ Υ. Thecase of a0 ≤ −1 is analogous.

LEMMA 89. If γ ∈ Υ then γn = ST−nγ ∈ Υ if and only if n = a0 = 〈γ+〉λ.

Proof. By definition, if γn ∈ Υ then Sγn+ = γ+ − nλ ∈ I∞q and hence γ+ ∈(

nλ− λ2 , nλ + λ

2

)⇒ n = a0. It is also clear that γa0 = Fqγ ∈ Υ.

LEMMA 90. Suppose that q is even. If γ = γ (ξ, η) ∈ Υ with a0 = 〈γ+〉λ ≥ 2intersects T a0−1L0 then the first return map is given as T (ξ, η) = Fq

K(ξ) (ξ, η) ∈P−1

(Σn(ξ)

)where K (ξ) and n (ξ) are as in Definition 77.

Proof. Consider Figures 6 and 8 showing the arcs around the point ρ. The pictureis symmetric with respect to <z = λ

2 and invariant under translation, so it appliesin the present case. After passing through T a0−1L0 the geodesic γ will intersect asequence of translates of the arcs χj and ωj which are the reflections of χj and ωj

in <z = λ2 exactly as in Lemma 76, except that it now passes through every arc.

Note, even the argument why γ intersects ωj and not its extension applies. Let wn

and An be as in Definition 61 except that now w0 ∈ T a0−1L0.Then w2j ∈ T a0−1ωj = T a0−1

(ST−1

)jSL0, for 0 ≤ j ≤ h and also w2j+1 ∈

T a0−1χj =(ST−1

)j+1L1, for 0 ≤ j ≤ h with corresponding maps A2j+1 =

(TS)j+1 T 1−a0 and A2j = (ST )j ST 1−a0 . Set γk := Akγ and ξk := Akξ for0 ≤ k ≤ 2h + 1. There are three cases when the point wk can define a return to Σ:

a) if γ2j+1 ∈ Υr and z = Pγ2j+1 ∈ Σ1,b) if γ2j−1 /∈ Υr but γ′2j−1 = TSγ2j−1 ∈ Υr with ξ′2j−1 ∈

(3λ2 , λ + 1

)and

z = Pγ′2j−1 ∈ Σ2, orc) ifγ2j ∈ Υr and z = Pγ2j ∈ Σ0.

According to Lemma 87 we get T a0−1ξ ∈ (λ2 , 1

]and cq (ξ) =

[a0; 1h−1, ah, ah+1, . . .

]

with ah = 2 or ah = 1 and ah+1 ≤ −1. Also λ2 =

[(−1)h]

and 1 = R =[(−1)h ,−2, (−1)h−1]∗.



Hence ξ2j+1 ∈ (−φj+1,−rh−j ] ⊆ Iq ⇒ γ2j+1 /∈ Υ for 1 ≤ j ≤ h − 1(cf. Remark 36). Also note, that A2h+1 = (TS)h+1 T 1−a0 =

(ST−1

)hST−a0 ,

therefore γ2h+1 = A2h+1γ = ST−1Fqhγ and hence γ2h+1 /∈ Υ unless ah = 1 in

which case γ2h+1 = Fqh+1γ ∈ Υr and Pγ2h+1 ∈ Σ1.

For γ′2j+1 = TSγ2j+1 = γ2j+3 we conclude γ′2j+1 /∈ Υ for 0 ≤ j ≤ h− 2. Forah = 1 we find γ′2h−1 = γ2h+1 ∈ Υr, but ξ′2h−1 < 0 and hence we do not get apoint of Σ2. If ah = 2 on the other hand then γ′2h−1 /∈ Υr but γ′2h+1 = TSγ2h+1 =Fq

hγ ∈ Υr and ξ′2h+1 > 3λ2 so Pγ′2h+1 ∈ Σ2.

For γ2j = Sγ2j−1 we get ξ2j ∈ S (−φj ,−rh+1−j) = (−λ− φj+1,−λ− rh−j)and therefore −3λ

2 < ξ2j < −λ2 for 0 ≤ j ≤ h − 1. Since T 1−a0η < −R we

have η2j ∈ S (TS)j (−∞,−R) = (φh−j , rj) that is η2j < 0. Hence γ2j /∈ Υr

for 0 ≤ j ≤ h − 1. Note, that F hq ξ > 2

λ implies that ξ2h = T−1ST−1F hq ξ ∈

T−1ST−1(

2λ ,∞)

=(−λ,−λ + λ

λ2−2

)for q > 4 and ξ2h ∈ (−∞,−λ) for

q = 4. In any case ξ2h < 0 and since η2h < r it is clear, that γ2h /∈ Υ.We conclude that the first return is given by w2h+1 and T (ξ, η) = F h+1

q (ξ, η) ∈P−1

(Σ1

)if ah = 1 and T (ξ, η) = F h

q (ξ, η) ∈ P−1(Σ2

)if ah = 2. This can be

written in the form T = FK(ξ)q ∈ P−1

(Σn(ξ)

)withK (ξ) and n (ξ) as in Definition

77.

LEMMA 91. For q even consider ξ ∈ (−λ2 , 1− λ

)with cq (ξ) =

[0; 1h−1, ah, ah+1, . . .

].

Then ah = 1 if and only if ξ < −λ3

λ2+4= −λ + 4λ

λ2+4.

Proof. By Lemma 87 we know either ah = 1 and ah+1 ≤ −1 or ah = 2. It is clearthat the boundary point between these two cases is given by ξ0 =

[0; 1h, (−1)h]

=

(ST )h (λ2

). Using (2) one can show that ξ0 = (ST )h (

λ2

)= (λ2−2)λ

2+λ

−λ λ2−2

= −λ3

4+λ2 .

The following corollary is easy to verify by estimating the intersection of γ(−r,

(a0 − 1

2

)λ)

and T a0−1L0. It implies that the case T = F hq does not occur for ξλ ≥ 3.

COROLLARY 92. Let q be even and suppose that γ = γ (ξ, η) ∈ Υ with a0 =ξλ ≥ 3. If γ intersects T a0−1L0 then cq (ξ) =

[a0; 1h, ah+1, . . .

]with ah+1 ≤

−1.

LEMMA 93. Let q be odd and suppose that γ ∈ Υ. Let l be the geodesic arc[ρ + λ, 1 + λ], i.e. the continuation of L3. Then γ does not intersect ±l outwards(i.e. in the direction from 0 to ±∞).

Proof. Take γ = γ (ξ, η) ∈ Υ and assume ξ > 0. Suppose that γ intersects l in theoutwards direction. Since γ can not intersect the geodesic TSL−1 =

[λ, λ + 2

λ

]more than once we have ξ ∈ (

λ + 1, λ + 2λ

)and because γ ∈ Υ we have −R ≤

η < −r. If w (ξ, η) is the intersection between γ and the line TL1 = 3λ2 + iR+

then =w (ξ, η) > =w (ξ,−r) ≥ =w (λ + 1,−r). To show that =w > =Tρ =



=ρ = sin πq it is enough to bound =w (λ + 1,−r) from below. By Lemma 69 we

have

=w (λ + 1,−r)2 =(

λ + 1− 3λ

2

)(3λ

2− r

)=

(1− λ

2

) (5λ

2−R

)

=(

1− λ

2

)(1 +

λ

2+ (2λ−R− 1)

)

= sin2 π

q+

(1− λ

2

)(2λ−R− 1) > sin2 π

q

since 2λ > R+1 and 1− λ2 > 0. Hence=w (ξ, η) > sin π

q and γ does not intersectl in the direction from 0 to ∞. An analogous argument for ξ < 0 concludes theLemma.

LEMMA 94. For q odd, let γ = γ (ξ, η) ∈ Υr be reduced with a0 = ξλ ≥ 2. Ifγ intersects T a0−1L0 then T (ξ, η) = F h+1

q (ξ, η) ∈ P−1(Σ3

).

Proof. Consider once more Figure 6 showing the arcs around ρ. Analogous tothe proof of Lemma 90 we have w2j ∈ T a0−1ωj = T a0−1

(ST−1

)jSL0, for

0 ≤ j ≤ h + 1 and w2j+1 ∈ T a0−1χj = T a0−1(ST−1

)j+1L1, for 0 ≤ j ≤ h

with the corresponding maps A2j+1 = (TS)j+1 T 1−a0 and A2j = (ST )j ST 1−a0 .Set γj := Ajγ and ξj := Ajξ. There are now four possibilities to produce a returnto Σ:

a) if γ2j+1 ∈ Υr and z = Pγ2j+1 ∈ Σ1,b) if γ2j−1 /∈ Υ but TSγ2j+1 ∈ Υr and z = PTSγ2j+1 ∈ Σ2,c) if γ2j ∈ Υr and z = Pγ2j ∈ Σ0,d) if γ2j /∈ Υ but T±1γ2j ∈ Υr and z = Pγ2j ∈ Σ±3.

We will see that most of these cases do not give a return. Since T 1−a0ξ ∈ (λ2 , 1

)Lemma 87 shows that cq (ξ) =

[a0; 1h, ah+1, . . .

]with ah+1 ≥ 2. Suppose also,

that c∗q (η) =[0; b1, b2, . . .

]∗. For the following arguments it is important to re-member that the action of TS on ∂H ∼= R∗ ∼= S1 is monotone as a rotation aroundρ.

Since γ2j+1 = (TS)j+1 T 1−a0γ we have ξ2j+1 ∈ (TS)j+1 (λ2 , 1

)= (−φ2j+2,−φ2j+1) ⊆

Iq and hence γ2j+1 /∈ Υ for 0 ≤ j ≤ h−1. Furthermore ξ2h+1 ∈ TS (−φ2h,−φ2h+1) =(−1, 0) and therefore |ξ2h+1| < 2

λ so that also γ2h+1 /∈ Υ.If γ′2j+1 = TSγ2j+1 = γ2j+3 then γ′2j+1 /∈ Υ for 1 ≤ j ≤ h − 1 and since

we have A2h+3 = (TS)h+2 T 1−a0 =(ST−1

)hST−a0 it is clear that γ2h+3 =

A2h+3γ = F h+1q γ ∈ Υr and we have a return at w2h+1 with z1 ∈ L2!

Since γ2j = (ST )j ST 1−a0γ obviously ξ2j ∈ S (TS)j (λ2 , 1

)= (S (−φ2j) , S (−φ2j−1)) =

−λ− (φ2j+2, φ2j+1) and hence −3λ2 < ξ2j < −λ

2 . But T 1−a0η < −r − λ = −R

and hence η2j ∈ (ST )j S (−∞,−R) =(φ2(h−j)+1, r2j+1

)for 0 ≤ j ≤ h (cf. Re-

mark 37) respectively η2h+2 ∈ ST (φ1, r2h+1) = ST (1− λ, r) = S (1, R) =(−1,−1/R) and therefore η2j < 0 and γ2j /∈ Υr for 0 ≤ j ≤ h + 1.



Finally, since T−1η2j < r2j+1−λ < −R and Tξ2j ∈ Iq it is clear that T±1γ2j /∈Υ for 0 ≤ j ≤ h + 1.

Acknowledgments. The authors appreciate also the many helpful discussions withTobias Muhlenbruch on different aspects of this paper.

REFERENCES

[1] R. L. Adler and L. Flatto. Cross section maps for geodesic flows. I. The modular surface. InErgodic theory and dynamical systems, II (College Park, Md., 1979/1980), volume 21 of Progr.Math., pages 103–161. Birkhauser Boston, Mass., 1982.

[2] R. L. Adler and L. Flatto. The backward continued fraction map and geodesic flow. ErgodicTheory Dynam. Systems, 4(4):487–492, 1984.

[3] R. L. Adler and L. Flatto. Cross section map for the geodesic flow on the modular surface.In Conference in modern analysis and probability (New Haven, Conn., 1982), volume 26 ofContemp. Math., pages 9–24. Amer. Math. Soc., Providence, RI, 1984.

[4] R. L. Adler and L. Flatto. Geodesic flows, interval maps, and symbolic dynamics. Bull. Amer.Math. Soc. (N.S.), 25(2):229–334, 1991.

[5] E. Artin. Ein mechanisches System mit quasiergodischen Bahnen. Hamb. Math. Abh., 3:170–177, 1924.

[6] D. Birkhoff. Quelques theoremes sur le mouvement des systemes dynamiques. Bull. Soc. Math.France, 40:305–323, 1912.

[7] R. Bowen and C. Series. Markov maps associated with Fuchsian groups. Inst. Hautes EtudesSci. Publ. Math., (50):153–170, 1979.

[8] R. M. Burton, C. Kraaikamp, and T. A. Schmidt. Natural extensions for the Rosen fractions.Trans. Amer. Math. Soc., 352(3):1277–1298, 2000.

[9] Paula Cohen and J. Wolfart. Modular embeddings for some nonarithmetic Fuchsian groups.Acta Arith., 56(2):93–110, 1990.

[10] K. Dajani, C. Kraaikamp, and W. Steiner. Metrical theory for α-Rosen fractions.Arxiv:math.NT/0702516v1, Feb 2007.

[11] M. Einsiedler and T. Ward. Ergodic theory: with a view towards number theory. In preparation.[12] D. Fried. Symbolic dynamics for triangle groups. Invent. Math., 125(3):487–521, 1996.[13] D. Fried. Reduction theory over quadratic imaginary fields. J. Number Theory, 110(1):44–74,

2005.[14] K. Grochenig and A. Haas. Backward continued fractions, Hecke groups and invariant measures

for transformations of the interval. Ergodic Theory Dynam. Systems, 16(6):1241–1274, 1996.[15] J. Hadamard. Les surfaces a courbures opposees et leurs lignes geodesiques. J. Math. Pures et

Appl., 5(4):27–73, 1898.[16] G. A. Hedlund. On the measure of the non-special geodesics on a surface constant negative

curvature. Proc. Natl. Acad. Sci. USA, 19:345–348, 1933.[17] G. A. Hedlund. On the metrical transitivity of the geodesics on closed surfaces of constant

negative curvature. Ann. of Math. (2), 35(4):787–808, 1934.[18] G. A. Hedlund. Two-dimensional manifolds and transitivity. Ann. of Math. (2), 37(3):534–542,

1936.[19] G. A. Hedlund. Fuchsian groups and mixtures. Ann. of Math. (2), 40(2):370–383, 1939.[20] E. Hopf. Fuchsian groups and ergodic theory. Trans. Amer. Math. Soc., 39(2):299–314, 1936.[21] E. Hopf. Statistik der geodatischen Linien in Mannigfaltigkeiten negativer Krummung. Ber.

Verh. Sachs. Akad. Wiss. Leipzig, 91:261–304, 1939.[22] A. Hurwitz. uber eine besondere Art der Kettenbruch-Entwickelung reeller Grossen. Acta

Math., 12:367–405, 1889.[23] S. Katok. Fuchsian Groups. The University of Chicago Press, 1992.



[24] S. Katok and I. Ugarcovici. Arithmetic coding of geodesics on the modular surface via continuedfractions. In European women in mathematics—Marseille 2003, volume 135 of CWI Tract,pages 59–77. Centrum Wisk. Inform., Amsterdam, 2005.

[25] S. Katok and I. Ugarcovici. Symbolic dynamics for the modular surface and beyond. Bull. Amer.Math. Soc. (N.S.), 44(1):87–132 (electronic), 2007.

[26] P. Koebe. Riemannsche mannigfaltigkeiten und nichteuklidische raumformen. iv: Verlaufgeodatischer linien. Sitzungsberichte Akad. Berlin, 1929:414–457, 1929.

[27] J. Lehner. Discontinuous groups and automorphic functions. Mathematical Surveys, No. VIII.American Mathematical Society, Providence, R.I., 1964.

[28] D. Lind and B. Marcus. An introduction to symbolic dynamics and coding. Cambridge Univer-sity Press, Cambridge, 1995.

[29] F. Lobell. Uber die geodatischen Linien der Clifford-Kleinschen Flachen. Math. Z., 30(1):572–607, 1929.

[30] G. A. Morse, M. Hedlund. Symbolic Dynamics. Amer. J. Math., 60(4):815–866, 1938.[31] H. M. Morse. A One-to-One Representation of Geodesics on a Surface of Negative Curvature.

Amer. J. Math., 43(1):33–51, 1921.[32] H. M. Morse. Recurrent geodesics on a surface of negative curvature. Trans. Amer. Math. Soc.,

22(1):84–100, 1921.[33] P. J. Myrberg. Ein Approximationssatz fur die Fuchsschen Gruppen. Acta Math., 57(1):389–

409, 1931.[34] H. Nakada. Metrical theory for a class of continued fraction transformations and their natural

extensions. Tokyo J. Math., 4(2):399–426, 1981.[35] H. Nakada. Continued fractions, geodesic flows and Ford circles. In Algorithms, fractals, and

dynamics (Okayama/Kyoto, 1992), pages 179–191. Plenum, New York, 1995.[36] J. Nielsen. Om geodætiske linier i lukkede mangfoldigheder med konstant negativ krumning.

Mat. Tidsskrift B, 1925:37–44, 1925.[37] D. Ornstein. Ornstein theory. Scholarpedia, 2008. 3(3):3957.[38] D. Ornstein and B. Weiss. On the Bernoulli nature of systems with some hyperbolic structure.

Ergodic Theory Dynam. Systems, 18(2):441–456, 1998.[39] D. S. Ornstein and B. Weiss. Geodesic flows are Bernoullian. Israel J. Math., 14:184–198, 1973.[40] H. Poincare. Memoire sur les courbes definies par une equation differentielle. J. Math. Pures

Appl., 3(8):251–296, 1882. Ch. V-IX.[41] J. G. Ratcliffe. Foundations of Hyperbolic Manifolds. Springer-Verlag, 1994.[42] D. Rosen. A class of continued fractions associated with certain properly discontinuous groups.

Duke Math. J., 21:549–563, 1954.[43] D. Rosen and T. A. Schmidt. Hecke groups and continued fractions. Bull. Austral. Math. Soc.,

46(3):459–474, 1992.[44] D. Rosen and C. Towse. Continued fraction representations of units associated with certain

Hecke groups. Arch. Math. (Basel), 77(4):294–302, 2001.[45] T. A. Schmidt. Remarks on the Rosen λ-continued fractions. In Number theory with an emphasis

on the Markoff spectrum (Provo, UT, 1991), volume 147 of Lecture Notes in Pure and Appl.Math., pages 227–238. Dekker, New York, 1993.

[46] P. Schmutz Schaller and J. Wolfart. Semi-arithmetic Fuchsian groups and modular embeddings.J. London Math. Soc. (2), 61(1):13–24, 2000.

[47] C. Series. Symbolic dynamics for geodesic flows. Acta Math., 146(1-2):103–128, 1981.[48] C. Series. The modular surface and continued fractions. J. London Math. Soc. (2), 31(1):69–80,

1985.



DIETER MAYER <[email protected]>: Institut fur Theor. Physik,TU Clausthal,Abt. Statistische Physik und Nichtlineare Dynamik, Arnold Sommerfeld Straße 6, 38678 Clausthal-Zellerfeld, Germany

FREDRIK STROMBERG <[email protected]>: Fachbereich Math-ematik, AG AGF, TU Darmstadt, Schloßgartenstraße 7, 64289 Darmstadt, Germany


€¦ · journal of modern dynamics web site: volume 3, no. 3, 2009, 1–46 symbolic dynamics for...

Documents