If, in addition, is connected, then it is a fundamental domain for Γ in H. The Fuchsian groupF Γ is called geometrically finite if there exists a convex fundamental region for Γ in H with finitely many sides. In this article we will use isometric fundamental regions (Ford fundamental regions), whose construction and existence we recall in the following. Let Γ := g Γ g. = ∞ { ∈ | ∞ ∞} a b denote the stabilizer subgroup of in Γ. Let g = c d ΓrΓ , which implies that c = 0. Let denote the Euclidean∞ norm on C. The∈isometric∞ sphere of g is defined6 as | · |
I(g) := z H g0(z) =1 = z H cz + d =1 . { ∈ | | | } { ∈ | | | } The exterior of I(g) is the set ext I(g) := z H cz + d > 1 , { ∈ | | | } and int I(g) := z H cz + d < 1 { ∈ | | | } is its interior. If the representative of g is chosen such that c> 0, then the isometric d 1 d 1 sphere I(g) is the complete geodesic segment with endpoints c c and c + c . If z = x + iy is an element of I(g), then the geodesic segment− (z−, ) is− contained 0 0 0 0 ∞ in ext I(g), and the geodesic segment (x0,z0) belongs to int I(g). Moreover, H = ext I(g) I(g) int I(g) ∪ ∪ is a partition of H into convex subsets such that ∂ ext I(g)= I(g)= ∂ int I(g). Let := := ext I(g) K KΓ g Γ Γ∞ ∈\\ denote the common part of the exteriors of all isometric spheres of Γ. We call a point z ∂gH a cuspidal point for Γ if Γ contains a parabolic element that stabilizes z. In∈ other words, z is called cuspidal if it is a representative of a cusp of Γ or, equivalently, of Y =Γ H. If is cuspidal for Γ, then \ ∞ 1 λ Γ = ∞ 0 1 for some λ> 0. For each r R, the set ∈ (r) := ,Γ(r) := (r, r + λ)+ iR>0 F∞ F∞ is a fundamental domain for Γ in H. The following proposition states the existence of isometric fundamental domains.∞ This proposition is well-known. For proofs in various generalities we refer to, e.g., [17, 22, 40, 35, 34]. Proposition 2.1. Let Γ be a geometrically finite Fuchsian group that has as cuspidal point. Then, for any r R, the set ∞ ∈ (r) := (r) F F∞ ∩ K is a convex fundamental domain for Γ in H with finitely many sides. SYMBOLIC DYNAMICS 2179
2.3. Symbolic dynamics. As before, let Γ be a Fuchsian group and set Y =Γ H. \ Let CS (“cross section”) be a subset of SY . Suppose that γ is a geodesic on Y . If
γ0(t) CS, then we say that γ intersects CS in t. Further, γ is said to intersect CS ∈ infinitelyc often in the future if we find a sequence (tn)n N withb tn as n ∈ → ∞ → ∞ andb γ0(tcn) CS for all n N.b Analogously,cγ is said to intersectb CS infinitely oftenc ∈ ∈ in the past if we find a sequence (tn)n N with tn as n and γ0(tn) CS ∈ → −∞ → ∞ ∈ for allb n N.c Let µ be a measure on the spaceb of geodesics on cY . The set CS is ∈ called a cross section (w.r.t. µ) for the geodesic flow Φ if b c (C1) µ-almost every geodesic γ on Y intersects CS infinitely often in the pastc and in the future, b (C2) each intersection of γ andb CS is discrete inc time: if γ0(t) CS, then there is ∈ ε> 0 such that γ0((t ε,t + ε)) CS = γ0(t) . − ∩ { } c c1 We call a subset U of Yba totally geodesic suborbifold ofbY if π− (U) is a totally geodesic submanifoldb of H. Let pr: SY c Y denoteb the canonical projection on → base points. If pr(CS)b is a totally geodesic suborbifold of Y and CS doesb not contain elements tangent to pr(CS), then CS automatically satisfies (C2). Suppose that CSc is a cross section for Φ. If, in addition, CS satisfiesc the property that each geodesic intersectingc CSc at all intersects it infinitely often in the past and in the future, thenc CS will be called a strongb cross section,c otherwise a weak cross section. Clearly, every weak crossc section contains a strong cross section. The first return mapc of Φ w.r.t. the strong cross section CS is the map
R: CS CS, v γv0(t ), b → 7→ 0 c where v SH with π(v)= v, π(γv)= γv, and ∈ c c b b t0 := min t> 0 γ0 (t) CS b b v ∈ n o is the first return time ofv ˆ or γv. This definition requires that t0 = t0(v) exists b c for each v CS, which will indeed be the case in our situation. For a weak cross ∈ section CS, the first return mapb can only be defined on a subset of CS. Inb general, this subsetb isc larger than the maximal strong cross section contained in CS. Supposec that CS is a strong cross section and let Σ be an at mostc countable c set. Decompose CS into a disjoint union α Σ CSα. To each v CS we assign the ∈ Z ∈ (two-sided infinite)ccoding sequence (an)n Z Σ defined by S∈ ∈ c cn b c an := α if and only if R (v) CSα. ∈ Note that R is invertible and let Λ be the set of all sequences that arise in this way. Then Λ is invariant under the left shift b c Z Z σ : Σ Σ , σ (an)n Z := ak+1. → ∈ k Suppose that the map CS Λ is also injective, which it will be in our case. Then we → have the inverse map Cod: Λ CS which maps a coding sequence to the element → in CS it was assigned to.c Obviously, the diagram
c R CS / CS c O O Codc cCod σ Λ / Λ 2180 ANKE D. POHL commutes. The pair (Λ, σ) is called a symbolic dynamics for Φ with alphabet Σ. If CS is only a weak cross section and hence R is only partially defined, then Λ also contains one- or two-sided finite coding sequences. b cLet CS0 be a set of representatives for the cross section CS, that is, CS0 is a subset of SH such that π 0 is a bijection CS0 CS. Relative to CS0, we define |CS → the map τ : CS ∂gH ∂gH by c → × c τ(v) := (γv( ),γv( )) c ∞ −∞ 1 where v := (π 0 )− (v). For some cross sections CS it is possible to choose CS0 |CS in such a way that τ is a bijectionb between CS and some subset DS of R R. × In this case the dynamicalb system (CS, R) is conjugatec to (DS, F ) by τ, where 1 F := τ R τ − is the induced selfmap on DSc (partially defined iffCS is only a ◦ ◦ weak cross section). Moreover, to constructc a symbolic dynamicsf efor Φ, one can starte with a decomposition of DS into pairwisef disjoint subsets Dα, α cΣ. Finally, let (Λ, σ) be a symbolic dynamics with alphabet Σ. Suppose∈b that we have a map i:Λ DS for somef DS R such that i (an)n Z e depends only on → ⊆ ∈ (an)n N , a (partial) selfmap F : DS DS, and a decomposition of DS into a ∈ 0 → disjoint union α Σ Dα such that ∈ S F i((an)n Z) Dα a1 = α ∈ ∈ ⇔ for all (an)n Z Λ. Then F , more precisely the triple F, i, (Dα)α Σ , is called a generating∈ function∈ for the future part of (Λ, σ). If such a generating∈ function exists, then the future part of a coding sequence is independent of the past part. Remark 2.2. We stated a rather general definition of cross sections. In particular, we did not require any regularity of the set CS nor restricted the admissible measures µ. Any additional requirements on CS or µ should be due to applications. In the following we restrict the measures toc an admissible set and construct subsets CS of SY which are simultaneous crossc sections for all admissible measures (see Theorem 8.15). For transfer operator approaches, any regularity of CS only plays an implicitc role. For these applications, the most important properties of a cross section are that all periodic geodesics intersect it and that it is conjugatec to a discrete dynamical system (DS, F ) with a (in some weak sense) piecewise smooth map F . In all hitherto existing examples of transfer operator approaches, the geodesics which tend to a cusp in thef futuree or past direction can be neglected. For this reason,e all measures are admissible which assign zero mass to this set of geodesics. We will see that the base sets pr(CS) of the constructed cross sections CS are always a finite family of complete geodesics and hence smooth. The weak cross sections are smooth subsets of SY . In contrast,c the strong cross sections are typicallyc fractal.
3. A sketch of the construction. In this section we provide a sketch of the construction of cross sections and symbolic dynamics, illustrated by examples. For further examples we refer to [21, 32, 38, 36]. As we will see already in this sketch, the cuspidal point plays a distinguished role. Throughout let Γ be a geometrically finite Fuchsian∞ group with as cuspidal point. ∞ Additional condition on Γ. To state the additionally required condition on Γ we need a few notions. The height of a point z H is defined to be ∈ ht(z) := Im z. SYMBOLIC DYNAMICS 2181
Let g = a b ΓrΓ with c chosen positive, and consider its isometric sphere c d ∈ ∞ d 1 I(g)= z H z + = . ∈ c c
Its point of maximal height
d 1 s = + i − c c is called the summit of I(g). Recall the set = ext I(g), K g ΓrΓ∞ ∈\ which is the subset of H contained in the exterior of all isometric spheres of Γ. We call an isometric sphere I of Γ relevant if I contributes nontrivially to the boundary of , that is, if I ∂ contains a submanifold of H of codimension 1. If the isometricK sphere I is∩ relevant,K then I ∂ is called its relevant part. An endpoint of the relevant part of a relevant isometric∩ K sphere is called a vertex of . From now on we impose the following condition on Γ: K For each relevant isometric sphere, its summit is contained in ∂ but not a (A) vertex of . K K π Example 3.1. (i) For n N, n 3, let λn := 2cos . The Hecke triangle group ∈ ≥ n Gn is the subgroup of PSL2(R) which is generated by 0 1 1 λ S := and T := n . 1− 0 n 0 1 The relevant isometric spheres are I(S) and its T k-translates, k Z. n ∈
K
3 λn λn 3 λn λn 0 λn λn − 2 − − 2 2 2
Figure 1. The set for G . K 5 (ii) Let a b PΓ (5) := PSL (Z) c 0 mod5 . 0 c d ∈ 2 ≡
The isometric spheres are the sets
d 1 Ic,d = z H z + = ∈ 5c 5c
2182 ANKE D. POHL
where c N and d Z. The set is given by (see Figure 2) ∈ ∈ K = z H z + d > 1 . K ∈ 5 5 d Z \∈
K
1 0 1 2 3 4 1 6 − 5 5 5 5 5 5
Figure 2. The set for PΓ (5). K 0
0 1 1 4 (iii) Let S := 1− 0 and T := [ 0 1 ], and denote by Γ the subgroup of PSL2(R) which is generated by S and T . The set is indicated in Figure 3. K
K
1 0 1 2 3 4 5 −
Figure 3. The set for the group Γ from Example 3.1(iii). K
The first cross section. The (first) cross section CS for the geodesic flow on Γ H will be defined with the help of a very specific tesselation of . For that we note\ K that there are two kinds of vertices of . Let v be ac vertex of . If v H, then we call v an inner vertex, otherwise v is calledK an infinite vertex.K Now we∈ construct a family of geodesic segments in the following way: Whenever v is an infinite vertex of , thenG the (complete) geodesic segment (v, ) belongs to this family. Moreover, wheneverK s is the summit of a relevant isometric∞ sphere, then the geodesic segment (s, ) belongs to this family. Let CS denote the set of unit tangent vectors on H which∞ are based on the geodesic segments in this family but which are not tangent to any of these geodesic segments.f In other words, CS consists of the unit tangent vectors whose base points are on any of these geodesic segments and which point to the right or the left, but not straight up or down.f Then CS := π(CS) is a weak cross section for the geodesic flow. By eliminating a certain set of vectors c f from it, we will also construct a strong cross section. SYMBOLIC DYNAMICS 2183
Construction of a set of representatives. The family of geodesic segments induces a tesselation of into hyperbolic triangles, quadrilateralG and strips. We call these tesselation elementsK precells in H. The union of certain finite subfamilies of these precells in H form isometric fundamental regions or even, if the union is connected, isometric fundamental domains. This relation is useful for the following two reasons. Pick such a subfamily of precells in H and let CSpre0 denote the set of unit tangent which are based on the vertical sides of these precells and point into them. Then the link to isometric fundamental regions yields that CSpre0 is a set of representatives for CS. This set, however, is not very useful for coding purposes. For this reason, we use another property of isometric fundamental regions to modify the set of representatives.c Isometric fundamental regions allow to define cycles of their sides as in the Poincar´eFundamental Polyhedron Theorem. The acting elements in these cycles are just the generating elements of the relevant isometric spheres which contribute to the boundary of the fundamental region. Using these cycles we glue translates of precells in H to ideal polyhedrons in H, that is, polyhedrons all of which vertices are in ∂gH. Even more, all vertices are in Γ. . At this point, Condition (A) is essential. We call these polyhedron cells in H.∞ All these cells in H have two vertical sides, and all these non-vertical sides are Γ-translates of vertical sides of some cells in H. These elements in Γ can be determined by an algorithm from the generators of the relevant isometric spheres. Lifting this construction to the unit tangent bundle SH and using this to re- distribute CSpre0 allows us to find a set of representatives CS0 such that CS0 is the disjoint union
CS0 = CSα0 α A [∈ for some finite index set A such that for each α A, the subset CSα0 is the set of unit tangent vectors based on a vertical side (a complete∈ geodesic segment) of some cell in H and pointing into this cell. The definition of precells in H and the construction of cells in H from precells is based on ideas in [44]. Our construction differs from Vulakh’s construction in three important aspects: We define three kinds of precells in H of which only one are precells in the sense of Vulakh. Finally, contrary to Vulakh, we extend the considerations to precells and cells in the unit tangent bundle. Example 3.2. The set
λn n := z H z > 1, Re z < F ∈ | | | | 2 is an isometric fundamental domain for the Hecke triangle group Gn (see Figure 4).
Hecke triangle groups have up to equivalence under (Gn) = Tn only one precell in H. It is indicated in Figure 5. ∞ h i Figure 6 provides an idea of the lifting and redistribution procedure. Discrete dynamical system on the boundary and associated transfer oper- ator families. The relation between isometric fundamental regions and the specific way of lifting to SH yields the following property of CS0. Whenever L is the side of some Γ-translate of some cell in H and L0 is the set of all unit tangent vectors based on that side pointing into this Γ-translate or all vectors pointing out of this Γ-translate, then there is a unique pair (α, g) A Γ such that L0 = g. CS0 . Both, ∈ × α 2184 ANKE D. POHL
n F
i
λn λn − 2 2
Figure 4. An isometric fundamental domain for G in H. F5 5
A
i λn + i
λ 0 n λn 2
Figure 5. The precell of G . A 5
Figure 6. Lifting to SH and defining CS0 for G5.
α and g can be determined algorithmically. Further, Γ. CS0 is just the union of all such L0. These properties allow to read off the induced discrete dynamical system F on the boundary of H. It is given by finitely many local diffeomorphisms of the form 1 (Iα Jα ) g− .(Iα Jα ) α g.(Iα Jα ) (Iα Jα ) α , 1 × 1 ∩ 2 × 2 ×{ 1} → 1 × 1 ∩ 2 × 2 ×{ 2} (x, y, α ) (g.x, g.y, α ), (3) 1 7→ 2 SYMBOLIC DYNAMICS 2185 where α , α A, Iα , Iα ,Jα ,Jα are intervals, and g Γ depends on α , α . 1 2 ∈ 1 2 1 2 ∈ 1 2 The associated transfer operator F,β with parameter β C is then defined on spaces of functions f on the domainL of F and given by ∈ f(u) F,βf(w)= β . L − F 0(u) u F 1(w) ∈ X | | A more explicit formula is provided in Section 10.
Example 3.3. Figure 7 shows the translates of CS0 for the Hecke triangle group G5. Here, we set λ 1 U := T S = n . n n 1− 0 For general Hecke triangle groups Gn, the picture is similar. If one restricts here to
1 T5− .CS0 S.CS0 CS0 U5.CS0 T5.CS0
1 4 1 2 4 2 T5− U5 .CS0 T5− U5 .CS0 U5 .CS0 U5 .CS0 1 3 3 T5− U5 .CS0 U5 .CS0 4 2 U5 S.CS0 U5 S.CS0 3 U5 S.CS0 1 4 1 3 1 2 3 2 T − U . = λ T − U . T − U . 0 U . U . U5. = λ5 5 5 ∞ − 5 5 5 ∞ 5 5 ∞ 5 ∞ 5 ∞ ∞
Figure 7. The shaded parts are translates of CS0 (in the unit tangent bundle) as indicated. the forward-direction of the geodesic flow, then the discrete dynamical system for Gn is F : R> rGn. R> rGn. , 0 ∞ → 0 ∞ given by the bijections 1 (gk.0,gk. )rGn. R> rGn. , x g− .x ∞ ∞ → 0 ∞ 7→ k for k =1,...,n 1 with − k gk := Un S.
The associated transfer operator F,β with parameter β C is then defined on L ∈ functions f : R> rGn. C and given by 0 ∞ → n 1 − β F,βf(x)= g0 (x) f(gk.x). L k k=1 X For the modular group PSL2(Z)= G3 the transfer operator becomes
2β x F,βf(x)= f(x +1)+(x + 1)− f . L x +1 The 1-eigenfunctions of this transfer operator are characterized by the functional equation (2). In [32], the relation between eigenfunctions of this transfer operator and period functions, as well as the relation to Mayer’s transfer operator and the factorization (1) are discussed in detail. 2186 ANKE D. POHL
The second cross section and the reduced discrete dynamical system. For some lattices Γ it may happen that in some of the local diffeomorphisms of the form (3) we have g = id. To avoid that and at the same time to be able to eliminate the bits α and α from the domain and range of F , we will shrink the cross { 1} { 2} section CS and the set of representatives CS0 in an algorithmic way to deduce a new cross section CSred with a set of representatives CSred0 . In terms of notions to be introducedc later, we will eliminate all vectors from CS with id in the label. A group that does not satisfy (A). The previous examples as well as those c from [21, 38, 36] show that there are several geometrically finite Fuchsian groups with as cuspidal point and which satisfy the condition (A). We now provide an example∞ of a geometrically finite Fuchsian group with as cuspidal point, which does not satisfy (A). ∞ Let Γ be the subgroup of PSL2(R) which is generated by the matrices 1 17 18 5 3 1 t := 11 , g := and g := . 0 1 1 11 −3 2 10 −3 − − Set 4 3 g := g g = , 3 1 2 3 −2 − let := 2 , 19 + iR+ and (see Figure 8) F∞ 11 11 1 1 := extI(g1) ext I(g1− ) ext I(g2) ext I(g1g2) ext I (g1g2)− . F F∞ ∩ ∩ ∩ ∩ ∩
s5 F s6
s1 s8
s3 s4 s2 s7 v2v3v4 v6
v1 v5 v7
Figure 8. The fundamental domain F We indicated the points 3 i 91 √24 v0 := v3 := 10 + 10 v6 := 55 + i 55 ∞2 19 √24 19 v1 := 11 v4 := 55 + i 55 v7 := 11 14 √24 v2 := 55 + i 55 v5 :=1 and the geodesic segments
s1 := [v0, v1] s4 := [v3, v4] s7 := [v6, v7] s2 := [v1, v2] s5 := [v4, v5] s8 := [v7, v0]. s3 := [v2, v3] s6 := [v5, v6] SYMBOLIC DYNAMICS 2187
Proposition 3.4. The set is a fundamental domain for Γ in H. F Proof. The sides of are s ,s ,s s ,s ,s ,s and s . Further we have the F 1 2 3 ∪ 4 5 6 7 8 side-pairings ts1 = s8, g1s2 = s7, g2s3 = s4 and g3s5 = s6, where tv0 = v0, tv1 = v7, g1v1 = v7, g1v2 = v6, g2v2 = v4, g2v3 = v3, g3v4 = v6 and g3v5 = v5. Now Poincar´e’s Fundamental Polyhedron Theorem (see e.g. [25]) yields that is F a fundamental domain for the group generated by t,g1,g2 and g3. This group is exactly Γ.
Proposition 3.5. Γ does not satisfy (A). Proof. Proposition 3.4 states that is a fundamental domain for Γ in H. Its shape and side-pairings yield that it is anF isometric fundamental domain. Therefore, the isometric sphere I(g1) is relevant and s1 is its relevant part. The summit of I(g1) is s := 3+i . One easily calculates that s int I(g ). Thus, Γ does not satisfy (A). 11 ∈ 2
In [44], Vulakh states that each geometrically finite subgroup of PSL2(R) for which is a cuspidal point satisfies (A). The previous example shows that this statement∞ is not correct. This property is crucial for the results in [44]. Thus, Vulakh’s constructions do not apply to such a huge class of groups as he claims.
4. Precells in H. Let Γ be a geometric finite Fuchsian group with as cuspidal point and which satisfies Condition (A). To avoid empty statements∞ suppose that Γ =Γ , which means that there are relevant isometric spheres. 6 In this∞ section we introduce the notion of precells in H and basal families of precells in H. Moreover, we discuss their relation to fundamental regions and study some of their properties. We recall that := ext I(g). K g ΓrΓ∞ ∈\ 4.1. The structure of . We start with a short discussion of the vertex structure of . K KThe set of all isometric spheres of Γ need not be locally finite. For example, g in the case of the modular group PSL2(Z), each neighborhood of 0 in H contains infinitely many isometric spheres. However, from Γ being geometrically finite, it follows immediately that the set of relevant isometric spheres is locally finite. In turn, the set of infinite vertices of has no accumulation points. Moreover, if v is an inner vertex of , then there areK exactly two distinct relevant isometric spheres such that v is a commonK endpoint of their relevant parts. If v is an infinite vertex, then two situations can occur. If v is an endpoint of the relevant parts of two distinct relevant isometric spheres, we call v a two-sided infinite vertex. Otherwise we call v a one-sided infinite vertex. Straightforward geometric arguments prove the following statement on the local situation at one-sided infinite vertices of . For this let K g pr : H r R, pr (z) := z ht(z)=Re z ∞ {∞} → ∞ − denote the geodesic projection from to ∂gH. For a,b R we set ∞ ∈ [a,b] if a b, a,b := ≤ h i ([b,a] otherwise. 2188 ANKE D. POHL
Proposition 4.1. Let v be a one-sided infinite vertex of . Then there exists K 1 a unique one-sided infinite vertex w of such that the strip pr− ( v, w ) H is 1 K ∞ h i ∩ contained in . In particular, pr− ( v, w ) does not intersect any isometric sphere K ∞ 1 h i in H, and, of all vertices of , pr− ( v, w ) contains only v and w. K ∞ h i For any one-sided infinite vertex v of , we call the interval v, w from Propo- sition 4.1 a boundary interval, and we callK w the one-sided infiniteh vertexi adjacent to v. Boundary intervals will be needed for the definition and investigation of strip precells defined below. Example 4.2. Recall the group Γ from Example 3.1(iii). The boundary intervals of are the intervals [1 + 4m, 3+4m] for each m Z. K ∈ 4.2. Precells in H and basal families. We now define the notion of precells in H. Definition 4.3. Let v be a vertex of . Suppose first that v is an inner vertex or a two-sided infinite vertex. Then thereK are (exactly) two relevant isometric spheres I1, I2 with relevant parts [a1, v] resp. [v,b2]. Let s1 resp. s2 be the summit of I1 resp. I2. By Condition (A), the summits s1 and s2 do not coincide with v. 1 If v is a two-sided infinite vertex, then define 1 to be the hyperbolic triangle with vertices v, s and , and define to be theA hyperbolic triangle with vertices 1 ∞ A2 v, s2 and . The sets 1 and 2 are the precells in H attached to v. Precells arising in this∞ way are calledA cuspidalA . If v is an inner vertex, then let be the hyperbolic quadrilateral with vertices A s1, v, s2 and . The set is the precell in H attached to v. Precells that are constructed in∞ this way areA called non-cuspidal. Suppose now that v is a one-sided infinite vertex. Then there exist exactly one relevant isometric sphere I with relevant part [a, v] and a unique one-sided infinite 1 vertex w other than v such that pr− ( v, w ) does not contain vertices other than v and w (see Proposition 4.1). Let s∞beh thei summit of I. Define 1 to be the hyperbolic triangle with vertices v, s and , and define 2 A 1 ∞ A to be the vertical strip pr− ( v, w ) H. The sets 1 and 2 are the precells in H attached to v. The precell∞ h is calledi ∩ cuspidal, andA isA called a strip precell. A1 A2 Example 4.4. The precells in H of the congruence group PΓ0(5) from Exam- ple 3.1(ii) are indicated in Figure 9 up to PΓ0(5) -equivalence. ∞
(v ) (v ) (v ) (v ) (v ) A 0 A 1 A 2 A 3 A 4
m1 m2 m3 m4 v1 v2 v3
v0 =0 v4 =1
Figure 9. Precells in H of PΓ0(5).
1We consider the boundary of the triangle in H to belong to it. SYMBOLIC DYNAMICS 2189
The inner vertices of are K 2k +1 √3 v = + i , k =1, 2, 3, k 10 10 and their translates under PΓ0(5) . The summits of the indicated isometric spheres are ∞ k i m = + , k =1,..., 4. k 5 5 The group PΓ0(5) has cuspidal as well as non-cuspidal precells in H, but no strip precells. Example 4.5. The precells in H of the group Γ from Example 3.1(iii) are up to Γ -equivalence one strip precell 1 and two cuspidal precells 2, 3 as indicated in∞ Figure 10. Here, v = 3, v =A 1, v = 1 and m = i. A A 1 − 2 − 3
A1 A2 A3 m
v1 v2 0 v3
Figure 10. Precells in H of Γ.
Condition (A) yields that the relation between different precells and between the precells and are well-structured. K Proposition 4.6. (i) If is a precell in H, then A 1 1 = pr− pr ( ) and ◦ = pr− pr ( ◦) . A ∞ ∞ A ∩ K A ∞ ∞ A ∩ K (ii) If two precells in H have a common point, then either they are identical or they coincide exactly at a common vertical side. (iii) The set is the essentially disjoint union of all precells in H, K = precell in H , K {A|A } and contains the disjoint[ union of the interiors of all precells in H, K ◦ precell in H . {A | A } ⊆ K Proof. We use the notation[ from Definition 4.3 to discuss the relation between precells and . Suppose firstK that is a non-cuspidal precell in H attached to the inner vertex A v. Condition (A) implies that the summits s1 and s2 are contained in the relevant parts of the relevant isometric spheres I1 and I2, respectively. Therefore they are contained in ∂ . Hence is the subset of with the two vertical sides [s , ] and K A K 1 ∞ [s2, ], and the two non-vertical ones [s1, v] and [v,s2]. The geodesic projection of from∞ is A ∞ pr ( )= Re s1, Re s2 . ∞ A h i 2190 ANKE D. POHL
Suppose now that is a cuspidal precell in H attached to the infinite vertex v. Then has two verticalA sides, namely [v, ] and [s, ], and a single non-vertical side, namelyA [v,s]. As for non-cuspidal precells∞ we find∞ that [v,s] is contained in the relevant part of some relevant isometric sphere, and hence is a subset of . The geodesic projection of from is A K A ∞ pr ( )= Re s, v . ∞ A h i 1 Suppose finally that is the strip precell pr− ( v, w ) H. Then is attached to the two vertices v andA w. It has the two vertical∞ h sidesi ∩ [v, ] andA [w, ] and no non-vertical ones. The geodesic projection of from is ∞ ∞ A ∞ pr ( )= v, w . ∞ A h i By Proposition 4.1, is contained in . From these observations, the statements follow easily. A K
Before we investigate the relation between precells in H and fundamental regions in Theorem 4.8 below, we state a few properties of isometric spheres. Their proofs are straightforward. Lemma 4.7. Let g ΓrΓ . ∈ ∞ 1 (i) We have g.I(g)= I(g− ). 1 (ii) If s is the summit of I(g), then g.s is the summit of I(g− ). 1 (iii) The geodesic projection pr (s) of the summit s of I(g) is the center g− . of I(g). ∞ ∞ 1 (iv) If I(g) is relevant with relevant part [a,b], then I(g− ) is relevant with relevant part g.[a,b] = [g.a, g.b]. Let Λ be a Fuchsian group. A subset of H is called a closed fundamental F region for Λ in H if is closed and ◦ is a fundamental region for Λ in H. If, in F F addition, ◦ is connected, then is said to be a closed fundamental domain for Λ in H. NoteF that if is a non-connectedF fundamental region for Λ in H, then can happen to be a closedF fundamental domain. F Let Aj j J be a family of real submanifolds (possibly with boundary) of H g { | ∈ } or H , and let n := max dim Aj j J . We call the union { | ∈ }
Aj j J [∈ essentially disjoint if for each i, j J, i = j, the intersection Ai Aj is contained in a real submanifold (possibly with∈ boundary)6 of dimension n ∩1. − Theorem 4.8. There exists a set j j J of precells in H, indexed by J, such {A | ∈ } that the (essentially disjoint) union j is a closed fundamental region for Γ j J A in H. The set J is finite and its cardinality∈ does not depend on the choice of the S specific set of precells. The set j j J can be chosen such that j J j is a {A | ∈ } ∈ A closed fundamental domain for Γ in H. In each case, the (disjoint) union ◦ S j J Aj is a fundamental region for Γ in H. ∈ S Proof. By Proposition 4.6(ii), the union of each family of pairwise different precells in H is essentially disjoint. Let r be the center of some relevant isometric sphere I. Let r := Re s for the summit s of some relevant sphere of Γ or let r := Re v for an SYMBOLIC DYNAMICS 2191 infinite vertex v of . Let λ> 0 be the unique positive number such that K 1 λ Γ = . ∞ 0 1 From Lemma 4.7 and Proposition 4.6 one easily deduces that
(r)= (r) = ([r, r + λ]+ iR>0) F F∞ ∩ K ∩ K decomposes into a finite set A := j j J of precells in H. Clearly, (r) is a closed fundamental domain. {A | ∈ } F
Let A2 := k k K be a set of precells in H such that := k K k is a closed fundamental{A | region∈ } for Γ in H. Proposition 4.6(iii) impliesF that ∈ A S = h. h Γ , A . K { A | ∈ ∞ A ∈ } Let k A2 and pick z k◦[. Then there exists hk Γ and jk J such that A ∈ ∈ A ∈ ∞ ∈ hk.z jk . Therefore hk. k◦ jk = . The Γ -invariance of shows that hk. k ∈ A A ∩ A 6 ∅ ∞ K A is a precell in H. Then Proposition 4.6(ii) implies that hk. k = j , and in turn A A k hk and jk are unique. We will show that the map ϕ: K J, k jk, is a bijection. → 7→ To show that ϕ is injective suppose that there are l, k K such that jk = jl =: j. 1 ∈ 1 Then hl. l = j = hk. k, hence hk− hl. l = k. In particular, hk− hl. l◦ k◦ = . A A A A A A1 ∩A 6 ∅ Since h K h◦ ◦ and ◦ is a fundamental region, it follows that hk− hl = id and ∈ A ⊆F F l = k. Thus, ϕ is injective. To show surjectivity let j J and z j◦. Then there S ∈ ∈ A 1 exists g Γ and k K such that g.z k. On the other hand, k = h− . j . ∈ ∈ ∈ A A k A k Hence hkg. ◦ j = . Since j and j are convex polyhedrons, it follows that Aj ∩ A k 6 ∅ A A k hkg. ◦ j = . Since (r) is a fundamental region and ◦, ◦ (r), we find Aj ∩ A k 6 ∅ F Aj Ajk ⊆F that hkg = id and j = jk. Hence, ϕ is surjective. It follows that #K = #J. It remains to show that the disjoint union P := k K k◦ is a fundamental region ∈ A for Γ in H. Obviously, P is open and contained in ◦. This shows that P satisfies F (F1) and (F2). Since ( )= for each precell inSH and K is finite, it follows that A◦ A P = = k = . Ak◦ A F k K k K [∈ [∈ Hence, P satisfies (F3) as well, and thus it is a fundamental region for Γ in H.
Each set A := j j J of precells in H, indexed by J, with the property that {A | ∈ } := j is a closed fundamental region is called a basal family of precells in F j J A H or a family∈ of basal precells in H. If, in addition, is connected, then A is called a connectedS basal family of precells in H or a connectedF family of basal precells in H. Example 4.9. Recall the Examples 3.2, 4.4 and 4.5. The set of precells in H {A} for Gn is a connected basal family of precells in H, as well as (v0),..., (v4) for PΓ (5), and , , for Γ. {A A } 0 {A1 A2 A3} The proof of Theorem 4.8 shows the following statements. Let 1 λ t := λ 0 1 with λ> 0 be such that Γ = tλ . ∞ h i Corollary 4.10. Let A be a basal family of precells in H.
(i) For each precell in H there exists a unique pair ( 0,m) A Z such that m A A ∈ × t . 0 = . λ A A 2192 ANKE D. POHL
m( ) (ii) For each A choose an element m( ) Z. Then tλ A . A is A ∈ A ∈ mA( ) A ∈ A A a basal family of precells in H. For each , the precell tλ . is of the same type as . A ∈ A (iii) The set is theA essentially disjoint union h. h Γ , A . K { A | ∈ ∞ A ∈ } 4.3. The tesselation of H by basal familiesS of precells. The following propo- sition is crucial for the construction of cells in H from precells in H. Note that the element g ΓrΓ in this proposition depends not only on and b but also on the choice∈ of the basal∞ family A of precells in H. In this sectionA we will use the proposition as one ingredient for the proof that the family of Γ-translates of all precells in H is a tesselation of H. Proposition 4.11. Let A be a basal family of precells in H. Let A be a basal precell that is not a strip precell, and suppose that b is a non-verticalA side ∈ of . Then there is a unique element g ΓrΓ such that b I(g) and g.b is the non-verticalA ∈ ∞ ⊆ side of some basal precell 0 A. If is non-cuspidal, then 0 is non-cuspidal, A ∈ A A and, if is cuspidal, then 0 is cuspidal. A A Proof. Let I be the (relevant) isometric sphere with b I. We will show first that there is a generator g of I such that g.b is a side of⊆ some basal precell. Then 1 g.b g.I(g)= I(g− ), which implies that g.b is a non-vertical side. Let⊆ h ΓrΓ be any generator of I, let s be the summit of I and v the vertex of that∈ is attached∞ to. Then b = [v,s]. Further, b is contained in the relevant partK of I =AI(h). By Lemma 4.7, the set h.b = [h.v, h.s] is contained in the relevant 1 part of the relevant isometric sphere I(h− ), the point h.v is a vertex of and h.s 1 K is the summit of I(h− ). Thus, there is a unique precell h with non-vertical side A h.b. By Corollary 4.10, there is a unique basal precell 0 and a unique m Z such that A ∈ m h = t . 0 = 0 + mλ. A λ A A m m Then tλ− h.b is a non-vertical side of 0, and tλ− h.b is contained in the relevant A 1 1 m m 1 part of the relevant isometric sphere I(h− ) mλ = I(h− tλ ) = I((tλ− h)− ). m − Clearly, also g := tλ− h is a generator of I. To prove the uniqueness of g, let k be any generator of I. Then there exists a unique n Z such that k = tnh. Thus, k.b = tnh.b = h.b + nλ and therefore ∈ λ λ k = h + nλ. Then A A m+n k = 0 + mλ + nλ = t . 0, A A λ A (m+n) (m+n) and tλ− k is a generator of I such that tλ− k.b is a side of some basal precell. Moreover, (m+n) m n m tλ− k = tλ− tλ− k = tλ− h = g. This shows the uniqueness. The basal precell 0 cannot be a strip precell, since it has a non-vertical side. Finally, is cuspidalA if and only if v is an infinite vertex. This is the case if and only A if gv is an infinite vertex, which is equivalent to 0 being cuspidal. This completes the proof. A
Lemma 4.12. Let be a precell in H. Suppose that S is a vertical side of . Then A A there exists a precell 0 in H such that S is a side of 0 and 0 = . In this case, A A A 6 A S is a vertical side of 0. A SYMBOLIC DYNAMICS 2193
Proposition 4.13. Let 1, 2 be two precells in H and let g1,g2 Γ. Suppose A A ∈ 1 that g1. 1 g2. 2 = . Then we have either g1. 1 = g2. 2 and g1g2− Γ , or g . Ag .∩ isA a6 common∅ side of g . and g A. , or gA. g . is∈ a∞ point 1 A1 ∩ 2 A2 1 A1 2 A2 1 A1 ∩ 2 A2 which is the endpoint of some side of g1. 1 and some side of g2. 2. If S is a 1A A common side of g1. 1 and g2. 2, then g1− .S is a vertical side of 1 if and only if 1 A A A g− .S is a vertical side of . 2 A2
Proof. W.l.o.g. g1 = id. Let A be a basal family of precells in H. Corollary 4.10 shows that we may assume that 1 A. Let S1 = [a1, ] and S2 = [a2, ] be the vertical sides of . If hasA non-vertical∈ sides, let∞ these be S = [∞b ,b ] A1 A1 3 1 2 resp. S = [b ,b ] and S = [b ,b ]. Lemma 4.12 shows that we find precells 0 3 1 2 4 3 4 A1 and 0 such that 0 = = 0 and S is a vertical side of 0 and S is a A2 A1 6 A1 6 A2 1 A1 2 vertical side of 0 . Proposition 4.11 shows that there exist (h , 0 ) Γ A resp. A2 3 A3 ∈ × (h , 0 ), (h , 0 ) Γ A such that h .S is a non-vertical side of 0 and h .S is 3 A3 4 A4 ∈ × 3 3 A3 4 4 a non-vertical side of 0 and h . = 0 and h . = 0 . Recall that each precell A4 3 A1 6 A3 4 A1 6 A4 in H is a convex polyhedron. Therefore 1 10 is a polyhedron with (a1, ) in its 1 A ∪ A ∞ interior, and 1 h3− . 30 is a polyhedron with (b1,b2) in its interior. Likewise for A ∪ 1 A 0 and h− . 0 . A1 ∪ A2 A1 ∪ 4 A4 Suppose first that ◦ g . = . By Corollary 4.10 there exists a pair (h, 0) A1 ∩ 2 A2 6 ∅ A ∈ Γ A such that 2 = h. 0. Then 1◦ g2h. 0 = . Since 0 is a convex ∞ × A A A ∩ A 6 ∅ A polyhedron, ◦ g h.( 0)◦ = . Now ( )◦ A is a fundamental region A1 ∩ 2 A 6 ∅ A A ∈ for Γ in H (see Theorem 4.8). Therefore, g2h = id and, by Proposition 4.6(ii), 1 S 1 = 0. Hence g2− Γ and 1 = g2. 2. b b A A ∈ ∞ A A Suppose now that ◦ g . = . If g . (a , ) = , then g . ( 0 )◦ = A1 ∩ 2 A2 ∅ 2 A2 ∩ 1 ∞ 6 ∅ 2 A2 ∩ A1 6 ∅ and the argument from above shows that 10 = g2. 2 and g2 Γ . From this it A A ∈ ∞ follows that S1 is a vertical side of 2. A 1 If g2. 2 (b1,b2) = , then g2. 2 h3− .( 30 )◦ = . As before, g2h3 Γ and A ∩1 6 ∅ A ∩ A 6 ∅ ∈ ∞ g . = h− . 0 . Then S is a non-vertical side of . The argumentation for 2 A2 3 A3 3 A2 g2. 2 (a2, ) = and g2. 2 (b3,b4) = is analogous. AIt remains∩ ∞ the6 case∅ that gA. ∩ intersects6 ∅ is an endpoint v of some side of . 2 A2 A1 A1 By symmetry of arguments, v is an endpoint of some side of 2. This completes the proof. A
We call a family Sj j J of polyhedrons in H a tesselation of H if { | ∈ } (T1) H = j J Sj , and ∈ (T2) if Sj Sk = for some j, k J, then either Sj = Sk or Sj Sk is a common S∩ 6 ∅ ∈ ∩ side or vertex of Sj and Sk.
Corollary 4.14. Let A be a basal family of precells in H. Then
Γ.A = g. g Γ, A { A | ∈ A ∈ } is a tesselation of H which satisfies in addition the property that if g1. 1 = g2. 2, then g = g and = . A A 1 2 A1 A2 Proof. Let := A . Theorem 4.8 states that is a closed fundamental F {A|A∈ } F region for Γ in H, hence g Γ g. = H. This proves (T1). Property (T2) follows S ∈ F directly from Proposition 4.13. Now let (g , ), (g , ) Γ A with g . = S 1 A1 2 A2 ∈ × 1 A1 g . . Then g . ◦ = g . ◦. Recalling that ◦ is a fundamental region for Γ in H 2 A2 1 A1 2 A2 F and that ◦, ◦ ◦, we get that g = g and = . A1 A2 ⊆F 1 2 A1 A2 2194 ANKE D. POHL
5. Cells in H. Let Γ be a geometrically finite subgroup of PSL2(R) of which is a cuspidal point and which satisfies (A). Suppose that the set of relevant isometric∞ spheres is non-empty. Let A be a basal family of precells in H. To each basal precell in H we assign a cell in H, which is an essentially disjoint union of certain Γ- translates of certain basal precells. More precisely, using Proposition 4.11 we define so-called cycles in A Γ in the same way as the cycles in the Poincar´eFundamental Polyhedron Theorem× are defined (see e.g. [25]). These are certain finite sequences of pairs ( ,h) A ΓrΓ , such that each cycle is determined up to a cyclic permutationA by∈ any pair× which∞ belongs to it. Moreover, if ( ,h ) is an element of some cycle, then h is an element in ΓrΓ assigned to byA PropositionA 4.11 (or h = id if is a stripA precell). Conversely,∞ if h is anA element assigned to by PropositionA A4.11, then ( ,h ) determines a cycleA in A ΓrΓ . A One of the crucial propertiesA A of each cell in H is that× it is a∞ convex polyhedron with non-empty interior of which each side is a complete geodesic segment. This fact is mainly due to the condition (A) of Γ. The other two important properties of cells in H are that each non-vertical side of a cell is a Γ-translate of some vertical side of some cell in H and that the family of Γ-translates of all cells in H is a tesselation of H. 5.1. Cycles in A Γ. Let A be a non-cuspidal precell in H. The definition of precells shows that× is attachedA ∈ to a unique (inner) vertex v of , and is the unique precell attachedA to v. Therefore we set (v) := . Further,K hasA two A A A non-vertical sides b1 and b2. Let k1( ), k2( ) be the two elements in ΓrΓ given { A A } ∞ by Proposition 4.11 such that bj I(kj ( )) and kj ( ).bj is a non-vertical side of some basal precell. Necessarily, the∈ isometricA spheresA I(k ( )) and I(k ( )) are 1 A 2 A different, therefore k1( ) = k2( ). The set k1( ), k2( ) is uniquely determined by Proposition 4.11, theA assignment6 A k ({ ) clearlyA dependsA } on the enumeration A 7→ 1 A of the non-vertical sides of . Now w := kj ( ).v is an inner vertex. Let (w) be the (unique non-cuspidal) basalA precell attachedA to w. Since one non-verticalA side 1 of (w) is kj ( ).bj , which is contained in the relevant isometric sphere I(kj ( )− ), A 1 A A and kj ( )− kj ( ).bj = bj is a non-vertical side of some basal precell, namely of , A A 1 A one of the elements in ΓrΓ assigned to (w) by Proposition 4.11 is kj ( )− . ∞ A A Construction 5.1. Let A be a non-cuspidal precell and suppose that = A ∈ A (v) is attached to the vertex v of . We assign to two sequences (hj ) of elements Ain ΓrΓ using the following algorithm:K A ∞ (step 1) Let v1 := v and let h1 be either k1( ) or k2( ). Set g1 := id, g2 := h1 and carry out (step 2). A A (step j) Set vj := gj.v and j := (vj ). Let hj be the element in ΓrΓ such 1 A A ∞ that hj ,hj− 1 = k1( j ), k2( j ) . Set gj+1 := hj gj. If gj+1 = id, then the − A A algorithm stops. If gj = id, then carry out (step j + 1). +1 6 Example 5.2. Recall the Hecke triangle group Gn and its basal family A = {A}n of precells in H from Example 3.2. The two sequences assigned to are (Un)j=1 1 n A and Un− j=1. The statements of Propositions 5.3-5.7 follow as in the Poincar´eFundamental Polyhedron Theorem, using Lemma 4.7 and elementary convex geometry. For this reason we omit the detailed proofs. SYMBOLIC DYNAMICS 2195
Proposition 5.3. Let = (v) be a non-cuspidal basal precell. A A (i) The sequences from Construction 5.1 are finite. In other words, the algorithm for the construction of the sequences always terminates. (ii) Both sequences have same length, say k N. ∈ (iii) Let (aj )j=1,...,k and (bj)j=1,...,k be the two sequences assigned to . Then they are inverse to each other in the following sense: For each j =1,...,kA we have 1 aj = bk− j+1. − (iv) For j =1,...,k set cj+1 := aj aj 1 a2a1, dj+1 := bjbj 1 b2b1 and c1 := − ··· − ··· id =: d1. Then k k 1 1 := c− . cj v = d− . dj v . B j A j A j=1 j=1 [ [ Further, both unions are essentially disjoint, and is the polyhedron with the 1 B 1 1 (pairwise distinct) vertices (in this order) c1− . , c2− . , ..., ck− . resp. 1 1 1 ∞ ∞ ∞ d− . , d− . , ..., d− . . 1 ∞ 2 ∞ k ∞ Definition 5.4. Let A be a non-cuspidal precell and suppose that is attached to the vertex v of .A Let ∈ h be one of the elements in ΓrΓ assignedA to by K A ∞ A Proposition 4.11. Let(hj )j ,...,k be the sequence assigned to by Construction 5.1 =1 A with h1 = h . For j = 1,...,k set g1 := id and gj+1 := hj gj. Then the (finite) A sequence ( (gj v),hj ) is called the cycle in A Γ determined by ( ,h ). A j=1,...,k × A A Let A be a cuspidal precell. Suppose that b is the non-vertical side of A ∈ A and let h be the element in ΓrΓ assigned to by Proposition 4.11. Let 0 be the (cuspidal)A basal precell with non-vertical∞ sideAh .b. Then the (finite) sequenceA 1 A ( ,h ), ( 0,h− ) is called the cycle in A Γ determined by ( ,h ). A A A A × A A Let A be a strip precell. Set h := id. Then ( ,h ) is called the cycle in A A A Γ determinedA ∈ by ( ,h ). A × A A Example 5.5. (i) Recall Example 5.2. The cycle in A Gn determined by n × ( ,Un) is ( ,Un) . A A j=1 (ii) Recall the group PΓ (5) and the basal family A = (v ),..., (v ) from 0 {A 0 A 4 } Example 4.4. The element in PΓ0(5) r PΓ0(5) assigned to (v0) is h := 4 1 ∞ A 5 −1 . The cycle in A PΓ0(5) determined by ( (v0),h) is − × A 1 ( (v ),h), ( (v ),h− ) . A 0 A 4 3 2 2 1 Further let h1 := h, h2 := 5 −3 and h3 := 5 −2 . The cycle in A PΓ0(5) determined by ( (v ),h ) is − − × A 1 1 ( (v ),h ), ( (v ),h ), ( (v ),h ) . A 1 1 A 3 2 A 2 3 (iii) Recall the group Γ and the basal family A = 1, 2, 3 from Example 4.5. The cycle in A Γ determined by ( ,S) is {A( ,SA ),A( },S) . × A2 A2 A3 Proposition 5.6. Let A be a non-cuspidal precell in H and suppose that h is one of the elementsA ∈ in Γ r Γ assigned to by Proposition 4.11. Let A ∞ A ( j ,hj ) be the cycle in A Γ determined by ( ,h ). Let j 1,...,k A j=1,...,k × A A ∈ { } and define the sequence ( 0,al) by Al l=1,...,k hl+j 1 for l =1,...,k j +1, al := − − (hl+j 1 k for l = k j +2,...,k, − − − 2196 ANKE D. POHL and
l+j 1 for l =1,...,k j +1, l0 := A − − A ( l+j 1 k for l = k j +2,...,k. A − − − Then a1 = hj is one of the elements in ΓrΓ assigned to j by Proposition 4.11 ∞ A and ( 0,al) is the cycle in A Γ determined by ( j ,hj ). Al l=1,...,k × A Proposition 5.7. Let A be a cuspidal precell in H and let h be the element A ∈ A1 in ΓrΓ assigned to by Proposition 4.11. Let ( ,h ), ( 0,h− ) be the cycle ∞ A 1 A A A A in A Γ determined by ( ,h ). Then h− is the element in ΓrΓ assigned to × A A 1 A ∞ 0 by Proposition 4.11 and ( 0,h− ), ( ,h ) is the cycle in A Γ determined by A 1 A A A A × ( 0,h− ). A A 5.2. Cells in H and their properties. Now we can define a cell in H for each basal precell in H. Construction 5.8. Let be a basal strip precell in H. Then we set A ( ) := . B A A Let be a cuspidal basal precell in H. Suppose that g is the element in ΓrΓ A 1 ∞ assigned to by Proposition 4.11 and let ( ,g), ( 0,g− ) be the cycle in A Γ determinedA by ( ,g). Define A A × A 1 ( ) := g− . 0. B A A ∪ A The set ( ) is well-defined because g is uniquely determined. Let BbeA a non-cuspidal basal precell in H and fix an element h in Γ r Γ A A ∞ assigned to by Proposition 4.11. Let ( j ,hj ) be the cycle in A Γ A A j=1,...,k × determined by ( ,h ). For j =1,...,k set g1 := id and gj+1 := hj gj. Set A A k 1 ( ) := g− . j . B A j A j=1 [ By Proposition 5.3, the set ( ) does not depend on the choice of h . The family B := ( ) A is calledB A the family of cells in H assigned to A.A Each element of B is{B calledA | A a cell ∈ in} H. Note that the family B of cells in H depends on the choice of A. If we need to distinguish cells in H assigned to the basal family A of precells in H from those assigned to the basal family A0 of precells in H, we will call the first ones A-cells and the latter ones A0-cells.
Example 5.9. Recall the Example 5.5. For the Hecke triangle group G5, Figure 11 shows the cell assigned to the family A = from Example 3.2. For the group {A} PΓ0(5), the family of cells in H assigned to A is indicated in Figure 12. Figure 13 shows the family of cells in H assigned to the basal family A of precells of Γ. In the following series of six propositions we investigate the structure of cells and their relations to each other. This will allow to show that the family of Γ-translates of cells in H is a tesselation of H, and it will be of interest for the labeling of the cross section. Proposition 5.10. Let be a non-cuspidal basal precell in H. Suppose that h is A A an element in ΓrΓ assigned to by Proposition 4.11 and let ( j ,hj ) ∞ A A j=1,...,k SYMBOLIC DYNAMICS 2197
( ) B A
i A λ5 + i 1 4 U − U − 5 A 5 A 2 3 U − U − 5 A 5 A
4 3 2 U U U U5 5 ∞ 5 ∞ 5 ∞ ∞ =0 = λ5
Figure 11. The cell ( ) for G . B A 5
( (v )) ( (v )) ( (v )) ( (v )) ( (v )) B A 0 B A 1 B A 2 B A 3 B A 4
0 1 2 3 4 1 5 5 5 5
Figure 12. The family of cells in H assigned to A for PΓ0(5).
( ) ( ) ( ) B A1 B A2 B A3
v1 v2 0 v3
Figure 13. The family of cells in H assigned to A for Γ.
be the cycle in A Γ determined by ( ,h ). For j = 1,...,k set g1 := id and × A A gj+1 := hj gj. Then the following assertions hold true. 1 (i) The set ( ) is the convex polyhedron with vertices (in this order) g1− . , 1 B A 1 ∞ g2− . ,..., gk− . . (ii) The∞ boundary of∞ ( ) consists precisely of the union of the images of the B A 1 vertical sides of j under g− , j =1,...,k. More precisely, if sj denotes the A j summit of I(hj ) for j =1,...,k, then k k+1 1 1 ∂ ( )= gj− .[sj , ] gj− .[hj 1.sj 1, ]. B A ∞ ∪ − − ∞ j=1 j=2 [ [ (iii) For each j = 1,...,k we have gj . ( ) = ( j ). In particular, the side 1 1 B A B A 1 [gj− . ,gj−+1. ] of ( ) is the image of the vertical side [ ,hj− . ] of ( j) ∞ 1 ∞ B A ∞ ∞ B A under gj− . 2198 ANKE D. POHL
(iv) Let be a basal precell in H and let h Γ such that h. ( )◦ = . Then A ∈ A∩B1 A 6 ∅ there exists a unique j 1,...,k such that h = g− and = j . In ∈ { } j A A particular,b is non-cuspidal and h. ( )= ( ). b A B A B A b Proof. By Proposition 5.3, ( ) is the polyhedron with vertices (in this order) 1 1 b 1 B A b g1− . , g2− . ,...,gk− . . Since each of its sides is a complete geodesic segment, ( ∞) is convex.∞ This∞ shows (i). The statement (ii) follows from the proof of PropositionB A 5.3. To prove (iii), fix j 1,...,k and recall from Proposition 5.6 the cycle ∈ { } ( 0,al) in A Γ determined by ( j ,hj ). For l = 1,...,k set c := id Al l=1,...,k × A 1 and c := a c . Then l+1 l l 1 gl+j 1gj− for l =1,...,k j +1 − cl = 1 − (gl+j k 1gj− for l = k j +2,...,k. − − − Hence k k 1 1 ( j)= c− . 0 = gj (clgj)− . 0 B A l Al Al l l [=1 [=1 k j+1 k − 1 1 = gj . gl−+j 1. l+j 1 gj. gl−+j k 1. l+j k 1 − A − ∪ − − A − − l=1 l=k j+2 [ [− k 1 = gj . g− . l = gj . ( ). l A B A l [=1 1 1 This immediately implies that the side [gj− . ,gj−+1. ] of ( ) maps to the side 1 1 1 ∞ ∞ B A gj.[g− . ,g− . ] = [ ,h− . ] of ( j), which is vertical. j ∞ j+1 ∞ ∞ j ∞ B A To prove (iv), fix z h. ( )◦. Then there exists l 1,...,k such that 1 ∈ A∩B A 1 ∈ { } z h. gl− . l. Let b := h. gl− . l. By Proposition 4.13 there are three possibilities∈ A ∩ for Ab. b A ∩ A 1 If b =b h. = gl− . l, then glh.b = l. Since and l are both basal, it follows A1 A A A A A that h = gl− and = l. Supposeb that v Ais theA vertex ofb to which bis attached. If b is a common side 1 K A of h. and gl− . l,b then, since z ( )◦, gl.b must be a non-vertical side of l (seeA (ii)). This impliesA that v b∈. BInA turn, there is a neighborhood U of v suchA ∈ that Ub ( ) and U h.( )◦ = . Hence, h.( )◦ ( )◦ = . Thus there exists ⊆B A ∩ A 6 ∅1 A ∩B A 6 ∅ j 1,...,k such that h.( )◦ gj− . j = . Proposition 4.13 implies that = j ∈{ 1} Ab ∩ A 6 ∅ b A A and h = gj− . b 1 b If b is a point, then b = z must be the endpoint of some side of g− . l. From l A z ( )◦ it follows that z = v. Now the previous argument applies. ∈B A 1 To show the uniqueness of j 1,...,k with = j and h = gj− , suppose ∈ { } A A 1 that there is p 1,...,k with = p and h = g− . Then gj = gp. By ∈ { } A A p Proposition 5.3(iv) j = p. The remaining parts of (ivb) follow from (iii). b Proposition 5.11. Let be a cuspidal basal precell in H which is attached to the vertex v of . SupposeA that g is the element in Γ r Γ assigned to by K 1 ∞ A Proposition 4.11. Let ( ,g), ( 0,g− ) be the cycle in A Γ determined by ( ,g). Then we have the followingA properties.A × A SYMBOLIC DYNAMICS 2199
1 (i) The set ( ) is the hyperbolic triangle with vertices v, g− . , . (ii) The boundaryB A of ( ) is the union of the vertical sides of ∞with∞ the images B A 1 A of the vertical sides of 0 under g− . A (iii) The sets g. ( ) and ( 0) coincide. In particular, the non-vertical side 1 B A B A 1 [v,g− . ] of ( ) is the image of the vertical side [g.v, ] of 0 under g− . ∞ B A ∞ A (iv) Suppose that is a basal precell in H and h Γ such that h. ( )◦ = . A 1 ∈ A∩B A 6 ∅ Then either h = id and = or h = g− and = 0. In particular, is A A A A A cuspidal and h.b ( )= ( ). b B A B bA b b Proposition 5.12. Let be a basal strip precell in H. Let be a basal precell in bA A H and h Γ such that h. ( )◦ = . Then h = id and = . ∈ A∩B A 6 ∅ Ab A Corollary 5.13. The map A B, ( ) is a bijection. b → A 7→ B A b The following proposition is implied by following the glueing procedure from precells to cells. The details are easily seen by straightforward deductions. Proposition 5.14. (1) Let be a non-cuspidal basal precell in H. Suppose that A (hj )j=1,...,k is a sequence in ΓrΓ assigned to by Construction 5.1. For ∞ A j = 1,...,k set g := id and gj := hj gj. Let 0 be a basal precell in H and 1 +1 A g Γ such that ( ) g. ( 0) = . Then we have the following properties. ∈ B A ∩ B A 6 ∅ (i) Either ( )= g. ( 0), or ( ) g. ( 0) is a common side of ( ) and B A B A B A ∩ B A B A g. ( 0). B A 1 (ii) If ( )= g. ( 0), then g = g− for a unique j 1,...,k . In particu- B A B A j ∈{ } lar, 0 is non-cuspidal. A (iii) If ( ) = g. ( 0), then 0 is cuspidal or non-cuspidal. If 0 is cuspidal B A 6 B A A A and k ΓrΓ is the element assigned to 0 by Proposition 4.11, then we ∈ ∞ 1 A have ( ) g. ( 0)= g.[k− . , ]. (2) Let be aB cuspidalA ∩ B basalA precell in∞H ∞which is attached to the vertex v of . SupposeA that h ΓrΓ is the element assigned to by Proposition 4.11. LetK ∈ ∞ A 0 be a basal precell in H and g Γ such that we have ( ) g. ( 0) = . AThen the following assertions hold∈ true. B A ∩ B A 6 ∅ (i) Either ( )= g. ( 0), or ( ) g. ( 0) is a common side of ( ) and B A B A B A ∩ B A B A g. ( 0). B A 1 (ii) If ( ) = g. ( 0), then either g = id or g = h− . In particular, 0 is cuspidal.B A B A A (iii) If ( ) = g. ( 0), then 0 is cuspidal or non-cuspidal or a strip precell. B A 6 B A A 1 If 0 is a strip precell, then [h− . , ] = ( ) g. ( 0). If 0 is a cuspidalA precell attached to the vertex∞ ∞w 6 ofB Aand∩ kB AΓ r Γ Ais the K ∈ ∞ element assigned to 0 by Proposition 4.11, then ( ) g. ( 0) is either A 1 B A1 ∩ B A [v, ] = g.[w, ] or [v, ] = g.[w, k− . ] or [v,h− . ] = g.[w, ] or ∞ 1 ∞ 1 ∞ 1 ∞ 1 ∞ ∞ [v,h− . ]= g.[w, k− . ] or [h− . , ]= g.[k− . , ]. If 0 is a non- ∞ ∞ ∞ ∞ ∞1 ∞ A cuspidal precell, then we have ( ) g. ( 0) = [h− . , ]. B A ∩ B A ∞ ∞ (3) Let be a basal strip precell in H. Let 0 be a basal precell in H and g Γ A A ∈ such that ( ) g. ( 0) = . Then the following statements hold. B A ∩ B A 6 ∅ (i) Either ( )= g. ( 0), or ( ) g. ( 0) is a common side of ( ) and B A B A B A ∩ B A B A g. ( 0). B A (ii) If ( )= g. ( 0), then g = id and = 0. B A B A A A (iii) If ( ) = g. ( 0), then 0 is a cuspidal or strip precell. If 0 is cuspidal B A 6 B A A A and k ΓrΓ is the element assigned to 0 by Proposition 4.11, then we ∈ ∞ 1 A have ( ) g. ( 0) = g.[k− . , ]. B A ∩ B A 6 ∞ ∞ 2200 ANKE D. POHL
Corollary 5.15. The family of Γ-translates of all cells provides a tesselation of H. In particular, if is a cell in H and S a side of , then there exists a pair B B ( 0,g) B Γ such that S = g. 0. Moreover, ( 0,g) can be chosen such that B 1 ∈ × B ∩ B B g− .S is a vertical side of 0. B 6. The base manifold of the cross sections. Let Γ be a geometrically finite subgroup of PSL (R) of which is a cuspidal point and which satisfies (A), and 2 ∞ suppose that there are relevant isometric spheres. In this section we define a set CS which will turn out to be a cross section for the geodesic flow on Y =Γ H w.r.t. to \ certain measures µ, which will be characterized in Section 8.1 below. Here we willc already see that CS satisfies Condition (C2) by showing that pr(CS) is a totally geodesic suborbifold of Y of codimension one and that CS is the set of unit tangent vectors based onc pr(CS) but not tangent to it. To achieve this,c we start at the other end. We fix a basal family A of precells in H andc consider the family B of cells in H assigned tocA. We define BS(B) to be the set of Γ-translates of the sides of these cells. Then we show that the set BS := BS(B) is in fact independent of the choice of A. We proceed to prove that BS is a totally geodesic submanifold of H of codimension one and define CS to be the set of unit tangent vectors based on BS but not tangent to it. Then CS := π(CS) is our geometric cross section and pr(CS) = BS := π(BS). This construction shows in particular that the set CS does not depend on the choice of Ac. For future purposes we already define the sets NC(cB) andc bd(B) and show that also these are independent of the choice of A. Throughoutc let 1 λ t = λ 0 1 with λ> 0 be such that Γ = tλ . ∞ h i Definition 6.1. Let A be a basal family of precells in H and let B be the family of cells in H assigned to A. For B let Sides( ) be the set of sides of . Then set B ∈ B B Sides(B) := Sides( ) B B B∈[ and define BS(B) := Γ. Sides(B)= g.S g Γ, S Sides(B) . { | ∈ ∈ } For B define bd( [) := ∂g and let[ NC( ) be the set of geodesics on Y which haveB a ∈ representativeB on H bothB endpoints ofB which are contained in bd( ). Further set B bd(B) := g. bd( ) B g Γ B [∈ B∈[ and NC(B) := NC( ). B B B∈[ Proposition 6.2. Let A and A0 be two basal families of precells in H and suppose that B resp. B0 are the families of the corresponding cells in H assigned to A resp. A0. There exists a unique map A Z, m( ) such that → A 7→ A m( ) ψ : A precells in H , t A . →{ } A 7→ λ A SYMBOLIC DYNAMICS 2201 is a bijection from A to A0. Then m( ) χ: B B0, ( ) t A . ( ) → B A 7→ λ B A is a bijection as well. Further we have that BS(B) = BS(B0), NC(B) = NC(B0) and bd(B) = bd(B0). Proof. This follows immediately from a combination of Corollary 4.10, Theorem 4.8, Corollary 5.13, Proposition 4.11 and Proposition 5.3. We set BS := BS(B), bd := bd(B) and NC := NC(B) for the family B of cells in H assigned to an arbitrary family A of precells in H. Proposition 6.2 shows that BS, bd and NC are well-defined. Proposition 6.3. The set BS is a totally geodesic submanifold of H of codimension one. Proof. The set BS is a disjoint countable union of complete geodesic segments, which is locally finite. Let CS denote the set of unit tangent vectors in SH that are based on BS but not tangent to BS. Recall that Y denotes the orbifold Γ H and recall the canonical \ projections π : H Y , π : SH SY from Section 2. Set BS := π(BS) and CS := π(CS). → → c c Proposition 6.4. The set BS is a totally geodesic suborbifold of Y of codimension one, CS is the set of unit tangent vectors based on BS but not tangent to BS and CS satisfies (C2). c c c c 1 Proof.c Since BS is Γ-invariant by definition, we see that BS = π− (BS). Therefore, BS is a totally geodesic suborbifold of Y of codimension one. Moreover, CS = 1 π− (CS) and hence CS is indeed the set of unit tangent vectors basedc on BS but notc tangent to BS. Finally, pr(CS) = BS. Therefore, CS satisfies (C2). c c c Remark 6.5. Let NIC be the set of geodesics on Y of which at least one endpoint c cg c g c is contained in π(bd). Here, π : H Γ H denotes the extension of the canonical g → \ projection H Y to H . In Section 8.1 we will show that CS is a cross section for the geodesic→ flow on Y w.r.t. any measure µ on the space of geodesics on Y for which µ(NIC) = 0. c We end this section with a short explanation of the acronyms. Obviously, CS stands for “cross section” and BS for “base of (cross) section”. Then bd stands for “boundary” in the sense of geodesic boundary, and bd( ) is the geodesic boundary of the cell . Moreover, which will become clearer in SectionB 8.2 (see Remark 8.32), NC standsB for “not coded” and NC( ) for “not coded due to the cell ”. Finally, NIC stands for “not infinitely often coded”.B B
7. Precells and cells in SH. Let Γ be a geometrically finite subgroup of PSL2(R) which satisfies (A). Suppose that is a cuspidal point of Γ and that the set of relevant isometric spheres is non-empty.∞ In this section we define the precells and cells in SH and study their properties. The purpose of precells and cells in SH is to get very detailed information about the set CS from Section 6 and its relation to the geodesic flow on Y , see Section 8.1. c 2202 ANKE D. POHL
Definition 7.1. Let U be a subset of H and z U. A unit tangent vector v at z ∈ is said to point into U if the geodesic γv determined by v runs into U, i. e., if there exists ε > 0 such that γv((0,ε)) U. The unit tangent vector v is said to point ⊆ along the boundary of U if there exists ε > 0 such that γv((0,ε)) ∂U. It is said to point out of U if it points into H rU. ⊆ Definition 7.2. Let be a precell in H. Define to be the set of unit tangent A A vectors that are based on and point into ◦. The set is called the precell in A A A SH corresponding to . If is attached to the vertexe v of , we call a precell in SH attached to v. A A e K A e Recall the projection pr: SH H on base points. → Remark 7.3. Let be a precell in H and the corresponding precell in SH. A A Since is a convex polyhedron with non-empty interior, pr( ) is the precell in H A A to which corresponds. e A e Lemma 7.4. Let , be two different precells in H. Then the precells and e A1 A2 A1 2 in SH are disjoint. A f Proof. This is an immediate consequence of Proposition 4.6(ii) and Definition 7.2. f
Definition 7.5. Let be a precell in H and the corresponding precell in SH. A A The set vb( ) of unit tangent vectors based on ∂ and pointing along ∂ is called A A A the visual boundary of . Further, vc( ) := e vb( ) is said to be the visual A A A ∪ A closure of .e A e e e e e
Figure 14. The precell in SH and its visual boundary for a non- cuspidal precell in H.
The next lemma is clear from the definitions. Lemma 7.6. Let be a precell in H and be the corresponding precell in SH. A A Then vc( ) is the disjoint union of and vb( ). A A e A We remark that the visual boundary and the visual closure of a precell in SH e e e A is a proper subset of the topological boundary resp. closure of in SH. A e Proposition 7.7. Let j j J be a basal family of precells in H and let {A | ∈ } b j j J be the set of corresponding precells in SH. Then there is a fundamental {A | ∈ } set for Γ in SH such that fF j vc j . (4) e A ⊆ F ⊆ A j J j J [∈ [∈ f e f SYMBOLIC DYNAMICS 2203
Moreover, pr( ) = j J j . If is a fundamental set for Γ in SH such that F ∈ A F j J vc( j ), thenS satisfies (4). Conversely, if j j J is a set, F ⊆ ∈ Ae F e {A | ∈ } H indexedS by J, of precells in S such that (4) holds for some fundamental set for e f e e F Γ in SH, then the family pr( j ) j J is a basal family of precells in H. { A | ∈ } e Proof. This is a slight extension of a similar statement in [21]. e Remark 7.8. Recall from Theorem 4.8 that each basal family of precells in H contains the same finite number of precells, say m. Proposition 7.7 shows that if k k K is a set of precells in SH, indexed by K, such that (4) holds for some {A | ∈ } fundamental set for Γ in SH, then #K = m. e F Definition 7.9. Let A := j j J be a basal family of precells in H. Then e {A | ∈ } the set A := j j J of corresponding precells in SH is called a basal family of precells in {SAH or| a∈family} of basal precells in SH. If A is a connected family of basal precellse infH, then A is said to be a connected family of basal precells in SH or a connected basal family of precells in SH. e Let A := j j J be a basal family of precells in H and let A := j j J be the corresponding{A | ∈ basal} family of precells in SH. {A | ∈ } e f Definition and Remark 7.10. We call two cycles c ,c in A Γ equivalent if 1 2 × there exists a basal precell A and elements g1,g2 ΓrΓ such that ( ,g1) is A ∈ ∈ ∞ A an element of c1 and ( ,g2) is an element of c2. Obviously, equivalence of cycles is an equivalence relationA (on the set of all cycles). If [c] is an equivalence class of cycles in A Γ, then each element ( ,h ) A Γ in any representative c of [c] is called a generator× of [c]. A A ∈ × Lemma 7.11. Let be a non-cuspidal basal precell in H and suppose that h is A A an element in ΓrΓ assigned to by Proposition 4.11. Let ( j ,hj) be ∞ A A j=1,...,k the cycle in A Γ determined by ( ,h ). If = l for some l 2,...,k , then × A A A A ∈{ } hl = h . Moreover, if A q := min l 1,...,k 1 l = ∈{ − } A +1 A exists, then q does not depend on the choice of h , k is a multiple of q, and A ( l q,hl q) = ( l,hl) for l 1,...,k q . A + + A ∈{ − } Proof. This is an obvious consequence of the definition of cycles. Definition 7.12. Let be a non-cuspidal basal precell in H and let h be an A A element in ΓrΓ assigned to by Proposition 4.11. Suppose that ( j ,hj ) ∞ A A j=1,...,k is the cycle in A Γ determined by ( ,h ). We set × A A cyl( ) := min l 1,...,k 1 l = k . A ∈{ − } A +1 A ∪{ } Lemma 7.11 shows that cyl( ) is well-defined. Moreover, it implies that cyl( ) does not depend on the choiceA of the generator ( ,h ) of an equivalence classA of cycles. For a cuspidal basal precell in H we setA A A cyl( ) :=3, A and for a basal strip precell in H we define A cyl( ) :=2. A 2204 ANKE D. POHL
Example 7.13. Recall Example 5.5. For the Hecke triangle group Gn and its basal precell in H we have = and hence cyl( ) = 1. In contrast, the basal A A A2 A precell (v1) in H of the congruence group PΓ0(5) appears only once in the cycle in A PΓA (5) and therefore cyl (v ) = 3. × 0 A 1 Construction and Definition 7.14.Set := j J j . Pick a fundamental set F ∈ A for Γ in SH such that F S j vc j , e A ⊆ F ⊆ A j J j J [∈ [∈ which is possible by Propositionf7.7. Fore each basalf precell A and each z A ∈ ∈F let Ez( ) denote the set of unit tangent vectors in vc( ) based at z. Fix A F ∩ A any enumeration of the index set J of A, say J = j ,...,jk . For z and { 1 } ∈ F l e1,...,k set e e ∈{ } l 1 − z j := Ez j and z j := Ez j r Ez j . F A 1 A 1 F A l A l A m m=1 [ Further sete e e e e
( ) := z( ) F A F A z [∈A for A. Recall from Propositione 7.7 that pr(e )= . Thus, A ∈ F F z( )= , (5) F A eF z A [∈F A∈[ and the union is disjoint. For each equivalencee classe of cycles in A Γ fix a generator and let S denote the set of chosen generators. Let ( ,h ) S. × Suppose that is a non-cuspidal precell in H andA letA v∈ be the vertex of to A K which is attached. Let ( j ,hj ) be the cycle in A Γ determined by A A j=1,...,k × ( ,h ). For j = 1,...,k set g1 := id and gj+1 := hj gj, and let sj be the summit A A of I(hj ). Further, for convenience, set h0 := hk and s0 := sk. In the following we partition certain ( j ) into k subsets. More precisely, we partition each element F A of the set ( j ) j =1,...,k into k subsets. Let j 1,..., cyl( ) . {F A e| } ∈{ A } For each z j◦ (hj 1.sj 1,gj.v] [gj .v, sj ) we pick any partition of z( j ) − − e ∈ A ∪ (1)∪ (k) F A into k non-empty disjoint subsets Wj,z ,...,Wj,z . (1) (2) (k) e For z [sj , ) we set Wj,z := z( j ) and Wj,z = ... = Wj,z := . ∈ ∞ F (1)A (2) ∅ (3) For z [hj 1.sj 1, ) we set Wj,z := , Wj,z := z( j ) and Wj,z = ... = ∈ − − ∞ e ∅ F A W (k) := . j,z ∅ For m 1,...,k and j 1,..., cyl( ) we set e ∈{ } ∈{ A } (m) j,m := W A j,z z j ∈A[ and e k 1 j ( ,h ) := gjgl− . l,l j+1 B A A A − l [=1 e e where the first part (l) of the subscript of l,l j+1 is calculated modulo cyl( ) and the second part (l j + 1) is calculated moduloA − k. A − e SYMBOLIC DYNAMICS 2205
Suppose that is a cuspidal precell in H. Let ( ,h ), ( ,h ) be the cycle A A1 1 A2 2 in A Γ determined by ( ,h ). Set g := h1 = h , g1 := id and g2 := h1 = g. Suppose× that v is the vertexA ofA to which A is attached, A and let s be the summit K of I(g). Let j 1, 2 . We partition ( j ) into three subsets as follows. ∈{ } F A For z j◦ (gj .v, gj.s) we pick any partition of z( j ) into three non-empty ∈ A ∪ (1) (2) (3) e F A disjoint subsets Wj,z , Wj,z and Wj,z . (1) (2) e (3) For z (gj .v, ) we set Wj,z := z( j ) and Wj,z = Wj,z := . ∈ ∞ (1) F(2)A (3) ∅ For z [gj .s, ) we set W = W := and W := z( j ). ∈ ∞ j,z j,z ∅ j,z F A For m 1, 2, 3 and j 1, 2 wee set ∈{ } ∈{ } (m) e j,m := W . A j,z z j ∈A[ Then we define e 1 1( ,h ) := 1,1 g− . 2,2, B A A A ∪ A 2( ,h ) := g. 1,2 2,1, Be A A eA ∪ A e 1 3( ,h ) := 1,3 g− . 2,3. Be A A A e ∪ e A Suppose that is a strip precell in H. Let v1, v2 be the two (infinite) vertices of A e e e to which is attached and suppose that v1 < v2. We partition ( ) into two Ksubsets as follows.A F A For z ◦ we pick any partition of z( ) into two non-empty disjointe subsets (1) ∈ A(2) F A Wz and Wz . (1) (2) For z (v1, ) we set Wz := z( e) and Wz := . For z (v2, ) we set (1) ∈ ∞(2) F A ∅ ∈ ∞ Wz := and Wz := z( ). For m∅ 1, 2 we defineF A e ∈{ } e (1) (2) 1( ,h ) := Wz and 2( ,h ) := Wz . B A A B A A z z [∈A [∈A The set S (“selection”)e of generators of the equivalencee classes of cycles in A Γ is called a set of choices associated to A. The family ×
BS := j( ,h ) ( ,h ) S, j =1,..., cyl( ) B A A A A ∈ A is called the family of cellsn in SH associated to A and S. The elementso in this family e e are subject to the choice of the fundamental set , the enumeration of J and some choices for partitions in unit tangent bundle. However,F all further applications of BS are invariant under these choices. This justifiese to call BS the family of cells in SH associated to A and S. Each element in BS is called a cell in SH. e e Example 7.15. For the Hecke triangle group G from Example 5.5 we choose e 5 S = ( ,U ) . Here we have k = 5 and cyl( ) = 1. The first figure in Figure 15 { A 5 } A indicates a possible partition of ( ) into the sets , , , ,..., , . The unit F A A1 1 A1 2 A1 5 tangent vectors in black belong to , , those in very light grey to , , those in A1 1 A1 2 light grey to , , those in middlee grey to , , and thosee ine dark greye to , . The A1 3 A1 4 A1 4 second figure in Figure 15 shows thee cell 1( ,U5) in SH. e e Be A e Example 7.16. For the group Γ from Examplee 3.1(iii) we choose as set of choices S = ( , id), ( ,S) (cf. Example 5.5). The cells in SH which arise from ( , id) { A1 A2 } A1 2206 ANKE D. POHL
( ) ( ,U ) F A B1 A 5 e e
Figure 15. A partition of ( ) and the cell ( ,U ) in SH. F A B1 A 5 e e ( , id) ( , id) B1 A1 B2 A1 e e
v1 v2 v1 v2
Figure 16. The cells in SH arising from ( , id). A1 are shown in Figure 16. The cells in SH arising from ( 2,S) are indicated in Figure 17. A
( ,S) ( ,S) ( ,S) B1 A2 B2 A2 B3 A2 e e e
v2 0 0 v3 v2 0
Figure 17. The cells in SH arising from ( , id). A1
Let S be a set of choices associated to A.
Proposition 7.17. The union e e is disjoint and a fundamental set for Γ in BS B SH. B∈ S e Proof. Construction 7.14 picks a fundamental set for Γ in SH and chooses a F family := z( ) z , A of subsets of it. Since the union in (5) is P {F A | ∈ F A ∈ } disjoint, is a partition of . Recall the notatione from Construction 7.14. One P e F e SYMBOLIC DYNAMICS 2207 considers the family
1 := z( j ) z , ( ,h ) S, j =1,..., cyl( ) . P F A ∈F A A ∈ A The elements of nare pairwise disjoint and each element of is containedo in . P1 e P P1 Hence, is a partition of . The next step is to partition each element of into P1 F P1 a finite number of subsets. Thus, is partitioned into some family of subsets F P2 of . Then each element We of is translated by some element g(W ) in Γ to get F P2 the family 3 := g(W ).W W e2 . Since is a fundamental set for Γ in SH, thee elementsP of { are pairwise| disjoint∈ P } and F is a fundamental set for Γ in SH. P3 P3 Now 3 is partitioned into certain subsets, saye into the subsets l, l L. Each P S Q ∈ e cell in SH is the union of the elements in some l( ) such that l( 1) = l( 2) if B Q B B 6 B 1 = 2. Therefore, the union BS is disjoint and a fundamental set Bfor6 Γe inB SH. B B ∈ e e S e e e e e For each BS set b( ) := pr( ) ∂ pr( ) and let CS0( ) be the set of unit B ∈ B B ∩ B B tangent vectors in that are based on b( ) but do not point along ∂ pr( ). e e B e e B e e B Example 7.18. Recall the congruence subgroup PΓ0(5) and its cycles in A PΓ0(5) e e 1 e× from Example 5.5. We choose S := (v4),h− , (v1),h1 as set of choices associated to A and set A A 1 := (v ),h− , := (v ),h , B1 B1 A 4 B4 B1 A 1 1 1 2 := 2 (v4),h− , 5 := 2 (v1),h1, Be Be A Be Be A 1 3 := 3 (v4),h− , 6 := 3 (v1),h1, Be Be A Be Be A as well as e e e e CS0 := CS0 j j B for j =1,..., 6. Figure 18 shows the sets CS j0 . e
CS20 CS40 CS60 CS50 CS30 CS10
0 1 2 3 4 1 5 5 5 5
Figure 18. The sets CSj0 .
Lemma 7.19. Let , be two basal precells in H and let g Γ such that A1 A2 ∈ g. vc( ) vc( ) = . A1 ∩ A2 6 ∅ Suppose that 1 = 2 or g = id. Then A 6 A 6 f f g. vc vc g. vb vb . A1 ∩ A2 ⊆ A1 ∩ A2 Moreover, suppose that is cuspidal or non-cuspidal and that there is a unit A1 tangent vector w vc( f) pointingf into a non-verticalf sidef S of such that ∈ A1 1 A1 gw vc( 2). Then g.w points into a non-vertical side S2 of 2 and g.S1 = S2. ∈ A f A f 2208 ANKE D. POHL
Proof. We have pr(g. vc( )) = g. pr(vc( )) = g. and pr(g. vb( )) = g.∂ , A1 A1 A1 A1 A1 and likewise for pr(vc( 2)) = 2 and pr(vb( 2)) = ∂ 2. From A f A f A A f g. vc( ) vc( ) = f A1 ∩ Af2 6 ∅ then follows that g. 1 2 = . By Proposition 4.13, either g. 1 = 2 and g Γ A ∩A 6 ∅ f f A A ∈ ∞ or g. 1 2 g.∂ 1 ∂ 2. Assume on the contrary that g. 1 = 2 with g Γ . SinceA ∩Aand⊆ areA ∩ basal,A Corollary 4.10 shows that g = idA andA = .∈ This∞ A1 A2 A1 A2 contradicts the hypotheses of the lemma. Hence g. 1 2 g.∂ 1 ∂ 2 and therefore A ∩ A ⊆ A ∩ A g. vc( ) vc( ) g. vb( ) vb( ). A1 ∩ A2 ⊆ A1 ∩ A2 Let 1 and w fulfill the assumptions of the Lemma. Further let γ be the geodesic determinedA by w. Then g.γfis the geodesicf determinedf byfg.w. By definition there exists ε> 0 such that γ((0,ε)) S and g.γ((0,ε)) . Then ⊆ 1 ⊆ A2 1 1 γ (0,ε) S g− . g− . . ⊆ 1 ∩ A2 ⊆ A1 ∩ A2 1 1 Since the sets 1 and g− . 2 intersect in more than one point and 1 = g− . 2, A A 1 A 6 1 A Proposition 4.13 states that g− . is a common side of and g− . . A1 ∩ A2 A1 A2 Necessarily, this side is S1. According to Proposition 4.13, g.S1 is a non-vertical side of . Thus, g.w points along the non-vertical side g.S of . A2 1 A2 Proposition 7.20. Let ( ,h ) S and suppose that is a non-cuspidal precell A A ∈ A in H. Let ( j ,hj ) be the cycle in A Γ determined by ( ,h ). For each A j=1,...,k × A A 1 m =1,..., cyl( ) we have b( m( ,h )) = (hm− . , ) and A B A A ∞ ∞ 1 pr m( ,h ) = ( m)◦ (hm− . , ). B Ae A B A ∪ ∞ ∞ 1 Moreover, CS0( m( ,h e)) is the set of unit tangent vectors based on (hm− . , ) B A A ∞ ∞ that point into ( m)◦, and pr(CS0( m( ,h ))) = b( m( ,h )). Be A B A A B A A Proof. We use the notation from Construction 7.14. Let j 1,..., cyl( ) and e e ∈ { A } z j . First we show that z( j ) = . For each choice of we have j ∈ A F A 6 ∅ F A ⊆ vc( j ). Remark 7.3 states that pr( j ) = j . Hence Ez( j ) j = . More F ∩ A e A A A e∩ A 6 ∅ f precisely, if j z denotes the set of unit tangent vectors based on z that point into e f A f e e ◦, then j = Ez( j ) j . The set j is non-empty, since j is convex Aj A z A ∩ A A z A with non-emptyf interior. Let k J such that k = j . Then ∈ A 6 A f e f f Ez( k) Ez( j ) vc k vc j vb k vb j , A ∩ A ⊆ A ∩ A ⊆ A ∩ A where the laste inclusione follows fromf Lemmaf7.19. Sincef j vb(f j ) = by Lemma 7.6, it follows that A ∩ A ∅ f f j Ez( k)= j Ez( j ) Ez( k) j vb j = . A ∩ A A ∩ A ∩ A ⊆ A ∩ A ∅ Hence f e f e e f f j z( j ). (6) A z ⊆ F A Let j 1,..., cyl( ) set j := j( ,h ) and ∈{ A } B Bf A Ae
Tj := j◦ (hj 1.sj 1,gj .v] [gj .v, sj ). Ae ∪ e − − ∪ SYMBOLIC DYNAMICS 2209
Let m 1,...,k . Then ∈{ } (m) pr j,m = pr W A j,z z j ∈A[ e = pr W (m) pr W (m) pr W (m) j,z ∪ j,z ∪ j,z z Tj z [sj , ) z [hj− .sj− , ) [∈ ∈ [∞ ∈ 1[ 1 ∞ Tj for m / 1, 2 , ∈{ } = Tj [sj , ) for m = 1, ∪ ∞ Tj [hj 1.sj 1, ) for m = 2. ∪ − − ∞ Note that necessarily k 3. Then ≥ k 1 j = gjgl− . l,l j+1 B A − l [=1 j 1 k e − e 1 1 1 1 = gj gl− . l,l j+1 gj gj− . j,1 gj gj−+1. j+1,2 gj gl− . l,l j+1. A − ∪ A ∪ A ∪ A − l l j [=1 =[+2 Since l j +1 1,e2 mod k for l e1,...,j 1 e j +2,...,k , it followse that − 6≡ ∈{ − }∪{ } j 1 − 1 1 pr j = gjgl− . pr l,l j+1 pr j,1 hj− . pr j+1,2 B A − ∪ A ∪ A l=1 [ e k e e e 1 gj gl− . pr l,l j+1 ∪ A − l=j+2 [ j 1 e k − 1 1 1 1 = gjg− .Tl Tj [sj , ) h− .Tj h− .[hj .sj , ) gjg− .Tl l ∪ ∪ ∞ ∪ j +1 ∪ j ∞ ∪ l l l j [=1 =[+2 k 1 1 = gj g− .Tl (h− . , ). l ∪ j ∞ ∞ l [=1 1 For the last equality we use that pr (sj ) = hj− . by Lemma 4.7. Hence sj is ∞ 1 1 ∞ 1 contained in the geodesic segment pr− (hj− . ) H = (hj− . , ), which shows ∞ ∞ ∩ ∞ ∞1 that the union of the two geodesic segments [sj , ) and [sj ,hj− . ) is indeed 1 ∞ ∞ (h− . , ). Proposition 5.10 implies that j ∞ ∞ 1 pr j = ( j )◦ (h− . , ). B B A ∪ j ∞ ∞ 1 This shows that b( j ) = (h− . , ). The set of unit tangent vectors in j based B j e∞ ∞ B on b( j) is the disjoint union B e e (1) 1 (2) D0 := W h− . W e j j,z ∪ j j+1,z z [sj , ) z [hj .sj , ) ∈ [∞ ∈ [ ∞ 1 = z( j ) h− . z( j ). F A ∪ j F A +1 z [sj , ) z [hj .sj , ) ∈ [∞ ∈ [ ∞ e e To show that CS0( j ) is the set of unit tangent vectors based on b( j ) that point B B into ( j)◦ we have to show that Dj0 contains all unit tangent vectors based on B A 1 [sj , ) that point intoe ◦, all unit tangent vectors based on (h− . ,se j ] that point ∞ Aj j ∞ 2210 ANKE D. POHL
1 into h− . ◦ and the unit tangent vector which is based at sj and points into j Aj+1 [sj ,gjv]. If w is a unit tangent vector based on [hj .sj , ) that points into j◦+1, 1 ∞ 1 A then, clearly, hj− .w is a unit tangent vector based on [sj ,hj− . ) that points into 1 ∞ h− . ◦ . Hence, (6) shows that D0 contains all unit tangent vectors of the first j Aj+1 j two kinds mentioned above. Let w be the unit tangent vector with pr(w) = sj which points into [sj ,gj.v]. Suppose first that w . Then w vc( j ) and therefore w Es ( j ). ∈ F ∈ A ∩ F ∈ j A Let k J with k = j . Assume on the contrary that w vc( k). Lemma 7.19 ∈ A 6 A ∈ A implies that [sj ,gj .v] is a non-verticale side off k, whiche is a contradiction.e Hence A w / Es ( k). Therefore, w s ( j ) and hence w D0 . f ∈ j A ∈ F j A ∈ j Suppose now that w / . Then there exists a unique g Γr id such that ∈ F ∈ { } gw e . Let be a basal precelle in H such that g.w vc( ) . Lemma 7.19 ∈ F A ∈ A 1∩ F shows that g.[sj ,gj.v] is a non-verticale side S of . Thus, g− . ( j)◦ = . A A∩B A 16 ∅ By Propositione 5.10(iv) there is a unique l 1,...,k suche thateg = glg− and ∈ { } j = l. Now [sj ,gj.v] is mapped by hj to the non-vertical side [hj .sj,gj .v] of A A +1 j . Thus, g = hj and = j . Then hj .w vc( ]j ). As before we see that A +1 A A +1 ∈ A +1 w D0 . Moreover, pr(CS0( j)) = b( j). ∈ j B B Analogously to Proposition 7.20 one proves the following two propositions. e e Proposition 7.21. Let ( ,h ) S and suppose that is a cuspidal precell in H. A A ∈ A 1 Let v be the vertex of to which is attached and let ( ,g), ( 0,g− ) be the cycle in A Γ determinedK by ( ,h A). Then A A × A A 1 b 1( ,h ) = (v, ), b 2( ,h ) = (g.v, ), b 3( ,h ) = (g− . , ) B A A ∞ B A A ∞ B A A ∞ ∞ and e e e pr 1( ,h ) = ( )◦ (v, ), B A A B A ∪ ∞ pr 2( ,h ) = ( 0)◦ (g.v, ), Be A A B A ∪ ∞ 1 pr 3( ,h ) = ( )◦ (g− . , ). Be A A B A ∪ ∞ ∞ Moreover, CS0( m( ,h )) is the set of unit tangent vectors based on b( m( ,h )) B A A e B A A that point into pr( m( ,h ))◦, and pr(CS0( m( ,h ))) = b( m( ,h )) for m = 1, 2, 3. e B A A B A A B A Ae e e e Proposition 7.22. Let ( ,h ) S and suppose that is a strip precell. Let v1, v2 be the two (infinite) verticesA ofA ∈to which is attachedA and suppose that v < v . K A 1 2 For m =1, 2 we have b( m( ,h )) = (vm, ) and B A A ∞ pr m( ,h ) = ( )◦ (vm, )= ◦ (vm, ). B Ae A B A ∪ ∞ A ∪ ∞ Moreover, CS0( m ( ,h )) is the set of unit tangent vectors based on (vm, ) that B eA A ∞ point into ( )◦, and pr(CS0( m( ,h ))) = b( m( ,h )). B A e B A A B A A Corollary 7.23. Let BS. Then := cl(pr( )) is a cell in H and b( ) a side B ∈ e B eB B of . Moreover, pr( )= ◦ b( ) and pr( )◦ = ◦. B B e Be∪ B B eB e The development of a symbolic dynamics for the geodesic flow on Y via the e e e family BS of cells in SH is based on the following properties of the cells in SH: It B uses that cl(pr( )) is a convex polyhedron of which each side is a complete geodesic segmente and thatB each side is the image under some element g Γ of thee complete ∈ e SYMBOLIC DYNAMICS 2211 geodesic segment b( 0) for some cell 0 in SH. It further uses that BS is a B B fundamental set for Γ in SH and that g. cl(pr( )) g Γ, BS is a tesselation { B | ∈ B ∈ } S of H. Moreover, onee needs that b( ) ise a vertical side of pr( ) and that CSe0( ) is B B B the set of unit tangent vectors based on b( ) thate point intoe pr(e )◦. It does not B B use that cl(pr( )) BS is thee set of all cells in H nor doese one need thate for { B | B ∈ } some cells 1, 2 BS one has cl(pr( 1)) =e (pr( 2)). This meanse that one has the freedomB toB performe ∈ e (horizontal)e translationsB B ofB single cells in SH by elements in Γ . Thee followinge e definition is motivatede by thise fact. We will see that in some situations∞ the family of shifted cells in SH will induce a symbolic dynamics which has a generating function for the future part while the symbolic dynamics that is constructed from the original family of cells in SH does not have one.
Definition 7.24. Each map T: BS Γ (“translation”) is called a shift map for → ∞ BS. The family BS,T :e= T . BS e B B B ∈ is called the family of cells in SH associated to A, S and T. Each element of BS,T is called a shifted cell in SHe. e e e e For each BS,T define b( ) := pr( ) ∂ pr( ) and let CS0( ) be the sete of B ∈ B B ∩ B B unit tangent vectors in that are based on b( ) but do not point along ∂ pr( ). e e B e e B e e B Let T be a shift map for BS. e e e Remark 7.25. The results of Propositions 7.17 and 7.20-7.22 remain true for BS T e , after the obvious changes. More precisely, the union BS,T is disjoint and a fun- damental set for Γ in SH, and if BS, then pr T( ). = T( ). pr( )e and B ∈ S B B B B b T( ). = T( ).b( ). Then CS0 T( ). is the set of unite tangent vectors based B B B B B B T T e e e e e e on ( ).b( ) that point into ( ). pr( )◦. eB e B e e B Be e 8. Geometrice e symbolic dynamics.e eLet Γ be a geometrically finite subgroup of PSL2(R) of which is a cuspidal point and which satisfies (A). Suppose that the set of relevant isometric∞ spheres is non-empty. Let A be a basal family of precells in H and denote the family of cells in H assigned to A by B. Suppose that S is a set of choices associated to A and let BS be the family of cells in SH associated to A and S. Fix a shift map T for BS and denote the family of cells in SH associated to A, S and T by BS,T. Recall the set CSe 0( ) for BS,T from Definition 7.24. We set e B B ∈ CS0 BS,eT := CS0 ande CS eBS,Te := π CS0 BS,T . e e B BS,T B∈[ e e c e e In Section 8.1, we will use the results from Section 7 to show that CS satisfies (C1) and hence is a cross section for the geodesic flow on Y w.r.t. certain measures µ. It will turn out that the measures µ are characterized by the conditionc that NIC (see Remark 6.5)be a µ-null set. 8.1. Geometric cross section. Recall the set BS from Section 6.
Lemma 8.1. Let BS,T. Then pr( ) is a convex polyhedron and ∂ pr( ) con- B ∈ B B sists of complete geodesic segments. Moreover, we have that pr( )◦ BS = and B ∩ ∅ ∂ pr( ) BS and pr(e e) BS = b( ) ande that b( ) is a connected componente of BS. B ⊆ B ∩ B B e e e e e 2212 ANKE D. POHL
Proposition 8.2. We have CS = CS(BS,T). Moreover, the union
CS0 BS,T = CS0 BS,T c c e B B ∈ [ is disjoint and CS0(BS,T) ise a set of representativese e foreCS.
Proof. We start by showing that CS(BS,T) CS. Let BS,T. Then there ex- e ⊆ cB ∈ ists a (unique) BS such that = T( ). . Lemma 8.1 shows that b( ) B1 ∈ B B1 B1 B1 is a connected component of BS.c Thee set CSc0( ) consistse e of unit tangent vec- B1 tors based on b(e ) whiche are not tangente toe it.e Therefore, CS0( ) CS. Nowe B1 B1 ⊆ b( ) = T( ). and CS0( ) = T( ). CS0( )e with T( ) Γ. Thus, we see B B1 B1 B B1 B1 B1 ∈ that π(b( )) eπ(BS) = BS and π(CS0( )) π(CS) = CS. Thise shows that B ⊆ B ⊆ CS(eBS,T) eCS.e e e e e ⊆ Conversely,e let v CS. Wec will show thate there is a unique c BS,T and a unique ∈ 1 B ∈ vc eCS0( ) suchc that π(v)= v. Pick any w π− (v). Remark 7.25 shows that the ∈ B ∈ set := b BcS,T is a fundamental set for Γ in SH.e Hencee there exists a P {B | B ∈ } 1 unique paire ( ,g) BS,T Γb such that v := g.w . Note that π− (CS) = CS. S B ∈ × ∈ B Thus, v CSe ande hencee pr(v) pr( ) BS. Lemma 8.1 shows that pr(v) b( ). ∈ 1 ∈ B ∩ ∈ B Therefore, v e π− (b(e )) . Since v CS, it doese not point along b(c). Hence ∈ B ∩ B ∈ B v does not point along ∂ pr( ), whiche shows that v CS0( ). This proves thate B ∈ B CS CS(BS,T). e e e ⊆ 1 To see the uniqueness of e and v suppose that w π−e(v). Let ( ,g ) B 1 ∈ B1 1 ∈ BcS,T cΓ bee the unique pair such that g1.w1 1. There exists a unique element × 1 ∈ B 1 h Γ such that h.w = w1.e Then g1hg− .v = g.w1 and v,g1hgb − .v e . Now ∈ 1 ∈ P e being a fundamental set shows that g hg− e= id, which proves that g .w = P 1 1 1 g h.w = g.w = v and = . This completes the proof. 1 B1 B Corollary 8.3. Let γ be a geodesic on Y which intersects CS in t. Then there is e e a unique geodesic γ on H which intersects CS0(BS,T) in t such that π(γ)= γ. b c Definition 8.4. Let γ be a geodesic on Y which intersects CS in γ (t ). If e 0 0 b s := min t>t0 γ0(t) CS b ∈ c b exists, we call s the first return timeof γ0(t0 ) and γ0(s) the next point of intersection b c of γ and CS. Let γ be a geodesic on H. If γ0(t) CS, then we say that γ intersects ∈ CS in t. If there is a sequence (tn)n Nb with limnb tn = and γ0(tn) CS for ∈ →∞ ∞ ∈ all n N,c then γ is said to intersect CS infinitely often in the future. Analogously, b ∈ if we find a sequence (tn)n N with limn tn = and γ0(tn) CS for all n N, then γ is said to intersect ∈CS infinitely→∞ often in the−∞ past. Suppose∈ that γ intersects∈ CS in t0. If s := min t>t γ0(t) CS 0 ∈ exists, we call s the first return timeof γ0(t0 ) and γ0(s) the next point of intersection of γ and CS. Analogously, we define the previous point of intersection of γ and CS resp. of γ and CS. c Remark 8.5. A geodesic γ on Y intersects CS if and only if some (and henceb any) 1 1 representative of γ on H intersects π− (CS). Recall that CS = π− (CS), and that CS is the set of unit tangentb vectors based onc BS but which are not tangent to BS. Since BS is a totallyb geodesic submanifoldc of H (see Proposition 6.3),c a geodesic γ SYMBOLIC DYNAMICS 2213 on H intersects CS if and only if γ intersects BS transversely. Again because BS is totally geodesic, the geodesic γ intersects BS transversely if and only if γ intersects BS and is not contained in BS. Therefore, a geodesic γ on Y intersects CS if and only if some (and thus any) representative γ of γ on H intersects BS and γ(R) BS. 6⊆ A similar argument simplifies the search for previousb and next points ofc intersec- tion. To make this precise, suppose that γ is ab geodesic on Y which intersects CS in γ0(t ) and that γ is a representative of γ on H. Then γ0(t ) CS. There is a next 0 0 ∈ point of intersection of γ and CS if and onlyb if there is a next point of intersectionc ofb γ and CS. The hypothesis that γ0(t0)b CS implies that γ(R) is not contained in ∈ BS. Hence each intersectionb ofcγ and BS is transversal. Then there is a next point of intersection of γ and CS if and only if γ((t , )) intersects BS. Suppose that 0 ∞ there is a next point of intersection. If γ0(s) is the next point of intersection of γ and CS, then and only then γ0(s) is the next point of intersection of γ and CS. In this case, s = min t>t γ(t) BS . { 0 | ∈ } Likewise, there was a previousb point of intersection of γ and CS ifb and onlyc if there was a previous point of intersection of γ and CS. Further, there was a previous point of intersection of γ and CS if and only if γ(( ,t )) intersectsc BS. If there −∞ 0 b was a previous point of intersection, then γ0(s) is the previous point of intersection of γ and CS if and only if γ0(s) was the previous point of intersection of γ and CS. Moreover, s = max tDefinition 8.7. Let BS,T and suppose that b( ) is the complete geodesic B ∈ B segment (a, ) with a R. We assign to two intervals I( ) and J( ) which are given as follows:∞ ∈e e B e B B e e e (a, ) if pr( ) z H Re z a , I := ∞ B ⊆{ ∈ | ≥ } B (( ,a) if pr( ) z H Re z a , −∞ Be ⊆{ ∈ | ≤ } e and e ( ,a) if pr( ) z H Re z a , J := −∞ B ⊆{ ∈ | ≥ } B ((a, ) if pr( ) z H Re z a . ∞ Be ⊆{ ∈ | ≤ } e We note that the combination of Remarke7.25 with Propositions 5.10(i) and 7.20, Propositions 5.11(i) and 7.21 resp. Proposition 7.22 shows that indeed each BS,T B ∈ gets assigned a pair I( ),J( ) of intervals. B B e e e e 2214 ANKE D. POHL
Lemma 8.8. Let BS,T. For each v CS0( ) let γv denote the geodesic on H B ∈ ∈ B determined by v. If v CS0( ), then (γv( ),γv( )) I( ) J( ). Conversely, ∈ B ∞ −∞ ∈ B × B if (x, y) I( ) Je( ),e then there exists a uniquee element v in CS0( ) such that ∈ B × B B (γv( ),γv( )) = (x, y). e e e ∞ −∞ e e e Let BS,T and g Γ. Suppose that I( ) = (a, ). Then B ∈ ∈ B ∞ (g.a, g. ) if g.a < g. , e e g.I = ∞ e ∞ B ((g.a, ] ( , g. ) if g. < g.a, ∞ ∪ −∞ ∞ ∞ and e (g.a, ] ( , g. ) if g.a < g. , g.J = ∞ ∪ −∞ ∞ ∞ B ((g.a, g. ) if g. < g.a. ∞ ∞ Here, the interval (b,e ] denotes the union of the interval (b, ) with the point ∞ ∞ ∂gH. Hence, the set I := (b, ] ( ,c) is connected as a subset of ∂gH. ∞The ∈ interpretation of I is more eluminating∞ ∪ −∞ in the ball model: Via the Cayley 1 transform the set ∂gH is homeomorphic to the unit sphere S . Let b0 := (b), C 1 C c0 := (c) and I0 := (I). Then I0 is the connected component of S r b0,c0 which containsC ( ). C { } C ∞ Suppose now that I( ) = ( ,a). Then B −∞ ( , g.a) (g.( ), ] if g.a < g.( ), g.I = e −∞ ∪ −∞ ∞ −∞ B ((g.( ), g.a) if g.( ) < g.a, −∞ −∞ and e (g.( ), g.a) if g.a < g.( ), g.J = −∞ −∞ B (( , g.a) (g.( ), ] if g.( ) < g.a. −∞ ∪ −∞ ∞ −∞ eα β Note that for g = γ δ we have h i αt + β α + βs α + βs g.( ) = lim = lim = lim = g. . −∞ t γt + δ s 0 γ + δs s 0 γ + δs ∞ &−∞ % & In particular, id .( )= . −∞ ∞ Let a,b R and set (a,b) := (min(a,b), max(a,b)) and ∈ + (a,b) := (max(a,b), ] ( , min(a,b)). − ∞ ∪ −∞ Proposition 8.9. Let BS,T and suppose that S is a side of pr( ). Then there B ∈ B exist exactly two pairs ( ,g ), ( ,g ) BS,T Γ such that S = gj .b( j). Moreover, B1 1 B2 2 ∈ × B g . cl(pr( )) = cl(pr( e)) ande g . cl(pr( )) cl(pr( )) = S or vicee versa. The 1 B1 B 2 B2 ∩ B union g . CS0( ) g .eCS0( )eis disjointe and equals the set of alle unit tangent 1 B1 ∪ 2 B2 vectors ine CS that aree based on S. Let ea,b ∂gH bee the endpoints of S. Then ∈ g1.I( 1) g1.Je( 1) = (a,b)+e (a,b) and g2.I( 2) g2.J( 2) = (a,b) (a,b)+ or viceB versa.× B × − B × B − × e e e e Proof. Let D0 denote the set of unit tangent vectors in CS that are based on S. By Lemma 8.1, S is a connected component of BS. Hence D0 is the set of unit tangent vectors based on S but not tangent to S. The complete geodesic segment S divides H into two open half-spaces H1,H2 such that H is the disjoint union H S H . Moreover, pr( )◦ is contained in H or H , say pr( )◦ H . Then 1 ∪ ∪ 2 B 1 2 B ⊆ 1 e e SYMBOLIC DYNAMICS 2215
D0 decomposes into the disjoint union D0 D0 where D0 denotes the non-empty 1 ∪ 2 j set of elements in D0 that point into Hj . For j =1, 2 pick vj D0 . Since CS0(BS,T) ∈ j is a set of representatives for CS = π(CS) (see Proposition 8.2), there exists a e uniquely determined pair ( j ,gj) BS,T Γ such that vj gj. CS0( j). We will B ∈ × ∈ B show that S = gj .b( j). Assumec on the contrary that S = gj .b( j). Since S and B 6 B gj.b( j ) are complete geodesice segments,e the intersection of S and gj .b(e j) in pr(vj ) B B is transversal. Recalle that S ∂ pr( ) and b( j ) ∂ pr( j ) ande that ∂ pr( 0)◦ = ⊆ B B ⊆ B B ∂ pr(e0) for each 0 BS,T. Then there exists ε> 0 such that Bε(pr(vj ))e pr( )◦ = B B ∈ ∩ B Bε(pr(vj )) H and e e e e ∩ 1 e e e e Bε pr(vj ) gj . pr j ◦ H = . ∩ B ∩ 1 6 ∅ Hence pr( )◦ gj . pr( j )◦ = . Proposition 5.14 in combination with Remark 7.25 B ∩ B 6 ∅ e shows that cl(pr( )) = gj . cl(pr( j )). But then e B e B ∂ pr( )= gj .∂ pr( j ), e e B B which implies that S = gj .b( j). This is a contradiction, and hence S = gj .b( j). B e e B Then Lemma 8.8 implies that gj .I( j ) gj .J( j ) equals (a,b)+ (a,b) or (a,b) B × B × − − × (a,b)+. On the other hand e e e e ∂gH ∂gH = γv( ),γv( ) v D0 = γv( ),γv( ) v D0 1 × 2 ∞ −∞ ∈ 1 −∞ ∞ ∈ 2 equals (a,b)+ (a,b ) or (a,b) (a,b )+. Therefore, again by Lemma 8.8, we have × − − × gj. CS0( j ) = D0 . This shows that the union g . CS0( ) g . CS0( ) is disjoint B j 1 B1 ∪ 2 B2 and equals D0. We havee cl(pr( )) H1 and g1. cl(pr( 1)) H1 withe S ∂ pr( )e g1.∂ pr( 1). Let z S. Then thereB ⊆ exists ε> 0 such thatB ⊆ ⊆ B ∩ B ∈ e e e e Bε(z) pr ◦ = Bε(z) H = Bε(z) g . pr ◦. ∩ B ∩ 1 ∩ 1 B1 Hence pr( )◦ g . pr( ) ◦ = . As above we find that cl(pr( )) = g . cl(pr( )). B ∩ 1 B1 e6 ∅ Be 1 B1 Finally, g2. cl(pr( 2)) H2 with e B ⊆e e e S g . cl pr H H H = S. e ⊆ 2 B2 ∩ 1 ⊆ 2 ∩ 1 Hence cl(pr( )) g . cl(pr( )) = S. B ∩ 2 B2 e Let BS,T and suppose that S is a side of pr( ). Let ( ,g ), ( ,g ) be the B ∈ e e B B1 1 B2 2 two elements in BS,T Γ such that S = gj .b( j) and g . cl(pr( )) = cl(pr( )) and × B 1 B1 B g2. cl(pr(e 2))e cl(pr( )) = S. Then we define e e e B ∩ e B e e e p ,S := ,g and n ,S := ,g . e Be B1 1 B B2 2
Remark 8.10. Let BS,T and suppose that S is a side of pr( ). We will show eB ∈ e e e B how one effectively finds the elements p( ,S) and n( ,S). Let e e B B e , k := p ,S and , k := n ,S . B1 1 B e B2 2 e B Suppose that is the (unique) element in BS such that T ( ). = and suppose 0 e e e e0 0 B B B B 1 further that j0 BS such that T( j0 ). j0 = j for j = 1, 2. Set S0 := T( 0)− .S. Be ∈ B B eB e e e B Then S0 is a side of pr( 0). For j =1, 2 we have B e e 1 e 1 e e 1 e S0 = T 0 − .S = T 0 − kj .b j = T 0 − kj T 0 .b 0 B e B B B Bj Bj e e e e e e 2216 ANKE D. POHL and
kj . cl pr j = kj T 0 . cl pr 0 . B Bj Bj e e e Moreover, cl(pr( )) = T( 0). cl(pr( 0)). Then k1. cl(pr( 1)) = cl(pr( )) is equiva- lent to B B B B B e e e e e 1 T 0 − k T 0 . cl pr 0 = cl pr 0 , B 1 B1 B1 B e e e e and k . cl(pr( )) cl(pr( )) = S is equivalent to 2 B2 ∩ B
e 1 e T 0 − k T 0 . cl pr 0 cl pr 0 = S0. B 2 B2 B2 ∩ B e e e 1e Therefore, ( 1, k1) = p( ,S) if and only if ( 10 , T( 0)− k1T( 10 )) = p( 0,S0), and B B 1 B B B B ( , k )= n( ,S) if and only if ( 0 , T( 0)− k T( 0 )) = n( 0,S0). B2 2 B B2 B 2 B2 B By Corollarye 7.23, thee sets 0 := cl(pr( e0)) ande 0 := cl(pr(e 0 ))e are A-cells B B Bj Bj ine H. Supposee first that 0 arisese frome the non-cuspidale e basal precell 0 in H. Then there is a unique elementB ( ,h ) Se such that for some eh ΓA the pair A A ∈ ∈ ( 0,h) is contained in the cycle in A Γ determined by ( ,h ). Necessarily, A × A A is non-cuspidal. Let ( j ,hj ) be the cycle in A Γ determined by A A j=1,...,k × ( ,h ). Then 0 = m for some m 1,..., cyl( ) and hence 0 = ( m) A A A A ∈ { A } B B A and 0 = m( ,h ). For j = 1,...,k set g1 := id and gj+1 := hj gj . Propo- B B A A sition 5.10(iii) states that ( m) = gm. ( ) and Proposition 5.10(i) shows that B A 1 B A1 S0 ise the geodesice segment [gmgj− . ,gmgj−+1. ] for some j 1,...,k . Then 1 1 ∞ ∞ ∈ { } gjg− .S0 = [ ,h− . ]. Let n 1,..., cyl( ) such that n j mod cyl( ). m ∞ j ∞ ∈ { A } ≡ A Then hn = hj by Lemma 7.11. Proposition 7.20 shows that b( n( ,h )) = A 1 1 1 B A [ ,hn− . ]= gj gm− .S0. We claim that ( j ( ,h ),gmgj− )= p( 0,S0). For this is A ∞ ∞ 1 B A B e remains to show that gmgj− . cl(pr( j ( ,h ))) = cl(pr( 0)). Proposition 7.20 shows B A eA B e that cl(pr( j ( ,h ))) = ( n) and Lemma 7.11 implies that ( n)= ( j). Let B A A B A e e B A B A1 v be the vertex of to which is attached. Then gj .v ( j )◦ and gmgj− gj .v = e K A 1 ∈B A gm.v ( m)◦. Therefore gmgj− . ( j )◦ ( m)◦ = . From Proposition 5.14(1) ∈B A 1 B A ∩B A 6 ∅ it follows that gmgj− . ( j ) = ( m). Recall that ( m) = cl(pr( 0)). Hence 1 B A B A B A B ( j( ,h ),gmgj− )= p( 0,S0) and B A A B e
e e 1 1 − T j( ,h ) . j ,h , T 0 gmgj− T j( ,h ) = p ,S . B A A B A A B B A A B e e e e e Analogously one proceeds if 0 is cuspidal or a strip precell. A Now we show how one determines ( , k ). Suppose again that 0 arises from B2 2 B the non-cuspidal basal precell 0 in H. We use the notation from the determination A of p( 0,S0). By Corollary 4.10 there ise a unique pair ( ,s) A Z such that B s s s 1 A ∈ × b( n( ,h )) tλ. = and tλ. = n. Then tλ− gjgm− .S0 is a side of the cell ( ) A B eA ∩ A 6 ∅ A 6 A b s B1 A in H. As before, we determine ( 3, k3) BS Γ such that k3.b( 3)= tλ− gj gm− .S0 e b bB ∈ ×1 B b and k3. cl(pr( 3)) = ( ). Recall that gj gm− .S0 is a side of ( j) = ( n). We B B A e e B Ae B A e b SYMBOLIC DYNAMICS 2217 have 1 s 1 s gmg− t k . cl pr cl pr 0 = gmg− t . ( ) ( m) j λ 3 B3 ∩ B j λ B A ∩B A 1 s 1 = gmg− t . ( ) gmg− . ( j ) e e j λ B Ab ∩ j B A 1 s = gmg− . t . ( ) ( j) j λ BbA ∩B A 1 s = gmg− . t . ( ) ( n) j λ B Ab ∩B A 1 1 = gmgj− g jgm− .S0 b = S0.
1 s Thus n( 0,S0) = ( ,gmg− t k ) and B B3 j λ 3 1 s 1 n ,S = T . , T 0 gmg− t k T − . e eB B3 B3 B j λ 3 B3 H If 0 arises from a cuspidale or stripe precelle e in , then the constructione of n( ,S) is analogous.B B e Proposition 8.11. Let γ be a geodesic on Y and suppose that γ intersects CS in γ0(t ). Let γ be the unique geodesic on H such that γ0(t ) CS0(BS,T) and 0 0 ∈ π(γ0(t )) = γ0(t ). Let b BS,T be the unique shifted cell in SH forb which we havec 0 0 B ∈ γ0(tb ) CS0( ). e 0 ∈ B (i) Thereb is a next pointe ofe intersection of γ and CS if and only if γ( ) does not e ∞ belong to ∂g pr( ). B (ii) Suppose that γ( ) / ∂g pr( ). Then there is a unique side S of pr( ) in- ∞ ∈ B B tersected by γ((et , )). Suppose that ( ,g) = n( ,S). The next point of 0 ∞ B1 B intersection is on g. CS0( ).e e B1 (iii) Let ( 0,h)= n( ,b( )). Then there wase a previouse point of intersection of γ B B B and CS if and only if γ( e ) / h.∂g pr( 0). −∞ ∈ B (iv) Supposee that γ( e )e/ h.∂g pr( 0). Then there is a unique side S of h. pr( 0) −∞ ∈ B 1 1 B intersected by γ(( ,t )). Let ( ,h− ek) = p( 0,h− S). Then the previous −∞ 0 B2 B point of intersection was on k. CSe 0( 2). e e B e Proof. We start by proving (i). Recall from Remark 8.5 that there is a next point e of intersection of γ and CS if and only if γ((t , )) intersects BS. Since γ0(t ) 0 ∞ 0 ∈ CS0( ), Proposition 7.20 resp. 7.21 resp. 7.22 in combination with Remark 7.25 B shows that γ0(t ) points into pr( )◦. Lemma 8.1 states that pr( )◦ BS = and 0 B B ∩ ∅ ∂ pr(e) BS. Hence γ((t , )) does not intersect BS if and only if γ((t , )) B ⊆ 0 ∞ 0 ∞ ⊆ pr( )◦. In this case, e e B e γ( ) cl g (pr( )) ∂gH = ∂g pr( ). e ∞ ∈ H B ∩ B Conversely, if γ( ) ∂ pr( ), then γ((t , )) pr( ) or γ((t , )) ∂ pr( ). g e 0 ◦ e 0 In the latter case,∞ Lemma∈ 8.1B shows that∞γ((t⊆, ))B BS. Hence,∞ if⊆ γ( )B 0 ∞ ⊆ ∞ ∈ ∂g pr( ), then γ((t , )) pr(e )◦. e e B 0 ∞ ⊆ B Suppose now that γ( ) / ∂g pr( ). The previous argument shows that the ∞ ∈ B geodesice segment γ((t , )) intersectse ∂ pr( ), say γ(t ) ∂ pr( ) with t (t , ). 0 ∞ B 1 ∈ B 1 ∈ 0 ∞ If there was an element t (t , )r et with γ(t ) ∂ pr( ), then there is a side 2 ∈ 0 ∞ { 1} 2 ∈ B S of pr( ) such that γ(R) = S, where thee equality followse from the fact that S is a completeB geodesic segment. But then, by Lemma 8.1e , γ(R) BS, which ⊆ e 2218 ANKE D. POHL contradicts to γ0(t ) CS. Thus, γ(t ) is the only intersection point of ∂ pr( ) 0 ∈ 1 B and γ((t , )). Since γ((t ,t )) pr( )◦, γ0(t ) is the next point of intersection of 0 ∞ 0 1 ⊆ B 0 γ and CS. Moreover, γ0(t ) points out of pr( ), since otherwise γ((t , )) woulde 1 B 1 ∞ intersect ∂ pr( ) which would lead toe a contradiction as before. Proposition 8.2 B states that there is a unique pair ( ,g) BS,eT Γ such that γ0(t ) g. CS0( ). B1 ∈ × 1 ∈ B1 Then γ0(t ) pointse into g. pr( )◦. Let S be the side of pr( ) with γ(t ) S. By 1 B1 B 1 ∈ Proposition 8.9, either g. cl(pr( ))e = cl(pr(e)) or g. cl(pr( )) cl(pr( )) = S.e In B1 B B1 ∩ B the first case, γ0(t1) points intoe pr( )◦, which is a contradiction.e Therefore e B e e e g. cl(pr( )) cl(pr( )) = S, Be1 ∩ B which shows that ( ,g)= n( ,S). This completes the proof of (ii). B1 B e e Let ( 0,h)= n( ,b( )). Since γ(t ) b( ) and γ0(t ) CS0( ), Proposition 8.9 B B B 0 ∈ B 0 ∈ B implies that γ(t ) e h.b( 0) ande γ0(t ) / h. CS0( 0). Since γ(R) h.∂ pr( 0), the 0 ∈ B 0 ∈ B 6⊆ B unit tangente vectoreγ0(te) points out of pr( 0e). Because the intersectione of γ(R) and 0 B h.b( 0) is transversal ande pr( 0) is a convex polyhedrone with non-empty interior,e B B γ((t0 ε,t0)) h. pr( )◦ = for each εe > 0. As before we find that there was a previouse− point∩ of intersectionB 6 ∅e of γ and CS if and only if γ(( ,t )) intersects −∞ 0 ∂ pr( 0) and that thise is the case if and only if γ( ) / h.∂g pr( 0). B −∞ ∈ B Suppose that γ( ) / h.∂g pr( 0). As before, there is a unique t 1 ( ,t0) −∞ ∈ B − ∈ −∞ suche that γ(t 1) h.∂g pr( 0). Let S be the side of h. pr( 0) withe γ(t 1) S. − ∈ B B − ∈ Necessarily, γ((t 1,t0) h. pr( 0)◦e, which shows that γ0(t 1) points into h. pr( 0)◦ − ⊆ B − B and that γ0(t 1) is the previouse point of intersection. Let ( 2,e k) BS,T Γ be the − B ∈ × unique pair such that γ0(t 1) k.e CS0( 2) (see Proposition 8.2). By Proposition e8.9, − ∈ B we have either k. cl(pr( )) = h. cl(pr( 0)) or k. cl(pr( )) e h. cl(pr(e 0)) = S. In B2 B B2 ∩ B the latter case, γ0(t 1) points out of eh. pr( 0)◦ which is a contradiction. Hence 1 − B 1 1 h− k. cl(pr( 2) = cl(pr(e 0)), which showse that ( 2,h− ek)= p( 0,h− Se). B B e B B Corollary 8.12.e Let γ bee a geodesic on Y and supposee that γ doese not intersect CS infinitely often in the future. If γ intersects CS at all, then there exists t R such ∈ that γ0(t) CS and γb((t, )) BS = . Analogously, supposeb that η is a geodesicc ∈ ∞ ∩ ∅ on Y which does not intersect CSb infinitely oftenc in the past. If η intersects CS at all, thenb therec exists bt R such thatc η0(t) CS and η(( ,t)) BSb = . ∈ c ∈ −∞ ∩ ∅ c Example 8.13. Recall the setting of Example 7.18. We considerb the two shift c c maps T id, and b b 1 ≡ 1 1 T2 1 := − and T2 j := id for j =2,..., 6. B 0 1 B For simplicity sete 1 := T2 1 . 1 and CS0e1 := T2 1 . CS10 . Further we set B− B B − B 1 0 2 1 3 2 4 1 g := ,e g := e e , g := e , g := , 1 5 1 2 5 −2 3 5 −3 4 5 −1 − − − 4 5 1 1 1 0 g := , g := , g := . 5 5 −6 6 0 1 7 −5 1 − − Figure 19 shows the translates of the sets CSj0 which are necessary to determine the location of the next point of intersection if the shift map is T1, and Figure 20 those if T2 is the chosen shift map. SYMBOLIC DYNAMICS 2219
CS20 CS40 CS60 CS50 g4.CS40 CS30 CS10 g6.CS20
g1.CS20 g2.CS50 g3.CS30 g4.CS60 g5.CS10 0 1 2 3 4 1 5 5 5 5
Figure 19. The translates of CS0 relevant for determination of the location of the next point of intersection for the shift map T1.
g7.CS40 CS0 1CS20 CS40 CS60 CS50 CS30 g6.CS20 −
g7.CS0 1 g1.CS20 g2.CS50 g3.CS30 g4.CS60 g4.CS0 1 − − 1 0 1 2 3 4 1 − 5 5 5 5 5
Figure 20. The translates of CS0 relevant for determination of the location of the next point of intersection for the shift map T2.
Recall the set bd from Section 6. The following characterization is now obvious. Proposition 8.14. Let γ be a geodesic on Y . (i) γ intersects CS infinitely often in the future if and only if γ( ) / π(bd). ∞ ∈ (ii) γ intersects CS infinitelyb often in the past if and only if γ( ) / π(bd). −∞ ∈ b c b Recall the set NIC from Remark 6.5. b c b Theorem 8.15. Let µ be a measure on the space of geodesics on Y . Then CS is a cross section w. r. t. µ for the geodesic flow on Y if and only if µ(NIC) = 0. c Proof. Proposition 6.4 shows that CS satisfies (C2). Let γ be a geodesic on Y . Then Proposition 8.14 implies that γ intersects CS infinitely often in the past and the future if and only if γ / NIC. Thisc completes the proof.b ∈ c Let denote the set of unit tangentb vectors to the geodesics in NIC and set E b CS := CSr . st E Corollary 8.16. Let µ be a measure on the space of geodesics on Y such that c c µ(NIC) = 0. Then CSst is the maximal strong cross section w. r. t. µ contained in CS. c 8.2. Geometric coding sequences and geometric symbolic dynamics. A c label of a unit tangent vector in CS or CS is a symbol which is assigned to this vector. The labeling of CS resp. CS is the assigment of the labels to its elements. The set of labels is commonly calledc the alphabet of the arising symbolic dynamics. We establish a labelingc of CS and CS in the following way: Let v CS0(BS,T) ∈ and suppose that BS,T is the unique shifted cell in SH such that v CS0( ). Let γ be the geodesicB ∈ on H determinedc by v. ∈ e B e e e 2220 ANKE D. POHL
Suppose first that γ( ) / ∂g pr( ). Proposition 8.11(ii) states that there is a ∞ ∈ B unique side S of pr( ) intersected by γ((0, )) and that the next point of inter- B ∞ section of γ and CS is on g. CS0( )e if ( ,g)= n( ,S). We assign to v the label B1 B1 B ( ,g). e B1 Suppose now that γ( ) ∂g pr(e ). Propositione e8.11(i) shows that there is no nexte point of intersection∞ of∈γ and CS.B Let ε be an abstract symbol which is not contained in Γ. Then we label v by εe(“end” or “empty word”). Let v CS. By Proposition 8.2 there is a unique v CS0(BS,T) such that ∈ 1 ∈ π(v)= v. We endow v and each element in π− (v) with the label of v. Theb followingc proposition is the key result for the determinatione of the set of labels. b b b
Proposition 8.17. Let BS,T and suppose that Sj , j = 1,...,k, are the sides B ∈ of pr( ). For j =1,...,k set ( j ,gj) := n( ,Sj ). Let v CS0( ) and suppose that B B B ∈ B γ is the geodesic determinede bye v. Then γ( ) gj .I( j) if and only if γ((0, )) ∞ ∈ B ∞ intersectse Sj . Moreover, if Sj e= b( ), thene gj.I( j ) I( ).e If Sj = b( ), then 6 B B ⊆ B B gj.I( j )= J( ). e B B e e e e Proof. This follows from Propositions 8.9 and 8.11 and Lemma 8.8 with similar e e arguments as in the proof of Proposition 8.9.
For BS,T let Sides( ) denote the set of sides of pr( ). B ∈ B B Corollary 8.18. Let BS,T. For each S Sides( ) set ( S,gS) := n( ,S). e e B ∈e ∈ B e B B Then I( ) is the disjoint union B e e e e e e I( )= I( ) ∂g pr( ) gS.I( S). B B ∩ B ∪ e e B S Sides([ )rb( ) e e e ∈ B B e Let Σ denote the set of labels. Corollary 8.19. The set Σ of labels is given by Σ= ε { } ∪ ( ,g) BS,T Γ 0 BS,T S Sides( 0)rb( 0): ( ,g)= n( 0,S) . ∪ B ∈ × ∃ B ∈ ∃ ∈ B B B B n o Moreover, Σ is finite. e e e e e e e e Proof. Note that for each 0 BS,T we have ∂g pr( 0) I( 0) = . Thus, ε is a B ∈ B ∩ B 6 ∅ label. Then the equality for Σ follows immediately from Corollary 8.18. Since BS,T is finite and each shifted celle in SeH has only finitely manye sides,e Σ is finite. e Example 8.20. For the Hecke triangle group Gn let A = , S = ( ,Un) and {A} { A } T id. Then BS,T = 1( ,Un) . Set := 1( ,Un). Then the set of labels is (see≡ Figure 7) {B A } B B A e Σ=e ε, ,U kS ke 1e,...,n 1 . B n ∈{ − } n o Example 8.21. Recall Example 8.13. If the shift map is T1, then the set of labels e is Σ= ε, ,g , , id , ,g , , id , ,g , , id , ,g , , id , B2 1 B4 B5 2 B6 B3 3 B5 B6 4 B3 1,g5 , 2,g6 , 4,g4 . e e e e e Be Be Be e e e SYMBOLIC DYNAMICS 2221
With the shift map T2, the set of labels is
Σ= ε, 4,g7 , 1,g7 , 2,g1 , 4, id , 5,g2 , 6, id , 3,g3 , B B− B B B B B 5, id, 6,g4, 3, id, 1,g4 , 2,g6 . e e Be Be Be Be− eB Definition and Remark 8.22. Let v CS0( BS T) and suppose that γ is the e e e, e e geodesic on H determined by v. Proposition∈ 8.11 implies that there is a unique sequence (tn)n J in R which satisfies the followinge properties: ∈ (i) J = Z (a,b) for some interval (a,b) with a,b Z and 0 (a,b), ∩ ∈ ∪ {±∞} ∈ (ii) the sequence (tn)n J is increasing, ∈ (iii) t0 = 0, (iv) for each n J we have γ0(tn) CS and ∈ ∈ γ0((tn,tn+1)) CS = and γ0((tn 1,tn)) CS = ∩ ∅ − ∩ ∅ where we set tb := if b< and ta := if a> . ∞ ∞ −∞ −∞ The sequence (tn)n J is said to be the sequence of intersection times of v (with respect to CS). ∈ 1 − Let v CS and set v := π 0 Be (v). Then the sequence of intersection ∈ |CS ( S,T) 1 times (w. r. t. CS) of v and of each w π− (v) is defined to be the sequence of intersectionb c times of v. ∈ b Now we define the geometricb coding sequences.b Definition 8.23. For each s Σ we set ∈ CSs := v CS v is labeled with s ∈ and c b c b 1 CSs := π− CSs = v CS v is labeled with s . ∈ Let v CS and let (t ) be the sequence of intersection times of v. Suppose that n n J c γ is the∈ geodesic on Y determined∈ by v. The geometric coding sequence of v is the sequencec (an)n J in Σ defined by b ∈ b b an := s if andb only if γ0(tn) CSs b ∈ for each n J. Let v ∈CS. The geometric coding sequence bof v is definedc to be the geometric coding sequence∈ of π(v).
Proposition 8.24. Let v CS0. Suppose that (tn)n J is the sequence of intersec- ∈ ∈ tion times of v and that (an)n J is the geometric coding sequence of v. Let γ be the geodesic on H determined by v∈. Suppose that J = Z (a,b) with a,b Z . ∩ ∈ ∪ {±∞} (i) If b = , then an Σr ε for each n J. ∞ ∈ { } ∈ (ii) If b< , then an Σr ε for each n (a,b 2] Z and ab 1 = ε. ∞ ∈ { } ∈ − ∩ − (iii) Suppose that an = ( n,hn) for n (a,b 1) Z and set B ∈ − ∩ g0 := h0 if b 2, e ≥ gn := gnhn for n [0,b 2) Z, +1 +1 ∈ − ∩ g 1 := id, − 1 g (n+1) := g nh−n for n [1, (a + 1)) Z. − − − ∈ − ∩ Then γ0(tn ) gn. CS0( n) for each n (a,b 1) Z. +1 ∈ B ∈ − ∩ e 2222 ANKE D. POHL
Proof. We start with some preliminary considerations which will prove (i) and (ii) and simplify the argumentation for (iii). Let n J and consider w := γ0(tn). The ∈ definition of geometric coding sequences shows that γ0(tn) CSa . Since CS is ∈ n the disjoint union k Γ k. CS0(BS,T) (see Proposition 8.2), there is a unique k Γ 1 ∈ 1 ∈ such that k− .w CS0(BS,T). The label of k− .w is an. Let η be the geodesic on H ∈1S 1 determined by k− .w. Note thateη(t) := k− .γ(t+tn) for each t R. The definition ∈ of labels shows that an e= ε if and only if there is no next point of intersection of η and CS. In this case γ0((tn, )) CS = and hence b = n + 1. This shows (i) and ∞ ∩ ∅ (ii). Suppose now that an = ( ,g). Then there is a next point of intersection of η B 1 and CS, say η0(s), and this is on g. CS0( ). Then k− .γ0(s + tn) g. CS0( ) and 1 B ∈ B k− .γ0((tn,s + tn)) CS = . Hencee tn = s + tn and γ0(tn ) kg. CS0( ). ∩ ∅ +1 +1 ∈ B Now we show (iii). Suppose that b 2.e Then v = γ0(t ) is labeled with ( e ,h ). ≥ 0 B0 0 Hence for the next point of intersection γ0(t1) of γ and CS we have e e γ0(t ) h . CS0 = g . CS0 . 1 ∈ 0 B0 0 B0 Suppose that we have already shown that e e γ0(tn ) gn. CS0 n +1 ∈ B for some n [0,b 1) Z and that b n+3. By( i) resp. (ii), γ0((tn+1, )) CS = ∈ − ∩ ≥ e ∞ ∩ 6 ∅ and hence γ0(tn+1) is labeled with ( n+1,hn+1). Our preliminary considerations show that B γ0(tn ) gnhn . CSe0 n = gn . CS0 n . +2 ∈ +1 B +1 +1 B +1 Therefore γ0(tn+1) gn. CS0( n) for each n [0,b 1) Z . ∈ B e ∈ − ∩ e Suppose that a 2. The element γ0(t 1) is labeled with ( 1,h 1). Since ≤ − − B− − γ0(t 1) k. CS0(BS,T) for somee k Γ, our preliminary considerations show that − ∈ ∈ γ0(t0) kh 1. CS0( 1). Because γ0(t0) = v CS0(BS,T), Propositione 8.2 implies ∈ −1 e B− ∈ that k = h−1 and − e 1e γ0(t0) CS0 1 = g 1. CS0 1 and γ0(t 1) h−1. CS0 BS,T = g 2. CS0 BS,T . ∈ B− − B− − ∈ − − Suppose that we have already shown that e e e e γ0(t (n 1)) g n. CS0 n and γ0(t n) g (n+1). CS0 BS,T − − ∈ − B− − ∈ − for some n [1, a) Z and suppose that a n 2. Then γ0(t n 1) exists and ∈ − ∩ e ≤− − − −e is labeled with ( n 1,h n 1). Since γ0(t n 1) h. CS0(BS,T) for some h Γ, we B− − − − − − ∈ ∈ know that γ0(t n) hh n 1. CS0( n 1). Therefore − e ∈ − − B− − e γ0(t n) hh (n+1). CS0 (n+1) g (n+1). CS0 BS,T . − ∈ − e B− ∩ − Proposition 8.2 implies that hh (n+1) = g (n+1) , γ0(t n) g (n+1). CS0( (n+1)) − e − − −e − and ∈ B 1 S T S T γ0(t (n+1)) g (n+1)h−(n+1). CS0 B , = g (n+2). CS0 B , . e − ∈ − − − Therefore γ0(tn+1) gn. CS0( n) for each n (a, 1] Z. This completes the e e proof. ∈ B ∈ − ∩ e Let Λ denote the set of geometric coding sequences and let Λσ be the subset of Λ which contains the geometric coding sequences (an)n (a,b) Z with a,b Z ∈ ∩ ∈ ∪ {±∞} for which b 2. Let Σall denote the set of all finite and one- or two-sided infinite sequences in≥ Σ. The left shift σ : Σall Σall, → σ (an)n J := ak+1 for all k J ∈ k ∈ SYMBOLIC DYNAMICS 2223 induces a partially defined map σ :Λ Λ resp. a map σ :Λσ Λ. Suppose that → → Seq: CS Λ is the map which assigns to v CS the geometric coding sequence of v. Recall→ the first return map R from Section∈ 2.3. Propositionc 8.25. The diagram b c b R CS / CS
Seq Seq c c σ Λ / Λ commutes. In particular, for v CS, the element R(v) is defined if and only if ∈ Seq(v) Λσ. ∈ c Proof. This follows immediatelyb from the definition of geometricb coding sequences, Propositionsb 8.11 and 8.24.
1 Set CSst := π− (CSst) and CSst0 (BS,T) := CSst CS0(BS,T) and let Λst denote the set of two-sided infinite geometric coding sequences.∩ c e e Remark 8.26. The set of geometric coding sequences of elements in CSst (or only CS0 (BS,T)) is Λ . Moreover, Λ Λσ. st st st ⊆ In the following we will show that (Λ , σ) is a symbolic dynamics for the geodesic e st flow on Φ. Elementary convex geometry proves the following lemma. Lemma 8.27. Suppose that x, y ∂ Hrbd, x