Convergence of spherical averages for actions of Fuchsian Groups
CCONVERGENCE OF SPHERICAL AVERAGESFOR ACTIONS OF FUCHSIAN GROUPS
ALEXANDER I. BUFETOV, ALEXEY KLIMENKO, AND CAROLINE SERIES
Abstract.
Pointwise convergence of spherical averages is proved for a measure-preservingaction of a Fuchsian group. The proof is based on a new variant of the Bowen-Seriessymbolic coding for Fuchsian groups that, developing a method introduced by Wroten,simultaneously encodes all possible shortest paths representing a given group element.The resulting coding is self-inverse, giving a reversible Markov chain to which methodspreviously introduced by the first author for the case of free groups may be applied.
MSC classification: 20H10, 22D40, 37A30Keywords: Ergodic theorem, pointwise convergence, Fuchsian group, spher-ical sums Introduction
Formulation of the main result.
Let G be a finitely generated group with asymmetric set of generators G . For g ∈ G , denote by | g | the length of the shortest wordin G representing g . Let S ( n ) be be the sphere of radius n in G : S ( n ) = { g ∈ G : | g | = n } Suppose that G acts on a probability space ( X, µ ) by measure-preserving transformations T g , g ∈ G . For a function f ∈ L ( X, µ ) consider spherical averages(1) S n ( f ) = 1 S ( n ) (cid:88) g ∈ S ( n ) f ◦ T g . The main result of this paper, Theorem A below, gives the almost sure conver-gence of spherical averages for measure-preserving actions of Fuchsian groups and for f ∈ L log L ( X, µ ).Let G be a Fuchsian group and let R be a fundamental domain for G . The images of R under the action of G induce a tessellation T R = { g R : g ∈ G } of the hyperbolic disc D . Following [10], we say that R has even corners if the geodesic extension of every sideof R is entirely contained in T R , more precisely in the union of boundaries of all domains g R ∈ T R .Let v ∈ D be a vertex of T R . If R has even corners, then the boundary of T R in asmall neighbourhood of v consists of n geodesic segments intersecting at v and dividingour neighbourhood into 2 n sectors. Write n = n ( v ) and let N ( R ) denote the number ofsides of R inside D . We need the following assumption on R . Assumption 1.1. (i) R has even corners,(ii) One of the following conditions holds for R : a r X i v : . [ m a t h . D S ] M a y ALEXANDER I. BUFETOV, ALEXEY KLIMENKO, AND CAROLINE SERIES • N ( R ) ≥ , • N ( R ) = 4 and either R is non-compact or R is compact and does not havetwo opposite vertices v, v (cid:48) such that n ( v ) = n ( v (cid:48) ) = 2 , • N ( R ) = 3 and R is non-compact. Let G be the set of all group elements mapping R to the domains of T R having acommon side with R . As is well known, G is a symmetric set of generators for G . Ourmain result is the following: Theorem A.
Let G be a non-elementary Fuchsian group G and let R be its fundamentaldomain satisfying Assumption 1.1. Let G act on a Lebesgue probability space ( X, µ ) bymeasure-preserving transformations. Let G be the set of generators of G mapping R tothe neighbouring domains. Denote by I G the sigma-algebra of sets invariant under allmaps T g g , g , g ∈ G . Then, for any function f ∈ L log L ( X, µ ) , as n → ∞ , we have S n ( f ) → E ( f |I G ) almost surely and in L . The condition that R have even corners is not as restrictive as it appears. In factit is clear that our result only depends on the generators G and the coding, and not onthe precise geometry of R . Thus Theorem A extends immediately to any presentationof a Fuchsian group for which one can find deformed group G (cid:48) which has a fundamentaldomain R (cid:48) with the same pattern of sides and side-pairings and even corners, see [18, 10]and [43] for a detailed discussion. The need to restrict to spheres of even radius can beseen by considering the action of the free group F on the two-element set { , } in whichboth generators of F act by interchanging the elements, in which case the value of S n ( f )depends on the parity of n . This seems to indicate that the condition on all relators havingeven length, as is implied by the even corner condition, may be essential.The Ces`aro convergence of the averages S n ( f ) is proven in [18] using the Bowen—Series Markovian coding [10], see also [3, 42, 43], in order to reduce the statement to theergodic theorem for Markov operators, cf. [12, 13]. To establish convergence of sphericalaverages themselves, and thus of powers of our Markov operator, we develop the approachfrom [14] for free groups. The argument of [14] relies on a symmetry condition for the cod-ing, which allows one to relate the Markov operator generated by the coding to its adjoint.The Bowen—Series coding of [10] is however not symmetric, and the main constructionof this paper is a new symmetric coding for Fuchsian groups.This new coding is constructed using a variant of the coding introduced by MatthewWroten [46], see also a related idea in [21] and [44]. Wroten’s idea is to code all possiblerepresentations of a group element as a shortest word simultaneously. Then the set of allpossible paths in the Markov chain can be inverted. It would be interesting to obtain asimilar coding for a more general hyperbolic groups. In particular, it is not clear to ushow to invert paths in the classical Cannon—Gromov coding [20, 29].We now briefly describe Wroten’s approach in our setting. Every shortest word in theFuchsian group G corresponds to a shortest path in the Cayley graph of G relative to G .This graph is embedded in D by sending g ∈ G to gO ∈ D , where O is some fixed base ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 3 point in int R . Vertices gO, hO are joined by an edge if and only if g − h ∈ G . If β isa shortest path in the Cayley graph, we refer to the sequence of regions traversed by theedges of β also as a shortest path. If g ∈ G then the thickened path [ g ] associated to g isby definition the collection of all those h R , h ∈ G which are traversed by some shortestpath from R to g R . Every domain h R ∈ [ g ] is endowed with its index , which equals thedistance in the Cayley graph from R to h R . The set of all domains with index k we willrefer to as the level of [ g ] and denote by [ g ] k .Now the coding works as follows. As above, let G be a Fuchsian group with theset of generators G associated to a fundamental domain R , so that the assumptions ofTheorem A hold. We will define a space of states Ξ = { X , . . . , X k } and a Ξ × Ξ transitionmatrix M = ( M ij ) such that M ij = 1 if transition from X i to X j is possible and M ij = 0otherwise. There is a subset Ξ S ⊂ Ξ of start states, and another subset Ξ F ⊂ Ξ of endstates.The states in Ξ represents how [ g ] k and [ g ] k +1 are attached to each other. It turns outthat every [ g ] k contains at most two fundamental domains and the domains from [ g ] k +1 are glued to the ones from [ g ] k across one, two or three sides, see Figure 7. We endow thisgeometrical configuration with some additional data to obtain a Markov chain generatingthickened paths; in particular, the data records the generators needed to carry out thegluing. We then prove that thickened paths from R to g R with | g | = n are in one-to-onecorrespondence with admissible sequences of length n in this Markov coding starting inΞ S and ending in Ξ F .The reversiblity or self-inverse symmetry property of the coding can be expressedas follows. We introduce two maps γ, ω : Ξ → G , closely related to the attaching mapsbetween [ g ] k and [ g ] k +1 , see Section 5.1. These maps satisfy relations given by Lemma 5.1.Follwing [14] we then construct Markov operators P and U on L ( X × Ξ) which as aconsequence of these relations satisfy P ∗ = U P U and U ∗ = U − = U . Hence we can applythe Alternierende Verfahren method similar to [14]. The reason for the symmetry conditionis that inverting a thickened path yields a thickened path and our coding preserves thissymmetry.Using symmetry, we next establish an inequality between P n and ( P ∗ ) k P k , which isthe base for maximal inequality in the Alternierende Verfahren scheme. For free groupswe have cU P n − ϕ ≤ ( P ∗ ) n P n ϕ for any nonnegative ϕ . In the case of Fuchsian groups theinequality is more complicated and in particular containing an error term A n ϕ , see (11) inSection 6 below. The underlying geometric meaning of this inequality, Lemma 7.6, is thatfor a majority of thickened paths the following holds. Consider a thickened path of length2 n + 1, let A and B be its end domains, and let D denote the central domain or domainsof level n + 1. Then there exists a domain C such that the thickened paths AD and AC coincide from their initial point up to some point at distance at most a fixed boundeddistance n from D , C and likewise the thickened paths DB and CD coincide except for atmost n terms from their initial points. The few thickened paths that cannot be embeddedto a triangle in this manner give rise to the error term A n ϕ in inequality (11). ALEXANDER I. BUFETOV, ALEXEY KLIMENKO, AND CAROLINE SERIES
Organization of the paper.
The paper is organized as follows. In the next sectionwe give some notation and definitions regarding Fuchsian groups and their fundamentaldomains.Section 3 is devoted to the central step in the proof of Theorem A. Here we give adescription of shortest paths and thickened paths in terms of their local structure (seeCorollary 3.16 and Proposition 3.23). To do this it is convenient to consider the classof shortest paths and the class of thickened paths simultaneously and to switch betweenthem when necessary. A transition from a thickened path to a collection of shortest pathsis immediate: every sequence of domains in the thickened path with growing indices is ashortest path. The inverse transition is obtained via the new convexification technique,which we also introduce in Section 3. We show that the thickened path between the endsof a given shortest path is the convexification of the latter, that is, a minimal convex unionof fundamental domains that contains this path. The convexification can be obtained in anumber of convexification steps , where each step adds several domains to remove concaveangle at one point of the path boundary. In Section 4 thickened paths are represented as a realizations of a topological Markovchain, which is based on the local description of thickened paths obtained in the previoissection. Using this Markov chain, in Section 5 we represent the spherical averages in termsof a Markov operator associated to this Markov coding. In the next section we provegeneral theorem on pointwise convergence of powers of a Markov operator (Theorem 6.6).Finally, in Section 7 we apply this theorem to the operator associated with our Markovcoding and conclude the proof ot Theorem A.1.3.
Historical remarks.
For two rotations of a sphere, convergence of spherical averageswas established by Arnold and Krylov [1], and a general mean ergodic theorem for actionsof free groups was proved by Guivarc’h [30].A first general pointwise ergodic theorem for convolution averages on a countablegroup is due to Oseledets [38] who relied on the martingale convergence theorem.The first general pointwise ergodic theorems for free semigroups and groups were givenby R.I. Grigorchuk in 1986 [26], where the main result is Ces`aro convergence of sphericalaverages for measure-preserving actions of a free semigroup and group. Convergence of theactual spherical averages for free groups was established by Nevo [33] for functions in L and Nevo and Stein [35] for functions in L p , p > L p , p >
1. Note that, as shown by Tao [45],whose argument is inspired by Ornstein’s counterexample [37], pointwise convergence ofspherical averages for functions in L does not hold even for actions of free groups.The method of Markov operators in the proof of ergodic theorems for actions of freesemigroups and groups was suggested by R. I. Grigorchuk [27, 28], J.-P. Thouvenot (oral We hope in future version to be able to replace the convexification process using methods derivedfrom [3].
ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 5 communication), and in [12]. In [14] pointwise convergence is proved for Markovian spheri-cal averages under the additional assumption that the Markov chain be reversible. The keystep in [14] is the triviality of the tail sigma-algebra for the corresponding Markov opera-tor; this is proved using Rota’s “Alternierende Verfahren” [41], that is to say, martingaleconvergence. The reduction of powers of the Markov operator to Rota’s “AlternierendeVerfahren” in [14] essentially relies on the reversibility of the Markov chain. In this paper,the proof of Theorem A is based on a general result on convergence of Markov operators,which is an extension of the result from [14], and its proof also goes along the same lines,see Section 6. Another result in this direction was obtained in [5]; it states the mean con-vergence for analogues of spherical averages for an arbitrary Markov chain satisfying verymild conditions. It is not known whether similar result holds for pointwise convergence.The study of Markovian averages is motivated by the problem of ergodic theorems forgeneral countable groups, specifically, for groups admitting a Markovian coding such asGromov hyperbolic groups [29] (see e.g. Ghys—de la Harpe [23] for a detailed discussionof the Markovian coding for Gromov hyperbolic groups). the first results on convergenceof spherical averages for Gromov hyperbolic groups, obtained under strong exponentialmixing assumptions on the action, are due to Fujiwara and Nevo [22]. For actions ofhyperbolic groups on finite spaces, an ergodic theorem was obtained by L. Bowen in [4].Ces`aro convergence of spherical averages for all measure-preserving actions of Markovsemigroups, and, in particular, Gromov hyperbolic groups, was established in [15, 17];earlier partial results were obtained in [11, 13]. In the special case of hyperbolic groups ashorter proof of this theorem, using the method of Calegari and Fujiwara [19], was latergiven by Pollicott and Sharp [39]. Using the method of amenable equivalence relations,Bowen and Nevo [6, 7, 8, 9] established ergodic theorems for “spherical shells” in Gromovhyperbolic groups. For further background see the surveys [34, 25, 16].1.4.
Acknowledgements.
The research of A. Bufetov on this project has received fund-ing from the European Research Council (ERC) under the European Union’s Horizon 2020research and innovation programme under grant agreement No 647133 (ICHAOS).2.
Definitions and notation
Tessellation and labelling.
Let G be a finitely generated non-elementary Fuchsiangroup acting in the hyperbolic disk D with fundamental domain R , which we assume tobe closed. We suppose R to be a finite-sided convex polygon with vertices contained in D ∪ ∂ D , such that the interior angle at each vertex is strictly less than π . By a side of R we mean the closure in D of the geodesic arc joining a pair of adjacent vertices. Weallow the infinite area case in which some adjacent vertices on ∂ D are joined by an arccontained in ∂ D ; we do not count these arcs as sides of R . Further we usually mean by vertices of R only vertices inside D . Sometimes it is convenient to count as vertices alsoends of sides that belong to ∂ D , this instances will be specified explicitly. Two sides are adjacent if they share a common vertex lying in D . ALEXANDER I. BUFETOV, ALEXEY KLIMENKO, AND CAROLINE SERIES
We assume that the sides of R are paired; that is, for each side s of R there is a(unique) element e ∈ G such that e ( s ) is also a side of R and the domains R and e ( R )are adjacent along e ( s ). (Notice that this includes the possibility that e ( s ) = s , in whichcase e is elliptic of order 2 and the side s contains the fixed point of e in its interior. Thecondition that the vertex angle is strictly less than π excludes the possibility that the fixedpoint of e is counted as a vertex of R .) ee − ee − e − R R e R Figure 1.
Labelling the sides of the fundamental domain R . The label e appears interior to R on the side of R adjacent to the region e − R .We denote by ∂ R the union of the sides of R , in other words, ∂ R is the part of theboundary of R inside the disk D . Each side of ∂ R is assigned with two labels, one interiorto R and one exterior, in such a way that the interior and exterior labels are mutuallyinverse elements of G . We label the side s ⊂ ∂R interior to R by e if e carries s to anotherside e ( s ) of R , while we label the same side exterior to R by e − , see Figure 1. With thisconvention, R and e − ( R ) are adjacent along the side s whose interior label is e , whilethe side e ( s ) has interior label e − .Let G be the set of labels on sides of R . The labelling extends to a G -invariantlabelling of all sides of the tessellation T R of D by images of R . (By a side of T R ,we mean a side of g R for some g ∈ G .) The conventions have been chosen in such away that if two regions g R , h R are adjacent along a common side s , then h − g ∈ G and the label on s interior to g R is h − g , while that on the side interior to h R is g − h .Suppose that O is a fixed basepoint in R and that γ is an oriented path in D from O to gO , g ∈ G , which avoids all vertices of T R , passing through in order adjacent regions R = g R , g R , . . . , g n R = g R . Then the labels of the sides crossed by γ , read in such away that if γ crosses from g i − R into g i R we read off the label e i = g − i − g i of the commonside interior to g i R , are in order e , e , . . . , e n so that g = e e . . . e n . This proves the wellknown fact that G generates G , see for example [2].As explained in the introduction, the fundamental domain R is said to have evencorners if for each side s of R , the complete geodesic in D which extends s is contained inthe sides of T R . This condition is satisfied for example, by the regular 4 g -gon of interiorangle π/ g whose sides can be paired with the standard generating set { a i , b i , i = 1 , . . . , g | Π[ a i b i ] } to form a surface of genus g . For further discussion on the even corner condition,see the references in the introduction. ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 7 A local description of thickened paths
In this section we describe the structure of shortest paths and thickened paths (asdefined in the introduction) between pairs of fundamental domains. This description isinvariant under interchange of direction and the ends of the path, and provides the basisfor a symmetric Markov coding constructed in the next section. Throughout this sectionwe assume that G and R satisfy assumptions of Theorem A.The argument is rather indirect. Namely, we define in local geometric terms a classof locally shortest paths , which is subsequently seen to coincide with the class of shortestpaths. Similarly, we consider a class of unions of fundamental domains constrained byexplicit local rules and then we prove that this class coincides with the class of thickenedpaths with indices defined above. As we have said in the introduction, the main tool inthis section is a convexification procedure that transforms a shortest path to a thickenedpath.3.1. Locally shortest paths.
Let R = ( R , . . . , R N ) be a path of fundamental domains ,i. e. R i and R i +1 share a common side s i , and the sides of R i common with R i − and R i +1 are different.We start by giving a definition of the boundary of a path of fundamental domains R ,which works well even if the path has self-tangencies or self-overlaps, and for paths withoutself-tangencies or self-overlaps this definition gives the usual boundary of the polygonaldomain (cid:83) i R i oriented counterclockwise.Each side s i is co-oriented (from R i to R i +1 ) and thus oriented, so we refer to theends of this side as the left and the right ends. The complement of s i − and s i in ∂ R i hastwo connected components. One connects the left ends of s i − and s i , the other connectsthe right ones, we denote them by ∂ L R i and ∂ R R i respectively. The boundary of the path R is the oriented closed curve in D that goes counter-clockwise along ∂ R \ s , then along ∂ R R , . . . , ∂ R R N − , ∂ R N \ s N − , ∂ L R N − , . . . , ∂ L R and meets with its beginning at theleft end of s . If l consecutive paths in this list are empty, there are l + 2 consecutivedomains touching the same vertex v (possibly v ∈ ∂ D ). We will call the angle in v ∈ D convex ( less than π ) if l + 2 < n ( v ), straight ( equal to π ) if l + 2 = n ( v ), concave ( largerthan π ) if l + 2 > n ( v ) (including the case when l + 2 > n ( v )). Let us also say that theangle is minimally concave if l + 2 = n ( v ) + 1 and minimally convex if l + 2 = n ( v ) − Definition 3.1.
A path R = ( R , . . . , R N ) of fundamental domains is called locallyshortest if the following holds. Consider any segment of ∂ R that lies inside D . Let v , v , . . . , v M be the consecutive vertices on this segment. Then the boundary angle inevery vertex v i is convex, straight, or minimally concave, and if the angles in v i and v j , i < j , are minimally concave then there is a vertex v k , i < k < j , where the boundary hasa convex angle.We shall see that a path is locally shortest if and only if it is shortest. We start bycheck the ‘if’ part. The ‘only if’ part is established till Corollary 3.21. ALEXANDER I. BUFETOV, ALEXEY KLIMENKO, AND CAROLINE SERIES
I u u (cid:48) R i R j Figure 2.
To the proof of Lemma 3.2.
Lemma 3.2.
1) Any shortest path is locally shortest.2) Let A and B be two fundamental domains. Take a ∈ A , b ∈ B and suppose that thegeodesic segment I = ab does not pass through vertices. Then the sequence R = ( R = A , R , . . . , R N = B ) of domains intersected by I is a locally shortest path.Proof.
1) Assume that a shortest path R has a non-minimally concave angle at a vertex v , that is, there are n ( v ) + 2 consecutive regions R i , . . . , R i + n ( v )+1 around v in our path.But then R is not the shortest since one can reach R i + n ( v )+1 from R i going around v inanother direction and it takes n ( v ) − n ( v ) + 1 steps for the originalpath.Similarly, let v = u, v , . . . , v k = u (cid:48) be a sequence of boundary vertices such that theangles at u and u (cid:48) are minimally concave, and those at v , . . . , v k − are straight. Let R i be the first domain in R touching u and R j be the last domain touching u (cid:48) . Then it takes j − i = n ( v ) + (cid:80) k − l =1 ( n ( v l ) −
1) + n ( v k ) steps to go from R i to R j along R . But one cansee that it takes only (cid:80) kl =0 ( n ( v l ) −
1) = j − i − uu (cid:48) . The original and the shorter paths are shown on Figure 2 by solid anddashed arrows respectively.2) Consider the 2 n ( v ) sectors around the vertex v . If the segment I intersects at least n ( v ) + 2 of them, then it has at least n ( v ) + 1 intersections with geodesic lines separatingthem, i. e. I intersects one of these n ( v ) lines twice, which is impossible.Similarly, assume that the boundary has two concave vertices u and u (cid:48) and the anglesat all vertices between them are straight. Let (cid:96) be a line connecting u and u (cid:48) . Then I should cross (cid:96) twice: once on each connected component of (cid:96) \ uu (cid:48) , see Figure 2, whichgives the desired contradiction. (cid:3) Convexification in terms of the boundary curve: a locally shortest pathdoes not touch itself.
In this subsection we prove the following statement.
Lemma 3.3.
Let R = {R , . . . , R N } be a locally shortest path. Then domains R i and R j can share a vertex v if and only if v is a common vertex of all R k for k = i, . . . , j . Lemma 3.3 implies that a locally shortest path R has no self-intersections, even bya side or a vertex on its boundary, so we may define the boundary of R as the usual ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 9 boundary of the domain (cid:83) i R i oriented in the counterclockwise direction rather than usethe definition from the previous subsection.The proof of Lemma 3.3 occupies the rest of the subsection. We start with thedefinition of a class of almost convex curves. Informally speaking, a closed curve is almostconvex if it satisfies the following. It goes along the sides of the tessellation, the tangentvector of our curve makes only one turn, in any vertex the tangent vector turns eitherto the left or exactly by one sector to the right and any two right turns have a left turnbetween them. In particular, the boundary of any locally shortest path R is an almostconvex curve. We proceed to the formal definition. Note that some care is needed in thecase of non-compact fundamental domains.Consider a curve Γ consisting of finitely many sides of the tessellation T R and arcs of ∂ D . Let v ∈ Γ be a vertex of T R , so v ∈ D is a final point of a side s ⊂ Γ and an initialpoint of a side s (cid:48) ⊂ Γ. Consider the sector to the left of Γ at v , that is, the sector at v sweptby a ray going from s to s (cid:48) in the clockwise direction. This sector covers several petals at v , an we will say that the angle of Γ at v is ( minimally ) convex , straight , or ( minimally ) concave if the number of covered petals satisfies the same inequalities as specified in thefirst paragraph of Subsection 3.1. Definition 3.4.
An oriented closed curve Γ ∈ D is called almost convex if it satisfies thefollowing conditions.(1) Γ consists of finitely many sides of the tessellation T R and arcs of ∂ D .(2) Let v ∈ D be a vertex of T R such that it is the final point of a side s ⊂ Γ andthe initial point of a side s (cid:48) ⊂ Γ. Then s (cid:54) = s (cid:48) and the angle of Γ at v is convex,straight, or minimally concave. Vertices where the angle of Γ is minimally concavewill be called right turns .(3) Let I = vv (cid:48) be a maximal geodesic segment in Γ, v, v (cid:48) ∈ D . Then v and v (cid:48) cannotbe right turns simultaneously.(4) Every arc of Γ lying on ∂ D goes in the counterclockwise direction. Also, let v bean isolated point of Γ ∩ ∂ D , thus v ∈ ∂ D is the final point of a side s ⊂ Γ and theinitial point of a side s (cid:48) ⊂ Γ. Then s (cid:48) lies to the left of s , that is, if w, w (cid:48) ∈ ∂ D bethe other ends of the geodesics containing s, s (cid:48) , then the points v, w, w (cid:48) appear on ∂ D in this clockwise circular order.(5) The curve Γ makes one turn in counterclockwise direction. That is, let γ : S → Γ be any parametrization of Γ spending nonzero time in each vertex joiningsides/arcs. Define a map δ : S → ∂ D as follows. If γ ( t ) lies inside a side s then δ ( t ) is the final point in the direction of Γ of the geodesic containing s ; if γ ( t )lies inside some arc on ∂ D , then δ ( t ) = γ ( t ).It remains to define δ for the intervals corresponding to the endpoints of sides andarcs of Γ. If γ ( t ) ≡ v ∈ D for t ∈ [ t , t ], define δ | [ t ,t ] ( t ) as a continuous functionjoining δ ( t −
0) and δ ( t + 0) so that all these δ ( t ) belongs to the sector betweenthe rays vδ ( t −
0) and vδ ( t + 0) that measures less than π . If γ ( t ) ≡ v ∈ ∂ D for t ∈ [ t , t ], define δ | [ t ,t ] ( t ) as a continuous function joining δ ( t −
0) and δ ( t + 0) in the counterclockwise direction. For the constructed map δ we requirethat deg δ = 1.A straightforward induction in N gives the following statement. Proposition 3.5.
The boundary ∂ R of a locally shortest path R = {R , . . . , R N } is analmost convex curve. Now we define a convexification procedure that transforms an almost convex curveinto a convex one (i.e. those without right turns).
Definition 3.6. A flower at a vertex v ∈ D is the union of all fundamental domains thathave v on their boundary; these domains will be called petals . wpv p (cid:48) w (cid:48) Figure 3.
The convexification step at v Definition 3.7.
Consider an almost convex curve. A convexification step at a vertex v with a right turn is defined as follows. Let s, s (cid:48) be the sides of the boundary beforeand after v . Consider the flower at v and modify the boundary as follows: at the start-ing point p of s turn right to the boundary of the flower of v and go counter-clockwisealong this boundary until reaching the final point p (cid:48) of s (cid:48) , then proceed along the originalpath. (Figure 3 shows the original and the resulting curves by dashed and dotted linesrespectively.) Proposition 3.8.
1) A convexification step at a vertex v of an almost convex curve Γ again yields an almost convex curve ˆΓ .2) After finitely many convexification steps the curve Γ becomes a curve without rightturns. Moreover, if Γ becomes a curve without right turns after consecutive convexificationsteps at the vertices v , . . . , v k , then the set { v , . . . , v k } coincides with the set A (Γ) of allvertices v of the curve Γ satisfying one of the following three conditions:(i) v has a right turn,(ii) v has a straight angle, and if I is a maximal geodesic arc in Γ containing v , one ofthe ends of I has a right turn,(iii) the angle of Γ at v is minimally convex, and if I − , I + are maximal geodesic arcs in Γ adjacent to v , then the other ends of both these arcs have right turns. ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 11
Proof.
1) The first four conditions in Definition 3.4 are clear. The last condition is obtainedas follows. The curve Γ around the added petals (i. e. the light gray area on Figure 3)is also almost convex. Join the curves Γ and Γ together and eliminate their commonsegment. So we obtain a curve ˆΓ with deg ˆ δ = deg δ + deg δ − δ .2) Clearly, if a convexification step can be performed in v then v ∈ A (Γ) (in fact, v should satisfy condition (i)). Also there are no possible steps for Γ if and only if A (Γ) = ∅ . Thus it remains to prove that A (ˆΓ) = A (Γ) \ { v } .First consider a vertex u ∈ ˆΓ in the added path. Then the sector to the left of ˆΓ at u covers one or two petals at u , so the angle of ˆΓ at u is at most straight. Moreover, let w and w (cid:48) be the vertices on the added path that are adjacent to p and p (cid:48) respectively. Thenfor u = w, w (cid:48) the sector covers one petal, hence the angle at u is convex. Therefore, allvertices between p and p (cid:48) do not belong to A (ˆΓ).Special care is needed when the added path from p to p (cid:48) contains at most two sides,i.e. either N ( R ) = 3 and there are one or two added petals, or N ( R ) = 4 and thereis one added petal. The latter case means that n ( v ) = 2, hence by condition (ii) ofAssumption 1.1 either w = w (cid:48) , p , or p (cid:48) lies on ∂ D , or R is compact and n ( w ) >
2, so (iii)fails for w . The former case N ( R ) = 3 is even simpler: for one added petal either p or p (cid:48) lies on ∂ D , for two added petals at least one of the following holds: w lies on ∂ D or both p and p (cid:48) lie on ∂ D . Therefore, in these cases we still have that the added vertices do notbelong to A (ˆΓ).For any vertex u with a right turn denote by I ( u ) the union of two maximal geodesicsegments of Γ adjacent to u , Let I (Γ) be the union of interiors of I ( u ) over all u with aright turn. Then the vertices with condition (i) or (ii) are those belonging to I (Γ), and thevertices with condition (iii) are those belonging to int clos I (Γ) \ I (Γ) and having minimallyconvex angle.After the convexification step at v the boundary angle is increased only at p and p (cid:48) ,where this angle is increased by one petal. Therefore, the only possible new vertices withconcave angle are p and p (cid:48) . If, say, the angle of ∂ ˆΓ at p is minimally concave then theangle of the angle of ∂ Γ at p is straight, so p ∈ int I ( v ) and ˆ I ( p ) is the union of [ p, w ] andthe part of I ( v ) lying on the other side of p with respect to v .On the other hand, for any vertex u (cid:54) = v we have I ( u ) ⊂ ˆ I ( u ). Assume that I ( u ) (cid:54) =ˆ I ( u ). Then the angle at an end z of I ( u ) for ˆΓ is larger than that for Γ. This is possibleonly if z is one of the vertices p and p (cid:48) . Moreover, as the angle at z of ˆΓ is greater by onepetal than that of Γ, the angle of Γ at z must be minimally convex. In this case we haveˆ I ( u ) = I ( u ) ∪ [ p, w ] if z = p and ˆ I ( u ) = I ( u ) ∪ [ p (cid:48) , w (cid:48) ] if z = p (cid:48) .Therefore, I (Γ) and I (ˆΓ) coincide on the common segment of the curves Γ and Γ (cid:48) between points p and p (cid:48) , so any vertex other than p and p (cid:48) belongs to A (Γ) and A (ˆΓ)simultaneously.To conclude the proof it remains to check that each of the points p and p (cid:48) belongs to A (Γ) and A (ˆΓ) simultaneously. Consider the point p . If p is internal for I ( v ), then p isalso internal for ˆ I ( p ), so p lies in both I (Γ) and I (ˆΓ). If p is the end of I ( v ), and not theend of another segment I ( u ), then p does not belong to ˆ I ( u ) for any u , so p does not lie in both A (Γ) and A (ˆΓ). Now suppose that p is the common end of I ( v ) and I ( u ). Assumefirst that the angle at p for Γ is convex but not minimally convex. The sector to the leftof ˆΓ in p is larger by one petal than those for Γ, so the angle of ˆΓ at p is convex, hence forˆΓ the vertex p is the end of only one segment ˆ I ( u ) = I ( u ), and p does not belong to A (Γ)and A (ˆΓ). Finally, if the angle of Γ at p is minimally convex, then p satisfies condition(iii) for Γ and condition (ii) for ˆΓ, so p belongs to both A (Γ) and A (ˆΓ). (cid:3) Recall the following statement for the affine plane: if a smooth closed curve γ alwaysturns to the left: Vol( ˙ γ, ¨ γ ) ≥
0, and ˙ γ makes one complete turn, then γ has no self-intersections. The next statement is a hyperbolic analogue of this fact. Proposition 3.9.
An almost convex curve Γ with no right turns has no self-intersections.Proof. Let us relax conditions on Γ: instead of condition 1 of Definition 3.4 we assumeonly that Γ consists of finitely many geodesic segments and arcs of ∂ D , we do not need therequirement that these segments are sides of T R . Condition 2 now says that the angles ofΓ in all vertices are not greater than π , and condition 3 is trivial.Assuming that Γ has self-intersections, we can perturb Γ inside the class specified inthe previous paragraph so that the intersection takes place either in an internal point oftwo arcs on ∂ D , or in an internal point of two geodesic segments belonging to differentlines. Denote this perturbed curve by the same symbol Γ, let γ and δ be defined as incondition 5 of Definition 3.4, and let ∆ : R → R be a lift of δ to the universal covers, thuswe have ∆( t + 1) = ∆( t ) + 2 π . Since Γ has no right turns, ∆ is nondecreasing.If v = γ ( t ) = γ ( t ), t < t < t + 1 is a common point of two arcs of Γ on ∂ D , then∆( t ) ≡ ∆( t ) (mod 2 π ), hence either ∆( t ) = ∆( t ) or ∆( t ) = ∆( t + 1). Both theseequalities lead to contradiction: in the former case we have ∆( t − ε ) < ∆( t ) for small ε > v = γ ( t ) = γ ( t ), t < t < t +1 is a common point of two geodesicsegments of Γ. Denote the lines containing these segments by (cid:96) i = a i b i ( i = 1 , (cid:96) (cid:54) = (cid:96) ,where a i and b i are the initial and the final points of these geodesic lines on ∂ D . As (cid:96) and (cid:96) intersect each other, the ends of (cid:96) and the ends of (cid:96) are interlaced in the cyclical orderon ∂ D . So we may assume that the order is a , a , b , b in counterclockwise direction; theother case is reduced to this one by the exchange of (cid:96) and (cid:96) .Consider the part γ ([ t , t ]) of our curve Γ. Note that δ ([ t , t ]) lies on the arc of ∂ D going from b to b in the counterclockwise direction, so δ ([ t , t ]) lies on the left half-disc H with respect to a b . Then one can inductively show that all consecutive segmentsand arcs of γ ([ t , t ]) belong to clos H , hence γ ([ t , t ]) ⊂ clos H . But γ ( t − ε ) lies in theright half-disc with respect to a b , so we get a contradiction. (cid:3) Proof of Lemma 3.3.
Assume the contrary and consider a minimal subpath R (cid:48) = {R i , . . . , R j } that still violates the conclusion of this lemma. Then ∂ R (cid:48) intersects itself, and, due to ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 13 minimality of R (cid:48) , any point of self-intersection belongs only to the first and the last do-mains in R (cid:48) . Hence this intersection is either a vertex or a side of ∂ R i and ∂ R j notadjacent to s i = R i ∩ R i +1 and s j − = R j − ∩ R j .Apply the convexification procedure to the boundary of {R i , . . . , R j } . Note that theconvexification step can be applied to no vertex v ∈ ∂ R i \ s i . Indeed, ∂ R (cid:48) has a convexangle at v , as there is only one petal in the sector to the left of ∂ R (cid:48) at v . Even if theangle at v is minimally convex, condition (iii) of Proposition 3.8 cannot hold there sinceat least one of the neighbours of v also belongs to ∂ R i \ s i and hence also has a convexboundary angle. The case N ( R ) = 3 needs special consideration: here if the only vertex v of ∂ R i \ s i lies in D , then one of its neighbours, i.e. the ends of s i , should lie on ∂ D ,hence condition (iii) does not hold for v . Therefore, the convexified curve still has aself-intersection, contradicting Proposition 3.9. (cid:3) Convexification in terms of domains: structure of path convexifications.
Keeping in mind that convexification does not give rise to self-intersections, consider theset of fundamental domains inside our curve during the convexification procedure. Initiallythe curve is ∂ R and the domains inside it are R , . . . , R N . The convexification step at v increases the set of domains inside the curve by the n ( v ) − v ,these petals are shown by the light gray area on Figure 3. The next lemma endows thiscollection of domains with indices. Note that properties 1–5 of this lemma hold for athickened path [ g ] when index k is assigned to domains from [ g ] k . Lemma 3.10.
Let R = ( R , . . . , R N ) be a locally shortest path. Then one can assign anindex from to N to every domain of its convexification in such a way that:1) every R i has index i ;2) the sequence of indices for consecutive domains bordering ∂ L [ R ] or ∂ R [ R ] is precisely { , . . . , N } ;3) if two domains have a common side, their indices differ by one;4) every domain with an index i borders domains with indices i − (provided i ≥ ) and i + 1 (provided i ≤ N − );5) only R has index , only R N has index N .Proof. Straightforward induction. The properties 1–5 hold for the initial collection ofdomains {R , . . . , R N } , and are preserved by convexification. Indeed, if a convexificationstep is performed at the vertex v , then property 2 implies that the domains bordering v have indices i, i + 1 , . . . , i + n ( v ), and the sides of the boundary adjacent to v are incidentto domains with indices i and i + n ( v ). Hence we may endow the added domains from theflower at v with the indices i + 1 , . . . , i + n ( v ) − (cid:3) Denote the convexification of the path R = ( R , . . . , R N ) by [ R ] and the union ofdomains in [ R ] with index i by [ R ] i . Since all convexification steps are performed onlyat the vertices of ∂ R , the set [ R ] i can contain at most three elements: one from theoriginal curve, one from the convexification step at a vertex from ∂ L R and one from the convexification at a vertex from ∂ R R . Let us show that in fact there are at most twoelements in every [ R ] i . Proposition 3.11.
Consider a sequence u = ( u , . . . , u k ) of adjacent vertices along theboundary of a locally shortest path R so that all u j belong to the set A ( R ) := A ( ∂ R ) ofvertices where convexification steps occur. The sequence u is assumed to be oriented from R to R N . Let u v be the border between R i and R i +1 with minimal possible i , u k v k bethe border between R j and R j +1 with maximal possible j . Then all vertices between v and v k do not belong to A ( R ) .Proof. We start with the following statement.
Claim 3.12.
Let u be a vertex on ∂ L R . Consider all domains R i , . . . , R i + a bordering u and denote opp( u ) = ∂ R R i +1 ∪ · · · ∪ ∂ R R i + a − . Then no internal vertices of opp( u ) belongto A ( R ) .Proof repeats the first part of the proof of item 2 in Proposition 3.8: opp( u ) is a part ofthe boundary of the flower at u , and we have seen there that a convexification step cannotbe applied at any of its internal vertices. (cid:3) This claim yields that no vertex v between v and v k can border three domains:otherwise opp( v ) is nonempty, so opp( v ) either contains an internal vertex u l , which thenfails to belong to A ( R ), or is non-compact (if N ( R ) = 3), cutting u into two sequences.Therefore, every vertex v between v and v k can satisfy only condition (ii) or (iii) ofProposition 3.8, and the former is possible only if n ( v ) = 2 and v is incident to twodomains in R .Further, there exists u j satisfying condition (i) and thus bordering at least threedomains in R . Then opp( u j ) is either non-compact or has internal vertex, hence it is notpossible to have a continuous sequence of vertices in A ( R ) between v and v k . Therefore,the only remaining case is when several vertices adjacent to v or to v k satisfy condition(ii). Let v be the vertex adjacent to v on ∂ R R and assume that v satisfy (ii) in Propo-sition 3.8. Due to Claim 3.12 this is possible only if u borders two domains (otherwise v is internal to opp( u )). Therefore, u satisfy (ii), n ( u ) = 2, and ∂ L R i +1 contains oneside. Then ∂ R R i +1 contains N ( R ) − N ( R ) ≥ v borders only one domainin R .Finally, assume N ( R ) = 4. Since u , . . . , u k contains a vertex satisfying condition (i)of Proposition 3.8, we have k ≥
1, i. e. R i +1 is a compact quadrilateral u u v v . Thus wearrive at a contradiction with condition (ii) in Assumption 1.1 as n ( u ) = n ( v ) = 2. (cid:3) Now consider a maximal sequence u = ( u , . . . , u k ) of adjacent vertices in A ( R ), it canjump between the left and the right boundaries. The previous proposition yields that suchsequence is uniquely defined by one its term. Namely, let u belong to the left boundaryof R . Then we should proceed along the left boundary while possible: for intermediatevertices there are no vertices from A ( R ) adjacent to them via sides s i . Reaching the last ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 15 vertex u s in the continuous sequence in A ( R ) ∩ ∂ L R it is possible to continue the sequence u only via the side s j incident to u s with maximal j . So if the other end u s +1 of s j belongsto A ( R ), we can proceed only along the right boundary in the same direction from R to R N , and j is the minimum of all k such that s k is incident to u s +1 , and so on. Proposition 3.13.
Let R be a locally shortest path.1) Let u = ( u , . . . , u k ) be a maximal sequence of adjacent vertices from A ( R ) . Let [ i, j +1] be the set of indices t such that R t is incident to a vertex from u . Then R i (resp., R j +1 )intersects u only by u (resp., u k ).2) Let u (cid:48) = ( u (cid:48) , . . . , u (cid:48) l ) be another such sequence, let [ i (cid:48) , j (cid:48) + 1] be the same segmentas above for u (cid:48) . Then the segments [ i, j + 1] and [ i (cid:48) , j (cid:48) + 1] are either non-intersecting orhave a common end. If, say, j + 1 = i (cid:48) , then R j +1 is incident to u k and u (cid:48) and they arenot adjacent.Proof.
1) If u and u lies on the same side of the boundary, then u borders at leasttwo domains, and only the last of them contains u . That is, R i , being the first domainincident to u , does not contain u . Similarly, if u and u lie on the different sides ofthe boundary, then u satisfies condition (i) of Proposition 3.8 and hence is incident to atleast three domains from R . Therefore, u u is a common side of the last two domains in R that are incident to u , hence u does not belong to the first domain R i incident to u .2) As we have seen above, all domains with indices from the segment [ i + 1 , j ] have novertices in A ( R ) except of those in u . Hence [ i + 1 , j ] ∩ [ i (cid:48) , j (cid:48) + 1] = ∅ . The last statementin the proposition is clear: if u k and u (cid:48) were adjacent, u and u (cid:48) might be joined into onesequence. (cid:3) The next step is to describe the geometry of a sequence u from the last proposition. Definition 3.14.
A curve going along sides of T R is called almost straight if the followingholds: in every vertex the angle is either straight or off by one sector to the left or to theright; and there are no two same-side turns with only straight angles between them. T T T T T T T T B B B B B B B B B E L C E R C E R E L D Figure 4.
An example of the adjacency graph for the domains in a non-trivial section. Dashed lines represents boundaries of domains, the curves u α,β are shown in bold. Letters at the bottom of the figure represent typesof states of the Markov coding defined in Section 4. Proposition 3.15.
Let u = ( u , . . . , u k ) and [ i, j + 1] be the same as in the previousproposition. Then the following holds.
1) The convexification steps at u , . . . , u k add exactly one domain with each index from i + 1 to j .2) The corresponding adjacency graph has the structure shown on Figure 4: there are twosequences of graph vertices R i → T i +1 → · · · → T j → R j +1 and R i → B i +1 → · · · → B j →R j +1 and there are several “crossings”, i.e. edges of the form T i → B i +1 or B i → T i +1 .3) Let u , − u be the two sides of R i adjacent to u , and u k u , k +1 be the two sides of R j +1 adjacent to u k . Then for any choice of α, β ∈ { , } the curve u α,β = ( u α − u u . . . u k u βk +1 ) is almost straight.Proof. We start with the proof of statements 1 and 2. Arrange the domains R i , . . . , R j +1 into the top and bottom rows as follows: all domains R s , i + 1 ≤ s ≤ j having verticesfrom A ( R ) only on their left (respectively, right) boundary are placed on the bottom row: B s := R s (respectively, on the top row: T s := R s ). Further, if a domain R s has verticesfrom A ( R ) on both parts of its boundary, then both ends of either R s − ∩ R s or R s ∩ R s +1 belong to u . If, say, u t u t +1 = R s ∩ R s +1 and u t belongs to the left boundary of R and u t +1 to the right one, we place R s to the bottom row and R s +1 to the top row, and viceversa. Finally, we place R i and R j +1 on the different rows from R i +1 and R j respectively.Let us define a “pit” as a series of already-defined domains T k , B k +1 , . . . , B l , T l +1 sothat T k +1 , . . . , T l are still undefined. This pit is “opened to the top”, there are “opened tothe bottom” pits with symmetric conditions.One can check that a convexification step preserves the following properties. (i) Everyopened to top (resp., bottom) pit corresponds to the maximal continuous sequence ofvertices on the left (resp., right) boundary where the convexification step is allowed butnot yet performed. The domains of the pit are exactly those that border vertices fromthis sequence. (ii) Every minimal cycle (i.e. those without edges inside) in the constructedadjacency graph corresponds to a vertex where the convexification step has been alreadyperformed. Domains of the cycle are exactly those bordering this vertex. The cycle hasthe form of either a trapezoid or a parallelogram.There are three possible cases when applying convexification step at vertex v belongingto a sequence associated to a pit T k , B k +1 , . . . , B l , T l +1 . Firstly, n ( v ) + 1 already defineddomains bordering v can comprise the whole pit. Then the convexification step at v addsthe remaining petals T k +1 , . . . , T l , and the pit is completely removed. Secondly, thesedomains can belong to the side of the pit, say, they are T k , B k +1 , . . . , B s +1 . Then wedenote the added domains as T k +1 , . . . , T s with the crossing edge T s → B s +1 at the end,and the pit is shortened. Finally, all these n ( v ) + 1 domains can belong to the bottom row: B r , . . . , B s +1 . Then the added domains are denoted as T r +1 , . . . T s and the pit is split intotwo, as well as the sequence of vertices where convexification step is not yet performed.Clearly, the statements (i) and (ii) above are preserved in any case.We proceed to statement 3. Edges of the cycle corresponding to the vertex u t cor-respond to sides between domains incident to u t . Crossings, which belong to two cycles,correspond to sides of the form u t u t +1 . One can see that if the cycle corresponding to u t is a parallelogram, then the sides u t − u t and u t u t +1 are n ( u t ) sectors apart, i. e. they form ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 17 the straight angle. Similarly, wide-bottom and wide-top trapezoids correspond to minimalleft and right turns respectively. As wide-top and wide-bottom trapezoids should inter-leave, we obtain that the curve u α,β is almost straight. The choice of α and β correspondsto the placement of R i and R j +1 on the top or the bottom rows. (cid:3) Corollary 3.16.
Consider the adjacency graph for the convexification of a locally shortestpath R . Then if [ R ] i and [ R ] j +1 , where j > i , contain one domain each and all [ R ] s , i +1 ≤ s ≤ j , contain more than one domain, then the adjacency graph for [ R ] i ∪· · ·∪ [ R ] j +1 is exactly what is obtained in the previous proposition.Remark . Clearly, if j = i in this corollary, then [ R ] i and [ R ] i +1 contain one domaineach, and the adjacency graph for [ R ] i ∪ [ R ] i +1 contains the only possible edge R i → R i +1 .3.4. Convexified paths and thickened paths.
The aim of this subsection is to charac-terize all locally shortest paths inside the convexification of a given locally shortest path.As a result we will obtain that all locally shortest paths between two domains have thesame length and hence they are indeed shortest.
Lemma 3.18.
Let [ R ] be a convexification of a locally shortest part R = ( R , . . . , R N ) .1) Let S = ( S , . . . , S M ) be a locally shortest path inside [ R ] such that S = R , S M = R N .Then the index in [ R ] of every domain S i equals i , and hence M = N .2) Let S = ( S , . . . , S N ) be any path inside [ R ] such that S i has index i in [ R ] . Then S islocally shortest and [ S ] = [ R ] .Proof. First of all, note that if R ] i = 1, then S contains the only domain R i with index i .Therefore, we can analyze only pieces of S inside each section described in Corollary 3.16.For an individual section Proposition 3.15 gives the structure of the adjacency graph. Let S i (cid:48) = R i , S i (cid:48) +1 , . . . , S j (cid:48) +1 = R j +1 be a part of S in this section.Consider firstly only the initial part of the path S going in the positive direction (i. e.with indices increasing). Since adjacent domains have indices differing by one, every S i inthis part of the path has the index i . Claim 3.19.
Assume that S goes along the cycle corresponding to a vertex v on the bottomrow, and that before that S goes only in positive direction. Consider the previous crossingwhere S goes from the top to the bottom row, let u be the vertex corresponding to the cycleafter this crossing. Then the part Γ u,v of the left boundary of S from the vertex previousto u to the vertex v is an almost straight curve, and either Γ u,v has no non-straight angles,or the first non-straight angle of Γ u,v is minimally concave. Moreover,(i) if the last crossing before the cycle corresponding to v is “top to bottom” then the lastnon-straight angle in Γ u,v is minimally concave;(ii) if this crossing is “bottom to top” then the last non-straight angle in Γ u,v is minimallyconvex or Γ u,v curve has no non-straight angles at all. Indeed, if v is adjacent to u , then either the cycle at u has the form of a wide-toptrapezoid, thus S goes through n ( u ) domains of this cycle, the angle at u is straight, andthis agrees with case (ii), or the cycle has the form of a parallelogram with n ( u ) + 1 of its a) · · · · · · b) · · · · · · c) · · ·· · · d) · · ·· · · e) f) Figure 5.
To the proof of Lemma 3.18. Shaded region is a cycle in theadjacency graph. Black vertices represent domains belonging to S , whitevertices represent other domains from [ R ].domains belonging to S , and this agrees with case (i). The inductive step from a cycle v to the next cycle v (cid:48) is considered similarly.Now let us pass to the proofs of the lemma statements.1) Suppose that at some moment the sequence S goes in the negative direction. As S cannot go forward and backward along the same edge, there are two possibilities: S goes either forward along the row and then backward along a crossing, or forward alonga crossing and then backward along the row. Assume that before this maneuver S goesalong the bottom row of the cycle corresponding to a vertex v . There are three possibilitiesfor the position of the crossing before this cycle: this crossing can go “bottom to top”,“top to bottom”, or it can be “top to bottom” and S goes along this crossing. So we getsix cases shown in Figure 5. In the cases shown as a), b), e), and f) the vertex v borders n ( v ) + 2 domains in S hence ∂ S is not an almost convex curve. In the remaining casesshown as c) and d) the angle of ∂ S at v is minimally concave, but by the last claim theprevious non-straight angle on ∂ S is also minimally concave, so again ∂ S fails to be almostconvex. Therefore, S goes in the positive direction, hence the index of S i in [ R ] equals i ,and the first statement is proven.2) Consider a part of S between two successive crossings, and let u and v be thevertices corresponding to the first and the last cycles between these crossings. Apply thelast claim to the vertex v and consider separately cases (i) and (ii). In the first case thecycle of v has the form of a wide-top trapezoid with n ( v ) of its domains lying in S , sothe boundary angle at v is straight. In the second case the cycle of v has the form ofa parallelogram, v is incident to n ( v ) + 1 domains of S and the boundary angle at v isminimally concave. Finally, if u = v then v is incident to n ( v ) + 1 domains of S andthe boundary angle at v is minimally concave. Therefore, in any case ∂ S contains thesequence of vertices from u to v , the angles of ∂ S at these vertices are minimally convex,straight, or minimally concave, the convex and the concave angles alternate, and the firstand the last non-straight angles are concave. Proposition 3.8 now yields that all verticesof this sequence from u to v belong to A ( S ). Consequently, the convexification procedurefor S includes the convexification steps at all vertices corresponding to the cycles in theadjacency graph for [ R ], thus [ S ] ⊃ [ R ]. The inverse inclusion follows from the fact ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 19 that [ S ] is a minimal convex union of fundamental domains containing S , and that [ R ] isconvex. (cid:3) Lemma 3.20.
Any two locally shortest paths R = ( R , . . . , R N ) and S = ( S , . . . , S M ) with R = S = A and R N = S M = B have the same convexification and the same length.Proof. Consider the intersection [ R ] ∩ [ S ] of their convexifications. It is a convex set. Takeany geodesic segment ab with a ∈ A and b ∈ B that does not pass through vertices. ByLemma 3.2 the sequence of domains intersected by ab is a locally shortest path T ⊂ [ R ] ∩ [ S ]going from A to B . Lemma 3.18 then states that [ R ] = [ T ] = [ S ] and that the lengths of R and S are both equal to those of T . (cid:3) Corollary 3.21.
All locally shortest paths are shortest.
Corollary 3.22.
Let R = ( R , . . . , R N ) be any (locally) shortest path from A = R to B = R N . Then the thickened path from A to B coincides with [ R ] and the set of all shortestpaths going from A to B is the set of all paths in [ R ] going in the positive direction.Proof. All shortest paths belong to [ R ] by the previous lemma, hence the thickened pathlies in [ R ]. On the other hand, every domain S j ⊂ [ R ] j can be included in a path in[ R ] going from A to B in the positive direction: we choose arbitrarily S j +1 adjacent to S j , then we choose S j +2 , . . . , S N , as well as S j − , S j − , . . . , S , this is possible due toLemma 3.10. (cid:3) We conclude this section with the converse of Proposition 3.15.
Proposition 3.23.
Let S be a family of fundamental domains with indices from to N satisfying properties 1–5 from Lemma 3.10. Denote by [ S ] i the union of domains withindex i and by A and B the only domains with indices and N respectively. Supposethat S is convex, and every section [ S ] i ∪ · · · ∪ [ S ] j +1 such that [ S ] i and [ S ] j +1 containonly one domain, and all intermediate [ S ] k contain at least two domains, has the structuredescribed in Proposition 3.15. Then S is a thickened path from A to B .Proof. Consider the path R = ( R , . . . , R N ) going from A to B along the left boundaryof S . All vertices of ∂ R belonging to the boundary of S has convex or straight angles,and we need to check that the vertices of the boundary of R that lies inside a nontrivialsection of S satisfy conditions for locally shortest path. But the part of ∂ R inside thenontrivial section is a curve of the form u α,β from Proposition 3.15. This curve is almoststraight hence ∂ R is an almost convex curve and R is a locally shortest path.Now we need to check that [ R ] = S . This is done exactly in the same way as in thesecond statement of Lemma 3.18. Finally, Corollary 3.22 shows that S is a thickened pathfrom A to B . (cid:3) The Markov coding
In this section we construct a Markov coding generating the set of thickened paths.This coding is based on the description of the structure of thickened paths in the localterms given in Propositions 3.15 and 3.23.
As it was stated in the introduction, states of this topological Markov chain describehow the ‘past’ level S − = [ S ] i of a thickened path is attached to its ‘future’ level S + =[ S ] i +1 . More specifically, a state of the Markov chain describes the arrangement of S − and S + up to the action of G . However, we have to endow these arrangements with additionaldata to construct our coding. This is done in Subsection 4.1.In the next subsection we define the transition matrix Π of the coding and showthat this coding indeed generates all thickened paths. Subsection 4.3 shows that theconstructed Markov chain has a time-reversing involution on the set of states. Finally, inSubsection 4.4 we will show that our Markov chain is strongly connected and aperiodic.4.1. States of the Markov chain.
As we have seen in Proposition 3.15, the adjacencygraph for the domains in S − ∪ S + has one of the following types: A . S − = S + = 1, and the graph contains the only possible edge from S − to S + . This corresponds to a trivial section of the thickened path described in Re-mark 3.17. B . S − = 1, S + = 2, and the graph contains both edges from S − to S + . This statestarts a nontrivial section from Proposition 3.15. C . S − = 2, S + = 2, and the edges join the left domain in S − to the left domainin S + and the right domain in S − to the right domain in S + . D . S − = 2, S + = 1, and the graph contains both edges from S − to S + . This stateends a nontrivial section. E . S − = 2, S + = 2, and the graph contains three edges, the two described fortype C , and one more. Namely, this type is subdivided into the type E L , wherethe third edge goes from the left domain in S − to the right domain in S + , andthe type E R , where it goes from the right domain in S + to the left domain in S + .The states of type E correspond to the transitions from one flower to the next oneinside a nontrivial section.This types are illustrated on Figure 4.The notation for each state of our Markov chain includes the type A . . . E of the statefrom the list above and the labels on the sides separating S − and S + . More precisely,these sides form a polygonal curve, which is co-oriented from S − to S + and thus oriented,and we read off the labels on the S + -side of the separating sides going from the left endof the separating curve to its right end.Clearly, the labels in the notation of the state should satisfy some restrictions, and toexpress these restrictions we introduce some notation regarding vertices, sides, and labels(see Figure 6). For any e ∈ G consider the side s e of R so that its label inside R is e .We co-orient this side from outside to inside of R , and the corresponding orientation of s e allows us to define for s e its left vertex v L ( e ) and its right vertex v R ( e ). Note that v L ( e )or v R ( e ) is undefined if the corresponding end of s e lies on ∂ D . The same notation v L,R ( s )will be used for the ends of a co-oriented side s of the tessellation T R .The labels e and e are called adjacent if the sides of R with these outgoing labelshave a common vertex adjacent, i. e. either e = e , or v L ( e − ) = v R ( e − ), or vice versa. ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 21 ee − l ( e ) r ( e ) v L ( e ) v R ( e ) e e Figure 6.
To the definitions of vertices v L ( e ), v R ( e ) and labels l ( e ), r ( e ).Labels e , e ∈ G are adjacent.Let us define maps l and r on the set of labels. Informally speaking, we do thefollowing: for e ∈ G we go around v L ( s e ) in the counterclockwise direction, then the nextside we crossed after s e has the label l ( e ) outside R . Similarly, going clockwise around v R ( s e ) we obtain r ( e ). Formally we define l ( e ) and r ( e ) as the labels such that v R ( l ( e ) − ) = v L ( e ), v L ( r ( e ) − ) = v R ( e ). Note that l ( e ) or r ( e ) is undefined if the corresponding end of s e lies on ∂ D . Definition 4.1.
The set (cid:98)
Ξ is the set of all possible arrangements of S − and S + up to theaction of G . Namely, it consists of the following elements (see Figure 7): • A ( e ): S − = S + = 1, and e is the label on the S + -side of the common side of S − and S + . • B ( e L , e R ): S − = 1, S + = 2, e L and e R are S + -labels on the common sides if S − with the left and the right domains in S + respectively. Since these sides of S − are adjacent, we have that v L ( e − L ) = v R ( e − R ). • C k ( e L , e R ): S − = S + = 2, and all four domains in S ± share a common ver-tex v . The label e L (respectively, e R ) is the S + -label on the common side ofthe left (respectively, right) domains in S − and S + , and the sector of the flowerat v between these two sides that contains S − consists of 2 k + 1 petals. Denote n ( e L , e R ) = n ( v ) = n ( v R ( e L )) = n ( v L ( e R )), then 1 ≤ k ≤ n ( e L , e R ) − l k +1 ( e − L ) = e R . • D ( e L , e R ): S − = 2, S + = 1, e L and e R are S + -labels on the common sides ofthe left and the right domain in S − with the domain S + . The adjacency conditiongives v R ( e L ) = v L ( e R ). • E L,R ( e L , e M , e R ): S − = S + = 2. The four domains in S − and S + do nothave a common vertex, and there are three sides separating them. The state E L represents the case when these sides form an N -shaped line, that is, the leftpast domain borders both future domains via sides with the S + -labels e L and e M , and the right past domain borders only the right future domain via the sidewith the label e R . Thus we have v L ( e − L ) = v R ( e − M ) and v R ( e M ) = v L ( e R ). The a) e b) e L e R c) e L e R d) e L e R e) e L e M e R Figure 7.
Configurations for states of the Markov coding:a) A ( e ), b) B ( e L , e R ), c) C ( e L , e R ), d) D ( e L , e R ), e) E R ( e L , e M , e R ).state E R is the same with left and right inverted: the boundary is N -shaped, and v R ( e L ) = v L ( e M ), v L ( e − M ) = v R ( e − R ).It is clear that every configuration of adjacent levels in a thickened path belong to theset (cid:98) Ξ. On the other hand, the set of all possible sequences of configurations cannot begenerated by a Markov chain. For example, for a vertex v with n ( v ) ≥ S ] i , [ S ] i +1 , [ S ] i +2 are consequent petals around v , say, in the counterclockwise direction.Then if e is the label on the future side of [ S ] i ∩ [ S ] i +1 , the label on the future side of[ S ] i +1 ∩ [ S ] i +2 is l ( e ), and we have that the transition A ( e ) → A ( l ( e )) is admissible. Onthe other hand, a long sequence A ( e ) → A ( l ( e )) → A ( l ( l ( e ))) → . . . is not admissible,since the respective sets [ S ] i are still the consecutive petals around a vertex v , and it isnot allowed that a thickened path has v on its boundary and contains more than n ( v )petals around v .To solve this problem we endow the states of type A with an additional information.This is based on the following statement. Proposition 4.2.
Let S be a thickened path. Let a vertex v ∈ ∂ S belongs to the boundariesof [ S ] k for k = i, . . . , j + 1 , where j > i . Then one of the following cases takes place: ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 23 (1) j = i + 2 and both pairs ([ S ] i , [ S ] i +1 ) , ([ S ] i +1 , [ S ] i +2 ) represent E -states,(2) for all k = i + 1 , . . . , j − the pair ([ S ] k , [ S ] k +1 ) represents a state of type A , for k = i it represents a state of type A or D , and for k = j it represents a state oftype A or B .Proof. Indeed, assume that a vertex v ∈ D belongs to three consecutive levels [ S ] k , [ S ] k +1 ,[ S ] k +2 of the thickened path, and S ] k +1 = 2. If, say, v belongs to ∂ L S , then ∂ L [ S ] k +1 consists of the only vertex v . Then v is compact and hence N ( R ) ≥
4. On the otherhand, the left domain T k +1 in the level [ S ] k +1 has at most two common sides with [ S ] k and at most two common sides with [ S ] k +2 , whence N ( R ) = 4, and each of T k +1 ∩ [ S ] k and T k +1 ∩ [ S ] k +2 contains two sides. Therefore, the states representing the pairs ([ S ] k , [ S ] k +1 ),([ S ] k +1 , [ S ] k +2 ) are of types E R and E L respectively. In particular, this means that v cannot belong to four consecutive levels of the thickened path, and we see that the firstcase in the conclusion of the lemma takes place. This case is illustrated on Figure 4, where T k +1 = T . The vertex v is not shown there, it is the common end of the sides crossingedges T → T and T → T of the adjacency graph.It remains to consider the case when S ] k = 1 for all k = i + 1 , . . . , j −
1, and itclearly implies the second case of the lemma conclusion. (cid:3)
If ([ S ] k , [ S ] k +1 ) forms a configuration A ( e ), one can specify four numbers i ± ,L and i ± ,R as follows: i − ,α (resp., i + ,α ), α ∈ { L, R } , is the number of m ≤ k (resp., m ≥ k + 1)such that [ S ] m contains v α ( s k ). If the vertex v α ( s k ) is not defined, we set i ± ,α = 1.Note that it is not possible to have i − ,L > i − ,R > ∂ L [ S ] k and ∂ R [ S ] k contain only a vertex. For N ( R ) ≥ N ( R ) = 3 theremaining vertex of [ S ] k lies on ∂ D , hence the previous state ([ S ] k − , [ S ] k ) cannot belongto any of types A, . . . , E . The same argument applies to i + ,L and i + ,R as well.The convexity of ∂ S at v α ( s k ), α = L, R , is now equivalent to i − ,α + i + ,α ≤ n ( v α ( s k )).Therefore, the configuration A ( e ) can be subdivided as follows, see Figure 8: • A ( e ): all four i ± ,L/R equal one. • A L [ i − , i + ]( e ): here i − ,L = i − , i + ,L = i + , i − ,R = i + ,R = 1, and the indices i ± should satisfy 3 ≤ i − + i + ≤ n ( v L ( e )). • A R [ i − , i + ]( e ): symmetric to the previous case; here 3 ≤ i − + i + ≤ n ( v R ( e )). • A LR [ i − , i + ]( e ): here i − ,L = i − , i + ,R = i + , and i + ,L = i − ,R = 1. The conditions onthe indices i ± are 2 ≤ i − ≤ n ( v L ( e )) −
1, 2 ≤ i + ≤ n ( v R ( e )) − • A RL [ i − , i + ]( e ): symmetric to the previous case; here 2 ≤ i − ≤ n ( v R ( e )) − ≤ i + ≤ n ( v L ( e )) − Remark . Notice that if N ( R ) = 3 and R has a compact side, some of these states maybe absent. Namely, let s be the only compact side of R , let g be its label outside of R . If([ S ] k , [ S ] k +1 ) has the form A ( g ), then [ S ] k +2 should contain at least one of the domainsadjacent to the sides of [ S ] k +1 , hence either i + ,L or i + ,R is greater than one. Similarly,either i − ,L > i − ,R >
1. This case needs special consideration in several statementsbelow, and we usually refer to it as “the special case from Remark 4.3”. a) A ( e ) b) A L [2 , e ) c) A LR [3 , e ) e e e d) Impossible “ A R [2 , e )”: i − + i + > n ( v R ( e )), thusconvexity in v R ( e ) fails e) Impossible subtype:both i + ,L and i + ,R are greater than 1 e e Figure 8.
Possible (a–c) and impossible (d–e) subtypes for type A states.Dark and medium gray domains are respectively the past and the futuredomain for the current state, light gray domains are other domains fromthe thickened path. Here n ( v L ( e )) = 4, n ( v R ( e )) = 3.Note that even in this case the list of A ... ( g )-states is not completely empty. Indeed,since s is the only compact side, it should be paired to itself: g = g − . Therefore, the endsof s are swapped by the action of g , hence n ( v L ( g )) = n ( v R ( g )) = n . Let α and β be theangles of R in the ends of s . Consider the flower around a vertex v ∈ D . Note that thesides incident to v are alternatingly compact and non-compact, and the angles betweenthese sides are alternatingly α and β . Therefore, nα + nβ = 2 π . On the other hand, thesum of angles in the hyperbolic triangle R is α + β < π . Consequently, n ≥
3, and forexample, the state A LR [2 , g ) is well-defined. Definition 4.4.
The set of states
Ξ of our Markov chain is the set of all states of types
B, C, D, E from the set (cid:98)
Ξ and of all subtypes of type A states enumerated in the previouslist. We denote the projection from Ξ to (cid:98) Ξ by π .Finally, let us define sets Ξ S , Ξ F ⊂ Ξ as follows:Ξ S = { A ( e ) , A L [1 , i + ]( e ) , A R [1 , i + ]( e ) , B ( e L , e R ) } , Ξ F = { A ( e ) , A L [ i − , e ) , A R [ i − , e ) , D ( e L , e R ) } , ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 25 where the parameters i ± , e , e L , e R admit all possible values. In the special case fromRemark 4.3 these definitions are amended as follows: if g = g − is the label on the compactside of R , we include A LR [2 , i + ]( g ) , A RL [2 , i + ]( g ) to Ξ S and A LR [ i − , g ) , A RL [ i − , g ) toΞ F instead of those A ... ( g ).4.2. The Markov coding.Definition 4.5.
The set of admissible transitions in our Markov coding is enumeratedin the following list. We denote by Π the Ξ × Ξ adjacency matrix for the correspondingtopological Markov chain and write j → j (cid:48) if the transition from j to j (cid:48) is admissibleaccordingly to this list. • A ( e ) → A ( e (cid:48) ) for e (cid:48) non-adjacent to e − ,A L [1 , i + ]( e (cid:48) ) for e (cid:48) non-adjacent to e − , any admissible i + ,A R [1 , i + ]( e (cid:48) ) for e (cid:48) non-adjacent to e − , any admissible i + ,B ( e L , e R ) for any e L , e R non-adjacent to e − . • If i + > A L [ i − , i + ]( e ) → A L [ i − + 1 , i + − l ( e )) ,A LR [ i − + 1 , j + ]( l ( e )) , (cid:12)(cid:12)(cid:12)(cid:12) if i + = 2for any admissible j + , B ( l ( e ) , r ( l ( e )) − ) , if i + = 2 . • A L [ i − , e ) → (the same cases as for A ( e )). • A RL [ i − , i + ]( e ) → (the same cases as for A L [1 , i + ]( e )). • The transitions for the A R - and A LR -states are similar with the exchange of leftand right. • B ( e L , e R ) → C ( r ( e L ) , l ( e R )) if n ( e L , e R ) ≥ n ( e L , e R ) = 2 the transitions for B ( e L , e R ) are the same as for C n ( e L ,e R ) − ( e L , e R )below. • C i ( e L , e R ) → C i +1 ( r ( e L ) , l ( e R )), for i < n ( e L , e R ) − • C n ( e L ,e R ) − ( e L , e R ) → D ( r ( e L ) , l ( e R )) ,E L ( r ( r ( e L ) − ) , r ( e L ) , l ( e R )) ,E R ( r ( e L ) , l ( e R ) , l ( l ( e R ) − )) . • D ( e L , e R ) → A ( e (cid:48) ) for e (cid:48) non-adjacent to e − L , e − R ,A L [1 , i + ]( e (cid:48) ) A R [1 , i + ]( e (cid:48) ) (cid:41) (cid:12)(cid:12)(cid:12)(cid:12) for e (cid:48) non-adjacent to e − L , e − R ,any admissible i + , B ( e (cid:48) L , e (cid:48) R )) (cid:12)(cid:12)(cid:12)(cid:12) for e (cid:48) L , e (cid:48) R either not adjacent to e − L , e − R ,or adjacent via a vertex v with n ( v ) > A L [2 , i + ]( l ( e L )) for any admissible i + ,A R [2 , i + ]( r ( e R )) for any admissible i + . • E L ( e L , e M , e R ) has the same set of transitions as B ( e L , e M ). • E R ( e L , e M , e R ) has the same set of transitions as B ( e M , e R ). Denote(2) P S → FN − = { ( j , . . . , j N − ) ⊂ Ξ N : j ∈ Ξ S , j N − ∈ Ξ F , Π j k ,j k +1 = 1 for k = 0 , . . . , N − } . We now show that this set is in 1:1-correspondence with the set of the thickened paths oflength N . Theorem 4.6.
Let S = ([ S ] , . . . , [ S ] N ) be a thickened path starting at R . Then thereexists a unique sequence of states j ∈ P S → FN − such that for each k the pair ([ S ] k , [ S ] k +1 ) represents the configuration π ( j k ) . Moreover, this mapping of thickened paths of length N starting in R to the set P S → FN − is a bijection.Proof.
1. Consider a thickened path S . Each pair ([ S ] k , [ S ] k +1 ) represents a uniqueconfiguration ˆ k ∈ (cid:98) Ξ. For every configuration of type A one can recover indices i ± ,L/R asdescribed above, thus arriving at the states j k with π ( j k ) = ˆ k . Note that if π ( j ) = A ( e )then the state j has i − ,L = i + ,R = 1, so j ∈ Ξ S . In the special case from Remark 4.3we need to amend these indices as follows: if π ( j ) = A ( g ), where g is the label on thecompact side, we have either i + ,L ≥ i + ,R ≥
2. In the former case we then set i − ,L = 2, i − ,R = 1 and in the latter we set i − ,R = 2, i − ,L = 1. This corresponds to the additionof the “virtual domain” S − to our thickened path. Note that this addition still yields athickened path.Now one can check that all transitions j k → j k +1 are admissible. There are threetypes of restrictions on the pair of states ( j k , j k +1 ) in the list of Definition 4.5.First, there are the restrictions on the configurations ˆ k , ˆ k +1 . For example, if S + =[ S ] k is a pair of petals around a vertex v , and there are more than one petals in the sectoraround v that is bounded by the sides in S − ∩ S + and contains S + , then every domain in S + has the adjacent domain in S ++ = [ S ] k +1 , hence S ++ is the next pair of petals insidethis sector, and the triple ( S − , S + , S ++ ) of level in the thickened path corresponds to thetransition C k → C k +1 with the appropriate conditions on the labels given in the list fromDefinition 4.5.Further, there are restrictions on the indices i ± of A -states. For example if j k is an A -state with i + ,L > S ] k +2 should contain the next petal at the vertex v L ( s k ) inthe counterclockwise direction after [ S ] k +1 . If i + ,L > j k +1 is again an A -state with i − increased by one and i + decreased by one. On the otherhand, if i + ,L = 2, it is possible that [ S ] k +2 contains not only the above-mentioned petal,but also the domain adjacent to [ S ] k +1 along the next side on its boundary.Finally, there are restrictions related to the convexity of ∂ S . Namely, we need tocheck these conditions for the boundary vertices v that are incident to at least three levelsin S . These cases are enumerated in Proposition 4.2. In the cases when the correspondingsequence of states contains A -states, the convexity is guaranteed by the inequalities on theindices i ± for these states, so we need to consider only the cases when ( j k , j k +1 ) have types( E L , E R ), ( E R , E L ), and ( D, B ). In the first two cases the convexity at v is guaranteed:as we have seen in the proof of Proposition 4.2, in this case N ( R ) = 4, R is compact, andthe vertex u opposite to v in T k has n ( u ) = 2, as u correspond to a cycle of four domains ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 27 in the adjacency graph. Therefore, n ( v ) ≥
3, while v is incident to three domains in S .The remaining case ( D, B ) is specially mentioned in the Definition 4.5: if [ S ] k − ∩ [ S ] k and [ S ] k ∩ [ S ] k +1 have a common vertex v in [ S ] k , then we require that n ( v ) > with indices i ± ,L/R inanother way than the one described in the previous part of the proof. Indeed, one can seefrom the set of transitions that the indices i − ,L/R for the state j k are uniquely defined bythe configurations π ( j k − ) , π ( j k ) and by the same indices for the state j k − (assuming j k − has type A ). Therefore, one can successively find these indices for all states starting from i − ,L/R ( j ) = 1 as implied by j ∈ Ξ S . Similarly, the indices i + ,L/R ( j k ) are successivelyfound starting from the end of the sequence: i + ,L/R ( j N − ) = 1.As above, the special case from Remark 4.3 needs a special consideration if ˆ = A ( g ).Then we have j = A ( e ), where e is a label on a non-compact side of R . Then either e = l ( g ) or e = r ( g ), and, say, in the former case we have i + ,L ( j ) ≥ i + ,R ( j ) = 1, hence j ∈ Ξ S implies i − ,L ( j ) = 1, i − ,R = 2. The latter case is considered in the same way.Now we can recover i − ,L/R ( j k ) successively in the same way as in the general case.3. It remains to show that every sequence j ∈ P S → FN − represents a thickened path S .We can inductively recover all [ S ] k starting with [ S ] = R . For example, if j k = B ( e L , e R ),then [ S ] k contains only one domain, so we take its sides s e L and s e R having the outsidelabels e L and e R , and [ S ] k +1 contains two domains that are adjacent to [ S ] k via the sides s e L and s e R . Our set of transitions guarantees that the construction of the next level is welldefined, for example, for E L -state we can construct the right future domain via the sideswith the labels e M and e R , and the result is the same. Moreover, one can check inductivelythat for A -states the indices i ± ,L , i ± ,R coincide with the corresponding numbers of domainsadjacent to the ends of the side separating the past and the future domain.Therefore, we can define the boundary of the sequence S = ([ S ] , . . . , [ S ] N ) in theway similar to the one from Subsection 3.1. As we have seen in the previous part of theproof, our set of transitions guarantees that the curve ∂ S is convex, i. e. always turns left.Hence the boundary ∂ S is not self-intersecting, and we apply Proposition 3.23 to showthat S is a thickened path. All assumptions of this proposition are clear except that thestructure of nontrivial sections is the one described in Proposition 3.15. The second itemthere is clear: the corresponding sequence of states starts with B -state, ends with D -state,and contains C - and E -states in between. Every C -state yields just two edges T i → T i +1 and B i → B i +1 , while E L -state (resp., E R -state) yields also a crossing B i → T i +1 (resp., T i → B i +1 ). Now the third item follows from the fact that in every flower there is one petalwith the minimal index, one petal with the maximal one, and these petals are opposite.The segments joining the centers of the adjacent flowers belong to the boundaries of thesetwo petals, hence the angle between these segments differs from the straight angle by notmore than one sector. Therefore, for every j ∈ P S → FN − we have constructed the thickenedpath S and it is clear that the sequence of states corresponding to S coincides with j . (cid:3) Time-reversing involution.
The Markov coding defined above has the followingproperty: the Markov chain with time reversed, that is, the Markov chain with the matrix Π T , is the same as the initial one with the states renamed. Namely, define the followinginvolution ι : Ξ → Ξ, which can be informally described as the one swapping the past andthe future domains for the state: A ( e ) ↔ A ( e − ) , A L [ i − , i + ]( e ) ↔ A R [ i + , i − ]( e − ) ,A LR [ i − , i + ]( e ) ↔ A RL [ i + , i − ]( e − ) , B ( e L , e R ) ↔ D ( e − R , e − L ) ,C k ( e L , e R ) ↔ C n ( e L ,e R ) − k − ( e − R , e − L ) ,E α ( e L , e M , e R ) ↔ E α ( e − R , e − M , e − L ) ( α = L, R ) . Proposition 4.7.
The involution ι maps the topological Markov chain with the adjacencymatrix Π to the same chain with reversed time, that is, Π ι ( j ) ι ( k ) = Π kj . Also, ι (Ξ S ) = Ξ F and vice versa.Proof. This follows directly from the definitions. (cid:3)
Properties of the Markov coding.
Let us recall the definitions of the followingproperties of a topological Markov chain.
Definition 4.8.
Let X and M be respectively the set of states and the adjacency matrixof a topological Markov chain. 1. The topological Markov chain ( X, M ) is called stronglyconnected if for any x, y ∈ X there exists a sequence z = x, z , . . . , z k = y such that z j → z j +1 is an admissible transition for any j .2. The topological Markov chain ( X, M ) is called aperiodic if there is no p > x → y one has τ ( y ) = τ ( x ) + 1, where τ is the map τ : X → Z /p Z .In this subsection we will show that our Markov coding (Ξ , Π) is strongly connectedand aperiodic. We start by constructing some paths, which will be useful in the consider-ations below.
Proposition 4.9.
Assume that R does not belong to the special case from Remark 4.3.Then there exists m ≤ such that for every e ∈ G there exists a locally shortest pathof fundamental domains T e = ( T e = R , T e , . . . , T em ) such that its boundary is convex,the corresponding sequence of states i → · · · → i m − = t ( e ) starts with i = A ( e ) ,and if the path T e is extended to any locally shortest path R − n , . . . , R − , T e , . . . , T em , and j − n → · · · → j m − is the sequence of states corresponding to its convexification, then j m − = t ( e ) .Similarly, there exists a locally shortest path H e = ( H e − m , . . . , H e − , H e = R ) with theconvex boundary, its final state is A ( e ) , and its initial state h ( e ) does not change if thepath is extended arbitrarily to the right.Proof. Observe that the convexification may affect the positive half of the extended pathonly if the convexification step applies to an end of the side s = T e ∩T e , and then possiblyto several adjacent boundary vertices with straight angles. Now the proof is presented onFigure 9. Dashed lines there show the maximal possible extent of the segments I ( u ) from ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 29 a) e ˆ e b) e ˆ e ∞ c) e ˆ e ∞ d) e ˆ e ≥ ≥ e) e ˆ e ≥ ≥ ≥ ≥ f) e ˆ e ≥ ≥ ≥ g) e ˆ e ∞∞ h) e ˆ e ∞ ∞ i) e ˆ e ∞≥ ≥ ≥ ≥ j) e ˆ e ∞ ∞≥ ≥ ≥ ≥ Figure 9. “Tail” paths T e from Proposition 4.9 and Remark 4.10. Do-mains T e are shaded gray. Dashed lines indicate boundary sides that canbe affected by convexification of an extended path. Numbers in circles show n ( . . . ) for the corresponding vertices, ∞ means that this vertex lies on ∂ D .Note that t ( e ) = A (ˆ e ) in all cases except f), and t ( e ) = A L [2 , e ) in thecase f).the proof of Proposition 3.8, where u is an end of s . Note that as these segments do notreach endpoints of s m − = T em − ∩ T em , the convexification does not add a new domainwith index m −
1, so π ( j m − ) = π ( i m − ) = A (ˆ e ). The indices i ± ,L , i ± ,R for this state arealso not changed, hence j m − = i m − .If N ( R ) ≥
5, it is sufficient to take m = 3 and construct the path T e so that ∂ L T e and ∂ R T e contain at least two sides each, see Fig. 9a).If N ( R ) = 4, and R is non-compact, one can construct T e as shown of Fig. 9b), c).If N ( R ) = 4, R is compact and has no opposite vertices with n ( . . . ) = 2, there arethe following three cases. Let s be the side of T e = e R opposite to s , we choose T e to be the domain on the other side of s . If both ends of s has n ( . . . ) ≥
3, we use thepath shown on Fig. 9d). Otherwise, if both ends of s has n ( . . . ) ≥
3, the same holds for both ends of s , the side of T e opposite to s , and we use the path from Fig. 9e). Theremaining case is when both s and s has ends with n ( . . . ) = 2. Then these ends lie onthe same (say, right) boundary, and the left ends of s , s , and s all have n ( . . . ) ≥ N ( R ) = 3 and each its side is non-compact, we can use the paths shown onFig. 9g), h).As for the second statement, the path H e is constructed by H e − j = T e − j , j = 0 , . . . , m .In particular, we have h ( e ) = ι ( t ( e − )). (cid:3) Remark . In the special case from Remark 4.3 we define the paths T e as shown onFig. 9i), j). Note that if g = g − is a label on the compact side, then all statements ofProposition 4.9 hold for these paths except that T g contains three, not two domains thatare incident to one of the vertices of R = T g .The following statement shows two important combinations of “head” and “tail” pathsfrom the previous proposition. Proposition 4.11.
Let H e , T e be the paths from Proposition 4.9 or Remark 4.10.1) For any e , ˆ e such that ˆ e (cid:54) = e − the path R = ( R − m , . . . , R m ) , where R j = H ˆ ej for j ≤ , R j = T ej for j ≥ , is locally shortest. Let k − m → · · · → k m − be the sequence of statescorresponding to its convexification. Then k − m = h (ˆ e ) , k m − = t ( e ) .2) For any e , ˆ e such that ˆ e (cid:54) = e − denote S j = H ˆ ej for j = − m, . . . , , S = e R . At mostone vertex w of S is shared with S − . Choose any side of S that is not incident to w andlet ˜ e be the label on this side outside of S . Denote S j +1 = e T ˜ ej for j = 0 , . . . , m . Thenthe path S = ( S j ) m +1 j = − m is locally shortest and the sequence of states corresponding to theconvexification of S starts with h (ˆ e ) and ends with t (˜ e ) .Proof.
1) The convexity of R may fail only at the vertices of R = R , and in a non-special case any vertex of R is adjacent to at most three domains: R − , R , R , thusthe boundary angle at any vertex is at most minimally concave. In the special case fromRemark 4.3 only one of the elements e, ˆ e can be equal to g , thus only one of paths H ˆ e , T e can have three domains adjacent to some vertex of R . Therefore, at most one vertex u of R can have at most four adjacent domains from R , but since n ( u ) ≥
3, the boundaryangle at u is again at most minimally concave.2) The convexity of S may fail only at the common vertices of S and S , since all othervertices are adjacent to only one of paths H ˆ e or e T ˜ e . Let u be a common vertex of S and S . Then u is incident either only to H ˆ e and S or only to S and e T ˜ e , hence the angle of ∂ S at u is greater that the angle of ∂ H ˆ e or e T ˜ e there, which is at most straight. Therefore,the angle of ∂ S at the common vertices of S and S is at most minimally concave. Finally,the common vertices of S and S cannot be joined by a straight segment of ∂ S as theyare joined by s = S ∩ S , which lies inside S , thus ∂ S is almost convex.The last part of both statements follows directly from Proposition 4.9. (cid:3) Lemma 4.12.
The topological Markov chain (Ξ , Π) introduced in Definition 4.5 is stronglyconnected. ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 31
Proof.
The scheme of the proof is the following. We will consider several cases, and inevery case we firstly choose the set Ω ⊂ Ξ with ι (Ω) = Ω, and then we prove the followingproperties:(i) for every j ∈ Ξ there exists a path along the arrows in the adjacency graph of theMarkov chain from j to some state k ∈ Ω;(ii) for every k , k ∈ Ω there exists a path from k to k .Let us denote by j (cid:32) k that there exists a path going from j to k . Observe that proper-ties (i) and (ii) imply strong connectivity. Namely, from (i) we have that for any states j, j (cid:48) ∈ Ξ there exist k, k (cid:48) ∈ Ω such that j (cid:32) k and ι ( j (cid:48) ) (cid:32) k (cid:48) . Applying the involution ι to the second of these relations, we get ι ( k (cid:48) ) (cid:32) j (cid:48) . Finally, (ii) yields k (cid:32) ι ( k (cid:48) ), and wehave j (cid:32) k (cid:32) ι ( k (cid:48) ) (cid:32) j (cid:48) .The scheme of the proof of property (i) is similar in all cases. Namely, we show thatthis property holds for the set Ω = { A ( e ) : e ∈ G } . First of all, we can reach an A -statefrom a state j via a series of states of the form C . . . CDA . Then we transform an A -stateto a state with smaller index i + , arriving eventually to a set k with π ( k ) = A (ˆ e ) and i + ,L/R ( k ) = 1. Then the state k can be followed by a state A ( e ) where e is any label notadjacent to ˆ e − .In the special case of Remark 4.3 we can decrease i + until it reaches 2. Then if ˆ e isnon-compact and, say, i + ,L = 2 we have l (ˆ e ) = g , where g is the label on the compactside, hence the state k can be followed by A LR [ i − , g ). It remains to consider a state k state with π ( k ) = A ( g ) and, say, i + ,R ( k ) = 2, i + ,L ( k ) = 1. The state k can be followedby A R [ i − , e ), where ˜ e = r ( g ), and, finally, by A (˜ e ), since the labels ˜ e and ˜ e − are notadjacent.In all cases below we have Ω ⊂ Ω , and if Ω (cid:54) = Ω to establish property (i) we willcheck that for every j ∈ Ω we have j (cid:32) k for some k ∈ Ω.1. Let N ( R ) ≥
5. Here we set Ω = Ω and it remains to check property (ii). Let usconstruct the paths T e shown on Figure 9a) in a uniform manner, namely, we choose thedomains T e , in such a way that ∂ R T e and ∂ L T e contain one side each. Denote by t ( e )the label such that t ( e ) = A ( t ( e )); t ( e ) is shown as ˆ e on Figure 9. Note that e (cid:55)→ t ( e ) isa bijection: for any e (cid:48) ∈ G we consider a pair of domains T and T comprising the state A ( e (cid:48) ), then we add domains T and T such that ∂ L T and ∂ R T contain one side each.Then if ( T , T ) corresponds the state A ( e ), we have e (cid:48) = t ( e ) and T e = h T , where h ∈ G is such that T = h − R .Now take any f, ˆ f ∈ G such that ˆ f (cid:54) = f − and denote e = t − ( f ), ˆ e − = t − ( ˆ f − ).Since t is a bijection, e (cid:54) = ˆ e − , and we may consider the path from the first part ofProposition 4.11 for these e, ˆ e . This path shows that A ( ˆ f ) (cid:32) A ( f ) for any f, ˆ f suchthat ˆ f (cid:54) = f − . Finally, to have A ( f ) (cid:32) A ( f − ) choose any e ∈ G \ { f, f − } and observethat A ( f ) (cid:32) A ( e ) (cid:32) A ( f − ).2. Let us assume that R has a side with both ends lying on ∂ D . Then we set ω = { e ∈ G : v L ( e ) and v R ( e ) are undefined } , Ω = { A ( e ) : e ∈ ω } . It is clear that ω − = ω , hence ι (Ω) = Ω. Moreover, A ( e ) (cid:32) A (ˆ e ) for any e ∈ G ,ˆ e ∈ ω , e (cid:54) = ˆ e − . In particular, for e ∈ G \ ω the last inequality holds, hence property (i) isestablished. To check property (ii) it is remains to show that A ( e ) (cid:32) A ( e − ) for any e ∈ ω . To do this we choose any f ∈ G \ { e, e − } , and note that A ( e ) → A ( f ) → A ( e − ).3. Let us assume that N ( R ) = 4, R is non-compact, and there are no sides with bothends lying on ∂ D . Then we set Ω = Ω L (cid:116) Ω R , where ω α = { e : v α ( e ) is undefined } , Ω α = { A ( e ) : e ∈ ω α } , α = L, R, thus ι (Ω L ) = Ω R and vice versa. Note that if s e is a compact side then the oppositeside of R is non-compact, hence A ( e ) (cid:32) A ( f ) ∈ Ω, and property (i) holds. We pass toproperty (ii). Note that the one of the following holds:(3) ω L = { e } , ω R = { e − } , or ω L = { e, f } , ω R = { e − , f − } . Assume that the second case takes place. Let ( S , S ) represents the state A ( e ).Denote s = S ∩ S , thus the left end of s lies on ∂ D . Let s be the side of S oppositeto s . Then s has its right end on ∂ D , and if g is the label on s outside S , then g (cid:54) = e − and g ∈ ω R . Therefore, g = f − and we have A ( e ) → A ( f − ). Similarly, we have(4) A ( e ) (cid:33) A ( f − ) and A ( f ) (cid:33) A ( e − ) . Now consider the side s (cid:48) of S adjacent to s via its left end, which lies on ∂ D . Thenthe outside label g (cid:48) of s (cid:48) belongs to ω L . Assume that g (cid:48) = f . Then A ( e ) (cid:32) A ( f ) and,similarly, A ( e ) (cid:33) A ( f ), A ( e − ) (cid:33) A ( f − ). Therefore, in this case property (ii)holds.It remains to consider the case g (cid:48) = e and the first case from (3). In both these casesthe side s (cid:48) has the outside label e . Consider the domain S adjacent to S via the side s . Consider the sides of S that have labels e ± . They do not include the side s andwe have the two cases shown on Figure 10a), b), where the domains S , S , S are shownin bold. Then we construct a path S joining A ( e ) to A ( e − ) as shown on these figures.Thus we have A ( e ) (cid:32) A ( e − ) and property (ii) is established, keeping in mind (4) forthe second case in (3).4. Let us assume that N ( R ) = 4 and R is compact. Here we set Ω = Ω . Definethe bijection τ : G → G as follows. For e ∈ G consider the side of R with the insidelabel e . Then the outside label on the opposite side of R equals τ ( e ). Choose m such that τ m = id. Then A ( e ) → A ( τ ( e )), and, moreover, A ( e ) → A ( τ ( e )) → · · · → A ( τ m − ( e )) → A ∗ ( e ) , where A ∗ ( e ) is any of the following states: A ( e ), A L [1 , e ) (provided n ( v L ( e )) ≥ A R [1 , e ) (provided n ( v R ( e )) ≥ e ∈ G such that n ( v L ( e )) , n ( v R ( e )) ≥
3. Then if e L = l ( e ), e R = r ( e ), e D = τ ( e ) one can see that { e L , e R , e D } = G \ { e − } , and A ( e ) → A ( e D ) , A ( e ) (cid:32) A α [1 , e ) → A α [2 , e α ) (cid:32) A ( e α ) , α = L, R.
ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 33 a) ∞ e e e ∞ e e e b) ∞ e e e ∞ e e e c) ∞ ∞ f f f f Figure 10.
To the proof of Lemma 4.12. The notation is similar to thatof Figure 9. The first and the last domains in the paths are shaded darkgray and light gray, the domain shown with dashed lines is included in thethickened path if the path without this domain is not convex.Therefore, A ( e ) (cid:32) A ( f ) for any f (cid:54) = e − . Note that n ( v L ( e − )) = n ( v R ( e )) and viceversa, hence the argument above can be applied to e − , and we have A ( e − ) (cid:32) A ( f ) for f (cid:54) = e .Applying the involution ι to the obtained relations one can see that A ( f ) (cid:33) A ( e )for f (cid:54) = e − and A ( f ) (cid:33) A ( e − ) for f (cid:54) = e . It remains to choose any f (cid:54) = e , e − andwrite A ( e ) (cid:33) A ( f ) (cid:33) A ( e − ).5. Finally, consider the special case from Remark 4.3. We setΩ = { A ( e ) : s e is not compact } , the property (i) was established in the beginning of the proof, and the proof of property (ii),namely, that A ( f ) (cid:32) A ( f − ), is shown on Figure 10c). (cid:3) Lemma 4.13.
The topological Markov chain (Ξ , Π) defined in Definition 4.5 is aperiodic.Proof. Suppose that our Markov chain has the period c , that is, an index τ ( i ) ∈ Z /c Z isassigned to every state i ∈ Ξ and all allowed transitions i → j satisfy τ ( j ) = τ ( i ) + 1.Take any e , e and choose ˆ e (cid:54) = e − , e − . Then using the paths from the first partof Proposition 4.11 we have τ ( t ( e s )) = τ ( h (ˆ e )) + 2 m − s = 1 ,
2. Therefore, τ ( t ( e ))is the same for all e , we denote it by τ t . Similarly, τ ( h ( e )) equals the same number τ h for all e , and τ t = τ h + 2 m −
1. On the other hand, any path from the second part ofProposition 4.11 yields τ t = τ h + 2 m . Therefore, 2 m − ≡ m (mod c ), whence c = 1. (cid:3) Corollary 4.14.
By Lemmas 4.12 and 4.13 our Markov chain (Ξ , Π) is strongly connectedand aperiodic, hence there exists N > such that all entries of the matrix Π N are positive. Spherical sums and Markov operator
In this section we express the spherical averages in terms of powers of a Markovoperator, see Lemma 5.5, and obtain an identity relating this Markov operator with itsadjoint, see Lemma 5.6.5.1.
Thickened paths and the sphere in the group.
Consider a state k ∈ Ξ and let P L,R , F L,R be the left/right past and future domains of some representation of k in thetessellation; if the state has only one past or future domain, we have P L = P R or F L = F R respectively. Then the maps γ, ω : Ξ → G are defined as follows. Let P L = h R , then(5) F L = hγ ( k ) − R , F R = hω ( k ) − R . Clearly, these definitions do not depend on the choice of a representation for k . Lemma 5.1.
The above-defined maps γ and ω satisfy the following identities:1) ω ( ι ( k )) = ω ( k ) − for any k ∈ Ξ ,2) γ ( k ) = ω ( j ) − γ ( ι ( j )) − ω ( k ) for any j, k ∈ Ξ such that k → j is an admissible transition.Proof.
1. Consider any representation of a state k in the tessellation, P L,R , F L,R beingits left/right past and future domains. Then the domains (cid:101) P L,R = F R,L and (cid:101) F L,R = P R,L represents the state ι ( k ). Therefore, if P L = h R , F R = g R , we have ω ( k ) − = h − g , ω ( ι ( k )) − = g − h .2. Consider six domains P L,R , F L,R , G L,R such that the domains P L,R , F L,R arethe past and the future domains for a representation of k and the domains F L,R , G L,R are the past and the future domains for a representation of j . Let P L = h R . Then F L = hγ ( k ) − R and G R = hγ ( k ) − ω ( j ) − R . Since G R,L and F R,L represent the state ι ( j ), we have F R = hγ ( k ) − ω ( j ) − γ ( ι ( j )) − R . On the other hand, F R = hω ( k ) − R . (cid:3) Lemma 5.2.
Consider the set P S → Fn − defined by (2) . and the map Φ : P S → Fn − → G , where Φ( j → · · · → j n − ) = ω ( j n − ) γ ( j n − ) . . . γ ( j ) . Then Φ is a bijection of P S → Fn − onto the set S n ( G ) = { g ∈ G : | g | = n } .Remark . Note that for j n − ∈ Ξ F there is only one future fundamental domain hence ω ( j n − ) = γ ( j n − ). A reason for the separate notation on the last step will be explainedlater. Proof.
Theorem 4.6 shows that sequences from P S → Fn − bijectively correspond to thickenedpaths S from R to g R with g ∈ S n ( G ).Take j ∈ P S → Fn − , let S be the corresponding thickened path and define h k ∈ G so that h j R is the left domain in [ S ] k . Then g = h n − ω ( j n − ) − = h n − γ ( j n − ) − ω ( j n − ) − = · · · = [ ω ( j n − ) γ ( j n − ) . . . γ ( j )] − , and it remains to use that g (cid:55)→ g − is a bijective map on the sphere S n ( G ). (cid:3) ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 35
Parry measure.
Let Π be the adjacency matrix of the topological Markov chaindescribed in Definition 4.5. Lemma 4.12 shows that this Markov chain is strongly con-nected. The Perron—Frobenius theorem then yields that the matrix Π has a unique (up toscaling) eigenvector h with nonnegative coordinates, that all coordinates of h are positive,and that the eigenvalue λ corresponding to h is greater than the absolute value of anyother eigenvalue of Π: (cid:88) j Π ij h j = λh i , h i > i. The eigenvalue λ is called the Perron—Frobenius (PF) eigenvalue and h is called the rightPerron—Frobenius eigenvector. The matrix P with entries p ij = h j λh i Π ij is stochastic and the corresponding Markov chain has the following property: the prob-ability of an admissible sequence of transitions depends only on the initial and the finalstates in this sequence and the number of steps:(6) p i i . . . p i n − i n = h i n λ n h i Π i i . . . Π i n − i n = h i n λ n h i . The Markov measure defined by the matrix P is called the Parry measure . Its stationarydistribution is(7) p i = α i h i , where α is the left PF eigenvector of Π: α Π = λα , normalized by αh = (cid:80) i α i h i = 1.The time-reversing involution on the set of states implies certain symmetries for theParry measure. Proposition 5.4.
Let an involution ι : Ξ → Ξ be such that Π ι ( j ) ι ( k ) = Π kj for all j, k ∈ Ξ .Then the transition probability matrix ( p ij ) and the stationary distribution ( p i ) of the Parrymeasure corresponding to the matrix Π satisfy the following equations: p ι ( j ) = p j , p ι ( j ) ι ( k ) = p k p kj p j for all j, k ∈ Ξ . Proof.
Let J be the matrix for the substitution ι . Then J = J T = J − , J Π J = Π T . Let,as above, λ be the Perron—Frobenius (PF) eigenvalue for Π and let α and h be the left andthe right PF eigenvectors for Π normalized by αh = 1. Then αJ is a left PF eigenvectorfor J Π J = Π T , whence ( αJ ) T = J α T is a right PF eigenvector for M . Therefore, J α T isproportional to h : α ι ( k ) = ch k . Now p ι ( j ) = α ι ( j ) h ι ( j ) = ch j · c α j = p j and p ι ( j ) ι ( k ) = Π ι ( j ) ι ( k ) h ι ( k ) λh ι ( j ) = Π kj c − α k λc − α j = Π kj h j λh k h k α k h j α j = p k p kj p j . (cid:3) Markov operator.
Recall that the group G acts on a Lebesgue probability space( X, µ ) by measure-preserving maps T g . We denote T g f := f ◦ T − g for any function f ∈ L p ( X, µ ). Denote (cid:101) S n ( f ) = (cid:88) | g | = n T − g f, then S n ( f ) = (cid:101) S n ( f ) (cid:101) S n (1) = (cid:80) | g | = n T − g f { g : | g | = n } , where S n ( f ) is defined by (1).Consider the probability space Y = Ξ × X with the product measure ν = p × µ . Here p ( { i } ) = p i , where p i is defined by (7). It is convenient to identify a function ϕ ∈ L ( Y, ν )with a tuple of functions ( ϕ i ) i ∈ Ξ , where ϕ i ( · ) = ϕ ( i, · ).Define the following operators P, U : L ( Y, ν ) → L ( Y, ν ):(8) (
P ϕ ) i = (cid:88) j p ij T − γ ( i ) ϕ j , ( U ϕ ) j = T − ω ( j ) ϕ ι ( j ) . It is clear that P and U are measure-preserving Markov operators. Lemma 5.5.
For any function f ∈ L ( X, µ ) define a function ϕ ( f ) ∈ L ( Y, ν ) by ( ϕ ( f ) ) j = h ι ( j ) f, j ∈ Ξ S , , otherwise . Then (9) (cid:101) S n ( f ) = λ n − (cid:88) j ∈ Ξ S h j ( P n − U ϕ ( f ) ) j . Proof.
Indeed, (cid:101) S n ( f ) = (cid:88) i ∈ Ξ S ,i n − ∈ Ξ F ,i ,...,i n − ∈ Ξ M i i . . . M i n − i n − T − ω ( i n − ) γ ( i n − ) ...γ ( i ) f = λ n − (cid:88) i ∈ Ξ S ,i n − ∈ Ξ F ,i ,...,i n − ∈ Ξ h i p i i . . . p i n − i n − h i n − T − γ ( i ) . . . T − γ ( i n − ) T − ω ( i n − ) f = λ n − (cid:88) i ∈ Ξ S h i (cid:18)(cid:88) i p i i T − γ ( i ) (cid:18) . . . (cid:18)(cid:88) i n − p i n − i n − T − γ ( i n − ) T − ω ( i n − ) (cid:18) χ Ξ F ( i n − ) h i n − f (cid:19)(cid:124) (cid:123)(cid:122) (cid:125) ( Uϕ ( f ) ) in − (cid:19) . . . (cid:19)(cid:19) = λ n − (cid:88) i h i ( P n − U ϕ ( f ) ) i . (cid:3) Dual operator.
Let us recall that for ϕ, ψ ∈ L ( Y, ν ) we have (cid:104) ϕ, ψ (cid:105) = (cid:88) k ∈ Ξ p k (cid:104) ϕ k , ψ k (cid:105) . A short computation shows that if an operator Q has the form( Qϕ ) i = (cid:88) j ∈ Ξ p ij T ij ϕ j , then ( Q ∗ ψ ) j = (cid:88) k ∈ Ξ p k p kj p j T ∗ kj ψ k . ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 37
Therefore, for P defined by (8) we have(10) ( P ∗ ψ ) j = (cid:88) k ∈ Ξ p k p kj p j T γ ( k ) ψ k . Lemma 5.6.
The Markov operators P and U defined by (8) satisfy the following identities: U = U − = U ∗ , P ∗ = U P U.
Proof.
This identities follow from the identities for maps γ and ω given in Lemma 5.1,from Proposition 5.4. For example, let us prove the second identity:( U P U ψ ) j = T − ω ( j ) ( P U ψ ) ι ( j ) = T − ω ( j ) (cid:88) l p ι ( j ) ,l T − γ ( ι ( j )) ( U ψ ) l = (cid:88) l p ι ( j ) ,l T − ω ( j ) T − γ ( ι ( j )) T − ω ( l ) ψ ι ( l ) = (cid:88) k p ι ( j ) ,ι ( k ) T ω ( j ) − γ ( ι ( j )) − ω ( k ) ψ k . For the last equality we substitute l = ι ( k ) and use the first identity in Lemma 5.1. Nowusing Proposition 5.4, the second identity in Lemma 5.1 and formula (10) one can see thatthe right-hand side equals ( P ∗ ψ ) j . (cid:3) General theorem on pointwise convergence
In this section, extending Theorem 1 in [14], we prove a general pointwise convergencetheorem for powers of a Markov operator. Let (
Z, η ) be a Lebesgue probability space,and let Q be a measure-preserving Markov operator on L ( Z, η ). We need the followingassumptions.
Assumption 6.1.
There exists a decomposition Q = V W , where V and W are measure-preserving Markov operators, so that Q ∗ = W V . Assumption 6.2.
For every n ∈ N the equation Q n ψ = ψ has only constant solutions in L ( Z, η ) . Assumption 6.3.
There exists m ∈ N such that the equation ( Q ∗ ) m Q m ψ = ψ has onlyconstant solutions in L ( Z, η ) . Assumption 6.4.
There exists a sequence of operators A n and constants C, K > and a, b, n ∈ N so that for all n ≥ n the following inequality holds for any nonnegative ϕ ∈ L ( Z, η ) : (11) W Q n − a ϕ ≤ C b (cid:88) j = − b ( Q ∗ ) n Q n + j ϕ + A n ϕ, where W is the operator from Assumption 6.1, and the operators A n : L ( Z, η ) → L ( Z, η ) map nonnegative functions into nonnegative ones, for any p ∈ [1 , ∞ ] map L p to itself, and (cid:107) A n (cid:107) L p ≤ α n , with (cid:80) ∞ n = n α n ≤ K .Remark . Applying V (cid:48) = QV to both sides of (11), we arrive at the inequality(11 (cid:48) ) Q n − a (cid:48) ϕ ≤ CV (cid:48) b (cid:88) j = − b ( Q ∗ ) n Q n + j ϕ + A (cid:48) n ϕ with the same estimates on the norms of the operators A (cid:48) n . We will use both (11) and(11 (cid:48) ). Theorem 6.6.
Let Q : L ( Z, η ) → L ( Z, η ) be a measure-preserving Markov operatoracting on a Lebesgue probability space ( Z, η ) and satisfying Assumptions 6.1–6.4. Thenfor every function ϕ ∈ L log L ( Z, η ) the sequence Q n ϕ converges almost surely and in L to (cid:82) Z ϕ dη as n → ∞ . The proof follows the scheme from [14] and occupies the rest of this section.6.1.
Space of trajectories.
The space of trajectories corresponding to Q is the space( Z , P Q ), where Z = Z Z with the usual Borel sigma-algebra B Z , and the measure P Q isgiven by the Ionescu Tulcea Extension Theorem, where all probability kernels are the sameand depend only on one preceding element of the trajectory. Namely, for z ∈ Z and ameasurable set A ⊂ Z we define P Q ( z, A ) = P Q,z ( A ) = Q [ A ]( z ) . Then the probability measure P Q is given as follows: P Q { z m ∈ A m , . . . , z n ∈ A n } = (cid:90) z m ∈ A m dν ( z m ) (cid:20)(cid:90) z m +1 ∈ A m +1 d P Q,z m ( z m +1 ) (cid:20) . . . (cid:20)(cid:90) z n ∈ A n d P Q,z n − ( z n ) (cid:21) . . . (cid:21)(cid:21) . By definition the shift map σ : Z → Z , ( σ ( z )) n = z n +1 preserves the measure P Q .The sigma-algebras F lk , k, l ∈ Z ∪{ + ∞ , −∞} are the minimal complete sigma-algebrassuch that all functions π j : z = ( z n ) (cid:55)→ z j are measurable for k ≤ j ≤ l . For brevity wedenote F nn = F n . Let us also recall that the tail sigma-algebra is defined as F tail = ∞ (cid:92) n =0 F ∞ n . For any function ϕ ∈ L ( Z, η ) we define the function ϕ ∈ L ( Z , P Q ) by the formula ϕ ( z ) = ϕ ( z ). We have(12) E ( ϕ |F − n −∞ )( z ) = E ( ϕ |F − n )( z ) = ( Q n ϕ )( z − n ) , E ( ϕ |F + ∞ n )( z ) = E ( ϕ |F n )( z ) = (( Q ∗ ) n ϕ )( z n ) . Mixing of the operator Q . We start by proving mixing for ˜ Q = Q m where m isdefined in Assumption 6.3. Lemma 6.7.
Let ˜ Q be a measure-preserving Markov operator on L ( Z, η ) such that theequation ˜ Q ∗ ˜ Qϕ = ϕ has only constant solutions in L ( Z, η ) . Then for any ϕ, ψ ∈ L ( Z, η ) we have (13) (cid:104) ˜ Q n ϕ, ψ (cid:105) = (cid:90) Z ˜ Q n ϕ · ψ dη → (cid:90) Z ϕ dη (cid:90) Z ψ dη as n → ∞ . ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 39
Proof.
The statement follows from the mixing of the shift map σ in the trajectory space( Z , P ˜ Q ). To obtain the latter we shall prove that σ has K -property: there exists a sub-sigma-algebra K of the Borel sigma-algebra B Z such that K ⊂ σ K , (cid:87) ∞ n =0 σ n K = B Z , (cid:84) ∞ n =0 σ − n K = { ∅ , Z } .By the Rokhlin—Sinai theorem (see [40], [24, Ch. 18]) the K -property is equivalent tothe triviality of the Pinsker sigma-algebra Π( σ ). Consider F − = F −∞ . Then σ F − ⊂ F − and (cid:87) k ∈ Z σ k F − = B Z . Thus Π( σ − ) ⊂ F − (see, e.g., Lemma 18.7.3 in [24]). Similarly, for F + = F ∞ one has Π( σ ) ⊂ F + . Therefore, Π( σ ) = Π( σ − ) ⊂ F − ∩ F + = F .We have proved that any Π( σ )-measurable function ϕ ∈ L ( Z , P ˜ Q ) depend only onthe zeroth coordinate: ϕ ( z ) = ϕ ( z ). More generally, ϕ ( z ) = ϕ k ( z k ). Formulas (12) yieldthat ϕ − = ˜ Qϕ and ϕ = ˜ Q ∗ ϕ − . Therefore, ϕ = ˜ Q ∗ ˜ Qϕ , and the assumption of thelemma yields that ϕ = const, thus Π( σ ) is trivial. (cid:3) Corollary 6.8.
The operator Q is also mixing, that is, (13) holds for Q instead of ˜ Q .Proof. The sequence ( (cid:104) Q n ϕ, ψ (cid:105) ) n ≥ is the union of the subsequences ( (cid:104) Q nm + r ϕ, ψ (cid:105) ) n ≥ .Each of them converges to the desired limit by Lemma 6.7 applied to the pair of functions( Q r ϕ, ψ ). (cid:3) Triviality of the tail sigma-algebra.
The next step is to prove that the tail sigma-algebra for Q is trivial. First, we prove that the tail sigma-algebra cannot be totallynontrivial , that is, it cannot contain infinitely many different sets (up to sets of measurezero).The proof follows that of Lemma 6 in [14], which is a version of the 0–2 law in theform of Kaimanovich [31]. Lemma 6.9.
For a measure-preserving Markov operator R on L ( Z, η ) the following holds.If the tail sigma-algebra of R is totally nontrivial then for any b ∈ N and any ε > thereexist nonnegative functions ϕ, ψ ∈ L ∞ ( Z, η ) with averages equal to such that (14) lim sup n →∞ (cid:104) ( R ∗ ) n + b ϕ, ( R ∗ ) n ψ (cid:105) L ( Z,η ) + · · · + (cid:104) ( R ∗ ) n − b ϕ, ( R ∗ ) n ψ (cid:105) L ( Z,η ) < ε. Proof.
Let ( Z , P R ) be the corresponding trajectory space. If F tail contains infinitely manysubsets, it contains a subset of arbitrarily small measure. Indeed, split Z = A (2)1 (cid:116) A (2)2 ,where A (2) i ∈ F tail have nonzero measure. Then at least one of these parts can be split intotwo sets of nonzero measure (otherwise F tail contains only finitely many sets, the unions ofsome of A (2) i ). Repeating this procedure, we get Z = A ( n )1 (cid:116) · · · (cid:116) A ( n ) n . Then the measureof at least one of A ( n ) j is not more than 1 /n .Take any set A ∈ F tail with P R ( A ) < / (2 b + 1). Then the set B = Z \ (cid:83) bs = − b σ s ( A )has positive measure. DenoteΦ( z ) = A ( z ) / P R ( A ) , Ψ( z ) = B ( z ) / P R ( B ) . Observe that Φ and Ψ are nonnegative, F tail -measurable, bounded by some constant M ,their expectations are equal to 1, and (Φ ◦ σ − j ) · Ψ = 0 for j = − b, . . . , b . Set ϕ k = E (Φ |F k −∞ ), ψ k = E (Ψ |F k −∞ ). Note that ϕ k ( z ) depends only on z k , soabusing notation we use the same symbol ϕ k for the corresponding function in L ( Z, η ).For example, we will write ϕ k ◦ σ j ( z ) = ϕ k ( z k + j ).Clearly, ϕ k and ψ k are nonnegative and bounded by M . Therefore, the martingaleconvergence theorem gives that ϕ k → Φ, ψ k → Ψ in L ( Z , P R ). Moreover, ϕ k ( z k − j ) = ϕ k ◦ σ − j → Φ ◦ σ − j . Hence E ( ϕ k ( z k − j ) |F tail ) → E (Φ ◦ σ − j |F tail ) = Φ ◦ σ − j ( z ) , E ( ψ k ( z k ) |F tail ) → Ψ( z )in L ( Z , P R ). Since all these functions are bounded by the same constant M , for large k we have that (cid:90) Z E ( ϕ k ( z k − j ) |F tail ) E ( ψ k ( z k ) |F tail ) d P R < ε b + 1 . By (12) we have E ( γ ( z k ) |F ∞ n + k ) = [( R ∗ ) n γ ]( z n + k ), whence for any j = − b, . . . , b (cid:90) Z [( R ∗ ) n + j ϕ k ]( z n + k ) · [( R ∗ ) n ψ k ]( z n + k ) dη = (cid:90) Z E ( ϕ k ( z k − j ) |F ∞ n + k ) · E ( ψ k ( z k )) |F ∞ n + k ) d P R → (cid:90) Z E ( ϕ k ( z k − j ) |F tail ) · E ( ψ k ( z k )) |F tail ) d P R < ε b + 1 as n → ∞ . Therefore, the functions ϕ k and ψ k for large k satisfy (14). (cid:3) Lemma 6.10.
Under the assumptions of Theorem 6.6 the tail sigma-algebra for Q ∗ cannotbe totally nontrivial.Proof. Assuming the contrary, the inequality (14) in Lemma 6.9 for R = Q ∗ yields thatfor some nonnegative functions ϕ, ψ with their averages equal to 1 and for all sufficientlylarge n we have(15) (cid:104) ( Q n + b + · · · + Q n − b ) ϕ, Q n ψ (cid:105) L ( Z,η ) < ε. On the other hand, by Assumption 6.4 the left-hand side of (15) is not less than1 C (cid:104) W Q n − a ϕ − A n ϕ, ψ (cid:105) = 1 C (cid:104) Q n − a ϕ, W ∗ ψ (cid:105) − C (cid:104) A n ϕ, ψ (cid:105) → C + 0 . Here we use Corollary 6.8 here, note that the average values of both ϕ and W ∗ ψ are equalto 1. Therefore, for large n the left-hand side of (15) is larger than 1 /C − ε , so taking ε < / C we arrive at a contradiction. (cid:3) Lemma 6.11.
Under the assumptions of Theorem 6.6 the tail sigma-algebra for Q cannotbe totally nontrivial.Proof. Consider the trajectory space ( Z , P ) for the infinite sequence . . . , V, W, V, W, . . . ofMarkov operators, that is, P ( z n +1 ∈ A | z n ) = V [ A ]( z n ) , P ( z n +2 ∈ A | z n +1 ) = W [ A ]( z n +1 ) . Then one can check that P ( z n +2 ∈ A | z n ) = V W [ A ]( z n ) = Q [ A ]( z n ) , P ( z n +1 ∈ A | z n − ) = Q ∗ [ A ]( z n − ) , ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 41 hence the projections π : z = ( z n ) (cid:55)→ ( z n ) and π : z = ( z n ) (cid:55)→ ( z n +1 ) map the trajectoryspace ( Z , P ) to the trajectory spaces for Q and Q ∗ respectively. Therefore, the trivialityof the tail sigma-algebras in the trajectory spaces for Q and Q ∗ is respectively equivalentto the triviality of sigma-algebras F tail , = (cid:92) n (cid:95) k ≥ n F k and F tail , = (cid:92) n (cid:95) k +1 ≥ n F k +1 respectively. Thus to prove the lemma it is sufficient to show that in ( Z , P ) the sigma-algebras F tail , F tail , , F tail , are trivial simultaneously. Obviously, the triviality of F tail implies the triviality of F tail ,j , j = 0 ,
1. Let us prove the converse.Consider any A ∈ F tail and check that, say, A ∈ F tail , . Indeed, A ∈ (cid:87) m ≥ n F m forevery n , and we can eliminate any finite number of F k with odd k from the set of m ’sthere:(16) A ∈ F n ∨ F n +2 ∨ · · · ∨ F n + s − ∨ (cid:95) m ≥ n + s ) F m . Consider the conditional probability P ( · | z n , z n +2 , . . . ) with respect to the sigma-algebra (cid:87) k ≥ n F k . As (16) shows, with respect to this conditional probability A depend onlyon “odd tail” (cid:87) k ≥ n + s F k +1 . But since the odd coordinates z n +1 , . . . , z n + s )+1 , . . . areindependent for the fixed even coordinates z n , . . . , z n + s ) , . . . , by Kolmogorov’s 0–1 Lawwe obtain that A is trivial with respect to this conditional probability, so A is measurablewith respect to (cid:87) k ≥ n F k , and hence A ∈ F tail , . (cid:3) Lemma 6.12.
Under the assumptions of Theorem 6.6 the tail sigma-algebra for Q istrivial.Proof. It remains to eliminate the case when F tail contains only finitely many differentsets. Assume that Z = A (cid:116) · · · (cid:116) A r , r >
1, where each A j ∈ F tail has no nontrivialsubsets belonging to F tail . The shift map σ interchanges these subsets, whence for A = A there exists n such that σ n A = A . As in Lemma 6.9, we define Φ = A / P Q ( A ) and ϕ k ( z k ) = ϕ k ( z ) = E (Φ |F k −∞ ). Then E (Φ |F k −∞ ) ◦ σ n = E (Φ ◦ σ n |F k + n −∞ ) = E (Φ |F k + n −∞ ) = ϕ k + n , hence ϕ k + n ( z k ) = ϕ k + n ◦ σ − n = E (Φ |F k −∞ ) == E ( E (Φ |F k + n −∞ ) |F k −∞ ) = E ( ϕ k + n ( z k + n ) |F k −∞ ) = [ Q n ϕ k + n ]( z k ) . Thus we arrive at the equation ϕ k + n ( z k ) = [ Q n ϕ k + n ]( z k ) and Assumption 6.2 impliesthat ϕ k + n is constant. Taking averages, we get E ( ϕ k + n ) = E (Φ) = 1, thus ϕ l ≡ l . But this contradicts to the convergence ϕ l → Φ (cid:54)≡
1, which was obtained in proof ofLemma 6.9. (cid:3)
Convergence.Lemma 6.13 (see [31]; [14, Propositions 4, 5]) . For a measure-preserving Markov oper-ator R on ( Z, η ) with the trivial tail sigma-algebra we have R n ϕ → (cid:82) Z ϕ dη , where theconvergence takes place in L for ϕ ∈ L ( Z, η ) and in L for ϕ ∈ L ( Z, η ) . Lemma 6.14 ([14, Lemma 8]) . For a measure-preserving Markov operator R on ( Z, η ) for any p > there exists a constant A p > such that for every nonnegative function ϕ ∈ L p ( Z, η ) we have (cid:13)(cid:13)(cid:13) sup n ≥ ( R ∗ ) n R n ϕ (cid:13)(cid:13)(cid:13) L p ≤ A p (cid:107) ϕ (cid:107) L p . Similarly, there exists a constant A log > such that for every nonnegative function ϕ ∈ L log L ( Z, η ) we have (cid:13)(cid:13)(cid:13) sup n ≥ ( R ∗ ) n R n ϕ (cid:13)(cid:13)(cid:13) L ≤ A log (cid:107) ϕ (cid:107) L log L . Remark . The following inequalities hold for any s ∈ Z :(17) (cid:13)(cid:13)(cid:13) sup( R ∗ ) n + s R n ϕ (cid:13)(cid:13)(cid:13) L p ≤ A p (cid:107) ϕ (cid:107) L p , (cid:13)(cid:13)(cid:13) sup( R ∗ ) n + s R n ϕ (cid:13)(cid:13)(cid:13) L ≤ A log (cid:107) ϕ (cid:107) L log L , the supremums here are taken over all n such that both n and n + s are nonnegative.Indeed, for s > R s ϕ and use inequality (cid:107) R s ϕ (cid:107) L log L ≤ (cid:107) ϕ (cid:107) L log L .For s < R ∗ ) | s | (cid:104) sup n ≥ ( R ∗ ) n R n ϕ (cid:105) ≥ sup n ≥ ( R ∗ ) n + | s | R n ϕ which yields the same equation for the L -norms of both sides. Note also that the L -normof left-hand side does not exceed the same norm for the function in the square bracketsthere. Hence A log (cid:107) ϕ (cid:107) L log L ≥ (cid:13)(cid:13)(cid:13) sup n ≥ ( R ∗ ) n R n ϕ (cid:13)(cid:13)(cid:13) ≥ (cid:13)(cid:13)(cid:13) sup n ≥ ( R ∗ ) n + | s | R n ϕ (cid:13)(cid:13)(cid:13) , and it remains to change n (cid:55)→ n + s . Proof of Theorem 6.6.
Combining (17) for R = Q and Assumption 6.4 in the form (11 (cid:48) ),for any nonnegative function ϕ ∈ L log L ( Z, η ) we obtain(18) (cid:13)(cid:13)(cid:13) sup n ≥ n Q n − a (cid:48) ϕ (cid:13)(cid:13)(cid:13) L ≤ C (cid:13)(cid:13)(cid:13) V (cid:48) (cid:16) sup n ≥ n b (cid:88) j = − b ( Q ∗ ) n Q n + j ϕ (cid:17)(cid:13)(cid:13)(cid:13) L + (cid:13)(cid:13) sup n ≥ n A (cid:48) n ϕ (cid:13)(cid:13) L ≤ (2 b + 1) A log C (cid:107) ϕ (cid:107) L log L + (cid:88) n ≥ n (cid:107) A (cid:48) n ϕ (cid:107) L ≤ B log (cid:107) ϕ (cid:107) L log L . Decomposing a function ϕ into its positive and negative parts we obtain (18) for all real-valued ϕ ∈ L log L ( Z, η ) with a larger B log . The same estimates hold for L p -norm.Now consider a real-valued function ϕ ∈ L ( Z, η ) with the zero average. Applying(18) to ( Q k ϕ ) we have (cid:13)(cid:13)(cid:13) sup m ≥ n + k Q m − a (cid:48) ϕ (cid:13)(cid:13)(cid:13) L = (cid:13)(cid:13)(cid:13) sup n ≥ n Q n +2 k − a (cid:48) ϕ (cid:13)(cid:13)(cid:13) L ≤ B (cid:107) Q k ϕ (cid:107) L . ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 43
Since the right-hand side tends to zero by Lemma 6.13, the sequence Q m − a (cid:48) ϕ tends tozero almost everywhere and in L as m → ∞ .We now extend pointwise convergence to all ϕ ∈ L log L . Namely, for a real-valuedfunction ϕ ∈ L log L ( Z, η ) with zero average consider ϕ (cid:48) ∈ L ( Z, η ) with zero average suchthat (cid:107) ϕ − ϕ (cid:48) (cid:107) L log L ≤ ε/B log . Then almost surely we havelim sup n →∞ | Q n − a (cid:48) ϕ ( z ) | ≤ lim sup n →∞ | Q n − a (cid:48) ϕ (cid:48) ( z ) | + lim sup n →∞ | Q n − a (cid:48) ( ϕ − ϕ (cid:48) )( z ) | . By convergence for functions in L , the first term in the right-hand side equals zero, while,by the maximal inequality, the second satisfies (cid:107) lim sup n →∞ | Q n − a (cid:48) ( ϕ − ϕ (cid:48) )( z ) |(cid:107) L ≤ B log (cid:107) ϕ − ϕ (cid:48) (cid:107) L log L ≤ ε. Therefore, lim sup | Q n − a (cid:48) ϕ ( z ) | ≤ δ outside of the set of measure less than ε/δ . Taking ε = 1 /l → δ = 1 /l (cid:48) → L follows from the same decomposition: (cid:107) Q n − a (cid:48) ϕ ( z ) (cid:107) L ≤ (cid:107) Q n − a (cid:48) ϕ (cid:48) (cid:107) L + (cid:107) Q n − a (cid:48) ( ϕ − ϕ (cid:48) )( z ) (cid:107) L , where the first term tends to zero even with L -norm instead of L , and the second termis less than (cid:107) ϕ − ϕ (cid:48) (cid:107) L ≤ ε/B log .Finally, combining the obtained convergence Q m − a (cid:48) ϕ → Qϕ in place of ϕ , we conclude that Q n ϕ → (cid:3) Proof of Theorem A
The proof of Theorem A in the case of trivial I G is based on Theorem 6.6. Namely,we denote(19) Q = P , V = P U, and W = U P, where P and U are defined in (8). Assumption 6.1 now holds due to Lemma 5.6. In thenext two subsections we will check that the remaining Assumptions 6.2, 6.3, and 6.4 holdfor the operators defined by (19), provided I G is trivial. Finally, in Subsection 7.3 wedeal with the case of nontrivial I G and conclude the proof of Theorem A.7.1. P k - and ( P ∗ ) k P k -invariant functions. To check Assumptions 6.2 and 6.3 we ex-press the equations from these assumptions in terms of the components ϕ j , j ∈ Ξ, of afunction ϕ . Proposition 7.1.
Let P be the Markov operator defined by (8) . Then the following holds.1) A function ϕ ∈ L ( Y, ν ) is a solution to the equation P k ϕ = ϕ if and only if for everyadmissible sequence i → i → · · · → i k of states we have (20) ϕ i = T − γ ( i ) . . . T − γ ( i k − ) ϕ i k .
2) If k ≥ N , where N is defined in Corollary 4.14, then a function ϕ ∈ L ( Y, ν ) is asolution for ( P ∗ ) k P k ϕ = ϕ if and only if for every admissible sequences i → i → · · · → i k and j → i → · · · → i k with i = j we have T − γ ( i ) . . . T − γ ( i k − ) ϕ i k = T − γ ( j ) . . . T − γ ( j k − ) ϕ j k . Proof.
1) The equation P k ϕ = ϕ is equivalent to ϕ i = ( P k ϕ ) i = (cid:88) i ,...,i k p i i . . . p i k − i k T − γ ( i ) . . . T − γ ( i k − ) ϕ i k , hence, since T g ’s are unitary,(21) (cid:107) ϕ i (cid:107) L ≤ (cid:88) i ,...,i k p i i . . . p i k − i k (cid:107) ϕ i n (cid:107) L . Multiplying these inequalities by p i and summing them up for all i ∈ Ξ we obtain (cid:88) i p i (cid:107) ϕ i (cid:107) L ≤ (cid:88) i k (cid:20) (cid:88) i ,...,i k − p i p i i . . . p i k − i k (cid:21) (cid:107) ϕ i k (cid:107) L = (cid:88) i k p i k (cid:107) ϕ i k (cid:107) L . Therefore, for each i inequality (21) is indeed an equality, and the vector ( (cid:107) ϕ i (cid:107) L ) i ∈ Ξ isa left eigenvector of the matrix Π k with nonnegative coordinates, where Π = ( p ij ). As theMarkov chain corresponding to Π is strongly connected and aperiodic, by the Perron—Frobenius theorem the vector ( (cid:107) ϕ i (cid:107) L ) i ∈ Ξ is proportional to (1 , . . . , ϕ i hasthe same norm.Finally, in the Hilbert space L ( X, µ ) the triangle inequality (21) reaches equality onlyif all nonzero summands are proportional to each other with positive coefficients, whence ϕ i = c · T − γ ( i ) . . . T − γ ( i n − ) ϕ i n . Calculating the L -norms of both sides, we get c = 1.2) Similarly, ( P ∗ ) k P k ϕ = ϕ yields ϕ j k = (cid:88) j ,...,j k − i ,...,i k (cid:20) p j p j j . . . p j k − j k p j k p j i p i i . . . p i k − i k × T γ ( j k − ) . . . T γ ( j ) T − γ ( i ) . . . T − γ ( i k − ) ϕ i k (cid:21) , The remaining is the same as in the first statement: ( (cid:107) ϕ i (cid:107) ) i ∈ Ξ is a left eigenvector of astochastic matrix with positive entries (indeed, for given i k , j k we choose j k − , . . . , j , j arbitrarily with p j s j s +1 >
0, then we can choose i , . . . , i k − since (Π k ) j i k > k ≥ N .Therefore, the L -norms of all ϕ i ’s are equal, and the same argument with the triangleinequality completes the proof. (cid:3) Lemma 7.2.
There exists k ∗ such that for any k ≥ k ∗ if a function ϕ ∈ L ( Y, ν ) satisfiesequalities (22) T − γ ( i ) . . . T − γ ( i k − ) ϕ i k = T − γ ( j ) . . . T − γ ( j k − ) ϕ j k for all admissible sequences i → i → · · · → i k , j → j → · · · → j k with i = j , then ϕ ( x, j ) does not depend on j : ϕ ( x, j ) = ϕ ◦ ( x ) , and ϕ ◦ ( x ) is G -invariant.Remark . If (22) holds for all pairs of sequences of a given length k , then it holds forany pair of sequences i → i → · · · → i k (cid:48) , j → j → · · · → j k (cid:48) of length k (cid:48) ≤ k with i = j . Indeed, append an arbitrary prefix i − ( k − k (cid:48) ) → · · · → i to these sequences and ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 45 apply (22) to the resulting sequences of length k . One can see that T − γ ( i − ( k − k (cid:48) )+1 ) . . . T − γ ( i ) cancels out and we arrive to (22) for the initial sequences of length k (cid:48) .Let us first deduce Assumptions 6.2 and 6.3 from Lemma 7.2. Corollary 7.4.
Assumptions 6.2 and 6.3 hold for the operator Q defined by (8) and (19) ,assuming I G is trivial.Proof.
1) Suppose that Q n ϕ = ϕ . Choose l such that k = 2 nl ≥ k ∗ . Then P k ϕ = ( Q n ) l ϕ = ϕ . Therefore, (22) holds, as both its sides are equal to T γ ( i ) ϕ i by (20). Lemma 7.2 thenimplies that all ϕ j are equal to the same function ϕ ◦ , where ϕ ◦ is G -invariant and henceconstant.2) Suppose that ( Q ∗ ) m Q m ϕ = ϕ , where m satisfies 2 m ≥ max( k ∗ , N ) and N is definedin the second part of Proposition 7.1. Then this proposition implies that (22) holds for ϕ with k = 2 m , so ϕ is constant. (cid:3) It remains to prove Lemma 7.2.
Proof of Lemma 7.2.
We will prove the statement of the lemma for k ∗ = 2 m , where m ≤ e ∈ G consider a path T e from thisproposition. Let i e → · · · → i em − = t ( e ) be the corresponding sequence of states. Denote ψ e = T − γ ( i e ) . . . T − γ ( i em − ) ϕ i em − = T g e ϕ i em − , where g e = γ ( i e ) − . . . γ ( i em − ) − is such that T em − = g e R .Take any e , e and choose ˆ e (cid:54) = e − , e − . Let R s ( s = 1 ,
2) be the path from the firstpart of Proposition 4.11 applied to e s and ˆ e , and let j s − m → · · · → j sm − be the sequenceof states corresponding to its convexification, thus j s − m = h ( e ), j sm − = t ( e s ). Applying(22) to these two sequences we get(23) T γ ( j − m +1 ) − ...γ ( j m − ) − ϕ t ( e ) = T γ ( j − m +1 ) − ...γ ( j m − ) − ϕ t ( e ) . Note that [ R s ] − m +1 = R s − m +1 = H ˆ e − m +1 , [ R s ] m − = R sm − = T e s m − . Therefore, if H ˆ e − m +1 = ˆ g R we have g e s = ˆ gγ ( j s − m +1 ) − . . . γ ( j sm − ) − , and (23) takes form T ˆ g − g e ϕ t ( e ) = T ˆ g − g e ϕ t ( e ) , or T ˆ g − ψ e = T ˆ g − ψ e . Thus all ψ e areequal to the same function ψ ◦ .Now take again any e , e , choose ˆ e (cid:54) = e − , e − and apply the same argument to thepaths from the second part of Proposition 4.11 for ˆ e , e s and ˜ e s . We obtain that T ˆ g − e ψ ˜ e = T ˆ g − e g ˜ e ϕ t (˜ e ) = T ˆ g − e g ˜ e ϕ t (˜ e ) = T ˆ g − e ψ ˜ e , Therefore, T e − e ψ ◦ = ψ ◦ , so ψ ◦ is G -invariant. (cid:3) Proof of Assumption 6.4.Lemma 7.5.
Assumption 6.4 holds for the operators defined by (8) and (19) . Namely, (11) holds for a = 6 , b = 2 and n being the maximum of n ( v ) over all vertices of R . The proof rests on the following geometric statement. Denote by P M the set of alladmissible sequences i = ( i → · · · → i M ) of states in the Markov chain. i i i n − a i n − a +1 j j j n − j n k k k n + s ) − k n + s ) ··· ··· ··· ··· ··· ··· g i g j g k Figure 11.
To Lemma 7.6
Lemma 7.6.
There exists a subset E N − ⊂ P N − with E N − = O ( λ N ) , where λ isthe Perron—Frobenius eigenvalue of the Markov chain, and the following holds for every i ∈ P N − \ E N − : there exists α ∈ { , , , } , β ∈ {− , , , } and paths j = ( j →· · · → j N − β + α − ) , k = ( k → · · · → k N + β + α − ) with the following properties.(i) j = k , j N − β + α − = ι ( i ) , k N + β + α − = i N − .(ii) Let R = ([ R ] , . . . , [ R ] N ) be any thickened path representing i . Construct repre-sentations Q = ([ Q ] , . . . , [ Q ] N − β + α ) , S = ([ S ] , . . . , [ S ] N + β + α ) of sequences j and k respectively, with [ Q ] N − β + α = [ R ] , [ S ] N + β + α = [ R ] N . Then [ Q ] = [ S ] .Finally, the mapping i (cid:55)→ ( j, k ) is injective.Remark . We slightly abuse our terminology here by applying the term “thickenedpath” to an arbitrary indexed set of domains generated by a sequence of states of theMarkov chain, i.e. without conditions on their first and last states. This will create noproblems, as all operations in the proof of this lemma will not affect states and n =max n ( v ) domains near each of the ends of the sequences i, j, k ).The statement of this lemma is illustrated by Figure 11. Every state from the se-quences i, j, k is represented by a straight arrow, while the past and the future domains ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 47 of the state are shown as the pairs of squares near the start and the end of this arrow.Other details of this figure, including numbering for elements of j , k , which is differentfrom that in the lemma statement, are discussed below when proving equality (27). Proof of Lemma 7.5 assuming Lemma 7.6.
The values of a and b from the statement ofLemma 7.5 are in fact b = max | β | , a = b + max α , where the possible values of α and β are described in Lemma 7.6. Thus for a nonnegative function ϕ ∈ L ( Y, ν ) we have(24) (
W Q n − a ϕ ) l = ( U P n − a +1 ϕ ) l = (cid:88) i ,...,i n − a +1 p ι ( l ) ,i p i i . . . p i n − a i n − a +1 T − ω ( l ) T − γ ( i ) T − γ ( i ) . . . T − γ ( i n − a ) ϕ i n − a +1 . The coefficient in a term of the last sum is nonzero if and only if the sequence ι ( l ) → i →· · · → i n − a +1 is admissible, and, as ( p ij ) is the matrix for the Parry measure, formula (6)yields(25) ( W Q n − a ϕ ) l ≤ (cid:101) C λ − n (cid:88) i ∈ P n − a +1 ,i = ι ( l ) T ω ( i ) T − γ ( i ) T − γ ( i ) . . . T − γ ( i n − a ) ϕ i n − a +1 . Similarly,(( Q ∗ ) n Q n + s ϕ ) l = (cid:88) j n − ,...,j ,k ,...,k n +2 s p j p l p j n − l p j n − j n − . . . p j j p j k p k k . . . p k n +2 s − k n +2 s × T γ ( j n − ) . . . T γ ( j ) T − γ ( j ) T − γ ( k ) . . . T − γ ( k n +2 s − ) ϕ k n +2 s , and (6) yields the estimate(26) (( Q ∗ ) n Q n + s ϕ ) l ≥ (cid:101) C λ − n (cid:88) j ∈ P n ,k ∈ P n +2 s ,j = k ,j n = l T γ ( j n − ) . . . T γ ( j ) T − γ ( k ) T − γ ( k ) . . . T − γ ( k n +2 s − ) ϕ k n +2 s . Here (cid:101) C is chosen in such a way that this inequality holds for any s with | s | ≤ b . ApplyLemma 7.6 to a sequence i from (25). There are O ( λ n ) sequences from E n − a +1 , andthe corresponding terms in (24) comprise ( A n ϕ ) l . We see that (cid:107) A n (cid:107) = O ( λ − n ), so theseries (cid:80) n (cid:107) A n (cid:107) converges.Suppose now that i / ∈ E n − a +1 . Then Lemma 7.6 provides paths ˜ ∈ P n − a − β + α and ˜ k ∈ P n − a + β + α . Denote γ = a − α + β ≥ t = (˜ − γ → · · · → ˜ ). Prepend this “left tail” t to ˜ and ˜ k to construct j and k . Now j ∈ P n , k ∈ P n +2 β , and one can see that the statements (i) and (ii) of Lemma 7.6 stillhold for j and k .Let us prove that the terms in (25) and (26) (for s = β ) corresponding to these i , j ,and k are equal. Indeed, l = ι ( i ) = j n and k n +2 s = i n − a +1 , so it remains to prove that(27) ω ( i ) γ ( i ) − γ ( i ) − . . . γ ( i − n − a ) (cid:124) (cid:123)(cid:122) (cid:125) g i = γ ( j n − ) . . . γ ( j ) (cid:124) (cid:123)(cid:122) (cid:125) g j γ ( k ) − . . . γ ( k n +2 s − ) − (cid:124) (cid:123)(cid:122) (cid:125) g k , where we define g i , g j , g k as shown.Consider the thickened paths R , Q , S representing i , j , k and satisfying statement (ii)in Lemma 7.6. Let h R be the right domain in [ R ] . Then by the definitions of γ ( · ) and ω ( · ) (see formula (5)) hω ( i ) R is the left domain in [ R ] , hω ( i ) γ ( i ) − R is the leftdomain in [ R ] , . . . , hg i R is the left domain in [ R ] n − a +1 .On the other hand, as [ R ] = [ Q ] n +1 and i = ι ( j n ), we have that [ R ] = [ Q ] n and h R is the left domain in [ Q ] n . The same argument as above gives us that hg j R is theleft domain in [ Q ] , which coincides with the left domain in [ S ] , so hg j g k R is the leftdomain in [ S ] n +2 s . But as [ S ] n +2 s +1 = [ R ] n − a +2 and k n +2 s = i n − a +1 , we obtainthat [ S ] n +2 s = [ R ] n − a +1 , and their left domains coincide. Therefore, hg i R = hg j g k R ,so (27) holds. This is shown on Figure 11: the curved arrows link the domains hg R , where g is an initial segment of either the left-hand or the right-hand side of (27); the shadedregions correspond to g = id (left), g = g j (top), g = g i = g j g k (right).Therefore, we have proved that for every term in the right-hand side of (25) exceptthose with i ∈ E n − a +1 there exists the equal term in the right-hand side of (26) for | s | ≤ b . Moreover, due to the last statement in Lemma 7.6 different sequences i yielddifferent pairs ( j, k ), so (cid:88) i ∈ P n − a +1 \ E n − a +1 ,i = ι ( l ) T g i ϕ i n − a +1 ≤ b (cid:88) s = − b (cid:88) j ∈ P n ,k ∈ P n +2 s ,j = k ,j n = l T g j g k ϕ k n +2 s . Combining this inequality with (25) and (26) we establish (11). (cid:3)
We are now passing to the proof of Lemma 7.6.
Proof of Lemma 7.6.
1. Take any i ∈ P N − . We start by choosing the correspondingvalue of β . Claim 7.8.
There exist β ∈ {− , , , } , a domain R ∗ ∈ [ R ] N − β , and a side s ⊂ ∂ R ∗ ∩ ∂ R . In the special case from Remark 4.3 it is also required that s is either non-compactor the angle of ∂ R at one of the ends of s is less than π .Proof. In fact, in a non-special case we may choose β ∈ {− , , } . Assume first that N ( R ) ≥
4. If for some β ∈ {− , , } there are two domains in [ R ] N − β , then among their2 N ( R ) sides there are at most six sides common with [ R ] N − β − or [ R ] N − β +1 , hence thereexists a side s ∈ ∂ [ R ] N − β ∩ ∂ R . It remains to consider the case when [ R ] N − β for all β contain one domain each. Then [ R ] N has two sides bordering [ R ] N ± , and we choose anyother side as s .Now consider a non-special case with N ( R ) = 3. Any side of R has an end on ∂ D , sothere are no polygonal chains of three sides in T R , and hence there are no states of type ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 49 E . Therefore, the previous argument works with the only amendment: if [ R ] N − β containstwo domains, then at most four of their six sides are common with the adjacent domainsfrom [ R ] N − β ± , and we may choose s to be any other side.In the special case two states of type E cannot be successive: indeed, between two E -states there are n ( v ) − C , and here n ( v ) ≥ v ∈ D .Therefore, if [ R ] N − β contains two domains, at most five of six sides are common with[ R ] N − β ± , so we choose s to be any other side. Note that two domains in [ R ] N − β has acommon vertex w ∈ D , and s is not adjacent to w , hence s is not compact. It remains toconsider the case when each of [ R ] N − β , β = − , , , s bethe side of R N − not common with R N − ∪ R N . We can set s = ˆ s except if ˆ s is compactand ∂ R has the straight angle in the end u of ˆ s incident to R N . In the latter case we mayassume that, say, ˆ s belong to ∂ L R . Since n ( u ) ≥ u is incident to R N and R N +1 and ∂ L R N = { u } . Then we choose s = ∂ R R N , which is non-compact. (cid:3)
2. The base for the construction of a thickened path Q is a path Q (1) , which canbe informally defined as follows: start with the path ([ R ] , . . . , [ R ] N − β ), replace [ R ] N − β by R ∗ from Claim 7.8, and then remove some domains from [ R ] N − β − , [ R ] N − β − , . . . ifnecessary to get a thickened path. This is not always possible, as we need to conserve thefirst n + 1 elements ([ R ] , . . . , [ R ] n ) of this path. · · · · · · · · ·· · ·R ∗ T N − β + q T N − β − p − B N − β + q +1 B N − β − p u ∗ Q (1) S (1) Figure 12.
The construction of Q (1) and S (1) . Black vertices in the ad-jacency graph represent domains in the union Q (1) ∪ S (1) , white verticesrepresent other domains in R . The boundary of the union Q (1) ∪ S (1) hasconvex or straight angles in all vertices except for the vertex u ∗ correspond-ing to the shaded cycle, but the boundaries of Q (1) and S (1) have at moststraight angles at u ∗ .More precisely, let β and R ∗ be given by Claim 7.8. If R ∗ is the only domain in [ R ] N − β ,we denote [ Q (1) ] t = [ R ] N − β − t , t = 0 , . . . , N − β , [ S (1) ] t = [ R ] N − β + t , t = 0 , . . . , N + β .Otherwise we may assume that R ∗ is the left domain in [ R ] N − β . Consider the structure ofthe corresponding section in the thickened path given by Proposition 3.15, so R ∗ = T N − β .Let B N − β − p − → T N − β − p , p ≥
0, be the last “bottom to top” crossing before R ∗ and T N − β + q → B N − β + q +1 , q ≥ R ∗ . Then we set[ Q (1) ] t = T N − β − t , t = 0 , . . . , p, [ R ] N − β − t , t = p + 1 , . . . , N − β, [ S (1) ] t = T N − β + t , t = 0 , . . . , q, [ R ] N − β + t , t = q + 1 , . . . , N + β. Let us show that the paths Q (1) and S (1) are thickened paths. Indeed, the convexity isclear for all vertices of their boundaries except for the vertices u , . . . , u l correspondingto the cycles in the adjacency graph between the crossings B N − β − p − → T N − β − p and T N − β + q → B N − β + q +1 . Denote by u ∗ = u l ∗ the vertex corresponding to the cycle in theadjacency graph containing both domains in [ R ] N − β . Then ∂ Q (1) (respectively, ∂ S (1) )contains the vertices u j only for j ≤ l ∗ (respectively, j ≥ l ∗ .Consider a vertex u j , j ≤ l ∗ . Then the cycle corresponding to u j is either a wide-bottom trapezoid, or a left-slanted parallelogram, or an incomplete cycle, that is, it con-tains domains from [ R ] N − β +1 ; the latter case takes place for j = l ∗ only. In all these casesthe path Q (1) contains at most half of the domains in this cycle, see Figure 12. Therefore,the angle of ∂ Q (1) at u j is at most straight.Later we will also need the following observation. Remark . The union Q (1) ∪ S (1) has only one concave angle on its boundary. Namely, ∂ ( Q (1) ∪ S (1) ) has a minimally concave angle at the above-defined vertex u ∗ . The convex-ification of Q (1) ∪ S (1) equals R .As we have to keep the first and the last n + 1 domains of the path R unchanged, werequire that p, q ≤ N − n −
3. If this inequality does not hold for p then the section givenby Proposition 3.15 goes from [ R ] N − β to the left beyond [ R ] n and all crossings here are“top to bottom”. Therefore, the types of the states in the sequence i n +1 → · · · → i N − β − have the following pattern:(28) . . . , E L , C, . . . , C, E L , C, . . . , C, E L , C, . . . , C, E L , . . . . Similarly, if R ∗ is the right domain in [ R ] N − β , these states follows the same pattern with E R in place of E L . Note that a sequence i n +1 → · · · → i N − β − following one of thesetwo patterns is uniquely defined by its final state i N − β − , Therefore, there are O ( λ N )sequences i such that the subsequence i n +1 → · · · → i N − β − follows any of these twopatterns.The other inequality q ≥ N − n − i N − β → · · · → i N − n − follows either the pattern (28) or the same pattern with E R in place of E L , and again thereare O ( λ N ) sequences i satisfying this inequality.Therefore, the conditions p, q ≤ N − n − O ( λ N ) sequences, which comprisethe set E (1)2 N − , a part of the set E N − from the lemma statement.3. Let s be the side of R ∗ provided by Claim 7.8, let e be the label on s inside of R ∗ = g ∗ R , and let v ± be the ends of s incident to [ R ] N − β ± . Add the “head” g ∗ H e from ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 51
Proposition 4.9 to Q (1) and S (1) :[ Q (1) ] t = [ S (1) ] t = g ∗ H et , t = − α, . . . , − . In the special case of Remark 4.3 if s is compact then ∂ R has a convex angle at an end v ∗ ∈ { v − , v + } of the side s . Then we choose H e to be either the path shown on Figure 9j)or its mirror image so that the end v ∗ is incident to the domains g ∗ H et for t = 0 , − , − s is incident to these domains for t = 0 , − Q (1) ] t ) N − βt = − α and ([ S (1) ] t ) N + βt = − α are “partially convexified paths”,that is, their boundaries are almost convex curves, and right turns may appear only at v − for the former path and at v + for the latter.Indeed, consider the former path. Then any vertex on v ∈ ∂ ([ Q (1) ] t ) N − βt = − α except for v = v − is incident to either ∂ ([ Q (1) ] t ) N − βt =0 or to g ∗ H e only, which are thickened paths, sotheir boundaries are convex. Let v = v − . Then the addition of g ∗ H e increases the numberof domains incident to v by at most one, hence the boundary angle at v , which is at moststraight for ([ Q (1) ] t ) N − βt =0 is at most minimally convex for ([ Q (1) ] t ) N − βt = − α . Similarly, in thespecial case if the added head is the one shown on Figure 9j) and v − = v ∗ , the number ofdomains incident to v − is increased by at most two when adding the head g ∗ H e , the angleof ∂ ([ Q (1) ] t ) N − βt =0 at v is convex, so the same angle for ∂ ([ Q (1) ] t ) N − βt = − α is at most minimallyconcave.Let Q (2) , S (2) be the convexifications of ([ Q (1) ] t ) N − βt = − α and ([ S (1) ] t ) N + βt = − α and let j (2) , k (2) be the corresponding sequences of states. Then Proposition 4.9 implies that j (2) − α = h ( e ) = k (2) − α .4. Let us show that j (2) N − β − = ι ( i ), k (2) N + β − = i N − This is true if the convexificationprocedure for ([ Q (1) ] t ) N + βt = − α and ([ S (1) ] t ) N + βt = − α does not change domains with t ≥ N − n − Q (1) ] t ) N − βt = − α starting from the end v − of s goes along a geodesic segment I of its boundary that reaches [ R ] n . Without lossof generality we suppose that s lies on the left boundary of R , and ∂ L [ R ] t ⊂ I for t = n + 1 , . . . , N − β − R ] t contains only one domain for t = n + 3 , . . . , N − β − R ] t = {T t , B t } . Then ∂ L [ R ] t is the boundary of T t minus its common sides with [ R ] t ± . Note that if ∂ L [ R ] t containstwo sides then the vertex between them is incident to only one domain in R , and theboundary angle there is convex, so ∂ L [ R ] t fails to belong to I . Therefore, T t has at least N ( R ) − R . This is impossible if N ( R ) ≥
6; if N ( R ) = 5, then ( i t − , i t ) = ( E R ( . . . ) , E L ( . . . )), hence there are no admissible i t − or i t +1 .Moreover, R should be compact since any vertex of T t lying on ∂ D belongs to ∂ R .It remains to consider the case of the compact domain R with N ( R ) = 4. Then T t hasat least three common sides with other domains, hence ( i t − , i t ) has types BE L , E L E L , CE L , E R C , E R E L , E R E R , or E R D . Let us show that ( i t − , i t ) cannot have types E L E L or E R E R . Assuming the former we have that the adjacency graph has a cycle T t − T t B t +1 B t ,which corresponds to a vertex u with n ( u ) = 2. Consider the common side of T t and T t +1 ,and let v be its left end when looking from T t . Then T t has three common sides with other domains from R , namely, with T t +1 , B t +1 , T t in the clockwise order starting from v . Inparticular, u and v are the opposite vertices of T t , hence n ( v ) ≥
3. On the other hand,the sides of T t +1 common with the other domains in R form a polygonal curve with threesides going from v in the counterclockwise direction. Therefore, v is incident to only twodomains in R , namely T t and T t +1 , whence the angle of ∂ R in v is convex. The case when( i t − , i t ) have types E R E R is considered in the same way.Therefore, we have shown that non- A states can appear in the sequence ( i t ) N − β − t = n +1 only as D or E R D at the beginning of this sequence or as B or BE L at its end.Now consider the sequence ( i t ) N − β − t = n +3 , which consists of A -states only. If all domains[ R ] t with t = n + 3 , . . . , N − β − I , then all states i t with t = 2 n + 3 , . . . , N − β − n − A L [ i − , i + ]( e ) with i − + i + = n ( v L ( e )). Moreover, the se-quence ( i t ) N − β − n − t =2 n +3 is uniquely defined by one its term i n +3 , since the only possibletransitions are A L [ i − , i + ]( e ) → A L [ i − + 1 , i + − l ( e )) , i + > ,A L [ i − , e ) → A L [1 , i (cid:48) + ]( l ( l ( e ) − )) . Here the former transition corresponds to taking the next petal around the same vertexon I , while the latter one means that we have arrived to a domain having a common side˜ s with I , thus we switch from taking petals around one end of ˜ s to that around the otherend.Therefore, the condition ∂ L [ R ] t ⊂ I for t = n + 1 , . . . , N − β − i → · · · → i N ), or O ( λ N ) possibilities for the entire sequence i . Denoteby E (2)2 N − the set of all sequences i such that the convexification procedure for either([ Q (1) ] t ) N − βt = − α or ([ S (1) ] t ) N + βt = − α reaches their domains with t ≥ N − n −
2. We have provedthat E (2)2 N − contains O ( λ N ) elements. Thus we define the set E N − = E (1)2 N − ∪ E (2)2 N − , and this set contains O ( λ N ) elements. The paths Q = ([ Q ] t ) N − β + αt =0 and S = ([ S ] t ) N + β + αt =0 from the lemma statement are now obtained from Q (2) and S (2) by a shift of numeration.5. It remains to check that the map i (cid:55)→ ( j, k ) is injective. We will show how torecover R from Q and S .Let i / ∈ E N − . Denote(29) Y = ([ Q (1) ] t ) N − βt =0 ∪ ([ S (1) ] t ) N + βt =0 ∪ g ∗ H e , where Q (1) , S (1) , and g ∗ are defined above in steps 2 and 3 of this proof. This is a Y -shaped union of fundamental domains, with the three terms of (29) joining at R ∗ .While Y is not a path, its boundary is an almost convex curve: right turns may occurin only three vertices, namely, the ends v ± of side s (see step 1) and the vertex u ∗ fromRemark 7.9. By the construction, the segments of A ( ∂ Y ) around vertices v − , v + , and u ∗ have no common points. Therefore, the convexification on each of these segments can beperformed independently, and the domains added are exactly the domains that are added ONVERGENCE OF SPHERICAL AVERAGES FOR ACTIONS OF FUCHSIAN GROUPS 53 when convexifying the path that is a union of the corresponding two terms in (29):[ Y ] = Y (cid:116) C v − (cid:116) C v + (cid:116) C u ∗ , where Q = (cid:2) ([ Q (1) ] t ) N − βt =0 ∪ g ∗ H e (cid:3) = (cid:0) ([ Q (1) ] t ) N − βt =0 ∪ g ∗ H e (cid:1) (cid:116) C v − , S = (cid:2) ([ S (1) ] t ) N + βt =0 ∪ g ∗ H e (cid:3) = (cid:0) ([ S (1) ] t ) N + βt =0 ∪ g ∗ H e (cid:1) (cid:116) C v + , R = (cid:2) ([ Q (1) ] t ) N − βt =0 ∪ ([ S (1) ] t ) N + βt =0 (cid:3) = (cid:0) ([ Q (1) ] t ) N − βt =0 ∪ ([ S (1) ] t ) N + βt =0 (cid:1) (cid:116) C u ∗ . Let us perform the convexification procedure for Y only for the segments of A ( ∂ Y ) around v ± . We arrive at the set ˆ Y = Q ∪ S with at most one non-convex vertex u ∗ . Note thatthe convexifications of Y and ˆ Y coincide, and they are equal to ˆ Y ∪ R by Remark 7.9.Therefore, R can be reconstructed from Q and S as follows. Let [ ˆ Y ] be the convexifi-cation of ˆ Y = Q ∪ S . Let J be a segment of ∂ [ ˆ Y ] going from [ Q ] N − β + α to [ S ] N + β + α andnot touching Q = S . Then the sequence of all domains in [ ˆ Y ] having common pointswith J is a locally shortest path, and its convexification coincide with R by Lemma 3.18.Formally speaking, to use this lemma we have to assume that [ Q ] N − β + α and [ S ] N + β + α contain one domain each. If this is not true, we append “caps” with states of types C . . . CD to j and/or k , consider ˆ Y and J for these elongations of the paths Q and S ,apply Lemma 3.18, and then remove the domains corresponding to the “caps” from thepath R . (cid:3) Conclusion of the proof of Theorem A.
Consider an ergodic decomposition ofthe measure µ with respect to the action of the subgroup of G generated by G = { g g : g , g ∈ G } . Note that in general, for arbitrary G -invariant measure ˜ µ , the operator P does not preserve the measure p × ˜ µ , but the operators Q, V, W defined by (19) do, asthey contain only terms of the form f ◦ T g ◦ T g for g , g ∈ G . Formula (9) then yields (cid:101) S n ( f ) = λ n − (cid:88) j ∈ Ξ S h j ( Q n − V ϕ ( f ) ) j . Note also that S (2 n ) equals the number of paths from Ξ S to Ξ F of the length 2 n , thus S (2 n ) = (cid:88) i ∈ Ξ S ,j ∈ Ξ F (Π n − ) ij = Cλ n − (1 + o (1)) , whence S n ( f ) = ˜ C (cid:88) j ∈ Ξ S h j ( Q n − V ϕ ( f ) ) j · (1 + o (1)) . Now we apply Theorem 6.6 to operators (19) acting on the space L ( Y, ˜ ν ), where ˜ ν = ˜ µ × p .Recall that we have checked Assumptions 6.2–6.4 for these operators in Corollary 7.4 andLemma 7.5. Therefore, we obtain that the following holds for ˜ µ -almost every x : • S n ( f )( x ) converges to some limit, which we denote as ˜ f ( x ). • ˜ f ( x ) = ˜ f ( T g g x ) for any g , g ∈ G . The second item results from the fact that ˜ f is constant ˜ µ -almost everywhere.Therefore, the set of x ∈ X such that these two conditions hold, has full measure withrespect to every convex combination of the ergodic measures, in particular, with respect tothe initial measure µ . Thus lim n →∞ S n ( f )( x ) exists µ -almost surely and is G -invariant.On the other hand, for every A ∈ I G one has (cid:90) A f dµ = (cid:90) A S n ( f ) dµ → (cid:90) A ˜ f dµ, whence ˜ f = E ( f |I G ). References [1] V. I. Arnold, A. L. Krylov. Equidistribution of points on a sphere and ergodic properties of solutionsof ordinary differential equations in a complex domain.
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Aix—Marseille Universit´e, CNRS, Centrale Marseille, I2M, UMR 7373,39 rue F. Joliot Curie Marseille France;Steklov Mathematical Institute of Russian Academy of Sciences,Gubkina str. 8, 119991, Moscow, Russia [email protected]
Alexey Klimenko
Steklov Mathematical Institute of Russian Academy of Sciences,Gubkina str. 8, 119991, Moscow, Russia;and National Research University Higher School of Economics,Usacheva str. 6, 119048, Moscow, Russia [email protected]
Caroline Series