aa r X i v : . [ m a t h . D S ] O c t MORSE CODING FOR A FUCHSIAN GROUP OF AFINITE COVOLUME
ARSENY EGOROV,
THE PENNSYLVANIA STATE UNIVERSITY
Abstract.
We consider a Fuchsian group Γ and the factor surface H / Γ, which has constant curvature − H (except for a subset ofa lower dimention), we obtain a fundamental domain D of Γ. Thiscan be done in different ways, but we restrict the choice to Dirichletdomains, which always are convex polygonal subsets of H . Givena generic geodesic on H , one can produce a so called geometricMorse code (or the cutting sequence) of the geodesic with respectto D . We prove that the set of Morse codes of all generic geodesicson H with respect to D forms a topological Markov chain, if andonly if D is an ideal polygon. Introduction
There have been several approaches to coding of geodesics on the hy-perbolic plane with respect to fundamental domains of Fuchsian groups.One approach was introduced by Morse [8], in which the plane is tesse-lated according to the chosen domain and the action of the group, thesides of the tesselation are labeled in a certain way, and the geodesicsare then coded using their intersections with the sides of the tessela-tion. Another approach was developed by Artin [1] and for coding ituses the endpoints of geodesics on the absolute, which are identifiedwith real numbers and are expanded into continued fractions.Artin’s approach was broadened to different types of continued frac-tions, e.g. [4], [7], and different Fuchsian groups (the original one wasused with the modular group), [2], [9, 10], but it requires some reduc-tion process (i.e. finding an appropriate geodesic among Γ-equivalentones) before actual encoding. Moreover, the coding process may bedefined differently for the same tessellation of the plane, which corre-sponds to different types of continued fractions. Being more syntheticthan Morse’s, this approach produces a set of sequences of integers,which is a topological Markov chain and is much easier to structurizethan just a “random” set of sequences.
Morse’s approach utilizes only the way the hyperbolic plane is tiledby fundamental domains of the group, which can be reproduced “in-ternally” on the factor surface H / Γ by marking curves on the surfacealong which it needs to be “cut” to obtain one of the fundamental do-mains on the plane. And Morse’s approach is defined uniquely once atessellation is chosen. So it is somewhat more natural and canonic thanArtin’s. We became interested, under what conditions Morse’s codingproduces a set of sequences, which is a topological Markov chain too.In [5] the case of the modular group
P SL (2 , Z ) is considered along witha slightly modified version of the coding, which uses integer numbersinstead of the group’s elements. Their result implies the results of thispaper for the case of the modular group and its standard fundamentaldomain, and is the primary motivation for this work.1.1. Fuchsian groups and Dirichlet domains.
The following defi-nitions provide us with basic terms:
Definition 1.
A group of orientation preserving isometries of the hy-perbolic plane H , acting discretely, is called a Fuchsian group.Given a Fuchsian group Γ, one can consider the factor space H / Γ.Since Γ acts discretely on H , the space is an orbifold. Definition 2.
If the volume of the factor surface H / Γ is finite, thegroup is said to have a finite covolume.
Definition 3.
Let Γ be a Fuchsian group, and let x ∈ H be a pointwhich is not fixed by any element of Γ except the identity. Then theDirichlet domain of Γ with respect to x is the set D x (Γ) = { y ∈ H :dist( y, x ) ≤ dist( y, gx ) ∀ g ∈ Γ } .A Dirichlet domain of a Fuchsian group is a fundamental domain ofthe group, [3, § ∂ ∞ H . In terms of the factorsurface H / Γ, it may have a finite number of singular points and cusps,but cannot have funnels.
Some agreements. • Given a Fuchsian group Γ and its Dirichlet domain D , we willrefer to the edges and vertices of the tessellation Γ D as just“edges” and “vertices”, because there are no edges or verticesconsidered in this paper other than those. ORSE CODING FOR A FUCHSIAN GROUP OF A FINITE COVOLUME 3 • We will call the vertices on ∂ ∞ H “infinite” and all the others— “finite”. • The geodesics are always directed. Then referring to the leftand right hand side of the plane with respect to a given geodesicmakes sense. If one geodesic crosses another one, we will say itis on the right from the other one, if its direction is from theleft to the right, otherwise we will say it is on the left from theother one. Clearly, if one geodesic is on the right from anotherone, then the latter one is on the left from the former one andvice versa.1.2.
Morse coding.
Consider a Fuchsian group Γ along with a Dirich-let domain D . D shares its edges with some of its images (“copies”)under the action of Γ. Let the images, sharing edges with D , be g D , . . . , g n D , where g , . . . , g n ∈ Γ. Let us label the edges of D by theelements g , . . . , g n , so that the edge shared with g i D is labeled by g i . This labeling is made inside the domain. Now we spread this la-beling to the rest of the tessellation: for every g ∈ Γ the edge of g D ,corresponding to the one labeled by g i inside D , is labeled by g i inside g D . g g − gg − D g g − γg D g g − gg D g g − Figure 1.
A geodesic crossing multiple fundamental domains.When the labeling is done, each edge on the plane is labeled on bothsides. It is easy to show, that each edge is labeled by elements inverseto each other from opposite sides.
ARSENY EGOROV
A generic geodesic on the hyperbolic plane avoids the vertices, finiteor infinite. Given a generic geodesic, one can write down the biinfinitesequence of labels on the edges, which the geodesic crosses on its way(each time the geodesic crosses an edge, one writes down the label onthe edge from the domain the geodesic is leaving). This sequence iscalled the geodesic’s cutting sequence. It was used by Morse in hispaper [8], and is also referred to as (geometric) Morse code of thegeodesic. According to [10], it can be traced back to [6].It is easy to see from the geometric construction, that in the cuttingsequence of a geodesic, an entry cannot be immediately followed by itsinverse from Γ. This fact follows from two observations: that everylabel is unique inside one domain, and that a geodesic cannot cross thesame edge twice in a row.1.3.
Main result.
We consider the set of cutting sequences of allgeneric geodesics on H with respect to a given Dirichlet domain D . Definition 4.
Let Σ be a no more than countable alphabet. Let Λ ⊆ Σ Z be a set of biinfinite sequences of elements of Σ. Assume there are k ∈ N and Λ k ⊆ Σ k +1 , such that Λ = { λ ∈ Σ Z : ( λ n , . . . , λ n + k ) ∈ Λ k , ∀ n ∈ Z } . Then Λ is called a k -step topological Markov chain.Let us say that Λ is a topological Markov chain, if it is a k -steptopological Markov chain for some k ∈ N .The elements of Λ k are called “allowed” ( k + 1)-tuples. Theorem 1.
Let Γ be a Fuchsian group with a finite covolume and aDirichlet domain D . Then the set of Morse codes of generic geodesicson H with respect to D is a topological Markov chain, if and only if D does not have finite vertices.Remark. The proof of sufficiency is essentially contained in [10, §
1, pp.603–604]: one needs to consider the tree, which connects the elementsof the Γ-orbit of the point, defining D , if and only if they are containedin copies of D , sharing a side. Every infinite sequence consisting of g , . . . , g n will then correspond to a branch of the tree, starting at theroot, and a point on ∂ ∞ H . Every biinfinite sequence will correspond totwo points on ∂ ∞ H and, thus, to a geodesic. One only needs to checkthis geodesic’s cutting sequence coincides with the original biinfinitesequence.Although this scheme produces a good proof, we present our ownproof of sufficiency, which seems to be slightly shorter and doesn’tneed too many details to be checked. ORSE CODING FOR A FUCHSIAN GROUP OF A FINITE COVOLUME 5 Auxiliary statements
Throughout this section Γ is a Fuchsian group of a finite covolume. D is a Dirichlet domain for Γ. Lemma 1 (There are plenty of finite vertices on the plane) . Assume D has a finite vertex. Let y ∈ H be some point. Consider the set Ξ ofdirections leading from y to all the finite vertices on the plane. Thenthe set is dense in the set of all directions at y (or, equivalently, betweenany two directions at y there is a direction from Ξ ).Proof. There are two cases: D has no infinite vertices and otherwise. Case . Suppose D has no infinite vertices, that is D is bounded. Con-sider η and η , two arbitrary directions at y . Let us prove there is adirection between η and η leading from y to a finite vertex.Let γ and γ be the two geodesics, passing through y and following η and η , respectively. Since the geodesics are not parallel, thereshould be a point z inside the angle formed by the two geodesics, suchthat dist( z, γ ∪ γ ) > diam( D ). Since z is contained in a copy of D ,that copy is entirely contained inside the angle as well. Since D has afinite vertex, so does the copy, containing z . Let V be a finite vertex ofthe copy. Clearly, V is inside the angle between γ and γ , too. Thus,the direction, leading from y to V , is between η and η . Case . Assume D has an infinite vertex. Since D has a finite vertextoo, D should have a side with both a finite and an infinite vertex. Letus denote the infinite vertex of the side by ξ ′ and the finite vertex by X ′ . By [3, Theorem 4.2.5], every infinite vertex is the fixed point ofsome parabolic element of Γ, therefore so is ξ ′ .Γ has a finite covolume, so by [3, Lemma 4.5.3], in the unit circlePoincare model the Euclidean diameters of sets g n D , g n ∈ Γ, go to 0, as n → ∞ . So if one considers any point η on ∂ ∞ H and a circle L of smallradius ε > g η ( ε ) ∈ Γ, suchthat g η ( ε ) D is inside L , and thus its infinite vertex, corresponding to ξ ′ is inside L , too. Since η and ε are arbitrary, we showed, that infinitevertices corresponding to ξ ′ in all copies of D are dense on ∂ ∞ H .Consider a parabolic element g ′ ∈ Γ, fixing ξ ′ . The points g ′ m X ′ , m ∈ Z , lie on an horocycle, passing through ξ ′ , and form a regularpolygon with infinite number of sides. Each of these points is a finitevertex. As m → ±∞ , the geodesics, going from y to g ′ m X ′ , convergeto the geodesic, going from y to ξ ′ . So the direction from y to ξ ′ is anaccumulation point of Ξ.A similar argument can be used for any infinite vertex of the form gξ ′ , g ∈ Γ. Thus Ξ has accumulation points at every direction from y ARSENY EGOROV to gξ ′ . As mentioned above, the set { gξ ′ : g ∈ Γ } is dense in ∂ ∞ H . Butit is the same as saying that the directions from y to gξ ′ , are dense inthe set of all directions at y . Hence Ξ is dense in the set of all directionsas well. (cid:3) Lemma 2.
If a geodesic passes near enough to a finite vertex, it inter-sects an edge, ending at the vertex. (There exists ε > , depending onlyon D , such that if a geodesic γ passes at a distance less than ε from afinite vertex X , then γ intersects one of the edges, ending at X .)Proof. Since D has a finite volume, it has a finite number of edges. Let ϕ be the greatest angle of D , and let a be the length of the shortest edgeof D . Choose ε to be such, that a right triangle with the hypothenuseof length a and an acute angle ϕ has the leg at that angle equal to ε .Suppose a geodesic γ passes closer than ε to a finite vertex X , con-sider the perpendicular from X to γ , let the base of it be point y . Theminimal angle between [ X, y ] and an edge ending at X is at most ϕ ,the length of [ X, y ] is less than ε , and the length of the edge formingthe minimal angle with [ X, y ] is at least a , so the edge has to cross γ . (cid:3) Lemma 3.
Suppose D has at least one finite vertex. Assume a geo-desic γ passes through a point y and then crosses edges h A , B i , . . . , h A n , B n i . Then there are geodesics, γ l and γ r , which pass through y ,cross the same edges and at least one more after that, such that thenew edge for γ l has a finite vertex on the right from γ l , and the newedge for γ r has a finite vertex on the left from γ r .Proof. It is enough to prove the statement for γ r . The proof for γ l isabsolutely the same.It is clear that we can rotate γ around y clockwise so that it stillintersects all of h A i , B i i . consider such a rotation of γ and call the newgeodsic ˆ γ r .According to Lemma 1, there is a direction from y between γ andˆ γ r , which leads to a finite vertex. Consider such a vertex X . Withoutloss of generality we may assume, that X is located behind h A n , B n i from y (since finite vertices constitute a discrete set on H , there is onlya finite number of finite vertices in the triangle bounded by γ , ˆ γ r , and h A n , B n i , while there is an infinite number of finite vertices in the anglebetween γ and ˆ γ r ).Now rotate ˆ γ r around y counterclockwise, so that the distance be-tween X and the new geodesic becomes less than ε (cf. Lemma 2),but X is still on the left. Call the new geodesic γ r . Clearly, it passesthrough y and crosses all of h A i , B i i . Since it is closer than ε to X , ORSE CODING FOR A FUCHSIAN GROUP OF A FINITE COVOLUME 7 it has to cross an edge ending at X , and the edge cannot be any of h A i , B i i , because X is behind all of them from y . (cid:3) Proof of Theorem 1
To prove that a set Λ ⊂ Σ Z is a topological Markov chain, we onlyneed to present a number k ∈ N and a set of allowed ( k + 1)-tuples Λ k .To prove otherwise, we need to show that given any k ∈ N onecan find a ( k + l + 1)-tuple λ , l >
0, such that no infinite sequence,containing λ , is in Λ, but every subsequence of λ of length k + 1 iscontained in some infinite sequence from Λ. Proof.
In our case the alphabet consists of the labels put on the edgesof D : Σ = { g , . . . , g n } , where n is the number of edges of D , and theset of infinite sequences Λ is the set of Morse codes of generic geodesicson H with respect to D .First we prove that if D has a finite vertex, the set of Morse codes,generated by D , is not a topological Markov chain.We want to prove that given any k ∈ N , there can be found a se-quence λ k + l = ( h , h , . . . , h k , h k +1 , . . . , h k + l ) of elements of Σ, for some l ∈ N , such that there exist generic geodesics γ ( i ) , for 0 ≤ i ≤ l , sothat Morse code of γ ( i ) contains ( h i , h i , . . . , h k + i ), 0 ≤ i ≤ l , but nogeodesic has its Morse code containing λ k + l as a subsequence.Fix some k ∈ N . Consider a finite vertex V . Consequently applyingLemma 3, we can find a geodesic γ , passing through V and crossingat least k edges, with the last edge having a finite vertex on the rightfrom γ . Without loss of generality we may assume that exactly the k -th edge that γ crosses after passing V has a finite vertex and thevertex is on the right from γ . Since there are only countably manyfinite vertices on the plane, we can choose γ in such a way, that itdoesn’t pass through any finite vertex other than V .Denote the endpoints of the first k edges γ crosses after passingthrough V , finite or infinite, by A , B ; . . . ; A k , B k with A i ’s being onthe left from γ .Let { γ ϕ } ≤ ϕ< π be the family of geodesics, passing through V , suchthat the angle between γ and γ ϕ in the clockwise direction is ϕ . It isclear that for all sufficiently small ϕ > k edges γ ϕ crossesafter it passes through V , are the same as for γ .Let ϕ ′ = inf { ϕ >
0, such that the first k edges, crossed by γ ϕ after it passes through V , are different from those of γ } . Obviously, ϕ ′ >
0. It is also clear that γ ϕ ′ should be a geodesic, intersecting all of h A , B i , h A , B i , . . . , h A k , B k ] within their relative interior or at theright endpoint and passing through at least one of the right endpoints. ARSENY EGOROV A V B D A h B A i h i B i A k h k B k γ γ ϕ γ ϕ ′ Figure 2.
The family of geodesics passing through V .The above construction of γ ϕ ′ can be thought of as a clockwise rota-tion of γ around V until it meets the first of B i ’s.Now consider the fundamental domain, which γ enters immediatelyafter it passes through V , name it D . Let the edges of D , that end at V , have the other endpoints denoted A and B , so that A is on theleft from γ .It can be easily seen, that by another small rotation of γ aroundone of its points other than V , one can find a generic geodesic γ b ,which intersects [ V, B i , h A , B i , . . . , h A k , B k ]. Let the label at [ V, B i outside D be h , and the labels on h A i , B i i on the side facing V be h i ,1 ≤ i ≤ k . Then the sequence ( h , h , . . . , h k ) is a part of Morse codeof γ b . Denote this sequence by λ (0) .On the other hand, rotating γ ϕ ′ clockwise around a point w behind h A k , B k ], it is possible to find another generic geodesic γ a , which inter-sects h A , V ], h A , B i , . . . , h A k , B k ].Due to Lemma 1, between γ a and γ ϕ ′ after they cross there shouldbe an infinite number of finite vertices. We could have chosen γ a sothat one of those finite vertices is close enough to γ a , so we can applyLemma 2. Let Z be such a finite vertex between the two geodesics,that an edge, ending at Z , crosses γ a .Let the edges, crossed by γ a after it crosses h A k , B k ] and up to themoment it crosses the edge, ending at Z , be labeled by h k +1 , . . . , h k + l ,and have endpoints A k +1 , B k +1 ; . . . ; Z , B k + l . This automaticallymeans that sequences λ ( i ) = ( h i , h i , . . . , h k + i ), 0 < i ≤ l , all areparts of Morse code of γ a . ORSE CODING FOR A FUCHSIAN GROUP OF A FINITE COVOLUME 9 A V B D A B A i B i A k B k wγ b γ ϕ ′ γ a Figure 3.
Geodesics γ ϕ ′ , γ b , and γ a . V B A i B i w ˜ γ γ ϕ ′ γ a ZB k + l Figure 4.
It is impossible for a geodesic to cross [
V, B i , h A , B i , . . . , h A k , B k ], and [ Z, B k + l i all at the same time.Let us show that the sequence ( h , h , . . . , h k , h k +1 , . . . , h k + l ) cannotbe a part of Morse code of any geodesic. Assume this sequence is Morsecode of some geodesic. Then there should one such geodesic ˜ γ thatcrosses [ V, B i , h A , B i , . . . , [ Z, B k + l i . But this is impossible, because[ V, B i and [ Z, B k + l i lie on the same side from γ ϕ ′ , and [ Z, B k + l i doesnot intersect γ ϕ ′ , while at least one of h A i , B i i , 1 ≤ i ≤ k , lies on theother side of γ ϕ ′ .So we proved, that ( h , h , . . . , h k + l ) is not a part of a geodesic’sMorse code, while ( h , . . . , h k ) is a part of Morse code of γ b , and( h i , . . . , h k + i ), 0 < i ≤ l , are parts of Morse code of γ a . This endsthe proof of necessity. Now assume that D does not have finite vertices. Then every edgeof D is a complete geodesic on H .Recall that set Σ consists of the labels put on the edges of D ,which are some elements of Γ. Since any edge is labeled by inverseelements from the opposite sides, and the same label does not ap-pear on different edges inside a single domain, two edges, crossedby a geodesic subsequently, cannot be labeled by inverse elements.Consider set Λ = { λ ∈ Σ Z : λ n +1 = λ − n , ∀ n ∈ Z } . If we intro-duce set Λ = { ( g, h ) ∈ Σ : g = h − } , set Λ can be written as { λ ∈ Σ Z : ( λ n , λ n +1 ) ∈ Λ , ∀ n ∈ Z } . It is clear, that Morse code of anygeodesic must be a subset of Λ. In order to prove that Morse codingof geodesics with respect to D is a topological Markov chain, we willshow, that every sequence from Λ can be realized as Morse code ofsome geodesic.Consider a sequence λ ′ ∈ Λ. Note that we can always find a sequenceof copies of D , ( . . . , D − , D , D , . . . ), such that D n and D n +1 share anedge, which is labeled by λ ′ n inside D n . To end the proof, we only needto find a geodesic, which crosses all the edges in the same order. Ifsuch a geodesic exists, it cannot cross any other edges in between ofthe ones just mentioned, since all domains are convex sets, and anytwo consecutive edges in the sequence belong to a same domain.Let the edge, shared by D n and D n +1 , have endpoints ξ n , ξ ′ n ∈ ∂ ∞ H .It is clear, that ξ n + l and ξ ′ n + l lie on the same side from geodesic ( ξ n , ξ ′ n ),for any n ∈ Z and l = 0 ∈ Z . Consider intervals [ ξ n , ξ ′ n ] on ∂ ∞ H , forall n = 0, chosen in such a way, that they do not contain ξ and ξ ′ .Then [ ξ n , ξ ′ n ], n = 0, form two nested sequences as n → ∞ and as n → −∞ , one on each side from geodesic ( ξ , ξ ′ ). Thus, there shouldbe points ξ ∞ , ξ −∞ ∈ ∂ ∞ H , such that ξ ∞ is on the other from ( ξ , ξ ′ )side of ( ξ n , ξ ′ n ), for all n >
0, and ξ −∞ is on the other from ( ξ , ξ ′ )side of ( ξ n , ξ ′ n ), for all n <
0. Obviously, ξ ∞ and ξ −∞ are two differentpoints. Consider the geodesic ( ξ −∞ , ξ ∞ ). It has to cross all of ( ξ n , ξ ′ n )and they can only be crossed in the order of increasing n . (cid:3) The proof of the sufficiency implies the following corollary.
Corollary 1.
If Morse coding, given by a Dirichlet domain of a Fuch-sian group, produces a topological Markov chain (topological Markovchain), it produces a 1-step topological Markov chain.
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