All finitely generated Kleinian groups of small Hausdorff dimension are classical Schottky groups
aa r X i v : . [ m a t h . G T ] D ec All finitely generated Kleinian groups of smallHausdorff dimension are classical Schottkygroups.
Yong HouPrinceton University
Abstract
This is the second part of our works on Hausdorff dimension ofSchottky groups [16]. In this paper we prove that there exists a uni-versal positive number λ >
0, such that up to finite index, any finitely-generated non-elementary Kleinian group with limit set of Hausdorffdimension < λ is a classical Schottky group. The proof rely on ourprevious works in [16], [15] which provide the foundation to the generalresult of this paper. Our result can also be considered as a converseto the well-known theorem of Doyle [11] and Phillips-Sarnak [23].
Contents Γ Actions on Hyperbolic Space 94 Relations of Trace, Fixed Points, and Hausdorff Dimension 115 Sufficient Conditions for k -generated 156 Degenerate types 21 Bounded Type IV 248 Double boundary degeneracy Type I 259 Single boundary degeneracy Type II 4410 Elliptical degeneracy Type III 4411 Non-Collapsing fixed points subspaces 4512 Collapsing fixed points types 4713 Collapsing fixed points I 4814 Collapsing fixed points II 5015 Extension from fixed Rank k to all ranks 6216 Proof of Main Theorem 65 Let x ∈ H be a point in the hyperbolic 3-space. For a given Kleinian group(finitely generated, torsion-free, discrete subgroup Γ of PSL(2 , C )), the limitset Λ Γ of Γ is defined as Λ Γ = Γ x ∩ ∂ H . Λ Γ is the minimal non-emptyclosed subset of ∂ H which is invariant under Γ . The group Γ is called non-elementary if Λ Γ contains more then two points. When Λ Γ = ∂ H then thegroup Γ is said to be of second kind, otherwise it is called first kind. The setΩ Γ = ∂ H − Λ Γ is called region of discontinuity of Γ, and it follows that Γacts properly discontinuously on Ω Γ . Given a collection of disjointed closed topological disks D i , D ′ i , ≤ i ≤ k in the Riemann sphere ∂ H = C with boundary curves ∂D i = ∆ i , ∂D ′ i =∆ ′ i . By definition ∆ i are closed Jordan curves in Riemann sphere ∂ H . Let { γ i } k ⊂ PSL(2 , C ) such that γ i (∆ i ) = ∆ ′ i and γ i ( D i ) ∩ D ′ i = ∅ , the group Γgenerated by { γ , ..., γ k } is a free Kleinian group of rank k, and Γ is calledmarked Schottky group with marking { γ , ..., γ k } . Equivalently, one can alsodefine rank k Schottky group to be the image of ρ, discrete faithful convex-compact representation of rank k free group F k into PSL(2 , C ) , with marking2iven by ρ. The equivalence of these two definitions simply follows from ˇSvarc-Milnor lemma and the classification of finitely generated purely loxodromicfree Kleinian groups.Given a finitely generated Kleinian group Γ it is called Schottky group ifit is a marked Schottky group for some markings. Whenever there exists a set { γ i , ..., γ k } of generators with all ∆ i , ∆ ′ i as circles then it is called a markedclassical Schottky group with classical markings { γ i , ..., γ k } , and { γ , ..., γ k } are called classical generators. A Schottky group Γ is called classical Schottkygroup if there exists a classical markings for Γ.Denote by J k ⊂ PSL(2 , C ) the space of all rank k Schottky groups up toconjugacy, and J k,o be the set of all rank k classical Schottky groups up toconjugacy. One can topologies and put analytic structure on J k as follows:Note that one can embed J k into C k − as (3 k − J k is infinite index covering space, in fact analytic cover, ofthe moduli space of genus k closed Riemann surfaces. Hence, J k can alsobe considered fine moduli space of genus k Riemann surfaces. The universalcovering space of J k is the Techim¨uller space.Equivalently, Schottky space J k can also be viewed as the quasiconformaldeformation space QF (Γ) of a fixed given rank k classical Schottky group Γ . The equivalency of these two definitions simply follows from representationdeformation theory of Klenian groups and the definition of Schottky groups.Note that QF (Γ) is independent of choice of Γ since given any two ranked k classical Schottky groups Γ , Γ ′ marked with { γ i } , { γ ′ i } , ≤ i ≤ k thereexists quasiconformal f with { γ ′ i } = { f ◦ γ i ◦ f } . The boundary ∂QF (Γ) ⊂ PSL(2 , C ) k contains points that are non-Schottky groups which are called geometrically infinite Kleinian groups of first kind (i.e. limit set is C ). TheseKleinian groups have limit set of Hausdorff dimension 2 . In fact, there areuncountable many such elements in ∂QF (Γ) . The set of classical Schottky groups is a non-empty and non-dense subsetof set of Schottky groups [19], also see [11].For a given Schottky group Γ, denote the Hausdorff dimension of Λ Γ by D Γ . Our main result is:
Theorem 1.1.
There exists a universal λ > , such that all finitely generatednon-elementary Kleinian group Γ with D Γ < λ are classical Schottky groups. Note that Theorem 1.1 holds up to finite index. Theorem 1.1 can beviewed as the converse to the result by Phillips-Sarnak [23] (for higher di-mensional Kleinian groups) and Doyle [11] which states that there exists a3niversal upper bound on the Hausdorff dimension of limit sets of all classi-cal Schottky groups. There are also some recent interesting works on boundsof Hausdorff dimensions of limit sets of Kleinian groups by Kapovich[17],Bowen[5].Let λ > J λk,o as the set of classicalSchottky groups of Hausdorff dimension < λ. Then by analyticity of Haus-dorff dimension function on Schottky space and Theorem 1.1, we have thefollowing simple observation:
Corollary 1.2. J λk,o is (3 k − -dimensional complex manifold. The proof is based on induction on rank k. It follows from main result of [16],we know Theorem 1.1 holds for k = 2 . We assume Theorem 1.1 is true for k − k > . The ideas and techniques used are mostly generalizationsof methods developed in [16].Note that for a given rank k Schottky group, when we speak of set ofgenerators or generating set, we always mean a set of k elements of freegenerators. Hence by definition it is always minimal. • Preliminary estimates : We state some preliminary estimates and someof it’s generalizations that are given in [16] on the locations of fixedpoints, centers of isometric circle of a given set of generators of aSchottky group. Detailed proofs of these estimates can be found in[16]. These estimates will give us a sufficient controls on how the fixedpoints of a set of generators move in terms of the Hausdorff dimen-sion of the limit set of the group. More precisely, when the Hausdorffdimension of rank k Schottky group Γ is small then, any set of k gen-erators will have either the norm of the trace exponentially large or itsfixed points degenerates into single point exponentially fast. The mainingredients of the proofs of these estimates rely on the results of [15].This is done in Section 3 and Section 4 . • Sufficient condition for classscial-ness : For a given sequence of rank k Schottky groups, we will derive a set of sufficient conditions on anysequence of set of generators with respective to Hausdorff dimensionsof its limit sets for it to contain classical Schottky group generators. Ifa given sequence of set of generators satisfies these conditions then, the4equence will contain a subsequence that will eventually be classicalSchottky groups.These sufficient conditions ensures that a given sequence of decreasingHausdorff dimensions will contain a subsequence of eventually classicalSchottky groups are given in Section 5 . For computational purpose,it turns out that the computations in this part are easier by usingthe unit ball model of the hyperbolic 3-space. However all statementsare made in the upper-half space of hyperbolic 3-space. Of courseall statements are model independent. In fact, this is our generalizedversion of conditions given in [16] to rank k. These conditions in thebasic form simply states that, if the norm of the traces of generatorsgrowth sufficiently fast compare to the degenerations of it’s fixed points,then all the Schottky disks will eventually be disjoint. • Obstructions to classical-ness : Let be a sequence of rank k Schottkygroups Γ n of decreasing Hausdorff dimensions D n → . Denote it’s setof generators by S Γ n , we will classify all obstructions for this sequenceto contain any classical Schottky group subsequence. It follows fromTheorem 3.1 and Proposition 4.1, there are at least k − D n . Hence we first assume the sequence is k − classical ofstandard form, i.e generators have k − | trace | . Then we define degenerate types
I, II, III, IV for the givensequence. These are the obstructions to existence of classical Schottkygroup. • Degenerate types becomes classical when Hausdorff siemsnion is suffi-ciently small : We show that for each of the degenerate types I, II, II,IV given as obstructions to the existence of classical Schottky group ina sequence of Schottky groups, when the Hausdorff dimension D Γ n → Non-collapsing fixed points subspaces : Based on degeneracies resolu-tions for sufficiently Hausdorff dimension, we then prove that Theorem1.1 holds for J k ( τ ) Schottky subspaces for τ > . Here the Schottkysubspace J k ( τ ) is defined as collection of Schottky groups up to con-jugation that have a generating sets S Γ with uniform lower bound on Z S Γ > τ (bounded separations of it’s fixed point sets). This is provedby induction on k and removal of obstructions to classical-ness whenHausdorff dimension is small. • Removing uniform lower fixed points bound constraint : We classify pos-sible types of fixed points collapses into Collapsing fixed points I andCollapsing fixed points II. Given a sequence of Γ n with D Γ n → , weshow that both types of collapsing fixed points I and II we must haveeither, uniform lower bound on D Γ n + Z Γ n (sum of Hausdorff dimensionsand fixed points bounds), or Γ n will contain a subsequence of classicalSchottky groups. This is done by analyzing fixed points distributionsbased on estimates of isometric circles centers of generating sets whenthe Hausdorff dimensions is sufficiently small. • Theorem 1.1 holds for fixed rank : Based on the removability of theconstraint of uniform lower bound Z Γ > τ and result on J k ( τ ) , weshow that Theorem 1.1 is true for rank k whenever k − • Rank extension : Finally we prove rank-extension based on fixed rankresult. We show that if Theorem 1.1 holds for rank k, i.e. there ex-ists bounds λ k then for sufficiently small Hausdorff dimension, λ k mustbe independent on k. This is proved by using Rank-length growth es-timates Lemma 3.4. The Rank-length estimates is growth estimatesthat relate the rank number k and norms of the traces with respect toother generators with respect to Hausdorff dimension. Section 2 is used to define and lists global notations that will be used through-out the paper. In Section 3 , we state a strong form of Theorem 1.1 given in[15] for rank k free groups acting on hyperbolic space. Note that the originalresults of [15] are state for pinched negatively curved manifolds. This willbe used for generators selection processes, see Corollary 3.2. In Section 4 , given any sequence of rank k Schottky groups Γ n with bounded Z Γ n (see6ection 2), we shall prove estimates that will enable us to bounds locationsof fixed points of a given sequence of generators of Γ n in terms of the Haus-dorff dimensions D Γ n of Γ n . These estimates will be used in the generatorsset selection procedure. In Section 5, we will derive sufficient conditions fora given set of k generators to be a classical generators. This is done in thehyperbolic ball space B . Sections 6 to 11, given a sequence of rank k Schottkygroups which is k − k to finitely generated by using already established resultsin earlier sections. Section 16 , we finish the proofs of our main theorem bysimple standard topological arguments.ACKNOWLEDGEMENTSThe author would like to express appreciation to Benson Farb, PeterSarnak for giving me opportunity to discuss this work in details. I wouldalso like to express gratitude to Peter Shalen, Marc Culler, Dick Canary formany thoughts on this paper. Finally, I like to express sincere appreciationto the referee. Let Γ ⊂ PSL(2 , C ) be a rank k Schottky group generated by < α , β , ..., β k > with α having fixed points 0 , ∞ . Note that by our choice here, all generatingset are minimal , i.e generating set of k elements for given rank k Schottkygroup. Asssume that γ ∈ Γ is a loxodromic element having fixed points = ∞ . For a γ we write in matrix form as γ = (cid:0) a bc d (cid:1) , with det( γ ) = 1 . We will setthe following notations and definitions throughout the rest of the paper.
Notation 1. • Denote the critical exponent of Γ by D Γ , which is equal the Hausdorffdimension of Λ Γ (see Section 3). • R γ the radius of isometric circles of γ . • η γ = − dc , ζ γ = ac . We will define two different ways to denotes the two fixed points of γ : { z γ,l , z γ,u } the two fixed points of γ in C with | z γ,l | ≤ | z γ,u | , and { z γ, − , z γ, + } the two fixed points with z γ, ± given by quadratic formulawith subscripts ± corresponding to ± p tr ( γ ) − . Note that we alwaystake the principle branch for the square roots of complex numbers. • L γ is the axis of γ. • T γ is the translation length of γ. • Z β i := min {| z β i , − − z β i , + | , | z βi, − − z βi, + |} . Geometrically, Z β i measurehow far apart do the fixed points of β i are in the conformal boundaryˆ C of H . This will be used subsequently for rate of convergence of fixedpoints with respect to norm of the trace. For instance it is use forestimates in Section 4 . • Z <α ,β i > := min { Z β i , | z β i , + | , | z β i , − | , | z β i , + | − , | z β i , − | − } for all 2 ≤ i ≤ k. whenever we need to emphases that β i is element of the generating setwe will also denote Z <α ,β i > by Z <α ,β ,...,β k > to explicitly write out thegenerators set.Geometrically, Z <α ,β i > or Z <α ,β ,...,β k > measures relative position ofthe fixed points of β i with respect to the fixed points of α in theconformal boundary ˆ C . These relative positions will be used to makeselection of generating sets. • For ǫ > , we say that Z Γ > ǫ, if there exists a generating set <α , β , ..., β k > of Γ such that Z <α ,β ,...,β k > > ǫ. • Given any two sequences of real numbers { p n , q n } , we use the notation p n ≍ q n iff, there exists σ > σ − < lim inf p n q n ≤ lim sup p n q n <σ. • Given two real numbers a ≤ b , set D a,b := { z ∈ C | a ≤ | z | ≤ b } . • Given a loxodromic γ, denote by D γ, ± , Schottky disks of γ. Set D γ := D γ, + ∪ D γ, − . Note that Schottky disks are defined to be circular diskssuch that γ (interior( D γ, + )) ∩ γ (interior( D γ, − )) = ∅ . Also note that itis true that any loxodromic element always have Schottky disks (seeProposition 5.5), but distinct elements generally don’t have disjointedSchottky disks. In addition, Schottky disks may not always be isometric8ircle disks, since not all loxodromic elements have disjoint isometriccircles. But for sufficiently large norm of the trace and control on Z Γ once can always take the Schottky disks to be isometric circular disks(disks bounded by isometric circles). Notation 2.
Let { γ n } ⊂ PSL(2 , C ) be a sequence of loxodromic transfor-mations. Let { p n } be a sequence of complex numbers, and { q n } a sequenceof positive real numbers. We write: | z γ n , ± − p n | < q n if there exists N such that for every n > N we have at least one of thefollowing holds,(i) | z γ n , + − p n | < q n , (ii) | z γ n , − − p n | < q n . Γ Actions on Hyperbolic Space
Let Γ be a finitely generated non-elementary Kleinian group, the criticalexponent of Γ is defined as the unique positive number δ Γ such that thePoincar´e series of Γ given by P γ ∈ Γ e − s dist( x,γx ) is divergent if s < δ Γ andconvergent if s > δ Γ . If the Poincar´e series diverges at s = δ Γ then Γ issaid to be divergent. Bishop-Jones showed that δ Γ ≤ D Γ for all analyticallyfinite non-elementary Kleinian groups Γ. In-fact, if Γ is topologically tame( H / Γ homeomorphic to the interior of a compact manifold-with-boundary)then δ Γ = D Γ . Hence it follows from Ian Agol [1], and independently DannyCalegari and Dave Gabai [8] proof of tameness conjecture that δ Γ = D Γ forall finitely generated non-elementary Kleinian groups. Note that Agol’s proofof tamness conjecture extends to pinched negatively curved manifolds.The critical exponent and hence Hausdorff dimension is a geometricallyrigid object in the sense that a decrease in D Γ corresponds to a decrease in“geometric complexity” of the group. Our following results are indicationsof geometric complexity.We state the following theorem from [15], which provides the relationbetween the negatively curved free group action and the critical exponent.9 heorem 3.1 (Hou[15]) . Let Γ be a free non-elementary Kleinian group ofrank k with free generating set S , and x ∈ H then X γ ∈S
11 + exp( D Γ dist( x, γx )) ≤ . In particular we have at least k − distinct elements { γ i j } ≤ j ≤ k − of S that satisfies dist( x, γ i j x ) ≥ log(3) / D Γ , and at least one element γ i j with dist( x, γ i j x ) ≥ log(2 k − / D Γ . The following is a useful corollary of Theorem 3.1, stated here for thecase of Γ is a free group of rank 2.
Corollary 3.2.
Let S = { γ , γ } be a generating set for a free non-elementaryKleinian group Γ . Let x ∈ H . Then dist( x, γ x ) ≥ D Γ log (cid:18) e D Γ dist( x,γ x ) + 3 e D Γ dist( x,γ x ) − (cid:19) Corollary 3.3.
Let S = { γ , γ } be a generating set for a free non-elementaryKleinian group Γ . Let x ∈ H . Let m be any integer. Then at least one ofthe elements γ ′ of S ′ = { γ m γ , γ m +11 γ } satisfies dist( x, γ ′ x ) ≥ log 3 / D Γ . We will need a much stronger estimates which will relates both the rankof Γ and elements of generators. This is needed in order to prove extensiontheorem from fixed rank to arbitrary rank.
Lemma 3.4 (Rank-length growth estimate) . Let S = { γ , γ , ..., γ k } be agenerating set for a free non-elementary Kleinian group Γ of rank k . Let x ∈ H . Then for every i there exists some j = i such that, dist( x, γ j x ) ≥ D Γ log (cid:18) (2 k − e D Γ dist( x,γ i x ) + (2 k − e D Γ dist( x,γ i x ) − (cid:19) . Proof.
Assume otherwise, X γ ∈S
11 + e D Γ dist( x,γx ) > X ≤ l ≤ k −
11 + (cid:16) (2 k − e D Γ dist( x,γix ) +(2 k − e D Γ dist( x,γix ) − (cid:17) + 11 + e D Γ dist( x,γ i x ) > ( e D Γ dist( x,γ i x ) − k − e D Γ dist( x,γ i x ) + 1)(2 k −
2) + 11 + e D Γ dist( x,γ i x ) > , which is a contradiction to Theorem 3.1.10 Relations of Trace, Fixed Points, and Haus-dorff Dimension
The relationships of fixed points of generating sets of a given sequence ofSchottky groups Γ n and the Hausdorff dimensions of Λ Γ n will be stated inthis section. Proofs of these can be found in [16] which is stated for generalrank k. The main is idea is to find a relationship between the distribution functionof the fixed points of one of the generators in terms of the translation lengthgrowth of the other generators and the Hausdorff dimension of its limit set.By obtaining these types of relationships we will be able to construct a newset of generators of the given generating set of generators so that it will havepredetermined distribution of its fixed points. These newly constructed setof generators will be a crucial part in our proof of the main theorem.Let { Γ n } be a sequence of rank k Schottky groups with D n → α ,n , β i,n ∈ PSL(2 , C ) , ≤ i ≤ k in H with α ,n = (cid:0) λ ,n λ − ,n (cid:1) , | λ ,n | > β i,n = (cid:0) a i,n b i,n c i,n d i,n (cid:1) . Throughout this section we assume that there exists
M > T α ,n < M for all n . Set D n = D Γ n . Let tr = trace . First we will state Corollary 3.2 in the trace form.
Proposition 4.1. [16] Suppose there exists ∆ > such that Z β i,n > ∆ . There exists ρ > depending on ∆ , such that | tr( β i,n ) | > ρ (cid:18) | λ ,n | D n + 3 | λ ,n | D n − (cid:19) D n for large n .Proof. Let T i,n be the translation length of β i,n and R i,n = dist( L α ,n , L β i,n ) . Let x n be a point on axis of α ,n which is the nearest point of L α ,n to L β i,n . Bytriangle inequality, T i,n ≥ dist( x n , β i,n x n ) − R i,n . Here the triangle inequalityis: dist( x n , β i,n x n ) ≤ dist( β i,n x n , β i,n y n ) + dist( β i,n y n , y n ) + dist( x n , y n ) , where y n is the closest point on L β i,n to L α ,n . Note for sufficiently large T i,n , we have for some positive constant c > , | tr ( β i,n ) | > ce T i,n . Now for large n, from Corollary 3.2, | tr ( β i,n ) | > c (cid:18) | λ ,n | D n + 3 | λ ,n | D n − (cid:19) D n (cid:16) e − L α ,n , L βi,n ) (cid:17) . T i,n ≥ dist( x n , β i,n x n ) − R i,n , and using Corollary 3.2 for e dist( x n ,β i,n x n ) . The second factor of right hand side comes from the equality e − R i,n = e − L α ,n , L βi,n ) . Now Z β i,n > ∆ implies that dist( L α ,n , L β i,n ) < M for some M > . Hence the result follows.We will also need to state the rank k form of the above estimates thatrelates the rank of Γ as well. Proposition 4.2.
Let Γ n be a sequence of rank k free group generated by < γ ,n , ..., γ k,n > with γ i,n having fixed points , ∞ . Suppose there exists ∆ > such that Z <γ ,n ,...,γ k,n > > ∆ . There exists j = i , and ρ > depending on ∆ ,such that | tr( γ j,n ) | > ρ (cid:18) (2 k − | λ γ i,n | D n + (2 k − | λ γ i,n | D n − (cid:19) D n for large n . Corollary 4.3.
We can state Proposition 4.2 with condition Z <γ ,n ,...,γ k,n > droped as follows, | tr ( γ j,n ) | > c (cid:18) (2 k − | λ γ i,n | D n + (2 k − | λ γ i,n | D n − (cid:19) D n (cid:16) e − L γi,n , L γj,n ) (cid:17) . Remark 4.1.A . Without assuming bounds on Z β i,n we can state aboveProposition 4.1 as, | tr ( β i,n ) | > c (cid:18) | λ ,n | D n + 3 | λ ,n | D n − (cid:19) D n (cid:16) e − L α ,n , L βi,n ) (cid:17) . If dist( L α ,n , L α ,n β i,n ) < ǫ then there exists δ > | tr( β i,n ) | > ρ (cid:18) | λ ,n | D n + 3 | λ ,n | D n − (cid:19) D n for large n . 12ollowing lemmas and corollaries are given in [16] as Lemma 4 .
2, Corollary4 . .
6. Note that all these results in [16] are originally stated for generalrank k. The proofs are verbatim to the proofs give in [16] with obviousnotation modifications, hence omitted.The results are estimates of convergence rates of fixed points of all gen-erators of Γ n in terms of its limit set Λ n Hausdroff dimension. Set, ω ( k i,n , l i,n ) = (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n ,n β i,n α li,n ,n , ± − ζ α ki,n ,n β i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n ,n β i,n α li,n ,n , ∓ − η α ki,n ,n β i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) . Note that each given k i,n , l i,n provides a Nielsen transformation of the gener-ators. We will determine what k i,n , l i,n are needed during the proof in laterpart of the paper. The idea is that, eventually we will need to choose somecollection of generators which are Nielsen transformations of α ,n , β i,n basedon the values of k i,n , l i,n . Determination on the constraints of k i,n , l i,n willrequire extensive analysis of both fixed points and centers of circles of gener-ators. Here the definition of ω ( k i,n , l i,n ) is to measures the distance betweenfixed points and centers of circles of generators relative to k i,n , l i,n . Indices k i,n , l i,n are introduced to determine relationship between fixedpoints convergence rates with respect to Nielsen transformation of β i,n by α ,n . Proofs of following statements can be found in [16].
Lemma 4.4. [16] Suppose there exists ∆ > , M > such that Z <α ,n ,β i,n > > ∆ and T α ,n < M for all n. Let k i,n , l i,n be any integers such that T α ,nki,n ,T α ,nli,n < M . Then there exists a constant ρ > such that, ω ( k i,n , l i,n ) < ρ | tr( β n ) || λ l n − k n n | for large n. Note that in general one could have T α i,n → , which implies that wecould have k i,n , l i,n → ∞ . In fact, this is one of the degenerate cases that wewill consider in Section 7 . Corollary 4.5. [16] Suppose there exists ∆ > , M > such that Z <α ,n ,β i,n > > ∆ and T α n < M for all n. Let k i,n , l i,n be any integers such that T α ,nki,n ,T α ,nli,n < M . Then for any δ > there exists ǫ > such that if D n < ǫ then, ω ( k i,n , l i,n ) < δ ( | λ ,n | − . emark . Note that condition Z <α ,n ,β i,n > > ∆ in Lemma 4.4 can bereplaced with conditions | tr( β i,n ) | → ∞ and | tr( β i,n ) | ≤ | c i,n | . Corollary 4.7. [16] Suppose there exists ∆ > , M > such that Z <α ,n ,β i,n > > ∆ and T α ,n < M for all n. Let k i,n , l i,n to be any integers such that T α ,nki,n ,T α ,nli,n < M . Then there exists constants δ, ρ > such that, if | tr( α k i,n ,n β i,n α l i,n ,n ) |→ ∞ then, ω ( k i,n , l i,n ) < ρ | tr( α k i,n ,n β i,n α l i,n ,n ) || tr( β i,n ) | for all D n < δ . Lemma 4.8. [16] Suppose there exists ∆ > , M > such that Z <α ,n ,β i,n > > ∆ and T α ,n < M for all n. Let k i,n , l i,n to be any integers such that T α ,nki,n ,T α ,nli,n < M . Then there exists a constant σ , σ > such that, σ | tr( α k i,n ,n β i,n α l i,n ,n ) || tr( β i,n ) | ≥ | z α ki,n ,n β i,n α li,n ,n , + − z α ki,n ,n β i,n α li,n ,n , − | ≥ σ | tr( α k i,n ,n β i,n α l i,n ,n ) || tr( β i,n ) | , for all n sufficiently large. Although next Lemma is not used in the rest of the paper but we include ithere to demonstrate relations between fixed points and Hausdorff dimensions.
Lemma 4.9. [16] Suppose there exists ∆ > , M > such that Z <α ,n ,β i,n > > ∆ and M − < T α ,n < M for all n. Let k i,n , l i,n to be any integers such that T α ,nki,n , T α ,nli,n < M . Then for at least one J ∈ { , } , and any inte-gers k ′ i,n , l ′ i,n with | ( k i,n − k ′ i,n ) | + | ( l i,n − l ′ i,n ) | = J , we have | z α k ′ i,n ,n β i,n α l ′ i,n ,n , + − z α k ′ i,n ,n β i,n α l ′ i,n ,n , − | ≥ σ D n | tr( β i,n ) | , for all n sufficiently large. The following observation can be seen from the proof of Lemma 4.4 [16].For a more rough upper bound of fixed points of α k i,n ,n β i,n α l i,n ,n to ζ α ki,n ,n β i,n α li,n ,n , and η α ki,n ,n β i,n α li,n ,n , then we can relax the conditions in Lemma 4.4 and stateas follows, Remark 4.2.B . Suppose that | z β i,n , + − z β i,n , − | < c and | c i,n λ l i,n − k i,n ,n | → ∞ . Then { z α ki,n ,n β i,n α li,n ,n , + , z α ki,n ,n β i,n α li,n ,n , − } → { ζ α ki,n ,n β i,n α li,n ,n , η α ki,n ,n β i,n α li,n ,n } . Sufficient Conditions for k -generated Given a sequence of Schottky groups with decreasing Hausdorff diemnsions,we will state and prove a set of conditions that will be sufficient for the givensequence to contain a subsequence which will be classical Schottky groupsfor small enough Hausdorff dimensions.Denote the unit ball model of hyperbolic 3 − space by ( B , dist B ) . Set π : B −→ H the stereographic hyperbolic isometry. Let Γ n to be a sequences of k − generated Schottky groups.For a loxodromic element α of PSL(2 , C ) acting on the unit ball B modelof hyperbolic 3-space, denote by S α,r and S α − ,r the isometric spheres ofEuclidean radius r of α. We set λ π ∗ ( α ) as the multiplier of π ∗ ( α ) in the upperspace model H . For
R > , set C R as the circle in C about origin of radius R. We will denote the north pole of ∂ B by e = (0 , , . Proposition 5.1. [16] Let α be a loxodromic element of PSL(2 , C ) acting on B , with axis passing through the origin and fixed points on north and southpoles. Then π ( S α,r ∩ ∂ B ) , π ( S α − ,r ∩ ∂ B ) maps to C λπ ∗ ( α ) , C λ π ∗ ( α ) . Lemma 5.2.
Let < α ,n , β ,n , ..., β k,n > be generators for Schottky groups Γ n in the upper-half space model H with | tr( β i,n ) | → ∞ and α n = (cid:0) λ ,n λ − ,n (cid:1) , | λ ,n | > . Suppose one of the following set of conditions holds:there exists Λ > , such that for large n, we have | λ ,n | < Λ and, • | λ ,n | − < | z β i,n ,l | ≤ | z β i,n ,u | < | λ ,n | , and • lim inf n (cid:26) | z β i,n ,u | − | λ ,n | ) | tr( β i,n ) | , | z β i,n ,l | − | λ ,n | − ) | tr( β i,n ) | (cid:27) = 0 , • | z β i,n, ± − z β i ′ ,n, ± || tr( β i,n ) | → ∞ ; | tr( β i,n ) | ≤ | tr( β i ′ ,n ) | , i = i ′ . or there exists κ > and for large n, we have | λ i,n | > κ and, • κ − < | z β ,n ,l | ≤ | z β ,n ,u | < κ, and lim inf n (cid:26) | z β i,n ,u | − κ ) | tr( β i,n ) | , | z β i,n ,l | − κ − ) | tr( β ,n ) | (cid:27) = 0 , • | z β i,n, ± − z β i ′ ,n, ± || tr( β i,n ) | → ∞ ; | tr( β i,n ) | ≤ | tr( β i ′ ,n ) | , i = i ′ . then there exists a subsequence such that for m large, π − < α ,n m , β ,n m , ..., β k,n m >π are classical generators for Γ n m in the unit ball model B .Proof. Suppose that there exists some subsequence < α ,n i , β ,n i , ..., β k,n i > which satisfies the first given set of above conditions. Let us first assume forlarge i, min ≤ j ≤ k {| z β j,ni,u − z β j,ni,l |} > δ > . Let r i , ρ j,i be the Euclidean radii of isometric spheres of generator, π − α ,n i π and π − β j,n i π respectively for 2 ≤ j ≤ k. Since that 4 cosh( T π − β j,ni π ) ≥| tr ( π − β j,n i π ) | implies that there exists some c ′ > e T π − βj,niπ ≥ c ′ | tr ( π − β j,n i π ) | . By ρ − j,i = cosh dist( o, L π − β j,ni π ) sinh( T π − β j,ni π ) ([3] p175),we have that ρ − j,i ≥ sinh( T π − β j,ni π ) so for large i there exists some c > ρ − j,i ≥ ce T π − βj,niπ . Therefore there exists some δ > ρ j,i ≤ δ | tr( β j,n i ) | − for large i. Since for z, w ∈ C , | π − ( z ) ′ | = 2 | z + e | and , | π − ( z ) − π − ( w ) | = | π − ( z ) ′ | / | π − ( w ) ′ | / | z − w | . This implies for large i, and x j,i ∈ C | z βj,ni,u | and y j,i ∈ C | λ ,ni | ,ρ j,i | π − x j,i − π − y j,i | ≤ δ | x j,i + e || y j,i + e || tr( β j,n i ) || x j,i − y j,i | . Since | λ ,n i | < Λ , there exists δ > , such that | x j,i + e || y j,i + e | < δ , andlim i ρ j,i | π − x j,i − π − y j,i | ≤ lim i δ δ | tr( β j,n i ) || x j,i − y j,i | = 0 for 2 ≤ j ≤ k. Similarly there exists δ > , such that for w j,i ∈ C | z βj,ni,l | and z j,i ∈ C | λ ,ni | − , lim i ρ j,i | π − w j,i − π − z j,i | ≤ lim i δ δ | tr( β j,n i ) || w j,i − z j,i | = 0 for 2 ≤ j ≤ k. δ > , such that for j = i and f j,i ∈ C | z βj,ni,u | ∪ C | z βj,ni,l | and g j ′ ,i ∈ C | z βj ′ ,ni ,u | ∪ C | z βj ′ ,ni ,l | and | tr( β j,i ) | ≤ | tr( β j ′ ,i ) | , lim i ρ j,i + ρ j ′ ,i | π − f j,i − π − g j ′ ,i | ≤ ρ j,i | π − f j,i − π − g j ′ ,i |≤ lim i δ δ | tr( β j,n i ) || f j,i − g j ′ ,i | = 0 for 2 ≤ j ≤ k. Therefore it follows for large i and Proposition 5.1 we have the isometricspheres S π − α ,ni π,r i , S π − α − ,ni π,r i , S π − β j,ni π,ρ i , S π − β − j,ni π,ρ i and S π − β j,ni π,ρ i , S π − β − j,ni π,ρ i ,S π − β j ′ ,ni π,ρ i , S π − β − j ′ ,ni π,ρ i are disjoint for all 2 ≤ j < j ′ ≤ k. Remark . We see from above proof that without the assumption of | λ j,n i | < Λ then we can’t put bounds on | x j,i + e || y j,i + e | . But since | x j,i + e || y j,i + e | ≤ ( | z β j,ni ,u | +1)( | λ ,n i | +1) for x j,i ∈ C | z βni,u | , y j,i ∈ C | λ ,ni | and | x j,i + e || y j,i + e | ≤ ( | z β j,ni ,l | + 1)( | λ ,n i | − + 1) for x j,i ∈ C | z βj,ni,l | , y j,i ∈ C | λ ,ni | − . Therefore wecan restate the condition as, • | λ ,n | − < | z β j,n ,l | ≤ | z β j,n ,u | < | λ ,n | , and • lim inf n (cid:26) ( | z β j,n ,u | + 1)( | λ ,n | + 1)( | z β j,n ,u | − | λ ,n | ) | tr( β j,n ) | , ( | z β j,n ,l | + 1)( | λ ,n | − + 1)( | z β j,n ,l | − | λ ,n | − ) | tr( β j,n ) | (cid:27) = 0 , • | z β j,n, ± − z β j ′ ,n, ± || tr( β j,n ) | → ∞ ; | tr( β j,n ) | ≤ | tr( β j ′ ,n ) | , j = j ′ . Next assume for some j we have that | z β j,ni ,u − z β j,ni ,l | → . Then withthis assumption we can do a much sharper estimate of the lower bounds oncosh dist( J, L β j,ni ) , which is the distance between the point J on the vertical J -axis and the translation axis of β j,n i in the space H . Eventhough a muchweaker lower bounds will be sufficient for our case.For any given two points h = ( z , θ ) , h = ( z , θ ) ∈ H recall that thehyperbolic distance is provided by equation,cosh dist( h , h ) = | z − z | + | θ − θ | θ θ + 1 . | z β j,ni ,u − z β j,ni ,l | → , also < | z β j,ni ,l | ≤ | z β j,ni ,u | < Λ we can esti-mate cosh dist( J, L β j,ni ) by using the above formula for ( z , θ ) ∈ L α ,ni and( z , θ ) ∈ L β j,ni . Since for large i we have, | z − z | ≥ | z β j,ni ,l | , | θ − θ | ≥|| z β j,ni ,l | − ( | z β j,ni ,u − z β j,ni ,l | ) | , θ θ ≤ | z β j,ni ,u || z β j,ni ,u − z β j,ni ,l | . Hence forlarge i we have,cosh dist( J, L β j,ni ) ≥ | z β j,ni ,l | + ( | z β j,ni ,l | − ( | z β j,ni ,u − z β j,ni ,l | )) | z β j,ni ,u || z β j,ni ,u − z β j,ni ,l | + 1 . Since | Λ | − < | z β j,ni ,l | ≤ | z β j,ni ,u | < Λ and | z β j,ni ,u − z β j,ni ,l | → , we have forlarge i there exits σ > J, L β j,ni ) ≥ σ | z β j,ni ,u − z β j,ni ,l | . Also by Λ − < | z β j,ni ,l | ≤ | z β j,ni ,u | < Λ and | λ ,n i | < | Λ | , there exists σ ′ > | π − ( z β j,ni ,u ) ′ || π − ( z β j,ni ,l ) ′ | > σ ′ . Since, | π − z β j,ni ,u − π − z β j,ni ,l | = | π − ( z β j,ni ,u ) ′ || π − ( z β j,ni ,l ) ′ || z β j,ni ,u − z β j,ni ,l | , we have | π − z β j,ni ,u − π − z β j,ni ,l | ≥ σ ′ | z β j,ni ,u − z β j,ni ,l | , From ρ − j,i = cosh dist( o, L π − β j,ni π ) sinh( T π − β j,ni π ) and the above estimates,implies that for i large, there exists δ > ρ j,i ≤ δ | π − z β j,ni ,u − π − z β j,ni ,l || tr( β j,n i ) | − . Hence there exists δ > x j,i ∈ C | z βj,ni,u | , y j,i ∈ C | λ ,ni | , lim i ρ j,i | π − x j,i − π − y j,i | ≤ lim i δ | π − z β j,ni ,u − π − z β j,ni ,l || tr( β j,n i ) || x j,i − y j,i | = 0 . Similarly there exists δ > , such that for w j,i ∈ C | z βj,ni,l | and z j,i ∈ C | λ ,ni | − , lim i ρ j,i | π − w j,i − π − z j,i | ≤ lim i δ | π − z β j,ni ,u − π − z β j,ni ,l || tr( β j,n i ) || w j,i − z j,i | = 0 . As before there exists δ > , such that for j = i and f j,i ∈ C | z βj,ni,u | ∪ C | z βj,ni,l | and g j ′ ,i ∈ C | z βj ′ ,ni ,u | ∪ C | z βj ′ ,ni ,l | and | tr( β j,i ) | ≤ | tr( β j ′ ,i ) | , lim i ρ j,i + ρ j ′ ,i | π − f j,i − π − g j ′ ,i | ≤ ρ j,i | π − f j,i − π − g j ′ ,i |≤ lim i δ δ | tr( β j,n i ) || f j,i − g j ′ ,i | = 0 for 2 ≤ j ≤ k.
18t follows from above calculations and Proposition 5.1 one have that for large i, S π − α ,ni π,r ,i , S π − α − ,ni π,r ,i are disjointed from S π − β j,ni π,ρ j,i , S π − β − j,ni π,ρ j,i also since | tr( β j,n i ) | → ∞ implies S π − β j,ni π,ρ j,i and S π − β − j,ni π,ρ j,i are disjointedfor i is large enough, and additionally we have S π − β j,ni π,ρ j,i ∪ S π − β − j,ni π,ρ j,i and S π − β j ′ ,ni π,ρ j,i ∪ S π − β − j ′ ,ni π,ρ j,i are also disjointed for 2 ≤ j < j ′ ≤ k. Thereforewe get first part of our lemma.For our second part of the lemma it is easily seen it can be proved in thesame way as first part.
Remark . During the course of our proof we realize that when | λ ,n | → , then we can certainly relax our first set of conditions for 2 ≤ j ≤ k in theabove lemma: • | λ ,n | − < | z β j,n ,l | ≤ | z β j,n ,u | < | λ ,n | , and • lim inf n (cid:26) | z β j,n ,l − z β j,n ,u | ( | z β j,n ,u | − | λ ,n | ) | tr( β j,n ) | , | z β j,n ,l − z β j,n ,u | ( | z β j,n ,l | − | λ ,n | − ) | tr( β j,n ) | (cid:27) = 0 . Next we consider when | tr( β j,n ) | < C for all n. In this case we can’t usethe isometric circles but instead we can employ the following Proposition toget disjoint circles for β i,n . Proposition 5.5.
For any given loxodromic transformation γ with fixedpoints = 0 , ∞ and mutiplier λ γ we can find disjoint circles S o,r , S o ′ ,r ′ of center o radius r and center o ′ radius r ′ respectively such that, γ ( interior ( S o,r )) ∩ interior ( S o ′ ,r ′ ) = ∅ , and r + r ′ = | z γ, + − z γ, − | | λ γ || λ γ | − . Note that since | λ γ | > , so by this equality for r + r ′ we have a upper boundas, r + r ′ < | z γ, − − z γ, + | | λ γ | +1 | λ γ |− . Proof.
We first conjugates γ into Mobius transformation γ ′ have fixed points { , ∞} . First we look at circles S , | λ γ | − , S , | λ γ | . The transformation φ ( x ) = x − x +1 maps the fixed points of γ ′ consists of { , ∞} to {− , } as fixed points19espectively. Additionally it also maps S , | λ γ | − , S , | λ γ | to S z ,r , S z ′ ,r ′ respec-tively. Now we will use basic formulas to determine the values of z , z ′ , r , r ′ (see page 91 of [21]). By [21] we have, r = (cid:12)(cid:12)(cid:12)(cid:12) −| λ γ | − − −| λ γ | − + 1 − | λ γ | − − | λ γ | − + 1 (cid:12)(cid:12)(cid:12)(cid:12) = 2 | λ γ || λ γ | − r ′ = (cid:12)(cid:12)(cid:12)(cid:12) −| λ γ | − −| λ γ | + 1 − | λ γ | − | λ γ | + 1 (cid:12)(cid:12)(cid:12)(cid:12) = 2 | λ γ || λ γ | − r + r ′ = 4 | λ γ || λ γ | − . The distance between the centers is, | z − z ′ | = (cid:12)(cid:12)(cid:12)(cid:12) −| λ γ | − − −| λ γ | − + 1 − −| λ γ | − −| λ γ | + 1 (cid:12)(cid:12)(cid:12)(cid:12) = 2 | λ γ | + 1 | λ γ | − . Since ( | λ γ | + 1) − | λ γ | = ( | λ γ | − > , implies S z ,r , S z ′ ,r ′ are disjoint. Byconjugating φγ ′ φ − with ψ ( x ) = x + z γ, + + z γ, − z γ, + − z γ, − we map the fixed points {− , } to n z γ, − z γ, + − z γ, − , z γ, + z γ, + − z γ, + o . Since ψ ( x ) is a translation transformation (i.e eu-clidian isometry), the circles are then mapped to circles S z ,r , S z ′ ,r ′ whichhave same radius as before and also preserves the disjointness as before. Thenby conjugating ψφγφ − ψ − with θ ( x ) = z γ, + − z γ, − sends n z γ, − z γ, + − z γ, − , z γ, + z γ, + − z γ, + o to points { z γ, − , z γ, + } and also maps the circles to S z ,r , S z ′ ,r ′ . We have that r , r ′ = | z γ, + − z γ, − | | λ γ || λ γ | − and also the disjointness is preserved. Also wehave γ = θψφγ ′ φ − ψ − θ − . Therefore we get that the sum of the radius ofour disjointed circles are given by, r + r ′ = | z γ, − − z γ, + | | λ γ || λ γ | − γ we will denote the set of disjoint disks D γ, + , D γ, − , suchthat D γ, + = ∂ S o,r , D γ, + = ∂ S o ′ ,r ′ with the convention that z γ, + ∈ D γ, + and z γ, − ∈ D γ, − . We also set R γ = | z γ, − − z γ, + | | λ γ |− for rest of the sections. Proposition 5.6.
Given a sequence < α n , β n > with D n → such that | tr( α n ) | < C. Assume that α n have fixed points , ∞ and | λ α n | − ≤ | η β n | ≤| ζ β n | ≤ | λ α n | then either < α n , β n > or < α n , α − n β n > is classical Schottkygroup for large n. Proof.
This is a direct consequence Theorem 1 . . k − k − k .To do so we will first state the k − Proposition 5.7.
Given a sequence of k − -generators < α ,n , ..., α k − ,n > with D n → such that | tr( α ,n ) | < C. Assume that α ,n have fixed points , ∞ and | λ α ,n | − ≤ | η α j,n | ≤ | ζ α j,n | ≤ | λ α ,n | then, < α ,n , α a ,n α ,n , ..., α a k − ,n α j,n > is classical Schottky group for large n and some a j ∈ { , } . Denote G α n ; β n = inf {| z β n , ± − z | : z ∈ D α n , ± } . Note that for a sequenceof generators < α n , β n > with z α n , ± , z β n , ± ∈ C and z β n , ± D α n , ± . Supposethat G α n ; β n R − α n → ∞ , then < α n , β n > generates a classical Schottky groupfor large n. This section we classify all possible behaviors such that a given sequence Γ n ofSchottky groups with D Γ n → obstructions that a sequence of Γ n to containany classical Schottky group will be called degeneracies. Analyzing degeneracies rely on the analysis of the behaviors of centers ofisometric circles of generators. To do this analysis, we must characterize allpossible dynamics (here dynamic refers to behaviors with respect to decreas-ing Hausdorff dimension) of sequence of generators according to behaviors oftheir centers of circles. To do so we first consider dichotomized behaviors andthen utilize them to show general result. More precisely, we show that eachof the dichotomized degenerate behavior will lead into a classical Schottky21roup for sufficiently large n. The dichotomization of degeneracies are intofour types as: TypeI, Type II, Type III, Type IV given in the following.In addition, each of these types of degeneracies will have several sub-typeswhich are stated later on separately.
Definition 6.1.
For τ >
0, we set: J k ( τ ) := { [Γ] ∈ J k | Z Γ > τ, ∃ for some Γ ∈ [Γ] } . Recall that J k denotes Schottky space of rank k . Here [Γ] denotes equiva-lence class of Γ , and J k ( τ ) is just collection of Schottky groups that have somegenerating set with uniform lower bounds on Z Γ by τ under conjugation.Let us assume that there exists a sequence { [Γ n ] } ⊂ J k ( τ ) of nonclassicalSchottky groups with strictly decreasing D n →
0. Set Γ n = < α ,n , α ,n , ..., α k,n > with that Z <α ,n ,α ,n ,...,α k,n > > τ . We arrange the generators so | tr( α i,n ) | ≤| tr( α i +1 ,n ) | . There are two possibilities: • ( I ) There exists a subsequence such that, | tr( α ,n ) | → ∞• ( II ) | tr( α ,n ) | < M , for some M > . Case ( I ) is trivial. Since | tr( α i,n j ) | → ∞ as n → ∞ for all 1 ≤ i ≤ k , it follows from Z <α i,nj ,...,α k,nj > > τ , there must exists N such that <α i,n j , ..., α i,n j > becomes classical Schottky groups for j > N . A contradic-tion.Now we consider case ( II ).We work in upper space model H . Conjugate < α ,n , ..., α k,n > by a Mobiustransformation into α ,n = (cid:0) λ α ,n λ − α ,n (cid:1) with | λ α ,n | >
1. Denote α j,n = (cid:0) a j,n b j,n c j,n d j,n (cid:1) . Since | tr( α ,n ) | < M implies | λ α ,n | < M ′ for some M ′ > , it follows from Proposition 4.1 and D n →
0, we have | tr( α j,n ) | → ∞ . Byreplacing α i,n with α − i,n if necessary, we can assume | ζ α i,n | ≤ | η α i,n | , ≤ j ≤ k. Since Z <α ,n ,...,α k,n > > τ , there exists ∆ , ∆ , ∆ , ∆ > < | z α j,n ,l | ≤ | z α j,n ,u | < ∆ , and ∆ < | z α j,n ,l − z α j,n ,u | < ∆ . It follows fromLemma 4.4, ∆ < | ζ α j,n | ≤ | η α j,n | < ∆ , and ∆ < lim n | ζ α j,n − η α j,n | < ∆ . For each n and i ≥
2, choose integers k i,n , l i,n such that:1 ≤ | ζ α i,n λ k i,n ,n | < | λ ,n | , ≤ | η α i,n λ l i,n ,n | < | λ ,n | . We consider the generating set < α ,n , ..., α k i,n ,n α i,n α l i,n ,n , ..., α k k,n ,n α k,n α l k,n ,n > .Denote these new generators by β i,n for i ≥ .
22y passing to subsequence and considering inverses if necessary, we havecan assume | η β i,n λ l i,n ,n | ≤ | ζ β i,n λ k i,n ,n | . Definition 6.2 (standard form) . Given a set of generators S Γ = { α , ..., α k } of Schottky group Γ, we say S Γ is of standard form if (up to conjugation byMobius transformation), z α , ± = { , ∞} and1 ≤ | ζ α i | < | λ α | , ≤ | η α i | < | λ α | , ≤ i ≤ k. Definition 6.3 ( k − . Given Γ rank- k Schottky group we say Γis k − classical if every rank k − G is a classical Schottkygroup.In Section 11.2 we will show that, given a sequence of Γ n Schottky groupwhich is k − D Γ n →
0, we can always choose a subsequenceof generators S Γ n such that it’s k − k − Definition 6.4.
Let Γ n be a sequence of Schottky groups with generatingset S Γ n = < α ,n , α ,n , ..., α k,n > such that α ,n have fixed points 0 , ∞ . We say S Γ n is a normal sequence of generating sets if (up to conjugation by Mobiustransformation), all isometric circles of α j,k are strictly bounded between 1and | λ α ,n | for 2 ≤ j ≤ k and large n . Proposition 6.5 (Normal sequence) . Let Γ n be a sequence of Schottky groupswhich is k − classical with D Γ n → . Then there exists a normal subsequenceof generating set S Γ n (we use same index for subsequence ).Proof. Since Γ n is k − S n so that every pair α ,n , α j,n is classical for 1 ≤ j ≤ k. We assume bysame convention that α ,n is of minimal | trace | . Conjugate α ,n to have fixedpoints 0 , ∞ and denote the new sequence by the same notation. It followsfrom Corollary 4.3 and Remark 4, isometric circles of α j,n have radius → ≤ j ≤ k. Since all fixed points are strictly bounded between 1 and | λ α ,n | , the result follows.The behaviors of elements of our sequence of generators are done accord-ing to location of their centers of circles of α j,n for 2 ≤ j ≤ k. Lets assume we23ave a sequence of generating sets which is k − α j,n for 2 ≤ j ≤ k to circles of α ,n . Definition 6.6 (Degenerate types) . Using above notations, let Γ n be a se-quence of rank k Schottky groups which is k − D n →
0. Let S n be a sequence of generating sets which is k − • Type I: Double boundary degeneracy || ζ α i,n λ k i,n ,n | − | η α i,n λ l i,n ,n || → . • Type II: Single boundary degeneracy || ζ α i,n λ k i,n ,n | − | λ ,n | | → . • Type III: Elliptical degeneracy | ζ α i,n λ k i,n ,n | → . • Type IV: Bounded || ζ α i,n λ k i,n ,n | − | η α i,n λ l i,n ,n || > C, and || ζ α i,n λ k i,n ,n | − | λ ,n | | > C. In the next four sections we will show that each of these degenerate typeswill lead to classical Schottky group for large n. We show this type of degeneracy is classical for large n. We use the same notation index for subsequences. For large n there exists1 < | λ | < | λ α ,n | , σ <
1, such that max i | ζ i,n λ k i,n ,n | → σ | λ | . Let ψ n be theMobius transformation that fixes { , ∞} defined by ψ ( x ) = x √ σ | λ | , x ∈ C . Let1 ≤ χ ≤ σ | λ | be given as χ = min i lim n | η i,n λ l i,n ,n | . Then min i | η ψβ i,n ψ − | → √ σ | λ | and max i | ζ ψβ i,n ψ − | → √ σ | λ | . Let | tr( β i ∗ ,n ) | = min | tr( β i,n ) | . It followsfrom Lemma 4.4, for n large,max { max i | z ψβ i,n ψ − , ∓ − η ψβ i,n ψ − | , max i | z ψβ i,n ψ − , ± − ζ ψβ i,n ψ − |} < ρ | tr( β i ∗ ,n ) | . Hence there exists ρ ′ , ρ ′′ , ρ ′′′ > (cid:12)(cid:12) z ψβ i,n ψ − , + − z ψβ i,n ψ − , − (cid:12)(cid:12) > (cid:12)(cid:12)(cid:12)(cid:12) |√ σ | λ | − χ √ σ | λ | (cid:12)(cid:12)(cid:12)(cid:12) − ρ ′ | tr( β i ∗ ,n ) | , and (cid:12)(cid:12)(cid:12)(cid:12) z ψβ i,n ψ − , + − z ψβ i,n ψ − , − (cid:12)(cid:12)(cid:12)(cid:12) > ρ ′′ (cid:12)(cid:12)(cid:12)(cid:12) | √ σ | λ | χ − √ σ | λ | (cid:12)(cid:12)(cid:12)(cid:12) − ρ ′′′ | tr( β i ∗ ,n ) | . This implies that there exist ∆ > Z ψβ i,n ψ − > ∆ . Hence byapplying Proposition 4.1 to the generators < ψα ,n ψ − , ψβ i,n ψ − > , implies | tr( ψβ i,n ψ − ) | → ∞ . Set κ = q σ +12 | λ | . Then for sufficiently large n we have κ < | λ ,n | , κ − < | z ψβ i,n ψ − ,l | ≤ | z ψβ i,n ψ − ,u | < κ. And obviously,lim n κ − | z ψβ i,n ψ − ,u | ) | tr( ψβ i,n ψ − ) | = 0 , lim n | z ψβ i,n ψ − ,l | − κ − ) | tr( ψβ i,n ψ − ) | = 0 . Also by κ ( S Γ n ) → ∞ , | z β i,n , ± − z β j,n , ± || tr( β i,n ) | → ∞ , for | tr( β i,n ) | ≤ | tr( β j,n ) | Therefore, < ψα ,n ψ − , ψβ ,n ψ − , ..., ψβ k,n ψ − > satisfies the second set ofconditions of Lemma 5.2, and so by Lemma 5.2, these will be classical gen-erators for large n , a contradiction. We show this type of degeneracy is classical for large n. This case we need to further break into two possibilities based on whether α ,n converging into a elliptical element or a nonellipitical element.By passing to a subsequence if necessary, we have two possibilities: • Type I : | λ α ,n | − • Type I : There exists λ >
1, such that | λ α ,n | ≥ | λ | for large n .25 .1 Type I In the degeneration into elliptical case, we need to consider whether thecollapsing of fixed points is at must faster rate relative to collapsing of fixedpoints to circles: ( i ) , ( ii ).Here we either have( i ) lim sup n (cid:12)(cid:12)(cid:12)(cid:12) | ζ α ki,n ,n β i,n α li,n ,n | − | λ α ,n | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ζ α ki,n ,n β i,n α li,n ,n − η α ki,n ,n β i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) − < ∞ or( ii ) lim inf n (cid:12)(cid:12)(cid:12)(cid:12) | ζ α ki,n ,n β i,n α li,n ,n | − | λ α ,n | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ζ α ki,n ,n β i,n α li,n ,n − η α ki,n ,n β i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) − → ∞ . ( ii )In this subcase, we show that under appropriate Nielsen transformation andconjugation, we can reduce to case ( i ) . Hence it would be enough just toconsider subcase ( i ), which is done next.Take Mobius transformation ψ n defined by, ψ i,n ( x ) = xζ αki,nn βi,nαli,n ,n . Set ψ n ( x ) be the ψ i ∗ ,n with maximal | ψ i,n | , ≤ i ≤ k. We conjugate α k i,n ,n β i,n α l i,n ,n by ψ n . Consider ( ψ n α k i,n ,n β i,n α l i,n ,n ψ − n ) − . Since by factor out ζ α ki,n ,n β i,n α li,n ,n λ − ,n in ( ii ),lim inf n | ζ α ki,n ,n β i,n α li,n ,n | | λ ,n | − (cid:12)(cid:12)(cid:12)(cid:12) | − | ζ α ki,nn β i,n α li,n ,n | − | λ ,n | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ζ − α ki,n ,n β i,n α li,n ,n λ ,n − λ ,n (cid:12)(cid:12)(cid:12)(cid:12) − → ∞ . lim inf n (cid:12)(cid:12)(cid:12)(cid:12) | − | ζ − α ki,n ,n β i,n α li,n ,n λ n | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ζ − α ki,n ,n β i,n α li,n ,n λ ,n − λ ,n (cid:12)(cid:12)(cid:12)(cid:12) − → ∞ . Since ζ l i,n − α ki,n ,n β i,n α li,n ,n λ ,n = η α ki,nn β i,n α − ,n and ζ ( ψ n α ki,n ,n β i,n α li,n − ,n ψ − n ) − = η ψ n α ki,n ,n β i,n α li,n − ,n ψ − n we have,lim inf n (cid:12)(cid:12)(cid:12)(cid:12) | − | η α ki,n ,n β i,n α li,n − ,n | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) η α ki,n ,n β i,n α li,n − ,n − λ ,n (cid:12)(cid:12)(cid:12)(cid:12) − → ∞ , giving,lim sup n (cid:12)(cid:12)(cid:12)(cid:12) | ζ ( ψ n α ki,n ,n β i,n α li,n − ,n ψ − n ) − | − | λ ,n | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ζ ( ψ n α ki,n ,n β ,n α li,n − ,n ψ − n ) − − (cid:12)(cid:12)(cid:12)(cid:12) − < ∞ . ψ n α k i,n ,n β i,n α l i,n − ,n ψ − n ) − satisfies ( i ). Hence replacing the gen-erators if necessary we can always assume the generators satisfy ( i ). Andwithout lost of generality we will assume that < α ,n , ..., α k i,n ,n β i,n α l i,n ,n , ... > satisfies ( i ) . ( i ) . Here we must consider whether the collapsing is bounded from below or not:( i ) , ( i ) . In this case, we have either:( i ) lim sup n (cid:12)(cid:12)(cid:12)(cid:12) | ζ α ki,n ,n β i,n α li,n ,n | − | λ α ,n | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ζ α ki,n ,n β i,n α li,n ,n − η α ki,n ,n β i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) − > δ > , or( i ) lim sup n (cid:12)(cid:12)(cid:12)(cid:12) | ζ α ki,n ,n α i,n α li,n ,n | − | λ α ,n | (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) ζ α ki,n ,n β i,n α li,n ,n − η α ki,n ,n β i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) − = 0 . First we present the proof for ( i ) . ( i ) . Lemma 8.1.
There exists c > such that, dist( L α ,n , L α ki,n ,n α i,n α li,n ,n ) < log( c | λ ,n | − . Proof.
We first show that | tr( α i,n ) | ( | λ ,n | − →
0. From Proposition 4.1 wehave,lim 1 | tr( α i,n ) | ( | λ ,n | − ≤ lim ρ (cid:18) | λ ,n | D n − | λ ,n | D n + 3)( | λ ,n | − D n (cid:19) D n and for large n, we have | λ ,n | D n − < | λ ,n | − ρ ′ > | tr( α i,n ) | ( | λ ,n | − ≤ lim ρ ′ ( | λ ,n | − − D n D n = 0 .
27t follows from Lemma 4.4, (cid:12)(cid:12)(cid:12)(cid:12) ζ α ki,n ,n α i,n α li,n ,n − η α ki,n ,n α i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) − ρ ′ | tr( α i,n ) | − ≤ (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n ,n α i,n α li,n ,n , − − z α ki,n ,n α i,n α li,n ,n , + (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) ζ α ki,n ,n α i,n α li,n ,n − η α ki,n ,n α i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) + ρ ′′ | tr( α i,n ) | − Since 1 ≤ | ζ α ki,n ,n α i,n α li,n ,n | < | λ ,n | , we have (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n ,n α i,n α li,n ,n , − − z α ki,n ,n α i,n α li,n ,n , + (cid:12)(cid:12)(cid:12)(cid:12) | λ ,n | − ≥ (cid:12)(cid:12)(cid:12)(cid:12) ζ α ki,n ,n α i,n α li,n ,n − η α ki,n ,n α i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) | λ ,n | − − ρ ′ | tr( α i,n ) | ( | λ ,n | − . By the condition of ( i ) we have, (cid:12)(cid:12)(cid:12)(cid:12) ζ α ki,n ,n α i,n α li,n ,n − η α ki,n ,n α i,n α li,n ,n (cid:12)(cid:12)(cid:12)(cid:12) | λ ,n | − | ζ α ki,n ,n α i,n α li,n ,n − η α ki,n ,n α i,n α li,n ,n || λ ,n | − | ζ α ki,n ,n α i,n α li,n ,n | + | ζ α ki,n ,n α i,n α li,n ,n | − > M + 1 , for some M > . Hence for large n there exists κ > (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n ,n α i,n α li,n ,n , − − z α ki,n ,n α i,n α li,n ,n , + (cid:12)(cid:12)(cid:12)(cid:12) | λ ,n | − > M + 1 − ρ ′ | tr( α i,n ) | ( | λ ,n | − > κ. For the upper bounds we have, | z α ki,n ,n α i,n α li,n ,n , − − z α ki,n ,n α i,n α li,n ,n , + | < | λ ,n | + ρ ′′ | tr( α i,n ) | − . Note thatdist( L α ,n , L α ki,n ,n α i,n α li,n ,n ) = inf (cid:26) dist( h , h ) | h ∈ L α ,n , h ∈ L α ki,n ,n α i,n α li,n ,n (cid:27) . Set h j = ( z j , θ j ); j = 1 , z , θ ) = (0 , | ζ α ki,n ,n α i,n α li,n ,n ,l | + 12 | z α ki,n ,n α i,n α li,n ,n ,u − z α ki,n ,n α i,n α li,n ,n ,l | ) , ( z , θ ) = ( 12 ( z α ki,n ,n α i,n α li,n ,n ,u + z α ki,n ,n α i,n α li,n ,n ,l ) , | z α ki,n ,n α i,n α li,n ,n ,u − z α ki,n ,n α i,n α li,n ,n ,l | ) . By Lemma 4.4,1 − σ | tr( α i,n ) | − < | z α ki,n ,n α i,n α li,n ,n ,l | ≤ | z α ki,n ,n α i,n α li,n ,n ,u | ≤ | λ ,n | + σ | tr( α i,n ) | − | z α ki,n ,n α i,n α li,n ,n ,u − z α ki,n ,n α i,n α li,n ,n ,l | we have,cosh dist( L α ,n , L α ki,n ,n α i,n α li,n ,n ) ≤ | z α ki,n ,n α i,n α li,n ,n ,u + z α ki,n ,n α i,n α li,n ,n ,l | + | z α ki,n ,n α i,n α li,n ,n ,l | | z α ki,n ,n α i,n α li,n ,n ,u − z α ki,n ,n α i,n α li,n ,n ,l | ( | z α ki,n ,n α i,n α li,n ,n ,l | + | z α ki,n ,n α i,n α li,n ,n ,u − z α ki,n ,n α i,n α li,n ,n ,l | ) +1 < | λ ,n | + ρ ′′′ | tr( α i,n ) | − + κ ( | λ ,n | −
1) + 1 κ ( | λ ,n | − , ρ ′′′ > . This last inequality implies the Lemma.
Lemma 8.2. lim n | tr( α k i,n ,n α i,n α l i,n ,n ) | ( | λ ,n | −
1) = 0
Proof.
It follows from Proposition 4.1 and Lemma 8.1, there exists ρ > (cid:12)(cid:12)(cid:12) tr( α k i,n ,n α i,n α l i,n ,n ) (cid:12)(cid:12)(cid:12) ≥ ρ (cid:18) ( | λ ,n | D n + 3)( | λ ,n | − D n | λ ,n | D n − (cid:19) D n . hence we have,lim 1 | tr( α k i,n ,n α i,n α l i,n ,n ) | ( | λ ,n | − ≤ lim ρ ′ (cid:18) | λ ,n | D n − | λ ,n | D n + 3)( | λ ,n | − D n (cid:19) D n . Since | λ ,n | D n − < | λ ,n | − n we have,lim 1 | tr( α k i,n ,n α i,n α l i,n ,n ) | ( | λ ,n | − ≤ lim ρ ′′ ( | λ ,n | − − D n D n = 0 . For large n by condition ( i ) , we have δ ( | λ ,n | − < | λ ,n | −| ζ α ki,n ,n α i,n α li,n ,n | then by Lemma 4.4, || z α ki,n ,n α i,n α li,n ,n ,u | − | ζ α ki,n ,n α i,n α li,n ,n || < χ | tr( α ki,n ,n α i,n α li,n ,n ) | forsome χ > | λ ,n | − | z α ki,n ,n α i,n α li,n ,n ,u | > | λ ,n | − | ζ α ki,n ,n α i,n α li,n ,n | − χ | tr( α k i,n ,n α i,n α l i,n ,n ) | > δ ( | λ ,n | − − χ | tr( α k i,n ,n α i,n α l i,n ,n ) | . ǫ n = min ≤ i ≤ k δ ( | λ ,n | − − χ (cid:12)(cid:12)(cid:12)(cid:12) tr( α ki,n ,n α i,n α li,n ,n ) (cid:12)(cid:12)(cid:12)(cid:12) ! . z u,n = max ≤ i ≤ k (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n ,n α i,n α li,n ,n ,u (cid:12)(cid:12)(cid:12)(cid:12) . Define Mobius transformations by ψ n ( x ) = (cid:16) ǫ n z u,n (cid:17) λ − ,n ( x ) . Then, | λ i,n | − | z ψ n α ki,n ,n α i,n α li,n ,n ψ − n ,u | = | λ i,n | − (1 + ǫ n z u,n ) | λ ,n | − | z α ki,n ,n α i,n α li,n ,n ,u |≥ ( | λ ,n | − | z α ki,n ,n α i,n α li,n ,n ,u | − ǫ n | ) | λ ,n | − > ( ǫ n − ǫ n | λ ,n | − = ǫ n | λ ,n | − . Also by Lemma 4.4 we have, || z α ki,n ,n α i,n α li,n ,n ,l | − | η α ki,n ,n α i,n α li,n ,n || < χ ′ | tr( α i,n ) | . Since | η α ki,n ,n α i,n α li,n ,n | ≥ ≤ i ≤ k this implies, | z ψα ki,n ,n α i,n α li,n ,n ψ − ,l | − | λ ,n | − = (1 + ǫ n z u,n ) | z α ki,n ,n α i,n α li,n ,n ,l || λ ,n | − − | λ ,n | − = | λ ,n | − | z α ki,n ,n α i,n α li,n ,n ,l | − ǫ n | z α ki,n ,n α i,n α li,n ,n ,l | z u,n > | λ ,n | − ǫ n | z α ki,n ,n α i,n α li,n ,n ,l | z u,n − χ ′ | tr( α i,n ) | > | λ ,n || tr( α i,n ) | | tr( α i,n ) | ǫ n | z α ki,n ,n α i,n α li,n ,n ,l | z u,n − χ ′ > | λ ,n || tr( α i,n ) | δ | tr( α i,n ) | ( | λ ,n | − | z α ki,n ,n α i,n α li,n ,n ,l | − χ | z α ki,n ,n α i,n α li,n ,n ,l | z u,n − χ ′ By Lemma 8.2 and above inequality we have, | z ψ n z αki,n ,n αi,nαli,n ,n ,l ψ − n ,l |−| λ ,n | − > | tr( α k i,n ,n α i,n α l i,n ,n ) | ( | z α ki,n ,n α i,n α li,n ,n ,l |−| λ ,n | − ) → ∞ and | tr( α k i,n ,n α i,n α l i,n ,n ) | ( | λ ,n |−| z ψ n α ki,n ,n α i,n α li,n ,n ,u | ) → ∞ . Hence, • | λ ,n | − < | z ψ n α ki,n ,n α i,n α li,n ,n ψ − n ,l | ≤ | z ψ n α ki,n ,n α i,n α li,n ,n ψ − n ,u | < | λ ,n | , and30 lim n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z ψnαki,n ,n αi,nαli,n ,n ψ − n ,u (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −| λ ,n | !(cid:12)(cid:12)(cid:12)(cid:12) tr (cid:18) α ki,n ,n α i,n α li,n ,n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) lim n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z ψnαki,n ,n αi,nαli,n ,n ψ − n ,l (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −| λ ,n | − !(cid:12)(cid:12)(cid:12)(cid:12) tr (cid:18) α ki,n ,n α i,n α li,n ,n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) . Also by κ ( S Γ n ) → ∞ , implies that (cid:12)(cid:12)(cid:12)(cid:12) z ψ n ;1 α ki,n ,n α i,n α li,n ,n ψ − n ;1 , ± − z ψ n ;1 α kj,n ,n α j,n α lj,n ,n ψ − n ;1 , ± (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:18) ǫ n z u,n (cid:19) λ − ,n (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n ,n α i,n α li,n ,n , ± − z α kj,n ,n α j,n α lj,n ,n , ± (cid:12)(cid:12)(cid:12)(cid:12) → ∞ , for i = j, ≤ i, j ≤ k and (cid:12)(cid:12)(cid:12) tr (cid:16) α k i,n ,n α i,n α l i,n ,n (cid:17)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) tr (cid:16) α k j,n ,n α j,n α l j,n ,n (cid:17)(cid:12)(cid:12)(cid:12) . Hence the generators < ψ n α ,n ψ − n , ψ n α k ,n ,n α ,n α l ,n ,n ψ − n , ..., ψ n α k k,n ,n α k,n α l k,n ,n ψ − n > satisfies conditions of Lemma 5.2 for large n. For future reference we set theMobius transformations defined earlier as ψ n ;1 ( x ) . Next we present the proof for ( i ) . This case requires a bit more compu-tation. This is due to the fact that we need to address the possibility thatcollapsing of fixed points imply that the distance between axises of β i,n and α ,n will go to ∞ . So we can’t conclude that | tr( β i,n ) | → ∞ . Hence we willuse different set of circles given by Proposition 5 . ( i ) . There exists | ρ α ki,n − ,n α i,n α li,n ,n | → ζ α ki,n − ,n α i,n α li,n ,n ρ α ki,n − ,n α i,n α li,n ,n = η α kn − n β n . If lim sup n | ρ α ki,n − ,n α i,n α li,n ,n − | >
0, then (with the same indexnotation for subsequence there exists a subsequence) such thatlim inf n | ζ α ki,n − ,n α i,n α li,n ,n − η α ki,n ,n α i,n α li,n ,n | > . This implies by Lemma 4.4, lim inf n | z α ki,n − ,n α i,n α li,n ,n , + − z α ki,n ,n α i,n α li,n ,n , − | > . Hence by Proposition 4.1, there exists ρ >
0, such that for large n , | tr( α k i,n − ,n α i,n α l i,n ,n ) | ≥ ρ (cid:18) | λ ,n | D n + 3 | λ ,n | D n − (cid:19) D n .
31n particular, we have | tr( α k i,n − ,n α i,n α l i,n ,n ) | → ∞ . Notelim n | ρ α ki,n − ,n α i,n α li,n ,n | − | λ ,n | − . This can be seen as follows: since | λ ,n | − | ζ α ki,n ,n α i,n α li,n ,n | ≤ | λ ,n | − | λ ,n | −| ζ αki,n ,n αi,nαli,n ,n || λ ,n | − → ≥ | λ ,n | −| ζ αki,n ,n αi,nαli,n ,n || λ ,n | − > ǫ > . Assume that the latter inequality holds. This is equivalent to ( i ) andwe follows the same idea used in ( i ) . Set Mobius transformations ψ n ;2 ( x ) = λ − ,n (1 − ǫ n ) − x, with ǫ n = ǫ ( | λ ,n | − | λ ,n | . Then, | λ ,n | − | ζ ψ n ;2 α ki,n ,n α i,n α li,n ,n ψ − n ;2 | = | λ ,n | − | λ − ,n (1 − ǫ n ) − ζ α ki,n ,n α i,n α li,n ,n | = | λ ,n | − (1 − ǫ n ) − (cid:18) | λ ,n | (1 − ǫ n ) − | ζ α ki,n ,n α i,n α li,n ,n | (cid:19) Since | λ ,n | − | ζ α ki,n ,n α i,n α li,n ,n | > ǫ ( | λ ,n | −
1) we have, | λ ,n |−| ζ ψ n ;2 α ki,n ,n α i,n α li,n ,n ψ − n ;2 | > | λ ,n | − (1 − ǫ n ) − ( ǫ ( | λ ,n | − −| λ ,n | ǫ n ) = ǫ ( | λ ,n | − | λ ,n | (1 − ǫ n ) . Since that ǫ n → , it follows from last inquality that for large n we have, | λ ,n |−| ζ ψn ;2 αki,n − ,n αi,nαli,n ,n ψ − n ;2 || λ ,n | − > ǫ ′ > . And | η ψ n ;2 α ki,n ,n α i,n α li,n ,n ψ − n ;2 | ≥ | λ ,n | − (1 − ǫ n ) − we have, | η ψ n ;2 α ki,n ,n α i,n α li,n ,n ψ − n ;2 | − | λ n | − | λ | − ǫ | λ n | (1 − ǫ n ) > ǫ ′′ > . Since | tr( α k i,n ,n α i,n α l i,n ,n )( | λ ,n | − | → ∞ , it follows that for large n , • | λ ,n | − < | η ψ n ;2 α ki,n ,n α i,n α li,n ,n ψ − n ;2 | ≤ | ζ ψ n ;2 α ki,n ,n α i,n α li,n ,n ψ − n ;2 | < | λ ,n | , and • n (cid:12)(cid:12)(cid:12)(cid:12) tr (cid:18) α ki,n ,n α i,n α li,n ,n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) η ψn ;2 αki,n ,n αi,nαli,n ,n ψ − n ;2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −| λ ,n | − ! (cid:12)(cid:12)(cid:12)(cid:12) tr (cid:18) α ki,n ,n α i,n α li,n ,n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ζ ψn ;2 αki,n ,n αi,nαli,n ,n ψ − n ;2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −| λ ,n | ! κ ( S Γ n ) → ∞ , implies that (cid:12)(cid:12)(cid:12)(cid:12) z ψ n ;2 α ki,n ,n α i,n α li,n ,n ψ − n ;2 , ± − z ψ n ;2 α kj,n ,n α j,n α lj,n ,n ψ − n ;2 , ± (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12) λ − ,n (1 − ǫ n ) − (cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n ,n α i,n α li,n ,n , ± − z α kj,n ,n α j,n α lj,n ,n , ± (cid:12)(cid:12)(cid:12)(cid:12) → ∞ , for i = j, ≤ i, j ≤ k and (cid:12)(cid:12)(cid:12) tr (cid:16) α k i,n ,n α i,n α l i,n ,n (cid:17)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) tr (cid:16) α k j,n ,n α j,n α l j,n ,n (cid:17)(cid:12)(cid:12)(cid:12) . Hence by Lemma 4.4, < ψ n ;2 α ,n ψ − n ;2 , ..., ψ n ;2 α k k,n ,n α k,n α l k,n ,n ψ − n ;2 > satisfiesLemma 5.2.If the former holds then | λ ,n | ( | η α ki,n ,n α i,n α li,n ,n |−| ζ α ki,n ,n α i,n α li,n ,n λ − ,n | ) = | λ ,n | ( | η α ki,n ,n α i,n α li,n ,n |−| ζ α ki,n − ,n α i,n α li,n ,n | )= | λ ,n | ( | ζ α ki,n − ,n α i,n α li,n ,n ρ α ki,n − ,n α i,n α li,n ,n | − | ζ α ki,n − ,n α i,n α li,n ,n | )= | λ ,n | | ζ α ki,n − ,n α i,n α li,n ,n | ( | ρ α ki,n − ,n α i,n α li,n ,n | − n | ρ αki,n − ,n αi,nαli,n ,n |− | λ ,n | − = 0 , and | tr( α k i,n − ,n α i,n α l i,n ,n )( | λ ,n | − | → ∞ , it follows that for large n , • | λ ,n | − < | ζ α ki,n − ,n α i,n α li,n ,n | ≤ | η α ki,n − ,n α i,n α li,n ,n | < | λ ,n | , and • n (cid:12)(cid:12)(cid:12)(cid:12) tr (cid:18) α ki,n − ,n α i,n α li,n ,n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ζ αki,n − ,n αi,nαli,n ,n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −| λ ,n | − ! (cid:12)(cid:12)(cid:12)(cid:12) tr (cid:18) α ki,n − ,n α i,n α li,n ,n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) η αki,n − ,n αi,nαli,n ,n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −| λ ,n | ! • (cid:12)(cid:12)(cid:12)(cid:12) z α ki,n − ,n α i,n α li,n ,n , ± − z α kj,n − ,n α j,n α lj,n ,n , ± (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) tr (cid:16) α k i,n − ,n α i,n α l i,n ,n (cid:17)(cid:12)(cid:12)(cid:12) → ∞ ; i = j Hence by Lemma 4.4, < α ,n , ..., α k k,n − ,n α k,n α l k,n ,n > satisfies Lemma 5.2.Consider the case that, ρ α ki,n − ,n α i,n α li,n ,n → α ,n , α k i,n − ,n α i,n α l i,n ,n and α k i,n ,n α i,n α l i,n ,n . To do so we do some two generators estimates. Fix a i thatsatisfies ρ α ki,n − ,n α i,n α li,n ,n →
1. We first normalize so that η α ki,n ,n α i,n α li,n ,n = 1 . Since our estimates will be in terms of traces so for simplicity we can take l i,n = 0 and set η α ki,n − ,n α i,n α li,n ,n = 1 . by conjugation with a desired Mobiustransformation.Let f i,n be the Mobius transformations given by, f i,n ( x ) = 1 σ i,n z α ki,n − ,n α i,n , − x − z α ki,n − ,n α i,n , − z α ki,n − ,n α i,n , + x − z α ki,n − ,n α i,n , − , x ∈ C . where σ i,n = z αki,n − ,n αi,n, − z αki,n − ,n αi,n, + − z αki,n − ,n αi,n, − . Let φ i,n be the Mobiustransformations defined by, φ i,n ( x ) = η − f i,n α ki,n ,n α i,n f − i,n x. Then ζ φ i,n f i,n α ki,n ,n α i,n f − i,n φ − i,n is given by, ζ φ i,n f i,n α ki,n ,n α i,n f − i,n φ − i,n = − z α ki,n − ,n α i,n , − − z α ki,n − ,n α i,n , + α k i,n ,n α i,n ( z α ki,n − ,n α i,n , − ) − z α ki,n − ,n α i,n , + α k i,n ,n α i,n ( z α ki,n − ,n α i,n , + ) − z α ki,n − ,n α i,n , − To see this we do a simple computation. Set ˜ α i,n = φ n ψ n α k i,n − ,n α i,n α l i,n ,n ψ − n φ − n and ˜ β i,n = φ n ψ n α k i,n ,n α i,n α l i,n ,n ψ − n φ − n . Writing in matrix ˜ β i,n = (cid:0) ˜ a n ˜ b n ˜ c n ˜ d n (cid:1) . Notethat by our choice of φ n , we have η ˜ β i,n = 1 , so ˜ c n = − ˜ d n and ζ ˜ β i,n = ˜ a n − ˜ d n . Bystraight forward matrix multiplications we have,˜ a n = − z α kn − n β n , − λ k n n a n − z α kn − n β n , − λ k n n b n + z α kn − n β n , − z α kn − n β n , + ( z α kn − n β n , − λ − k n n c n + λ − k n n d n )˜ a n = (cid:18) − z αkn − n βn, − ( λ knn a n z αkn − n βn, − + λ knn b n ) z αkn − n βn, − λ − knn c n + λ − knn d n + z α kn − n β n , − z α kn − n β n , + (cid:19) ( z α kn − n β n , − λ − k n n c n + λ − k n n d n ) − = − z α kn − n β n , − ( α k n n β n ( z α kn − n β n , − ) − z α kn − n β n , + )( z α kn − n β n , − λ − k n n c n + λ − k n n d n ) − d n = z α kn − n β n , − z α kn − n β n , + λ k n n a n + z α kn − n β n , − λ k n n b n − z α kn − n β n , − ( z α kn − n β n , + λ k n n c n λ − k n n d n )˜ d n = (cid:18) z αkn − n βn, − ( λ knn a n z αkn − n βn, + + λ knn b n ) z αkn − n βn, + λ − knn c n + λ − knn d n − z α kn − n β n , − (cid:19) ( z α kn − n β n , + λ − k n n c n + λ − k n n d n ) − = z α kn − n β n , − ( α k n n β n ( z α kn − n β n , + ) − z α kn − n β n , − )( z α kn − n β n , + λ − k n n c n + λ − k n n d n ) − Now by η n = 1 we have, z α kn − n β n , + λ − k n n c n + λ − k n n d n z α kn − n β n , − λ − k n n c n + λ − k n n d n = 1 − z α kn − n β n , + − z α kn − n β n , − . Since ζ ˜ β i,n = ˜ a n − ˜ d n , the formula for ζ ˜ β i,n follows from above equations for ˜ a n and ˜ d n . Let ˜ λ n , denote the multiplier of ˜ α ,n . First, we need to get a estimate of the growth of | tr( ˜ β i,n ) | in terms of | tr( β i,n ) | . Note that | tr( ˜ β i,n ) | = | tr( α k i,n ,n α i,n α l i,n ,n ) | . Remark 6.A . There exists σ, σ ′ > σ ′ | tr( α k i,n ,n α i,n α l i,n ,n ) | > | tr( ˜ β i,n ) | > σ | tr( α k i,n ,n α i,n α l i,n ,n ) | ( | λ ,n | −
1) for n large . In fact we only need the lower bound for | tr( ˜ β i,n ) | . Proof.
Since, | λ ,n | − | ζ α ki,n ,n α i,n α li,n ,n − | + | ζ α ki,n ,n α i,n α li,n ,n | − | λ ,n | | ζ α ki,n ,n α i,n α li,n ,n − | = | ζ α ki,n ,n α i,n α li,n ,n | − | ζ α ki,n ,n α i,n α li,n ,n − | ≤ , and by condition ( i ) , | ζ αki,n ,n αi,nαli,n ,n |−| λ ,n | | ζ αki,n ,n αi,nαli,n ,n − | → | ζ α ki,n ,n α i,n α li,n ,n − || λ ,n | − > ǫ > , for large n. η α ki,n ,n α i,n α li,n ,n = 1, and by Lemma 4.8, more precisely by equation (1)in the proof of Lemma 4.8 and | tr( α i,n ) | ≍ | c i,n | , | tr( α k i,n ,n α i,n α l i,n ,n ) || tr( α i,n ) | > σ ′ | ζ α ki,n ,n α i,n α li,n ,n − η α ki,n ,n α i,n α li,n ,n | > σ ′ ǫ ( | λ ,n | − , for large n. The upper bound is trivial.We define a fuction of generators α i,n , α ,n as f ( α i,n , α ,n ) = | ζ ˜ αi,n − || ˜ λ ,n | − . In next Lemma, we will prove ( i ) by: if lim j f ( α i,n j , α ,n j ) = 0 then wehave classical Schottky generators. In Lemma 7 .
6, we will prove ( i ) for thecase f ( α i,n j , α ,n j ) > M for some M > . Hence we will be done for ( i ) . Lemma 8.3.
Assume that there exists a subsequence such that, lim j f ( α i,n j , α ,n j ) = 0 . Then S j = < α k i,nj − ,n j α i,n j α l i,nj ,n j , ..., α k i,nj ,n j α i,n j α l i,nj ,n j , ..., α k k,nj ,n j α k,n j α l k,nj ,n j > clas-sical Schottky generators for large n. i.e with generator α ,n j replaced by α k i,nj − ,n j α i,n j α l i,nj ,n j and rest generators remains the same.Proof. We will show that S j satisfies conditions of Lemma 5.2 with Remark5.4.Since by Remark 6.A, σ ′ | tr( α i,n ) | > | tr( ˜ β i,n j ) | > σ | tr( α i,n j ) | ( | λ ,n j | − . In particular | tr( ˜ β i,n j ) | → ∞ , we havelim j (cid:12)(cid:12)(cid:12) z ˜ β i,nj , + − z ˜ β i,nj , − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ζ ˜ β i,nj − (cid:12)(cid:12)(cid:12) = lim j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q | tr( ˜ β i,n j ) | − β i,n j ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1Since | tr( ˜ β i,n j ) | → ∞ implies that the isometric circles of ˜ β i,n j are disjointfor large n. Since ζ ˜ β i,nj , η ˜ β i,nj are centers of these isometric circles and soby disjointness, the radius of these isometric circles must be < | ζ ˜ βi,n − η ˜ βi,nj | for large j. In addition, each isometric circle contains one of the fixed point z ˜ β i,nj ,l , z ˜ β i,nj ,u . By our convention z ˜ β i,nj ,l , z ˜ β i,nj ,u are contained within the iso-metric circles with centers ζ ˜ β i,nj , η ˜ β i,nj respectively. Note that η ˜ β i,nj = 1 . n j we have, | z ˜ β i,nj ,l − ζ ˜ β i,nj || ζ ˜ β i,nj − | , | z ˜ β i,nj ,u − || ζ ˜ β i,nj − | < . From these bounds we have,lim j | z ˜ β i,nj ,l − z ˜ β i,nj ,u | ( | z ˜ β i,nj ,u | − | ˜ λ i,n j | ) | tr( ˜ β i,n j ) | = lim j (cid:12)(cid:12)(cid:12) ζ ˜ β i,nj − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( | z ˜ β i,nj ,u | −
1) + (1 − | ˜ λ i,n j | ) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) tr( ˜ β i,n j ) (cid:12)(cid:12)(cid:12) ≤ lim j (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) | ˜ λ i,nj |− ζ ˜ βi,nj − (cid:12)(cid:12)(cid:12)(cid:12) − (cid:19) | tr( ˜ β i,n j ) | = 0Similarly we have,lim j | z ˜ β i,nj ,l − z ˜ β i,nj ,u | ( | z ˜ β i,nj ,l | − | ˜ λ i,n j | − ) | tr( ˜ β i,n j ) | = lim j (cid:12)(cid:12)(cid:12) ζ ˜ β i,nj − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( | z ˜ β i,nj ,u | − | ζ ˜ β i,nj | ) + ( | ζ ˜ β i,nj | −
1) + (1 − | ˜ λ i,n j | − ) (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) tr( ˜ β i,n j ) (cid:12)(cid:12)(cid:12) ≤ lim j (cid:18) | ˜ λ i,nj | (cid:12)(cid:12)(cid:12)(cid:12) | ˜ λ i,nj |− ζ ˜ βi,nj − (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) | ζ ˜ βi,nj |− ζ ˜ βi,nj − (cid:12)(cid:12)(cid:12)(cid:12) − (cid:19) | tr( ˜ β i,n j ) | = 0Hence by Remark 5.4, ˜ α n j , ˜ β n j have disjoint circles for large n j which impliesthat α i,n j , β i,n j have disjoint cricles. By κ ( S Γ nj ) → ∞ we have | z α ki,nj − ,nj α i,nj α li,nj ,nj , ± − z α kj,nj ,nj α j,nj α lj,nj ,nj , ± || tr( α k i,nj ,n j α i,n j α l i,nj ,n j ) | → ∞ , for i = j. This implies < α k i,nj − ,n j α i,n j α l i,nj ,n j , ..., α k i,nj ,n j α i,n j α l i,nj ,n j , ..., α k k,nj ,n j α k,n j α l k,nj ,n j > is classical Schottky group for large n j . We are done for the subcase inf n f ( α i,n , α ,n ) = 0 . Next we will deal theremain subcase of
M < f ( α i,n , α ,n ) for some M > . First we need toestimate the norm of the trace of Nielsen transformed generators ˜ β i,n withrespect to the rate the generators ˜ α ,n degenerates into elliptical elements,i.e. | ˜ λ ,n | decreases to 1. The required estimate is given by next Lemma.37 emma 8.4. Assume that there exists a
M > such that, M < f ( α i,n , α ,n ) . Then there exists N , σ > such that, | ˜ λ ,n | D n − > σ | tr( ˜ β i,n ) | D n D n − , for n > N . Proof.
Use matrix representation we can write, ˜ β i,n = (cid:0) ˜ a i,n ˜ b i,n ˜ c i,n ˜ d i,n (cid:1) . Note that η ˜ β i,n = 1 . So ˜ c i,n = − ˜ d i,n we have, ˜ β i,n = (cid:0) ˜ a i,n ˜ b i,n − ˜ d i,n ˜ d i,n (cid:1) . By | ζ ˜ β i,n | ≤ M < | ζ ˜ βi,n − || ˜ λ ,n | − , we have, M ( | ˜ λ ,n | − < (cid:12)(cid:12)(cid:12) ζ ˜ β i,n − (cid:12)(cid:12)(cid:12) < M ′ . By | tr( ˜ β i,n ) | → ∞ and as in the proof of Lemma 8.3 we have,lim n (cid:12)(cid:12)(cid:12) z ˜ β i,n , + − z ˜ β i,n , − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ζ ˜ β i,n − (cid:12)(cid:12)(cid:12) = lim n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q | tr( ˜ β i,n ) | − β i,n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1Since | z ˜ β i,n , + − z ˜ β i,n , − | = (cid:12)(cid:12)(cid:12) ˜ d i,n (cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12)q tr ˜ β i,n − (cid:12)(cid:12)(cid:12)(cid:12) it follows that we have forsome σ, σ ′ > n that, σ | tr( ˜ β i,n ) | < | ˜ d i,n | < σ ′ | tr( ˜ β i,n ) | ( | ˜ λ ,n | − − . Let e be the Euler number. If there exists a subsequence such that lim j | ζ ˜ β i,nj | = e − , then 1 − e − < | z ˜ β i,nj , + − z ˜ β i,nj , − | < e − for large n j . Otherwise welet m n > m n + 1 ≤ | ˜ λ ,n | ≤ m n . From this definition we have, lim n | ˜ λ ,n | m n = lim n (1 + m − n ) m n = e. Thenthere exists
N, δ > n > N we have, | ζ ˜ α mn ,n ˜ β i,n − η ˜ α mn ,n ˜ β i,n | ≤ (cid:12)(cid:12)(cid:12) | ζ ˜ β i,n || ˜ λ ,n | m n + 1 (cid:12)(cid:12)(cid:12) ≤ δ ( e + 1)and, | ζ ˜ α mn ,n ˜ β i,n − η ˜ α mn ,n ˜ β i,n | ≥ (cid:12)(cid:12)(cid:12) | ζ ˜ β i,n || ˜ λ ,n | m n − (cid:12)(cid:12)(cid:12) ≥ δ ( e − . κ > κ − < | z ˜ α mn ,n ˜ β i,n , + − z ˜ α mn ,n ˜ β i,n , − | < κ. Therefore by setting m n = 0 for the subsequence with lim j | ζ ˜ β i,nj | = e − , wecan always assume that for large nκ − < | z ˜ α mn ,n ˜ β i,n , + − z ˜ α mn ,n ˜ β i,n , − | < κ. Since | z ˜ α mn ,n ˜ β i,n , + − z ˜ α mn ,n ˜ β i,n , − | = (cid:12)(cid:12)(cid:12)(cid:12)q tr ( ˜ α m n ,n ˜ β i,n ) − (cid:12)(cid:12)(cid:12)(cid:12) | ˜ d i,n ˜ λ m n ,n | , we have σ ′′ | tr( ˜ β i,n ) | < | tr( ˜ α m n ,n ˜ β i,n ) | < σ ′′′ | tr( ˜ β i,n ) | ( | ˜ λ ,n | − . Since κ − < | z ˜ α mn ,n ˜ β i,n , + − z ˜ α mn ,n ˜ β i,n , − | < κ, we have dist( L ˜ α ,n , L ˜ α mn ,n ˜ β i,n ) < δ forsome δ > . By Remark 4.1.A of Proposition 4.1 applied to < ˜ α ,n , ˜ α m n ,n ˜ β i,n > we have, | tr( ˜ α m n ,n ˜ β i,n ) | > ρ | ˜ λ ,n | D n + 3 | ˜ λ ,n | D n − ! D n and above bound for | tr( ˜ α m n ,n ˜ β i,n ) | we have, | tr( ˜ β i,n ) | > ρσ ′′′− ( | ˜ λ ,n | − | ˜ λ ,n | D n + 3 | ˜ λ ,n | D n − ! D n > ρ ′ | ˜ λ ,n | − (cid:16) | ˜ λ ,n | D n − (cid:17) D n > ρ ′′ (cid:16) | ˜ λ ,n | D n − (cid:17) − D n The last inequality implies that, | ˜ λ ,n | − > | ˜ λ ,n | D n − > ρ ′′′ | tr( ˜ β i,n ) | D n D n − . Next we are ready to take care the subcase when f ( α i,n j , α ,n j ) > M. roposition 8.5. Suppose that there exists a
M > such that, M < f ( α i,n , α ,n ) . Then < α ,n , α k ,n − ,n α ,n α l ,n ,n , ..., α k k,n − ,n α k,n α l k,n ,n > are classical generators forlarge n. To prove Proposition 8.5 when lim sup n | tr( α k i,n − ,n β i,n ) | < ∞ we use dis-joint non-isometric circles for α k i,n − ,n β i,n based on the following Lemma. Proof. (Proposition 8.5) First assume that lim sup n | tr( α k i,n − ,n α i,n α l i,n ,n ) | < ∞ . Let S o i,n ,r i,n , S o ′ i,n ,r ′ i,n be the disjoint circles for α k i,n − ,n α i,n α l i,n ,n given byProposition 5.5. We will show that lim n r n + r ′ n | λ ,n | − = 0 . Note thatlim n r i,n + r ′ i,n | λ ,n | − ≤ lim n | z α ki,n − ,n α i,n α li,n ,n , + − z α ki,n − ,n α i,n α li,n ,n , − | | ˜ λ i,n | ( | ˜ λ i,n | − | λ ,n | − | tr( α k i,n − ,n α i,n α l i,n ,n ) | < C for some C > ≤ lim n | z α ki,n − ,n α i,n α li,n ,n , + − z α ki,n − ,n α i,n α li,n ,n , − | C ( | ˜ λ i,n | − | λ ,n | − n | tr( α k i,n − ,n α i,n α l i,n ,n ) | < ∞ and lim sup n | λ ,n | k i,n − < ∞ , wehave by Lemma 4.8, | z α ki,n − ,n α i,n α li,n ,n , + − z α ki,n − ,n α i,n α li,n ,n , − | ≍ | tr( α i,n ) | . By Proposition 4.1, | tr( α i,n ) | D n > ρ D n | λ ,n | D n + 3 | λ ,n | D n − > ρ D n | λ ,n | − . The last inequality follows from | λ ,n | D n − < | λ ,n | − n. By our assumptiom that
M < | ζ ˜ βi,n − || ˜ λ i,n | − , and Lemma 8.4 we have,lim n r i,n + r ′ i,n | λ ,n | − ≤ lim n (4 ρ D n ) − M ′ | tr( α i,n ) | − | tr( α i,n ) | D n | tr( ˜ β i,n ) | D n − D n since | tr( ˜ β i,n ) | < σ | tr( α i,n ) | we have, < lim n (4 ρ D n ) − σ D n − D n M ′ | tr( α i,n ) | D n (6 − D n ) − − D n = 040ow the circles S o i,n ,r i,n contains one of z α ki,n − ,n α i,n α li,n ,n , − , z α ki,n − ,n α i,n α li,n ,n , + and S o ′ i,n ,r ′ i,n contains the other fixed point, and since | η α kn − n β n | → | ζ α kn − n β n | → , it follows from Lemma 4.4, we must have S o i,n ,r i,n , S o ′ i,n ,r ′ i,n contained in theregion between | λ ,n | and | λ ,n | for large n. Hence we have classical generatorsfor large n. Now if there exists a subsequence such that | tr( α k i,n − ,n α i,n α l i,n ,n ) | → ∞ , then < α ,n , ..., α k k,n − ,n α k,n α l k,n ,n > satisfies conditions of Lemma 5.2.Hence the case ( i ) is complete. Next we consider Type I . This is rela-tively straightforward and parallel to case Type I but we provide anothershort proof in the following. For the Type I it can be seen that we can certainly follow similarly procedureas given in our proof of the Type I and just simply do some modifications.In fact many of our previous estimates will be became much simpler by | λ ,n | > λ hence all estimates that requires bounds on ( | λ ,n | − − aretrivial. To avoid much repetition of our previous proof of the Type I wetherefore present a alternative and somewhat much shorter proof of Type I . Lemma 8.6.
Let α n , β n be a sequence of loxodromic transformations withfixed points of α n be , ∞ . Suppose | z β n , + − z β n , − | → and C − < | z β n , ± | < C. Then there exists δ > such that dist( L α n , L β n ) < log (cid:18) δ | z β n , + − z β n , − | (cid:19) . Proof.
By using hyperbolic distance formula, and since | z β n , + − z β n , − | → | z β n , ± | → . we have for large n ,cosh dist( L α n , L β n ) < | z β n ,u | + ( | z β n ,u + z β n ,l | ) ( | z β n ,l | + | z β n ,u − z β n ,l | )( | z β n ,u − z β n ,l | ) + 1 < ρ | z β n ,u − z β n ,l | , for some ρ > roof. (Type I ) Suppose that there is a subsequence (use same index forsubsequence) such that | tr( α k i,n ,n α i,n α l i,n ,n ) | → ∞ . Then by our assumption oftype I , | ζ α ki,n ,n α i,n α li,n ,n | → | λ ,n | > λ > n , • λ − < | η α ki,n ,n α i,n α li,n ,n | ≤ | ζ α ki,n ,n α i,n α li,n ,n | < λ, and • n (cid:12)(cid:12)(cid:12)(cid:12) tr (cid:18) α ki,n ,n α i,n α li,n ,n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) η ψn ;2 αki,n ,n αi,nαli,n ,n ψ − n ;2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −| λ ,n | − ! (cid:12)(cid:12)(cid:12)(cid:12) tr (cid:18) α ki,n ,n α i,n α li,n ,n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ζ ψn ;2 αki,n ,n αi,nαli,n ,n ψ − n ;2 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) −| λ ,n | ! By Lemma 4.4, < α ,n , α k ,n ,n α ,n α l ,n ,n , ..., α k k,n ,n α k,n α l k,n ,n > satisfies the secondset of conditions of Lemma 5.2, hence classical.Otherwise we have | tr( α k i,n ,n α i,n α l i,n ,n ) | < C for some C > . Since | ζ α ki,n ,n α i,n α li,n ,n |→ η α ki,n ,n α i,n α li,n ,n → , by Lemma 4.4 we have | z α ki,n ,n α i,n α li,n ,n , ± | → . Nowby Remark 4.1.A and | tr( α k i,n ,n α i,n α l i,n ,n ) | < C we must have dist( L α ,n , L α ki,n ,n α i,n α li,n ,n ) → ∞ . This implies that | z α knn β n , + − z α knn β n , − | → . More precisely we have,Let S o,r i,n , S o ′ ,r ′ i,n be the circles given by Proposition 5.5. Proposition 8.7. | tr( α k i,n ,n α i,n α l i,n ,n ) | < C for some C > . Then we musthave ( r i,n + r ′ i,n ) → . Proof.
First note that as we have showed | tr( α k i,n ,n α i,n α l i,n ,n ) | < C implies, | z α ki,n ,n α i,n α li,n ,n , + − z α ki,n ,n α i,n α li,n ,n , − | → | z α ki,n ,n α i,n α li,n ,n , ± | → . Set ψ i,n ( x ) = x − z αki,n ,n αi,nαli,n ,n , + x − z αki,n ,n αi,nαli,n ,n , − . Let λ α ki,n ,n α i,n α li,n ,n be the mutiplier of ψ i,n α k i,n ,n α i,n α l i,n ,n ψ − i,n . By Remark 4.1.A applied to ψ i,n < α i,n , α k i,n ,n α i,n α l i,n ,n > ψ − i,n we have, | λ α ,n | > | λ α ki,n ,n α i,n α li,n ,n | D n + 3 | λ α ki,n ,n α i,n α li,n ,n | D n − D n (cid:18) e − L ψi,nα ,nψ − i,n , L ψi,nαki,n ,n αi,nαli,n ,n ψ − i,n ) (cid:19) . { z ψ i,n α ,n ψ − i,n , + , z ψ i,n α ,n ψ − i,n , − } = { , z αki,n ,n αi,nαli,n ,n , + z αki,n ,n αi,nαli,n ,n , − } we have, | z ψ i,n α ,n ψ − i,n , ± | → , and (cid:12)(cid:12)(cid:12) z ψ i,n α ,n ψ − i,n , + − z ψ i,n α ,n ψ − i,n , − (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − z α ki,n ,n α i,n α li,n ,n , + z α ki,n ,n α i,n α li,n ,n , − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) → . By Lemma 8.6 we have for large n ,dist( L ψ i,n α ,n ψ − i,n , L ψ i,n α ki,n ,n α i,n α li,n ,n ψ − i,n ) < log δ | z ψ i,n α ,n ψ − i,n , + − z ψ i,n α ,n ψ − i,n , − | ! . This implies that for large n , | λ α ,n | > δ − | λ α ki,n ,n α i,n α li,n ,n | D n + 3 | λ α ki,n ,n α i,n α li,n ,n | D n − D n | z ψ i,n α ,n ψ − i,n , + − z ψ i,n α ,n ψ − i,n , − | > δ − (cid:18) | λ α ki,n ,n α i,n α li,n ,n | D n + 3 (cid:19) D n | z ψ i,n α ,n ψ − i,n , + − z ψ i,n α ,n ψ − i,n , − || λ α ki,n ,n α i,n α li,n ,n | − > δ − (cid:18) | λ α ki,n ,n α i,n α li,n ,n | D n + 3 (cid:19) D n ( r n + r ′ n )2 C ′ | z α ki,n ,n α i,n α li,n ,n , − | The last inequality follows from Proposition 5.5 and | λ α ki,n ,n α i,n α li,n ,n | < C ′ . Thesecond inequality in the above calculations follows from that | λ α ki,n ,n α i,n α li,n ,n |
10 Elliptical degeneracy Type III
We show this type of degeneracy is classical by reducing to type I for large n. Since 1 ≤ | η α ki,n ,n α i,n α li,n ,n | ≤ | ζ α ki,n ,n α i,n α li,n ,n | , Type III implies Type I. In thiscan trivially reduce to Type I. 44ow we have completed the proof for Types I, II, III, IV. next we willlink them together and consider the general case without any restrictions on α ,n .
11 Non-Collapsing fixed points subspaces
The main result in this section is the following, when we have a sequence ofΓ n of Schottky groups which is k − Z Γ n > τ for some τ > D Γ n → n will contain a subsequence which will be classicalSchottky groups for large enough n. Given a set of generators < α , ..., α k > we define Nielsen maps F i,l,m for integers 1 ≤ i ≤ k and natural number l, m as: F i,l,m α j,k = α li α j α mi . First we show that given Γ n sequence of k − D n →
0, we can always choose generators so that the k − α ,n for n sufficiently large. Lemma 11.1.
Let Γ n be a sequence of rank k Schottky group which is k − classical and D Γ n → . Then we can find a subsequence of generating set S n of Γ n , up to Mobius conjugation, such that { α ,n , ..., α k,n } ⊂ S n is a sequenceof k − classical generators and, S n is of standard form.Proof. Since Γ n is k − D Γ n →
0, it follows from Prop 6.4,we have a normal subsequence of generating sets (we using same index forsubsequence) < α ,n , α ,n , ..., α k,n > of Γ n . As before there exists integers l ,j,n , m ,j,n such that we can put these sequence of generating sets in tostandard form by F ,l ,j,n ,m ,j,n applied to α j,n for 2 ≤ j ≤ k. We can assumethat the all centers of isometric circles of F ,l ,j,n ,m ,j,n α j,n are strictly boundedaway from | λ ,n | . Note that the new sequence will not necessarily be k − n is k − k − α ,n . Infact, there exists F i,l i,j,n ,m i,j,n such that elements of this classical generatorswill be of the form F i,l i,j,n ,m i,j,n α j,n for i = 1 and integers l i,j,n , m i,j,n dependson indices i, j, n. Note that if it’s not of standard form then, by repeat-ing this process for a fixed n , we have generating set with generators with | trace | → ∞ . The generators constructed will have leading term of the form45 i,l i,j,n ,m i,j,n F ,l ,j,n ,m ,j,n where i = 1 and some integers l i,j,n , m i,j,n . In partic-ular continue this process, we have generators of arbitrarily small isometriccircles. And since we only terminate when we have elements of leading termof the form F ,l ,j,n ,m ,j,n F i,l i,j,n ,m i,j,n becomes k − F ,l ,j,n ,m i,j,n F i,l ,j,n ,m i,j,n is distinct for the repetition. This impliesall fixed points of these k − < F i,l i,j,n ,m i,j,n α ′ j,n > for i ≥ k − < α ′ j,n > is also gener-ators of Γ which must also of fixed points and isometric centers arbitrarilyclose, and < F i,l i,j,n ,m i,j,n α ′ j,n > is classical, we must have fixed points of < α ′ j,n > contained within isometric circles of F i,l i,j,n ,m i,j,n α ′ j,n . Since isomet-ric circle centers of α ′ j,n are bounded within 1 , | λ ,n | and strictly away from | λ ,n | , we must have isometric centers of F i,l i,j,n ,m i,j,n α ′ j,n also bounded ar-bitrarily close to 1 and away from | λ ,n | for sufficiently many repetitions.Hence we can find a Mobius map (scaling) such that, fixes 0 , ∞ and moves allisometric centers of F i,l i,j,n ,m i,j,n α ′ j,n to be bounded within 1 , | λ ,n | . Thereforethe repetition process must terminate with k − α ′ j,n withcenters bounded within 1 , | λ ,n | , hence a generating set of k − n sequence of k − D n → , Lemma11.1 provides generators of standard form with k − n. However note that, the circles of the k − α ,n , hence they don’t necessarily constitute classical gener-ating set for Γ n . This is due to the fact we have degenerate types I, II, III,IV. Theorem 11.2.
Let J k be the rank k Schottky space. For each τ > thereexists a ν > such that, { [Γ] ∈ J k ( τ ) | D Γ ≤ ν, Γ is k − classical , for some Γ ∈ [Γ] } ⊂ J k,o . Proof.
We proof by contradiction. It follows from argument before for ( I )we only need to assume that | α ,n | < C for some C > . Now it follows from Lemma 11.1, we have sequence of generating sets S n of Γ n such that it is k − α j,n ,46or 2 ≤ j ≤ k are classical generators and have centers bounded within 1and | λ α ,n | (standard form), the only issues that prevents < α ,n , ..., α k,n > been a classical generators are degeneracies described earlier. Hence, we havereduced the general case to degenerate type I, II, III, IV. Now it follows fromthat degenerate types are all in fact classical generators for D Γ n sufficientlysmall, i.e large n , which gives our result.
12 Collapsing fixed points types
This section and following sections are devoted in proving Theorem 12.1.The Theorem 12.1 will enable us to remove the constraint on Z Γ n . that wasplaced in the Theorem 11.2. The idea of the proof is based on analysis of thebehavior of the set of fixed points of generators. This should be compared tothe previous section where analysis of centers of isometric circles were used.Recall Theorem 11.2 states that a given Schottky group Γ is classicalSchottky group if: Z <α ,...,α k > > τ and Hausdorff dimension D Γ is sufficientlysmall and Γ is k − k − D Γ we have the following dichotomy: • Z <α ,n ,...,α j,n > > c for some c > , • is classical Schottky group.Therefore, combining with Theorem 11.2, we have the proof that Schottkygroups of fixed rank k which is k − Theorem 12.1.
There exists c > such that: Let Γ n be a sequence of rank k Schottky groups that is k − classical with D n → . Then for sufficientlylarge n , there exists a subsequence (using same index) Γ n with generating set < α ,n , α ,n , ..., α k,n > such that: • Z <α ,n ,...,α k,n > > c or, • < α ,n , α ,n , ..., α k,n > is classical Schottky groups for large n. The idea of the proof of Theorem 12.1 is to use Theorem 11.2 and show bycontradiction. The proof involves analysis of collapsing of fixed points given47elow as Collapsing I and II. First we pick an sequence of generating sets S Γ n of Γ n which is k − S Γ n we will examine when S Γ n dosenot satisfies conditions of Theorem 11.2. In each situation where Γ n J k ( τ )we can always obtain new sequence of generating sets such that it will leadsto contradictions.The proof will occupy next two sections of Collapsing fixed points I, II. Proof.
We prove by contradiction. Suppose there exists Γ n a sequence ofSchottky groups which is k − < α ,n , ..., α k,n > of Γ n we have Z <α ,n ,...,α k,n > → n , by replacing, < α ,n , ..., α k,n > with < α ,n , α m ,n ,n α ,n , ..., α m k,n ,n α k,n >, for sufficiently large m k,n if necessary, we can always assume that every Γ n is generated by generators with | tr( α ,n ) | ≤ | tr( α k,n ) | and | tr( α k,n ) | ≥ log 3 D n .We use H the upper space model. Conjugating with Mobius transfor-mations it is sufficient to assume α ,n with fixed points 0 , ∞ and multiplier λ ,n and α i,n , i ≥ | λ ,n | − < | η α i,n | ≤
1. Note that as before we denoteby | z α i,n ,l | ≤ | z α i,n ,u | as the two fixed points of α i,n by z α i,n ,l , z α i,n ,u . Wheneverwe have α i,n in matrix form then we always assume | a i,n | ≤ | d i,n | otherwisejust replace α i,n by α − i,n .Note that by Lemma 11.1, we can assume S Γ n satisfies is k − S Γ n to be the generatorsset.We when Z Γ n →
0, which we say
Collapsing fixed points.
This can bedivided into possible collapsing types: • Collapsing fixed points I: | z α i,n ,u − z α i,n ,l | → i ≥ . • Collapsing fixed points II: z α i,n ,l → i ≥ . After we analyzed both Collapsing fixed points I, II we will then address thegeneral situation. We will address these collapsing fixed points in next twosections.
13 Collapsing fixed points I
We show that in this collapsing situation we have further additional I andI such that: 48 I implies Z Γ n > c and, • I implies classical.Where Collapsing fixed points I is dichotomized into:I : lim inf n | λ ,n | = 1 . I : There some exists λ > | λ ,n | > λ. Let us first analyze subclass I which is straightforward. Since | z α i,n ,l − z α i,n ,u | → | λ ,n | > λ > , we have λ < | z α i,n ,l | ≤ | z α i,n ,u | <λ for large n. Hence by < α ,n , ..., α k,n > is k − n. In this case we will need to perform transformations on α i,n , i ≥ λ ,n degenerates into 1 , i.e. α ,n collapsing into an ellipticalelement. This is done by divid the rate of degeneracy of λ ,n into smallerand smaller intervals, and then transform the α i,n , i ≥ | λ ,n | strictlydecreasing to 1. For large sufficiently enough n , we can choose a sequenceof positive integers m n which depends on index n so that 1 + m n +1 ≤| λ ,n | ≤ m n . Set ζ α i,n = a i,n c i,n , η α i,n = − d i,n c i,n , and since | tr( α i,n ) | → ∞ and | √ tr ( α i,n ) − c i,n | = | z α i,n , + − z α i,n , − | → , we have | tr( α i,n ) | < | c i,n | thenfrom Lemma 4.4 and Remark 4.6 it follows that, (cid:12)(cid:12)(cid:12) z α imn ,n α i,n α imn ,n , ± − ζ i,n λ im n i,n (cid:12)(cid:12)(cid:12) ≤ ρ | λ im n i,n || tr( α i,n ) | ≤ ρ e i | tr( α i,n ) | (cid:12)(cid:12)(cid:12) z α imn ,n α i,n α imn ,n , ∓ − η i,n λ im n i,n (cid:12)(cid:12)(cid:12) ≤ ρ | λ im n i,n || tr( α i,n ) | ≤ ρ e i | tr( α i,n ) | n. By Lemma 4.4 and Remark 4.6 and the conditionthat | z α i,n ,l − z α i,n ,u | → | ζ n | , | η n | →
1. We have || z α imn ,n α i,n α imn ,n , ± | − | λ im n ,n || + || z α imn ,n α i,n α imn ,n , ∓ | − | λ im n ,n || → . By the definition of m n we must have | λ im n n | → e i , | λ im n n | → e i . Hence itfollows that | z α imn ,n α i,n α imn ,n , ± | → e and also | z α imn ,n α i,n α imn ,n , ∓ | → . From thisit follows that there exists c > Z <α ,n ,...,α imn ,n α i,n α imn ,n ,...,α kmn ,n α k,n α kmn ,n > > c for sufficiently large n.
14 Collapsing fixed points II
We show in this collapsing situation we have: • II implies Z Γ n > c and, • II implies classical for n .Where Collapsing fixed points II is further dichotomized into:II : lim inf n | λ ,n | < Λ for some Λ > . II : lim inf n | λ ,n | → ∞ .We also assume that | ζ i,n | ≤ | η i,n | as before, otherwise we just simply replaceby it’s inverse. In this case we prove that there are integers l i,n with, Z <α ,n ,α l ,n ,n β ,n ,...,α lk,n ,n β k,n > > c for some c > . By taking a subsequence we can just assume we have | λ ,n | ≤ Λ for largeenough n . First we consider α ,n and then move to α k,n inductively as follows.Choose positive integers l ,n to be the smallest such that e ≤ | ζ ,n λ l ,n ,n | . Since | λ ,n | ≤ Λ , we must have some σ > e < | ζ ,n λ l ,n ,n | < σ . emma 14.1. With above notations, there exists < ǫ < e and n > N ǫ such that e − ǫ < | z α l ,n ,n α ,n , + | , | z α l ,n ,n α ,n , − | < σ + ǫ . To show this Lemma 14.1 in the following we can use Remark 4.2.B ofLemma 4.4.
Proof.
Note that since | z α ,n , − − z α ,n , + | < ǫ for some ǫ > n, and also | tr( α ,n ) | → ∞ , we have | c ,n | → ∞ . By Remark 4.2.B we have | z α ,n ,u − η ,n | → , hence for large n we have Λ − < | η ,n | < δ. Let us show that | z α l ,n ,n α ,n , + − z α l ,n ,n α ,n , − | 6→ ∞ . Suppose this is not true.Denote by ρ ,n the center of the circle having z α l ,n ,n α ,n , + and z α l ,n ,n α ,n , − aspair of antipodal points on the boundary sphere. By | ζ α l ,n ,n α ,n + η α l ,n ,n α ,n | <σ + 1 + ǫ ′ for some ǫ ′ > n, and ρ ,n = z αl ,n ,n α ,n, + + z αl ,n ,n α ,n, − , and z α l ,n ,n α ,n , + + z α l ,n ,n α ,n , − = ζ α l ,n ,n α ,n + η α l ,n ,n α ,n we get | ρ ,n − σ ′ | < κ for some positive number κ > σ ′ = σ + 1 + ǫ ′ . Since | ρ ,n − σ ′ | ≤| ρ ,n | + | σ ′ | < σ +1+ ǫ ′ + σ + 1 + ǫ ′ we can take κ = ( σ + 1 + ǫ ′ ) . Hencewe have that dist( L α ,n , L α l ,n ,n α ,n ) < δ for some δ > . By Remark 4.1.Awe have | tr( α l ,n ,n α ,n ) | → ∞ , and | ζ α l ,n ,n α ,n − η α l ,n ,n α ,n | = | tr( α l ,n ,n α ,n ) || c ,n λ − l ,n ,n | , we get | tr( α l ,n ,n α ,n ) | ≍ | c ,n λ − l ,n ,n | . But this imples that | z α l ,n ,n α ,n , + − z α l ,n ,n α ,n , − | < C for some C > , hence a contradiction.Note that if | z α l ,n ,n α ,n , + − z α l ,n ,n α ,n , − | → | z α l ,n ,n α ,n , ± | → | ζ α l ,n ,n α ,n + η α l ,n ,n α ,n | . Since ( e − − δ ) < | ζ α l ,n ,n α ,n + η α l ,n ,n α ,n | < σ ′ for large n. Thisimplies that ( e − − δ ) < | z α l ,n ,n α ,n , ± | < σ ′ for large n. Now if c < | z , + − z α l ,n ,n α ,n , − | < c ′ for some c, c ′ > , then it follows fromRemark 4.1.A we get | tr( α l ,n ,n α ,n ) | → ∞ and it imples that | c ,n λ − l ,n ,n | → ∞ . So by Remark 4.2.B we get { z α l ,n ,n α ,n , + , z α l ,n ,n α ,n , − } → { ζ α l ,n ,n α ,n , η α l ,n ,n α ,n } which gives the lemma.Next we proceed down to generator α i +1 ,n and do the same above proce-dure as we have done for α i,n with i ≥ . More precisely, we choose l i +1 ,n tobe the smallest integer such that e i +1) + | z α li,n ,n α ,n ,u | ≤ | ζ i +1 ,n λ l i +1 ,n ,n | . Then51t follows from above proof we have, e i +1) + | z α li,n ,n α ,n ,u | − ǫ < | z α li +1 ,n ,n α i +1 ,n , + | , | z α li +1 ,n ,n | α i +1 ,n | < σ i +1 + ǫ, i ≥ . Now for these transformed generators we have the following two possibilitiesaccording to the collapsing or the non-collapsing fixed points respectively as, • II a : lim inf (cid:12)(cid:12)(cid:12)(cid:12) z α l ,n ,n α ,n , + − z α l ,n ,n α ,n , − (cid:12)(cid:12)(cid:12)(cid:12) → . • II b : lim inf n (cid:12)(cid:12)(cid:12)(cid:12) z α l ,n ,n α ,n , + − z α l ,n ,n α ,n , − (cid:12)(cid:12)(cid:12)(cid:12) > b , we have Z <α ,n ,α l ,n ,n β ,n ,...,α lk,n ,n α k,n > > c , for some c > n .Hence suppose we have II a . Passing to a subsequence if necessary we canassume that | z α li +1 ,n ,n α i +1 ,n , + − z α li +1 ,n ,n α i +1 ,n , − | → . If | λ ,n | → m i,n , m ′ i,n as defined in Collaps-ing fixed points I , such that e i ) + | z α lk,n ,n α k,n ,u | ≤ | η i,n λ l i,n +2 m i,n ,n | < e M i,n ,e M i,n +2 ≤ | ζ i,n λ l i,n +2 m i,n ,n | < e M ′ i,n , and also κ i,n < (cid:12)(cid:12)(cid:12)(cid:12) z α li,n + mi,n ,n α i,n , + − z α li,n + mi,n ,n α i,n , − (cid:12)(cid:12)(cid:12)(cid:12) < κ ′ i,n , for some 0 < κ i,n , κ ′ i,n , M i,n , M ′ i,n . By Remark 4.1.A, | tr( α l i,n + m i,n ,n α i +1 ,n ) | →∞ . Hence by Remark 4.2.B we have { z α li,n + mi,n ,n α i,n , + , z α li,n + mi,n ,n α i,n , − } → { ζ α li,n + mi,n ,n α i,n , η α αli,n + mi,n ,n αi,n } . This implies that e i + | z α lk,n ,n α k,n ,u | ≤ | z α li,n + mi,n ,n α i,n , ± | < e M i,n and e M i,n +2 < | z α li,n + mi,n ,n α i,n , ∓ | < e M ′ i,n . Hence there exists c >
0, such that Z <α ,n ,...,α li,n + mi,n ,n α i,n ,...,α lk,n ,n α k,n > > c for large n. If λ ,n | > c > m i,n . Then δ < | z α li,n + ˜ mi,n ,n α i,n , + − z α li,n + ˜ mi,n ,n α i,n , − | < δ for some 0 < δ , δ . And it follows from Remark 4.1.A and Remark 4.2.B wehave that, Z <α ,n ,...,α li,n + ˜ mi,n ,n α i,n ,...,α lk,n ,n α k,n > > c for large n. Take a subsequence of α ,n if necessary we can assume that | λ ,n | is strictlyincreasing to. Let l i,n ≥ | ζ i,n λ l i,n ,n | ≤ | ζ α i,n λ l i,n ,n | = 1 and lim sup | ζ α i,n λ l i,n ,n − η α i,n | 6 = 0 then we have asubsequence such that,lim inf j Z <α ,nj ,...,α li,nj +2( i − nj α i,nj α − i − nj ,...,α lk,nj +2( k − nj α k,nj α − k − nj > > . Suppose lim sup ζ α i,n λ l i,n ,n = 1 then, denote by < α ,n j , ..., α l k,nj ,n j α k,n j > the subsequence of < α ,n , ..., α l k,n ,n α k,n > with lim j ζ α i,nj λ l i,nj ,n j = 1.If lim sup | tr( α l ,nj i,n j α i,n j ) | = ∞ then by passing to a subsequence if nec-essary, for large j , α l i,nj ,n j α i,n j will have disjointed isometric circles. Since | tr( α i,n j ) | → ∞ , z α i,nj ,l → , so | z α i,nj , + − z α i,nj , − | < c for c > , we have | c i,n j | → ∞ . By Remark 4.2.B, lim j min {| η α i,nj − z α i,nj , − | , | η α i,nj − z α i,nj , + |} →
0. Since ζ α li,nj ,nj α i,nj → | tr( α l ,nj ,n j α i,n j ) | → ∞ we have, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ζ α li,nj ,nj α i,nj − η α li,nj ,nj α ,nj (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) tr( α l i,nj ,n j α ,n j ) (cid:12)(cid:12)(cid:12) | λ − l i,nj ,n j c i,n j | → | λ − l i,nj ,n j c i,n j | → ∞ . Also (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z α li,nj ,nj α i,nj , + − z α li,nj ,nj α i,nj , − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)q tr ( α l i,nj ,n j α i,n j ) − (cid:12)(cid:12)(cid:12)(cid:12) | λ − l i,nj ,n j c i,n j | → . Hence it follows from Remark 4.2.B, there exists a κ > j we have, κ − < | z α li,nj ,nj α i,nj ,l | ≤ | z α li,nj ,nj α i,nj ,u | < κ . And since | λ ,n j | → ∞ ,we can choose Mobius transformations ψ i such that S j = ψ j < α ,n j , ..., α l i,nj ,n j α i,n j , ..., α l i,nj ,n j α i,n j > ψ − j satisfies Lemma 5.2. And since it is k − S j generates classical Schottky groups for large j. α i,n , i ≥ | tr( α i +1 ,n ) | ≤ | tr( α i,n ) | . If we havethat lim sup | tr( α l k,nj ,n j α k,n j ) | < ∞ then set φ j be Mobius transformations sothat φ j α l k,nj ,n j α k,n j φ − j have fixed points 0 , ∞ and fixed points z φ j α ,nj φ − j , ± be z α lk,nj α k,nj , ± . Then it follows that z φ j α i,nj φ − j ,l →
1. Hence we can reducedthis case to Collapsing fixed points I which has already been considered.If lim sup | ζ α i,n λ l i,n ,n | <
1, then we have: • II a : lim inf | ζ α i,n λ l i,n +21 ,n | = 1 . • II b : lim inf | ζ α i,n λ l i,n +21 ,n | > . a We arrange so that < α ,n j , ..., α l k,nj ,n j α k,n j > be a subsequence of such thatlim j | ζ α k,nj λ l k,nj +21 ,n j | →
1. If sup j | tr( α l k,nj +11 ,n j α k,n j ) | < ∞ , then we conjugate α ,n j , α l k,nj ,n j α k,n j to ˆ α k,j = φ j α ,n j φ − j , and ˆ α ,j = φ j α l k,nj +11 ,n j α ,n j φ − j with ˆ α ,i have fixed points 0 , ∞ and ˆ α k,j have z α lk,nj +11 ,nj α k,nj , ± . Since sup j | ˆ λ ,j | < ∞ ,it follows that, if z ˆ α i,j ,l → < ˆ α ,j , ..., ˆ α k,j > , falls under Collapsingfixed points I, and if z ˆ β i ,l → < ˆ α ,j , ..., ˆ α k,j > falls under Collapsingfixed points II Otherwise there exists ǫ > ǫ < | z ˆ α i,j ,l | < − ǫ, hence as in Collapsing fixed points I we can choose integers N i,n such that Z < ˆ α ,j ,..., ˆ α Ni,j ,j ˆ α i,j ˆ α − Ni,n ,j ,..., ˆ α Nk,j ,j ˆ α k,j ˆ α − Nk,n ,j > > c for some c > . Now if we assume that sup j | tr( α l i,nj +11 ,n j α i,n j ) | = ∞ then for large j, sincethe radius of isometric circles are R α li,nj +11 ,nj α i,nj = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z αli,nj +11 ,nj αi,nj ,u − z αli,nj +1 i,nj αi,nj ,l (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) | tr( α li,nj +11 ,nj α i,nj ) | and the distance between the centers is | ζ α li,nj +11 ,nj α ,nj − η α li,nj +11 ,nj α i,nj | = | tr( α li,nj +11 ,nj α i,nj ) || c i,nj λ − li,nj − ,nj | , | z α li,nj +11 ,nj α i,nj ,u − z α li,nj +11 ,nj α i,nj ,l | = | r tr ( α li,nj +11 ,nj α i,nj ) − || c i,nj λ − li,nj − ,nj | we have,lim j | ζ α li,nj +11 ,nj α i,nj − η α li,nj +11 ,nj α i,nj | R α li,nj +11 ,nj α i,nj = lim i | tr( α l i,nj +11 ,n j α i,n j ) | | q tr ( α l i,nj +11 ,n j α i,n j ) − | > δ | tr( α l i,nj +11 ,n j α i,n j ) | for some δ > . Hence | ζ α li,nj +11 ,nj α i,nj − η α li,nj +11 ,nj α i,nj | > R α li,nj +11 ,nj α i,nj for large j. From this it implies that α l i,nj +11 ,n j α i,n j have disjointed isometric circles forlarge j. By Lemma 4.4 we have η α i,nj → . And since η α li,nj +11 ,ni α i,nj = η α i,nj , implies that η α li,nj +11 ,nj α i,nj → . Note that if inf j | ζ α li,nj +11 ,nj α i,nj − | > | ζ α li,nj +11 ,nj α i,nj | → , then by | tr( α l i,nj +11 ,n j α i,n j ) | → ∞ we have | − z α li,nj +11 ,nj α i,nj ,l | , | ζ α li,nj +11 ,nj α i,nj − z α li,nj +11 ,nj α i,nj ,u | → , and inf j | z α li,nj +11 ,nj α i,nj ,u − z α li,nj +11 ,nj α i,nj ,l | > . Hence for large j there exits ǫ > − ǫ < (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z α li,nj +11 ,nj α i,nj ,l (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z α li,nj +11 ,nj α i,nj ,u (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ǫ. Hence we can choose integers N i,n such that,inf j Z <α ,nj ,...,α li,ni + Ni,nj +11 ,nj α i,nj α − Ni,nj ,nj ,...,α lk,ni + Nk,nj +11 ,nj α k,nj α − Nk,nj ,nj > > . Now suppose that ζ α li,nj +11 ,nj α i,nj → . Then we have | ζ α li,nj +11 ,nj α i,nj − η α li,nj +11 ,nj α i,nj | → . Hence | tr( α li,nj +11 ,nj α i,nj ) || c i,nj λ − li,nj − ,nj | → . Since | tr( α l i,nj +11 ,n j α i,n j ) | → ∞ , which gives | z α kni +1 ni β ni ,u − z α kni +1 ni β ni ,l | = q tr ( α l ni +1 n i β n i ) − | c n i λ − l ni − n i | → . R α li,nj +11 ,nj α i,nj → . Since inf j | λ ,n j | > c > , we have for large j that the isometric circles of α l i,nj +11 ,n j α i,n j disjointed and lies between c − and c. In particular, c − < | z α li,nj +11 ,nj α i,nj ,l | ≤ | z α li,nj +11 ,nj α i,nj ,u | < c. Set, S j = < α ,n j , ..., α l i,nj +11 ,n j α i,n j , ..., α l k,nj +11 ,n j α k,n j > .S j satisfies Lemma 5.2, hence classical. b First we define a new sequence of < ˜ α ,n , ..., ˜ α k,n > as follows: Consider S ′ n = < α ,n , ..., α l i,n ,n α i,n , ..., α l k,n ,n α k,n > . We arrange < α l ,n ,n α ,n , ..., α l k,n ,n α k,n > so that | tr( α l i +1 ,n ,n α i +1 ,n ) | ≤ | tr( α l i,n ,n α i,n ) . If | tr( α l k,n ,n α k,n ) | ≥ | tr( α ,n ) | , thenset ˜ α i,n = α i,n , i ≥
1. Otherwise, let φ n be the Mobius map so that φ n α l k,n ,n α k,n φ − n have fixed points 0 , ∞ , and φ n α ,n φ − n have z φ n α ,n φ − n , ± = z α k,n , ± . Set α ,n ;1 = φ n α l k,n ,n α k,n φ − n , α i,n ;1 = φ n α i,n φ − n , i = k. We define integer l i,n ;1 with respectto < α ,n ;1 , ..., α k,n ;1 > the same way as we defined l i,n before.Suppose that | tr( α l k,n ;1 ,n ;1 α k,n ;1 ) | ≥ | tr( α ,n ;1 ) | then we set ˜ α i,n = α i,n ;1 , i ≥ α ,n = α ,n ;1 . Otherwise, we repeat this construction to get a sequence < α ,n ; m , ..., α k,n ; m > . By construction for a each n , either there exists a m such that | tr( α l k,n ; m ,n ; m α k,n ; m ) | ≥ | tr( α ,n ; m ) | or we have | tr( α l k,n ; m ,n ; m α k,n ; m ) | ≥| tr( α ,n ; m ) | for all m. Assume the latter, since α ,n,m +1 = φ n,m α l k,n ; m ,n ; m α k,n ; m φ − n,m we have | tr( α ,n ; m +1 ) | < | tr( α ,n ; m ) | for all m. If lim m | tr( α ,n ; m ) | = 0 thentake m n to be the first integer m with with | tr( α ,n ; m ) | < n . If lim m | tr( α ,n ; m ) | > m n to be the first integer m with | tr( α ,n ; m +1 ) | > | tr( α ,n ; m ) | − n . If the former holds, we set m n to be the first integer m with | tr( α l k,n ; m ,n ; m α k,n ; m ) | ≥| tr( α ,n ; m ) | . Hence there exists a m n , such that either | tr( α l k,n ; mn ,n ; m n α k,n ; m n ) | > | tr( α ,n ; m n ) | − n , or | tr( α ,n ; m n ) | < n . Note that by construction arrange-ment, we have | tr( α l i,n ; m ,n ; m α i,n ; m ) | ≥ | tr( α l i +1 ,n ; m ,n ; m α i +1 ,n ; m ) | , i ≥ . This alsoimplies that | tr( α l i,n ; mn ,n ; m n α i,n ; m n ) | > | tr( α ,n ; m n ) | − n , i ≥ | tr( α l k,n ; mn ,n ; m n α k,n ; m n ) | > | tr( α ,n ; m n ) | − n . We define ˜ α ,n = α ,n ; m n , ˜ α i,n = α i,n,m n , i ≥ . For S ′ n = < α ,n ; m , ..., α l k,n ; m ,n ; m α k,n ; m >, note that S ′ n is also of k − < ˜ α ,n , ..., ˜ α k,n > . If lim inf n | tr( ˜ α ,n ) | < ∞ , wechoose a subsequence with | tr( ˜ α ,n i ) | < c for all large i and some c >
0. Let p i,j , ≤ i ≤ k be a sequence of least positive integers such that | tr( ˜ α p i,j ,n j ˜ α i,n j ) | > D nj . We conjugate ˜ α p i,j ,n j ˜ α i,n j by ψ j that fixes 0 , ∞ andmax ≤ i ≤ k { z ψ j ˜ α pi,j ,nj ˜ α i,nj ψ − j ,u } = 1 . Set ¯ α ,j = ψ j ˜ α ,n j ψ − j , ¯ α i,j = ψ j ˜ α p i,j ,n j ˜ α i,n j ψ − j . By construction, if z ¯ α i,j ,l → < ¯ α ,j , ..., ¯ α k,j > satisfies Collapsing fixedpoints II and if z ¯ α i,j ,l → < ¯ α ,j , ..., ¯ α k,j > satisfies Collapsing fixedpoints II. Otherwise there exists ǫ > ǫ < | z ¯ α i,j ,l | < − ǫ then wecan chose N i > Z < ¯ α ,j ,..., ¯ α Ni ,j ¯ α i,j ¯ α − Ni ,j ,..., ¯ α Nk ,j ¯ α k,j ¯ α − Nk ,j > > c forsome c > . Hence in either case, we are done.Now suppose that lim inf n | tr( ˜ α ,n ) | = ∞ . Since | tr( ˜ α i,n ) | ≥ | tr( ˜ α ,n ) | , i ≥ < ˜ α ,n , ..., ˜ α k,n > satisfies case II b , otherwisewe are done. We define ˜ l i,n = l i,n ; m n , i ≥ . Set β ,n = ˜ α ,n , β i,n = ˜ α ˜ l i,n ,n ˜ α i,n . Since | tr( ˜ α i,n ) | ≥ | tr( ˜ α ,n ) | and | z ˜ α i,n , − − z ˜ α i,n , + | ≤ δ n with δ n → | λ ,n | − ≤ | z ˜ α i,n ,u | ≤ , z ˜ α ,n ,l → | tr( α l i,n ; mn ,n ; m n α i,n ; m n ) | > | tr( α ,n ; m n ) | − n , which implies lim n | tr( β i,n ) || ˜ λ ,n | > ǫ for ǫ > λ ,n is the multiplier of β ,n , and | η β i,n − ζ β i,n | < δ | ˜ λ ,n | − , for some δ > . Since η β i,n β ,n = η β i,n ˜ λ − ,n and η β ,n β i,n = η β i,n , we have | η β i,n β ,n − ˜ λ − ,n | = | η β i,n ˜ λ − ,n − ˜ λ − ,n | < δ | ˜ λ ,n | − and | η β ,n β i,n − ζ β i,n | = | η β i,n − ζ β i,n | < δ | ˜ λ ,n | − , for large n. Now we have reduced II b to the collapsing behaviors of η β i,n , ζ β i,n thereare four cases that needs to be considered as follows:(I) lim j | η ˜ α i,nj | − | ˜ λ − ,n j || ˜ λ ,n j | − , | ζ ˜ α i,nj ˜ λ l i,nj ,n j | − | ˜ λ − ,n j || ˜ λ ,n j | − < C . (II) lim j | η ˜ α i,nj | − | ˜ λ − ,n j || ˜ λ ,n j | − < C , lim j | ζ ˜ α i,nj ˜ λ l i,nj ,n j | − | ˜ λ − ,n j || ˜ λ ,n j | − = ∞ . (III) lim j | η ˜ α i,nj | − | ˜ λ − ,n j || ˜ λ ,n j | − = ∞ , lim j | ζ ˜ α i,nj ˜ λ l i,nj ,n j | − | ˜ λ − ,n j || ˜ λ ,n j | − < C . n | η ˜ α i,n | − | ˜ λ − ,n || ˜ λ ,n | − , | ζ ˜ α i,n ˜ λ l i,n ,n | − | ˜ λ − ,n || ˜ λ ,n | − = ∞ . Consider (I) :In this case, we will produce a set of classical generators for sufficiently large n j by using condition (I), which is the transformed set of generators from theoriginal set of generators.Since | ζ ˜ α i,nj ˜ λ l i,nj ,n j | > | ˜ λ ,n j | − and our assumption and ζ β ,nj β i,nj = ˜ λ ,n j ζ β i,nj , we have for large j , 1 < | ζ β ,nj β i,nj | < ( C + 1) , ≤ i ≤ k. Since η β i,nj = η ˜ α ˜ li,nj ,nj ˜ α i,nj = η ˜ α i,nj and | ˜ λ ,n j | − < | η ˜ α i,nj | ≤ < | η β i,nj ˜ λ ,n j | . By our assumption we have Similarly, 1 < | η β i,nj β − ,nj | < ( C + 1) . Since | ˜ λ ,n j | → ∞ it follows that there exists κ > − κ − < | η β ,nj β i,nj β − ,nj | , | ζ β ,nj β i,nj β − ,nj | < κ for large i. By Lemma4.4 with Remark 4.6 and | tr( β i,n j ) | > ǫ | ˜ λ ,n j | for large j we have for some ρ , ρ > − κ − − ρ | ˜ λ ,n j | − < | z β ,nj β i,nj β − ,nj ,l | , | z β ,nj β i,nj β − ,nj ,u | < κ + ρ | ˜ λ ,n j | − . Hence there exists κ ′ > κ ′− < | z β ,nj β i,nj β − ,nj ,l | ≤ | z β ,nj β i,nj β − ,nj ,u | < κ ′ . Consider, S j = < β ,n j , ..., β ,n j β i,n j β − ,n j , ..., β ,n j β k,n j β − ,n j > . Since | tr( β ,n j β i,n j β − ,n j ) | → ∞ and by k − S j satisfies thesecond set of conditions of Lemma 5.2, hence classical. Consider (IV) :In this case, we will produce a set of classical generators for sufficiently large n. By using condition (IV), the generators are conjugate to the original setof generators. 58here exists 0 < ρ n → ∞ with ρ n < | ˜ λ ,n | , such that | ζ ˜ α i,n ˜ λ l i,n ,n | −| ˜ λ ,n | − > ρ n | ˜ λ ,n | − . Let χ i,n be Mobius transformations defined by χ n ( x ) = ˜ λ ,n √ ρ n x. We will show that χ n < β ,n , ..., β i,n , ..., β k,n > χ − n satisfies Remark5.3 of Lemma 5.2.If we have that 1 − δ < | η β i,n | ≤ > δ > | ˜ λ ,n |√ ρ n (1 − δ ) < | η χ n β i,n χ − n | < | ˜ λ ,n |√ ρ n . Otherwise we have, √ ρ n | ˜ λ ,n | + 1 | ˜ λ ,n |√ ρ n < | η χ n β i,n χ − n | < | ˜ λ ,n |√ ρ n . Similarly, by condition of Collapsing fixed points II we have | ζ ˜ α i,n ˜ λ l i,n ,n | < . This gives, √ ρ n | ˜ λ ,n | + 1 | ˜ λ ,n |√ ρ n < | ζ χ n β i,n χ − n | < | ˜ λ ,n |√ ρ n . By Lemma 4.4 with Remark 4.6, | z β i,n , ± − ζ β i,n | < C | tr( β i,n ) | − and | z β i,n , ∓ − η β i,n | < C | tr( β i,n ) | − for some C > . Since | tr( β i,n ) | > ǫ | ˜ λ ,n | for large n we have, | z χ n β i,n χ − n , ± − η χ n β i,n χ − n | < Cǫ | ˜ λ ,n |√ ρ n , | z χ n β i,n χ − n , ∓ − ζ χ n β i,n χ − n | < Cǫ | ˜ λ ,n |√ ρ n . Hence, | z χ n β i,n χ − n ,u | < | ˜ λ ,n |√ ρ n + 1 | ˜ λ ,n |√ ρ n + Cǫ | ˜ λ ,n |√ ρ n and , | z χ n β i,n χ − n , l | > √ ρ n | ˜ λ ,n | − | ˜ λ ,n |√ ρ n − Cǫ | ˜ λ ,n |√ ρ n . We have | ˜ λ ,n | − < | z χ n β i,n χ − n ,l | ≤ | z χ n β i,n χ − n ,u | < | ˜ λ ,n | for large n. By above estimates for fixed points of χ n β i,n χ − n , and | tr( β i,n ) | > ǫ | ˜ λ ,n |
59e have,( | z χ n β i,n χ − n ,u | + 1)( | ˜ λ ,n | + 1) | tr( β i,n ) | ( | ˜ λ ,n | − | z χ n β i,n χ − n ,u | ) < | ˜ λ ,n | √ ρ n + | ˜ λ ,n |√ ρ n + | ˜ λ ,n | + δǫ | ˜ λ ,n | (1 − ρ n − | ˜ λ ,n | √ ρ n − Cǫ | ˜ λ ,n | √ ρ n ) < | ˜ λ ,n | + √ ρ n | ˜ λ ,n | + δ √ ρ n | ˜ λ ,n | ǫ √ ρ n (1 − ρ n − | ˜ λ ,n | √ ρ n − Cǫ | ˜ λ ,n | √ ρ n )by ρ n < | ˜ λ ,n | we have some δ ′ > < δ ′ ǫ √ ρ n (1 − ρ n − | ˜ λ ,n | √ ρ n − Cǫ | ˜ λ ,n | √ ρ n ) → . For the other part of the conditions of Remark 5.3 we have:If | z χ n β i,n χ − n ,l | < M then,( | z χ n β i,n χ − n ,l | + 1)( | ˜ λ ,n | − + 1) | tr( β i,n ) | ( | z χ n β i,n χ − n ,l | − | ˜ λ ,n | − ) < ( M + 1)( | ˜ λ ,n | − + 1) ǫ | ˜ λ ,n | ( √ ρ n | ˜ λ ,n | − | ˜ λ ,n |√ ρ n − Cǫ | ˜ λ ,n |√ ρ n − | ˜ λ ,n | ) < M ′ ǫ ( √ ρ n − √ ρ n − Cǫ √ ρ n − < M ′ ǫ ′′ √ ρ n → . Otherwise we have | z χ n β i,n χ − n ,l | → ∞ and,( | z χ n β i,n χ − n ,l | + 1)( | ˜ λ ,n | − + 1) | tr( β i,n ) | ( | z χ n β i,n χ − n ,l | − | ˜ λ ,n | − ) < δ ′′ ǫ | ˜ λ ,n | → , for some δ ′′ > . Set S n = χ n < β ,n , ..., β i,n , ..., β k,n > χ − n , S ′ n = < ˜ α ,n , ..., ˜ α i,n , ..., ˜ α k,n > . Since S ′ n is k − β i,n = α ˜ l i,n ,n α i,n , we have S n satisfies, • | ˜ λ ,n | − < | z χ n β i,n χ − n ,l | ≤ | z χ n β i,n χ − n ,u | < | ˜ λ ,n |• lim n ( ( | z χ n β i,n χ − n ,u | + 1)( | ˜ λ ,n | + 1) | tr( β i,n ) | ( | ˜ λ ,n | − | z χ n β i.n χ − n ,u | ) , ( | z χ n β i,n χ − n ,l | + 1)( | ˜ λ ,n | − + 1) | tr( β i,n ) | ( | z χ n β i,n χ − n ,l | − | ˜ λ ,n | − ) ) = 060onditions of Remark 5.3. Consider (II) :In this case, we will produce a set of classical generators for sufficiently large n. Using condition (II), the set of generators are conjugate to the transformedfrom the original set generators.Since | ˜ λ ,n j | − < | η ˜ α i,nj | ≤ , and by our condition we have 1 < | η ˜ α i,nj ˜ λ ,n j | ≤ C . As in (IV) and using same notations similarly we have, | ˜ λ ,n |√ ρ n < | η χ n β i,nj β − ,nj χ − n | < | ˜ λ ,n |√ ρ n (1 + C ) , √ ρ n | ˜ λ ,n | + 1 | ˜ λ ,n |√ ρ n < | ζ χ n β i,n χ − n | < | ˜ λ ,n |√ ρ n . Now by same argument as in (IV) we have, χ n < β ,n j , ..., β i,n β − ,n j , ..., β k,n j β − ,n j > χ − n are classical Schottky groups for large j. Consider (III) :Finally in this case, we will produce a set of classical generators which areconjugate to the transformed set of generators from the original set.As in (I) we have, 1 < | ζ β ,nj β i,nj | < (1 + C ) . And also as in (IV) and (II)we also have, | ˜ λ ,n |√ ρ n < | ζ χ n β ,nj β i,nj β − ,nj χ − n | < | ˜ λ ,n |√ ρ n (1 + C ) , √ ρ n | ˜ λ ,n | + 1 | ˜ λ ,n |√ ρ n < | ζ χ n β ,nj β i,n β − ,nj χ − n | < | ˜ λ ,n |√ ρ n . Then by same arguments as in (IV) we have, χ n < β ,n j , ..., β ,n j β i,n β − ,n j , ..., β ,n j β k,n j β − ,n j > χ − n are classical Schottky groups for large j. Hence we have completed the proof for all cases (I), (II), (III), (IV).Now from above proofs of (I)-(IV), suppose we have some generatorsthat satisfies some of (I)-(IV) then, there exists a ,n , ..., a k,n , b ,n , ..., b k,n ∈{ , − , } such that, χ n < β ,n j , ..., β a i,nj ,n j β i,n β b i,nj ,n j , ..., β a k,nj ,n j β k,n j β b k,nj ,n j > χ − n , is classical Schottky group. 61 k Completing proof of Theorem 12.1 : For general case, we combine resultsof Collapsing fixed points I, II to complete the proof as follows. Given asequence Γ n which is k − D n →
0, it follows that if Z Γ n isnot bounded below by some constant c > Z Γ n below. In particular, suppose both collapsing I and II existsfor all sequence S n of generating set of Γ n . If Γ n is sequence of non-classicalSchottky groups with D n → Z Γ n . Hence we have completedproof Theorem 12.1.
Theorem 14.2 (Rank-k classical Schottky) . There exists ν k > such thatany rank k non-classical Schottky group Γ must have D Γ > ν k . Proof.
We proof by induction. By Main theorem of [16], there exists ν > < ν . Let k > . Hence by induction and Theorem 11.2 and Thereom 12.1, there exists ν k > D Γ < ν k is classical Schottky group.
15 Extension from fixed Rank k to all ranks This is section we will prove that our result on fixed rank k can be extendedto all finitely generated Γ . Define D k as: D k = sup {D ′ k | any rank k nonclassical Schpottky group Γ must have D Γ ≥ D ′ k } . Lemma 15.1.
There exists a positive number δ > and K a positive integersuch that, given any Γ k +1 rank k + 1 Schottky group with the property thatall rank
N < k + 1 subgroups of Γ k +1 are classical schottky groups then, Γ k +1 is classical Schottky group provided that D Γ k +1 < δ and k > K . The idea of the proof is induction on rank. To do so we use Proposition4.2, which say that with respect to α ,n there must exists some generatorsuch that the norms of its trace growth exponentially with respect to the62ank k. So assuming that we have all rank k of “small” Hausdorff dimensionsare classical Schottky groups then we will show that the additional generatorof the “ k + 1” element will have disjointed Schottky circles from rest of the“ k ” elements when the rank is sufficiently large. Proof.
Let { Γ k +1 } be a sequence of Schottky groups each Γ k +1 is of rank k + 1 , such that any rank N that is < k + 1 is classical. We will show thatfor sufficently large k, Γ k +1 is classical for small D k +1 . In order to prove Γ k +1 is eventually classical we will need to use theestimates given by Proposition 4.2. We denote generators set of Γ k +1 as S Γ k +1 = < α ,k +1 , ..., α k +1 ,k +1 > . We can assume that D Γ k +1 → , otherwisewe are done with the lemma.If necessary, for i ≥ α i,k +1 by α l i,k +1 ,k +1 α i,k +1 α m i,k +1 ,k +1 such that | λ ,k +1 | − ≤ | ζ α li,k +11 ,k +1 α i,k +1 α mi,k +11 ,k +1 | , | η α li,k +11 ,k +1 α i,k +1 α mi,k +11 ,k +1 | ≤ , we can assume that all | λ ,k +1 | − ≤ | ζ α i,k +1 | , | η α i,k +1 | ≤ , i ≥ . For each n ≤ k + 1 , we choose and arrange the geneartors as follows. Fol-lows from Corollary 4.3 there exists a generators that satisfies the estimatesstated in Corollary 4.3 with respect to λ α ,k +1 . Set α k +1 ,k +1 that satisfies theestimates provided by Corollary 4.3. We arrange α k,k +1 simiarly by usingCorollary 4.3 on < α ,k +1 , ..., α k,k +1 > . And we arrange generators this wayfor all generators down to α n +1 ,k +1 and in particular we have, | tr( α n +1 ,k +1 ) | > c (2 n − | λ α ,k +1 | D Γ k +1 + (2 n + 1) | λ α ,k +1 | D Γ k +1 − ! D Γ k +1 (cid:16) e − dist( L α ,k +1 , L αn +1 ,k +1 ) (cid:17) . Let Γ n ; k +1 ⊂ Γ k +1 be generated by S Γ n ; k +1 := < α ,k +1 , ..., α n,k +1 > . Since D Γ n ; k +1 → k → ∞ , and by assumption that it is k − S Γ n ; k +1 such thatit is k − γ N +1 ,k +1 , ..., γ k +1 ,k +1 are disjoint fromcircles C | λ α ,k +1 | − , C | λ α ,k +1 | . Suppose that, | z γ m,k +1 , − − z γ m,k +1 , + | ( | λ α ,k +1 | − > C C > , m > N. by Lemma 8.6 and above inequality and assumethat D Γ k +1 < then,1 | tr( γ m,k +1 ) | ( | λ α ,k +1 | − < ρ ′ | λ α ,k +1 | D Γ k +1 − m − | λ α ,k +1 | D Γ k +1 + (2 m + 1) ! D Γ k +1 ( | λ α ,k +1 | − − < ρ ′ | λ α ,k +1 | D Γ k +1 − m − | λ α ,k +1 | D Γ k +1 + (2 m + 1))( | λ α ,k +1 | − D Γ k +1 ! D Γ k +1 since | λ α ,k +1 | D Γ k +1 − | λ α ,k +1 | − D Γ k +1 < | λ α ,k +1 | − | λ α ,k +1 | − D Γ k +1 < < ρ ′′ (cid:18) m −
1) + (2 m + 1) (cid:19) → , for m → ∞ . for small D Γ k +1 and large m. If | z γm,k +1 , − − z γm,k +1 , + | ( | λ α ,k +1 | − → | λ ,k +1 | > C ′ forsome C ′ > . In this case we have the same result as to Proposition 8.7 that r γ m,k +1 + r ′ γ m,k +1 → . To see this we can simply reproduce the same proof asgiven before. More precisely, we can assume that | tr( γ m,k +1 ) | < C ′′ for some C ′′ > r γ m,k +1 + r ′ γ m,k +1 < M ′ | z γ m,k +1 , − | (cid:16) (2 m − | λ γ m,k +1 | D Γ k +1 + (2 m + 1) (cid:17) D Γ k +1 < M ′′ (2 m + 1) → , for D Γ k +1 ≤ . For | λ α ,k +1 | → f ( γ m,k +1 , α ,k +1 ) → f ( γ m,k +1 , α ,k +1 ) > M for some M > . For f ( γ m,k +1 , α ,k +1 ) → , by repeating the same technique we can showthat S ′′ Γ k +1 with α ,k +1 replaced by α − ,k +1 γ m,k +1 in S ′ Γ k +1 generates a classical64chottk group for large m. Similarly for f ( γ m,k +1 , α ,k +1 ) > M, we have ˆ S Γ k +1 with γ m,k +1 replaces by α − ,k +1 γ m,k +1 generates a classical Schottky group forlarge m. Corollary 15.2. inf k D k > . Proof.
Let δ > , K be given by Lemma 15.1 then let D = min { δ, D K } thenany nonclassical Schottky group Γ must have D Γ > D .
16 Proof of Main Theorem
Theorem 16.1.
There exists ǫ > such that all finitely generated Schottkygroup Γ with D Γ < ǫ is a classical Schottky group.Proof. This follows from Theorem 11.2 and Theorem 12.1.
Proof of Theorem 1.1.
The final argument that we need to show is thatKleinian groups with limit set of small enough Hausdorff dimension are Schot-tky groups. This is a well established fact and detailed arguments can befound in [16]. For completeness we give a quick review here.Let Γ be a non-elementary finitely generated Kleinian group. By Selberglemma we can assume Γ is a torsion-free non-elementary finited generatedKleinian group. D Γ < implies Γ is geoemtrically finite convex cocompactof second kind. Since Ω Γ / Γ consists of finitely many compact compressibleRiemann surfaces, we can decomposes Γ along compression disks and afterfinitely many steps we left with topological balls. Hence H / Γ is a handlebody.E-mail: [email protected]
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