# Projections in Moduli Spaces of the Kleinian Groups

Authors: Hala Alaqad; Jianhua Gong; Gaven Martin
Journal: Abstract and Applied Analysis
 (2022)

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## Abstract

A two-generator Kleinian groupf,g can be naturally associated with a discrete group f,ϕ with the generator ϕ of order two and where f,ϕfϕ−1=f,gfg−1⊂f,g,f,ϕ:f,gfg−1=2. This is useful in studying the geometry of the Kleinian groups since f,g will be discrete only if f,ϕ is, and the moduli space of groups f,ϕ is one complex dimension less. This gives a necessary condition in a simpler space to determine the discreteness of f,g. The dimension reduction here is realised by a projection of principal characters of the two-generator Kleinian groups. In applications, it is important to know that the image of the moduli space of Kleinian groups under this projection is closed and, among other results, we show how this follows from Jørgensen’s results on algebraic convergence.


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## Body

## 1. Introduction

It is shown in [1] that every two-generator Kleinian group f,g is determined uniquely up to conjugacy by its principal character, i.e., the triple of complex parameters
(1)γ,β,β~=γf,g,βf,βg.Here,(2)γ=γf,g=trf,g−2,β=βf=tr2f−4,β~=βg=tr2g−4,where f,g=fgf−1g−1 is the multiplicative commutator. Thus, the space of the two-generator Kleinian groups f,g can be identified as a subspace of the three complex dimensional space ℂ3. We define D=γ,β,β~:γ,β,β~istheprincipalcharacterofaKleiniangroup via the mapping
(3)f,g↦γf,g,βf,βg.Things are set up so the groupId↦0,0,0∈ℂ3, where Id is the identity of the group.Applying [1, Lemma 2.29], given a two-generator Kleinian group f,g with principal character γ,β,β~ with γ≠β, there is a two-generator discrete group Γϕ=f,ϕ with principal character γ,β,−4, i.e., ϕ is elliptic of order two. However, we can waive the condition γ≠β (see Lemma 8 below).If the groupΓϕ is Kleinian (i.e., not virtually abelian), then there are a series of inequalities on the entries of the principal character which need to hold. One such is Jørgensen’ inequality, γ+β≥1. Conversely, it is often easier to establish new discreteness conditions of two-generator groups with one generator of order two since we know of many polynomial trace identities that yield a polynomial semigroup action on the associated moduli space, and the dynamics of this action can be used to investigate discreteness of groups. Let us give one such example leading to Jørgensen’s inequality to motivate our subsequent results. We note
(4)trf,gfg−1−2=trf,g−2trf,gfg−1+2−tr2f,which we write as γf,gfg−1=γγ−β. Following [2], we write pβz=zz−β,pβ0z=z, and pβnz=pβ∘pβ∘⋯∘pβz. Repeated application of this identity followed by projection yields the sequence
(5)γ,β,−4↦pβγ,β,−4↦⋯↦pβnγ,β,−4↦⋯giving a sequence of traces of commutators pβnγn=0∞ in the original discrete group Γϕ, where the series of commutators is
(6)f,g,f,f,g,f,f,f,g,⋯.It is usually not too difficult to prove such sequences cannot be infinite and contain convergent subsequences. Jørgensen’s inequality is implied by the assertion thatpβnγ↛0; hence, γ>1+β. Every such trace identity yields a new inequality and the semigroup structure is clearly seen in general by
(7)γ,β,−4↦pβγ,β,−4↦qβpβγ,β,−4,where for example, qβz=z1+β−z2 from the trace identity
(8)trf,gfg−1fg−2=trf,g−2tr2f−trf,g−12,γf,gfg−1fg=γ1+β−γ2.In another direction, the ergodicity of the polynomial semigroup acting in the complement of the Riley slice,R=intγ,0,−4:f,gisdiscreteandfreelygeneratedbyfandg is used to prove the supergroup density and the existence of the unbounded Nielsen classes of generating pairs [3]. Note that this is not the usual definition of the Riley slice but is known to be equivalent.In order to further these studies, we establish a number of useful results in this article. Dynamically, the elementary subgroups are sinks for the polynomial semigroup action on a slice and are isolated within the subset corresponding to discrete groups (apart from trivial abelian and dihedral examples). Thus, after we observe elementary properties of a two-generator discrete group with principal characterγ,β,−4 in §3, we list the possible principal characters in the first table in §4 representing. Then, we prove the two complex dimensional slice space D2=γ,β:thereisβ~soγ,β,β~isaprincipalcharacterofaKleiniangroup is closed in §5. Then, D2 is the projection of the three complex dimensional moduli space D to ℂ2. Applications of this set being closed are in the generalizations of Jørgensen’s inequality quantifying the isolated nature of the elementary groups in the moduli space of discrete groups [4].


## 2. Preliminaries

LetMo¨b+ℂ¯ be the Möbius group of the normalized orientation preserving Möbius transformations on the Riemann sphere ℂ¯:(9)Mo¨b+ℂ¯=az+bcz+d:a,b,c,d∈ℂandad−bc=1,and let PSL2,ℂ be the projective special linear topological quotient group:
(10)PSL2,ℂ=SL2,ℂ/±Id,where SL2,ℂ=abcd:a,b,c,d,∈ℂandad−bc=1. Then, these two groups are topologically isomorphic, Mo¨b+ℂ¯≅PSL2,ℂ and so there are at least two different ways of thinking about groups throughout this paper, as either subgroups of Mo¨b+ℂ¯ or subgroups of PSL2,ℂ. Each of these groups has its own topology; however, the topological isomorphism shows us that the concept of discreteness is the same.A subgroupΓ of Mo¨b+ℂ¯ is called a Kleinian group if it is a nonelementary discrete group (see [5, 6]). Here, a group is elementary if it is virtually abelian, i.e., a finite extension of an abelian group. The elementary groups are classified [5, Section 5.1]. A subgroup of Mo¨b+ℂ¯ is nonelementary if the limit set, LΓ=thesetofaccumulationpointsofagenericorbitΓz=fz:f∈Γ, contains infinitely many points. Notice from [7] that a group Γ is Kleinian if and only if every two-generator subgroup of Γ is discrete. It is this characterization of the Kleinian groups that motivates a focus on two-generator groups.The principal character of a representation of the two-generator groupf,g conveniently encodes various other geometric quantities. One of the most important is the complex hyperbolic distanceδ+iθ between two axes axf and axg of the two elliptic or loxodromic Möbius transformations f and g, where axf is the hyperbolic line in the hyperbolic 3-space ℍ3 joining the fixed points of f and g in ℂ¯, δ is the hyperbolic distance between those two axes, and θ is the holonomy of the transformation whose axis contains the common perpendicular between those two axes and moves axf to axg. Thus,
(11)sinh2δ+iθ=4γββ~.The parameterβf describes the conjugacy class of the Möbius transformation in which f lies.Lemma 1.
If an elementf of Mo¨b+ℂ¯ is not the identity, then
(a)
f is parabolic if and only if βf=0.(b)
f is elliptic if and only if βf∈−4,0.(c)
f is loxodromic if and only if βf∈ℂ∖−4,0.In addition, the parameterβf gives the order of each elliptic element.Lemma 2.
If an elementf of Mo¨b+ℂ¯ is not the identity, then f is elliptic of order p if and only if
(12)βf=−4sin2kπp,for1≤k<pandk,p=1.We use the following easy lemma: a group is Kleinian if and only if it isvirtually Kleinian, i.e., has a Kleinian subgroup of finite index.Lemma 3.
LetH be a finite index subgroup of Γ. Then, Γ is Kleinian if and only if H is Kleinian.The proof follows directly from the fact that ifH has finite index in Γ, then it has a coset decomposition
(13)Γ=ϕ1H∪ϕ2H∪⋯∪ϕnH,where ϕ1=Id,ϕ2,⋯,ϕn∈Γ for some n. Thus, any sequence gn contains a subsequence gnk lying in one coset.


## 3. Slicing Spaces of Two-Generator Discrete Groups

Suppose thatf,g is two-generator discrete subgroup of Mo¨b+ℂ¯ with γ=γf,g≠0 and γf,g=γ≠β=βf. Then, Gehring and Martin [1] state that there exist elliptics ϕ and ψ of order two such that Γϕ=f,ϕ and Γψ=f,ψ are discrete with the related two-generator groups principal characters γ,β,−4 and β−γ,β,−4, respectively.Notice thatγf,g=0 implies f and g have a common fixed point on ℂ¯ and hence f,g is elementary (see [5], Theorem 4.3.5 and [1], Identity (1.5)). Thus, the condition γ≠0 is necessary for Kleinian groups. We now consider the other restriction γ≠β.We start with our observation of the properties of a two-generator discrete group with the principal characterγ,β,−4, which has one generator of order two.Lemma 4.
LetΓ=f,g be a discrete group with principal character γ,β,−4. If γ=β, then Γ is an elementary discrete group, which is isomorphic to a dihedral group Dp, for some p∈1,2,⋯,∞.Proof.
There are two cases to consider. Iff is parabolic, then γ=β=0 by Lemma 1, and hence, f and g have a common fixed point which we may assume by conjugacy is ∞∈ℂ¯. Then, Γ is a subgroup of the upper-triangular matrices, where
(14)f=1x01andg=±iy0∓i,and the result follows. We may now assume f and g have the form
(15)f=λ001λandg=αμδν,whereλ≠0,±1,αν−μδ=1.
Computing the commutator,(16)f,g=αν−λ2μδαλ2μ−αμ1λ2δν−δναν−1λ2μδ,and hence, γ=trf,g−2=−μδλ−1/λ2, and β=tr2f−4=λ−1/λ2. The assumption γ=β now implies that μδ=−1. Since αν−μδ=1,αν=0. Now, βg=−4, so trg=0, which gives α+ν=0. Then, α=ν=0 and
(17)g=0μ−1μ0.
Thus,gz=−μ2/z interchanges the fixed points 0 and ∞ of f. So Γ has a finite orbit 0,∞, and hence, it is an elementary discrete group, [5]. Computation shows the conjugate gfg−1=f−1:(18)gfg−1=1λ00λ=f−1.
Thus,Γ is a dihedral group Dp, for some p∈ℕ∪∞ from Theorem 9.We recall that the Riemann sphereℂ¯=ℂ∪∞ is the boundary of the hyperbolic 3-space ℍ3.Lemma 5.
LetΓ=f,g be a discrete group with the principal character γ,−4,−4, then Γ is isomorphic to one of the following elementary discrete groups:
(a)
The dihedral groupD∞, if axf∩axg=∅ or axf∩axg≠∅ in ℂ.(b)
A dihedral groupDp for some p∈ℕ, if axf∩axg≠∅ in ℍ3.(c)
A parabolic group, ifaxf∩axg=∞.Proof.
(1)
axf∩axg=∅: Let α be the common perpendicular between two axes axf and axg. Since f and g are elliptic of order two, they are the rotations of order two about their axes axf and axg in ℍ3, respectively. And each of f and g interchanges the ending points of α and fixes α setwise; hence, Γ is elementary. Furthermore, the product fg fixes the ending points of α and fixes α setwise, so the axis axfg=α. It follows that fg is loxodromicNext, from the Fricke identity (see [8]), we have
(19)γf,g=βf+βg+βfg−trftrgtrfg+8.Sinceβf=βg=−4,trf=0, and f2=Id, the previous identity (19) gives γf,fg=βfg, and by Lemma 4, f,fg is a dihedral group. Notice that fg is loxodromic with infinite order; hence, f,g=f,fg≅D∞.(2)
axf∩axg≠∅ in ℍ3: Since f and g have a common fixed point (say x0), Γ is elementary. Since Γ is discrete, the axes of elliptics both of order two must intersect at x0 an angle θ=kπ/n for some k and n≥2, the product fg is a rotation about axfg with the oriented angle 2θ=2kπ/p for any n≥2 and some p∈ℕ, where axfg is perpendicular to axf and axg and passing through x0. Therefore, fg is elliptic of order p. Notice that f and g are rotations of order two and their rotation axes axf and axg meet in ℍ3¯, we have gfg−1=f−1. It follows that f,g≅Dp(3)
axf∩axg≠∅ in ℂ¯: We may assume by conjugacy that ∞ is the common fixed point of f and g in ℂ¯. Then, f and g fix ∞, and hence, axf and axg are vertical hyperbolic lines. Notice that both f and g are rotations with the angles π at distinct centers. Since the sum of two rotation angles is 2π, the product fg is a translation, and hence, fg is parabolic fixing ∞. Thus, the classification of elementary groups [5, Section 5.1], f,g is a group of the Euclidean similarities of ℂ
We next note the following elementary result.Lemma 6.
LetΓ=f,g be a two-generator group with principal character γ,β,−4, then Γg=f,gfg−1 is an index two subgroup of f,g:Γ=Γg∪gΓg.Then, we have the following.Lemma 7.
Suppose thatf,g is a two-generator Kleinian group with principal character γ,β,β~. If β~≠−4, then
(20)Fixf∩Fixgfg−1=∅.Proof.
Sincef,g is a two-generator Kleinian group, γ≠0, and hence, f and g cannot share any fixed points, so Fixf∩Fixg=∅. Since gFixf=Fixgfg−1, there are two cases for us to consider.IfFixf has one element, say Fixf=z0, then Fixgfg−1=gz0. But gz0≠z0. Thus, Fixf∩Fixgfg−1=∅.IfFixf has two elements, say Fixf=z1,z2, then Fixgfg−1=gz1,gz2 and
(21)gz1≠z1,gz2≠z2.IfFixf∩Fixgfg−1≠∅, then by [5, Theorem 5.1.2] f and gfg−1 cannot share exactly one fixed point, and hence, Fixf=Fixgfg−1=gFixf. Thus, the only possibility is that g interchange the fixed points of f, so gz1=z2 and gz2=z1, which implies that g2 fixes z1 and z2 but they are not fixed by g. We conclude that g2 has at least three fixed points and this implies that g2 is the identity, g has order two and β~=−4.In [1, Remark 2.6] it is shown that if f,g is a two-generator discrete group, then there are two elliptic elements ϕ and ψ of order two such that
(22)ϕfϕ−1=gfg−1andψfψ−1=gf−1g−1.In particular, iff is elliptic or loxodromic, then the axes axϕ and axψ meet at right angles to one and other, and they bisect the common perpendicular between the axes axf and axgfg−1.Lemma 8.
Letf,g be a Kleinian group with principal character γ,β,β~. Then, there are two elliptic elements ϕ and ψ of order two such that Γϕ=f,ϕ and Γψ=f,ψ are discrete ℤ2-extension of Γg with principal characters γ,β,−4 and β−γ,β,−4, respectively.Proof.
NoteΓg=f,gfg−1 is a subgroup of both Γϕ=f,ϕ and Γψ=f,ψ. Suppose that the principal characters for Γϕ and Γψ are γ1,β,−4 and γ2,β,−4, respectively. Using [1, Lemma 2.1]
(23)γγ−β=γf,gfg−1=γf,ϕfϕ−1=γ1γ1−β,γγ−β=γf,gfg−1=γf,ψfψ−1=γ2γ2−β.
We have two quadratic equations(24)γ12−βγ1−γγ−β=0,γ22−βγ2−γγ−β=0.
Solving these two quadratic equations,γ1=γ, or β−γ and γ2=γ, or β−γ. As [9, Identity (6.9)] shows both possibilities occur, so
(25)γ1,γ2=γ,β−γ.
Thus, after relabeling, the principal characters forΓϕ and Γψ are γ,β,−4 and β−γ,β,−4, respectively. This completes the proof.


## 4. Exceptional Set of Principal Characters

Now, the natural question is to determine whether two discrete groupsΓϕ=f,ϕ and Γψ=f,ψ in Lemma 8 are actually Kleinian. Since both Γϕ and Γψ are ℤ2-extension of Γg, by Lemma 3, we only need to decide if Γg is a Kleinian subgroup. First, we list the possible principal characters in Tables 1–3 representing two-generator elementary discrete groups. The set of principal characters occurring in these three tables is called the exceptional set of principal characters.Table 1
Principal Characters: 2,p.
psin2θγGroupsPrincipal characters323−2A4−2,−3,−4313-1S4−1,−3,−433−565−32A55−32,−3,−433+56−3+52A5−3+52,−3,−431-3D3−3,−3,−4412-1S4−1,−2,−441−2D4−2,−2,−455−5105−32A55−32,5−52,−455−510−1A5−1,−5+52,−455+510−1A5−1,5−52,−455+510−3+52A5−3+52,−5+52,−4515−52D55−52,5−52,−451−5+52D5−5+52,−5+52,−4Table 2
Principal Characters: 3,p.
psin2θγGroupsPrincipal characters349−1A5−1,−3,−3389-2A4−2,−3,−3423−1S4−1,−2,−3510−25155−32A55−32,5−52,−3510−2515−1A5−1,−5+52,−3510+2515-1A5−1,5−52,−3510+2515−3+52A5−3+52,−5+52,−3Table 3
Principal Characters: 4,4 and 5,5.
p,psin2θγGroupParameters4,41-1S4−1,−2,−25,5455−3/2A55−3/2,5−5/2,5−5/25,545-1A5−1,−5+5/2,5−5/25,545−3+5/2A5−3+5/2,−5+5/2,−5+5/2Theorem 9.
LetG be an elementary discrete group of Mo¨b+ℂ¯, then G is isomorphic to one of the following groups, where p∈1,2,⋯,∞.
A cyclic groupℤp:(26)a:ap=Id.
A dihedral groupDp≅ℤp⋉ℤ2:(27)a,b:ap=b2=Id,bab−1=a−1.
The groupℤp×ℤ⋉ℤ2 or ℤp×ℤ:(28)a,b,c:aba−1b−1=bp=c2=Id,cac−1=a−1,cbc−1=b−1.
A Euclidean translation groupℤ×ℤ.
A Euclidean triangle groupΔ2,3,6, or Δ2,4,4, or Δ3,3,3:(29)a,b:ap=bq=abr=Id,1p+1q+1r=1.
A finite spherical triangle groupA4=Δ2,3,3, or S4=Δ2,3,4, or A5=Δ2,3,5:(30)a,b:ap=bq=abr=Id,1p+1q+1r>1.Letf and g be elliptic of order p∈2,3,4,5 in Mo¨b+ℂ¯, and let θ be the angle subtended at the origin between axf and axg, and hence, the hyperbolic distance between those two axes is δ=0. Then, sin2θ can be computed by using [9, Lemmas 6.19, 6.20, and 6.21], the parameters βg and βf∈−3,−4,5−5/2,−5+5/2, and the commutator parameter is γf,g=−βfβg/4sin2θ, see [1]. With these elementary observations, some spherical trigonometry, and Theorem 9 about the classification of the elementary discrete groups, we obtain the following Tables 1 and 2, which list the possible principal characters (including the cases in the thesis [10]) of two-generator elementary discrete groups with nonzero commutator parameters γf,g≠0. They are either the dihedral groups (if γf,g=βf and βg=−4 or the finite spherical triangle groups A4=Δ2,3,3, S4=Δ2,3,4, and A5=Δ2,3,5.Notice that the angle between intersecting axes of elliptics of order4 in a discrete group is always either 0 when they meet on the Riemann sphere ℂ¯ or π/2. This yields the additional parameter −1,−2,−2 for the elementary group S4 when generated by two elements of order 4. Furthermore, the angle between intersecting axes of elliptics of order 5 in a discrete group is either arcsin2/5 or its complement arcsin−2/5. After possibly taking powers of the generator of order 5, the three additional principal characters can be obtained in the following Table 3.The classification of the elementary discrete groups (Theorem9) plays an important role in listing the tables above. We make the following additional remarks.
(i)
The axes of elliptics both of order two can intersect at an anglekπ/p for any k and p≥2 giving the dihedral group Dp with principal character −4sin2kπ/p,−4,−4 (see Lemma 5).(ii)
The axes of elliptics elements of orders2 and p, p≥3, can meet at right angles. In this case, the principal character of dihedral group Dp is −4sin2π/p,−4sin2π/p,−4.(iii)
The axes of elliptics of orderp and q (p≤q) in a discrete group meet on ℂ¯, i.e., meeting with angle 0, if and only if p,q∈2,2,2,3,2,4,2,6,3,3,3,6,4,4,6,6.For all of these groupsγf,g=0. In particular, the Euclidean triangle groups Δ2,3,6, Δ3,3,3, Δ2,4,4, and cyclic groups have γf,g=0. Also, the Euclidean translation groups have γ=0.Lemma 10.
Letf,g be a Kleinian group with the principal character γ,β,−4 that is not one of those exceptional groups listed in Table 1. Then, the subgroup Γg=f,gfg−1 is Kleinian.Proof.
We need to show that the discrete subgroupΓg of the Kleinian group f,g can not be elementary. Let γ=γf,g and β=βf. By Lemma 8 there is an elliptic h of order two such that Γh=f,h is a discrete group containing Γg with index two and the principal character for Γh is γ,β,−4. By the hypothesis, γ,β,−4 is not one of those exceptional groups listed in Table 1; then, Γh is not a finite spherical triangle group A4,S4, and A5. Thus, Γg can only be elementary if Γg=γγ−β=0, i.e., γ=0 or γ=β. Since f,g is Kleinian, γ≠0. So it can only be γ=β, and hence, it is the dihedral group. In this case, since f is not of order two, gfg−1=f±1. In the case gfg−1=f, it gives gFixf=Fixgfg−1=Fixf, and hence, g is elliptic of order two (Lemma 7) fixing or interchanging the fixed points of f in ℂ¯. In the case gfg−1=f−1,g might be a power of f. In either case f,g is not Kleinian, a contradiction.The following theorem guarantees thatΓg is a Kleinian group if f is not elliptic of order p≤6.Theorem 11.
Suppose thatf,g is a Kleinian group. If f is not elliptic of order p∈2,3,4,6, then the subgroup Γg=f,gfg−1 is Kleinian.Proof.
a Suppose that g is elliptic of order two. Then, Γg is a subgroup of index two in Γ and Lemma 3 ensures that Γg is also Kleinian. b Suppose that the order of g is not 2. We need to show that Γg is nonelementary. If Γg is elementary, then (see [5], Section 5.1]).
(i)
Γg is elementary of type I; each nontrivial element is elliptic. Since the order of f is not p≤6 or p=5, neither is the order of gfg−1. If γf,gfg−1≠0, then from Tables 1–3, Γg is neither the finite spherical triangle groups A4,S4, and A5, nor the dihedral groups D3,D4, and D5. If γf,gfg−1=0, then γ=β, and Fixf∩Fixgfg−1≠∅. On the other hand, by Lemma 7, Fixf∩Fixgfg−1=∅, a contradiction.(ii)
SupposeΓg is an elementary group of type II; then, Γg is conjugate to a subgroup of Mo¨bℂ¯ fixing ∞ whose every element is parabolic of the form z↦z+b, b∈ℂ. Thus, the group Γg is abelian, and hence, gFixf=Fixgfg−1=Fixf, a contradiction to Γ being Kleinian.(iii)
SupposeΓg is an elementary group of type III; then, both f and gfg−1 are elliptic or both are loxodromic. In either case Γg is abelian and as above this is a contradiction.
We have shown thatΓg cannot be elementary if g does not have order two. Hence, in all two cases, Γg is a Kleinian group.Notice that a Kleinian groupf,g with f elliptic of order p=5 or p≥7 and βg≠−4 cannot have γ=β. If so, then f and gfg−1 have a common fixed point. But this discrete group cannot be a Euclidean triangle group as they do not have elliptics of order 5 or greater than 6. Thus, f and gfg−1 have two common fixed points and the same trace. Hence, gfg−1=f±1 and g must fix or interchange the fixed points of f. The first is not possible, and the second shows g2=Id, the identity.


## 5. Projections and Principal Characters

Our methods to establish new inequalities rely on a fundamental result concerning spaces of finitely generated Kleinian groups. Namely they are closed in the topology of algebraic convergence—a result originally due to Jørgensen [7]. We state this now.Definition 12.
A sequence ofn-generator subgroups Γj=fj,1,fj,2,⋯,fj,n is said to be convergent algebraically to a n-generator subgroup Γ=f1,f2,⋯,fn in Mo¨bℂ¯ if for each k=1,2,⋯,n, the sequence of the corresponding generators fj,k converges uniformly to fk in the spherical metric of ℂ¯.Theorem 13.
(Jørgensen).
That the map back is an isomorphism was proved by Jørgensen in [7]. There is a higher dimensional analogue to this Jørgensen’s theorem, but additional hypotheses are necessary—for instance giving a uniform bound on the torsion in a sequence or asking that the limit set be in geometric position (see [11], Proposition 5.8).Now, we show the sliceD2 is closed. Notice that we are free to normalize a sequence of two-generator groups by conjugation and to pass to subsequences. First show a preliminary result.Theorem 14.
The slice
D∗=γ,β,−4:γ,β,−4istheprincipalcharacterofaKleiniangroup is closed in the two complex dimensional space ℂ2 in the usual topology.Proof.
Suppose thatγj,βj,−4 is a sequence of principal characters of Kleinian groups, say fj,gj. We proceed by considering two cases: up to subsequence, fi is parabolic for all i or not.
Case (a). Suppose fj is parabolic for all j. Conjugate each Γj so that the first generator is now represented by the matrix fj=1101. Then, fj is constant and it remains to show that the sequence gj also converges. Since trgj=0, we suppose the matrix for second generator is
(31)gj=ajbjcj−aj.
We calculate thatγj=cj2, and hence, cj⟶c. Jørgensen’s inequality gives γj≥1 for all j, and hence, cj⟶c≠0.
Set(32)hj=1−aj+i1+cjcj01andhjgjhj−1=i1+cj1ci−i1+cj.
Sincefj commutes with hj,fj is left unchanged under conjugation. The result follows by Theorem 13.
Case (b). We now suppose that fj is not parabolic for all j. By conjugation, we may assume that fj is represented by the matrix
(33)fj=λj001λj, where λj≠0,±1. Thus, the parameter βj=λj−1/λj2, which gives the quadratic equation
(34)λj2−βjλj−1=0,and hence, λj=βj±βj+4/2. Since βj⟶β,λi converges to λ=β±β+4/2. Noting that λ≠0,±1 for either choice of λ, we conclude that fj converges to the element f represented by the matrix
(35)fj⟶f=λ001λandβj⟶β=λ−1λ2.
It remains to show that the order two elementsgj also converge. Note first that γj=−βjbjcj and so we may assume that bjcj≠0. As in the previous case, the matrix for gj can be written as
(36)gj=aj−1+aj2cjcj−aj,so that aj≠±i and 1+aj2≠0. Consider the conjugacy of the group fj,gj by the following diagonal matrix,
(37)ϕj=icj1+aj200−i1+aj2cj,and then, gj (replaced by the conjugacy ϕjgjϕj−1) has the form
(38)gj=aj1−1+aj2−aj.
Sincefj commutes with ϕj,fj is left unchanged under the conjugation.
The commutator is now(39)fj,gj=fjgjfj−1gj−1=−aj2+λj21+aj2−aj+λj2ajaj1+aj2λj2−aj1+aj21+aj2λj2−aj2.
Hence, the parameterγj for the group fj,gj is
(40)γj=aj2λj2−12+λj2−12λj2.
At this point we would like to showλj↛1, but this can actually happen in limits of Dehn surgeries—see the following example. Thus, we have two subcases:
Case (b) (i) Suppose λj↛1
Solving foraj yields
(41)aj2=γjλj2−λj2−12λj2−12⟶γλ2−λ2−12λ2−12.
Leta2=γλ2−λ2−12/λ2−12, then gj converges to the element g of PSL2,ℂ represented by the matrix
(42)g=a1−1+a2−a,and γj⟶γf,g=γ. Applying Theorem 13, the limit group Γ=f,g is Kleinian.
Case (b) (ii) Suppose λj⟶1. This is equivalent to βj⟶0. For each j, we may conjugate fj and gj so that fj has a fixed point at ∞ and gj has a fixed point at 0. Thus, fj is upper triangular and gj is lower triangular. We may further conjugate by a diagonal matrix to achieve the upper entry in the matrix representing fj is 1. With this normalization we have
(43)fj=λj101λjandgj=i0cj−i, and λj⟶1. We calculate that
(44)γj=γfj,gj=cjcj+2iλj2−1λj.
It follows thatcj⟶±γ, for one choice of sign. Thus, fj⟶f and gj⟶g(45)f=1101andg=i0γ−i.
This group is nonelementary by Theorem13 so γ≠0. One can see this since as soon as βj<1/2, Jørgensen’s inequality gives γj>1/2. This completes the proof.Example 1.
We next give an example to show the last case can occur. That isβj⟶0 and γj⟶γ (and γ=1). Let
(46)f=1101andh=10a∞1.
Then,f,h is (a representation of) the two bridge figure of eight knot group if a∞=1+i3/2. The relator in this group is
(47)hfh−1fh=fhf−1hf.
If we performp,0 Dehn surgery on the complement of this knot, we obtain a Kleinian group generated by two elliptics of order p, fp, and hp. Then, up to conjugacy fp and hp have the form
(48)fp=eiπ/p10e−iπ/pandhp=eiπ/p0ape−iπ/p.
Then, the relator (47) holds if and only if
(49)ap=123−2cos2πp−−1−4cos2πp+2cos4πpor its conjugate. Thus, fp,hp is the group obtained from this orbifold Dehn surgery. As such it is discrete and nonelementary. Then, as p⟶∞,
(50)βfp=βhp=−4sin2πp⟶0=βf,γfp,hp=ap−2+ap+2cos2πp⟶a∞2=γf,g.
Asβfp=βhp, we find an involution gp such that hp=gpfpgp−1 with γfp,gp=γfp,hp.We next have the following lemma.Lemma 15.
Letγj,βj,β~j be a sequence of principal characters of the two-generator Kleinian groups fj,gj, and let γ,β,β~ be the principal character of two-generator group f,g. Suppose that γj,βj converges to γ,β and f is not elliptic of order p∈2,3,4,6. Then, γ≠0 and γ≠β.Proof.
Sincef is not elliptic of order p∈2,3,4,6, fj is likewise not elliptic of these orders for all but finitely many j. Applying Theorem 11, fj,gjfjgj−1 is a sequence of Kleinian groups with corresponding principal characters γjγj−βj,βj,βj which converge to γγ−β,β,β. Now, fj,gjfjgj−1 is a Kleinian group generated by two elements of the same trace. Thus, γjγj−βj≠0, and hence, γj≠0 and γj≠βj.
Now we may apply a result by Cao [12, Theorem 5.1] which gives a universal lower bound on the commutator parameter for such groups.
(51)γjγj−βj≥0.198.
Thus,limj⟶∞γjγj−βj=γγ−β≥0.198. It follows that γ≠0 and γ≠β.Theorem 16.
The two complex dimensional space
D2=γ,β:γ,β,β~istheprincipalcharacterofaKleiniangroupforsomeβ~ is closed in the two complex dimensional space ℂ2 in the usual topology.Proof.
Suppose thatγj,βj is a sequence in D2 with limit γ,β in ℂ2. By the definition of D2, there is a sequence β~j so that γj,βj,β~j is the principal character of a Kleinian group. Then, we project to the principal character γj,βj,−4 of fj,ϕj. If infinitely many of these groups are Kleinian, the result we seek follows from Theorem 14. After passing to a subsequence, we are left with the case fj,ϕj is elementary for every j.
Case (i). Suppose fj is not elliptic of order p∈2,3,4,6 for infinitely many j. Then, Lemma 15 tells us that γj∉0,βj, and hence, fj,ϕj is Kleinian, so this case cannot occur
Case (ii). We may now suppose, after passing to a subsequence, that each fj is elliptic of order p∈2,3,4,5,6, βfj=−4sin2kπ/p=β, k,p=1. fj,ϕj is a discrete elementary group generated by elliptics of orders p and 2. The order of any such group is less than that of A5. If infinitely many of these groups are finite, then in fact fj,ϕ is elliptic of order 1,2,3,4,5, or 6. Thus, γj lies in a finite set of values. Again, after a subsequence, we have γj=γ. Then, γj,βj,β~j=γ,β,β~j and the latter group is Kleinian by hypothesis. Thus, γ,β∈D2
Case (iii). We are left with the situation that fj,ϕj is generated by elliptics of orders p and 2, infinite and elementary for each j. The classification tells us this only happens for the Euclidean triangle groups and any such group has a common fixed point. Then, γfj,ϕj=0=γfj,gj which is not possible as fj,gj is a Kleinian group.



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*Source: 6311193-2022-05-06.xml*