# Infinite Paths of Minimal Length on Suborbital Graphs for Some Fuchsian Groups

Authors: Khuanchanok Chaichana; Pradthana Jaipong
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

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## Abstract

In this study, we work on the Fuchsian groupHm where m is a prime number acting on mℚ^ transitively. We give necessary and sufficient conditions for two vertices to be adjacent in suborbital graphs induced by these groups. Moreover, we investigate infinite paths of minimal length in graphs and give the recursive representation of continued fraction of such vertex.


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## Body

## 1. Introduction

TheHecke group, Hλ, introduced by Hecke in [1], is the group generated by the two Möbius transformations(1)Rz=−1z,Sz=z+λ,where λ is a real number such that λ=λq=2cosπ/q and q is an integer greater than 2 or λ>2. When q=3, Hλ3=H1 is the modular group PSL2,ℤ. If q=4,6, it is known, see [2], that Hλq=Hm, where m=2,3 consists of all transformations of the following two types:(2)Tz=az+bmcmz+d,a,b,c,d∈ℤ,ad−bcm=1,(3)Tz=amz+bcz+dm,a,b,c,d∈ℤ,adm−bc=1.However, Rosen [3] showed that the above two transformations need not to be in Hλ if λ≠1,2,3. He proved that T is an element of Hλ if and only if a/c is a finite λ-fraction, i.e.,(4)where r0 is an integer and ri is a positive integer for i=1,…,n. Later, Keskin [4] presented the Fuchsian group Hm for a squarefree positive integer m, which consists of all mappings of the forms (2) and (3).In 1991, Jones et al. [5] studied the modular group H1 by applying the idea of the suborbital graphs for a permutation group, introduced in [6]. Later, Akbas [7] proved his conjecture stating that a suborbital graph for the modular group is a forest if and only if it contains no triangles. They completely characterized circuits in suborbital graphs for the modular group. These studies lead us to explore infinite paths in suborbital graphs for the Fuchsian group Hm.In [8], Yayenie gave a remark stating that the Fuchsian group Hm acts transitively on the set mℚ^ if and only if m is either 1 or prime. So, we study on the Fuchsian group Hm where m is a prime number. We divide this work into four sections. In Section 2, we will show the way to construct a suborbital graph G∞,u/vm for the group Hm and we give the edge conditions for two vertices that are joined in the graph. In Sections 3 and 4, we investigate vertices on the infinite path of minimal length in the graph G∞,u/vm for two cases: m,v=1 and m,v=m, respectively; here, ordered pairs denote greatest common divisors. Finally, we represent each vertex on the infinite path of minimal length by recurrence relations and determine the limit point of the sequence of the vertices.


## 2. Suborbital Graphs forHm

Letℚ^=ℚ∪∞ and m be a squarefree positive integer, every element of mℚ^ can be represented as a fraction x/ym with x,y∈ℤ and x,y=1. We represent ∞ as 1/0m=−1/0m. Hm acts on mℚ^ naturally by(5)T⋅xym=Txym,T
∈
H
m
,
x
/
y
m
∈
m
ℚ
^
. The following remark was given by Yayanie in [8], regarding the Fuchsian group Hm acting transitively on vertices mℚ^.Remark 1.
[8]. The Fuchsian group Hm acts transitively on the set mℚ^ if and only if m is either 1 or prime.
From now on, it will be assumed thatm is prime. So, Hm acts transitively on mℚ^. We will give a construction of suborbital graphs for the Fuchsian group Hm. Let Hm act on mℚ^×mℚ^ by(6)gα,β=gα,gβ.
g
∈
H
m
,
α
,
β
∈
m
ℚ
^. The orbits of this action are called suborbitals of Hm. The suborbital containing α,β is denoted by Oα,β. We can form a suborbital graph Gα,β whose vertices are the elements of mℚ^, and there is a directed edge from λ to δ if γ,δ∈Oα,β, denoted by γ⟶δ. We can see that Oβ,α is also a suborbital such that Oβ,α=Oα,β or Oβ,α∩Oα,β=∅. In the latter case, Gβ,α is just Gα,β with reversed arrows and we call Gβ,α and Gα,β paired suborbital graphs. In the case Gβ,α=Gα,β, the graph consists of pairs of oppositely directed edges, and we replace each pair with an undirected edge for convenience. We call the graph self-paired.
SinceHm acts on mℚ^ transitively, each suborbital contains a pair ∞,u/vm for some u/vm∈mℚ^. The following two theorems are valid for prime number m and we can use the same technique in the proofs of Theorem 1 and 2 in [9], which were stated for m=2 and 3.Theorem 1.
Letu and v be relatively prime and m prime. If m,v=1, then, there exists a directed edge from r/sm to x/ym in G∞,u/vm if and only if ry−sx=±v and either.(i)
x≡±urmodv,y≡±usmodv and ms or(ii)
x≡±murmodv,y≡±musmodv and my.Theorem 2.
Letu and v be relatively prime and m prime. If m,v=m, then, there exists a directed edge from r/sm to x/ym in G∞,u/vm if and only if either.(i)
ry−sx=±v,ms and x≡±urmodv,y≡±usmodv or(ii)
ry−sx=±v/m and x≡±urmodv/m,y≡±usmodv.
Here, the choice of signs forx and y are always the same. By using Theorems 1 and 2, we obtain the following two corollaries that characterize a self-paired graph.Corollary 1.
Letu and v be relatively prime and m prime such that m,v=1. Then the suborbital graph G∞,u/vm is self-paired if and only if mu2+1≡0modv.Corollary 2.
Letu and v be relatively prime and m prime such that m,v=m. Then the suborbital graph G∞,u/vm is self-paired if and only if u2+1≡0modv.
Next, we will show the existence of an integerk such that mu2+kmu+1≡0modv.Lemma 1.
Letu and v be relatively prime and m prime such that m,v=1. Then there exist integers k and l with 1<k,l≤v such that mu2+kmu+1≡0modv and mu2−lmu+1≡0modv.Proof.
Sincem,v=u,v=1, we have mu,v=1. Then, there exists an integer x such that mux≡1modv. So mux−mu2−1≡−mu2−1modv. Taking k=x−mu2−1, it is seen that mu2+kmu+1≡0modv is satisfied. Note that k and l are uniquely determined.Theorem 3.
Letu and v be relatively prime and m prime such that m,v=1. Suppose that mu2+kmu+1≡0modv and mu2−lmu+1≡0modv where k,l are integers such that 1<k,l≤v. If G∞,u/vm is self-paired, then k=l=v; otherwise, l=v−k.Proof.
Sincemu2+kmu+1≡0modv and mu2−lmu+1≡0modv, kmu≡−lmumodv. As u,v=1=m,v, we obtain that k≡−lmodv. So, there exists an integer y such that k+l=vy and then 2<vy≤2v since 1<k,l≤v. Hence, y=1 or 2. Assume that G∞,u/vm is self-paired. By Corollary 1, we have mu2+1≡0modv which implies that ku≡0modv and −lu≡0modv. Since v∤u, vk, and vl. From 1<k,l≤v, we get k=v=l, so y=2. For the case y=1, we have l=v−k.Lemma 2.
Letu and v be relatively prime and m prime. Then there exist integers k and l with 1<k,l≤v such that u2+ku+1≡0modv and u2−lu+1≡0modv.Theorem 4.
Letu and v be relatively prime and m prime. Suppose that u2+ku+1≡0modv and u2−lu+1≡0modv where k,l are integers such that 1<k,l≤v. If G∞,u/vm is self-paired, then k=l=v; otherwise, l=v−k.


## 3. Infinite Path of Minimal Length inG∞,u/vm Where m,v = 1

Letv1,v2,v3,…,vn be vertices of the suborbital graph G∞,u/vm, we call the configurations(7)v1⟶v2⟶⋯⟶vn,v1⟶v2⟶v3⟶⋯,a path and an infinite path, respectively. If vi⟶>vj (or vi⟵<vj) and there is no vertex which has greater (or smaller) value than vj joined with the vertex vi, then vj is the farthest vertex which can be joined with the vertex vi. The path v1⟶v2⟶⋯⟶vn is called of minimal length if and only if vi↮vj, where i<j−1,i∈1,2,3,…,n−2,j∈3,4,5,…,n and vi+1 must be the farthest vertex which can be joined with the vertex vi.In this section, we focus on the infinite path of minimal length in the suborbital graphG∞,u/vm where m,v=1. By the choice of prime, m, and remark 1, we obtain transitivity. Thus, we can map the first edge of any infinite path to the edge ∞⟶u/vm. We start investigating vertices in the infinite path of minimal length by determining the farthest vertex which can be joined with the vertex u/vm.Theorem 5.
Letu and v be relatively prime and m prime such that m,v=1, and let k,l be the integers uniquely determined in Theorem 6. Then, we have the following results in G∞,u/vm:(i)
The farthest vertices which can be joined with u/vm on the right and the left are(8)u+1/kmvm,u−1/lmvm,respectively. No nearest vertex exists.(ii)
The farthest vertices which can be joined with u+1/km/vm and u−1/lm/vm are(9)u+1/km−1/kvm,u−1/lm−1/lvm,respectively. No nearest vertex exists.(iii)
The farthest vertices which can be joined withu+1/km−1/k/vm and u−1/lm−1/l/vm are(10)u+1/km−1/k−1/km/vm,u−1/lm−1/l−1/lm/vm,respectively. No nearest vertex exists.Proof.
i For the right side of u/vm, we assume that there exists an edge u/vm⟶x/ym in G∞,u/vm and u/vm<x/ym. We can write x/ym in the form(11)xym=uvm+vxvym−uyvym=u+vx−uy/yvm.
With this and the fact thatuy<vx, we can replace x/ym with(12)u+t/svm=su+tsvm,where t/s is in ℚ+. Let d be the greatest common divisor of su+t and sv; then, we get su+t/d,sv/d=1 and(13)uvm⟶su+t/dsv/dm.
Theorem 1 gives the conditions when this edge exists. Sincem,v=1, we have m∤v so case i in Theorem 1 cannot happen. Then, we thus consider case ii. In this case, we have su+t/d≡±mu2modv and usv/d−vsu+t/d=±v.
Ifsu+t/d≡mu2modv and usv/d−vsu+t/d=v, then t=−d which implies su−d/d≡mu2modv. As mu2+kmu+1≡0modv, we have su−d/d≡−kmu−1modv, that is, su≡−dkmumodv. Since u,v=1, we get s≡−dkmmodv. In other words, s=−dkm−vz for some integer z. Thus,(14)ts=−d−dkm−vz=1km+vz/d.
We will find the largest value oft/s by defining a function f:ℝ+∪0⟶ℝ,(15)fz=u+1/km+vz/dvm.
The derivative off is f′z=−d/dkm+vz2m, which is negative for every nonnegative z. This implies that the maximum occurs at z=0 and maximum value is(16)u+1/kmvm=kmu+1kmvm.
By Theorem 1, it now suffices to show thatkmu+1/kmv is an irreducible fraction. Thus,(17)u+1/kmvm,is a vertex in G∞,u/vm and is the farthest one joined with u/vm. We also see that(18)limz⟶∞u+1/km+vz/vzvm=uvm.
This implies that there is no such nearest point joined with the vertexu/vm.
Ifsu+t/d≡−mu2modv and usv/d−vsu+t/d=−v, then t=d, which implies su+d/d≡−mu2modv. We have mu2+kmu+1≡0modv. This implies that su+d/d≡kmu+1modv; that is, su≡dkmumodv. The fact that u,v=1 implies that s≡dkmmodv. Therefore, s=dkm+vz for some z in ℕ∪0. Hence,(19)ts=ddkm+vz=1km+vz/d.
The proof is similar to the previous case. Next, we will consider the left side ofu/vm. Assume that there exists an edge(20)uvm⟶xym,in G∞,u/vm and u/vm>x/ym. We can replace x/ym with(21)u−p/qvm=qu−pqvm,where p/q is in ℚ+. Let c be the greatest common divisor of qu−p and qv; then, we get qu−p/c,qv/c=1 and(22)uvm⟶qu−p/cqv/cm.
By Theorem 1, casei cannot happen. So, we will consider case ii. Then, we have qu−p/c≡±mu2modv and uqv/c−vqu−p/c=±v.
Ifqu−p/c≡mu2modv and uqv/c−vqu−p/c=v, then p=c which implies qu−c/c≡mu2modv. As mu2−lmu+1≡0modv, we have qu−c/c≡lmu−1modv; that is, qu≡clmumodv. Since u,v=1, we get q≡clmmodv. In other words, q=clm+vz for some integer z. Thus,(23)pq=cclm+vz=1lm+vz/c.
We define a functionf:ℝ+∪0⟶ℝ,(24)fz=u−1/lm+vz/cvm.
The derivative off is f′z=c/clm+vz2m, which is positive for every nonnegative z. This implies that the minimum occurs at z=0 and minimum value is(25)u−1/lmvm=lmu−1lmvm.
By Theorem 1, it now suffices to show thatlmu−1/lmv is an irreducible fraction. Thus,(26)u−1/lmvm,is a vertex in G∞,u/vm and is the farthest one joined with u/vm. We also see that(27)limz⟶∞u−1/lm+vz/cvm=uvm.
This implies that there is no such nearest point joined with the vertexu/vm.
Ifqu−p/c≡−mu2modv and uqv/c−vqu−p/c=−v, then p=−c which implies qu+c/c≡−mu2modv. We have mu2−lmu+1≡0modv. This implies that qu+c/c≡−lmu+1modv; that is, qu≡−clmumodv. The fact that u,v=1 implies that q≡−clmmodv. Therefore, q=−clm−vz for some z in ℕ∪0. Hence,(28)pq=−c−clm−vz=1lm+vz/c.
This case is done by using a similar argument to that of the previous case.Corollary 3.
Letu and v be relatively prime and m prime such that m,v=1. If there are integers k,l such that mu2+kmu+1≡0modv and mu2−lmu+1≡0modv, then(29)φz=−umz+mu2+kmu+1/v−vz+u+km,ψz=−umz+mu2−lmu+1/v−vz+u−lm,are elements of Hm. Moreover, φi+1u/vm,ψi+1u/vm are the farthest vertices which can be joined with φiu/vm and ψiu/vm, respectively, where i∈ℕ.Corollary 4.
Letu and v be relatively prime and m prime. If mu2+kmu+1≡0modv where 1<k≤v, then there is an infinite path of minimal length:(30)10m⟶uvm⟶u+1/kmvm⟶u+km−1/k/1vm⟶u+1/km−1/k−1/kmvm⟶⋯,whose vertices are in the set(31)V1≔∪n=1∞u+Tn0vm:Tn=t0t1…tn,t0z=z,tiz=−1−km+zm;i∈1,2,⋯,n∪∞,uvminG∞,uvm.Corollary 5.
Letu and v be relatively prime and m prime. If mu2−lmu+1≡0modv where 1<l≤v, then there is an infinite path of minimal length:(32)⋯⟵u−1/lm−1/l−1/lmvm⟵u−1/lm−1/lvm⟵u−1/lmvm⟵uvm⟵10m,whose vertices are in the set(33)V2≔∪n=1∞u−Tn0vm:Tn=t0t1…tn,t0z=z,tiz=−1−lm+zm;i∈1,2,…,n∪∞,uvminG∞,uvm.


## 4. Infinite Path of Minimal Length inG∞,u/vm Where m,v=m

In the previoussection, we provided the existence of infinite path of minimal length in the suborbital graph G∞,u/vm where m,v=1. We find that the existence property is also valid for the suborbital graph G∞,u/vm where m,v=m in very close analogy.Theorem 6.
Letu and v be relatively prime and m prime such that m,v=m, and let k,l be the integers uniquely determined in Theorem 8. Then we have the following results in G∞,u/vm:(i)
The farthest vertices which can be joined withu/vm on the right and the left are(34)u+1/kvm,u−1/lvm,respectively. No nearest vertex exists.(ii)
The farthest vertices which can be joined withu+1/k/vm andu−1/l/vm are(35)u+1/k−1/kvm,u−1/l−1/lvm,respectively. No nearest vertex exists.(iii)
The farthest vertices which can be joined withu+1/k−1/k/vm andu−1/l−1/l/vm are(36)u+1/k−1/k−1/kvm,u−1/l−1/l−1/lvm,respectively. No nearest vertex exists.Corollary 6.
Letu and v be relatively prime and m prime such that m,v=m. If there are integers k,l such that u2+ku+1≡0modv and u2−lu+1≡0modv, then(37)λz=−uz+u2+ku+1/vmmz+u+k,γz=−uz+u2−lu+1/vm−v/mmz+u−l,are elements of Hm. Moreover, λi+1u/vm,γi+1u/vm are the farthest vertices which can be joined with λiu/vm and γiu/vm, respectively, where. i=1,2,…Corollary 7.
Letu and v be relatively prime and m prime. If u2+ku+1≡0modv where 1<k≤v, then there is an infinite path of minimal length:(38)10m⟶uvm⟶u+vm⟶u+1/k−1/kvm⟶u+1/k−1/k−1/kvm⟶⋯,whose vertices are in the set(39)W1≔∪n=1∞u+Tn0vm:Tn=t0t1…tn,t0z=z,tiz=−1−k+z;i∈1,2,…,n∪∞,uvminG∞,uvm.Corollary 8.
Letu and v be relatively prime and m prime. If u2−lu+1≡0modv where 1<l≤v, then there is an infinite path of minimal length:(40)⋯⟵u−1/l−1/l−1/lvm⟵u−1/l−1/lvm⟵u−1/lvm⟵uvm⟵10m,whose vertices are in the set(41)W2≔∪n=1∞u−Tn0vm:Tn=t0t1…tn,t0z=z,tiz=−1−l+z;i∈1,2,⋯,n⋃∞uvminG∞,uvm.


## 5. Continued Fractions and Recurrence Relations

From results in Sections3 and 4, we have that any vertex on the infinite path of minimal length can be represented by a continued fraction expansion. As a continued fraction is related to recurrence relations, we use them to investigate vertices on the infinite path of minimal length. We conclude this section by finding the limit point of the sequence of the vertices.Letann∈ℕ,bnn∈ℕ0 be sequences of complex numbers with an≠0 for n≥1 and tmn∈ℕ0,Tmn∈ℕ0 be sequences of Möbius transformations defined as follows:(42)t0z=b0+z,tnz=anbn+z,forn∈ℕ,T0z=t0z,Tnz=Tn−1tnz,forn∈ℕ.We considerT00,T10,T20,T30 and so on and form a continued fraction of the form(43)For convenience, we denote this by(44)b0+a1b1+a2b2+a3b3+⋯+anbn.In [10], the nth numerator An and the nth denominator Bn of a continued fraction as in (43) are defined by the recurrence relations(45)AnBn=bnAn−1Bn−1+anAn−2Bn−2,n=1,2,3,…,with initial conditions(46)A−1=1,B−1=0,A0=b0,B0=1.For a given sequenceznn∈ℕ0, Tnzn can be written as(47)Tnzn=An+An−1znBn+Bn−1zn,n=0,1,2,…,and then(48)Tn0=AnBn.Now we consider infinite paths in suborbital graphG∞,u/vm. For the case when m,v=1, the infinite path of minimal length for the right direction in Corollary 4 gives an=−1/m and bn=−k for n≥1. By recurrence relations in (45), we obtain Bn=−mAn+1, and then, we have a vertex on this path:(49)u+Tn0vm=An+1mu−AnAn+1mvm.Similarly, we have a vertex on the infinite path of minimal length for the left direction is(50)u−Tn0vm=An+1lu+AnAn+1lvm.Theorem 7.
Ifm,v=1 and k≥2, then we have(51)An=−1mn21−n∑t=1nkm+km2−4mn−tkm−km2−4mt−1.Proof.
From the recurrence relation, we have(52)mAn+2+kmAn+1+An=0,with A−1=1 and A0=0. The characteristic equation for the relation (53) is(53)mx2+kmx+1=0,which gives two roots(54)x1=−km+km2−4m2m,x2=−km−km2−4m2m.
Then, any solution of (9) have the form(55)An=α−km+km2−4m2mn+β−km−km2−4m2mn.
By using the initial conditions, we have(56)A0=α+β=0,A1=α−km+km2−4m+β−km−km2−4m=−2,which implies(57)α=−1km2−4m,β=1km2−4m.
Hence, we get(58)An=−12mn1km2−4mkm+km2−4mn−km−km2−4mn.
Sincekm+km2−4mn−km−km2−4mn is equal to(59)2km2−4m∑t=1nkm+km2−4mn−tkm−km2−4mt−1,then we obtain(60)An=−1mn21−n∑t=1nkm+km2−4mn−tkm−km2−4mt−1.Theorem 8.
Ifm,v=1 and l≥2, then we have(61)An=−1mn21−n∑t=1nlm+lm2−4mn−tlm−lm2−4mt−1.
Next, we consider casem,v=m. By Corollary 7, we have an=−1,bn=−k for n≥1. So we get Bn=An+1 from recurrence relations in (45). Then, a vertex on the infinite path of minimal length for the right direction is(62)u+Tn0vm=An+1u−AnAn+1vm.
Likewise, a vertex on the infinite path of minimal length for the left direction is(63)u−Tn0vm=An+1u+AnAn+1vm.Theorem 9.
Ifm,v=m and k≥2, then we have(64)An=−1nn,k=2−1n21−n∑t=1nk+k2−4n−tk−k2−4,k>2.Theorem 10.
Ifm,v=m and l≥2, then we have(65)An=−1nn,l=2,−1n21−n∑t=1nl+l2−4n−tl−l2−4t−1,l>2.
Having characterized the vertices on the infinite path of minimal length, we investigate the limit point of this path by using the Śleszyński–Pringsheim theorem.Theorem 11.
[10] (Śleszyński–Pringsheim). The continued fraction(66)a1b1+a2b2+a3b3+⋯+anbn+⋯converges to some valued f with f≤1 if(67)bn≥1+an,n≥1.Corollary 12.
The sequence of the vertices of infinite path of minimal length (30) converges to(68)2mu+mk−mk2−4⋅m2mvm.Proof.
Since we havean=−1/m and bn=−k where m is prime and k≥2, we get bn≥1+an for all n≥1. By Theorem 11, the continued fraction(69)1km−1k−1km−⋯converges to f with f≤1; that is, limn⟶∞Tn0=f. As we know(70)Tn0=1km−mTn−10,Tn0km−mTn−10=1, and since limn⟶∞Tn0=limn⟶∞Tn−10, we have fkm−mf=1. Hence, mf2−mkf+1=0 and(71)f=mk±m2k2−4m2m.
Ask≥2 and f≤1, we get f=mk−m2k2−4m/2m. Therefore, we obtain that the sequence of the vertices of infinite path of minimal length (30) converges to(72)u+mk−m2k2−4m/2mvm=2mu+mk−mk2−4⋅m2mvm.Corollary 13.
The sequence of the vertices of infinite path of minimal length (32) converges to(73)2mu−ml+ml2−4⋅m2mvm.Corollary 14.
The sequence of the vertices of infinite path of minimal length (38) converges to(74)2u+k−k2−42vm.Corollary 15.
The sequence of the vertices of infinite path of minimal length (40) converges to(75)2u−l+l2−42vm.
We observe that the limit points in Corollaries 23 and 24 are not in the setmℚ^, but the limit points in Corollaries 25 and 26 will be in the set 2 if k=l=2.



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