On Dirichlet fundamental domainsfor Fuchsian groups

（フックス群のディリクレ基本領域について）

理学研究科

数物系専攻

梅本 悠莉子

Contents

Abstract

1. Hyperbolic geometry1.1. The hyperbolic plane1.2. Isometries of the hyperbolic plane1.3. Geodesics in the hyperbolic plane1.4. The hyperbolic area and the Gauss-Bonnet formula1.5. The disk model for the hyperbolic plane

2. Fuchsian groups2.1. The topological group PSL(2,R)2.2. Commutative Fuchsian groups2.3. Properly discontinuous groups

3. Dirichlet fundamental domains3.1. Fundamental domains3.2. Dirichlet domains3.3. Structure of Dirichlet domains

4. The Siegel’s theorem and cofinite Fuchsian groups

5. Triangle groups5.1. Dirichlet domains for cofinite Fuchsian groups5.2. Triangle groups5.3. Main theorem5.4. Generalizations

Acknowlegements

References

Abstract

This thesis presents Dirichlet fundamental domains for Fuchsian groupsmainly referring to the book written by Katok([4]). Fuchsian groups arediscrete subgroups of PSL(2,R), and they act on the hyperbolic plane H.We will see in the third chapter that any Fuchsian group possesses a nicefundamental domain which is called a Dirichlet domain. In the last chapter,we consider Dirichlet domains for triangle groups.

1 Hyperbolic geometry

1.1 The hyperbolic plane

Let C be the complex plane. The hyperbolic plane is the metric spaceconsisting of the upper half-plane

H = {z ∈ C | Im(z) > 0}

endowed with a metric ρ defined below. We shall use the usual notationsfor the real and imaginary parts of z = x+ iy ∈ C, Re(z) = x, Im(z) = y.

To define the hyperbolic metric ρ, we first define the hyperbolic lengthof a curve in H.

Definition 1. (hyperbolic length)Let I = [0, 1] and γ : I → H be a piecewise differentiable curve

γ = {z(t) = x(t) + iy(t) ∈ H | t ∈ I}.

Then its hyperbolic length h(γ) is given by

h(γ) =

∫γ

1

Im(z)|dz| =

∫ 1

0

∣∣∣dz(t)dt

∣∣∣ dty(t)

=

∫ 1

0

√(dx(t)dt )2 + (dy(t)dt )2

y(t)dt.

An application of the chain rule shows that this hyperbolic length isindependent of the parametrization of γ.

Definition 2. (hyperbolic distance)The hyperbolic distance ρ(z, w) between two points z, w ∈ H is defined

by the formulaρ(z, w) = inf h(γ),

where the infimum is taken over all γ joining z to w in H.

1

The above ρ is non-negative, symmetric, and satisfies the triangle in-equality

ρ(z, w) ≤ ρ(z, ξ) + ρ(ξ, w),

i.e., is a distance function on H.We will see in a later section that the among the curves joining z and w,

the one with shortest hyperbolic length (i.e., a geodesic) is a straight lineor a semicircle orthogonal to the real axis R = {z ∈ C | Im(z) = 0}.

γ

z

wγ

z

w

Fig. 1:

1.2 Isometries of the hyperbolic plane

The set of fractional linear (or Mobius) transformations of the Rie-mann sphere C onto itself of the form{

z 7→ az + b

cz + d|(a bc d

)∈ SL(2,R)

}where SL(2,R) is the unimodular group, i.e., a, b, c, d ∈ R, ad − bc = 1,forms a group, and it is called PSL(2,R). The product of two transforma-tions of PSL(2,R) corresponds to the product of corresponding matrices ofSL(2,R) and the inverse corresponds to the inverse matrix. Each transfor-mation T ∈ PSL(2,R) is represented by a pair of matrices ±g ∈ SL(2,R)and thus, PSL(2,R) is isomorphic to SL(2,R)/{±12}, where 12 is the 2× 2identity matrix. The identity transformation in PSL(2,R) will be denotedby Id.

Remark 1. PSL(2,R) contains all fractional linear transformations of theform z 7→ az+b

cz+d with a, b, c, d ∈ R and ∆ = ad − bc > 0 since by dividing

the numerator and denominator by√∆ we obtain a new matrix for it of

determinant 1. In particular, PSL(2,R) contains all transformations of theform z 7→ az + b (a, b ∈ R, a > 0), and the transformation z 7→ −1/z.

2

Theorem 1. PSL(2,R) acts on H by homeomorphisms.

Proof. First we show that any transformation of PSL(2,R) maps H ontoitself. Let T ∈ PSL(2,R), and w = T (z) = az+b

cz+d . Then

w =(az + b)(cz + d)

|cz + d|2=

ac|z|2 + adz + bcz + bd

|cz + d|2,

so that

Im(w) =w − w

2i=

z − z

2i|cz + d|2=

Im(z)

|cz + d|2. (1)

Therefore Im(z) > 0 implies Im(w) > 0. The theorem now follows from thecontinuity of T (z) and its inverse. �

A transformation of H onto itself is called an isometry if it preserves thehyperbolic distance on H and it is homeomorphism. The set of all isometriesof H forms a group; we shall denote it by Isom(H).

Theorem 2. PSL(2,R) ⊂ Isom(H), i.e., any T ∈ PSL(2,R) satisfies

ρ(z, w) = ρ(T (z), T (w))

for any two points z, w ∈ H.

Proof. It will be enough to show that, if γ : I → H is a piecewise differ-entiable curve in H then for any T ∈ PSL(2,R) we have h(T (γ)) = h(γ).Suppose γ : I → H is given by z(t) = x(t) + iy(t), and w(t) = T (z(t)) =u(t) + iv(t). We have

dw

dz=

a(cz + d)− c(az + b)

(cz + d)2=

1

(cz + d)2.

By (1) v = y|cz+d|2 , and hence |dwdz | =

vy . Thus

h(T (γ)) =

∫ 1

0

|dwdz |dtv(t)

=

∫ 1

0

|dwdzdzdt |dt

v(t)=

∫ 1

0

|dzdt |dty(t)

= h(γ).

�

Theorem 3. Any transformation of PSL(2,R) is conformal.

Proof. Any T (z) = az+bcz+d ∈ PSL(2,R) is conformal on the complex plane

since for any z ∈ C, T ′(z) = 1(cz+d)2

= 0. �

3

1.3 Geodesics in the hyperbolic plane

Definition 3. (geodesic)Let z, w be two points in H and γ be a piecewise differentiable curve

joining them. Then if γ satisfies the equation

ρ(z, w) = h(γ),

γ will be called a geodesic on H.

Lemma 1. ([2], Lemma 2.4.) If two points z, w ∈ H are located on thesame vertical line, say z = x0+ iy0, w = x0+ iy1, the straight line γ0 joiningz and w has the shortest hyperbolic length among all piecewise differentiablecurves joining z and w. In addition, the hyperbolic length of any other curvejoining z and w has strictly larger hyperbolic length, and

ρ(z, w) = h(γ0) = log

∣∣∣∣y1y0∣∣∣∣ .

Proof. Assuming y0 ≤ y1 without loss of generality, let us first compute thehyperbolic length h(γ0). Parametrize this line segment by

t 7→ x0 + it, y0 ≤ t ≤ y1.

Then,

h(γ0) =

∫ y1

y0

1

tdt = log

y1y0

.

Now, consider a piecewise differentiable curve γ joining z and w, whichis parametrized by

t 7→ x(t) + iy(t), 0 ≤ t ≤ 1.

Its hyperbolic length is

h(γ) =

∫ 1

0

√(dx(t)dt )2 + (dy(t)dt )2

y(t)dt ≥

∫ 1

0

|dy(t)dt |y(t)

dt ≥∫ 1

0

dy(t)dt

y(t)dt

= logy(1)

y(0)= log

y1y0

= h(γ0).

In addition, for the first term to be equal to the last one, the above twoinequalities must be equalities. Equality in the first inequality requires thatthe function x(t) is constant, while equality in the second one implies thaty(t) is weakly increasing. This shows that the curve γ is equal to the curveγ0 if h(γ) = h(γ0). �

4

Theorem 4. Let γ be a piecewise differentiable curve joining two pointsz, w ∈ H. Then γ is a geodesic segment on H if and only if γ is a part of asemicircle or a straight line orthogonal to the real axis R.

zw

γ

a b

A(z)

A(w)

A(γ)

0

Fig. 2:

Proof. Suppose γ is a part of a semicircle. Let a, b ∈ R, a < b be end pointsof the semicircle. The transformation A : z 7→ − z−a

z−b ∈ PSL(2,R) moves γto the imaginary axis. Since A is an isometric and by Lemma 1,

ρ(z, w) = ρ(A(z), A(w)) = h(A(γ)) = h(γ).

Conversely, if γ is a geodesic, i.e., ρ(z, w) = h(γ). We can choose thetransformation A in PSL(2,R) which moves z, w to the imaginary axis.Then ρ(A(z), A(w)) = h(A(γ)), and so A(γ) is a geodesic connecting A(z)and A(w). So A(γ) is a straight line by Lemma 1, and then γ is a semicircleor a straight line orthogonal to the real axis R by conformality of A. �

By Theorem 4, any two points z, w ∈ H can be joined by a uniquegeodesic. We will denote it by [z, w].

Corollary 1. If z and w are two distinct points in H, then

ρ(z, w) = ρ(z, ξ) + ρ(ξ, w)

if and only if ξ ∈ [z, w].

5

1.4 The hyperbolic area and the Gauss-Bonnet formula

Definition 4. (hyperbolic area)For a subset A ⊆ H we define µ(A), the hyperbolic area of A, by

µ(A) =

∫A

dxdy

y2

if this integral exists.

Theorem 5. The hyperbolic area is invariant under all transformations inPSL(2,R) : if A ⊆ H, µ(A) exists, and T ∈ PSL(2,R) then µ(T (A)) =µ(A).

Proof. Let z = x+ iy,

T (z) =az + b

cz + d(a, b, c, d ∈ R, ad− bc = 1),

and w = T (z) = u + iv. Then using the Cauchy-Riemann equations wecalculate the Jacobian

∂(u, v)

∂(x, y)=

∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x=

(∂u

∂x

)2

+

(∂v

∂x

)2

=

∣∣∣∣dTdz∣∣∣∣ = 1

| cz + d |4.

Thus

µ(T (A)) =

∫T (A)

dudv

v2=

∫A

∂(u, v)

∂(x, y)

dxdy

v2

=

∫A

1

|cz + d|4|cz + d|4

y2dxdy = µ(A)

using (1). �

Let H = H ∪ R ∪ {∞} be the Euclidean closure of H. A hyperbolicn-sided polygon is a closed set of H bounded by n hyperbolic geodesicsegments. If two line segments intersect in H, then the point of their inter-section is called a vertex of the polygon, and if line segments intersect inR ∪ {∞}, then the point of their intersection is called a ideal vertex.

In Figure 3 we illustrate four types of hyperbolic triangles depending onwhether 0, 1, 2, or 3 vertices of the triangle are ideal vertices.

The Gauss-Bonnet formula shows that the hyperbolic area of a hyper-bolic triangle depends only on its angles.

6

Fig. 3:

Theorem 6. (Gauss-Bonnet)Let ∆ be a hyperbolic triangle with angles α, β, γ. Then

µ(∆) = π − α− β − γ.

Proof. Case 1.We suppose first that at least one of the vertices of ∆, say γ, belongs

to R ∪ {∞}, and hence γ is equal to zero. If it belongs to the real axis,by applying a transformation in PSL(2,R) we can map this vertex to ∞without altering the hyperbolic area and the angles. Now we get the trianglewhose two sides are vertical geodesics and one side is a semicircle orthogonalto the real axis. By applying transformations of the form z 7→ z+k (k ∈ R),z 7→ λz (λ > 0) we can assume that the semicircle has center 0 and radius1. These transformations will not change the area of ∆, by Theorem 5.Since each angle of ∆ will be preserved by conformality, the angles AOCand BOD are equal to α and β (Fig. 4). So the vertical geodesics throughA and B are the lines x = cos(π − α) and x = cosβ, respectively. We nowcalculate

µ(∆) =

∫∆

dxdy

y2=

∫ cosβ

cos(π−α)dx

∫ ∞

√1−x2

dy

y2=

∫ cosβ

cos(π−α)

dx√1− x2

.

Make the substitution x = cos θ(0 ≤ θ ≤ π); then

µ(∆) =

∫ β

π−α

− sin θdθ

sin θ= π − α− β.

Case 2.Suppose ∆ has no vertices in R∪{∞}. If ∆ has vertices A,B,C, we can

apply a transformation in PSL(2,R) to map A,B onto the imaginary axis.Let draw a vertical geodesic from C to ∞. Then we have the situation of

7

the right picture in Figure 4; θ denotes ∠BC∞. Here ∆ = ∆1 −∆2 where∆1 has vertices A,C,∞ and ∆2 has vertices B,C,∞. Then

µ(∆) = µ(∆1)−µ(∆2) = π−α− (γ+θ)−{π−θ− (π−β)} = π−α−β−γ.

A

B

Cα

β

γθ

0

BA

C D0

Δ

αβ

α β

Fig. 4:

�

1.5 The disk model for the hyperbolic plane

We shall now describe a model of the hyperbolic geometry in the unit disc:

U = {z ∈ C, |z| < 1}.

Let I = [0, 1] and γ∗ : I → U be a piecewise differential curves γ∗ = {w(t) =u(t) + iv(t) ∈ H | t ∈ I}. Then its hyperbolic length h∗(γ∗) is given by

h∗(γ∗) =

∫γ∗

2

1− |w|2|dw| =

∫ 1

0

2

1− |w(t)|2

∣∣∣∣dw(t)dt

∣∣∣∣ dt.The hyperbolic distance ρ∗(p, q) between two points p, q ∈ U is defined bythe formula

ρ∗(p, q) = infh∗(γ∗),

where the infimum is taken over all piecewise differentiable curve γ∗ joiningp and q in U .

The map

f(z) =zi+ 1

z + i(2)

8

is a 1-1 map of H onto U , thus ρ∗ satisfies the equation

ρ∗(p, q) = ρ(f−1(p), f−1(q))(p, q ∈ U ),

i.e., f is an isometry from (H, ρ) onto (U , ρ∗). In the model U geodesics aresegments of Euclidean circles orthogonal to the principal circle C(0; 1) ={z ∈ C | |z| = 1} and its diameters.

We shall change from one model to the other as each has its own partic-ular advantage.

Let U = U ∪ C(0; 1) be the Euclidean closure of U .

9

2 Fuchsian groups

2.1 The topological group PSL(2,R)

Besides being a group, PSL(2,R) is also a topological space in which atransformation z 7→ az+b

cz+d can be identified with the point (a, b, c, d) ∈ R4.More precisely, as a topological space, SL(2,R) can be identified with thesubset of R4,

{(a, b, c, d) ∈ R4 | ad− bc = 1}.

Definition 5. A subgroup Γ of Isom(H) is called discrete if the inducedtopology on Γ is a discrete topology, i.e., if Γ is a discrete set in the topo-logical space Isom(H).

There are three types of elements in PSL(2,R) = {z 7→ T (z) = az+bcz+d | ad−

bc = 1} distinguished by the value of its trace: Tr(T ) = |a + d|. IfTr(T ) < 2, T is called elliptic; if Tr(T ) = 2, T is called parabolic; and ifTr(T ) > 2, T is called hyperbolic.

T ∈ PSL(2,R) is hyperbolic if and only if its corresponding matrix is

conjugate in SL(2,R) to a matrix

(λ 00 1/λ

), λ = 1, T is parabolic if and

only if its matrix is conjugate in SL(2,R) to a unique matrix

(1 10 1

), and

T is elliptic if and only if its matrix is conjugate in SL(2,R) to a unique

matrix

(cos θ sin θ− sin θ cos θ

).

The fixed points are found by solving

z =az + b

cz + d(a, b, c, d ∈ R, ad− bc = 1),

and we see that a hyperbolic transformation has two fixed points in R∪{∞},a parabolic transformation has one fixed point in R ∪ {∞}, and elliptictransformation has a pair of complex conjugate fixed points, and therefore,one fixed point in H.

2.2 Commutative Fuchsian groups

Definition 6. (Fuchsian group)A discrete subgroup of PSL(2,R) is called a Fuchsian group.

Remark 2. 1. Any subgroup of a Fuchsian group is a Fuchsian group.

10

2. Any Fuchsian group at most a countable group.

Before we give some examples of Fuchsian groups, let us identify discretesubgroups of one-dimensional topological groups: R, the additive group ofreal numbers, and S1, the multiplicative group of complex numbers of mod-ulus 1.

Lemma 2. 1. Any non-trivial discrete subgroup of R is an infinite cyclicgroup.

2. Any discrete subgroup of S1 is a finite cyclic group.

Theorem 7. 1. All hyperbolic and parabolic cyclic subgroups of PSL(2,R)are Fuchsian groups.

2. An elliptic cyclic subgroup of PSL(2,R) is a Fuchsian group if andonly if it is finite.

Definition 7. (commutator)For g, h ∈ PSL(2,R), [g, h] = g ◦ h ◦ g−1 ◦ h−1 ∈ Γ is called the com-

mutator of g and h.

Remark 3. For g, h ∈ PSL(2,R), g and h are commutative if and only if[g, h] = Id.

Theorem 8. Two non-identity elements g, h ∈ PSL(2,R) are commutativeif and only if they have the same fixed-point set, i.e.,

Fix(g) = Fix(h).

Theorem 9. Every commutative Fuchsian group is cyclic.

Proof. By Theorem 8 , all non-identity elements in a commutative Fuchsiangroup Γ have the same fixed-point set. So all elements of Γ−{Id} belong tothe same type. Suppose Γ−{Id} consists of hyperbolic elements fixing p, q ∈R∪{∞}. There exists A ∈ PSL(2,R) such that A(p) = 0, A(q) = ∞. ThenAΓA−1 is a discrete subgroup of {z 7→ λz | λ > 0} ∼= R+. So AΓA−1 shouldbe cyclic by Lemma 2. Suppose Γ − {Id} consists of parabolic elementsfixing r ∈ R∪{∞}. There exists B ∈ PSL(2,R) such that B(r) = ∞. ThenBΓB−1 is a discrete subgroup of {z 7→ z + k | k ∈ R} ∼= R. So BΓB−1

should be cyclic. Suppose Γ−{Id} consists of elliptic elements fixing s ∈ H.There exists C ∈ PSL(2,R) such that C(s) = i. Then (fC)Γ(fC)−1 is adiscrete subgroup of {z 7→ eiθz | θ ∈ [0, 2π)} ∼= S1 where f(z) = zi+1

z+i (see

§1.5 (2)). So (fC)Γ(fC)−1 should be cyclic. �

11

2.3 Properly discontinuous groups

Let G be a subgroup of Isom(H).

Definition 8. (G-orbit)For z ∈ H, the set G(z) = {g(z) | g ∈ G} is called the G-orbit of the

point z.

Definition 9. (locally finite)The G-orbit of z ∈ H is called locally finite if, for arbitrary compact

set K ⊂ H, there exist only finitely many g ∈ G such that g(z) ∈ K.

Definition 10. (properly discontinuously)We say that a group G acts properly discontinuously on H if for all

z ∈ H, G(z), the G-orbit of z is locally finite.

Theorem 10. Let Γ be a subgroup of PSL(2,R). Then Γ is a Fuchsiangroup if and only if Γ acts properly discontinuously on H.

Proof. We first show that a Fuchsian group acts properly discontinuously onH. Let z ∈ H and K be a compact subset of H. Then {T ∈ Γ | T (z) ∈ K} ={T ∈ PSL(2,R) | T (z) ∈ K} ∩ Γ is a finite set because it is the intersectionof a compact and discrete set, and hence Γ acts properly discontinuously.Conversely, suppose Γ acts properly discontinuously, but it is not a discretesubgroup of PSL(2,R). Choose a point s ∈ H not fixed by any non-identityelement of Γ: such points exist by Remark 2.2. As we are assuming thatΓ is not discrete, there exists a sequence {Tk} of distinct elements of Γsuch that Tk → Id as k → ∞. Hence Tk(s) → s as k → ∞ and as s isnot fixed by any non-identity element of Γ, {Tk(s)} is a sequence of pointsdistinct from s. Hence every closed hyperbolic disc centered at s containsinfinitely many points of the Γ-orbit of s, and hence Γ does not act properlydiscontinuously. �

Corollary 2. Let Γ be a subgroup of PSL(2,R). Then Γ acts properlydiscontinuously on H if and only if for all z ∈ H, Γ(z), the Γ-orbit of z, isa discrete subset of H.

Proof. Suppose Γ acts properly discontinuously on H, hence each Γ-orbitis locally finite, hence a discrete set of H. Conversely, suppose Γ does notact properly discontinuously on H and hence by Theorem 10 is not discrete.Repeating the argument in the proof of Theorem 10, we construct a sequence{Tk(s)} of points distinct from s such that Tk(s) → s, hence the Γ-orbit ofthe point s is not discrete. �

12

3 Dirichlet fundamental domains

3.1 Fundamental domains

We are going to be concerned with fundamental domains of mainly Fuchsiangroups, however it is convenient to give a definition in a slightly more generalsituation as follows.

Definition 11. (fundamental domain)Let Γ be a discrete group of isometries of the hyperbolic plane H. A

closed region F ⊆ H (i.e. a closure of a non-empty open set F ◦, called theinterior of F ) is defined to be a fundamental domain for Γ if

1.∪

T∈Γ T (F ) = H,

2. F ◦ ∩ T (F ◦) = ∅ for all T ∈ Γ− {Id}.

The set ∂F = F −F ◦ is called the boundary of F . The family {T (F ) | T ∈Γ} is called the tessellation of H.

We shall prove in § 3.2 that any Fuchsian group possesses a nice (con-nected and hyperbolically convex) fundamental domain. Now we give anexample in the simplest situation.

Example 1. Let Γ be the cyclic group generated by the transformationz 7→ 2z. Then the semi-annulus shown in the left picture of Figure 5 isa fundamental domain for Γ. Also we can find other fundamental domain ;an arbitrary small perturbation of the lower semicircle determines a pertur-bation of the upper semicircle, and gives yet another fundamental domainshown in the right picture of Figure 5.

Fig. 5:

Theorem 11. Let Γ be a discrete group of isometries of the hyperbolic planeH, and let F1 and F2 be two fundamental domains for Γ, and µ(F1) < ∞.Suppose that the boundaries of F1 and F2 have zero hyperbolic area. Thenµ(F2) = µ(F1).

13

Proof. We have µ(F ◦i ) = µ(Fi), i = 1, 2. Now

F1 ⊇ F1 ∩ (∪T∈Γ

T (F ◦2 )) =

∪T∈Γ

(F1 ∩ T (F ◦2 )).

Since F ◦2 is the interior of a fundamental domain, the sets F1 ∩ T (F ◦

2 ) aredisjoint, and hence

µ(F1) ≥∑T∈Γ

µ(F1 ∩ T (F ◦2 )) =

∑T∈Γ

µ(T−1(F1) ∩ F ◦2 ) =

∑T∈Γ

µ(T (F1) ∩ F ◦2 ).

Since F1 is a fundamental domain∪T∈Γ

T (F1) = H,

and therefore ∪T∈Γ

(T (F1) ∩ F ◦2 ) = F ◦

2 .

Hence ∑T∈Γ

µ(T (F1) ∩ F ◦2 ) ≥ µ(

∪T∈Γ

(T (F1) ∩ F ◦2 )) = µ(F ◦

2 ) = µ(F2).

Interchanging F1 and F2, we obtain µ(F2) ≥ µ(F1). Hence µ(F2) = µ(F1).�

Theorem 12. Let Γ be a discrete group of isometries of the hyperbolic planeH, and Λ be a subgroup of Γ of index n. If

Γ = ΛT1 ∪ ΛT2 ∪ ... ∪ ΛTn

is a decomposition of Γ into Λ-cosets and if F is a fundamental domain forΓ then

1. F1 = T1(F ) ∪ T2(F ) ∪ ... ∪ Tn(F ) is a fundamental domain for Λ,

2. if µ(F ) is finite and the hyperbolic area of the boundary of F is zerothen µ(F1) = nµ(F ).

Proof. We shall prove that F1 is a closed region. To prove this, we shall showthe following statement; if A,B ⊆ H be two closed regions, then the unionA ∪B is also a closed region. By the definition of the closed region, A = Uand B = V for some open sets U, V ⊆ H. Then A ∪ B = U ∪ V = U ∪ V ,and so the statement follows.

14

Let z ∈ H. Since F is a fundamental domain for Γ, there exists w ∈ Fand T ∈ Γ such that z = T (w). We have T = STi for some S ∈ Λ and somei, 1 ≤ i ≤ n. Therefore

z = STi(w) = S(Ti(w)).

Since Ti(w) ∈ F1, z is in the Λ-orbit of some point of F1. Hence the Λ-imagesof F1 is H.

Now suppose that z ∈ F ◦1 and that S(z) ∈ F ◦

1 , for S ∈ Λ. We needto prove that S = Id. Let ϵ > 0 be so small that Bρ(z; ϵ) (the openhyperbolic disc of radius ϵ centered at z) is contained in F ◦

1 . ThenBϵ(z) has a non-empty intersection with exactly k of the images of F ◦ underT1, ..., Tn, where 1 ≤ k ≤ n. Suppose these images are Ti1(F

◦), ..., Tik(F◦).

Let Bϵ(S(z)) = S(Bϵ(z)) have a non-empty intersection with Tj(F◦) say,

1 ≤ j ≤ n. It follows that Bρ(z; ϵ) has a non-empty intersection withS−1Tj(F

◦) so that S−1Tj = Til where 1 ≤ l ≤ k. Hence

ΛTj = ΛS−1Tj = ΛTil ,

so that Tj = Til and S = Id. Hence F ◦1 contains precisely one point of each

Λ-orbit. So F1 is a fundamental domain for Λ.The second statement follows, as µ(T (F )) = µ(F ) for all T ∈ PSL(2,R),

and µ(Ti(F ) ∩ Tj(F )) = 0 for i = j. �

3.2 Dirichlet domains

Definition 12. (perpendicular bisector)A perpendicular bisector of the geodesic segment [z1, z2] is the geodesic

through w, the mid-point of [z1, z2] orthogonal to [z1, z2].

Remark 4. 1. When the geodesic segment [z1, z2] is given, the perpen-dicular bisector of [z1, z2] is determined uniquely.

2. The perpendicular bisector of the geodesic segment [z1, z2] is equal toa line given by

{z ∈ H | ρ(z, z1) = ρ(z, z2)}.

Definition 13. (Dirichlet domain)Let Γ be an arbitrary Fuchsian group and let p ∈ H be not fixed by

any element of Γ− {Id}. We shall denote the perpendicular bisector of thegeodesic segment [p, T (p)] where T ∈ Γ− {Id} by Lp(T ), i.e.,

Lp(T ) = {z ∈ H | ρ(z, p) = ρ(z, T (p))},

15

and denote the hyperbolic half-plane containing p by Hp(T ), i.e.,

Hp(T ) = {z ∈ H | ρ(z, p) ≤ ρ(z, T (p))}.

We define the Dirichlet domain for Γ centered at p to be the set

Dp(Γ) =∩

T∈Γ−{Id}

Hp(T ).

T(p)

p

Fig. 6:

Remark 5.

Dp(Γ) = {z ∈ H | ρ(z, p) ≤ ρ(z, T (p)) for all T ∈ Γ}= {z ∈ H | ρ(z, p) ≤ ρ(T (z), p) for all T ∈ Γ}

Theorem 13. The Dirichlet domain Dp(Γ) is a hyperbolically convex fun-damental domain for Γ.

Proof. Let z ∈ H, and Γ(z) be its Γ-orbit. Since Γ(z) is a discrete set, thereexists z0(= z) ∈ Γ(z) with the smallest ρ(z0, p). Then ρ(z0, p) ≤ ρ(T (z0), p)for all T ∈ Γ, and by Remark 5 z0 ∈ Dp(Γ). Thus Dp(Γ) contains at leastone point from every Γ-orbit.

Next we show that if z1, z2 are in the interior of Dp(Γ), they cannot liein the same Γ-orbit. If ρ(z, p) = ρ(T (z), p) for some T ∈ Γ − {Id}, thenρ(z, p) = ρ(z, T−1(p)) and hence z ∈ Lp(T

−1). Then either z ∈ Dp(Γ) or zlies on the boundary of Dp(Γ); hence if z is in the interior of Dp(Γ), ρ(z, p) <ρ(T (z), p) for all T ∈ Γ− {Id}. If two points z1, z2 lie in the same Γ-orbit,this implies ρ(z1, p) < ρ(z2, p) and ρ(z2, p) < ρ(z1, p), a contradiction. Thusthe interior of Dp(Γ) contains at most one point in each Γ-orbit . Being anintersection of closed half-planes, Dp(Γ) is closed and hyperbolically convex.Thus Dp(Γ) is path-connected, hence connected. �

16

Example 2. Let Γ be the modular group PSL(2,Z). The point ki(k > 1) isnot fixed by any non-identity element of PSL(2,Z), so choose p = ki, wherek > 1. Then

F = {z ∈ H | |z| ≥ 1, |Re(z)| ≤ 1/2},

illustrated in Figure 7 is the Dirichlet domain for Γ centered at p.

p=ki

i

-1 1-1/2 1/20

Fig. 7:

3.3 Structure of Dirichlet domains

Dirichlet domains for Fuchsian groups are bounded by geodesics in H andpossibly by segments of the real axis R. Recall that Dp(Γ) is a closed subsetof H, but it may not be a closed subset of H. We call the closure of Dp(Γ) in

H the Euclidean closure of Dp(Γ) and denote it by Dp(Γ). We also define

the Euclidean boundary of Dp(Γ); ∂oDp(Γ) : ∂oDp(Γ) = Dp(Γ) − Dp(Γ)which obviously belongs to R ∪ {∞}.

Proposition 1. Let Γ be a Fuchsian group. Then {Lp(T ) | T ∈ Γ−{Id}} islocally finite, i.e., for arbitrary compact set K ⊂ H, there exists only finitelymany Lp(T ) such that Lp(T ) ∩K = ∅.

Theorem 14. Let Γ be a Fuchsian group. Then the Dirichlet tessellation{T (Dp(Γ)) | T ∈ Γ} is locally finite.

Proof. Let F = Dp(Γ) where p is not fixed by any element of Γ − {Id}.Let a ∈ F , and let K ⊂ H be a compact neighborhood of a. Suppose thatK ∩ Ti(F ) = ∅ for some infinite sequence T1, T2, ... of distinct elements ofΓ. Let σ = supz∈K ρ(p, z). Since H ∈ z 7→ ρ(p, z) ∈ R is a continuous

17

function about z, this function is bounded on the compact set K, so σ =supz∈K ρ(p, z) is finite. Let wj ∈ K ∩ Tj(F ). Then wj = Tj(zj) for zj ∈ F ,and by the triangle inequality,

ρ(p, Tj(p)) ≤ ρ(p, wj) + ρ(wj , Tj(p))

= ρ(p, wj) + ρ(zj , p)

≤ ρ(p, wj) + ρ(wj , p) (as zj ∈ Dp(Γ))

≤ 2σ.

Thus the infinite set of points T1(p), T2(p), ... belongs to the compacthyperbolic ball with center p and radius 2σ, but this contradicts the properlydiscontinuous action of Γ.

�

Definition 14. (vertex, ideal vertex)The point u ∈ ∂Dp(Γ) is called a vertex of Dp(Γ) if there exist two

geodesics Lp(T1), Lp(T2) such that their intersection in H is only the pointu, or u is a fixed point of an elliptic element of order 2 in Γ. If ∂oDp(Γ)has a component with one point, it is called an ideal vertex.

Definition 15. (side, free side)If, for some T ∈ Γ − {Id}, the intersection of Dp(Γ) with Lp(T ) is

a segment of a geodesic, this segment is called a side of Dp(Γ). If thissegment contains an elliptic vertex of order 2 as an inner point, say v, weshall consider the segment as two sides which have v as an end point of bothsides. If ∂oDp(Γ) has a component with more than two points, it is called afree side.

Remark 6. Since {Lp(T ) | T ∈ Γ−{Id}} is locally finite by Proposition 1,vertices of Dp(Γ) do not accumulate in H.

Proposition 2. If a side of Dp(Γ) has an end point in H, then that pointis a vertex. And if, for some S ∈ Γ − {Id}, the intersection of Lp(S) withDp(Γ) is one point, then this point is also a vertex.

Proof. We shall show the first statement. For some T ∈ Γ− {Id}, Lp(T ) ∩Dp(Γ) is a geodesic having an end point u ∈ H. Suppose that u is not avertex of Dp(Γ), in other words, T is the only element of Γ such that Lp(Γ)contains u. Since {Lp(g) | g ∈ Γ − {Id}} is locally finite, for some ϵ > 0,

only finitely many g ∈ Γ − {Id} satisfy the condition Bρ(u; ϵ) ∩ Lp(g) = ∅(Fig. 8). So if we choose sufficiently small ϵ′, ϵ > ϵ′ > 0, only T satisfies the

18

u

p

ρ

ρ

p

Fig. 8:

condition Bρ(u; ϵ′) ∩ Lp(g) = ∅. Then u ∈ (Bρ(u; ϵ

′) ∩ Lp(Γ)) ⊂ Dp(Γ), andthis contradicts the fact u is an end point.

The second statement can be proved similarly. �

Proposition 3. A vertex of Dp(Γ) is an end point of two sides.

Proof. Let u ∈ ∂Dp(Γ) be a vertex and Lp(T1) ∩ Lp(T2) = {u} for somedistinct T1, T2 ∈ Γ − {Id}. Since {Lp(g) | g ∈ Γ − {Id}} is locally finite,only finitely many Lp(g) go through the point u.

u

ρ

p 1

p 2

p 3g

u

p 1

p 2

p 3g

Fig. 9:

Let us call them Lp(g1), Lp(g2), ..., Lp(gl) where g1 = T1, g2 = T2 (Fig.9). Again since {Lp(g) | g ∈ Γ− {Id}} is locally finite, for sufficiently small

ϵ > 0, the situation Bρ(u; ϵ) ∩ Lp(g) = ∅ is satisfied by g1, ..., gl ∈ Γ. Nowsince l ≥ 2, u is an end point of two sides.

19

�

Proposition 4. Each point z ∈ ∂Dp(Γ) is a vertex or an inner point of theside.

Proof. Since z ∈ ∂Dp(Γ), there exists T ∈ Γ − {Id} such that z ∈ Lp(T ).If such T exists more than two, z is a vertex. If such T exists only one,for sufficiently small ϵ > 0, only Lp(T ) intersect with Bρ(z; ϵ). So z ∈(Lp(T ) ∩Bρ(z; ϵ)) ⊂ Dp(Γ), and then z is an inner point of the side.

p

p

ρT

z

z

Fig. 10:

�

Corollary 3. If Dp(Γ) is compact, then Dp(Γ) is a convex hyperbolic polygonbounded by finitely many sides.

Definition 16. (congruent)We shall say that two points of H, z1 and z2, are congruent if z2 =

T (z1) for some T ∈ Γ.

Remark 7. Since Dp(Γ) is a fundamental domain, if two distinct pointsz1, z2 ∈ Dp(Γ) are congruent, z1, z2 ∈ ∂Dp(Γ).

Proposition 5. For arbitrary z ∈ ∂Dp(Γ), there exists some T ∈ Γ− {Id}such that T (z) ∈ ∂Dp(Γ).

Proof. If z ∈ ∂Dp(Γ) is an elliptic vertex of order 2, this statement is obvi-ously true. So let us remove this case. By the property of Dp(Γ) as a funda-mental domain and the fact that the Dirichlet tessellation {T (Dp(Γ)) | T ∈Γ} is locally finite, only finitely many elements of T ∈ Γ−{Id}, say T1, ..., Tl

satisfy the condition Bρ(z; 1)∩Ti(Dp(Γ)) = ∅ (i = 1, ..., l) . For each natural

20

number n, choose gn ∈ {T1, ..., Tl} such that Bρ(z; 1/n) ∩ gn(Dp(Γ)) = ∅,and pick up a point zn from this intersection. Then the sequence {zn} con-verges to z. Since some Tk is infinitely contained in the sequence {gn} as aterm, the subsequence {znν} ⊂ Bρ(z; 1/n) ∩ Tk(Dp(Γ)) also converges to z.Since Tk(Dp(Γ)) is a closed set, then z ∈ Tk(Dp(Γ)) and so T−1

k (z) ∈ Dp(Γ).If T−1

k (z) ∈ Dp(Γ)◦, then z ∈ Tk(Dp(Γ)

◦) = (Tk(Dp(Γ)))◦. So for some

ϵ > 0, Bρ(z; ϵ) ⊂ (Tk(Dp(Γ)))◦. On the other hand, Bρ(z; ϵ) ∩Dp(Γ)

◦ = ∅because z ∈ ∂Dp(Γ). Then (Tk(Dp(Γ)))

◦ ∩Dp(Γ)◦ = ∅, a contradiction. So

T−1k (z) ∈ ∂Dp(Γ). �

Theorem 15. (side paring)For an arbitrary side s1 of Dp(Γ), there exists a side s2 and a transfor-

mation T ∈ Γ− {Id} uniquely such that s2 = T (s1).

Proof. Since s1 is a side, there exists a unique T ∈ Γ such that the in-tersection of Lp(T ) with Dp(Γ) is s1. Suppose z is an inner point of theside s1, then ρ(p, z) = ρ(T (p), z) and so T−1(z) ∈ Lp(T

−1). Since z ∈∩g∈Γ−{Id,T}Hp(g)

◦, ρ(p, z) < ρ(g(p), z) for all g ∈ Γ − {Id, T}. Then

T−1(z) ∈∩

h∈Γ−{Id,T−1}Hp(h)◦. So T−1(z) ∈ Lp(T

−1)∩Dp(Γ), and T−1(z)

is an inner point of the side Lp(T−1) ∩Dp(Γ).

Since {Lp(g) | g ∈ Γ − {Id}} is locally finite, we can choose ϵ > 0such that Lp(T ) is the only perpendicular bisector intersecting Bρ(z; ϵ) andLp(T

−1) is the only perpendicular bisector intersecting Bρ(T−1(z); ϵ). Since

T is a homeomorphism and Dp(Γ) is a fundamental domain, the shaded partof Bρ(T

−1(z); ϵ) is moved to the shaded part of Bρ(z; ϵ) (Fig.11). So suchT cannot exists more than two, then the theorem follows.

z

-1

ρ ρ-1

p p-1

p

T

Fig. 11:

21

�

Corollary 4. Vertices are only congruent to vertices, while inner points ofsides are only congruent to inner points of sides.

Proof. We can find this from the proof of Theorem 15. �

Corollary 5. If the Dirichlet domain Dp(Γ) is compact, Dp(Γ) is a convexhyperbolic polygon whose number of sides is even.

Definition 17. (cycle)Choose a vertex or an ideal vertex u of Dp(Γ) and we shall call the set

of all vertices of Dp(Γ) congruent to u a cycle including u.

Remark 8. If the two vertices u, v are in the same cycle, each hyperbolicdistance from p to each point is the same, i.e.,

ρ(p, u) = ρ(p, v)

by Remark 5.

Definition 18. (elliptic vertex, elliptic cycle)If a vertex u ∈ ∂Dp(Γ) is a fixed point of some elliptic element of Γ,

u is called an elliptic vertex, and the cycle including such u is called anelliptic cycle.

Remark 9. Every element of an elliptic cycle is an elliptic vertex.

Theorem 16. There is a one-to-one correspondence between the ellipticcycles of Dp(Γ) and the conjugacy classes of non-trivial maximal ellipticcyclic subgroups of Γ.

Proof. For an elliptic cycle C, choose an arbitrary vertex u ∈ C. Then thestabilizer Stab(u) = {T ∈ Γ | T (u) = u} is an elliptic cyclic subgroup of Γ,and the Γ-conjugacy class of Stab(u) corresponds to C. �

Definition 19. (accidental vertex)If a vertex u ∈ ∂Dp(Γ) is not a fixed point of Γ, u is called an accidental

vertex.

Theorem 17. Let F be a Dirichlet domain for Γ. Let {v1, ..., vt} be allcongruent vertices of F , and θ1, ..., θt be the internal angles of them. Letm be the order of the stabilizer of v1 in Γ of one of these vertices. Thenθ1 + θ2 + ...+ θt = 2π/m.

22

Remark 10. 1. As F is locally finite, there are only finitely many ver-tices in a cycle.

2. As the stabilizers of two points in a congruent set are conjugate sub-groups of Γ, they have the same order.

3. If a vertex is not a fixed point, we have m = 1 and θ1 + ...+ θt = 2π.

Proof. (of Theorem 17)Let

H = {Id, S, S2, ..., Sm−1}

be the stabilizer of v1 in Γ. Then each Sr(F )(0 ≤ r ≤ m − 1) has a vertexat v1 whose angle is θ1. For each k ∈ {1, ..., t} there exist Tk such thatTk(vk) = v1. Then the set of all elements which map vk to v1 is HTk, a cosetwhich has m elements, so the SrTk(F ) have v1 as a vertex with an angle ofθk. On the other hand, if a domain A(F )(A ∈ Γ) has v1 as a vertex, thenA−1(v1) ∈ F , hence A−1(v1) = vi for some i, 1 ≤ i ≤ t. Thus A ∈ HTi, andA(F ) has been included in the above description. So we have mt domainssurrounding v1. These domains are distinct, for if SrTk(F ) = SqTl(F ), thenSrTk = SqTl, and hence k = l and r = q. We conclude then that

m(θ1 + ...+ θt) = 2π.

�

Theorem 18. (generators)Let {Ti} be the subset of Γ consisting of those elements which pair the

sides of some fixed Dirichlet domain F . Then {Ti} is a set of generators forΓ (see [4], Theorem 3.5.4).

23

4 The Siegel’s theorem and cofinite Fuchsian groups

Definition 20. (cofinite)A Fuchsian group Γ is called cofinite if there exists a Dirichlet domain

Dp(Γ) whose hyperbolic area is finite.

Remark 11. If Γ is confinite and µ(Dp(Γ)) < ∞, then Theorem 11 impliesthat for every fundamental domain F for Γ such that µ(∂F ) = 0,

µ(F ) = µ(Dp(Γ)).

Theorem 19. (Siegel’s Theorem)([4], [7] §3.4, Theorem 1 )If Γ is cofinite then Dp(Γ) is a convex hyperbolic polygon.

Proof. Let us adopt the unit disk model. We shall prove that any Dirichletdomain F = Dp(Γ) has finitely many sides. Since the vertices of F areisolated, any compact subset K ⊂ H contains only finitely many vertices.This takes care of the case in which F is compact. Now suppose that F isnot compact.

Since F cannot has a free side, it will be enough to show that F hasonly finitely many vertices and ideal vertices. First, we shall prove thatthe number of ideal vertices is finite. Let us take any N ideal vertices andname them v1, ...vN in the positive direction. Now we join v1 to v2, v2 tov3,..., vN to v1 by means of geodesic segments, and we obtain a hyperbolicpolygon F1 bounded by N geodesics and contained inside F such that itsideal vertices are v1, ...vN . Then we can divide F1 into N − 2 hyperbolictriangles by joining v1 to v3, ..., vN−1 through geodesic segments (see Fig.12). By Theorem 6 (Gauss-Bonnet) the hyperbolic area of each triangle isπ, and so

(N − 2)π = µ(F1) ≤ µ(F ) < ∞.

Thus N is bounded from above, and the assertion follows.Now we shall prove that F has finitely many vertices. The main ingre-

dient of the proof is an estimation of the angles ω at vertices of the domainF . More precisely, we are going to prove that∑

ω

(π − ω) ≤ µ(F ) + 2π (3)

where the sum is taken over all vertices of F . Let t ∈ ∂F be a vertex. Itfollows that if, beginning at t, we traverse the potion of the boundary of Fin a definite direction, we encounter;

24

p

F1

Fig. 12:

1. a finite number of vertices of F and then the rim of the unit disk, or

2. an infinite sequence of vertices, or

3. finitely many vertices and then the starting point t.

We call the connected portion of the boundary of F on both sides of tthe boundary component R determined by t. Let ..., z−1, z0, z1, ... denotethe sequence where z0 = t, finite or infinite to either side, of vertices of Fassociated with R traversed in the positive direction. In the third case Ris a closed hyperbolic polygon, in view of the convexity of F , encloses allthe Dirichlet domain F ; then F is surely compact, and of course, a convexhyperbolic polygon. From now on we can exclude the third case in our proof.

Now we join the vertices zk(k = 0,±1, ...) on R to the center p of F bymeans of geodesics. If zk, zk+1 are two successive vertices of F , then the threepoints zk, zk+1, p are vertices of a hyperbolic triangle ∆k with correspondingangles αk, βk, γk which, in view of the convexity of F , is entirely containedin F (see Fig. 13). If zk or zk+1 is an ideal vertex, then αk or βk is 0.

By Theorem 6 (Gauss-Bonnet), the hyperbolic area of the triangle ∆k is

µ(∆k) = π − αk − βk − γk(k = 0,±1, ...).

Summation over a connected portion of R yields

n∑k=m

µ(∆k) =

n∑k=m

(π − αk − βk − γk) (m < n). (4)

If zk is not an ideal vertex of F , then αk+βk−1 = ωk is the correspondingvertex angle of F and, in the view of the convexity of F , we have

0 < ωk = αk + βk−1 < π. (5)

25

pα

β

γk-1 k-1

k-1

α

β

k

k

γkΔk-1

Δk

kz

k-1z

k+1z

Fig. 13:

Upon introduction of the ωk, (4) becomes

n∑k=m

µ(∆k) +n∑

k=m

γk =n∑

k=m+1

(π − ωk) + π − αm − βn (m < n). (6)

Now we choose n as large as possible in the first case and let n go to∞ in the second case. Similarly, we choose m as small as possible in thefirst case and let m go to −∞ in the second case. Since all the triangles ∆k

belong to F and do not overlap,

n∑k=m

µ(∆k) ≤ µ(F ). (7)

Because the summands on the left are all positive, each of the infinite sumsassociated with the transitions to the limit converges, and the same is trueof

n∑k=m

γk ≤ 2π. (8)

By (5), αm < π, βn < π, so that the αm are bounded for m → −∞ and theβn are bounded for n → ∞. Again by (5), the term π−ωk of the sum on theright-hand side of (6) are all positive. This fact and the boundedness of itspartial sums (implied by (7) and (8)) assume its convergence for m → −∞

26

and n → ∞. Let us denote these limit α−∞ and β∞. On the other hand,if there is a minimal m or maximal n, then αm = 0 or βn = 0. Now weinvestigate the case when there is no maximal n and prove that, in thatcase,

β∞ ≤ π/2. (9)

Let rk = ρ(p, zk). If zk is on Lp(Tk), then, we have the inequality

2rk ≥ ρ(p, Tk(p)). (10)

p

zk+1

zk

T(p)k

p k

Fig. 14:

Since the Tk(p) have no limit point in U , |Tk(p)| → 1 for k → ∞and, as a result, rk → ∞. This means that rk+1 > rk for infinitely manynatural numbers k. We now show that in a hyperbolic triangle, too, thelarger angle is opposite the larger side, so that αk > βk. To this end, weconstruct the perpendicular bisector M of the side from zk to zk+1. Sinceρ(p, zk+1) > ρ(p, zk), zk and p lie on the same side of M , and, consequently,M meets the side of the triangle joining p to zk+1 at an interior point η (seeFig. 15). Reflection in M carries the hyperbolic triangle with vertices zk,zk+1, η into itself so that the angle λ at the vertex zk is equal to the angleβk at zk+1.

On the other hand, we have λ < αk, and so αk > βk. Since αk+βk+γk <π, we have, surely, 2βk < αk+βk < π, and βk is an acute angle for infinitelymany natural k. This implies, for m → ∞, the assertion (9). If there is nominimal m, then we show, analogously, that rk > rk+1 for infinitely manynegative k, and this implies βk > αk, 2αk < π and

α−∞ ≤ π/2.

27

p

zk+1

zk

M

η

λ

βk

αk

Fig. 15:

If we use n not only as a maximal n but also as ∞, and m not only as aminimal m but also as −∞, we have, invariably,

π − αm − βn ≥ 0,

and (6) becomes

n∑k=m+1

(π − ωk) ≤n∑

k=m

µ(∆k) +n∑

k=m

γk

Now we consider the finite or countably infinite class of boundary compo-nents of F . Since each vertex ω of F is included in one of the three casesand all the triangles ∆ω belong to F and do not overlap, then∑

ω

(π − ω) ≤∑ω

µ(∆ω) +∑ω

γω ≤ µ(F ) + 2π.

where the sum is taken over all vertices of F . So (3) follows.Now we are going to prove, using this estimate, that the number of

vertices which lie a finite distance from the point p is finite. Let a be avertex and a(1) = a, a(2),...,a(n) all vertices congruent to a. If we denote theangle at vertex a(i) by ω(i), we have by theorem:

ω(1) + ω(2) + ...+ ω(n) = 2π, (11)

if a is not a fixed point for any T ∈ Γ− {Id}; and

ω(1) + ω(2) + ...+ ω(n) = 2π/m, (12)

if a is a fixed point of order m ≥ 2. Since ω(i) < π for each cycle of the type(11), we have n ≥ 3, and hence

n∑i=1

(π − ω(i)) = (n− 2)π > π. (13)

28

Comparing (13) with (3) we conclude that the number of cycles where a isnot a fixed point for any T ∈ Γ − {Id} is finite. For each cycle of the type(12) we have

n∑i=1

(π − ω(i)) = (n− 2/m)π > π/3. (14)

Comparing (14) with (3) we conclude that the number of elliptic cycles oforder≥ 3 is finite. Finally, any elliptic fixed point of order 2 is also finite.Thus we have proved that F has only finitely many vertices, and the theoremfollows. �

Definition 21. (parabolic vertex, parabolic cycle)If an ideal vertex u ∈ ∂0Dp(Γ) is a fixed point of some parabolic element

of Γ, u is called a parabolic vertex, and the cycle including such u is calleda parabolic cycle.

Theorem 20. If Γ is cofinite and has non-compact Dirichlet domain Dp(Γ),then ideal vertices of Dp(Γ) are parabolic vertices.

Theorem 21. If Γ is cofinite there is a one-to-one correspondence betweenthe parabolic cycles of Dp(Γ) and the conjugacy classes of non-trivial maxi-mal parabolic cyclic subgroups of Γ.

Proof. For a parabolic cycle C, choose an arbitrary vertex u ∈ C. ThenStab(u) is a parabolic cyclic subgroup of Γ, and the Γ-conjugacy class ofStab(u) corresponds to C. �

By Theorem 16 and Theorem 21, we can see the number of elliptic cyclesand parabolic cycles is invariant under Γ.

Definition 22. (signature)Let Γ be a cofinite Fuchsian group, and let g be the genus of the Rie-

mannian surface H/Γ (∼= Dp(Γ)/Γ). Let n be the sum of the number ofelliptic cycles and parabolic cycles of Γ, and m1, ...,mn is the order of thestabilizer of any element of each cycle. Then we say that Γ has signature(g;m1, ...,mn).

29

5 Triangle groups

5.1 Dirichlet domains for cofinite Fuchsian groups

Theorem 22. ([1], Theorem 10.5.1)Let Γ be a cofinite Fuchsian group with signature (g;m1, ...,mn), and let

Dp(Γ) be the Dirichlet polygon with center p with N sides. Then

1. N ≤ 12g + 4n− 6.

2. If n = 0, then N ≥ 4g.

3. If n ≥ 1, then N ≥ 4g + 2n− 2.

Proof. Suppose that Dp(Γ) has elliptic or parabolic cycles C1, ..., Cn andaccidental cycles Cn+1, ..., Cn+A: either (but not both) of these sets of cyclesmay be absent. In general, we let |C| denote the number of points in thecycle C.

Now

|Cj | ≥ 1 if 1 ≤ j ≤ n; |Cj | ≥ 3 if n < j ≤ n+A,

andn+A∑j=1

|Cj |.

Thus0 ≤ A ≤ (N − n)/3.

Euler’s formula yields

2− 2g = 1− (N/2) + n+A (15)

and the inequalities in 1 and 3 follow by eliminating A. The inequality in2 follows from (15) by putting n = 0 and observing that as n = 0, we haveA ≥ 1. �

In particular, for a cofinite Fuchsian group Γ with signature (0;m1,m2,m3),the number of sides of Dp(Γ) is 6 or 4.

30

5.2 Triangle groups

Theorem 23. (Poincare theorem)([3], THEOREME)Let P be a finite sided hyperbolic polygon of finite area. If every vertex of

P has an interior angle of the form 2πk , 2 ≤ k ≤ ∞, and is not a hyperbolic

fixed point, then the set of side pairing transformations generate a Fuchsiangroup Γ and P is a fundamental polygon of Γ.

Triangle Fuchsian group([4], Section 4.4)Let mi be a positive integer or ∞ (i = 1, 2, 3) such that

1

m1+

1

m2+

1

m3< 1,

and△ be a hyperbolic triangle with vertices v1, v2, v3, angles π/m1, π/m2, π/m3 at these vertices, and sides M1,M2,M3 opposite to these vertices. LetRi be the hyperbolic reflection in the geodesic containing Mi (i = 1, 2, 3).By means of Poincare theorem, the side pairing transformations RiRj andRiRk of △ ∪ Ri(△) ({i, j, k} = {1, 2, 3}) generate a Fuchsian group with afundamental polygon △∪Ri(△), which is called a triangle group.

It is known that a cofinite Fuchsian group is with signature (0;m1,m2,m3)if and only if it is a triangle group. ([1], Theorem 10.6.4.)

M

M

1

M23

πm1

πm2π

m3

Fig. 16:

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5.3 Main theorem

Theorem 24. Let △ be a hyperbolic triangle with vertices v1, v2, v3, interiorangles π/m1, π/m2, π/m3 at these vertices, and sides M1,M2,M3 opposite tothese vertices. Let Ri be the hyperbolic reflection in the geodesic containingMi (i = 1, 2, 3) and consider the triangle group Γ generated by R1R2, R2R3

and R3R1 (This group is the same to the group generated by two transforma-tions RiRj and RjRk, i, j, k = {1, 2, 3}). Then the Dirichlet domain Dp(Γ)of Γ centered at p is determined as follows:

1. If p is on the interior of Mi, then Dp(Γ) is a quadrilateral; moreprecisely

Dp(Γ) = Hij ∩Hji ∩Hik ∩Hki

where {i, j, k} = {1, 2, 3}, and Hij denotes Hp(RiRj).

2. If p is in the interior of ∆, then Dp(Γ) is a hexagon; more precisely

Dp(Γ) = H12 ∩H21 ∩H23 ∩H32 ∩H31 ∩H13.

πm1

πm2π

m3

p

12

31

21

32

13

Δ

πm1

πm2π

m3

πm3

πm2

πm1

Δ

L

LL

L21

12

13

31

p

Fig. 17:

Proof. Since △∪ Ri(△) is a fundamental domain for Γ, for every Dirichletdomain, µ(Dp(Γ)) = µ(△∪Ri(△)) = 2µ(△) by Theorem 5.(about Case 1) We can find easily that

Dp(Γ) ⊂ Hij ∩Hji ∩Hik ∩Hki.

Hij ∩Hji ∩Hik ∩Hki is a quadrilateral bounded by Lij , Lji, Lik, Lki whereLij denotes Lp(RiRj). On the other hand,

Hij ∩Hji ∩Hik ∩Hki = △∪Ri(△),

32

soµ(Hij ∩Hji ∩Hik ∩Hki) = µ(△) + µ(Ri(△)) = 2µ(△).

ThenDp(Γ) = Hij ∩Hji ∩Hik ∩Hki.

(about Case 2) First, we shall consider the case that △ is contained inH. We can find easily that

Dp(Γ) ⊂ H12 ∩H21 ∩H23 ∩H32 ∩H31 ∩H13.

H12∩H21∩H23∩H32∩H31∩H13 is a hexagon bounded by L12, L21, L23, L32, L31,L13 where Lij denotes Lp(RiRj). At the same time, we can find that theangles of the hexagon is like a figure in Fig. 18.

p

12

31

21

32

13

θθ

4

5

6

4

5

6θ

θ

Fig. 18:

Proposition 6. (Ceva’s Theorem for Hyperbolic Triangles ([5], p.81))If X is a point not on any side of a hyperbolic triangle ABC such that

AX and BC meet in D, BX and CA in E, and CX and AB in F (Fig.19),

sinhAF

sinhFB

sinhBD

sinhDC

sinhCE

sinhEA= 1.

Conversely, if D lies on the side BC, E on CA and F on AB such that theabout equation holds, then AD, BE and CF are concurrent.

33

A

BC

D

EF

x

Fig. 19:

Let us extend geodesic segments [v1, p], [v2, p], [v3, p] and name pointsA,B,C,D,E, F like a left figure in Fig. 20. Then we shall divide angles∠A,∠B,∠C like a right figure in Fig. 20 and name points D′, E′, F ′.

pθ

θ4

5

6θθ 4

5

6

D

EF

A

2

B

3

C2

B

A

3

C

Fig. 20:

By the Sine rule in △ACF , △BCF , △BAD, △CAD, △CBE, △ABE,

sinhAF

sin θ6=

sinhCF

sin∠A ,sinhFB

sin θ5=

sinhCF

sin∠B ,sinhBD

sin θ2=

sinhAD

sin∠B ,

sinhDC

sin θ1=

sinhAD

sin∠C ,sinhCE

sin θ4=

sinhBE

sin∠C ,sinhEA

sin θ3=

sinhBE

sin∠A ,

and thensinhAF

sinhFB

sinhBD

sinhDC

sinhCE

sinhEA=

sin θ2 sin θ4 sin θ6sin θ1 sin θ3 sin θ5

.

Sosin θ2 sin θ4 sin θ6sin θ1 sin θ3 sin θ5

= 1 (16)

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by Ceva’s theorem. Similarly

sinhAF ′

sinhF ′B

sinhBD′

sinhD′C

sinhCE′

sinhE′A=

sin θ1 sin θ3 sin θ5sin θ2 sin θ4 sin θ6

.

So by (16)sinhAF ′

sinhF ′B

sinhBD′

sinhD′C

sinhCE′

sinhE′A,= 1

and so AD′, BE′ and CF ′ are concurrent by Ceva’s theorem.

θθ 4

5

6

A

3

2

B

Cp

12

31

21

32

13

θθ

4

5

6

4

5

6θ

θ

Fig. 21:

Then the area of (H12∩H21 ∩H23∩H32 ∩H31∩H13)−△ (a left picturein Fig. 21) is equal to that of △ (a right figure in Fig. 21) because of thecongruence of triangles. So µ(H12 ∩H21 ∩H23 ∩H32 ∩H31 ∩H13) = 2µ(△)and then Dp(Γ) = H12 ∩H21 ∩H23 ∩H32 ∩H31 ∩H13.

Now we shall consider the case that △ has ideal vertices. We cannotuse Ceva’s theorem directly because the hyperbolic length of the side oneof whose end point is an ideal vertex is infinite. Let us adopt the unit diskmodel. For each j ∈ {1, 2, 3}, we shall consider the sequence {vjn} such thatvjn converges to vj and the hyperbolic triangle △n = △v1nv2nv3n includesp as an interior point (Fig. 22).

Then △n converges to △ and so each interior angle of △n converges toeach interior angle of △. So µ(△n) also converges to µ(△) by Theorem 6(Gauss-Bonnet). On the other hand, since Linjn converges to Lij where i, j ∈{1, 2, 3}, i = j, Linjn∩Linkn converges to Lij∩Lik where i, j, k ∈ {1, 2, 3}, i =j = k. So the hexagon Hn = H1n2n ∩H2n1n ∩H2n3n ∩H3n2n ∩H3n1n ∩H1n3n

converges to the hexagon H = H12 ∩ H21 ∩ H23 ∩ H32 ∩ H31 ∩ H13 andso µ(Hn) converges to µ(H ) by Theorem 6 (Gauss-Bonnet). On the other

35

v

v

v12

3

L12

L13

L23

L32 L31

L21

v1n v2n

v3n

L nn

L nn

L nn

L nn

L nnL nn

212

1323

32p p

Fig. 22:

hand, µ(Hn) = 2µ(△n) for each n, and then it must be µ(H ) = 2µ(△). SoDp(Γ) = H .

�

Finally we remark that for an arbitrary point p ∈ H, there exists g ∈ Γsuch that g(p) ∈ △ ∪ Ri(△) because △ ∪ Ri(△) is a fundamental polygonfor Γ. If g(p) is in the interior of △ or Ri(△), Dp(Γ) is a hexagon, and ifg(p) is on the interior of the side of △ or Ri(△), Dp(Γ) is a quadrilateral.

5.4 Generalizations

We can extend this main theorem to other geometries; Euclidean and spher-ical triangle groups ([5]). We will try to extend to higher dimensional cases;Simplex reflection groups ([6]).

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Acknowlegements

I would like to thank my supervisor Associate Prof. Yohei Komori forhaving suggested this topic, for his sensible advice and comments.

My thanks also go to Prof. Yoichi Imayoshi for listening to my talk inthe weekly seminal and advising me.

I am also grateful to Katsuya Shakudo for listening to my weekly talkand advising me.

References

[1] Alan F. Beardon, The Geometry of Discrete Groups, SpringerGTM 91.

[2] Francis Bonahon, Low-Dimensional Geometry, AMS,STML/IAS/PARK CITY 49.

[3] Georges de Rham, Sur les polygones generateurs de groupes fuch-siens, Enseign. Math, 17, 1971, 49-61. Quart. J. Math. Oxford Ser.(2) 33, 1982, 451-461.

[4] Svetlana Katok, Fuchsian Groups, Chicago Lectures in Mathemat-ics.

[5] Minoru Nakaoka, Introduction to hyperbolic geometry (Japanese),Science-sha.

[6] John G. Ratcliffe, Foundations of Hyperbolic Manifolds, SpringerGTM 149.

[7] C.L.Siegel, Topics in complex function theory, Volume 2, Wiley-Interscience 25.

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