The Arithmetic of Quaternion Algebra

90 CHAPTER 4. APPLICATIONS TO ARITHMETIC GROUPS
3. The group ¯ Γ is of finite type, and generated by the homographies {¯gi, 1 ≤
i ≤ n}. It comes from that {¯gF |¯g ∈ ¯ Γ} forms a pavement of H. If ¯g ∈ ¯ Γ, it
exists ¯g ′ belonging to the group generated by these ¯gi such that ¯gF = ¯g ′ F ,
hence ¯g = ¯g ′ . by Using again an argument of pavement we see:
4. A cycle of F being a equivalent class of vertices of F in H ∪ R ∪ ∞ modulo
¯Γ,The sum of the angles around the vertices of a cycle is of the form 2π/q
where q is an integer great than 1, or q = ∞.
Definition 4.7. A cycle is said to be
hyperbolic if q = 1,
elliptic of order q if q > 1, q �= ∞,
parabolic if q = ∞.
The angle 2π/q is the angle of cycle. eq denotes the number of cycles of angle
2π/q.
Definition 4.8. A point of H ∪ R ∪ ∞ is said to be elliptic of order q (rep.
parabolic or a point) for ¯ Γ if it is a double point of an elliptic homography of
order q (resp. parabolic) of ¯ Γ.
It is easy to show that the elliptic cycles of order q constitute a system of
representatives modulo ¯ Γ of the elliptic points of order q. It is the same for
the parabolic points . The interior of F contains no any elliptic point, no any
parabolic. The union of H and the points of ¯ Γ is denoted by H ∗ .
Searching cycles. We look for the cycles in such a way: Let A, B, C, ... be the
vertices of F in H ∗ when we run along the boundary of F in a sense given in
advance. In order to find the cycle of A, we run along the edge AB = C1 and
then the edge which is congruent to A ′ B ′ = g1(C1) in the chosen sense. It
remains that B ′ = A2, and runs along the next edge C2, then the edge which
is congruent to g2(C2) with its end point A3,... till to that when we find again
A = Am. Integer m is the length of the cycle.
EXAMPLE:
1)The fundamental domain of modular group P SL(2, Z):
A picture here !!!
a point {∞}, a cycle {A, C} of order 3, a cycle B of order 2. The group is
generated by the homographies z ↦→ z + 1 and z ↦→ −1/z.
2)
A picture here!!!
In the example given by this figure, we have two points {A, B, E} and ∞,
and two cycles of order 2: {B}, {D}.
Lemma 4.2.9. The number of elliptic cycles of order q is equal to the half
of the number of conjugate classes of Γ of the characteristic polynomial X 2 −
2 cos(2π/aq)X + 1, where a is the index of the center in Γ.
Proof. The two numbers defined in (1), (2) are equal to eq:
(1) The number of the equivalent classes modulo ¯ Γ of the set Eq = {z ∈
H| elliptic of order q} = {z ∈ H| ¯ Γz is cyclic of order q}, where ¯ Γz is the isotropic
group of z in ¯ Γ.