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Deﬁnition. Let X = H∗ /Γ. A point x = Γ · τ is called an elliptic point of order 2
(resp, of order 3) if τ ∈ Γ(1) · i (resp. τ ∈ Γ(1) · ρ) and Γτ = 1.
Theorem 8.4. The genus of H∗ /Γ is equal to
g =1+ µΓ
r2
r3
r∞
−
−
−
,
12
4
3
2 where µΓ is the index of Γ/Γ ∩ (±1) in Γ(1)/(±1), r2 is the number of elliptic points of
Γ of order 2, r3 is the number of elliptic points of Γ of order 3, and r∞ is the number
of cusps of Γ.
Proof. Notice ﬁrst that the number µΓ is equal to the degree of the meromorphic
function X (Γ) → X (Γ(1)) ∼ P1 (C) deﬁned by the j function j : H → C. In fact, the
=
number of the points in the preimage of a general z ∈ C is equal to the number of
Γorbits in H contained in a Γ(1)orbit. Applying (8.22), we have
X
2g − 2 = −2µ +
(ex (j ) − 1) =
x∈X −2µ + X X (ex (j ) − 1) + j (x)=j (i) (ex (j ) − 1) + X (ex (j ) − 1). j (x)=∞ j (x)=j (ρ) We have (µ − r2 )/2 points over j (i) with ex (j ) = 2 and (µ − r3 )/3 points over j (ρ)
with ex (j ) = 3. Also by (8.21), the sum of indices of cusps is equal to µ. This gives
2g − 2 = −2µ + (µ − r2 )/2 + 2(µ − r3 )/3 + (µ − r∞ ),
hence
g =1+ µ
r2
r3
r∞
−
−
−
.
12
4
3
2 We shall concentrate on the special subgroups Γ of Γ(1) introduced earlier. They
are the principal congruence subgroup Γ(N ) of level N and
„
«
αβ
Γ0 (N ) = {
∈ SL(2, Z) : N γ }.
γδ
Obviously
Γ(N ) ⊂ Γ0 (N ).
Lemma 8.5. ( µ0,N := µΓ0 (N ) 1
N3
2 Q pN (1 − p−2 ) if N > 2,
if N = 2,
Y
= [Γ(1) : Γ0 (N )] = N
(1 + p−1 ), µN := µΓ(N ) = 6 (8.23) pN where p denotes a prime number.
Proof. This easily follows from considering the action of the group SL(2, Z/N ) on the
set (Z/N )2 . The isotropy subgroup of the vector (1, 0) is isomorphic to the group
„
«
a
b
of Γ0 (N )/Γ(N ) ⊂ SL(2, Z/N ). It consists of matrices of the form
. The
0 a−1
number of invertible elements a in the ring Z/N is equal to the value of the Euler ...
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This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.
 Fall '09
 ONTONKONG

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