MA342J: Introduction to Modular Forms
Tutorial 3, February 23
Modular forms of higher levels
1. Let
MN
na
=
c
b
d
a, b, c, d ∈ Z , ad − bc = N
o
be the set of integer matrices with determinant N . The group SL(2, Z)
acts on this space by multiplication from the left. Prove that every orbit
for its action has a unique representative of the form
a b
a, d > 0, ad = N, 0 ≤ b < d .
0 d
(It follows that there are exactly σ1 (N ) SL(2, Z)-orbits in MN .)
2. Prove that if f (z) ∈ Mk (Γ0 (M )) than f (N z) ∈ Mk (Γ0 (M N )).
Remark: One can use previous exercise to prove property (iii), i.e.
boundnes in all cusps. Namely, prove that if f is such that
f |k g = O(1) when Im(z) → ∞
for every g ∈ SL(2, Z), than for any m ∈ M
N thefunction f |k m also has
N 0
this property. To get f (N z) we take m =
.
0 1
3. For a subgroup Γ0 ⊂ Γ and a cusp [α] ∈ Γ0 \P 1 (Q) we defined the width
h([α]) to be the ramification index of the point [α] in the covering
X(Γ0 ) → X(Γ) .
One can prove the following formula for the width of a cusp:
h([α]) = [Stabα (Γ) : Stabα (Γ0 )] .
Question 1. Find Stab∞ (Γ).
Question 2. Prove that the r.h.s. of the above formula is independent of
the choice of α in the same Γ0 -orbit.
Question 3. Let Γ0 be a normal subgroup, e.g. Γ0 = Γ(N ). Prove that
all cusps have equal width.
In the case Γ0 = Γ(N ) one easily sees that h([∞]) = N . Since
(
Q
1 3
−2
) if N > 2
p|N (1 − p
2N
[Γ : Γ0 (N )] =
6
if N = 2
we conclude that this group has
(
Q
1 2
−2
N
)
p|N (1 − p
ν∞ = 2
3
1
if N > 2
if N = 2
cusps. Also we proved in class that this group has no elliptic points, i.e.
ν2 = ν3 = 0. Therefore we get from the genus formula that the modular
curve X(Γ0 (N )) for N > 3 has genus
g(N ) = 1 +
N −6 2Y
N
(1 − p−2 ) .
24
p|N
N
ν∞
g
2
3
0
3
4
0
4
6
0
5
12
0
6
12
1
7
24
3
8
24
5
2
9
36
10
10
36
13
11
60
26