8Modular forms for subgroups of SL2(ℤ)

III Modular Forms and L-functions

8.1 Definitions
For the rest of the course, we are going to look at modular forms defined on
some selected subgroups of SL
2
(Z).
We fix a subgroup Γ
Γ(1) of finite index. For Γ(1), we defined a modular
form to a holomorphic function
f
:
H C
that is invariant under the action of
Γ(1), and is holomorphic at infinity. For a general subgroup Γ, the invariance
part of the definition works just as well. However, we need something stronger
than just holomorphicity at infinity.
Before we explain what the problem is, we first look at some examples. Recall
that we write
¯
Γ for the image of Γ in PSL
2
(Z).
Lemma.
Let Γ
Γ(1) be a subgroup of finite index, and
γ
1
, ··· , γ
i
be right
coset representatives of
¯
Γ in Γ(1), i.e.
Γ(1) =
d
a
i=1
¯
Γγ
i
.
Then
d
a
i=1
γ
i
D
is a fundamental domain for Γ.
Example. Take
Γ
0
(p) =

a b
c d
SL
2
(Z) : b 0 (mod p)
Recall there is a canonical map
SL
2
(
Z
)
SL
2
(
F
p
) that is surjective. Then Γ
0
(
p
)
is defined to be the inverse image of
H =

a 0
b c

SL
2
(F
p
).
So we know
(Γ(1) : Γ
0
(p)) = (SL
2
(F
q
) : H) =
|SL
2
(F
q
)|
|H|
= p + 1,
where the last equality follows from counting. In fact, we have an explicit choice
of coset representatives
SL
2
(F
p
) =
a
bF
p
0
∗ ∗
1 b
0 1
a
0
∗ ∗
1
1
Thus, we also have coset representatives of Γ
0
(p) by
T
b
=
1 b
0 1
: b F
p
S =
0 1
1 0

.
For example, when
p
= 3, then
b {
0
,
1
,
+1
}
. Then the fundamental domain
is
D
T
1
DT
1
D
SD
So in defining modular forms, we’ll want to control functions as
z
0 (in
some way), as well as when
y
. In fact, what we really need is that the
function has to be holomorphic at all points in
P
1
(
Q
) =
Q {∞}
. It happens
that in the case of Γ(1), the group Γ(1) acts transitively on
Q {∞}
. So by
invariance of
f
under Γ(1), being holomorphic at
ensures we are holomorphic
everywhere.
In general, we will have to figure out the orbits of
Q{∞}
under Γ, and then
pick a representative of each orbit. Before we go into that, we first understand
what the fundamental domain of Γ looks like.
Definition (Cusps). The cusps of Γ (or
¯
Γ) are the orbits of Γ on P
1
(Q).
We would want to say a subgroup of index
n
has
n
many cusps, but this
is obviously false, as we can see from our example above. The problem is that
we should could each cusp with “multiplicity”. We will call this the width. For
example, in the fundamental domain above
D
p =
T
1
DT
1
D
SD
0
In this case, we should count
p
=
three times, and
p
= 0 once. One might
worry this depends on which fundamental domain we pick for Γ. Thus, we will
define it in a different way. From now on, it is more convenient to talk about
¯
Γ
than Γ.
Since
Γ(1)
acts on
P
1
(
Q
) transitively, it actually just suffices to understand
how to define the multiplicity for the cusp of
. The stabilizer of
in
Γ(1)
is
Γ(1)
=
±
1 b
0 1
: b Z
.
For a general subgroup
Γ Γ(1)
, the stabilizer of
is
Γ
=
Γ Γ(1)
. Then
this is a finite index subgroup of Γ(1)
, and hence must be of the form
Γ
=
±
1 m
0 1

for some m 1. We define the width of the cusp to be this m.
More generally, for an arbitrary cusp, we define the width by conjugation.
Definition
(Width of cusp)
.
Let
α Q {∞}
be a representation of a cusp of
Γ. We pick g Γ(1) with g() = α. Then γ(α) = α iff g
1
γg() = . So
g
1
Γ
α
g = (g
1
Γg)
=
±
1 m
α
0 1

for some m
α
1. This m
α
is called the width of the cusp α (i.e. the cusp Γα).
The g above is not necessarily unique. But if g
0
is another, then
g
0
= g
±1 n
0 ±1
for some n Z. So m
α
is independent of the choice of g.
As promised, we have the following proposition:
Proposition. Let Γ have ν cusps of widths m
1
, ··· , m
ν
. Then
ν
X
i=1
m
i
= (
Γ(1) :
¯
Γ).
Proof. There is a surjective map
π :
¯
Γ \ Γ(1) cusps
given by sending
¯
Γ · γ 7→
¯
Γ · γ().
It is then an easy group theory exercise that |π
1
([α])| = m
α
.
Example. Consider the following subgroup
Γ = Γ
0
(p) =

a b
c d
: c 0 (mod p)
.
Then we have
Γ(1) : Γ
0
(p)
=
Γ(1) : Γ
0
(p)
= p + 1.
We can compute
Γ
=
1,
1 1
0 1

= Γ(1)
.
So m
= 1. But we also have
Γ
0
=
1,
1 0
p 1

,
and this gives
m
0
=
p
. Since
p
+ 1 =
p
= 1, these are the only cusps of Γ
0
(
p
).
Likewise, for Γ
0
(p), we have m
= p and m
0
= 1.
Equipped with the definition of a cusp, we can now define a modular form!
Definition
(Modular form)
.
Let Γ
SL
2
(
Z
) be of finite index, and
k Z
. A
modular form of weight k on Γ is a holomorphic function f : H C such that
(i) f |
k
γ = f for all γ Γ.
(ii) f is holomorphic at the cusps of Γ.
If moreover,
(iii) f vanishes at the cusps of Γ,
then we say f is a cusp form.
As before, we have to elaborate a bit more on what we mean by (ii) and (iii).
We already know what it means when the cusp is
(i.e. it is Γ
). Now in
general, we write our cusp as Γα = Γg() for some g Γ(1).
Then we know
¯
Γ
α
= g
±
1 m
0 1

g
1
.
This implies we must have
1 m
0 1
or
1 m
0 1
g
1
Γ
α
g.
Squaring, we know that we must have
1 2m
0 1
g
1
Γ
α
g.
So we have
f |
k
g|
k
(
1 2m
0 1
) = f |
k
g.
So we know
(f |
k
g)(z + 2m) = (f |
k
g)(z).
Thus, we can write
f |
k
g =
˜
f
g
(q) =
X
nZ
(constant)q
n/2m
=
X
nQ
2mnZ
a
g,n
(f)q
n
,
where we define
q
a/b
= e
2πiaz/b
.
Then f is holomorphic at the cusp α = g() if
a
g,n
(f) = 0
for all n < 0, and vanishes at α if moreover
a
g,0
(f) = 0.
Note that if I Γ, then in fact
1 m
0 1
g
1
Γ
α
g.
So the q expansion at α is actually series in q
1/m
.
There is a better way of phrasing this. Suppose we have
g
(
) =
α
=
g
0
(
),
where g
0
GL
2
(Q)
+
. Then we have
g
0
= gh
for some h GL
2
(Q)
+
such that h() = . So we can write
h = ±
a b
0 c
where a, b, d Q and a, d > 0.
Then, we have
f |
k
g
0
= (f |
k
g)|
k
h
=
X
nQ
2mnZ
a
g,n
(f)q
n
|
k
±
a b
0 d
= (±i)
k
X
n
a
g,n
(f)q
an/d
e
2πbn/d
.
In other words, we have
f |
k
g =
X
n0
c
n
q
rn
for some positive r Q. So condition (ii) is equivalent to (ii’):
(ii’) For all g GL
2
(Q)
+
, the function f |
k
g is holomorphic at .
Note that (ii’) doesn’t involve Γ, which is convenient. Likewise, (iii) is
consider to
(iii’) f |
k
g vanishes at for all g GL
2
(Q)
+
.
We can equivalently replace GL
2
(Q)
+
with SL
2
(Z).
Modular form and cusp forms of weight
k
on Γ form a vector space
M
k
(Γ)
S
k
(Γ).
Recall that for Γ = Γ(1) = SL
2
(Z), we knew M
k
= 0 if k is odd, because
f |
k
(I) = (1)
k
f.
More generally, if
I
Γ, then
M
k
= 0 for all odd
k
. But if
I 6∈
Γ, then
usually there can be non-zero forms of odd weight.
Let’s see some examples of such things.
Proposition.
Let Γ
Γ(1) be of finite index, and
g G
=
GL
2
(
Q
)
+
. Then
Γ
0
=
g
1
Γ
g
Γ(1) also has finite index in Γ(1), and if
f M
k
(Γ) or
S
k
(Γ), then
f |
k
g M
k
0
) or S
k
0
) respectively.
Proof.
We saw that (
G,
Γ) has property (H). So this implies the first part. Now
if γ Γ
0
, then gγg
1
Γ. So
f |
k
gγg
1
= f f |
k
g|
k
γ = f |
k
g.
The conditions (ii’) and (iii’) are clear.
This provides a way of producing a lot of modular forms. For example, we
can take Γ = Γ(1), and take
g =
N 0
0 1
.
Then it is easy to see that Γ
0
= Γ
0
(
N
). So if
f
(
z
)
M
k
(Γ(1)), then
f
(
Nz
)
M
k
0
(
N
)). But in general, there are lots of modular forms in Γ
0
(
N
) that
cannot be constructed this way.
As before, we don’t have a lot of modular forms in low dimensions, and there
is also an easy bound for those in higher dimensions.
Theorem. We have
M
k
(Γ) =
(
0 k < 0
C k = 0
,
and
dim
C
M
k
(Γ) 1 +
k
12
(Γ(1) : Γ).
for all k > 0.
In contrast to the case of modular forms of weight 1, we don’t have explicit
generators for this.
Proof. Let
Γ(1) =
d
a
i=1
Γγ
i
.
We let
f M
k
(Γ),
and define
N
f
=
Y
1id
f |
k
γ
i
.
We claim that
N
f
M
kd
(Γ(1)), and
N
f
= 0 iff
f
= 0. The latter is obvious by
the principle of isolated zeroes.
Indeed, f is certainly holomorphic on H, and if γ Γ(1), then
N
f
|
k
γ =
Y
i
f |
k
γ
i
γ = N
f
.
As f M
k
(Γ), each f |
k
γ
i
is holomorphic at .
If k < 0, then N
f
M
kd
(Γ(1)) = 0. So f = 0.
If
k
0, then suppose
dim M
k
(
G
)
> N
. Pick
z
1
, ··· , z
N
D \ {i, ρ}
distinct. Then there exists 0 6= f M
k
(Γ) with
f(z
1
) = ··· = f (z
N
) = 0.
So
N
f
(z
1
) = ··· = N
f
(z
N
) = 0.
Then by our previous formula for zeros of modular forms, we know
N
kd
12
.
So dim M
k
(Γ) 1 +
kd
12
.
If k = 0, then M
0
(Γ) has dimension 1. So M
0
(Γ) = C.