8Modular forms for subgroups of SL2(ℤ)

III Modular Forms and L-functions

8.3 Examples of modular forms
We now look at some examples of (non-trivial) modular forms for different
subgroups. And the end we will completely characterize
M
2
0
(4)). This seems
like a rather peculiar thing to completely characterize, but it turns out this
understanding M
2
0
(4)) can allow us to prove a rather remarkable result!
Eisenstein series
At the beginning of the course, the first example of a modular form we had was
Eisenstein series. It turns out a generalized version of Eisenstein series will give
us more modular forms.
Definition (G
r,k
). Let k 3. Pick any vector r = (r
1
, r
2
) Q
2
. We define
G
r,k
(z) =
X
0
mZ
2
1
((m
1
+ r
1
)z + m
2
+ r
2
)
k
,
where
P
0
means we omit any m such that m + r = 0.
For future purposes, we will consider r as a row vector.
As before, we can check that this converges absolutely for
k
3 and
z H
.
This obviously depends only on r mod Z
2
, and
G
0,k
= G
k
.
Theorem.
(i) If γ Γ(1), then
G
r,k
|
k
γ = G
rγ,k
.
(ii) If Nr Z
2
, then G
r,k
M
k
(Γ(N)).
Proof.
(i) If g GL
2
(R)
+
and u R
2
, then
1
(u
1
z + u
2
)
k
|
k
g =
(det g)
k/2
((au
1
+ cu
2
)z + (bu
1
+ du
2
))
k
=
(det g)
k/2
(v
1
z + v
2
)
k
,
where v = n · g. So
G
r,k
|
k
γ =
X
0
m
1
(((m + r)
1
γ)z + ((m + r)γ)
2
)
k
=
X
m
0
1
((m
0
1
+ r
0
1
)z + m
0
2
+ r
0
2
)
k
= G
rγ,k
(z),
where m
0
= mγ and r
0
= rγ.
(ii)
By absolute convergence,
G
r,k
is holomorphic on the upper half plane.
Now if
Nr Z
2
and
γ
Γ(
N
), then
Nrγ Nr
(
mod N
). So
rγ r
(mod Z
2
). So we have
G
r,k
|
k
γ = G
rγ,k
= G
r,k
.
So we get invariance under Γ(
N
). So it is enough to prove
G
r,k
is holo-
morphic at cusps, i.e.
G
r,k
|
k
γ
is holomorphic at
for all
γ
Γ(1). So it is
enough to prove that for all r, G
r,k
is holomorphic at .
We can write
G
r,k
=
X
m
1
+r
1
>0
+
X
m
1
+r
1
=0
+
X
m
1
+r
1
<0
!
1
((m
1
+ r
1
)z + m
2
+ r
2
)
k
.
The first sum is
X
m
1
+r
1
>0
=
X
m
1
>r
1
X
m
2
Z
1
([(m
1
+ r
1
)z + r
2
] + m
2
)
k
.
We know that (
m
1
+
r
1
)
z
+
r
2
H
. So we can write this as a Fourier series
X
m
1
>r
1
X
d1
(2π)
k
(k 1)!
d
k1
e
2πr
2
d
q
(m
1
+r
1
)d
.
We now see that all powers of q are positive. So this is holomorphic.
The sum over m
1
+ r
1
= 0 is just a constant. So it is fine.
For the last term, we have
X
m
1
+r
1
<0
=
X
m
1
<r
1
X
m
2
Z
(1)
k
((m
1
r
1
)z r
2
m
2
)
k
,
which is again a series in positive powers of q
m
1
r
1
.
ϑ functions
Our next example of modular forms is going to come from
ϑ
functions. We
previously defined a ϑ function, and now we are going to call it ϑ
3
:
ϑ
3
(z) = ϑ(z) =
X
nZ
e
πin
2
z
= 1 + 2
X
n1
q
n
2
/2
.
We proved a functional equation for this, which was rather useful.
Unfortunately, this is not a modular form. Applying elements of Γ(1) to ϑ
3
will give us some new functions, which we shall call ϑ
2
and ϑ
4
.
Definition (ϑ
2
and ϑ
4
).
ϑ
2
(z) =
X
nZ
e
πi(n+1/2)
2
z
= q
1/8
X
nZ
q
n(n+1)/2
= 2q
1/8
X
n0
q
n(n+1)/2
ϑ
4
(z) =
X
nZ
(1)
n
e
πin
2
z
= 1 + 2
X
n1
(1)
n
q
n
2
/2
.
Theorem.
(i) ϑ
4
(z) = ϑ
3
(z ±1) and θ
2
(z + 1) = e
πi/4
ϑ
2
(z).
(ii)
ϑ
3
1
z
=
z
i
1/2
ϑ
3
(z)
ϑ
4
1
z
=
z
i
1/2
ϑ
2
(z)
Proof.
(i) Immediate from definition, e.g. from the fact that e
πi
= 1.
(ii)
The first part we’ve seen already. To do the last part, we use the Poisson
summation formula. Let
h
t
(x) = e
πt(x+1/2)
2
= g
t
x +
1
2
,
where
g
t
(x) = e
πtx
2
.
We previously saw
ˆg
t
(y) = t
1/2
e
πy
2
/t
.
We also have
ˆ
h
t
(y) =
Z
e
2πixy
g
t
x +
1
2
dx
=
Z
e
2πi(x1/2)y
g
t
(x) dx
= e
πiy
ˆg
t
(y).
So by the Poisson summation formula,
ϑ
2
(it) =
X
nZ
h
t
(n) =
X
nZ
ˆ
h
t
(n) =
X
nZ
(1)
n
t
1/2
e
πn
2
/t
= t
1/2
ϑ
4
i
t
.
There is also a
ϑ
1
, but we have
ϑ
1
= 0. Of course, we did not just invent
a funny name for the zero function for no purpose. In general, we can define
functions
ϑ
j
(
u, z
), and the
ϑ
functions we defined are what we get when we set
u = 0. It happens that ϑ
1
has the property that
ϑ
1
(u, z) = ϑ
1
(u, z),
which implies ϑ
1
(z) = 0.
We now see that the action of
SL
2
(
Z
) send us between
ϑ
2
,
ϑ
3
and
ϑ
4
, up to
simple factors.
Corollary.
(i) Let
F =
ϑ
4
2
ϑ
4
3
ϑ
4
4
.
Then
F (z + 1) =
1 0 0
0 0 1
0 1 0
F, z
2
F
1
z
=
0 0 1
0 1 0
1 0 0
F
(ii) ϑ
4
j
M
2
(Γ) for a subgroup Γ
Γ(1) of finite index. In particular,
ϑ
4
j
|
z
γ
is
holomorphic at for any γ GL
2
(Q)
+
.
Proof.
(i) Immediate from the theorem.
(ii)
We know
Γ(1)
=
hS, T i
, where
T
=
±(
1 1
0 1
)
and
S
=
±
0 1
1 0
. So by (i),
there is a homomorphism ρ : Γ(1) GL
3
(Z) and ρ(I) = I with
F |
2
γ = ρ(γ)F,
where
ρ
(
γ
) is a signed permutation. In particular, the image of
ρ
is finite,
so the kernel Γ = ker ρ has finite index, and this is the Γ we want.
It remains to check holomorphicity. But each
ϑ
j
is holomorphic at
.
Since
F |
2
γ
=
ρ
(
γ
)
F
, we know
ϑ
4
j
|
2
is a sum of things holomorphic at
,
and is hence holomorphic at .
It would be nice to be more specific about exactly what subgroup
ϑ
4
j
is
invariant under. Of course, whenever
γ
Γ, then we have
ϑ
4
j
|
2
γ
=
ϑ
4
j
. But in
fact ϑ
4
j
is invariant under a larger group.
To do this, it suffices to do it for
ϑ
4
=
ϑ
4
3
, and the remaining follow by
conjugation.
We introduce a bit of notation
Notation. We write
W
N
=
0 1
N 0
Note that in general, W
N
does not have determinant 1.
Theorem.
Let
f
(
z
) =
ϑ
(2
z
)
4
. Then
f
(
z
)
M
2
0
(4)), and moreover,
f |
2
W
4
=
f.
To prove this, we first note the following lemma:
Lemma. Γ
0
(4) is generated by
I, T =
1 1
0 1
, U =
1 0
4 1
= W
4
1 1
0 1
W
1
4
.
Four is special. This is not a general phenomenon.
Proof. It suffices to prove that Γ
0
(4) is generated by T and U = ±(
1 0
4 1
).
Let
γ = ±
a b
c d
Γ
0
(4).
We let
s(γ) = a
2
+ b
2
.
As
c
is even, we know
a
1 (
mod
2). So
s
(
γ
)
1, and moreover
s
(
γ
) = 1 iff
b = 0, a = ±1, iff γ = T
n
for some n.
We now claim that if
s
(
γ
)
6
= 1, then there exists
δ {T
±1
, U
±1
}
such that
s(γδ) < s(γ). If this is true, then we are immediately done.
To prove the claim, if s(γ) 6= 1, then note that |a| 6= |2b| as a is odd.
If
|a| < |
2
b|
, then
min{|b±a|} < |b|
. This means
s
(
γT
±1
) =
a
2
+(
b±a
)
2
<
s(γ).
If
|a| > |
2
b|
, then
min{|a ±
4
b|} < |a|
, so
s
(
γU
±1
) = (
a ±
4
b
)
2
+
b
2
<
s(γ).
Proof of theorem. It is enough to prove that
f |
2
T = f |
2
U = f.
This is now easy to prove, as it is just a computation. Since
ϑ
(
z
+ 2) =
ϑ
(
z
), we
know
f |
2
T = f(z + 1) = f(z).
We also know that
f |
2
W
4
= 4(4z)
2
f
1
4z
=
1
4z
2
ϑ
1
2z
4
= f(z),
as
ϑ
1
z
=
z
i
1/2
ϑ(z).
So we have
f |
2
U = f |
2
W
4
|
2
T
1
|
2
W
4
= (1)(1)f = f.
We look at Γ
0
(2) and Γ
0
(4) more closely. We have
Γ
0
(2) =

a b
2c d

=
γ Γ(1) : γ =
1
0 1
mod 2
.
We know
|SL
2
(
Z/
2
Z
)
|
= 6, and so (Γ(1) : Γ
0
(2)) = 3. We have coset representa-
tives
I,
0 1
0 1
,
1 0
1 1
.
We also have a map
Γ
0
(2) Z/2Z
a b
2c d
7→ c,
which one can directly check to be a homomorphism. This has kernel Γ
0
(4). So
we know
0
(2) : Γ
0
(4)) = 2, and has coset representatives
I,
1 0
2 1
So
Γ
0
(2) =
T, ±
1 0
2 1
= W
2
1 1
0 1
W
1
2
.
We can draw fundamental domains for Γ
0
(2):
and Γ
0
(4):
We are actually interested in Γ
0
(2) and Γ
0
domains look “dual”.
Consider
g(z) = E
2
|
2
(
2 0
0 1
) E
2
= 2E
2
(2z) E
2
(z).
E
2
(z) = 1 24
X
n1
σ
1
(n)q
n
= z
2
E
2
1
z
12
2πiz
.
Proposition. We have g M
2
0
(2)), and g|
2
W
2
= g.
Proof. We compute
g|
2
W
2
=
2
(2z)
2
g
1
2z
=
1
z
2
E
2
1
z
2
(2z)
2
E
2
1
2z
= E
2
(z) +
1
2πiz
2
E
2
(2z) +
12
2πi · 2z
= g(z).
We also have
g|
2
T = g(z + 1) = g(z),
and so
g|
2
(
1 0
2 1
) = g|
2
W
2
T
1
W
1
2
= g.
Moreover,
g
is holomorphic at
, and hence so is
g|
2
W
2
=
g
. So
g
is also
holomorphic at 0 =
W
2
(
). As
has width 1 and 0 has width 2, we see
that these are all the cusps, and so
g
is holomorphic at the cusps. So
g
M
2
0
(2)).
Now we can do this again. We let
h = g(2z) =
1
2
g|
2
(
2 0
0 1
) = 2E
2
(4z) E
2
(2z).
Since g M
2
0
(2)), this implies h M
2
0
(4)) M
0
0
(2)).
The functions
g
and
h
are obviously linearly independent. Recall that we
have
dim M
2
0
(4)) 1 +
k(Γ(1) : Γ
0
(4))
12
= 2.
So the inequality is actually an equality. We have therefore shown that
Theorem.
M
2
0
(4)) = Cg Ch.
Recall we also found an
f
(
z
) =
ϑ
(2
z
)
4
M
2
0
(4)). So we know we must
have
f = ag + bh
for some constants a, b C.
It is easy to find what
a
and
b
are. We just write down the
q
-expansions. We
have
f = ϑ(2z)
4
= (1 + 2q + 2q
4
+ ···)
4
= 1 + 8q + 24q
2
+ 32q
3
+ ···
g = 2E
2
(2z) E
2
(z)
= 1 + 24
X
n1
σ
1
(n)(q
n
2q
2n
)
= 1 + 24q + 24q
2
+ 96q
3
+ ···
h = g(2z)
= 1 + 24q
2
+ ···
By looking at the first two terms, we find that we must have
f =
1
3
g +
2
3
h =
1
3
(4E
2
(4z) E
2
(z)) = 1 + 8
X
k1
σ
1
(n) 4σ
1
n
4

q
n
,
where σ
1
n
4
= 0 if
n
4
6∈ Z.
But recall that
f =
X
nZ
q
n
2
!
4
=
X
a,b,c,dZ
q
a
2
+b
2
+c
2
+d
2
=
X
nN
r
4
(n)q
n
,
where
r
4
(
n
) is the number of ways to write
n
as a sum of 4 squares (where order
matters). Therefore,
Theorem (Lagrange’s 4-square theorem). For all n 1, we have
r
4
(n) = 8
σ
1
(n) 4σ
1
n
4

= 8
X
d|n4-d
d.
In particular, r
4
(n) > 0.
We could imagine ourselves looking at other sums of squares. For example,
ϑ
2
(2
z
)
2
, which turns out to be an element of
M
1
1
(4)),
one can get a similar formula for the number of ways of writing
n
as a sum of 2
squares.
We can also consider higher powers, and then get approximate formulae for
r
2k
(
n
), because the appropriate Eisenstein series no longer generate
M
k
. There
may be a cusp form as well, which gives an error term.
In general, if we have
γ =
a b
Nc d
Γ
0
(N),
then we find
W
N
γW
1
N
d c
Nb a
Γ
0
(N).
So
W
N
normalizes the group Γ
0
(
N
). Then if
f M
k
0
(
N
)), then
f |
k
W
N
M
k
0
(N)), and this also preserves cusp forms.
Moreover, we have
f |
k
W
2
N
= f |
k
N 0
0 N
= f,
as I Γ
0
(N). So
M
k
0
(N)) = M
k
0
(N))
+
M
k
0
(N))
,
where we split into the (
±
1)-eigenspaces for
W
N
, and the cusp forms decompose
similarly. This
W
N
is the Atkin-Lehner involution. This is the “substitute” for
the the operator S =
0 1
1 0
in Γ(1).