4The modular group

III Modular Forms and L-functions 4 The modular group
We now move on to study the other words that appear in the title of the course,
namely modular forms. Modular forms are very special functions defined on the
upper half plane
H = {z C : Im(z) > 0}.
The main property of a modular form is that they transform nicely under
obius transforms. In this chapter, we will first try to understand these obius
transforms. Recall that a matrix
γ =
a b
c d
GL
2
(C)
acts on C by
z 7→ γ(z) =
az + b
cz + d
.
If we further restrict to matrices in
GL
2
(
R
), then this maps
C \R
to
C \R
, and
R {∞} to R {∞}.
We want to understand when this actually fixes the upper half plane. This
is a straightforward computation
Im γ(z) =
1
2i
az + b
cz + d
a¯z + b
c¯z + d
=
1
2i
|cz + d|
2
= det(γ)
Im z
|cz + d|
2
.
Thus, we know
Im
(
γ
(
z
)) and
Im
(
z
) have the same sign iff
det
(
γ
)
>
0. We write
Definition (GL
2
(R)
+
).
GL
2
(R)
+
= {γ GL
2
(R) : det γ > 0}.
This is the group of obius transforms that map H to H.
However, note that the action of
GL
2
(
R
)
+
on
H
is not faithful. The kernel
is given by the subgroup
R
×
· I = R
×
·
1 0
0 1
.
Thus, we are naturally led to define
Definition (PGL
2
(R)
+
).
PGL
2
(R)
+
=
GL
2
(R)
+
R
×
· I
.
There is a slightly better way of expressing this. Now note that we can obtain
any matrix in
GL
2
(
R
+
), by multiplying an element of
SL
2
(
R
) with a unit in
R
.
So we have
PGL
2
(R)
+
=
SL
2
(R)/I} PSL
2
(R).
What we have is thus a faithful action of
PSL
2
(
R
) on the upper half plane
H
.
From IA Groups, we know this action is transitive, and the stabilizer of
i
=
1
is SO(2)/I}.
In fact, this group
PSL
2
(
R
) is the group of all holomorphic automorphisms
of H, and the subgroup SO(2) SL
2
(R) is a maximal compact subgroup.
Theorem. The group SL
2
SL
2
(R) = KAN = NAK,
where
K = SO(2), A =

r 0
0
1
r

, N =

1 x
0 1

Note that this implies a few of our previous claims. For example, any
z = x + iy C can be written as
z = x + iy =
1 x
0 1
y 0
0
1
y
· i,
using the fact that K = SO(2) fixes i, and this gives transitivity.
Proof.
This is just Gram–Schmidt orthogonalization. Given
g GL
2
(
R
), we
write
ge
1
= e
0
1
, ge
2
= e
0
2
,
By Gram-Schmidt, we can write
f
1
= λ
1
e
0
1
f
2
= λ
2
e
0
1
+ µe
0
2
such that
kf
1
k = kf
2
k = 1, (f
1
, f
2
) = 0.
So we can write
f
1
f
2
=
e
0
1
e
0
2
λ
1
λ
2
0 µ
Now the left-hand matrix is orthogonal, and by decomposing the inverse of
λ
1
λ
2
0 µ
, we can write g =
e
0
1
e
0
2
as a product in KAN.
In general, we will be interested in subgroups Γ
SL
2
(
R
), and their images
¯
Γ in Γ PSL
2
(R), i.e.
¯
Γ =
Γ
Γ I}
.
We are mainly interested in the case Γ = SL
2
(Z), or a subgroup of finite index.
Definition (Modular group). The modular group is
PSL
2
(Z) =
SL
2
(Z)
I}
.
There are two particularly interesting elements of the modular group, given
by
S = ±
0 1
1 0
, T = ±
1 1
0 1
.
Then we have
T
(
z
) =
z
+ 1 and
S
(
z
) =
1
z
. One immediately sees that
T
has
infinite order and S
2
= 1 (in PSL
2
(Z)). We can also compute
T S = ±
1 1
1 0
and
(T S)
3
= 1.
The following theorem essentially summarizes the basic properties of the modular
group we need to know about:
Theorem. Let
D =
z H :
1
2
Re z
1
2
, |z| > 1
{z H : |z| = 1, Re(z) 0}.
1
2
1
1
2
1
ρ = e
πi/3
i
Then
D
is a fundamental domain for the action of
¯
Γ
on
H
, i.e. every orbit
contains exactly one element of D.
The stabilizer of
z D
in Γ is trivial if
z 6
=
i, ρ
, and the stabilizers of
i
and
ρ
are
¯
Γ
i
= hSi
=
Z
2Z
,
¯
Γ
ρ
= hT Si
=
Z
3Z
.
Finally, we have
¯
Γ = hS, T i = hS, T Si.
In fact, we have
¯
Γ = hS, T | S
2
= (T S)
3
= ei,
but we will neither prove nor need this.
The proof is rather technical, and involves some significant case work.
Proof.
Let
¯
Γ
=
hS, T i
¯
Γ
. We will show that if
z H
, then there exists
γ
¯
Γ
such that γ(z) D.
Since
z 6∈ R
, we know
Z
+
Zz
=
{cz
+
d
:
c, d Z}
is a discrete subgroup of
C. So we know
{|cz + d| : c, d Z}
is a discrete subset of
R
, and is in particular bounded away from 0. Thus, we
know
Im γ(z) =
Im(z)
|cz + d|
2
: γ =
a b
c d
¯
Γ
is a discrete subset of
R
>0
and is bounded above. Thus there is some
γ
¯
Γ
with
Im γ
(
z
) maximal. Replacing
γ
by
T
n
γ
for suitable
n
, we may assume
|Re γ(z)|
1
2
.
We consider the different possible cases.
If |γ(z)| < 1, then
Im Sγ(z) = Im
1
γ(z)
=
Im γ(z)
|γ(z)|
2
> Im γ(z),
which is impossible. So we know
|γ
(
z
)
|
1. So we know
γ
(
z
) lives in the
closure of D.
If Re(γ(z)) =
1
2
, then T γ(z) has real part +
1
2
, and so T (γ(z)) D.
If
1
2
< Re
(
z
)
<
0 and
|γ
(
z
)
|
= 1, then
|Sγ
(
z
)
|
= 1 and 0
< Re Sγ
(
z
)
<
1
2
,
i.e. Sγ(z) D.
So we can move it to somewhere in D.
We shall next show that if
z, z
0
D
, and
z
0
=
γ
(
z
) for
γ
¯
Γ
, then
z
=
z
0
.
Moreover, either
γ = 1; or
z = i and γ = S; or
z = ρ and γ = T S or (T S)
2
.
It is clear that this proves everything.
To show this, we wlog
Im(z
0
) =
Im z
|cz + d|
2
Im z
where
γ =
a b
c d
,
and we also wlog c 0.
Therefore we know that |cz + d| 1. In particular, we know
1 Im(cz + d) = c Im(z) c
3
2
since z D. So c = 0 or 1.
If c = 0, then
γ = ±
1 m
0 1
for some
m Z
, and this
z
0
=
z
+
m
. But this is clearly impossible. So we
must have m = 0, z = z
0
, γ = 1 PSL
2
(Z).
If
c
= 1, then we know
|z
+
d|
1. So
z
is at distance 1 from an integer.
As z D, the only possibilities are d = 0 or 1.
If d = 0, then we know |z| = 1. So
γ =
a 1
1 0
for some a Z. Then z
0
= a
1
z
. Then
either a = 0, which forces z = i, γ = S; or
a = 1, and z
0
= 1
1
z
, which implies z = z
0
= ρ and γ = T S.
If d = 1, then by looking at the picture, we see that z = ρ. Then
|cz + d| = |z 1| = 1,
and so
Im z
0
= Im z =
3
2
.
So we have z
0
= ρ as well. So
+ b
ρ 1
= ρ,
which implies
ρ
2
(a + 1)ρ b = 0
So ρ = 1, a = 0, and γ = (T S)
2
.
Note that this proof is the same as the proof of reduction theory for binary
What does the quotient
¯
Γ \ N
look like? Each point in the quotient can be
identified with an element in
D
. Moreover,
S
and
T
identify the portions of
the boundary of
D
. Thinking hard enough, we see that the quotient space is
homeomorphic to a disk.
An important consequence of this is that the quotient Γ
\H
has finite invariant
measure.
Proposition. The measure
dµ =
dx dy
y
2
is invariant under
PSL
2
(
R
). If Γ
PSL
2
(
Z
) is of finite index, then
µ
\H
)
<
.
Proof. Consider the 2-form associated to µ, given by
η =
dx dy
y
2
=
idz d¯z
2(Im z)
2
.
We now let
γ =
a b
c d
SL
2
(R).
Then we have
Im γ(z) =
Im z
|cz + d|
2
.
Moreover, we have
dγ(z)
dz
=
a(cz + d) c(az + b)
(cz + d)
2
=
1
(cz + d)
2
.
Plugging these into the formula, we see that η is invariant under γ.
Now if
¯
Γ PSL
2
(
Z
) has finite index, then we can write
PSL
2
(
Z
) as a union
of cosets
PSL
2
(Z) =
n
a
i=1
¯γγ
i
,
where n = (PSL
2
(Z) :
¯
Γ). Then a fundamental domain for
¯
Γ is just
n
[
i=1
γ
i
(D),
and so
µ(
¯
Γ \ H) =
X
µ(γ
i
D) = (D).
So it suffices to show that µ(D) is finite, and we simply compute
µ(D) =
Z
D
dx dy
y
2
Z
x=
1
2
x=
1
2
Z
y=
y=
2/2
dx dy
y
2
< .
It is an easy exercise to show that we actually have
µ(D) =
π
3
.
We end with a bit terminology.
Definition
(Principal congruence subgroup)
.
For
N
1, the principal congru-
ence subgroup of level N is
Γ(N) = {γ SL
2
(Z) : γ I (mod N)} = ker(SL
2
(Z) SL
2
(Z/NZ)).
Any Γ
SL
2
(
Z
) containing some Γ(
N
) is called a congruence subgroup, and its
level is the smallest N such that Γ Γ(N)
This is a normal subgroup of finite index.
Definition
0
(N), Γ
1
(N)). We define
Γ
0
(N) =

a b
c d
SL
2
(Z) : c 0 (mod N)
and
Γ
1
(N) =

a b
c d
SL
2
(Z) : c 0, d 1 (mod N)
.
We similarly define Γ
0
(
N
) and Γ
1
(
N
) to be the transpose of Γ
0
(
N
) and Γ
1
(
N
)
respectively.
Note that “almost all” subgroups of
SL
2
(
Z
) are not congruence subgroups.
On the other hand, if we try to define the notion of congruence subgroups in
higher dimensions, we find that all subgroups of
SL
n
(
Z
) for
n >
2 are congruence!