6Hecke operators

III Modular Forms and L-functions

6.1 Hecke operators and algebras
Recall that for f : H C, γ GL
2
(R)
+
and k Z, we defined
(f |
k
γ)(z) = (det γ)
k/2
j(γ, z)
k
f(γ(z)),
where
γ =
a b
c d
, j(γ, z) = cz + d.
We then defined
M
k
= {f : f |
k
γ = f for all γ Γ(1) + holomorphicity condition}.
We showed that these are finite-dimensional, and we found a basis. But there
is more to be said about modular forms. Just because we know polynomials
have a basis 1
, x, x
2
, ···
does not mean there isn’t anything else to say about
polynomials!
In this chapter, we will figure that
M
k
has the structure of a module for
the Hecke algebra. This structure underlies the connection with arithmetic, i.e.
Galois representations etc.
How might we try to get some extra structure on
M
k
? We might try to see
what happens if we let something else in
GL
2
(
R
)
+
act on
f
. Unfortunately, in
general, if
f
is a modular form and
γ GL
2
(
R
)
+
, then
g
=
f |
k
γ
is not a modular
form. Indeed, given a δ Γ(1), then it acts on g by
g|
k
δ = f |
k
γδ = (f |
k
γδγ
1
)γ
and usually
γδγ
1
6∈
Γ(1). In fact the normalizer of Γ(1) in
GL
2
(
R
)
+
is generated
by Γ(1) and aI for a R
.
It turns out we need to act in a smarter way. To do so, we have to develop
quite a lot of rather elementary group theory.
Consider a group
G
, and Γ
G
. The idea is to use the double cosets of Γ
defined by
ΓgΓ = {γgγ
0
: γ, γ
0
Γ}.
One alternative way to view this is to consider the right multiplication action of
G
, hence Γ on the right cosets Γ
g
. Then the double coset Γ
g
Γ is the union of
the orbits of Γg under the action of Γ. We can write this as
ΓgΓ =
a
iI
Γg
i
for some g
i
gΓ G and index set I.
In our applications, we will want this disjoint union to be finite. By the
orbit-stabilizer theorem, the size of this orbit is the index of the stabilizer of Γ
g
in Γ. It is not hard to see that the stabilizer is given by Γ
g
1
Γ
g
. Thus, we
are led to consider the following hypothesis:
Hypothesis (H): For all g G, (Γ : Γ g
1
Γg) < .
Then (
G,
Γ) satisfies (H) iff for any
g
, the double coset Γ
g
Γ is the union of
finitely many cosets.
The important example is the following:
Theorem.
Let
G
=
GL
2
(
Q
), and Γ
SL
2
(
Z
) a subgroup of finite index. Then
(G, Γ) satisfies (H).
Proof. We first consider the case Γ = SL
2
(Z). We first suppose
g =
a b
c d
Mat
2
(Z),
and det g = ±N, N 1. We claim that
g
1
Γg Γ Γ(N),
from which it follows that
(Γ : Γ g
1
Γg) < .
So given
γ
Γ(
N
), we need to show that
gγg
1
Γ, i.e. it has integer coefficients.
We consider
±N·gγg
1
=
a b
c d
γ
d b
c a
a b
c d
d b
c a
NI 0 (mod N).
So we know that
gγg
1
must have integer entries. Now in general, if
g
0
GL
2
(
Q
),
then we can write
g
0
=
1
M
g
for
g
with integer entries, and we know conjugating by
g
and
g
0
give the same
result. So (G, Γ) satisfies (H).
The general result follows by a butterfly. Recall that if (
G
:
H
)
<
and
(
G
:
H
0
)
<
, then (
G
:
H H
0
)
<
. Now if Γ
Γ(1) =
SL
2
(
Z
) is of finite
index, then we can draw the diagram
Γ(1)
g
1
Γ(1)g
Γ
Γ(1) g
1
Γ(1)g
g
1
Γg
Γ g
1
Γ(1)G Γ(1) g
1
Γg
Γ g
1
Γg
finite
finite
finite
finite
Each group is the intersection of the two above, and so all inclusions are of finite
index.
Note that the same proof works for GL
N
(Q) for any N.
Before we delve into concreteness, we talk a bit more about double cosets.
Recall that cosets partition the group into pieces of equal size. Is this true for
double cosets as well? We can characterize double cosets as orbits of Γ
×
Γ acting
on G by
(γ, δ) · g = γgδ
1
.
So G is indeed the disjoint union of the double cosets of Γ.
However, it is not necessarily the case that all double cosets have the same
size. For example
|
Γ
e
Γ
|
=
|
Γ
|
, but for a general
g
,
|
Γ
g
Γ
|
can be the union of
many cosets of Γ.
Our aim is to define a ring
H
(
G,
Γ) generated by double cosets called the
Hecke algebra. As an abelian group, it is the free abelian group on symbols
g
Γ] for each double coset
g
Γ]. It turns out instead of trying to define a
multiplication for the Hecke algebra directly, we instead try to define an action
of this on interesting objects, and then there is a unique way of giving
H
(
G,
Γ)
a multiplicative structure such that this is a genuine action.
Given a group
G
, a
G
-module is an abelian group with a
Z
-linear
G
-action.
In other words, it is a module of the group ring
ZG
. We will work with right
modules, instead of the usual left modules.
Given such a module and a subgroup Γ G, we will write
M
Γ
= {m M : = m for all γ Γ}.
Notation. For g G and m M
Γ
, we let
m|gΓ] =
n
X
i=1
mg
i
, ()
where
ΓgΓ =
n
a
i=1
Γg
i
.
The following properties are immediate, but also crucial.
Proposition.
(i) m|gΓ] depends only on ΓgΓ.
(ii) m|gΓ] M
Γ
.
Proof.
(i) If g
0
i
= γ
i
g
i
for γ
i
Γ, then
X
mg
0
i
=
X
i
g
i
=
X
mg
i
as m M
Γ
.
(ii) Just write it out, using the fact that {Γg
i
} is invariant under Γ.
Theorem.
There is a product on
H
(
G,
Γ) making it into an associative ring,
the Hecke algebra of (
G,
Γ), with unit
e
Γ] = [Γ], such that for every
G
-module
M, we have M
Γ
is a right H(G, Γ)-module by the operation ().
In the proof, and later on, we will use the following observation: Let
Z
\G
]
be the free abelian group on cosets
g
]. This has an obvious right
G
-action by
multiplication. We know a double coset is just an orbit of Γ acting on a single
coset. So there is an isomorphism between
Θ : H(G, Γ) Z \ G]
Γ
.
given by
gΓ] 7→
X
g
i
],
where
ΓgΓ =
a
Γg
i
.
Proof. Take M = Z \ G], and let
ΓgΓ =
a
Γg
i
ΓhΓ =
a
Γh
j
.
Then
X
i
g
i
] M
Γ
,
and we have
X
i
g
i
]|hΓ] =
X
i,j
g
i
h
j
] M
Γ
,
and this is well-defined. This gives us a well-defined product on
H
(
G,
Γ).
Explicitly, we have
gΓ] · hΓ] = Θ
1
X
i,j
g
i
h
j
]
.
It should be clear that this is associative, as multiplication in
G
is associative,
and [Γ] = [ΓeΓ] is a unit.
Now if M is any right G-module, and m M
Γ
, we have
m|gΓ]|hΓ] =
X
mg
i
|hΓ] =
X
mg
i
h
j
= m([ΓgΓ] · hΓ]).
So M
Γ
is a right H(G, Γ)-module.
Now in our construction of the product, we need to apply the map Θ
1
. It
would be nice to have an explicit formula for the product in terms of double
cosets. To do so, we choose representatives S G such that
G =
a
gS
ΓgΓ.
Proposition. We write
ΓgΓ =
r
a
i=1
Γg
i
ΓhΓ =
s
a
j=1
Γh
j
.
Then
gΓ] · hΓ] =
X
kS
σ(k)[ΓkΓ],
where σ(k) is the number of pairs (i, j) such that Γg
i
h
j
= Γk.
Proof. This is just a simple counting exercise.
Of course, we could have taken this as the definition of the product, but we
have to prove that this is independent of the choice of representatives
g
i
and
h
j
,
and of S, and that it is associative, which is annoying.