9Hecke theory for Γ0(N)

III Modular Forms and L-functions 9 Hecke theory for Γ
0
(N)
Note that it is possible to do this for other congruence subgroups. The key case
is
Γ
1
(N) =

a b
c d
SL
2
(Z) : c 0 (mod N), d, a 1 (mod N)
The map
a b
c d
7→ d mod N
is a homomorphism Γ
0
(
N
)
(
Z/nZ
)
×
, and
the kernel is Γ
1
(N).
So we can describe
S
k
1
(N)) =
M
χ
\
(Z/NZ)
×
S
k
1
(N), χ),
where f S
k
1
(n), χ) if
f |
k
a b
c d
= χ(d)f for all
a b
c d
Γ
0
(N).
Of course,
S
k
1
(
N
)
, χ
trivial
) =
S
k
0
(
N
)). In general, everything we can do for
Γ
0
(N) can be done for S
k
1
(N), χ).
But why not study Γ(N) itself? We can check that
1 0
0 N
Γ(N)
1 0
0 N
1
Γ
1
(N
2
).
So we can go from modular forms on Γ(N ) to ones on Γ
1
(N
0
).
For various representation-theoretic reasons, things work much better on
Γ
1
(N).
Last time, we used in various places the matrix
W
N
=
0 1
N 0
.
Then we decomposed
S
k
0
(N)) = S
k
2
(N))
+
S
k
0
(N))
,
according to the
±
-eigenspaces of the operator
W
N
. A while ago, we proved
some theorem about the function equation of L-functions.
Theorem. Let f S
k
0
(N))
ε
, where ε = ±1. Then define
L(f, s) =
X
n1
a
n
n
s
.
Then L(f, s) is am entire function, and satisfies the functional equation
Λ(f, s) = (2π)
s
Γ(s)L(f, s) = ε(N)
k/2
Λ(f, k s).
Proof. We have f |
k
W
N
= εf, and then we can apply our earlier result.
This is a rather remarkable thing. We didn’t really use much about the
properties of f .
Now we want to talk about Hecke operators on Γ
0
(
N
). These are a bit more
complicate. It is much better to understand these in terms of representation
theory instead, but that is outside the scope of the course. So we will just state
the relevant definitions and results.
Recall that a modular form of level 1 is defined by the
q
-expansion, and if
what we have is in fact a Hecke eigenform, then it suffices to know the Hecke
eigenvalues, i.e. the values of
a
p
. We describe this as having “multiplicity one”.
Theorem
(Strong multiplicity one for
SL
2
(
Z
))
.
Let
f, g S
k
(Γ(1)) be normal-
ized Hecke eigenforms, i.e.
f|T
p
= λ
p
f λ
p
= a
p
(f)
g|T
p
= µ
p
g µ
p
= a
p
(g).
Suppose there exists a finite set of primes
S
such that such that for all
p 6∈ S
,
then λ
p
= µ
p
. Then f = g.
Note that since the space of modular forms is finite dimensional, we know
that the modular forms can only depend on finitely many of the coefficients. But
this alone does not give the above result. For example, it might be that
a
2
(
f
) is
absolutely crucial for determining which modular form we are, and we cannot
miss it out. The strong multiplicity one theorem says this does not happen.
Idea of proof. We use the functional equations
Λ(f, k s) = (1)
k/2
Λ(f, s)
Λ(g, k s) = (1)
k/2
Λ(g, s)
So we know
L(f, k s)
L(f, s)
=
L(g, k s)
L(g, s)
.
Since these are eigenforms, we have an Euler product
L(f, s) =
Y
p
(1 λ
p
p
s
+ p
k12s
)
1
,
and likewise for g. So we obtain
Y
p
1 λ
p
p
sk
+ p
2sk1
1 λ
p
p
s
+ p
k12s
=
Y
p
1 µ
p
p
sk
+ p
2sk1
1 µ
p
p
s
+ p
k12s
.
Now we can replace this
Q
p
with
Q
pS
. Then we have some very explicit
rational functions, and then by looking at the appropriate zeroes and poles, we
can actually get λ
p
= µ
p
for all p.
This uses L-functions in an essential way.
The reason we mention this here is that a naive generalization of this theorem
does not hold for, say, Γ
0
(
N
). To even make sense of this statement, we need to
say what the Hecke operators are for Γ
0
(
N
). We are going to write the definition
in a way as short as possible.
Definition (Hecke operators on Γ
0
(N)). If p - N , we define
T
p
f = p
k
2
1
f |
k
p 0
0 1
+
p1
X
k=0
f |
k
1 b
0 p
!
which is the same as the case with Γ(1).
When p | N , then we define
U
p
f = p
k
2
1
p1
X
n=0
f |
k
1 b
0 p
.
Some people call this T
p
instead, and this is very confusing.
We can compute the effect on
q
-expansions as follows when
p - N
, then
we have
a
n
(T
p
f) = a
np
(f) + p
k1
a
n/p
(f),
where the second term is set to 0 if p - n. If p | N, then we have
a
n
(U
p
f) = a
np
(f).
Proposition. T
p
, U
p
send S
k
0
(N)) to S
k
0
(N)), and they all commute.
Proof. T
p
, U
p
do correspond to double coset actions
Γ
0
(N)
1 0
0 p
Γ
0
(N) =
(
Γ
0
(N)
p 0
0 1
q
`
b
Γ
0
(N)
1 b
0 p
p - N
`
b
Γ
0
(N)
1 b
0 p
p | N
.
Commutativity is checked by carefully checking the effect on the
q
-expansions.
However, these do not generate all the Hecke operators. For example, we
have W
N
!
Example. Consider S
12
0
(2)). This contains f = ∆(z) and
g = f |
12
(
2 0
0 1
) = 2
6
∆(2z) = ∆ |
12
W
2
,
using the fact that
|
k
0 1
1 0
= ∆.
So the matrix of W
2
on span{f, g} is
0 1
1 0
.
We can write out
f =
X
τ(n)q
n
= q 24q
2
+ 252q
3
1472q
4
+ ···
g = 2
6
X
τ(n)q
2n
= 2
6
(q
2
+ 24q
4
+ ···)
So we find that
U
2
g = 2
6
f.
It takes a bit more work to see what this does on f. We in fact have
U
2
f =
X
τ(2n)q
n
= 24q 1472q
4
+ ··· = 24f 32g.
So in fact we have
U
2
=
24 64
32 0
.
Now
U
2
and
W
2
certainly do not commute. So the Hecke algebra is not com-
mutative. In fact, generates a two-dimensional representation of the Hecke
algebra.
This makes life much worse. When we did Hecke algebras for Γ(1), all our
representations are 1-dimensional, and we can just work with linear spans. Now
everything has higher dimensional, and things go rather wrong. Similarly, we
can consider ∆(
dz
)
S
12
0
(
N
)) for any
d | N
, and this gives a whole bunch of
things like this.
This turns out to be the only obstruction to the commutativity of the action
of the Hecke algebra. We know S
k
0
(N)) contains
{f(dz) : f S
k
0
(M)), dM | N, M 6= N}.
We let
S
k
0
(
N
))
old
be the span of these. These are all the forms that come
from a smaller level.
Now
S
k
0
(
N
)) has an inner product! So the natural thing to do is to
consider the orthogonal complement of
S
k
0
(
N
))
old
, and call it
S
k
0
(
N
))
new
.
Theorem
(Atkin–Lehner)
.
The Hecke algebra
H
(
G,
Γ
0
(
N
)) fixes
S
k
0
(
N
))
new
and
S
k
0
(
N
))
old
, and on
S
k
0
(
N
))
new
, it acts as a commutative subalgebra
of the endomorphism ring, is closed under adjoint, and hence is diagonalizable.
Moreover, strong multiplicity one holds, i.e. if
S
is a finite set of primes, and we
have
{λ
p
:
p 6∈ S}
given, then there exists at most one
N
1 and at most one
f S
k
0
(N), 1)
new
(up to scaling, obviously) for which
T
p
f = λ
p
f for all p - N, p 6∈ S.