8Modular forms for subgroups of SL2(ℤ)

III Modular Forms and L-functions As promised earlier, we define an inner product on the space of cusp forms.
We let
f, g S
k
(Γ). Then the function
y
k
f
(
z
)
g(z)
is Γ-invariant, and is
bounded on
H
, since
f
and
g
vanish at cusps. Also, recall that
dx dy
y
2
is an
GL
2
(R)
+
-invariant measure. So we can define
hf, gi =
1
v(Γ)
Z
Γ\H
y
k
f(z)g(z)
dx dy
y
z
C,
where
R
Γ\H
means we integrate over any fundamental domain, and
v
(Γ) is the
volume of a fundamental domain,
v(Γ) =
Z
ΓH
dx dy
y
2
= (Γ(1) :
¯
Γ)
Z
D
dx dy
y
2
.
The advantage of this normalization is that if we replace Γ by a subgroup Γ
0
of finite index, then a fundamental domain for Γ
0
is the union of (
¯
Γ
:
¯
Γ
0
) many
fundamental domains for Γ. So the expression (
) is independent of Γ, as long
as both f, g S
k
(Γ).
This is called the Petersson inner product.
Proposition.
(i) h·, ·i is a Hermitian inner product on S
k
(Γ).
(ii) h·, ·i
is invariant under translations by
GL
2
(
Q
)
+
. In other words, if
γ GL
2
(Q)
+
, then
hf |
k
γ, g|
k
γi = hf, gi.
(iii) If f, g S
k
(Γ(1)), then
hT
n
f, gi = hf, T
n
gi.
This completes our previous proof that the
T
n
can be simultaneously diago-
nalized.
Proof.
(i)
We know
hf, gi
is
C
-linear in
f
, and
hf, gi
=
hg, fi
. Also, if
hf, fi
= 0,
then
Z
Γ\H
y
k2
|f|
2
dx dy = 0,
but since
f
is continuous, and
y
is never zero, this is true iff
f
is identically
zero.
(ii) Let f
0
= f |
k
γ and g
0
= g|
k
γ S
k
0
), where Γ
0
= Γ γ
1
Γγ. Then
y
k
f
0
¯g
0
= y
k
(det γ)
k
|cz + d|
2k
· f(γ(z))
g(γ(z)) = (Im γ(z))
k
f(γ(z))g(γ(z)).
Now Im γ(z) is just the y of γ(z). So it follows that Then we have
hf
0
, g
0
i =
1
v
0
)
Z
D
Γ
0
y
k
f ¯g
dx dy
y
2
γ(z)
=
1
v
0
)
Z
γ(D
Γ
0
)
y
k
f ¯g
dx dy
y
2
.
Now
γ
(
D
Γ
0
) is a fundamental domain for
γ
Γ
0
γ
1
=
γ
Γ
γ
1
Γ, and note that
v
0
) = v(γΓ
0
γ
1
) by invariance of measure. So hf
0
, g
0
i = hf, gi.
(iii)
Note that
T
n
is a polynomial with integer coefficients in
{T
p
:
p | n}
. So it
is enough to do it for n = p. We claim that
hT
p
f, gi = p
k
2
1
(p + 1)hf |
k
δ, gi,
where δ Mat
2
(Z) is any matrix with det(δ) = p.
Assuming this, we let
δ
a
=
1
Mat
2
(Z),
which also has determinant p. Now as
g|
k
p 0
0 p
= g,
we know
hT
p
f, gi = p
k
2
1
(p + 1)hf |
k
δ, gi
= p
k
2
1
(p + 1)hf, g |
k
δ
1
i
= p
k
2
1
(p + 1)hf, g |
k
δ
a
i
= hf, T
p
gi
To prove the claim, we let
Γ(1)
p 0
0 1
Γ(1) =
a
0jp
Γ(1)δγ
i
for some γ
i
Γ(1). Then we have
hT
p
f, gi = p
k
2
1
*
X
j
f |
k
δγ
j
, g
+
= p
k
2
1
X
j
hf |
k
δγ
j
, g |
k
γ
j
i
= p
k
2
1
(p + 1)hf |
k
δ, gi,
using the fact that g|
k
γ
j
= g.