8Modular forms for subgroups of SL_{2}(ℤ)

III Modular Forms and L-functions

8.2 The Petersson inner product

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

k−2

|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

(Γ

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(Γ

0

)

Z

D

Γ

0

y

k

f ¯g

dx dy

y

2

γ(z)

=

1

v(Γ

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(Γ

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

= pδ

−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

0≤j≤p

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.