III Modular Forms and L-functions - Modular forms of level 1

5Modular forms of level 1

III Modular Forms and L-functions

5.1 Basic definitions

We can now define a modular form. Recall that we have SL

(Z) = Γ(1).

Definition

(Modular form of level 1)

A holomorphic function

H → C

is a

modular form of weight k ∈ Z and level 1 if

(i) For any

γ =



a b

c d



∈ Γ(1),

we have

f(γ(z)) = (cz + d)

f(z).

(ii) f is holomorphic at ∞ (to be defined precisely later).

What can we deduce about modular forms from these properties? If we take

γ = −I, then we get

f(z) = (−1)

f(z).

So if

is odd, then

f ≡

0. So they only exist for even weights. If we have even

weights, then it suffices to consider

Γ = hS, T i. Since

f(z) 7→ (cz + d)

−k

f(γ(z))

is a group action of Γ(1) on functions on

, it suffices to check that

is invariant

under the generators S and T . Thus, (i) is equivalent to

f(z + 1) = f(z), f(−1/z) = z

f(z).

How do we interpret (ii)? We know

-periodic. If we write

2πiz

, then

we have

z ∈ H

iff 0

< |q| <

1, and moreover, if two different

give the same

then the values of

on the two

agree. In other words,

(

) only depends on

and thus there exists a holomorphic function

f(q) on {0 < |q| < 1} such that

f(e

2πiz

) = f(z).

Explicitly, we can write

f(q) = f



2πi

log q



By definition,

is a holomorphic function on a punctured disk. So we have a

Laurent expansion

f(q) =

∞

n=−∞

(f)q

called the Fourier expansion or

-expansion of

. We say

is meromorphic

(resp. holomorphic) at ∞ if

f is meromorphic (resp. holomorphic) at q = 0.

In other words, it is meromorphic at

∞

(

) = 0 for

sufficiently negative,

and holomorphic if

(

) = 0 for all

n ≥

0. The latter just says

(

) is bounded

as Im(z) → ∞.

The following definition is also convenient:

Definition

(Cusp form)

A modular form

is a cusp form if the constant term

(f) is 0.

We will later see that “almost all” modular forms are cusp forms.

In this case, we have

f =

n≥1

(f)q

From now on, we will drop the ˜, which should not cause confusion.

Definition

(Weak modular form)

A weak modular form is a holomorphic form

on H satisfying (i) which is meromorphic at ∞.

We will use these occasionally.

The transformation rule for modular forms seem rather strong. So, are there

actually modular forms? It turns out that there are quite a lot of modular forms,

and remarkably, there is a relatively easy way of listing out all the modular

forms.

The main class (and in fact, as we will later see, a generating class) of modular

forms is due to Eisenstein. This has its origin in the theory of elliptic functions,

but we will not go into that.

Definition (Eisenstein series). Let k ≥ 4 be even. We define

(z) =

m,n∈Z

(m,n)6=(0,0)

(mz + n)

(m,n)∈Z

(mz + n)

Here the

denotes that we are omitting 0, and in general, it means we

don’t sum over things we obviously don’t want to sum over.

When we just write down this series, it is not clear that it is a modular form,

or even that it converges. This is given by the following theorem:

Theorem. G

is a modular form of weight

and level 1. Moreover, its

expansion is

(z) = 2ζ(k)





1 −

n≥1

k−1

(n)q





, (1)

where

(n) =

1≤d|n

Convergence of the series follows from the following more general result. Note

that since z 6∈ R, we know {1, z} is an R-basis for C.

Proposition. Let (e

, ··· , e

) be some basis for R

. Then if r ∈ R, the series

m∈Z

+ ··· + m

−r

converges iff r > d.

Proof. The function

) ∈ R

7→



i=1



is a norm on

. As any 2 norms on

are equivalent, we know this is equivalent

to the sup norm k · k

∞

. So the series converges iff the corresponding series

m∈Z

kmk

−r

∞

converges. But if 1

≤ N ≤ Z

, then the number of

m ∈ Z

such that

kmk

∞

is (2N + 1)

− (2N − 1)

∼ 2

d−1

. So the series converges iff

N≥1

−r

d−1

converges, which is true iff r > d.

Proof of theorem.

Then convergence of the Eisenstein series by applying this

∼

. So the series is absolutely convergent. Therefore we can simply

compute

(z + 1) =

m,n

(mz + (m + n))

= G

(z).

Also we can compute



−



m,n

(−m + nz)

= z

(z).

satisfies the invariance property. To show that

is holomorphic, and

holomorphic at infinity, we’ll derive the q-expansion (1).

Lemma.

∞

n=∞

(n + w)

(−2πi)

(k − 1)!

∞

d=1

k−1

2πidw

for any w ∈ H and k ≥ 2.

There are (at least) two ways to prove this. One of this is to use the series

for the cotangent, but here we will use Poisson summation.

Proof. Let

f(x) =

(x + w)

We compute

f(y) =

∞

−∞

−2πixy

(x + w)

dx.

We replace this with a contour integral. We see that this has a pole at

−w

. If

y > 0, then we close the contour downwards, and we have

f(y) = −2πi Res

z=−w

−2πiyz

(z + w)

= −2πi

(−2πiy)

k−1

(k − 1)!

2πiyw

y ≤

0, then we close in the upper half plane, and since there is no pole, we

have

f(y) = 0. So we have

∞

n=−∞

(n + w)

n∈Z

f(n) =

d∈Z

f(d) =

(−2πi)

(k − 1)!

d≥1

k−1

2πidw

by Poisson summation formula.

Note that when we proved the Poisson summation formula, we required

to decrease very rapidly at infinity, and our

does not satisfy that condition.

However, we can go back and check that the proof still works in this case.

Now we get back to the Eisenstein series. Note that since

is even, we can

drop certain annoying signs. We have

(z) = 2

n≥1

+ 2

m≥1

n∈Z

(n + mz)

= 2ζ(k) + 2

m≥1

(2πi)

(k − 1)!

d≥1

k−1

= 2ζ(k) + 2

(2πi)

(k − 1)!

n≥1

k−1

(n)q

Then the result follows from the fact that

ζ(k) = −

(2πi)

So we see that G

is holomorphic in H, and is also holomorphic at ∞.

It is convenient to introduce a normalized Eisenstein series

Definition (Normalized Eisenstein series). We define

(z) = (2ζ(k))

−1

(z)

= 1 −

n≥1

k−1

(n)q

(m,n)=1

m,n∈Z

(mz + n)

The last line follows by taking out any common factor of

m, n

in the series

defining G

Thus, to figure out the (normalized) Eisenstein series, we only need to know

the Bernoulli numbers.

Example. We have

, B

−1

, B

−1

, B

−631

2730

, B

Using these, we find

= 1 + 240

(n)q

= 1 − 504

(n)q

= 1 + 480

(n)q

= 1 − 264

(n)q

= 1 +

65520

691

(n)q

= 1 − 24

(n)q

We notice that there is a simple pattern for k ≤ 14, except for k = 12.

For more general analysis of modular forms, it is convenient to consider the

following notation:

Definition (Slash operator). Let



a b

c d



= γ ∈ GL

(R)

, z ∈ H,

and f : H → C any function. We write

j(γ, z) = cz + d.

We define the slash operator to be

(f |

γ)(z) = (det γ)

k/2

j(γ, z)

−k

f(γ(z)).

Note that some people leave out the

det γ

k/2

factor, but if we have it, then

whenever γ = Ia, then

f |

γ = sgn(a)

which is annoying. In this notation, then condition (i) for

to be a modular

form is just

f |

γ = f

for all γ ∈ SL

(Z).

To prove things about our j operator, it is convenient to note that





= j(γ, z)



γ(z)



. (∗)

Proposition.

(i) j(γδ, z) = j(γ, δ(z))j(δ, z) (in fancy language, we say j is a 1-cocycle).

(ii) j(γ

−1

, z) = j(γ, γ

−1

(z))

−1

(iii) γ

ϕ 7→ f |

is a (right) action of

(

)

on functions on

. In

other words,

f |

γ |

δ = f |

(γδ).

Note that this implies that if if Γ ≤ GL

(R)

and Γ = hγ

, ··· , γ

i then

f |

γ = f ⇐⇒ f |

= f for all i = 1, ··· , m.

The proof is just a computation.

Proof.

(i) We have

j(γδ, z)



γδ(z)



= γδ





= j(δ, z)γ



δ(z)



= j(δ, z)j(γ, δ(z))





(ii) Take δ = γ

−1

(iii) We have

((f |

γ)|

δ)(z) = (det δ)

k/2

j(δ, z)

−k

(f |

γ)(δ(z))

= (det δ)

k/2

j(δ, z)

−k

(det γ)

k/2

j(γ, δ(z))

−k

f(γδ(z))

= (det γδ)

k/2

j(γδ, z)

−k

f(γδ(z))

= (f |

γδ)(z).

Back to the Eisenstein series.

arise naturally in elliptic functions, which

are coefficients in the series expansion of Weierstrass ℘ function.

There is another group-theoretic interpretation, which generalizes in many

ways. Consider

Γ(1)

∞





1 n

0 1



: n ∈ Z



≤ Γ(1) = SL

(Z),

which is the stabilizer of ∞. If

δ = ±



1 n

0 1



∈ Γ(1)

∞

then we have

j(δγ, z) = j(δ, γ(z))j(γ, z) = ±j(γ, z).

So j(γ, z)

depends only on the coset Γ(1)

∞

γ. We can also check that if

γ =



a b

c d



, γ =





∈ Γ(1),

then Γ(1)

∞

γ = Γ(1)

∞

iff (c, d) = ±(c

, d

Moreover, gcd(c, d) = 1 iff there exists a, b such that



a b

c d



= 1.

We therefore have

(z) =

γ∈Γ(1)

∞

\Γ(1)

j(γ, z)

−k

where we sum over (any) coset representatives of Γ(1)

∞

We can generalize this in two ways. We can either replace

with some other

appropriate function, or change the groups.