Math 726: L-functions and modular forms Fall 2011

Lecture 11 : Hecke Operators and Hecke theory

Instructor: Henri Darmon Notes written by: Celine Maistret

We begin with an aside on Question 1 of assignment 1 :

If ρ = Ind_{K}^{ℚ}1_{K} then we have the following facts :

• ρ can be identified with the permutation representation of G_{ℚ} acting on Hom(K, ℚ̄) which is a finite set of cardinality d = [K : ℚ]. It can therefore be viewed as the permutation representation of G_{ℚ} acting on the roots of the polynomial F_{α}(x), where F_{α}(x) is the monic characteristic polynomial of any primitive element α ∈ K.

• If p is prime, we can choose α ∈ K such that F_{α}(x) ∈ ℚ[x] ∩ ℤ_{p}[x] has p-integral coefficients and ℤ_{p}[x]/F_{α}(x) = O_{K} ⊗ ℤ_{p}

Now, in the ring of integers, factor F_{α}(x) mod p :

F_{α} = F_{1}^{e_{1}}....F_{r}^{e_{r}} with deg(F_{j}) = f_{j}, (p) = ℘_{1}^{e_{1}}....℘_{r}^{e_{r}} and N(℘_{j}) = p^{f_{j}}. This determines the factorisation of p in O_{K}

Using this, we can see how the inertia group at p permutes among themselves the roots of F_{α} that reduce to a common root mod p. Hence V^{I_{p}} can be identified with the permutation representation of G_{𝔽_{p}} acting on the roots of F_{1}....F_{r} mod p.

Therefore, det(1 − xF rob_{p}) = (1 − x^{f_{1}})(1 − x^{f_{2}})...(1 − x^{f_{r}}) so that the Euler factor at p of L(s, ρ) is

(1 − p^{−sf_{1}})(1 − p^{−sf_{2}})...(1 − p^{−sf_{r}})

but N(℘_{i}) = p^{f_{i}} hence

L(s, ρ) = Π_{℘|p}(1 − N_{℘}^{−s})^{−1}

Q.E.D.

Back to Hecke Operators in case of Modular Forms of Level 1.

A Modular Form of weight k can be viewed as a homogeneous function on Π , the space of lattices in ℂ, in the following way :

f ↦ F_{f}(Λ) = ω_{2}^{−k}f(ω_{1}/ω_{2})

where ω_{1}, ω_{2} are a set of integral generators of the lattice Λ, and were chosen such that their ratio belongs to the upper half plane.

Note that we want such functions F_{f} to be Modular/Cuspidal homogeneous functions on Π, ie we want to control the growth conditions such that when we consider the other direction we have :

F ↦ f_{F} (τ) = F (ℤ + τℤ)

with f_{F} holomorphic at τ and has the right behavior at ∞.

Once these conditions hold, we get a bijection between the two sets.

We can now define the Hecke Operator more naturally in terms of homogeneous functions on Π :

Definition 1. For all n ≥ 1, the Hecke Operator T_{n} acting on the space of homogeneous functions of weight k on Π is defined by :

(T_{n}F)(Λ) = n^{k−1} ∑_{Λ′⊂Λ_{[Λ/Λ′]=n}} F (Λ′)

It is clear from this definition that the image of a homogeneous function of weight k on lattices is still a homogeneous function of weight k on lattices. But we want to check that T_{n}F preserves the image of the space of modular forms and hence that T_{n}F induces an action on modular forms.

In order to do this, we make a previsional definition of T_{n} acting on M_{k}(SL_{2}(ℤ)) and S_{k}(SL_{2}(ℤ)) :

Definition 2. F_{T_{n}f} = T_{n}F_{f}

Using Definition 2 we derive a precise formula :

T_{n}f(z) = n^{k−1} ∑_{γ∈SL_{2}(ℤ)\M_{n}} f(γz)(cz + d)^{−k}

Where M_{n} is the set of matrices in M_{2}(ℤ) of determinant n, on which SL_{2}(ℤ) acts by left multiplication.

This is motivated by the fact that, in order to produce all the sublattices of (1,z) of index n, we apply to z various möbius tranformations corresponding to matrices of determinant n.

It is now clear that T_{n} preserves holomorphicity, it remains to show that it also preserves the growth properties.

Theorem 3. 1) If f = ∑_{m=0}^{∞} a(m)q^{m} belongs to M_{k}(SL_{2}(ℤ)), then

T_{n}(f)(q) = ∑_{m=0}^{∞} ∑_{d|(m,n)} d^{k−1}a(\frac{nm}{d^{2}})q^{m}  (1)

In particular if n = p prime, then

T_{p}(f)(q) = ∑_{m=0}^{∞} a(mp)q^{m} + p^{k−1} ∑_{m=0}^{∞} a(m)q^{mp}

2)

T_{n}T_{m} = ∑_{d|(m,n)} d^{k−1}T_{\frac{mn}{d^{2}}}

In particular, 1) shows that T_{n} preserves Modular and Cusp forms, and 2) implies :

T_{n}T_{m} = T_{nm} if (m, n) = 1
T_{p}T_{p^{t}} = T_{p^{t+1}} + p^{k−1}T_{p^{t−1}}

In order to prove Theorem 3 we need the following lemma :

Lemma 1. Every orbit SL_{2}(ℤ) (\begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix}) ∈ SL_{2}(ℤ) \ M_{n} has a unique representative of the

form (\begin{matrix}a & b \\ 0 & d \end{matrix}) with ad = n ; a, d > 0 ; 0 ≤ b < d.

Proof : Let gcd(a′, c′) = g, and let α = (\begin{matrix}∗ & ∗ \\ −\frac{c′}{g} & \frac{a′}{g} \end{matrix} ) ∈ SL_{2}(ℤ)

Then

α (\begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix}) = (\begin{matrix}∗ & ∗ \\ 0 & ∗ \end{matrix}) with determinant = n

= (\begin{matrix}a & ∗ \\ 0 & d \end{matrix}) with ad = n; ad > 0

Here we can assume by eventually multipying by (\begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix}) that a, d > 0

Also since (\begin{matrix}1 & t \\ 0 & 1 \end{matrix}) (\begin{matrix}a & b \\ 0 & d \end{matrix}) = (\begin{matrix}a & b + td \\ 0 & d \end{matrix})

there exists a unique t ∈ ℤ such that b + td ∈ [0, d)

Hence SL_{2}(ℤ) (\begin{matrix}a′ & b′ \\ c′ & d′ \end{matrix}) = SL_{2}(ℤ) (\begin{matrix}a & b \\ 0 & d \end{matrix}) with ad = n ; a, d > 0 ; 0 ≤ b < d. □

Proof of theorem 3 :

1) We can rewrite (1) as

T_{n}(f)(z) = n^{k−1} ∑_{γ=(\begin{matrix}a & b \\ c & d \end{matrix})∈SL_{2}ℤ\M_{n}}(cz + d)^{−k}f(\frac{az + b}{cz + d})

which, using the lemma :

= n^{k−1} ∑_{a,d>0_{ad=n}} ∑_{b=0}^{d−1} d^{−k}f(\frac{az + b}{d})

By bringing in the Fourier expansion

= n^{k−1} ∑_{a,d>0_{ad=n}} ∑_{b=0}^{d−1} ∑_{m=0}^{∞} d^{−k}a(m)e^{2πim( \frac{az+b}{d})}

= n^{k−1} ∑_{a,d>0_{ad=n}} ∑_{b=0}^{d−1} ∑_{m=0}^{∞} d^{−k}a(m)q^{ma/d}(e^{2πim/d})^{b}

if we rearrange

= n^{k−1} ∑_{a,d>0_{ad=n}} d^{−k} ∑_{m=0}^{∞} a(m)q^{ma/d}(∑_{b=0}^{d−1}(e^{2πim/d})^{b})

Here we note that

∑_{b=0}^{d−1}((e^{2πim/d})^{b}) = d if d|m and 0 otherwise

hence

= n^{k−1} ∑_{a,d>0_{ad=n}} d^{1−k} ∑_{d|m=0}^{∞} q^{am/d}

by changing variables m → md

= ∑_{a,d>0_{ad=n}} a^{k−1} ∑_{m=0}^{∞} a(dm)q^{am}

finally we group terms

= ∑_{m=0}^{∞} ∑_{a|(m,n)} a^{k−1}a(\frac{mn}{a^{2}})q^{m}

= ∑_{m=0}^{∞} ( ∑_{d|(m,n)} d^{k−1}a(\frac{mn}{d^{2}}))q^{m}

2) Exercise □