Math 726: L-functions and modular forms Fall 2011

Week 3, lecture 8: More on Modular Forms

Instructor: Henri Darmon Notes written by: Luiz Kazuo Takei

Recall

We are studying modular forms with respect to a congruence group Γ, i.e., a subgroup Γ ≤ SL_{2}(ℤ) such that Γ ⊇ Γ(N) for some integer N.

Recall our notations:

modular forms M_{k}(Γ)
∪
cusp forms S_{k}(Γ).

More on Modular Forms

Claim 1. (to be justified) Cusp forms (of an appropriate type) play the same role for 2dimensional representations of G_{ℚ} as do Dirichlet characters for 1-dimensional representations of G_{ℚ}.

From now on, we assume

(\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) ∈ Γ.

This implies that, for any f ∈ S_{k}(Γ),

f(z) = ∑_{n=1}^{∞} a_{n}e^{2πinz}.

We can then define the L-function attached to f by

L(f, s) = ∑_{n=1}^{∞} a_{n}n^{−s}.

Question 1. Where does this sum converge?

In order to answer this question, we will introduce some basic results about Möbius transformations and modular forms. In what follows, we assume f ∈ S_{k}(Γ).

Lemma 1. y (\frac{az+b}{cz+d}) = \frac{y(z)}{(cz+d)(c\overline{z}+d)} for all ( \begin{matrix}a & b \\ c & d \end{matrix}) ∈ SL_{2}(ℝ).

Proof.

y (\frac{az + b}{cz + d}) = \frac{1}{2i} [\frac{az + b}{cz + d} − \overline{(\frac{az + b}{cz + d})}] = \frac{1}{2i} [\frac{(ad − bc)(z − \overline{z})}{(cz + d)(c\overline{z} + d)}] = \frac{y(z)}{(cz + d)(c\overline{z} + d)}.

Corollary 1. y^{k}f(z)\overline{f(z)} is invariant under Γ. Hence, so is y^{k/2}|f(z)|.

Lemma 2. There exists a constant C_{f} (depending on f) such that

|f(x + iy)| < C_{f}y^{−k/2}

for all z = x + iy ∈ H.

Proof. Since f is a cusp form, the values of y^{k/2}|f(z)| are uniformly bounded in a neighborhood of all the cusps. Hence, there exists C_{f} such that

y^{k/2}|f(z)| < C_{f}.

Proposition 1. There exists a constant C′_{f} such that

|a_{n}| < C′_{f}n^{k/2}.

Proof. Let z = x + iy ∈ H. Then

f(z) = ∑_{n=1}^{∞} a_{n}e^{2πin(x+iy)} = ∑_{n=1}^{∞} a_{n}e^{−2πny}e^{2πinx}.

By Fourier inversion,

a_{n}e^{−2πny} = ∫_{0}^{1} f(x + iy)e^{−2πinx}dx.

So

|a_{n}|e^{−2πny} ≤ ∫_{0}^{1} |f(x + iy)|dx ≤ C_{f}y^{−k/2}.

And this implies that

|a_{n}| ≤ C_{f}y^{−k/2}e^{2πny}.

Notice that for each y > 0 we have a bound for |a_{n}|. Choosing y = \frac{k}{4πn} will minimize the function on the right hand side and we obtain

|a_{n}| ≤ C′_{f}n^{k/2},

where C′_{f} = C_{f}k^{−k/2}(4π)^{k/2}e^{k/2}.

Corollary 2. L(f, s) converges absolutely for Re(s) > 1 + k/2. Definition 1. The Mellin transform of f is

M(f)(s) = ∫_{0}^{∞} f(it)t^{s} \frac{dt}{t}.

Proposition 2. M(f)(s) = (2π)^{−s}Γ(s)L(f, s). Proof. Just like in the proof of proposition 1 in lectures 5 and 6, we have that

M(f)(s) = ∫_{0}^{∞} f(it)t^{s} \frac{dt}{t} = ∫_{0}^{∞} ∑_{n=1}^{∞} a_{n}e^{−2πnt}t^{s} \frac{dt}{t} = ∑_{n=1}^{∞} ∫_{0}^{∞} a_{n}e^{−2πnt}t^{s} \frac{dt}{t}.

Applying the change of variables u = 2πnt yields

M(f)(s) = ∑_{n=1}^{∞} ∫_{0}^{∞} a_{n}e^{−u}u^{s}(2πn)^{−s} \frac{du}{u}.

Finally, we can rearrange terms and obtain

M(f)(s) = (2π)^{−s} (∑_{n=1}^{∞} a_{n}n^{−s}) ∫_{0}^{∞} e^{−u}u^{s} \frac{du}{u}.

Remark 1. If ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) ∈ Γ, then M_{k}(Γ) = \{0\} when k is odd.

Theorem 2. Assume f ∈ S_{k}(SL_{2}(ℤ)). Then L(f, s) extends to an analytic function on ℂ and, if

Λ(f, s) := (2π)^{−s}Γ(s)L(f, s),

then

Λ(f, s) = (−1)^{k/2}Λ(f, k − s).

Proof.

Λ(f, s) = M(f)(s) = ∫_{0}^{∞} f(it)t^{s} \frac{dt}{t} = ∫_{0}^{1} f(it)t^{s} \frac{dt}{t} + ∫_{1}^{∞} f(it)t^{s} \frac{dt}{t}.

Applying the change of variables t ↔ 1/t for the first integral gives us

∫_{1}^{∞} f(i/t)t^{−s} \frac{dt}{t}.

Now, since

f(i/t) = f(−1/it) = (it)^{k}f(it),

the first integral becomes

i^{k} ∫_{1}^{∞} f(it)t^{k−s} \frac{dt}{t} = (−1)^{k/2} ∫_{1}^{∞} f(it)t^{k−s} \frac{dt}{t}.

Hence

Λ(f, s) = (−1)^{k/2} ∫_{1}^{∞} f(it)t^{k−s} \frac{dt}{t} + ∫_{1}^{∞} f(it)t^{s} \frac{dt}{t},

which implies that Λ(f, s) has an analytic continuation to ℂ. Moreover, we can also readily see that

Λ(f, s) = (−1)^{k/2}Λ(f, k − s).

Remark 2. If f ∈ S_{k}(SL_{2}(ℤ)) then the sign in the functional equation for L(f, s) is

\{ \begin{matrix}1, & if k ≡ 0 (mod 4) \\ −1, & if k ≡ 2 (mod 4). \end{matrix}

In particular, if k ≡ 2 (mod 4), then L(f, k/2) = 0.

Question 2. Where are the zeros of L(f, s) on Re(s) > 1 + k/2?

Questions 3. (i) The proof of analytic continuation works well when ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ∈ Γ. What about other Γ (e.g. Γ_{0}(N), Γ_{1}(N))?

(ii) When does L(f, s) admit an Euler product factorization? And if so, what does it look like?

The key tool for these questions is

Hecke operators .