Les concepts de base des formes modulaires

Rephrasing http://www.math.umn.edu/~garrett/m/v/basic_rankin_selberg.pdf

- Take $\Gamma_0(N)$ a congruence subgroup of $SL_2(\mathbb{Z})$ containing $\scriptstyle\begin{pmatrix}1&1\\ 0 &1\end{pmatrix}$.

- Decompose $\Gamma_0(N) = P \times \Gamma_0(N)/P$ where $P = \{\scriptstyle\begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}$ $,n \in \mathbb{Z}\}$.

- Let $$E_s(z) = \sum_{\gamma \in \Gamma_0(N)/P} Im(\gamma z)^s, \qquad\qquad Re(s) > 1$$
By definition, it is $\Gamma_0(N)/P$ invariant, and we will show below it is also $P$ invariant. Therefore $E_s$ is $\Gamma_0(N)$ invariant.

- We'll show below that $\pi ^{-s} \Gamma(s) \zeta(2s) E_s(z)$ is invariant by $s \to 1-s$ and that $E_s$ is $\Gamma_0(N)$ invariant not only for $Re(s) > 1$.

- Take two modular forms $f \in S_k(\Gamma_0(N)),g\in M_l(\Gamma_0(N))$, and let
$$\langle f E_s,g\rangle = \int_{\Gamma_0(N)\setminus \mathbb{H}} f(z) E_s(z)\overline{g(z)} y^{k+l}\frac{dx dy}{y^2}$$
The idea is that $\varphi(z) =f(z)\overline{g(z)}y^{k+l}$ is $\Gamma_0(N) $ invariant. Since $\frac{dx dy}{y^2}$ is a $SL_2(\mathbb{Z})$ invariant measure, thus also $\Gamma_0(N)$ invariant, we can now write that
$$\int_{P\setminus \mathbb{H}} \psi(z)\varphi(z)\frac{dx dy}{y^2} = \sum_{\gamma \in \Gamma_0(N)/P}\int_{\Gamma_0(N)\setminus \mathbb{H}} \psi(\gamma z)\varphi(z)\frac{dx dy}{y^2} $$
and taking $\psi(z) = (Im(z))^s $ :
$$\int_{P\setminus \mathbb{H}} (Im(z))^s\varphi(z)\frac{dx dy}{y^2} = \int_{\Gamma_0(N)\setminus \mathbb{H}} \sum_{\gamma \in \Gamma_0(N)/P}(Im(\gamma z))^s\varphi(z)\frac{dx dy}{y^2} = \langle f E_s,g\rangle$$
- The LHS is much easier to compute : $\displaystyle\int_{P \setminus \mathbb{H}} = \int_0^\infty \int_0^1 $ and since $f(z) = \sum_{n\ge 1} a_n e^{2i \pi n z},g(z) = \sum_{n\ge 0} b_n e^{2i \pi n z}$ we get

$$\langle f E_s,g\rangle =\int_0^\infty \int_0^1 \sum_{n \ge 1,m \ge 0}a_n \overline{b_m} e^{-2\pi y (n+m)}e^{-2\pi(n+m)}y^{k+l+s-2}dxdy$$
$$=\sum_{n \ge 1,m \ge 0}a_n \overline{b_m}\int_0^\infty y^{s+k+l-2}e^{-2\pi(n+m)y}(\int_0^1 e^{2i \pi (n-m)x}dx)dy$$
$$ =\sum_{n \ge 1}a_n \overline{b_n}\int_0^\infty y^{s+k+l-2}e^{-4\pi n y}dy = \Gamma(s+k+l-1)(4\pi)^{k+l-s+1}\sum_{n \ge 1}a_n \overline{b_n} n^{k+l-s+1}$$

By construction $\pi ^{-s} \Gamma(s) \zeta(2s) \langle f E(.,s),g\rangle$ is invariant under $s \to 1-s$, so that $a_n \overline{b_n} n^{k+l}$ are the coefficients of a modular form.

---

About the needed properties of the real Eisenstein series $E_s(z)$ :

- Let $\scriptstyle\begin{pmatrix}a & b\\ c & d \end{pmatrix},\begin{pmatrix}e & f\\ c & d \end{pmatrix}$ $\in \Gamma_0(N)$, then $\scriptstyle\begin{pmatrix}e & f\\ c & d \end{pmatrix}\begin{pmatrix}a & b\\ c & d \end{pmatrix}^{-1} = \begin{pmatrix}e & f\\ c & d \end{pmatrix}\begin{pmatrix}d & -b\\ -c & a \end{pmatrix} = \begin{pmatrix}1 & af-eb\\ 0 & 1 \end{pmatrix}$ $ \in P$


- $$E_s(z) = \sum_{\gamma \in \Gamma_0(N)/P} (Im(\gamma z))^s = \sum_{\gamma \in \Gamma_0(N)/P} \frac{Im(z)}{|cz+d|^{2s}} =\sum_{(c,d), gcd(Nc,d) = 1} \frac{Im(z)}{|cNz+d|^{2s}} $$
Then look at the modified series
$$\tilde{E}_s(z) = \sum_{(c,d) \ne (0,0)} \frac{Im(z)}{|cNz+d|^{2s}} = \sum_{f=1}^\infty \sum_{(c,d), gcd(Nc,d) = 1} \frac{Im(z)}{f^{2s}|cNz+d|^{2s}} = \zeta(2s)E_s(z)$$
$\tilde{E}_s(z)$ is clearly $P$ invariant, therefore $E_s(z)$ is $P$ invariant too.

- It is also easier to put in an integral form. Let $u= (x,y)$ and $\phi(u)= e^{-\pi |u|^2}$ and $\Theta(g) = \sum_{c,d} \phi((c,d)g)$ for some matrix $ g \in GL_2(\mathbb{R})$ and
$$F_s(g) = \int_0^\infty t^{s-1} (\Theta(tg)-1)dt = \pi^{-s}\Gamma(s)\sum_{(c,d) \ne (0,0)} |(c,d)g|^{-2s}$$
choosing $g = \scriptstyle\begin{pmatrix}1&Nx \\ 0 & 1\end{pmatrix}\begin{pmatrix}\sqrt{Ny}&0 \\ 0 & 1/\sqrt{Ny} \end{pmatrix}$ we have $(c,d)g = (\sqrt{Ny}c,\frac{Ncx+d}{\sqrt{Ny}})$ and $|(c,d)g|^2 = \frac{N^2y^2 c^2+ (Ncx+d)^2}{Ny} = \frac{|cN(x+iy)+d|^2}{Ny}$ so that $ F_s(g) = \pi^{-s}N^{s}\Gamma(s) \tilde{E}_s(x+iy)$

- The bi-dimensional Dirac comb $\text{I}\Pi(x,y) = \sum_{c,d} \delta(x-c)\delta(y-d)$ is its own Fourier transform, as $\phi((x,y))= e^{-\pi(x^2+y^2)}$

- And we have the following effect of a linear transformation $g \in GL_2(\mathbb{R})$ on the Fourier transform :
$$\mathcal{F}[h(gx)](\xi) = \int_{\mathbb{R}^n} e^{2i \pi \xi^T x} h(gx)dx = \int_{\mathbb{R}^n} e^{2i \pi \xi^T g^{-1} y} h(y)dg^{-1}y = \frac{1}{\det(g)}\mathcal{F}[h(x)](g^{-T}\xi) $$


- Hence the Poisson summation formula applies
$$\Theta(tg) = \langle \text{I}\Pi,\phi((x,y)tg)\rangle =\langle \text{I}\Pi(x,y),\frac{1}{t}\phi((x,y)g^{-T}/t)\rangle
= \frac{1}{t}\sum_{c,d} \phi((c,d)g^{-T}/t) = \frac{1}{t}\Theta(g^{-T}/t)$$

where $g^{-1} =\scriptstyle \begin{pmatrix}1/\sqrt{Ny}&0 \\ 0 & \sqrt{Ny} \end{pmatrix}\begin{pmatrix}1&-Nx \\ 0 & 1\end{pmatrix}$ and $g^{-T} = \scriptstyle\begin{pmatrix}1&0 \\ -Nx & 1\end{pmatrix}\begin{pmatrix}1/\sqrt{Ny}&0 \\ 0 & \sqrt{Ny} \end{pmatrix}$ .


Note that $(c,d)g^{-T} = (c-dNx,d)\scriptstyle\begin{pmatrix}1/\sqrt{Ny}&0 \\ 0 & \sqrt{Ny} \end{pmatrix}$ $ = (\frac{c-dNx}{\sqrt{Ny}},\sqrt{Ny}d)$, $|(c,d)g^{-T}|^2 = \frac{(c-dNx)^2+ d^2N^2y^2}{N y} = \frac{|Nd(x+iy)-c|^2}{Ny}$ and hence
$$\int_0^\infty t^{s-1} (\Theta(g^{-T}t)-1)dt = \pi^{-s}\Gamma(s)\sum_{(c,d) \ne (0,0)}\frac{N^s y^s}{|Nd(x+iy)-c|^2} = \pi^{-s}N^s\Gamma(s) \tilde{E}_{s}(z)$$
Proving that (for this marticular matrix $g$) : $\ \Theta(g^{-T}t) = \Theta(gt)$ and

- $\displaystyle\int_0^1 t^{s-1} (\Theta(gt)-1)dt=\int_0^1 t^{s-1} (\frac{1}{t}\Theta(g^{-T}/t)-1)dt =\int_1^\infty t^{-s-1} (t\Theta(g^{-T}t)-1)dt$ $\displaystyle =\frac{1}{s-1}-\frac{1}{s}+\int_1^\infty t^{(1-s)-1} ( \Theta(g^{-T}t)-1)dt
= \frac{1}{s-1}-\frac{1}{s}+\int_1^\infty t^{(1-s)-1} ( \Theta(gt)-1)dt$
i.e.
$$\int_0^\infty t^{s-1} (\Theta(gt)-1)dt=\frac{1}{s-1}-\frac{1}{s}+\int_1^\infty (t^{s-1}+t^{(1-s)-1} ) ( \Theta(gt)-1)dt$$
and we get the desired formula
$$\Lambda(s,z)=\pi^{-s}\Gamma(s) \tilde{E}_{s}(z)= \pi^{-s}\Gamma(s)\zeta(2s) E_{s}(z)$$
$$\Lambda(1-s,z)=\Lambda(s,z)$$