Let $E(z,s):=\pi^{-s}\Gamma (s) \sum_{(m,n)=1}\frac{y^s}{|mz+n|^{2s}}$ be the real-analytic Eisenstein series.

It satisfies the functional equation $E(z,s)=E(z,1-s)$ with two poles at $s=0,1$.

The method I know to prove this is to calculate the Fourier coefficients individually. They are either divisor functions or Riemann zeta function. Each satisfies the functional equation.

However, is there more conceptual way to prove this function equation without looking at individual coefficients?


1 Answer 1


As usual, the functional equation on the Dirichlet series side comes from a theta function on the modular side.

Using the poisson summation formula we show $$\vartheta_z(x) = \sum_{(c,d) \in \mathbb{Z}^2} \exp(-\pi x \frac{|cz+d|^2}{|\Im(z)|}) = x^{-1} \vartheta_z(1/x)$$

Then let $$E_z(s) = \sum_{\gamma \in SL_2(\mathbb{Z})} \Im(\gamma(z))^{s}=\sum_{(c,d) \in \mathbb{Z}^2, gcd(c,d)=1} (\frac{|cz+d|^{2}}{\Im(z)})^{-s}$$ (with $gcd(0,d) = d$). It comes from a Mellin transform $$\Gamma(s)\pi^{-s}\zeta(2s)E_z(s) = \Gamma(s)\pi^{-s}\sum_{c,d \in \mathbb{Z}^2 \setminus(0,0)} (\frac{|cz+d|^2}{|\Im(z)|})^{-s}\\ = \int_0^\infty (\vartheta_z(x)-1) x^{s-1}dx = \int_1^\infty (\vartheta_z(x)-1) x^{s-1}dx+\int_1^\infty (\vartheta_z(1/x)-1) x^{-s-1}dx \\ = \frac{1}{s-1}+\frac{1}{-s }+\int_1^\infty (\vartheta_z(x)-1) (x^{s-1}+x^{-s})dx$$ Which proves the functional equation.

Let $h(u) = e^{-\pi \|u\|^2}, u \in \mathbb{R}^2$ which is its own Fourier transform. For some self-adjoint matrix $B \in GL_2(\mathbb{R})$ let $g(u)= e^{-\pi u^T B u} = h(B^{1/2}u)$ then $\widehat{g}(u) = \frac{1}{|\det(B)|^{1/2}} h((B^{-1/2})^Tu)$. Then apply the Poisson summation formula to $$\theta_B(x) = \sum_{n \in \mathbb{Z}^2} e^{-\pi x (u^T Bu)}= \sum_{n \in \mathbb{Z}^2} h((x^{1/2} I)B n)\\=\sum_{n \in \mathbb{Z}^2} \frac{x^{-1}}{|\det(B)|^{1/2}} h((x^{-1/2}I)B^{-T/2}n)= \frac{x^{-1}}{|\det(B)|^{1/2}}\theta_{B^{-1}}(1/x)$$

Finally $|cz+d|^2 = (c,d)B{\scriptstyle\begin{pmatrix} c \\ d \end{pmatrix}}$ where $B = {\scriptstyle\begin{pmatrix} |z|^2 & \Re(z) \\ \Re(z) & 1\end{pmatrix}}, \det(B) = \Im(z)^2, B^{-1} = \frac{1}{\Im(z)^2} {\scriptstyle\begin{pmatrix} 1 & -\Re(z) \\ -\Re(z) & |z|^2\end{pmatrix}}$ so that $(c,d)B^{-1}{\scriptstyle\begin{pmatrix} c \\ d \end{pmatrix}} = \frac{|c-dz|^2}{\Im(z)^2}$ and $\theta_{B^{-1}}(x) =\theta_B(\frac{x}{\Im(z)^2})$ and $\vartheta_z(x) = \theta_B(\frac{x}{|\Im(z)|})$.

(it is very possible there are some typos)

  • $\begingroup$ Also it is clear this might work for $\sum_{\gamma \in \Gamma_0(N)} \Im(\gamma(z))^s$ and the twisted version $ \sum_{\gamma \in \Gamma_0(N)} \Im(\gamma(z))^s \chi(d)$ for some Dirichlet character $\chi$, but I never could do it properly. $\endgroup$
    – reuns
    Jul 31, 2017 at 1:37

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