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Fourier expansions of Eisenstein series as a Poincare series for the Fuchsian group

Mathematics Asked by LWW on January 7, 2022

In Miyake’s book, Modular Forms, Ch 2.6, thm 2.6.9, there is a statement which relate to Fourier expansion of the Eisenstein series.

Let $Gamma$ be a Fuchsian group, $chi$ a character of $Gamma$ of finite order, and $k$ an integer. We suppose $chi(-1)=(-1)^k$ if $-1 in Gamma$. Let $Lambda$ be a subgroup of $Gamma$, and $phi$ be a meromorphic function of $mathbb{H}$ which is a seed function for a the Poincare series. We write the Poincare series by

$$
F(z)=F_k(z;phi,chi,Lambda,Gamma)=sum_{gamma in Lambda setminus Gamma} overline{chi(gamma)} (phi|_k gamma)(z).
$$

Let $x$ be a cusp and $sigma in SL_2(mathbb{R})$ satisfies $sigma(x)=infty$, and $h$ be a cusp width of $x$. Take $phi_m(z)=j(sigma,z)^{-k}e^{2pi i m sigma z/h}$, $Lambda=Gamma_x$, and take $chi$ which satisfies

$$
chi(gamma)j(sigma gamma sigma^{-1},z)^k =1
$$

for $gamma in Gamma_x$. If $m=0$, then $F(z)=F_k(z;phi_0,chi,Gamma_x,Gamma)$ is called the Eisenstein series.

Theorem. Suppose $kgeq 3$. If $m=0$, then $F(z)=F_k(z;phi_0,chi,Gamma_0,Gamma) in M_k(Gamma,chi).$ It has the Fourier expansion at $x$ of the form
$$
(F|_k sigma^{-1})(z)=1+sum_{n=1}^{infty} a_n e^{2pi i n z/h},
$$

and vanishes at all cusps which are inequivalent to $x$.

But I don’t know why the Fourier coefficient $a_0$ is $1$.

Here is my attempt.

$$
a_0=frac{1}{h}int_{z_0}^{z_0+h} (F|_k sigma^{-1})(z) dz
$$

$$
=frac{1}{h}int_{z_0}^{z_0+h}F_k(z;1,chi^{sigma},sigmaGamma_x sigma^{-1} ,sigma Gamma sigma^{-1}) dz,
$$

where $chi^{sigma}(sigma gamma sigma^{-1})=chi(gamma)$, and the above integral is

$$
=frac{1}{h}int_{z_0}^{z_0+h}sum_{gamma in Gamma_x setminus Gamma } chi(gamma)^{-1}j(sigma gamma sigma^{-1},z)^{-k} dz.
$$

The last integral must be 1, and if $gamma in Gamma_x$, we know that $chi(gamma)j(sigma gamma sigma^{-1},z)^k=1$. But I don’t know about the other $gamma$.

One Answer

It seems that you started off well enough. Maybe for clarity, note that we may get rid of $sigma$, assume that $x = infty$ and what one really wants to show is that for all characters $chi$ of finite order, which acts trivially on $Gamma_{infty}$ we have $$ I = int_{z_0}^{z_0+h} sum_{1 ne gamma in Gamma_{infty} backslash Gamma} chi(gamma) j(gamma, z)^{-k} dz = 0 $$

One could do that directly (see below), but in fact Miyake already does that in the course of proving Theorem 2.6.7. I would first like to draw your attention to condition (v), right before the theorem. Note that for $phi = 1$, if $sigma^{-1} = left( begin{array}[cc] aa & b \ c & d end{array} right)$, then we get different bounds when $c = 0$ and when $c ne 0$. Namely, for $c = 0$, we have $varepsilon = 0$, but for $c ne 0$, we have the much stronger $varepsilon = k$.

Now, consider again the functions $phi_{alpha}$ in the proof of Theorem 2.6.7. In this case, $alpha$ runs over a set of representatives for $Gamma_{infty} backslash Gamma / Gamma_{infty}$ .

Note that the lower left entry of a matrix (the one we call $c$) is the same for all elements of the double coset. Moreover, an element with $c = 0$ must be in $Gamma_{infty}$. Therefore, for any nontrivial $alpha$, we have $c ne 0$, and the stronger bound, implying that for any element $alpha beta$ in the double coset, this is the case, and so by the same proof as in that of Theorem 2.6.7, we see that $phi_{alpha}$ vanishes at $infty$ for all $alpha ne Gamma_{infty}$.

Since $F(z) = sum_{alpha} phi_{alpha} (z)$ (see 2.6.6), we see that at $infty$ the value of $F$ coincides with that of $phi_1 = 1$.

(*) If you would really like to evaluate the integral, here is one way to proceed: begin{align*} I &= int_{Gamma_{infty} backslash mathbb{R}} sum_{1 ne alpha in Gamma_{infty} backslash Gamma / Gamma_{infty}} sum_{beta in Gamma_{infty} alpha backslash Gamma} chi(alpha beta) j(alpha beta, z)^{-k} dz \ &= sum_{1 ne alpha in Gamma_{infty} backslash Gamma / Gamma_{infty}} chi(alpha) int_{Gamma_{infty} backslash mathbb{R}} sum_{beta in (alpha^{-1} Gamma_{infty} alpha cap Gamma_{infty}) backslash Gamma_{infty}} j(alpha, beta z)^{-k} dz \ &= sum_{1 ne alpha in Gamma_{infty} backslash Gamma / Gamma_{infty}} chi(alpha) int_{(alpha^{-1} Gamma_{infty} alpha cap Gamma_{infty}) backslash mathbb{R}} j(alpha, z)^{-k} dz = 0. end{align*}

Here we have used that $chi(beta) = 1$ for $beta in Gamma_{infty}$, that $j(alpha beta, z) = j(alpha, beta z) j(beta, z)$, that $j(beta, z) = 1$ for $beta in Gamma_{infty}$, that the integral $int j(alpha,z)^{-k} dz$ converges for $alpha notin Gamma_{infty}$, and that the sum $sum chi(alpha)$ vanishes. This only works when $chi$ is non-trivial due to convergence issues, but it gives a rough idea of what one should do.

Answered by assaferan on January 7, 2022

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