# 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 $$gamma$$.

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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