\documentclass[12pt]{article} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsthm} \usepackage{amsmath} \usepackage{graphicx} \usepackage{color} \usepackage[colorlinks]{hyperref} \usepackage{newlfont} \usepackage{mathrsfs} \usepackage[active]{srcltx} % SRC Specials for DVI Searching \begin{document} \title{\bf{Fourier Coefficients in Vector Valued Modular Forms }} \author{Pepe Gimenez} \date{\today} \maketitle \noindent {\setlength{\baselineskip}% {1.5\baselineskip}blah blah blah \begin{align} F_1(\tau)& = \sum_{n=\mu_1}^{\infty}a_n(1)q^\frac{n}{N_1}, \quad \mu_1 \epsilon\mathbb{Z}, \quad q=e^{2\pi i \tau}\\ & = q^{m_1}\sum_{n=\beta_1}^{\infty}a_n(1)q^n, \quad \beta_1 \epsilon\mathbb{Z}, \quad 0 \leqslant m_1< 1, \quad q=e^{2\pi i \tau} \end{align} blah blah blah \begin{equation}\label{EqTeddy} \rho:\Gamma\rightarrow GL_2(\mathbb{C}) \end{equation} blah blah blah (\ref{EqTeddy}) is a representation \begin{equation} \rho (V)=\left(% \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{array}% \right) \end{equation} \begin{equation} V=\left(% \begin{array}{cc} a & b \\ c & d \\ \end{array}% \right), \quad V\epsilon SL_2(\mathbb{Z}) \end{equation} \begin{align}\label{EqtransFj} (c\tau+d)^{r}F_1(V\tau)& =b_{11}F_1(\tau)+ b_{12}F_2(\tau)\\ (c\tau+d)^{r}F_2(V\tau)& =b_{21}F_1(\tau)+ b_{22}F_2(\tau) \end{align} Let $x=e^{2 \pi i \tau}$, and define \begin{equation} q^{-m_1}F_1(\tau)=f_1(x)=\sum_{n=\beta_1}^{\infty}a_n(1)x^n \end{equation} The functions $f_j(x)$ are analytic inside the unit circle except at zero where there is a pole. \begin{equation} a_m(j)=\frac{1}{2\pi i}\int_C \frac {f_j(x)} {x^{m+1}}dx \end{equation} where C could be the circle $|x|=e^{2\pi N^{-2}}$ \begin{equation} a_m(j)=\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\frac{1}{2\pi i}\int_{\xi_{h,k}}\frac {f_j(x)} {x^{m+1}}dx \end{equation} where $\xi_{h,k}$ is the Farey arc. We can make the change of variable \begin{equation}\label{EqCV} x=e^{-2\pi N^{-2}+ 2\pi i\frac{h}{k} + 2\pi i \varphi}, \quad -\theta_{h,k}^{'}\leq \varphi \leq \theta_{h,k}^{''} \end{equation} \begin{equation}\label{Eqamj} a_m(j)=e^{-2\pi N^{-2}m}\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}e^{-2\pi i m \frac{h}{k}}\int_{-\theta_{h,k}^{'}}^{\theta_{h,k}^{''}}f_j(e^{2\pi i \frac{h}{k}-2\pi(N^{-2}-i \varphi)})e^{-2\pi i m \varphi}d\varphi \end{equation} From now on we will consider $c>0$ in $V$and we write \begin{equation} -i(c\tau + d)=z, \quad \tau=\frac {i z}{c}-\frac {d}{c}, \quad \Re(z)>0 \end{equation} and choose the modular transformation \begin{equation} V=\left(% \begin{array}{cc} a & b \\ c & d \\ \end{array}% \right)=\left(% \begin{array}{cc} h' & -\frac {hh'+1}{k} \\ k & -h \\ \end{array}% \right) \end{equation} Where $h'$ is a solution of \begin{equation} hh'\equiv -1 \pmod {k} \end{equation} We then have \begin{equation}\label{Eqdefz} \tau=\frac{i z}{k}+\frac{h}{k},\quad V\tau=\tau'=\frac{i}{kz}+\frac{h'}{k} \end{equation} And therefore by (\ref{EqtransFj}) \begin{align} F_1(V\tau)=\varepsilon(V)z^{-r}(b_{11}F_1(\tau)+ b_{12}F_2(\tau))\\ F_2(V\tau)=\varepsilon(V)z^{-r}(b_{21}F_1(\tau)+ b_{22}F_2(\tau)) \end{align} Let \begin{equation} \begin{array}{cc} M_1^1(V)= \frac{\varepsilon^{-1}(V)(-i(c\tau + d))^{r}}{b_{11}} & M_2^1(V)=-\frac{b_{12}}{b_{11}} \\ M_1^2(V)=-\frac{b_{21}}{b_{22}} & M_2^2(V)=\frac{\varepsilon^{-1}(V)(-i(c\tau + d))^{r}}{b_{22}} \\ \end{array} \end{equation} Then, provided that $b_{11},b_{22}\neq 0$ \begin{align} e^{2\pi i m_1\tau}f_1(x)& = F_1\left(\tau\right)\\ & =\frac{\varepsilon^{-1}(V)(-i(c\tau + d))^{r}F_1(V\tau)-b_{12}F_2(\tau)}{b_{11}}\\ & =M_{1}^1(V)F_1(V\tau)+M_{2}^1(V)F_2(\tau) \end{align} \begin{align*} e^{2\pi i m_2\tau}f_2(x)& =F_2\left(\tau\right)\\ & =\frac{\varepsilon^{-1}(V)(-i(c\tau + d))^{r}F_2(V\tau)-b_{21}F_1(\tau)}{b_{22}}\\ & =M_{1}^2(V)F_1(\tau)+M_{2}^2(V)F_2(V\tau) \end{align*} Therefore \begin{align*} e^{2\pi i m_1\tau}f_1(x)& =M_{1}^1(V)F_1(V\tau)+M_{2}^1(V)(M_{1}^2(V)F_1(\tau)+M_{2}^2(V)F_2(V\tau))\\ & =M_{1}^1(V)F_1(V\tau)+M_{2}^1(V)(M_{1}^2(V)(e^{2\pi i m_1\tau}f_1(x))+M_{2}^2(V)F_2(V\tau)) \end{align*} \begin{equation} (1-M_{2}^1(V)M_{1}^2(V))e^{2\pi i m_1\tau}f_1(x)=M_{1}^1(V)F_1(V\tau)+M_2^1(V)M_{2}^2(V)F_2(V\tau)) \end{equation} \begin{align*} e^{2\pi i m_2\tau}f_2(x)& =M_{1}^2(V)(M_{1}^1(V)F_1(V\tau)+M_{2}^1(V)F_2(\tau))+M_{2}^2(V)F_2(V\tau)\\ & =M_{1}^2(V)(M_{1}^1(V)F_1(V\tau)+M_{2}^1(V)(e^{2\pi i m_2\tau}f_2(x)))+M_{2}^2(V)F_2(V\tau) \end{align*} \begin{equation} (1-M_{1}^2(V)M_{2}^1(V))e^{2\pi i m_1\tau}f_2(x)=M_{1}^2(V)M_{1}^1(V)F_1(V\tau)+M_{2}^2(V)F_2(V\tau) \end{equation} \begin{align} M_1^2(V)M_1^1(V)=-\frac{b_{21}\varepsilon^{-1}(V)(-i(c\tau + d))^{r}}{b_{11}b_{22}} \end{align} \begin{align} M_2^1(V)M_2^2(V)=-\frac{b_{12}\varepsilon^{-1}(V)(-i(c\tau + d))^{r}}{b_{11}b_{22}} \end{align} \begin{align} M_2^1(V)M_1^2(V)=\frac{b_{12}b_{21}}{b_{11}b_{22}} \end{align} \begin{align} 1-M_2^1(V)M_1^2(V)& =\frac{b_{11}b_{22}-b_{12}b_{21}}{b_{11}b_{22}}\\ & =\frac{|\rho(V)|}{b_{11}b_{22}} \end{align} therefore, provided $|\rho(V)|\neq 0$. Question: Can we always choose a $V$ Such that $|\rho(V)|\neq 0$ and $b_{11},b_{22}\neq 0$? \begin{align} f_1(x)=\frac{e^{-2\pi i m_1 \tau}\varepsilon^{-1}(V)z^{r}}{|\rho(V)|}\left(b_{22}F_1(V\tau)- b_{12}F_2(V\tau)\right) \end{align} \begin{align} f_2(x)=\frac{e^{-2\pi i m_2 \tau}\varepsilon^{-1}(V)z^{r}}{|\rho(V)|}\left( b_{11}F_2(V\tau)-b_{21}F_1(V\tau)\right) \end{align} And \begin{align*} \{ f(x)\}^t & = \left( f_1(x),f_2(x)\right)\\ & = \left(% \begin{array}{cc} e^{-2\pi i m_1\tau} & o \\ o & e^{-2\pi i m_2\tau} \\ \end{array}% \right)\left(% \begin{array}{c} F_1(\tau) \\ F_2(\tau) \\ \end{array}% \right)\\ & =\varepsilon^{-1}(V)z^{r}\rho(V)^{-1} \left(% \begin{array}{cc} e^{-2\pi i m_1\tau} & 0 \\ 0 & e^{-2\pi i m_2\tau} \\ \end{array}% \right)F(V\tau)\\ & =\varepsilon^{-1}(V)z^{r}\rho(V)^{-1} \left(% \begin{array}{cc} e^{-2\pi i m_1\tau} & 0 \\ 0 & e^{-2\pi i m_2\tau} \\ \end{array}% \right)\left(% \begin{array}{cc} e^{2\pi i m_1V\tau} & 0 \\ 0 & e^{2\pi i m_2V\tau} \\ \end{array}% \right)f(V\tau)\\ & =\varepsilon^{-1}(V)z^{r}\rho(V)^{-1}A(m_1,m_2)B (m_1,m_2,V)f(e^{-\frac{2\pi }{k z}}e^{\frac{2\pi i h' }{k}}) \end{align*} where by (\ref{Eqdefz}) \begin{align} A(m_1,m_2)& =\left(% \begin{array}{cc} e^{2\pi m_1\left(\frac{z-i h}{k}\right)} & 0 \\ 0 & e^{2\pi m_2\left(\frac{z-i h}{k}\right)} \\ \end{array}% \right)\\ & =\left(% \begin{array}{cc} e^{2\pi m_1\frac{z}{k}} & 0 \\ 0 & e^{2\pi m_2\frac{z}{k}} \\ \end{array}% \right)\left(% \begin{array}{cc} e^{-2\pi i m_1\frac{h}{k}} & 0 \\ 0 & e^{-2\pi i m_2\frac{h}{k}} \\ \end{array}% \right) \end{align} and \begin{align}B (m_1,m_2,V)& = \left(% \begin{array}{cc} e^{-\frac{2\pi m_1}{k z}}e^{\frac{2\pi i h' m_1}{k}} & 0 \\ 0 & e^{-\frac{2\pi m_2}{k z}}e^{\frac{2\pi i h' m_2}{k}} \\ \end{array}% \right)\\ & = \left(% \begin{array}{cc} e^{-2\pi m_1\frac{1}{kz}} & 0 \\ 0 & e^{-2\pi m_2\frac{1}{kz}} \\ \end{array}% \right)\left(% \begin{array}{cc} e^{2\pi i m_1\frac{h'}{k}} & 0 \\ 0 & e^{2\pi i m_2\frac{h'}{k}} \\ \end{array}% \right) \end{align} Therefore if we restrict $h'$ to $0 \leq h'< k$, then $h'$ is unique, and we can set \begin{equation} \Omega_{h,k}=\varepsilon^{-1}(V)\rho(V)^{-1}\left(% \begin{array}{cc} e^{2\pi i m_1\frac{h'-h}{k}} & 0 \\ 0 & e^{2\pi i m_2\frac{h'-h}{k}} \\ \end{array}% \right) \end{equation} and \begin{equation} \Psi_k(z)=z^{r} \left(% \begin{array}{cc} e^{2\pi m_1\frac{z-1/z}{k}} & 0 \\ 0 & e^{2\pi m_2\frac{z-1/z}{k}} \\ \end{array}% \right) \end{equation} Then \begin{equation}\label{Eqtrans} f(e^{-2\pi\frac{z-ih}{k}})^t=\Omega_{h,k}\Psi_k(z)f(e^{2\pi i \frac{h'}{k}}e^{-2\pi\frac{1}{kz}}) \end{equation} Now let $a_m$ be the vector of the m-th Fourier coefficients \begin{equation} a_m^t=(a_m(1),a_m(2)) \end{equation} then by Cauchy integral formula, \begin{equation} a_m^t=\frac{1}{2\pi i}\int_C \frac{f(x)^t}{x^{m+1}}dx, \end{equation} where $C$ could be the circle $|x|=e^{2\pi N^{-2}}$. We understand that \begin{align} \int_C\frac{f(x)^t}{x^{m+1}}dx& =\int_C\left(% \begin{array}{cc} \frac{f_1(x)}{x^{m+1}}, & \frac{f_2(x)}{x^{m+1}} \\ \end{array}% \right)dx\\ & =\left(% \begin{array}{cc} \int_C\frac{f_1(x)}{x^{m+1}}dx, & \int_C\frac{f_2(x)}{x^{m+1}}dx \\ \end{array}% \right) \end{align} \begin{equation} a_m^t=\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\frac{1}{2\pi i}\int_{\xi_{h,k}}\frac {f(x)^t} {x^{m+1}}dx \end{equation} where $\xi_{h,k}$ is the Farey arc. As we did in (\ref{Eqamj}), we can make the change of variable (\ref{EqCV}) to get \begin{equation} a_m^t=e^{-2\pi N^{-2}m}\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}e^{-2\pi i m \frac{h}{k}}\int_{-\theta_{h,k}^{'}}^{\theta_{h,k}^{''}}f(e^{2\pi i \frac{h}{k}-2\pi(N^{-2}-i \varphi)})^t e^{-2\pi i m \varphi}d\varphi \end{equation} Now by (\ref{Eqtrans}), where $z=k(N^{-2}-i \varphi)$ \begin{equation}\begin{split} a_m^t=e^{2\pi N^{-2}m}\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\Omega_{h,k}e^{-2\pi i h\frac{m}{k}}\int_{-\theta_{h,k}^{'}}^{\theta_{h,k}^{''}}\Psi_k(k(N^{-2}-i \varphi))\\ f\left(e^{\frac{2\pi}{k}\left(ih'-k^{-1}(N^{-2}-i \varphi)^{-1}\right)}\right)^t e^{-2\pi i m \varphi}d\varphi \end{split}\end{equation} \section{An Estimation} We will show now that $f(x)$ in the neighborhood of $x=0$ is dominated by the principal part $P(x)$, where \begin{align} P(x)=\left(% \begin{array}{c} \sum_{m=1}^{-min(\beta_1,\beta_2)}a_m (1)x^{-m} \\ \sum_{m=1}^{-min(\beta_1,\beta_2)}a_m (2)x^{-m} \\ \end{array}% \right), && D(x)=f(x)-P(x) \end{align} \begin{align} a_m& =\left(% \begin{array}{c} a_m(1) \\ a_m(2) \\ \end{array}% \right)\\ & =\left(% \begin{array}{c} Q_m(1)+R_m(1) \\ Q_m(2)+R_m(2) \\ \end{array}% \right)\\ & =\left(% \begin{array}{c} Q_m(1) \\ Q_m(2) \\ \end{array}% \right)+\left(% \begin{array}{c} R_m(1) \\ R_m(2) \\ \end{array}% \right)\\ & = Q_m + R_m \end{align} where \begin{equation}\begin{split} Q_m^t=e^{2\pi N^{-2}m}\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\Omega_{h,k}e^{-2\pi i h\frac{m}{k}}\int_{-\theta_{h,k}^{'}}^{\theta_{h,k}^{''}}\Psi_k(k(N^{-2}-i \varphi))\\ P\left(e^{\frac{2\pi}{k}\left(ih'-k^{-1}(N^{-2}-i \varphi)^{-1}\right)}\right)^t e^{-2\pi i m \varphi}d\varphi \end{split}\end{equation} \begin{equation}\begin{split} R_m^t=e^{2\pi N^{-2}m}\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\Omega_{h,k}e^{-2\pi i h\frac{m}{k}}\int_{-\theta_{h,k}^{'}}^{\theta_{h,k}^{''}}\Psi_k(k(N^{-2}-i \varphi))\\ D\left(e^{\frac{2\pi}{k}\left(ih'-k^{-1}(N^{-2}-i \varphi)^{-1}\right)}\right)^t e^{-2\pi i m \varphi}d\varphi \end{split}\end{equation} We are going to make an estimate of $R_m$. From the theory of Farey fractions we have \begin{align} \frac{1}{2kN} \leq \theta_{h,k}^{'} \leq \frac{1}{kN}, && \frac{1}{2kN} \leq \theta_{h,k}^{''} \leq \frac{1}{kN} \end{align} and therefore since $k \leq N$, we find for $-\theta_{h,k}^{'} \leq \varphi \leq \theta_{h,k}^{''}$ \begin{equation} \Re\left(k(N^{-2}-i \varphi) \right)=kN^{-2}, \end{equation} \begin{equation} \Re\left(\frac{1}{k(N^{-2}-i \varphi) }\right)=\frac{N^{-2}}{k(N^{-4}+ \varphi^2)}\geq \frac{N^{-2}}{k(N^{-4}+ k^{-2}N^{-2})}=\frac{k}{k^2N^{-2}+1}\geq\frac{k}{2}, \end{equation} \begin{equation} |k(N^{-2}-i \varphi)|=k(N^{-4}+ \varphi^2)^{\frac{1}{2}}\leq (k^2 N^{-4}+ N^{-2})^{\frac{1}{2}} \leq 2^{\frac{1}{2}}N^{-1} \end{equation} \begin{equation} |e^{\frac{2\pi m_j}{k}\left(k(N^{-2}-i \varphi))-1/(k(N^{-2}-i \varphi)\right)}|\leq e^{\frac{2\pi m_j}{k}\left( kN^{-2} -\frac{k}{2}\right)}=e^{2\pi m_j N^{-2}}e^{-\pi m_j} \end{equation} Therefore \begin{eqnarray*} \lefteqn{|\Psi_k(k(N^{-2}-i \varphi))| } \\ & & \leq |(k(N^{-2}-i \varphi))|^{r}\sqrt{|e^{\frac{2\pi m_1}{k}\left(k(N^{-2}-i \varphi))-1/(k(N^{-2}-i \varphi)\right)}|^2+|e^{\frac{2\pi m_2}{k}\left(k(N^{-2}-i \varphi))-1/(k(N^{-2}-i \varphi)\right)}|^2} \\ & & \leq 2^{\frac{r}{2}}N^{-r}\sqrt{2}e^{2\pi \alpha N^{-2}}e^{-\pi \beta } \end{eqnarray*} where $\alpha= max\{m_1,m_2\}$ and $\beta=min \{m_1,m_2\}$ Also, using the fact that if $a,b>0$, then $(a+b)^2=a^2+2ab+b^2 \geq a^2 +b^2$ \begin{eqnarray*} \lefteqn{|D\left(e^{\frac{2\pi}{k}\left(ih'-k^{-1}(N^{-2}-i \varphi)^{-1}\right)}\right)^t|}\\ && \leq \sum_{m=0}^\infty |a_m|e^{-2\pi m \Re\left(k^{-1}(N^{-2}-i \varphi)^{-1} \right)}\\ && \leq \sum_{m=0}^\infty |a_m|e^{-\pi m}\\ && = \sum_{m=0}^\infty\sqrt{|a_m(1)|^2+|a_m(2)|^2}e^{-\pi m}\\ && \leq \sum_{m=0}^\infty\left(|a_m(1)|+|a_m(2)| \right)e^{-\pi m}\\ && = \sum_{m=0}^\infty|a_m(1)|e^{-\pi m}+\sum_{m=0}^\infty|a_m(2)|e^{-\pi m} \end{eqnarray*} the last equality holds since it converges absolutely since $|e^{-\pi}|\leq 1$ Using these results we have \begin{eqnarray*} \lefteqn{|\Psi_k(k(N^{-2}-i \varphi))D\left(e^{\frac{2\pi}{k}\left(ih'-k^{-1}(N^{-2}-i \varphi)^{-1}\right)}\right)^t|}\\ && \leq 2^{\frac{r}{2}}N^{-r}\sqrt{2}e^{2\pi \alpha N^{-2}}e^{-\pi \beta }\left(\sum_{m=0}^\infty|a_m(1)|e^{-\pi m}+\sum_{m=0}^\infty|a_m(2)|e^{-\pi m}\right)\\ && =CN^{-r}e^{2\pi \alpha N^{-2}} \end{eqnarray*} where \begin{equation} C=2^{\frac{r}{2}}\sqrt{2}e^{-\pi \beta }\left(\sum_{m=0}^\infty|a_m(1)|e^{-\pi m}+\sum_{m=0}^\infty|a_m(2)|e^{-\pi m}\right) \end{equation} Which is finite since $|e^{-\pi}|\leq 1$ and both series are convergent inside the unit circle \begin{eqnarray*} \lefteqn{|\Omega_{h,k}|} \\ && =|\varepsilon^{-1}(V)\rho(V)^{-1}\left(% \begin{array}{cc} e^{2\pi i m_1\frac{h'-h}{k}} & 0 \\ 0 & e^{2\pi i m_2\frac{h'-h}{k}} \\ \end{array}% \right)| \end{eqnarray*} In order to bound $|\Omega_{h,k}|$ we have to discuss some things about $\rho(V)$. Let $L(V)$, be the Eichler length of $V$, with respect to the generators $S$ and $T$ of $\Gamma$. Namely, we write $V$ as a product \begin{equation} V=\pm V_1\ldots V_L \end{equation} where each $V_j$ is equal to either $S$ or $T^{n_j}$ for some integer $n_j$, no two consecutive $V_j$ are both equal to $S$ or a power of $T$, and where $L$ is minimal. we have that \begin{equation} L(V) \leq n_1\log\mu(V)+n_2 \end{equation} Where $\mu (V)= {h'}^2+\left( \frac{hh'+1}{k}\right)^2+k^2+h^2$ and $n_1$, $n_2$ are constants independent of $V$ Now we have $\rho(V)= \rho(V_1)\ldots\rho(V_L)$, so that \begin{equation} |\rho_{jm}(V)|\leq \sum_{1 \leq m1,\ldots,m_{L-1} \leq p} |\rho_{jm_1}(V_1)|.|\rho_{m_1m_2}(V_2)|\ldots |\rho_{m_{L-1}m}(V_L)| \end{equation} Let $K_1$ be a constant such that it is a bound for all elements in the matrix $\rho (S)$. Basically $|\rho_{lm} (S)| \leq K_1$ for all $1 \leq m,l \leq p$. Also if $V_j=T^{n_j}$ then $|\rho_{lm}(V)|=\delta_{l,m}$ (Kronecker delta) because the Vector valued form is reduced. Also since there are $p^{L-1}$ summands we have \begin{equation} |\rho_{jm}(V)| \leq p^{L(V)-1}K_1^{L(V)} \end{equation} Therefore, since we have $p^2$ elements in $\rho(V)$ \begin{eqnarray*} \lefteqn{|\rho(V)|}\\ &&\leq \sqrt{p^2\left(p^{L(V)-1}K_1^{L(V)}\right)^2}\\ &&=\left(pK_1\right)^{L(V)}\\ &&\leq \left(pK_1\right)^{n_1\log\mu(V)+n_2}\\ &&\leq C_2 \mu(V)^\xi\\ &&=C_2 \left({h'}^2+\left( \frac{hh'+1}{k}\right)^2+k^2+h^2\right)^\xi\\ &&\leq C_3 k^{2\xi}\\ &&\leq C_3N^{2\xi} \end{eqnarray*} Where $\xi= n_1\log(pK_1)$, and $C_2$, $C_2$ are a constants independent of $V$ Therefore and since in our case $p=2$ \begin{eqnarray*} \lefteqn{|\Omega_{h,k}|} \\ && =|\varepsilon^{-1}(V)\rho(V)^{-1}\left(% \begin{array}{cc} e^{2\pi i m_1\frac{h'-h}{k}} & 0 \\ 0 & e^{2\pi i m_2\frac{h'-h}{k}} \\ \end{array}% \right)|\\ && \leq C_3N^{2\xi}\sqrt{2}\\ &&=C_4N^{2\xi} \end{eqnarray*} thus we have, \begin{eqnarray*} \lefteqn{|R_m|}\\ && \leq e^{2\pi N^{-2}m}\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}C_4N^{2\xi}\int_{-\theta_{h,k}^{'}}^{\theta_{h,k}^{''}}CN^{-r}e^{2\pi \alpha N^{-2}}d\varphi\\ &&\leq C_5e^{2\pi N^{-2}(m+\alpha)}N^{-r+2\xi}\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\int_{-\theta_{h,k}^{'}}^{\theta_{h,k}^{''}}d\varphi\\ &&=C_5e^{2\pi N^{-2}(m+\alpha)}N^{-r+2\xi} \end{eqnarray*} And therefore we can conclude that $|R_m|=0$ for $r>2\xi$ \section{A convergent series for $a_m$} We now evaluate $Q_m$, under the condition \begin{equation} m+max\{m_1,m_2\}>0 \end{equation} todavia no se muy bien porque, parece ser que sera necesario para la convergencia de cierta integral Now if we make the substitution \begin{equation} \omega=N^{-2}-i\varphi \end{equation} We have, \begin{equation}\begin{split} Q_m^t=\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\Omega_{h,k}e^{-2\pi i h\frac{m}{k}}\frac{1}{i}\int_{N^{-2}-\theta_{h,k}^{''}}^{N^{-2}+\theta_{h,k}^{'}}\Psi_k(k\omega) P\left(e^{\frac{2\pi}{k}\left(ih'-k^{-1}\omega^{-1}\right)}\right)^t e^{2\pi m \omega}d\omega \end{split}\end{equation} Now since \begin{align} P(x)=\left(% \begin{array}{c} \sum_{\nu=1}^{-min(\beta_1,\beta_2)}a_{-\nu} (1)x^{-\nu} \\ \sum_{\nu=1}^{-min(\beta_1,\beta_2)}a_{-\nu} (2)x^{-\nu} \\ \end{array}% \right), &&\Psi_k(z)=z^{r} \left(% \begin{array}{cc} e^{2\pi m_1\frac{z-1/z}{k}} & 0 \\ 0 & e^{2\pi m_2\frac{z-1/z}{k}} \\ \end{array}% \right) \end{align} Then \begin{equation} \Psi_k(k\omega)=k^{r} \omega^{r} \left(% \begin{array}{cc} e^{2\pi m_1\omega}e^{-\frac{2\pi m_1}{k^2\omega}} & 0 \\ 0 & e^{2\pi m_2\omega}e^{-\frac{2\pi m_2}{k^2\omega}} \\ \end{array}% \right) \end{equation} then \begin{eqnarray*} \lefteqn{Q_m}\\ &&\begin{split} =\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\Omega_{h,k}e^{-2\pi i h\frac{m}{k}}\frac{1}{i}\int_{N^{-2}-\theta_{h,k}^{''}}^{N^{-2}+\theta_{h,k}^{'}}k^{r} \omega^{r} \left(% \begin{array}{cc} e^{2\pi m_1\omega}e^{-\frac{2\pi m_1}{k^2\omega}} & 0 \\ 0 & e^{2\pi m_2\omega}e^{-\frac{2\pi m_2}{k^2\omega}} \\ \end{array}% \right)\\ \left(% \begin{array}{c} \sum_{\nu=1}^{-min(\beta_1,\beta_2)}a_{-\nu} (1)e^\frac{2\pi i h' \nu}{k}e^\frac{2\pi \nu}{k^2\omega} \\ \sum_{\nu=1}^{-min(\beta_1,\beta_2)}a_{-\nu} (2)e^\frac{2\pi i h' \nu}{k}e^\frac{2\pi \nu}{k^2\omega} \\ \end{array}% \right) e^{2\pi m \omega}d\omega \end{split}\\ &&\begin{split} =\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\Omega_{h,k}e^{-2\pi i h\frac{m}{k}}k^{r}\\ \left(% \begin{array}{c} \sum_{\nu=1}^{-min(\beta_1,\beta_2)}e^\frac{2\pi i h' \nu}{k}\frac{1}{i}\int_{N^{-2}-\theta_{h,k}^{''}}^{N^{-2}+\theta_{h,k}^{'}} \omega^{r}a_{-\nu} (1)e^\frac{2\pi (\nu-m_1)}{k^2\omega}e^{2\pi (m+m_1) \omega} d\omega \\ \sum_{\nu=1}^{-min(\beta_1,\beta_2)}e^\frac{2\pi i h' \nu}{k}\frac{1}{i}\int_{N^{-2}-\theta_{h,k}^{''}}^{N^{-2}+\theta_{h,k}^{'}} \omega^{r}a_{-\nu} (2)e^\frac{2\pi (\nu-m_2)}{k^2\omega}e^{2\pi (m+m_2) \omega} d\omega\\ \end{array}% \right) \end{split}\\ &&=\sum_{\tiny{\begin{array}{c} h,k \\ 0 \leq h < k \leq N\\ (h,k)=1 \\ \end{array}}}\Omega_{h,k}e^{-2\pi i h\frac{m}{k}}k^{r}\sum_{\nu=1}^{-min(\beta_1,\beta_2)}e^\frac{2\pi i h' \nu}{k}\left(% \begin{array}{cc} a_{-\nu} (1) & 0 \\ 0 & a_{-\nu} (2) \\ \end{array}% \right)I_k(m,\nu) \end{eqnarray*} where \begin{eqnarray*} \lefteqn{I_k(m,\nu)}\\ &&=\left(% \begin{array}{c} \frac{1}{i}\int_{N^{-2}-i\theta_{h,k}^{''}}^{N^{-2}+i\theta_{h,k}^{'}} \omega^{r}e^\frac{2\pi (\nu-m_1)}{k^2\omega}e^{2\pi (m+m_1) \omega} d\omega \\ \frac{1}{i}\int_{N^{-2}-i\theta_{h,k}^{''}}^{N^{-2}+i\theta_{h,k}^{'}} \omega^{r}e^\frac{2\pi (\nu-m_2)}{k^2\omega}e^{2\pi (m+m_2) \omega} d\omega \\ \end{array}% \right)\\ &&=\left(% \begin{array}{c} I_k^1(m,\nu) \\ I_k^2(m,\nu) \\ \end{array}% \right) \end{eqnarray*} Now we cut the complex plane from $0$ to $-\infty$ along the negative real axis, and consider the path shown in the figure below \includegraphics[scale=0.3]{path.eps} Then we can write \begin{eqnarray*} \lefteqn{I_k^j(m,\nu)}\\ &&=\frac{1}{i}\int_{-\infty}^{(0+)}-\frac{1}{i}\int_{-\infty}^{-\varepsilon}-\frac{1}{i}\int_{-\varepsilon}^{-\varepsilon-i\theta_{h,k}^{''}}-\frac{1}{i}\int_{-\varepsilon-i\theta_{h,k}^{''}}^{N^{-2}-i\theta_{h,k}^{''}}-\frac{1}{i}\int_{N^{-2}+i\theta_{h,k}^{'}}^{-\varepsilon+i\theta_{h,k}^{'}}-\frac{1}{i}\int_{-\varepsilon+i\theta_{h,k}^{'}}^{-\varepsilon}-\frac{1}{i}\int_{-\varepsilon}^{-\infty}\\ &&=L_k^j(m,\nu)-J_1^j-J_2^j-J_3^j-J_4^j-J_5^j-J_6^j \end{eqnarray*} Where the integrand in all the integrals is \begin{equation} \omega^{r}e^\frac{2\pi (\nu-m_j)}{k^2\omega}e^{2\pi (m+m_j) \omega} \end{equation} We will also assume that $0<\varepsilon