The convergence of Eisenstein series of weight zero - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:16:17Z http://mathoverflow.net/feeds/question/9885 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9885/the-convergence-of-eisenstein-series-of-weight-zero The convergence of Eisenstein series of weight zero Alex 2009-12-27T14:42:38Z 2009-12-28T07:43:26Z <p>Consider Eisenstein series of weight zero, i.e.</p> <p>$E_{\mathfrak{a}}(z,\ s,\ \chi) = \sum_{ \gamma \in \Gamma_{\mathfrak{a}} \backslash \Gamma } \bar{\chi}(\gamma) (Im\sigma_{\mathfrak{a}}^{-1} \gamma z)^s$, </p> <p>where $\chi$ is a multiplier system of weight zero ( $\chi\ :\ \Gamma \rightarrow \mathbb{C}^*$ is a group homomorphism) singular at cusp $\mathfrak{a}$. Then my first question is that why this series converges absolutely in $Re(s)>1$?</p> <p>My second question is how to calculate the following summation:</p> <p>$\sum_{d\ (mod c)}\ \epsilon_d(\frac{c}{d})$, where $\gamma =$ $[ \begin{pmatrix} a &amp; b\ c &amp; d \end{pmatrix} ]$ $\in \Gamma_0(4)$, $(\frac{c}{d})$ is the extended quadratic residue symbol and $c = b^2.$</p> http://mathoverflow.net/questions/9885/the-convergence-of-eisenstein-series-of-weight-zero/9940#9940 Answer by Idoneal for The convergence of Eisenstein series of weight zero Idoneal 2009-12-28T06:26:02Z 2009-12-28T06:26:02Z <p>It seems you have started reading something from the middle.</p> <p>Hint for the first one: Do it for $SL_2(\mathbb{Z})$ first. Note that </p> <p>$Im \frac{az+b}{cz+d} = y/|cz+d|^2$.</p>