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Consider Eisenstein series of weight zero, i.e.

$ 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 $,

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 $Re(s)>1$?

My second question is how to calculate the following summation:

$ \sum_{d\ (mod c)}\ \epsilon_d(\frac{c}{d}) $, where $ \gamma = $ $\[ \begin{pmatrix} a & b\\ c & d \end{pmatrix} \]$ $\in \Gamma_0(4) $, $(\frac{c}{d})$ is the extended quadratic residue symbol and $c = b^2. $

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closed as unclear what you're asking by quid, Stefan Kohl, José Figueroa-O'Farrill, Alexey Ustinov, Yoav Kallus Oct 26 '15 at 20:16

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Any chance you could flesh out that question a bit? Specifically, can you define more explicitly all the notation that you use (and not just some of it) - it might be obvious to you since you're familiar with it, but to someone else learning from a different book, perhaps not. (e.g. what does $\epsilon$ mean, what is $\sigma_{a}$, etc). – Vinoth Dec 28 '09 at 1:36
I voted to close. If you provide more background and explain why you're interested in this question, it might not seem like you're just trying to get someone else to do your work for you. – Qiaochu Yuan Dec 28 '09 at 8:10
Well, I think that my problem is not fitful or complete. Actually, my problem comes from Automorphic Form, and you guys can refer to this place: – Alex Dec 31 '09 at 14:07

It seems you have started reading something from the middle.

Hint for the first one: Do it for $SL_2(\mathbb{Z})$ first. Note that

$Im \frac{az+b}{cz+d} = y/|cz+d|^2$.

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I got it! Thanks – Alex Dec 31 '09 at 14:17

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