Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is an exercise about Eisenstein Series $G_2(\tau)$, to prove that $$(G_2[\gamma]_2)(\tau)=G_2(\tau)-{2\pi i\over\tau}$$ where $\gamma=\begin{pmatrix} 0&1\\ -1 &0 \end{pmatrix}$

I just cannot show that $$\sum_{d\in\mathbb{Z}}\sum_{c\ne0}\frac{1}{(c\tau+d)(c\tau+d+1)}=-{2\pi i\over \tau}$$ I tried $$\lim_{N\to\infty}\sum_{d=-N}^{N-1}\sum_{c\ne0}\frac{1}{(c\tau+d)(c\tau+d+1)}=\lim_{N\to\infty}\sum_{c\ne0}({1\over c\tau-N}-{1\over c\tau+N})$$ and to use the equality $$\pi\cot\pi N/\tau={\tau\over N}-\tau\sum_{c=1}^{\infty}({1\over c\tau-N}-{1\over c\tau+N})$$ but $\displaystyle\lim_{N\to\infty}\pi\cot\pi N/\tau$ doesn't exist.

share|improve this question
    
What is $\tau$ supposed to be? The infinite series $\sum_{d\in\mathbb{Z}}\sum_{c\ne0}\frac{1}{(c\tau+d)(c\tau+d+1)}$ does not converge if $\tau \in \{0,1\}$. And if it were to converge for some $\tau \in \mathbb{R}$, the answer could not be a complex number. –  Hans Engler Apr 3 '13 at 22:29
    
@HansEngler The book says $\tau\in\mathbb{c}\cup\{\infty\}$ –  Urukec Apr 3 '13 at 22:39
    
$\tau$ is almost always meant to represent two complex numbers, say $\frac{\omega_1}{\omega_2}$, whose ratio is not real and is contained in the upper half plane –  Brent J Apr 4 '13 at 0:52
    
@ Brent: So the right hand side of this identity is analytic. How about the left hand side? Is the double series expected to converge in an open set of the complex plane? If so, it should be possible to continue it analytically. (This is just a reality check. I don't actually know anything about Eisenstein series.) –  Hans Engler Apr 4 '13 at 16:39
    
@HansEngler Not sure why you're asking me because I didn't ask the question, although I may know a solution. Anyways, an Eisenstein series, like any modular form, is holomorphic at the cusp. In fact, it is absolutely convergent to a holomorphic function of $\tau$. I believe it has an analytic continuation to the whole complex plane with a simple pole at $\pm\frac{1}{2}$ –  Brent J Apr 4 '13 at 17:10
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.