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Let $ \theta_z(t) = \sum \limits_{m,n\in\mathbb{z}}e^{-\pi Q_z(m,n)t}$ where $Q_z(m,n)=y^{-1}|mz+n|^2$.

I need to prove that $\theta_z(t)=t^{-1}\theta_z(t^{-1})$.

Now, looking that up I know that there is a thing called "theta functions" and that $\theta_z(t)$ is such a function, and that this functional equation is a general property of theta functions.

However, I do not have this to rely on for this calculation. That is, I'm trying to prove this explicitly for this function, and I imagine there's a better way than proving myself the theorem for the general case.

I tried using the generalized version of Poisson's summation formula, for that I defined the function $f(x,y) = e^-\pi Q_z (x,y)t$.

From this I immediately attain using Poissin's summation formula (and the easy to prove fact that $(\mathbb{Z}\times\mathbb{Z})^{\vee}=\mathbb{Z}\times\mathbb{Z}$) that $\theta_z(t) = \sum \limits_{m,n\in\mathbb{Z}}f(m,n) = \sum \limits_{m,n\in\mathbb{z}}\hat{f}(m,n)$.

The only "missing link" is to calculate $\hat{f}$, which I wasn't able to do.

I won't type in any of my attempted calculations because they're long, cumbersome and frankly - pretty useless.

I would, however, appreciate either a pointer to calculating this transform, or maybe if any of you think that my proposed strategy for this calculation is bad - a better way to do it?

Thanks in advance.


Edit: the $y$ in the definition of $Q_z$ is the imaginary part of $z$. It doesn't really matter though, because $z$ can be treated as a constant in this context, it only comes to play in a later part of solving the problem I'm tackling (this is just a part in proving that the Eisenstein series has a meromorphic continuation)


Edit 2: I think the problem is with choosing $f$, I'll try working with $f(w)=e^{-\pi w\bar{w}\frac{t}{y}}$. It's Fourier transform is trivial, but it's harder to calculate the lattice and dual lattice.


Yup, that seemed to work

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    what is $y$ in the definition of $Q_z(m,n)$?2012-07-28
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    $y$ is the imaginary part of $z=x+iy$. It doesn't matter, though, because $z$ is a constant.2012-07-28

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