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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 $Q_z(m,n)$? – robjohn Jul 28 '12 at 10:25
$y$ is the imaginary part of $z=x+iy$. It doesn't matter, though, because $z$ is a constant. – Shai Deshe Jul 28 '12 at 10:26

We have $Q(m,n):=\frac{1}{y}|mz+n|^2$. Choose a basis of $\Bbb R^2$ orthonormal w.r.t. $Q$ so that with the base-change matrix $U$ and coordinates $\mathbf{r}=U\mathbf{s}$ we have $Q(U\mathbf{s})=\|\mathbf{s}\|^2$ and therefore

$$\begin{array}{c l}\widehat{e^{-\pi Q(\cdot)}}(\mathbf{m})& = \int_{\Bbb R^2}e^{-\pi Q(\mathbf{r})t}e^{2\pi i\langle\mathbf{m},\mathbf{r}\rangle}d\mathbf{r} \\ & =\int_{\Bbb R^2}e^{-\pi Q(U\mathbf{s})t}e^{2\pi i \langle\mathbf{m},U\mathbf{s}\rangle}\left|\frac{\partial \mathbf{r}}{\partial\mathbf{s}}\right|d\mathbf{s} \\ & = \int_{\Bbb R^2}e^{-\pi \|\mathbf{s}\|^2}e^{2\pi i\langle U^T\mathbf{m},\mathbf{s}\rangle}|\det U|d\mathbf{s} \\ & = e^{-\pi \|U^T \mathbf{m}\|^2}=e^{-\pi Q(\mathbf{m})}.\end{array}$$

Note the orthonormality implies $UU^T=I$ and so $| \det U|=1$. Also see these notes on $\Theta$ and lattices.

share|cite|improve this answer
Apparently this is what you did in edit #2. – anon Jul 28 '12 at 13:49

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