Consider the function: $$ f\big(x,y ; \,t \big) = \big(x^2 - y^2\big)\,e^{-\pi\,(x^2 + y^2)\,t} $$ If I could find a Fourier transform, then perhaps I could write the statement of Poisson summation in this case: $$ \sum_{m,n \in \mathbb{Z}} f(m,n) = \sum_{m,n \in \mathbb{Z}} \widehat{f}(m,n) $$
In a clumsy way I could write: $$ \widehat{f}(m,n) = \int_0^\infty dx\int_0^\infty dy \; \left[ e^{2\pi i \,(\,mx + ny\,)} \big(x^2 - y^2\big)\,e^{-\pi\,(x^2 + y^2)\,t} \right] $$ This does not factor into an integral of $dx$ times an integral of $dy$ because of the joint $x^2 - y^2 $. I believe the integrals behave nicely, so I have that:
$$ \sum_{(m,n) \in \mathbb{Z}^2} \big(m^2 - n^2\big)\,e^{-\pi\,(m^2 + n^2)\,t} = \frac{1}{\sqrt{t}}\sum_{(m,n) \in \mathbb{Z}^2} \big(m^2 - n^2\big)\,e^{-\pi\,(m^2 + n^2)\,/\,t} \tag{$\ast$}$$
It doesn't seem right.
Therefore I am ransacking my old Fourier analysis textbooks. It writes it in one variable but the extension to several variables is clear. Let $f(x) \in \mathcal{S}(\mathbb{R})$ be an element of Schwartz class, e.g. $f(x) = x^2 \, e^{-\pi x^2 \,t}$.
- $f(x+h) \longrightarrow \hat{f}(\xi) e^{2\pi i h \xi}$
- $f(x) e^{-2\pi i xh} \longrightarrow \hat{f}(\xi+h)$
- $f(\delta x) \longrightarrow \delta^{-1}\hat{f}(\delta^{-1}\xi)$
- $f'(x) \longrightarrow 2\pi i \xi \hat{f}(\xi)$
- $-2\pi i x f(x) \longrightarrow \frac{d}{d\xi}\hat{f}(\xi)$ The last property looks like the one I want because if I know $$f_0(x) = e^{-2\pi i \, x^2 t} \longrightarrow \hat{f}_0(t) = e^{-2\pi i \, x^2 /t}$$ Then I could take the second derivative and get another Fourier transform formula: $$f_0(x) = x^2 \, e^{-2\pi i \, x^2 t} \longrightarrow \frac{d^2}{dt^2}\hat{f}_0(t) = \frac{d^2}{dt^2} \left[ e^{-2\pi i \, x^2 /t} \right]$$ That might lead to the formula $(\ast)$ that I have above.
In this case $(*)$ is true. Both sides equal zero, as you see by swapping the summation variables.
In general, if $\phi(x,y)$ is a homogeneous harmonic polynomial (a spherical function) and $f(x,y)=\phi(x,y)\exp(-\pi(x^2+y^2))$ then $\hat f$ is a multiple of $f$.