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Taking a modular form such that we have Fricke involution:

$\sum_{n=1} a_n e^{-\pi nx^2} = \frac{A}{x^k} \sum_{n=1} a_n e^{-\pi \frac{n}{x^2}}$ [1]

I would like to know if there exists results on possible formula with $a_n$ coeficient which will be like the classical Poisson summation formula (it works for Dirichlet L functions with the twisted Poisson summation formula so why not for modular forms for some functions?)

More specifically, for example, does following equality holds?:

$\sum_{n=1} a_n e^{-(z+ \sqrt{\pi n} x)^2 } = \frac{A}{x^k} \sum_{n=1} a_n \widehat{f_z}(\frac{1}{x})$ [2]

(Where $f_z(x)=e^{-(z+ \sqrt{\pi n}x)^2}$ and Fourier transform is done on variable x: to have an iso Poisson formula)

Note that in z=0 equation [2] becomes the formula [1]. I thought this can be demonstrate by showing that both side also satisfy same partial differential equation (but partial differential equation are slightly different...).

So my question is; do you have any reference on the subject? Is there some fucntions for which an equation like [2] can hold? for example as an extension of [1]?

(Originally I would like to have asymptotic of $\sum_{n=1}a_n e^{-(z+ \sqrt{\pi n}x)^2}$ in x near 0.)

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1 Answer 1

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The functional equation of the Riemann Zeta function is equivalent to the Poisson summation formula. This should be adoptable to the setting of automorphic $L$-functions. I am not sure where the abelian Fourier Analysis should happen here, probably in higher rank though(?)

Here is a reference, which is even more general than what you ask for: Why is the functional equation of the Riemann zeta function equivalent to the Poisson summation formula?

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Thanks, the answer to my question is as you say in [Cohen_H.]_Number_theory_vol.2._Analytic_and_modern, page 178. The generalisation of Poisson Summation formula I was looking for is explained in details. –  Bertrand Sep 13 '13 at 8:43

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