I was wondering if there exists a Poisson Summation formula (like the one existing with primitive character) for imprimitive Dirichlet characters ?

For a primitive Dirichlet character $\chi$ we have:

$$ \sum\limits_{n=-\infty}^{\infty}\chi(n) f\bigg(\frac{n}{q}x\bigg) =\frac{K}{x} \sum\limits_{n=-\infty}^{\infty} \overline{\chi(n)} \hat{f}\bigg(\frac{n}{x}\bigg)$$

But for imprimitive characters ?

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    $\begingroup$ It would help greatly if you stated which version of the Poisson summation formula for primitive characters you are referring to. $\endgroup$ – Daniel Loughran Feb 4 '15 at 16:31
  • $\begingroup$ You are right, it is done. $\endgroup$ – Bertrand Feb 5 '15 at 12:23
  • $\begingroup$ You might be interested in a preprint of Daileda and Jones, where they show that by modifying the way in which one extends primitive characters to imprivitive characters (in particular, by making a choice other than $\chi(n)=0$ for $n$ not coprime to $q$ -- and, iirc, by choosing it so that the Gauss sum is well-behaved), these new imprimitive characters behave nicely analytically. It's available here: olemiss.edu/working/ncjones/primitivity9.pdf . $\endgroup$ – rlo Feb 5 '15 at 19:18
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    $\begingroup$ Ignorant question: do you have a reference for the formula in the question? specifically, what is the value of $K$? $\endgroup$ – kodlu Jul 3 '19 at 4:13

The reason we can get that (twisted) Poisson summation formula in the first place is that in the primitive case you can interpolate the character to a smooth real function via Gauss sums.

In the imprimitive case this is not the case anymore, and you can't get a function nice enough to anything that resembles a Poisson formula to hold.

Of course nothing is lost, since imprimitive characters are induced by primitive ones, and for example (since you have used the Dirichlet series tag), we have:

$$L(\chi,s)=\prod_{\substack{ p|m \\ p\nmid f }} (1-\chi (p)p^{-s})L(\chi ',s)$$

which gives you functional equation and analytic continuation of Dirichlet series for imprimitive $\chi'$. This kind of induced-character argument bypasses the need for Poisson summation in any case I can think of.

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  • $\begingroup$ Thanks for the answer, it confirms what I though. $\endgroup$ – Bertrand Feb 5 '15 at 15:41
  • $\begingroup$ I might quibble that the issue is that the "finite" (or maybe finite-adelic, or p-adic, etc) Fourier transform of (the naively natural way to present) a non-primitive character does not produce a character... not quite $\overline{\chi}$, for example. But the procedure is still coherent. $\endgroup$ – paul garrett Jul 3 '19 at 1:29

Yes, of course.

Poisson summation formula has nothing to do with characters. If χ is any periodic function, all you need to do is to replace χ̅ by the discrete Fourier Transform of χ.

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  • $\begingroup$ I agree that an all-too-common presentation of things makes (the application of) Poisson summation appear to be (partly) about characters... which it is not. Nevertheless, in an Iwasawa-Tate adelic context, it becomes more clear that the Schwartz-Bruhat functions are "dummies", and the (Hecke/idele-class) character is really "the thing". $\endgroup$ – paul garrett Jul 3 '19 at 1:31

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