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(Tuesday, Sept 5:) For a number field $Fˣ$ and a number ring $Oˣ$ it is common to define:

$Z(f,χ) = ʃ_{Fˣ} f(x) χ(x) dˣ x$

$g(ω,ψ) = ʃ_{Oˣ} ω(x) ψ(x) dˣ x$

where $dˣx$ is the multiplicative Haar measure on $Fˣ$ and $Oˣ$ respectively.

In the nonarchimedian setting we can consider the characteristic function $𝟙_{O}$. I'm trying to get an understanding of why the non-Archimedean case often entails integrating against $𝟙_{O}$ and what the analogue of $𝟙_{O}$ is for Archimedean fields $ℝ$ and $ℂ$. It seems like this would be analogue of $e^{-x^2}$ somehow, but I can't quite see the similarity.

Perhaps $𝟙_{ℤₚ} : ℚₚ⭢ℂ$ is its own Fourier transform?

Z(f,χ) specializes when we take particular values of $f$. What makes $𝟙_{O}$ (for nonarchimedian) and $e^{-x^2}$ the most common choices?

Another kind of zeta function is to do with exp(Σ Tr(φⁿ)xⁿ⁄n) (note the power series of log(1 - t φ) is $Σ xⁿ⁄n$, and compare with det •exp = exp • tr for finite dimensional Hilbert spaces). This can be understood as a modification of the characteristic polynomial.

Is there an easy way of seeing which of these latter kind of zeta functions should arise specifically from taking particular values of $f$ and $ω$ in $Z$ and $g$?

Maybe there are some fairly straightforward equations relating the derived characteristic polynomial and multiplicative fourier transform, but as it is I can't quite understand why these should be related. Or maybe there is a toy example featuring the ordinary characteristic polynomial.

Wed Sep 6: There are the product rules for both of these kinds of zeta functions, which makes me interested in the case for a local field. From the comments below one might consider a Schwartz-Bruhat function on a locally compact local field which is its own Fourier transform. So let's specialize $f(x)$ in $Z(f,χ)$ to be a Schwartz-Bruhat function which is its own Fourier dual in what proceeds, so that $λχ↦Z(f,χ)$ gives the multiplicative Fourier dual of $f$.

Note also that 1/(1-x) is the Hasse-Weil zeta function of a point. $𝟙_{ℕ} : ℤ ⭢ ℂ$ can be compared with the $𝟙_{ℤₚ} : ℚₚ ⭢ ℂ$. Mainly I am confused about why this Hasse-Weil zeta function gets precomposed with an exponential function, and I suspect this is to do with the map $ℤₚ ⊗_{𝔽ₚ} 𝔽ₚˢᵉᵖ ⭢ 𝔽ₚˢᵉᵖ$.

I find it straightforward to consider the function $Π_{i=1}^{n} Det( 1-t * Hⁱ Frbₚ )$ on 𝔽ₚˢᵉᵖ for positive characteristic as well, which produces 1/(1-x) for the variety Spec(𝔽ₚˢᵉᵖ). But the thing that is confusing me is precomposing the Hasse Weyl 1/(1-s) with the exponential function

It's common to precompose the zeta function for a positive characteristic variety with an exponential function $p^{-s}$ to get $Z_{V,p}(p^{-s})$. I was thinking that this must correspond to the map $ℤₚ ⭢ 𝔽ₚˢᵉᵖ$ somehow. Is it true that $Z_{V,p}(p^{-s})$ is the Hasse Weil zeta function for the lift of a positive characteristic variety in the way of Hensel's lemma and Newton's method in positive characteristic? So, I want this recomposition with $p⁻ˢ$ to correspond to the canonical map $ℤₚ ⊗_{𝔽ₚ} 𝔽ₚˢᵉᵖ ⭢ 𝔽ₚˢᵉᵖ$ or a similar map into $𝔽ₚˢᵉᵖ$, but I'm not quite sure how this goes.

So what I want to ask is this:

for a finite extension of ℤₚ, do we get $Z_{Spec(A ⊗ 𝔽ₚˢᵉᵖ,p)} • p⁻ˢ = Z_{Spec(A),p}$?

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    $\begingroup$ The indicator function of $\mathbb{Z}_p$ is a Schwartz-Bruhat function that is its own Fourier transform. The archimedean analogue of this is $e^{-\pi x^2}$ for $\mathbb{R}$ and $e^{-2\pi z\overline{z}}$ for $\mathbb{C}$. $\endgroup$ Commented Sep 5, 2023 at 4:35
  • $\begingroup$ @PeterHumphries thanks and feel free to leave that as an answer. Any other links or insights are much appreciated. I'm quite glad of this fourier reflexivity for $1_{\mathbb{Z}_{p}}$. $\endgroup$
    – user30211
    Commented Sep 5, 2023 at 4:55
  • $\begingroup$ It seems that, after Peter Humphries had given a short but informative answer to the original question in the comments, the original post was edited to add a second (only very distantly related) question onto the end of the original one. If you have something new to ask, then please ask a new question. $\endgroup$ Commented Sep 6, 2023 at 6:50

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