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I am working on Tate's thesis, and I have some problems with computations, yet the result seems to be a good natural motivation for introducing the arithmetic conductor of a character.

Let $F$ be a non-archimedean local field, and $\psi$ a non-trivial additive character. Define the $\psi$-Fourier transform for a locally constant and compactly supported function on $F$ to be $$\widehat{\Phi}(x) = \int_F \Phi(y)\psi(xy)dy$$

We normalize the involved Haar measure so that the Fourier inversion formula holds, that is to say $$\widehat{\widehat{\Phi}}(x) = \Phi(-x)$$

I would like to understand why this implies that the ring of integers $\mathcal{O}$ have volume $q^{c(\psi)/2}$?

Recall that $c(\psi)$ is defined by $\psi$ trivial on $\mathfrak{p}^{c(\psi)}$ and not on $\mathfrak{p}^{c(\psi)-1}$, where $\mathfrak{p}$ is the maximal ideal of $\mathcal{O}$. I tried manipulating the integrals splitting into classes modulo $\mathfrak{p}^{c(\psi)}$, but I cannot make it work properly.

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    $\begingroup$ Venkataramana gave a great answer, but let me point out that the statement is also part of Corollary 3 in Chapter VII-2 of Weil: Basic number theory (which is a great source for these matters anyways). $\endgroup$
    – GH from MO
    Feb 27, 2017 at 15:00
  • $\begingroup$ I looked at Tate's thesis, where he makes this computation for one specific character $\psi$. From this, the general case can easily be derived. $\endgroup$ Mar 7, 2017 at 8:43

1 Answer 1

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Apply the Fourier Inversion Formula to the characteristic function $\Phi(x) = \chi_\mathcal{O}(x)$ of the ring $\mathcal{O}$ of integers in $F$. The Fourier transform is the integral $\widehat{\Phi }(x)=\int_\mathcal{O} \psi (xy)dy$. This integral is zero if and only if $y\mapsto \psi (xy)$ is a non-trivial character on the additive group $\mathcal{O}$ (orthogonality relation for characters). If $\psi (xy)$ is the trivial character, then this means that $x\in {\mathfrak p}^{c(\psi)}=\pi ^{-m}\mathcal{O}$ ($m=-c(\psi)$). We then have $$ \widehat{\Phi}(x)= \mathrm{vol}(\mathcal{O}) \chi_{\pi^{-m}\mathcal{O}}(x).$$ Taking Fourier transforms on both sides of this equation, and bearing in mind that the measure is so chosen that Fourier inversion holds, we get
$$\chi_{\mathcal{O}}(x)=\Phi (x)= \Phi (-x)= \widehat{\widehat{\Phi}}(x) = \mathrm{vol}(\mathcal{O})\mathrm{vol}(\pi^{-m}\mathcal{O})\chi_\mathcal{O}(x).$$ We then get the equality $$\mathrm{vol}(\mathcal{O})\mathrm{vol}(\pi^{-m}\mathcal{O})=1, \quad\text{i.e.}, \quad \mathrm{vol}(\mathcal{O})=q^{-m/2}.$$ This is what was to be proved.

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  • $\begingroup$ Great! Thanks for such a limpid a comprehensive answer! $\endgroup$ Feb 27, 2017 at 14:45
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    $\begingroup$ No problem severus. And @GH from MO, thank you for the editing. $\endgroup$ Feb 27, 2017 at 14:58

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