%&latex % January 12, 2000 \documentclass[11pt]{amsart} \usepackage{amsmath,amssymb,amscd} \usepackage[mathscr]{eucal} \newcommand{\Aut}{{\mathrm{Aut} }} \newcommand{\End}{{\mathrm{End} }} \newcommand{\GL}{{\mathbf{GL} }} \newcommand{\Gal}{{\mathrm{Gal}\, }} \newcommand{\G}{{\mathbf G}_{m}} \newcommand{\Hom}{{\mathrm{Hom} }} \newcommand{\Tr}{{\mathrm{Tr}\, }} \newcommand{\Q}{\mathbf Q} \newcommand{\Z}{\mathbf Z} \newcommand{\R}{\mathbf R} \newcommand{\C}{\mathbf C} \newcommand{\F}[1]{\mathbf{F}_{#1}} \newcommand{\Fbar}[1]{\overline{\mathbf{F}}_{#1}} \newcommand{\fr}[1]{\,{\rm Frob}_{\, #1}} \newcommand{\q}[1]{\mathbf{Q}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\msr}[1]{\mathscr{#1}} \newcommand{\mbf}[1]{\mathbf{#1}} \newcommand{\mfr}[1]{\mathfrak{#1}} \newcommand{\mr}[1]{\mathrm{#1}} \newcommand{\ok}[1]{O_{#1}} \newcommand{\re}[1]{{\rm Re} (#1)} \newcommand{\im}[1]{{\rm Im} (#1)} \theoremstyle{plain} \newtheorem{thm}{Theorem}[section] \newtheorem{lem}[thm]{Lemma} \newtheorem{pro}[thm]{Proposition} \newtheorem{cor}[thm]{Corollary} \theoremstyle{definition} \newtheorem{Def}[thm]{Definition} %\theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \newtheorem{exa}[thm]{Example} \begin{document} \title{Lecture 36} \author{Michael Broome} \address{Louisiana State University \\ Department of Mathematics \\ Baton Rouge, LA 70803, USA} \date{\today} \email{broome@math.lsu.edu} \maketitle \section{Section 3: Characters of Local Fields} We have previously shown that $K^{\wedge}\simeq K$ for the additive group of a local field. Also, $\eta\rightarrow X_{\eta}$ where $\eta\in K$ and $X$ is any one fixed nontrivial character. By definition, $X_{\eta}(\zeta)=X(\eta\,\zeta)$.\newline For $\R$, $\lambda :\R\rightarrow\R/\Z$, defined by $\lambda(x)=-x$, is a continuous homomorphism, and the basic character is defined by $X(\zeta)=e^{2\pi i\lambda(\zeta)}=e^{-2\pi i\zeta}$.\newline For $\C$, the basic character is defined by $X(\zeta)=e^{2\pi i\lambda(Tr_{\C/\R}(\zeta))}=e^{-2\pi i(\zeta+\bar{\zeta})}$.\newline If $K$ is $p$-adic, the continuous homomorphism is $\lambda_{p} :\Q_{p}\rightarrow\Q/\Z$ with the basic character $X(\zeta)=e^{2\pi i\lambda_{p}(\zeta)}$.\newline For $x\in\Q_{p}$, $x$ can be written as \[ x=\frac{a_{\nu}}{p^{\nu}}+\frac{a_{\nu-1}}{p^{\nu-1}}+\cdots+\frac{a_{1}}{p} +a_{0}+a_{1}p_{1}+\cdots \] where $\frac{a_{\nu}}{p^{\nu}}+\frac{a_{\nu-1}}{p^{\nu-1}}+\cdots+\frac{a_{1}}{p}$ is the polar part and the remaining terms are elements in $\Z_{p}$.\newline Now define $\lambda_{p}(x)$ to be the following: \[ \frac{a_{-\nu}}{p^{\nu}}+\cdots+\frac{a_{-1}}{p}\in \Z[\frac{1}{p}]. \] \[ \begin{matrix} &&\Q_{p}&&&\\ &\lambda_{p}&\downarrow&&\searrow&\\ &&&&&\Q_{p}/\Z_{p}\\ &&&&\swarrow&\\ \stackrel{\bigoplus}{p}\Q_{p}/\Z_{p}&=&\Q/\Z&&& \end{matrix} \] For $K/\Q_{p}$, the basic character is $X(\zeta)=e^{2\pi i\lambda_{p}(Tr_{K/\Q_{p}}(\zeta))}$.\newline\newline Here is some notation for convenience. $\Lambda=\lambda\circ$Tr. Thus, $X(\zeta)=e^{2\pi i\Lambda(\zeta)}$.\newline Recall our study of the Fourier transform:\newline Given $f:G\rightarrow\C$, the Fourier transform is $\hat{f}:\hat{G}\rightarrow\C$. Also, we have the following: \[ \begin{matrix} \hat{f}(\chi)=\int_{G}f(g)\overline{\chi(g)}dg\quad and\\\\ f(g)=\int_{\hat{G}}\hat{f}(x)\chi(g)d\chi \end{matrix} \] where $dg$ is Haar measure and $d\chi$ is the Haar measure on the dual group.\newline\newline If $G\simeq\hat{G}$ defined by $g\mapsto\chi g$, then if we rewrite the Fourier transform, we get the following: \[ \hat{f}(g)=\int_{G}f(s)\overline{\chi_{g}(s)}ds. \] We would like for the following to be true: \[ f(g)=\int_{G}\hat{f}(s)\chi_{g}(s)ds. \] In order to do this, we have to scale the Haar measure, $ds$. You choose a self-dual Haar measure with a constant $c>0$.\newline If $K=\R$, the self dual Haar measure is the usual Lebesgue measure.\newline If $K=\C$, the self dual Haar measure is $dx=2|dz\wedge d\bar{z}|$, that is twice the usual Lebesgue measure.\newline If $K$ is $p$-adic, the self dual Haar measure is $dx$,which is the measure in which the volume is as below. \[ vol(O_{K})=\int_{O_{K}}dx=(N\vartheta)^{-1/2} \] where $\vartheta=\vartheta_{K/\Q_{p}}=$the different ideal$=p^{r}$. \begin{Def} \label{D:IV.2.1} $K$ is a $p$-adic field. The inverse different \[ \begin{matrix} \vartheta^{-1}=\vartheta_{K/\Q_{p}}^{-1} =\{y\in K:Tr_{K/\Q_{p}}(xy)\in\Z_{p} \,for \,all \,x\in O_{K}\} =\{y:Tr(yO_{K})\subseteq\Z_{p}\}. \end{matrix} \] \end{Def} Note that $O_{K}\subseteq\vartheta^{-1}=p^{-r}.$ \begin{thm} \label{T:IV.2.1} Now, we define the Fourier transforms as follows: \[ \hat{f}(y)=\int_{K}f(x)e^{-2\pi i\Lambda(xy)}dx.\quad Then, f(x)=\int_{K}\hat{f}(y)e^{2\pi i\Lambda(xy)}dy \] with $dx$ chosen as the self dual Haar measure. We also know $\hat{f}(y)=c\cdot f(x)$ and $\hat{\hat{f}}(x)=f(-x)$. \end{thm} \begin{proof} We know that the two sides differ by a constant multiple $c>0$. For $c=1$, we need to only verify the Fourier transform of one $f$. \begin{itemize} \item[(1)]For $K=\R$, $f(x)=e^{-\pi|x|^{2}}=e^{-\pi x^{2}}$ ,and we need to verify $\hat{f}(y)=e^{-\pi y^{2}}$. Now, \[ \hat{\hat{f}}(x)=e^{-\pi x^{2}}=e^{-\pi(-x)^{2}}=f(-x). \] We have proven previously that \[ \int_{-\infty}^{\infty}e^{-ax^{2}+2bx}dx=\Big(\frac{\pi}{a}\Big)^{1/2}e^{b^{2}/a}\quad,Re(a)>0. \] Therefore, \[ \begin{matrix} \hat{f}(y)&=\int_{-\infty}^{\infty}e^{-\pi x^{2}}e^{2\pi ixy}dx\\ &=\int e^{-\pi x^{2}+2\pi iyx}\\ &=(\frac{\pi}{\pi})^{1/2}e^{(\frac{-\pi^{2}y^{2}}{\pi})} &=e^{-\pi y^{2}}. \end{matrix} \] So, we let $a=\pi$ and $b=\pi iy$ in the previous expression, and we are done. \item[(2)]For $K=\C$, $f(z)=e^{-2\pi|z|^{2}}$, and we need to verify $\hat{f}(w)=e^{-2\pi|w|^{2}}$. Now, \[ \hat{f}(w)=2\int_{\C}e^{-2\pi|z|^{2}}e^{4\pi i(wz+\overline{wz})}. \] Let $z=x+iy$ adn $u+iv$.Then, \[ \hat{f}(w)=2\int_{\C}e^{-2\pi(x^2+y^2)}e^{4\pi i(u^2-v^2)}\newline =2\int_{\R}e^{-2\pi x^2 +4\pi iu^2}\int_{\R}e^{-2\pi y^2 -4\pi iv^2} \] Let $a=2\pi$ in both integrals, and let $b=2\pi iu$ for the first integral and $b=2\pi iv$ for the second integral. Then, using the statement proven earlier, we get\newline \[ \hat{f}(w)=2\Big(\frac{\pi}{2\pi}\Big)^{1/2}\Big(\frac{-4\pi^2u^2}{2\pi}\Big) \Big(\frac{\pi}{2\pi}\Big)^{1/2}\Big(\frac{-4\pi^2v^2}{2\pi}\Big) \] \[ =2\Big(\frac{1}{\sqrt{2}}\Big)e^{-2\pi u^2}\Big(\frac{1}{\sqrt{2}}\Big)e^{-2\pi v^2}=e^{-2\pi(u^2+v^2)}=e^{-2\pi|w|^2}. \] So, we are done in this case. \item[(3)]If $K$ is $p$-adic, then $f$ is the characteristic function of $O_{K}$ where the value of $f(x)$ is 1 if $f(x)\in O_{K}$ and 0 otherwise.\newline \begin{lem} \label{L:IV.2.2} $\hat{f}=(N\vartheta)^{-1/2}g$ and $\hat{g}=(N\vartheta)^{1/2}f$. \end{lem} If we assume the above lemma, then \[ \hat{\hat{f}}=(N\vartheta)^{-1/2}(N\vartheta)^{1/2}f=f(x). \] So, $f(-x)=f(x)$. \end{itemize} \end{proof} \begin{proof} This is the proof of the preceding lemma.\newline \[ \hat{f}=\int_{K}f(x)e^{-2\pi i\Lambda(xy)}dx=\int_{O_{K}}e^{-2\pi i\Lambda(xy)}dx \] since $f(x)$ is either 0 or 1 if $f(x)$ is on $O_{K}$.\newline Now, assume that $y\in\vartheta^{-1}\Rightarrow Tr(yO_{K})\subseteq\Z_{p}\iff\lambda\circ Tr(yO_{K})=0\iff\Lambda(yx)=0\,for\,all\,x\in O_{K}$.Therefore, if $y\in\vartheta^{-1}$, then \[ \hat{f}(y)=\int_{O_{K}}dx=(N\vartheta)^{-1/2}.\newline \] Now, assume $y\notin\vartheta^{-1}$. Then there exists $x_{0}\in O_{K}$ such that $Tr(yx_{0})\notin\Z_{p}\iff\Lambda(yx_{0})\ne 0$. So, if $y\notin\vartheta^{-1}$, then \[ \hat{f}(y)=\int_{O_{K}}e^{-2\pi i\Lambda(y(x+x_{0}))}dx\newline =e^{-2\pi i\Lambda(yx_{0})}\int_{O_{K}}e^{-2\pi i\Lambda(yx)}dx.\newline \] Note that $e^{-2\pi i\Lambda(yx_{0})}\ne 1$. Thus, \[ \hat{f}(y)=e^{-2\pi i\Lambda(yx_{0})}\hat{f}(y)\Rightarrow\hat{f}(y)=0\,for\,y\notin\vartheta^{-1}. \] A similar argument holds true for the second half of the lemma. \end{proof} \bibliographystyle{amsplain} \providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace} \begin{thebibliography}{10} \bibitem{sL} S. Lang, \emph{Algebraic Number Theory}, Second Edition, Graduate Texts in Mathematics, {\bf 110}, Springer - Verlag, 1994. \end{thebibliography} \end{document}