%&latex \NeedsTeXFormat{LaTeX2e} [1994/12/01] \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}} %========== I added a couple of commands \newcommand{\ok}[1]{\mathcal{O}_{#1}} \newcommand{\norm}[1]{\left\Vert #1\right\Vert} \newcommand{\abs}[1]{\left\vert #1\right\vert} \newcommand{\set}[1]{\left\{ #1\right\}} \newcommand{\gal}{\mathcal G} \newcommand{\ba}{\mathbf a} \newcommand{\bb}{\mathbf b} \newcommand{\bt}{\mathbf t} %=========== \newcommand{\re}[1]{{\rm Re} (#1)} \newcommand{\im}[1]{{\rm Im} (#1)} \newcommand{\dirlim}{\lim\put(-15,-6){$\leftarrow$}} \newcommand{\nequiv}{\put(5,0){/}\equiv} \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 44} \author{Uroyoan R. Walker} \address{Louisiana State University, Department of Mathematics, Baton Rouge, LA 70803, USA} \date{May 11, 2000} \email{walker@math.lsu.edu} \maketitle \setcounter{section}{3} \section{Tate's Thesis} \subsection{Global Functional Equation} We will write the idele $J=J_k$ as a direct product by embedding the positive reals $\R_+$ in it, namely, \begin{eqnarray*} \R_+ &\longrightarrow & J\\ t &\longmapsto & \left( t^{1/N},\ldots ,t^{1/N},1,1,\ldots\right)\end{eqnarray*} where $N$ is the number of archimedean components, $t^{1/N}$ is a component in every archimedean place $v$, and 1 at the non-archimedean places. So $$J\cong\R_+\times J^0$$ In particular, every idele $\ba$ can be written uniquely as a product $\ba =\bt\bb$ with $\bt\in\R_+$ and $\bb\in J^0$. Hence $$\int_J{f(a)c(a)d^\times a}=\int_{\R_+} \left( \int_{J^0}f(\bt\bb )c(\bt\bb )d^\times \bb\right) \frac{dt}{t}$$ \noindent Now we state two theorems that we will use in the proof of the global functional equation. \begin{thm}[Riemann-Roch]If $f(x)$ satisfies \begin{itemize} \item $f(x)$ is continuous and in $\mr L_1(A)$. \item $\sum _{\alpha\in k}{f(\ba (x+\alpha ))}$ is convergent for ideles $\ba$ and adeles $x$ uniformly for these variables ranging over compact subsets of the idele and adele groups respectively. \item $\sum _{\alpha\in k}{\hat f(\ba\alpha )}$ is convergent for all ideles $\ba$. \end{itemize}Then: $$\frac{1}{\norm\ba}\sum _{\alpha\in k}{\hat f\left(\frac{\alpha}{\ba}\right)} =\sum _{\alpha\in k}{f(\ba\alpha )}$$ \end{thm}\begin{thm}[Poisson Formula]Let $f$ be continuous and in $\mr L_1(A)$. Assume that $\sum _{\alpha\in k}{\abs{f(x+\alpha )}}$ is uniformly convergent for $x$ in a compact subset of $A$, and that $\sum _{\alpha\in k}{\hat f(\alpha )}$ is convergent. Then: $$\sum _{\alpha\in k}{\hat f(\alpha )}=\sum _{\alpha\in k}{f(\alpha )}$$\end{thm}\noindent To get global zeta functions, we consider functions $f$ satisfying: \\ $\Z 1$. $f(x)$ and $\hat f(x)$ are continuous and in $\mr L_1(A)$.\\ $\Z 2$. The sums $\sum _{\alpha\in k}{f(\ba (x+\alpha ))}$ and $\sum _{\alpha\in k}{\hat f(\ba (x+\alpha ))}$ are both convergent, absolutely and uniformly for ideles $\ba$ and adeles $x$ ranging over compact subsets of the idele and adele groups respectively.\\ $\Z 3$. The functions $f(\ba )\norm \ba^\sigma$ and $\hat f(\ba )\norm \ba^\sigma$ are in $\mr L_1(J)$ for $\sigma >1$.\vskip.2cm\noindent Note that if a function $f$ satisfies $\Z 1-\Z 3$, then $f$ satisfies the hypothesis of the Riemann-Roch theorem. For each one of these functions $f$ satisfying $\Z 1-\Z 3$, we define, for quasi-characters $c$ with $\re s>1$, a zeta function $$\zeta (f,c)=\int{f(a)c(a)d^\times a}$$ the integral taken over the idele group. If $c(a)=\chi (a)\norm a^s$, then $$\zeta (f,c)=\zeta (f,\chi ,s)=\int{f(a)\chi (a)\norm a^sd^\times a}$$ Notice that by fixing $\chi$, our function $\zeta$ becomes a function of $s$, and a holomorphic one, by $\Z 3$, in $\re s>1$. \begin{thm} By analytic continuation we may extend the definition of any zeta function $\zeta (f,c)$ to the domain of all quasi-characters of $J/k^\times$. The extended function is single valued and holomorphic, except at $c(a)=1$ and $c(a)=\norm a$, where it has simple poles with residues $-\kappa f(0)$ and $+\kappa\hat f(0)$ respectively, where $\kappa =$ volume of a multiplicative fundamental domain of $J_k^0 \mod k^\times$. We have the functional equation $$\zeta (f,c)=\zeta (\hat f, \hat c)$$ where $\hat c(a)=\norm ac^{-1}(a)$. \end{thm}\proof We have $$\zeta (f,c)=\int_{\norm a\geqslant 1}{f(a)c(a)d^\times a}+\int_{\norm a\leqslant 1}{f(a)c(a)d^\times a}$$ The second integral clearly converges, so we have to work with the first one. We'll prove a more precise statement: \begin{thm}We have $$\zeta (f,c)=\int_{\norm a\geqslant 1}{f(a)c(a)d^\times a}+\int_{\norm a\geqslant 1}{\hat f(a)\hat c(a)d^\times a}+\delta_\chi\left[ \frac{\kappa\hat f(0)}{s-1}- \frac{\kappa f(0)}{s}\right]$$ where $\delta_\chi$ is 1 or 0 according as the character $\chi$ induced by $c$ on $J^0$ is trivial or not in which case $s$ is the unique complex number such that $$c(a)=\norm a^s$$ The two integrals are convergent for all $c$, and uniformly in every strip $$\sigma_0\leqslant\re c\leqslant\sigma_1$$ \end{thm}\noindent It is clear that the integrals converge uniformly in any given strip, so replace $(f,c)$ with $(\hat f,\hat c)$ on the right hand side. Recall $$\hat{\hat f}(0)=f(-0)=f(0)\quad\mbox{and}\quad\hat{\hat f}(a)=f(-a)$$ so changing variables in the second integral we have, $c(-1)$ coming out of the integral. But, by assumption, $c$ is trivial on $k^\times$. Hence the right hand side is invariant under the transformation $(f,c)\mapsto(\hat f,\hat c)$, we're done\endproof\noindent Now we want to finish the proof by working with the integral taken over $\norm a<1$. First we write the ideles as $J\cong\R_+\times J^0$, and let $E$ be a fundamental domain for the lattice $J^0/k^\times$. Hence for any given (fixed) $t$, \begin{eqnarray*}& & \int_{J^0}{f(tb)c(tb)d^\times b}+f(0)\int_E{c(tb)d^\times b}\\ &=& \sum_{\alpha\in k^\times}{\int_{\alpha E}{f(tb)c(tb)d^\times b}}+f(0)\int_E{c(tb)d^\times b}\\ &=& \sum_{\alpha\in k^\times}{\int_E{f(\alpha tb)c(tb)d^\times b}}+f(0)\int_E{c(tb)d^\times b}\enskip\mbox{since $c(k^\times )=1$ and $d^\times$ is trans. inv.}\\ &=& \int_E{\sum_{\alpha\in k}{f(\alpha tb)c(tb)d^\times b}}\quad\mbox{by $\Z 2$}\\ &=& \int_E{\sum_{\alpha\in k}{\hat f\left(\frac{\alpha}{tb}\right)\frac{1}{\norm{tb}}c(tb)d^\times b}}\quad\mbox{by Riemann-Roch}\end{eqnarray*}\noindent On the other hand, \begin{eqnarray*}& & \int_{J^0}{\hat f(t^{-1}b)\hat c(t^{-1}b)d^\times b}+\hat f(0)\int_E{\hat c(t^{-1}b)d^\times b}\\ &=& \int_{J^0}{\hat f((bt)^{-1})\hat c((bt)^{-1})d^\times b}+\hat f(0)\int_E{\hat c((bt)^{-1})d^\times b}\quad\mbox{by $b\mapsto b^{-1}$ and trans. inv} \\ &=& \sum_{\alpha\in k^\times}{\int_{\alpha E}{\hat f((bt)^{-1})\hat c((bt)^{-1})d^\times b}}+\hat f(0)\int_E{\hat c((bt)^{-1})d^\times b}\\ &=& \sum_{\alpha\in k^\times}{\int_E{\hat f(\alpha (bt)^{-1})\hat c((bt)^{-1})d^\times b}}+\hat f(0)\int_E{\hat c((bt)^{-1})d^\times b}\\ &=& \int_E{\sum_{\alpha\in k}{\hat f(\alpha (bt)^{-1})\hat c((bt)^{-1})d^\times b}}\quad\mbox{by $\Z 2$}\\ &=& \int_E{\sum_{\alpha\in k}{\hat f\left(\frac{\alpha}{tb}\right)c^{-1}((bt)^{-1})\norm{tb}^{-1}d^\times b}}\\ &=& \int_E{\sum_{\alpha\in k}{\hat f\left(\frac{\alpha}{tb}\right)c(tb)\frac{1}{\norm{tb}}d^\times b}}\end{eqnarray*}\noindent So $$\int_{J^0}{f(tb)c(tb)d^\times b}+f(0)\int_E{c(tb)d^\times b}=\int_{J^0}{\hat f(t^{-1}b)\hat c(t^{-1}b)d^\times b}+\hat f(0)\int_E{\hat c(t^{-1}b)d^\times b}$$\noindent But $c(tb)=c(t)c(b)=t^sc(b)$. Hence $$\int_E{c(tb)d^\times b}=t^s\int_E{c(b)d^\times b}=\left\{\begin{array}{ll}\kappa t^s & c(a)=\norm a^s\\ 0 & \mbox{if $c$ is non-trivial on $J^0$}\end{array}\right.$$\noindent where $\kappa =\frac{2^{r_1}(2\pi )^{r_2}hR}{\omega d_\kappa ^{1/2}}$ is the volume of the fundamental domain $E$\footnote{for notation see theorem VII.6 \S 3 on Serge Lang's {\it Algebraic Number Theory\/}.}. Integrating now from $t=0$ to $t=1$, and on RHS apply the transformation $t\mapsto t^{-1}$ so the limits of integration are now 1 to $\infty$. Make this substitution and the theorem follows at once.\endproof \end{document}