%&latex \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} \setcounter{section}{4} % %------------------------------------------- % Apr. 28, Friday, note by Changheon Kang %------------------------------------------- % % Chapter IV. Section 4. % We were talking about a Haar measure on (the multiplicative group) $K^\times$ $$d^\ast x = \frac{dx}{|x|}$$ In the $\mfr{p}$-adic case, we choose a normalization $$d^\ast x =\frac{N\mfr{p}}{N\mfr{p}-1}\frac{dx}{|x|}$$ %--------------------- % Lemma 4.1 %--------------------- \begin{lem}\label{L:IV.4.1} Let $K$ be a $\mfr{p}$-adic field. Then $$\mathrm{vol}(U_K)\stackrel{\rm def}{=}\int_{U_K}d^\ast x=(N\mfr{d})^{-\frac{1}{2}}.$$ \end{lem} \begin{proof} Consider $\int_{U_K}dx$. \begin{eqnarray*} \int_{U_K}dx = \int_{\ok{K}-\mfr{p}_K} dx &=& \int_{\ok{K}}dx-\int_{\mfr{p}_K} dx \\ &=& \mu(\ok{K})-\mu(\pi\ok{K}) \;\;\;\mbox{($\pi$ : a primeelement of $K$)}\\ &=& \mu(\ok{K})-|\pi|\mu(\ok{K}) \\ &=& \mu(\ok{K})(1-N\mfr{p}^{-1}) \\ &=&\left(N\mfr{d}\right)^{-\frac{1}{2}}\left(\frac{N\mfr{p}-1}{N\mfr{p}}\right) \end{eqnarray*} So, $$\int_{U_K}\frac{dx}{|x|}=\int_{U_K}dx=\left(N\mfr{d}\right)^{-\frac{1}{2}}\left(\frac{N\mfr{p}-1}{N\mfr{p}}\right)$$ Thus, we have $$\int_{U_K}d^\ast x=\int_{U_K}\frac{N\mfr{p}}{N\mfr{p}-1}\frac{dx}{|x|}=\left(N\mfr{d}\right)^{-\frac{1}{2}}$$ \end{proof} Let $f\in\mc{S}(K)$. we defined a local zeta function $$\zeta(f,c)=\int_{K^\times}f(y)c(y)d^\ast y$$ with a quasi-character $c\in \Omega(K)$. Since we can write $c(y)=c_0(y)|y|^s$ with some $c_0\in \widehat{U}_K$ and $s\in\C$, the local zeta function can be written as $$\zeta(f,c_0,s)=\int_{K^\times}f(y)c_0(y)|y|^sd^\ast x.$$ %--------------------- % Proprosition 4.1 / Definition of uniform convergence on compact subsets %--------------------- \begin{pro}\label{P:IV.4.1} Let $G$ be a locally compact group (not necessarly abelian) with a Haar measure $dx$. Let $E\subset\C$ be an open set. Let $f:G\times E\rightarrow\C$ be a continuous function such that \begin{itemize} \item [(1)] $\forall x\in G$, $s \mapsto f(x,s)$ is holomorphic on $E$ \item [(2)] $\forall s\in E$, $x \mapsto f(x,s)$ and $x\mapsto f_s(x,s)$ are integrable with respect to $dx$, where $f_s(x,s)$ is the derivative of $f(x,s)$ with respect to $s$. \end{itemize} Define the function $$g(s)=\int_Gf(x,s)dx \;\;\mbox{($s\in E$).}$$ Suppose the integral defining $g$ converges uniformly on compact subsets of $E$, that is, for any compact subset $D\subset E$ and any $\epsilon>0$ there is a compact set $C=C(D,\epsilon)\subset G$ such that $$\left|\int_{G-C}f(x,s)dx\right|<\epsilon\;\;\;\mbox{ for all }s\in K.$$ Then $g(s)$ is holomorphic in $E$. \end{pro} \begin{proof}\footnote{{\sl Analytic Number Theory}, Larry J. Goldstein} \end{proof} %--------------------- % Proprosition 4.2 %--------------------- \begin{pro}\label{P:IV.4.2} The integral for $\zeta(f,c_0,s)$ converges uniformly on compact subsets of the region $\re{s}>0$. Therefore, it is a holomorphic function of $s$ in that region. \end{pro} \begin{proof} Assume that $K=\C$. Then $f\in\mc{S}(K)$ is rapidly decreasing. Take a compact subset $D\subset\{s:\re{s}>0\}$ and choose $\delta>0$ such that $\re{s}\geq\delta$ for all $s\in D$. Since $f\in\mc{S}(K)$ is rapidly decreasing, \begin{eqnarray*} &\exists& A>0 \;\;\mbox{ such that } \left|f(re^{i\theta})\right|\leq Ar^{-3\delta} \;\;\mbox{ if } r\geq 1 \\ \mbox{and }&\exists& B>0 \;\;\mbox{ such that } \left|f(re^{i\theta})\right|\leq B \;\;\;\;\;\;\mbox{ if } r\leq 1. \end{eqnarray*} Then \begin{eqnarray*} \left|\zeta(f,c_0,s)\right|&\leq&\int_{\C^\times}|f(z)c_0(z)||z|_\C^\sigma d^\ast z \\ &=&2\int_0^{2\pi}\int_0^\infty|f(re^{i\theta})|r^{2\sigma-1}dr d\theta \;\;\;\mbox{($\sigma=\re{s}$).} \end{eqnarray*} Choose $0