\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 42} \maketitle We now finish the theorem in the p-adic case. We will use the following. \begin{pro} \label{P:compact} If G is a compact group and $\chi : G \to T$ is a non-trivial character, then $(1, \chi)_G = \int_{G} \chi(y) \, dy = 0$. \end{pro} \begin{rem} \label{R:nicef} { \em Let $c_0$ be a character of $U_K$ with conductor $\frak p^n$ such that $c_0(\pi) = 1$ and $c_n = \{x \mapsto c_n(x)|x|^s$ , $s \in \Bbb C \}$. We will evaluate $\zeta(f_n$, $c_n|\cdot|^s)$ for a well-chosen $f_n$. Define $$ f_n(x) = \left \{\begin{array}{ll} e^{2{\pi}i{\Lambda}(x)} & \mbox{if $x \in \frak D^{-1}\frak p^{-n}$} \\ 0 & \mbox{if $x \not\in \frak D^{-1}\frak p^{-n}$} \end{array} \right. $$ } \end{rem} A property of $f_n$ is the following. \begin{lem} \label{L:fhat} $$ \hat f_n(x) = \left \{\begin{array}{ll} (N\frak D)^{1/2}(N\frak p)^n & \mbox{if $x \equiv 1 \bmod \frak p^n$} \\ 0 & \mbox{if $x \not\equiv 1 \bmod \frak p^n$ } \end{array} \right. $$ \end{lem} \begin{proof} We have $$ \begin{aligned} \hat f_n(x) & = \int_{K} f_n(y) e^{-2{\pi}i{\Lambda}(xy)} \, dy \\ & = \int_{\frak D^{-1} \frak p^{-n}} e^{2{\pi}i{\Lambda}(y)} e^{-2{\pi}i{\Lambda}(xy)} \, dy \\ & = \int_{\frak D^{-1} \frak p^{-n}} e^{-2{\pi}i{\Lambda}((x-1)y)} \, dy. \end{aligned} $$ If $x-1 \in \frak p^n$ and $y \in {\frak D^{-1} \frak p^{-n}}$, then $y(x-1) \in \frak D^{-1}$ and so $\hat f_n(x) = \int_{\frak D^{-1} \frak p^{-n}} 1 \, dy = N(\frak D^{-1} \frak p^{-n})^{-1} \int_{\mathcal{O}_K} \, dy = (N\frak D)^{1/2} (N\frak p^n)$. If $x-1 \not\in \frak p^n$ and $y \in {\frak D^{-1} \frak p^{-n}}$, then $y(x-1) \not\in \frak D^{-1}$. Thus $y \mapsto e^{-2{\pi}i{\Lambda}((x-1)y)}$ is a non-trivial character of ${\frak D^{-1} \frak p^{-n}}$ and we conclude by Proposition \ref{P:compact}. \end{proof} Now, let us first handle the unramified case: $n=0$. In this case, $c_0 = 1$ and $f_0 = \chi_{\frak D^{-1}}$ (i.e., the characteristic function on $\frak D^{-1}$). We compute $$ \zeta(f_0, |\cdot|^s) = \int_{K} f_0(y)|y|^s \, d^{*}y = \int_{\frak D^{-1}} |y|^s \, d^{*}y. $$ Recall the annulus, $A_{\nu} = \{x \in K: |x|= N\frak p^{-\nu} \}$. Let $\frak D = \frak p^d$. Then we may express $\frak D^{-1}$ as a disjoint union, namely $$ \frak D^{-1} = \bigcup_{\nu = -d}^{\infty} A_{\nu} $$ and so $$ \begin{aligned} \zeta(f_0, |\cdot|^s) & = \sum_{\nu = -d}^{\infty} \int_{A_{\nu}} |y|^s \, d^{*}y \\ & = \sum_{\nu = -d}^{\infty} N\frak p^{-{\nu}s} \int_{A_{\nu}} \, d^{*}y \\ & = \sum_{\nu = -d}^{\infty} N\frak p^{-{\nu}s} \int_{\pi^{\nu}U_K} \, d^{*}y \\ & = \sum_{\nu = 1}^{\infty} N\frak p^{-{\nu}s} \int_{U_K} \, d^{*}y \\ & = (N\frak D)^{-1/2} \sum_{\nu = -d}^{\infty} (\frac{1}{N\frak p^s})^{\nu} \\ & = (N\frak D)^{-1/2} \frac{N\frak p^{ds}}{1 - N\frak p^{-s}}. \end{aligned} $$ As $\hat f_0$ is $(N\frak D)^{1/2}$ times $\chi_{\frak D}$, we have $$ \begin{aligned} \zeta(\hat f_0, \hat{|\cdot|^s}) & = \zeta(\hat f_0, |\cdot|^{1-s}) \\ & = (N\frak D)^{1/2} \int_{\frak D} |y|^{1-s} \, d^{*}y \\ & = \sum_{\nu = 0}^{\infty} N\frak p^{-{\nu}(1-s)} \\ & = \frac{1}{1 - N\frak p^{s-1}}. \end{aligned} $$ For the ramified case, $n > 0$, we have $$ \begin{aligned} \zeta(f_n, c_{n}|\cdot|^s) & = \int_{\frak D^{-1} \frak p^{-n}} e^{2{\pi}i{\Lambda}(y)} c_{n}(y)|y|^s \,d^{*}y \\ & = \sum_{\nu = -d-n}^{\infty} N\frak p^{-{\nu}s} \int_{A_{\nu}} e^{2{\pi}i{\Lambda}(y)} c_{n}(y) \, d^{*}y \\ \end{aligned} $$ We will use the following Lemma. \begin{lem} \label{L:zero} $\int_{A_{\nu}} e^{2{\pi}i{\Lambda}(y)} c_{n}(y) \, d^{*}y = 0$ for $\nu > -d-n$. \end{lem} \begin{proof} There are two cases. \\ Case 1: $\nu \geq -d$. Then $A_{\nu} \subset \frak D^{-1}$ and thus $e^{2{\pi}i{\Lambda}(y)} = 1$ on $A_{\nu}$. So we have $$ \int_{A_{\nu}} c_{n}(y) \, d^{*}y = \int_{U_K} c_{n}(y{\pi^{\nu}}) \, d^{*}y = 0 $$ since $c_{n}(y)$ is ramified and thus non-trivial on the subgroup $U_K$. \\ Case 2: $-d > \nu > -d-n$. In this case, we can express $A_{\nu}$ as a disjoint union of sets of the form $y_0 + \frak D^{-1} = y_0 + \frak p^{-d} = y_0(1 + \frak p^{-d-\nu})$. On these sets, $\Lambda$ is constant and $$ \int_{y_0 + \frak D^{-1}} e^{2{\pi}i{\Lambda}(y)} c_{n}(y) \, d^{*}y = e^{2{\pi}i{\Lambda}(y_0)} \int_{y_0 + \frak D^{-1}} c_{n}(y) \, d^{*}y $$ which equals $0$ by Proposition \ref{P:compact} as we are integrating over a multiplicative subgroup $1+\frak p^{-d-\nu}$ which has non-trivial character $c_{n}(y)$. \end{proof} Now from Lemma \ref{L:zero}, we have $$ \zeta(f_n, c_n|\cdot|^s) = N\frak p^{(d+n)s} \int_{A_{-d-n}} e^{2{\pi}i{\Lambda}(y)} c_{n}(y) \, d^{*}y. $$ Let $\{\epsilon\}$ be the coset representatives for $U_K/<1+\frak p^n>$. Then we have the disjoint union, $$ A_{-d-n} = {\pi}^{-d-n}U_K = {\pi}^{-d-n} \bigcup_{\epsilon} \epsilon(1+\frak p^n) = \bigcup_{\epsilon} (\epsilon{\pi}^{-d-n} + \frak D^{-1}) $$ On each of these disjoint sets, $e^{2{\pi}i{\Lambda}(y)}$ and $c_{n}(y)$ are constant. To see this, note that $e^{2{\pi}i{\Lambda}(\epsilon{\pi}^{-d-n} + \frak D^{-1})} = e^{2{\pi}i{\Lambda}(\epsilon{\pi}^{-d-n})} e^{2{\pi}i{\Lambda}(\frak D^{-1})} = e^{2{\pi}i{\Lambda}(\epsilon{\pi}^{-d-n})} \cdot 1$. Also, recall $c_{n}(\pi) = 1$ and $c_{n}(\epsilon) = c_{n}(\epsilon^{'})$ if and only if $\epsilon = \epsilon^{'}$. Therefore $c_{n}(\epsilon{\pi}^{-d-n}) = c_{n}(\pi)^{-d-n} \cdot c_{n}(\epsilon) = c_{n}(\epsilon)$. So, $$ \zeta(f_n, c_{n}|\cdot|^s) = N\frak p^{(d+n)s}\left(\sum_{\epsilon} c_{n}(\epsilon) e^{2{\pi}i{\Lambda}(\epsilon \pi^{-d-n})} \right) \int_{1+\frak p^n} d^{*}y. $$ Let us now compute $\zeta(\hat f_n, \hat{c_{n}|\cdot|^s}) = \zeta(\hat f_n, c_{n}^{-1}|\cdot|^{1-s})$. As $\hat f_n$ is $(N\frak D)^{1/2} N\frak p^{n} \chi_{1+\frak p^n}$ (on $\chi_{1+\frak p^n}$, $c_{n}^{-1}(y)|y|^{1-s} = 1$), we have $$ \zeta(\hat f_n, \hat{c_{n}|\cdot|^s}) = (N\frak D)^{1/2}N\frak p^n \int_{1+\frak p^n} \, d^{*}y. $$ \end{document}