%&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{\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 33} \author{Uroyoan R. Walker} \address{Louisiana State University, Department of Mathematics, Baton Rouge, LA 70803, USA} \date{April 12, 2000} \email{walker@math.lsu.edu} \maketitle \setcounter{section}{3} \section{Tate's Thesis} \subsection{Overview of Zeta Functions} \begin{thm}[Riemann]\label{1.1} $\zeta(s)=\displaystyle{\sum _{n=1}^\infty {\frac{1}{n^s}}}$ where $\re s>1$, extends to an entire meromorphic function of $s$. It has a simple pole at $s=1$ with residue $1$, and no other poles. Moreover, if we define $$Z(s)=\pi ^{-s/2}\Gamma (s/2)\zeta (s)$$ then we have the functional equation $Z(s)=Z(1-s)$. \end{thm} \noindent So, in particular we must have, $\zeta (s)=\frac{1}{s-1}+$ holomorphic function of $s$. Let's recall that $\Gamma (s)=\int _0^\infty {e^{-t}t^s\,\frac{dt}{t}}\quad\mbox{ where }\re s>0$ and $\frac{dt}{t}$ is Haar measure on $\R _+^\times$. One of the main properties of the Gamma Function is that $\Gamma (s+1)=s\,\Gamma (s)$. We also have, as can be readily checked, $\Gamma (1)=1$. \begin{eqnarray*}\Gamma (s+1) &=& \int _0^\infty {e^{-t}t^s\, dt}\\ &=& -t^se^{-t}\Bigr] _{t=0}^\infty +s\int _0^\infty {e^{-t}t^{s-1}\, dt} \quad\mbox{doing integration by parts}\\ &=& s\,\Gamma (s)\end{eqnarray*}\noindent If $n\in\Z _+$, then $\Gamma (n+1)=n!$. This follows by induction since $\Gamma (1)=1$ and assuming the result true for all $k\leqslant n$, we get: \begin{eqnarray*}\Gamma (n+1) &=& n\,\Gamma (n)\quad\mbox{ by the above argument}\\ &=& n(n-1)\,\Gamma (n-1)\quad\mbox{ by induction } \\ &\vdots& \\ &=& n!\end{eqnarray*} \noindent \underline{Fact}: $\Gamma (s)$ extends meromorphically to the entire complex $s$-plane with simple poles at $0,\, -1,\, -2,\ldots$. \vskip.6cm\noindent Weierstrass Product $$\frac{1}{\Gamma (z)}=ze^{\gamma z}\prod _{n=1}^\infty {(1+\frac{z}{n})e^{-z/n}}$$ \noindent where $\gamma =\displaystyle{\lim _{N\rightarrow\infty}{\left( 1+\frac{1}{2}+\cdots +\frac{1}{N}-\ln (N)\right)}}\approx 0.57712\ldots$ is the Euler-Mascherioni constant. \\ \noindent Main idea of Riemann's proof:\vskip.5cm\noindent First, take the Gamma Function and make a change of variables, $t\rightarrow \pi n^2t=u$ where $n\in\Z$. So, $du=\pi n^2\, dt$, hence $dt=\frac{du}{\pi n^2}$. Thus, we get $$\int _0^\infty {e^{-n^2\pi t}t^{s/2}\,\frac{dt}{t}}=\left(\frac{1}{\pi n^2}\right)^{s/2}\int _0^\infty {e^{-u}u^{s/2}\,\frac{du}{u}}$$ so $$\int _0^\infty {e^{-n^2\pi t}t^{s/2}\,\frac{dt}{t}} =\pi ^{-s/2}\frac{1}{n^s}\,\Gamma (s/2)$$ \noindent Now, summing over $n$ we get \begin{eqnarray*}\int _0^\infty {\left(\sum _{n=1}^\infty {e^{-\pi n^2t}}\right)t^{s/2}\,\frac{dt}{t}} &=& \pi ^{-s/2}\,\Gamma (s/2)\,\zeta (s)\\ &=& Z(s)\end{eqnarray*}\noindent Notice that the parenthesized term inside the integral equals $\frac{1}{2}\left[\theta (t)-1\right]$, where $\theta (t)$ is the $\theta$-series. We'll work with a modified version of the above, which we'll call $\varphi$. We define $$\varphi (s)=\int _1^\infty {t^{s/2}\left[\theta (t)-1\right]\,\frac{dt}{t}}+\int _0^1{\left[\theta (t)-\frac{1}{\sqrt t}\right]t^{s/2}\,\frac{dt}{t}}$$ \begin{pro}\label{1.3}$\varphi (s)$ is an entire function of $s$.\end{pro}\begin{lem}\label{1.2}Let $f(t)$ be a continuous function on $[1,\infty )$ and suppose that $\abs {f(t)}\leqslant K\cdot e^{-Ct}$ for $t\gg 1$ and some $C>0$. Then $$\psi (s)=\int _1^\infty {t^sf(t)\, dt}$$ is an entire function of $s$ and $\psi '(s)=\int _1^\infty {\ln t\cdot t^sf(t)\, dt}\quad$ just differentiate under integral sign with respect to $s$.\end{lem}\noindent We will use this fact without providing a proof. What you must show is that no matter what $s$ is, the integral converges. \proof (of Proposition 1.3) Enough to show that each integral in $\varphi (s)$ is an entire function of $s$. By lemma 1.2, to check the first integral, we must show that $f(t)=\theta (t)-1$ satisfies $\abs{f(t)}\leqslant K\, e^{-Ct}$ for some $C>0$ and $t\gg 1$. But this is more or less clear if you look at the $\theta$-series. We have: \begin{eqnarray*}f(t) &=& 2\sum _{n=1}^\infty {e^{-\pi n^2t}}\\ &=& 2\left( e^{-\pi t}+e^{-4\pi t}+e^{-9\pi t}+\cdots \right)\end{eqnarray*}\noindent Which is exponentially decaying. To check the second integral, we proceed as follows: We have: $$\int _0^1{\left[\theta (t)-\frac{1}{\sqrt t}\right]t^{s/2}\,\frac{dt}{t}}$$ \noindent Recall that we showed, $\theta (t)=\frac{1}{\sqrt t}\,\theta (\frac{1}{t})$. We use this to get:\begin{eqnarray*}\int _0^1{\left[\theta (t)-\frac{1}{\sqrt t}\right]t^{s/2}\,\frac{dt}{t}} &=& \int _0^1{\left[\frac{1}{\sqrt t}\theta (\frac{1}{t})-\frac{1}{\sqrt t}\right]t^{s/2}\,\frac{dt}{t}}\\ &=& \int _0^1{\left[ \theta (\frac{1}{t})-1\right] t^{\frac{s-1}{2}}\,\frac{dt}{t}} \end{eqnarray*}\noindent Now make the substitution, $u=\frac{1}{t}$, so we have $\frac{dt}{t}=-\frac{du}{u}$. Thus, \begin{eqnarray*}\int _0^1{\left[\theta (t)-\frac{1}{\sqrt t}\right]t^{s/2}\,\frac{dt}{t}} &=& \int _0^1{\left[ \theta (\frac{1}{t})-1\right] t^{\frac{s-1}{2}}\,\frac{dt}{t}}\\ &=& \int _1^\infty {\left[\theta (u)-1\right]\,u^{\frac{1-s}{2}}\, \frac{du}{u}}\end{eqnarray*}\noindent Now we are in the situation of lemma 1.2, so we are done.\endproof\noindent Next time we'll finish the proof of the functional equation. \end{document} ¬