\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 31} \maketitle Last time we stated the following theorem. Let us now prove it. \begin{thm} \label{T:Pois} (Poisson summation) Choose a Haar measure $\mu$ on G such that $\mu(G/\Gamma) = 1$. Then $$ \sum_{\gamma \in \Gamma} f(\gamma) = \sum_{\hat \gamma \in {\Gamma}^{\bot}} \hat f(\hat \gamma). $$ for all "sufficiently regular" f. \end{thm} \begin{proof} Let us first note that for any well-behaved function $\phi : G \to \Bbb C$, we have by Fubini's theorem, $$ \int_{G} \phi(g) \, d\mu(g) = \int_{G/\Gamma} \sum_{\gamma \in \Gamma} \phi(g + \gamma) \, d\mu(\dot g) \hspace{.10in} \mbox{*}. $$ Let $f: G \to \Bbb C$ be given (i.e well-behaved. This means $f \in C_{0}(G, \Bbb C)$ ) and define $F: G \to \Bbb C$ by $$ F(g) = \sum_{\gamma \in \Gamma} f(g + \gamma). $$ Note that $F(g + \gamma) = F(g)$, $\forall \gamma \in \Gamma$. Thus we can regard F as a function on the coset space i.e. $G/\Gamma \to \Bbb C$. Now take its Fourier transform i.e. consider $\hat F: \hat{(G/\Gamma)}$ $\to \Bbb C$. So $$ \begin{aligned} \hat F(\hat \gamma) & = \int_{G/\Gamma} F(\dot g)\hat \gamma(\dot g) \, d\dot\mu(\dot g) \\ & = \int_{G/\Gamma} \sum_{\gamma \in \Gamma} f(g + \gamma) \hat\gamma(\dot g) \, d\dot\mu(\dot g) \\ & = \int_{G/\Gamma} \sum_{\gamma \in \Gamma} f(g + \gamma)\hat\gamma(g + \gamma) \, d\dot\mu(\dot g) \\ & = \int_{G} f(g) \hat\gamma(g) \, d\mu(g) \hspace{.05in} \mbox{by *} \\ & = \hat f(\hat\gamma). \end{aligned} $$ Thus $$ \hat F(\hat\gamma) = \hat f(\hat\gamma). \hspace{.05in} \hspace{.05in} \mbox{**} $$ Now do Fourier Inversion for $G/\Gamma$ and $\Gamma^{\bot}$. Thus we have $$ \begin{aligned} F(g) & = \sum_{\gamma \in \Gamma} f(g + \gamma) \\ & = \sum_{\hat\gamma \in {\Gamma}^{\bot}} \hat F(\hat\gamma)\hat\gamma(-g) \\ & = \sum_{\hat\gamma \in {\Gamma}^{\bot}} \hat f(\hat\gamma)\hat\gamma(-g) \hspace{.05in} \mbox{by **} \end{aligned} $$ Taking $g = 0$ and so $\hat\gamma(0) = 1$, we obtain $\sum_{\gamma \in \Gamma} f(\gamma) = \sum_{\hat\gamma \in {\Gamma}^{\bot}} \hat f(\hat\gamma)$. \end{proof} \begin{rem} \label{R:theta} {\em Let us consider an important application of Poisson summation, the transformation formula for $\theta$-series. Define $$ \theta(t) = \sum_{n \in \Bbb Z} e^{-{\pi}tn^2} $$ where $t \in \Bbb R$ and $t > 0$. Note that $\theta(t)$ converges. Here $G = \Bbb R$, $\Gamma = \Bbb Z$, $\mu = $ Lebesgue measure, and $\mu(\Bbb R/\Bbb Z) = 1$. We have seen that $\hat G = \Bbb R$. For each $y \in \Bbb R$, we associate a character, ${\rm X}_{y}(x) = e^{-2{\pi}ixy}$. Now what is ${\Bbb Z}^{\bot}$? Well, $$ {\Bbb Z}^{\bot} = \{y \in \Bbb R: {\rm X}_{y}(\Bbb Z) = 1\} = \{y \in \Bbb R: e^{-2{\pi}iyn} = 1, \forall n \in \Bbb Z \} = \Bbb Z. $$ Now, let $f(x) = e^{-{\pi}tx^2}$. We {\rm claim} that $\hat f(y) = \frac{1}{\sqrt{t}}e^{\frac{-{\pi}y^2}{t}}$. Assuming this claim, let us do Poisson summation to get $$ \sum_{n \in \Bbb Z} f(n) = \sum_{m \in \Bbb Z} \hat f(m). $$ And so $$ \theta(t) = \sum_{n \in \Bbb Z} e^{-{\pi}tn^2} = \frac{1}{\sqrt{t}} \sum_{m \in \Bbb Z} e^{\frac{-{\pi}m^2}{t}} = \frac{1}{\sqrt{t}} \theta(\frac{1}{t}). $$ } \end{rem} \begin{thm} \label{T:Jac} $\theta(t) = \frac{1}{\sqrt{t}} \theta(\frac{1}{t})$. \end{thm} To prove the {\rm claim} in the above Remark, we will need the following. \begin{lem} \label{L:stuff} Let $a$,$b \in \Bbb C$ where Re$(a) > 0$. Then $$ \int_{\Bbb R} e^{-ax^2 + 2bx} \, dx = (\frac{\pi}{a})^{1/2}e^{\frac{b^2}{a}} $$ where Re$(\frac{\pi}{a})^{1/2} > 0$. \end{lem} Assuming this lemma, let us prove the {\em claim}. So $$ \begin{aligned} \hat f(y) & = \int_{\Bbb R} f(x)e^{-2{\pi}ixy} \,dx \\ & = \int_{\Bbb R} e^{-{\pi}tx^2}e^{-2{\pi}ixy} \,dx \\ & = \int_{\Bbb R} e^{-{\pi}tx^2 - 2{\pi}ixy} \,dx \\ & = (\frac{1}{t})^{1/2}e^{\frac{-{\pi}y^2}{t}} \hspace{.05in} \mbox{by Lemma \ref{L:stuff} where $a = {\pi}t$, $b = -i{\pi}y$}. \end{aligned} $$ Now let us see the lemma. \begin{proof} Recall $J = \int_{0}^{\infty} e^{-x^2} \,dx$. So $J^2 = \int_{0}^{\infty} e^{-x^2} \,dx \int_{0}^{\infty} e^{-y^2} \,dy = \int_{0}^{\infty}\int_{0}^{\infty} e^{-(x^2 + y^2)} \,dxdy $. Using polar coordinates, we have $$ \int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty} re^{-r^2} \,drd\theta = \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \,d\theta = \frac{\pi}{4}. $$ So $J = \frac{\sqrt{\pi}}{2}$ and thus $\int_{\Bbb R} e^{-x^2} \,dx = \sqrt{\pi}$. Now $$ \begin{aligned} \int_{\Bbb R}e^{-ax^2 + 2bx} \,dx & = \int_{\Bbb R}e^{-a(x - \frac{b}{a})^2 + \frac{b^2}{a}} \,dx \hspace{.05in} \mbox{by completing the square} \\ & = e^{\frac{b^2}{a}}\int_{\Bbb R} e^{-a(x - \frac{b}{a})^2} \,dx \\ & = e^{\frac{b^2}{a}}\int_{\Bbb R} e^{-au^2} \,du \hspace{.05in} \mbox{letting $u = x - \frac{b}{a}$} \\ & = e^{\frac{b^2}{a}} \frac{1}{\sqrt{a}}\int_{\Bbb R} e^{-w^2} \,dw \hspace{.05in} \mbox{letting $w = u\sqrt{a}$} \\ & = (\frac{\pi}{a})^{1/2}e^{\frac{b^2}{a}} \end{aligned} $$ \end{proof} \end{document}