%&latex % January 12, 2000 \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 29} \maketitle Recall that if $G$ is a locally compact group, then we have defined the {\sl Haar measure} $\mu$, a left-invariant Borel measure on $G$. We could also require that $\mu$ is right-invariant, that is $$ \mu(U) = \mu(Ug), \quad \forall g \in G, \ \ U \subseteq G, \ \ U \ \mbox{measurable}.$$ In general, left and right-Haar measures do not coincide. If the right-Haar measure equals the right-Haar measue then we say that $G$ is {\sl unimodular}. \begin{exa} \begin{itemize} \item [] \item [$(1)$] If $G$ is a commutative group, then $G$ is unimodular. \item [$(2)$] If $G$ is compact, then $G$ is unimodular. \item [$(3)$] If $G$ is a semi-simple (reductive) Lie group, then $G$ is unimodular. \end{itemize} \end{exa} \begin{exa} Let $G = \left\{g=\begin{pmatrix} a & b \\ 0 & \frac{1}{a} \end{pmatrix}:\ a>0, b \in \R\ \right\}.$ Then the differential form $dg=\frac{da}{a^{2}} \wedge db$ gives a left-invariant Haar-measure,$$ d(g_{0}g)=dg,$$ which is not right-invariant, as $$ d(gg_{0})= \frac{1}{a_{0}^{2}}dg.$$ \end{exa} Let $\alpha: G \longrightarrow G$ be a topological group automorphism. Then $$ \mu(\alpha(U))=\Delta(\alpha)\cdot\mu(U),\quad \Delta(\alpha) \in \R_{+},$$ or equivalently $$\int_{G}f(\alpha^{-1}x)d\mu(x) = \Delta(\alpha)\int_{G}f(x)d\mu(x).$$ The constant $\Delta(\alpha)$, which is independent of $\mu$, is often called the modulus of the automorphism $\alpha$. We now consider an analog of Fubini's theorem. Let $G$ be a locally compact abelian group with closed subgroup $G^{\prime} \le G$. Then $G^{\prime\prime} \cong G/G^{\prime}$ is also locally compact and we have Haar measures $\mu,\mu^{\prime},\mu^{\prime\prime}$ on $G,G^{\prime},G^{\prime\prime}$, such that $$\int_{G}f(x)d\mu(x)=\int_{G^{\prime\prime}}\left(\int_{G^{\prime}}f(x+y)d\mu^{\prime}(y) \right)d\mu^{\prime\prime}(\dot{x}),$$ where $\dot{x}$ is the coset represented by $x+y$. \begin{exa} Let $K$ be a locally compact field of characteristic 0 and let $\mu$ be the Haar measure for $(K, +)$. Then $\forall a\in K^{\times},$ the map $x \mapsto ax$ is an automorphism of $(K,+)$ and thus has modulus as follows. \end{exa} \begin{lem} \label{2.1} The Haar measure for the above example is $$\mu(a) = |a| = \ \ \mbox{the normalized absolute value of \ } a.$$ \end{lem} \begin{proof} If $K=\R$, then $\mu= \ $Lebesgue measure and $|a|=|a|_{\R}$. If $K=\C$, then $\mu= \ $Lebesgue measure $dx \wedge dy$ and $\mu(a) = |a|_{\C}^{2}= \ $the normalized absolute value of $a$. If $K$ is $p$-adic, we consider two cases. First suppose $a\in \ok{K}$ and thus $a\ok{K} \subseteq \ok{K}$ . Let $\dot{x}_{i} \in \ok{K}/a\ok{K}$ be coset representatives, that is $\dot{x}_{i} = x_{i} + a\ok{K}$. Then $$\mu(\ok{K})=\sum_{i} \mu(x_{i} + a\ok{K})= \#\left(\ok{K}/a\ok{K}\right)\cdot\mu(a\ok{K}),$$ as $\mu$ being a Haar measure implies that $\mu(x_{i} + a\ok{K})=\mu(a\ok{K}), \ \forall i.$ Therefore, $$\frac{\mu(a\ok{K})}{\mu(\ok{K})} = \frac{1}{\#\left(\ok{K}/a\ok{K}\right)}=\left( \frac{1}{N{\frak{p}}} \right) ^{ord_{\frak{p}}(a)}=|a|_{\frak{p}}.$$ If $a \notin \ok{K},$ then $a^{-1} \in \ok{K}$ and we are reduced to the first case. Thus, $\mu(a\ok{K})=|a|_{\frak{p}}\cdot \mu(\ok{K})$. \end{proof} \section{Fourier Analysis} Let $G$ be a locally compact, abelian group and recall that $\hat{G}$ is the group of characters of $G$, $i.e.$ the functions $\chi:G\longrightarrow T$. The idea of Fourier Analysis is to write an ``arbitrary'' function $f:G\longrightarrow \C$ as a linear combination of characters $$f=\sum_{\chi\in \hat{G}}C_{\chi}\cdot\chi, \quad C_{\chi} \in \C, \quad \mbox{(the ``Fourier coefficients'')}.$$ Recall $L^{2}(G,\mu)=\{f:G\rightarrow\C\ :\ \mbox{square-summable with respect to a Haar measure} \ \mu\}$. Similarly we can define $L^{2}(\hat{G},\hat{\mu}).$ These are both Hilbert spaces with inner products $$(f_{1},f_{2})_{G} = \int_{G} f_{1}(x)\overline{f_{2}(x)} d\mu(x)\quad\mbox{and}\quad(\varphi_{1},\varphi_{2})_{\hat{G}} = \int_{\hat{G}} \varphi_{1}(\chi)\overline{\varphi_{2}(\chi)} d\hat{\mu}(\chi),$$ respectively. \begin{pro} \label{3.1} Let $G$ be a compact group. Then the different characters of $G$ are orthogonal in $L^{2}(G,\mu)$. \end{pro} \begin{proof} Let $\chi_{1} \ne \chi_{2} \in \hat{G}.$ We will show that $(\chi_{1},\chi_{2})_{G}=0.$ By definition, \begin{align*} (\chi_{1},\chi_{2})_{G} &= \int_{G}\chi_{1}(s)\overline{\chi_{2}(s)}d\mu(s)\quad \mbox{which, by a change of variable,} \\ &= \int_{G}\chi_{1}(s+t)\overline{\chi_{2}(s+t)}d\mu(s+t),\ t\in G \quad \mbox{and, as } \mu \mbox{ is a Haar measure,} \\ &=\int_{G}\chi_{1}(s)\chi_{1}(t)\overline{\chi_{2}(s)}\,\overline{\chi_{2}(t)}d\mu(s)\\ &=\chi_{1}(t)\chi_{2}(t)^{-1}\int_{G}\chi_{1}(s)\chi_{2}(s)d\mu(s)\\ &= \chi_{1}(t)\chi_{2}(t)^{-1}\cdot(\chi_{1},\chi_{2})_{G} \end{align*} Simplifying, we have $$\left( 1-\frac{\chi_{1}(t)}{\chi_{2}(t)}\right)(\chi_{1},\chi_{2})_{G}.$$ Now, $\chi_{1}\ne\chi_{2}\implies \exists t\in G$ such that $\chi_{1}(t) \ne \chi_{2}(t).$ Therefore ${\left( 1-\frac{\chi_{1}(t)}{\chi_{2}(t)}\right) \ne 0,}$ and thus $(\chi_{1},\chi_{2})_{G} =0.$ \end{proof} \end{document}