%&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{\norm}[1]{\Vert #1 \Vert}
\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}
\centerline{\bf{Lecture 43}}
\vskip .4cm
\section{Abstract restricted direct products}

Let $\mfr{p}$ be any set of indices and consider, for each $\mfr{p}$, a commutative locally compact group $G_{\mfr{p}}$ with a given compact subroup $H_{\mfr{p}}$ for all but finitely many $\mfr{p}$'s. Let $G$ be the restricted direct product of the $G_{\mfr{p}}$ with respect to $H_{\mfr{p}}$.
If $c$ is any quasi-character of $G$, denote by $c_{\mfr{p}}$ the restriction of $c$ to $G_{\mfr{p}}$ (i.e. $c_{\mfr{p}}(a_{\mfr{p}})=c(1,1,...,a_{\mfr{p}},...)$) which is clearly a quasi-character of $G_{\mfr{p}}$.
\begin{lem} \label{lem:IV.5.1}  $c_{\mfr{p}}$ is trivial on $H_{\mfr{p}}$ for almost all $\mfr{p}$ and for any $a\in G$ we have:
\[
c(a)=\prod_{\mfr{p}}c_{\mfr{p}}(a_{\mfr{p}})
\]
almost all factors in the product being equal to 1.
\end{lem} 
\noindent{\em Proof}: Let $U$ be a neighborhood of 1 in the complex plane which does not contain any multiplicative subgroup other than $\{1\}$. This neighborhood can be constructed because if a non-trivial multiplicative subgroup of $\C$ contains an element of absolute value different than 1, then that subgroup contains numbers of arbitrarily large or small absolute value, and we can avoid this situation by chosing $U$ inside of a ball of small radius centered at 1. If the same subgroup contains a complex number other than 1 on the unit torus $T$, then it contains all powers of that number, hence at least a complex number in the half-plane $Re(z)\leq 0$, and this situation can be avoided in the same manner. Since $c$ is continuous, let $N=\prod_{\mfr{p}}N_{\mfr{p}}$ be a neighborhood of 1 in $G$ such that $c(N)\subseteq U$. Select the set $S$ such that it contains all $\mfr{p}$ for which $N_{\mfr{p}}\neq H_{\mfr{p}}$ along with all $\mfr{p}$ for which $H_{\mfr{p}}$ is not defined. If we denote by 
\[
G^S=\{a\in G:\ a_{\mfr{p}}=1,\forall \mfr{p}\in S,\ a_{\mfr{p}}\in H_{\mfr{p}},\forall \mfr{p}\notin S\}
\] 
then clearly $G^S\subseteq N$ and thus $c(G^S)\subseteq U$. Since $U$ contains no non-trivial subgroups, $c(G^S)=1$, and then necessarily $c_{\mfr{p}}(H_{\mfr{p}})=1,\ \forall \mfr{p}\notin S$. \\
To prove the product formula, fix $a\in G$. By enlarging $S$ if necessary, we can assume that $a_{\mfr{p}}\in H_{\mfr{p}},\ \forall \mfr{p}\notin S$. Thus we can write $a=\prod_{\mfr{p}\in S}a_{\mfr{p}}\cdot a^S$, with $a^S\in G^S$. Since $c$ is multiplicative and trivial on $G^S$, we can write
\[
c(a)=\prod_{\mfr{p}\in S}c_{\mfr{p}}(a_{\mfr{p}})\cdot c(a^S)=\prod_{\mfr{p}\in S}c_{\mfr{p}}(a_{\mfr{p}}).
\]
Finally, we know that outside $S$ $c_{\mfr{p}}(a_{\mfr{p}})=1$, and the equality follows.
\\
\par
In fact one can show a converse of this result:
\begin{lem} \label{lem:IV.5.2}  Let $c_{\mfr{p}}$ be a quasi-character of $G_{\mfr{p}}$ for each $\mfr{p}$, such that $c_{\mfr{p}}$ is trivial on $H_{\mfr{p}}$ for almost all $\mfr{p}$. If we define 
\[
c(a)=\prod_{\mfr{p}}c_{\mfr{p}}(a_{\mfr{p}})
\]
then $c$ is a quasi-character of $G$.
\end{lem} 
\noindent{\em Proof}: The multiplicativity of $c$ is obvious. Let 
\[
S=\{\mfr{p}:c_{\mfr{p}}(H_{\mfr{p}})\not= 1\}
\]
and define $s=|S|$. Fix a neighborhood $U$ of 1 in the complex plane, and a neighborhood $V$ of 1 such that $V^s\subset U$ (the map $x\rightarrow x^s$ is continuous). Since the maps $c_{\mfr{p}}$ are continuous, we can find a neighborhood $N_{\mfr{p}}$ of 1 in $G_{\mfr{p}}$ such that $c_{\mfr{p}}(N_{\mfr{p}})\subset V$ for $\mfr{p}\in S$. For $\mfr{p}\notin S$ define $N_{\mfr{p}}=H_{\mfr{p}}$. Then $c(\prod_{\mfr{p}} N_{\mfr{p}})\subset V^s\subset U$, so $c$ is continuous.
\begin{rem} $c$ is a character if and only if all $c_{\mfr{p}}$ are characters.
\end{rem} 
\begin{thm} \label{thm:IV.5.1} The restricted direct product of the groups $\hat{G}_{\mfr{p}}$ relative to the subgroups $H_{\mfr{p}}^{\bot}$ is naturally isomorphic, topologically and algebraically, to $\hat{G}$. 
\end{thm}
\noindent{\em Proof}: Lemma \ref{lem:IV.5.1}{} and Lemma \ref{lem:IV.5.2}{} show that the map $c\rightarrow \prod_{\mfr{p}}c_{\mfr{p}}$ is in fact an algebraic isomorphism between the two groups. We only need to check that this is also a topological isomorphism. It is enough to check that this isomorphism preserves the open neighborhoods of 1. Let $c$ be a character of $G$, close to 1 (as a character). Then we can find a large compact set $B\subset G$ of the form $\prod_{\mfr{p}}B_{\mfr{p}}$, where $B_{\mfr{p}}\subset G_{\mfr{p}}$ is compact for all $\mfr{p}$, $B_{\mfr{p}}=H_{\mfr{p}}$ for almost all $\mfr{p}$, such that $c(B)$ is close to 1 in $\C$. We have seen that $c_{\mfr{p}}(H_{\mfr{p}})=1$ for almost all $\mfr{p}$ (the only sugroups of $\C$ that are close to 1 are trivial - see the proof of Lemma IV.3.1), so $c_{\mfr{p}}(B_{\mfr{p}})=1$ for almost all $\mfr{p}$ (to be more precise, this happens for all $\mfr{p}$ for which $B_{\mfr{p}}=H_{\mfr{p}}$). For the remaining $\mfr{p}$,  $c_{\mfr{p}}(B_{\mfr{p}})$ will be just close to 1 (because $c(B)$ is close to 1). But then $c_{\mfr{p}}$ is close to 1 in $\hat{G}_{\mfr{p}}$ for a finite number of $\mfr{p}$, and $c_{\mfr{p}}\in H_{\mfr{p}}^{\bot}$ for the remaining $\mfr{p}$, which simply means that $c$ is close to 1 in the restricted direct product of the $\hat{G}_{\mfr{p}}$.\\
The converse is essentially reversing the logical implications, so we omit the formal proof.
\\
\par Suppose now that we have chosen a Haar measure $da_{\mfr{p}}$ on each $G_{\mfr{p}}$ such that $Vol(H_{\mfr{p}})=1$ for almost all $\mfr{p}$. We would like to define a Haar measure $da$ on $G$ for which, in some sense, $da=\prod_{\mfr{p}\in S}da_{\mfr{p}}$. To accomplish this fix a set $S$ (containing all $\mfr{p}$ for which $H_{\mfr{p}}$ is not defined), and define $G_S=(\prod_{\mfr{p}\in S}G_{\mfr{p}})\times G^S$. Note that $G_S=\prod_{\mfr{p}\in S}G_{\mfr{p}}\times \prod_{\mfr{p}\notin S}H_{\mfr{p}}$ is a direct product of locally compact groups, all but finitely many of which are compact. So with the product topology $G_S$ is locally compact. Thus we can consider the product measure on $G_S$. Since the topology on $G$ is the smallest topology in which each $G_S$ is open, there exists a unique Haar measure on $G$ that induces the product measure on each $G_S$, and let us denote this measure by $\prod_{\mfr{p}}da_{\mfr{p}}$. \\
\begin{pro} \label{prop:IV.5.1} If $f$ is a function on $G$ then
\[
\int f(a)da=lim_S\int_{G_S}f(a)da
\]
where either:\\
(1) $f$ is measurable, $f(a)\geq 0$, and we accept $+\infty$ as a value for the integrals; or\\
(2) $f\in L_1(G)$, in which case the values of the integrals are complex numbers.
\end{pro} 
\noindent{\em Proof}: $\int f(a)da$ is the limit of $\int_B f(a)da$, for larger and larger compact subsets $B$ in $G$. Every compact $B$ is contained in a certain $G_S$. Then by using characteristic functions we get in case (1):
\[
f(a)=lim_B f(a)\cdot \chi_B\leq lim_{G_S} f(a)\cdot \chi_{G_S}\leq f(a)  
\]
and the equality follows by integration. In case (2) the statement follows by a similar argument involving absolute values. 

\section{Global additive duality} 

Throughout this section $K$ denotes an algebraic number field. Recall that the adele group of $K$, denoted by $\mfr{A}_K$, is the locally compact group obtained as a restricted direct product of the local fields $G_{\mfr{p}}=K_{\mfr{p}}$ (regarded as additive groups) with respect to the subgroups $H_{\mfr{p}}=O_{\mfr{p}}$, given at the non-archimedian places. \\
If we combine Theorem \ref{thm:IV.5.1}{}, and Proposition \ref{prop:III.2.1}{} we note that $\hat{\mfr{A}}_K$ is the restricted direct product of the groups $\hat{K}_{\mfr{p}}\simeq K_{\mfr{p}}$ relative to the subgroups $O_{\mfr{p}}^{\bot}$, for $\mfr{p}$ non-archimedian. The isomorphisms are algebraic and topologic. We have also noticed that $O_{\mfr{p}}^{\bot}\simeq \mfr{D_{\mfr{p}}}^{-1}$, where $\mfr{D_{\mfr{p}}}$ is the different of $K_{\mfr{p}}$. But $\mfr{D_{\mfr{p}}} = O_{\mfr{p}}$ for almost all $\mfr{p}$, which means that $\hat{\mfr{A}}_K\simeq \mfr{A}_K$, algebraically and topologically. Moreover, according to Proposition \ref{prop:III.2.1}{} the identification given by the above isomorphism is:
\[
\alpha=(...,\alpha_{\mfr{p}},...)\rightarrow [x=(...,x_{\mfr{p}},...)\rightarrow \prod_{\mfr{p}}exp(2\pi i\Lambda_{\mfr{p}}(\alpha_{\mfr{p}}x_{\mfr{p}}))]
\]
To summarize:
\begin{thm} \label{thm:IV.6.1} The adele group is self-dual. We identify any adele $\alpha$ with the character $x\rightarrow e^{2\pi i \Lambda(\alpha x)}$, where $\Lambda(y)=\sum_{\mfr{p}}\Lambda_{\mfr{p}}(y_{\mfr{p}})$.
\end{thm}
We define the Schwartz space of the adele ring:
\[
\mfr{S}(\mfr{A}_K)=\{\sum_{i=1}^n f_i: f_i=f_{\infty}^{(i)}\times \prod_{\mfr{p}}f_{\mfr p}^{(i)}\}
\]
where:\\
\hskip 1cm $f_{\infty}^{(i)}\in \mfr{S}(\R^{r_1+2r_2})$, (as usually, $r_1$= the number of real embeddings, and $r_2$= the number of pairs of complex embeddings),\\
\hskip 1cm $f_{\mfr{p}}^{(i)}\in \mfr{S}(K_{\mfr{p}})$, for all $\mfr{p}$ non-archimedian,\\
\hskip 1cm $f_{\mfr{p}}^{(i)}=\chi_{O_{\mfr{p}}}$, for almost all $\mfr{p}$ non-archimedian.\\
It turns out that every function in $\mfr{S}(\mfr{A}_K)$ is continuous.\\
Now the measure on $\mfr{A}_K$ is $dx=\prod_{\mfr{p}}dx_{\mfr{p}}$, where $dx_{\mfr{p}}$ was chosen like in section III.2 (i.e. to be self-dual). \\
Fact: If each $dx_{\mfr{p}}$ is self-dual then $dx$ is self-dual either. \\
The consequence of this observation is:
\begin{thm} \label{thm:IV.6.2} If for a function $f\in \mfr{S}(\mfr{A}_K)$ we define the Fourier transform
\[
\hat{f}(y)=\int f(x)e^{2\pi i \Lambda(yx)}dx
\]
then $\hat{f}\in \mfr{S}(\mfr{A}_K)$ and the following inversion formula holds:
\[
f(x)=\int \hat{f}(y)e^{2\pi i \Lambda(xy)}dy
\] 
where $dx$ is chosen as above.
\end{thm}  
\par Let us investigate now the following problem: given an adele $\alpha\in \mfr{A}_K$, is $x\rightarrow \alpha \cdot x$ and automorphism of $\mfr{A}_K$? Certainly this is a continuous endomorphism of $\mfr{A}_K$. Here is the complete answer of the question:\\
\begin{lem} \label{lem:IV.6.1} The map $x\rightarrow \alpha \cdot x$ is an automorphism of $\mfr{A}_K$ if and only if $\alpha\in \mfr{I}_K$.
\end{lem}
\noindent{\em Proof}: It is enough to find $\beta \in \mfr{A}_K$ such that $\alpha\cdot \beta=(...,1,...)=1$. If such a $\beta$ exists, then the map $x\rightarrow \beta \cdot x$ is the inverse of the given map, so this endomorphism is actually an automorphism. Conversely, if the map is an automorphism then it is surjective, so we can find such a $\beta$. Hence the map is an automorphism iff we can find $\beta \in \mfr{A}_K$ such that $\alpha\cdot \beta=1$. But the multiplication on $\mfr{A}_K$ is component-wise, so in order for such a $\beta$ to exist we need $\alpha_{\mfr{p}}\not= 0$, $\alpha_{\mfr{p}}^{-1}\in O_{\mfr{p}}$ for almost all $\mfr{p}$, and $\beta_{\mfr{p}}=\alpha_{\mfr{p}}^{-1}$. In other words $\alpha_{\mfr{p}}$ is a unit in $O_{\mfr{p}}$ for almost all $\mfr{p}$ and is non-zero for the remaining $\mfr{p}$. But this means that $\alpha$ is an idele.\\
\par Let us consider an idele $\alpha$. Almost all components $\alpha_{\mfr{p}}$ are units in $O_{\mfr{p}}$, so the product $\prod_{\mfr{p}}|\alpha_{\mfr{p}}|$ is really a finite product. We define the size of the idele $\alpha$:
\[
|\alpha|=\prod_{\mfr{p}}|\alpha_{\mfr{p}}|
\] 
and the following equality is true:
\begin{lem} \label{lem:IV.6.2} For an idele $\alpha$:
\[
d(\alpha\cdot x)=|\alpha|\cdot dx.
\]
\end{lem}
\noindent{\em Proof}: We will use the following fact: if $B=\prod_{\mfr{p}}B_{\mfr{p}}$ is a compact neighborhood of 0 in the adele ring then $\int_{B}dx=\prod_{\mfr{p}}\int_{B_{\mfr{p}}}dx_{\mfr{p}}$ (remember that $dx=\prod_{\mfr{p}}dx_{\mfr{p}}$). Then: 
\[
\int_{\alpha B} dx = \prod_{\mfr{p}} \int_{\alpha_{\mfr{p}} B_{\mfr{p}}} dx_{\mfr{p}}.
\]
We have seen that $d(\alpha_{\mfr{p}}x_{\mfr{p}})=|\alpha_{\mfr{p}}|dx_{\mfr{p}}$, and this proves the equality. 
\end{document}