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\begin{document}

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%
%---------------------------------------------
%   Mar. 24, Friday, note by Changheon Kang
%---------------------------------------------
%
%    Chapter III. Fourier Analysis on Locally Compact Abelian Groups%
%    Section 1. Pontryagin Duality
%---------------------------------------------
\section{Pontryagin Duality}

$G$ = a locally compact abelian group.

\begin{Def}\label{D:III.1.1}
 A character of $G$ is a continuous homomorphism
$$\chi:G\rightarrow T=\{z\in\C : |z|=1\}$$
\end{Def}

Let $\dual{G}$ be the set of all characters $\chi:G\rightarrow T$ of $G$.
Then $\dual{G}$ is an abelian group (called the {\it character group} of $G$)
by defining the product $\chi_1\chi_2$ as follows:
$$(\chi_1\chi_2)(g)=\chi_1(g)\chi_2(g)\;\;\mbox{($g\in G$)}.$$

$\dual{G}$ can be given a topology as follows:
It's enough to describe the neighborhoods of the unit character
$\mathbf{1}\in\dual{G}$ ($\mathbf{1} : G \rightarrow \{1\}$).
Let $U\subset T$ be an open neighborhood of $1\in T$. Let $K\subset G$ be any compact set in $G$. Then a basic open neighborhood of the unitcharacter is
$$C(K,U)=\{\chi: \chi(K)\subseteq U\}\subset\dual{G}.$$

\begin{rem} With respect to this topology,
\begin{itemize}
\item [(1)] $\dual{G}$ is a locally compact abelian group.
\item [(2)] If $G$ is discrete, then $\dual{G}$ is compact, and if $G$ is
compact, then $\dual{G}$ is discrete.
\end{itemize}
\end{rem}

%-----------------------------------
%   Examples
%-----------------------------------
\begin{exa}

(1) $G=\Z$ then $\dual{G}=T$: $\chi\in\dual{G}$ can be written as
$$\chi=\chi_t \;\mbox{ defined by $\chi_t(n)=t^n\;$ ($t\in T$).}$$

(2) $G=T$ then $\dual{G}=\Z$: $\chi\in\dual{G}$ can be written as
$$\chi=\chi_n \;\mbox{ defined by $\chi_n(t)=t^n\;$ ($n\in\Z$).}$$

(3) $G=\Z/n$ then $\dual{G}=U_n=\{\zeta\in\C : \zeta^n=1\}$, the
set of all $n$-th roots of unity: $\chi\in\dual{G}$ can be written as
$$\chi=\chi_\zeta \;\mbox{ defined by
$\chi_\zeta([i]_n)=\zeta^i\;$ ($\zeta\in U_n$).}$$

(4) $G=(\R,+)$ then $\dual{G}=\R$: $\chi\in\dual{G}$ can be written as
$$\chi=\chi_r \;\mbox{ defined by $\chi_r(x)=e^{2\pi ixr}\;$ ($r\in\R$).}$$

(5) $G=\Z_p$.

Note that $\Z_p$ is compact and abelian, whence $\dual{G}$ is discrete.
Since we can write $$\Z_p=\lim_\leftarrow\Z/p^i,$$ we have
$$\dual{\Z_p}=\dual{\lim_\leftarrow\Z/p^i}=\lim_\rightarrow\dual{\Z/p^i}  =\lim_\rightarrow U_{p^i} = \bigcup_i U_{p^i} = U_{p^\infty}$$

(Q) If $G=\Q_p$, $A_K$, or $J_K$ then what is $\dual{G}$ ?

\end{exa}

%---------------------------------------------------
%    Theorem III.1.1 [Pontryagin Duality Theorem]
%---------------------------------------------------

\begin{thm}[Pontryagin Duality Theorem]\label{T:III.1.1}
 Let $G$ be a locally compact abelian group. The natural map $$G
\rightarrow \dual{\dual{G}}$$ which sends $g\in G$ to the character $\chi_g:\dual{G}\rightarrow T$ by the formula
$$\chi_g(\psi) = \psi(g)\;\;\mbox{ for }\; \psi\in\dual{G}$$
is an isomorphism of topological groups.
\end{thm}

\begin{proof}
\end{proof}

Note that if $G$ is locally compact then so is $G/H$ for a closed
subgroup $H\subset G$.
$$\dual{(G/H)}=H^\perp\stackrel{\rm def}{=}\{\chi:G\rightarrow T |
\chi(H)=1\}.$$


%-----------------------------------
%     Section 2. Haar Measures
%-----------------------------------
\section{Haar Measures}

%-----------------------------------
%    Theorem III.2.1 [Haar]
%-----------------------------------
\begin{thm}[Haar]\label{T:III.2.1}
 Let $G$ be a locally compact group. Then there is a Borel measure $\mu$ on $G$ that is invariant under the left translation in $G$:
$$\mu(xU)=\mu(U)\;\;\mbox{ for all }\; x\in G \;\mbox{ and any
measurable set }U\subset G.$$
(If $G$ is additive, we can replace $xU$ by $x+U$.)

Moreover, $\mu$ is unique up to scalar multiples.
\end{thm}

We call $\mu$ in the theorem above {\it a left Haar measure}.

\begin{rem}
\begin{itemize}
\item [(1)] For any Haar measure $\mu$, if $D\subset G$ is compact then $\mu(D)<\infty$.
In particular, if $G$ itself is compact, we can choose $\mu$ so
that $\mu(G)=1$.
\item [(2)] In terms of integral, $\mu(xU)=\mu(U)$ can be written as
$$\int_U f(xy)d\mu(y)=\int_U f(y)d\mu(y)\;\mbox{ or just $d\mu(xy)=d\mu(y)$.}$$
\end{itemize}
\end{rem}

%-----------------------------------
%   Examples
%-----------------------------------
\begin{exa}

(1) $G=(\R,+)$ : $\mu =$ Lebesgue measure $\lambda$ is a Haar measure.
Clearly, $\lambda(x+U)=\lambda(U)$.

(2) $G=$ a finite group : $\mu(g)=c_0$ ($c_0$ a constant) is a Haar measure.
Then $\mu(U)= c_0 \cdot\#U$. Conveniently we take $c_0 =
\frac{1}{\#G}$ so that $\mu(G)=1$.

(3) If $G=\GL(2,\R)$ then $\mu(A)=\frac{\lambda(A)}{|A|^2}$ is a Haar measure.

(4) If $G=\Z_p$ then we can choose a Haar measure such that $\mu(\Z_p)=1$.
Consider $\mu(p\Z_p)$.
Note $p\Z_p \subset \Z_p$ is open.

Since we can write $$\Z_p=\bigcup_{i=0}^{p-1} (i+p\Z_p),$$ we have
$$1=\mu(\Z_p)=\sum_{i=0}^{p-1}\mu(i+p\Z_p)=\sum_{i=0}^{p-1}\mu(p\Z_p)=p\mu(p\Z
_p).$$
Hence $\mu(p\Z_p)=\frac{1}{p}.$

What about $\mu(p^\nu\Z_p)=$ ?

\end{exa}

%-----------------------------

\end{document}