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\begin{document}

\title{Lecture 35}

\author{Soad Ahmad}


\date{\today}

\email{sahmad02@yahoo.com}

\maketitle


\begin{pro}   Let $K$ be a local field. Then $K$ is isomorphic to
$K\hat{\ } $ algebraically and topologically. More precisely, let
$\xi \longmapsto\chi(\xi)$ be any nontrivial character of $K$. For
any $\eta\in K$ define
\[\chi_{\eta}(\xi)=\chi(\eta\xi).\]
This is a character and the assignment
$\eta\longmapsto\chi_{\eta}$ is an isomorphism  of $K$ to $K\hat{\
}$.
\end{pro}
\begin{proof}
  1) since $\xi\longmapsto\eta\xi$ is a continuous
  homomorphism of $(K,+)\longrightarrow(K,+)$ it's clear that
  $\xi\longmapsto\chi(\eta\xi)$ is a character.

  
  2) Suppose that $\chi(\eta\xi)=1$  $\forall\xi\in K$. Since
  $\chi\neq 1$ this implies that $\eta K\subsetneqq K$ hence
  $\eta=0$.
  Therefore,
\[K\longrightarrow K\hat{\ } \mbox{ is a surjective homomorphism}\]
3) If $\chi(\eta\xi)=1$  $\forall\eta ,$ then $K\xi\subsetneqq K$
and hence $\xi=0 .$

So the characters of the form $\chi_{\eta}$ are dense in $K\hat{\
}$. This is because, if we let $C=\{\chi_{\eta}:\eta\in
K\}\subseteq K\hat{\ }$ then $\overline{C}\subseteq K\hat{\ }$ is
a closed subgroup. Consider \[ \left(K\hat{\
}/\overline{C}\right)\hat{\
}=\overline{C}^{\perp}=C^{\perp}\]where $C^{\perp}=\{\psi\in
K\hat{\ }\hat{\ }:\psi(C)=1\}.$

    $=\{\xi\in K:\xi(C)=1\}$

    $=\{\xi \in K: \chi(\eta\xi)=1\forall \eta\}$

    $=\{0\}$

$\therefore \hat{C}=K\hat{\ }.$

4) $\eta\longmapsto\chi_{\eta}$ is bicontinous :

Let $B=\{\xi\in K:|\xi|\leq M\}.$ $B$ is closed and bounded hence
 compact .

 If $\eta$ is near $0$ in $K$, then $\eta B$ is near $0$ in $K$
 and $\chi(\eta B)$ is near 1 in T , so $ \chi_{\eta}$ is near 1
 in $K\hat{\ }$ and hence it's continuous.

 Now let $\xi_0 \in K$ be any element with $\chi(\xi_0)\neq1.$
 Suppose $\chi_{\eta}$ is near 1 in the dual $K\hat{\ }$, then
 $\chi_{\eta}(B)=\chi(\eta B)$ is near 1 in $T$. In fact we can
 assume it's even closer than  $\chi(\xi_0)$.

 5) By 4, $\{\chi_\eta:\eta\in K\}$ is locally compact and hence
 is complete closed subset of $K\hat{\ }.$ But it's  dense in $K\hat{\
 }$ hence it's all of it.

 \end{proof}

Special Cases:

1) $K=\R:$

   $\chi(\xi)=\exp(2\pi i\xi)$

  and

   $\R \simeq \R\hat{\ }$

   $\eta\longmapsto$ the character $\chi :\xi\longmapsto \exp(2\pi
   i\eta\xi).$

 \

 2) $K=\C$

     $\chi(\xi)=\exp(Tr_{\C/\R}(\xi)).$

 \

 3) $K=p-adic$:

    $\chi(\xi)=\chi(Tr_{K/\Q_p}(\xi)$  this works where $\chi$ is  non trivial
    character of $\Q_p.$

\end{document}