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% Class notes on April 24, 2000 Lecture Notes no4 prepared for Dr. Hoffman's 
%Number Theory Class.

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\begin{document}

\title{Lecture 37}

\author{Cem Guneri}

\address{Louisiana State University \\
Department of Mathematics \\
Baton Rouge, LA 70803, USA}

\date{\today}

\email{guneri@math.lsu.edu}
\maketitle

\section{Quasi-Characters Of Local Fields}
\begin{Def} Let $K$ be a local field. A quasi-character of $K$ is a continuous 
homomorphism 
\[\omega : K^* \longrightarrow \C^*\]

We say that $\omega$ is unramified if it is identically one on the units
\[U_{K}=O_{K}^*=\{x\in K^*; \hspace{2pt} |x|=1\}\]

We let $\Omega(K)$ be the set of quasi-characters of $K$.
\end{Def}

\begin{pro} The unramified quasi-characters are all of the form
\[ x \longmapsto |x|^s\]
where $s\in \C$ and uniquely determined if $K$ is archimedean and unique upto 
adding integer multiples of $\frac{2\pi i}{\log Np}$ if $K$ is p-adic.
\end{pro}

\proof There is no doubt that all such maps are continuous homomorphisms 
(thinking multiplicatively of course) from $K^*$ to $\C^*$ and they are identically 
one on $U_{K}$. So all such maps are unramified quasi-characters. In case of an 
archimedean local field (i.e: $\R$ or $\C$) these maps are obviously unique for every 
$s\in \C$. In the nonarchimedean case (i.e; p-adic) we know that values 
of $|.|_{p}$ are $q^{\Z}\subset \R_{+}^*$ (which is isomorphic to $\Z$) where 
$q=Np$. Then we take the $s^{th}$ power to $\C^*$ to obtain a quasi-character. 
Now
\[q^s=q^{s'} \iff e^{s \log(q)} = e^{s' \log(q)}\]

But the equality onthe right hand side is true upto the constant 
$\frac{2\pi i}{\log Np}$.

Now conversely let $\omega$ be any quasi-character. We have the following two 
epimorphisms:
\[K:archimedean \Longrightarrow |.|:K^* \longrightarrow \R_{+}^*\]
\[K:nonarchimedean \Longrightarrow |.|_{p}:K^* \longrightarrow \Z\]
and both of these have $U_{K}$ as their kernel. Therefore we have the following 
descriptions for $K^*$:
\[K:archimedean \Longrightarrow K^* \simeq \R_{+}^* \times U_{K}\]
\[K:nonarchimedean \Longrightarrow K^* \simeq \Z \times U_{K}\]

Hence an unramified quasi-character is a function of the form 
\[x \longmapsto \omega_{1}(|x|)\]
where $\omega_{1}$ is a continuous quasi-character of group of values. 
(i.e; $\R_{+}^*$ or $\Z$) If we consider the archimedean case, for instance, then 
the continuous homomorphisms from $\R_{+}^*$ to $\C^*$ are of the form 
$t\mapsto t^s$ and this is the form we were looking for.
\endproof

\begin{thm} For a local field $K$, the quasi-characters are maps of the form
\[ \omega: \hspace{2pt} x \longmapsto \tilde \omega(u) |x|^s\]
where $s\in \C$ as in Prop 3.1., $\tilde \omega$ is a character of the group 
of units and $u$ is the unit part of $x\in K^*$ if we write $x=\pi^{\nu}u$ for some 
prime element $\pi$. 
(i.e: $\tilde \omega \in \widehat{U_{K}}$)
\end{thm}

\proof Since $\tilde \omega$ is a character, any map $\omega$ of this form is a 
continuous homomorphism from $K^*$ to $\C^*$ and hence a quasi-character. 
Conversely let $\omega \in \Omega(K)$ be any quasi-character of $K$. Consider 
it's restriction to $U_{K}\subset K^*$ and call it $\tilde \omega$. Since $U_{K}$ 
is compact $|\tilde \omega(U_{K})|$ is a compact subgroup of $\R_{+}^*$. But 
only such subgroup of $\R_{+}^*$ is $\{1\}$ and hence $|\tilde \omega(U_{K})|=1$. 
But then $\tilde \omega$ is a map from $U_{K}$ to $T\subset \C$ and it is a 
continuous homomorphism since it comes from a quasi-character $\omega$. That means 
$\tilde \omega$ is a character of $U_{K}$.  

Note that 
\[x \longmapsto \frac{\omega(x)}{\tilde \omega(u)}\]
is an unramified quasi-character of $K^*$ since it is identically one on $U_{K}$. 
(Note that $u$ is again the unit part of $x$ with respect to some prime element)
Therefore this map is of the form $x \longmapsto |x|^s$ by Prop 3.1. But this 
implies that $\omega \in \Omega(K)$ is of the form that we wanted.
\endproof

By the last theorem we see that a quasi-character of $K$ is actually a product 
of a character of $U_{K}$ and an unramified quasi-character of $K$.

\begin{note} Given a quasi-character $\omega$ of $K$. The complex absolute value of 
$\omega(x)$ for any $x\in K^*$ can be computed as follows:
\[|\omega(x)|=|\tilde \omega(u) |x|^s|=|\tilde \omega(u)|||x|^s|=|x|^{Re(s)}\]

Observe that even though the choice of $u$ depends on the prime element you work 
with, the exponent $s\in\C$ comes from the unramified quasi-character 
$x\longmapsto \frac{\omega(x)}{\tilde \omega(u)}$ and in Prop 3.1. we showed a 
certain uniqueness property of this complex number $s$. Hence in both the 
archimedean and the nonarchimedean cases, real part of $s$ will be well defined.

This observation makes it possible to define the so called exponent of the 
quasi-character $\omega$ of $K$ as $\sigma=Re(\omega):=Re(s)$. 
\end{note}

\begin{exa} Let's compute $\Omega(K)$ for real and complex numbers and say few words 
in case of p-adic numbers.

$K=\R$: In this case $U_{K}=\{\pm{1}\}$ and $\widehat{U_{K}}=\{id,sgn\}$, where 
$sgn(-1)=-1$. Then
\[\Omega(\R)=\{x\mapsto (sgn)^m|x|^s; \hspace{2pt} m=0,1 \hspace{2pt} \mbox{and}
\hspace{2pt} s\in \C\} \] 

Since $m$ has only two possible values, we can say that $\Omega(\R)$ consists 
of two copies of $\C$.

$K=\C$: In this case $U_{K}=T=\{e^{i\theta}\}$ and $\widehat{U_{K}}=\Z$ where 
the basic characters of $U_{K}$ are $\chi_m(t)=t^m$. But then 
\[\Omega(\C)=\{x\mapsto \chi_m\biggl(\frac{x}{|x|}\biggr)|x|^s; \hspace{2pt} m\in \Z 
\hspace{2pt} \mbox{and} \hspace{2pt} s\in \C\}\]

Thinking as in the first example we can say that $\Omega(\C)$ is countably many 
copies of $\C$.

$K=\Q_p$: In this case 
\[U_{K}=\Z_p^*=\lim_{\gets}\bigl(\Z/p^n\bigr)^*\]

Every continuous character of $\Z_p^*$ factors through $\bigl(\Z/p^n\bigr)^*$ 
for some $n\in \Z_{+}$. Therefore it is enough to understand continuous 
characters of $\bigl(\Z/p^n\bigr)^*$. But for this we have the following 
decomposition due to Gauss:
\[\bigl(\Z/p^n\bigr)^* \simeq \Z/(p-1) \times \Z/p^{n-1} \hspace{7pt} \mbox{if}
\hspace{2pt} p\neq 2\]
\[\bigl(\Z/2^n\bigr)^* \simeq \Z/2 \times \Z/2^{n-2} \hspace{7pt} \mbox{if} 
\hspace{2pt} p=2\]

Now one can use the knowledge on the characters of the objects on the right hand 
side to answer the original problem of finding $\Omega(\Q_p)$.


\end{exa}






\end{document}