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\begin{document}

\title{Lecture 41}

\author{Meg Onoda}

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

\date{\today}

\email{onoda@math.lsu.edu}

\maketitle

\section{Lecture 41}
Last time, we stated Theorem 4.2 and started proving the theorem.
We are going to verify the case for $K=\R, K=\C, and K=
\mathfrak{p}-adic field.$ For each equation class $C$ of
quasi-characters, we choose one nice $f_C \in \mathfrak{S}(K)$. we
compute $\zeta$, explicitly. They will obviously have analytic
continuations.\\
\[
\frac{\zeta(f_C, C)}{\zeta(\widehat{f_C}, \hat{C})} = \frac{\zeta
(f,C)}{\zeta (\hat{f},C)}
\]
Let $K=\R.$ We have two kinds of quasi-characters.
\[
C_0 = \lbrace {x \longrightarrow |x|^s} \rbrace
\]
\[
C_1 = \lbrace {x \longrightarrow sgn(x) |x|^s} \rbrace
\]
We choose functions to work with as,
\[
f_0(x) = e^{-\pi x^2}
\]
\[
f_1(x) = \chi e^{-\pi x^2} \]
 Hence we have,
\[
\hat{f_0} (y) = e^{-\pi y^2} = f_0
\]
\[
 \hat{f_1} (y) = i\cdot f_1 (y)
 \]
 The $\zeta-fuction$.
 \[
 \begin{array}{lll}
 \zeta (f_0, ||^s) &=& \int_{\R ^x} f_0 (y) |y|^s d^x y\\
 &=& 2 \int_0^{+\infty} e^{-\pi y^2} \cdot y^s \frac{dy}{y}\\
 &=& \pi ^{-s/2} \Gamma { \mbox{}\hfill
 \displaystyle s \atopwithdelims() \displaystyle 2
 \hfill\mbox{}}
 \end{array}
 \]
% ${ \mbox{}\hfill
% \displaystyle s \atopwithdelims() \displaystyle 2
%\hfill\mbox{}}$
Let $k = \C$.\\ Equivalent class of quasi-characters is\\
\[
c_n = \lbrace z \longrightarrow \left( \frac{z}{|z|} \right)^n
\cdot |z|^s,   s\in \C ,    n\in \Z \rbrace
\]
The corresponding $\zeta -function$ is
\[
f_n(z) = \left\{ \begin{array}{ll}
 \bar{z}^n e^{-2\pi z \bar{z}} & \mbox { for $n\geq \Z$};\\
 z^n e^{-2\pi z \bar{z}}& \mbox{ for $n\leq \Z$}.\end{array}
 \right.
\]


We have\\
\[
\hat f_n = i^{|n|} f_{-n} for all n.
\]
The $\zeta -functions$. For $y=re^{i\theta}$, we have\\
\[
\begin{array}{ll}
\mbox f_n (y) = r^{|n|} e^{-in \theta} e^{-2 \pi r^2}&
|y|^s = r^{2s}  \\
\mbox c_n(y) = e^{in\theta} & dy = \frac{2rdrd\theta}{r^2} \\
\end{array}
\]


Therefore,\\
\[
\begin{array}{lll}
\zeta (f_n, c_n||^s) &=& \int f_n(y)c_n(y) |y| dy =
\int_0^{+\infty} \int_0^{2\pi} r^{2(s-1)+|n|} e^{-2\pi r^2}2r dr
d\theta\\
 &=& 2\pi \int_0^{\infty} (r^2)^{(s-1)+\frac{|n|}{2}}e^{-2\pi r^2}
 d(r^2) = (2\pi)^{(1-s) +\frac{|n|}{2}} \Gamma \left( 1-s +
 \frac{|n|}{2}\right)\\
\end{array}
\]
Let $K=p-adic field.$\\
 $\lambda (x) = \lambda ( Tr_{K/\Q_p} (x) ) \in \Q/\Z$\\
 $x=\tilde{x} \cdot \pi ^\nu$, where $\tilde{x} \in U_k, and \nu \in
 \Z$\\
 $|x| = \mathcal{N} \mathfrak{p}^{-\nu}$ \\
 $\int_{O_K} dx = (\mathcal{N} \vartheta)^{-1/2}$ so,
 $\int_{U_K} d^x x = (\mathcal{N} \vartheta)^{-1/2}$.\\

 %\begin{array}{lll}
 %$\vartheta ^{-1} &=& \lbrace x\in K:\Lambda (xy) \con 0 mod\Z, for
 %all y\in O_K \rbrace\\
  %&=& \lbrace x \in K: e^{2\pi i \Lambda (xy)} = 1, for all y \in
  %O_K \rbrace$.
 %\end{array}

If $G$ is a compact group, and $\chi: G \longrightarrow T$ is a
non-trivial character. Then $\int_G \chi (y) dy = 0.$ Hence, every
quasi-character of $K$ can be written in the form, $c_0:U_K
\longrightarrow T$ defined as $ y \longrightarrow c_0(y)\cdot
|y|^s$.\\ Note that as a group, $K^{\times} = U_K \otimes \langle
\pi \rangle \sim U_K \otimes \Z.$ We can view $c_0$ as a character
of K trivial on $\pi$, so $c_0 (\pi) = 1$. And each $c_0$ factors
through a finite quotient,\\
\[
\begin{array}{ccc}
U_K & \longrightarrow & U_K/{1 + \mathfrak{p}^n }\\
    & \searrow        & \downarrow \\
    &                 &  T\\
\end{array}
\]

The largest $n$ for which you have such factorization is called
conductor of $c_0 ( or c)$. Another way of saying this is $c_0(1 +
\mathfrak{p}^n) \cong 1$. Each equivalent class of
quasi-characters of $K$ is uniquely given by a character of $U_K$.
\end{document}