Commit 04dcb144 authored by Han-Miru Kim's avatar Han-Miru Kim
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Reduction theory, finished program

parent 677e760c
#+TITLE: Dirichlet L-functions - README
#+AUTHOR: Han-Miru Kim
#+PROPERTY: header-args :tangle ./header.tex
* About
This file contains general information about the lecture notes aswell as the configuration for my Seminar notes.
** Structure
This file is an Emacs Org Mode file which is edited and then /tangled/ with the ~header.tex~ file, which I do not touch manually.
The benefit of this is that comments, which would have been written in plaintext inside ~header.tex~ can now be written using Org Mode Formatting, allowing for literate configuration.
For example, I can use headings, use formatted text, insert tables, source code blocks or even images, which would not be possible in LaTeX comments.
I structure my notes as follows:
Instead of a ~main.tex~ file, I have a ~main.org~ file which I then export to latex via ~M-x org-export-dispatch l o~.
Since I have just started using Emacs, my configuration does not quite allow me to take notes in real-time.
This means I still take my notes in Vim by writing ~.tex~ files and putting them in the ~./files~ directory, from where they will be sourced and can be compiled using my existing vim-tex setup.
** How to use
After changing the configuration, place the point at the top of this document (~g g~ in evil-normal mode) and do ~C-c C-c~ to refresh. Then execute ~M-x org-babel-tangle~, which will update ~header.tex~.
It should output ~Tangled n code blocks from README.org~
To use source code blocks without having them exported to ~header.tex~, use the ~:tangle no~ option.
** TODO
- I plan on trying out a language-agnostic solution where I can export this document to ~.html~ format and upload it on my webpage.
- As of now, I use different configuration files for every lecture. This means that there is alot of code redundancy as the configuration only differs by about 5% (if at all) between different lecture notes.
But since the README.org files are only a couple kilobytes in size each, I should be fine.
* Package Imports
** Font Settings
#+BEGIN_SRC latex
\usepackage[utf8]{inputenc}
\usepackage{lmodern} % Latin Modern
\usepackage{fontenc}[t1]
\usepackage{microtype}
#+END_SRC
** Content
#+BEGIN_SRC latex
\usepackage{caption}
\usepackage{enumerate}
\usepackage{hyperref}
%\usepackage{wrapfig}
\usepackage{array}
\newcolumntype{L}{>{$}l<{$}} % math mode l
\usepackage[framemethod=TikZ]{mdframed}
\usepackage[customcolors,norndcorners]{hf-tikz}
#+end_src
** Page Settings
#+BEGIN_SRC latex
\usepackage{geometry}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}
\hoffset = -0.45 in
\voffset = -0.3 in
\textwidth = 500pt
\textheight = 650pt
\setlength{\headheight}{25.6pt}
\setlength{\headwidth}{500pt}
\setlength{\parindent}{0pt}
\marginparwidth = 0pt
\fancyhead[L]{\rightmark}
\fancyhead[R]{\today}
\fancyfoot[L]{Han-Miru Kim}
\fancyfoot[C]{\href{mailto:kimha@ethz.ch}{kimha@ethz.ch}}
\fancyfoot[R]{\thepage}
#+END_SRC
** Math
The ~amsmath~ and ~amssymb~ form a basic collection of mathematic packages.
Insert more explanations here.
#+BEGIN_SRC latex
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{dsfont}
\usepackage{amsfonts}
\usepackage{mathdots}
\usepackage{mathrsfs}
\usepackage{bm} % bold math
\usepackage{empheq} % colored equations
\usepackage[most]{tcolorbox}
\usepackage{mathtools}
\usepackage{faktor} % Quotients
\usepackage{xcolor}
\usepackage{stmaryrd}
\let\tempphi\phi
\let\phi\varphi % curly
\let\varphi\phi % ugly
#+END_SRC
The last three lines swap the commands ~\phi~ and ~\varphi~.
I do this becuase curly one (now ~\phi~) looks much than the default.
** Graphics
#+BEGIN_SRC latex
\usepackage{graphicx}
\usepackage{tikz}
\usetikzlibrary{cd}
#+END_SRC
** Formatting
Packages for sectioning and positioning.
#+BEGIN_SRC latex
\usepackage{sectsty}
\sectionfont{\LARGE}
\subsectionfont{\Large}
\subsubsectionfont{\large}
\paragraphfont{\large}
%\usepackage{adjustbox} %\adjustbox{scale=2,center}{}
%\usepackage{paracol}
\usepackage{float}
#+END_SRC
* Macros
** Symbols
~\DeclareMathOperator~ defines commands that print out with the standard text-font over the italic math-fonts used for variables etc.
#+BEGIN_SRC latex
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\IS}{\mathbb{S}}
\newcommand{\ID}{\mathbb{D}}
\newcommand{\IT}{\mathbb{T}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\K}{\mathbb{K}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\IP}{\mathbb{P}}
\DeclareMathOperator{\charac}{char}
\DeclareMathOperator{\degree}{deg}
\newcommand{\del}{\partial}
\newcommand{\es}{\text{\o}}
#+END_SRC
** Delimiters
#+BEGIN_SRC latex
\DeclarePairedDelimiter\abs{\vert}{\vert}
\DeclarePairedDelimiter\ceil{\lceil}{\rceil}
\DeclarePairedDelimiter\floor{\lfloor}{\rfloor}
\DeclarePairedDelimiter\Norm{\vert\vert}{\vert\vert}
\DeclarePairedDelimiter\scal{\langle}{\rangle}
\DeclarePairedDelimiter\dbrack{\llbracket}{\rrbracket}
#+END_SRC
** Math Operators
#+BEGIN_SRC latex
% Algebra
\newcommand{\iso}{\cong}
\newcommand{\mono}{\rightarrowtail}
\newcommand{\epi}{\twoheadrightarrow}
\DeclareMathOperator{\Bil}{Bil}
\DeclareMathOperator{\ev}{ev}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Ker}{Ker}
\DeclareMathOperator{\Image}{Im}
\DeclareMathOperator{\spn}{span}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\rang}{rang}
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\End}{End}
\DeclareMathOperator{\Eig}{Eig}
\DeclareMathOperator{\Hau}{Hau}
\DeclareMathOperator{\trace}{tr}
\DeclareMathOperator{\rot}{rot}
\DeclareMathOperator{\ad}{ad}
\DeclareMathOperator{\Mat}{Mat}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\lcd}{lcd}
\DeclareMathOperator{\ggT}{ggT}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\sgn}{sgn}
\DeclareMathOperator{\Alg}{-Alg}
\renewcommand{\mod}{\mathrm{mod }}
% Cat
\newcommand*{\Top}{\text{\sffamily Top}}
\newcommand*{\Htpy}{\text{\sffamily Htpy}}
\newcommand*{\Grp}{\text{\sffamily Grp}}
\renewcommand*{\Vec}{\text{\sffamily Vec}}
\newcommand*{\Set}{\text{\sffamily Set}}
\newcommand*{\Ring}{\text{\sffamily Ring}}
\DeclareMathOperator{\Nat}{Nat}
%\newcommand{\coprod}{\amalg}
% Analysis
\DeclareMathOperator{\grad}{grad}
\DeclareMathOperator{\graph}{graph}
\DeclareMathOperator{\supp}{supp}
% Formatting
\newsavebox\MBox
\newcommand\cunderline[2][red]{ % Colored underline
{\sbox\MBox{$#2$}%
\rlap{\usebox\MBox}\color{#1}\rule[-1.2\dp\MBox]{\wd\MBox}{0.5pt}}
}
#+END_SRC
* Environments
** Default Environemnts
Changing the behaviour of the ~enumerate~ environment to use letters over roman numerals.
#+BEGIN_SRC latex
\renewcommand{\labelenumi}{(\alph{enumi})}
#+END_SRC
** Theorem
Define the environments using the ~amsthm~ package.
They are easy to set up and respect the section numbering, but I prefer the fancy coloured boxes using ~mdframed~.
#+BEGIN_SRC latex
%\newtheorem{thm}{Theorem}[section]
%\newtheorem{cor}{Corollary}[thm]
%\newtheorem{lem}[thm]{Lemma}
%\newtheorem{prop}{Proposition}[thm]
%\theoremstyle{definition}
%\newtheorem{dfn}[thm]{Definition}
%\newtheorem{rem}[thm]{Remark}
%\newtheorem{note}[thm]{Note}
%\newtheorem{ex}[thm]{Example}
#+END_SRC
** Math boxes
From https://tex.stackexchange.com/questions/20575/attractive-boxed-equations:
Requires the the ~empheq~ and ~tcolorbox~ packages.
Makes the cool blue equation boxes.
#+BEGIN_SRC latex
\newtcbox{\bluebase}[1][]{%
nobeforeafter, math upper, tcbox raise base,
enhanced, colframe=black,
colback=blue!20, boxrule=0.5pt,
sharp corners,
#1
}
#+END_SRC
You can then write the equations as follows
#+BEGIN_SRC latex :tangle no
\begin{empheq}[box=\bluebase]{align*}
\nabla \cdot E = \frac{\rho}{\epsilon_0}
\end{empheq}
#+END_SRC
** Fancy Boxes
*** Setup
Taken from this [texblog post](https://texblog.org/2015/09/30/fancy-boxes-for-theorem-lemma-and-proof-with-mdframed/)
Setup numbering system and define a function that creates environment.
~\newenvironment~ takes one optional and three necessary arguments.
- The environment name (that which is put inside ~\begin{###}~)
- The Title name which is printed on the page
- The colour of the box. I use colors in the format ~blue!50!green!50~ etc.
#+BEGIN_SRC latex
\newcounter{theo}[section]
\renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}
\newcommand{\defboxenv}[3]{
\newenvironment{#1}[1][]{
\refstepcounter{theo}
\ifstrempty{##1} { % no Title
\mdfsetup{
frametitle={
\tikz[baseline=(current bounding box.east), outer sep=0pt]
\node[anchor=east,rectangle,fill=#3]
{\strut #2~\thetheo};
}
}
}{ % else: with Title
\mdfsetup{
frametitle={
\tikz[baseline=(current bounding box.east),outer sep=0pt]
\node[anchor=east,rectangle,fill=#3]
{\strut #2~\thetheo:~##1};
}
}
} % either case
\mdfsetup{
linecolor=#3,
innertopmargin=0pt,
linewidth=2pt,
topline=true,
frametitleaboveskip=\dimexpr-\ht\strutbox\relax,
}
\begin{mdframed}[]\relax
}{ % post env command
\end{mdframed}
}
}
#+END_SRC
*** Environments
#+BEGIN_SRC latex
\defboxenv{dfn}{Definition}{orange!50}
\defboxenv{edfn}{(Definition)}{orange!20}
\defboxenv{thm}{Theorem}{blue!40}
\defboxenv{lem}{Lemma}{blue!20}
\defboxenv{prop}{Proposition}{orange!30}
\defboxenv{cor}{Corollary}{blue!30}
\defboxenv{xmp}{Example}{green!50!blue!30!white}
\defboxenv{exr}{Exercise}{black!30}
\defboxenv{rem}{Remark}{red!20}
#+END_SRC
\section{Dirichlet Series}
\begin{dfn}[Dirichlet Seires]
\begin{align*}
\sum_{n=1}^{\infty}a_n e^{- \lambda_n s}
\end{align*}
for $\lambda_1 < \lambda2 < \ldots$ and $\lambda_n \to \infty$.
\end{dfn}
For power series, there is a convergence radius, $\abs{z} < R$ converges.
\begin{thm}[]
For dirichlet seires, the series convergences for all $\text{Re}(s) > \sigma_0$ for some $\sigma_0 \in R$.
\textbf{Abscissa of convergence}.
\end{thm}
\textbf{Notation:} Given a dirichlet series $\sum_{n=1}^{\infty}a_n e^{-\lambda_n s}$, we define
\begin{align*}
A(N)
&:=
\sum_{n=1}^{N}a_n \in \C\\
A(M,N)
&:=
\sum_{n=M}^{N}a_n \in \C
\\
S(M,N)
&:=
\sum_{n=M}^{N}a_nn^{-s}
: \C|_{\sigma_0} \to \C
\end{align*}
with the convention that $A(M,N) = 0$ for $M > N$.
We can determine the abscissa of convergence by the coefficients.
\begin{thm}[]
If $\sum_{n=1}^{\infty}a_n$ diverges, then the abscissa of convergence is
\begin{empheq}[box=\bluebase]{align*}
\sigma_0 = \limsup_{N \to \infty} \frac{\log \abs{A(N)}}{\lambda_N}
\end{empheq}
Then $\sigma_1 \leq \sigma_0 + 1$.
\end{thm}
\begin{thm}[]
Let $\sum a_n n^{-s}$ be a Dirichlet series with abscisse of convergence $\sigma_0$ and let $\sigma_1 (\geq \sigma_0)$ be the abscisse of convergence of $\sum \abs{a_n}n^{-s}$.
\end{thm}
For power series, we know that the radius of convergence is determined by the singularity with smallest absolute value.
For ordinary dirichlet series, we can show that there is a singularity at the boundary.
\begin{thm}[Landau]
Let $\sum_{n=1}^{\infty}a_N n^{-s}$ be an ordinary Dirichlet series with non-negtive real coefficients and abscissa $\sigma_0$.
Then
\begin{align*}
f(s) = \sum_{n=1}^{\infty}a_n n^{-s} qtq
\end{align*}
has a singularity at $s = \sigma_0$.
\end{thm}
\section{Gamma Function}
$s = \sigma + \tau i$.
\begin{align*}
\Gamma(s) = \int_{0}^{\infty}t^{s-1}e^{-t}dt, \quad \text{for} \quad \sigma > 0
\end{align*}
Is holomorphic with poles at $0,-1,-2,\ldots$.
$\frac{1}{\Gamma(s)}$ is holomorphic everywhere with zeros at $0,-1,-2,\ldots$.
\begin{align*}
\zeta(s) = \frac{1}{\Gamma(s)} \int_0^{\infty}\frac{t^{s-1}}{e^{t}-1} dt \quad \text{for} \quad \sigma > 0
\end{align*}
\section{Rieman Zeta Function}
\begin{align*}
\psi(s) = \left(
1 - \frac{1}{2^{s}} + \frac{1}{3^{s}} + \ldots
\right)
\end{align*}
\begin{align*}
\psi(s) = (1 - 2^{1-s})\zeta(s)
\end{align*}
So the only singularities are the zeros of $(1 - 2^{1-s})$.
\begin{align*}
\sigma(1 - 2n) = - \frac{B_{2n}}{2n}
\end{align*}
$\zeta(2\N) = 0$.
\section{Dirichlet Characters}
\begin{dfn}[]
A \textbf{fundamental discrimant} (Grundzahl) is an integer $D$ with
\begin{align*}
D = 1 (\mod 4), D \text{ not a square} \quad \text{or} \quad D = 4m, m \text{ not a square }, m = 2 \text{ or }3 (\mod 4)
\end{align*}
\end{dfn}
\begin{dfn}[]
Let $\chi$ be a dirichlet charater $(\mod N)$.
The \textbf{Dirichlet $L$-series} associated to this character is the series
\begin{align*}
L(s,\chi) = \sum_{n=1}^{\infty} \frac{\chi(n)}{n^{s}}
\end{align*}
Since $\abs{\chi(n)} \leq 1$, this series converges absolutely for $\sigma > 1$.
\end{dfn}
\section{Reduction theory}
In previous Chapters, we found the number of equivalence classes of forms $h(D)$ aswell as the number of inequivalent forms representing $n$.
Goals
\begin{enumerate}
\item Fix $D$. We want to find a finite set of forms with discrimanant $D$ such that every equivalence class has at least one representant.
\item Find an algorithm to determine if two forms are equivalent.
\item Givn Find a finite set of representations of anumber with a given form,
\item if two representations of a number are equivalent.
\end{enumerate}
In chapter 8, we saw that given any form $ax^{2} + bxy + cy^{2}$, we can use the transformation
\begin{align*}
S_n = \begin{pmatrix}
n & 1\\
-1 & 0
\end{pmatrix}
:
ax^{2} + bxy + cy^{2}
\mapsto
(an^{2} - bn + c)x^{2} + (2an - b)xy + ay^{2}
\end{align*}
to obtain a Form $ax^{2} + bxy + cy^{2}$ with
\begin{align*}
- \abs{a} < b \leq \abs{a} \leq \abs{c}
\end{align*}
\begin{dfn}[]
A positive definite form $ax^{2} + bxy + cy^{2}$ is said to be \textbf{reduced}, if
\begin{align*}
-a < b \leq a < c \quad \text{or} \quad 0 \leq b \leq a = c
\end{align*}
\end{dfn}
Then, every positive definite form is equivalent to a reduced form.
Now, we claim that reduced positive definite forms are pairwise inequivalent.
This answers question (b):
Two forms are equivalent, if and only if they lead to the same reduced form.
First note that for a reduced form $f$ with $(x,y) \neq (0,0)$ one has
\begin{align*}
f(x,y) = ax^{2} + bxy + cy^{2} \geq a(x^{2} - \abs{xy} + y^{2}) \geq a
\end{align*}
\begin{dfn}[]
Let $f = ax^{2} + bxy + cy^{2}$.
We define an action on the set of forms
\begin{align*}
TF = S_n f \quad \text{for} \quad n \in \Z \text{ with }n > \overline{f} > n-1
\end{align*}
where $\overline{f} = \frac{b + \sqrt{D}}{2a}$
\begin{align*}
S_n = \begin{pmatrix}
n & 1\\
-1 & 0
\end{pmatrix}
:
ax^{2} + bxy + cy^{2}
\mapsto
(an^{2} - bn + c)x^{2} + (2an - b)xy + ay^{2}
\end{align*}
\end{dfn}
\textbf{Notation:} We notate a form $f(x,y) = ax^{2} + bxy + cy^{2}$ as a triple $[a,b,c]$
Warning: the notation $\overline{f}$ is not used in the book.
\begin{dfn}[]
An indefinite form $ax^{2} + bxy + cy^{2}$ is \textbf{reduced}, if
\begin{align*}
a > 0, c > 0, b > a + c
\end{align*}
\end{dfn}
\begin{thm}[]
Let $D > 0$, not a square.
Then there are only finitely many reduced forms with discriminant $D$.
Every form with discriminant $D$ can be reduced with finitely many transformations.
The transformation of a reduced form is reduced and so forms cycles in the set of reduced forms.
Every equivalence of reduced forms is obtained by iteration of $T$.
In particular, two reduced forms are equivalent if and only if they belong to the same cycle.
\end{thm}
\begin{xmp}[]
Use \texttt{bqf.hs} to show cycle of $f = x^{2} - 6 xy + 3y^{2}$.
\end{xmp}
\section{Binary quadratic forms}
\begin{dfn}[]
A \textbf{binary quadratic form} is an expression of the form
\begin{empheq}[box=\bluebase]{align*}
f(x,y) = ax^{2} + bxy + cy^{2}
\end{empheq}
where $a,b,c$ are fixed \emph{coefficients} and $x,y$ are variables.
\end{dfn}
We will assume that $a,b,c \in \Z$ and that the number of variables $x,y$ is always two, so we will drop the word ``binary''.
We will also omit the word ``quadratic'' aswell and speak of ``forms'', meaning binary quadratic forms.
The main question is whether the equation
\begin{align}\label{eq:bqf}
f(x,y) = ax^{2} + bxy + cy^{2} = n \quad \text{for} \quad x,y \in \Z
\end{align}
has solutions or not.
If we plot the graph of $z = f(x,y)$, we would expect that transformations such as flipping $x,y$ or doing $90$-degree rotations should preserve the structure of the set of solutions to \ref{eq:bqf}.
Let $\begin{pmatrix}
\alpha & \beta\\
\gamma & \delta
\end{pmatrix}$ with integer coefficients and determinant $1$, i.e. $\begin{pmatrix}
\alpha & \beta\\
\gamma & \delta
\end{pmatrix}
\in \text{SL}(2,\Z)$.
With the substitution
\begin{align*}
\begin{pmatrix}
x\\
y
\end{pmatrix}
\mapsto \begin{pmatrix}
x'\\
y'
\end{pmatrix}
=
\begin{pmatrix}
\alpha & \beta\\
\gamma & \delta
\end{pmatrix}
\begin{pmatrix}
x\\
y
\end{pmatrix}
=
\begin{pmatrix}
\alpha x + \beta y\\
\gamma x + \delta y
\end{pmatrix}
\end{align*}
equation \ref{eq:bqf} turns into
\begin{align*}
f(x',y')
&=
a' x^{2} + b'' xy + c'y^{2}
\\
&= a(\alpha x + \beta y)^{2} + b(\alpha x + \beta y)(\gamma x + \delta y) + c(\gamma x + \delta y)^{2}
\\
&= a'x^{2} + b'xy + c'y^2
\end{align*}
where $a',b',c'$ is given by
\begin{align*}
a' &= a \alpha^{2} + b \alpha \gamma + c \gamma^{2}\\
b' &= 2 a \alpha \beta + b(\alpha \delta + \beta \gamma) + 2 c \gamma \delta\\
c' &= a \beta^{2} + b \beta \delta + c \delta^{2}
\end{align*}
\begin{dfn}[]
Two quadratic forms $f(x,y) =ax^{2} + bxy + cy^{2}$ and $f'(x,y) = a'x^{2} + b'xy + c'y^{2}$ are \textbf{equivalent}, if there exist $\begin{pmatrix}
\alpha & \beta\\
\gamma & \delta
\end{pmatrix}
\in \text{SL}(\Z)$ such that
\begin{align*}
a' &= a \alpha^{2} + b \alpha \gamma + c \gamma^{2}\\
b' &= 2 a \alpha \beta + b(\alpha \delta + \beta \gamma) + 2 c \gamma \delta\\
c' &= a \beta^{2} + b \beta \delta + c \delta^{2}
\end{align*}
\end{dfn}
One quickly verifies that this defines an equivalence relation on all binary quadratic forms, as $\text{SL}_2(\Z)$ forms a group, so matrix multiplication and inverses exist.
\begin{prop}[]
Given a form $ax^{2} + bxy + cy^{2}$, the \textbf{discriminant}
\begin{empheq}[box=\bluebase]{align*}
D = b^{2} - 4ac
\end{empheq}