The lecture notes are based on the number theory topics course on 3 Feb, 2016.
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
\documentclass[a4paper]{article}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{tikz-cd}
\usepackage{mathrsfs}
\usepackage[colorinlistoftodos]{todonotes}
\usepackage{enumitem}
\usepackage{yfonts}
\usepackage{ dsfont }
\title{Modular forms of half integral weights, noncongruence subgroups, metaplectic groups}
\author{Yang Li}
\date{\today}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lem}[thm]{Lemma}
\newtheorem{defn}[thm]{Definition}
\newtheorem{eg}[thm]{Example}
\newtheorem{ex}[thm]{Exercise}
\newtheorem{conj}[thm]{Conjecture}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{claim}[thm]{Claim}
\newtheorem{rmk}[thm]{Remark}
\newcommand{\ie}{\emph{i.e.} }
\newcommand{\cf}{\emph{cf.} }
\newcommand{\into}{\hookrightarrow}
\newcommand{\dirac}{\slashed{\partial}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\LieT}{\mathfrak{t}}
\newcommand{\T}{\mathbb{T}}
\newcommand{\A}{\mathds{A}}
\begin{document}
\maketitle
\begin{abstract}
The lecture notes are based on the number theory topics course on 3 Feb, 2016.
\end{abstract}
\section{modular forms of half integral weights}
Let $\Gamma \subset SL2(Z)$ be a finite index subgroup. Let k be an integer. Recall a weight k, level $\Gamma$ modualr form is a holomorphic function on the upper half plane satisfying the funcitonal equation:
$f(\frac{a\tau+b}{c\tau+d})=(c\tau+d)^kf(\tau)$ for $\gamma \in \Gamma$
\begin{defn}
Half integral weight modular forms are holomorphic functions on the upper half plane with the modified functional equation:
$f(\gamma \tau)=\epsilon(\gamma)(c\tau+d)^(k/2)f(\tau)$ for $\gamma \in \Gamma$
where $\epsilon$ is some root of unity and the square root is chosen in some half plane.
\end{defn}
\begin{eg}
$\theta(\tau)=\sum{exp(2\pi i n^2 \tau)}$
$\Gamma(8)$=congruence subgroup mod 8,
then
$\theta(\gamma(\tau))=\begin{cases}
\theta(\tau) & c=0\\
(\frac{c}{d})(c\tau+d)^{1/2}\theta(\tau) &c>0
\end{cases}$
where $(\frac{c}{d})$ is the Legendre symbol.
\end{eg}
\begin{ex}
For all N, there exist $\gamma \in \Gamma(N)$, such that the Legende symbol $(\frac{c}{d})=-1$ for $\gamma= \left( \begin{array}{cc}
\ a & b \\
c & d \\
\end{array} \right) $
\end{ex}
For integral weight forms the transformation law is simple:
$j(\gamma, \tau)=(c\tau+d)^k$
then $j(\gamma_1\gamma_2, \tau)=j(\gamma_1, \gamma_2\tau)j(\gamma_2, \tau)$
so $j(\gamma, \tau)$ is a multiplier system.
But $(c\tau+d)^1/2$ is not a multiplier system.
\section{The metaplectic group}
\begin{defn}
$Mp_2(\R)=\{(g,\phi)\space | g\in SL2(\R), \phi: H \mapsto \C, \phi^2=c\tau+d\}$
\end{defn}
We see $Mp_2(\R)$ has a natural covering map to $SL2(\R)$. $Mp_2(\R)$ is a Lie group but not the real points of an algebraic group; in particular it cannot be realised by a matrix representation.
The group law is given by:
$(g,\phi)*(g\prime, \phi\prime)=(gg\prime, \tau \mapsto \phi(g\prime \tau)\phi\prime(\tau))$
Recall the $\theta$ function satisfies some functional equation. This means the factor of automorphy forms a multiplier system. This fact is equivalent to:
The covering map $Mp_2(\R)\mapsto SL_2(\R)$ splits on $\Gamma(8)$ with the splitting given by $(\frac{c}{d})(c\tau+d)^{1/2}$
\begin{rmk}
The way to prove this is indeed a multiplier system: either use the fact that the theta function is nonzero, or use quadratic reciprocity.
\end{rmk}
\section{Congruence subgroup problem for $SL_n$}
Question: if $\Gamma \subset SL(O_K)$ has finite index, where K is a number field, is $\Gamma$ a congruence subgroup?
Here the congruence subgroup means the coefficients of the matrix equals the identity matrix mod the ideal (n).
\begin{eg}
For $SL_2(Z)$, the answer is no.
Take $\Gamma \subset SL_2(\Z)$ small enough so that $\Gamma$ is not torsion free. Then $\Gamma$ is a free group, so there is a surjection $\Gamma \mapsto \Z$.
Let $\hat{\Gamma}=\varprojlim{\Gamma/\Upsilon}$
$\Upsilon$ has finite index in $\Gamma$.
Let $ \bar{\Gamma}=\varprojlim{\Gamma/\Gamma(n)}$.
The hom from $\Gamma$ to $\Z$ extends to $\hat{\Gamma}\mapsto \hat{\Z}$.
$\bar{\Gamma}$ is the closure of $\Gamma$ in $SL_2(\A_f)$.
Since $SL_2$ is semisimple, the commutator map is surjective, $[sl_2, sl_2]\mapsto sl_2$.
So $[\bar{\Gamma},\bar{\Gamma}]$ is open in $SL_2(\A_f)$, since $\bar{\Gamma}$ is open in $SL_2(\A_f)$. So $[\bar{\Gamma},\bar{\Gamma}]$ has finite index in $\bar{\Gamma}$.
Hence there is no hom $\bar{\Gamma}\mapsto \hat{\Z}$ apart from 0.
There is $1\mapsto C \mapsto \hat{\Gamma}\mapsto \bar{\Gamma}\mapsto 1$.
C is called the congruence kernel.
\end{eg}
\begin{thm}
The theorem of Bass-Milnor-Serre says that if n is greater or equal to 3, and the number field K has a real place, then every subgroup of finite index in $SL_n(O_K)$ is a congruence subgroup.
\end{thm}
If K is totally complex there will be a noncongruence subgroup.
Let K be totally complex, and contains an n-th root of unity. We can define the n-th power Legendre symbol on K, as follows:
Let $a\in K$, p=prime ideal in $O_K$, p does not divide na, then
$a^{\frac{Np-1}{n}}$=some n-th root of unity mod p.
Define the Legendre symbol $(\frac{a}{p})$ to be the n-th root of 1.
For a general ideal coprime to na, define the Legendre symbol by the product law.
Define $\Gamma(n^2)$ to be the congruence subgroup in $SL_2(O_K)$ mod the ideal $(n^2)$.
Define a map $\kappa: \Gamma(n^2) \mapsto \mu_n$
\\
$
\left( \begin{array}{cc}
\ a & b \\
c & d \\
\end{array} \right) \mapsto \begin{cases}
(\frac{c}{d}) & c \neq 0 \\
1 & c=0 \\
\end{cases}
$
\begin{thm}
Kubata: $\kappa$ is a hom, and its kernel is a noncongruence subgroup.
\end{thm}
\begin{ex}
Prove this.
\end{ex}
Bass-Milnor-Serre extended the $ \kappa$ to $SL_m(O_K, n^2)$.
$\kappa$ gives an isomorphism between the congruence kernel and $\mu_n$ as long as n is the total number of roots of unity in K.
This means every subgroup of finite index in $SL_m(O_K, n^2)$ contains some $\Gamma(N)\cap ker(\kappa)$. (If either m is at least 3 or [K:$\Q$]) is at least 4).
\begin{rmk}
Kubata's exercise is equivalent to the reciprocity formula for the Legendre symbol in K, ie the Artin reciprocity law for Kummer extensions of K.
\end{rmk}
\section{Digression on K theory}
Before going on, define the K2 group of a field. Let K be any infinite field. The group $SL_m(K)$ is perfect for m at least 3, meaning it is equal to its own commutator subgroup.
Hence $SL_m(K)$ has a universal central extension.
$ 1 \mapsto K2(K) \mapsto St_m(K) \mapsto SL_m(K) \mapsto 1$
Here K2(K) is defined to be the kernel. It does not depend on m as long as m is at least 3.
We recall what it means to be a universal central extension:
for any Abelian group A, the central extensions of the form
$ 1 \mapsto A \mapsto ? \mapsto SL_m(K) \mapsto 1$
are in bijective correspondence with the hom set
$Hom(K2(K),A)$
where the correspondence is given by the obvious morphism of extension sequences.
For a field K, the group K2(K) is calculated by Matsumoto as follows (giving a presentaion of K2(K)):
$K2(K)=K^*\otimes_\Z K^* /<a\otimes 1-a, a\in K\setminus \{0,1\} > $
We will write $\{a,b\}$ for the image of the tensor $a \otimes b$ in K2(K).
\begin{rmk}
In terms of matrices this means:
$[ \widetilde{ diag(a, a^{-1}, 1, \dots, 1)}, \widetilde{diag(b, b^{-1}, 1, \dots, 1)} ] \in K_2(K)$
Notice we need at least 3*3 matrices for this to make sense. The \textasciitilde means taking the preimage in $St_m(K)$.
\end{rmk}
We also get an extension sequence for $SL_2$:
$ 1 \mapsto K2(K) \mapsto something \mapsto SL_2(K) \mapsto 1$
by taking the middle term to be the preimage of $SL_2(K)$ in $St_3(K)$.
This extension is easy to describe: here is a inhomogeneous 2-cocycle.
$ \sigma(g,h)=\{X(gh)/X(g), X(gh)/X(h) \}, g,h\in SL_2(K)$
$ X(\left( \begin{array}{cc}
\ a & b \\
c & d \\
\end{array} \right) )=\begin{cases}
c & c \neq 0 \\
d & c=0 \\
\end{cases} $
This satisfies the cocycle relation.
$ \sigma(g_1 g_2, g_3)\sigma(g_1,g_2)=\sigma(g_1, g_2 g_3)\sigma(g2,g3)$
\begin{rmk}
The cocycle condition is equivalent to the associativity of the group law on $SL_2(K) \times K_2(K)$.
\end{rmk}
\begin{ex}
Show $\sigma$ is a 2-cocycle.
(Need properties of $\{a,b\}) $:
the bilinearity of the tensor and the relation
$\{x,1-x\})=1 $ for $x \neq 1$.
\end{ex}
\section{Hilbert symbol, metaplectic group again}
Let $\Q_p$=either a p-adic field or the real numbers. Define for $a,b\in \Q_p$
$(a,b)_p=\begin{cases}
1 & ax^2+by^2=1 \textrm{has a solution in }\Q_p \\
-1 & \textrm{if not} \\
\end{cases}$
For the real number case,
$(a,b)=\begin{cases}
1 & a>0 \textrm{ or } b>0 \\
-1 & a,b<0\\
\end{cases}$
The (a,b) is called the Hilbert symbol and it satisfies the bilinear relations and the property that $(x,1-x)=1 $ for $x \neq 1$.
In other words the Hilbert symbol is a hom
$K_2(\Q_p)\mapsto \{1,-1\}$. In fact it is the only nontrivial such.
For the real number case we get a central extension of $SL_2(\R)$ which reproduces our $Mp_2(\R)$. This is a unique connected double cover.
Note: if G=Lie group, then G is homotopic to the maximal compact subgroup. In the case of $SL_2(\R)$, the maximal compact subgroup is the circle, so the first fundamental group is $\Z$, hence there is a unique connected double cover.
The quadratic reciprocity can be stated as:
$a, b \in \Q^*, \prod_{ \textrm{p prime or infinity} }{(a,b)_p}=1 $
For each prime we have a central extension
$ 1 \mapsto \mu_2 \mapsto \widetilde{SL_2(\Q_p)} \mapsto SL_2(\Q_p) \mapsto 1$
defined by the relavent two-cycle $\sigma_p$.
We can put these together to obtain an adelic version:
$ 1 \mapsto \mu_2 \mapsto \widetilde{SL_2(\A)} \mapsto SL_2(\A) \mapsto 1$
where $\sigma_{\A}=\prod{\sigma_p^\prime}$,
and $\sigma_p^\prime $ is cohomologous to $\sigma_p$.
By the Hilbert symbol version of the reciprocity law, the cocycle $\sigma_{\A}$ splits on $SL_2(\Q)$.
It turns out if p is odd, then $\sigma_p$ splits on $SL_2(\Z_p)$ and $\sigma_2$ splits on $SL_2(\Z_2, 4)$. $\sigma_{\A}$ will split on $U=\prod_{p odd}{SL_2(\Z_p)\times SL_2(\Z_2,4)}$.
Now on $\Gamma(4)$ we have two different splittings of almost the same extension (the difference between the two extensions is $\sigma_{\infty})$.
If we divide one splitting by another, we get a map $\kappa: \Gamma(4)\mapsto \mu_2$.
If these were two different splittings of the same cocycle, $\kappa$ would be a hom. But if they are not, then $\kappa$ is a splitting of $\sigma_{\infty})$, ie, $
\sigma_{\infty}(g,h)=\kappa(g)\kappa(h)/\kappa(gh)$.
\begin{rmk}
This is how we show $\kappa(\gamma)(c\tau+d)^{1/2}$ is a multiplier system. And when we work out what $\kappa$ is, we get
$\kappa(\left( \begin{array}{cc}
\ a & b \\
c & d \\
\end{array} \right))=(\frac{c}{d})$
\end{rmk}
\begin{eg}
If K is totally complex, then
$SL_2(K_{\infty})=SL_2(\C)^N,
K_\infty=K\otimes_\Q \R$
$SL_2(\C)$ is simply connected, ie, it has no nontrivial covering groups. Complex Hilbert symbols are 1.
So the extension
$ 1 \mapsto \mu_n \mapsto \widetilde{SL_2(\A)} \mapsto SL_2(\A) \mapsto 1$
splits on $SL_2(K)$ by reciprocity law, and also splits on $U\times SL_2(K_\infty)$.
$\Gamma(n^2)=SL_2(K)\cap (U\times SL_2(K_\infty))$.
On $\Gamma(n^2)$ we have two splittings of the same extension.
Dividing one extension by another, we get a hom $\kappa: \Gamma(n^2)\mapsto \mu_n$.
This is exactly the same $\kappa$ we had before. $ker(\kappa)$ is a noncongruence subgroup.
\end{eg}
\begin{rmk}
metaplectic forms are automorphic forms on $\hat{G(\A)}$ for any reductive G over a number field.
\end{rmk}
\end{document}