(click to close)

Please wait

(click to close)

(click to close)

Author:

liyang

License:

Creative Commons CC BY 4.0
^{(?)}

Abstract:

The lecture notes are based on the number theory topics course on 3 Feb, 2016.

Tags:

` ````
\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}
```

Something went wrong...

Our gallery is the easiest way to put your LaTeX templates, examples and articles online. Just create a free project and use the publish menu. It only takes a couple of clicks!

The LaTeX templates, examples and articles in the Overleaf gallery all come from our amazing community.

New content is added all the time. Follow us on twitter for the highlights!