%% LyX 2.0.6 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[british]{article}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage{geometry}
\geometry{verbose,tmargin=3cm,bmargin=3cm,lmargin=2cm,rmargin=2cm}
\usepackage{babel}
\usepackage{rotating}
\usepackage{textcomp}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[all]{xy}
\PassOptionsToPackage{normalem}{ulem}
\usepackage{ulem}
\usepackage[unicode=true]
 {hyperref}
\usepackage{breakurl}

\makeatletter

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands.
%% Because html converters don't know tabularnewline
\providecommand{\tabularnewline}{\\}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
\usepackage{enumitem}		% customizable list environments
\newlength{\lyxlabelwidth}      % auxiliary length 
  \theoremstyle{plain}
  \newtheorem*{thm*}{\protect\theoremname}
\theoremstyle{plain}
\newtheorem{thm}{\protect\theoremname}[section]
  \theoremstyle{definition}
  \newtheorem{defn}[thm]{\protect\definitionname}
  \theoremstyle{remark}
  \newtheorem*{notation*}{\protect\notationname}
  \theoremstyle{definition}
  \newtheorem*{example*}{\protect\examplename}
 \newcommand\thmsname{\protect\theoremname}
 \newcommand\nm@thmtype{theorem}
 \theoremstyle{plain}
 \newtheorem*{namedtheorem}{\thmsname}
 \newenvironment{namedthm}[1][Undefined Theorem Name]{
   \ifx{#1}{Undefined Theorem Name}\renewcommand\nm@thmtype{theorem*}
   \else\renewcommand\thmsname{#1}\renewcommand\nm@thmtype{namedtheorem}
   \fi
   \begin{\nm@thmtype}}
   {\end{\nm@thmtype}}
  \theoremstyle{plain}
  \newtheorem{lem}[thm]{\protect\lemmaname}
  \theoremstyle{plain}
  \newtheorem{cor}[thm]{\protect\corollaryname}
  \theoremstyle{remark}
  \newtheorem*{note*}{\protect\notename}
  \theoremstyle{remark}
  \newtheorem*{rem*}{\protect\remarkname}
  \theoremstyle{remark}
  \newtheorem*{claim*}{\protect\claimname}
% labeling-like list based on enumitem's description list with
% mandatory second argument (label-pattern):
\newenvironment{elabeling}[2][]%
{\settowidth{\lyxlabelwidth}{#2}
\begin{description}[font=\normalfont,style=sameline,
leftmargin=\lyxlabelwidth,#1]}
{\end{description}}
  \theoremstyle{definition}
  \newtheorem*{xca*}{\protect\exercisename}
 \theoremstyle{definition}
 \newtheorem*{defn*}{\protect\definitionname}
  \theoremstyle{plain}
  \newtheorem*{fact*}{\protect\factname}
  \theoremstyle{plain}
  \newtheorem{prop}[thm]{\protect\propositionname}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\newcommand{\xyR}[1]{
 \xydef@\xymatrixrowsep@{#1}}
\newcommand{\xyC}[1]{
  \xydef@\xymatrixcolsep@{#1}}
\date{}
\setcounter{section}{-1}

\makeatother

  \providecommand{\claimname}{Claim}
  \providecommand{\corollaryname}{Corollary}
  \providecommand{\definitionname}{Definition}
  \providecommand{\examplename}{Example}
  \providecommand{\exercisename}{Exercise}
  \providecommand{\factname}{Fact}
  \providecommand{\lemmaname}{Lemma}
  \providecommand{\notationname}{Notation}
  \providecommand{\notename}{Note}
  \providecommand{\propositionname}{Proposition}
  \providecommand{\remarkname}{Remark}
  \providecommand{\theoremname}{Theorem}
\providecommand{\theoremname}{Theorem}

\begin{document}
\global\long\def\zz{\mathbb{Z}}
\global\long\def\im{\operatorname{im}}
\global\long\def\re{\operatorname{re}}
\global\long\def\rr{\mathbb{R}}
\global\long\def\cc{\mathbb{C}}
\global\long\def\vv{\mathbb{V}}
\global\long\def\SL{\operatorname{SL}}
\global\long\def\ord{\operatorname{ord}}
\global\long\def\Spec{\operatorname{Spec}}
\global\long\def\qq{\mathbb{Q}}
\global\long\def\pp{\mathbb{P}}
\global\long\def\Hom{\operatorname{Hom}}
\global\long\def\id{\operatorname{id}}
\global\long\def\gcd{\operatorname{gcd}}
\global\long\def\lcm{\operatorname{lcm}}
\global\long\def\Sym{\operatorname{Sym}}
\global\long\def\ff{\mathbb{F}}
\global\long\def\nn{\mathbb{N}}
\global\long\def\hh{\mathbb{H}}
\global\long\def\Ann{\operatorname{Ann}}
\global\long\def\End{\operatorname{End}}
\global\long\def\sgn{\operatorname{sgn}}
 \global\long\def\rk{\operatorname{rk}}
\global\long\def\Br{\operatorname{Br}}
 \global\long\def\Hasse{\operatorname{Hasse}}
\global\long\def\GL{\operatorname{GL}}
\global\long\def\Coll{\operatorname{Col}}
 \global\long\def\Tr{\operatorname{Tr}}
\global\long\def\aa{\mathbb{A}}
\global\long\def\ld{\operatorname{in}_{<}}
\global\long\def\ob{\operatorname{Ob}}
\global\long\def\mor{\operatorname{mor}}
\global\long\def\ext{\operatorname{Ext}}
\global\long\def\tor{\operatorname{Tor}}
\global\long\def\cok{\operatorname{coker}}
\global\long\def\ilim{\varprojlim}



\title{Further Representation Theory}


\author{Alex Bartel\\
Notes by Florian Bouyer}

\maketitle
\begin{center}
Copyright (C) Bouyer 2014. 
\par\end{center}

\begin{center}
Permission is granted to copy, distribute and/or modify this document 
\par\end{center}

\begin{center}
under the terms of the GNU Free Documentation License, Version 1.3 
\par\end{center}

\begin{center}
or any later version published by the Free Software Foundation; 
\par\end{center}

\begin{center}
with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts. 
\par\end{center}

\begin{center}
A copy of the license can be found at \href{http://www.gnu.org/licenses/fdl.html}{http://www.gnu.org/licenses/fdl.html}
\par\end{center}

\tableofcontents{}

\newpage{}

\textbf{Aims:} 
\begin{itemize}
\item More character theory
\item A bridge between representations and modules
\item Representations of finite groups over fields (of characteristic $0$)
that are not algebraically closed.
\end{itemize}

\section*{Applications to Representation Theory}
\begin{thm*}[Burnside]
 If $\left|G\right|=p^{\alpha}q^{\beta}$ where $p,q$ are primes,
then $G$ is soluble. 
\end{thm*}

\begin{thm*}[Frobenius]
 If $H\leq G$ is such that $gHg^{-1}\cap H=\{1\}$ $\forall g\in G\setminus H$.
Then $\exists N\lhd G$ such that $N\cap H=\{1\}$ and $NH=G$, i.e.,
$G=N\rtimes H$.
\end{thm*}
Idea of Proof: Define $N=G\setminus\left(\cup_{g\in G}gHg^{-1}\cup\{1\}\right)$.
We then use representation theory to prove that $N$ is a normal subgroup
of $G$.
\begin{thm*}
If $G=S_{n}$, $1\in G$ has the most square roots among all $g\in S_{n}$.

More generally, can express the square root counting function through
characters.
\end{thm*}
\newpage{}


\section{Revision}


\subsection{Representations}

Let $G$ be a finite group.
\begin{defn}
A \emph{representation of $G$ over a field $K$} is a $K$-vector
space $V$ together with a group homomorphism $\rho:G\to\GL(V):=$\{invertible
linear maps $V\to V\}$. Suppose $\dim V=n$ and $v_{1},\dots,v_{n}$
s a basis of $V$, such a choice identifies $\GL(V)$ with $\GL_{n}(K)=\{$invertible
$n\times n$ matrices over $K$\}.

If $(V_{1},\rho_{1}),(V_{2},\rho_{2})$ are two representations of
$G$ over $K$, a \emph{homomorphism} $\phi:(V_{1},\rho_{1})\to(V_{2},\rho_{2})$
is a vector space homomorphism $\phi:V_{1}\to V_{2}$ such that for
all $v\in V_{1},g\in G$ $\phi(\rho_{1}(g)\cdot v)=\rho_{2}(g)\cdot\phi(v)$.\end{defn}
\begin{notation*}
~
\begin{itemize}
\item Sometimes just say ``$V$ is a representation'' when the map $\rho$
is understood.
\item Write $^{g}v$ or $g\cdot v$ instead of $\rho(g)\cdot v$. 
\end{itemize}
\end{notation*}
\begin{defn}
If $V$ is a representation, a \emph{subrepresentation} is a a subvector
space $W\subset V$ such that $G\cdot W=W$. We denote it $W\leq V$.
\end{defn}
We have the obvious notion of $V/W$ as a representation and the usual
isomorphism theorems. (In particular kernels and images of homomorphism
are subrepresentations.)
\begin{example*}
Let $G=C_{2}=\left\langle g\right\rangle $, let $V$ be of dimension
of $2$, with basis $v_{1},v_{2}$. We could have $\rho:g\mapsto\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}$. $W_{1}=\left\langle \begin{pmatrix}1\\
1
\end{pmatrix}\right\rangle $ is a subrepresentation (one can easily see that it is invariant under
$\rho$). The other subrepresentation is $W_{2}=\left\langle \begin{pmatrix}1\\
-1
\end{pmatrix}\right\rangle $.
\end{example*}
If $W_{1},W_{2}\leq V$ we say $V=W_{1}\oplus W_{2}$ if this is true
on the level of vector spaces. 
\begin{defn}
A representation is \emph{indecomposible} if it's not a direct sum
of proper subrepresentation. 

A representation is \emph{irreducible} if it is non-zero and has no
proper non-zero subrepresentation.\end{defn}
\begin{example*}
Let $G=C_{p}=\left\langle g\right\rangle $ (where $p$ is a prime).
Let $K=\ff_{p}$, $V$ is $2$ dimensional. Let the representation
be $G\to\GL_{2}(\ff_{p})$, defined by $g\mapsto\begin{pmatrix}1 & 1\\
0 & 1
\end{pmatrix}$. This is not irreducible (i.e, reducible) since $W=\left\langle \begin{pmatrix}1\\
0
\end{pmatrix}\right\rangle $ is invariant under $G$ (and $W\leq V$). But $V$ is indecomposible,
since there is no other proper subrepresentation. 
\end{example*}

\begin{example*}
~
\begin{itemize}
\item Given any group $G$ and any field $K$, $G\to\GL_{1}(K)=K^{*}$ defined
by $g\mapsto1$. This is called the \emph{trivial representation}
denoted $\mathbb{I}$. 
\item \textbullet{} Given any group $G$, $X$ a finite $G$-set (i.e.,
$G$ acts on $X$ by permutations) with $\left|X\right|=n$. Take
an $n$-dimensional vectors space $V$ over any field $K$, with a
basis$\{v_{x}:x\in X\}$. Let $G$ act on $V$ by $g\cdot v_{x}=v_{g(x)}$.
This representation is denoted by $K[X]$. 
\item Important special case: $X=G$, $G$ acts by left multiplication.
The resulting representation, $K[G]$, is called the \emph{regular
representation}. 
\end{itemize}
\end{example*}
\begin{namedthm}[Schur's Lemma]
 Let $G$ be a group, $V_{1},V_{2}$ be two irreducible representations.
Any homomorphism $V_{1}\to V_{2}$ is either $0$, or an isomorphism. \end{namedthm}
\begin{lem}
Any irreducible representation $V$ of $G$ over $K$ is isomorphic
to a quotient of the regular representation $K[G]$.\end{lem}
\begin{proof}
Take any $v\in V\setminus\{0\}$, define a map $K[G]\to V$ by $g\mapsto g\cdot v$.
This is a homomorphism of representations, it is not the zero map,
so it is onto. So we are done by the first isomorphism theorem. \end{proof}
\begin{thm}[Maschke]
 Suppose $\mbox{char}K\nmid\left|G\right|$. Given any $W_{1}\leq V$,
representations of $G/K$. Then there exists a representation $W_{2}\leq V$
such that $V=W_{1}\oplus W_{2}$. \end{thm}
\begin{cor}
Every irreducible representation $V$ (in the case $\mbox{char}K\nmid\left|G\right|$)
is isomorphic to a subrepresentation of $K[G]$. 
\end{cor}

\subsection{Modules}
\begin{defn}
An \emph{algebra} $A$ \emph{over a field} $K$ is a ring (with $1$)
that is also a $K$-vector space, such that $(x\cdot\alpha)\cdot(y\cdot\beta)=(xy)\cdot(\alpha\cdot\beta)$
for all $x,y\in K,\;\alpha,\beta\in A$.

Equivalently, $A$ is a ring with $K\subset Z(A)$.\end{defn}
\begin{example*}
~
\begin{itemize}
\item $\cc$ is a $\cc$-algebra, but it is also an $\rr$-algebra
\item If $A$ is any $K$-algebra, then the ring of $n\times n$ matrices
over $A$, denoted $M_{n}(A)$, is also a $K$-algebra. 
\item $\hh=\left\langle \rr\cdot1+\rr\cdot\underline{i}+\rr\cdot\underline{j}+\rr\cdot\underline{k}\mid ij=k,jk=i,ki=j,ij=-k,kj=-i,ik=-j,i^{2}=j^{2}=k^{2}=-1\right\rangle $
is an $\rr$-algebra.
\item If $G$ is a group, $K$ is a field, the group algebra $K[G]$ is
a vector space spanned by vectors $v_{g}$, $g\in G$, with multiplication
$v_{g}\cdot v_{j}=v_{gh}$. 
\end{itemize}
\end{example*}
\begin{defn}
If $A$ is a $K$-algebra, a \emph{left} $A$\emph{-module} is an
abelian group $(M,+)$ with a map $A\times M\to M$ such that
\begin{itemize}
\item $a\times(m_{1}+m_{2})=a\times m_{1}+a\times m_{2}$
\item $(a_{1}+a_{2})\times m=a_{1}\times m+a_{2}\times m$
\item $0_{A}\times m=0_{M}$
\item $1_{A}\times m=m$
\item $(a_{1}\cdot a_{2})\times m=a_{1}\times(a_{2}\times m)$
\end{itemize}
Equivalently, the map $A\to\End(M)=\Hom(M,M)$ defined by $a\mapsto(m\mapsto a\times m)$
is a ring homomorphism. 
\end{defn}
\textbf{Moral}: $K[G]$-modules are the same as representations of
$G$ over $K$. 

We have the obvious notions of homomorphisms of modules, submodules,
quotients, isomorphisms theorems, etc.
\begin{example*}
Any algebra $A$ can be thought as a module over itself: $M=A$ and
$a\times m=a\cdot m$. This is called the \emph{left regular module}
of $A$.

The left regular module of $K[G]$ is the same as the regular representation
of $G$ over $K$. \end{example*}
\begin{defn}
A module $M$ is \emph{simple} if $M\neq0$ and there exists no proper
non-zero submodules.

A module $M$ is \emph{semi-simple} if it's a direct sum of simple
modules.\end{defn}
\begin{namedthm}[Schur's Lemma]
 If $M_{1},M_{2}$ are simple $A$-modules, then any homomorphisms
$M_{1}\to M_{2}$ is either the $0$ map, or an isomorphism.

In particular, if $M$ is simple, then $\End(M)$ is a Division Ring
(i.e., every non zero elements has a two-sided inverse)\end{namedthm}
\begin{note*}
A submodule of the left regular module of $A$ is nothing but a left
ideal.\end{note*}
\begin{namedthm}[Maschke's Theorem]
 The left regular module of $K[G]$ is semi-simple, when $\mbox{char}K\nmid\left|G\right|$.\end{namedthm}
\begin{thm}[Artin - Wedderburn]
 Any algebra whose regular module is semi-simple is isomorphic to
$\oplus_{i}M_{n_{i}}(D_{i})$ where $D_{i}$ are division rings. 
\end{thm}
Hence, if $\mbox{char}K\nmid\left|G\right|$ we have $K[G]\cong\oplus_{i}M_{n_{i}}(D_{i})$,
where $D_{i}$ are division algebras over $K$. 
\begin{rem*}
~
\begin{enumerate}
\item $M_{n}(D)$ is really semi-simple. $I_{i}=\begin{pmatrix}0 & \cdots & 0 & * & 0 & \cdots & 0\\
\vdots &  & \vdots & \vdots & \vdots &  & \vdots\\
0 & \cdots & 0 & * & 0 & \cdots & 0
\end{pmatrix}$ is clearly a left ideal, $M_{n}(D)=\oplus I_{i}$ as a module.

\begin{claim*}
$I_{i}$ is simple.\end{claim*}
\begin{proof}
If $U\leq I_{i}$, $v\in U$ is non-zero, without loss of generality
$v=\begin{pmatrix}\alpha\\
\vdots
\end{pmatrix}$, $\alpha\in D^{*}$. Then 
\begin{eqnarray*}
\begin{pmatrix}1 & 0 & \cdots & 0\\
0\\
\vdots &  & 0\\
0
\end{pmatrix}\begin{pmatrix} & \alpha\\
0 & * &  & 0\\
 & \vdots\\
 & *
\end{pmatrix} & = & \begin{pmatrix} & \alpha\\
0 & 0 &  & 0\\
 & \vdots\\
 & 0
\end{pmatrix}\in U\\
\begin{pmatrix}\alpha^{-1} & 0 & \cdots & 0\\
0\\
\vdots &  & 0\\
0
\end{pmatrix}\begin{pmatrix} & \alpha\\
0 & * &  & 0\\
 & \vdots\\
 & *
\end{pmatrix} & = & \begin{pmatrix} & 1\\
0 & 0 &  & 0\\
 & \vdots\\
 & 0
\end{pmatrix}\in U
\end{eqnarray*}
So $U=I_{i}$. \end{proof}
\begin{claim*}
The $I_{i}$ are all pairwise isomorphic.
\end{claim*}
\item ~

\begin{cor}
Then number of irreducible representations (up to isomorphism) of
$G$ over $K$ (equivalently simple $K[G]$-modules) is equal to the
number of conjugate classes of elements of $G$\end{cor}
\begin{proof}
Compute $\dim_{K}Z(K[G])$ on both sides. On the left had side $Z=\left\langle \sum_{h\in G}hgh^{-1}|g\in G\right\rangle $,
so $\dim_{K}Z$ is precisely the number of conjugacy classes of elements
of $G$. On the right hand side $Z=\left\langle (0,\dots,0,I_{n_{i}},0,\dots0)\right\rangle $
where $I_{n_{i}}\in M_{n_{i}}(D_{i})$, so $\dim_{K}Z$ equals the
number of distinct isomorphism classes of simple modules (one for
each $i$).
\end{proof}
\item Suppose $A$ is a semi-simple algebra. $S$ is a simple $A$-module.
Then Schur's Lemma says $\End_{A}(S)=D$ is a division algebra. Put
$M=\underset{n\mathrm{\, copies}}{\underbrace{S\oplus\dots\oplus S}}$.
Then $\End_{A}(M)=M_{n}(D)$. (Each endomorphism of $M$ is determined
by the image of $(0,\dots,0,s,0,\dots,0)$ (in the $i$th place),
which is determined by projections to all components). The Wedderburn
isomorphism comes by identifying $A$ (actually $A^{\mathrm{op}}$)
with its endomorphism ring. 
\item A decomposition $A=\oplus_{i}M_{n_{i}}(D_{i})$ corresponds to writing
$1=\sum e_{i}$, where $e_{i}$ are non-zero orthogonal primitive
central idempotent.

\begin{elabeling}{00.00.0000}
\item [{idempotent}] $e_{i}^{2}=e_{i}$
\item [{central}] $e_{i}a=ae_{i}$ for all $a\in A$.
\item [{orthogonal}] $e_{i}e_{j}=0$ if $i\neq j$
\item [{primitive}] $e_{i}$ is not a sum of non-zero orthogonal central
idempotent elements.
\end{elabeling}

If $A=\oplus M_{n_{i}}(D_{i})$, then $1=(1,\dots,1)=(1,0,\dots,0)+\dots+(0,\dots,0,1)$.
Conversely if $1=\sum e_{i}$, then $U_{i}=e_{i}A$ gives $A=\oplus U_{i}$.
Since they are orthogonal, we have $U_{i}\cap U_{j}=\{0\}$, since
they are central idempotent, $U_{i}$ are ideals, since they are primitive
$U_{i}$ are isotypical ($S\oplus\dots\oplus S$), and since their
sum are $1$, $A=\sum U_{i}$.

\end{enumerate}
\end{rem*}
\begin{example*}
If $G\cong C_{3}$, then $\cc[G]\cong\cc\oplus\cc\oplus\cc$. On the
other hand $\rr[G]\cong\rr\oplus\cc$. 

To see the second statement, consider $e_{1}=\frac{1}{3}(1+g+g^{2})$.
We have $e_{1}^{2}=\frac{1}{9}(1+g+g^{2})(1+g+g^{2})=\frac{1}{9}(3+3g+3g^{2})=e_{1}$.
The ideal $e_{1}\rr[G]$ had $\rr$-dimension $1$, so it is isomorphism
to $\rr$. 

Consider $e_{2}=1-e_{1}=\frac{2}{3}-\frac{1}{3}g-\frac{1}{3}g^{2}$.
The ideal $e_{2}\rr[G]$ is generated (as $\rr$-vector space) by
$\alpha=e_{2}$ and $\beta=\frac{1}{\sqrt{3}}(g-g^{2})$. We have$\beta^{2}=\frac{1}{3}(g-g^{2})^{2}=\frac{1}{3}(g^{2}-2+g)=-\alpha$.
So $\alpha\mapsto1$, $\beta\mapsto i$ is an isomorphism $e_{2}\rr[G]\to\cc$. 
\end{example*}

\subsection{Characters}

If $\rho:G\to\GL_{n}(\cc)$ is a representation, the corresponding
\emph{character} $\chi_{\rho}(g)=\Tr(\rho(g))$. 
\begin{thm}
Let $\rho_{1},\rho_{2}$ be two representations, then $\rho_{1}\cong\rho_{2}$
if and only if $\chi_{\rho_{1}}=\chi_{\rho_{2}}$.\end{thm}
\begin{rem*}
~
\begin{itemize}
\item $\chi_{\rho_{1}\oplus\rho_{2}}=\chi_{\rho_{1}}+\chi_{\rho_{2}}$. 
\item Recall: the number of distinct irreducible representations of $G$
over $\cc$ is equal to the number of conjugacy classes of elements.
The character table of $G$ is a square table, \\
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline 
 & $1$ & $g_{2}$ & $g_{3}$ & $\dots$ & $\dots$ & $g_{k}$\tabularnewline
\hline 
\hline 
$\mathbb{I}$ & $1$ & $1$ & $1$ & $\dots$ & $\dots$ & $1$\tabularnewline
\hline 
$\chi_{1}$ & $\dim\chi_{1}$ & $\chi_{2}(g_{2})$ & $\chi_{2}(g_{3})$ & $\dots$ & $\dots$ & $\chi_{2}(g_{k})$\tabularnewline
\hline 
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\ddots$ & $\ddots$ & $\vdots$\tabularnewline
\hline 
$\chi_{k}$ & $\dim\chi_{k}$ & $\chi_{k}(g_{2})$ & $\chi_{k}(g_{3})$ & $\dots$ & $\dots$ & $\chi_{k}(g_{k})$\tabularnewline
\hline 
\end{tabular}\\
Character are class functions, i.e., constant on conjugacy classes
and the irreducible characters span the vector space of class functions. 
\end{itemize}
\end{rem*}
\begin{thm}[Schur's Lemma in disguise, Row Orthogonality]
 If $\chi_{1},\chi_{2}$ are irreducible characters of $G$, then
the inner product 
\[
\left\langle \chi_{1},\chi_{2}\right\rangle =\frac{1}{\left|G\right|}\cdot\sum_{g\in G}\chi_{1}(g)\overline{\chi_{2}(g)}=\begin{cases}
1 & \mathrm{if}\,\chi_{1}=\chi_{2}\\
0 & \mathrm{if}\,\chi_{1}\neq\chi_{2}
\end{cases}.
\]

\end{thm}
We also have column orthogonality. (Exercise: try to derive it using
Row Orthogonality)

An arbitrary $G$ representation $\rho$ is a sum $\rho=\sum_{i}n_{i}\rho_{i}$
where $\rho_{i}$ are distinct irreducible representations and $n_{i}\in\zz$.
So $\chi_{\rho}=\sum n_{i}\chi_{\rho_{i}}$, and from the Row Orthogonality
Theorem, $n_{i}=\left\langle \chi_{\rho},\chi_{\rho_{i}}\right\rangle $,
i.e.,$\chi_{\rho}=\sum\left\langle \chi_{\rho},\chi_{\rho_{i}}\right\rangle \chi_{\rho_{i}}$. 
\begin{rem*}
The inner product can be defined for arbitrary class functions, so
the theorem says that irreducible characters form an orthonormal basis
of the space of class functions. 
\end{rem*}

\subsubsection*{New characters / representations from old ones}

If $N\lhd G$, then any group homomorphism from $G/N$ induces a group
homomorphism from $G$. So we can lift representations and characters
from quotients. 
\begin{example*}
Let $G\cong S_{3}$ and $N\cong C_{3}$. Then $G/N\cong C_{2}$. Let
$\epsilon:C_{2}=\left\langle g\right\rangle \to\cc^{*}=\GL_{1}(\cc)$
be defined by $g\mapsto-1$ \\
\begin{tabular}{|c|c|c|c|}
\hline 
 & $1$ & $(12)$ & $(123)$\tabularnewline
\hline 
\hline 
$\mathbb{I}$ & $1$ & $1$ & $1$\tabularnewline
\hline 
$\epsilon$ & $1$ & $-1$ & $1$\tabularnewline
\hline 
$\rho$ & $2$ & $0$ & $-1$\tabularnewline
\hline 
\end{tabular}\\
Where the last row was worked out using the dimension (sum of the
dimension squared need to equal $\left|G\right|$) and column orthogonality.
\end{example*}
Another way to get new representations from old ones is using \uline{restriction}.
Any group homomorphism $G\to X$ (where $X$ is anything) restricts
to a group homomorphism $H\to X$ for any $H\leq G$. We will write
this $\mbox{Res}_{G/H}\rho$ or $\rho\downarrow_{H}^{G}$. 

We also have \uline{induction}: Let $H\leq G$ and $\rho:H\to\GL(V)$.
Take a set of coset representatives $\{g_{1}H,\dots,g_{n}H\}$ for
$G/H$. Define a new vector space $W=\oplus_{g_{i}}\underset{\cong V\mathrm{\, as\, v.s.}}{\underbrace{g_{i}\cdot V}}$.
For any $g\in G$ and for each $g_{i}$, write (uniquely) $g\cdot g_{i}=g_{j}h$,
with $h\in H$. Let $g$ act on $W$ by $g\cdot(g_{i}v)=g_{j}\rho(h)(v)$.
This defines a representation of $G$ on $W$, ( note that $\dim W=\dim V\cdot\left|G/H\right|$).
We write this as $\mbox{Ind}_{G/H}\rho$, or $\rho\uparrow_{H}^{G}$. 

The character of $\rho\uparrow_{H}^{G}$ is 
\[
\chi\uparrow_{H}^{G}(x)=\frac{1}{\left|H\right|}\sum_{g\in G}\chi^{0}(gxg^{-1})\;\mathrm{where\;}\chi^{0}(y)=\begin{cases}
\chi(y) & \mathrm{if}\, y\in H\\
0 & \mathrm{if}\, y\notin H
\end{cases}.
\]

\begin{namedthm}[Frobenius reciprocity]
 If $H\leq G$, $\chi$ is a character of $G$ and $\tau$ is a character
of $H$, then
\[
\left\langle \chi,\tau\uparrow_{H}^{G}\right\rangle _{G}=\left\langle \chi\downarrow_{H}^{G},\tau\right\rangle _{H}.
\]

\end{namedthm}
More functorial statement of Frobenius reciprocity is the following:

\emph{If $H\leq G$, $\rho:H\to\GL(V)$ and $\rho':G\to\GL(V')$,
then there is a natural isomorphism $\Hom_{G}\left(\rho',\rho\uparrow_{H}^{G}\right)\cong\Hom_{H}\left(\rho'\downarrow_{H}^{G},\rho\right)$.}

This works over any field!


\subsection*{Some Properties of Characters}
\begin{itemize}
\item Character values are sum of roots of unities, more specifically if
$g\in G$ has order $n$, $\chi$ is $d$-dimensional, then $\chi(g)$
is the sum of $d$ $n$th root of unity.


In particular, $\chi(g)$ is an algebraic integer, i.e., roots of
monic polynomial with integer coefficients.


Also, it follows that $\left|\chi(g)\right|\leq\left|\chi(1)\right|$,
with equality if and only if the matrix corresponding to $g$ is in
fact scalar (independent of basis on the vector space). Furthermore
$\chi(g)=\chi(1)$ if and only if $g\mapsto I_{n}$ (the identity
matrix). Hence define $\ker\chi=\{g\in G|\chi(g)=\chi(1)\}$. Define
the centre, $Z(\chi)=\{g\in G|\left|\chi(g)\right|=\chi(1)\}$.

\item There exists a bijection between irreducible characters $\chi$ of
$G$ with $\ker\chi\geq N\lhd G$ and irreducible characters of $G/N$
lifted to $G$. 


All normal subgroups of $G$ are obtained as intersections of $\ker\chi$
for suitable irreducible character $\chi$. Also $Z(G)=\cap_{\chi\in\mathrm{Irr}(G)}Z(\chi)$

\item Recall that $\cc[G]=\oplus_{\rho_{i}\mathrm{\, irr}}\rho_{i}^{\oplus\dim\rho_{i}}$,
so for any irreducible $\chi$ we have $\left\langle \cc[G],\chi\right\rangle _{G}=\dim\chi$,
hence $\left|G\right|=\sum_{\chi\in\mathrm{Irr}(G)}\left(\dim\chi\right)^{2}$
\item Let $G'=\left\langle ghg^{-1}h^{-1}|g,h\in G\right\rangle \lhd G$.
This is called the \emph{derived subgroup} or \emph{commutator subgroup}.
It is the unique minimal normal subgroup with abelian quotient, i.e.,
if $N\lhd G$ is such that $G/N$ is abelian then $N\geq G'$. It
is easy to see that $G'=\cap_{\dim\chi=1}\ker\chi$. 
\item If $\phi$ is any character and $\chi$ is a $1$-dimensional character,
then $\phi\otimes\chi(g)=\phi(g)\cdot\chi(g)$ is also a character
(check!)\end{itemize}
\begin{example*}
~
\begin{enumerate}
\item Cyclic groups, $C_{n}=\left\langle g\right\rangle $, of order $n$.
All irreducible characters are $1$-dimensional, $\chi_{k}:g\mapsto e^{\frac{2\pi i}{n}k}$
for $k=0,\dots,n-1$. Let $\zeta=e^{\frac{2\pi1}{n}}$\\
\begin{tabular}{|c|c|c|c|c|c|}
\hline 
 & $1$ & $g$ & $g^{2}$ & $\dots$ & $g^{n-1}$\tabularnewline
\hline 
\hline 
$\mathbb{I}$ & $1$ & $1$ & $1$ & $\dots$ & $1$\tabularnewline
\hline 
$\chi_{1}$ & $1$ & $\zeta$ & $\zeta^{2}$ & $\dots$ & $\zeta^{n-1}$\tabularnewline
\hline 
$\chi_{2}$ & $1$ & $\zeta^{2}$ & $\zeta^{4}$ & $\dots$ & $\zeta^{2(n-1)}$\tabularnewline
\hline 
$\vdots$ &  &  &  &  & \tabularnewline
\hline 
\end{tabular}
\item Abelian groups, $A=C_{n_{1}}\times\dots\times C_{n_{r}}=\left\langle g_{1}\right\rangle \times\dots\times\left\langle g_{r}\right\rangle $.
Then all irreducible characters are $1$-dimensional, $\chi_{k_{1}\dots k_{r}}:g_{j}\mapsto e^{\frac{2\pi i}{n_{j}}k_{j}}$
for $0\leq k_{j}\leq n_{j}-1$. 
\item Non-abelian group of order $8$: 

\begin{itemize}
\item $G_{1}=D_{8}=\left\langle \sigma,\tau\mid\sigma^{2}=\tau^{4}=1,\sigma\tau\sigma=\tau^{-1}\right\rangle $
\item $G_{2}=Q_{8}=\left\langle -1,i,j,k\mid(-1)^{2}=1,i^{2}=j^{2}=k^{2}=-1,ij=k,jk=i,ki=j,ji=-k,kj=-i,ik--j\right\rangle $.
\end{itemize}

First look at $G_{1}/G_{1}'\cong C_{2}\times C_{2}$ where $G_{1}'=\left\langle \tau^{2}\right\rangle $.
So we can easily lift the characters of $G_{1}/G_{1}'$ to $G_{1}$.
\\
\begin{tabular}{|c|c|c|c|c|c|}
\hline 
 & $1$ & $\sigma$ & $\tau$ & $\tau^{2}$ & $\sigma\tau$\tabularnewline
\hline 
\hline 
$\mathbb{I}$ & $1$ & $1$ & $1$ & $1$ & $1$\tabularnewline
\hline 
$\epsilon_{1}$ & $1$ & $1$ & $-1$ & $1$ & $-1$\tabularnewline
\hline 
$\epsilon_{2}$ & $1$ & $-1$ & $1$ & $1$ & $-1$\tabularnewline
\hline 
$\epsilon_{3}$ & $1$ & $-1$ & $-1$ & $1$ & $1$\tabularnewline
\hline 
$\chi$ & $2$ & $0$ & $0$ & $-2$ & $0$\tabularnewline
\hline 
\end{tabular}\\
$G_{2}$ is left as an exercise, but they do have the same character
table (but $G_{1}\ncong G_{2}$) 

\item $G=S_{4}$. First recall that the character table for $S_{3}$ is\\
 %
\begin{tabular}{|c|c|c|c|}
\hline 
 & $1$ & $(12)$ & $(123)$\tabularnewline
\hline 
\hline 
$\mathbb{I}$ & $1$ & $1$ & $1$\tabularnewline
\hline 
$\epsilon$ & $1$ & $-1$ & $1$\tabularnewline
\hline 
$\rho$ & $2$ & $0$ & $-1$\tabularnewline
\hline 
\end{tabular}\\
and recall that $S_{4}/V_{4}\cong S_{3}$. Hence we can lift the three
characters of $S_{3}$ into $S_{4}$. Then we use the dimensional
formula to find that the last two characters must be $3$ dimensional,
so can complete using column orthogonality and the fact that $\epsilon\otimes\chi_{1}$
must be a character. \\
\begin{tabular}{|c|c|c|c|c|c|}
\hline 
 & $1$ & $(12)$ & $(123)$ & $(1234)$ & $(12)(34)$\tabularnewline
\hline 
\hline 
$\mathbb{I}$ & $1$ & $1$ & $1$ & $1$ & $1$\tabularnewline
\hline 
$\epsilon$ & $1$ & $-1$ & $1$ & $-1$ & $1$\tabularnewline
\hline 
$\rho$ & $2$ & $0$ & $-1$ & $0$ & $2$\tabularnewline
\hline 
$\chi_{1}$ & $3$ & $1$ & $0$ & $-1$ & $-1$\tabularnewline
\hline 
$\chi_{2}$ & $3$ & $-1$ & $0$ & $1$ & $-1$\tabularnewline
\hline 
\end{tabular}
\end{enumerate}
\end{example*}
\newpage{}


\section{Mackey's Formula and Applications}

Let $H\leq G$, and $\rho$ is a representation of $H$, what is $\rho\uparrow^{G}\downarrow_{H}$? 
\begin{defn}
Let $H,K\leq G$, a \emph{double coset} is a set of the form $KgH=\{kgh|k\in K,h\in H\}=\cup_{k\in K}kgH=\cup_{h\in H}Kgh$. 

$K\backslash G/H$ is the set of double cosets (or, by slight abuse
of notation, the set of double coset representative).\end{defn}
\begin{note*}
$Kg_{1}H=Kg_{2}H$ if and only if $g_{2}\in Kg_{1}H$. 
\end{note*}
\textbf{Warning:} Different double cosets can have different size.
\begin{example*}
Let $G=S_{3}$ and $H=K=\left\langle (12)\right\rangle $. 
\begin{itemize}
\item $H\cdot1\cdot K=H$, size is 2.
\item $H\cdot(123)\cdot K=\{(123),(12)(123)(12)=(132),(123)(12)=(23),(12)(123)=(13)\}$,
size is 4.
\end{itemize}
\end{example*}
\begin{namedthm}[Mackey's Formula]
 Let $H,K\leq G$ and $\rho$ a representation of $H$ over any field
$L$. Then 
\[
\rho_{H}\uparrow^{G}\downarrow_{K}=\bigoplus_{g\in K\backslash G/H}\,^{g}\rho\downarrow_{K\cap gHg^{-1}}\uparrow^{K}
\]
where $^{g}\rho(ghg^{-1})=\rho(h)$ for all $h\in H$. \end{namedthm}
\begin{proof}
(Not Examinable) Let $V$ be the vector space corresponding to $\rho$,
then $\rho\uparrow^{G}$ is represented on $W=\oplus_{g\in G/H}gV$.
Now $G$ acts transitively on $G/H$, but $K$ may not. Suppose 
\[
\underset{\cup=Kg_{1}H}{\underbrace{g_{1}H,\dots,g_{r_{1}}H}},\underset{\cup=Kg_{r_{1}+1}H}{\underbrace{g_{r_{1}+1}H,\dots,g_{r_{2}}H}},\dots
\]
For a giving $k\in K$, there exists $g_{n}H$ such that $kg_{1}H=g_{n}H$
if and only if $g_{n}\in kg_{1}H$. So $g_{1}V\oplus\dots\oplus g_{r_{1}}V$
is a $K$-subrepresentation of $W$. By direct calculation, we see
that this is isomorphism to $\,^{g_{1}}\rho\downarrow_{K\cap g_{1}Hg_{1}^{-1}}\uparrow^{K}$\end{proof}
\begin{example*}
Take $\rho=\mathbb{I}_{H}$, then $\rho\uparrow^{G}=L[G/H]$ where
$L$ is the field as above. Now $\rho\uparrow^{G}\downarrow_{K}=\oplus_{g\in K\backslash G/H}\mathbb{I}_{K\cap gHg^{-1}}\uparrow^{K}=\oplus_{g\in K\backslash G/H}L[K/K\cap gHg^{-1}]$. 
\end{example*}
Check: The orbit of $K$ acting on $G/H$ are in bijection with $K\backslash G/H$. 


\subsection{Application I: principal series representation of $\GL_{2}(\ff_{p})$}

Let $G=\GL_{2}(\ff_{p})$ (where $p\in\zz$ is any prime). We have
the subgroups $B=\left\{ \begin{pmatrix}* & *\\
0 & *
\end{pmatrix}\right\} \leq G$, $T=\left\{ \begin{pmatrix}* & 0\\
0 & *
\end{pmatrix}\right\} \leq B$ and $U=\left\{ \begin{pmatrix}1 & *\\
0 & 1
\end{pmatrix}\right\} \leq B$. Note that $\left|U\right|=p$, $\left|T\right|=(p-1)^{2}$, $\left|B\right|=(p-1)^{2}p$
and $\left|G\right|=(p^{2}-1)(p^{2}-p)=(p-1)^{2}p(p+1)$. We have
$T\cong\ff_{p}^{*}\times\ff_{p}^{*}$, $U\cong(\ff_{p},+)\lhd B$.
We have $B/U\cong\ff_{p}^{*}\times\ff_{p}^{*}$, in fact $B=U\rtimes T$
(i.e, $U\lhd B$, $T\leq B$, $U\cap T=\left\{ 1\right\} $ and $UT=B$). 

Let $\chi_{1},\chi_{2}:\ff_{p}^{*}\to\cc^{*}$ be two irreducible
characters, then define $\tau=\chi_{1}\otimes\chi_{2}:B\to\cc^{*}$
by $\begin{pmatrix}a & b\\
0 & d
\end{pmatrix}\mapsto\chi_{1}(a)\cdot\chi_{2}(d)$. (Note $U\leq\ker\tau$). 
\begin{thm}
$\tau\uparrow_{B}^{G}$ (which has dimension $p+1$) is either
\begin{itemize}
\item irreducible if $\chi_{1}\neq\chi_{2}$ or
\item (one dimensional)$\oplus$ irreducible if $\chi_{1}=\chi_{2}$. 
\end{itemize}
\end{thm}
\begin{proof}
Recall that a character is irreducible if and only if $\left\langle \tau,\tau\right\rangle _{G}=1$.
We have 
\begin{eqnarray*}
\left\langle \tau,\tau\right\rangle _{B} & \underset{\mathrm{Frob\, rec}}{=} & \left\langle \chi_{1}\otimes\chi_{2}\uparrow_{B}^{G}\downarrow_{B},\chi_{1}\otimes\chi_{2}\right\rangle _{B}\\
 & \underset{\mathrm{Mackey}}{=} & \sum_{g\in B\backslash G/B}\left\langle \,^{g}\left(\chi_{1}\otimes\chi_{2}\right)\downarrow_{B\cap gBg^{-1}}\uparrow^{B},\chi_{1}\otimes\chi_{2}\right\rangle _{B}\\
 & \underset{\mathrm{Frob\, rec}}{=} & \sum_{g\in B\backslash G/B}\left\langle \,^{g}\left(\chi_{1}\otimes\chi_{2}\right)\downarrow_{B\cap gBg^{-1}},\chi_{1}\otimes\chi_{2}\downarrow_{B\cap gBg^{-1}}\right\rangle _{B\cap gBg^{-1}}
\end{eqnarray*}

\begin{claim*}
$B\backslash G/B=\{B\cdot1\cdot B,B\cdot\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}\cdot B\}$\end{claim*}
\begin{proof}
It is enough to show that any $\begin{pmatrix}a & b\\
c & d
\end{pmatrix}\in G$ with $c\neq0$ is of the form $X\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}Y$ with $X,Y\in B$. Let $X=\begin{pmatrix}x & y\\
0 & z
\end{pmatrix}$ and $Y=\begin{pmatrix}w & v\\
0 & u
\end{pmatrix}$, we compute that $X\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}Y=\begin{pmatrix}x & y\\
0 & z
\end{pmatrix}\begin{pmatrix}0 & u\\
w & v
\end{pmatrix}=\begin{pmatrix}yw & ux+vy\\
zw & vz
\end{pmatrix}$. For $a,b,c,d$ with $c\neq0$ and $ad-bc\neq0$, we can solve $x,y,z,w,u,v$. 
\end{proof}
Going back to the equality above, we have
\begin{eqnarray*}
\left\langle \tau,\tau\right\rangle  & = & \left\langle \,^{1}(\chi_{1}\otimes\chi_{2})\downarrow_{B\cap1\cdot B\cdot1}\uparrow^{B},\chi_{1}\otimes\chi_{2}\right\rangle _{B}\\
 &  & +\left\langle \,^{\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}}(\chi_{1}\otimes\chi_{2})\downarrow_{B\cap\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}B\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}},\chi_{1}\otimes\chi_{2}\downarrow_{B\cap\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}B\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}}\right\rangle \\
 & = & 1+\left\langle \underset{\begin{pmatrix}a & b\\
c & d
\end{pmatrix}\mapsto\chi_{1}(d)\chi_{2}(a)}{\underbrace{\,^{\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}}(\chi_{1}\otimes\chi_{2})}}\downarrow_{T},\chi_{1}\otimes\chi_{2}\downarrow_{T}\right\rangle _{T}\\
 & = & 1+\begin{cases}
0 & \mathrm{if}\,\chi_{1}\neq\chi_{2}\\
1 & \mathrm{if\,\chi_{1}=\chi_{2}}
\end{cases}\\
 & = & \begin{cases}
1 & \mathrm{if\,\chi_{1}\neq\chi_{2}}\\
2 & \mathrm{if}\,\chi_{1}=\chi_{2}
\end{cases}.
\end{eqnarray*}
We deduce that if $\chi_{1}\neq\chi_{2}$, then $\tau$ is irreducible
and otherwise it is the sum of two distinct irreducible.
\begin{claim*}
If $\chi_{1}=\chi_{2}$, then $\left\langle \tau,\chi_{1}\circ\det\right\rangle _{G}=1$. \end{claim*}
\begin{proof}
We have 
\[
\left\langle \tau,\chi_{1}\circ\det\right\rangle _{G}=\left\langle \underset{\chi_{1}(a)\chi_{2}(d)}{\underbrace{\chi_{1}\otimes\chi_{2}}},\underset{\chi_{1}(ad)}{\underbrace{(\chi_{1}\circ\det)}}\downarrow_{B}\right\rangle _{B}=1
\]

\end{proof}
So if $\chi_{1}=\chi_{2}$, then $\tau=\chi_{1}\circ\det\oplus$ (a
$p$-dimensional irreducible character)\end{proof}
\begin{example*}
If $\chi_{1}=\chi_{2}=\mathbb{I}$, then $\tau\cong\cc[G/B]=\mathbb{I}+($\uline{Steinberg
representation}) 
\end{example*}

\subsection{Application II: Semi-direct products by Abelian groups}

Let $G\lhd N$, such that $N\cap H=\{1\}$, $NH=G$. Then $G=N\rtimes H$
(called semi-direct product). Note that this implies that for any
$g\in G$ there exists unique $n\in N,h\in H$ such that $g=nh$.
So as sets $G\leftrightarrow N\times H$. Under this bijection, $(n_{1},h_{1})\cdot(n_{2},h_{2})=(n_{1}\underset{\in N}{\underbrace{(h_{1}n_{2}h_{1}^{-1})}},h_{1}h_{2})$.
We have that $H$ acts on $N$ by conjugation, i.e., $\,^{h}n=hnh^{-1}$.
This defines a map $H\to\mbox{Aut}(N)$. So $G$ is uniquely determined
by $N,H$ and the map $\phi:H\to\mbox{Aut}(N)$. Conversely, given
$N,H$ and $\phi$, we can construct the group $G$ defined by as
a set $N\times H$, with $(n_{1},h_{1})\cdot(n_{2},h_{2})=(n_{1}\phi(h_{1})\cdot n_{2},h_{1}h_{2})$
\begin{example*}
~
\begin{itemize}
\item $D_{2n}=C_{n}\rtimes C_{2}$ with $\phi:C_{2}=\left\langle \sigma\right\rangle \to\mbox{Aut}(C_{n})$
is defined by $\sigma\mapsto(\tau\mapsto\tau^{-1})$.
\item If $\phi:H\to\mbox{Aut}(N)$ is defined by $h\mapsto\id$, then we
get the direct product, $G=N\times H$. 
\end{itemize}
\end{example*}
\textbf{Caution}: We can get things that are isomorphic to $N\times H$
even if $\phi$ is non-trivial. 

If $N\lhd G$, then $G$ acts on the irreducible characters of $N$
by $\,^{g}\chi(n)=\chi(g^{-1}ng)$. Note that $N$ acts trivially
on its own characters, so we get a well-defined action of $G/N$ on
$\mathrm{Irr}(N)$. 
\begin{rem*}
$\,^{g}\chi$ is certainly a class function (if $\chi$ is) but why
is it a character, i.e., what is the corresponding representation?
If $\rho:N\to\GL(V)$ is the representation attached to $\chi$, then
$\,^{g}\rho:N\to\GL(V)$ is defined by $n\mapsto\rho(g^{-1}ng)$.
Note that the latter definition make sense over any fields.
\end{rem*}

\begin{rem*}
If $N\lhd G$ and $H\cong G/N$, then in general $G\ncong N\rtimes H$. 
\begin{example*}
Let $G=C_{4}$, $N\cong C_{2}$, $G/N\cong C_{2}$ but $G\ncong C_{2}\rtimes C_{2}$. 
\end{example*}
\end{rem*}
Now suppose $G=A\rtimes H$ where $A$ is an Abelian group. We will
completely describe $\mbox{Irr}(G)$. Let $\chi\in\mbox{Irr}(A)$
(hence $\chi$ is one dimensional), let $S_{\chi}:=\mbox{Stab}_{H}(\chi)=\{h\in H|\chi(h^{-1}ah)=\chi(a)\,\forall a\in A\}$.
We extend $\chi$ to $A\rtimes S_{\chi}$ by $\chi(a\cdot s)=\chi(a)$.
(Check that this is a $1$-dimensional character of $A\rtimes S_{\chi}$).
Take any irreducible character $\rho$ of $S_{\chi}\cong(A\rtimes S_{\chi})/A$,
thought of as a character of $A\rtimes S_{\chi}$, and define $\tau_{\chi,\rho}=\chi\otimes\rho\uparrow^{G}$.
\begin{thm}
\label{thm:2.3}~
\begin{itemize}
\item $\tau_{\chi,\rho}$ are all irreducible,
\item All irreducible characters of $G$ are of this form,
\item $\tau_{\chi,\rho}=\tau_{\chi',\rho'}$ if and only if $\chi,\chi'$
lie in the same orbit under the $H$-action (i.e., there exists $h\in H$
such that $S_{\chi}=hS_{\chi'}h^{-1}$) and $\rho=\,^{h}\rho'$.
\end{itemize}
\end{thm}
\begin{proof}
Left as an exercise\end{proof}
\begin{example*}
We describe the characters of $D_{2n}\cong C_{n}\rtimes C_{2}$ in
this way. If $C_{n}=\left\langle g\right\rangle $, and $C_{2}=\left\langle h\right\rangle $,
then we know the characters $\chi_{k}:C_{n}\to\cc^{*}$ are defined
by $g\mapsto e^{2\pi ik/n}$. Also $\,^{h}\chi_{k}(g)=\chi_{k}(h^{-1}gh)=\chi_{k}(g^{-1})=e^{-2\pi ik/n}$,
so 
\[
\mbox{Stab}_{\chi_{k}}(C_{2})=\begin{cases}
C_{2} & \mathrm{if\,}e^{-2\pi ik/n}=e^{2\pi ik/n}\iff k=0,n/2\\
1 & \mathrm{otherwise}
\end{cases}.
\]
So if $k\neq0,n/2$ then $\chi_{k}\uparrow^{G}$ is irreducible. If
$k=0,n/2$ then $\chi_{k}$ extends (in two ways) to a $1$-dimensional
character of $G$. Also $\chi_{k}\uparrow^{G}=\chi_{k'}\uparrow^{G}$
is and only if $k=\pm k'$. \end{example*}
\begin{xca*}
~
\begin{itemize}
\item We describe the characters of $S_{4}\cong V_{4}\rtimes S_{3}$ (where
$V_{4}\cong C_{2}\times C_{2}$ is the Klein group)
\item Suppose $H\leq S_{n}$, $A$ is any (Abelian) group. Consider $G=\underset{n-\mathrm{times}}{(\underbrace{A\times\dots\times A})}\rtimes H=:A\wr H$,
the \emph{wreath product}. Describe $\mbox{Irr}(G)$ using Theorem
\ref{thm:2.3}. 

\begin{rem*}
$\mbox{Syl}_{p}(S_{p^{2}})\cong C_{p}\wr C_{p}$. (prove it!)
\end{rem*}
\end{itemize}
\end{xca*}

\subsection{Application III: Clifford Theory (induction from and restriction
to normal subgroups)}
\begin{thm}[Clifford]
Let $G$ be any finite group, $N\lhd G$, $F$ any fields, and $\rho$
any irreducible representations of $G$ over $F$, (equivalently a
simple $F[G]$-module). Then $\rho\downarrow_{N}=\oplus\tau_{i}^{\oplus e}$,
where $\tau_{i}$ are simple $F[N]$-modules, that form a \uline{single
orbit} under the $G$-action. \end{thm}
\begin{proof}
Let $\tau_{1}$ be a simple quotient submodule of $\rho\downarrow_{N}$.
So then $\Hom_{N}(\rho\downarrow_{N},\tau_{1})$ is non-trivial. By
Frobenius reciprocity, $\Hom_{G}(\rho,\tau_{1}\uparrow^{G})$ is non-trivial.
Equivalently, $\rho$ is isomorphic to a submodule of $\tau_{1}\uparrow^{G}$.
So $\rho\downarrow_{N}$ is a submodule of 
\begin{eqnarray*}
\tau_{1}\uparrow^{G}\downarrow_{N} & = & \bigoplus_{N\backslash G/N}\,^{g}\tau_{1}\downarrow_{N\cap gNg^{-1}}\uparrow^{N}\\
 & = & \bigoplus_{g\in G/N}\,^{g}\tau_{1}
\end{eqnarray*}
As $\tau_{1}$ is simple, all $\,^{g}\tau_{1}$ are simple. Using
the exercise below, we see that $\rho\downarrow_{N}=\oplus_{\mathrm{some\,}g}\,^{g}\tau_{1}$.
Note that $\Hom_{N}(\rho\downarrow_{N},\tau_{1})=\Hom_{N}(\,^{g}\rho\downarrow_{N},\,^{g}\tau_{1})$.
If $\rho\downarrow_{N}\cong\oplus\tau_{i}^{\oplus e_{i}}$ then $\Hom_{N}(\rho\downarrow_{N},\tau_{i})\cong D^{\oplus e_{i}}$,
where $D=\End_{N}(\tau_{i})$. So $e_{i}$ are the same for all the
distinct conjugates of $\tau_{1}$. \end{proof}
\begin{xca*}
Submodules of semisimple modules are semisimple.
\end{xca*}

\begin{xca*}
Prove the above theorem over $\cc$, using characters.\end{xca*}
\begin{example*}
~
\begin{itemize}
\item Let $G=S_{3}$ and $N=C_{2}$. Consider the $2$-dimensional irreducible
representation, $\rho$, over $\cc$. Then $\rho\downarrow_{N}=\chi\oplus\overline{\chi}$
where $\chi$ is defined by $(123)\mapsto e^{2\pi i/3}$. Note $\,^{(12)}\chi=\overline{\chi}$. 
\item Let $G=S_{n}$ and $N=A_{n}$. 
\end{itemize}
\end{example*}
Now we want to translate Clifford's theorem into a statement about
induction.
\begin{example*}
$G=C_{3}\rtimes C_{4}=\left\langle x,y|x^{3}=y^{4}=1,yxy^{-1}=x^{-1}\right\rangle $,
where $C_{4}$ acts on $C_{3}$ through the quotient $C_{4}/C_{2}$.
Let $N=C_{3}$, $\chi$ a non-trivial $1$-dimensional character of
$C_{3}$. We investigate $\chi\uparrow^{G}$ in two steps. Consider
$\chi\uparrow^{G}=\chi\uparrow^{C_{3}\rtimes C_{2}=C_{6}}\uparrow^{G}=(\tau\oplus\tau')\uparrow^{G}$
where $\tau$ and $\tau'$ are distinct irreducible $\mbox{Irr}(C_{6})$.
Both $\tau\uparrow^{G}$ and $\tau'\uparrow^{G}$ are irreducible
by the 2nd exercise sheet\end{example*}
\begin{defn*}
Let $N\lhd G$ and $\chi$ an irreducible character of $N$. The \emph{inertia}
subgroup of $\chi$ in $G$ is $I_{G}(\chi)=\mbox{Stab}_{G}(\chi)=\{g\in G|\,^{g}\chi=\chi)=\{g\in G|\chi(g^{-1}ng)=\chi(n)\,\forall n\in N\}$.\end{defn*}
\begin{thm}
Let $N\lhd G$, $\chi\in\mbox{Irr}(N)$, $T=I_{G}(\chi)\geq N$. Let
$\tau$ be an irreducible summand of $\chi\uparrow^{\tau}.$
\begin{enumerate}
\item $\rho=\tau\uparrow^{G}$ is irreducible
\item $\tau\to\tau\uparrow^{G}$ is a bijection between the distinct irreducible
summand of $\chi\uparrow^{T}$ and those of $\chi\uparrow^{G}$
\item $\rho\downarrow_{T}=\tau+$\emph{(stuff that is disjoint from $\chi\uparrow^{T}$)}.
By disjoint we mean $\psi\in\mbox{Irr}(T)$ such that $\left\langle \psi,\chi\uparrow^{T}\right\rangle =0$.
\item $\left\langle \rho\downarrow_{N},\chi\right\rangle =\left\langle \tau\downarrow_{N},\chi\right\rangle $. 
\end{enumerate}
\end{thm}
\begin{proof}
First note that $\tau\downarrow_{N}=e\cdot\chi$, and hence $\,^{g}\tau\downarrow_{N}=e\cdot\,^{g}\chi$.
If $g\notin T$ then $\,^{g}\chi\neq\chi$, hence $\left\langle \tau\downarrow_{N},\,^{g}\tau\downarrow_{N}\right\rangle =0$
if $g\notin T$. Now compute
\begin{eqnarray*}
\left\langle \tau\uparrow^{G},\tau\uparrow^{G}\right\rangle _{G} & = & \left\langle \tau\uparrow^{G}\downarrow_{T},\tau\right\rangle _{T}\\
 & = & \sum_{g\in T\backslash G/T}\left\langle \,^{g}\tau\downarrow_{T\cap gTg^{-1}}\uparrow^{T},\tau\right\rangle _{T}\\
 & = & 1+\sum_{\mathrm{some}g\notin T}\left\langle \,^{g}\tau\downarrow_{T\cap gTg^{-1}},\tau\downarrow_{T\cap gTg^{-1}}\right\rangle \quad\quad T\cap gTg^{-1}\geq N\\
 & = & 1+0
\end{eqnarray*}


This proves $1$. We have that $2.$ and $4.$ follows from $3.$~.To
prove $3.$~, 
\begin{eqnarray*}
\rho\downarrow_{T} & = & \tau\uparrow^{G}\downarrow_{T}\\
 & = & \sum_{T\backslash G/T}\,^{g}\tau\downarrow_{T\cap gTg^{-1}}\uparrow^{T}\\
 & = & \tau+\sum_{\mathrm{some\,}g\notin T}\,^{g}\tau\downarrow_{T\cap gTg^{-1}}\uparrow^{T}
\end{eqnarray*}

\end{proof}

\subsection{Application IV: Frobenius groups}
\begin{thm}[Frobenius]
 \label{thm:2.6}Suppose $H\leq G$ is such that $H\cap gHg^{-1}=\{1\}$
for all $g\notin H$ \emph{($H$ }is called a \emph{Frobenius complement)}.
Then there exists $N\lhd G$ such that $G=N\rtimes H$. 
\end{thm}
To prove this we will need several lemma. Define 
\[
N=G\backslash\left(\cup_{g}gHg^{-1}\right)\cup\{1\}.
\]

\begin{lem}
Let $N$ be defined as above, $\left|N\right|=\frac{\left|G\right|}{\left|H\right|}$,
also is $M\lhd G$ intersect $H$ trivially, then $M\subset N$\end{lem}
\begin{proof}
The second part is by definition of $N$. For the first part
\begin{eqnarray*}
\left|N\right| & = & \left|G\right|-\frac{\left|G\right|}{\left|H\right|}\left(\left|H\right|-1\right)\\
 & = & \left|G\right|-\left|G\right|+\frac{\left|G\right|}{\left|H\right|}
\end{eqnarray*}
\end{proof}
\begin{lem}
\label{lem:2.8}Let $G$ and $H$ be as in Theorem \ref{thm:2.3}.
Let $\theta$ be a class function on $H$ with $\theta(1)=0$. Then
$\theta\uparrow^{G}\downarrow_{H}=\theta$. \end{lem}
\begin{proof}
By Machke we have 
\begin{eqnarray*}
\theta\uparrow^{G}\downarrow_{H} & = & \sum_{g\in H\backslash G/H}\,^{g}\theta\downarrow_{H\cap gHg^{-1}}\uparrow^{H}\\
 & = & \theta+\sum_{g\notin H}\,^{g}\theta\downarrow_{H\cap gHg^{-1}}\uparrow^{H}\\
 & = & \theta+\sum0_{\{1\}}\uparrow^{H}\\
 & = & \theta
\end{eqnarray*}

\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:2.6}]
 Motivation: if $\chi\in\mbox{Irr}(G)$ is such that $\ker\chi\supseteq N$,
then $\chi\downarrow_{H}$ is irreducible. We want to recover $\chi$
from $\chi\downarrow_{H}$. 

Let $\mathbb{I}_{H}\neq\phi\in\mbox{Irr}(H)$. Define $\theta_{\phi}=\phi-\phi(1)\mathbb{I}_{H}$,
hence $\theta_{\phi}(1)=0$. Note that 
\begin{eqnarray*}
\left\langle \theta_{\phi}\uparrow^{G},\mathbb{I}_{G}\right\rangle _{G} & = & \left\langle \theta_{\phi},\mathbb{I}_{H}\right\rangle _{H}\\
 & = & -\phi(1).
\end{eqnarray*}
Let us set $\chi_{\phi}=\phi_{\psi}\uparrow^{G}+\phi(1)\mathbb{I}_{G}$,
hence $\left\langle \chi_{\phi},\mathbb{I}_{G}\right\rangle _{G}=0$.
Furthermore
\begin{eqnarray*}
\left\langle \chi_{\phi},\chi_{\phi}\right\rangle +\phi(1)^{2} & = & \left\langle \theta_{\phi}\uparrow^{G},\theta_{\phi}\uparrow^{G}\right\rangle _{G}\\
 & = & \left\langle \theta_{\phi},\theta_{\phi}\uparrow^{G}\downarrow_{H}\right\rangle _{H}\\
 & \underset{\mathrm{Lemma\,\ref{lem:2.8}}}{=} & \left\langle \theta_{\phi},\theta_{\phi}\right\rangle _{H}\\
 & \underset{\mathrm{def\,\theta_{\phi}}}{=} & \underset{=1}{\underbrace{\left\langle \phi,\phi\right\rangle }}+\phi(1)^{2}.
\end{eqnarray*}
Hence $\left\langle \chi_{\phi},\chi_{\phi}\right\rangle _{G}=1$
and is irreducible. Now $\theta_{\phi}$ is the difference of two
characters, therefore so is $\theta_{\phi}\uparrow^{G}$, and hence
so is $\chi_{\rho}$. But $\left\langle \chi_{\phi},\chi_{\phi}\right\rangle =1$,
hence $\pm\chi_{\phi}\in\mbox{Irr}(G)$. But also, 
\begin{eqnarray*}
\chi_{\phi}\downarrow_{H} & = & \theta_{\phi}\uparrow^{G}\downarrow_{H}+\theta(1)\cdot\mathbb{I}_{G}\downarrow_{H}\\
 & = & \theta_{\phi}+\phi(1)\cdot\mathbb{I}_{G}\\
 & = & \theta.
\end{eqnarray*}
So $\chi_{\theta}\in\mbox{Irr}(G)$. 

Define 
\[
M=\bigcap_{\mathbb{I}_{H}\neq\phi\in\mathrm{Irr}(H)}\ker(\chi_{\phi})\lhd G.
\]

\begin{claim*}
$M\cap H=\{1\}$ (and hence $M\subseteq N$)
\end{claim*}

Indeed, if $h\in H$, then $\chi_{\phi}(h)=\phi(h)$. So $H\cap M=\cap_{\phi\in\mathrm{Irr}(H)}\ker\phi=\{1\}$
\begin{claim*}
$N\subseteq M$. 
\end{claim*}

If $n\notin gHg^{-1}$$ $, then 
\begin{eqnarray*}
\chi_{\phi}(n) & = & \theta_{\phi}\uparrow^{G}(n)+\phi(1)\cdot\mathbb{I}_{G}\\
 & = & \phi\uparrow^{G}(n)-\phi(1)\cdot\mathbb{I}\uparrow^{G}(n)+\phi(1)\cdot\mathbb{I}_{G}\\
 & \underset{\mathrm{def\, of\, irreducible\, char}}{=} & 0+0+\phi(1)\\
 & = & \chi_{\phi}(1)
\end{eqnarray*}
for all $\mathbb{I}_{H}\neq\phi\in\mathrm{Irr}(H)$. Hence $n\in M$.

So $N=M\lhd G$ and we are done.
\end{proof}
\newpage{}


\section{Tensor Products, Frobenius - Schur indicators and much more}

Let $G$ be a finite group and $K$ be any field.

\textbf{Motivation}: If $\chi,\phi$ are characters of $G$, then
so is $\chi+\phi$. But what about $\chi\cdot\phi$?
\begin{defn}
Let $V$ and $W$ be vector spaces (over $K$). The \emph{tensor product
$V\otimes W$ }is the vector space spanned by ``symbols'' $v\otimes w$
with $v\in V,w\in W$, with relations 
\begin{itemize}
\item $(kv)\otimes w=v\otimes(kw)=k(v\otimes w)$
\item $(v_{1}+v_{2})\otimes w=v_{1}\otimes w+v_{2}\otimes w$
\item $v\otimes(w_{1}+w_{2})=v\otimes w_{1}+v\otimes w_{2}$
\end{itemize}
\end{defn}
\begin{fact*}
If $v_{1},\dots,v_{n}$ is a basis for $V$ and $w_{1},\dots,w_{m}$
a basis of $W$, then $v_{i}\otimes w_{j}$ for $1\leq i\leq n$,$1\leq j\leq m$
is a basis for $V\otimes W$. \end{fact*}
\begin{prop}
Tensor products have the following properties:
\begin{itemize}
\item $(V\otimes W)\otimes U\cong V\otimes(W\otimes U)$
\item $(V\oplus U)\otimes W\cong V\otimes W\oplus U\otimes W$
\end{itemize}
\end{prop}
\begin{proof}
Check that: 
\begin{itemize}
\item $(v\otimes w)\otimes u\mapsto v\otimes(w\otimes u)$
\item $(u,v)\otimes w\mapsto(v\otimes w,u\otimes w)$
\end{itemize}
are isomorphisms.
\end{proof}
If $V,W$ are $G$-representation, then $G$ acts on $V\otimes W$
via $g\cdot(v\otimes w)=g\cdot v\otimes g\cdot w$. Suppose that $g$
is represented by $A=(a_{ij})_{1\leq i,j\leq n}$ on $V$ with respect
to $v_{1},\dots,v_{n}$, and it is represented by $B=(b_{ij})_{1\leq i,j\leq m}$
on $W$ with respect to $w_{1},\dots,w_{m}$. Then $g\cdot(v_{i}\otimes w_{k})=gv_{i}\otimes gw_{k}=\left(\sum a_{ij}v_{j}\right)\otimes\left(\sum b_{kl}w_{l}\right)=\sum_{j,l}a_{ij}b_{kl}(v_{j}\otimes w_{l})$.
So with respect to the basis $v_{1}\otimes w_{1},\dots,v_{1}\otimes w_{m},v_{2}\otimes w_{1},\dots,v_{n}\otimes w_{n}$
of $V\otimes W$, $g$ is represented by
\[
\left(\begin{array}{c|c|c|c}
a_{11}B & a_{12}B & \dots & a_{1n}B\\
\hline \vdots &  & \\
\hline a_{n1}B &  &  & a_{nn}B
\end{array}\right)=:A\otimes B
\]

\begin{example*}
Let $G=S_{3}$ and take $\rho$ to be the standard representation,
that is $\rho$ is defined by 
\begin{eqnarray*}
(123) & \mapsto & \begin{pmatrix}-1 & -1\\
1 & 0
\end{pmatrix}\\
(12) & \mapsto & \begin{pmatrix}1 & 1\\
0 & -1
\end{pmatrix}.
\end{eqnarray*}
The basis of $\rho\otimes\rho$ is $v_{1}\otimes v_{1},v_{1}\otimes v_{2},v_{2}\otimes v_{1},v_{2}\otimes v_{2}$.
So $\rho\otimes\rho$ is defined by
\begin{eqnarray*}
(123) & \mapsto & \left(\begin{array}{cc|cc}
1 & 1 & 1 & 1\\
-1 & 0 & -1 & 0\\
\hline -1 & -1 & 0 & 0\\
1 & 0 & 0 & 0
\end{array}\right)\\
(12) & \mapsto & \left(\begin{array}{cc|cc}
1 & 1 & 1 & 1\\
0 & -1 & 0 & -1\\
\hline 0 & 0 & -1 & -1\\
0 & 0 & 0 & 1
\end{array}\right)
\end{eqnarray*}

\end{example*}
If $V,W$ are complex representations with characters $\chi,\phi$
respectively then the character $\tau$ of $V\otimes W$ is
\begin{eqnarray*}
\tau(g) & = & \sum_{1\leq i\leq n,1\leq k\leq m}a_{ii}b_{kk}\\
 & = & \left(\sum_{i}a_{ii}\right)\cdot\left(\sum_{k}b_{kk}\right)\\
 & = & \chi(g)\cdot\phi(g).
\end{eqnarray*}



\subsubsection*{Aside: Duals and homomorphism spaces}
\begin{defn}
Let $V$ be a representation of $ $ over $K$. The \emph{dual} representation
is $V^{*}=\{f:V\to K|f(v+\alpha w)=f(v)+\alpha f(w)\forall\alpha\in K,v,w\in V\}$
with $G$ action on $V^{*}$ by $(g\cdot f)(v)=f(g^{-1}v)$, i.e.,
$g\cdot f:v\mapsto f(g^{-1}v)\in K$\end{defn}
\begin{lem}
If $V$ is a complex representation with character $\chi$, then the
character of $V^{*}$ is $\overline{\chi}$.\end{lem}
\begin{proof}
Let $v_{1},\dots,v_{n}$ be a basis of $V$. Take the dual basis $V^{*}$
to be $f_{1},\dots,f_{n}$ such that $f_{i}(v_{j})=\delta_{ij}=\begin{cases}
1 & i=j\\
0 & i\neq j
\end{cases}$. Assume without loss of generality that $g\in G$ is represented
by $\begin{pmatrix}\alpha_{1}\\
 & \ddots\\
 &  & \alpha_{n}
\end{pmatrix}$ with respect to $v_{1},\dots,v_{n}$. Then check that $g$ in represented
by $\begin{pmatrix}\alpha_{1}^{-1}\\
 & \ddots\\
 &  & \alpha_{n}^{-1}
\end{pmatrix}$ with respect to $f_{1},\dots,f_{n}$. Since $\alpha_{i}$ are roots
of unity, $\alpha_{i}^{-1}=\overline{\alpha_{i}}$.\end{proof}
\begin{cor}
$V\cong V^{*}$ (as representations over $\cc$) if and only if $\chi$
is $\rr$-valued.\end{cor}
\begin{defn}
If $V,W$ are representations of $G$ over $K$, then $\Hom_{K}(V,W)$
is a $G$-representation via $(g\cdot f)(v)=g\cdot f(g^{-1}v)$.\end{defn}
\begin{lem}
If $V$ and $W$ are representations over $\cc$ with characters $\chi,\phi$
respectively then the character $\tau$ of $\Hom_{K}(V,W)$ is $\overline{\chi}\cdot\phi$. \end{lem}
\begin{proof}
Use matrices with respect to basis $f_{ik}:v_{j}\mapsto\delta_{ij}w_{k}$. 
\end{proof}
In particular, over $\cc$, $V\otimes W\cong\Hom(V^{*},W)$ (by comparing
characters)
\begin{lem}
Over any field $K$, $V\otimes_{K}W\cong\Hom_{K}(V^{*},W)$.\end{lem}
\begin{proof}
Check that $V\otimes W\to\Hom(V^{*},W)$ defined by $v\otimes w\mapsto(f\mapsto f(v)\cdot w)$
is an isomorphism (of $G$-representations).\end{proof}
\begin{rem*}
The fixed subspace of $\Hom(V,W)$ under the $G$-action is 
\[
\Hom_{G}(V,W)=\{f:V\to W\mathrm{\, linear}|g\cdot f(v)=f(g\cdot v)\forall v\in V,g\in G\}
\]

\end{rem*}
\textbf{Assume}: For the rest of this chapter that the characteristic
of $K$ is $0$
\begin{notation*}
$V^{\otimes n}=\underset{n\mathrm{\, times}}{\underbrace{V\otimes V\otimes\dots\otimes V}}$\end{notation*}
\begin{example*}
Going back to the case that $G=S_{3}$, $\rho$ the standard representation.
The character table is\\
\begin{tabular}{|c|c|c|c|}
\hline 
 & $1$ & $(123)$ & $(12)$\tabularnewline
\hline 
\hline 
$\rho^{\otimes2}$ & $4$ & $1$ & $0$\tabularnewline
\hline 
\end{tabular}

\begin{eqnarray*}
\left\langle \rho^{\otimes2},\mathbb{I}\right\rangle  & = & \frac{1}{6}(4+2)=1\\
\left\langle \rho^{\otimes2},\mathrm{sign}\right\rangle  & = & \frac{1}{6}(4+2)=1\\
\left\langle \rho^{\otimes2},\rho\right\rangle  & = & 1
\end{eqnarray*}
Hence we have that $\rho^{\otimes2}=\mathbb{I}+\epsilon+\rho$.
\end{example*}
$V^{\otimes n}$ carries an action of $S_{n}$, $\sigma(v_{1}\otimes\dots\otimes v_{n})=v_{\sigma^{-1}(1)}\otimes\dots\otimes v_{\sigma^{-1}(n)}$
for all $\sigma\in S_{n},v_{i}\in V$. This action commutes with the
$G$-action. (So we get an action of $G\times S_{n}$).
\begin{claim*}
If $\rho_{1},\dots,\rho_{k}$ is a complete set of irreducible $K$-representations
of $S_{n},$then $V^{\otimes n}=\oplus_{i=1}^{k}V_{(\rho_{i})}^{\otimes n}$
as $G$-representations. \end{claim*}
\begin{proof}
If $g\in G$, then (for $t\in V^{\otimes n}$) $t\mapsto g\cdot t$
is a homomorphism of $S_{n}$-representation. So if $t\in V_{(\rho_{i})}^{\otimes n}$,
then the projection of $g\cdot t$ to any $V_{(\rho_{j})}^{\otimes n}$
for $j\neq i$ is $0$ by Schur's lemma. \end{proof}
\begin{example*}
Let $v_{1},\dots,v_{n}$ be a basis for $V$. We consider $S_{2}$
and hence $V^{\otimes2}$, which has basis $v_{i}\otimes v_{j}$ for
$1\leq i,j\leq n$.
\begin{itemize}
\item $V_{(\mathbb{I})}^{\otimes2}$ has basis $v_{i}\otimes v_{j}+v_{j}\otimes v_{i}$
for $1\leq i\leq j\leq n$. 
\item $V_{(\mathrm{sign})}^{\otimes2}$ has basis $v_{i}\otimes v_{j}-v_{j}\otimes v_{i}$
for $1\leq i<j\leq n$. 
\end{itemize}
So in part, $V_{(\mathbb{I})}^{\otimes2}$ has dimension $\frac{n(n+1)}{2}$,
$V_{(\mathrm{sign)}}^{\otimes2}$ has dimension $\frac{n(n-1)}{2}$. \end{example*}
\begin{defn}
$V_{(\mathbb{I})}^{\otimes2}$ is called the \emph{symmetric square
of $V$}, written $S^{2}V$. $V_{(\mathrm{sign)}}^{\otimes2}$ is
called the \emph{alternating square} \emph{of $V$, }written $\wedge^{2}V$.\end{defn}
\begin{lem}
The characters of $S^{2}V$ and $\wedge^{2}V$ are 
\begin{eqnarray*}
\chi_{S^{2}V}(g) & = & \frac{1}{2}\left(\chi(g)^{2}+\chi(g^{2})\right)\\
\chi_{\wedge^{2}V}(g) & = & \frac{1}{2}\left(\chi(g)^{2}-\chi(g^{2})\right)
\end{eqnarray*}
\end{lem}
\begin{proof}
Let $g\in G$, take a basis of $V$ to be $v_{1},\dots,v_{n}$ such
that $g=\mathrm{diag}(\alpha_{1},\dots,\alpha_{n})$ with respect
to that basis. Then $g\cdot(v_{i}\otimes v_{j}+v_{j}\otimes v_{i})=\alpha_{i}\alpha_{j}(v_{i}\otimes v_{j}+v_{j}\otimes v_{i})$.
So 
\begin{eqnarray*}
\chi_{S^{2}V}(g) & = & \sum_{1\leq i\leq j\leq n}\alpha_{i}\alpha_{j}\\
 & = & \frac{1}{2}\left(\chi(g)^{2}+\chi(g^{2})\right).
\end{eqnarray*}
A similar calculation shows that 
\begin{eqnarray*}
\chi_{S^{2}V}(g) & = & \sum_{1\leq i<j\leq n}\alpha_{i}\alpha_{j}\\
 & = & \frac{1}{2}\left(\chi(g)^{2}-\chi(g^{2})\right).
\end{eqnarray*}
\end{proof}
\begin{rem*}
$S^{2}\chi+\wedge^{2}\chi=\chi^{2}$. \end{rem*}
\begin{defn}
Let $\chi$ be an irreducible character of $G$, the \emph{Frobenius
- Schur} \emph{indicator} of $\chi$ is 
\[
s_{2}(\chi):=\frac{1}{\left|G\right|}\sum_{g\in G}\chi(g^{2}).
\]
\end{defn}
\begin{thm}
$s_{2}(\chi)\in\{0,1,-1\}$ and $s_{2}(\chi)=0$ if and only if $\chi\neq\overline{\chi}$.\end{thm}
\begin{proof}
Note that $\chi(g^{2})=S^{2}\chi(g)-\wedge^{2}\chi(g)$, so $s_{2}(\chi)=\left\langle S^{2}\chi,\mathbb{I}\right\rangle _{G}-\left\langle \wedge^{2}\chi,\mathbb{I}\right\rangle _{G}$.
\begin{claim*}
$\left\langle \underset{\chi^{2}}{\underbrace{S^{2}\chi+\wedge^{2}\chi}},\mathbb{I}\right\rangle _{G}=0$
or $1$
\end{claim*}

Let $V$ be the vector space attached to $\chi$. Then $\left\langle V^{\otimes2},\mathbb{I}\right\rangle =\dim\left((V\otimes V)^{G}\right)=\Hom_{G}(V^{*},V)$
(where $(-)^{G}$ are elements fixed by $G$.) By Schur's lemma, these
$G$-homomorphism are $1$-dimensional if $V^{*}\cong V$ (i.e., if
$\chi\cong\overline{\chi}$) and $0$ otherwise. 

\end{proof}
So what does this $\pm1$ mean for $s_{2}$ of real-valued characters?
\begin{example*}
Let $G=S_{3}$ and $\chi$ the standard character, i.e., $\chi(1)=2,\chi((123))=-1$
and $\chi((12))=0$. Now $s_{2}(\chi)=\frac{1}{6}(2+3\cdot2+2\cdot(-1))=1$.
\end{example*}

\subsubsection*{Pairings on vector spaces}
\begin{defn}
Let $V$ be a vector space over a field $K$. A \emph{pairing} on
$V$ is a bilinear map $\left\langle \,,\,\right\rangle :V\times V\to K$. 
\end{defn}
Given a paring, we get a linear map $V\to V^{*}$ defined by $v\mapsto(w\mapsto\left\langle v,w\right\rangle )$.
Conversely, given a homomorphism $\phi:V\to V^{*}$, we can define
a pairing by $\left\langle v,w\right\rangle =\phi(v)(w)$. These operations
are inverses to each other.
\begin{defn}
A pairing is non-degenerate if, given $v\in V$, $\left\langle v,w\right\rangle =0\forall w\in V$
then $v=0$. (This is equivalent to ``right non-degenerate'' for
finite dimensional vector spaces.)
\end{defn}
In the language of $\phi:V\to V^{*}$, this is equivalent to $\phi$
being an isomorphism.
\begin{defn}
$\left\langle \,,\,\right\rangle $ is \emph{symmetric }if $\left\langle v,w\right\rangle =\left\langle w,v\right\rangle $
for all $v,w\in V$

It is \emph{alternating} if $\left\langle v,w\right\rangle =-\left\langle w,v\right\rangle $
for all $v,w\in V$. 

Let $G$ act on $V$. We say that $\left\langle \,,\,\right\rangle $
is $G$\emph{-invariant} if $\left\langle gv,gw\right\rangle =\left\langle v,w\right\rangle $
$\forall v,w\in V,\forall g\in G$.
\end{defn}
This is equivalent to $\phi:V\to V^{*}$ being a $G$-homomorphism.
So we have a bijection between $\Hom_{G}(V,V^{*})$ and $G$-invariant
pairings on $V$. 

If $U,V,W$ are vector spaces, then bilinear maps $U\times V\to W$
are ``the same things as'' linear maps $U\otimes V\to W$ in the
following sense: there is a canonical map $U\times V\to U\otimes V$
defined by $(u,v)\mapsto u\otimes v$, and given any bilinear map
\[
\xymatrix{U\times V\ar[r]\ar[rd] & U\otimes V\ar@{-->}[d]^{\exists!}\\
 & W
}
\]
so that the diagram commutes.

In particular, bilinear maps $V\times V\to K$ correspond canonically
to maps $V\otimes V\to K$, and the set of $G$-invariant pairings
on $V$ is in bijection with $\Hom_{G}(V\otimes V,\mathbb{I})$. 
\begin{itemize}
\item The pairing is symmetric if the map $V\otimes V\to K$ is $0$ on
$\wedge^{2}V$, i.e., such pairings correspond to maps $V\otimes V/\wedge^{2}V\cong S^{2}V\to K$.
\item The pairing is alternating if the map $V\otimes V\to K$ is $0$ on
$S^{2}V$, i.e., such pairings correspond to maps $V\otimes V/S^{2}V\cong\wedge^{2}V\to K$. 
\item $V\cong V^{*}$ if and only if there exists a $G$-invariant non-degenerate
pairing on $V$. Conversely, given $f:V\to V^{*}$, take $\left\langle u,v\right\rangle =f(u)v$
if and only if $\chi=\overline{\chi}$. If $V\cong V^{*}$, then $\Hom_{G}(V,V^{*})$
is $1$-dimensional
\item If $\left\langle \,,\,\right\rangle $ is a non-degenerate $G$-invariant
pairings, we can write it 
\[
\left\langle u,v\right\rangle =\frac{1}{2}\underset{\left\langle \,,\,\right\rangle _{s}}{\underbrace{\left(\left\langle u,v\right\rangle +\left\langle v,u\right\rangle \right)}}+\frac{1}{2}\underset{\left\langle \,,\,\right\rangle _{a}}{\underbrace{\left(\left\langle u,v\right\rangle -\left\langle v,u\right\rangle \right)}}
\]
We cannot have both $\left\langle \,,\,\right\rangle _{s}$ and $\left\langle \,,\,\right\rangle _{a}$
non-degenerate, since they would have to be multiple of each other.
Another way of saying this: since $V\otimes V=S^{2}V\oplus\wedge^{2}V$
we either have 

\begin{itemize}
\item $\dim(\Hom_{G}(S^{2}V,\mathbb{I}))=1$ and $\Hom_{G}(\wedge^{2}V,\mathbb{I})=0$
or
\item $\dim(\Hom_{G}(S^{2}V,\mathbb{I}))=0$ and $\Hom_{G}(\wedge^{2}V,\mathbb{I})=1$
\end{itemize}
\item There exists a symmetric $G$-invariant, non-degenerate pairing on
$V$ if and only if $\Hom_{G}(S^{2}V,\mathbb{I})\neq0$, if and only
if $V\cong V^{*}$ and $\Hom_{G}(\wedge^{2}V,\mathbb{I})=0$. 


Explicitly, if $f:S^{2}V\to\mathbb{I}$, construct the pairing by
$ $$\left\langle u,v\right\rangle =f(u\otimes v+v\otimes u)$

\item There exists an alternating $G$-invariant, non-degenerate pairing
on $V$ if and only if $\Hom_{G}(\wedge^{2}V,\mathbb{I})\neq0$, if
and only if $V\cong V^{*}$ and $\Hom_{G}(S^{2}V,\mathbb{I})=0$. \end{itemize}
\begin{thm}
Let $V$ be an irreducible complex representation of $G$.
\begin{enumerate}
\item There exists a non-degenerated $G$-invariant pairing on $V$ if and
only if $V\cong V^{*}$ if and only if $\chi\cong\overline{\chi}$
\item There exists a symmetric non-degenerate $G$-invariant pairing on
$V$ if and only if there exists a basis on $V$ with respect to which
$G$ is represented by real matrices.
\item There exists an alternating non-degenerate $G$-invariant pairing
on $V$ if and only if $\chi$ is real-valued, but $V$ cannot be
defined over $\rr$ (in the above sense)
\end{enumerate}
\end{thm}
\begin{example*}
Let $G=D_{10}=\left\langle \tau,\sigma|\tau^{2}=\sigma^{5}=1,\tau\sigma\tau=\sigma^{-1}\right\rangle $.
Rotations and reflections of the $5$-gon gives the following representation:
$\tau\mapsto\begin{pmatrix}1 & 0\\
0 & -1
\end{pmatrix}$, $\sigma\mapsto\begin{pmatrix}e^{2\pi i/5} & 0\\
0 & e^{-2\pi i/5}
\end{pmatrix}.$ But with respect to ``the right'' basis, this can be written as
$\tau\mapsto\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}$, $\sigma\mapsto\begin{pmatrix}\cos\frac{2\pi}{5} & \sin\frac{2\pi}{5}\\
-\sin\frac{2\pi}{5} & \cos\frac{2\pi}{5}
\end{pmatrix}$\end{example*}
\begin{thm*}[Part 2]

\begin{enumerate}
\item holds if and only if $s_{2}(\chi)=\left\langle S^{2}\chi,\mathbb{I}\right\rangle -\left\langle \wedge^{2}\chi,\mathbb{I}\right\rangle \neq0$
\item holds if and only if $s_{2}(\chi)=1$
\item holds if and only if $s_{2}(\chi)=-1$
\end{enumerate}
\end{thm*}
\begin{proof}[Partial proof]
 Suffices to show 2.~(as we already know 1.~). We will only prove
the implication $\Leftarrow$.

Suppose that $V$ is definable over $\rr$. This means that if $V$is
regarded as an $\rr$-vector space (of twice its dimension over $\cc$),
then $V=W+iW$, where $W$ is invariant under $G$. Let $(\,,\,)$
be any positive-definite pairing on $W$. Define $\left\langle \,,\,\right\rangle _{1}$
on $W$ by $\left\langle u,v\right\rangle _{1}=\frac{1}{\left|G\right|}\sum_{g\in G}(g\cdot u,g\cdot v)$.
This is clearly $G$-invariant and positive-definite.

Define $\left\langle u,v\right\rangle _{2}=\left\langle u,v\right\rangle _{1}+\left\langle v,u\right\rangle _{1}$.
Then this is still $G$-invariant and positive-definite, furthermore
it is symmetric. Define $\left\langle u,v\right\rangle _{3}$ on $V$
by $\left\langle u+iu',v+iv'\right\rangle _{3}=\left\langle u,v\right\rangle _{2}-\left\langle u',v'\right\rangle _{2}+i\left(\left\langle u,v'\right\rangle _{2}+\left\langle u',v\right\rangle _{2}\right)$,
this is the required pairing. 

For the implication ``$\Rightarrow$'' see for example Serre, chapter
2, Theorem 31, or Curtis - Reiner, Vol II, Section 73.13.
\end{proof}

\subsubsection*{Application}

Define $r_{2}(g)=\#\{h\in G|h^{2}=g\}$. First observe: $h\mapsto xhx^{-1}$
gives a bijection between square roots of $h^{2}$ and those of $xh^{2}x^{-1}$.
So $r_{2}$ is a class function. Thus $r_{2}=\sum_{\chi\in\mathrm{Irr}(G)}\alpha_{\chi}\cdot\chi$
for $\alpha_{\chi}\in\cc$. Now
\begin{eqnarray*}
\alpha_{\chi} & = & \left\langle r_{2},\chi\right\rangle _{G}\\
 & = & \frac{1}{\left|G\right|}\sum_{g\in G}r_{2}(g)\chi(g)\\
 & = & \frac{1}{\left|G\right|}\sum_{g\in G}\#\{h\in G|h^{2}=g\}\chi(g)\\
 & = & \frac{1}{\left|G\right|}\sum_{g\in G}\sum_{h\in G,h^{2}=g}\chi(g)\\
 & = & \frac{1}{\left|G\right|}\sum_{g\in G}\sum_{h\in G,h^{2}=g}\chi(h^{2})\\
 & = & \frac{1}{\left|G\right|}\sum_{h\in G}\chi(h^{2})\\
 & = & s_{2}(\chi)\in\{-1,0,1\}.
\end{eqnarray*}
Hence $r_{2}=\sum_{\rho\mathrm{\, irr\, reps\, realisable\, over\,}\rr}\chi_{\rho}-\sum_{\mathrm{self\, dual\, irr\, reps\, not\, realisable\, over\,\rr}}\chi_{\rho}.$
\begin{cor}
Let $G$ be an abelian group, then $r_{2}$ takes its maximum at the
identity element.\end{cor}
\begin{proof}
$r_{2}(g)=\left|\sum_{\mathrm{real}\,\chi}\chi(g)\right|\leq\sum_{\mathrm{real\,}\chi}\left|\chi(g)\right|\leq\sum_{\mathrm{real}\,\chi}\chi(1)=r_{2}(1)$.
\end{proof}
Similarly for dihedral groups, symmetric groups, alternating groups,
and for all groups that don't have $\chi$ with $s_{2}(\chi)=-1$. 
\begin{rem*}
One can define higher Frobenius - Schur indicators:
\[
s_{k}(\chi):=\frac{1}{\left|G\right|}\sum_{g\in G}\chi(g^{k}),\, k\in\nn.
\]
For $k\geq3$, these are unbounded as $G$ varies (hint for proof:
consider the Heisenberg group of order $p^{3}$, i.e., $\begin{pmatrix}1 & * & *\\
0 & 1 & *\\
0 & 0 & 1
\end{pmatrix}\subset\GL_{3}(\ff_{p})$)
\end{rem*}
To finish our discussion of $\rr[G]$-modules, we should talk about
Wedderburn components. 

Recall: $R[G]\cong\oplus_{i}M_{n_{i}}(D_{i})$ where $D_{i}$ are
division algebras. In fact $D_{i}=\End_{\rr[G]}(\rho_{i})$, where
$\rho_{i}$ are the distinct simple $\rr[G]$-modules.
\begin{fact*}
The only associative division algebras over $\rr$ are $\rr,\cc,\hh$. \end{fact*}
\begin{thm}
Let $\rho$ be a complex irreducible representation of $G$.
\begin{enumerate}
\item If $\rho\neq\rho^{*}$, then $\rho\oplus\rho^{*}$ is realisable over
$\rr$, is simple as an $\rr[G]$-module and the corresponding Wedderburn
block is isomorphic to $M_{n_{i}}(\cc)$
\item If $\rho$ is realisable over $\rr$, then the Wedderburn component
is isomorphic to $M_{n_{i}}(\rr)$
\item If $\rho\cong\rho^{*}$ but not realisable over $\rr$ (i.e., $\rho$
is \emph{simpletic} or \emph{quaternion}) then $\rho\oplus\rho$ is
realisable over $\rr$, it is simple and the corresponding Wedderburn
block is isomorphic to $M_{n_{i}}(\hh)$. 
\end{enumerate}
\end{thm}
\begin{proof}[Proof (sketch)]
 In cases $1$.~ and $3$.~, to prove realisability over $\rr$,
we construct a symmetric non-degenerate, $G$-invariant pairing.

E.g., in case $3$.~ let $[\,,\:]$ be a $G$-invariant non-degenerate
alternating pairing on $\rho$. Define $\left\langle \,,\,\right\rangle $
on $\rho\oplus\rho$ by $\left\langle (u_{1},v_{1}),\left(u_{2},v_{2}\right)\right\rangle =\left[u_{1},v_{2}\right]-\left[v_{1},u_{2}\right]$.
We can see that this is symmetric.

Case 1.~ is omitted

To find the corresponding Wedderburn component, notice that $\rr,\cc$
and $\hh$ have different dimensions over $\rr$. So just need to
know $\dim_{\rr}\End_{\rr[G]}(-)$, which we use the following lemma
for.
\begin{lem}
If $\tau$ is an $\rr[G]$-module, $\tau\otimes_{\rr}\cc$ the corresponding
$\cc[G]$-module. Then $\dim_{\rr}\End_{\rr[G]}(\tau)=\dim_{\cc}\End_{\cc[G]}(\tau\otimes_{\rr}\cc)$
\end{lem}
We can calculate the following:
\begin{enumerate}
\item $\rho\neq\rho^{*}$, $\left\langle \rho\oplus\rho^{*},\rho\oplus\rho^{*}\right\rangle =2$
hence $\End_{\rr[G]}=\cc$
\item $\left\langle \rho,\rho\right\rangle _{G}=1$ hence $\End_{\rr[G]}=\rr$
\item $\left\langle \rho\oplus\rho,\rho\oplus\rho\right\rangle _{G}=4$
hence $\End_{\rr[G]}=\hh$.
\end{enumerate}
\end{proof}
Recall that $V^{\otimes n}\cong\oplus_{\chi\in\mathrm{Irr}(S_{n})}V_{(\chi)}^{\otimes n}$.
In general, if $V$ is a $\cc[G]$-module, $V=\oplus_{\chi\in\mathrm{Irr}(G)}V_{(\chi)}$,
to find $V_{(\chi)}\subset V$, use idempotent:
\begin{itemize}
\item $e_{\chi}=\frac{1}{\left|G\right|}\sum_{g\in G}\overline{\chi(g)}\cdot g\in\cc[G]$
- a primitive central idempotent. 
\end{itemize}
Hence $V_{(\chi)}=e_{\chi}\cdot V=\{e_{\chi}\cdot v|v\in V\}$. 
\begin{example*}
Let $G=S_{3},V=\cc[S_{3}]$, $\chi$ be the the standard character
(2-dimensional).
\[
e_{\chi}=\frac{1}{6}\left(2\cdot\id-(123)-(132)\right)\in\cc[G],
\]
e.g, $e_{\chi}\cdot\id=e_{\chi}$
\begin{eqnarray*}
e_{\chi}(12) & = & \frac{1}{6}\left(2\cdot(12)-(23)-(13)\right)\\
e_{\chi}(13) & = & \frac{1}{6}(2\cdot(13)-(12)-(23))
\end{eqnarray*}
etc, we get a $4$ dimensional subrepresentation of $\cc[S_{3}]\cong\rho^{\oplus2}$
(where $\rho$ is the standard representation)

\newpage{}
\end{example*}

\section{Permutation representation, monomial representation, induction theorems}

\textbf{Recall}: Let $X$ be a finite $G$-set, i.e., $X=\{1,\dots,n\}$,
and there is a group homomorphism $G\to S_{n}$. Then $\cc[X]$ is
the associated permutation representation.

Philosophy: these are easy, so we want to express pother representations
in terms of these. 

\textbf{Recall}: If $X=G/H$, then $\cc[X]\cong\mathbb{I}_{H}\uparrow^{G}$.
In particular, $\left\langle \cc[X],\mathbb{I}\right\rangle _{G}=\left\langle \mathbb{I}_{H}\uparrow^{G},\mathbb{I}\right\rangle _{G}=\left\langle \mathbb{I}_{H},\mathbb{I}\right\rangle _{H}=1$.
So we can write $\cc[X]\cong1\oplus\rho$. When is $\rho$ irreducible?
\begin{lem}
If $X$ is transitive, i.e., $\forall x,y\in X$ there exists $g\in G$
such that $g\cdot x=y$, define $H=\mbox{Stab}_{G}(X)$ for a fixed
$x\in X$. Then $X\cong G/H$, i.e., there is a bijection of sets
that commutes with the $G$-action.\end{lem}
\begin{proof}
Define $X\to G/H$ by $g\cdot x\mapsto g\cdot H$. 
\begin{itemize}
\item This is well-defined and one to one:
\begin{eqnarray*}
g\cdot x=g'\cdot x,g,g'\in G & \iff & g^{-1}gx=g^{-1}g'x\\
 & \iff & g^{-1}g'\in\mathrm{Stab}_{G}(x)=H\\
 & \iff & g\cdot H=g'\cdot H
\end{eqnarray*}

\item Surjective by Orbit-Stabiliser: 
\[
\left|G/H\right|=\frac{\left|G\right|}{\left|H\right|}=\left|\mathrm{Orbit}(x)\right|=\left|X\right|
\]

\item An isomorphism of $G$-sets: $g(hx)=(gh)x\mapsto(gh)H=g(hH)$
\end{itemize}
\end{proof}
\begin{rem*}
$\mbox{Stab}_{G}(g\cdot x)=g\mbox{Stab}_{G}(x)g^{-1}$. In particular
$G/H\cong G/(gHg^{-1})$
\end{rem*}
An arbitrary set $X$ can be written as a union of orbits, $X=\coprod_{i=1}^{r}G/H_{i}$.
Then $\cc[X]=\oplus_{i=1}^{r}\cc\left[G/H_{i}\right]$ and 
\begin{eqnarray*}
\left\langle \cc[X],\mathbb{I}\right\rangle _{G} & = & \sum_{i=1}^{r}\left\langle \cc[G/H_{i}],\mathbb{I}\right\rangle _{G}\\
 & = & r\\
 & = & \mbox{number of orbit of}\, X\,\mbox{under}\, G.
\end{eqnarray*}

\begin{lem}
Let $X$ be a transitive set, $X\cong G/H$ and $\chi$ be the permutation
character. Then $\left\langle \chi,\chi\right\rangle _{G}=$ the number
of orbits on $X$ under the action of $H$\end{lem}
\begin{proof}
Let number of orbits under $H$ is
\begin{eqnarray*}
\left\langle \chi\downarrow_{H},\mathbb{I}_{H}\right\rangle _{H} & = & \left\langle \chi,\mathbb{I}_{H}\uparrow^{G}\right\rangle _{G}\\
 & = & \left\langle \chi,\chi\right\rangle _{G}
\end{eqnarray*}
 \end{proof}
\begin{cor}
$\cc[G/H]\cong\mathbb{I}\oplus\rho$ with $\rho$ irreducible if and
only if $H$ acts transitively on the non-trivial cosets. We say that
$X=G/H$ is \emph{doubly transitive}.\end{cor}
\begin{example*}
~
\begin{itemize}
\item $S_{n}$, $n\geq2$, acts doubly transitively on $\{1,\dots,n\}$,
so we get an $(n-1)$-dimensional irreducible character $\chi$. E.g.,
$n=4$, 
\begin{eqnarray*}
\chi'((123)) & = & \chi((123))-\mathbb{I}((123))\\
 & = & \underset{\#\mathrm{fixed\, pts}}{1}-1=0
\end{eqnarray*}

\item $G=\GL_{2}(\ff_{p})$ acts doubly transitively on the $(p+1)$ lines
through $0$ in $\left(\ff_{p}\right)^{2}$; e.g. the stabiliser of
$\left\langle (1,0)\right\rangle $ is $\begin{pmatrix}* & *\\
0 & *
\end{pmatrix}$, and that acts transitively on the remaining $p$ lines $\left\langle (a,1)\right\rangle $,
$a\in\ff_{p}$.


So we get a $p$-dimensional irreducible representation, $\cc[G/H]-\mathbb{I}$. 

\end{itemize}
\end{example*}
\begin{namedthm}[Artin's Induction]
 Let $\chi$ be a $\qq$-valued character (i.e., $\chi(g)\in\qq$
for all $g\in G$). Then $\chi=\sum_{H\leq G}\frac{a_{H}}{\left[N_{G}(H):H\right]}\cdot\mathbb{I}_{H}\uparrow^{G}$,
where the sum runs over representatives over conjugacy classes of
cyclic subgroups and $a_{H}\in\zz$ and $N_{G}(H)=\left\{ g\in G|gHg^{-1}=H\right\} $.\end{namedthm}
\begin{proof}
The idea is to express $\chi$ as linear combinations of characteristic
functions $\Phi_{x}(y)=\begin{cases}
1 & \mathrm{if\,}\left\langle x\right\rangle \sim\left\langle y\right\rangle \\
0 & \mathrm{else}
\end{cases}$, and then express $\Phi_{x}$ as linear combinations of $\mathbb{I}_{H}\uparrow^{G}$. 
\begin{lem}
$\chi(x)=\chi(y)$ whenever $\left\langle x\right\rangle \sim\left\langle y\right\rangle $,
i.e., $\chi$ is a $\qq$-linear combinations of characteristics functions
of $\Phi_{x}$.\end{lem}
\begin{proof}
Suppose $\left\langle x\right\rangle \sim\left\langle y\right\rangle $,
i.e., $x$ is a conjugate to some $y^{m}$, where $m$ is coprime
to the $\ord(x)=\ord(y)=n$. Up to replacing $x$ by a conjugate,
let $x=y^{m}$. We can diagonalise $x$ and $y$ in the representation
corresponding to $\chi$, $x=\mathrm{diag}(\epsilon_{1}^{m},\dots,\epsilon_{d}^{m})$
and $y=\mathrm{diag}(\epsilon_{1},\dots,\epsilon_{d})$, where $\epsilon_{i}$
are $n$th-roots of unity. There is an automorphism $\sigma$ of $\qq(e^{2\pi i/n})$
such that for any $n$th-root of unity $\epsilon$, $\sigma(\epsilon)=\epsilon^{m}$.
So $\sigma(\chi(y))=\sigma(\sum\epsilon_{i})=\sum\epsilon_{i}^{m}=\chi(x)\in\qq$.
But $\sigma|_{\qq}=\id$, hence $\chi(x)=\chi(y)$
\end{proof}
Let $H_{1},\dots,H_{s}$ be representatives of conjugacy classes of
cyclic subgroups, $\Phi_{i}(g)=\begin{cases}
1 & \mathrm{if\,}\left\langle g\right\rangle \sim H_{i}\\
0 & \mathrm{else}
\end{cases}$. By the above lemma, $\mathbb{I}_{H_{i}}\uparrow^{G}=\sum_{j=1}^{s}b_{ij}\Phi_{j}$.
So we want to invert $B=(b_{ij})$. First note that the $\Phi_{i}$
are orthogonal with respect to $\left\langle \,,\,\right\rangle _{G}$:
\begin{eqnarray*}
\left\langle \Phi_{i},\Phi j\right\rangle  & = & \frac{1}{\left|G\right|}\sum_{g\in G}\Phi_{i}(g)\Phi_{j}(g)\\
 & = & \begin{cases}
0 & i\neq j\\
\frac{1}{\left|G\right|}\phi\left(\left|H_{i}\right|\right)\cdot\frac{\left|G\right|}{\left|N_{G}(H_{i})\right|}=\frac{\phi\left(\left|H_{i}\right|\right)}{\left|N_{G}(H_{i})\right|} & i=j
\end{cases}
\end{eqnarray*}
By definition, $\Phi_{j}\downarrow_{H_{i}}$ is identically $0$ unless
$H_{j}\leq_{G}H_{i}$ in this case $\Phi_{j}\downarrow_{H_{i}}=1$
on the $\phi(\left|H_{j}\right|)$ generators of $H_{j}$ in $H_{i}$.
Now
\begin{eqnarray*}
b_{ij}\cdot\frac{\phi\left(\left|H_{i}\right|\right)}{\left|N_{G}(H_{j})\right|} & = & \left\langle \mathbb{I}_{H_{i}}\uparrow^{G},\Phi_{j}\right\rangle _{G}\\
 & = & \left\langle \mathbb{I}_{H_{i}},\Phi_{j}\downarrow_{H_{i}}\right\rangle _{H_{i}}\\
 & = & \frac{1}{\left|H_{i}\right|}\cdot\phi\left(\left|H_{j}\right|\right)\mathrm{if\,}H_{j}\leq_{G}H_{i}.
\end{eqnarray*}
Hence $b_{ij}=\frac{\left|N_{G}(H_{j})\right|}{\left|H_{i}\right|}$
if $H_{j}\leq_{G}H_{i}$ and otherwise $b_{ij}=0$. 

Now order the $H_{i}$ by size, then we have established that $B$
is triangular, integer entries, and in the $i$th row, all entries
are divisible by $\left[N_{G}(H_{i}):H_{i}\right],$because if $H_{j}\leq H_{i}$
then $\left|N_{G}(H_{i})\right|\big|\left|N_{G}(H_{j})\right|$. It
follows that $B$ is invertible, with denominators in the $i$th row
of the inverse dividing $\left[N_{G}(H_{i}):H_{i}\right]$. \end{proof}
\begin{rem*}
It is still an open question, how ``bad'' these denominators can
get, e.g., we do not know for what groups $G,$ any $\qq$-valued
$\chi$ can be written as $\sum_{H\leq G,\mathrm{cyclic}}c_{H}\mathbb{I}_{H}\uparrow^{G}$
with $c_{H}\in\zz$. This is possible in $S_{n}$.\end{rem*}
\begin{example*}
Let $G=C_{p}\times C_{p}$. There are $p+1$ cyclic subgroups of order
$p$, denote them $H_{1},\dots,H_{p+1}$. Any irreducible, non-trivial
character $\chi$ factors through a unique $G/H_{i}$. In fact, $\mathbb{I}_{H_{i}}\uparrow^{G}=\sum_{\ker\chi_{ij}\geq H_{i}}\chi_{ij}$
with $\chi_{i1}=\mathbb{I}$. After solving the system of linear equations,
we find that $\mathbb{I}_{\{1\}}\uparrow^{G}-\sum_{i}\mathbb{I}_{H_{i}}\uparrow^{G}=-p\cdot\mathbb{I}$.\end{example*}
\begin{cor}
The number of irreducible $\qq[G]$-modules is equal to conjugacy
classes of cyclic subgroups. \end{cor}
\begin{proof}
This corollary also depends on the theory of Schur indices, which
we will cover later.\end{proof}
\begin{cor}
Two $\qq[G]$-modules $V_{1},V_{2}$ are isomorphic if and only if
$\dim V_{1}^{H}=\dim V_{2}^{H}$ for all cyclic $H\leq G$. \end{cor}
\begin{proof}
Exercise\end{proof}
\begin{example*}
Let $G=S_{3}$, 
\begin{itemize}
\item $\mathbb{I}_{\{1\}}\uparrow^{G}=\mathbb{I}\oplus\epsilon\oplus\rho^{\oplus2}$,
where $\epsilon$ is the sign representation and $\rho$ is the standard
representation
\item $\mathbb{I}_{C_{2}}\uparrow^{G}=\mathbb{I}\oplus\rho$
\item $\mathbb{I}_{C_{3}}\uparrow^{G}=\mathbb{I}\oplus\epsilon$
\end{itemize}
Now $\phi=\mathbb{I}$, $-\mathbb{I}_{\{1\}}\uparrow^{G}+2\cdot\mathbb{I}_{C_{2}}\uparrow^{G}+\mathbb{I}_{C_{3}}\uparrow^{G}=2\cdot\mathbb{I}$. 
\end{example*}

\begin{example*}
Let $G=C_{p}\rtimes C_{p-1}$, then $-\mathbb{I}_{\{1\}}\uparrow^{G}+(p-1)\mathbb{I}_{C_{p-1}}\uparrow^{G}+\mathbb{I}_{C_{p}}\uparrow^{G}=(p-1)\cdot\mathbb{I}_{G}$.
Prove this as an exercise\end{example*}
\begin{rem*}
Even if one was allowed to use $\mathbb{I}_{H}\uparrow^{G}$ for all
$H\leq G$, one would still have denominators\end{rem*}
\begin{example*}
Let $G=Q_{8}\times C_{3}$ and let $\rho$ be the standard representations
of $Q_{8}$, $\chi$ a $1$-dimensional non-trivial character of $C_{3}$.
Then $\rho\otimes\left(\chi\otimes\overline{\chi}\right)$ is a representation
that can be defined over $\qq$, but it is not a $\zz$-linear combination
of $\mathbb{I}_{H}\uparrow^{G}$ for all $H\leq G$, but twice that
is.\end{example*}
\begin{defn}
A group is called $p$\emph{-quasi-elementary} if it's of the form
$G=C\rtimes P$ where $C$ is cyclic and $P$ a $p$-group (i.e.,
order $p^{n}$ for some $n$).

Without loss of generality, we can assume $p\nmid\left|G\right|$. \end{defn}
\begin{namedthm}[Solomon Induction]
 There exists $a_{H}\in\zz$ for $H\leq G$ quasi-elementary subgroups
such that $\mathbb{I}=\sum_{H\leq G}a_{H}\mathbb{I}_{H}\uparrow^{G}$.
\end{namedthm}

\begin{namedthm}[Brauer's Induction Theorem]
 Let $\phi\in\mathrm{Irr}(G)$. Then there exists $a_{H,\lambda}\in\zz$
for $H$ of the form $H=C\times P$ (with $C$ cyclic and $P$ a $p$-group,
these are called elementary groups), such that $\phi=\sum a_{H,\lambda}\lambda\uparrow^{G}$,
where $\lambda$ are $1$-dimensional characters of elementary subgroups.
\end{namedthm}
We will first deduce Brauer from Solomon, to do so we will use for
the first time the ring structure of the ring of class functions. 
\begin{defn}
We define $R(G)=\left\langle \mathrm{Irr}(G)\right\rangle _{\zz}=\left\{ \sum_{\chi\in\mathrm{Irr}(G)}a_{\chi}\cdot\chi\big|a_{\chi in}\zz\right\} $.

For any family of subgroups of $G$, $\mathcal{H}$, we define $I_{\mathcal{H}}(G)=\left\{ \sum_{H\in\mathcal{H},\lambda\in\mathrm{Irr}(H)}a_{H,\lambda}\lambda_{H}\uparrow^{G}\big|a_{H,\lambda}\in\zz\right\} $.\end{defn}
\begin{lem}
Let $H\leq G$, $\phi$ a class function of $H$, $\psi$ a class
function of $G$. Then $\phi_{H}\uparrow^{G}\cdot\psi=\left(\phi\cdot\psi\downarrow_{H}\right)\uparrow^{G}$.\end{lem}
\begin{proof}
Just do it. (exercise)\end{proof}
\begin{cor}
$I_{\mathcal{H}}(G)$ is an ideal in $R(G)$\end{cor}
\begin{proof}
If $\phi=\sum a_{H,\lambda}\lambda_{H}\uparrow^{G}\in I_{\mathcal{H}}(G)$,
$\psi\in\mathrm{Irr}(G)$, then $\psi\cdot\phi=\sum a_{H,\lambda}\psi\lambda_{H}\uparrow^{G}=\sum a_{H,\lambda}\left(\psi\downarrow_{H}\cdot\lambda\right)\uparrow^{G}\in I_{\mathcal{H}}(G)$. 
\end{proof}
Now let $\mathcal{H}=\left\{ C\times P\leq G\big|C\mathrm{\, cyclic,}\, P\mathrm{\, a\,}p\,\mathrm{group}\right\} $.
We will prove Brauer if we can show
\begin{itemize}
\item $\mathbb{I}\in I_{\mathcal{H}}(G)=:I(G)$
\item All elementary groups are $\mathrm{M}$-groups, i.e., every irreducible
characters is monomial, i.e., induced from a $1$-dimensional character.
(This is left as an exercise)\end{itemize}
\begin{thm}
$\mathbb{I}\in I(G)$. \end{thm}
\begin{proof}[Proof (assuming Solomon)]
 We do this by induction on $\left|G\right|$. We can use elementary
groups $G$ as our base case.

Assume that the theorem holds for all proper subgroups of $G$, i.e.,
for all $H\leq G$, $\mathbb{I}_{H}=\sum_{U\leq H\mathrm{\, elem,}\lambda\in\mathrm{Irr}(U)}a_{U,\lambda}\lambda_{U}\uparrow^{H}$.
Then it is enough to show that 
\[
\mathbb{I}=\sum_{H\lneq G,\lambda\in\mathrm{Irr}(H)}b_{H,\lambda}\lambda_{H}\uparrow^{G},\,\,\,\,\,\,\,(\dagger)
\]
because then, each $\lambda=\sum_{U\leq H\mathrm{\, elem}}a_{u,\mu}^{(\lambda)}\mu\uparrow^{H}$
and $\mathbb{I}_{G}=\sum\sum b_{H,\lambda}a_{U,\mu}^{(\lambda)}\mu\uparrow^{H}\uparrow^{G}$.
If $G$ is not quasi-elementary, then Solomon shows $(\dagger)$. 

So we are left with proving the statement 
\[
\mathbb{I}=\sum_{H\lneq G,\lambda\in\mathrm{Irr}(H)}b_{H,\lambda}\lambda_{H}\uparrow^{G},\,\: b_{H,\lambda}\in\zz
\]
 for $G=C\rtimes P$, where $P$ acts non-trivially on $C$ by conjugation,
so that $G\neq C\times P$. Let $Z=Z_{C}(P)=\{x\in C|xpx^{-1}=p\forall p\in P\}$.
Since $G\neq C\times P$, $Z\neq C$. Set $E=ZP\neq G$. 

We have $\mathbb{I}_{E}\uparrow^{G}=\mathbb{I}_{G}+\Xi$. It is enough
to show that any irreducible summand of $\Xi$ is induced from a proper
subgroup. Let $\xi$ be an irreducible summand of $\Xi$. Let $\chi$
be an irreducible summand of $\xi\downarrow_{C}$. Let $S=\mathrm{Stab}_{G}(\chi)$.
Recall that if $\iota$ is an irreducible summand of $\chi\uparrow^{S}$,
then $\iota\uparrow^{G}$ is irreducible and all summands of $\chi\uparrow^{G}$
are of this form (in particular this is true for $\xi$). So we now
just need to know that $S\neq G$, i.e., that $\chi$ is not invariant
under the $G$-action. To do so, we use the following lemma.
\begin{lem}
Let $G=C\rtimes P$, $p\nmid\left|C\right|$, $\chi\in\mathrm{Irr}(C)$,
$Z=Z_{C}(P)$ and assume $Z\in\ker\chi=:K$. If $\chi$ is invariant
in $G$, then $\chi=\mathbb{I}_{C}$. 
\begin{proof}
$\chi$ is a faithful character on $C/K$, so for $\chi$ to be invariant,
$G$ has to preserve each coset $cK$. But if $P$ acts on $cK$,
then the number of points moved is divisible by $p$ (by Orbit - Stabiliser).
But $p\nmid\left|C\right|$ so $p\nmid\left|K\right|$, hence $p\nmid\left|cK\right|$.
So there is at least one point in $cK$ that is normalised by $P$,
i.e., $cK\cap Z\neq\emptyset$. But $Z\subseteq K$, so $cK=K$ for
all $c$, i.e., $\ker\chi=K=C$. 
\end{proof}
\end{lem}
In our situation, $Z\lhd G$, so by Clifford, $Z\subset\ker(\mathbb{I}_{E}\uparrow^{G})$,
so in particular $Z\subset\ker\chi$. We claim that $\chi\neq\mathbb{I}$:
note 
\begin{eqnarray*}
\left\langle \mathbb{I}_{G}\downarrow_{C}+\Xi\downarrow_{C},\mathbb{I}\right\rangle  & = & \left\langle \mathbb{I}_{E}\uparrow^{G}\downarrow_{C},\mathbb{I}\right\rangle \\
 & = & \left\langle \oplus_{E\backslash G/C}\mathbb{I}\downarrow\R\uparrow,\mathbb{I}\right\rangle \,\,\,\mathrm{but\,}E\backslash G/C=E\cdot1\cdot C\\
 & = & 1
\end{eqnarray*}
So $\left\langle \xi\downarrow_{C},\mathbb{I}\right\rangle =0$, so
$\chi\neq\mathbb{I}$. Hence $\chi$ is not invariant. \end{proof}
\begin{lem}[Banashewski]
 Let $S$ be a finite set, and $R$ be a rng (i.e., a ring which
does not necessarily contain $1$) of functions $f:S\to\zz$. Then
either $R\ni\mathbb{I}_{S}$ or there exists $s\in S$ and a prime
$p$ such that $p|f(x)\forall x\in R$. \end{lem}
\begin{proof}
Suppose there is no such $x,p$. Then for any $x\in S$, $\gcd\left\{ f(x)|f\in R\right\} =1$.
So there exists $x\in R$ such that $f_{x}(x)=1$. Consider $\prod_{x\in S}(f_{x}-\mathbb{I}_{S})\equiv0$
on $S$. So expanding the product gives an expression for $\mathbb{I}_{S}$
as a linear combination of products of $f_{x}\in R$.\end{proof}
\begin{defn}
We define $P_{\mathcal{H}}(G)=\left\{ \sum_{H\in\mathcal{H}}a_{H}\mathbb{I}_{H}\uparrow^{G}|a_{H}\in\zz\right\} $.\end{defn}
\begin{lem}
Suppose that $\mathcal{H}$ is closed under taking subgroups, i.e.,
$H\in\mathcal{H}$ implies $U\in\mathcal{H}$ for all $U\leq H$.
Then $P_{\mathcal{H}}(G)$ is a rng. \end{lem}
\begin{proof}
Either use 
\begin{eqnarray*}
\mathbb{I}_{H}\uparrow^{G}\cdot\mathbb{I}_{H}\uparrow^{G} & = & \left(\mathbb{I}_{H}\cdot\mathbb{I}_{H}\uparrow^{G}\downarrow_{H}\right)\uparrow^{G}\\
 & = & \sum_{H\backslash G/H'}\mathbb{I}\downarrow_{H\cap gH'g^{-1}}\uparrow^{G}\in P_{\mathcal{H}}(G)
\end{eqnarray*}
or note that if $v_{1},\dots,v_{n}$ is a permutation basis of $V$,
$w_{1},\dots,w_{m}$ is a permutation basis of $W$, then $v_{i}\otimes w_{j}$
is a permutation basis. and the point stabiliser of $v_{i}\otimes w_{j}=\mathrm{Stab}(v_{i})\cap\mathrm{Stab}(w_{j})\in\mathcal{H}$
if one of the other stabilisers was in $\mathcal{H}$. 
\end{proof}
We want to use Banaschewski's lemma to conclude that if $\mathcal{H}=\{$quasi-elementary
subgroups$\}$, then $\mathbb{I}_{G}\in P_{\mathcal{H}}(G)$. 
\begin{lem}
For any prime $p$, any $x\in G$, there exists a quasi-elementary
$H\leq G$ such that $p\big\nmid\mathbb{I}_{H}\uparrow^{G}(x)$. \end{lem}
\begin{proof}
For $x\in G$, write $\left\langle x\right\rangle =C_{p}\times C_{p'}$,
where $C_{p}$ is a $p$-group and $C=C_{p'}$ has order not divisible
by $p$. Let $N=N_{G}(C)$, let $P$ be a $p$-Sylow group in $N$
containing $C_{p}$. Set $H=C\rtimes P$.
\begin{claim*}
$p\nmid\mathbb{I}_{H}\uparrow^{G}(x)$
\end{claim*}

Indeed, 
\begin{eqnarray*}
\mathbb{I}_{H}\uparrow^{G}(x) & = & \#\left\{ gH\in G/H\big|xgH=gH\right\} \\
 & = & \#\left\{ gH\in G/H\big|g^{-1}xg\in H\right\} .
\end{eqnarray*}
If $g^{-1}xg\in H$, then $g^{-1}Cg=C$. So
\begin{eqnarray*}
\mathbb{I}_{H}\uparrow^{G}(x) & = & \#\left\{ gH\in G/H\big|g^{-1}Cg=C\,\mathrm{and}\, g^{-1}xg\in H\right\} \\
 & = & \#\left\{ gH\in N/H\big|g^{-1}xg\in H\right\} .
\end{eqnarray*}
The action of $\left\langle x\right\rangle $ on $N/H$ factors through
$\left\langle x\right\rangle /C$, i.e., $C$ acts trivially on $N/H$:
indeed $C\lhd N$ and $C\leq H$, so $c\cdot nH=n\cdot c'\cdot H=nH$.
But $\left\langle x\right\rangle /C$ is a $p$-group, so the number
of elements of $N/H$ that are \uline{not} fixed by $\left\langle x\right\rangle /C$
is a multiple of $p$. So 
\begin{eqnarray*}
\mathbb{I}_{H}\uparrow^{G}(x) & \equiv & \left|N/H\right|\mod p\\
 & \not\equiv & 0\mod p
\end{eqnarray*}
because $H/C$ is a $p$-Sylow of $N/C$.

\end{proof}
\begin{rem*}
There is a counterpart to Brauer's theorem, which is called ``Brauer's
characterisation of Characters'':
\begin{thm}
A class function $\phi$ of $G$ is a $\zz$-linear combination of
characters ($\phi\in R(G)$) if and only if $\phi\downarrow_{H}\in R(G)$
for all $H\leq G$ elementary subgroups. \end{thm}
\begin{proof}[Idea of Proof]
 Define $R_{\mathcal{H}}(G)=\left\{ \mathrm{class\, functions\,\phi\mathrm{\, of\,}}G|\phi\downarrow_{H}\in R(H)\,\forall H\in\mathcal{H}\right\} $.
Note that $I_{\mathcal{H}}(G)\subset R_{\mathcal{H}}(G)$ is an ideal
(exercise). But $\mathbb{I}\in I_{\mathcal{H}}(G)$, so $I_{\mathcal{H}}(G)=R_{\mathcal{H}(G)}.$
\end{proof}
\end{rem*}
For consequences, see Isaacs, chapter on Brauer's Theorem


\section{Rationality questions, Schur indices.}
\begin{defn}
Let $M$ be a $K[G]$-module, $F\subset K$ a subfield. We say that
$M$ is \emph{realisable} over $F$ if there exists an $F[G]$-module
$M_{F}$ such that $M_{F}\otimes_{F}K\cong M$. 

In the language of representations: In the language of representations,
this means that we can find a $K$-basis on $M$ such that all $g\in G$
are represented by matrices with entries in $F$ with respect to this
basis. \end{defn}
\begin{example*}
Consider $G=S_{3}$. We have $\rho\colon(123)\mapsto\begin{pmatrix}e^{2\pi i/3} & 0\\
0 & e^{4\pi i/3}
\end{pmatrix},(12)\mapsto\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}$. But we can change this basis in such a way that $\rho$ becomes
$(123)\mapsto\begin{pmatrix}0 & -1\\
1 & -1
\end{pmatrix}$, $(12)\mapsto\begin{pmatrix}0 & 1\\
1 & 0
\end{pmatrix}$. So the $K[G]$-representation $\rho$ ($K=\qq(e^{2\pi i/3})=\qq(\sqrt{-3})$)
is realisable over $\qq$. 
\begin{itemize}
\item In fact, $\rho$ is induced from $\chi:(123)\mapsto e^{2\pi i/3}$,
$\left\langle (123)\right\rangle \cong C_{3}\lhd S_{3}$. This $\chi$
is definitely \uline{not} realisable over $\qq$, since $\GL_{1}(K)$
is commutative, so change of basis doesn't do anything to $\chi((123))$.


More generally, the character of a representation is independent of
basis, so if $\rho$ realisable over $F$, then we need $F\supseteq\qq(\chi_{\rho})$,
where $\qq(\chi_{\rho})$ is the field generated over $\qq$ by $\chi_{\rho}(g)$
for all $g\in G$.

\end{itemize}
Let $G=Q_{8}=\left\langle x,y|x^{4}=1,x^{2}=y^{2},yxy^{-1}=x^{-1}\right\rangle $.
Let $\rho:x\mapsto\begin{pmatrix}i & 0\\
0 & -i
\end{pmatrix}$, $y\mapsto\begin{pmatrix}0 & 1\\
-1 & 0
\end{pmatrix}$, this is a $K[G]$-representation where $K=\qq(i)$. We already know
that $\rho$ can not be defined over $\rr$, so it is certainly not
realisable over $\qq$. But we have exactly two copies of $\rho$
inside $\cc[G]$, which is realisable over $\qq$. The other summands
in $\cc[G]$ are $1$-dimensional, all realisable over $\qq$. So
$\qq[G]/$(all $1$-dimensional subrepresentations) $\cong\rho^{\oplus2}$.
So $\rho^{\oplus2}$ is realisable over $\qq$. \end{example*}
\begin{defn}
Let $K\subset\cc$, $\rho$ an irreducible (complex) representation
of $G$, the \emph{Schur index}, $m_{K}(\rho)$, of $\rho$ over $K$
is the the smallest integer $m$ such that there exists an irreducible
$K[G]$-representation $\tau$ with $\left\langle \tau,\rho\right\rangle =m$.
Equivalently, it's the unique integer $m$ such that $m|\left\langle \rho,\tau\right\rangle $
for all $K[G]$-module $\tau$. \end{defn}
\begin{example*}
We have:
\begin{itemize}
\item $m_{\qq}(\mathrm{standard\, representation\, of\,}Q_{8})=2$
\item $m_{\qq}(\mathrm{standard\, representation\, of\,}S_{3})=1$. 
\item $m_{\qq}(\chi:(123)\mapsto e^{2\pi i/3})=1$, although $\chi$ is
not realisable over $\qq$. (Note that $\chi+\overline{\chi}$ is
definable over $\qq$)
\end{itemize}
\end{example*}
\begin{defn}
A representation over $K$ is said to be \emph{absolutely irreducible}
if it is irreducible over $\cc$. 

A field $K\subset\cc$ is called a \emph{splitting field} of $G$
if every irreducible $K[G]$-representation is absolutely irreducible,
equivalently if every complex $G$-representation is realisable over
$K$. \end{defn}
\begin{lem}
Let $\chi$ be an irreducible character of $G$, $F\subset\cc$ such
that $F(\chi)=F$, i.e., $\chi$ takes values in $F$. Let $\tau$
be an irreducible $F[G]$-representations such that $\left\langle \tau,\chi\right\rangle \neq0$.
Then $\tau\otimes\cc=m_{F}(\chi)\cdot\chi$. \end{lem}
\begin{proof}
The element $e_{\chi}=\frac{1}{\left|G\right|}\sum_{g\in G}\chi(g)g^{-1}\in F[G]$.
So $e_{\chi}\cdot\tau=\tau$. But for any complex representation $V$,
$e_{\chi}\cdot V\cong\chi^{\oplus n}$, for some $n$. But this $n$
has to be $m_{F}(\chi)$. 
\end{proof}
If $\chi_{\rho}$ is a character, why is $\chi_{\rho}^{\sigma}$ a
character for all $\sigma\in\mbox{Gal}(\qq(\chi_{\rho})/\qq)=\mbox{Gal}(\overline{\qq}/\qq)/\mbox{Gal}(\overline{\qq}/\qq(\chi_{\rho}))$?
We have that if $\rho:G\to\GL_{n}(\overline{\qq})$ is defined by
$g\mapsto(a_{ij})$, then $\rho^{\sigma}:G\to\GL_{n}(\overline{\qq})$
is defined by $g\mapsto(a_{ij}^{\sigma})$. Now if $\sigma$ fixes
$\qq(\chi_{\rho})$, then by definition $\chi_{\rho^{\sigma}}=\chi_{\rho}$.
Hence $\rho^{\sigma}\cong\rho$.
\begin{thm}
Let $K\subset\cc$ be arbitrary, $\rho\in\mathrm{Irr}(G)$, $\tau$
a simple $K[G]$-module such that $\left\langle \tau,\rho\right\rangle \neq0$.
Then 
\[
\tau\otimes\cc=m_{K}(\rho)\sum_{\sigma\in\mathrm{Gal}(K(\chi_{\rho})/K)}\rho^{\sigma}.
\]
\end{thm}
\begin{proof}
Let $F=K(\chi_{\rho})$, let $\psi$ be a simple $F[G]$-module such
that $\left\langle \tau,\psi\right\rangle \neq0$. So by the lemma,
$\psi\otimes\cc=m_{F}(\rho)\cdot\rho$. Let $\sigma\in\mathrm{Gal}(F/K)$.
Since $\tau^{\sigma}=\tau$ (as $\tau$ is a $K[G]$-module), 
\begin{eqnarray*}
\left\langle \tau,\psi^{\sigma}\right\rangle  & = & \left\langle \tau^{\sigma},\psi^{\sigma}\right\rangle \\
 & = & \frac{1}{\left|G\right|}\sum_{g\in G}\chi_{\tau}^{\sigma}\overline{\chi_{\psi}^{\sigma}}\\
 & = & \left(\frac{1}{\left|G\right|}\sum_{g\in G}\chi_{\tau}\overline{\chi_{\psi}}\right)^{\sigma}\\
 & = & \left\langle \tau,\psi\right\rangle ^{\sigma}\\
 & = & \left\langle \tau,\psi\right\rangle .
\end{eqnarray*}
So each Galois conjugate occurs with equal multiplicity inside $\tau$,
i.e., $\tau\otimes F=\alpha\sum\psi^{\sigma}+$stuff that is not Galois
conjugate to $\psi$. So $\tau\otimes\cc=\alpha m_{F}(\rho)\sum_{\sigma\in\mathrm{Gal}(F/K)}\rho^{\sigma}+$stuff
that has nothing to do with $\rho$. In particular $m_{F}(\rho)|m_{K}(\rho)$.
We just need to prove now that $m_{F}(\rho)\sum\rho^{\sigma}$ is
realisable over $K$. 

Let $V$ be the $F$-vector space on which $\psi$ is represented.
Regarding $F$ as a $\left|\mathrm{Gal}(F/K)\right|$-dimensional
$K$-vector space, we can think of $V$ as a $\left|\mathrm{Gal(F/K)}\right|\cdot\dim_{F}V$-dimensional
vector space over $K$. Inspecting the action of $G$ on $V$ with
respect to this $K$-basis, one can see that $V\otimes_{K}F=\sum_{\sigma\in\mbox{Gal}(F/K)}\psi^{\sigma}$,
which is realisable over $K$.\end{proof}
\begin{example*}
Let $G=C_{7}\rtimes C_{9}$, where $C_{9}$ acts on $C_{7}$ through
$C_{9}/C_{3}\cong C_{3}$. Let $\chi\in\mathrm{Irr}(C_{7})$ be faithful,
extended trivially to $S_{\chi}\cong C_{3}$. Let $\tau$ be a faithful
character of $C_{3}\cong C_{7}\times C_{3}/C_{7}$, $\rho=\left(\chi\otimes\tau\right)\uparrow^{G}\in\mathrm{Irr}(G)$.
The field $\qq(\chi_{\rho})$ is the degree $4$ subfield of $\qq(\zeta_{21})$,
call it $F$. In particular, $\sum_{\sigma\in\mathrm{Gal}(F/\qq)}\rho^{\sigma}$
takes values in $\qq.$ But $m_{\qq}(\rho)=3$. So $\sum_{\sigma\in\mathrm{Gal}(F/\qq)}\rho^{\sigma}$
is not a character of a $\qq[G]$-module, but $3$ times it is (and
it's a \uline{simple} $\qq[G]$-module). \end{example*}
\begin{cor}
We have $m_{K}(\rho)=m_{K(\chi_{\rho})}(\rho)$. \end{cor}
\begin{thm}
Let $e=\mathrm{exponent}(G)=\min\{n\in\nn|g^{n}=\id\forall g\in G\}$.
Let $F=\qq(\zeta_{e})$. Then $F$ is a splitting field for $G$. \end{thm}
\begin{proof}
Let $\chi\in\mathrm{Irr}(G)$. Write 
\[
\chi=\sum_{H\leq G\mathrm{\, elem},\lambda\in\mathrm{Irr}(H)\,1\mathrm{-dim}}a_{H,\lambda}\lambda_{H}\uparrow^{G}
\]
with $a_{H,\lambda}\in\zz$. Clearly $\lambda$ can be realised over
$F$, therefore so can $\lambda\uparrow^{G}$. But $m_{F}(\chi)|\left\langle \chi,\mathrm{every\,}F[G]\mathrm{-module}\right\rangle $,
so $m_{F}(\chi)|\left\langle \mathrm{RHS},\chi\right\rangle =\left\langle \chi,\chi\right\rangle =1$.
So $m_{F}(\chi)=1$, so $\chi$ is realisable over $F$. \end{proof}
\begin{rem*}
~
\begin{itemize}
\item Schur indices can be arbitrarily large: let $p,q$ be primes, $q\equiv1\mod p$
and $q\not\equiv1\mod p^{2}$. Let $G=C_{q}\rtimes C_{p^{2}}$ where
$C_{p^{2}}$ acts on $C_{q}$ through $C_{p^{2}}/C_{p}\cong C_{p}$.
Then using the same notation as in the above example $\chi\otimes\psi$
is a faithful irreducible character of $C_{q}\times C_{p}$. Now let
$p=\left(\chi\otimes\psi\right)\uparrow^{G}$, then $m_{\qq}(\rho)=p$
(this claim can not be proven with techniques learn in this course)
\item Schur indices don't behave very well under induction, restriction
and tensor products (in part, if $G=H\times K$, $\chi\in\mathrm{Irr}(H)$,
$\psi\in\mathrm{Irr}(K)$, $m_{\qq}(\chi\otimes\psi)\neq m_{\qq}(\chi)\cdot m_{\qq}(\psi)$).
\item Schur indices are hard to calculate. There is an algorithm, based
on another induction theorem (Witt - Berman). But it's difficult to
understand Schur indices in natural families. 
\item There is no unique minimal field of definition for $\rho\in\mathrm{Irr}(G)$. 

\begin{example*}
Let $G=Q_{8}$ and $\rho$ be the standard representation. We know
that $\rho$ is not realisable over $\qq$. But it is realisable over
$\qq(i)$. In fact $\rho$ is realisable over $K$ if and only if
$-1=x^{2}+y^{2}$ for some $x,y\in K$. In particular, $\rho$ is
realisable in infinitely many quadratic fields.
\end{example*}
\item Note: $m_{\qq}(\rho)=1$ if and only if $\sum_{\sigma\in\mathrm{Gal}(\qq(\chi_{\rho})/\qq)}\rho^{\sigma}$
is realisable over $\qq$, if and only if $\rho$ is realisable over
$\qq(\chi_{\rho})$. 
\end{itemize}
\end{rem*}

\subsection{Schur indices and Artin - Wedderburn}

Recall: $K[G]\cong\oplus_{i}M_{n_{i}}(D_{i})$ with $n_{i}\in\nn$,
and $D_{i}$ are division algebras over $K$. These $D_{i}$ are isomorphic
to $\End_{G}(A_{i})$ as $A_{i}$ ranges over simple non-isomorphic
$K[G]$-modules. Each $D_{i}$ contains $K$ in its centre, $Z=Z(D_{i})$.
Also $Z$ is a field, but $D_{i}$ can also contain bigger fields. 
\begin{example*}
Let $D=\hh=\left\langle 1,i,j,k\right\rangle _{\rr}$. Note that $\zz\cong\rr=\left\langle 1\right\rangle $,
but $\left\langle 1,i\right\rangle _{\rr}\cong\cc$. 
\end{example*}
We will show the following theorem:
\begin{thm}
\label{thm:Proof later} If $\chi\in\mathrm{Irr}(G)$, $K=\qq(\chi)$,
$M$ a simple $K[G]$-module with $\left\langle M,\chi\right\rangle \neq0$,
i.e., $M\otimes\cc=m_{K}(\chi)\cdot\chi$, and let $D=\End_{K[G]}(M)$.
If $F\subset D$ is a maximal subfield of $D$, then $\chi$ can be
realised by a representation over $F$. Moreover $[F:K]=m_{K}(\chi)$,
and $\chi$ cannot be realised over any smaller degree extension. \end{thm}
\begin{lem}
Let $\rho:G\to\GL_{n}(F)$ be an irreducible representation over $F$.
Then the following are equivalent:
\begin{enumerate}
\item $\rho$ is absolutely irreducible
\item The centraliser of $\rho(G)$ in $M_{n}(F)$ is just scalar matrices.
\end{enumerate}
\end{lem}
\begin{proof}
Note that the centraliser of $\rho(G)$ in $M_{n}(F)$ is $\End_{G}(\rho)\cong F$. 
\begin{elabeling}{00.00.0000}
\item [{1.~$\Rightarrow$2.~}] Suppose that $\rho$ is absolutely irreducible,
let $M\in M_{n}(F)$ commute with $\rho(g)$ for all $g\in G$. Let
$L/F$ be an extension in which $M$ has an eigenvalue, $\lambda$.
Then $M-\lambda I$ commutes with $\rho(g)$ for all $g\in G$. But
$\rho\otimes L$ is still simple, so Schur's lemma implies $M-\lambda I$
is either $0$ or invertible. But, its singular, so $M-\lambda I=0$,
so $M=\lambda I$. 
\item [{$2.$~$\Rightarrow$1.~}] Suppose that $\rho$ is not absolutely
irreducible. Let $L/F$ be a \uline{finite} Galois extension of
$F$ such that $\rho\otimes L\cong\rho_{1}\oplus\rho_{2}$, so with
respect to a suitable $L$-basis, $\rho\to\left(\begin{array}{c|c}
*\\
\hline  & *
\end{array}\right)$ . Let $X$ be the change of basis matrix form the original basis
to this one. Any matrix of the form $A_{\lambda_{1}\lambda_{2}}=\left(\begin{array}{c|c}
\lambda_{1}I_{d_{1}}\\
\hline  & \lambda_{2}I_{d_{2}}
\end{array}\right)$ commutes with $\rho(G)$, note $\lambda_{1},\lambda_{2}\in L$. So
$XA_{\lambda_{1},\lambda_{2}}X^{-1}$ commutes with $\im(\rho\to M_{n}(F))$.
Also $\overline{A}=\sum_{\sigma\in\mbox{Gal}(L/F)}\left(XA_{\lambda_{1}\lambda_{2}}X^{-1}\right)^{\sigma}\in M_{n}(F)$
commutes with $\im(\rho\to M_{n}(F))$. Exercise: by varying $\lambda_{1},\lambda_{2}$,
we can arrange $\overline{A}$ to not be scalar. 
\end{elabeling}
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:Proof later}]
 Since $F\subset\End_{G[K]}(M)$, $M$ can be thought as a vector
space over $F$, and this $F$-commutes with the $G$-action, so $M$
becomes an $F[G]$-module. But $\End_{F[G]}(M)$ is the centraliser
of $F[G]$ in $D$. But by the maximality of $F$, $\End_{F[G]}(M)\cong F$,
so $M$ is an absolutely simple $F[G]$-module. So $M\otimes_{F}\cc$
has character $\chi$. 

It remains to compute $[F:K]$. Regarding $F$ as a $K$-vector space,
$M$ becomes a $K[G]$-module of $K$-dimension equal to $\dim_{F}(M)\cdot[F:K]$.
But also, this dimension is $m_{K}(\chi)\cdot\chi(1)$; on the other
hand, $M$ is an absolutely simple $F[G]$-module, so $\dim_{F}(M)=\chi(1)$.
So $m_{K}(\chi)=[F:K]$. Also, if $L/K$ is a degree $d$-extension
such that $\chi$ is realised by a simple $L[G]$-module $M'$ then
regarding $L$ as a $K$-vector space, $M'$ becomes a $K[G]$-module
with $\dim_{K}(M')=\dim_{L}(M')\cdot[L:K]=[L:K]\cdot\chi(1)$. So
by definition, $m_{K}(\chi)|[L:K]$. \end{proof}
\begin{rem*}
~
\begin{itemize}
\item There is a local to global theory of Schur indices: If $D$ is a division
algebra over a number field $K$, one defines $m_{\mathfrak{p}}(D)=m(D\otimes K_{\mathfrak{p}})$
for all place (finite and infinite) $\mathfrak{p}$ of $K$, and $m(D)=\lcm(m_{\mathfrak{p}}(D))$. 
\item If $p$ is an odd prime, and $G$ is a $p$-group, then $m_{\qq}(\chi)=1$
for all $\chi\in\mathrm{Irr}(G)$. 


If $p=2$ and $G$ is a $2$-group, then $m_{\qq}(\chi)=1$ or $2$,
and it's $1$ if and only if $\chi$ is realisable over $\rr$. 

\item $m(\chi)^{2}\big|\left|G\right|$
\end{itemize}
\end{rem*}
For more details see Curtis - Reiner, Vol II. 

\newpage{}


\section{Examples}
\begin{enumerate}
\item Let $G=C_{2}\wr C_{3}=\left(C_{2}\times C_{2}\times C_{2}\right)\rtimes C_{3}$.
The conjugacy classes: 

\begin{itemize}
\item $1=((0,0,0),\id)$
\item $((1,0,0),\id)$, this has 3 elements
\item $((1,1,0),\id)$, this has $3$ elements
\item $((1,1,1),\id)$, this has $1$ element
\item $((0,0,0),(123))$, this has $4$ elements
\item $\left((0,0,0),(132)\right)$, this has $4$ elements
\item $((1,0,0),(123))$, this has $4$ elements
\item $((1,0,0),(132))$, this has $4$ elements
\end{itemize}

All irreducible characters are $(\chi\otimes\psi)\uparrow^{G}$, with
$\chi\in\mathrm{Irr}(C_{2}^{3})$ and $\psi\in\mathrm{Irr}(\mathrm{Stab}_{C_{3}}(\chi))$.
Let $\epsilon$ denote the sign representation.
\begin{itemize}
\item $\chi=\mathbb{I}$. Then $\mathrm{Stab}(\chi)=C_{3}$, so we get $3$
irreducible characters corresponding to the irreducible characters
of $C_{3}$.
\item $\chi=(\epsilon,\mathbb{I},\mathbb{I})$, then $\mbox{Stab}(\chi)=\left\{ 1\right\} $.
Hence $\chi\uparrow^{G}$ is irreducible.
\item $\chi=(\epsilon,\epsilon,\mathbb{I})$, then $\mbox{Stab}(\chi)=\left\{ 1\right\} $.
Hence $\chi\uparrow^{G}$ is irreducible.
\item $\chi=(\epsilon,\epsilon,\epsilon)$, then $\mbox{Stab}(\chi)=C_{3}$.
So $\chi\otimes\psi$ is irreducible for $\psi\in\mathrm{Irr}(C_{3})$.
\end{itemize}

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline 
 & \begin{sideways}
$1$
\end{sideways} & \begin{sideways}
$((1,1,1),\id)$
\end{sideways} & \begin{sideways}
$((1,0,0),\id)$
\end{sideways} & \begin{sideways}
$((1,1,0),\id)$
\end{sideways} & \begin{sideways}
$((0,0,0),(123))$
\end{sideways} & \begin{sideways}
$((0,0,0),(132))$
\end{sideways} & \begin{sideways}
$((1,0,0),(123))$
\end{sideways} & \begin{sideways}
$((1,0,0),(132))$
\end{sideways}\tabularnewline
\hline 
\hline 
$\chi_{1}=\mathbb{I}$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$\tabularnewline
\hline 
$\chi_{2}$ & $1$ & $1$ & $1$ & $1$ & $e^{2\pi i/3}$ & $e^{4\pi i/3}$ & $e^{2\pi i/3}$ & $e^{4\pi i/3}$\tabularnewline
\hline 
$\chi_{3}$ & $1$ & $1$ & $1$ & $1$ & $e^{4\pi i/3}$ & $e^{2\pi i/3}$ & $e^{4\pi i/3}$ & $e^{2\pi i/3}$\tabularnewline
\hline 
$\chi_{4}$ & $1$ & $-1$ & $-1$ & $1$ & $1$ & $1$ & $-1$ & $-1$\tabularnewline
\hline 
$\chi_{5}$ & $1$ & $-1$ & $-1$ & $1$ & $e^{2\pi i/3}$ & $e^{4\pi i/3}$ & $-e^{2\pi i/3}$ & $-e^{4\pi i/3}$\tabularnewline
\hline 
$\chi_{6}$ & $1$ & $-1$ & $-1$ & $1$ & $e^{4\pi i/3}$ & $e^{2\pi i/3}$ & $-e^{4\pi i/3}$ & $-e^{2\pi i/3}$\tabularnewline
\hline 
$\chi_{7}$ & $3$ & $-3$ & $1$ & $-1$ & $0$ & $0$ & $0$ & $0$\tabularnewline
\hline 
$\chi_{8}$ & $3$ & $3$ & $-1$ & $-1$ & $0$ & $0$ & $0$ & $0$\tabularnewline
\hline 
\end{tabular}
\par\end{center}


We see that $s_{2}(\chi_{2,3,5,6})=0$, $s_{2}(\mathbb{I})=1$, $s_{2}(\chi_{4})=1$,
and $s_{2}(\chi_{7,8})=1$ because the corresponding representations
are inductions of rational representations. 


So for example, the number of square roots of 
\begin{eqnarray*}
((1,1,1),\id) & = & \sum_{\chi\in\mathrm{Irr}}s_{2}(\chi)\cdot\chi((1,1,1),\id)\\
 & = & 1-1-3+3=0.
\end{eqnarray*}
Indeed, there are no elements of order $4$. 


We could explicate Brauer's induction theorem, e.g., $\chi_{7}=\mathrm{Ind}_{c_{2}^{3}}.$
But for $\chi_{1\dots6}$, need to induce various characters of elementary
subgroups and look for linear relationships.

\item Let $G=\mbox{PSL}_{2}(\ff_{7})=\SL_{2}(\ff_{7})/\left\{ \pm\id\right\} $.
This is a simple group of order $168$. This has conjugacy classes
are $\id,\begin{pmatrix}5 & 0\\
0 & 3
\end{pmatrix},\begin{pmatrix}0 & 1\\
-1 & 0
\end{pmatrix},\begin{pmatrix}4 & 3\\
2 & 0
\end{pmatrix},\begin{pmatrix}1 & 1\\
0 & 1
\end{pmatrix},\begin{pmatrix}5 & 5\\
1 & 4
\end{pmatrix}$. Let $B=\left\langle \begin{pmatrix}5 & 0\\
0 & 3
\end{pmatrix},\begin{pmatrix}1 & 1\\
0 & 1
\end{pmatrix}\right\rangle \cong C_{7}\rtimes C_{3}$. Using Machke, you can check that if $\chi$ is a one-dimensional
non-trivial character of $B$, then $\chi\uparrow^{G}$ is irreducible
(8-dimensional). (To compute $ $$\left\langle \chi\uparrow^{G},\chi\uparrow^{G}\right\rangle =\left\langle \chi,\chi\uparrow^{G}\downarrow_{B}\right\rangle =\left\langle \chi,\oplus_{B\backslash G/B}\,^{g}\chi\downarrow_{B\cap\,^{g}B}\right\rangle $). 


$G$ acts doubly-transitively on lines in $\left(\ff_{7}\right)^{2}$.
($\SL_{2}$ acts doubly transitively, because, the line $\left\langle (1,0)\right\rangle $
is stabilised by $\begin{pmatrix}1 & 1\\
0 & 1
\end{pmatrix}$, which permutes the lines $\left\langle (a,1)\right\rangle $; and
$\pm\id$ acts trivially). Hence the (permutation character $-\mathbb{I}$)
is irreducible. ($7$-dimensional).


Also, $\mbox{PSL}_{2}(\ff_{7})\cong\GL_{3}(\ff_{2})\cong\mbox{PSL}_{3}(\ff_{2})$,
which acts on non-zero vectors in $\left(\ff_{2}\right)^{3}$. So
we get an irreducible $7-1=6$-dimensional character. By $\left|G\right|=\sum\left(\dim\chi\right)^{2}$,
there are $2$ $3$-dimensional irreducible characters. 


These can be obtained by column orthogonality. It then turns out that
$s_{2}(3$-dimensionals$)=0$. The $7$-dimensional character is the
character of a $\qq[G]$-module$/\qq[G]$-submodule, so the representation
is defined over $\qq$. Similarly, the $6$-dimensional one. The $8$-dimensional
is, a priory, defined over $\qq\left(\sqrt{-3}\right)=\qq\left(e^{2\pi i/3}\right)$.
But on the other hand, it is real-valued, and $s_{2}(\chi\uparrow^{G})=1$
(check, but only if you feel adventurous). From general theory of
Schur indices (beyond the scope of this course), $m_{\qq}(\chi\uparrow^{G})=1$. 

\item Let $G=S_{7}$. The conjugacy classes are: $\id,(12),(12)(34),(12)(34)(56),\dots$
(there are $15$ of them).


Obvious Characters: $\mathbb{I}$, sign, $\chi=$ (standard permutation
character $-\mathbb{I}$) which is $6$ dimensional, and $\chi\otimes\mathrm{sign}$.


Consider $\chi^{\otimes2}=S^{2}\chi\oplus\wedge^{2}\chi$, note that
$S^{2}\chi=\mathbb{I}+\dots$, hence it is still reducible. We have
$\wedge^{2}\chi(g)=\frac{1}{2}\left(\chi(g)^{2}-\chi(g^{2})\right)$,
so we can explicitly work out its values, and hence can calculate
that $\left\langle \wedge^{2}\chi,\wedge^{2}\chi\right\rangle =1$,
so $\wedge^{2}\chi$ is irreducible. This also gives us $\wedge^{2}\chi\otimes\mathrm{sign}$
for free.


It turns out that $\wedge^{k}\chi(g)$ is $(-1)^{k}$ times the coefficient
of $x^{k}$ in $\det(\rho(g)-x\cdot\id)$. So again, we can explicatively
calculate $\wedge^{3}\chi(g)$ for all $g$, and find that $\wedge^{3}\chi$
is irreducible. 

\item Generalised quaternions: $Q_{2^{n+2}}=\left\langle c,x|c^{2^{n}}=x^{2},xcx^{-1}=c^{-1}\right\rangle $
(so $\ord(c)=2^{n+1},\ord(x)=4)$. Hence $\left|Q_{2^{n+2}}\right|=2^{n+2}$.


Let $n=2$, so consider $G=Q_{2^{4}}$: We have the conjugacy classes: 


\begin{center}
\begin{tabular}{|c|c|c|}
\hline 
Representative & Size & Order\tabularnewline
\hline 
\hline 
$\id$ & $1$ & $1$\tabularnewline
\hline 
$x^{2}$ & $1$ & $2$\tabularnewline
\hline 
$x$ & $4$ & $4$\tabularnewline
\hline 
$c^{2}$ & $2$ & $4$\tabularnewline
\hline 
$cx$ & $4$ & $4$\tabularnewline
\hline 
$c$ & $2$ & $8$\tabularnewline
\hline 
$c^{3}$ & $2$ & $8$\tabularnewline
\hline 
\end{tabular}
\par\end{center}


Let $G'=\left\langle c^{2}\right\rangle $, $G/G'\cong C_{2}\times C_{2}$,
so we get 4 1-dimensional characters.


\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 
 & $\id$ & $x^{2}$ & $x$ & $c^{2}$ & $cx$ & $c$ & $c^{3}$\tabularnewline
\hline 
\hline 
$\mathbb{I}$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$ & $1$\tabularnewline
\hline 
$\chi_{1}$ & $1$ & $1$ & $-1$ & $1$ & $1$ & $-1$ & $1$\tabularnewline
\hline 
$\chi_{2}$ & $1$ & $1$ & $-1$ & $1$ & $-1$ & $1$ & $1$\tabularnewline
\hline 
$\chi_{3}$ & $1$ & $1$ & $1$ & $1$ & $-1$ & $-1$ & $-1$\tabularnewline
\hline 
$\tau_{1}$ & $2$ & $2$ & $0$ & $-2$ & $0$ & $0$ & $0$\tabularnewline
\hline 
$\tau_{2}$ & $2$ & $-2$ & $0$ & $0$ & $0$ & $\sqrt{2}$ & $-\sqrt{2}$\tabularnewline
\hline 
$\tau_{3}$ & $2$ & $-2$ & $0$ & $0$ & $0$ & $-\sqrt{2}$ & $\sqrt{2}$\tabularnewline
\hline 
\end{tabular}We then calculate an induced character from a non-abelian quotient
of order $8$ (hence either $Q_{8}$ or $D_{8}$, but they have the
same character table)


We now compute the Frobenius - Schur indicator: $s_{2}(\chi_{i})=1$,
$s_{2}(\tau_{1})=\frac{1}{16}\sum_{g\in G}\tau_{1}(g^{2})=\frac{1}{16}\left(2+2+2\cdot2+4\cdot2+\dots\right)>0$,
hence $s_{2}(\tau_{1})=1$. This actually shows that $G/\left\langle c^{4}\right\rangle \cong D_{8}$
(and not $Q_{8})$, and $\tau_{1}$ is realisable over $\qq$. Next
we compute $s_{2}(\tau_{2})=s_{2}(\tau_{3})=-1$. So $\tau_{2},\tau_{3}$
are nor realisable over $\rr$ (in particular not over $\qq\left(\sqrt{2}\right)$).
In particular, $m_{\qq}(\tau_{2,3})>1$. Let $H=\left\langle c\right\rangle \cong C_{8}$,
\begin{eqnarray*}
\left\langle \tau_{2}\downarrow_{H},\left(c\mapsto e^{2\pi i/8}\right)\right\rangle _{H} & = & \frac{1}{\left|H\right|}\sum_{k=0}^{7}\tau_{2}(c^{k})\cdot e^{2\pi ik/8}\\
 & = & 1\\
 & = & \left\langle \tau_{2},\left(c\mapsto e^{2\pi i/8}\right)\uparrow^{G}\right\rangle _{G}.
\end{eqnarray*}
Hence $\tau_{2}=\left(c\mapsto e^{2\pi i/8}\right)\uparrow^{G}$.
So we can define $\tau_{2}$ over $\qq\left(e^{2\pi i/8}\right)$.
Note that $[\qq\left(e^{2\pi i/8}\right):\qq\left(\sqrt{2}\right)]=2$,
so $m_{\qq}(\tau_{2})=2$. Similarly for $\tau_{3}$. This last calculation
also verifies Brauer's induction theorem for $\tau_{2}$ and $\tau_{3}$. 


Let us verify Artin's Induction: 
\begin{itemize}
\item The cyclic subgroups are: $1,Z=\left\langle x^{2}\right\rangle ,C_{1}=\left\langle c^{2}\right\rangle ,C_{2}=\left\langle x\right\rangle ,C_{3}=\left\langle cx\right\rangle ,C_{4}=\left\langle c\right\rangle $. 
\item So we get $\mathbb{I}_{\left\{ 1\right\} }\uparrow^{G}=1+\chi_{1}+\chi_{2}+\chi_{3}+\chi_{4}+2\cdot(\tau_{1}+\tau_{2}+\tau_{3})$
\item $\mathbb{I}_{Z}\uparrow^{G}=1+\chi_{1}+\chi_{2}+\chi_{3}+2\tau_{1}$
\item etc
\end{itemize}
\end{enumerate}
\newpage{}


\section{Symmetric Groups}

Motivation: 
\begin{enumerate}
\item Artin - Wedderburn: $\cc[G]\cong\oplus_{i=1}^{t}M_{n_{i}}(\cc)$.
It's clear what the simple modules of the RHS look like (they are,
as submodules of the regular module, columns). If $e_{i}=\left(0,\dots,0,I_{n_{i}},0,\dots,0\right)\in\oplus M_{n_{i}}(\cc)$,
then $e_{i}\cdot\oplus M_{n_{i}}(\cc)$ is a direct sum of $n_{i}$
mutually isomorphic simple modules. By finding that $e_{i}=\frac{\chi_{i}(1)}{\left|G\right|}\sum_{g\in G}\chi_{i}(g^{-1})g\in\cc[G]$,
one finds the ``$\chi_{i}$''-isotypical block $M_{n_{i}}(\cc)$
as a submodule of $\cc[G]$. For example:

\begin{itemize}
\item $\mathbb{I}=\left\langle \sum_{g\in G}g\right\rangle \leq\cc[G]$. 
\item $G=S_{3}$, $\chi$ the standard representation, so $e_{\chi}=\frac{2}{6}\left(2\id-(123)-(132)\right)\in\cc[S_{3}],$
and $e_{\chi}\cc[G]$ is $4$-dimensional. 
\end{itemize}

But if we knew what $f_{i}=\begin{pmatrix}1\\
 & 0\\
 &  & \ddots\\
 &  &  & 0
\end{pmatrix}\in M_{n_{i}}(\cc)$ looks like as an element of $\cc[G]$, then $\cc[G]f_{i}$ would
give us a simple summand. 

\item We know that the number of conjugacy classes of elements of $G$ is
the same as the number of isomorphism classes of simple $\cc[G]$-modules.
But there is, in general, no canonical bijection between these two
sets. 
\end{enumerate}

\subsection{Young Diagrams}

Conjugacy classes of $S_{n}\leftrightarrow$cycle types $\leftrightarrow$
partitions $\lambda=(\lambda_{1},\dots,\lambda_{k})$ of $n$ such
that $\lambda_{i}\in\nn$, $\lambda_{1}\geq\lambda_{2}\ge\dots\geq\lambda_{k}$
and $\sum\lambda_{i}=n$. 

The associated Young diagram is

\begin{tabular}{|c|c|c|c|cccc}
\cline{1-5} 
 &  &  &  & \multicolumn{1}{c|}{} &  &  & $\lambda_{1}$ boxes\tabularnewline
\cline{1-5} 
 &  &  &  &  &  &  & $\lambda_{2}$ boxes\tabularnewline
\cline{1-4} 
\multicolumn{1}{c}{} & \multicolumn{1}{c}{$\vdots$} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{$\vdots$} &  &  &  & $\vdots$\tabularnewline
\cline{1-3} 
 &  &  & \multicolumn{1}{c}{} &  &  &  & $\lambda_{k}$ boxes\tabularnewline
\cline{1-3} 
\end{tabular}

For example

\begin{tabular}{|c|c|c|c|c}
\hline 
 &  &  &  & \multicolumn{1}{c|}{}\tabularnewline
\hline 
 &  &  &  & \tabularnewline
\cline{1-4} 
 &  &  &  & \tabularnewline
\cline{1-4} 
 & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \tabularnewline
\cline{1-1} 
 & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \tabularnewline
\cline{1-1} 
\end{tabular}which corresponds to $5+4+4+1+1=15$.

The conjugate partition is obtained by reflecting the Young diagram
in the \textbackslash{} diagonal. A Young tableau is a numbering of
the boxes of a Young Diagram by numbers from $1$ to $n$. $S_{n}$
acts on the Young tableau of any given Young diagram. Fix any tableau
of the Young diagram corresponding to $\lambda$ (e.g., the obvious
one), define $P_{\lambda}=\{g\in S_{n}|g\mathrm{\, fixes\, each\, row\, of\, the\, tableau}\}$.

For example

$\lambda=$%
\begin{tabular}{|c|c|c}
\hline 
$1$ & $2$ & \multicolumn{1}{c|}{$3$}\tabularnewline
\hline 
$4$ & $5$ & \tabularnewline
\cline{1-2} 
$6$ & \multicolumn{1}{c}{} & \tabularnewline
\cline{1-1} 
$7$ & \multicolumn{1}{c}{} & \tabularnewline
\cline{1-1} 
\end{tabular}, then $P_{\lambda}=\left\langle (123),(45),(12)\right\rangle \cong S_{3}\times S_{2}\times S_{1}\times S_{1}$. 

We also define $Q_{\lambda}=\left\{ g\in S_{n}|g\mathrm{\, fixes\, each\, columns\, of\, the\, tableau}\right\} $.
Define: 
\begin{itemize}
\item $a_{\lambda}=\sum_{g\in P_{\lambda}}g\in\cc[S_{n}]$
\item $b_{\lambda}=\sum_{g\in Q_{\lambda}}\sgn(g)\cdot g\in\cc[S_{n}]$, 
\item $c_{\lambda}=a_{\lambda}\cdot b_{\lambda}$. This is the Young symmetriser
corresponding to $\lambda.$ \end{itemize}
\begin{example*}
Let $G=S_{3}$ and $\lambda=(2,1)$, i.e., %
\begin{tabular}{|c|c|}
\hline 
$1$ & $2$\tabularnewline
\hline 
$3$ & \multicolumn{1}{c}{}\tabularnewline
\cline{1-1} 
\end{tabular}. Then $P_{\lambda}=\left\langle (12)\right\rangle $, $Q_{\lambda}=\left\langle (13)\right\rangle $,
$a_{\lambda}=\id+(12)$, $b_{\lambda}=\id-(13)$ and hence $c_{\lambda}=\id+(12)-(13)-(123)$. \end{example*}
\begin{thm}
$V_{\lambda}:=\cc[G]\cdot c_{\lambda}$ is a simple $S_{n}$-module,
only depending on $\lambda$ up to isomorphism. $V_{\lambda}=V_{\mu}$
if and only $\lambda=\mu$, and all simple modules are isomorphic
to some $V_{\lambda}$. 
\end{thm}
Reference for the proof: Fulton, Harris: Representation Theory: a
first course. 
\begin{example*}
Continuing from above, $G=S_{3}$ and $\lambda=(2,1)$, we get $\cc[G]c_{\lambda}=\left\langle c_{\lambda},(13)\cdot c_{\lambda}\right\rangle :=\left\langle v_{1},v_{2}\right\rangle $.
Then see how $S_{3}$ acts on $v_{1},v_{2}$:
\begin{itemize}
\item $(13)v_{1}=v_{2}$, $(13)v_{2}=v_{1}$
\item $(123)v_{1}=(123)+(23)-(12)-(132)=-v_{1}-v_{2}$, $(123)v_{2}=v_{1}$. 
\end{itemize}
\end{example*}

\begin{example*}
Let $G=S_{n}$:
\begin{itemize}
\item Consider $\lambda=(n)$. Then we have $P_{\lambda}=S_{n}$, $Q_{\lambda}=\{1\}$,
$a_{\lambda}=\sum_{g\in S_{n}}g=c_{\lambda}$. So $V_{\lambda}\cong\mathbb{I}$.
\item Consider $\lambda=(1,\dots,1)$. Then we have $P_{\lambda}=\left\{ 1\right\} ,Q_{\lambda}=S_{n}$,
$b_{\lambda}=\sum_{g\in S_{n}}\sgn(g)\cdot g=c_{\lambda}$. So $V_{\lambda}\cong\mathrm{sign}$. 
\item Consider $\lambda=(n-1,1)$. Then we have $P_{\lambda}=S_{n-1}=\mathrm{stab}_{S_{n}}(n)\leq S_{n}$,
$Q_{\lambda}=\left\langle (1n)\right\rangle $. So $a_{\lambda}=\sum_{g\in\mathrm{Stab}(n)}g$,
$b_{\lambda}=\id-(1n)$, hence $c_{\lambda}=\sum_{g\in\mathrm{Stab}(n)}g-\sum_{g\in\mathrm{Stab}(n)\cdot(1n)}g$.
After some work, you should find that $V_{\lambda}$ is the standard
representation of $S_{n}$ ($\left(n-1\right)$-dimensional)
\item More generally: $\lambda=(n-i,1,\dots,1)$, then $V_{\lambda}=\wedge^{i}V_{(n-1,1)}$. 
\end{itemize}
\end{example*}

\subsection{Frobenius's Formula}

Let $\chi_{\lambda}$ be the character of $V_{\lambda}$. Let $g\in S_{n}$,
$i_{1}=$ number of $1$-cycles in $g$, $i_{2}=$ number of $2$-cycles
of $g$, $\dots$ , $i_{n}=$ number of $n$-cycles of $g$. As usual
$\lambda=(\lambda_{1},\dots,\lambda_{k})$. Consider the following
symmetric function in $x_{1},\dots,x_{k}$:
\begin{itemize}
\item $P_{j}(\underbar{x})=x_{1}^{j}+\dots+x_{k}^{j}$, $1\leq j\leq n$
\item $\Delta(\underbar{x})=\prod_{i<j}(x_{i}-x_{j})$
\end{itemize}
Set $l_{1}=\lambda_{1}+(k-1)$, $l_{2}=\lambda_{2}+(k-2)$, $\dots$
, $l_{k}=\lambda_{k}$. 
\begin{thm}[Frobenius]
 $\chi_{\lambda}(g)$ is the coefficient of $x_{1}^{l_{1}}\dots x_{k}^{l_{k}}$
in 
\[
\Delta(\underbar{x})\prod_{j=1}^{n}P_{j}(\underbar{x})^{i_{j}}.
\]
\end{thm}
\begin{example*}
Let $G=S_{3}$, $\lambda=(2,1)$.
\begin{itemize}
\item $g=(12)$, then we have $i_{1}=1$, $i_{2}=1$, $i_{3}=0$, $l_{1}=2+2-1=3$,
$l_{2}=1+2-2=1$. We have $\Delta(\underbar{x})=x_{1}-x_{2}$, so
$\chi_{\lambda}(g)$ is the coefficient of $x_{1}^{3}x_{2}$ in $(x_{1}-x_{2})(x_{1}+x_{2})^{1}(x_{1}^{2}+x_{2}^{2})^{1}=(x_{1}^{4}-x_{2}^{4})$.
So $\chi_{\lambda}(g)=0$.
\item $g=(123)$, then we have $i_{1}=0$, $i_{2}=0$, $i_{3}=1$, $l_{1}=3$,
$l_{2}=1$. So $\chi_{\lambda}(g)$ is the coefficient of $x_{1}^{3}x_{2}$
in $(x_{1}-x_{2})(x_{1}^{3}+x_{2}^{3})=x_{1}^{4}+x_{1}x_{2}^{3}-x_{2}x_{1}^{3}-x_{2}^{4}$.
So $\chi_{\lambda}(g)=-1$. 
\end{itemize}
\end{example*}
Note that with manipulation, we can find that 
\[
\dim\chi_{\lambda}=\frac{n!}{l_{1}!\dots l_{k}!}\cdot\prod_{i<j}(l_{1}-l_{j}).
\]


Another dimension formula: Define the look length of a box in a Young
Diagram is the number of boxes to the right and underneath it (counting
the box itself once). E.g.,

\begin{tabular}{|c|c|c|c}
\hline 
$7$ & $4$ & $3$ & \multicolumn{1}{c|}{$1$}\tabularnewline
\hline 
$5$ & $2$ & $1$ & \tabularnewline
\cline{1-3} 
$2$ & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \tabularnewline
\cline{1-1} 
$1$ & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \tabularnewline
\cline{1-1} 
\end{tabular}
\begin{thm}
~
\[
\dim\chi_{\lambda}=\frac{n!}{\prod\mathrm{hook\, length\, in\,}\lambda}
\]

\end{thm}
E.g., with the example above, we find that $\chi_{\lambda}=\frac{9!}{2\cdot2\cdot3\cdot4\cdot5\cdot7}=\frac{6\cdot8\cdot9}{2}=216$. 
\begin{rem*}
Since $c_{\lambda}\in\qq[G]$, all simple $\cc[S_{n}]$-modules are
realisable over $\qq$. 

\newpage{}
\end{rem*}

\section{Revision Quiz Session}


\subsubsection*{True or False}
\begin{itemize}
\item If $H\leq G$, $\chi\in\mathrm{Irr}(G)$, then all irreducible summands
of $\chi\downarrow_{H}$ have the same dimension.


No: Let $H=S_{3}$, we are looking for $G\geq S_{3}$ and $\chi\in\mathrm{Irr}(G)$,
such that $\left\langle \chi,\mathrm{stand\uparrow^{G}}\right\rangle =\left\langle \chi\downarrow_{S_{3}},\mathrm{stand}\right\rangle \neq0\neq\left\langle \chi\downarrow_{S_{3}},\mathrm{a\,}1\mathrm{-dimensional\, char}\right\rangle =\left\langle \chi,1\mathrm{-dim}\uparrow^{G}\right\rangle $.
Let $\rho$ be the standard representation and $\tau$ the $1$-dimensional
representation, then we want $\left\langle \rho\uparrow^{G},\tau\uparrow^{G}\right\rangle \neq0$.
We have
\begin{eqnarray*}
\left\langle \rho\uparrow^{G},\tau\uparrow^{G}\right\rangle  & = & \left\langle \rho\uparrow^{G}\downarrow_{S_{3}},\tau\right\rangle \\
 & = & \sum_{S_{3}\backslash G/S_{3}}\left\langle \,^{g}\rho\downarrow_{S_{3}\cap\,^{g}S_{3}}\uparrow^{S_{3}},\tau\right\rangle 
\end{eqnarray*}
If $G=S_{6}$, $S_{3}=\mathrm{Stab}(4,5,6)$, there exists a $g\in G$
such that $\,^{g}S_{3}=\mathrm{Stab}(1,2,3)$, so $S_{3}\cap\,^{g}S_{3}=\left\{ 1\right\} $,
so $\rho\downarrow_{S_{3}\cap\,^{g}S_{3}}=\mathbb{I}+\mathbb{I}$. 

\item If $H=N\lhd G$, $\chi\in\mathrm{Irr}(G)$, then all irreducible summand
of $\chi\downarrow_{H}$ have the same dimension.


Yes: (Clifford).

\item If $H\leq G$, $\chi\in\mathrm{Irr}(H)$, then all irreducible summand
of $\chi\uparrow^{G}$ have the same dimension.


No: Consider the regular representation with any groups.

\item If $H=N\lhd G$, $\chi\in\mathrm{Irr}(H)$, then all irreducible summand
of $\chi\uparrow^{G}$ have the same dimension.


No: by the above reasoning.

\item If $\chi$ is realisable over $K$, then so are all irreducible summands
of $\chi\downarrow_{H}$.


No: Let $G=S_{3}$, $H=C_{3}$, $\chi$ be the standard representation.

\item If $\chi$ is realisable over $K$, then so are all irreducible summands
of $\chi\uparrow^{G}$.


No: Take the regular representation.

\item If $\chi_{H}\uparrow^{G}$ is realisable over $K$, then so is $\chi$. 


No: Let $\chi\in\mathrm{Irr}(C_{3})$ be non-trivial and $G=S_{3}$. 

\item If $F/K$ is an extension of fields, then $m_{F}(\chi)\leq m_{K}(\chi)$. 


Yes: By definition of $m(\chi)$. (Also using Tower laws we get $m_{F}(\chi)\big|m_{K}(\chi)$)

\end{itemize}
Recall: If $F/K$ is a finite field extension, $M$ an $F[G]$-module,
by thinking of $F$ as a $K$-vector space, we can think of $M$ as
a $K[G]$-module of $K$-dimension equal to $\dim_{F}(M)\cdot[F:K]$. 

A completely different instance of the same ``philosophy'': Let
$C$ be a curve over $\qq$, e.g., $C:y^{2}=x^{3}+x$. We can think
of this as a curve over $\qq[i]$. So now, write $x=u+iv$ and $y=w+iz$,
with $u,v,w,z\in\qq$. So $C:(w+iz)^{2}=(u+iv)^{3}+(u+iv)$. By equating
real and imaginary parts, we get two equations in four unknowns over
$\qq$. So we now have a surface over $\qq$. 
\end{document}
