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chap4.tex
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%!TEX TS-program = xelatex
%!TEX encoding = UTF-8 Unicode
\documentclass[11pt]{report}
\usepackage[margin = 1in]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{mleftright}
\usepackage{enumitem}
\usepackage{textcomp, gensymb}
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\usepackage{bbding}
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\usepackage{extarrows}
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\usepackage{fancyhdr} % Header and Footer formatting
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% \usepackage{mathpazo}
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% Header and Footer Information
\lhead{\small\emph{Baitian Li}}
\chead{}
\rhead{\textsc{Representation Theory of Finite Groups}}
\lfoot{\today}
\cfoot{}
\rfoot{\thepage\ of \pageref{LastPage}} % Counts the pages.
\makeatletter % This provides a total page count as \ref{NumPages}
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%\lineskiplimit=-\maxdimen\relax
\usepackage{amsthm} % This will create the Problem environment
\newtheorem*{lemma}{Lemma}
\newtheorem*{theorem}{Theorem}
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\setlist[enumerate, 1]{label = {\arabic*.}}
% \setlist[enumerate, 2]{label = {(\alph*)}}
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itemsep = 0em}
\newcommand{\bbR}{\mathbb R}
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\newcommand{\bbQ}{\mathbb Q}
\newcommand{\bbC}{\mathbb C}
\newcommand{\Id}{\mathit{Id}}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\End}{End}
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\newcommand{\ang}[1]{\langle #1 \rangle}
\newcommand{\Ang}[1]{\left\langle #1 \right\rangle}
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\begin{document}
\setcounter{chapter}{3}
\chapter{Character Theory and the Orthogonality Relations}
\begin{exercise}
Let $\varphi\colon G \to \GL(U)$, $\psi\colon G \to \GL(V )$ and
$\rho\colon G \to \GL(W )$ be representations of a group $G$. Suppose that
$T\in \Hom_G(\varphi, \psi)$ and $S\in \Hom_G(\psi, \rho)$.
Prove that $ST \in \Hom_G(\psi, \rho)$.
\begin{proof}
For each $g$, $ST \varphi_g = S \psi_g T = \rho_g ST$, so $ST\in \Hom_G(\psi, \rho)$.
\end{proof}
\end{exercise}
\begin{exercise}
Let $\varphi$ be a representation of a group $G$ with character $\chi_\varphi$. Prove that
$\chi_\varphi(g^{-1}) =\overline{\chi_\varphi(g)}$.
\begin{proof}
$\varphi$ is equivalent to some unitary representation $\rho$, then we have $\chi_\varphi = \chi_\rho$.
Then we have $\chi_\rho(g^{-1}) = \Tr(\rho_{g^{-1}}) = \Tr(\rho_g^*) = \overline{\Tr \rho_g} = \overline{\chi_\rho(g)}$.
\end{proof}
\end{exercise}
\begin{exercise}
Let $\varphi\colon G \to \GL(V )$ be an irreducible representation. Let
\[ Z(G) = \{ a\in G \mid ag = ga, \forall g\in G\} \]
be the center of $G$. Show that if $a \in Z(G)$, then $\varphi(a) = \lambda I$ for some $\lambda \in \bbC^*$.
\begin{proof}
Since $a\in Z(G)$, we have $\varphi_a \varphi_g = \varphi_g \varphi_a$ for all $g$, so
$\varphi_a\in \Hom_G(\varphi, \varphi)$. Then we have $\varphi_a = \lambda I$, since
$\varphi_a\in \GL(V)$, we have $\lambda \neq 0$.
\end{proof}
\end{exercise}
\begin{exercise}
Let $G$ be a group. Show that $f \colon G \to \bbC$ is a class function if and
only if $f(gh) = f(hg)$ for all $g, h \in G$.
\begin{proof}
Since this is for all $g, h$, we substitude by $g = h^{-1}x$, the statement
is equivalent to $f(h^{-1}xh) = f(x)$ for all $x, h\in G$. Which is the definition
of class function.
\end{proof}
\end{exercise}
\begin{exercise}
For $v = (c_1, \dots , c_m) \in (\bbZ/2\bbZ)^m$, let $\alpha(v) = \{i \mid c_i = [1]\}$. For
each $Y \subseteq \{1, \dots, m\}$, define a function $\chi_Y\colon (\bbZ/2\bbZ)^m \to \bbC$ by
\[ \chi_Y (v) = (-1)^{|\alpha(v) \cap Y|}. \]
\begin{enumerate}
\item Show that $\chi_Y$ is a character.
\begin{proof}
We only need to prove that $\chi_Y$ is a representation, this can be verified by definition.
\end{proof}
\item Show that every irreducible character of $(\bbZ/2\bbZ)^m$ is of the form $\chi_Y$ for some
subset $Y$ of $\{1, \dots , m\}$.
\begin{proof}
Since $(\bbZ/2\bbZ)^m$ is abelian, the irreducible representation must have degree $1$.
Thus such character $\chi$ is a representation. Let $Y = \{ i \mid e_i = ([0], \dots, [1], \dots, [0]) = -1 \}$,
where the $i$th position of $e_i$ is $[1]$.
Then $\chi$ agrees with $\chi_Y$ on $\{ e_i \}$, which generates the whole $(\bbZ/2\bbZ)^m$.
Then we have $\chi = \chi_Y$.
\end{proof}
\item Show that if $X, Y \subseteq \{1, \dots , m\}$, then $\chi_X(v)\chi_Y (v) = \chi_{X\Delta Y}(v)$ where
$X \Delta Y = X\cup Y \smallsetminus (X\cap Y)$ is the symmetric difference.
\begin{proof}
Can be verified by definition.
\end{proof}
\end{enumerate}
\end{exercise}
\begin{exercise}
Let $\sgn\colon S_n \to \bbC^*$ be the representation given by
\[ \sgn(\sigma) = \begin{cases}
1 & \sigma \text{ is even}\\
-1 & \sigma \text{ is odd.}
\end{cases} \]
Show that if $\chi$ is the character of an irreducible representation of $S_n$ not equivalent
to $\sgn$, then
\[ \sum_{\sigma \in S_n} \sgn(\sigma) \chi(\sigma) = 0. \]
\begin{proof}
Since $\sgn$ has degree $1$, it is irreducible and $\chi_{\sgn} = \sgn$.
The equation directly follows from $\ang{\chi, \chi_{\sgn}} = 0$, which is implied by
the first orthogonality relations.
\end{proof}
\end{exercise}
\begin{exercise}
Let $\varphi\colon G \to \GL_n(\bbC)$ and $\rho\colon G \to \GL_m(\bbC)$ be representations.
Let $V = M_{mn}(\bbC)$. Define $\tau\colon G \to \GL(V )$ by $\tau_g(A) = \rho_g A \varphi^T_g$.
\begin{enumerate}
\item Show that $\tau$ is a representation of $G$.
\begin{proof}
Clearly $\tau_g$ is a linear mapping, and $\tau_1(A) = \rho_1 A \varphi_1^T = A$, and
$\tau_g \tau_h(A) = \tau_g(\rho_h A \varphi_h^T) = \rho_{gh}A\varphi_{gh}^T = \tau_{gh}(A)$.
Thus $\tau_g \in \GL(V)$ for all $g$, and $\tau$ is a homomorphism, therefore $\tau$ is a representation.
\end{proof}
\item Show that
\[ \tau_g E_{kl} = \sum_{i,j} \rho_{ik}(g) \varphi_{jl}(g) E_{ij}. \]
\begin{proof}
By definition.
\end{proof}
\item Prove that $\chi_\tau(g) = \chi_\rho(g)\chi_\varphi(g)$.
\begin{proof}
\[ \chi_\tau(g) = \Tr(\tau_g) = \sum_{i, j} (\tau_g E_{ij})_{ij}
=\sum_{i,j} \rho_{ii}(g) \varphi_{jj}(g) = \chi_\rho(g) \chi_\varphi(g). \qedhere \]
\end{proof}
\item Conclude that the pointwise product of two characters of $G$ is a character of $G$.
\begin{proof}
The product character $\chi_\rho(g) \chi_\varphi(g)$ is given by the representation $\tau$.
\end{proof}
\end{enumerate}
\end{exercise}
\begin{exercise}
Let $\alpha\colon S_n \to \GL_n(\bbC)$ be the standard representation from
Example 3.1.9.
\begin{enumerate}
\item Show that $\chi_\alpha(\sigma)$ is the number of fixed points of $\sigma$, that is, the number of
elements $k \in \{1, \dots , n\}$ such that $\sigma(k) = k$.
\begin{proof}
\[ \chi_\alpha(\sigma) = \Tr (\alpha(\sigma)) = \sum_{i} \ang{e_i, e_{\sigma(i)}}. \qedhere \]
\end{proof}
\item Show that if $n = 3$, then $\ang{\chi_\alpha, \chi_\alpha} = 2$ and hence $\alpha$ is not irreducible.
\begin{proof}
We have $\ang{\chi_\alpha, \chi_\alpha} = \frac 1{6} \sum_{\sigma} \|\chi_\alpha(\sigma)\|^2$.
\begin{itemize}
\item For $\sigma = 1$, $\norm{\chi_\alpha(\sigma)}^2 = 9$.
\item For $\sigma = (1\ 2), (2\ 3), (3\ 1)$, $\norm{\chi_\alpha(\sigma)}^2 = 1$.
\item Otherwise, $\norm{\chi_\alpha(\sigma)}^2 = 0$.
\end{itemize}
So the sum is $9 + 3\cdot 1 = 12$, we have $\ang{\chi_\alpha, \chi_\alpha} = 2$.
\end{proof}
\end{enumerate}
\end{exercise}
\begin{exercise}
Let $\chi$ be a non-trivial irreducible character of a finite group $G$. Show
that
\[ \sum_{g\in G} \chi(g) = 0. \]
\begin{proof}
Since the trivial representation $1$ is irreducible, we have a irreducible character $\chi_1(g) = 1$
for all $g$. Since $\chi\neq \chi_1$, by the first orthogonality relations we have
$\ang{\chi, \chi_1} = 0$.
\end{proof}
\end{exercise}
\begin{exercise}
Let $\varphi\colon G\to H$ be a surjective homomorphism and let
$\psi \colon H\to \GL(V )$ be an irreducible representation. Prove that $\psi \circ \varphi$ is an irreducible
representation of $G$.
\begin{proof}
Suppose $W\leq V$ is $G$-invariant, then for each $h\in H$, since $\varphi$ is surjective, we have
$\varphi(g) =h$, thus $\psi_h W = (\psi \circ \varphi)_g W \leq W$. We must have $W = 0$ or $W = V$.
\end{proof}
\end{exercise}
\setcounter{exercise}{12}
\begin{exercise}
Let $G$ be a group and let $G'$ be the commutator subgroup of $G$. Let $\varphi: G \to G/G'$ be the
canonical homomorphism given by $\varphi(g) = gG'$. Prove
that every degree one representation $\rho\colon G \to \bbC^*$ is of the form $\psi \circ \varphi$ where
$\psi\colon G/G' \to \bbC^*$ is a degree one representation of the abelian group $G/G'$.
\begin{proof}
Let $N = \ker \rho$, since $G/N \cong \rho(G) \subset \bbC^*$ is commutative, we have
$G' \leq N$. Then $\rho$ induces the homomorphism $\psi$ such that $\rho = \psi \circ \varphi$
naturally.
\end{proof}
\end{exercise}
\begin{exercise}
Show that if $G$ is a finite group and $g$ is a non-trivial element
of $G$, then there is an irreducible representation $ϕ$ with $\varphi(g) \neq I$.
\begin{proof}
Consider the regular representation $L\colon G \to \GL(\bbC G)$, since $L_g e = g$,
we have $L_g \neq I$. Consider the decomposition of representation
$L \sim d_1\varphi^{(1)} \oplus \dots \oplus d_s\varphi^{(s)}$, suppose $\varphi^{(i)}_g$
is identity for all $i$, we must have $L_g= I$, so we must have some $\varphi^{(i)}$
such that $\varphi^{(i)}_g \neq I$.
\end{proof}
\end{exercise}
\begin{exercise}
This exercise provides an alternate proof of the orthogonality
relations for characters. It requires Exercise 3.5. Let $\varphi\colon G \to \GL_m(\bbC)$ and
$\psi\colon G \to \GL_n(\bbC)$ be representations of a finite group $G$.
Let $V = M_{mn}(\bbC)$ and define a representation $\rho\colon G \to \GL(V )$ by $\rho_g(A) = \varphi_g A \psi_{g}^*$.
\begin{enumerate}
\item Prove that $\chi_\rho(g) = \chi_\varphi(g)\overline{\chi_\psi(g)}$ for $g \in G$.
\begin{proof}
Follows from Exercise 4.7.
\end{proof}
\item Show that $V^G = \Hom_G(\psi, \varphi)$.
\begin{proof}
$A \in V^G \iff \varphi_g A \psi_g^* = A \iff \varphi_g A = A \psi_g \iff A \in \Hom_G(\psi, \varphi)$.
\end{proof}
\item Deduce $\dim \Hom_G(\psi, \varphi) = \ang{\chi_\varphi, \chi_\psi}$ using Exercise 3.5.
\begin{proof}
\[ \ang{\chi_\varphi, \chi_\psi} = \frac 1{|G|} \sum_{g\in G} \chi_\varphi(g)\overline{\chi_\psi(g)}
= \frac 1{|G|} \sum_{g\in G} \chi_\rho(g)
\xlongequal{\text{exercise 3.5}} \dim V^G = \dim \Hom_G(\psi, \varphi). \qedhere \]
\end{proof}
\item Deduce using Schur's lemma that if $\varphi$ and $\psi$ are irreducible representations, then
\[ \ang{\chi_\varphi, \chi_\psi} = \begin{cases}
1 & \varphi \sim \psi,\\
0 & \varphi \nsim \psi.
\end{cases} \]
\begin{proof}
Schur's lemma states that,
if $\varphi \nsim \psi$, we have $\Hom_G(\psi, \varphi) = 0$, otherwise
$\Hom_G(\psi, \varphi) = \{ \lambda I \mid \lambda \in \bbC \}$.
\end{proof}
\end{enumerate}
\end{exercise}
\end{document}