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\begin{flashcard}[finite-abelian-has-G-irred-C-reps-prop]
\begin{proposition*}
  If $G$ is a finite abelian group then $G$ has
  exactly \cloze{$|G|$ \gls{irred}} $\CC$-\glspl{rep}.
\end{proposition*}

\begin{proof}
  \cloze{
  We saw \lastlecture\ that all \gls{irred}
  \glspl{rep} of an abelian group have
  \gls{rep_deg} $1$.
  \[ S = \{\text{\gls{simple} \glspl{rep} of
  $G$}\} / \sim \leftrightarrow \Hom(G,
  \CC^\times) \]
  Moreover, if $G = H \times K$, then
  \[ \Hom(G, \CC^\times) \leftrightarrow \Hom(H,
  \CC^\times) \times \Hom(K, \CC^\times) \qquad
  \varphi \mapsto (\varphi|_H, \varphi|_K) \]
  (need $\CC^\times$ is abelian to get
  surjectivity). Now by structure theorem for
  finite abelian groups, $G \simeq C_{n_1} \times
  \cdots \times C_{n_r}$ for some $n_1,\ldots n_r
  \in \NN$. So by a simple induction argument we
  can reduce to the case where $G$ is cyclic, $G
  \cong C_n = \langle x \rangle$ say.

  Then $\rho \in \Hom(G, \CC^\times)$ is
  determined by $\rho(x)$ and $\rho(x)^n = 1$,
  i.e. $\rho(x)$ is an $n$-th root of unity.
  Moreover, for $j = 0, \ldots, n - 1$, $x^k
  \mapsto e^{2\pi ijk/n}$ defines a
  \glsref[rep_deg]{$1$-dimensional}
  \gls{rep} of $G$.
  }
\end{proof}
\end{flashcard}

\begin{lemma*}
  \refsteplabel[Lemma]{lec7_sum_is_zero_lemma}
  If $(\rho_1, V_1)$ and $(\rho_2, V_2)$ are
  non-\gls{rep_iso}
  \glsref[rep_deg]{$1$-dimensional} \glspl{rep} of
  a finite group $G$, then
  \[ \sum_{g \in G} \rho_1(g^{-1}) \rho_2(g) = 0 \]
  (Note $\rho_1(g^{-1}) = \rho_1(g)^{-1}) =
  \ol{\rho_1(g)}$ since $\rho_1(g)^j = 1$ for some
  $j$ if $k = \CC$).
\end{lemma*}

\begin{proof}
  We've seen that $\Hom_k(V_1, V_2)$ is a
  \gls{rep} of $G$ via $g \cdot \varphi =
  \rho_2(g) \circ \varphi \circ
  \varphi_1(g^{-1})$. Moreover,
  \[ \sum_{g \in G} g \cdot \varphi \in
  \GHom_G(V_1, V_2) = 0 \]
  by our \nameref{lec5_key_idea} from \lecture{5} and
  \nameref{schurs_lemma}.
  Pick an \gls{rep_iso} $\varphi \in \Hom_k(V_1,
  V_2)$ and then
  \[ 0 = \sum_{g \in G} \rho_2(g) \varphi
  \rho_1(g^{-1}) = \left( \sum_{g \in G}
  \rho_1(g^{-1}) \rho_2(g) \right) \varphi \]
  So as $\varphi$ is injective, we're done.
\end{proof}

\begin{flashcard}[isotypic-component-defn]
\begin{definition*}[isotypic component]
  \glsnoundefn{iso_component}{$W$-isotypic
  component}{$W$-isotypic-components}
  \cloze{If $V$ is a \gls{comp_red} \gls{rep} of a
  group $G$ and $W$ is any \gls{simple} \gls{rep}
  of $G$, the \emph{$W$-isotypic component} of $V$
  is the smallest \gls{subrep} of $V$ containing
  all \glspl{subrep} of $V$ \gls{rep_iso} to $W$.}
\end{definition*}
\end{flashcard}
\vspace{-1em}

This exists since if $(V_i)_{i \in I}$ are
\glspl{subrep} of $V$ containing all
\glspl{subrep} of $V$ \gls{rep_iso} to $W$ then
$\bigcap_{i \in I} V_i$ is another (or we can
simply take the vector space sum of all
\glspl{subrep} \gls{rep_iso} to $W$).

\begin{flashcard}[unique-isotypical-decomp-defn]
\begin{definition*}[unique isotypical
  decomposition]
  \glsnoundefn{unique_iso_decomp}{unique
  isotypical decomposition}{unique isotypical
  decomposition}
  \glsnoundefn[unique_iso_decomp]{iso_decomp}
  {isotypical decomposition}{isotypical
  decompositions}
  \cloze{
  We say $V$ has a \emph{unique isotypical
  decomposition} if $V$ is the direct sum of its
  \isocomp{W} as $W$ goes over all \gls{simple}
  \glspl{rep} of $G$ (up to
  \glsref[rep_iso]{isomorphism}).
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[unique-iso-decomp-existence]
\begin{corollary*}
  If $G$ is a finite abelian group, then every
  $\CC$-\gls{rep} of $V$ has a
  \gls{unique_iso_decomp}.
\end{corollary*}

\begin{proof}
  \cloze{
  For each \glsref[rep_iso]{homomorphism}
  $\theta_i : G \to \CC^\times$, $i = 1, \ldots,
  |G|$ (i.e. each \gls{simple} \gls{rep} of $G$)
  we can define
  \[ W_i = \{v \in V \mid g \cdot v = \theta_i(g)
  v ~\forall g \in G\} \]
  the \isocomp{\theta_i} of $V$. Since $V$ is
  \gls{comp_red}, $V = \sum_{i = 1}^{|G|} W_i$. We
  need to show that if $\sum_{i = 1}^{|G|} w_i =
  0$ with $w_i \in W_i$ for each $i$, then $w_i =
  0$ for each $i$. But
  \[ \sum_{i = 1}^{|G|} w_i = 0 \implies 0 = g
  \cdot \sum_{i = 1}^{|G|} w_i = \sum_{i =
  1}^{|G|} \theta_i(g) w_i ~\forall g \in G \]
  Then for each $j$,
  \[ \sum_{i = 1}^{|G|} \left( \sum_{g \in G}
  \theta_j(g^{-1}) \theta_i(g) \right) w_i = 0 \]
  But by \hyperref[lec7_sum_is_zero_lemma]{the
  previous Lemma},
  \[ \LHS = \sum_{g \in G} \theta_j(g^{-1}) \theta_j(g)
  w_j = |G|w_j \]
  Thus $w_j = 0$ as required.
  }
\end{proof}
\end{flashcard}

This proof also works when $\CC$ is replace by any
other algebraically closed field with
characteristic $0$.

You will extend this to all finite groups on
\es{2}.

\newpage
\section{Characters}

Summary so far: We want to classify all
\glspl{rep} of a given (finite) group $G$. We've
seen when $G$ is finite and $\characteristic k =
0$ then every \gls{rep} decomposes as $V \cong
\bigoplus_{i = 1}^r n_i V_i$ with $V_1, \ldots,
V_r$ \gls{simple} and pairwise non-\gls{rep_iso}
and $n_i \ge 0$.

Moreover, if $k = \ol{k}$ then $n_i = \dim
\GHom_G(V_i, V)$. Next we want to discuss how to
classify \gls{irred} ($\CC$-)\glspl{rep} of a
finite group and understand how to \emph{compute}
the $n_i$ given $V$. We'll do both by character
theory.

\subsection{Definitions}

\begin{flashcard}[char-of-rep-defn]
\begin{definition*}[Character]
  \glspropdefn{char_rep}{character}{\gls{rep}}
  \glssymboldefn{char_rep}{\chi}{character of
  a \gls{rep}}
  \cloze{
  Given a \gls{rep} $\rho : G \to \GL(V)$, the
  \emph{character} of $\rho$ is the function $G
  \to k$, $\chi = \chi_\rho = \chi_V : G \to k$,
  $g \mapsto \Trace \rho(g)$.
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

Since for matrices $\Trace (BA) = \Trace(AB)$, the
\gls{char_rep} does not depend on a choice of basis
\[ \Trace(PAP^{-1}) = \Trace(AP^{-1} P) = \Trace(A) \]
Similarly, \gls{rep_iso} \glspl{rep} have the same
\gls{char_rep}.

\begin{example*}
  Let $G = D_6 = \langle s, t : s^2 = t^3 = e, sts
  = t^{-1} \rangle$, the dihedral group of order
  $6$. Let $G \to \GL_2(\RR)$ be the action of $G$
  by symmetries of a triangle. To compute
  $\chichar_\rho$ we just need to know the
  eigenvalues of each matrix $\rho(g)$. Each
  reflection (element $st^i$) has eigenvalue $1,
  -1$ so $\chichar(st^i) = 0$ for all $i$. The
  eigenvalues of the non-trivial rotation must be
  non-trivial cube roots of $1$ and sum to be a
  real number. Thus $\chichar(t) = \chichar(t^2) =
  e^{2\pi i / 3} + e^{4\pi i / 3} = -1$. Also,
  $\chichar(e) = 2$.
\end{example*}

\begin{proposition*}
  Let $(\rho, V)$ be a \gls{rep} of $G$. Then
  \begin{enumerate}[(i)]
    \item $\chichar_V(e) = \dim V$
    \item $\chichar_V(g) = \chichar_V(hgh^{-1})
      ~\forall g, h \in G$
    \item If $W$ is another \gls{rep}, then
      $\chichar_{V \oplus W} = \chichar_V +
      \chichar_W$
    \item If $V$ is unitary then
      $\chichar_V(g^{-1}) = \ol{\chichar_V(g)}
      \forall g \in G$
  \end{enumerate}
\end{proposition*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\chichar(e) = \Trace \id_V = \dim V$
    \item $\rho(hgh^{-1}) = \rho(h) \rho(g)
      \rho(h^{-1})$ thus $\rho(g)$ and
      $\rho(hgh^{-1})$ are conjugate in $\GL(V)$
      so have the same trace.
    \item Clear
    \item By choosing a basis we may view $\rho$
      as a homomorphism $G \to U(n)$. Then
      $\rho(g^{-1}) = \rho(g)^{-1} =
      \ol{\rho(g)}^\top$. So $\Trace \rho(g^{-1})
      = \ol{\Trace(\rho(g))}$ since $\Trace$ is
      transpose invariant. \qedhere
  \end{enumerate}
\end{proof}

The \glsref[char_rep]{characters} contain very little data: an
element of $k$ for each conjugacy class in $G$.
But when $G$ is finite and $k = \CC$, it contains
all we need to reconstruct $V$ up to
\gls{rep_isom}.

\begin{flashcard}[class-function-defn]
\begin{definition*}[Class function]
  \glsnoundefn{cls_fn}{class function}{class functions}
  \glssymboldefn{cls_fn_space}{space of class
  functions}{Cg}
  \cloze{
  A function $f : G \to k$ is a \emph{class
  function} if $f(hgh^{-1}) = f(g) ~\forall g, h
  \in G$. We'll write $\mathcal{C}_G$ for the
  $k$-vector space of class functions.

  Notice that if $O_1, \ldots O_r$ are the
  conjugacy classes in $G$ then the indicator
  functions
  \[ \mathbbm{1}_{O_i} : G \to k, \qquad g \mapsto
  \begin{cases}
    1 & g \in O_i \\
    0 & g \not\in O_i
  \end{cases} \]
  form a basis for $\mathcal{C}_G$. So $\dim
  \mathcal{C}_G = \#\text{conjugacy classes}$.
  }
\end{definition*}
\end{flashcard}