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\textbf{Question:} Suppose
\[ V \simeq \bigoplus_{i \in I} V_i \simeq
\bigoplus_{j \in J} W_j \]
with $V_i$, $W_j$ are \gls{irred} \glspl{rep}
of $G$. Can these decompositions be different?

\newpage
\section{Schur's Lemma}

Recall that if $V$ is a vector space then $\Aut(V)
= \GL_{\dim V}(k)$.

\begin{flashcard}[schurs-lemma]
\begin{theorem*}[Schur's Lemma]
  \refstepcounter{customtheorem}
  \label{schurs_lemma}
  \cloze{
  Suppose $V$ and $W$ are \gls{irred}
  \glspl{rep} of a group $G$. Then
  \begin{enumerate}[(i)]
    \item Every $\varphi \in \GHom_G(V, W)$ is
      either $0$ or an isomorphism;
    \item if $k$ is algebraically closed then
      $\dim \GHom_G(V, W)$ is $0$ or $1$.
  \end{enumerate}
  }
\end{theorem*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item \cloze{Note if $\varphi \in \GHom_G(V,
      W) \setminus \{0\}$ then $\ker \varphi \lneq
      V$ is a \Ginv\ subspace. So as $V$ is
      \gls{irred}, $\ker \varphi = 0$. Similarly
      $0 \neq \Im\varphi \le W$ is a \Ginv{}
      subspace so $\Im \varphi = W$. Then by
      \nameref{reps_first_iso}, $\varphi$ is an
      isomorphism.}
    \item \cloze{Suppose $\varphi_1, \varphi_2 \in
      \GHom_G(V, W) \setminus \{0\}$. By (i),
      $\varphi_1$ is an isomorphism such that
      $\varphi_1^{-1} \circ \varphi_2 \in
      \GHom_G(V, V)$. But then as $k$ is
      algebraically closed, every element of
      $\Hom_k(V, V)$ has an eigenvalue. In
      particular, $\exists \lambda \in k$ such
      that $\ker(\varphi_1^{-1} \circ \varphi_2 -
      \lambda \id_V) \neq 0$. But $\varphi_1^{-1}
      \circ \varphi_2 - \lambda \id_V \in
      \GHom_G(V, V)$. So $\ker(\varphi_1^{-1}
      \circ \varphi_2 - \lambda \id_V)$ is
      \Ginv\ so is equal to $V$ (since $V$
      \gls{irred}), i.e. $\varphi_1^{-1} \circ
      \varphi_2 = \lambda \id_V$ and $\varphi_2 =
      \lambda \varphi_1$, i.e. $\GHom_G(V, W) = k
      \varphi_1$.}
      \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

This says that in particular, $\GHom_G(V, V)
= k$ (such that $k = \ol{k}$), so an \gls{irred}
\gls{rep} is
rigid in the same sense that a 1-dimensional
vector space is rigid since their automorphism
groups are the same.

\begin{flashcard}[G-linear-maps-decomp-identities]
\begin{proposition*}
  If $V, V_1, V_2$ are \glspl{rep} of $G$ then
  \begin{enumerate}[(a)]
    \item $\GHom_G(V, V_1 \oplus V_2) \simeq
      \cloze{\GHom_G(V, V_1) \oplus \GHom_G(V,
      V_2)}$
    \item $\GHom_G(V_1 \oplus V_2, V) \simeq
      \cloze{\GHom_G(V_1, V) \oplus \GHom_G(V_2,
      V)}$
  \end{enumerate}
\end{proposition*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(a)]
    \item \cloze{There are natural \Glin\ inclusion
      maps $\iota_i : V_i \to V_1 \oplus V_2$ for
      $i = 1, 2$. These induce by post-composition
      \Glin\ maps $\Hom_k(V, V_i) \to
      \Hom_k(V, V_1 \oplus V_2)$, $f \mapsto
      \iota_i \circ f$. Together these give a
      linear isomorphism
      \[ \Hom_k(V, V_1) \oplus \Hom_k(V, V_2) \to
      \Hom_k(V, V_1 \oplus V_2) \qquad (f_1, f_2)
      \mapsto \iota_1 f_1 + \iota_2 f_2 \]
      Since $\iota_1, \iota_2$ are \Glin,
      this is an \gls{itt_map}
      \[ g \cdot (\iota_1 f_1 + \iota_2 f_2) =
      \iota_1 \circ (g \cdot f_1) + \iota_2 \circ
      (g \cdot f_2) \]
      Since in general if $\varphi : U \to W$ is
      an \gls{itt_map} between \glspl{rep}
      of $G$ it induces an isomorphism on
      $G$-fixed points since $g \cdot (\varphi(u))
      = \varphi(u) \iff g \cdot u = u$ ($\varphi$
      injective). It follows that there is an
      induced isomorphism as in (a).}
    \item \cloze{Since the natural projection maps $\pi_i
      : V_1 \oplus V_2 \to V_i$, $(v_1, v_2)
      \mapsto v_i$ for $i = 1, 2$ are also
      \Glin\ and induce a \Glin
      isomorphism
      \[ \Hom_k(V_1, V) \oplus \Hom_k(V_2, V) \to
      \Hom_k(V_1 \oplus V_2, V), \qquad (f_1, f_2)
      \mapsto f_1 \circ \pi_1 + f_2 \circ \pi_2 \]
      and again taking $G$-invariants gives the
      result.} \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{corollary*}
  If $V \simeq \bigoplus_{i \in I} V_i$ and $W
  \simeq \bigoplus_{j \in J} W_j$ then
  \[ \GHom_G(V, W) = \bigoplus_{i \in I}
  \bigoplus_{j \in J} \GHom_G (V_i, W_j) \]
\end{corollary*}

\begin{proof}
  This follows from the previous proposition and a
  simple induction argument.
\end{proof}

\begin{corollary*}
  If $k = \ol{k}$ and $V \simeq \bigoplus_{i =
  1}^r V_i$ is a decomposition of $V$ as a
  \gls{rep_dir_sum} of \gls{simple} \glspl{rep} then for each
  \gls{simple} \gls{rep} $W$ of $G$
  \[ |\{i : V_i \simeq W\}| = \dim \GHom_G(V, W) =
  \dim \GHom_G(W, V) .\]
  and does not depend on the choice of
  decomposition.
\end{corollary*}

\begin{proof}
  By the last result $\dim \GHom_G(V, W) =
  \bigoplus_{i = 1}^r \GHom_G(V_i, W)$ so $\dim
  \GHom_G(V, W) = \sum_{i = 1}^r \dim \GHom_G(V_i,
  W)$ and similarly $\dim \GHom_G(W, V) = \sum_{i
  = 1}^r \dim \GHom_G(W, V_i)$. Thus it suffices
  to show
  \[ \dim \GHom_G(V_i, W) = \dim \GHom_G(W,
  V_i) = \begin{cases}
    1 & W \simeq V_i \\
    0 & W \not\simeq V_i
  \end{cases} \]
  This is part of the statement of
  \nameref{schurs_lemma} when $k = \ol{k}$.
\end{proof}

\textbf{Exercise for enthusiasts:} Give a version
of this corollary when $k \neq \ol{k}$.

\textbf{Important question:} How can we compute
these numbers $\dim \GHom_G(V, W)$?
Note our
\hyperref[lecture5_last_remark]{final remark last
lecture} may help us at least when
$\characteristic(k) = 0$ since it said $\dim V^G =
\frac{1}{|G|} \sum_{g \in G} \Trace \rho(g)$. So we
need to understand these traces for the
\gls{rep} $\Hom_k(V, W)$.

\begin{flashcard}[abelian-irreducible-complex-has-degree-1]
\begin{corollary*}
  If $G$ is an abelian group then every
  \gls{irred} $\CC$-\gls{rep} of $G$ has
  \gls{rep_deg} $1$.
\end{corollary*}

\begin{proof}
  \cloze{
  Let $(\rho, V)$ be a complex \gls{irred}
  \gls{rep} of $G$. For each $g \in G$,
  $\rho(g)\rho(h) = \rho(h)\rho(g)$ for all $h \in
  G$. So $\rho(g) \in \GHom_G(V, V) = \CC \id_V$
  by \nameref{schurs_lemma}. Now if $v \in V
  \setminus \{0\}$ then $\rho(g) \langle v \rangle
  \le \langle v \rangle ~\forall g \in G$ so
  $\langle v \rangle$ is a \Ginv\ subspace
  of $V$, so $\langle v \rangle = V$ since $V$ is
  \gls{irred}.
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[finite-faithful-irred-implies-cyclic-centre]
\begin{corollary*}
  If \cloze{$G$ is a \fcemph{finite} group with a
  \fcemph{\gls{ff_rep}} \fcemph{\gls{irred}}
  \gls{rep} over an
  \fcemph{algebraically closed} field $k$}, then $Z(G)$ is cyclic.
\end{corollary*}

\begin{proof}
  \cloze{
  Let $(\rho, V)$ be a \gls{ff_rep} \gls{irred}
  \gls{rep} of $G$ and $z \in Z(G)$. Then
  $\rho(z)\rho(g) = \rho(g)\rho(z)$ for all $g \in
  G$, i.e. $\varphi(z) \in \Hom_G(V, V) = k \id_V$
  by \nameref{schurs_lemma}. We can write $\rho(z)
  = \lambda_z \id_V$ for $\lambda_z \in k$, and
  $\lambda_{z_1 z_2} = \lambda_{z_1} \cdot
  \lambda_{z_2}$ for all $z_1, z_2 \in Z(G)$, and
  $\lambda_e = 1$. So $Z(G) \to k^\times$, $z
  \mapsto \lambda_z$ is a \gls{ff_rep} \gls{rep}
  of $Z(G)$ since $V$ is \gls{ff_rep}, i.e. $Z(G)$ is
  isomorphic to a finite subgroup of $k^\times$,
  and any such subgroup is cyclic.
  }
\end{proof}
\end{flashcard}

\begin{example*}
  $G = C_4 = \langle x \rangle$. \repdim{1}
  $\CC$-\glspl{rep} of $G$ are given by
  \begin{center}
    \begin{tabular}{c|c|c|c}
      $1$ & $x$ & $x^2$ & $x^3$ \\
      \hline
      $1$ & $1$ & $1$ & $1$ \\
      $1$ & $i$ & $-1$ & $-i$ \\
      $1$ & $-1$ & $1$ & $-1$ \\
      $1$ & $-i$ & $-1$ & $i$
    \end{tabular}
  \end{center}
\end{example*}

\begin{example*}
  $C_2 \times C_2 = \langle x \rangle \times
  \langle y \rangle$. \repdim{1}
  $\CC$-\glspl{rep} of $G$ are given by
  \begin{center}
    \begin{tabular}{c|c|c|c}
      $1$ & $x$ & $y$ & $xy$ \\
      \hline
      $1$ & $1$ & $1$ & $1$ \\
      $1$ & $1$ & $-1$ & $-1$ \\
      $1$ & $-1$ & $1$ & $-1$ \\
      $1$ & $-1$ & $-1$ & $1$
    \end{tabular}
  \end{center}
\end{example*}
