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\begin{flashcard}[sub-representation-defn]
\begin{definition*}[Subrepresentation]
  \glsnoundefn{subrep}{subrepresentation}{subrepresentations}
  \glsadjdefn{Ginv}{$G$-invariant}{vector subspace}
  \cloze{
  Suppose $\rho : G \to \GL(V)$ is a
  \gls{rep} and $W \le V$ is a $k$-linear
  subspace. Then we say $W$ is
  \emph{$G$-invariant} if $\rho(g)(W) \subset W$
  for all $g \in G$.

  In this case we can define a \gls{rep}
  $(\rho_W, W)$ via
  \[ \rho_W(g)(w) = \rho(g)(w) \qquad \forall g
  \in G, w \in W .\]
  We call $(\rho_W, W)$ a \emph{subrepresentation}
  of $W$.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[proper-subrep]
\begin{definition*}[Proper Subrepresentation]
  \glsadjdefn{proper_subrep}{proper}{subrepresentation}
  \cloze{
  If $W \neq 0$ and $W \neq V$ we say $W$ is a
  \emph{proper} \gls{subrep}.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[irreducible-rep]
\begin{definition*}[Irreducible Representation]
  \glsadjdefn{irred}{irreducible}{\gls{rep}}
  \glsadjdefn[irred]{simple}{simple}{\gls{rep}}
  \cloze{
  We say $V \neq 0$ is \emph{irreducible} or
  \emph{simple} if $V$ has no \gls{proper_subrep}
  \glspl{subrep}.
  }
\end{definition*}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
  \item Any \repdim{1} \gls{rep} of any
    group is \gls{irred}.
  \item If $G = C_2$, $\rho : G \to \GL_2(k)$,
    with
    \[ \rho(-1) =
    \begin{pmatrix}
      -1 & 0 \\
      0 & 1
    \end{pmatrix}
    \]
    ($\characteristic k \neq 2$). Then $(\rho, k^2)$ has
    exactly 2 proper \glspl{subrep}, namely
    \[ \left\langle
    \begin{pmatrix}
      1 \\
      0
    \end{pmatrix}
    \right\rangle, \left\langle
    \begin{pmatrix}
      0 \\
      1
    \end{pmatrix}
    \right\rangle \]
    \begin{proof}
      It is easy to see that two given
      \glspl{subrep} are \Ginv,
      since the vectors are
      $\rho(-1)$-eigenvectors. Conversely, any
      \gls{proper_subrep} \gls{subrep} must have
      \gls{rep_dim} 1, so is spanned by an
      eigenvector of $\rho(-1)$ and the
      eigenspaces of $\rho(-1)$ are those
      described above.
    \end{proof}
  \item If $G = C_2$ then any \gls{simple}
    \gls{rep} of $G$ has
    \gls{rep_dim} $1$.
    \begin{proof}
      Suppose $\rho : G \to \GL(V)$ is an
      \gls{irred} \gls{rep} of $G$. The
      minimal polynomial of $\rho(-1)$ divides
      $X^2 - 1 = (X - 1)(X + 1)$ since
      $\rho(-1)^2 = \id_V$. So it has a linear
      factor and $\rho(-1)$ has a nonzero eigenvector
      $v$. Then $\rho(-1) \langle v \rangle
      \subseteq \langle v \rangle$ but also
      $\rho(1) \langle v \rangle \subseteq
      \langle v \rangle$. So $\langle v \rangle$
      is a \Ginv\ subspace of $V$. As $V$
      is \gls{irred} and $\langle v \rangle \neq
      0$, $\langle V \rangle = V$ has dimension
      1.
    \end{proof}

    Note we see that if $\characteristic k \neq 2$
    there are precisely 2 \gls{simple} \glspl{rep}
    of $C_2$ up to \gls{rep_isom} and only 1 if
    $\characteristic k = 2$.
  \item If $G = D_6$, then every complex
    \gls{irred} \gls{rep} has \gls{rep_deg} $\le 2$.
    \begin{proof}
      Let $\rho : G \to \GL(V)$ be an \gls{irred}
      \gls{rep} of $G$. Let $r$ be a
      non-trivial rotation in $G$ and $s$ a
      reflection in $G$ so $r^3 = e = s^2$, $srs =
      r^{-1}$ and $\{r, s\}$ generate $G$.

      Since $\rho(r)^3 = \id_V$, the minimal
      polynomial of $\rho(r)$ divides $X^3 - 1$,
      so $\rho(r)$ has an eigenvector $v$ with
      eigenvalue $\lambda$ such that $\lambda^3 =
      1$.

      Consider $W \defeq \langle v, \rho(s) v
      \rangle \le V$, so $\dim W \le 2$. Now
      $\rho(s)\rho(s)v = \rho(s^2)v = v$, and
      $\rho(r)\rho(s)v = \rho(s)\rho(r)^{-1}v =
      \lambda^{-1} \rho(s)v$, so $\rho(s)W \le W$
      and $\rho(R)W \le W$. Since $r$ and $s$
      generate $G$, it follows $W$ is a
      \Ginv\ subspace of $V$.

      So if $V$ is \gls{irred}, must have $W = V$,
      so $\dim V \le 2$ as required.
    \end{proof}

    \textbf{Exercise:} Show that there are
    precisely 3 \gls{irred} \glspl{rep} of
    $D_6$ up to \gls{rep_isom}: two of
    \gls{rep_deg} 1 and
    one of \gls{rep_deg} 2. (Hint: consider proof above
    and split into cases for each value of
    $\lambda$).
\end{enumerate}

\begin{flashcard}[quotient-rep]
\begin{definition*}[Quotient Representation]
  \cloze{
  If $(\rho, V)$ is a \gls{rep} of $G$ and $W
  \le V$ is a \Ginv\ \gls{subrep} then
  we can define a \emph{quotient representation}
  $(\rho_{V / W}, V / W)$ via
  \[ \rho_{V / W}(g)(v + W) = \rho(g)(v) + W .\]
  (This is well-defined since $\rho(g)(W) \subset
  W \forall g \in G$ means that the choice of
  coset representative doesn't matter).
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

We're going to start dropping the $\rho$ now where
it doesn't cause confusion.

\begin{flashcard}[G-linear-map]
\begin{definition*}[$G$-linear map]
  \glsadjdefn{Glin}{$G$-linear}{linear map}
  \glssymboldefn{GHom}{HomG}{HomG}
  \cloze{
  If $(\rho, V)$ and $(\sigma, W)$ are two
  \glspl{rep} of $G$ we say a $k$-linear map
  $\varphi : V \to W$ is \emph{$G$-linear} if
  $\varphi \circ g = g \circ \varphi$ for all $g
  \in G$, i.e. $\varphi \circ \rho(g) = \sigma(g)
  \circ \varphi$. We write
  \[ \Hom_G(V, W) = \{\varphi \in \Hom_k(V, W)
  \mid \text{$\varphi$ is $G$-linear}\} \]
  a $k$-vector subspace of $\Hom_k(V, W)$.
  }
\end{definition*}
\end{flashcard}

\begin{enumerate}[(1)]
  \item $\varphi \in \Hom_k(V, W)$ is an
    \gls{itt_map} if and only if $\varphi$ is
    a bijection and $\varphi \in \Hom_G(V, W)$,
    since $\varphi \circ \rho(g) = \sigma(g)
    \circ \varphi \iff \varphi \circ \rho(g)
    \circ \varphi^{-1} = \sigma(g)$.
  \item If $W \le V$ is a $G$-\gls{subrep}
    then the natural inclusion map
    \[ \iota : W \to V, w \mapsto w \in
    \GHom_G(W, V) \]
    and the natural projection map
    \[ \pi : V \to V / W, v \mapsto v + W \in
    \GHom_G(V, V / W) .\]
  \item Recall that $\Hom_k(V, W)$ is a
    \gls{rep} of $G$ via $g \cdot \varphi =
    g \circ \varphi \circ g^{-1}$ for $g \in G$,
    $\varphi \in \Hom_k(V, W)$ and $\varphi \in
    \GHom_G(V, W)$ if and only if $g \cdot
    \varphi = \varphi$ for all $g \in G$.
    \begin{lemma*}
      \vspace{0.5em}
      If $U$, $V$ and $W$ are \glspl{rep} of
      a group $G$ with $\varphi_1 \in \Hom_k(V,
      W)$, $\varphi_2 \in \Hom_k(U, V)$ then
      \[ g \cdot (\varphi_1 \circ \varphi_2) =
      (g \circ \varphi_1) \circ (g \circ
      \varphi_2) \forall g \in G .\]
      In particular, if:
      \begin{itemize}
        \item $\varphi_1 \in \GHom_G(V,
          W)$ then $g \cdot (\varphi_1 \circ
          \varphi_2) = \varphi_1 \circ (g \cdot
          \varphi_2)$ for all $g \in G$
        \item $\varphi_2 \in \GHom_G(U, V)$ then
          $g \cdot (\varphi_1 \circ \varphi_2) =
          (g \cdot \varphi_1) \circ \varphi_2$
          for all $g \in G$
        \item $\varphi_1 \in \GHom_G(V, W)$ and
          $\varphi_2 \in \GHom_G(U, V)$
      \end{itemize}
      Then $\varphi_1 \circ \varphi_2 \in
      \GHom_G(U, W)$.
    \end{lemma*}
    \begin{proof}
      With the notation in the statement,
      \[ (g \cdot \varphi_1) \circ (g \circ
      \varphi_2) = (g \circ \varphi_1 \circ
      g^{-1}) \circ (g \circ \varphi_2 \circ
      g^{-1}) = g \circ (\varphi_1 \circ
      \varphi_2) \circ g^{-1} = g \cdot (\varphi_1
      \circ \varphi_2) \]
      Everything else follows immediately.
    \end{proof}
\end{enumerate}

\begin{flashcard}[first-iso-thm-for-reps]
\begin{lemma*}[First Isomorphism for
  Representations]
  \refstepcounter{customlemma}
  \label{reps_first_iso}
  \cloze{
  Suppose $V$ and $W$ are two \glspl{rep} of
  $G$ and $\varphi \in \Hom_G(V, W)$. Then
  \begin{enumerate}[(i)]
    \item $\ker \varphi$ is a \gls{subrep} of
      $V$;
    \item $\Image \varphi$ is a \gls{subrep} of
      $W$;
    \item the linear isomorphism $\ol{\varphi} : V
      / \ker \varphi \to \Image \varphi$ given by the
      first isomorphism theorem for vector spaces
      is an \gls{itt_map}. Thus $V / \ker
      \varphi$ and $\Image \varphi$ are
      \gls{rep_iso} as \glspl{rep}.
  \end{enumerate}
  }
\end{lemma*}

\begin{proof}
  \cloze{
  \begin{enumerate}[(i)]
    \item If $v \in \ker \varphi$ and $g \in G$
      then $\varphi(g \cdot v) = g \cdot
      \varphi(v) = g \cdot 0 = 0$. So $g \cdot v
      \in \ker \varphi$.
    \item If $w = \varphi(v) \in \Image \varphi$
      and $g \in G$ then $g \cdot w = g \cdot
      \varphi(v) = \varphi(g \cdot v) \in \Image
      \varphi$.
    \item We know $\ol{\varphi}$ is given by the
      formula $\ol{\varphi}(v + \ker \varphi) =
      \varphi(v)$. Then $g \circ \ol{\varphi}(v +
      \ker \varphi) = g \cdot \varphi(v) =
      \varphi(g \cdot v) = \ol{\varphi}(g(v +
      \ker\varphi))$.
  \end{enumerate}
  }
\end{proof}
\end{flashcard}

\begin{proposition*}
  Suppose $\rho : G \to \GL(V)$ and $W \le V$ is a
  subspace. Then the following are equivalent
  \begin{enumerate}[(i)]
    \item $W$ is a \gls{subrep} of $V$.
    \item There exists a basis of $V$ such that
      $v_1, \ldots, v_r$ is a basis of $W$ and
      each $\rho(g)$ with respect to this basis is
      block upper triangular:
      \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/14721604682511ee.png}
      \end{center}
    \item For every basis $v_1, \ldots, v_d$ of
      $V$ such that $v_1, \ldots, v_r$ is a basis
      of $W$ each $\varphi(g)$ with respect to the
      basis is block upper triangular as in (ii).
  \end{enumerate}
\end{proposition*}

\begin{proof}
  Linear Algebra Example Sheet last year.
\end{proof}
