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\begin{enumerate}[(1)]
  \setcounter{enumi}{5}
  \item \glssymboldefn{kX}{f : X to k}{kX}
    Given a finite set $X$ we can form the
    vector space
    \[ kX \defeq \{f : X \to k\} \]
    with pointwise operations. This has a basis
    $\langle \delta_x : x \in X \rangle$ where
    $\delta_x(y) = \delta{xy}$ for $y \in X$.
    If $f \in kX$ then $f = \sum_{x \in X} f(x)
    \delta_x$.

    If a group $G$ acts on $X$, we can define
    \begin{align*}
      \rho : G
      &\to \Aut(kX)
      \rho(g)(f)(x)
      &= f(g^{-1} x)
      \qquad \forall f \in kX, g \in G, x \in X
    \end{align*}
    If is easy to check $\rho(g)$ is linear for
    all $g \in G$ and $\rho(e) = \id_{kX}$. So it
    suffices to show $\rho(gh) = \rho(g) \rho(h)
    ~\forall g, h \in G$. To show this, note that
    for all $g, h \in G$, $f \in kX$, $x \in X$,
    we have
    \begin{align*}
      \rho(gh)(f)(x)
      &= f(h^{-1} g^{-1} x) \\
      &= \rho(h)(f)(g^{-1} x) \\
      &= \rho(g)\rho(h)(f)(x)
    \end{align*}
    as required. Note that for $g \in G$, $x, y
    \in G$,
    \[ (\rho(g)\delta_x)(y) = \delta_x(g^{-1} y) =
    \delta_{x, g^{-1}y} = \delta_{gx, y} =
    \delta_{gx}(y) \]
    So by linearity $\rho(g) \left( \sum f(x)
    \delta_x \right) = \sum_{x \in X} f(x)
    \delta_{gx}$.
    \glsnoundefn{reg_rep}{regular
    representation}{regular representations}
  \item In particular if $G$ is finite then $G$
    acts on itself by left multiplication $G
    \times G \to G$, $(g, h) \mapsto gh$. This
    induces a \gls{rep} of $G$ on $kG$,
    called the \emph{regular representation}. If
    $g \in G$ then $\rho(g)(\delta_e) = \delta_g$
    so $\rho(g) = e \iff g = e$. So the regular
    representation is always \gls{ff_rep}.
  \item If $(\rho, V)$ is a \gls{rep} of $G$
    we can define a \gls{rep} $\rho^*$ of $G$
    on $V^*$ as follows
    \[ \rho^*(g)(\theta)(v) = \theta(g^{-1} v)
    \qquad \forall g \in G, \theta \in V^*, v \in
    V \]
    $\rho^*(g)$ can be viewed as the adjoint of
    $\rho(g)^{-1}$ and recall that with respect to
    a pair of dual bases for $V$ and $V^*$, the
    matrix of the adjoint of a linear map is the
    transpose of the matrix of the map. So if $V =
    k^d$ so $\rho : G \to \GL_d(k)$ then
    $\rho^*(g) = (\rho(g)^{-1})^\top$. This is a
    homomorphism because $\GL_d(k) \to \GL_d(k)$,
    $A \mapsto (A^{-1})^\top$ is a homomorphism.
  \item More generally, if $(\rho, V)$ and
    $(\sigma, W)$ are two \glspl{rep} of $G$
    then $(\tau, \Hom_k(V, W))$ is a
    \gls{rep} of $G$ as follows
    \[ \tau(g)(\alpha) = \sigma(g) \circ \alpha
    \circ \rho(g)^{-1} \qquad \forall g \in G,
    \alpha \in \Hom_k(V, W) \]
    \textbf{Exercise:} check the details (this is
    on Example Sheet 1).

    Note that if $W = k$ is the \gls{triv_rep}
    then we recover the previous
    example. Moreover if $V = k^n$, $W = k^m$ with
    the standard bases (so $\Hom_k(V, W) =
    \Mat_{m, n}(k)$) then $\tau(g)(A) = \sigma(g)
    A \rho(g)^{-1}$ for all $g \in G$, $A \in
    \Mat_{m, n}(k)$.
  \item If $\rho : G \to \GL(V)$ is a
    \gls{rep} (of $G$) and $\theta : H \to G$ is a
    group homomorphism then $\rho\theta : H \to
    \GL(V)$ is a \gls{rep} off $H$. If $H \le
    G$ and $\theta$ is the inclusion map then we
    call this the \emph{restriction of $\rho$ to
    $H$}.
\end{enumerate}

\subsection{The Category of Representations}

If $\rho : G \to \GL(V)$ is a \gls{rep} and
$\varphi : V \to W$ is a homomorphism of vector
spaces then $\sigma : G \to \GL(W)$ defined by
$\sigma(g) = \varphi \circ \rho(g) \circ
\varphi^{-1}$ for all $g \in G$.

\begin{flashcard}[isomorphic-representations-defn]
\begin{definition*}[Isomorphic Representations]
  \glsadjdefn{rep_iso}{isomorphic}{\gls{rep}}
  \glsnoundefn{rep_isom}{isomorphism}{isomorphisms}
  \glsverbdefn{itts}{intertwines}{representation}
  \glsnoundefn[itts]{itt_map}{intertwining
  map}{intertwining maps}
  \cloze{
  We say that $\rho : G \to \GL(V)$ and $\sigma :
  G \to \GL(W)$ are \emph{isomorphic
  representations} if there exists $\varphi : V
  \to W$ a $k$-linear isomorphism such that
  \[ \sigma(g) = \varphi \circ \rho(g) \circ
  \varphi^{-1} \qquad \forall g \in G \]
  We say $\varphi$ \emph{intertwines} $\rho$ and
  $\sigma$.
  }
\end{definition*}
\end{flashcard}

Note that:
\begin{enumerate}[(1)]
  \item $\id_V$ \gls{itts} $\rho$ and $\rho$.
  \item If $\varphi$ \gls{itts} $\rho$ and
    $\sigma$ then $\varphi^{-1}$ \gls{itts}
    $\sigma$ and $\rho$.
  \item If $\varphi$ \gls{itts} $\rho$ and
    $\sigma$ and $\varphi'$ \gls{itts} $\sigma$
    and $\tau$, then $\varphi' \circ \varphi$
    \gls{itts} $\rho$ and $\tau$.
\end{enumerate}
Therefore this definition of \gls{rep_isom} is an
equivalence relation.

Since every vector space is \gls{rep_iso} to $k^d$
for some $d \ge 0$, every \gls{rep} is
\gls{rep_iso} to a matrix \gls{rep}.

If $\rho, \sigma : G \to \GL_d(k)$ are
matrix \glspl{rep} of the same \gls{rep_deg} then an
\gls{itt_map} from $\rho$ to $\sigma$ is an
invertible matrix $P \in \GL_d(k)$ such that
\[ \sigma(g) = P \rho(g) P^{-1} \qquad \forall g
\in G \]
Thus matrix \glspl{rep} are \gls{rep_iso}
precisely if they represent the same family of
maps with respect to different bases.

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item If $G = \{e\}$ then $(\rho, V)$ and
      $(\sigma, W)$ are \gls{rep_iso} if and only if
      $\dim V = \dim W$.
    \item If $G = (\ZZ, +)$, then $(\rho, V)$ and
      $(\sigma, W)$ are \gls{rep_iso} if and only if
      there are bases for $V$ and $W$ such that
      $\rho(1)$ and $\sigma(1)$ are the same
      matrix. So
      \[ \{\text{\glspl{rep} of $(\ZZ, +)$}\} /
      \sim \leftrightarrow \{\text{conjugacy
      classes of invertible matrices}\} \]
      If $k = \CC$ the RHS is classified by Jordan
      Normal Form (more generally rational
      canonical form).
    \item If $G = C_2 = (\{\pm 1\}, \cdot)$ then
      \[ \{\text{\glspl{rep} of $C_2$}\} /
      \sim \leftrightarrow \{\text{conjugacy
      classes of matrices $A$ such that $A^2 = I$}\} \]
      Since the minimal polynomial of $A$ in RHS
      divides $X^2 - 1 = (X - 1)(X + 1)$ (which has
      distinct roots if characteristic of $k$ is
      not 2), every such matrix is conjugate to a
      diagonal matrix and all diagonal entries are
      $1$ or $-1$.

      \textbf{Exercise:} Show that there are
      precisely $n + 1$ \gls{rep_isom} classes of
      \glspl{rep} of $C_2$ of degree $n$
      (for any field of characteristic not equal
      to 2).
    \item If $G$ acts on sets $X$ and $Y$ and
      there is a bijection $f : X \to Y$ such that
      $g \cdot (f(x)) = f(g \cdot x)$ for all $g
      \in G$, $x \in X$, then $f$ induces an
      \gls{rep_isom} of \glspl{rep} $\tilde{f} :
      kX \to kY$, $\tilde{f}(\theta)(y) =
      \theta(f^{-1} y)$.

      \textbf{Exercise:} check this.
  \end{enumerate}
\end{example*}