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% 01/11/2023 11AM

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item \glsref[tens_prod]{Tensor product} of
      \glspl{rep} defined last time is a
      generalisation of the \gls{tens_prod} of a
      \gls{rep} and a \repdim{1} \gls{rep}
      previously defined.
    \item If $X$ and $Y$ are finite sets with
      $G$-action
      \begin{align*}
        \alpha_{X \times Y} : kX \tprod kY
        &\to kX \times Y \\
        \delta_x \tvprod \delta_y
        &\mapsto \delta_{(x, y)}
      \end{align*}
      is an \gls{itt_map}.
  \end{enumerate}
\end{remark*}

\begin{flashcard}[character-ring-defn]
\begin{definition*}[Character ring]
  \glsnoundefn{char_ring}{character
  ring}{character rings}
  \glssymboldefn{char_ring}{$R(G)$}{$R(G)$}
  \cloze{
  The \emph{character ring} $R(G)$ of a group $G$
  is defined by
  \[ R(G) \defeq \{\chichar_1 - \chichar_2 \mid
  \chichar_1, \chichar_2 \text{are
  \glsref[char_rep]{characters} of
  $G$}\} \subset \mathcal{C}_G \]
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

Since $\chichar_{V_1 \oplus V_2} =
\chichar_{V_1} + \chichar_{V_2}$, $\charring(G)$ is a
subgroup of $\mathcal{C}_G$ under $+$. Since
$\indicator{G}$ is a \gls{char_rep}, $\charring(G)$ contains
the multiplicative $1$ in $\mathcal{C}_G$.

Since $\chichar_{V_1 \tprod V_2} = \chichar_{V_1}
\cdot \chichar_{V_2}$,
\[ (\chichar_{V_1} - \chichar_{V_2}) \cdot
(\chichar_{W_1} - \chichar_{W_2}) = \chichar_{(V_1
\tprod W_1) \oplus (V_2 \tprod W_2)} -
\chichar_{(V_2 \tprod W_1) \oplus (V_1 \tprod
W_2)} \in \charring(G) \]
$\charring(G)$ is closed under $\cdot$ and so
$\charring(G)$ is a subring of $\mathcal{C}_G$.

\textbf{Observation:} If $(\rho, V)$ is a
\gls{rep} of $G$ and $(\sigma, W)$ is a \gls{rep}
of another group $H$, then
\begin{align*}
  \rho \trprod \sigma : G \times H
  &\to \GL(V \tprod W) \\
  (g, h)
  &\mapsto \rho(g) \tvprod \sigma(h)
\end{align*}
is a \gls{rep} of $G \times H$, by parts (i) and
(ii) in \nameref{the_last_lemma_last_time_l11}.
Moreover,
\[ (\chichar_V \otimes \chichar_W)(g, h) =
\chichar_{V \tprod W} (g, h) = \chichar_V(g)
\chichar_W(h) \]
by part (iii) of the
\hyperref[the_last_lemma_last_time_l11]{same
lemma}. Thus
\begin{align*}
  \charring(G) \times \charring(H)
  &\to \charring(G \times H) \\
  (\chichar_V, \chichar_W)
  &\mapsto \chichar_V \otimes \chichar_W
\end{align*}
defines a $\ZZ$-bilinear map.

The construction of $V \tprod W$ as a \gls{rep} of
$G$ from last time can be viewed as the case $G =
H$ in the construction followed by restriction
along
\begin{align*}
  G
  &\to G \times G \\
  g
  &\mapsto (g, g)
\end{align*}

\begin{flashcard}[chars-of-cartesian-group-prod-prop]
\begin{proposition*}
  Suppose $G$ and $H$ are finite groups, and
  $(\rho_1, V_1), \ldots, (\rho_r, V_r)$ are the
  \gls{irred} $\CC$-\glspl{rep} of $G$ and
  $(\sigma_1, W_1), \ldots, (\sigma_s, W_s)$ are
  all the \gls{irred} $\CC$-\glspl{rep} of $H$.

  For each $1 \le i \le r$, $1 \le j \le s$,
  $(\rho_i \trprod \sigma_j, V_i \tprod W_j)$ is a
  \gls{irred} $\CC$-\gls{rep} of $G \times H$.
  Moreover, all \gls{irred} $\CC$-\glspl{rep} of
  $G \times h$ arise in this way.
\end{proposition*}
\fcscrap{
We've seen this when $G$, $H$ are abelian before
since all these \glspl{rep} have degree $1$ in
this case.
}
\begin{proof}
  \cloze{
  Let $\chichar_1, \ldots, \chichar_r$ be the
  \glsref[char_rep]{characters} of $\rho_1,
  \ldots, \rho_r$, and $\psi_1, \ldots, \psi_s$
  the \glsref[char_rep]{characters} of $\sigma_1,
  \ldots, \sigma_s$. The \gls{char_rep} of $\rho_i
  \trprod \sigma_j$ is $(\chichar_i \otimes
  \psi_j)(g, h) = \chichar_i(g) \psi_j(h)$. Then
  \begin{align*}
    \langle \chichar_i \otimes \psi_j, \chichar_k
    \otimes \psi_l \rangle_{G \times H}
    &= \frac{1}{|G \times H|} \sum_{(g, h) \in G
    \times H} \ol{\chichar_i(g) \psi_j(h)}
    \chichar_k(g) \psi_l(h) \\
    &= \left( \frac{1}{|G|} \sum_{g \in G}
    \ol{\chichar_i(g)} \chichar_k(g) \right)
    \left( \frac{1}{|H|} \sum_{h \in H}
    \ol{\psi_j(g)} \psi_l(g) \right) \\
    &= \Gip\langle \chichar_i, \chichar_k
    \rangle_G \langle \psi_j, \psi_l \rangle_H \\
    &= \delta_{ik} \delta_{jl}
  \end{align*}
  So the $\chichar_i \otimes \psi_j$ are pairwise
  distinct and irreducible. Now
  \[ \sum_{i, j} (\dim V_i \tprod W_j)^2 = \left(
  \sum_i ()\dim V_i)^2 \right) \left( \sum_j (\dim
  W_j)^2 \right) = |G| |H| = |G \times H| \qedhere \]
  }
\end{proof}
\end{flashcard}

\textbf{Question:} If $V, W$ are \gls{irred}
\glspl{rep} of $G$, can $V \tprod W$ be an
\gls{irred} \gls{rep} of $G$? We've seen that if
$\dim V = 1$ or $\dim W = 1$ then yes. Typically
the answer is no.

\begin{example*}
  $G = S_3$
  \begin{center}
    \begin{tabular}{c|c|c|c}
       & $1$ & $(123)$ & $(12)$ \\
      \hline
      $\indicator{}$ & $1$ & $1$ & $1$ \\
      $\eps$ & $1$ & $1$ & $-1$ \\
      $V$ & $2$ & $-1$ & $0$
    \end{tabular}
  \end{center}
  clear that $\indicator{} \trprod W \simeq W$
  always. $\eps \trprod \eps \cong \indicator{}$,
  $\eps \trprod V = V$. Also, $V \trprod V$ has
  \gls{char_rep} $\chichar_V^2$.
  \[ \chichar_V^2(e) = 2^2 = 4, \quad \chichar_V^2
  ((123)) = (-1)^2 = 1, \quad \chichar_V^2((12)) =
  0^2 = 0 \]
  $\chichar_V^2 = \chichar_V + \eps +
  \indicator{}$.

  In general if $\chichar_1, \ldots, \chichar_r$
  are all \gls{irred}
  \glsref[char_rep]{characters} of $G$ and $1 \le
  i, j \le r$ then
  \[ \chichar_i \chichar_j = \sum_{k = 1}^r a_{i,
  j}^k \chichar_k \]
  for some $a_{i, j^k} \in \NN_0$. These numbers
  $a_{i, j}^k$ determine the ring structure on
  $\charring(G)$ since $\charring(G) =
  \bigoplus_{i = 1}^r \ZZ \chichar_i$ as a group
  under $+$.

  In fact, $V \tprod V, V \tprod V \tprod V,
  \ldots$ are never \gls{irred} if $\dim V > 1$.
\end{example*}

\subsection{Symmetric and exterior powers}

For any vector space $V$ we can define
\[ \sigma_V \defeq \sigma : V \tprod V \to V
\tprod V \qquad \sigma(v \tvprod w) = w \tvprod v ~\forall v,
w \in V \]
(Exercise: prove that $V \times V \to V \tprod V$,
$(v, w) \mapsto w \tvprod v$ is bilinear).

Notice $\sigma^2 = \id_{V \tprod V}$, so if
$\characteristic k \neq 2$, then $\sigma$
decomposes $V \tprod V$ into eigenspaces.
\begin{align*}
  \ssquare V
  &\defeq \{a \in V \tprod V \st \sigma(a) = a\}
  &&\text{the \emph{symmetric square} of $V$} \\
  \asquare V
  &\defeq \{a \in V \tprod V \st \sigma(a) = a\}
  &&\text{the \emph{exterior / alternating square}
  of $V$}
\end{align*}
In fact $V \tprod V = \ssquare V \oplus \asquare V$
is the \gls{iso_decomp} of $V \tprod V$ as a
\gls{rep} of $C_2$. Suppose for now that
$\characteristic k \neq 2$.

\begin{lemma*}
  Suppose $v_1, \ldots, v_m$ is a basis for $V$.
  \begin{enumerate}[(i)]
    \item $\ssquare V$ has basis $v_i v_j \defeq \half
      (v_i \tvprod v_j + v_j \tvprod v_i)$ for $1
      \le i \le j \le m$ ($v_j v_i = v_i v_j$ if
      $i > j$ allowed).
    \item $\asquare V$ has basis $v_i \wedge v_j
      \defeq \half (v_i \tvprod v_j - v_j \tvprod
      v_i)$ for $1 \le i < j \le m$ ($v_j \wedge
      v_i = -v_i \wedge v_j$ if $i \ge j$ allowed)
  \end{enumerate}
  Thus $\dim \ssquare V = \half m(m + 1)$ and $\dim
  \asquare V = \half m(m - 1)$.
\end{lemma*}

\begin{proof}
  It is easy to check:
  \begin{enumerate}[(i)]
    \item $v_i v_j \in \ssquare V$ for all $i, j$.
    \item $v_i \wedge v_j \in \asquare V$ for all
      $i, j$.
    \item The union of the claimed bases spans $V
      \tprod V$ and has size $m^2 = \dim V \tprod
      V$
  \end{enumerate}
  So it follows from this that $V \tprod V =
  \ssquare V
  \oplus \asquare V$. Everything else follows.
\end{proof}

You might want to ponder \es[11]{2} in this
context.

\begin{proposition*}
  Let $(\rho, V)$ be a \gls{rep} of $G$.
  \begin{enumerate}[(i)]
    \item $V \tprod V = \ssquare V \oplus \asquare V$
      as a direct sum of \glspl{rep} of $G$.
    \item For $g \in G$ such that $\rho(g)$ is
      diagonalisable (for convenience, but there
      exists a slightly more complicated proof not
      using this condition),
      \begin{align*}
        \chichar_{\ssquare V} (g)
        &= \half (\chichar_V(g)^2 +
        \chichar_V(g^2)) \\
        \chichar_{\asquare V}
        &= \half (\chichar_V(g)^2 -
        \chichar_V(g^2))
      \end{align*}
  \end{enumerate}
\end{proposition*}

\begin{proof}
  For (i) we need that $\ssquare V$ and $\asquare V$
  are \Ginv, i.e. if $a \in V \tprod V$ such that
  $\sigma(a) = \lambda a$ for $\lambda = \pm 1$,
  then $\sigma(ga) = \lambda g a ~\forall g \in
  G$. For this it suffices to show that $\rho(g)$
  and $\sigma$ commute $\forall g \in G$, i.e.
  $\sigma \in \GHom_G(V \tprod V, V \tprod V)$.
  BUt $\sigma \circ g(v \tvprod w) = \sigma( gv
  \tvprod gw)) = gw \tvprod gv = g(w \tvprod v) =
  g \sigma(v \tvprod w) ~\forall g \in G$.

  To prove (ii) it suffices to compute
  $\chichar_{\ssquare V}$ since $\text{sum of }\RHS =
  \chichar_V^2 = \chichar_{V \tprod V}$. Let $v_1,
  \ldots, v_m$ form a basis for $V$ such that
  $\rho(g) v_i = \lambda_i v_i$ for $1 \le i \le
  m$.
  \[ g(v_i \lambda_i) = \lambda_i \lambda_j v_i
  v_j \]
  So
  \[ \chichar_{\ssquare V}(g)
  = \sum_{1 \le i \le j \le m} \lambda_i \lambda_j
  \]
  and
  \[ \chichar_V(g)^2 + \chichar_V(g^2)
  = \left( \sum_i \lambda_i \right)^2
    + \sum_j (\lambda_j)^2
  = 2 \sum \lambda_i^2
    + 2 \sum_{i < j} \lambda_i \lambda_j
  = 2 \chichar_{\ssquare V}(g) \qedhere \]
\end{proof}

\textbf{Exercise:} Prove the formula for
$\chichar_{\asquare V}$ directly.