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% 03/11/2023 11AM

Recall
\begin{align*}
  \chichar_{\ssquare V} (g)
  &= \half (\chichar_V(g)^2 + \chichar_v(g^2)) \\
  \chichar_{\asquare V}(g)
  &= \half (\chichar_V(g)^2 - \chichar_V(g^2))
\end{align*}

\begin{center}
  \begin{tabular}{c|c|c|c|c|c}
    $S_4$ & $e$ & $(12)(34)$ & $(123)$ & $(12)$ & $(1234)$ \\
    \hline
    $\indicator{}$ & $1$ & $1$ & $1$ & $1$ & $1$ \\
    $\eps$ & $1$ & $1$ & $1$ & $-1$ & $-1$ \\
    $\chichar_3$ & $3$ & $-1$ & $0$ & $1$ & $-1$ \\
    $\eps\chichar_3$ & $3$ & $-1$ & $0$ & $-1$ & $1$ \\
    $\chichar_5$ & $2$ & $2$ & $-1$ & $0$ & $0$ \\
    \hline
    $\chichar_3^2$ & $9$ & $1$ & $0$ & $1$ & $1$ \\
    $\chichar_3(g^2)$ & $3$ & $3$ & $0$ & $3$ & $-1$ \\
    $\ssquare \chichar_3$ & $6$ & $2$ & $0$ & $2$ & $0$ \\
    $\asquare \chichar_3$ & $3$ & $-1$ & $0$ & $-1$ & $1$
  \end{tabular}
\end{center}
since $e^2 = e = ((12)(34))^2 = (12)^2$,
$(123)^2 = (132)$, $(1234) = (13)(24)$.

Thus $\ssquare \chichar_3 = \indicator{} +
\chichar_3 + \chichar_5$, $\asquare \chichar_3 =
\eps \chichar_3$. So given $1, \eps,
\chichar_3$, we can construct the remaining
characters from $\ssquare \chichar_3$ and
$\asquare \chichar_3$. More generally, for any
vector sapce $V$ we may consider
\glssymboldefn{tpower}{$V^{\otimes
n}$}{$V^{\otimes n}$}
\[ V^{\tprod n} = \ub{V \tprod \cdots \tprod
V}_{\text{$n$ times}} = V \tprod V^{\tprod (n -
1)} \]
for $n \ge 1$ ($V^{\tprod 0} = k$, $V^{\tprod 1}
= V$). Then for any $w \in S_n$, we can define
an (invertible) linear map
\begin{align*}
  \sigma(w) : \tpower V^n
  &\to \tpower V^n \\
  v_1 \tvprod \cdots \tvprod v_n
  &\mapsto v_{w^{-1}(1)} \tvprod \cdots \tvprod
  v_{w^{-1}(n)}
\end{align*}
for $v_1, \ldots, v_n \in V$.

\textbf{Exercise:} Show that this defines a
\gls{rep} of $S_n$ on $\tpower V^n$ and that if
$V$ is a \gls{rep} of $G$ then the $G$-action and
$S_n$-action on $\tpower V^n$ commute.

Thus we can decompose $\tpower V^n$ as a \gls{rep}
of $S_n$ into \glsref[iso_component]{isotypic
components} if $\characteristic k = 0$ and each
will be a \Ginv\ subspace of $\tpower V^n$.

\begin{definition*}
  \glsnoundefn{symm_pow}{symmetric power}{N / A}
  \glsnoundefn{alt_pow}{alternating power}{exterior
  power}
  \glssymboldefn{symm_pow}{$S^n V$}{$S^n V$}
  \glssymboldefn{alt_pow}{$\Lambda^n V$}{$\Lambda^n V$}
  If $V$ is a vector space
  \begin{enumerate}[(i)]
    \item The $n$-th \emph{symmetric power} of $V$
      is
      \[ S^n V \defeq \{a \in \tpower V^n \st
      \sigma(w)(a) = a ~\forall w \in S_n\} \]
    \item The $n$-th \emph{alternating / exterior
      power} of $V$ is
      \[ \Lambda^n V \defeq \{a \in V \st
      \sigma(w)(a) = \eps(w) a ~\forall w \in
      S_n\} \]
  \end{enumerate}
\end{definition*}
\vspace{-1em}

Note that for $n \ge 3$,
\[ \spow{n}{V} \oplus \apow{n}{V} = \{a \in
\tpower V^n \st \sigma(w) a = a ~\forall w \in
A_n\} \subsetneq \tpower V^n \]
We also define the following notation for $v_1,
\ldots, v_n \in V$
\begin{align*}
  v_1 v_2 \cdots v_n
  &= \frac{1}{n!} \sum_{w \in S_n} v_{w(1)}
  \tvprod \cdots \tvprod v_{w(n)} \\
  v_1 \anwedge \cdots \anwedge v_n
  &= \frac{1}{n!} \sum_{w \in S_n} \eps(w)
  v_{w(1)} \tvprod \cdots \tvprod v_{w(n)}
\end{align*}
(for $\characteristic k = 0$).

\textbf{Exercise:} Show that if $v_1, \ldots, v_d$
is a basis for $V$ then
\[ v_{i_1} \cdots v_{i_n} \st 1 \le i_1 \le i_2
\le \cdots \le i_n \le d\} \]
is a basis for $\spow{n}{V}$, and
\[ v_{i_1} \anwedge \cdots \anwedge v_{i_n} \st 1 \le i_1 \le i_2
\le \cdots \le i_n \le d\} \]
is a basis for $\apow{n}{V}$. Hence given a basis
for $V$ with respect to which $\rho(g)$ is
diagonal, compute $\chichar_{\spow{n}{V}}(g)$ and
$\chichar_{\apow{n}{V}}(g)$ in terms of the
eigenvlaues of $\rho(g)$.

For any vector space $V$, $\apow{\dim V} V \simeq
k$ and $\apow{n}{V} = 0$ for $n > \dim V$.

\textbf{Exercise:} Show that if $(\rho, V)$ is a
\gls{rep} of $G$ then $\apow{\dim V} V \simeq \det
\rho$ as a \gls{rep} of $G$.

\begin{definition*}
  Given a vector space $V$ we can define the
  \emph{tensor algebra} of $V$
  \[ TV \defeq \bigoplus_{n \ge 0} \tpower V^n \]
  as an infinite dimensional vector space. Then
  $TV$ is a (non-commutative) graded ring with
  product
  \[ (v_1 \tvprod \cdots \tvprod v_r) \cdot (w_1
  \tvprod \cdots \tvprod w_n) = v_1 \tvprod \cdots
  \tvprod v_r \tvprod w_1 \tvprod \cdots \tvprod w_s
  \in \tpower V^{(r + s)} \]
  For $v_1 \tvprod \cdots v_r \in \tpower V^r$,
  $w_1 \tvprod \cdots \tvprod w_r \in \tpower
  V^r$ with graded quotient map. The
  \emph{symmetric algebra} of $V$
  \[ SV \defeq \frac{TV}{\{x \tvprod y - y \tvprod
  x \st x, y \in V\}} \]
  and \emph{exterior algebra} of $V$
  \[ \Lambda V \defeq \frac{TV}{\{x \tvprod y + y
  \tvprod x \st x, y \in V\}} \]
\end{definition*}
\vspace{-1em}

One can show $\symmalg V \simeq \bigoplus_{n \ge
0} \spow{n}{V}$ and $\altalg V \simeq \bigoplus_{n
\ge 0} \apow{n}{V}$ in characteristic $0$. This
can be seen via
\[ x_1 \cdots x_n \mapsfrom [x_1 \tvprod \cdots
\tvprod x_n] \qquad x_1 \awedge \cdots \awedge x_n
\mapsfrom [x_1 \tvprod \cdots \tvprod x_n] \]
$\symmalg V$ is a commutative graded ring and
$\altalg V$ is a graded commutative ($x \in
\apow{r} V$ and $y \in \apow{s}{V}$ then $x
\awedge y = (-1)^{rs} y \awedge x$) ring.

\subsection{Duality}

Recall that $\mathcal{C}_G$ has a $*$-operation
given by $f^*(g) = f(g^{-1})$ for all $f \in
\mathcal{C}_G$, $g \in G$. This also restricts to
$\charring(G)$. Recall also if $(\rho, V)$ is a
\gls{rep} of $G$ then the \emph{dual rep} $(\rho^*,
V^*)$ is defined by
\[ \rho^*(g)(\theta)(v) = \theta(\rho(g^{-1})(v))
\qquad \forall v \in V, g \in G, \theta \in V^* \]

\begin{lemma}
  $\chichar_{V^*} = \chichar_V^*$.
\end{lemma}

\begin{proof}
  If $\rho(G)$ is represented by $A$ with respect
  to a bais $v_1, \ldots, v_d$ for $V$ and
  $\eps_1, \ldots, \eps_d$ is the dual basis for
  $V^*$, then $\rho(g^{-1}) v_i = \sum_j
  (A^{-1})_{ji} v_j$ So
  \[ \rho^*(g)(\eps_k)(v_i) = \eps_k(\rho^{-1}(g)
  v_i) = \eps_k \left( \sum_j (A^{-1})_{ji} v_j
  \right) = (A^{-1})_{ki} \]
  and
  \[ \rho^*(g)(\eps_k) = \sum_j (A^{-1})^\top_{jk}
  \eps_j \]
  i.e. $\rho^*(g)$ is represented by
  $(A^{-1})^\top$ with respect to this dual basis.
  Taking traces gives $\chichar_{\rho^*}(g) =
  \chichar_\rho (g^{-1}) = \chichar_\rho^*(g)$.
\end{proof}

We say $V$ is self-dual if $V \simeq V^*$ as
\glspl{rep} of $G$. When $G$ is finite and $k =
\CC$ then $V$ is self-dual if and only if
$\chichar_V^* = \chichar_V$ which happens if and
only if $\chichar_V(g) \in \RR ~\forall g \in G$
since $\chichar_V^* = \ol{\chichar_V}$ in this
case.

\begin{example*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item $G = \langle x \rangle \simeq C_3$ and
      $V = \CC$, $\rho : G \to \CC^\times$, $x^j
      \mapsto e^{2\pi ij/3}$. Then $\rho^*(x^j) =
      e^{-2\pi ij/3}$ and $V$ is not self-dual.
    \item $G = S_n$ since $[g]_{S_n} =
      [g^{-1}]_{S_n} ~\forall g \in S_n$, every
      \gls{rep} of $S_n$ is self-dual.
    \item Permutation \glspl{rep} are always
      self-dual.
  \end{enumerate}
\end{example*}
\vspace{-1em}

We now have various ways of building \glspl{rep}
of a group $G$.
\begin{itemize}
  \item permutation \glspl{rep}.
  \item restrict \glspl{rep} of $H$ to $G$ along
    homomorphisms $\theta : G \to H$.
  \item tensor products.
  \item $\spow{n}{V}$ and $\apow{n}{V}$.
  \item decomposition of \glspl{rep} into
    \gls{irred} components.
  \item character theoretically, e.g. row / column
    orthogonality in \gls{char_table}.
\end{itemize}
One more next time related to restriction from $G$
to $H$ for $H \le G$ called induction.