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$\chichar_0 = 0$ chere $0$ is the $k$-vector space
of dimension $0$, so $\GL(0) = \{\id_0\}$.
$\chichar_k = \indicator{G}$ where $k$ is the
\gls{triv_rep}.

\textbf{Goal:} If $V$, $W$ are \glspl{rep}, build
a \gls{rep} $V \otimes W$ such that $\chichar_{V
\otimes W} = \chichar_V \cdot \chichar_W$.

\subsection{Tensor Products}

Suppose $V$ and $W$ are vector spaces over $k$
with bases $v_1, \ldots, v_m$ and $w_1, \ldots,
w_n$ respectively. We can view $V \oplus W$ as the
set of pairs $(v, w)$ with $v \in V$, $w \in W$
under pointwise operation or as the vector space
with basis $v_1, \ldots, v_m, w_1, \ldots, w_n$.

\begin{flashcard}[tensor-prod-defn]
\begin{definition*}
  \glsnoundefn{tens_prod}{tensor product}{tensor
  products}
  \glssymboldefn{tensor_product}{$\otimes$}{$\otimes$}
  The \emph{tensor product} \cloze{$V \otimes W$ of $V$
  and $W$ is the $k$ vector space with basis $w_i
  \otimes w_i$ for $1 \le i \le m$ and $1 \le j \le
  n$ (so $\dim V \otimes W = (\dim V)(\dim W)$).}
\end{definition*}
\end{flashcard}

\begin{example*}
  If $X$ and $Y$ are finite then $kX \tensprod kY$
  has basis $\delta_x \tensvecprod \delta_y$ for $x \in
  X$, $y \in Y$ and $\alpha_{X \times Y} : kX
  \tensprod kY \to kX \times Y$, $\delta_x
  \tensvecprod
  \delta_y \mapsto \delta_{(x, y)}$ extends to an
  isomorphism of vector spaces.
  \begin{notation*}
    If $v = \sum \lambda_i v_i \in V$ and $w = \sum
    \mu_j w_j \in W$ then
    \[ v \tensvecprod w \defeq \sum_{i, j} \lambda_i
    \mu_j v_i \tensvecprod w_j \in V \tensprod W .\]
  \end{notation*}
  So under $\alpha_{X \times Y}$,
  \[ \alpha_{X \times Y} (f \tensvecprod g)(x, y) =
  f(x)g(y) \]
  Note that in general, not every element of $V
  \tensprod W$ can be written in the form $V
  \tensprod
  W$. For example, $v_1 \tensvecprod w_1 + v_2
  \tensvecprod
  w_2$. The smallest number of summands needed is
  called the \emph{rank} of the tensor.
\end{example*}

\begin{lemma*}
  The map $V \times W \to V \tensprod W$, $(v, w)
  \mapsto v \tensvecprod w$ is bilinear.
\end{lemma*}

\begin{proof}
  We should prove that if $x, x_1, x_2 \in V$ and
  $y, y_1, y_2 \in W$ and $u_1, u_2 \in k$ then
  \begin{align*}
    (u_1 x_1 + u_2 x_2) \tensvecprod y
    &= u_1 (x_1 \tensvecprod y) + u_2 (x_2
    \tensvecprod y) \\
    x \tensvecprod (u_1 y_1 + u_2 y_2)
    &= u_1 (x \tensvecprod y_1) + u_2 (x
    \tensvecprod y_2)
  \end{align*}
  We'll do the second and then appeal to symmetry.
  We write $x = \sum \lambda_i v_i$, $y_k = \sum
  \mu_j^k w_j$ for $k = 1, 2$. Then
  \begin{align*}
    x \tensvecprod (u_1 y_1 + u_2 y_2)
    &= \sum_{i, j} \lambda_i (u_1 \mu_j^1 + u_2
    \mu_j^2) (v_i \tensvecprod w_j) \\
    u_1 (x \tensvecprod y_1) + u_2(x \tensvecprod
    y_2)
    &= \sum_{i, j} u_1 \lambda_i \mu_j^1 (v_i
    \tensvecprod w_j) + \sum_{i, j} u_2 \lambda_I
    \mu_j^2 (v_i \tensvecprod w_j)
  \end{align*}
  These are equal.
\end{proof}

\textbf{Exercise:} Show that given $U$, $V$ and
$W$ there is a one to one correspondence
\[ \{\text{linear maps $V \tensprod W \to U$}\}
\to \{\text{bilinear maps $V \times W \to U$}\} \]
given by precomposition with the bilinear map $(v,
w) \mapsto v \tensvecprod w$.

\begin{lemma*}
  If $x_1, \ldots, x_m$ is any basis for $V$ and
  $y_1, \ldots, y_n$ is any basis for $W$ then
  $x_i \tensvecprod y_j$ with $1 \le i \le m$ and
  $1 \le j \le n$ is a basis for $V \tensprod W$.
  Thus the definition of $V \tensprod W$ does not
  depend on the choice of basis.
\end{lemma*}

\begin{proof}
  It suffices to show that the given set spans $V
  \tensprod W$ since it has size $mn$. But if $v_i
  = \sum_r A_{ri} x_r$ and $w_J = \sum_s V_{sj}
  y_s$ then
  \[ v_i \tensvecprod w_j = \sum_{r, s} (A_{ri}
  B_{sj})  (x_r \tensvecprod y_s) \]
  But $\{v_i \tensvecprod w_j\}_{1 \le i \le m, 1
  \le j \le n}$ span $V \tensvecprod V \tensprod
  W$ so we're done.
\end{proof}

\begin{remark*}[for enthusiast's]
  In fact we could've defined $V \tensprod W$ in a
  basis independent way. Let $F$ be the (infinite
  dimensional) vector space basis $\langle v
  \tensvecprod w \st v \in V, w \in W \rangle$ and
  $R$ the subspace spanned by elements
  \begin{align*}
    x \tensvecprod (u_1 y_1 + u_2 y_2) - u_1 (x
    \tensvecprod y_1) - u_2 (x \tensvecprod y_2)
    \\
    (u_1 v_1 + u_2 v_2) \tensvecprod y - u_1 (x_1
    \tensvecprod y) - u_2(x_2 \tensvecprod y)
  \end{align*}
  for all $x, x_1, x_2 \in V$, $y, y_1, y_2 \in
  W$, $u_1, u_2 \in k$. Then let $V \tensprod W =
  F / R$.
\end{remark*}

\textbf{Exercise:} Show that for vector spaces
$U$, $V$ and $W$ there is a natural (basis
independent) isomorphism
\[ (U \oplus V) \tensprod W \to (U \tensprod W)
\oplus (V \tensprod W) .\]

\begin{flashcard}[lin-map-tens-prod]
\begin{definition*}
  \glssymboldefn{tens_lin_prod}{$\otimes$}{$\otimes$}
  Suppose $V$ and $W$ are vector spaces with bases
  $v_1, \ldots, v_m$ and $w_1, \ldots, w_n$ and
  $\varphi : V \to V$ and $\psi : W \to W$ are
  linear maps. We can define
  \begin{align*}
    \varphi \otimes \psi
    &: \cloze{V \tensprod W \to V \tensprod W}
    (\varphi \otimes \psi)
    &\cloze{(v_i \tensvecprod w_j) = \varphi(v_i
    \tensvecprod \psi(w_j)}
  \end{align*}
\end{definition*}
\end{flashcard}

\begin{example*}
  If $\varphi$ is represented by $A$ and $\psi$ is
  represented by $B$ with respect to given bases,
  then if we order $V_i \tprod W_j$
  lexicographically (i.e. $v_1 \tvprod w_1$, $v_1
  \tvprod w_2$, $v_1 \tvprod w_3$, \ldots, $v_1
  \tvprod w_n$, $v_2 \tvprod w_1$, \ldots, $v_m
  \tvprod w_n$), then $(\varphi \tlprod \psi)$ is
  representede by the block matrix
  \[
  \begin{pmatrix}
    A_{11} B & A_{12} B & \cdots & \\
    A_{21} B & A_{22} B & \cdots & \\
    \vdots & \vdots & \ddots & \\
     &  &  & A_{mm} B
  \end{pmatrix} \]
  Since
  \[ (\varphi \tlprod \psi) (v_i \tvprod w_j) =
  \left( \sum_k A_{ki} v_k \right) \tvprod \left(
  \sum_l B_{lj} w_l \right) = \sum_{k, l} A_{ki}
  B_{lj} v_k \tvprod w_l \]
\end{example*}

\begin{lemma*}
  The linear map $\varphi \tlprod \psi$ does not
  depend on the basis. Indeed,
  \[ (\varphi \tlprod \psi)(v \tvprod w) =
  \varphi(v) \tvprod \psi(w) \qquad \forall v \in
  V, w \in W .\]
\end{lemma*}

\begin{proof}
  Writing $v = \sum \lambda_i v_i$ and $w = \sum
  \mu_j w_j$,
  \begin{align*}
    (\varphi \tlprod \psi)(v \tvprod w)
    &= (\varphi \tlprod \psi) \left( \sum_{i, j}
    \lambda_i \mu_j v_i \tvprod w_j \right) \\
    &= \sum_{i, j} \lambda_i \mu_j \varphi(v_i)
    \tvprod \psi(w_j) \\
    &= \varphi(v) \tvprod \psi(w)
  \end{align*}
\end{proof}

\begin{remark*}
  The proof really says that $V \times W \to V
  \tprod W$, $(v, w) \mapsto v \tvprod w$ is bilinear
  and $\varphi \tlprod \psi$ is the corresponding
  linear map $V \tprod W \to V \tprod W$ from an
  earlier exercise.
\end{remark*}

\begin{lemma*}
  \refsteplabel[the last lemma last time]{the_last_lemma_last_time_l11}
  Suppose $\varphi, \varphi_1, \varphi_2 \in
  \Hom_k(V,V)$ and $\varphi, \varphi_1, \varphi_2
  \in \Hom_k(W, W)$. Then
  \begin{enumerate}[(i)]
    \item $(\varphi_1 \varphi_2) \tlprod (\psi_1
      \psi_2) = (\varphi_1 \tlprod
      \psi_1)(\varphi_2 \tlprod \psi_2) \in
      \Hom_k(V \tprod W, V \tprod W)$.
    \item $\id_V \tlprod \id_W = \id_{V \tprod W}$
    \item $\Trace(\varphi \tlprod \psi) =
      \Trace(\varphi) \Trace(\psi)$.
  \end{enumerate}
\end{lemma*}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Given $v \in V$, $w \in W$ by the last
      lemma we can compute
      \begin{align*}
        (\varphi_1 \varphi_2 \tlprod \varphi_1
        \varphi_2)(v \tvprod w)
        &= \varphi_1 \varphi_2(v) \tvprod
        \varphi_1 \varphi_2(w) \\
        &= (\varphi_1 \tlprod \psi_1)(\varphi_2(v)
        \tvprod \psi_2(w)) \\
        &= (\varphi_1 \tlprod \psi_1)(\varphi_2
        \tlprod \psi_2)(v \tvprod w)
      \end{align*}
      We're done since all maps and linear and
      $\{v \tvprod w\}$ spans $V \tprod W$.
    \item Is clear.
    \item By earlier example it suffices to see
      \[ \Trace
      \begin{pmatrix}
        A_{11}B & & & \\
         & A_{22} B & & \\
         & & \ddots & \\
         & & & A_{nn} B
      \end{pmatrix}
      = \sum_{i, j} B_{ii} A_{jj} = (\Trace
      A)(\Trace B) \qedhere \]
  \end{enumerate}
\end{proof}

\begin{flashcard}[rep-tens-prod-defn]
\begin{definition*}
  \glssymboldefn{tens_rep_prod}{$\otimes$}{$\otimes$}
  Given two \glspl{rep} $(\rho, V)$ and $\sigma,
  W)$ of a group $G$ we can define a \gls{rep}
  $(\rho \otimes \sigma, V \tprod W)$ via
  \[ (\rho \otimes \sigma)(g) = \cloze{\rho(g) \tvprod
  \sigma(g)} \]
\end{definition*}
\end{flashcard}
\vspace{-1em}

This is a \gls{rep} by parts (i) and (ii) of the
last lemma, and $\chichar_{\rho \trprod \sigma} =
\chichar_\rho \cdot \chichar_\sigma$ by part
(iii).