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\section{Arithmetic Properties of Characters}

We'll assume $G$ is finite and $k = \CC$.

\subsection{Arithmetic results}
\label{subsec7_1}

The following facts will be proved in
\courseref[Number Fields]{NF} next term.

\begin{definition*}
  $x \in \CC$ is an \emph{algebraic integer} if it
  is a root of a monic polynomial with integer
  coefficients.
\end{definition*}

\subsubsection*{Facts}

\begin{enumerate}[(1)]
  \refsteplabel[Fact (1)]{lec17_fact1}
  \item The algebraic integers form a subring
    $\algO$ of $\CC$.
    \glssymboldefn{algO}{$O$}{$O$}
  \refsteplabel[Fact (2)]{lec17_fact2}
  \item Any subring of $\CC$ that is finitely
    generated as an (additive) abelian group is
    contained in $\algO$.
  \refsteplabel[Fact (3)]{lec17_fact3}
  \item If $x \in \algO \cap \QQ$ then $x
    \in \ZZ$ (see )
\end{enumerate}
For (1) and (2), see
\href{http://www.dpmms.cam.ac.uk/study/IB/GroupsRings\%2BModules/2022-2023/Example\%20sheet\%204.pdf}{GRM
Example Sheet 4, Q13 from 2023}. For (3), see
\href{http://www.dpmms.cam.ac.uk/study/IA/Numbers\%2BSets/2021-2022/numset3_2021.pdf}{Numbers
\& Sets Example Sheet 3, Q3 from 2021}.

\begin{lemma*}
  If $\chichar$ is a \gls{char_rep} of $G$ then
  $\chichar(g) \in \algO$ for all $g \in G$.
\end{lemma*}

\begin{proof}
  We know that $\chichar(g)$ is a sum of $n$-th
  roots of unity ($n = |G|$, say). Each such
  $n$-th root of unity satisfies $X^n - 1$ and so
  lies in $\algO$. So $\chichar(g) \in
  \algO$ by \nameref{lec17_fact1}.
\end{proof}

\subsubsection*{The group algebra}

We now want to make the $k$-vector space $kG$ into
a ring. There are two sensible ways to do this.
One is by pointwise multiplication making $kG$ a
commutative ring.
\begin{flashcard}[convolution-product]
More usefully for us right now
is the \emph{comvolution product}
\glsnoundefn{conv_prod}{convolution product}{N/A}
\glssymboldefn{conv_ring}{$kG$}{$kG$}
\[ (f_1 f_2)(g) = \cloze{\sum_{x \in X} f_1(gx)
f_2(x^{-1}) = \sum_{\substack{x, y \in G \\ xy =
g}} f_1(x) f_2(y)} .\]
\cloze{that makes $kG$ into a (usually) non-commutative
ring.}
\end{flashcard}

We can verify that $\delta_g \delta_h =
\delta_{gh}$ for all $g, h \in G$. So we can
rephrase the product as
\[ \left( \sum{g \in G} \lambda_g \delta_g \right)
\left( \sum_{h \in G} \mu_H \delta_h \right) =
\sum_{k \in G} \left( \sum{\substack{g, h \in G \\
gh = k}} \lambda_j \mu_h \right) \delta_k .\]
From now on we'll have this product in mind when
we view $\convring kG$ as a ring. A (finitely generated)
$\convring kG$-module is ``the same'' as a \gls{rep} of
$G$. Given a \gls{rep} $(\rho, V)$ of $G$ we make
$V$ into a (finitely generated) $\convring kG$-module via
\[ f \cdot v = \sum_{g \in G} f(g) \rho(g) v
\qquad \forall v \in V, f \in \convring kG .\]
Conversely, given a finitely generated $\convring kG$-module
$M$, the underlying $k$-vector space is a
\gls{rep} of $G$ via
\[ \rho(g)(m) = \delta_{g \cdot m} \qquad \forall
m \in M, g \in G .\]
Moreover, under this correspondence \Glin{} maps
correspond to $\convring kG$-module homomorphisms.

\textbf{Exercise:} Suppose $kX$ is a permutation
\gls{rep} of $G$. Calculate the action of $f \in
\convring kG$ on $kX$ under the correspondence.

\glssymboldefn{conv_center}{$Z(kG)$}{$Z(kG)$}
It will prove useful to study $Z(\convring kG)$, the
\emph{centre} of $\convring kG$; that is the subring of
$\convring kG$
consisting of elements $f \in \convring kG$ such that $fh =
hf$ for all $h \in \convring kG$. This is because
for $f \in \convZ(\convring kG)$,
\[ \sum_{g \in G} f(g) \rho(g) \in \GHom_G(V, V) \]
for every \gls{rep} $(\rho, V)$ of $G$.

\begin{flashcard}[center-kG-characterisation-lemma]
\begin{lemma*}
  Suppose $f \in \convring kG$. Then $f \in
  \convZ(\convring kG)$ if and only if \cloze{$f \in
  \mathcal{C}_G$ the space of \glspl{cls_fn}. In
  particular, $\dim_k \convZ(\convring kG) =
  \#\text{conjugacy classes in $G$}$.}
\end{lemma*}

\begin{proof}
  \cloze{\begin{align*}
    f \in \convring kG
    &\iff fh = hf \quad \forall h \in \convring kG
    \\
    &\iff f\delta_g = \delta_g f \quad \forall g
    \in G \\
    &\iff \delta_{g^{-1}} f \delta_g = f
    &&\text{(since $\delta_e = 1$ and
    $\delta_{g^{1}} \delta_g = \delta_e$)}
  \end{align*}
  But
  \begin{align*}
    (\delta_{g^{-1}} f \delta_g)(x)
    &= \sum_{g \in G} (\delta_{g^{-1}} f)(xy^{-1})
    \delta_g(y) \\
    &= (\delta_{g^{-1}} f)(xg^{-1}) \\
    &= f(gxg^{-1}) \qquad \forall g \in G
  \end{align*}
  So $f \in \convZ(\convring kG)$ if and only if
  $f \in \mathcal{C}_G$ as required.}
\end{proof}
\end{flashcard}

\begin{remark*}
  The multiplication on $\convZ(\convring kG)$ and
  $\mathcal{C}_G$ will not be the same even though
  their $k$-vector space structures are the same
  even though both are commutative.
\end{remark*}

\begin{notation*}
  \glssymboldefn{clsum}{$C_i$}{$C_i$}
  Given $g \in G$ define the \emph{class sum}
  \[ C_{[g]_G}(x) = \begin{cases}
    1 & x \in [g]_G \\
    0 & x \not\in [g]_G
  \end{cases} \]
  Then if $[g_1]_G, \ldots, [g_r]_G$ is a list of
  conjugacy classes in $G$ we write
  \[ C_i \defeq C_{[g_i]_G} .\]
  We used to write $\indicator{[g_i]_G}$ for
  $C_i$. We have switched to draw attention to the
  different multiplication.
\end{notation*}

\begin{flashcard}[clsum-product-prop]
\begin{proposition*}
  \[ \clsum_i \clsum_j = \cloze{\sum_{l = 1}^r a_{ij}^l
  \clsum_l} \]
  where
  \[ a_{ij}^l = \cloze{|\{(x, y) \in [g_1]_G \times
  [g_j]_G \st xy = g_l\} \in \ZZ} .\]
  \cloze{The $a_{ij}^l$ are called the \emph{structure
  constants} of $\convZ(\convring kG)$.}
\end{proposition*}

\begin{proof}
  \cloze{Since $\convZ(\convring kG)$ is a ring,
  \[ \clsum_i \clsum_j = \sum_{l = 1}^r a_{ij}^l
  C_l \]
  for some $a_{ij}^l \in k$. But we explicitly
  compute
  \begin{align*}
    a_{ij}^l
    &= (\clsum_i \clsum_j)(g_l) \\
    &= \sum_{x, y \in G} \clsum_i(x) \clsum_j(y)
    \\
    &= |\{(x, y) \in [g_i]_G \times [g_j]_G \st xy
    = g_l\}|
  \end{align*}
  as required.}
\end{proof}
\end{flashcard}

SUppose now that $(\rho, V)$ is an \gls{irred}
\gls{rep} of $G$. Then we've seen that if $z \in
\convZ(\convring kG)$, then
\[ z : V \to V, \quad zv = \sum_{g \in G} z(g)
\rho(g) \in \GHom_G(V, V) = k\id_V \]
($k$ algebraically closed). So we get a
$k$-algebra homomorphism
\[ \omega_\rho : \convZ(\convring kG) \to k \]
where $z \in \convring \convZ(kG)$ acts by
$\omega_{\rho(z)} \id_V$ on $V$. Taking traces we
see
\begin{align*}
  \label{lec17_l206_eq}
  (\dim V_i) \homega_{\rho(z)}
  &= \sum_{g \in G} z(g) \chichar_\rho(g) \\
  \implies \homega_\rho(z)
  &= \sum_{g \in G} \frac{z(g)
  \chichar_\rho(g)}{\chichar_\rho(e)} \\
  \implies \homega_\rho(\clsum_i)
  &= \frac{\chichar_\rho(g_i)}{\chichar_\rho(e)}
  |[g_i]_G| \tag{\dag}
\end{align*}
We now see that $\homega_\rho$ only depends on
$\chichar_\rho$ so we can write
$\homega_{\chichar_\rho} = \homega_\rho$.

\begin{lemma*}
  The values $\homega_{\chichar}(\clsum_i) \in
  \algO$ for all \gls{irred}
  \glsref[char_rep]{characters} $\chichar$.
\end{lemma*}
\vspace{-1em}

Note that htis is not immediately clear as
$\frac{1}{\chichar(e)} \not\in \algO$ for
$\chichar(e) \neq 1$.

\begin{proof}
  Since $\homega_{\chichar}$ is an algebra
  homomorphism,
  \[ \label{lec17_l233_eq} \homega_{\chichar}(\clsum_i)
  \homega_{\chichar}(\clsum_j) = \sum_{l = 1}^r
  a_{ij}^l \homega_{\chichar}(\clsum_l) \tag{$*$} \]
  so the subring of $\CC$ generated by the
  $\homega_{\chichar}(\clsum_i)$ is a finitely
  generated abelian group under $+$. By
  \nameref{lec17_fact2} in \cref{subsec7_1}, it
  follows that ecah $\homega_{\chichar}(\clsum_i)
  \in \algO$.
\end{proof}

\begin{lemma*}
  \[ a_{ij}^l = \frac{|G|}{|\clsum_G(g_i)|
  |\clsum_G(g_j)|} \sum_{\chichar \in \Irr(G)}
  \frac{\chichar(g_i) \chichar(g_j)
  \chichar(g_l^{-1})}{\chichar(e)} .\]
  In particular, $a_{ij}^l$ is deterined by the
  character table.
\end{lemma*}

\begin{proof}
  By \eqref{lec17_l233_eq} and \eqref{lec17_l206_eq}
  \[ \frac{\chichar(g_i)}{\chichar(e)} |[g_i]_G|
  \frac{\chichar(g_j)}{\cancel{\chichar(e)}}
  |[g_j]_G| = \sum_{l = 1}^r \frac{a_{ij}^l
  \chichar(g_l) |[g_l]_G|}{\cancel{\chichar(e)}} \]
  TODO
\end{proof}