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\ihead{Marginalia to ``Structure theory for the group algebra ...''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{\href{https://eudml.org/doc/287582}{\textbf{Structure theory for the
group algebra of the symmetric group, with applications to polynomial
identities for the octonions}}}

\textit{Murray R. Bremner, Sara Madariaga, Luiz A. Peresi}

Comment. Math. Univ. Carolin. \textbf{57},4 (2016), pp. 413--452.

version of 5 December 2016

\textbf{Errata by Darij Grinberg}

\bigskip
\end{center}

\setcounter{section}{5}

\section*{Marginalia}

The following are my comments on specific places in the paper
\textquotedblleft\href{https://eudml.org/doc/287582}{Structure theory for the
group algebra of the symmetric group, with applications to polynomial
identities for the octonions}\textquotedblright\ by Murray R. Bremner, Sara
Madariaga, Luiz A. Peresi (Comment. Math. Univ. Carolin. 57,4 (2016), pp.
413--452). Very few of them are corrections (there is barely anything wrong in
the paper); most of them are additional details and steps that have been
omitted from the proofs.

\begin{itemize}
\item \textbf{page 414:} You write that the matrices obtained by restricting
$\phi$ to $S_{n}$ \textquotedblleft have entries in $\left\{  0,\pm1\right\}
$\textquotedblright. If you are talking about the matrices $R^{\lambda}\left(
p\right)  $ from Definition 1.49, then I don't see why this is true (and I
suspect it is not).

\item \textbf{page 415, Definition 1.5:} \textquotedblleft A \textbf{Young
tableau} $T^{\lambda}$\textquotedblright\ should be \textquotedblleft A
\textbf{Young tableau} $T$\textquotedblright. (The \textquotedblleft$\lambda
$\textquotedblright\ superscript is unnecessary and confusing; you just call
it \textquotedblleft$T$\textquotedblright\ afterwards.)

\item \textbf{page 416, Definition 1.9:} It would be good to point out that
this action of $S_{n}$ on the set $\left\{  \text{tableaux of shape }%
\lambda\right\}  $ is free and transitive. (This is being used tacitly further below.)

\item \textbf{page 417, Remark 1.14:} This is perhaps a bit out of place: You
have yet to use the notation $hvT$ at this point!

\item \textbf{page 417, proof of Proposition 1.15:} In the second paragraph of
this proof, replace \textquotedblleft obtaining tableaux $T^{\lambda^{\prime}%
}\succ T^{\mu^{\prime}}$ where $\lambda^{\prime}$ and $\mu^{\prime}$ are
partitions of $n-n_{1}$\textquotedblright\ by \textquotedblleft obtaining
tableaux $T^{\lambda^{\prime}}$ and $T^{\mu^{\prime}}$ whose shapes
$\lambda^{\prime}$ and $\mu^{\prime}$ are partitions of $n-n_{1}$ satisfying
$Y^{\lambda^{\prime}}\succ Y^{\mu^{\prime}}$ \textquotedblright.

\item \textbf{page 419, proof of Lemma 1.19:} It would be helpful to point out
that the third equality sign in the displayed equation relies on the facts
that $\epsilon\left(  v\right)  ^{-1}=\epsilon\left(  v\right)  $ (because
$\epsilon\left(  v\right)  \in\left\{  1,-1\right\}  $) and that $\epsilon$ is
a group homomorphism.

\item \textbf{page 419, proof of Proposition 1.20:} Remove \textquotedblleft
Since $\epsilon\left(  p\right)  =\epsilon\left(  p^{-1}\right)
$\textquotedblright. (You don't use the fact that $\epsilon\left(  p\right)
=\epsilon\left(  p^{-1}\right)  $ here.)

\item \textbf{page 419, Definition 1.21:} Replace \textquotedblleft in lex
order\textquotedblright\ by \textquotedblleft in some fixed order chosen in
such a way that the standard tableaux of shape $\lambda$ will be $T_{1}%
,T_{2},\ldots,T_{d_{\lambda}}$ in lex order\textquotedblright. Indeed, you
don't need all the $n!$ tableaux $T_{1},T_{2},\ldots,T_{n!}$ to be in lex
order; but you will later want the standard tableaux of shape $\lambda$ to be
$T_{1},T_{2},\ldots,T_{d_{\lambda}}$ in lex order. If you list all the $n!$
tableaux of shape $\lambda$ in lex order, then (in general) the first
$d_{\lambda}$ tableaux in your list will not be the standard tableaux of shape
$\lambda$.

\item \textbf{page 420, Corollary 1.24:} After \textquotedblleft be the
standard tableaux\textquotedblright, add \textquotedblleft of shape $\lambda
$\textquotedblright.

\item \textbf{page 421, proof of Proposition 1.25:} At the beginning of this
proof, add \textquotedblleft Again, set $H_{i}=H_{T_{i}}$ and $V_{i}=V_{T_{i}%
}$; thus, $D_{i}=H_{i}V_{i}$.\textquotedblright.

\item \textbf{page 421, proof of Proposition 1.25:} After \textquotedblleft%
$hD_{i}^{2}v=\left(  hH_{i}\right)  V_{i}H_{i}\left(  V_{i}v\right)
=\epsilon\left(  v\right)  H_{i}V_{i}H_{i}V_{i}=\epsilon\left(  v\right)
D_{i}^{2}$\textquotedblright, add \textquotedblleft$=\epsilon\left(  v\right)
\sum_{p\in S_{n}}x_{p}p$\textquotedblright\ (in order to make the step to the
next equality clearer).

\item \textbf{page 421, proof of Proposition 1.25:} Replace \textquotedblleft
On the left side of (7) take $p=\iota$, on the right side take $p=hv$, and
compare coefficients\textquotedblright\ by \textquotedblleft Comparing
coefficients of $hv$ on both sides of (7), we obtain\textquotedblright.

\item \textbf{page 421, proof of Proposition 1.25:} Replace \textquotedblleft
Setting $p=q$ on both sides, we obtain
\[
x_{q}tqq^{-1}tq=\epsilon\left(  q^{-1}tq\right)  x_{q}q,
\]
and this simplifies to $x_{q}q=-x_{q}q$\textquotedblright\ by
\textquotedblleft Comparing coefficients of $q$ on both sides of this
equation, we obtain $x_{q}=\epsilon\left(  q^{-1}tq\right)  x_{q}$ (since the
only $p\in S_{n}$ satisfying $tpq^{-1}tq=q$ is $q$), and this simplifies to
$x_{q}=-x_{q}$\textquotedblright.

\item \textbf{page 421, proof of Proposition 1.25:} After \textquotedblleft
Combining the results of the two cases,\textquotedblright, add
\textquotedblleft we obtain that $D_{i}^{2}=\sum_{h\in G_{H}\left(
T_{i}\right)  }\sum_{v\in G_{V}\left(  T_{i}\right)  }x_{\iota}\epsilon\left(
v\right)  hv$, since Lemma 1.12 shows that any permutation of the form $hv$
can be written in this form in exactly one way. In view of $D_{i}=\sum_{h\in
G_{H}\left(  T_{i}\right)  }\sum_{v\in G_{V}\left(  T_{i}\right)  }%
\epsilon\left(  v\right)  hv$, this rewrites as\textquotedblright.

\item \textbf{page 422, proof of Proposition 1.25:} The equality sign in
\textquotedblleft$\sum_{h,v}\epsilon\left(  v\right)  \operatorname*{trace}%
\left(  hv\right)  =\operatorname*{trace}\left(  I_{\mathbb{F}S_{n}}\right)
$\textquotedblright, again, relies on Lemma 1.12. (Indeed, Lemma 1.12 shows
that the only pair $\left(  h,v\right)  $ satisfying $hv=\iota$ is $\left(
\iota,\iota\right)  $.)

\item \textbf{page 422, Definition 1.26:} It is worth saying that
$E_{i}^{\lambda}$ will be denoted by $E_{i}$ when $\lambda$ is clear from the context.

\item \textbf{page 422, \S 1.5:} After Corollary 1.27, I suggest adding
another corollary (which is being tacitly used in the proof of Lemma 1.29):

\textbf{Corollary 1.27a.} Let $\lambda\vdash n$. Let $i,j\in\left\{
1,2,\ldots,n!\right\}  $. Then, $f_{i}=f_{j}$ and $c_{i}=c_{j}$ and
$E_{j}=s_{ji}E_{i}s_{ij}$ and $E_{i}s_{ij}=s_{ij}E_{j}$.

\textit{Proof of Corollary 1.27a.} From (5), we obtain $D_{j}=s_{ji}%
D_{i}s_{ij}$, so that $\mathbb{F}S_{n}D_{j}=\mathbb{F}\underbrace{S_{n}s_{ji}%
}_{=S_{n}}D_{i}s_{ij}=\mathbb{F}S_{n}D_{i}s_{ij}\cong\mathbb{F}S_{n}D_{i}$ as
vector spaces (since $s_{ij}\in\mathbb{F}S_{n}$ is invertible). Hence,
$\dim\left(  \mathbb{F}S_{n}D_{j}\right)  =\dim\left(  \mathbb{F}S_{n}%
D_{i}\right)  $. In other words, $f_{j}=f_{i}$ (since the numbers $f_{i}$ and
$f_{j}$ are defined to be $\dim\left(  \mathbb{F}S_{n}D_{i}\right)  $ and
$\dim\left(  \mathbb{F}S_{n}D_{j}\right)  $, respectively). In other words,
$f_{i}=f_{j}$. This yields $n!/f_{i}=n!/f_{j}$. In other words, $c_{i}=c_{j}$
(since the numbers $c_{i}$ and $c_{j}$ are defined to be $n!/f_{i}$ and
$n!/f_{j}$, respectively). Finally, the definition of $c_{i}$ yields
$c_{i}=n!/f_{i}$, so that $\dfrac{f_{i}}{n!}=\dfrac{1}{c_{i}}$. But the
definition of $E_{i}$ yields $E_{i}=\underbrace{\dfrac{f_{i}}{n!}}_{=\dfrac
{1}{c_{i}}}D_{i}=\dfrac{1}{c_{i}}D_{i}$. Likewise, $E_{j}=\dfrac{1}{c_{j}%
}D_{j}$. Hence,%
\[
E_{j}=\underbrace{\dfrac{1}{c_{j}}}_{\substack{=\dfrac{1}{c_{i}}\\\text{(since
}c_{i}=c_{j}\text{)}}}\underbrace{D_{j}}_{=s_{ji}D_{i}s_{ij}}=\dfrac{1}{c_{i}%
}\cdot s_{ji}D_{i}s_{ij}=s_{ji}\underbrace{\left(  \dfrac{1}{c_{i}}%
D_{i}\right)  }_{=E_{i}}s_{ij}=s_{ji}E_{i}s_{ij}.
\]
Hence,%
\[
\underbrace{s_{ij}}_{=\left(  s_{ji}\right)  ^{-1}}\underbrace{E_{j}}%
_{=s_{ji}E_{i}s_{ij}}=\underbrace{\left(  s_{ji}\right)  ^{-1}s_{ji}}_{=\iota
}E_{i}s_{ij}=E_{i}s_{ij},
\]
so that $E_{i}s_{ij}=s_{ij}E_{j}$. This completes the proof of Corollary
1.27a. $\blacksquare$

\item \textbf{page 422, \S 1.5:} I would also add another corollary (which is
being tacitly used in the proof of Lemma 1.32):

\textbf{Corollary 1.27b.} Let $\lambda\vdash n$. Let $i\in\left\{
1,2,\ldots,n!\right\}  $. Then, $D_{i}\neq0$ and $f_{i}\neq0$ and $E_{i}\neq0$.

\textit{Proof of Corollary 1.27b.} The definition of $D_{i}$ yields
\begin{align*}
D_{i}  &  =\underbrace{H_{T_{i}}}_{=\sum_{h\in G_{H}\left(  T_{i}\right)  }%
h}\underbrace{V_{T_{i}}}_{=\sum_{v\in G_{V}\left(  T_{i}\right)  }%
\epsilon\left(  v\right)  v}=\left(  \sum_{h\in G_{H}\left(  T_{i}\right)
}h\right)  \left(  \sum_{v\in G_{V}\left(  T_{i}\right)  }\epsilon\left(
v\right)  v\right) \\
&  =\sum_{h\in G_{H}\left(  T_{i}\right)  }\sum_{v\in G_{V}\left(
T_{i}\right)  }\epsilon\left(  v\right)  hv=\sum_{\left(  h,v\right)  \in
G_{H}\left(  T_{i}\right)  \times G_{V}\left(  T_{i}\right)  }\epsilon\left(
v\right)  hv.
\end{align*}
The group elements $hv$ on the right hand side of this equality are all
distinct (by the second sentence of Lemma 1.12); thus, the sum $\sum_{\left(
h,v\right)  \in G_{H}\left(  T_{i}\right)  \times G_{V}\left(  T_{i}\right)
}\epsilon\left(  v\right)  hv$ has no cancellations and therefore is nonzero.
In other words, $D_{i}\neq0$. Hence, the left ideal $\mathbb{F}S_{n}D_{i}$ is
nonzero, and thus $\dim\left(  \mathbb{F}S_{n}D_{i}\right)  >0$. The
definition of $f_{i}$ now yields $f_{i}=\dim\left(  \mathbb{F}S_{n}%
D_{i}\right)  >0$. Hence, $f_{i}\neq0$. Now, the definition of $E_{i}$ yields
$E_{i}=\dfrac{f_{i}}{n!}D_{i}\neq0$ (since $f_{i}\neq0$ and $D_{i}\neq0$).
This proves Corollary 1.27b. $\blacksquare$

\item \textbf{page 422, proof of Lemma 1.29:} At the beginning of the proof,
add the following sentence: \textquotedblleft Again, set $H_{i}=H_{T_{i}}$ and
$V_{i}=V_{T_{i}}$; thus, $D_{i}=H_{i}V_{i}$.\textquotedblright.

\item \textbf{page 422, proof of Lemma 1.29:} I would replace this proof with
the following more detailed version:

\textquotedblleft Again, set $H_{i}=H_{T_{i}}$ and $V_{i}=V_{T_{i}}$; thus,
$D_{i}=H_{i}V_{i}$. Define $H_{j}$ and $V_{j}$ likewise, so that $D_{j}%
=H_{j}V_{j}$.

First, assume that $s_{ji}=vh$ for some $h\in G_{H}\left(  T_{i}\right)  $ and
$v\in G_{V}\left(  T_{i}\right)  $. Then, Lemma 1.19 yields $hH_{i}=H_{i}$.
Now, from $D_{i}=H_{i}V_{i}$, we obtain $hD_{i}=\underbrace{hH_{i}}_{=H_{i}%
}V_{i}=H_{i}V_{i}=D_{i}$. Multiplying this by $\dfrac{f_{i}}{n!}$, we obtain
$hE_{i}=E_{i}$ (since $E_{i}=\dfrac{f_{i}}{n!}D_{i}$). Also, Lemma 1.19 yields
$V_{i}v=\epsilon\left(  v\right)  V_{i}$. Now, from $D_{i}=H_{i}V_{i}$, we
obtain $D_{i}v=H_{i}\underbrace{V_{i}v}_{=\epsilon\left(  v\right)  V_{i}%
}=\epsilon\left(  v\right)  \underbrace{H_{i}V_{i}}_{=D_{i}}=\epsilon\left(
v\right)  D_{i}$. Multiplying this by $\dfrac{f_{i}}{n!}$, we obtain
$E_{i}v=\epsilon\left(  v\right)  E_{i}$ (since $E_{i}=\dfrac{f_{i}}{n!}D_{i}$).

Now, Corollary 1.27a yields $E_{j}=s_{ji}E_{i}s_{ij}$, so that%
\begin{align*}
E_{i}E_{j}  &  =E_{i}\left(  \underbrace{s_{ji}}_{=vh}E_{i}s_{ij}\right)
=\underbrace{E_{i}v}_{=\epsilon\left(  v\right)  E_{i}}\underbrace{hE_{i}%
}_{=E_{i}}s_{ij}=\epsilon\left(  v\right)  \underbrace{E_{i}E_{i}%
}_{\substack{=\left(  E_{i}\right)  ^{2}=E_{i}\\\text{(by Corollary 1.27)}%
}}s_{ij}=\epsilon\left(  v\right)  E_{i}s_{ij}\\
&  =\xi_{ij}E_{i}s_{ij}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\xi
_{ij}=\epsilon\left(  v\right)  \right)  .
\end{align*}


Second, assume that $s_{ji}\neq vh$ for any $h\in G_{H}\left(  T_{i}\right)  $
and $v\in G_{V}\left(  T_{i}\right)  $. Thus, $\left(  s_{ji}\right)
^{-1}\neq\left(  vh\right)  ^{-1}$ for any $h\in G_{H}\left(  T_{i}\right)  $
and $v\in G_{V}\left(  T_{i}\right)  $. In other words, $s_{ij}\neq
h^{-1}v^{-1}$ for any $h\in G_{H}\left(  T_{i}\right)  $ and $v\in
G_{V}\left(  T_{i}\right)  $ (since $\left(  s_{ji}\right)  ^{-1}=s_{ij}$ and
$\left(  vh\right)  ^{-1}=h^{-1}v^{-1}$). Equivalently, $s_{ij}\neq hv$ for
any $h\in G_{H}\left(  T_{i}\right)  $ and $v\in G_{V}\left(  T_{i}\right)  $
(since $G_{H}\left(  T_{i}\right)  $ and $G_{V}\left(  T_{i}\right)  $ are
subgroups of $S_{n}$ and thus invariant under inversion). Hence, Lemma 1.16
(applied to $T=T_{j}$ and $p=s_{ij}$) shows that there exist two distinct
numbers $k,\ell$ that lie in the same row of $T_{j}$ and in the same column of
$s_{ij}T_{j}$. In view of $s_{ij}T_{j}=T_{i}$, this shows that $k$ and $\ell$
lie in the same column of $T_{i}$; therefore, the transposition $t=\left(
k,\ell\right)  $ satisfies $t\in G_{V}\left(  T_{i}\right)  $. Hence,
$V_{i}t=-V_{i}$. But $k$ and $\ell$ lie in the same row of $T_{j}$; thus,
$t\in G_{H}\left(  T_{j}\right)  $ and therefore $tH_{j}=H_{j}$. Now,%
\[
\underbrace{D_{i}}_{=H_{i}V_{i}}\underbrace{D_{j}}_{=H_{j}V_{j}}=H_{i}%
V_{i}H_{j}V_{j}=H_{i}\underbrace{V_{i}t}_{=-V_{i}}\underbrace{tH_{j}}_{=H_{j}%
}V_{j}=-\underbrace{H_{i}V_{i}}_{=D_{i}}\underbrace{H_{j}V_{j}}_{=D_{j}%
}=-D_{i}D_{j},
\]
so that $D_{i}D_{j}=0$. Since $E_{i}=\dfrac{f_{i}}{n!}D_{i}$ and $E_{j}%
=\dfrac{f_{j}}{n!}D_{j}$, this entails $E_{i}E_{j}=0=\xi_{ij}E_{i}s_{ij}$
(since $\xi_{ij}=0$). This completes the proof of Lemma
1.29.\textquotedblright

\item \textbf{page 423, proof of Lemma 1.32:} After \textquotedblleft and so
$\xi_{ij}=0$\textquotedblright, I would add \textquotedblleft(since Corollary
1.27b yields $E_{i}\neq0$, and thus $E_{i}s_{ij}\neq0$)\textquotedblright.

\item \textbf{page 423, proof of Lemma 1.32:} Replace \textquotedblleft and so
Lemma 1.29 gives $E_{i}=\xi_{ii}E_{i}$, hence $\xi_{ii}=1$\textquotedblright%
\ by the simpler argument \textquotedblleft and so the definition of $\xi
_{ii}$ yields $\xi_{ii}=\epsilon\left(  \iota\right)  =1$\textquotedblright.

\item \textbf{page 423, proof of Proposition 1.33:} \textquotedblleft Using
Proposition 1.22\textquotedblright\ $\rightarrow$ \textquotedblleft Using
Corollary 1.27a (specifically, the $E_{i}s_{ij}=s_{ij}E_{j}$
part)\textquotedblright.

\item \textbf{page 423, Corollary 1.36:} Please say that your definition of
\textquotedblleft subalgebra\textquotedblright\ does not require that the
unity of the subalgebra equals the unity of the algebra! (This is far from standard.)

\item \textbf{page 423:} After Corollary 1.36, I would add another corollary
for later use:

\textbf{Corollary 1.36a.} Let $\lambda,\mu\vdash n$ be distinct. Then,
$N^{\lambda}N^{\mu}=0$.

\textit{Proof.} It suffices to show that $E_{i}^{\lambda}s_{ij}^{\lambda}%
E_{k}^{\mu}s_{k\ell}^{\mu}=0$ for any $i,j\in\left\{  1,2,\ldots,d_{\lambda
}\right\}  $ and $k,\ell\in\left\{  1,2,\ldots,d_{\mu}\right\}  $. So let us
consider such $i,j$ and $k,\ell$. Proposition 1.23 yields $D_{j}^{\lambda
}D_{k}^{\mu}=0$ (since $\lambda\neq\mu$). But the definitions of
$E_{j}^{\lambda}$ and $E_{k}^{\mu}$ yield $E_{j}^{\lambda}=\dfrac{f_{j}}%
{n!}D_{j}^{\lambda}$ and $E_{k}^{\mu}=\dfrac{f_{k}}{n!}D_{k}^{\mu}$. But
Corollary 1.27a yields $E_{i}s_{ij}^{\lambda}=s_{ij}^{\lambda}E_{j}^{\lambda}$
and therefore
\[
\underbrace{E_{i}^{\lambda}s_{ij}^{\lambda}}_{=s_{ij}^{\lambda}E_{j}^{\lambda
}}E_{k}^{\mu}s_{k\ell}^{\mu}=s_{ij}^{\lambda}\underbrace{E_{j}^{\lambda}%
}_{=\dfrac{f_{j}}{n!}D_{j}^{\lambda}}\underbrace{E_{k}^{\mu}}_{=\dfrac{f_{k}%
}{n!}D_{k}^{\mu}}s_{k\ell}^{\mu}=\dfrac{f_{j}}{n!}\cdot\dfrac{f_{k}}{n!}%
s_{ij}^{\lambda}\underbrace{D_{j}^{\lambda}D_{k}^{\mu}}_{=0}s_{k\ell}^{\mu
}=0.
\]
This proves Corollary 1.36a. $\blacksquare$

\item \textbf{page 423:} After Lemma 1.37, I would add another corollary for
later use:

\textbf{Corollary 1.37a.} For any partition $\lambda\vdash n$ and any two
$d_{\lambda}\times d_{\lambda}$-matrices $B$ and $C$, we have%
\[
\alpha^{\lambda}\left(  B\right)  \alpha^{\lambda}\left(  C\right)
=\alpha^{\lambda}\left(  B\mathcal{E}^{\lambda}C\right)  .
\]


\textit{Proof.} Let $\lambda\vdash n$, and let $B$ and $C$ be two $d_{\lambda
}\times d_{\lambda}$-matrices $B$ and $C$. Then, we can write $B$ and $C$ in
the forms $B=\left(  b_{ij}\right)  $ and $C=\left(  c_{ij}\right)  $. Hence,
$B=\left(  b_{ij}\right)  =\sum_{i,j}b_{ij}E_{ij}$ and $C=\left(
c_{ij}\right)  =\left(  c_{k\ell}\right)  =\sum_{k,\ell}c_{k\ell}E_{k\ell}$.
Hence,%
\begin{align*}
&  \alpha^{\lambda}\left(  \underbrace{B}_{=\sum_{i,j}b_{ij}E_{ij}}\right)
\alpha^{\lambda}\left(  \underbrace{C}_{=\sum_{k,\ell}c_{k\ell}E_{k\ell}%
}\right) \\
&  =\alpha^{\lambda}\left(  \sum_{i,j}b_{ij}E_{ij}\right)  \alpha^{\lambda
}\left(  \sum_{k,\ell}c_{k\ell}E_{k\ell}\right)  =\sum_{i,j}b_{ij}\sum
_{k,\ell}c_{k\ell}\underbrace{\alpha^{\lambda}\left(  E_{ij}\right)
\alpha^{\lambda}\left(  E_{k\ell}\right)  }_{\substack{=\alpha^{\lambda
}\left(  E_{ij}\mathcal{E}^{\lambda}E_{k\ell}\right)  \\\text{(by Lemma
1.37)}}}\\
&  =\sum_{i,j}b_{ij}\sum_{k,\ell}c_{k\ell}\alpha^{\lambda}\left(
E_{ij}\mathcal{E}^{\lambda}E_{k\ell}\right)  =\alpha^{\lambda}\left(
\underbrace{\left(  \sum_{i,j}b_{ij}E_{ij}\right)  }_{=B}\mathcal{E}^{\lambda
}\underbrace{\left(  \sum_{k,\ell}c_{k\ell}E_{k\ell}\right)  }_{=C}\right) \\
&  =\alpha^{\lambda}\left(  B\mathcal{E}^{\lambda}C\right)  .
\end{align*}
This proves Corollary 1.37a. $\blacksquare$

\item \textbf{page 423, proof of Proposition 1.38:} After \textquotedblleft%
$\sum_{\mu\vdash n}\alpha^{\mu}\left(  A^{\mu}\right)  =0$\textquotedblright,
add \textquotedblleft for some matrices $A^{\mu}=\left(  a_{ij}^{\mu}\right)
_{i,j=1,2,\ldots,d_{\mu}}$\textquotedblright.

\item \textbf{pages 423--424, proof of Proposition 1.38:} I think this whole
proof would become clearer if rewritten as follows:

\textquotedblleft Assume that $\sum_{\mu\vdash n}\sum_{i,j=1}^{d_{\mu}}%
a_{ij}^{\mu}E_{i}^{\mu}s_{ij}^{\mu}=0$ for some family of scalars $a_{ij}%
^{\mu}\in\mathbb{F}$. We shall show that $a_{ij}^{\mu}=0$ for all $\mu$ and
$i,j$.

Fix a partition $\lambda$. Let $A$ be the $d_{\lambda}\times d_{\lambda}%
$-matrix $\left(  a_{ij}^{\lambda}\right)  _{i,j=1,2,\ldots,d_{\lambda}}$. We
shall show that $A=0$.

Fix $u,v\in\left\{  1,2,\ldots,d_{\lambda}\right\}  $. We have $E_{u}%
^{\lambda}\underbrace{\sum_{\mu\vdash n}\sum_{i,j=1}^{d_{\mu}}a_{ij}^{\mu
}E_{i}^{\mu}s_{ij}^{\mu}}_{=0}=0$, so that%
\begin{align*}
0  &  =E_{u}^{\lambda}\sum_{\mu\vdash n}\sum_{i,j=1}^{d_{\mu}}a_{ij}^{\mu
}E_{i}^{\mu}s_{ij}^{\mu}=\sum_{\mu\vdash n}\sum_{i,j=1}^{d_{\mu}}a_{ij}^{\mu
}E_{u}^{\lambda}E_{i}^{\mu}s_{ij}^{\mu}=\sum_{i,j=1}^{d_{\lambda}}%
a_{ij}^{\lambda}E_{u}^{\lambda}E_{i}^{\lambda}s_{ij}^{\lambda}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Proposition 1.23 yields that }E_{u}^{\lambda}E_{i}^{\mu}=0\text{
whenever }\lambda\neq\mu\text{,}\\
\text{and this entails that all addends }a_{ij}^{\mu}E_{u}^{\lambda}E_{i}%
^{\mu}s_{ij}^{\mu}\text{ with }\lambda\neq\mu\text{ vanish}%
\end{array}
\right) \\
&  =\sum_{i,j=1}^{d_{\lambda}}a_{ij}\underbrace{E_{u}E_{i}}_{\substack{=\xi
_{ui}E_{u}s_{ui}\\\text{(by Lemma 1.29)}}}s_{ij}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{from now on, we are omitting the
superscripts }\lambda\right) \\
&  =\sum_{i,j=1}^{d_{\lambda}}a_{ij}\xi_{ui}E_{u}\underbrace{s_{ui}s_{ij}%
}_{=s_{uj}}=\sum_{i,j=1}^{d_{\lambda}}a_{ij}\xi_{ui}E_{u}s_{uj}.
\end{align*}
Multiplying both sides of this equality with $E_{v}$ on the right, we obtain%
\begin{align*}
0  &  =\left(  \sum_{i,j=1}^{d_{\lambda}}a_{ij}\xi_{ui}E_{u}s_{uj}\right)
E_{v}=\sum_{i,j=1}^{d_{\lambda}}a_{ij}\xi_{ui}E_{u}s_{uj}\underbrace{E_{v}%
}_{\substack{=E_{v}s_{vv}\\\text{(since }s_{vv}=\iota\text{)}}}\\
&  =\sum_{i,j=1}^{d_{\lambda}}a_{ij}\xi_{ui}\underbrace{\left(  E_{u}%
s_{uj}\right)  \left(  E_{v}s_{vv}\right)  }_{\substack{=\xi_{jv}E_{u}%
s_{uv}\\\text{(by Proposition 1.33)}}}=\sum_{i,j=1}^{d_{\lambda}}a_{ij}%
\xi_{ui}\xi_{jv}E_{u}s_{uv}=\left(  \sum_{i,j=1}^{d_{\lambda}}\xi_{ui}%
a_{ij}\xi_{jv}\right)  E_{u}s_{uv}.
\end{align*}
Since $s_{uv}\in S_{n}$ is invertible, we can cancel $s_{uv}$ from this
equality and obtain%
\[
0=\left(  \sum_{i,j=1}^{d_{\lambda}}\xi_{ui}a_{ij}\xi_{jv}\right)  E_{u}.
\]
Since $E_{u}\neq0$ (by Corollary 1.27b), we thus obtain
\[
\sum_{i,j=1}^{d_{\lambda}}\xi_{ui}a_{ij}\xi_{jv}=0
\]
(since $\sum_{i,j=1}^{d_{\lambda}}\xi_{ui}a_{ij}\xi_{jv}$ is a scalar).

But $A=\left(  a_{ij}^{\lambda}\right)  _{i,j=1,2,\ldots,d_{\lambda}}=\left(
a_{ij}\right)  _{i,j=1,2,\ldots,d_{\lambda}}$ (since we are omitting the
superscript $\lambda$) and $\mathcal{E}^{\lambda}=\left(  \xi_{ij}\right)
_{i,j=1,2,\ldots,d_{\lambda}}$ (by the definition of $\mathcal{E}^{\lambda}$).
Hence, $\sum_{i,j=1}^{d_{\lambda}}\xi_{ui}a_{ij}\xi_{jv}$ is the $\left(
u,v\right)  $-th entry of the matrix $\mathcal{E}^{\lambda}A\mathcal{E}%
^{\lambda}$. Thus, we have showed that the $\left(  u,v\right)  $-th entry of
the matrix $\mathcal{E}^{\lambda}A\mathcal{E}^{\lambda}$ is $0$ (since we have
showed that $\sum_{i,j=1}^{d_{\lambda}}\xi_{ui}a_{ij}\xi_{jv}=0$).

Forget that we fixed $u,v$. We thus have proved that the $\left(  u,v\right)
$-th entry of the matrix $\mathcal{E}^{\lambda}A\mathcal{E}^{\lambda}$ is $0$
for each $u,v\in\left\{  1,2,\ldots,d_{\lambda}\right\}  $. In other words,
all entries of the matrix $\mathcal{E}^{\lambda}A\mathcal{E}^{\lambda}$ are
$0$. In other words, $\mathcal{E}^{\lambda}A\mathcal{E}^{\lambda}=0$. Since
$\mathcal{E}^{\lambda}$ is invertible (by Lemma 1.32), we thus obtain $A=0$.
Thus, all entries of $A$ are $0$. In other words, $a_{ij}^{\lambda}=0$ for any
$i,j\in\left\{  1,2,\ldots,d_{\lambda}\right\}  $ (since the entries of $A$
are $a_{ij}^{\lambda}$). Since we have proved this for any $\lambda\vdash n$,
we thus conclude that all our scalars $a_{ij}^{\mu}$ are $0$. This proves
Proposition 1.38. $\blacksquare$\textquotedblright

\item \textbf{page 424, proof of Corollary 1.40:} \textquotedblleft by
Proposition 1.23\textquotedblright\ $\rightarrow$ \textquotedblleft by
Proposition 1.38\textquotedblright.

\item \textbf{page 424, Remark at the end of \S 1.6:} You give the reference
[35, \S 5.1.4, Theorem A]. Here are a few alternative references for proofs of
the equality $\sum_{\lambda}d_{\lambda}^{2}=n!$:

\begin{itemize}
\item Proposition 1.3.3 in Marc A. A. van Leeuwen, \textit{The
Robinson-Schensted and Sch\"{u}tzenberger algorithms, an elementary approach},
version 25 Nov 2011.\newline\url{http://www-math.univ-poitiers.fr/~maavl/}

\item Corollary 8.5 in Richard P. Stanley, \textit{Algebraic Combinatorics:
Walks, Trees, Tableaux, and More}, Undergraduate Texts in Mathematics,
Springer 2013.\newline\url{http://www-math.mit.edu/~rstan/algcomb/index.html}

(This book also has a second edition; the equality still is Corollary 8.5 in it.)

\item Theorem 2.6.5 part 3. in Bruce E. Sagan, \textit{The Symmetric Group:
Representations, Combinatorial Algorithms, and Symmetric Functions}, 2nd
edition, Springer 2001.
\end{itemize}

\item \textbf{page 424, \S 1.7:} \textquotedblleft We prove that the map
$\psi$ in (1) is\textquotedblright\ $\rightarrow$ \textquotedblleft We shall
now construct the map $\psi$ in (1), and prove that it is\textquotedblright.

\item \textbf{page 425, Proposition 1.43:} It is worth saying that the
expression \textquotedblleft$\delta_{\lambda\mu}\delta_{jk}U_{i\ell}^{\lambda
}$\textquotedblright\ is understood to be $0$ if $\lambda=\mu$ (even if
$U_{i\ell}^{\lambda}$ is undefined in this case).

\item \textbf{page 425, proof of Proposition 1.43:} After \textquotedblleft If
$\lambda=\mu$ then\textquotedblright, add \textquotedblleft the definitions of
$U_{ij}$ and $U_{k\ell}$ and Lemma 1.37a yield (with the notation
$\mathcal{E}$ being used for $\mathcal{E}^{\lambda}$)\textquotedblright.

\item \textbf{page 425, proof of Proposition 1.43:} Replace \textquotedblleft
orthogonality of Proposition 1.23\textquotedblright\ by \textquotedblleft
orthogonality of Corollary 1.36a\textquotedblright.

\item \textbf{page 425, proof of Theorem 1.45:} Here is a more detailed
version of this proof:

\textquotedblleft Proposition 1.43 shows that the map $\psi$ is an
$\mathbb{F}$-algebra homomorphism. It remains to prove that $\psi$ is
bijective. The Remark at the end of \S 1.6 shows that $\sum_{\lambda
}d_{\lambda}^{2}=n!$; in other words, $\dim M=\dim\left(  \mathbb{F}%
S_{n}\right)  $. Hence, $\psi$ is an $\mathbb{F}$-linear map between two
$\mathbb{F}$-vector spaces of the same (finite) dimension. Thus, if $\psi$ is
injective, then $\psi$ is bijective. Therefore, it will suffice to show that
$\psi$ is injective.

Recall that $M=\bigoplus\limits_{i=1}^{r}M_{d_{i}}\left(  \mathbb{F}\right)
$. Let $\gamma:M\rightarrow M$ be the $\mathbb{F}$-linear map that sends each
$\left(  A_{1},A_{2},\ldots,A_{r}\right)  \in M$ to $\left(  A_{1}\left(
\mathcal{E}^{\lambda_{1}}\right)  ^{-1},A_{2}\left(  \mathcal{E}^{\lambda_{2}%
}\right)  ^{-1},\ldots,A_{r}\left(  \mathcal{E}^{\lambda_{r}}\right)
^{-1}\right)  \in M$. This map $\gamma$ is well-defined (since the $r$
matrices $\mathcal{E}^{\lambda_{1}},\mathcal{E}^{\lambda_{2}},\ldots
,\mathcal{E}^{\lambda_{r}}$ are all invertible) and injective (since the $r$
matrices $\left(  \mathcal{E}^{\lambda_{1}}\right)  ^{-1},\left(
\mathcal{E}^{\lambda_{2}}\right)  ^{-1},\ldots,\left(  \mathcal{E}%
^{\lambda_{r}}\right)  ^{-1}$ are all invertible). Moreover, it is easy to see
that $\psi=\alpha\circ\gamma$. (Indeed, by linearity, it suffices to show that
$\psi\left(  E_{ij}^{\lambda}\right)  =\left(  \alpha\circ\gamma\right)
\left(  E_{ij}^{\lambda}\right)  $ for all $\lambda\vdash n$ and
$i,j\in\left\{  1,2,\ldots,d_{\lambda}\right\}  $. But this is easy to check,
since the definition of $\psi$ yields%
\begin{align*}
\psi\left(  E_{ij}^{\lambda}\right)   &  =U_{ij}^{\lambda}=\alpha^{\lambda
}\left(  E_{ij}^{\lambda}\left(  \mathcal{E}^{\lambda}\right)  ^{-1}\right)
=\alpha\underbrace{\left(  0,0,\ldots,0,E_{ij}^{\lambda}\left(  \mathcal{E}%
^{\lambda}\right)  ^{-1},0,0,\ldots,0\right)  }_{=\gamma\left(  0,0,\ldots
,0,E_{ij}^{\lambda},0,0,\ldots,0\right)  =\gamma\left(  E_{ij}^{\lambda
}\right)  }\\
&  =\alpha\left(  \gamma\left(  E_{ij}^{\lambda}\right)  \right)  =\left(
\alpha\circ\gamma\right)  \left(  E_{ij}^{\lambda}\right)  .
\end{align*}
Thus, $\psi=\alpha\circ\gamma$ is proven.)

Now, the maps $\alpha$ and $\gamma$ are both injective (indeed, the map
$\alpha$ is injective by Corollary 1.40). Hence, their composition
$\alpha\circ\gamma$ is injective. In other words, the map $\psi$ is injective
(since $\psi=\alpha\circ\gamma$). As we have seen above, this completes the
proof of Theorem 1.45. $\blacksquare$

\item \textbf{page 427:} After \textquotedblleft Our next goal is to compute
explicitly the algebra homomorphism $\phi$\textquotedblright, add
\textquotedblleft inverse to $\psi$\textquotedblright.

\item \textbf{page 427:} Replace \textquotedblleft Proposition 1.22 and Lemma
1.29\textquotedblright\ by \textquotedblleft Lemma 1.29 and Corollary 1.27a
(specifically, the $E_{i}s_{ij}=s_{ij}E_{j}$ part of it)\textquotedblright.

\item \textbf{page 428:} I'd replace \textquotedblleft The Wedderburn
decomposition of $\mathbb{F}S_{n}$ shows that\textquotedblright\ by
\textquotedblleft The surjectivity of $\psi$ in Theorem 1.45 shows
that\textquotedblright\ (this is more concrete).

\item \textbf{page 428, Definition 1.49:} This definition tacitly uses the
fact that the $r_{ij}^{\lambda}\left(  p\right)  $ are uniquely determined by
$p$, $\lambda$, $i$ and $j$. This follows from the fact that the family
$\left(  U_{ij}^{\lambda}\right)  $ (with $\lambda$ ranging over all
partitions of $n$ and with $i,j$ ranging over $\left\{  1,2,\ldots,d_{\lambda
}\right\}  $ each) is a basis of the $\mathbb{F}$-vector space $\mathbb{F}%
S_{n}$. (And this fact follows from Theorem 1.45, since the family $\left(
E_{ij}^{\lambda}\right)  $ forms a basis of $M$ and is sent to the family
$\left(  U_{ij}^{\lambda}\right)  $ by the map $\psi$.)

\item \textbf{page 428, Lemma 1.50:} It is worth saying that Lemma 1.50 is a
consequence of Proposition 1.43.

\item \textbf{page 429, proof of Proposition 1.51:} Replace both
\textquotedblleft$\alpha$\textquotedblright s in this proof by
\textquotedblleft$\alpha^{\lambda}$\textquotedblright. (It is dangerous to
omit the superscript on an $\alpha$, since $\alpha$ already has a different
meaning given to it in Definition 1.39.)

\item \textbf{page 429, proof of Proposition 1.51:} Remove the
\textquotedblleft write $\mathcal{E}=A_{\iota}^{\lambda}$
and\textquotedblright\ part of the first sentence of the proof. Instead, at
the beginning of the proof, I'd add \textquotedblleft The matrix $A_{\iota
}^{\lambda}$ is the matrix $\mathcal{E}^{\lambda}$ from Definition 1.31, and
thus is invertible (by Lemma 1.32). We shall omit the superscripts $\lambda$,
so we write $A_{p}$ for $A_{p}^{\lambda}$, and we write $\mathcal{E}$ for
$\mathcal{E}^{\lambda}=A_{\iota}^{\lambda}=A_{\iota}$\textquotedblright.

\item \textbf{page 429, proof of Proposition 1.51:} Replace \textquotedblleft
We have\textquotedblright\ by \textquotedblleft Thus, $E_{ii}\mathcal{E}^{-1}$
is the $d_{\lambda}\times d_{\lambda}$-matrix whose $i$-th row has entries
$\eta_{i1},\eta_{i2},\ldots,\eta_{id_{\lambda}}$ while all other rows are $0$.
Therefore, the definition of $\alpha^{\lambda}$ yields%
\[
\alpha^{\lambda}\left(  E_{ii}\mathcal{E}^{-1}\right)  =\sum_{k=1}%
^{d_{\lambda}}\eta_{ik}E_{i}s_{ik}.
\]
Similarly,%
\[
\alpha^{\lambda}\left(  E_{jj}\mathcal{E}^{-1}\right)  =\sum_{\ell
=1}^{d_{\lambda}}\eta_{j\ell}E_{j}s_{j\ell}.
\]
Hence,\textquotedblright.

\item \textbf{page 429, proof of Proposition 1.51:} The second-to-last
equality sign in the long (displayed) computation relies on the equality%
\[
\sum_{\ell=1}^{d_{\lambda}}\eta_{j\ell}E_{i}s_{i\ell}=U_{ij},
\]
which is not completely obvious. Here is how it can be proved: The matrix
$\left(  \mathcal{E}^{\lambda}\right)  ^{-1}=\mathcal{E}^{-1}$ has entries
$\eta_{ij}$. Thus, $E_{ij}^{\lambda}\left(  \mathcal{E}^{\lambda}\right)
^{-1}$ is the $d_{\lambda}\times d_{\lambda}$-matrix whose $i$-th row has
entries $\eta_{j1},\eta_{j2},\ldots,\eta_{jd_{\lambda}}$ while all other rows
are $0$. Therefore, the definition of $\alpha^{\lambda}$ yields%
\[
\alpha^{\lambda}\left(  E_{ij}^{\lambda}\left(  \mathcal{E}^{\lambda}\right)
^{-1}\right)  =\sum_{\ell=1}^{d_{\lambda}}\eta_{j\ell}E_{i}s_{i\ell}.
\]


Now, the definition of $U_{ij}$ yields%
\[
U_{ij}=\alpha^{\lambda}\left(  E_{ij}^{\lambda}\left(  \mathcal{E}^{\lambda
}\right)  ^{-1}\right)  =\sum_{\ell=1}^{d_{\lambda}}\eta_{j\ell}E_{i}s_{i\ell
}.
\]
Thus, $\sum_{\ell=1}^{d_{\lambda}}\eta_{j\ell}E_{i}s_{i\ell}=U_{ij}$ is proven.

\item \textbf{page 429, proof of Proposition 1.51:} The last equality sign in
the long (displayed) computation relies on the equality%
\[
\sum_{k=1}^{d_{\lambda}}\eta_{ik}\xi_{kj}^{p}=\left(  A_{\iota}^{-1}%
A_{p}\right)  _{ij},
\]
which is not completely obvious. Here is how it can be proved: We have
$A_{\iota}=\mathcal{E}$, so that $A_{\iota}^{-1}=\mathcal{E}^{-1}$. Thus, the
entries of the matrix $A_{\iota}^{-1}$ are the entries of the matrix
$\mathcal{E}^{-1}$, which are the scalars $\eta_{ij}$ (by the definition of
$\eta_{ij}$). On the other hand, the entries of the matrix $A_{p}%
=A_{p}^{\lambda}$ are $\xi_{ij}^{p}$ (by the definition of $A_{p}^{\lambda}$).
Hence, the $\left(  i,j\right)  $-th entry of the matrix $A_{\iota}^{-1}A_{p}$
is $\sum_{k=1}^{d_{\lambda}}\eta_{ik}\xi_{kj}^{p}$ (by the definition of the
product of two matrices). In other words, $\left(  A_{\iota}^{-1}A_{p}\right)
_{ij}=\sum_{k=1}^{d_{\lambda}}\eta_{ik}\xi_{kj}^{p}$. Thus, $\sum
_{k=1}^{d_{\lambda}}\eta_{ik}\xi_{kj}^{p}=\left(  A_{\iota}^{-1}A_{p}\right)
_{ij}$ is proven.

\item \textbf{page 429, proof of Proposition 1.51:} \textquotedblleft
Therefore $r_{ij}^{\lambda}\left(  p\right)  $\textquotedblright%
\ $\rightarrow$ \textquotedblleft Therefore, by Lemma 1.50 (and because
$U_{ij}\neq0$), we obtain $r_{ij}^{\lambda}\left(  p\right)  $%
\textquotedblright.

\item \textbf{page 429:} It is worth explaining why exactly Proposition 1.51
provides an explicit way of computing the homomorphism $\phi$ in (11). Indeed,
each $p\in S_{n}$ satisfies%
\begin{align*}
&  \psi\left(  R^{\lambda_{1}}\left(  p\right)  ,R^{\lambda_{2}}\left(
p\right)  ,\ldots,R^{\lambda_{r}}\left(  p\right)  \right) \\
&  =\sum_{\lambda\vdash n}\psi\left(  \underbrace{R^{\lambda}\left(  p\right)
}_{\substack{=\sum_{i=1}^{d_{\lambda}}\sum_{j=1}^{d_{\lambda}}r_{ij}^{\lambda
}\left(  p\right)  E_{ij}^{\lambda}\\\text{(by the definition of }R^{\lambda
}\left(  p\right)  \text{)}}}\right)  =\sum_{\lambda\vdash n}\underbrace{\psi
\left(  \sum_{i=1}^{d_{\lambda}}\sum_{j=1}^{d_{\lambda}}r_{ij}^{\lambda
}\left(  p\right)  E_{ij}^{\lambda}\right)  }_{\substack{=\sum_{i=1}%
^{d_{\lambda}}\sum_{j=1}^{d_{\lambda}}r_{ij}^{\lambda}\left(  p\right)
U_{ij}^{\lambda}\\\text{(by the definition of }\psi\text{)}}}\\
&  =\sum_{\lambda\vdash n}\sum_{i=1}^{d_{\lambda}}\sum_{j=1}^{d_{\lambda}%
}r_{ij}^{\lambda}\left(  p\right)  U_{ij}^{\lambda}%
=p\ \ \ \ \ \ \ \ \ \ \left(  \text{by (13)}\right)
\end{align*}
and therefore $\left(  R^{\lambda_{1}}\left(  p\right)  ,R^{\lambda_{2}%
}\left(  p\right)  ,\ldots,R^{\lambda_{r}}\left(  p\right)  \right)
=\psi^{-1}\left(  p\right)  =\phi\left(  p\right)  $ (since $\psi^{-1}=\phi$).
Hence, by computing the matrices $R^{\lambda}\left(  p\right)  $ for all
$\lambda\vdash n$, we can obtain an explicit formula for $\phi\left(
p\right)  $.
\end{itemize}


\end{document}