\documentclass[12pt,final,notitlepage,onecolumn,german]{article}%
\usepackage[all,cmtip]{xy}
\usepackage{lscape}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{color}
\usepackage{amsmath}%
\setcounter{MaxMatrixCols}{30}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{CSTFile=LaTeX article (bright).cst}
%TCIDATA{Created=Sat Mar 27 17:33:36 2004}
%TCIDATA{LastRevised=Monday, August 16, 2010 13:11:40}
%TCIDATA{}
%TCIDATA{}
%TCIDATA{}
%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\definecolor{violet}{RGB}{143,0,255}
\definecolor{forestgreen}{RGB}{34, 100, 34}
\voffset=-2.5cm
\hoffset=-2.5cm
\setlength\textheight{24cm}
\setlength\textwidth{15.5cm}
\begin{document}
\begin{center}
\textbf{Group Characters, Symmetric Functions, and the Hecke Algebras}
\textit{David M. Goldschmidt}
published by the American Mathematical Society 1993
\textbf{Errata list} by Darij Grinberg
\bigskip
\end{center}
\section*{Chapter 1}
\begin{itemize}
\item It would not harm to explain that whenever you say "module", you usually
mean "right module". People are often used to left modules instead.
\item \textbf{1.4 (Rieffel-Wedderburn):} The arrow in "natural map
$B\mapsto\operatorname*{End}\nolimits_{D}\left( I\right) $" should be an
$\rightarrow$ arrow rather than an $\mapsto$ arrow.
\item \textbf{Between 1.4 and 1.5:} You write: "We remark that if $I$ is a
minimal right ideal above then $D$ must be a division ring. For if
$0\neq\varphi\in D$, then $\ker\left( \varphi\right) $ and
$\operatorname{im}\left( \varphi\right) $ are both right ideals of $A$,
[...]". Actually they are right ideals of $B$, not of $A$.
\item \textbf{Between 1.5 and 1.6:} You write: "In particular, $\dim
_{D}\left( eB\right) =n$ so $eB$ is a minimal right ideal." Could you
explain the word "so" in this sentence? Because for me, the real reason why
$eB$ is a minimal right ideal is that $B\cong D^{n\times n}$ acts transitively
on the set $eB\diagdown0\cong D^{n}\diagdown0$ (since, just as in linear
algebra, any nonzero row vector can be mapped to any other nonzero row vector
of the same size by a suitably chosen square matrix).
\end{itemize}
\section*{Chapter 2}
\begin{itemize}
\item \textbf{Page 5 and some times further in the book:} Your notation
concerning matrix rings and general linear groups is not 100\% consistent: the
notations $M_{n}\left( \mathbb{C}\right) $ and $M\left( n,\mathbb{C}%
\right) $ are used for one and the same thing (the ring of $n\times n$
matrices over $\mathbb{C}$); the notations $\mathbf{GL}_{n}\left(
\mathbb{C}\right) $ and $\mathbf{GL}\left( n,\mathbb{C}\right) $ are used
for one and the same thing (the multiplicative group of units of $M_{n}\left(
\mathbb{C}\right) $).
\item \textbf{The long formula on page 5:} In the formula%
\[
\left( \sum_{g\in G}\alpha\left( g\right) g\right) \left( \sum_{h\in
G}\beta\left( h\right) h\right) =\sum_{g,h}\alpha\left( g\right)
\beta\left( h\right) gh=\sum_{x}\sum_{g}\alpha\left( x\right) \beta\left(
xg^{-1}\right) x,
\]
the term $\beta\left( xg^{-1}\right) $ should be $\beta\left(
x^{-1}g\right) $.
\item \textbf{Proof of (2.1):} In the sentence "[...] where $P$ is the matrix
of the permutation $g\rightarrow g^{-1}$ ", the $\rightarrow$ arrow should be
a $\mapsto$ arrow.
\item \textbf{Between (2.5) and (2.6):} You write: "Indeed, the same argument
shows more: the class sums are a basis for the integral group ring
$\mathbb{Z}G\subseteq\mathbb{C}G$." I believe you meant this the right way,
but it is just asking to be misunderstood: of course, the class sums are a
basis for the \textit{center of} the integral group ring, not for the group
ring itself.
\item \textbf{Proof of (2.8):} You write: "Let $\mathbf{X}$ be the $s\times s$
matrix whose $\left( i,j\right) $ entry is $\chi_{i}\left( x_{j}\right)
$." Here, $\chi_{i}\left( x_{j}\right) $ should be replaced by $\chi
_{j}\left( x_{i}\right) $ (otherwise the rest of the proof doesn't work -
unless I have miscalculated $\mathbf{X}^{\ast}D\mathbf{X}$.)
\item \textbf{Page 10, between (2.8) and (2.9):} In the formula%
\[
\phi=\sum_{\chi\in\operatorname*{Irr}\left( g\right) }\left( \chi
,\phi\right) \chi,
\]
the $\operatorname*{Irr}\left( g\right) $ should be an $\operatorname*{Irr}%
\left( G\right) $.
\end{itemize}
\section*{Chapter 3}
\begin{itemize}
\item \textbf{(3.2):} In the statement of (3.2), you write: "The functions
$\omega_{i}:\mathbf{Z}\left( \mathbb{C}G\right) \rightarrow\mathbb{C}$ are
algebra homomorphisms whose values are algebraic integers." To be precise,
"values" should be "values at elements of $\mathbf{Z}\left( \mathbb{Z}%
G\right) $" here.
\end{itemize}
\section*{Chapter 4}
\begin{itemize}
\item \textbf{Proof of 4.4:} You write: "In particular, the set of products
$\left\{ x_{i}h_{ij}\ \mid\ 1\leq i\leq t,\ 1\leq j\leq T_{i}\right\} $ is a
set of right coset representatives for $K$ in $G$." The $T_{i}$ in $\left\{
x_{i}h_{ij}\ \mid\ 1\leq i\leq t,\ 1\leq j\leq T_{i}\right\} $ should be a
$t_{i}$ (with lowercase $t$).
\item \textbf{4.5 (Frobenius):} In the statement of 4.5, it would be better to
make the conditions more precise: $H\subseteq G$ is supposed to be a subgroup
of $G$ (not just some arbitrary subset), and "for $g\in G\diagdown H$" should
be "for all $g\in G\diagdown H$" (and not just for \textit{some} $g\in
G\diagdown H$).
\item \textbf{Between (4.8) and (4.9):} You write: "This observation,
originally due to Burnside, is useful in certain enumeration problems."
Indeed, this observation is known to the whole world as Burnside's lemma, so
the mention of Burnside is appropriate - but it should also be mentioned that
Burnside was \textit{not} the original author of this lemma.\footnote{cf. the
historical remarks on
\texttt{http://en.wikipedia.org/wiki/Burnside\%27s\_lemma}}
\end{itemize}
\section*{Chapter 5}
\begin{itemize}
\item \textbf{Between (5.2) and (5.3):} You write: "We first argue that
$A=\mathbf{C}_{G}\left( A\right) $, for if not then $\mathbf{C}_{G}\left(
A\right) \diagup A$ is a proper normal subgroup of the $p$-group $G\diagup A$
and therefore meets the center of $G\diagup A$ nontrivially." Here, the word
"proper" should be "nontrivial", in my opinion.
\item \textbf{Between (5.3) and (5.4):} You write: "it suffices by induction
to show that every nonlinear character $\chi$ of $G$ is induced from a proper
subgroup of $G$." Here, "character" should be replaced by "irreducible character".
\item \textbf{Between (5.4) and (5.5):} You write: "and let $\mathcal{B}%
\left( G;\mathcal{H}\right) $ be the set of permutation characters $\left\{
1_{H}^{G}\ \mid\ H\in\mathcal{H}\right\} $". This is wrong - you don't want
the \textit{set} of these characters, but you want the \textit{abelian group
generated by this set}.
\item \textbf{(5.5):} Maybe it wouldn't hurt to mention that "ring" means "not
necessarily unital ring" here.
\item \textbf{Proof of (5.6):} On the last line of page 20, you write
$I_{g}=R$. The $R$ should be a $\mathbb{Z}$ here.
\item \textbf{Page 21, before (5.7):} You write: "It is clear that any
subgroup of a quasi-elementary group is itself quasi-elementary". But why is
this clear? The shortest proof I can think of is nontrivial\footnote{The proof
mainly consists of showing the following lemma:
\par
\textbf{Lemma.} Let $G$ be a finite group. Then, $G$ is quasi-elementary if
and only if the subset $\left\{ g\in G\ \mid\ \operatorname*{ord}g\text{ is
prime to }p\right\} $ of $G$ is a cyclic subgroup for some prime $p$. Here,
$\operatorname*{ord}g$ denotes the order of the element $g$ in $G$.
\par
\textit{Proof of the Lemma.} $\Longrightarrow$: Assume that $G$ is
quasi-elementary. Then, there is some prime $p$ such that $G$ is a semidirect
product $PC$ of some $p$-subgroup $P$ of $G$ with some cyclic normal subgroup
$C$ of $G$ of order prime to $p$. Thus, $\left\vert G\right\vert =\left\vert
PC\right\vert =\left\vert P\right\vert \cdot\left\vert C\right\vert $, so that
$\left\vert G\diagup C\right\vert =\left\vert G\right\vert \diagup\left\vert
C\right\vert =\left\vert P\right\vert \cdot\left\vert C\right\vert
\diagup\left\vert C\right\vert =\left\vert P\right\vert $ is a power of $p$,
so that $G\diagup C$ is a $p$-group. Now, let $g\in G$ be some element such
that $\operatorname*{ord}g$ is prime to $p$. Then, the order of the element
$\overline{g}$ of the quotient group $G\diagup C$ is prime to $p$ as well
(because the order of $\overline{g}$ in $G\diagup C$ divides the order of $g$
in $G$). But the order of the element $\overline{g}$ of the quotient group
$G\diagup C$ must be a power of $p$ (since $G\diagup C$ is a $p$-group).
Hence, the order of the element $\overline{g}$ of the quotient group $G\diagup
C$ is $1$ (because it is both prime to $p$ and a power of $p$), and thus
$\overline{g}=1$, so that $g\in C$. We have thus shown that every element
$g\in G$ such that $\operatorname*{ord}g$ is prime to $p$ must lie in $C$.
Therefore, $\left\{ g\in G\ \mid\ \operatorname*{ord}g\text{ is prime to
}p\right\} \subseteq C$. Combining this with $C\subseteq\left\{ g\in
G\ \mid\ \operatorname*{ord}g\text{ is prime to }p\right\} $ (which is clear
because $\operatorname*{ord}g$ is prime to $p$ for every element $g\in C$,
since the order of $C$ is prime to $p$), we obtain $\left\{ g\in
G\ \mid\ \operatorname*{ord}g\text{ is prime to }p\right\} =C$, so that
$\left\{ g\in G\ \mid\ \operatorname*{ord}g\text{ is prime to }p\right\} $
is a cyclic subgroup. This proves the $\Longrightarrow$ direction of the
Lemma.
\par
$\Longleftarrow$: Assume that the subset $\left\{ g\in G\ \mid
\ \operatorname*{ord}g\text{ is prime to }p\right\} $ of $G$ is a cyclic
subgroup for some prime $p$. Denote this subset $\left\{ g\in G\ \mid
\ \operatorname*{ord}g\text{ is prime to }p\right\} $ by $C$. Clearly, $C$ is
a normal subgroup of $G$ (it is in fact a characteristic subgroup). Now we are
going to show that the quotient group $G\diagup C$ is a $p$-group. In fact,
assume that it is not. Then, $\left\vert G\diagup C\right\vert $ is not a
power of $p$, so there must be a prime $q\neq p$ such that $q\mid\left\vert
G\diagup C\right\vert $. Thus, by Cauchy's theorem, there exists an element of
$G\diagup C$ which has order $q$. Let $\overline{g}$ be this element (where
$g\in G$). Then, $\overline{g}^{q}=1$, so that $g^{q}\in C$. By the definition
of $C$, this yields that $\operatorname*{ord}\left( g^{q}\right) $ is prime
to $p$, and thus $\operatorname*{ord}g$ is prime to $p$ as well (since
$\operatorname*{ord}g\mid q\cdot\operatorname*{ord}\left( g^{q}\right) $ and
since $q$ is prime to $p$), so that $g\in C$ and thus $\overline{g}=1$. This
contradicts to our fact that the order of $\overline{g}$ is $q$. This
contradiction shows that our assumption was wrong, and thus $G\diagup C$ is a
$p$-group. Thus, $\left\vert G\diagup C\right\vert $ is some power of $p$
dividing $\left\vert G\right\vert $. Now let $P$ be a Sylow $p$-subgroup of
$G$. Then, $\left\vert P\right\vert \geq\left\vert G\diagup C\right\vert $
(since $\left\vert G\diagup C\right\vert $ is some power of $p$ dividing
$\left\vert G\right\vert $, while $\left\vert P\right\vert $ is the greatest
power of $p$ dividing $\left\vert G\right\vert $). Besides, $P\cap C=1$ (since
$P$ is a $p$-group, while all elements of $C$ have order prime to $p$) and
obviously $\left\vert P\right\vert \geq\left\vert G\diagup C\right\vert $
yields $\left\vert P\right\vert \cdot\left\vert C\right\vert \geq\left\vert
G\diagup C\right\vert \cdot\left\vert C\right\vert =\left\vert G\right\vert $,
so that $\left\vert PC\right\vert =\dfrac{\left\vert P\right\vert
\cdot\left\vert C\right\vert }{\left\vert P\cap C\right\vert }\geq
\dfrac{\left\vert G\right\vert }{1}=\left\vert G\right\vert $ and thus $PC=G$.
Hence, $G$ is a semidirect product of $P$ and $C$. This shows that $G$ is
quasi-elementary, and therefore the $\Longleftarrow$ direction of the Lemma is
proven as well.}.
\item \textbf{Proof of (5.7):} You write: "Let $P$ be a Sylow $p$-subgroup of
$N=\mathbf{N}_{G}\left( C\right) $ containing $g$". But a $p$-group cannot
contain $g$ in general (the order of $g$ is not always a power of $p$). I
guess you want $P$ to be a Sylow $p$-subgroup of $N=\mathbf{N}_{G}\left(
C\right) $ containing $g^{n}$ (this is possible and this leads to $g\in H$ afterwards).
\item \textbf{Proof of (5.7):} You write: "Namely, choose coset
representatives $\left\{ x_{1},...,x_{t}\right\} $ for $H$ in $G$." I would
say "\textit{right} coset representatives" here to be more precise.
\item \textbf{Proof of (5.8):} On the fourth line from the bottom of page 20,
you write "[...] where $\lambda$ is a linear character of some $H\in
\mathcal{H}$." I think the $\mathcal{H}$ here is supposed to mean
$\mathcal{E}$.
\item \textbf{Between (5.10) and 5.11:} In the formula%
\[
1_{G}=1_{N}^{G}-\sum_{i>0}a_{i}\lambda_{i}^{G},
\]
the minus sign should be a plus sign.
\item \textbf{5.11:} In the statement of Brauer's theorem 5.11, maybe you
should replace "power of $p$" by "nontrivial power of $p$" for better clarity.
\item \textbf{Page 23:} You write: "Since cyclic groups are direct products of
cyclic groups of prime power order, $H$ is of the form $P\times Q$ where $P$
is a $p$-group and $\left\vert Q\right\vert \not \equiv 0\left(
\operatorname{mod}p\right) $." But I think this follows directly from $H$
being elementary - where are you using the fact that cyclic groups are direct
products of cyclic groups of prime power order?
\end{itemize}
\section*{Chapter 6}
\begin{itemize}
\item \textbf{Page 27:} You write: "In a slight departure from usual
terminology, we will mean by a partition of $\Omega$ an ordered collection of
pairwise disjoint nonempty subsets $\mathcal{P}=\left\{ \mathcal{P}%
_{1},\mathcal{P}_{2},...,\mathcal{P}_{r}\right\} $ such that [...]".
Actually, I would propose to write $\left( \mathcal{P}_{1},\mathcal{P}%
_{2},...,\mathcal{P}_{r}\right) $ instead of $\left\{ \mathcal{P}%
_{1},\mathcal{P}_{2},...,\mathcal{P}_{r}\right\} $ here, because $\left\{
\mathcal{P}_{1},\mathcal{P}_{2},...,\mathcal{P}_{r}\right\} $ looks too much
like "the set with elements $\mathcal{P}_{1},\mathcal{P}_{2},...,\mathcal{P}%
_{r}$" which denies any ordering on the partitions, while you want the
partitions to be ordered.
\item \textbf{Proof of (6.2):} On the second line from the bottom of page 28,
a full stop is missing after $\mathcal{C}\left( T\right) =\mathcal{Q}$ (but
maybe it is just missing on the scanned version of the book).
\item \textbf{Page 29:} On the third line before Gale-Ryser's theorem (6.3),
you write: "It is clear that if $\lambda\leq\mu$, then $\lambda\ll\mu$." I
think it's the other way round: if $\lambda\ll\mu$, then $\lambda\leq\mu$.
\item \textbf{Proof of (6.4):} You write: "We say that $\sigma$ is a
$k$\textit{-cycle} if $k_{1}=k$ and $k_{2}=1$." To be completely rigorous
here, I would replace this by "We say that $\sigma$ is a $k$\textit{-cycle} if
$k_{1}=k$ and $k_{2}=1$ or $r=1$".
\item \textbf{Proof of (6.4):} You write: "The usual notation for a $k$-cycle
$\sigma$ is $\left( m_{0}m_{1}...m_{k-1}\right) $ where $m_{i}^{\sigma
}=m_{i+1}$ ($0\leq i\mathcal{C}_{1}$." Here,
$\mathcal{P}>\mathcal{C}_{1}$ should be $\mathcal{P}>\mathcal{R}_{1}$ instead.
\item \textbf{Last line of page 39:} The $T_{\mathcal{Q}}$ here should be
$\mathcal{Q}$ (or, more precisely, "standard Specht vectors which are smaller
than $T_{\mathcal{Q}}$" should be "standard Specht vectors of standard
tableaux whose row partitions are smaller than $\mathcal{Q}$").
\item \textbf{Proof of (8.9):} Two lines above the statement of (8.10), you
speak of "the map $T\rightarrow T^{\prime}$ ". The $\rightarrow$ arrow here
should be a $\mapsto$ arrow.
\item \textbf{Proof of (8.9):} One line above the statement of (8.10), you
write: "the disjoint union of the standard tableaus of type $\lambda_{j}$
($1\leq j\leq s$)". The $\lambda_{j}$ here should be $\lambda^{\left(
j\right) }$ instead.
\end{itemize}
\section*{Chapter 9}
\begin{itemize}
\item \textbf{(9.1):} When you write "the $e_{r}$ are algebraically
independent", you mean only the $e_{r}$ for $r>0$ (although you have defined
$e_{0}$ as well). I think this is worth a mention.
\item \textbf{(9.3):} Similarly, when you write "the $h_{r}$ are algebraically
independent", you mean only the $h_{r}$ for $r>0$ (although you have defined
$h_{0}$ as well).
\end{itemize}
\section*{Chapter 10}
\begin{itemize}
\item \textbf{Between (10.3) and (10.4):} "Note that if $\alpha_{j}<0$ for
some $j$, then the $j$th column of $\mathbf{H}_{\alpha}$ is zero, so we define
$a_{\alpha}=0$ if any $\alpha_{j}=0$." Of course, you mean $\alpha_{j}<0$ when
you write $\alpha_{j}=0$ here.
\item \textbf{(10.6):} You write: "In particular, the Schur functions of
degree $n$ are a $Z$-basis for $\Lambda^{n}$." The $Z$ here should be a
boldface $\mathbb{Z}$.
\item \textbf{(10.8):} In this formula, $\sum\limits_{\lambda}h_{\lambda
}\left( z\right) $ should be $\sum\limits_{k}h_{k}\left( z\right)
$\textbf{.}
\item \textbf{(10.11):} It seems to me that the condition $\mu<\lambda$ should
be $\mu>\lambda$ here.
\end{itemize}
\section*{Chapter 11}
\begin{itemize}
\item \textbf{Proof of (11.2):} You refer to the "Frobenius reciprocity
(3.1)". But the Frobenius reciprocity was (4.1), not (3.1).
\item \textbf{Proof of (11.2):} In the formula%
\[
\mathbf{ch}\left( fg\right) =\left( f\sharp g^{S^{n+m}},\rho\right)
=\left( f\sharp g,\rho_{S^{n}\times S^{m}}\right) =\dfrac{1}{n!m!}\sum
_{x,y}f\left( x\right) g\left( y\right) \rho\left( x,y\right) ,
\]
the term $f\sharp g^{S^{n+m}}$ should be $\left( f\sharp g\right) ^{S^{n+m}%
}$.
\item \textbf{Between (11.3) and (11.4):} When you say "and moreover $\left(
\left[ \lambda\right] ,\psi_{\mu}\right) =0$ for $\mu<\lambda$", it seems
to me that you mean $\mu>\lambda$ instead of $\mu<\lambda$.
\item \textbf{Between (11.3) and (11.4):} You write: "It now follows easily
from (7.2) that $\left[ \lambda\right] =\chi_{\lambda}$ [...]". But this
doesn't seem that easy to me. The simplest argument I can come up with is the
following: Since $\left[ \lambda\right] $ is an irreducible character of
$S^{n}$, there exists some partition $\rho\left( \lambda\right) $ of $n$
such that $\left[ \lambda\right] =\chi_{\rho\left( \lambda\right) }$. Thus
we have defined a map $\rho$ from the set of all partitions of $n$ to itself.
This map $\rho$ is injective (since for any two partitions $\lambda$ and $\mu$
such that $\rho\left( \lambda\right) =\rho\left( \mu\right) $, we have
$\left[ \lambda\right] =\chi_{\rho\left( \lambda\right) }=\chi
_{\rho\left( \mu\right) }=\left[ \mu\right] $ and thus $1=\left( \left[
\lambda\right] ,\left[ \lambda\right] \right) =\left( \left[
\lambda\right] ,\left[ \mu\right] \right) =\left\langle s_{\lambda}%
,s_{\mu}\right\rangle =\delta_{\lambda,\mu}$, so that $\lambda=\mu$), and thus
a permutation of the set of partitions of $n$ (since this set is finite).
Since $\left( \psi_{\lambda},\chi_{\rho\left( \lambda\right) }\right)
=\left( \chi_{\rho\left( \lambda\right) },\psi_{\lambda}\right) =\left(
\left[ \lambda\right] ,\psi_{\lambda}\right) =1\neq0$, we have $\rho\left(
\lambda\right) \gg\lambda$ for every $\lambda$ by (7.2). Since $\gg$ is a
partial order and $\rho$ is a permutation, this yields $\rho\left(
\lambda\right) =\lambda$ for every $\lambda$, and thus $\left[
\lambda\right] =\chi_{\rho\left( \lambda\right) }=\chi_{\lambda}$, qed. Is
there some simpler argument that I fail to see?
\item \textbf{Between (11.4) and (11.5):} You write: "Moreover, since
$\lambda_{1}+r-1\leq n$ with equality iff $\lambda_{2}=\lambda_{3}%
=\cdots=\lambda_{r}=1$, there can be at most one term equal to $\left[
n\right] $ in the expansion of any determinant of the form (11.3), and it
occurs in the expansion of exactly one such determinant [...]". I would add
"(for fixed $r$)" here (because if we do not fix $r$, then it can occur in the
expansion of \textit{more than one} such determinant).
\item \textbf{Between (11.4) and (11.5):} In the determinant%
\[
\det\left[
\begin{array}
[c]{ccccc}%
\left[ n-r+1\right] & 1 & 0 & \cdots & 0\\
\left[ n-r+2\right] & \left[ 1\right] & 1 & \cdots & 0\\
& & \left[ 1\right] & \ddots & \vdots\\
\vdots & \vdots & & \ddots & 1\\
\left[ n\right] & \left[ r-2\right] & & \cdots & \left[ 1\right]
\end{array}
\right] ,
\]
the $\left[ r-2\right] $ should be $\left[ r-1\right] $.
\item \textbf{Page 54:} You write: "In general, we need to evaluate a
determinant of the form $\det\left( f_{i}\left( \mu_{j}\right) \right) $
where $f_{i}$ is a monic polynomial of degree $i$." Here I would rather say
"of degree $r-i$", because otherwise you have to label the rows from $0$ to
$n-1$ rather than from $1$ to $n$ which is a bit unusual.
\end{itemize}
\section*{Chapter 12}
\begin{itemize}
\item \textbf{Last line of page 55:} Here you write $h_{ij}\left(
\lambda^{\prime}\right) =h_{i,j}\left( \lambda\right) $. The notations
$h_{ij}$ and $h_{i,j}$ denote one and the same thing; it would be best to
decide for one of them throughout the text (I personally favor $h_{i,j}$
because it is less ambiguous).
\item \textbf{Last line of page 56:} Here, "for any integer $\left\langle
a\right\rangle $" should be "for any integer $a$".
\item \textbf{Between (12.4) and (12.5):} On page 57, you write: "[...] we may
as well assume that there is some index $i\geq k$ such that $\mu_{i}>\mu
_{k}-m>\mu_{i+1}$, [...]". For the sake of completeness, it should be added
here that $\mu_{r+1}$ is supposed to mean $-1$.
\item \textbf{Page 57, one line above the picture of the Young diagram:} You
write: "and $\lambda^{\left( k\right) }=0$ for all other $k$". The equation
$\lambda^{\left( k\right) }=0$ is supposed to mean "$\lambda^{\left(
k\right) }$ is not a partition". The same mistake is repeated three lines
below the picture of the Young diagram.
\item \textbf{After 12.6:} You notice correctly that the Murnaghan-Nakayama
formula generalizes (8.9). I would add that it also generalizes (11.5).
\end{itemize}
\section*{Chapter 13}
\begin{itemize}
\item \textbf{First sentence of Chapter 13:} "In this section we define the
Hecke algebra (of type $A_{n-1}$) and prove that it is isomorphic to the group
algebra $\mathbb{Q}\left[ t\right] S^{n}$." But I don't think it is
isomorphic to $\mathbb{Q}\left[ t\right] S^{n}$. Maybe it becomes isomorphic
when tensored with an appropriate field.
\item \textbf{(13.1):} There are two mistakes here: First, $1\leq i\leq n$
should be $1\leq ij
\end{array}
\right. $.\newline In order to obtain a proof of (13.2'), it is enough to
read the proof of (13.2) with the following changes:\newline- replace every
$g_{i}$ by $\gamma_{i}$;\newline- replace $\left( t-1\right) wg_{n-1}+tw$ by
$w$;\newline- replace $H_{n}$ by $\Gamma$;\newline- read "word of the form"
instead of "$\mathbb{Q}\left[ t\right] $-linear combination of words of the
form";\newline- read "consists of" instead of "is spanned by".\newline Now,
(13.2') yields $\left\vert \Gamma\right\vert \leq n\left\vert \left\langle
\gamma_{1},\gamma_{2},...,\gamma_{n-2}\right\rangle \right\vert $, and thus by
induction $\left\vert \Gamma\right\vert \leq n\left( n-1\right)
\cdots1=n!=\left\vert S_{n}\right\vert $, so that the group homomorphism
$P:\Gamma\rightarrow S_{n}$ must be bijective (since it is surjective), and
thus $\Gamma\cong S_{n}$. This proves (13.1).
\item \textbf{The definition of a standard word (directly above (13.3)):} You
write: "Inductively, we define $w\in H_{n}$ to be a \textit{standard word} if
it is of the form $w_{1}g_{1,i}$ for some $i\geq0$, where $w_{1}$ is a
standard word in $\left\langle g_{1},...,g_{n-2}\right\rangle $." This is
slightly ambiguous - namely, if we would blindly follow this definition, we
would believe that a standard word in $\left\langle g_{1},...,g_{n-2}%
\right\rangle $ means a word of the form $w_{2}g_{1,i}$ for some $i\geq0$,
where $w_{2}$ is a standard word in $\left\langle g_{1},...,g_{n-3}%
\right\rangle $. But this makes no sense ($g_{1,i}$ is not in $\left\langle
g_{1},...,g_{n-2}\right\rangle $ at all). Instead, a standard word in
$\left\langle g_{1},...,g_{n-2}\right\rangle $ means a word of the form
$w_{2}g_{2,j}$ for some $j\geq1$. It wouldn't hurt to warn the reader about
this pitfall.
\item \textbf{Proof of (13.3):} In order to prove that the standard words
$w_{\sigma}$ are linearly independent, you write: "Moreover, if there were a
relation%
\[
\sum_{\sigma}p_{\sigma}\left( t\right) w_{\sigma}=0
\]
with $p_{\sigma}\left( t\right) \in\mathbb{Q}\left[ t\right] $ and
$\gcd\left\{ p_{\sigma}\left( t\right) \right\} =1$, [...]". But why can
you assume that $\gcd\left\{ p_{\sigma}\left( t\right) \right\} =1$ here?
If the $p_{\sigma}\left( t\right) $ have a common factor and you want to
cancel it from them, you need the Hecke algebra to be torsionfree; is this
trivial or have you silently proven this somewhere?
\item \textbf{First line of page 64:} You write: "is really not necessary in
the sequel". But don't you use (13.6) on page 69?
\item \textbf{(13.7):} In the statement of (13.7), you write: "If $\sigma\in
S^{n}$ fixes $\left\{ 1,...,k\right\} $ and $\rho\in S^{n}$ fixes $\left\{
k+1,...,n\right\} $, [...]". However, as the statement (iii) shows, you want
it exactly the other way round: you want $\sigma$ to fix $\left\{
k+1,...,n\right\} $ (so that $\sigma\in S^{k}$) and $\rho$ to fix $\left\{
1,...,k\right\} $ (so that $\rho\in S^{n-k}$).
\end{itemize}
\section*{Chapter 14}
\begin{itemize}
\item \textbf{Proof of (14.1):} You write: "namely we define%
\[
\tau_{n}\left( wg_{1,i}\right) =s\tau_{n-1}\left( wg_{2,i}\right)
\]
where $w$ is a standard word in $H_{n-1}$". Here you should add "and $i>0$",
because for $i=0$ this is wrong (and there is no need to define $\tau
_{n}\left( wg_{1,i}\right) $ for $i=0$, because $wg_{1,i}\in H_{n-1}$ for
$i=0$ and $\tau_{n}$ is supposed to extend $\tau_{n-1}$).
\item \textbf{Between (14.3) and (14.4):} You write: "So we may assume that
$w=w_{1}g_{1,j}$ in (14.3) for some standard word $w_{1}$ in $H_{n-1}$ and
some $j\geq2$." Here, $j\geq2$ should be $j\geq1$.
\item \textbf{The end of page 68:} When you write "Now if we define $\theta$
by the equation $\tau\left( \theta g_{i}\right) =\tau\left( \theta
^{-1}g_{i}^{-1}\right) $, [...]", it would be nice to add that the $\tau$
that you are using here is actually a base extended version of the $\tau$ that
you have defined before (namely, it is extended to a map $\tau:\mathbb{Q}%
\left( s,t,\theta\right) \rightarrow\mathbb{Q}\left( s,t,\theta\right) $).
\item \textbf{First line of page 69:} I think $\pi_{n}:B_{n}\rightarrow H_{n}$
should be $\pi_{n}:B_{n}\rightarrow H_{n}\otimes\mathbb{Q}\left(
s,t,\theta\right) $ here.
\item \textbf{Example (about the trefoil) on page 69:} In this example, you
obtain the formula%
\[
\widehat{\tau}\left( g_{1}^{3}\right) =\left( \theta s\right) \theta
^{3}\tau\left( \left( t-1\right) g_{1}^{2}+tg_{1}\right) =\theta^{4}%
s^{3}\left[ \left( t-1\right) ^{2}s+\left( t-1\right) t+ts\right] .
\]
Where does the $s^{3}$ come from?
\item \textbf{Proof of (14.9):} The last equation on page 70,%
\[
e_{n}g_{n-1}=\left( e_{n-1}+g_{n-1}\rho_{n-1}\right) g_{n-1}=te_{n},
\]
should be%
\[
e_{n}g_{n-1}=\left( e_{n-1}+e_{n-1}g_{n-1}\rho_{n-1}\right) g_{n-1}=te_{n}.
\]
\item \textbf{Proof of (14.9):} In the equation%
\[
e_{n}g_{n-i}=e_{n-1}\rho_{n}g_{n-1}=te_{n-1}\rho_{n}=te_{n}%
\]
(this is the last equation in the proof), the $g_{n-1}$ should be $g_{n-i}$.
\item \textbf{Page 71:} Here you say: "From (14.9) we see that $e_{n}H_{n}$ is
a one-dimensional right ideal [...]". But speaking about dimension makes sense
only over a field.
\end{itemize}
\end{document}