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\ihead{Errata to ``Mind your $P$ and $Q$-symbols''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Mind your $P$ and $Q$-symbols}

\textit{Geordie Williamson}

\url{http://web.archive.org/web/20250801094939/https://people.mpim-bonn.mpg.de/geordie/Hecke.pdf}

version of 2004-09-01

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

The following are corrections and comments on the honours thesis
\textquotedblleft Mind your $P$ and $Q$-symbols\textquotedblright\ by Geordie Williamson.

Note that I have myself only read Chapters 1, 2 (sans the proof of Lemma
2.8.1) and 3 (sans the proof of Proposition 3.4.1) carefully. The rest of the
text has only been skimmed, and most errors reported in these later chapters
have been found by GPT-5.5 (but verified by myself, except for a few where I
point it out). As for the appendices, I have not given them much scrutiny at all.

\setcounter{section}{7}

\section{Errata}

\begin{enumerate}
\item \textbf{page v:} It is not quite correct that \textquotedblleft there is
no one source that explains why the Hecke algebra of the symmetric group is a
cellular algebra\textquotedblright. In fact this is proved in Mathas's [24,
Theorem 3.20] through the Murphy cellular basis. What you seemingly intended
to say here is that no source explains why the Kazhdan--Lusztig basis is cellular.

\item \textbf{page vi:} The reference to \textquotedblleft Lusztig
[19]\textquotedblright\ is wrong; it should probably cite [21].

\item \textbf{page 1, just before Proposition 1.1.2:} \textquotedblleft the
number of simple transpositions\textquotedblright\ $\rightarrow$
\textquotedblleft the number $m$ of simple transpositions\textquotedblright.
(This is to clarify that the transpositions are counted with multiplicity --
e.g., the expression $s_{1}s_{2}s_{1}$ contains three, not two, simple transpositions.)

\item \textbf{page 2, proof of Theorem 1.1.5:} It should be said that you WLOG
assume that $i<j$ here.

\item \textbf{page 2, proof of Theorem 1.1.5:} After \textquotedblleft We
first verify that $\operatorname{Im}\left(  \varphi\right)  \subset N\left(
wt\right)  $.\textquotedblright, add \textquotedblleft Let $\left(
p,q\right)  \in N\left(  w\right)  $.\textquotedblright.

\item \textbf{page 2, proof of Theorem 1.1.5:} \textquotedblleft either $p=i$
and $q=j$\textquotedblright\ should be \textquotedblleft either $p=i$ or
$q=j$\textquotedblright.

\item \textbf{page 3, proof of Theorem 1.1.5:} The first paragraph on this
page (where you prove injectivity of $\varphi$) is confusing and unnecessary
complicated. Instead you can argue as follows: The value $\varphi\left(
p,q\right)  $ of the map $\varphi$ always belongs to $N\left(  t\right)  $ in
the case when $t\left(  p\right)  >t\left(  q\right)  $ (since $\left(
p,q\right)  \in N\left(  t\right)  $ in this case), but never belongs to
$N\left(  t\right)  $ in the case when $t\left(  p\right)  <t\left(  q\right)
$ (since $t\left(  t\left(  p\right)  \right)  =p<q=t\left(  t\left(
q\right)  \right)  $ entails that $\left(  t\left(  p\right)  ,t\left(
q\right)  \right)  \notin N\left(  t\right)  $ in this case). Hence, two
values $\varphi\left(  p_{1},q_{1}\right)  $ and $\varphi\left(  p_{2}%
,q_{2}\right)  $ of the map $\varphi$ cannot be equal unless they come from
the same case (i.e., unless the two inequalities $t\left(  p_{1}\right)
<t\left(  q_{1}\right)  $ and $t\left(  p_{2}\right)  <t\left(  q_{2}\right)
$ either both hold or both fail). But two values of $\varphi$ coming from the
same case cannot be equal unless the inputs are equal. Thus, $\varphi$ is injective.

\item \textbf{page 3, proof of Theorem 1.1.5:} The last paragraph of this
proof can be simplified: The extra consideration of the $k=m$ case is
unnecessary. Indeed, $u_{m+1}$ is defined as an empty product and thus equals
the identity permutation $id$. Hence, $u_{m+1}\left(  i\right)  <u_{m+1}%
\left(  j\right)  $ (because $i<j$). But $u_{1}\left(  i\right)  =w\left(
i\right)  >w\left(  j\right)  =u_{1}\left(  j\right)  $. Hence, there exists
some $k\in\left\{  1,2,\ldots,m\right\}  $ such that $u_{k}\left(  i\right)
>u_{k}\left(  j\right)  $ but $u_{k+1}\left(  i\right)  <u_{k+1}\left(
j\right)  $. From here, proceed as you do.

\item \textbf{page 3, Corollary 1.1.7:} Replace \textquotedblleft$w\in
Sym$\textquotedblright\ by \textquotedblleft$w\in Sym_{n}$\textquotedblright.

\item \textbf{page 4, proof of Lemma 1.2.2:} \textquotedblleft Hence the
exchange condition\textquotedblright\ $\rightarrow$ \textquotedblleft Hence
the left-hand version of the exchange condition\textquotedblright.

\item \textbf{page 5, proof of Proposition 1.2.1:} In (1.2.9), replace
\textquotedblleft$u_{2}$\textquotedblright\ by \textquotedblleft$u_{1}%
$\textquotedblright.

\item \textbf{page 6, proof of Corollary 1.2.3:} Replace \textquotedblleft%
$u_{n}$\textquotedblright\ by \textquotedblleft$u_{2}$\textquotedblright\ on
the first line of page 6.

\item \textbf{page 6, proof of Theorem 1.2.4:} \textquotedblleft by relation
(i)\textquotedblright\ $\rightarrow$ \textquotedblleft by relation
(1.2.11a)\textquotedblright.

\item \textbf{page 7, proof of Lemma 1.3.1:} Replace \textquotedblleft$u_{n}%
$\textquotedblright\ by \textquotedblleft$u_{m}$\textquotedblright\ twice in
the last paragraph of this proof.

\item \textbf{page 7, before Proposition 1.3.2:} \textquotedblleft$i_{1}%
,i_{2},\ldots i_{s}$\textquotedblright\ should be \textquotedblleft%
$i_{1},i_{2},\ldots,i_{s}$\textquotedblright.

\item \textbf{page 7, before Proposition 1.3.2:} \textquotedblleft$i_{1}%
<i_{2}\cdots<i_{s}$\textquotedblright\ should be \textquotedblleft$i_{1}%
<i_{2}<\cdots<i_{s}$\textquotedblright.

\item \textbf{page 7, proof of Proposition 1.3.2:} \textquotedblleft and
$t_{m}\in T$ such that $w_{m-1}=vt_{m}$\textquotedblright\ should be
\textquotedblleft and $t_{m-1}\in T$ such that $w_{m-1}=w_{m}t_{m-1}=vt_{m-1}%
$\textquotedblright.

\item \textbf{page 8, proof of Lemma 1.3.4:} \textquotedblleft If
$xr<yr$\textquotedblright\ should rather be \textquotedblleft If $xr\leq
yr$\textquotedblright.

\item \textbf{page 8, proof of Lemma 1.3.4:} \textquotedblleft subexpression
or\textquotedblright\ $\rightarrow$ \textquotedblleft subexpression
of\textquotedblright.

\item \textbf{page 8, \S 1.4:} \textquotedblleft\textit{right descent set} of
a permutation $w\in Sym_{n}$\textquotedblright\ $\rightarrow$
\textquotedblleft\textit{right descent set} of $w$\textquotedblright\ (since
you already have said \textquotedblleft Given a permutation $w\in Sym_{n}%
$\textquotedblright\ at the beginning of this sentence).

\item \textbf{page 8, \S 1.4:} \textquotedblleft$\mathcal{L}\left(  s_{1}%
s_{2}\right)  =s_{1},\ \mathcal{R}\left(  s_{1}s_{2}\right)  =s_{2}%
$\textquotedblright\ should be \textquotedblleft$\mathcal{L}\left(  s_{1}%
s_{2}\right)  =\left\{  s_{1}\right\}  ,\ \mathcal{R}\left(  s_{1}%
s_{2}\right)  =\left\{  s_{2}\right\}  $\textquotedblright\ (you forgot the set-braces).

\item \textbf{page 8, Lemma 1.4.3:} The \textquotedblleft$\supset
$\textquotedblright\ signs should be \textquotedblleft$\supseteq
$\textquotedblright\ signs. Likewise in the proof.

\item \textbf{page 10, \S 2.1:} After \textquotedblleft A \textit{partition}
is a weakly decreasing sequence $\lambda=\left(  \lambda_{1},\lambda
_{2},\ldots\right)  $\textquotedblright, add \textquotedblleft of nonnegative
integers\textquotedblright.

\item \textbf{page 10, \S 2.1:} In \textquotedblleft where $m=l\left(
\lambda\right)  $ and $\lambda_{i}=0$ for $i>l\left(  \lambda\right)
$\textquotedblright, replace the \textquotedblleft and\textquotedblright\ by
\textquotedblleft since\textquotedblright. (The \textquotedblleft$\lambda
_{i}=0$ for $i>l\left(  \lambda\right)  $\textquotedblright\ part is not a
requirement but a consequence of $m=l\left(  \lambda\right)  $.)

\item \textbf{page 10, \S 2.1:} \textquotedblleft\textit{Ferrer's
diagram}\textquotedblright\ $\rightarrow$ \textquotedblleft\textit{Ferrers
diagram}\textquotedblright.

\item \textbf{page 10, \S 2.2:} In the description of the bumping algorithm,
it is worth explaining that we regard a tableau $T$ of shape $\lambda$ as
containing infinitely many rows, with all but the first $l\left(
\lambda\right)  $ of them being empty. Thus, if an entry is bumped out of the
last nonempty row of $T$, then it is inserted into the next row, which is
empty, and finds rest in that row (which is no longer nonempty). Thus, the new
box created is the first box in the (formerly empty) $\left(  l\left(
\lambda\right)  +1\right)  $-st row.

\item \textbf{page 11, \S 2.2:} It should be said that the new box is counted
as part of the bumping route (despite the definition of the bumping route
sounding like it isn't).

\item \textbf{page 11, \S 2.2:} \textquotedblleft Similarly $\left(
i,j\right)  $ is \textit{strictly below} $\left(  k,l\right)  $ if $i<k$ and
\textit{weakly below} if $i\leq k$\textquotedblright\ $\rightarrow$
\textquotedblleft Similarly $\left(  i,j\right)  $ is \textit{strictly below}
$\left(  k,l\right)  $ if $i>k$ and \textit{weakly below} if $i\geq
k$\textquotedblright.

Also, it is worth saying that \textquotedblleft below\textquotedblright\ and
\textquotedblleft to the left\textquotedblright\ mean \textquotedblleft weakly
below\textquotedblright\ and \textquotedblleft weakly to the
left\textquotedblright\ unless qualified differently.

\item \textbf{page 11, Lemma 2.2.1:} This lemma is too crowded. To make it
clearer, I would break it up into three parts:

\begin{statement}
(a) The bumping route of $T\leftarrow x$ moves to the left. That is, if
$x_{i}$ and $x_{i+1}$ are (consecutive) elements in the bumping sequence then
$x_{i+1}$ lies weakly left of $x_{i}$ in $T\leftarrow x$.

(b) Furthermore, if $x<y$, then the bumping route of $T\leftarrow x$ is
strictly left of the bumping route of $\left(  T\leftarrow x\right)
\leftarrow y$, and the new box of $T\leftarrow x$ is strictly left and weakly
below the new box of $\left(  T\leftarrow x\right)  \leftarrow y$.

(c) Furthermore, if $x<y$, then the bumping route of $\left(  T\leftarrow
y\right)  \leftarrow x$ is weakly left of the bumping route of $T\leftarrow
y$, and the new box of $\left(  T\leftarrow y\right)  \leftarrow x$ is
strictly below and weakly left of the new box of $T\leftarrow y$.
\end{statement}

The three paragraphs of the proof correspond precisely to these three parts
(a), (b) and (c). And this correspondence is important, because the very first
sentence of the proof (\textquotedblleft If the new box of $T\leftarrow x$ is
in the first row then there is nothing to prove\textquotedblright) clearly
makes sense only for part (a), not for parts (b) and (c).

\item \textbf{page 11, proof of Lemma 2.2.1:} In the first paragraph of the
proof, replace \textquotedblleft$T_{i+1,y}$\textquotedblright\ by
\textquotedblleft$T_{i+1,j}$\textquotedblright\ twice (on the same line).

\item \textbf{page 11, proof of Lemma 2.2.1:} In the third paragraph of the
proof, replace \textquotedblleft$x_{1},\ldots x_{j}$\textquotedblright\ by
\textquotedblleft$x_{1},\ldots,x_{j}$\textquotedblright.

\item \textbf{page 11, proof of Lemma 2.2.1:} In the third paragraph of the
proof, replace \textquotedblleft and let $p$ be as above\textquotedblright\ by
\textquotedblleft and let $p$ be the minimum of $k$ and $j$\textquotedblright.

\item \textbf{page 11, proof of Lemma 2.2.1:} In the third paragraph of the
proof, replace \textquotedblleft bumping sequence of $T\left(  \leftarrow
y\right)  \leftarrow x$\textquotedblright\ by \textquotedblleft bumping route
of $\left(  T\leftarrow y\right)  \leftarrow x$\textquotedblright.

\item \textbf{page 12, proof of Lemma 2.2.2:} \textquotedblleft an an
element\textquotedblright\ $\rightarrow$ \textquotedblleft an
element\textquotedblright.

\item \textbf{page 12, \S 2.3:} In the displayed equation \textquotedblleft%
$\varnothing\leftarrow w=\left(  \ldots\left(  \left(  \varnothing\leftarrow
w_{1}\right)  \leftarrow w_{2}\right)  \ldots\right)  \leftarrow w_{n}%
$\textquotedblright, the \textquotedblleft$w_{n}$\textquotedblright\ should be
a \textquotedblleft$w_{i}$\textquotedblright.

\item \textbf{page 12, \S 2.3:} On the last line of this page, replace
\textquotedblleft the $w$ corresponds\textquotedblright\ by \textquotedblleft
that $w$ corresponds\textquotedblright.

\item \textbf{page 12, \S 2.3:} On the last line of this page, replace
\textquotedblleft Shensted\textquotedblright\ by \textquotedblleft
Schensted\textquotedblright.

\item \textbf{page 13, proof of Theorem 2.3.1:} \textquotedblleft pair of
tableau\textquotedblright\ $\rightarrow$ \textquotedblleft pair of
tableaux\textquotedblright.

\item \textbf{page 13, proof of Theorem 2.3.1:} This proof is missing a
noticeable part: You only showed that the reverse procedure undoes the
Robinson--Schensted correspondence, but it should also be proved that the
reverse procedure can be applied to a pair $\left(  P,Q\right)  $ of
same-shape standard tableaux (even if we don't know a-priori that this pair is
the image of a permutation under the Robinson--Schensted correspondence) and
can then be undone by the Robinson--Schensted correspondence. (Unless you
already know that the number of pairs $\left(  P,Q\right)  $ is $\leq
\left\vert Sym_{n}\right\vert $, which allows you to skip this step thanks to
the pigeonhole principle.) This requires checking, among other things, that
the reverse bumping algorithm, when applied to a tableau $S$ and an outside
corner $c$ of $S$, always produces a tableau $T$ and a number $x$ which
satisfy $T\leftarrow x=S$ and produce the new box $c$.

\item \textbf{page 14, first line:} \textquotedblleft we define $\left(
\lambda\cup\mu\right)  _{i}=$\textquotedblright\ should be \textquotedblleft
we define a new partition $\lambda\cup\mu$ by $\left(  \lambda\cup\mu\right)
_{i}=$\textquotedblright.

Also, \textquotedblleft for $1\leq i\leq\max\left\{  m,n\right\}
$\textquotedblright\ should be \textquotedblleft for $i\geq1$%
\textquotedblright\ (otherwise the formulation allows for nonzero parts beyond
$\max\left\{  m,n\right\}  $). Generally, it is better to use the infinite
form of partitions here (i.e., replace \textquotedblleft$\lambda=\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  $ and $\mu=\left(  \mu
_{1},\mu_{2},\ldots,\mu_{m}\right)  $\textquotedblright\ by \textquotedblleft%
$\lambda=\left(  \lambda_{1},\lambda_{2},\lambda_{3},\ldots\right)  $ and
$\mu=\left(  \mu_{1},\mu_{2},\mu_{3},\ldots\right)  $\textquotedblright), as
this will keep the letter $n$ free for its later use for the size of $\lambda$.

\item \textbf{page 14:} \textquotedblleft ommiting\textquotedblright%
\ $\rightarrow$ \textquotedblleft omitting\textquotedblright.

\item \textbf{page 14, Lemma 2.4.1:} Here you are writing saturated chains of
partitions $\varnothing=\lambda_{0}\subset\lambda_{1}\subset\cdots
\subset\lambda_{n}$ as $\lambda_{1}\subset\lambda_{2}\subset\cdots
\subset\lambda_{n}$ (that is, you are omitting $\lambda_{0}$). This is
perfectly valid ($\lambda_{0}$ is $\varnothing$ and thus is uninteresting) but
should probably be explained.

\item \textbf{page 14, Lemma 2.4.1:} The notation $w\left(  i\right)  $ here
is rather misleading, as it would normally mean the image of $i$ under $w$
(that is, the same as $w_{i}$).

\item \textbf{page 14, proof of Lemma 2.4.1:} \textquotedblleft we defined we
defined\textquotedblright\ $\rightarrow$ \textquotedblleft we
defined\textquotedblright.

\item \textbf{page 14, proof of Lemma 2.4.1:} \textquotedblleft be
adding\textquotedblright\ $\rightarrow$ \textquotedblleft by
adding\textquotedblright.

\item \textbf{page 14, proof of Lemma 2.4.1:} \textquotedblleft%
$Shape(P^{\left(  1\right)  )}$\textquotedblright\ should be \textquotedblleft%
$Shape\left(  P^{\left(  1\right)  }\right)  $\textquotedblright.

\item \textbf{page 15, first sentence:} It should be said that $\left(
i,j\right)  $ denotes the cell in the $i$-th row (counted from the bottom) and
the $j$-th column (counted from the left).

\item \textbf{page 15, first sentence:} It is also worth being explicit:
$w\left(  i,j\right)  $ is the word that remains if we start with $w_{1}%
w_{2}\ldots w_{j}$ and remove all letters larger than $i$.

\item \textbf{page 15:} \textquotedblleft row-bumping the $\left(  i,j\right)
^{th}$ partial permutation into the empty set\textquotedblright\ $\rightarrow$
\textquotedblleft row-bumping the $\left(  i,j\right)  ^{th}$ partial
permutation $w\left(  i,j\right)  $ into the empty tableau\textquotedblright.
Or you can just say $Shape\left(  \varnothing\leftarrow w\left(  i,j\right)
\right)  $, since you have that nice notation for it.

\item \textbf{page 15:} \textquotedblleft label them with the empty
set\textquotedblright\ $\rightarrow$ \textquotedblleft label them with the
empty tableau\textquotedblright.

\item \textbf{page 16, first paragraph:} An \textquotedblleft$\subset
$\textquotedblright\ sign is missing in \textquotedblleft$r\left(  n,0\right)
\subset r\left(  n,1\right)  \subset\cdots r\left(  n,n\right)  $%
\textquotedblright.

\item \textbf{page 16, proof of Lemma 2.5.1:} Add a period at the end of the
displayed relation.

\item \textbf{page 16, proof of Lemma 2.5.1:} In \textquotedblleft or
$w\left(  i+1,j\right)  =w_{1}w_{2}\ldots w_{l}\left(  \text{i+1}\right)
w_{l+1}\ldots w_{k}$\textquotedblright, the \textquotedblleft
i+1\textquotedblright\ should be \textquotedblleft$i+1$\textquotedblright%
\ (that is, in mathmode).

\item \textbf{page 16, proof of Lemma 2.5.1:} \textquotedblleft for some
$l\leq j$\textquotedblright\ $\rightarrow$ \textquotedblleft for some $l\leq
k$\textquotedblright.

\item \textbf{page 16, proof of Lemma 2.5.1:} At the very end,
\textquotedblleft$r\left(  i+1,j\right)  \subset r\left(  i,j\right)
$\textquotedblright\ should be \textquotedblleft$r\left(  i,j\right)  \subset
r\left(  i+1,j\right)  $\textquotedblright.

\item \textbf{page 16, possibility 2:} An easier and less confusing argument
would proceed as follows: \textquotedblleft In this case clearly $\mu=\lambda$
and $\rho=\nu$ but $\lambda\subset\nu$ and $\mu\subset\rho$. From
$\lambda\subset\nu$, we obtain $\nu=\lambda\cup\nu=\mu\cup\nu$ (since
$\lambda=\mu$) and thus $\rho=\nu=\mu\cup\nu$.\textquotedblright

\item \textbf{page 16, possibility 3:} An easier and less confusing argument
would proceed as follows: \textquotedblleft In this case clearly $\nu=\lambda$
and $\rho=\mu$ but $\lambda\subset\mu$ and $\nu\subset\rho$. From
$\lambda\subset\mu$, we obtain $\mu=\mu\cup\lambda=\mu\cup\nu$ (since
$\lambda=\nu$) and thus $\rho=\mu=\mu\cup\nu$.\textquotedblright

\item \textbf{page 17, possibility 4:} Replace \textquotedblleft$w_{k}%
$\textquotedblright\ by \textquotedblleft$w_{j}$\textquotedblright\ three
times in this argument.

\item \textbf{page 17, possibility 4:} After \textquotedblleft must be in the
same location as $i$ in $T\left(  \mu\right)  $\textquotedblright, I would add
\textquotedblleft(since $T\left(  \nu\right)  =T\left(  \lambda\right)
\leftarrow w_{j}$)\textquotedblright\ for clarity.

\item \textbf{page 17, possibility 4:} \textquotedblleft places $w_{k}$ in the
box previously occupied\textquotedblright\ $\rightarrow$ \textquotedblleft
bumps $i$ out of the box previously occupied\textquotedblright. (The entry
that bumps $i$ is not necessarily $w_{k}$.)

\item \textbf{page 17, possibility 4:} \textquotedblleft(the row number of $i$
in $\mu$)\textquotedblright\ $\rightarrow$ \textquotedblleft(the row number of
$i$ in $T\left(  \mu\right)  $)\textquotedblright.

\item \textbf{page 17, possibility 4:} I find the last two sentences of this
argument rather hard to follow. I would instead argue as follows:

By construction, $T\left(  \rho\right)  =T\left(  \mu\right)  \leftarrow
w_{j}$ and $T\left(  \nu\right)  =T\left(  \lambda\right)  \leftarrow w_{j}$.
Again by construction, the tableaux $T\left(  \lambda\right)  $ and $T\left(
\nu\right)  $ can be obtained from $T\left(  \mu\right)  $ and $T\left(
\rho\right)  $ (respectively) by deleting the entry $i$.

Note that there is a single box in $\rho\setminus\mu$ (since $T\left(
\rho\right)  =T\left(  \mu\right)  \leftarrow w_{j}$). This box does not
belong to $\mu$, and thus does not belong to $\nu$ (since $\mu=\nu$).

Let $\left(  s,s^{\prime}\right)  $ be the box of $T\left(  \mu\right)  $ that
contains the entry $i$. Thus, this box lies in the $s$-th row. Moreover, the
tableaux $T\left(  \mu\right)  $ and $T\left(  \lambda\right)  $ differ only
in the box $\left(  s,s^{\prime}\right)  $ (since $T\left(  \lambda\right)  $
can be obtained from $T\left(  \mu\right)  $ by deleting the entry $i$, which
lies in the box $\left(  s,s^{\prime}\right)  $). Hence, $\mu_{s}=\lambda
_{s}+1$, whereas $\mu_{t}=\lambda_{t}$ for all $t\neq s$.

Now, consider the bumping route of $T\left(  \mu\right)  \leftarrow w_{j}$
(which produces $T\left(  \rho\right)  $) and the bumping route of $T\left(
\lambda\right)  \leftarrow w_{j}$. If the former bumping route did not contain
the box $\left(  s,s^{\prime}\right)  $, then it would be identical with the
latter bumping route (since the tableaux $T\left(  \mu\right)  $ and $T\left(
\lambda\right)  $ differ only in the box $\left(  s,s^{\prime}\right)  $, and
thus the insertion of $w_{j}$ into both tableaux will proceed exactly
identically unless it hits this specific box); but this is impossible, since
the former bumping route contains the single box in $\rho\setminus\mu$
(because it produces the tableau $T\left(  \rho\right)  $) whereas the latter
bumping route does not (since this box does not belong to $\nu$). Hence, the
former bumping route must contain the box $\left(  s,s^{\prime}\right)  $. The
entry bumped from this box is, of course, $i$ (since this is the entry of
$T\left(  \mu\right)  $ in this box). After being bumped, this entry $i$ is
moved into the next (i.e., $\left(  s+1\right)  $-st) row, where it finds its
rest at the end of the row (it cannot bump any further entry, since $i$ is
larger than all other entries of $T\left(  \mu\right)  $). Thus, the new box
of $T\left(  \mu\right)  \leftarrow w_{j}$ lies in the $\left(  s+1\right)
$-st row. Since $T\left(  \mu\right)  \leftarrow w_{j}=T\left(  \rho\right)
$, this new box must be the single box in $\rho\setminus\mu$, and thus we
conclude that the single box in $\rho\setminus\mu$ lies in the $\left(
s+1\right)  $-th row. Hence, $\rho_{s+1}=\mu_{s+1}+1$, whereas $\rho_{t}%
=\mu_{t}$ for all $t\neq s+1$.

\item \textbf{page 17, after the four possibilities:} \textquotedblleft(since
$i$ is greater than all the $w_{i}$)\textquotedblright\ $\rightarrow$
\textquotedblleft(since $i$ is greater than each of $w_{1},w_{2},\ldots,w_{k}%
$)\textquotedblright\ (beware of overusing the letter $i$).

\item \textbf{page 17, after the four possibilities:} \textquotedblleft Hence
$\rho_{i}=\lambda_{i}$ when $i\neq1$ and $\rho_{1}=\lambda_{1}+1$%
\textquotedblright\ $\rightarrow$ \textquotedblleft Hence $\rho_{t}%
=\lambda_{t}$ when $t\neq1$ and $\rho_{1}=\lambda_{1}+1$\textquotedblright%
\ (again, beware of overusing the letter $i$).

\item \textbf{page 17, local rule 3:} \textquotedblleft then let $i$ be the
unique integer such that $\mu_{i}=\lambda_{i}+1$. Then, $\rho_{j}=\mu_{j}$ if
$j\neq i+1$ and $\rho_{i+1}=\mu_{i+1}+1$\textquotedblright\ $\rightarrow$
\textquotedblleft then let $s$ be the unique integer such that $\mu
_{s}=\lambda_{s}+1$. Then, $\rho_{t}=\mu_{t}$ if $t\neq s+1$ and $\rho
_{s+1}=\mu_{s+1}+1$\textquotedblright\ 

\item \textbf{page 17, local rule 4:} \textquotedblleft$\rho_{j}=\lambda_{j}$
if $j\neq1$\textquotedblright\ $\rightarrow$ \textquotedblleft$\rho
_{t}=\lambda_{t}$ if $t\neq1$\textquotedblright\ (you are overusing the letter
$j$ this time).

\item \textbf{page 17, \S 2.6:} The first paragraph is misadvertising Knuth
equivalence. When it comes to \textbf{deciding} whether two words have the
same $P$-symbol, the easiest way to proceed is to compute their $P$-symbols
using Robinson--Schensted insertion. Knuth equivalence does not provide an
efficient decision procedure (unless you mean the inefficient
\textquotedblleft map out the whole equivalence class by search and
backtracking\textquotedblright\ method).

\item \textbf{page 17, \S 2.6:} After \textquotedblleft three adjacent
elements\textquotedblright, I would add \textquotedblleft of $w$%
\textquotedblright\ (as opposed to three consecutive integers).

\item \textbf{page 18:} \textquotedblleft and write $u\equiv w$%
\textquotedblright\ $\rightarrow$ \textquotedblleft and write $v\equiv
w$\textquotedblright.

\item \textbf{page 18, proof of Proposition 2.6.1:} \textquotedblleft show
that elementary\textquotedblright\ $\rightarrow$ \textquotedblleft show that
an elementary\textquotedblright.

\item \textbf{page 18, proof of Proposition 2.6.1:} After \textquotedblleft
assume that $T$ has one row\textquotedblright, I would add \textquotedblleft%
(or zero rows, in which case we pretend that $T$ has one empty
row)\textquotedblright.

\item \textbf{page 18, proof of Proposition 2.6.1:} In the \textquotedblleft
seven possibilities\textquotedblright, it should be said that the
\textquotedblleft$t_{i-1}<$\textquotedblright\ part of the chain of
inequalities should be understood as void if $i=1$. Likewise, the case $m=0$
is counted towards possibility 1.

\item \textbf{page 18, proof of Proposition 2.6.1:} The \textquotedblleft
seven possibilities\textquotedblright\ are not a complete list. There are in
fact some further possibilities:

\begin{statement}
7a)\qquad$t_{m-1}<x<y<z<t_{m}$:%
\[
T\leftarrow zxy=%
\begin{tabular}
[c]{|c|ccccc}\cline{1-2}\cline{4-6}%
$t_{1}$ & $t_{2}$ & \multicolumn{1}{|c}{$\cdots$} &
\multicolumn{1}{|c}{$t_{m-1}$} & \multicolumn{1}{|c}{$x$} &
\multicolumn{1}{|c|}{$y$}\\\cline{1-2}\cline{4-6}%
$z$ &  &  &  &  & \\\cline{1-1}%
$t_{m}$ &  &  &  &  & \\\cline{1-1}%
\end{tabular}
\ \ =T\leftarrow xzy.
\]

\end{statement}

\begin{statement}
8)\qquad$t_{i-1}<x<y<t_{i}<\cdots<t_{m}<z$ and $i<m$:%
\[
T\leftarrow zxy=%
\begin{tabular}
[c]{|c|c|ccccc}\cline{1-2}\cline{4-5}\cline{7-7}%
$t_{1}$ & $t_{2}$ & $\cdots$ & \multicolumn{1}{|c}{$x$} &
\multicolumn{1}{|c}{$y$} & \multicolumn{1}{|c}{$\cdots$} &
\multicolumn{1}{|c|}{$z$}\\\cline{1-2}\cline{4-5}\cline{7-7}%
$t_{i}$ & $t_{j}$ &  &  &  &  & \\\cline{1-2}%
\end{tabular}
\ \ =T\leftarrow xzy\ \ \ \ \ \ \ \ \ \ \left(  t_{j}:=t_{i+1}\right)  .
\]

\end{statement}

\begin{statement}
8a)\qquad$t_{i-1}<x<y<t_{i}<\cdots<t_{m}<z$ and $i=m$ (thus $t_{m-1}%
<x<y<t_{m}<z$):%
\[
T\leftarrow zxy=%
\begin{tabular}
[c]{|c|c|cccc}\cline{1-2}\cline{4-6}%
$t_{1}$ & $t_{2}$ & $\cdots$ & \multicolumn{1}{|c}{$t_{m-1}$} &
\multicolumn{1}{|c}{$x$} & \multicolumn{1}{|c|}{$y$}\\\cline{1-2}\cline{4-6}%
$t_{m}$ & $z$ &  &  &  & \\\cline{1-2}%
\end{tabular}
\ \ =T\leftarrow xzy.
\]

\end{statement}

Furthermore, some of the possibilities need to be subdivided further:

\item \textbf{page 18, proof of Proposition 2.6.1:} In possibility 1, there
should not be a \textquotedblleft$t_{i}$\textquotedblright\ in the second row
of the tableau.

\item \textbf{page 18, proof of Proposition 2.6.1:} In possibility 5, the
definition of $t_{j}$ needs to be qualified slightly: We do indeed set
$t_{j}:=t_{i+1}$ if $i+1<k$; otherwise we set $t_{j}:=z$.

\item \textbf{page 19, proof of Proposition 2.6.1:} In possibility 7, you need
to require $i<m$, since the output looks different in the case $i=m$ (see
possibility 7a above).{}

\item \textbf{page 19, proof of Proposition 2.6.1:} \textquotedblleft In cases
1, 2 and 3\textquotedblright\ $\rightarrow$ \textquotedblleft In cases 1, 2,
3, 7a, 8 and 8a\textquotedblright.

\item \textbf{page 19, Lemma 2.6.3:} Add \textquotedblleft for any positive
integer $v$ not occurring in $T$\textquotedblright\ at the end of this lemma.

\item \textbf{page 20, proof of Lemma 2.6.3:} \textquotedblleft as
$r_{1},r_{2}\ldots r_{p}$\textquotedblright\ $\rightarrow$ \textquotedblleft
as $r_{1},r_{2},\ldots,r_{p}$\textquotedblright.

\item \textbf{page 20, proof of Lemma 2.6.3:} Missing $<$ sign in
\textquotedblleft$v=v_{1}<v_{2}\cdots<v_{r}$\textquotedblright.

\item \textbf{page 20, proof of Lemma 2.6.3:} For the equivalence
\textquotedblleft$r_{i}v_{i}\equiv v_{i+1}r_{i}^{\prime}$\textquotedblright%
\ to hold for all $i\leq r$, you need to set $v_{r+1}:=\varnothing$, and to
define $r_{i}$ or $r_{i}^{\prime}$ to be an empty word if the corresponding
tableau has no $i$-th row. Alternatively, you may want to state this
equivalence for $i\leq r-1$ only, and then add $r_{r}v_{r}\equiv r_{r}%
^{\prime}$ and $r_{i}=r_{i}^{\prime}$ for all $i>r$ as a last step. But it is
perhaps best to proceed differently: Set $v_{i}:=\varnothing$ for all $i>r$,
and then argue that $r_{i}v_{i}\equiv v_{i+1}r_{i}^{\prime}$ holds for all
$i\geq1$ (including the cases $i<r$ and $i\geq r$).

\item \textbf{page 20, \S 2.7:} After \textquotedblleft$i+1$ lies strictly
below and weakly left of $i$ in $P$\textquotedblright, it is worth adding
\textquotedblleft(actually the \textquotedblleft weakly left\textquotedblright%
\ part follows from the \textquotedblleft strictly below\textquotedblright%
\ part, since $P$ is a tableau)\textquotedblright. This is tacitly used in the
proof of Proposition 2.8.2.

\item \textbf{page 21, proof of Proposition 2.7.1:} \textquotedblleft row Row
Bumping Lemma\textquotedblright\ $\rightarrow$ \textquotedblleft Row Bumping
Lemma\textquotedblright.

\item \textbf{page 21:} After \textquotedblleft successively down each
column\textquotedblright, add \textquotedblleft(from the leftmost column to
the rightmost)\textquotedblright.

\item \textbf{page 21, Lemma 2.7.2:} I would replace \textquotedblleft
Suppose\textquotedblright\ by \textquotedblleft Let $P$ be a tableau such
that\textquotedblright.

\item \textbf{page 21, Lemma 2.7.2:} Replace the \textquotedblleft$\subset
$\textquotedblright\ sign by \textquotedblleft$\subseteq$\textquotedblright,
as I think (not sure about this!) that you use \textquotedblleft$\subset
$\textquotedblright\ for proper subsets.

\item \textbf{page 21, Lemma 2.7.2:} Add a period before \textquotedblleft
then\textquotedblright.

\item \textbf{page 21, proof of Lemma 2.7.2:} \textquotedblleft fill a
diagram\textquotedblright\ $\rightarrow$ \textquotedblleft fill the
diagram\textquotedblright\ (twice). Also, this is a bit misleading: You don't
want an arbitrary filling (such a filling would not be unique), but rather a
tableau. (You use this when you tacitly argue that the entries $1,2,\ldots
,c_{1}$ lie in the first column rather than further right.)

\item \textbf{page 21, proof of Lemma 2.7.2:} Again, replace the
\textquotedblleft$\subset$\textquotedblright\ signs by \textquotedblleft%
$\subseteq$\textquotedblright.

\item \textbf{page 21, proof of Lemma 2.7.2:} In \textquotedblleft%
$c_{1}+1,c_{1}+2,\ldots,c_{2}$\textquotedblright, replace \textquotedblleft%
$c_{2}$\textquotedblright\ by \textquotedblleft$c_{1}+c_{2}$\textquotedblright.

\item \textbf{page 21, proof of Lemma 2.7.2:} \textquotedblleft contained in
$\mathcal{D}\left(  S_{\lambda}\right)  $\textquotedblright\ $\rightarrow$
\textquotedblleft containing $\mathcal{D}\left(  S_{\lambda}\right)
$\textquotedblright.

\item \textbf{page 22, first paragraph:} In \textquotedblleft$\lambda
_{1}+\cdots+\lambda_{k}\trianglelefteq\mu_{1}+\cdots+\mu_{k}\trianglelefteq
\pi_{1}+\cdots+\pi_{k}$\textquotedblright, the \textquotedblleft%
$\trianglelefteq$\textquotedblright\ signs should be \textquotedblleft$\leq
$\textquotedblright.

\item \textbf{page 22, second paragraph:} \textquotedblleft in a partition or
tableau\textquotedblright\ $\rightarrow$ \textquotedblleft in a partition
$\lambda$ or tableau\textquotedblright\ (since you refer to $\lambda$ in the
same sentence).

\item \textbf{page 22, before Lemma 2.8.1:} After \textquotedblleft It turns
out that the dominance order\textquotedblright, add \textquotedblleft on the
partitions of a given integer $n$\textquotedblright\ (you are not saying
anything about partitions of different $n$'s here, even though you defined
dominance for them as well).

\item \textbf{page 22, proof of Lemma 2.8.1:} \textquotedblleft
operations\textquotedblright\ $\rightarrow$ \textquotedblleft
operation\textquotedblright.
\end{enumerate}

\begin{noncompile}
\textbf{page 23, first paragraph:} \textquotedblleft otherwise $\mu_{i}%
=\mu_{i-1}=\lambda_{i-1}\geq\lambda_{i}$\textquotedblright\ $\rightarrow$
\textquotedblleft otherwise $\mu_{i}=\mu_{i-1}\geq\lambda_{i-1}\geq\lambda
_{i}$\textquotedblright. (EDIT: This is a simplification but not necessary.)
\end{noncompile}

\begin{enumerate}
\item \textbf{page 23, second paragraph:} This argument does not work as
given. For example, let $\lambda=\left(  6,4,3,1\right)  $ and $\mu=\left(
5,5,2,2\right)  $. Then, $i=1$ and $j=4$ and therefore $\mu^{\prime}=\left(
6,5,2,1\right)  $. But we don't have $\mu^{\prime}\trianglelefteq\lambda$, and
thus we cannot apply the induction hypothesis.

One way to correct the argument is by changing the definition of $j$: We let
$j$ be the smallest integer $j>i$ such that $\lambda_{1}+\lambda_{2}%
+\cdots+\lambda_{j}\leq\mu_{1}+\mu_{2}+\cdots+\mu_{j}$. Then, it is easy to
see that $\lambda_{j}<\mu_{j}$ (otherwise, $j$ would not be the smallest) and
$\lambda_{j+1}\geq\mu_{j+1}$ (otherwise, we would have $\lambda_{1}%
+\lambda_{2}+\cdots+\lambda_{j+1}<\mu_{1}+\mu_{2}+\cdots+\mu_{j+1}$,
contradicting $\mu\trianglelefteq\lambda$), so that $\mu_{j+1}\leq
\lambda_{j+1}\leq\lambda_{j}<\mu_{j}$. Thus, $\left(  j,\mu_{j}\right)  $ is
an outside corner of $\mu$. Define $\mu^{\prime}$ in the way you suggest
(using this new $j$) and show that $\sum\left\vert \lambda_{k}-\mu_{k}%
^{\prime}\right\vert <n$ and $\mu^{\prime}\trianglelefteq\lambda$. Then use
the induction hypothesis.

In all of this, you should view $\lambda$ and $\mu$ as infinite sequences, so
that the case $l\left(  \mu\right)  >l\left(  \lambda\right)  $ disappears.

\item \textbf{page 23, proof of Proposition 2.8.2:} Replace the
\textquotedblleft$\subset$\textquotedblright\ sign by \textquotedblleft%
$\subseteq$\textquotedblright.

\item \textbf{page 23, proof of Proposition 2.8.2, Case 1:} \textquotedblleft
below\textquotedblright\ $\rightarrow$ \textquotedblleft strictly
below\textquotedblright.

\item \textbf{page 24, proof of Proposition 2.8.2, Case 2:} \textquotedblleft
below\textquotedblright\ $\rightarrow$ \textquotedblleft strictly
below\textquotedblright\ (three times).

\item \textbf{page 24, Note 2:} This is a bit imprecise: Skew shapes cannot be
arbitrary sets of tiles; if the shape is not a set difference of two Young
diagrams, then slides might not preserve the P-symbol of the reading word. I
don't know whether this makes the description of the construction here false.

\item \textbf{page 24, Note 2:} \textquotedblleft$w_{m}$\textquotedblright%
\ should be \textquotedblleft$w_{n}$\textquotedblright.

\item \textbf{page 24, Note 2:} \textquotedblleft
Sh\"{u}tzenberger\textquotedblright\ $\rightarrow$ \textquotedblleft
Sch\"{u}tzenberger\textquotedblright.

\item \textbf{page 26, \S 3.1, first paragraph:} \textquotedblleft algebra of
$A$\textquotedblright\ $\rightarrow$ \textquotedblleft algebra over
$A$\textquotedblright.

\item \textbf{page 26, \S 3.1:} \textquotedblleft in Theorem
1.2.11\textquotedblright\ $\rightarrow$ \textquotedblleft in Theorem
1.2.4\textquotedblright.

\item \textbf{page 26, \S 3.1:} \textquotedblleft are reduced
expression\textquotedblright\ $\rightarrow$ \textquotedblleft are reduced
expressions\textquotedblright.

\item \textbf{page 26, \S 3.1:} In the long(ish) computation, I would replace
\textquotedblleft$T_{i_{1}}T_{i_{2}}\ldots T_{i_{m}}^{2}$\textquotedblright%
\ by \textquotedblleft$T_{i_{1}}T_{i_{2}}\ldots T_{i_{m-1}}T_{i_{m}}^{2}%
$\textquotedblright\ to make it clearer.

\item \textbf{page 27, \S 3.2:} \textquotedblleft a ring
homomorphism\textquotedblright\ $\rightarrow$ \textquotedblleft an $A$-algebra
homomorphism\textquotedblright.

\item \textbf{page 27, \S 3.2:} I'd replace \textquotedblleft representations
afforded by $H_{n}\left(  q\right)  $\textquotedblright\ by \textquotedblleft
representations of $H_{n}\left(  q\right)  $\textquotedblright, since
\textquotedblleft afforded by\textquotedblright\ suggests modules to me (i.e.,
I would refer to a representation of $H_{n}\left(  q\right)  $ on some
$A$-module $M$ as a \textquotedblleft representation afforded by
$M$\textquotedblright, not \textquotedblleft by $H_{n}\left(  q\right)
$\textquotedblright).

\item \textbf{page 27, Lemma 3.3.1:} It would be best to add the extra
condition \textquotedblleft$\ell\left(  x\right)  \leq m$\textquotedblright%
\ under the summation sign (i.e., to replace the summation sign by
\textquotedblleft$\sum\limits_{\substack{x\leq w;\\\ell\left(  x\right)  \leq
m}}$\textquotedblright). This makes the lemma a little bit stronger (the proof
is very easy to adapt), and this extra strength is needed in the proof of
Proposition 3.3.2.

\item \textbf{page 28, proof of Lemma 3.3.1:} I'd replace \textquotedblleft%
(since $r_{1}r_{2}\ldots r_{i+1}<r_{1}r_{2}\ldots r_{i}$)\textquotedblright%
\ by the more detailed explanation \textquotedblleft(by (3.1.2), since
$r_{1}r_{2}\ldots r_{i+1}<r_{1}r_{2}\ldots r_{i}$)\textquotedblright.

\item \textbf{page 28, proof of Lemma 3.3.1:} It is also worth explaining how
you obtain the \textquotedblleft$qT_{r_{1}}\ldots\widehat{T_{r_{p}}}%
\ldots\widehat{T_{r_{q}}}\ldots T_{r_{m}}$\textquotedblright\ term. (Namely,
you rewrite $T_{r_{1}\ldots r_{i+1}}$ as $T_{r_{1}}\ldots\widehat{T_{r_{p}}%
}\ldots\widehat{T_{r_{q}}}\ldots T_{r_{i+1}}$ using the fact that $r_{1}%
\ldots\widehat{r_{p}}\ldots\widehat{r_{q}}\ldots r_{i+1}$ is a reduced
expression for $r_{1}r_{2}\ldots r_{i+1}$; then you multiply the extra factors
$T_{r_{i+2}},\ldots,T_{r_{m}}$ on the right of this.)

\item \textbf{page 28, proof of Proposition 3.3.2:} The long displayed
computation could use a bit more explanation: To get from the first line to
the second, you use the fact that $r_{m}r_{m-1}\ldots r_{1}$ is a reduced
expression for $w^{-1}$ (by Proposition 1.1.3) and thus we have $T_{w^{-1}%
}=T_{r_{m}}T_{r_{m-1}}\cdots T_{r_{1}}$.

\item \textbf{page 28, proof of Proposition 3.3.2:} On the last line of this
proof, \textquotedblleft$\left(  q^{-1}T_{r_{1}}\right)  \left(
q^{-1}T_{r_{1}}\right)  $\textquotedblright\ should be \textquotedblleft%
$\left(  q^{-1}T_{r_{1}}\right)  \left(  q^{-1}T_{r_{2}}\right)
$\textquotedblright.

\item \textbf{page 28, proof of Proposition 3.3.2:} The claim in the last
sentence of this proof is not quite obvious: Why cannot $T_{w}$ emerge from
any other subexpressions? This becomes clear once Lemma 3.3.1 is strengthened
as I suggested above (adding the $\ell\left(  x\right)  \leq m$ condition
under the summation sign), since then we see that any subexpression of length
$<m$ can only produce $T_{x}$ terms with $\ell\left(  x\right)  <m$.

\item \textbf{page 29:} \textquotedblleft Applying $^{\ast}$
yields\textquotedblright\ $\rightarrow$ \textquotedblleft Let $r=s_{j}\in S$.
Applying * to (3.1.2) yields\textquotedblright.

\item \textbf{page 29:} \textquotedblleft Now Lemma 1.4.2 shows
that\textquotedblright\ $\rightarrow$ \textquotedblleft Now Corollary 1.3.3
shows that\textquotedblright.

\item \textbf{page 29:} \textquotedblleft If is
straightforward\textquotedblright\ $\rightarrow$ \textquotedblleft It is
straightforward\textquotedblright.

\item \textbf{page 29, proof of Proposition 3.4.1:} In the first sentence, it
is not clear why you can extend $\iota$ multiplicatively like this (what if
two equal $T_{i_{1}}T_{i_{2}}\ldots T_{i_{m}}$s correspond to different
$\iota\left(  T_{i_{1}}\right)  \iota\left(  T_{i_{2}}\right)  \cdots
\iota\left(  T_{i_{m}}\right)  $'s?).

Instead I would proceed as follows:

Base-changing the standard $A$-algebra structure on $H_{n}\left(  q\right)  $
using the ring homomorphism $\overline{}:A\rightarrow A$, we obtain a new
$A$-algebra structure on $H_{n}\left(  q\right)  $, which is given explicitly
by%
\[
a\rightharpoonup h=\overline{a}\cdot h\ \ \ \ \ \ \ \ \ \ \text{for all }a\in
A\text{ and }h\in H_{n}\left(  q\right)
\]
(where the \textquotedblleft$\rightharpoonup$\textquotedblright\ symbol means
the action of $A$ on the new $A$-algebra $H_{n}\left(  q\right)  $, whereas
the \textquotedblleft$\cdot$\textquotedblright\ symbol means the original
action of $A$ on $H_{n}\left(  q\right)  $). Alternatively, if you view an
$A$-algebra as a ring $M$ equipped with a ring homomorphism from $A$ to its
center $Z\left(  M\right)  $, then this new $A$-algebra structure on
$H_{n}\left(  q\right)  $ is the composition $A\rightarrow A\rightarrow
H_{n}\left(  q\right)  $, where the left arrow is the involution $\overline
{}:A\rightarrow A$ and the right arrow is the ring homomorphism corresponding
to the original $A$-algebra structure on $H_{n}\left(  q\right)  $. Either
way, let us denote this new $A$-algebra structure on $H_{n}\left(  q\right)  $
by $\overline{H_{n}\left(  q\right)  }$. As a ring, it is still the old
$H_{n}\left(  q\right)  $, but the action of $A$ is different (in that any
polynomial $F\left(  q\right)  $ now acts as $\overline{F\left(  q\right)
}=F\left(  q^{-1}\right)  $).

Now, define an $A$-algebra homomorphism $\iota:H_{n}\left(  q\right)
\rightarrow\overline{H_{n}\left(  q\right)  }$ by setting $\iota\left(
T_{j}\right)  :=T_{j}^{-1}$ for each $1\leq j<n$. It is easy to see that this
homomorphism is well-defined, since the elements $T_{j}^{-1}$ in
$\overline{H_{n}\left(  q\right)  }$ satisfy the same relations (3.1.1) as the
elements $T_{j}$ in $H_{n}\left(  q\right)  $ (indeed, the first two relations
for the $T_{j}^{-1}$ follow immediately by inverting the corresponding
relations for the $T_{j}$; the third relation for the $T_{j}^{-1}$ is saying
that
\[
T_{j}^{-2}=\left(  q-1\right)  \rightharpoonup T_{j}^{-1}+q\rightharpoonup
T_{id}\ \ \ \text{in\ \ \ }\overline{H_{n}\left(  q\right)  },
\]
but this is equivalent to
\[
T_{j}^{-2}=\left(  q^{-1}-1\right)  T_{j}^{-1}+q^{-1}T_{id}%
\ \ \ \text{in\ \ \ }H_{n}\left(  q\right)  ,
\]
which can be easily derived from (3.1.1c)). Moreover, $\iota$ is an
$A$-algebra homomorphism from $H_{n}\left(  q\right)  $ to $\overline
{H_{n}\left(  q\right)  }$, thus a ring homomorphism from $H_{n}\left(
q\right)  $ to $H_{n}\left(  q\right)  $ (since $\overline{H_{n}\left(
q\right)  }=H_{n}\left(  q\right)  $ as rings). It is easy to see that
$\iota\left(  T_{w}\right)  =T_{w^{-1}}^{-1}$ for each $w\in Sym_{n}$ (indeed,
pick a reduced expression $s_{i_{1}}s_{i_{2}}\cdots s_{i_{m}}$ for $w$ and
observe that $T_{w}=T_{i_{1}}T_{i_{2}}\cdots T_{i_{m}}$ and $T_{w^{-1}%
}=T_{i_{m}}T_{i_{m-1}}\cdots T_{i_{1}}$, so that%
\begin{align*}
\iota\left(  T_{w}\right)   &  =\iota\left(  T_{i_{1}}T_{i_{2}}\cdots
T_{i_{m}}\right)  =\iota\left(  T_{i_{1}}\right)  \iota\left(  T_{i_{2}%
}\right)  \cdots\iota\left(  T_{i_{m}}\right) \\
&  =T_{i_{1}}^{-1}T_{i_{2}}^{-1}\cdots T_{i_{m}}^{-1}=\left(
\underbrace{T_{i_{m}}T_{i_{m-1}}\cdots T_{i_{1}}}_{=T_{w^{-1}}}\right)
^{-1}=T_{w^{-1}}^{-1}%
\end{align*}
). Since $\iota$ is an $A$-algebra homomorphism from $H_{n}\left(  q\right)  $
to $\overline{H_{n}\left(  q\right)  }$, this entails that%
\[
\iota\left(  \sum_{w\in Sym_{n}}F_{w}\left(  q\right)  T_{w}\right)
=\sum_{w\in Sym_{n}}F_{w}\left(  q\right)  \rightharpoonup T_{w^{-1}}%
^{-1}=\sum_{w\in Sym_{n}}\overline{F_{w}\left(  q\right)  }T_{w^{-1}}^{-1}%
\]
for any Laurent polynomials $F_{w}\left(  q\right)  \in A$. Finally, note that
$\iota$ is a ring homomorphism, thus a $\mathbb{Z}$-algebra homomorphism.
Recall that the $\mathbb{Z}$-algebra $H_{n}\left(  q\right)  $ is generated by
the $n+1$ elements $q,q^{-1},T_{1},T_{2},\ldots,T_{n-1}$ (since $A=\mathbb{Z}%
\left[  q,q^{-1}\right]  $). Since $\iota^{2}$ agrees with $id$ on each of
these $n+1$ generators (this is easily checked directly\footnote{Indeed,
$\iota$ sends each of these generators to its reciprocal, and thus $\iota^{2}$
must reciprocate it twice, but of course $\left(  x^{-1}\right)  ^{-1}=x$.}),
we thus conclude that $\iota^{2}=id$ (because $\iota$ and thus $\iota^{2}$ is
a $\mathbb{Z}$-algebra homomorphism). Hence, $\iota$ is an involution.

\item \textbf{page 30, proof of Proposition 3.4.2:} \textquotedblleft by Lemma
1.1.3\textquotedblright\ $\rightarrow$ \textquotedblleft by Proposition
1.1.3\textquotedblright.

\item \textbf{page 30, proof of Proposition 3.4.2:} In the first displayed
equation, replace \textquotedblleft$\iota\left(  T_{y}\right)  ^{\ast}%
$\textquotedblright\ by \textquotedblleft$\left(  \iota\left(  T_{y}\right)
\right)  ^{\ast}$\textquotedblright\ for greater clarity. A similar change is
worth doing in the second displayed equation.

\item \textbf{page 31, Theorem 3.5.1:} \textquotedblleft element $C_{w}%
$\textquotedblright\ $\rightarrow$ \textquotedblleft element $C_{w}\in
H_{n}\left(  q\right)  $\textquotedblright.

\item \textbf{page 31, after Theorem 3.5.1:} \textquotedblleft
basis\textquotedblright\ $\rightarrow$ \textquotedblleft basis of
$H_{n}\left(  q\right)  $\textquotedblright.

\item \textbf{page 31, after Theorem 3.5.1:} \textquotedblleft the linear map
sending $C_{w}$ to $T_{w}$\textquotedblright\ $\rightarrow$ \textquotedblleft
the linear map sending $T_{w}$ to $C_{w}$\textquotedblright\ (you don't know
yet that the $C_{w}$ form a basis, so you cannot define a linear map by its
action on the $C_{w}$).

\item \textbf{page 31, Proof of Uniqueness:} I would replace \textquotedblleft
our expression for $T_{y^{-1}}^{-1}$ in Section 3.3\textquotedblright\ by
\textquotedblleft our expression for $T_{y^{-1}}^{-1}$ from Proposition
3.3.2\textquotedblright\ for better clarity (\textquotedblleft
in\textquotedblright\ sounds like \textquotedblleft into\textquotedblright\ here).

\item \textbf{page 32, Proof of Uniqueness:} In (3.5.1), the \textquotedblleft%
$q_{x}^{-\tfrac{1}{2}}$\textquotedblright\ on the right-hand side should be
\textquotedblleft$q_{x}^{\tfrac{1}{2}}$\textquotedblright.

\item \textbf{page 32, Proof of Uniqueness:} In the paragraph after (3.5.1),
\textquotedblleft degree at least $1$\textquotedblright\ should be
\textquotedblleft degree at least $\dfrac{1}{2}$\textquotedblright.

\item \textbf{page 32, Proof of Existence:} \textquotedblleft$w$ lies in
S\textquotedblright\ $\rightarrow$ \textquotedblleft$w$ lies in $S$%
\textquotedblright\ (mathmode!).

\item \textbf{page 32, Proof of Existence:} \textquotedblleft there exists
$r\in R$\textquotedblright\ $\rightarrow$ \textquotedblleft there exists $r\in
S$\textquotedblright.

\item \textbf{page 32, Proof of Existence:} \textquotedblleft since
$C_{r}=q^{\tfrac{1}{2}}T_{r}-q^{\tfrac{1}{2}}T_{id}$\textquotedblright%
\ $\rightarrow$ \textquotedblleft since $C_{r}=q^{-\tfrac{1}{2}}%
T_{r}-q^{\tfrac{1}{2}}T_{id}$\textquotedblright.

\item \textbf{page 33, Proof of Existence:} \textquotedblleft and then
equating coefficients of $T_{x}$ in (3.5.3)\textquotedblright\ $\rightarrow$
\textquotedblleft and then equating coefficients of $T_{x}$ in Theorem
3.5.1\textquotedblright.

\item \textbf{page 34, Proof of Existence:} \textquotedblleft in the sum
$\sum\mu\left(  z,v\right)  q_{z}^{\tfrac{1}{2}}q_{w}^{\tfrac{1}{2}}P_{x,z}%
$\textquotedblright\ $\rightarrow$ \textquotedblleft in the sum $\sum
\mu\left(  z,v\right)  q_{z}^{-\tfrac{1}{2}}q_{w}^{\tfrac{1}{2}}P_{x,z}%
$\textquotedblright.

\item \textbf{page 35, proof of Theorem 3.6.1:} The formula for $T_{r}C_{r}$
is unnecessary: A strong induction needs no induction base. (The sum will just
be empty if you have $w=r$.)

\item \textbf{page 35, proof of Theorem 3.6.1:} Before the last long displayed
computation, replace \textquotedblleft Now:\textquotedblright\ by
\textquotedblleft Now, by (3.5.3) and (3.5.2), we have\textquotedblright.

\item \textbf{page 35, proof of Theorem 3.6.1:} After the last long displayed
computation, I would add \textquotedblleft(by (3.5.3) again)\textquotedblright.

\item \textbf{page 36, proof of Proposition 3.6.2:} In the third displayed
equation, it would make sense to replace \textquotedblleft$q_{y}^{-1}%
$\textquotedblright\ by \textquotedblleft$q_{y^{-1}}^{-1}$\textquotedblright%
\ (the two numbers are equal, but $q_{y^{-1}}^{-1}$ comes out more naturally
from the calculation). Same in the fourth displayed equation.

\item \textbf{page 37, Note 1:} \textquotedblleft for all primes
powers\textquotedblright\ $\rightarrow$ \textquotedblleft for all prime
powers\textquotedblright.

\item \textbf{page 37, Note 1:} Another reference for this proof is Theorem 11
in Daniel Bump's notes \textit{Hecke algebras} (
\url{http://sporadic.stanford.edu/bump/math263/hecke.pdf} ).

\item \textbf{page 37, Note 2:} I would replace \textquotedblleft%
$\operatorname*{End}\nolimits_{G}$\textquotedblright\ here by
\textquotedblleft$\operatorname*{End}\nolimits_{\mathbb{C}G}$%
\textquotedblright\ just to be on the safe side.

\item \textbf{page 37, Note 3:} Add a comma after \textquotedblleft then, in
fact\textquotedblright.

\item \textbf{page 37, Note 3:} Another proof of the Tits deformation theorem
can be obtained from: Richard Dipper and Gordon James, \textit{Blocks and
idempotents of Hecke algebras of general linear groups}, Proc. London Math.
Soc. (3) \textbf{54} (1987), pp. 57--82, Theorem 4.3 and \S 3.1 (vii). (The
former theorem entails that the Hecke algebra $H_{n}\left(  q\right)  $
decomposes as a direct product of matrix rings when it is semisimple, and that
the factors are indexed by the partitions of $n$. Then, \S 3.1 (vii) shows
that the dimensions of these factors are combinatorially given in terms of
standard tableaux and therefore independent of $q$.)

\item \textbf{pages 37--38, Note 5:} The condition for semisimplicity given
here is false. According to Theorem 4.3 in the paper by Dipper and James just
cited, the right condition for $H_{n}\left(  q\right)  $ to be semisimple
(over $\mathbb{C}$) is \textquotedblleft$q=1$ or $q^{2},q^{3},\ldots,q^{n}%
\neq1$ and $q\neq0$\textquotedblright\ (unless $n\leq2$, in which case the
\textquotedblleft$q\neq0$\textquotedblright\ part should be removed from the condition).

\item \textbf{page 38, Note 5:} \textquotedblleft if $e\left(  z\right)
<n$\textquotedblright\ $\rightarrow$ \textquotedblleft if $1<e\left(
z\right)  \leq n$\textquotedblright.

\item \textbf{page 38, Note 8:} Is the preprint [7] available somewhere?

\item \textbf{page 38, Note 8:} The reference to \textquotedblleft Lusztig
[19]\textquotedblright\ is wrong; it should probably cite [21].

\item \textbf{page 39, Proposition 4.1.1:} It is worth pointing out that the
$g$ can depend on $i_{k}$ (i.e., there isn't always one single $g$ for the
entire chain).

\item \textbf{page 39, proof of Proposition 4.1.1:} \textquotedblleft Hence
there exists a chain $i=i_{1},i_{2},\ldots i_{m}=j$\textquotedblright\ should
be \textquotedblleft Hence there exists a chain $i=i_{1},i_{2},\ldots
,i_{m+1}=j$\textquotedblright\ (the chain needs to go till $i_{m+1}$ in order
for $g_{m}$ to be accounted for).

\item \textbf{page 40:} In the first table on this page, the rows are indexed
$T_{id}$ and $T_{id}$. Clearly, the second row should be indexed $T_{s_{1}}$ instead.

\item \textbf{page 41, (4.2.1):} Replace \textquotedblleft$y\leq
x$\textquotedblright\ under the summation sign by \textquotedblleft%
$y\underset{L}{\leq}x$\textquotedblright.

\item \textbf{page 41, last line of the page:} Replace \textquotedblleft%
$r_{a}\left(  y,w\right)  $\textquotedblright\ by \textquotedblleft%
$r_{a}\left(  y,x\right)  $\textquotedblright\ inside the sum.

\item \textbf{page 42:} In the \textquotedblleft slightly more complicated
example\textquotedblright, replace \textquotedblleft$T_{s_{1}}C_{s_{1}}%
=-C_{s}$\textquotedblright\ by \textquotedblleft$T_{s_{1}}C_{s_{1}}=-C_{s_{1}%
}$\textquotedblright.

\item \textbf{page 42:} The second representing matrix in the middle of this
page should presumably be labelled \textquotedblleft$T_{s_{2}}\mapsto
$\textquotedblright\ rather than \textquotedblleft$T_{s_{1}}\mapsto
$\textquotedblright. Moreover, in the following sentence, the first two
matrices should instead be%
\[
T_{s_{1}s_{2}}\mapsto\left(
\begin{array}
[c]{cc}%
0 & -q^{1/2}\\
q^{3/2} & -q
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ T_{s_{2}s_{1}}\mapsto\left(
\begin{array}
[c]{cc}%
-q & q^{3/2}\\
-q^{1/2} & 0
\end{array}
\right)  .
\]


\item \textbf{page 43, Lemma 4.3.2:} Part (ii) needs the requirement $x\leq w$
(otherwise, $P_{x,w}$ would not exist).

\item \textbf{page 43, proof of Lemma 4.3.2:} In the first displayed equation
of this proof, \textquotedblleft$q_{rw}$\textquotedblright\ should be
\textquotedblleft$q_{w}$\textquotedblright, whereas \textquotedblleft$P_{x,z}%
$\textquotedblright\ should be \textquotedblleft$P_{rw,z}$\textquotedblright.

\item \textbf{page 43, proof of Lemma 4.3.2:} In the proof of part (ii), you
should first explain why we have $rw<w$ (since this allows you to apply
(4.3.1) and thus talk about $c_{x}$) and $rx\leq rw$ (in order for $P_{rx,rw}$
to exist). Indeed, $rw<w$ holds because we would otherwise have $x\leq w<rw$,
contradicting $x\not \leq rw$. Thus, Lemma 1.3.4 (or, more precisely, the
analogue of Lemma 1.3.4 where $r$ is multiplied from the left rather than from
the right\footnote{This analogue follows from the original Lemma 1.3.4 by
applying it to $x^{-1}$ and $y^{-1}$ instead of $x$ and $y$.}) yields $rx\leq
rw$ (since $rx<x$ and $rw<w$).

\item \textbf{page 43, proof of Lemma 4.3.2:} In the proof of part (iii), you
should first explain why we have $rx\leq w$ (so that $P_{rx,w}$ makes sense).
This can be shown as follows: By Lemma 1.3.1 (or, more precisely, the analogue
of Lemma 1.3.1 where $r$ is multiplied from the left rather than from the
right\footnote{This analogue follows from the original Lemma 1.3.1 by applying
it to $v^{-1}$ and $w^{-1}$ instead of $v$ and $w$.}), we have $rx\leq w$ or
$rx\leq rw$ (since $x\leq w$). In both cases, we obtain $rx\leq w$ (since
$rw<w$).

\item \textbf{page 44, (4.4.1):} Replace \textquotedblleft if $r\notin%
\mathcal{L}\left(  z\right)  $\textquotedblright\ by \textquotedblleft if
$r\notin\mathcal{L}\left(  w\right)  $\textquotedblright. On the other hand,
under the summation sign, replace \textquotedblleft$r\in\mathcal{L}\left(
w\right)  $\textquotedblright\ by \textquotedblleft$r\in\mathcal{L}\left(
z\right)  $\textquotedblright.

\item \textbf{page 45:} \textquotedblleft coeffcient\textquotedblright%
\ $\rightarrow$ \textquotedblleft coefficient\textquotedblright.

\item \textbf{page 45, Proposition 4.5.1 and the following paragraph:} Each of
the four chains of the form \textquotedblleft$x=x_{1}-x_{w}-\ldots-x_{m}%
=y$\textquotedblright\ should instead be \textquotedblleft$x=x_{1}%
-x_{2}-\ldots-x_{m}=y$\textquotedblright.

\item \textbf{page 46, Proposition 4.5.3:} Replace both \textquotedblleft%
$\supset$\textquotedblright\ signs by \textquotedblleft$\supseteq
$\textquotedblright\ signs (in order to be consistent with the rest of the
text). Same in the proof.

\item \textbf{page 46, Proposition 4.5.3:} The second part labelled
\textquotedblleft(i)\textquotedblright\ should be \textquotedblleft%
(ii)\textquotedblright.

\item \textbf{page 46, Note 3:} \textquotedblleft of part (ii) of Lemma
4.3.2\textquotedblright\ should be \textquotedblleft of part (iii) of Lemma
4.3.2\textquotedblright.

\item \textbf{page 48, first paragraph:} \textquotedblleft
non-trival\textquotedblright\ $\rightarrow$ \textquotedblleft
non-trivial\textquotedblright.

\item \textbf{page 48, \S 5.1:} The notations in the matrix ring example
(after the definition of a cellular algebra) are confusing: It is not a good
idea to denote the matrix ring by $R$, since it plays the role of $H$ (not of
$R$) in the definition of a cellular algebra. It is also not good to use $k$
for the base field, since $k$ appears as an index in some sums. I would denote
the base ring by $R$ (this fits with the definition of a cellular algebra) and
use $R^{n\times n}$ for the matrix rings.

(The same issue reappears on page 49 and perhaps later.)

\item \textbf{page 49, second-to-last displayed equation on the page:} Replace
\textquotedblleft$r_{a}\left(  P,P^{\prime}\right)  $\textquotedblright\ by
\textquotedblleft$r_{a}\left(  P^{\prime},P\right)  $\textquotedblright. Make
this same change on the line below (in the sentence \textquotedblleft Now,
axiom (C2) ...\textquotedblright).

\item \textbf{page 49:} \textquotedblleft the the left
module\textquotedblright\ $\rightarrow$ \textquotedblleft the left
module\textquotedblright.

\item \textbf{page 50, \S 5.2:} \textquotedblleft it to show\textquotedblright%
\ $\rightarrow$ \textquotedblleft is to show\textquotedblright.

\item \textbf{page 50, proof of Lemma 5.2.1:} \textquotedblleft Hence
$w^{0}s_{i}s_{i+1}<w^{0}s_{i}s_{i+1}s_{i}$\textquotedblright\ should be
\textquotedblleft Hence $w^{0}s_{i+1}s_{i}\leq w^{0}s_{i}s_{i+1}s_{i}%
$\textquotedblright.

\item \textbf{page 51:} After \textquotedblleft On the other
hand,\textquotedblright\ and before \textquotedblleft$\mathcal{R}\cap\left\{
s_{i},s_{i+1}\right\}  $ contains\textquotedblright, add \textquotedblleft
if\textquotedblright.

\item \textbf{page 52, proof of Proposition 5.2.3:} \textquotedblleft and
$\ldots zxy\ldots$ to $\ldots xzy\ldots$\textquotedblright\ $\rightarrow$
\textquotedblleft and $\ldots yzx\ldots$ to $\ldots yxz\ldots$%
\textquotedblright\ (otherwise you are listing the same transformation twice).

\item \textbf{page 52, Corollary 5.2.4:} A comma is missing in
\textquotedblleft$i_{1},i_{2},\ldots i_{m}$\textquotedblright.

\item \textbf{page 52, proof of Corollary 5.2.4:} Replace \textquotedblleft%
$K_{i_{k_{1}}}$\textquotedblright\ by \textquotedblleft$K_{i_{1}}%
$\textquotedblright\ in the last sentence of the proof.

\item \textbf{page 53:} \textquotedblleft transormations\textquotedblright%
\ $\rightarrow$ \textquotedblleft transformations\textquotedblright. Also,
\textquotedblleft tranformation\textquotedblright\ $\rightarrow$
\textquotedblleft transformation\textquotedblright.

\item \textbf{page 54, proof of Proposition 5.3.1:} \textquotedblleft and
$yr>r$\textquotedblright\ should be \textquotedblleft and $yr>y$%
\textquotedblright. On the same line, \textquotedblleft$yt<t$%
\textquotedblright\ should be \textquotedblleft$yt<y$\textquotedblright.

\item \textbf{page 55, proof of Proposition 5.3.1:} Replace \textquotedblleft%
$P_{x,w}=P_{xr,wr}$\textquotedblright\ by \textquotedblleft$P_{x,y}=P_{xr,yr}%
$\textquotedblright. In the same sentence, \textquotedblleft$\mu\left(
x,w\right)  =\mu\left(  xr,wr\right)  $\textquotedblright\ should be
\textquotedblleft$\mu\left(  x,y\right)  =\mu\left(  xr,yr\right)
$\textquotedblright.

\item \textbf{page 55, proof of Proposition 5.3.1:} In the degree estimates
following $x\not \prec z$, replace \textquotedblleft$\deg q_{z}^{\tfrac{1}{2}%
}q_{y}^{\tfrac{1}{2}}P_{x,y}$\textquotedblright\ by \textquotedblleft$\deg
q_{z}^{-\tfrac{1}{2}}q_{y}^{\tfrac{1}{2}}P_{x,z}$\textquotedblright.

\item \textbf{page 55, proof of Proposition 5.3.1:} In Case 2, replace
\textquotedblleft$yt<x$\textquotedblright\ by \textquotedblleft$yt<y$%
\textquotedblright.

\item \textbf{page 55, proof of Proposition 5.3.1:} In Case 2, replace
\textquotedblleft Hence either $xr<x$ and $yt>y$ or $xr<x$ and $yt>y$%
\textquotedblright\ by \textquotedblleft Hence either $xr<x$ and $yt>y$ or
$xr>x$ and $yt<y$\textquotedblright.

\item \textbf{page 55, proof of Proposition 5.3.1:} At the beginning of Case
2, replace \textquotedblleft$K_{i}\left(  y\right)  =xt$\textquotedblright%
\ should be \textquotedblleft$K_{i}\left(  y\right)  =yt$\textquotedblright.

\item \textbf{page 56, proof of Proposition 5.3.1:} \textquotedblleft and so
$\mu\left(  K_{i}\left(  x\right)  ),K_{i}\left(  y\right)  \right)
$\textquotedblright\ has a closing parenthesis too much.

\item \textbf{page 56, proof of Proposition 5.3.1:} \textquotedblleft neither
$x$ not $yr$\textquotedblright\ $\rightarrow$ \textquotedblleft neither $x$
nor $yr$\textquotedblright.

\item \textbf{page 57, proof of Corollary 5.3.2:} \textquotedblleft%
$\mathcal{L}\left(  K_{i}\left(  x_{j-1}\right)  \right)  \not \subseteq
\mathcal{L}\left(  K_{i}\left(  x_{j}\right)  \right)  $ for all
$i<m$\textquotedblright\ should be \textquotedblleft$\mathcal{L}\left(
K_{i}\left(  x_{j-1}\right)  \right)  \not \subseteq \mathcal{L}\left(
K_{i}\left(  x_{j}\right)  \right)  $ for all $j>1$\textquotedblright\ or,
equivalently (but more consistent with the rest of the proof),
\textquotedblleft$\mathcal{L}\left(  K_{i}\left(  x_{j}\right)  \right)
\not \subseteq \mathcal{L}\left(  K_{i}\left(  x_{j+1}\right)  \right)  $ for
all $j<m$\textquotedblright.

\item \textbf{page 57, \S 5.4:} \textquotedblleft the descent of a
tableau\textquotedblright\ $\rightarrow$ \textquotedblleft the descent set of
a tableau\textquotedblright.

\item \textbf{page 57, proof of Theorem 5.4.1:} In the sentence
\textquotedblleft Hence $K_{i}\left(  y\right)  $ is a well defined element
and $K_{i}\left(  x\right)  \underset{L}{\sim}K_{i}\left(  y\right)  $ by
Corollary 5.3.2\textquotedblright, replace each of the three \textquotedblleft%
$i$\textquotedblright s by \textquotedblleft$i_{1}$\textquotedblright.

\item \textbf{page 58, Corollary 5.4.3:} A comma is missing in
\textquotedblleft$i_{1},i_{2},\ldots i_{m}$\textquotedblright.

\item \textbf{page 60, proof of Proposition 5.5.2:} \textquotedblleft by the
shape\textquotedblright\ $\rightarrow$ \textquotedblleft be the
shape\textquotedblright.

\item \textbf{Page 60, end of the proof of Proposition 5.5.2.} This is
reported by GPT-5.5; I have not verified it myself:

\textquotedblleft The appeal to Corollary 5.4.3 does not by itself justify the
claim, since that corollary only asserts the existence of a suitable sequence
of change-of-label maps; it does not show that the particular inverse sequence
occurring here preserves equality of the $Q$-symbols.

A direct repair is: \textquotedblleft Since $\widehat{x}\sim_{L}\widehat{y}$,
repeated application of Corollary 5.3.2 along the inverse sequence
\[
K_{i_{m}},K_{i_{m-1}},\ldots,K_{i_{1}}%
\]
shows that $x\sim_{L}y$. Hence $Q(x)=Q(y)$ by Theorem 5.4.1.\textquotedblright%
\textquotedblright

\item \textbf{page 61, (5.6.1):} The \textquotedblleft$r\in\mathcal{L}\left(
z\right)  $\textquotedblright\ under the summation sign should say
\textquotedblleft$s_{i}\in\mathcal{L}\left(  z\right)  $\textquotedblright.
(Also, you can replace \textquotedblleft the multiplication formula given in
Section 4.4\textquotedblright\ by the more precise reference \textquotedblleft%
(4.4.3)\textquotedblright.)

\item \textbf{page 61, (5.6.2):} The \textquotedblleft$s_{i}\in\mathcal{L}%
\left(  w\right)  $\textquotedblright\ under the summation sign should say
\textquotedblleft$s_{i}\in\mathcal{L}\left(  z\right)  $\textquotedblright.

\item \textbf{page 61, between the last two displays:} \textquotedblleft%
$\mu\left(  \widehat{z},\widehat{x}\right)  =\mu\left(  z,w\right)
$\textquotedblright\ should be \textquotedblleft$\mu\left(  \widehat{z}%
,\widehat{w}\right)  =\mu\left(  z,w\right)  $\textquotedblright.

\item \textbf{page 62:} In the equality (5.6.4) and the next displayed
equality after it, the \textquotedblleft$C_{P^{\prime},P}^{\lambda}%
$\textquotedblright\ on the right-hand side should be \textquotedblleft%
$C_{P^{\prime},Q}^{\lambda}$\textquotedblright.

\item \textbf{page 62, Note 6:} \textquotedblleft
Garsian--McLarnan\textquotedblright\ should be \textquotedblleft
Garsia--McLarnan\textquotedblright.

\item \textbf{page 65:} \textquotedblleft Lusztig [19]\textquotedblright\ is
probably a mis-reference ([19] is by Knuth, not by Lusztig, and certainly is
not about the basis theorem).

\item \textbf{page 65:} \textquotedblleft use of Lemma 3.3.2\textquotedblright%
\ $\rightarrow$ \textquotedblleft use of Proposition 3.3.2\textquotedblright.

\item \textbf{page 65:} In the displayed equation between (A.1.2) and (A.1.3),
the sum $\sum\limits_{y<w}R_{y,w}T_{y}$ needs to be multiplied by
$q_{w}^{\tfrac{1}{2}}$. The same applies to the similar sum on the right-hand
side of (A.1.3).

\item \textbf{page 65:} In (A.1.3), the \textquotedblleft$\widetilde{T}_{w}%
$\textquotedblright\ inside the sum should be a \textquotedblleft%
$\widetilde{T}_{y}$\textquotedblright.

\item \textbf{page 65:} In \textquotedblleft there is a unique element
$C_{w}=\sum_{y\leq w}\epsilon_{y}\epsilon_{w}q_{w}^{\tfrac{1}{2}}%
q_{y}^{-\tfrac{1}{2}}\overline{P}_{y,w}\widetilde{T}_{w}$\textquotedblright,
the \textquotedblleft$\overline{P}_{y,w}$\textquotedblright\ should be a
\textquotedblleft$\overline{P_{y,w}}$\textquotedblright\ (the line should
cover the whole term), whereas the \textquotedblleft$\widetilde{T}_{w}%
$\textquotedblright\ should be \textquotedblleft$\widetilde{T}_{y}%
$\textquotedblright. Moreover, \textquotedblleft element\textquotedblright%
\ should be \textquotedblleft$\iota$-invariant element\textquotedblright.

Actually, the \textquotedblleft$\overline{P}_{y,w}$\textquotedblright$\mapsto
$\textquotedblleft$\overline{P_{y,w}}$\textquotedblright\ (and likewise for
$P_{x,w}$) change should be done throughout \S A.1.

\item \textbf{page 66, immediately after (A.1.4):} GPT-5.5 suggests to replace
\textquotedblleft$\widetilde{T}_{yr}+q^{1/2}\widetilde{T}_{y}$%
\textquotedblright\ by \textquotedblleft$\widetilde{T}_{yr}-q^{1/2}%
\widetilde{T}_{y}$\textquotedblright\ (thus the two alternatives should be
\[
\widetilde{T}_{yr}-q^{1/2}\widetilde{T}_{y}\qquad\text{or}\qquad
\widetilde{T}_{yr}-q^{-1/2}\widetilde{T}_{y}%
\]
). I have not checked this myself.

\item \textbf{page 66, formula (A.1.6):} GPT-5.5 suggests to replace
\textquotedblleft$h_{y}T_{y}-h_{y}(0)C_{y}$\textquotedblright\ by
\textquotedblleft$h_{y}\widetilde{T}_{y}-h_{y}(0)C_{y}$\textquotedblright. I
have not checked this myself.

\item \textbf{page 67:} \textquotedblleft the outside corner
(2,3)\textquotedblright\ $\rightarrow$ \textquotedblleft the inside corner
$\left(  2,3\right)  $\textquotedblright.

\item \textbf{page 67:} \textquotedblleft from the first row to the
third\textquotedblright\ $\rightarrow$ \textquotedblleft from the first column
to the third\textquotedblright.

Likewise, \textquotedblleft second two rows\textquotedblright\ $\rightarrow$
\textquotedblleft second two columns\textquotedblright.

This said, it is not quite clear what \textquotedblleft this
procedure\textquotedblright\ means (i.e., how it can be generalized from this example).

\item \textbf{page 67:} In the table, replace \textquotedblleft$\mathcal{R}%
\left(  x_{1}\right)  =\left\{  s_{1},s_{4}\right\}  $\textquotedblright\ by
\textquotedblleft$\mathcal{R}\left(  x_{1}\right)  =\left\{  s_{1},s_{2}%
,s_{4}\right\}  $\textquotedblright\ in the first row, replace
\textquotedblleft$\mathcal{R}\left(  x_{2}\right)  =\left\{  s_{1},s_{2}%
,s_{4}\right\}  $\textquotedblright\ by \textquotedblleft$\mathcal{R}\left(
x_{2}\right)  =\left\{  s_{2},s_{4}\right\}  $\textquotedblright, replace
\textquotedblleft$\mathcal{R}\left(  x_{3}\right)  $\textquotedblright\ by
\textquotedblleft$\mathcal{R}\left(  x_{4}\right)  $\textquotedblright\ in the
second-to-last row, and replace \textquotedblleft$\mathcal{R}\left(
x_{4}\right)  $\textquotedblright\ by \textquotedblleft$\mathcal{R}\left(
x_{5}\right)  $\textquotedblright\ in the last row.

\item \textbf{page 67:} \textquotedblleft For $1<i\leq4$ we
have\textquotedblright\ $\rightarrow$ \textquotedblleft For $1<i\leq5$ we
have\textquotedblright.

\item \textbf{pages 67--68: use of Lemma 4.3.2 (i):} The proof uses right
multiplication: $x_{i+1}=x_{i}s_{i}$ (not $s_{i}x_{i}$) on page 67, and
$w_{i+1}=w_{i}s_{i}$ (not $s_{i}w_{i}$) on page 68. Thus, it is not Lemma
4.3.2 (i) being used here, but its right-hand version. (This was reported by
GPT-5.5 and has not been checked.)

\item \textbf{page 67, proof of Proposition A.2.1:} After \textquotedblleft at
the end of the $k^{th}$ column\textquotedblright, I would add
\textquotedblleft(which was possibly empty in $\lambda$)\textquotedblright.

\item \textbf{page 68:} The definition of the permutation $w$ has some
mistakes; as it stands, it produces $w=610435$ when applied to the example
given before the proof.

GPT-5.5 suggests replacing it by the following definition:%
\[
w\left(  i\right)  =%
\begin{cases}
c_{k}^{\prime}, & \text{if }i=c_{j-1}^{\prime}+1;\\
w_{\lambda}\left(  i\right)  -1, & \text{if }c_{j}^{\prime}<i\leq
c_{k}^{\prime};\\
w_{\lambda}\left(  i\right)  , & \text{otherwise.}%
\end{cases}
\]
That is, $w=\zeta^{-1}w_{\lambda}$, where $\zeta$ is the cycle $\left(
c_{j}^{\prime},c_{j}^{\prime}+1,c_{j}^{\prime}+2,\ldots,c_{k}^{\prime}\right)
$. GPT-5.5 also suggests that the $q$ should be defined by $q=c_{k-1}^{\prime
}$ rather than $q=c_{k-1}^{\prime}-1$.

(This said, I have not checked the rest of the proof.)

\item \textbf{page 68, proof of Proposition A.2.1:} GPT-5.5 claims that the
sentence proving the descent condition is malformed. It can be replaced by:

For every $p\leq i\leq q$, the entry $c_{k}^{\prime}$ occupies position $i$ in
$w_{i}$. Hence, for $p\leq i<q$, we have
\[
w_{i}(i)=c_{k}^{\prime}>w_{i}(i+1),\ \ \ \ \ \ \ \ \ \ \text{whereas}%
\ \ \ \ \ \ \ \ \ \ w_{i+1}(i)<w_{i+1}(i+1)=c_{k}^{\prime}.
\]
Therefore
\[
s_{i}\in\mathcal{R}(w_{i})\setminus\mathcal{R}(w_{i+1}).
\]


This also removes the phrase \textquotedblleft which is the greater than or
equal to\textquotedblright.

\item \textbf{page 68, proof of Proposition A.2.1:} The notation $Shape$ is
defined for tableaux, not for permutations, but in this proof it is used for
permutations. Of course, the intended meaning is clear: If $w$ is a
permutation, then $Shape\left(  w\right)  $ means $Shape\left(  P\left(
w\right)  \right)  =Shape\left(  Q\left(  w\right)  \right)  $.

\item \textbf{page 68, proof of Proposition A.2.1:} If $\lambda=\lambda
_{1},\lambda_{2},\ldots,\lambda_{m}=\mu$, then there are only $m-1$ raising
operations. Hence the pairs $x_{i}\leq_{R}y_{i}$ should be indexed by $1\leq
i<m$, and the chain should be
\[
x_{1}\leq_{R}y_{1}\sim_{LR}x_{2}\leq_{R}y_{2}\sim_{LR}\cdots\sim_{LR}%
x_{m-1}\leq_{R}y_{m-1}.
\]
There should be no pair $x_{m}\leq_{R}y_{m}$.

\item \textbf{page 68, proof of Proposition A.2.1:} Equal Robinson--Schensted
shape implies two-sided cell equivalence, not necessarily left-cell
equivalence. Thus replace \textquotedblleft$x\sim_{L}x_{1}$\textquotedblright%
\ by \textquotedblleft$x\sim_{LR}x_{1}$\textquotedblright. With the corrected
indexing (see the previous correction), the conclusion becomes
\[
x\sim_{LR}x_{1}\leq_{LR}y_{m-1}\sim_{LR}y,
\]
which is still sufficient to prove $x\leq_{LR}y$.
\end{enumerate}


\end{document}