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\ihead{Errata to ``Polynomial representations of $\operatorname{GL}_n$''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Polynomial Representations of $\operatorname{GL}_{n}$}

\textit{J. A. Green}

\href{https://doi.org/10.1007/3-540-46944-3}{2nd edition, Springer 2007}

\textbf{Errata and addenda by Darij Grinberg}
\end{center}

\setcounter{section}{13}

\section{Corrections}

This is a list of corrections to \textquotedblleft Polynomial Representations
of $\operatorname*{GL}\nolimits_{n}$\textquotedblright\ by J. A. Green (2nd
edition 2007). Most of it concerns the Appendix on Schensted correspondence
and Littelmann paths by K. Erdmann, J.A. Green and M. Schocker
(\href{https://link.springer.com/content/pdf/bbm:978-3-540-46959-9/1}{freely
available from SpringerLink}). I have read almost everything in this Appendix
up until the end of (D.7). After the list of corrections, I give an
alternative proof of the Littlewood--Richardson rule that avoids any
representation theory (see Section \ref{sec.LRproof} below).

\subsection{Corrections to \S A}

\begin{enumerate}
\item \textbf{page 74, (A.2b):} It is worth mentioning that what is called a
\textquotedblleft standard\textquotedblright\ tableau here is often called a
\textquotedblleft semistandard\textquotedblright\ tableau in more
combinatorial-minded sources.

\item \textbf{page 75, \S A.3:} Add \textquotedblleft Let $i=i_{1}i_{2}\ldots
i_{r}\in I\left(  n,r\right)  $ be a word.\textquotedblright\ after
\textquotedblleft we need some preliminary definitions\textquotedblright.

\item \textbf{page 75, (A.3a):} The equality $h_{c}^{i}\left(  t\right)
=\omega\left(  i_{1}\right)  +\cdots+\omega\left(  i_{t}\right)  $ is actually
true for all $t\in\left\{  0,1,\ldots,r\right\}  $, including $t=0$.

\item \textbf{page 75, (A.3b):} Again, this is actually true for all
$t\in\left\{  0,1,\ldots,r\right\}  $, including $t=0$.

\item \textbf{page 76, (A.3e):} It is worth saying that $M^{i}=0$ is
equivalent to $q=0$ (since the least $t\in\left\{  0,1,\ldots,r\right\}  $
satisfying $h_{c}^{i}\left(  t\right)  =0$ is clearly $0$).

\item \textbf{page 76, (A.3f):} It is worth saying that $M^{i}=h_{c}%
^{i}\left(  r\right)  $ is equivalent to $\overline{q}=r$ (since the greatest
$t\in\left\{  0,1,\ldots,r\right\}  $ satisfying $h_{c}^{i}\left(  t\right)
=h_{c}^{i}\left(  r\right)  $ is clearly $r$). Hence, if we don't have
$M^{i}=h_{c}^{i}\left(  r\right)  $, then we don't have $\overline{q}=r$, and
thus we have $\overline{q}<r$, so that speaking of $s_{\overline{q}+1}$ makes sense.

\item \textbf{page 76, (A.3):} There is an urgent need for examples here! Here
is one:

Let $n=5$ and $c=2$ and $r=10$ and $i=4232133253$. Then, the height function
$h_{c}^{i}=h_{2}^{i}$ sends the numbers $0,1,\ldots,10$ to
$0,0,1,0,1,1,0,-1,0,0,-1$, respectively. Hence, $M=M_{c}^{i}=1$ and
$q=q_{c}^{i}=2$ and $\overline{q}=\overline{q}_{c}^{i}=5$. Thus,
\begin{align*}
\widetilde{f}_{c}\left(  i\right)   &  =4\underline{3}%
32133253\ \ \ \ \ \ \ \ \ \ \text{and}\\
\widetilde{e}_{c}\left(  i\right)   &  =423213\underline{3}253,
\end{align*}
where we have underlined the unique entry of $i$ that got changed.

Now let $j=\widetilde{f}_{c}\left(  i\right)  =4332133253$. Then, the height
function $h_{c}^{j}=h_{2}^{j}$ sends the numbers $0,1,\ldots,10$ to
$0,0,-1,-2,-1,-1,-2,-3,-2,-2,-3$, respectively. Hence, $M=M_{c}^{j}=0$ and
$q=q_{c}^{j}=0$ and $\overline{q}=\overline{q}_{c}^{j}=1$. Thus,
\begin{align*}
\widetilde{f}_{c}\left(  j\right)   &  =\infty\text{ (since }M_{c}%
^{j}=0\text{)}\ \ \ \ \ \ \ \ \ \ \text{and}\\
\widetilde{e}_{c}\left(  j\right)   &  =4\underline{2}32133253=i,
\end{align*}
which confirms (A.3g)(5).

\item \textbf{page 76, (A.3g) (6):} It is worth saying that%
\[
h_{c}^{i\mid j}\left(  t\right)  =%
\begin{cases}
h_{c}^{i}\left(  t\right)  , & \text{if }t\leq r;\\
h_{c}^{i}\left(  r\right)  +h_{c}^{i}\left(  t-r\right)  , & \text{if }t\geq r
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for all }t\in\left\{  0,1,\ldots,r+s\right\}
\]
(note that the case $t=r$ is covered by both cases here), and therefore%
\begin{align*}
M_{c}^{i\mid j}  &  =\max\left\{  M_{c}^{i},\ h_{c}^{i}\left(  r\right)
+M_{c}^{j}\right\}  \ \ \ \ \ \ \ \ \ \ \text{and}\\
q_{c}^{i\mid j}  &  =%
\begin{cases}
q_{c}^{i}, & \text{if }M_{c}^{i}\geq h_{c}^{i}\left(  r\right)  +M_{c}^{j};\\
h_{c}^{i}\left(  r\right)  +q_{c}^{j}, & \text{if }M_{c}^{i}<h_{c}^{i}\left(
r\right)  +M_{c}^{j}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\overline{q}_{c}^{i\mid j}  &  =%
\begin{cases}
\overline{q}_{c}^{i}, & \text{if }M_{c}^{i}>h_{c}^{i}\left(  r\right)
+M_{c}^{j};\\
h_{c}^{i}\left(  r\right)  +\overline{q}_{c}^{j}, & \text{if }M_{c}^{i}\leq
h_{c}^{i}\left(  r\right)  +M_{c}^{j}.
\end{cases}
\end{align*}
These facts are easy to verify, and are the reason why the formulas for
$\widetilde{f}_{c}\left(  i\mid j\right)  $ and $\widetilde{e}_{c}\left(
i\mid j\right)  $ hold.

\item \textbf{page 79, Definition:} Replace \textquotedblleft$y_{m,\lambda
_{m-1}}$\textquotedblright\ by \textquotedblleft$y_{m-1,\lambda_{m-1}}%
$\textquotedblright.

\item \textbf{page 79:} \textquotedblleft proved in \S C.4\textquotedblright%
\ $\rightarrow$ \textquotedblleft proved in \S C.6\textquotedblright.
\end{enumerate}

\subsection{Corrections to \S B}

\begin{enumerate}
\item \textbf{page 80, \S B.1:} What is called a \textquotedblleft%
\textit{standard}\textquotedblright\ tableau here is usually called a
\textquotedblleft\textit{semistandard}\textquotedblright\ tableau in texts on combinatorics.

\item \textbf{page 83, Step 1:} Replace \textquotedblleft($<x_{1,\mu\left(
1\right)  +1}=\infty$)\textquotedblright\ by \textquotedblleft($<u_{1,\mu
\left(  1\right)  +1}=\infty$)\textquotedblright.

\item \textbf{page 83:} It should be said that the notation $\mu_{a}$ is used
as synonym for $\mu\left(  a\right)  $.

\item \textbf{page 84, Definition of z:} It should be explained why $z\leq n$
(otherwise, the vector $\varepsilon_{z}$ in (B.3d) is undefined; and earlier
even, the $k\left(  a\right)  $ in the inductive step would be undefined if
$a=z+1$).

There is an easy explanation: We WLOG assume that $z>1$ (since the claim is
obvious for $z=1$). Then, $x_{z}=u_{z-1,k\left(  z-1\right)  }$ is an entry of
the tableau $u$, and hence belongs to $\underline{n}$. Thus, $x_{z}\leq n$.

We have $1\leq x_{1}<x_{2}<\cdots<x_{z}$ by the definition of the $x_{a}$.
Thus, $x_{a}\geq a$ for each $a$ (by induction on $a$). In particular,
$x_{z}\geq z$, so that $z\leq x_{z}\leq n$, qed.

\item \textbf{page 84, (B.3d):} \textquotedblleft We shall show in
(B.5b)\textquotedblright\ $\rightarrow$ \textquotedblleft We shall show in
(B.5a)\textquotedblright.

\item \textbf{page 85:} Replace \textquotedblleft we define $x_{a}%
:=k_{a-1,k\left(  a-1\right)  }$\textquotedblright\ by \textquotedblleft we
define $x_{a}:=u_{a-1,k\left(  a-1\right)  }$\textquotedblright.

\item \textbf{page 85:} Replace \textquotedblleft between the entries
$u_{a,k\left(  a-1\right)  }$\textquotedblright\ by \textquotedblleft between
the entries $u_{a,k\left(  a\right)  -1}$\textquotedblright.

\item \textbf{page 85:} It would be good to introduce another piece of
standard terminology: The sequence of places
\[
\left(  \left(  1,k\left(  1\right)  \right)  ,\ \left(  2,k\left(  2\right)
\right)  ,\ \ldots,\ \left(  z,k\left(  z\right)  \right)  \right)
\]
(these are the places in which $P$ differs from $U$) is called the
\textit{bumping route} (or \textit{bumping path}) of the insertion
$U\leftarrow x_{1}$.

\item \textbf{page 85, (B.4a):} Replace \textquotedblleft$P=P\left(  i\right)
$\textquotedblright\ by just \textquotedblleft$P$\textquotedblright. (There is
no \textquotedblleft$i$\textquotedblright\ involved yet.)

\item \textbf{page 85, (B.4a):} Remove the comma in \textquotedblleft The row
$\left(  z\right)  $ of $P$, is\textquotedblright.

\item \textbf{page 86, (B.4a):} Remove the comma in \textquotedblleft$3$ in
row $\left(  1\right)  $ of $U$, is bumped\textquotedblright.

\item \textbf{page 86, last paragraph:} Remove the comma in \textquotedblleft%
$P\left(  i_{1}i_{2},\cdots i_{r}\right)  $\textquotedblright.

\item \textbf{page 88, proof of (B.5b):} \textquotedblleft then by
(B.5a)\textquotedblright\ $\rightarrow$ \textquotedblleft then by
(B.3d)\textquotedblright.

\item \textbf{page 89, proof of (B.5b):} It took me a while to understand how
\textquotedblleft$u_{a,k\left(  a+1\right)  }\leq x_{a}$\textquotedblright\ is
obtained here. Namely: Combining $k\left(  a+1\right)  \leq k\left(  a\right)
$ with $k\left(  a+1\right)  =h\neq k\left(  a\right)  $, we obtain $k\left(
a+1\right)  <k\left(  a\right)  $, hence $k\left(  a+1\right)  \leq k\left(
a\right)  -1$. By the row-standardness of $U$, we thus have $u_{a,k\left(
a+1\right)  }\leq u_{a,k\left(  a\right)  -1}\leq x_{a}$ by (B.3c).

\item \textbf{page 90, Step 1 of the construction of the extrusion sequence:}
I would replace \textquotedblleft$w_{z}:=p_{z,\lambda_{z}}$\textquotedblright%
\ by \textquotedblleft$w_{z}:=p_{z,l\left(  z\right)  }=p_{z,\lambda_{z}}%
$\textquotedblright, in order to make it clear where the \textquotedblleft%
$w_{a+1}:=p_{a+1,l\left(  a+1\right)  }$\textquotedblright\ in the Inductive
step comes from.

\item \textbf{page 90, Step 1 of the construction of the extrusion sequence:}
Remove the comma in \textquotedblleft the entry in $P$, at the
place\textquotedblright.

\item \textbf{page 90, Step 2 of the construction of the extrusion sequence:}
Worth reminding the reader that $p_{k,\lambda_{k}+1}$ is understood to be
$\infty$, and that $p_{k,0}$ is understood to be $0$ for each $k$.

\item \textbf{page 91, proof of (B.6h):} The appeal to analogy in the last
sentence of this proof is a bit of a stretch, so let me explain in some detail
how the standardness of $U$ is proved:

First, we show that%
\begin{equation}
l\left(  1\right)  \geq l\left(  2\right)  \geq\cdots\geq l\left(  z\right)  .
\label{p91.1}%
\end{equation}
This is an analogue of (B.3f), and is proved rather similarly:

[\textit{Proof of (\ref{p91.1}):} Let $a\in\left\{  1,2,\ldots,z-1\right\}  $.
We must prove that $l\left(  a\right)  \geq l\left(  a+1\right)  $. Since $P$
is column-standard, we have $p_{a,l\left(  a+1\right)  }<p_{a+1,l\left(
a+1\right)  }=w_{a+1}$ (by the definition of $w_{a+1}$). But $l\left(
a\right)  $ is the largest $l\in\left\{  1,2,\ldots,\lambda_{a}\right\}  $
such that $p_{a,l}<w_{a+1}$. Hence, from $p_{a,l\left(  a+1\right)  }<w_{a+1}%
$, we obtain $l\left(  a\right)  \geq l\left(  a+1\right)  $. Thus,
(\ref{p91.1}) is proved.]

Next, we observe that
\begin{equation}
w_{1}<w_{2}<\cdots<w_{z}, \label{p91.2}%
\end{equation}
because each $a\in\left\{  1,2,\ldots,z-1\right\}  $ satisfies $w_{a}%
=p_{a,l\left(  a\right)  }<w_{a+1}$ (by (B.6f)).

Now, we can show that $U$ is standard:

First, we shall show that $U$ is row-standard, i.e., that $u_{a,h-1}\leq
u_{a,h}$ for all adjacent pairs $\left(  a,h-1\right)  ,\ \left(  a,h\right)
$ of places in any row $\left(  a\right)  $ of $\left[  \mu\right]  $.

[\textit{Proof:} This is very similar to the proof of (i) in the proof of
(B.5b). Let $\left(  a,h-1\right)  $ and $\left(  a,h\right)  $ be two
adjacent places in any row of $\left[  \mu\right]  $. Since $P$ is
row-standard, we have $p_{a,h-1}\leq p_{a,h}$. But the only entry in row $a$
that differs between $U$ and $P$ is $u_{a,l\left(  a\right)  }\neq
p_{a,l\left(  a\right)  }$. Thus, our claim $u_{a,h-1}\leq u_{a,h}$ follows
from $p_{a,h-1}\leq p_{a,h}$, unless $l\left(  a\right)  $ is one of $h-1$ and
$h$. So it remains to consider these two cases $l\left(  a\right)  =h-1$ and
$l\left(  a\right)  =h$. If $l\left(  a\right)  =h-1$, then we must prove that
$u_{a,l\left(  a\right)  }\leq u_{a,l\left(  a\right)  +1}$; but this follows
from $u_{a,l\left(  a\right)  }=w_{a+1}\leq p_{a,l\left(  a\right)  +1}$ (by
(B.6f)) and $p_{a,l\left(  a\right)  +1}=u_{a,l\left(  a\right)  +1}$. If
$l\left(  a\right)  =h$, then we must prove that $u_{a,l\left(  a\right)
-1}\leq u_{a,l\left(  a\right)  }$; but this follows from $u_{a,l\left(
a\right)  -1}=p_{a,l\left(  a\right)  -1}\leq p_{a,l\left(  a\right)
}<w_{a+1}$ (by (B.6f)) and $w_{a+1}=u_{a,l\left(  a\right)  }$. In either
case, $u_{a,h-1}\leq u_{a,h}$ is proved.]

Thus, we have shown that $U$ is row-standard. It remains to prove that $U$ is
column-standard, i.e., that if $\left(  a-1,h\right)  $ and $\left(
a,h\right)  $ are two adjacent places in the same column of $\left[
\mu\right]  $, then%
\begin{equation}
u_{a-1,h}<u_{a,h}. \label{p91.3}%
\end{equation}


[\textit{Proof:} Let $\left(  a-1,h\right)  $ and $\left(  a,h\right)  $ be
two adjacent places in the same column of $\left[  \mu\right]  $. If $h\neq
l\left(  a\right)  $ and $h\neq l\left(  a-1\right)  $, then $u_{a,h}=p_{a,h}$
and $u_{a-1,h}\neq p_{a-1,h}$, then (\ref{p91.3}) follows from $p_{a-1,h}%
<p_{a,h}$, which holds because $P$ is column-standard.

Now consider the case when $h=l\left(  a\right)  $ and $h\neq l\left(
a-1\right)  $. Then, $l\left(  a\right)  =h\neq l\left(  a-1\right)  $.
Moreover, $l\left(  a-1\right)  \geq l\left(  a\right)  $ (by (\ref{p91.1})),
so that $l\left(  a-1\right)  >l\left(  a\right)  $ (since $l\left(
a-1\right)  \neq l\left(  a\right)  $). Now, from $h=l\left(  a\right)  $, we
obtain $p_{a,h}=p_{a,l\left(  a\right)  }=w_{a}<w_{a+1}$ (by (\ref{p91.2})).
But the column-standardness of $P$ yields $p_{a-1,h}<p_{a,h}$. Since $h\neq
l\left(  a-1\right)  $, we have $u_{a-1,h}=p_{a-1,h}<p_{a,h}<w_{a+1}%
=u_{a,l\left(  a\right)  }=u_{a,h}$ (since $l\left(  a\right)  =h$). Thus,
(\ref{p91.3}) is proved in the case when $h=l\left(  a\right)  $ and $h\neq
l\left(  a+1\right)  $.

Now consider the case when $h\neq l\left(  a\right)  $ and $h=l\left(
a-1\right)  $. Thus, $h=l\left(  a-1\right)  \geq l\left(  a\right)  $ (by
(\ref{p91.1})). Combining this with $h\neq l\left(  a\right)  $, we find
$h>l\left(  a\right)  $. Therefore, $h\geq l\left(  a\right)  +1$, so that
$p_{a,h}\geq p_{a,l\left(  a\right)  +1}$ (since $P$ is row-standard).
Moreover, (B.6f) yields $p_{a,l\left(  a\right)  +1}\geq w_{a+1}>w_{a}$ (by
(\ref{p91.2})). Now, $h\neq l\left(  a\right)  $, so that $u_{a,h}=p_{a,h}\geq
p_{a,l\left(  a\right)  +1}>w_{a}=u_{a-1,l\left(  a-1\right)  }=u_{a-1,h}$
(since $l\left(  a-1\right)  =h$). That is, $u_{a-1,h}<u_{a,h}$. Thus,
(\ref{p91.3}) is proved in the case when $h\neq l\left(  a\right)  $ and
$h=l\left(  a+1\right)  $.

Finally, consider the case when $h=l\left(  a\right)  $ and $h=l\left(
a-1\right)  $. Thus, $u_{a-1,h}=u_{a-1,l\left(  a-1\right)  }=w_{a}<w_{a+1}$
(by (\ref{p91.2})) and $u_{a,h}=u_{a,l\left(  a\right)  }=w_{a+1}$, so that
$u_{a-1,h}<w_{a+1}=u_{a,h}$. Thus, (\ref{p91.3}) is proved in the case when
$h=l\left(  a\right)  $ and $h=l\left(  a+1\right)  $.

Hence, we have proved (\ref{p91.3}) in all four cases.]

We thus have shown that $U$ is column-standard. Since $U$ is also
row-standard, we thus know that $U$ is standard.

\item \textbf{page 91, (B.6j):} Remove the comma at the end of the insertion sequence.

\item \textbf{page 92, proof of (i):} Instead of \textquotedblleft%
$x_{a}=p_{a,k\left(  a\right)  }<w_{a+1}\leq u_{a,k\left(  a\right)
+1}=p_{a,k\left(  a\right)  +1}$\textquotedblright, it would be logically
clearer to write \textquotedblleft$p_{a,k\left(  a\right)  }=x_{a}%
<x_{a+1}=w_{a+1}=x_{a+1}\leq u_{a,k\left(  a\right)  +1}=p_{a,k\left(
a\right)  +1}$\textquotedblright.

\item \textbf{page 92, proof of (i):} Replace \textquotedblleft$\mathsf{E}%
\left(  \mathsf{J}\left(  \left(  \mu,U,V\right)  ,x_{1}\right)  \right)
))$\textquotedblright\ by \textquotedblleft$\mathsf{E}\left(  \mathsf{J}%
\left(  \left(  \mu,U,V\right)  ,x_{1}\right)  \right)  $\textquotedblright.

\item \textbf{page 92, proof of (i):} To prove that $\mathsf{E}\left(  \left(
\lambda,P,Q\right)  \right)  =\left(  \left(  \mu,U,V\right)  ,x_{1}\right)
$, we can proceed as follows: Each $a\in\left\{  1,2,\ldots,z\right\}  $
satisfies
\begin{align*}
u_{a,l\left(  a\right)  }  &  =u_{a,k\left(  a\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }l\left(  a\right)  =k\left(
a\right)  \text{ by (B.6m)}\right) \\
&  =x_{a+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (B.3b)}\right) \\
&  =w_{a+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (B.6m) again}\right)  .
\end{align*}
Thus, when the entries of $P$ in the places $\left(  a,l\left(  a\right)
\right)  $ get replaced by $w_{a+1}$ in the construction of $\mathsf{E}\left(
\left(  \lambda,P,Q\right)  \right)  $, they merely revert to their original
values $u_{a,l\left(  a\right)  }$ that they used to have in $U$. All the
other entries of $P$ were already equal to the respective entries of $U$
(since $l\left(  a\right)  =k\left(  a\right)  $ for all $a$), and remain so.
Hence, the tableau $P$ reverts back to $U$ in the construction of
$\mathsf{E}\left(  \left(  \lambda,P,Q\right)  \right)  $. Moreover, the
dominant weight $\lambda$ reverts back to $\mu$ (since $\lambda$ was
$\mu+\varepsilon_{z}$ and now becomes $\left(  \mu+\varepsilon_{z}\right)
-\varepsilon_{z}=\mu$), and the tableau $Q$ reverts back to $V$ (since it
loses the entry $r$ in cell $\left(  z,\lambda_{z}\right)  $). Hence,
$\mathsf{E}\left(  \left(  \lambda,P,Q\right)  \right)  =\left(  \left(
\mu,U,V\right)  ,x_{1}\right)  $.

\item \textbf{page 92, proof of (ii):} This proof relies on the tacit
assumption that the $z$ at the beginning of the extrusion sequence (B.6k) is
the same as the $z$ at the end of the insertion sequence (B.6o). This is
indeed the case, but it is not obvious from the get-go, so the notation needs
to be changed and some justification added. In more detail:

After \textquotedblleft Let (B.6k) be the extrusion sequence which defines
$\mathsf{E}\left(  \left(  \lambda,P,Q\right)  \right)  =\left(  \left(
\mu,U,V\right)  ,w_{1}\right)  $ (see (B.6g)).\textquotedblright, add
\textquotedblleft Thus, $\mu=\lambda-\varepsilon_{z}$.\textquotedblright.

Replace both \textquotedblleft$z$\textquotedblright s in (B.6o) by
\textquotedblleft$x$\textquotedblright s.

In (B.6p), add \textquotedblleft and $a\leq x$\textquotedblright\ after
\textquotedblleft$w_{a}=x_{a}$\textquotedblright.

After \textquotedblleft holds for some $a$\textquotedblright, add
\textquotedblleft$<z$\textquotedblright.

In the proof of (B.6p), replace \textquotedblleft unique element of $\left\{
1,2,,\ldots,z\right\}  $\textquotedblright\ by \textquotedblleft unique
element of $\left\{  1,2,\ldots,\lambda_{a}\right\}  $\textquotedblright%
\ (this is a separate typo).

After \textquotedblleft However this proves that $l\left(  a\right)  =k\left(
a\right)  $, from (B.3c).\textquotedblright, add: \textquotedblleft Thus,
$k\left(  a\right)  =l\left(  a\right)  \in\left\{  1,2,\ldots,\lambda
_{a}\right\}  =\left\{  1,2,\ldots,\mu_{a}\right\}  $ (because $a\neq z$ and
therefore $\lambda_{a}=\mu_{a}$). This entails $k\left(  a\right)  \neq\mu
_{a}+1$. Thus, by (B.3b), the insertion sequence (B.6o) does not stop at
$k\left(  a\right)  $; in other words, $a<z^{\prime}$. Hence, $a+1\leq
z^{\prime}$.

In the sentence \textquotedblleft Consequently $w_{a+1}=p_{a,l\left(
a\right)  }=p_{a,k\left(  a\right)  }=x_{a+1}$ (see (B.3b); we are here using
the insertion of $x_{1}$ into $\left(  \mu,U,V\right)  $)\textquotedblright,
replace both \textquotedblleft$p$\textquotedblright s by \textquotedblleft%
$u$\textquotedblright s.

Replace \textquotedblleft Now we can prove, by induction on $a$%
\textquotedblright\ by \textquotedblleft Thus, we have proved (B.6p) by
induction on $a$. As a side-result, we have obtained\textquotedblright.

After (B.6r), add \textquotedblleft In particular, this shows that $l\left(
z\right)  =k\left(  z\right)  $, so that $k\left(  z\right)  =l\left(
z\right)  =\lambda_{z}=\mu_{z}+1$. Therefore, by (B.3b), the insertion
sequence (B.6o) stops at $k\left(  z\right)  $, meaning that we have
$z^{\prime}=z$.\textquotedblright.

\item \textbf{page 94:} Remove the colon at the end of the second display of
the page. (This is the display saying \textquotedblleft$\mathsf{J}%
_{r-s}:T\left(  n,r-s-1\right)  \times I\left(  n,s+1\right)  \longrightarrow
T\left(  n,r-s\right)  \times I\left(  n,s\right)  :$\textquotedblright.)
\end{enumerate}

\subsection{Corrections to \S C}

\begin{enumerate}
\item \textbf{page 96, \S C.1:} \textquotedblleft in exactly one place; see
(A.3g)(2)\textquotedblright\ should reference (A.3g)(3) rather than (A.3g)(2).

\item \textbf{page 97, after (C.2b):} It is worth mentioning that the Knuth
unwinding $KY$ is also known as the \textit{reading word} or the \textit{row
word} of $Y$.

\item \textbf{page 97, proof of part (i) of Proposition (C.2c):} In
\textquotedblleft Now suppose that $m>1$ and that Proposition (C.2c)
holds\textquotedblright, replace \textquotedblleft(C.2c)\textquotedblright\ by
\textquotedblleft(C.2c)(i)\textquotedblright. (Part (ii) won't be proved until
a while later.)

\item \textbf{page 98, Diagram C.1:} All the \textquotedblleft$0$%
\textquotedblright s here should be \textquotedblleft$\infty$%
\textquotedblright s. The same applies to Diagram C.2.

\item \textbf{page 99:} \textquotedblleft given the element $x_{a}=y_{a,t}%
$\textquotedblright\ should just be \textquotedblleft given the element
$x_{a}$\textquotedblright, since the equality $x_{a}=y_{a,t}$ is not obvious
at this point.

The right time to state $x_{a}=y_{a,t}$ is after proving (2). Indeed, once (2)
has been proved, we can show that%
\begin{equation}
x_{a}=y_{a,t}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ k\left(
a\right)  =t\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ z\geq a
\label{p99.1}%
\end{equation}
for each $a\in\left\{  1,2,\ldots,\beta_{t}\right\}  $ (where $z$ is as in
(B.3c)). This is proved by induction on $a$: For $a=1$, it is clear (at least
$x_{1}=y_{1,t}$ and $z\geq1$ are clear; but $k\left(  1\right)  =t$ follows
from $y_{1,t-1}\leq y_{1,t}=x_{1}<y_{2,t}$). Now fix an $a\in\left\{
2,3,\ldots,\beta_{t}\right\}  $ and assume as induction hypothesis that
(\ref{p91.1}) holds for $a-1$ instead of $a$; in other words, assume that%
\[
x_{a-1}=y_{a-1,t}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ k\left(
a-1\right)  =t\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ z\geq a-1.
\]
We must prove that (\ref{p91.1}) holds for $a$; that is, we must prove that
\[
x_{a}=y_{a,t}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ k\left(
a\right)  =t\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ z\geq a.
\]
For this purpose, we note that the definition of insertion yields
$x_{a}=u_{a-1,k\left(  a-1\right)  }=u_{a-1,t}$ (since $k\left(  a-1\right)
=t$). But we have $U=X\left[  t-1\right]  $, so that $u_{a-1,t}=y_{a,t}$ and
$u_{a,t-1}=y_{a,t-1}$ and $u_{a,t}=y_{a+1,t}$. Hence, $x_{a}=u_{a-1,t}%
=y_{a,t}$. Furthermore, the equation (2) on page 99 can be rewritten as
$u_{a,t-1}\leq x_{a}<u_{a,t}$ (since $u_{a,t-1}=y_{a,t-1}$ and $u_{a,t}%
=y_{a+1,t}$ and $x_{a}=y_{a,t}$). But $k\left(  a\right)  $ is defined to be
the unique element of $\left\{  1,2,\ldots,\mu_{a}+1\right\}  $ such that
$u_{a,k\left(  a\right)  -1}\leq x_{a}<u_{a,k\left(  a\right)  }$. Comparing
this with $u_{a,t-1}\leq x_{a}<u_{a,t}$, we see that $t$ fits this
description, so that we have $k\left(  a\right)  =t$. It remains to show that
$z\geq a$. To do so, we observe that the $t$-th column of $U$ has $\beta
_{t}-1$ entries. Thus, it has an $\left(  a-1\right)  $-st entry (since
$a\leq\beta_{t}$ and thus $a-1\leq\beta_{t}-1$). In other words, $\mu
_{a-1}\geq t$. But the definition of $z$ yields $k\left(  z\right)  =\mu
_{z}+1$. If we had $z=a-1$, then we could rewrite this as $k\left(
a-1\right)  =\mu_{a-1}+1>\mu_{a-1}\geq t$, which would contradict $k\left(
a-1\right)  =t$. Hence, $z\neq a-1$. Combined with $z\geq a-1$, this leads us
to $z>a-1$ and thus to $z\geq a$. Thus, the induction step is complete, and we
have proved (\ref{p91.1}) for all $a\in\left\{  1,2,\ldots,\beta_{t}\right\}
$.

This confirms all the values claimed in Table C.1. To ensure that this is a
complete analysis of the Schensted insertion $X\left[  t-1\right]  \leftarrow
x_{1}$, we need to show that $z=\beta_{t}$ (meaning that there is no further
bumping after the $\beta_{t}$-th row). But this is easy: The $t$-th column of
$U$ has $\beta_{t}-1$ entries, but the $\left(  t-1\right)  $-st column of $U$
has $\beta_{t-1}\geq\beta_{t}$ entries. Thus, the $\beta_{t}$-th row of $U$
spans the first $t-1$ columns but no further. In other words, $\mu_{\beta_{t}%
}=t-1$. But applying (\ref{p91.1}) to $a=\beta_{t}$, we obtain $k\left(
\beta_{t}\right)  =t=\mu_{\beta_{t}}+1$ (since $\mu_{\beta_{t}}=t-1$). This
shows that $z=\beta_{t}$ (since $z$ is characterized as the first positive
integer for which $k\left(  z\right)  =\mu_{z}+1$).

\item \textbf{page 100, (C2.h):} The tableau $Y$ needs to be standard (i.e.,
semistandard, if we use the combinatorialists' lingo) in order for this to
hold. While this normally goes without saying (semistandardness is assumed by
default according to \S B.1), this might be worth a reminder, since a
non-standard tableau has just been constructed a few lines above.

\item \textbf{page 101, Table C.2:} The tableau $X\leftarrow y_{1,1}$ is not
an $\left(  \lambda^{\ast}+\varepsilon_{1}\right)  $-tableau, as claimed here,
but rather an $\left(  \lambda^{\ast}+\varepsilon_{\beta_{1}}\right)  $-tableau.

\item \textbf{page 101, Table C.2:} Again, it should be explained that the
\textquotedblleft$0$\textquotedblright s actually stand for $\infty$ (or
should just be disregarded completely). The same applies to Diagram C.3.

\item \textbf{page 103, definition of basic moves:} This is all good, but the
rules in (C.3c) and (C.3d) are somewhat hard to memorize. Here is a more
memorable way to restate the definition:

\begin{itemize}
\item Given three letters $x,y,z\in\underline{n}$, we say that the letter $z$
is \textit{numerically intermediate} between $x$ and $y$ if and only if
$\min\left\{  x,y\right\}  \leq z\leq\max\left\{  x,y\right\}  $.

\item A \textit{basic move of type }$K^{\prime}$ is a transformation that
swaps two consecutive letters $x$ and $y$ in a word under the condition that a
letter $z$ numerically intermediate between $x$ and $y$ lies immediately to
the right of $y$, and that we have neither $x\leq y\leq z$ nor $y\leq x\leq z$
(that is, the three letters don't form a weakly increasing subword before or
after the move).

\item A \textit{basic move of type }$K^{\prime\prime}$ is a transformation
that swaps two consecutive letters $x$ and $y$ in a word under the condition
that a letter $z$ numerically intermediate between $x$ and $y$ lies
immediately to the left of $x$, and that we have neither $z\leq x\leq y$ nor
$z\leq y\leq x$ (that is, the three letters don't form a weakly increasing
subword before or after the move).
\end{itemize}

So the strict inequalities $a<b$ in (C.3c) and $b<c$ in (C.3d) are preventing
our three relevant letters from forming a weakly increasing subword before or
after the move.

\item \textbf{page 104, the paragraph above (C3.h):} Remove the comma in
\textquotedblleft of the tableau $U$, is identical to the\textquotedblright.

\item \textbf{pages 104--106, proof of (C3.f) (including all items
(C3.g)--(C3.m)):} This proof tries too hard to micromanage the entries of the
tableau during the transformation. As a result, the proof is hard to follow
and suffers from some minor imprecisions (e.g., Proposition (C3.m) (ii)
requires $z>1$, and the proof of Proposition (C3.m) (i) presumes that
$k\left(  1\right)  <\mu_{1}$, which is not guaranteed).

A cleaner proof (although following essentially the same idea) can be made by
breaking up the Knuth unwindings of $U$ and $P$ row by row (rather than letter
by letter). Here is this proof:\footnote{That said, the notation defined in
(C3.g) is used elsewhere, so it is not worth skipping.}

We let $I\left(  n\right)  $ be the set of all (finite) words with entries in
$\underline{n}$. Thus, $I\left(  n\right)  =\bigsqcup\limits_{r\geq0}I\left(
n,r\right)  $.

On the set $I\left(  n\right)  $, we define a binary relation $\equiv$ by as
follows: Two words $i,j\in I\left(  n\right)  $ shall satisfy $i\equiv j$ if
and only if there is a finite sequence of words $i\left(  1\right)  ,i\left(
2\right)  ,\ldots,i\left(  s\right)  $ such that $i\left(  1\right)  =i$ and
$i\left(  s\right)  =j$ and each consecutive pair of words $i\left(
\sigma-1\right)  ,\ i\left(  \sigma\right)  $ is connected by a basic move of
type $K^{\prime}$ or $K^{\prime\prime}$. In other words, $i\equiv j$ if and
only if the word $i$ can be transformed into $j$ by a sequence of basic moves.

Clearly, the relation $\equiv$ is an equivalence relation (it is symmetric
since the basic moves are symmetric). Moreover, it is \textit{monoidal},
meaning that if two words $i,j\in I\left(  n\right)  $ satisfy $i\equiv j$,
then any further word $k\in I\left(  n\right)  $ satisfies $i\mid k\equiv
j\mid k$ and $k\mid i\equiv k\mid j$. (This is because any basic move that can
be applied to a word $w\in I\left(  n\right)  $ can also be applied to any
appearance of $w$ as a segment in a longer word.) These properties of the
relation $\equiv$ will be used without explicit mention.

We have the following two easy properties of the relation $\equiv$:

\begin{statement}
\textbf{(C3.g') Lemma.} If $b_{1},b_{2},\ldots,b_{s},x\in\underline{n}$ are
letters such that $s\geq1$ and $x<b_{1}\leq b_{2}\leq\cdots\leq b_{s}$, then
$b_{1}b_{2}\ldots b_{s}x\equiv b_{1}xb_{2}b_{3}\ldots b_{s}$.
\end{statement}

\textit{Proof.} We induct on $s$. The base case ($s=1$) is trivial, since it
just claims $b_{1}x\equiv b_{1}x$. For the induction step, we fix $s>1$ and
assume that $b_{1}b_{2}\ldots b_{s-1}x\equiv b_{1}xb_{2}b_{3}\ldots b_{s-1}$
has already been proved. We must now show that $b_{1}b_{2}\ldots b_{s}x\equiv
b_{1}xb_{2}b_{3}\ldots b_{s}$. But we have $x<b_{1}\leq b_{2}\leq\cdots\leq
b_{s}$ and thus $x<b_{s-1}\leq b_{s}$. Hence, we can transform the word
$b_{1}b_{2}\ldots b_{s}x$ into the word $b_{1}b_{2}\ldots b_{s-1}xb_{s}$ by a
basic move of type $K^{\prime}$ (applied to the last three letters). Thus,%
\[
b_{1}b_{2}\ldots b_{s}x\equiv\underbrace{b_{1}b_{2}\ldots b_{s-1}x}_{\equiv
b_{1}xb_{2}b_{3}\ldots b_{s-1}}b_{s}\equiv b_{1}xb_{2}b_{3}\ldots b_{s-1}%
b_{s}=b_{1}xb_{2}b_{3}\ldots b_{s}.
\]
This completes the induction step. Thus, (C3.g') is proved.

\begin{statement}
\textbf{(C3.h') Lemma.} If $b_{1},b_{2},\ldots,b_{s},x\in\underline{n}$ are
letters such that $s\geq1$ and $b_{1}\leq b_{2}\leq\cdots\leq b_{s}<x$, then
$b_{1}b_{2}\ldots b_{s-1}xb_{s}\equiv xb_{1}b_{2}\ldots b_{s}$.
\end{statement}

\textit{Proof.} We induct on $s$. The base case ($s=1$) is trivial, since it
just claims $xb_{1}\equiv xb_{1}$. For the induction step, we fix $s>1$ and
assume that $b_{1}b_{2}\ldots b_{s-2}xb_{s-1}\equiv xb_{1}b_{2}\ldots b_{s-1}$
has already been proved. We must now show that $b_{1}b_{2}\ldots b_{s-1}%
xb_{s}\equiv xb_{1}b_{2}\ldots b_{s}$. But we have $b_{1}\leq b_{2}\leq
\cdots\leq b_{s}<x$ and thus $b_{s-1}\leq b_{s}<x$. Hence, we can transform
the word $b_{1}b_{2}\ldots b_{s-1}xb_{s}$ into the word $b_{1}b_{2}\ldots
b_{s-2}xb_{s-1}b_{s}$ by a basic move of type $K^{\prime\prime}$ (applied to
the last three letters). Thus,%
\[
b_{1}b_{2}\ldots b_{s-1}xb_{s}\equiv\underbrace{b_{1}b_{2}\ldots
b_{s-2}xb_{s-1}}_{\equiv xb_{1}b_{2}\ldots b_{s-1}}b_{s}\equiv xb_{1}%
b_{2}\ldots b_{s-1}b_{s}=xb_{1}b_{2}\ldots b_{s}.
\]
This completes the induction step. Thus, (C3.h') is proved.

Now, we are ready for the \textit{proof of (C3.f):} Let $P=\left(
p_{a,b}\right)  _{\left(  a,b\right)  \in\left[  \lambda\right]  }$ be the
tableau $U\leftarrow x$. Write the tableau $U$ as $U=\left(  u_{a,b}\right)
_{\left(  a,b\right)  \in\left[  \mu\right]  }$. For any $a\in\underline{n}$,
we let $P_{a}$ be the word $p_{a,1}p_{a,2}\ldots p_{a,\lambda_{a}}$ (obtained
by reading the row $a$ of $P$ from left to right), and we let $U_{a}$ be the
word $u_{a,1}u_{a,2}\ldots u_{a,\mu_{a}}$ (obtained by reading the row $a$ of
$U$ from left to right). Note that the word $P_{a}$ is empty if $\lambda
_{a}=0$, and likewise for $U_{a}$. By the definition of Knuth unwinding, we
have $KU=U_{n}\mid U_{n-1}\mid\cdots\mid U_{1}$ and $KP=P_{n}\mid P_{n-1}%
\mid\cdots\mid P_{1}$.

Now, set $x_{1}=x$ and consider the \textquotedblleft
parameters\textquotedblright\ of the insertion $U\leftarrow x_{1}$ as defined
in (B.3b): the number $z$ (which is the number for which $\lambda
=\mu+\varepsilon_{z}$), the numbers $k\left(  1\right)  ,k\left(  2\right)
,\ldots,k\left(  z\right)  $ (which describe in which positions $U$ differs
from $P$) and the letters $x_{1},x_{2},\ldots,x_{z}$ (which are the bumped
entries). We note that the tableau $P=U\leftarrow x_{1}$ differs from $U$ only
in its first $z$ rows (since the bumping ends in the $z$-th row). Thus,
$P_{a}=U_{a}$ for each $a>z$. Hence,
\begin{equation}
P_{n}\mid P_{n-1}\mid\cdots\mid P_{z+1}=U_{n}\mid U_{n-1}\mid\cdots\mid
U_{z+1}. \label{p104.4}%
\end{equation}


Furthermore, row $z$ of $P=U\leftarrow x_{1}$ is just row $z$ of $U$ with a
new entry $x_{z}$ adjoined at its end (since (B.3d) says $p_{z,\mu_{z}%
+1}=x_{z}$). Hence,%
\begin{equation}
P_{z}=U_{z}\mid x_{z}. \label{p104.5}%
\end{equation}
Now we claim the following:

\begin{statement}
\textbf{(C3.i')} For any $a\in\underline{z-1}$, we have%
\begin{equation}
x_{a+1}\mid P_{a}\equiv U_{a}\mid x_{a}. \label{p104.6}%
\end{equation}

\end{statement}

\textit{Proof.} Let $a\in\underline{z-1}$. We have $P_{a}=p_{a,1}p_{a,2}\ldots
p_{a,\lambda_{a}}$ and $U_{a}=u_{a,1}u_{a,2}\ldots u_{a,\mu_{a}}$. From
$a\in\underline{z-1}$, we obtain $a\neq z$ and thus $\lambda_{a}=\mu_{a}$
(since $\lambda=\mu+\varepsilon_{z}$ differs from $\mu$ only in its $z$-th
entry). From (B.3d), we know that row $a$ of $P$ differs from row $a$ of $U$
only in its $k\left(  a\right)  $-th entry, which is $p_{a,k\left(  a\right)
}=x_{a}$ instead of $u_{a,k\left(  a\right)  }=x_{a+1}$. Hence, $p_{a,j}%
=u_{a,j}$ for all $j\neq k\left(  a\right)  $.

From (B.3), we know that $x_{1}<x_{2}<x_{3}<\cdots$, so that $x_{a}%
<x_{a+1}=u_{a,k\left(  a\right)  }\leq u_{a,k\left(  a\right)  +1}\leq
u_{a,k\left(  a\right)  +2}\leq\cdots\leq u_{a,\mu_{a}}$ (since $U$ is
row-standard). Hence, (C3.g') (applied to $x=x_{a}$ and $\left(  b_{1}%
,b_{2},\ldots,b_{s}\right)  =\left(  u_{a,k\left(  a\right)  },u_{a,k\left(
a\right)  +1},\ldots,u_{a,\mu_{a}}\right)  $) yields%
\begin{align}
&  u_{a,k\left(  a\right)  }u_{a,k\left(  a\right)  +1}\ldots u_{a,\mu_{a}%
}x_{a}\nonumber\\
&  \equiv\underbrace{u_{a,k\left(  a\right)  }}_{=x_{a+1}}x_{a}u_{a,k\left(
a\right)  +1}u_{a,k\left(  a\right)  +2}\ldots u_{a,\mu_{a}}\nonumber\\
&  =x_{a+1}x_{a}u_{a,k\left(  a\right)  +1}u_{a,k\left(  a\right)  +2}\ldots
u_{a,\mu_{a}}. \label{p104.7}%
\end{align}


From (B.3c), we know that $u_{a,k\left(  a\right)  -1}\leq x_{a}$. Since $U$
is row-standard, we have $u_{a,1}\leq u_{a,2}\leq\cdots\leq u_{a,k\left(
a\right)  -1}\leq x_{a}<x_{a+1}$. Hence, (C3.h') (applied to $x=x_{a+1}$ and
$\left(  b_{1},b_{2},\ldots,b_{s}\right)  =\left(  u_{a,1},u_{a,2}%
,\ldots,u_{a,k\left(  a\right)  -1},x_{a}\right)  $) yields%
\begin{align}
&  u_{a,1}u_{a,2}\ldots u_{a,k\left(  a\right)  -1}x_{a+1}x_{a}\nonumber\\
&  \equiv x_{a+1}u_{a,1}u_{a,2}\ldots u_{a,k\left(  a\right)  -1}x_{a}.
\label{p104.8}%
\end{align}


Now, we recall that%
\begin{align}
P_{a}  &  =p_{a,1}p_{a,2}\ldots p_{a,\lambda_{a}}=p_{a,1}p_{a,2}\ldots
p_{a,\mu_{a}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\lambda_{a}=\mu
_{a}\right) \nonumber\\
&  =\underbrace{p_{a,1}p_{a,2}\ldots p_{a,k\left(  a\right)  -1}%
}_{\substack{=u_{a,1}u_{a,2}\ldots u_{a,k\left(  a\right)  -1}\\\text{(since
}p_{a,j}=u_{a,j}\text{ for all }j\neq k\left(  a\right)  \text{)}%
}}\underbrace{p_{a,k\left(  a\right)  }}_{=x_{a}}\underbrace{p_{a,k\left(
a\right)  +1}p_{a,k\left(  a\right)  +2}\ldots p_{a,\mu_{a}}}%
_{\substack{=u_{a,k\left(  a\right)  +1}u_{a,k\left(  a\right)  +2}\ldots
u_{a,\mu_{a}}\\\text{(since }p_{a,j}=u_{a,j}\text{ for all }j\neq k\left(
a\right)  \text{)}}}\nonumber\\
&  =u_{a,1}u_{a,2}\ldots u_{a,k\left(  a\right)  -1}x_{a}u_{a,k\left(
a\right)  +1}u_{a,k\left(  a\right)  +2}\ldots u_{a,\mu_{a}}. \label{p104.9}%
\end{align}
On the other hand,%
\begin{align*}
U_{a}  &  =u_{a,1}u_{a,2}\ldots u_{a,\mu_{a}}\\
&  =u_{a,1}u_{a,2}\ldots u_{a,k\left(  a\right)  -1}u_{a,k\left(  a\right)
}u_{a,k\left(  a\right)  +1}\ldots u_{a,\mu_{a}},
\end{align*}
so that%
\begin{align*}
U_{a}\mid x_{a}  &  =u_{a,1}u_{a,2}\ldots u_{a,k\left(  a\right)
-1}\underbrace{u_{a,k\left(  a\right)  }u_{a,k\left(  a\right)  +1}\ldots
u_{a,\mu_{a}}x_{a}}_{\substack{\equiv x_{a+1}x_{a}u_{a,k\left(  a\right)
+1}u_{a,k\left(  a\right)  +2}\ldots u_{a,\mu_{a}}\\\text{(by (\ref{p104.7}%
))}}}\\
&  \equiv\underbrace{u_{a,1}u_{a,2}\ldots u_{a,k\left(  a\right)  -1}%
x_{a+1}x_{a}}_{\substack{\equiv x_{a+1}u_{a,1}u_{a,2}\ldots u_{a,k\left(
a\right)  -1}x_{a}\\\text{(by (\ref{p104.8}))}}}u_{a,k\left(  a\right)
+1}u_{a,k\left(  a\right)  +2}\ldots u_{a,\mu_{a}}\\
&  \equiv x_{a+1}\underbrace{u_{a,1}u_{a,2}\ldots u_{a,k\left(  a\right)
-1}x_{a}u_{a,k\left(  a\right)  +1}u_{a,k\left(  a\right)  +2}\ldots
u_{a,\mu_{a}}}_{\substack{=P_{a}\\\text{(by (\ref{p104.9}))}}}=x_{a+1}\mid
P_{a}.
\end{align*}


This proves (\ref{p104.6}).

Now, we have%
\begin{align*}
KP  &  =P_{n}\mid P_{n-1}\mid\cdots\mid P_{1}\\
&  =\underbrace{P_{n}\mid P_{n-1}\mid\cdots\mid P_{z+1}}_{\substack{=U_{n}\mid
U_{n-1}\mid\cdots\mid U_{z+1}\\\text{(by (\ref{p104.4}))}}}\mid
\underbrace{P_{z}}_{\substack{=U_{z}\mid x_{z}\\\text{(by (\ref{p104.5}))}%
}}\mid P_{z-1}\mid P_{z-2}\mid P_{z-3}\mid\cdots\mid P_{1}\\
&  =U_{n}\mid U_{n-1}\mid\cdots\mid U_{z+1}\mid U_{z}\mid\underbrace{x_{z}\mid
P_{z-1}}_{\substack{\equiv U_{z-1}\mid x_{z-1}\\\text{(by (\ref{p104.6}%
)}\\\text{for }a=z-1\text{)}}}\mid P_{z-2}\mid P_{z-3}\mid\cdots\mid P_{1}\\
&  \equiv U_{n}\mid U_{n-1}\mid\cdots\mid U_{z+1}\mid U_{z}\mid U_{z-1}%
\mid\underbrace{x_{z-1}\mid P_{z-2}}_{\substack{\equiv U_{z-2}\mid
x_{z-2}\\\text{(by (\ref{p104.6})}\\\text{for }a=z-2\text{)}}}\mid P_{z-3}%
\mid\cdots\mid P_{1}\\
&  \equiv U_{n}\mid U_{n-1}\mid\cdots\mid U_{z+1}\mid U_{z}\mid U_{z-1}\mid
U_{z-2}\mid\underbrace{x_{z-2}\mid P_{z-3}}_{\substack{\equiv U_{z-3}\mid
x_{z-3}\\\text{(by (\ref{p104.6})}\\\text{for }a=z-3\text{)}}}\mid\cdots\mid
P_{1}\\
&  \equiv\cdots\\
&  \equiv\underbrace{U_{n}\mid U_{n-1}\mid\cdots\mid U_{z+1}\mid U_{z}\mid
U_{z-1}\mid U_{z-2}\mid U_{z-3}\mid\cdots\mid U_{1}}_{=U_{n}\mid U_{n-1}%
\mid\cdots\mid U_{1}=KU}\mid\underbrace{x_{1}}_{=x}\\
&  =KU\mid x=w.
\end{align*}
In view of $P=U\leftarrow x$, we can rewrite this as $K\left(  U\leftarrow
x\right)  \equiv w$. In other words, $K\left(  U\leftarrow x\right)  $ can be
transformed into $w$ by a sequence of basic moves. This proves (C3.f).

\item \textbf{page 106, proof of the \textquotedblleft only
if\textquotedblright\ part of Knuth's theorem:} The shorthands $P_{s-1}$ and
$P_{s}$ for $P_{s-1}\left(  i\right)  $ and $P_{s}\left(  i\right)  $ should
be explained. Maybe it is better to just drop the \textquotedblleft$\left(
i\right)  $\textquotedblright\ argument altogether (since $i$ is fixed here),
and simply write $P_{s}$ for $P\left(  i_{1}\ldots i_{s}\right)  $.

Furthermore it would be nice to remind the reader that $P_{s}=P_{s-1}%
\leftarrow i_{s}$ by the construction of the P-symbol.

\item \textbf{page 110, item (ii):} Remove the comma in \textquotedblleft the
first row of $W$, is $\widetilde{W}=\widetilde{U}\leftarrow y$%
\textquotedblright.

\item \textbf{page 110:} When you say \textquotedblleft(The case $p=\mu_{1}+2$
only occurs when $k=\mu_{1}+1$.)\textquotedblright, I think you mean
\textquotedblleft$l=\mu_{1}+1$\textquotedblright\ rather than
\textquotedblleft$k=\mu_{1}+1$\textquotedblright.

\item \textbf{page 111:} In the second paragraph of this page (line 8),
replace \textquotedblleft such that $s\geq l$\textquotedblright\ by
\textquotedblleft such that $s>l$\textquotedblright. Also remove the comma
that follows these words.

\item \textbf{page 111:} In the second paragraph of this page (line 12),
replace \textquotedblleft to $W^{\prime\prime}=W\leftarrow b$%
\textquotedblright\ by \textquotedblleft to $W^{\prime\prime}=W\leftarrow
c$\textquotedblright.

\item \textbf{page 112, Case 1:} Remove the comma after \textquotedblleft of
$V$ with $s>k$\textquotedblright.

\item \textbf{page 112, Case 1:} Replace \textquotedblleft bumps the same
letter $y=v_{1,l}=w_{1,l}$\textquotedblright\ by \textquotedblleft bumps the
same letter $y$ from place $\left(  1,l\right)  $\textquotedblright\ (this
letter is the $\left(  1,l\right)  $-th entry of $V\leftarrow c$ and
$W\leftarrow a$, and it does equal $w_{1,l}$, but it is not always the same as
$v_{1,l}$).

\item \textbf{page 112, Table C.8:} Not pictured on this table is the case
$l=p$. In this case, there is no letter $c$ in the first row any more (since
it has been bumped into the lower part of the tableau under the guise of $z$),
but otherwise the picture looks the same (in particular, the entry $b$
occupies the $l=p$-th place in row ($1$)).

\item \textbf{page 113, Case 1:} The proof of $x\leq y<z$ in the first
paragraph of this page looks questionable to me. I would rather argue as follows:

By definition, $y$ is the $l$-th letter in the $1$-st row of $V\leftarrow c$.
Meanwhile, $c$ is the $p$-th letter in this row. Since $l\leq p$, we thus
obtain $y\leq c$ (since $V\leftarrow c$ is row-standard). Furthermore,
$c<u_{1,p}=z$. Hence, $y\leq c<z$.

By definition, $y$ is the $l$-th letter in the $1$-st row of $W\leftarrow a$.
Since the $1$-st rows of $W\leftarrow a$ and $W$ differ only in their $k$-th
letter (because $a$ bumps out the $k$-th letter of the $1$-st row of $W$),
this entails that $y$ is the $l$-th letter in the $1$-st row of $W$ as well
(since $k<l$ and thus $l\neq k$). In other words, $y=w_{1,l}$. But $x=w_{1,k}%
$. Since $W$ is row-standard, we have $w_{1,k}\leq w_{1,l}$ (since $k<l$), and
thus $x=w_{1,k}\leq w_{1,l}=y$. Hence, altogether, $x\leq y<z$.

\item \textbf{page 113, Case 2:} Better to replace the first \textquotedblleft%
$=$\textquotedblright\ sign in \textquotedblleft bumps the letter
$w=v_{1,k+1}=u_{1,k+1}$\textquotedblright\ by a \textquotedblleft%
$:=$\textquotedblright\ sign, since the letter \textquotedblleft%
$w$\textquotedblright\ (albeit with subscripts) has already been used for a
different purpose. Or replace this letter \textquotedblleft$w$%
\textquotedblright\ by an unused letter (such as \textquotedblleft%
$t$\textquotedblright).

\item \textbf{page 113, Case 2:} Here you write \textquotedblleft note that
$w_{1,p-1}=u_{1,p-1}\leq a<c=w_{1,p}$\textquotedblright. It is perhaps clearer
if this is replaced by \textquotedblleft note that $p=k$ and thus
$w_{1,p-1}=w_{1,k-1}=u_{1,k-1}\leq a<c=w_{1,p}$\textquotedblright.

\item \textbf{page 113, Case 2:} Replace \textquotedblleft From $v_{1,k}=a\leq
b<c=v_{1,k+1}$\textquotedblright\ by \textquotedblleft Let $\left(
v\leftarrow c\right)  _{r,s}$ denote the $\left(  r,s\right)  $-th letter of
$V\leftarrow c$. Then, from $\left(  v\leftarrow c\right)  _{1,k}=a\leq
b<c=\left(  v\leftarrow c\right)  _{1,k+1}$\textquotedblright.

\item \textbf{page 113, Case 2:} Replace \textquotedblleft From $w_{1,k}=a\leq
b<c<u_{1,k}\leq u_{1,k+1}=w_{1,k+1}$\textquotedblright\ by \textquotedblleft
Let $\left(  w\leftarrow a\right)  _{r,s}$ denote the $\left(  r,s\right)
$-th letter of $W\leftarrow a$. Then, from $\left(  w\leftarrow a\right)
_{1,k}=a\leq b<c<u_{1,p}=u_{1,k}\leq u_{1,k+1}=\left(  w\leftarrow a\right)
_{1,k+1}$\textquotedblright.

\item \textbf{page 114, the paragraph below (C.5a):} In \textquotedblleft the
set $\left\{  0\right\}  \cup\left\{  h_{c}^{KP}:t\in\left[  \lambda\right]
\right\}  $\textquotedblright, replace the \textquotedblleft$h_{c}^{KP}%
$\textquotedblright\ by \textquotedblleft$h_{c}^{KP}\left(  t\right)
$\textquotedblright. Moreover, it is worth pointing out that this set can also
be written as $\left\{  h_{c}^{KP}\left(  t\right)  :t\in\left[
\lambda\right]  \cup\left\{  0\right\}  \right\}  $.

\item \textbf{page 114, footnote }$^{3}$\textbf{:} After \textquotedblleft
view theorem (C.5b)\textquotedblright, add \textquotedblleft and Proposition
(C.2c)\textquotedblright.

\item \textbf{page 115, (C.5b):} Replace \textquotedblleft such that $h\left(
\left(  a,b\right)  \right)  =M$\textquotedblright\ by \textquotedblleft such
that $h^{KP}\left(  \left(  a,b\right)  \right)  =M$\textquotedblright.

\item \textbf{page 115, Table C.11:} I would replace the \textquotedblleft%
$>c+1$\textquotedblright\ in row $a+1$ and column $b+1$ by a \textquotedblleft%
$\geq c+1$\textquotedblright\ (see below for why).

\item \textbf{page 115, proof of (C.5b):} It is not clear to me why
\textquotedblleft Since $a\geq c+1$, entries in row ($a+1$) to the right of
column ($b$) are all $>c+1$\textquotedblright. But I do understand why these
entries are all $\geq c+1$ (since $p_{a+1,b}=c+1$ and since $P$ is
row-standard). Fortunately, this is sufficient for the proof (with a few
little modifications, listed below).

\item \textbf{page 115, proof of (C.5b):} In the equality \textquotedblleft%
$Y^{\ast}=\sum_{1\leq x\leq b^{\prime}}\omega\left(  p_{a,x}\right)
$\textquotedblright, the \textquotedblleft$b^{\prime}$\textquotedblright%
\ under the summation sign should be \textquotedblleft$b^{\prime}%
-1$\textquotedblright. Moreover, add a comma after this equality.

\item \textbf{page 115, proof of (C.5b):} Replace \textquotedblleft But for
$b+1\leq x\leq\lambda_{a+1}$ all the entries $p_{a+1,x}$ are $>c+1$, hence all
the summands $\omega\left(  p_{a+1,x}\right)  =0$, therefore $Y=0$%
\textquotedblright\ by \textquotedblleft However, all the entries $p_{a+1,x}$
with $b+1\leq x\leq\lambda_{a+1}$ are $\geq c+1$ and thus satisfy
$\omega\left(  p_{a+1,x}\right)  \in\left\{  0,-1\right\}  $; hence, $Y\leq
0$\textquotedblright.

\item \textbf{page 115, proof of (C.5b):} Replace \textquotedblleft%
$h^{KP}\left(  a,b\right)  =h^{KP}\left(  a+1,b^{\prime}-1\right)
$\textquotedblright\ by \textquotedblleft$h^{KP}\left(  a,b\right)  \leq
h^{KP}\left(  a+1,b^{\prime}-1\right)  $. Since $h^{KP}\left(  a,b\right)  =M$
is the largest of all $h^{KP}\left(  t\right)  $, we thus conclude that
$h^{KP}\left(  a,b\right)  =h^{KP}\left(  a+1,b^{\prime}-1\right)
$\textquotedblright.

\item \textbf{page 116, after the proof of (C.5b):} Add the sentence
\textquotedblleft In the situation of Theorem (C.5b), we denote the tableau
$\widetilde{P}$ by $\widetilde{f}_{c}P$. Thus, $K\left(  \widetilde{f}%
_{c}P\right)  =\widetilde{f}_{c}\left(  KP\right)  $.\textquotedblright.

\item \textbf{page 116, (C.5d):} \textquotedblleft two set\textquotedblright%
\ $\rightarrow$ \textquotedblleft two sets\textquotedblright.

\item \textbf{page 116, Proposition B:} One of the claims made here is that if
$\widetilde{f}_{c}\left(  i\right)  \neq\infty$, then $\widetilde{f}%
_{c}\left(  KP\left(  i\right)  \right)  \neq\infty$. But the converse claim
should also be made: If $\widetilde{f}_{c}\left(  KP\left(  i\right)  \right)
\neq\infty$, then $\widetilde{f}_{c}\left(  i\right)  \neq\infty$. (This is
used later on, in the proof of (D.1g).)

Both claims follow from Lemma (C.6c). (Indeed, by repeatedly applying Lemma
(C.6c), we see that if two words $i$ and $j$ in $I\left(  n,r\right)  $ are
connected by a sequence of basic moves, then the statements $\widetilde{f}%
_{c}\left(  i\right)  \neq\infty$ and $\widetilde{f}_{c}\left(  j\right)
\neq\infty$ are equivalent. Now it just remains to apply this to $j=KP\left(
i\right)  $.)

\item \textbf{page 116, Lemma (C.6b):} Add \textquotedblleft Let $c\in\left\{
1,2,\ldots,n-1\right\}  $.\textquotedblright\ at the beginning of the lemma.

\item \textbf{page 117, proof of Lemma (C.6b):} After the first sentence, I
would add the following: \textquotedblleft Thus, $j_{k}=z$ and $j_{k+1}=x$.
The definition of a height function yields
\begin{align*}
h_{c}^{i}\left(  k+1\right)   &  =h_{c}^{i}\left(  k\right)  +\omega
_{c}\left(  i_{k+1}\right)  =h_{c}^{i}\left(  k\right)  +\omega_{c}\left(
z\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
h_{c}^{i}\left(  k-1\right)   &  =h_{c}^{i}\left(  k\right)  -\omega
_{c}\left(  i_{k}\right)  =h_{c}^{i}\left(  k\right)  -\omega_{c}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
h_{c}^{j}\left(  k+1\right)   &  =h_{c}^{j}\left(  k\right)  +\omega
_{c}\left(  j_{k+1}\right)  =h_{c}^{j}\left(  k\right)  +\omega_{c}\left(
x\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
h_{c}^{j}\left(  k-1\right)   &  =h_{c}^{j}\left(  k\right)  -\omega
_{c}\left(  j_{k}\right)  =h_{c}^{j}\left(  k\right)  -\omega_{c}\left(
z\right)
\end{align*}
(where $\omega_{c}$ is short for $\omega_{c,c+1}$).\textquotedblright

\item \textbf{page 117, proof of (ii) in the proof of Lemma (C.6b):} This
proof can be redone in a nicer way (at least to my taste):

\textit{Proof of (ii):} By definition,
\begin{equation}
M^{i}=\max\left\{  h_{c}^{i}\left(  \nu\right)  \ \mid\ \nu\in\left\{
0,1,\ldots,r\right\}  \right\}  . \label{p117.1}%
\end{equation}
We shall now show that
\begin{equation}
M^{i}=\max\left\{  h_{c}^{i}\left(  \nu\right)  \ \mid\ \nu\in\left\{
0,1,\ldots,r\right\}  \setminus\left\{  k\right\}  \right\}  \label{p117.2}%
\end{equation}
holds as well. Indeed, assume the contrary. Thus, $M^{i}$ is the maximum of
the set $\left\{  h_{c}^{i}\left(  \nu\right)  \ \mid\ \nu\in\left\{
0,1,\ldots,r\right\}  \right\}  $, but not the maximum of the (only slightly
smaller) set $\left\{  h_{c}^{i}\left(  \nu\right)  \ \mid\ \nu\in\left\{
0,1,\ldots,r\right\}  \setminus\left\{  k\right\}  \right\}  $. In other
words, $M^{i}$ is the largest of all the numbers $h_{c}^{i}\left(  0\right)
,\ h_{c}^{i}\left(  1\right)  ,\ \ldots,\ h_{c}^{i}\left(  r\right)  $, but
ceases to be so if we remove the number $h_{c}^{i}\left(  k\right)  $ from
this list. Obviously, the only way this can happen is if $M^{i}$ equals
$h_{c}^{i}\left(  k\right)  $ but is larger than all the other numbers in the
list $h_{c}^{i}\left(  0\right)  ,\ h_{c}^{i}\left(  1\right)  ,\ \ldots
,\ h_{c}^{i}\left(  r\right)  $. So we must be in this exact situation. Thus,
$M^{i}=h_{c}^{i}\left(  k\right)  $ but
\begin{equation}
M^{i}>h_{c}^{i}\left(  \nu\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }\nu\neq
k. \label{p117.3}%
\end{equation}
In particular, from (\ref{p117.3}), we obtain $M^{i}>h_{c}^{i}\left(
k+1\right)  $ and $M^{i}>h_{c}^{i}\left(  k-1\right)  $ (here we use the fact
that $k\geq1$). Thus, $h_{c}^{i}\left(  k\right)  =M^{i}>h_{c}^{i}\left(
k+1\right)  =h_{c}^{i}\left(  k\right)  +\omega_{c}\left(  i_{k+1}\right)  $,
so that $\omega_{c}\left(  i_{k+1}\right)  <0$ and thus $i_{k+1}=c+1$ (by the
definition of $\omega_{c}$). Likewise, we can show (using $M^{i}>h_{c}%
^{i}\left(  k-1\right)  $) that $i_{k}=c$. Thus, the word $j$ is obtained from
$i$ by swapping the consecutive letters $i_{k}=c$ and $i_{k+1}=c+1$. Moreover,
this swap must be a basic move. If this basic move is of type $K^{\prime}$,
then we must have $k\geq2$ and $i_{k}<i_{k-1}\leq i_{k+1}$ (since
$i_{k}=c<c+1=i_{k+1}$), so that $i_{k-1}\in\left(  \underbrace{i_{k}}%
_{=c},\underbrace{i_{k+1}}_{=c+1}\right]  =\left(  c,c+1\right]  $ and thus
$i_{k-1}=c+1$, whence%
\begin{align*}
M  &  =h_{c}^{i}\left(  k\right)  =h_{c}^{i}\left(  k-2\right)  +\omega
_{c}\left(  \underbrace{i_{k-1}}_{=c+1}\right)  +\omega_{c}\left(
\underbrace{i_{k}}_{=c}\right) \\
&  =h_{c}^{i}\left(  k-2\right)  +\underbrace{\omega_{c}\left(  c+1\right)
}_{=-1}+\underbrace{\omega_{c}\left(  c\right)  }_{=1}\\
&  =h_{c}^{i}\left(  k-2\right)  +\left(  -1\right)  +1=h_{c}^{i}\left(
k-2\right)  ,
\end{align*}
and this contradicts (\ref{p117.3}). Thus, this basic move cannot be of type
$K^{\prime}$. Hence, it must be of type $K^{\prime\prime}$. Thus, we must have
$k\leq r-2$ and $i_{k}\leq i_{k+2}<i_{k+1}$ (since $i_{k}=c<c+1=i_{k+1}$), so
that $i_{k+2}\in\left[  \underbrace{i_{k}}_{=c},\underbrace{i_{k+1}}%
_{=c+1}\right)  =\left[  c,c+1\right)  $ and thus $i_{k+2}=c$, whence%
\begin{align*}
M  &  >h_{c}^{i}\left(  k+2\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{p117.3})}\right) \\
&  =h_{c}^{i}\left(  k\right)  +\omega_{c}\left(  \underbrace{i_{k+1}}%
_{=c+1}\right)  +\omega_{c}\left(  \underbrace{i_{k+2}}_{=c}\right) \\
&  =h_{c}^{i}\left(  k\right)  +\underbrace{\omega_{c}\left(  c+1\right)
}_{=-1}+\underbrace{\omega_{c}\left(  c\right)  }_{=1}\\
&  =h_{c}^{i}\left(  k\right)  +\left(  -1\right)  +1=h_{c}^{i}\left(
k\right)  ,
\end{align*}
and this contradicts $M=h_{c}^{i}\left(  k\right)  $. This contradiction shows
that our assumption was false. Thus, (\ref{p117.2}) is proved.

The same argument, applied to the word $j$ instead of $i$, shows that
\begin{equation}
M^{j}=\max\left\{  h_{c}^{j}\left(  \nu\right)  \ \mid\ \nu\in\left\{
0,1,\ldots,r\right\}  \setminus\left\{  k\right\}  \right\}  \label{p117.4}%
\end{equation}
(because we have not used the assumption $i_{k}<i_{k+1}$ in our above argument).

But the right hand sides of the equalities (\ref{p117.2}) and (\ref{p117.4})
are equal (since (i) shows that $h_{c}^{i}\left(  \nu\right)  =h_{c}%
^{j}\left(  \nu\right)  $ for all $\nu\neq k$). Hence, so are the left hand
sides. In other words, $M^{i}=M^{j}$.

\item \textbf{page 117, proof of Lemma (C.6b):} After the proof of (ii), I
would add another claim: \textquotedblleft Next we claim that

(iii) we have $q^{i}\geq k$ if and only if $q^{j}\geq k$.

Indeed, if $q^{i}<k$, then (i) yields $h_{c}^{j}\left(  q^{i}\right)
=h_{c}^{i}\left(  q^{i}\right)  =M^{i}=M^{j}$ by (ii), and thus $q^{j}\leq
q^{i}$ by the minimality of $q^{j}$; but this entails $q^{j}\leq q^{i}<k$.
Hence, $q^{i}<k$ implies $q^{j}<k$. Similarly, the converse is true. Hence, we
have $q^{i}<k$ if and only if $q^{j}<k$. By taking the contrapositive, we
obtain (iii).\textquotedblright

\item \textbf{page 117, proof of Lemma (C.6b):} I struggle to understand how
you prove part (a) of the lemma. So here is my proof of part (a):

\textquotedblleft Now, for the proof of (a), suppose that $q^{i}\notin\left\{
k,k+1\right\}  $. We must show that $q^{j}=q^{i}$.

We have $q^{i}\neq k$ (since $q^{i}\notin\left\{  k,k+1\right\}  $). Thus, (i)
shows that $h_{c}^{j}\left(  q^{i}\right)  =h_{c}^{i}\left(  q^{i}\right)
=M^{i}$ (by the definition of $q^{i}$). Thus, $h_{c}^{j}\left(  q^{i}\right)
=M^{i}=M^{j}$ by (ii). Therefore, $q^{i}\geq q^{j}$, by the minimality of
$q^{j}$.

If we also have $q^{j}\neq k$, then the same argument (with the roles of $i$
and $j$ interchanged) shows that $q^{j}\geq q^{i}$, and thus $q^{j}=q^{i}$
(since $q^{i}\geq q^{j}$). Hence, we are done in this case. Thus, from now on,
we WLOG assume that $q^{j}=k$.

Hence, $k=q^{j}$ is the first place where the height function $h_{c}^{j}$
reaches its maximum (by the definition of $q^{j}$). Hence, $h_{c}^{j}\left(
k-1\right)  <h_{c}^{j}\left(  k\right)  $. In other words, $\omega_{c}\left(
z\right)  >0$ (since $h_{c}^{j}\left(  k-1\right)  =h_{c}^{j}\left(  k\right)
-\omega_{c}\left(  z\right)  $). Hence, $z=c$. Thus, $x<z=c$, so that
$\omega_{c}\left(  x\right)  =0$. Now, (i) yields $h_{c}^{j}\left(
k+1\right)  =h_{c}^{i}\left(  k+1\right)  $, whence
\begin{align*}
h_{c}^{i}\left(  k+1\right)   &  =h_{c}^{j}\left(  k+1\right)  =h_{c}%
^{j}\left(  \underbrace{k}_{=q^{j}}\right)  +\underbrace{\omega_{c}\left(
x\right)  }_{=0}=h_{c}^{j}\left(  q^{j}\right) \\
&  =M^{j}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }M^{j}\right)
\\
&  =M^{i}.
\end{align*}
Hence, $q^{i}\leq k+1$ (by the minimality of $q^{i}$). Since $q^{i}%
\notin\left\{  k,k+1\right\}  $, we can conclude that $q^{i}<k$, which
contradicts $q^{i}\geq q^{j}=k$. This contradiction shows that the case we are
considering (that is, when $q^{j}=k$) is impossible, and the proof of (i) is
complete.\textquotedblright

\item \textbf{page 117, proof of Lemma (C.6b):} In the proof of (b), you write
\textquotedblleft In fact, by (i), $q^{j}=k$\textquotedblright. To me, this is
unclear, so I'd rather say \textquotedblleft But (iii) shows that $q^{j}\geq
k$, so that $q^{j}=k$\textquotedblright.

\item \textbf{page 117, proof of Lemma (C.6b):} In the proof of (c), you write
\textquotedblleft Then $x=c$, and $q^{j}\geq k$, by (ii)\textquotedblright.
Again, it is not clear to me how you obtain $q^{j}\geq k$ here, but you can
obtain it immediately from (iii).

\item \textbf{page 117, proof of Lemma (C.6b):} In the proof of (c), after
\textquotedblleft Then $z>c+1$\textquotedblright, I would add
\textquotedblleft(since $z>x=c$)\textquotedblright.

\item \textbf{page 117, Lemma (C.6c):} Add \textquotedblleft Let $c\in\left\{
1,2,\ldots,n-1\right\}  $.\textquotedblright\ at the beginning of the lemma.

\item \textbf{page 117, Lemma (C.6c):} Replace \textquotedblleft If
$\widetilde{f}_{c}\left(  i\right)  \neq\infty$, then $\widetilde{f}%
_{c}\left(  j\right)  \neq\infty$, and $\widetilde{f}_{c}\left(  j\right)  $
is obtained from $\widetilde{f}_{c}\left(  i\right)  $ by a basic
move\textquotedblright\ by \textquotedblleft Then, the inequality
$\widetilde{f}_{c}\left(  i\right)  \neq\infty$ holds if and only if
$\widetilde{f}_{c}\left(  j\right)  \neq\infty$. Moreover, if it holds, then
$\widetilde{f}_{c}\left(  j\right)  $ is obtained from $\widetilde{f}%
_{c}\left(  i\right)  $ by a basic move\textquotedblright.

In fact, in this form, the statement is symmetric in $i$ and $j$, which is
used in the proof of the lemma.

\item \textbf{page 117, Lemma (C.6c):} The last claim of this lemma
(\textquotedblleft There is a corresponding statement (and proof), with
$\widetilde{e}_{c}$ replacing $\widetilde{f}_{c}$\textquotedblright) is quite
an imposition on the reader, as it requires constructing analogues of both
Lemma (C.6b) and Lemma (C.6c) and their already rather patience-demanding proofs.

An easier way to prove this claim would be to use the results of \S D.3 (which
are completely independent), specifically Lemma (D.3e). This can be done as
follows. First, we show a simple property of the operator $C$:

\begin{statement}
\textbf{(D.3i') Lemma.} Let $i,j\in I\left(  n,r\right)  $ be two words such
that $j$ is obtained from $i$ by a basic move. Then, the word $C\left(
j\right)  $ is obtained from $C\left(  i\right)  $ by a basic move as well.
\end{statement}

\textit{Proof.} If $i\ K^{\prime}\ j$, then $i=\left(  \ldots,b,c,a,\ldots
\right)  $ and $j=\left(  \ldots,b,a,c,\ldots\right)  $ for some letters
$a<b\leq c$, and thus
\begin{align*}
C\left(  i\right)   &  =\left(  \ldots,\ n+1-a,\ n+1-c,\ n+1-b,\ \ldots
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
C\left(  j\right)   &  =\left(  \ldots,\ n+1-c,\ n+1-a,\ n+1-b,\ \ldots
\right)  ,
\end{align*}
which shows that $C\left(  i\right)  \ K^{\prime\prime}\ C\left(  j\right)  $
(since $a<b\leq c$ entails $n+1-c\leq n+1-b<n+1-a$). Similarly, if
$i\ K^{\prime\prime}\ j$, then $C\left(  i\right)  \ K^{\prime}\ C\left(
j\right)  $. In both cases, Lemma (D.3i') is proved.

\begin{statement}
\textbf{(D.3j') Lemma.} The analogue of Lemma (C.6c) for $\widetilde{e}_{c}$
instead of $\widetilde{f}_{c}$ holds. In other words:

Let $c\in\left\{  1,2,\ldots,n-1\right\}  $. Let $i,j\in I\left(  n,r\right)
$, and suppose $j$ is obtained from $i$ by a basic move. Then, the inequality
$\widetilde{e}_{c}\left(  i\right)  \neq\infty$ holds if and only if
$\widetilde{e}_{c}\left(  j\right)  \neq\infty$. Moreover, if it holds, then
$\widetilde{e}_{c}\left(  j\right)  $ is obtained from $\widetilde{e}%
_{c}\left(  i\right)  $ by a basic move.
\end{statement}

\textit{Proof.} Lemma (D.3i') shows that the word $Cj$ is obtained from $Ci$
by a basic move. Hence, Lemma (C.6c) (applied to $n-c$, $Ci$ and $Cj$ instead
of $c$, $i$ and $j$) shows that $\widetilde{f}_{n-c}\left(  Ci\right)
\neq\infty$ holds if and only if $\widetilde{f}_{n-c}\left(  Cj\right)
\neq\infty$, and furthermore that the word $\widetilde{f}_{n-c}\left(
Cj\right)  $ is obtained from $\widetilde{f}_{n-c}\left(  Ci\right)  $ by a
basic move (if these words are not $\infty$).

For convenience, set $C\left(  \infty\right)  :=\infty$. Lemma (D.3e) yields
$C\left(  \widetilde{e}_{c}\left(  i\right)  \right)  =\widetilde{f}%
_{n-c}\left(  Ci\right)  $ and similarly $C\left(  \widetilde{e}_{c}\left(
j\right)  \right)  =\widetilde{f}_{n-c}\left(  Cj\right)  $. Thus, we have the
following chain of equivalences:%
\begin{align*}
\left(  \widetilde{e}_{c}\left(  i\right)  \neq\infty\right)  \  &
\Longleftrightarrow\ \left(  C\left(  \widetilde{e}_{c}\left(  i\right)
\right)  \neq\infty\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }C\text{ is bijective}\\
\text{and sends }\infty\text{ to }\infty
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \widetilde{f}_{n-c}\left(  Ci\right)
\neq\infty\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }C\left(
\widetilde{e}_{c}\left(  i\right)  \right)  =\widetilde{f}_{n-c}\left(
Ci\right)  \right) \\
&  \Longleftrightarrow\ \left(  \widetilde{f}_{n-c}\left(  Cj\right)
\neq\infty\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\widetilde{f}_{n-c}\left(  Ci\right)  \neq\infty\text{ holds}\\
\text{if and only if }\widetilde{f}_{n-c}\left(  Cj\right)  \neq\infty
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  C\left(  \widetilde{e}_{c}\left(  j\right)
\right)  \neq\infty\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }C\left(
\widetilde{e}_{c}\left(  j\right)  \right)  =\widetilde{f}_{n-c}\left(
Cj\right)  \right) \\
&  \Longleftrightarrow\ \left(  \widetilde{e}_{c}\left(  j\right)  \neq
\infty\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }C\text{ is bijective}\\
\text{and sends }\infty\text{ to }\infty
\end{array}
\right)  .
\end{align*}
In other words, $\widetilde{e}_{c}\left(  i\right)  \neq\infty$ holds if and
only if $\widetilde{e}_{c}\left(  j\right)  \neq\infty$. Moreover, if it
holds, then $\widetilde{f}_{n-c}\left(  Ci\right)  \neq\infty$ and
$\widetilde{f}_{n-c}\left(  Cj\right)  \neq\infty$ hold as well (by the above
equivalence), and thus the word $\widetilde{f}_{n-c}\left(  Cj\right)  $ is
obtained from $\widetilde{f}_{n-c}\left(  Ci\right)  $ by a basic move (as we
have shown above). In other words, the word $C\left(  \widetilde{e}_{c}\left(
j\right)  \right)  $ is obtained from $C\left(  \widetilde{e}_{c}\left(
i\right)  \right)  $ by a basic move (since $C\left(  \widetilde{e}_{c}\left(
i\right)  \right)  =\widetilde{f}_{n-c}\left(  Ci\right)  $ and $C\left(
\widetilde{e}_{c}\left(  j\right)  \right)  =\widetilde{f}_{n-c}\left(
Cj\right)  $). Hence, Lemma (C.3i') shows that the word $C\left(  C\left(
\widetilde{e}_{c}\left(  j\right)  \right)  \right)  $ is obtained from
$C\left(  C\left(  \widetilde{e}_{c}\left(  i\right)  \right)  \right)  $ by a
basic move as well. Since $C\left(  C\left(  \widetilde{e}_{c}\left(
j\right)  \right)  \right)  =\widetilde{e}_{c}\left(  j\right)  $ (because
$C^{2}=\operatorname*{id}$) and $C\left(  C\left(  \widetilde{e}_{c}\left(
i\right)  \right)  \right)  =\widetilde{e}_{c}\left(  i\right)  $ (similarly),
we can rewrite this as follows: The word $\widetilde{e}_{c}\left(  j\right)  $
is obtained from $\widetilde{e}_{c}\left(  i\right)  $ by a basic move. Thus,
Lemma (D.3j') is proved.

\item \textbf{page 118, proof of Lemma (C.6c):} Before going into the cases,
it is worth reminding the reader that $\widetilde{f}_{c}\left(  i\right)  $ is
defined to be the word $i$ with the letter $c$ at position $q^{i}$ replaced by
$c+1$. Hence, $\widetilde{f}_{c}\left(  i\right)  $ differs from $i$ only in
position $q^{i}$.

\item \textbf{page 118, Case (a):} After \textquotedblleft The claim follows
directly if $y$ is not changed, either.\textquotedblright, add
\textquotedblleft Thus we restrict ourselves to the case when it is changed.
Then, $y=c$, and $q^{i}$ is either $k-1$ or $k+2$ (depending on the type of
the basic move).\textquotedblright.

\item \textbf{page 118, Case (a):} After \textquotedblleft we get $h_{c}%
^{j}\left(  k\right)  =h_{c}^{j}\left(  k-1\right)  +1$\textquotedblright, add
\textquotedblleft$=h_{c}^{j}\left(  q^{j}\right)  +1$ (since $k-1=q^{i}=q^{j}%
$)\textquotedblright.

\item \textbf{page 118, Case (a):} After \textquotedblleft we get $h_{c}%
^{i}\left(  k\right)  =h_{c}^{i}\left(  k+2\right)  $\textquotedblright, add
\textquotedblleft$=h_{c}^{i}\left(  q^{i}\right)  $ (since $k+2=q^{i}%
$)\textquotedblright.

\item \textbf{page 118, Case (c):} Replace \textquotedblleft since otherwise
$h_{c}^{i}\left(  k+2\right)  =h_{c}^{i}\left(  k\right)  +1$%
\textquotedblright\ by \textquotedblleft since otherwise $y=c$ (since $y\geq
x=c$) and therefore $h_{c}^{i}\left(  k+2\right)  =h_{c}^{i}\left(  k\right)
+1$ (because $z>c+1$) in contradiction to the fact that $h_{c}^{i}\left(
k\right)  =h_{c}^{i}\left(  q^{i}\right)  =M^{i}$ is the largest value of
$h_{c}^{i}$\textquotedblright.

\item \textbf{page 118, Case (c):} After \textquotedblleft since otherwise
$h_{c}^{i}\left(  k-2\right)  =h_{c}^{i}\left(  k\right)  $\textquotedblright,
add \textquotedblleft in contradiction to the fact that $k=q^{i}$ is the first
time the function $h_{c}^{i}$ attains its maximum value\textquotedblright.
\end{enumerate}

\subsection{Corrections to \S D}

\begin{enumerate}
\item \textbf{page 121:} Here is a \textit{proof of Lemma (D.1a):}

(i) Fix $i\in I\left(  n,r\right)  $ and $c\in\left\{  1,2,\ldots,n-1\right\}
$. We have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \widetilde{f}_{c}\left(  i\right)  =\infty\right) \\
&  \Longleftrightarrow\ \left(  M_{c}^{i}=0\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (A.3e)}\right) \\
&  \Longleftrightarrow\ \left(  h_{c}^{i}\left(  t\right)  \leq h_{c}%
^{i}\left(  0\right)  \text{ for all }t\in\left\{  0,1,\ldots,r\right\}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }M_{c}%
^{i}\text{ is defined as the maximum value of }h_{c}^{i}\right) \\
&  \Longleftrightarrow\ \left(  h_{c}^{i}\left(  t\right)  \leq h_{c}%
^{i}\left(  0\right)  \text{ for all }t\in\left\{  1,2,\ldots,r\right\}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the inequality }h_{c}^{i}\left(  t\right)  \leq h_{c}^{i}\left(
0\right) \\
\text{automatically holds for }t=0
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  h_{c}^{i}\left(  t\right)  \leq0\text{ for all
}t\in\left\{  1,2,\ldots,r\right\}  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }h_{c}^{i}\left(  0\right)  =0\right) \\
&  \Longleftrightarrow\ \left(  \#\left\{  \nu\leq t:i_{\nu}=c\right\}
-\#\left\{  \nu\leq t:i_{\nu}=c+1\right\}  \leq0\text{ for all }t\in\left\{
1,2,\ldots,r\right\}  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }h_{c}^{i}\left(  t\right)  =\#\left\{  \nu\leq t:i_{\nu
}=c\right\}  -\#\left\{  \nu\leq t:i_{\nu}=c+1\right\} \\
\text{for all }t\in\left\{  0,1,\ldots,r\right\}  \text{ (this is a
restatement of (A.3b))}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \#\left\{  \nu\leq t:i_{\nu}=c\right\}
\leq\#\left\{  \nu\leq t:i_{\nu}=c+1\right\}  \text{ for all }t\in\left\{
1,2,\ldots,r\right\}  \right)  .
\end{align*}
This proves Lemma (D.1a) (i).

(ii) Fix $i\in I\left(  n,r\right)  $ and $c\in\left\{  1,2,\ldots
,n-1\right\}  $. For each $s\in\left\{  1,2,\ldots,r\right\}  $, we have%
\begin{align}
&  \underbrace{h_{c}^{i}\left(  r\right)  }_{\substack{=\omega\left(
i_{1}\right)  +\omega\left(  i_{2}\right)  +\cdots+\omega\left(  i_{r}\right)
\\\text{(by the definition of height function)}}}-\underbrace{h_{c}^{i}\left(
s-1\right)  }_{\substack{=\omega\left(  i_{1}\right)  +\omega\left(
i_{2}\right)  +\cdots+\omega\left(  i_{s-1}\right)  \\\text{(by the definition
of height function)}}}\nonumber\\
&  =\left(  \omega\left(  i_{1}\right)  +\omega\left(  i_{2}\right)
+\cdots+\omega\left(  i_{r}\right)  \right)  -\left(  \omega\left(
i_{1}\right)  +\omega\left(  i_{2}\right)  +\cdots+\omega\left(
i_{s-1}\right)  \right) \nonumber\\
&  =\omega\left(  i_{s}\right)  +\omega\left(  i_{s+1}\right)  +\cdots
+\omega\left(  i_{r}\right) \nonumber\\
&  =\#\left\{  \nu\geq s:i_{\nu}=c\right\}  -\#\left\{  \nu\geq s:i_{\nu
}=c+1\right\}  \label{p121.2}%
\end{align}
(by the definition of $\omega=\omega_{c,c+1}$). Now, we have the following
chain of logical equivalences:%
\begin{align*}
&  \ \left(  \widetilde{e}_{c}\left(  i\right)  =\infty\right) \\
&  \Longleftrightarrow\ \left(  M_{c}^{i}=h_{c}^{i}\left(  r\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (A.3f)}\right) \\
&  \Longleftrightarrow\ \left(  h_{c}^{i}\left(  t\right)  \leq h_{c}%
^{i}\left(  r\right)  \text{ for all }t\in\left\{  0,1,\ldots,r\right\}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }M_{c}%
^{i}\text{ is defined as the maximum value of }h_{c}^{i}\right) \\
&  \Longleftrightarrow\ \left(  h_{c}^{i}\left(  t\right)  \leq h_{c}%
^{i}\left(  r\right)  \text{ for all }t\in\left\{  0,1,\ldots,r-1\right\}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since the inequality
}h_{c}^{i}\left(  t\right)  \leq h_{c}^{i}\left(  r\right)  \text{ holds
automatically for }t=r\right) \\
&  \Longleftrightarrow\ \left(  h_{c}^{i}\left(  s-1\right)  \leq h_{c}%
^{i}\left(  r\right)  \text{ for all }s\in\left\{  1,2,\ldots,r\right\}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have
substituted }s-1\text{ for }t\right) \\
&  \Longleftrightarrow\ \left(  h_{c}^{i}\left(  r\right)  -h_{c}^{i}\left(
s-1\right)  \geq0\text{ for all }s\in\left\{  1,2,\ldots,r\right\}  \right) \\
&  \Longleftrightarrow\ \left(  \#\left\{  \nu\geq s:i_{\nu}=c\right\}
-\#\left\{  \nu\geq s:i_{\nu}=c+1\right\}  \geq0\text{ for all }s\in\left\{
1,2,\ldots,r\right\}  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{p121.2}%
)}\right) \\
&  \Longleftrightarrow\ \left(  \#\left\{  \nu\geq s:i_{\nu}=c\right\}
\geq\#\left\{  \nu\geq s:i_{\nu}=c+1\right\}  \text{ for all }s\in\left\{
1,2,\ldots,r\right\}  \right)  .
\end{align*}
This proves Lemma (D.1a) (ii).

\item \textbf{page 122:} It is worth saying that the operators $W$, $B$ and
$C$ are known as \textit{complementation}, \textit{reversal} and
\textit{reverse-complementation}.

\item \textbf{page 122, proof of (D.1c):} \textquotedblleft Prove similarly
that $i\in\mathsf{T}$ implies $C\left(  i\right)  \in\Upsilon$%
\textquotedblright\ should be \textquotedblleft Prove similarly that
$i\in\Upsilon$ implies $C\left(  i\right)  \in\mathsf{T}$. Thus, $C$ restricts
to a map $\mathsf{T}\rightarrow\Upsilon$ and also to a map $\Upsilon
\rightarrow\mathsf{T}$. These two maps are mutually inverse, since
$C^{2}=\operatorname*{id}$.\textquotedblright.

\item \textbf{page 122, (D.1e):} I would replace \textquotedblleft column $t$
of $Z_{\lambda}$\textquotedblright\ by \textquotedblleft column $t$ of
$\left[  \lambda\right]  $\textquotedblright, so as to avoid an impression of
a circular definition.

\item \textbf{page 123, proof of Theorem (D.1g):} \textquotedblleft Let
$c\in\left\{  1,2,\ldots,n\right\}  $\textquotedblright\ should be
\textquotedblleft Let $c\in\left\{  1,2,\ldots,n-1\right\}  $%
\textquotedblright.

\item \textbf{page 123, proof of Theorem (D.1g):} \textquotedblleft From
\S A.3\textquotedblright\ could better be \textquotedblleft From
(A.3f)\textquotedblright.

\item \textbf{page 123, proof of Theorem (D.1g):} Remove the \textquotedblleft%
$=M_{c}^{i}$\textquotedblright\ part from \textquotedblleft Hence
$h_{c}^{KP\left(  i\right)  }\left(  r\right)  =h_{c}^{i}\left(  r\right)
=M_{c}^{i}$\textquotedblright.

\item \textbf{page 123, proof of Theorem (D.1g):} Instead of \textquotedblleft
We can calculate $M_{c}^{KP\left(  i\right)  }=M_{c}^{KT_{\lambda}}$ easily;
it is $\lambda_{c}-\lambda_{c+1}$, and it is attained at the last place
$\left(  1,\lambda_{1}\right)  $ of $KP\left(  i\right)  $. Therefore the
maximum $M_{c}^{i}$ of $h_{c}^{i}$ is also attained at the last place of
$i$\textquotedblright, it would be clearer to say: \textquotedblleft The
height function $h_{c}^{KP\left(  i\right)  }=h_{c}^{KT_{\lambda}}$ starts out
at $0$, then decreases by $1$ for each place in the $\left(  c+1\right)  $-st
row of $\lambda$, then increases by $1$ for each place in the $c$-th row of
$\lambda$; then it remains constant with the value $\lambda_{c}-\lambda_{c+1}$
ever after. Since $\lambda_{c}\geq\lambda_{c+1}$, we conclude that its maximum
value is $\lambda_{c}-\lambda_{c+1}=h_{c}^{KP\left(  i\right)  }\left(
r\right)  =h_{c}^{i}\left(  r\right)  $. In other words, $M_{c}^{KP\left(
i\right)  }=h_{c}^{i}\left(  r\right)  $. Hence, $M_{c}^{i}=M_{c}^{KP\left(
i\right)  }=h_{c}^{i}\left(  r\right)  $ as well\textquotedblright.

\item \textbf{page 123, proof of Theorem (D.1g):} You write: \textquotedblleft
We know $\widetilde{e}_{c}\left(  KP\left(  j\right)  \right)  =KP\left(
\widetilde{e}_{c}\left(  j\right)  \right)  $, by Proposition B, hence
$\widetilde{e}_{c}\left(  KP\left(  j\right)  \right)  =\infty$ for all
$c$\textquotedblright. This uses a variation of Proposition B instead of
Proposition B proper (Proposition B is about $\widetilde{f}_{c}$, not about
$\widetilde{e}_{c}$). But it is easier to replace the argument by a different
one: \textquotedblleft Let $c\in\left\{  1,2,\ldots,n-1\right\}  $. From
$j\in\mathsf{T}$, we obtain $\widetilde{e}_{c}\left(  j\right)  =\infty$ and
thus $M_{c}^{j}=h_{c}^{j}\left(  r\right)  $ by (A.3f). But $M_{c}^{KP\left(
j\right)  }=M_{c}^{j}$ by Lemma (C.6b) and Proposition (C.3p), and furthermore
$h_{c}^{KP\left(  j\right)  }\left(  r\right)  =h_{c}^{j}\left(  r\right)  $
since the letters of the word $KP\left(  j\right)  $ are a permutation of the
letters of $j$ (and both words have length $r$, so that the value of the
height function at $r$ is accounting for all their letters). Thus,
$h_{c}^{KP\left(  j\right)  }\left(  r\right)  =h_{c}^{j}\left(  r\right)
=M_{c}^{j}=M_{c}^{KP\left(  j\right)  }$, and therefore $\widetilde{e}%
_{c}\left(  KP\left(  j\right)  \right)  =\infty$ by (A.3f).\textquotedblright.

\item \textbf{page 124, proof of Theorem (D.1g):} After \textquotedblleft
Hence all entries of the first row of $P\left(  j\right)  $ are equal to
$1$\textquotedblright, I would add \textquotedblleft(since $P\left(  j\right)
$ is row-standard)\textquotedblright.

\item \textbf{page 124, proof of Theorem (D.1g):} When you say
\textquotedblleft Next consider the last entry, $t$ say, in the $s^{\text{th}%
}$ row of $P\left(  j\right)  $\textquotedblright, you should explain that you
are arguing by strong induction on $s$ (since you use the induction hypothesis
a few sentences later). So I would instead write: \textquotedblleft We shall
now show that all entries of the $s^{\text{th}}$ row of $P\left(  j\right)  $
are equal to $s$ for all $s\in\left\{  1,2,\ldots,n\right\}  $. We will prove
this by strong induction on $s$. So we assume this is proved for all entries
of rows $1,2,\ldots,s-1$, and we consider the last entry, $t$ say, in the
$s^{\text{th}}$ row of $P\left(  j\right)  $.\textquotedblright.

\item \textbf{page 124, proof of Theorem (D.1g):} After \textquotedblleft it
is constant on the letters of rows $1$ up to $s-1$\textquotedblright, add
\textquotedblleft(by the induction hypothesis, since $s-1<t-1$%
)\textquotedblright.

\item \textbf{page 124, proof of Theorem (D.1g):} Remove \textquotedblleft,
say $x$,\textquotedblright\ (you never use the notation $x$).

\item \textbf{page 124, proof of Theorem (D.1g):} After \textquotedblleft This
is a contradiction.\textquotedblright, I would add \textquotedblleft\ Hence,
$t\leq s$. Since $P\left(  j\right)  $ is row-standard, this shows that all
entries in the $s$-th row of $P\left(  j\right)  $ are $\leq s$. Since
$P\left(  j\right)  $ is column-standard, they cannot be $<s$, and thus they
are all $=s$.\textquotedblright.

\item \textbf{page 124, proof of Theorem (D.1g):} You write: \textquotedblleft
Using Lemma (C.6b) and Proposition B, as in the proof of (i), it is quite easy
to see that $i\in\Upsilon$\textquotedblright.

I think this needs more details (it is indeed similar to the corresponding
part of (i), but not completely analogous):

We want to show that $i\in\Upsilon$. Let $c\in\left\{  1,2,\ldots,n-1\right\}
$. We must prove that $\widetilde{f}_{c}\left(  i\right)  =\infty$.
Equivalently (by (A.3e)), we must prove that $M_{c}^{i}=0$. But Proposition
(C.3p) shows that the words $i$ and $KP\left(  i\right)  $ are connected by a
sequence of basic moves; hence, Lemma (C.6b) shows that their height functions
have the same maximum: $M_{c}^{i}=M_{c}^{KP\left(  i\right)  }$. Thus, it
remains to show that $M_{c}^{KP\left(  i\right)  }=0$. In other words, it
remains to show that all values of the height function $h_{c}^{KP\left(
i\right)  }$ are $\leq0$. In other words, it remains to show that
$h_{c}^{KP\left(  i\right)  }\left(  t\right)  \leq0$ for each $t\in\left\{
0,1,\ldots,r\right\}  $.

So let us show this. Fix $t\in\left\{  0,1,\ldots,r\right\}  $. The
construction of the tableau $Z_{\lambda}$ ensures that each appearance of $c$
in $Z_{\lambda}$ is followed by an appearance of $c+1$ one box further south
in the same column (since the entries of each column are $k+1,k+2,\ldots,n$
for some $k$). In the Knuth unwinding $KZ_{\lambda}$, the latter $c+1$ appears
earlier than the former $c$ (since it lies further south in $Z_{\lambda}$).
Thus, among the first $t$ letters of $KZ_{\lambda}$, there must be at least as
many $\left(  c+1\right)  $'s as there are $c$'s (since each column that
contributes a $c$ to these first $t$ letters must also contribute a $c+1$ to
them). In other words,%
\begin{align*}
&  \left(  \text{number of }\left(  c+1\right)  \text{'s among the first
}t\text{ letters of }KZ_{\lambda}\right) \\
&  \geq\left(  \text{number of }c\text{'s among the first }t\text{ letters of
}KZ_{\lambda}\right)  .
\end{align*}
But (A.3b) yields%
\begin{align*}
h_{c}^{KZ_{\lambda}}\left(  t\right)   &  =\left(  \text{number of }c\text{'s
among the first }t\text{ letters of }KZ_{\lambda}\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \text{number of }\left(  c+1\right)  \text{'s
among the first }t\text{ letters of }KZ_{\lambda}\right) \\
&  \leq0
\end{align*}
(by the preceding sentence). Since $P\left(  i\right)  =Z_{\lambda}$, we can
rewrite this as $h_{c}^{KP\left(  i\right)  }\left(  t\right)  \leq0$. This is
precisely what we needed to show. Thus, $i\in\Upsilon$ is proved.

\item \textbf{page 124, (D.1i):} The \textquotedblleft$\mathsf{Q}\left(
\lambda\right)  $\textquotedblright\ here means the tableau $\mathsf{Q}%
^{\left(  \lambda\right)  }$ defined in (C.2h). This should be said.

\item \textbf{page 124, proof of Proposition (D.1i):} Here is this proof in a
bit more detail:

From (D.1d), we have $i^{\lambda}=KT_{\lambda}$, and thus $P\left(
i^{\lambda}\right)  =P\left(  KT_{\lambda}\right)  =T_{\lambda}$ (by (C.2c)
(i)) and $Q\left(  i^{\lambda}\right)  =Q\left(  KT_{\lambda}\right)
=\mathsf{Q}^{\left(  \lambda\right)  }$ (by (C.2h)). On the other hand,
$P\left(  i^{\mathsf{Q}\left(  \lambda\right)  }\right)  =T_{\lambda}$ (by
(D.1g) (i) and (D.1h)) and $Q\left(  i^{\mathsf{Q}\left(  \lambda\right)
}\right)  =\mathsf{Q}\left(  \lambda\right)  $ (by (D.1h)). Hence, $P\left(
i^{\lambda}\right)  =T_{\lambda}=P\left(  i^{\mathsf{Q}\left(  \lambda\right)
}\right)  $ and $Q\left(  i^{\lambda}\right)  =\mathsf{Q}^{\left(
\lambda\right)  }=\mathsf{Q}\left(  \lambda\right)  =Q\left(  i^{\mathsf{Q}%
\left(  \lambda\right)  }\right)  $. Since any word $i$ is uniquely determined
by the pair $\left(  P\left(  i\right)  ,\ Q\left(  i\right)  \right)  $ (by
(B.6a)), we thus obtain $i^{\lambda}=i^{\mathsf{Q}\left(  \lambda\right)  }$.
The same argument, using $Z_{\lambda}$ and $i_{\lambda}$ instead of
$T_{\lambda}$ and $i^{\lambda}$, shows that $i_{\lambda}=i_{\mathsf{Q}\left(
\lambda\right)  }$.

\item \textbf{page 124, \S D.2:} \textquotedblleft(see
(A.4a))\textquotedblright\ should be \textquotedblleft(see
(A.4c))\textquotedblright.

\item \textbf{page 125, proof of Proposition (D.2b):} At the very beginning of
this proof, add the following: \textquotedblleft By Remark (A.3g) (5), the
cases $\widetilde{f}_{c}\left(  i\right)  =j$ and $\widetilde{e}_{c}\left(
i\right)  =j$ can be transformed into each other by swapping $i$ with $j$.
Thus, it suffices to handle one of them.\textquotedblright.

\item \textbf{page 125, proof of Proposition (D.2b):} \textquotedblleft By
Proposition B (see (C.6b))\textquotedblright\ should be \textquotedblleft By
Proposition B (see (C.6a))\textquotedblright.

\item \textbf{page 125, proof of Proposition (D.2b):} After \textquotedblleft
Now take $P=P\left(  i\right)  $ in Theorem (C.5b)\textquotedblright, add
\textquotedblleft(this is applicable, since $\widetilde{f}_{c}\left(
KP\left(  i\right)  \right)  =KP\left(  j\right)  \neq\infty$ guarantees that
$M_{c}^{KP\left(  i\right)  }\neq0$)\textquotedblright.

\item \textbf{page 125, proof of Proposition (D.2b):} After \textquotedblleft
Therefore $KP\left(  j\right)  =K\widetilde{P}$\textquotedblright, add
\textquotedblleft, and thus $P\left(  j\right)  =\widetilde{P}$ since any
tableau is uniquely determined by its Knuth unwinding (see (C.2c)
(i))\textquotedblright.

\item \textbf{page 125, proof of Proposition (D.2b):} After \textquotedblleft
If $q<r$, then $j^{\prime}=\widetilde{f}_{c}\left(  i^{\prime}\right)
$\textquotedblright, add \textquotedblleft(since $q<r$ shows that
$M^{i^{\prime}}=M^{i}$ and thus $q^{i^{\prime}}=q^{i}=q$, so that the
construction of $\widetilde{f}_{c}\left(  i^{\prime}\right)  $ from
$i^{\prime}$ changes the same $c$ as the construction of $\widetilde{f}%
_{c}\left(  i\right)  $ from $i$)\textquotedblright.

\item \textbf{page 125, proof of the \textquotedblleft only
if\textquotedblright\ part of Theorem 1:} Replace \textquotedblleft$i\left(
1\right)  ,\ \left(  2\right)  ,\ \ldots,\ i\left(  s\right)  $%
\textquotedblright\ by \textquotedblleft$i\left(  1\right)  ,\ i\left(
2\right)  ,\ \ldots,\ i\left(  s\right)  $\textquotedblright.

\item \textbf{page 125, last paragraph:} It is worth reminding that $\infty$
does not count as a word (and thus is not included in $\mathcal{S}\left(
w\right)  $).

\item \textbf{page 125, last paragraph:} After \textquotedblleft would be an
element of $\mathcal{S}\left(  w\right)  $ of size $S-1$.\textquotedblright%
\ (the last words on page 125), add \textquotedblleft Thus, $w^{\prime}%
\in\mathsf{T}$.\textquotedblright.

\item \textbf{page 126, first paragraph:} After \textquotedblleft But then
Theorem (D.1g)\textquotedblright, add \textquotedblleft(i)\textquotedblright.

\item \textbf{page 126, proof of Proposition (D.2d):} I would say a few words
about how the implication (1) $\Longrightarrow$ (3) in part (iii) is proven
(all the other claims do indeed follow from the things above). Namely, we
proceed similarly to the proof of the implication (1) $\Longrightarrow$ (2):
Let $w\in I\left(  Q,\approx\right)  $. Define $\mathcal{S}\left(  w\right)  $
to be the set of all words of the form $\widetilde{f}_{c_{1}}\widetilde{f}%
_{c_{2}}\cdots\widetilde{f}_{c_{t}}\left(  w\right)  $, where $c_{1}%
,c_{2},\ldots,c_{t}$ are arbitrary elements of $\left\{  1,2,\ldots
,n-1\right\}  $ (again, we allow $t$ to be $0$). This set $\mathcal{S}\left(
w\right)  $ is finite (being a subset of the finite set $I\left(  n,r\right)
$), and thus there exists an element $w^{\prime}$ of this set with largest
size. This element $w^{\prime}$ must then satisfy $\widetilde{f}_{c}\left(
w^{\prime}\right)  =\infty$ for all $c\in\left\{  1,2,\ldots,n-1\right\}  $
(since otherwise, $\widetilde{f}_{c}\left(  w^{\prime}\right)  $ would have
even larger size than $w^{\prime}$), and thus lies in $\Upsilon$. By
Proposition (D.2b), it satisfies $Q\left(  w^{\prime}\right)  =Q\left(
w\right)  =Q$ (since $w\in I\left(  Q,\approx\right)  $), so that $w^{\prime
}=i_{Q}$ (by Theorem (D.1g) (ii), since $w^{\prime}\in\Upsilon$). Hence,
$i_{Q}=w^{\prime}=\widetilde{f}_{c_{1}}\widetilde{f}_{c_{2}}\cdots
\widetilde{f}_{c_{t}}\left(  w\right)  $ for some $c_{1},c_{2},\ldots,c_{t}$
(since $w^{\prime}\in\mathcal{S}\left(  w\right)  $). Therefore,
$w=\widetilde{e}_{c_{t}}\cdots\widetilde{e}_{c_{2}}\widetilde{e}_{c_{1}%
}\left(  i_{Q}\right)  $. Thus, the implication (1) $\Longrightarrow$ (3) is proved.

\item \textbf{page 126, Weights:} \textquotedblleft Remember (see (A.3g)(3),
or \S 3.1)\textquotedblright\ should be \textquotedblleft Remember (see
(A.3g)(4), or \S 3.1)\textquotedblright.

\item \textbf{page 127, last line:} \textquotedblleft We shall see in
\S D.4\textquotedblright\ $\rightarrow$ \textquotedblleft We shall see in
\S D.8\textquotedblright.

\item \textbf{page 127, proof of Proposition (D.2f) (ii):} \textquotedblleft
For each $t\in\underline{n}$\textquotedblright\ should be \textquotedblleft
For each positive integer $t$\textquotedblright\ (the tableau $Z_{\lambda}$
can have more than $n$ columns).

\item \textbf{page 127, proof of Proposition (D.2f) (iii) and (iv):}
\textquotedblleft By (D.2d) we know that\textquotedblright\ should perhaps be
\textquotedblleft By (D.2d) (iii) we know that\textquotedblright.

\item \textbf{page 127, proof of Proposition (D.2f) (iii) and (iv):}
\textquotedblleft From (A.3g)(3)\textquotedblright\ should be
\textquotedblleft From (A.3g)(4)\textquotedblright.

\item \textbf{page 127, proof of Proposition (D.2f) (iii) and (iv):}
\textquotedblleft Therefore $i\trianglelefteq i^{Q}$\textquotedblright\ should
be \textquotedblleft Therefore $\mathsf{wt}\left(  i\right)  \trianglelefteq
\mathsf{wt}\left(  i^{Q}\right)  $\textquotedblright.

\item \textbf{page 128, between (D.3b) and (D.3c):} I would replace
\textquotedblleft and $n-i_{\nu}=n-c$\textquotedblright\ by \textquotedblleft
and $n-i_{\nu}+1=n-c+1$\textquotedblright\ (after all, the \textquotedblleft%
$+1$\textquotedblright s are in the formula that you are rewriting).

\item \textbf{page 128, between (D.3b) and (D.3c):} After \textquotedblleft So
(D.3b) gives\textquotedblright, add \textquotedblleft(upon substutituting
$\nu$ for $r-\rho+1$ and rewriting the conditions)\textquotedblright.

\item \textbf{page 128, after (D.3c):} \textquotedblleft for every subset
$\Pi$ of $\left\{  1,\ldots,s\right\}  $\textquotedblright\ should be
\textquotedblleft for every subset $\Pi$ of $\left\{  1,\ldots,t\right\}
$\textquotedblright.

\item \textbf{page 128, Lemma (D.3e):} It should be pointed out that you set
$C\infty:=\infty$, so that the statement makes sense even if some of the
operators yield $\infty$.

\item \textbf{page 128, proof of Lemma (D.3e):} \textquotedblleft in the
vertical line $x=r$\textquotedblright\ should be \textquotedblleft in the
vertical line $x=r/2$\textquotedblright\ (at least if reflection is understood
in the sense of elementary geometry).

\item \textbf{pages 128--129, proof of Lemma (D.3e):} The shorthands
\textquotedblleft$h$\textquotedblright\ and \textquotedblleft$\widetilde{h}%
$\textquotedblright\ for $h_{c}^{i}$ and $h_{n-c}^{Ci}$ are unnecessary and
only confusing in this proof; they should both be replaced by
\textquotedblleft$h_{c}^{i}$\textquotedblright\ and \textquotedblleft%
$h_{n-c}^{Ci}$\textquotedblright\ everywhere they are used.

\item \textbf{page 129, proof of Lemma (D.3e):} Remove the comma before
\textquotedblleft becomes the first maximum of\textquotedblright.

\item \textbf{page 129, proof of Lemma (D.3e):} After \textquotedblleft
assumes its maximum at place $r-\overline{q}$ for the first
time.\textquotedblright, add \textquotedblleft In other words, $q_{n-c}%
^{Ci}=r-\overline{q}$.\textquotedblright.

\item \textbf{page 129, proof of Lemma (D.3g):} \textquotedblleft the weight
of $C\left(  i^{Q}\right)  $ is $\left(  n^{\lambda_{1}},\left(  n-1\right)
^{\lambda_{2}},\ldots\right)  $\textquotedblright\ should be \textquotedblleft
the weight of $C\left(  i^{Q}\right)  $ is $\left(  \lambda_{n},\ldots
,\lambda_{1}\right)  $. In other words, the weight of $i_{R}$ is $\left(
\lambda_{n},\ldots,\lambda_{1}\right)  $ (since $i_{R}=C\left(  i\right)
=C\left(  i^{Q}\right)  $)\textquotedblright.

\item \textbf{page 129, proof of Lemma (D.3g):} This whole proof is
confusingly phrased and awkward with its unnecessary separation of two cases
that are really two parts of the same argument. Here is how I would write it:

Let $i\in I\left(  n,r\right)  $ have shape $\lambda$. We must show that $Ci$
has shape $\lambda$ as well.

Let $Q=Q\left(  i\right)  $, so that $Q\in\mathcal{Q}\left(  \lambda\right)
$. By the definition of $i^{Q}$, we have $Q\left(  i^{Q}\right)  =Q$ and
$i^{Q}\in\mathsf{T}$. The latter yields $Ci^{Q}\in C\left(  \mathsf{T}\right)
=\Upsilon$ by Lemma (D.1c). But each word $w\in\Upsilon$ has the form
$w=i_{R}$ for some standard tableau $R$ with entries $1,2,\ldots,r$ (indeed,
if we set $R:=Q\left(  w\right)  $, then $i_{R}$ is defined as the unique word
in $\Upsilon$ whose $Q$-symbol is $R$; but this word must be precisely $w$
because $w\in\Upsilon$ and $Q\left(  w\right)  =R$). Hence, the word $Ci^{Q}$
has this form (since $Ci^{Q}\in\Upsilon$). In other words, $Ci^{Q}=i_{R}$ for
some standard tableau $R$. Consider this $R$.

By Proposition (D.2f) (i), we know that $\mathsf{wt}\left(  i^{Q}\right)
=\lambda$. Hence, $\mathsf{wt}\left(  Ci^{Q}\right)  =\left(  \lambda
_{n},\ldots,\lambda_{1}\right)  $. In other words, $\mathsf{wt}\left(
i_{R}\right)  =\left(  \lambda_{n},\ldots,\lambda_{1}\right)  $, since
$C\left(  i^{Q}\right)  =i_{R}$. But Proposition (D.2f) (ii) shows that
$\mathsf{wt}\left(  i_{R}\right)  =\left(  \mu_{n},\ldots,\mu_{1}\right)  $,
where $\mu$ is the shape of $R$. Comparing these, we find $\left(  \lambda
_{n},\ldots,\lambda_{1}\right)  =\left(  \mu_{n},\ldots,\mu_{1}\right)  $,
hence $\lambda=\mu$. Thus, $\lambda$ is the shape of $R$ as well.

From Proposition (D.2d) (iii), we see that there are $c_{1},c_{2},\ldots
,c_{t}\in\left\{  1,2,\ldots,n-1\right\}  $ such that $i=\widetilde{f}_{c_{1}%
}\cdots\widetilde{f}_{c_{t}}\left(  i^{Q}\right)  $. Hence, by Lemma (D.3e),
we conclude that $Ci=\widetilde{e}_{n-c_{1}}\cdots\widetilde{e}_{n-c_{t}%
}\left(  Ci^{Q}\right)  =\widetilde{e}_{n-c_{1}}\cdots\widetilde{e}_{n-c_{t}%
}\left(  i_{R}\right)  $ (since $Ci^{Q}=i_{R}$). Since the operators
$\widetilde{e}_{c}$ preserve the shape of a word (because Proposition (D.2b)
shows that they even preserve its $Q$-symbol), we thus conclude that the shape
of $Ci$ is the shape of $i_{R}$. Hence, the shape of $Ci$ is $\lambda$, since
the shape of $i_{R}$ is $\lambda$. This completes the proof.

\item \textbf{page 130, definition of the Littelmann algebra:} It is worth
saying that algebras (and thus subalgebras) are not required to be unital here.

\item \textbf{page 130, (D.4b):} \textquotedblleft spanned the
set\textquotedblright\ should be \textquotedblleft spanned by the
set\textquotedblright.

\item \textbf{page 130, proof of Proposition (D.4e):} \textquotedblleft the
basis $\left\{  v_{i}:i\in I\right\}  $\textquotedblright\ should be
\textquotedblleft the basis $\left\{  v_{i}:i\in I\left(  n,r\right)
\right\}  $\textquotedblright.

\item \textbf{page 130, proof of Proposition (D.4e):} It is worth saying that
fact (iv) follows by induction on $w$ using fact (iii).

\item \textbf{page 130, proof of Proposition (D.4e):} After the final formula
on this page, add \textquotedblleft Thus, fact (iv) shows that $D_{S}\in L$,
since fact (ii) shows that each $D_{Z\left(  c\right)  }$ and each
$D_{Y\left(  c\right)  }$ lies in $L$.\textquotedblright.

\item \textbf{page 131, proof of Proposition (D.4e):} Replace
\textquotedblleft$\mathsf{End}_{F}\left(  S,S\right)  $\textquotedblright\ by
\textquotedblleft$\mathsf{End}_{F}\left(  S\right)  $\textquotedblright.

\item \textbf{page 131, \S D.5:} \textquotedblleft By Proposition B, this is
an $L$-submodule of $V^{\otimes r}$\textquotedblright\ should be
\textquotedblleft By Proposition (D.2b), this is an $L$-submodule of
$V^{\otimes r}$\textquotedblright.

\item \textbf{page 131, \S D.5:} After \textquotedblleft For $z=\sum_{i}%
\xi_{i}v_{i}\in V^{\otimes r}$\textquotedblright, add \textquotedblleft(with
$\xi_{i}\in F$)\textquotedblright.

\item \textbf{page 132, Lemma (D.5b):} \textquotedblleft lies in the
set\textquotedblright\ should be \textquotedblleft is a subset of
\textquotedblright.

\item \textbf{page 132, Corollary (D.5c):} \textquotedblleft lies
in\textquotedblright\ should be \textquotedblleft is a subset of
\textquotedblright.

\item \textbf{page 132, the paragraph below Corollary (D.5c):}
\textquotedblleft lies in\textquotedblright\ should be \textquotedblleft is a
subset of \textquotedblright.

\item \textbf{page 132, definition of an }$L$\textbf{-module:} After
\textquotedblleft so that $x\left(  ym\right)  =\left(  xy\right)
m$\textquotedblright, add \textquotedblleft and $\left(  x+y\right)
m=xm+ym$\textquotedblright. Also require that $\left(  \lambda x\right)
m=\lambda\left(  xm\right)  $ for all $\lambda\in F$ and $x\in L$ and $m\in M$.

\item \textbf{page 133, after (D.5f):} \textquotedblleft stated in
(A.3g)(4)\textquotedblright\ should be \textquotedblleft stated in
(A.3g)(5)\textquotedblright.

\item \textbf{page 133, (D.5g):} \textquotedblleft such that $\mathsf{sz}%
\left(  i\right)  \leq\mathsf{sz}\left(  j\right)  $ implies that $i\leq
j$\textquotedblright\ should be \textquotedblleft such that $\mathsf{sz}%
\left(  i\right)  <\mathsf{sz}\left(  j\right)  $ implies that $i\leq
j$\textquotedblright\ (otherwise, the total order would not exist, since
different words can have the same size).

\item \textbf{page 133, (D.5g):} Replace \textquotedblleft upper
triangular\textquotedblright\ and \textquotedblleft lower
triangular\textquotedblright\ by \textquotedblleft strictly upper
triangular\textquotedblright\ and \textquotedblleft strictly lower
triangular\textquotedblright, respectively.

\item \textbf{page 133, Corollary (D.5h):} Remove the comma after
\textquotedblleft subalgebra of $L$\textquotedblright.

\item \textbf{page 133, (D.5j):} This argumentation is a bit sloppy here. It
has been shown that the $L$-module $V^{\otimes r}$ is a direct sum of finitely
many simple $L$-modules (indeed, this follows from Lemma (D.5i) and from
$V^{\otimes r}=\bigoplus\limits_{\lambda\in\Lambda^{+}\left(  n,r\right)
}\ \ \bigoplus\limits_{Q\in\mathcal{Q}\left(  \lambda\right)  }M_{Q}$). In
other words, $V^{\otimes r}$ is completely reducible as an $L$-module. But
this does not immediately yield that $L$ itself is completely reducible as an
$L$-module, which would be necessary to apply [11, Theorem (25.2), page 164].
Instead, the complete reducibility of $L$ can be justified as follows:

The $L$-module $V^{\otimes r}$ is faithful (since $L$ is defined as an
$F$-algebra of endomorphisms of $V^{\otimes r}$). Hence, the canonical
$F$-algebra morphism $L\rightarrow\operatorname*{End}\nolimits_{F}\left(
V^{\otimes r}\right)  $ is injective. Moreover, this morphism is a morphism of
$L$-representations, if we equip $L$ with the left regular $L$-module
structure and equip $\operatorname*{End}\nolimits_{F}\left(  V^{\otimes
r}\right)  $ with the $L$-module structure given by post-action (i.e., the
$L$-action given by $\left(  \ell\alpha\right)  \left(  v\right)  =\ell
\cdot\alpha\left(  v\right)  $ for all $\ell\in L$ and $\alpha\in
\operatorname*{End}\nolimits_{F}\left(  V^{\otimes r}\right)  $ and $v\in
V^{\otimes r}$). Hence, the left regular $L$-module $L$ can be identified with
an $L$-submodule of $\operatorname*{End}\nolimits_{F}\left(  V^{\otimes
r}\right)  $. But the $L$-module $\operatorname*{End}\nolimits_{F}\left(
V^{\otimes r}\right)  $ is isomorphic to $\bigoplus\limits_{i\in I\left(
n,r\right)  }V^{\otimes r}$ (by the isomorphism that sends each $\alpha
\in\operatorname*{End}\nolimits_{F}\left(  V^{\otimes r}\right)  $ to the
family $\left(  \alpha\left(  v_{i}\right)  \right)  _{i\in I\left(
n,r\right)  }\in\bigoplus\limits_{i\in I\left(  n,r\right)  }V^{\otimes r}$),
and thus is completely reducible (since $V^{\otimes r}$ is completely
reducible as an $L$-module, and since complete reducibility is inherited by
finite direct sums). Hence, the left regular $L$-module $L$ is completely
reducible as well (since any submodule of a completely reducible $L$-module is
completely reducible -- see [11, (15.2)] for a proof). Now, [11, Theorem
(25.2)] really does show that $L$ is semisimple.

\item \textbf{page 135, proof of (D.7c):} You write: \textquotedblleft But
$KP\left(  i\right)  $ equals $KP\left(  KP\left(  i\right)  \right)  $, hence
its $Q$-symbol is $\mathsf{Q}^{\left(  \lambda\right)  }$ (see
(C.2i))\textquotedblright. It would be easier to argue that Proposition (C.2h)
yields $Q\left(  KP\left(  i\right)  \right)  =\mathsf{Q}^{\left(
\lambda\right)  }$.

\item \textbf{page 136, (D.7e):} In this diagram, both \textquotedblleft%
$\widetilde{f}_{c\left(  i\right)  }$\textquotedblright s should be
\textquotedblleft$\widetilde{f}_{c}\left(  i\right)  $\textquotedblright s.

\item \textbf{page 136, after (D.7e):} You write: \textquotedblleft In the
same way, one has a diagram like (D.7e), with $\widetilde{e}_{c}$ replacing
$\widetilde{f}_{c}$\textquotedblright. This requires the $\widetilde{e}_{c}%
$-analogue of Proposition B, which has never been stated explicitly and which
would need to be proved. Here is a quick way to do this:

\begin{statement}
\textbf{(D.7e') Proposition B'.} Let $i\in I\left(  n,r\right)  $ and
$c\in\left\{  1,2,\ldots,n-1\right\}  $. Then, $\widetilde{e}_{c}\left(
i\right)  \neq\infty$ holds if and only if $\widetilde{e}_{c}\left(  KP\left(
i\right)  \right)  \neq\infty$. Moreover, if these two inequalities hold, then
$\widetilde{e}_{c}\left(  KP\left(  i\right)  \right)  =KP\left(
\widetilde{e}_{c}\left(  i\right)  \right)  $.
\end{statement}

\textit{Proof.} We distinguish the cases $\widetilde{e}_{c}\left(  i\right)
\neq\infty$ and $\widetilde{e}_{c}\left(  i\right)  =\infty$:

\begin{itemize}
\item \textit{Case 1:} We have $\widetilde{e}_{c}\left(  i\right)  \neq\infty
$. Then, we must show that $\widetilde{e}_{c}\left(  KP\left(  i\right)
\right)  =KP\left(  \widetilde{e}_{c}\left(  i\right)  \right)  $ and
$\widetilde{e}_{c}\left(  KP\left(  i\right)  \right)  \neq\infty$.

Indeed, $\widetilde{e}_{c}\left(  i\right)  $ is a word (since $\widetilde{e}%
_{c}\left(  i\right)  \neq\infty$). Let us denote this word by $j$. Thus,
$j=\widetilde{e}_{c}\left(  i\right)  $, so that $\widetilde{f}_{c}\left(
j\right)  =i\neq\infty$. Therefore, (C.6a)\ (applied to $j$ instead of $i$)
yields $\widetilde{f}_{c}\left(  KP\left(  j\right)  \right)  \neq\infty$ and
$\widetilde{f}_{c}\left(  KP\left(  j\right)  \right)  =KP\left(
\widetilde{f}_{c}\left(  j\right)  \right)  $. The latter equality entails
\begin{align*}
KP\left(  j\right)   &  =\widetilde{e}_{c}\left(  KP\left(
\underbrace{\widetilde{f}_{c}\left(  j\right)  }_{=i}\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widetilde{e}_{c}\text{ undoes
}\widetilde{f}_{c}\right) \\
&  =\widetilde{e}_{c}\left(  KP\left(  i\right)  \right)  .
\end{align*}
Hence, $\widetilde{e}_{c}\left(  KP\left(  i\right)  \right)  =KP\left(
\underbrace{j}_{=\widetilde{e}_{c}\left(  i\right)  }\right)  =KP\left(
\widetilde{e}_{c}\left(  i\right)  \right)  \neq\infty$. This proves (D.7e')
in Case 1.

\item \textit{Case 2:} We have $\widetilde{e}_{c}\left(  i\right)  =\infty$.
Then, we must show that $\widetilde{e}_{c}\left(  KP\left(  i\right)  \right)
=\infty$ as well.

The word $KP\left(  i\right)  $ can be obtained from $i$ by a sequence of
basic moves (by Proposition (C.3p)). Each basic move might change the height
function of the word, but leaves the maximum of this height function unchanged
(by the first claim in Lemma (C.6b)). Hence, the height functions of the words
$KP\left(  i\right)  $ and $i$ have the same maximum. In other words,
$M_{c}^{KP\left(  i\right)  }=M_{c}^{i}$. Moreover, $h_{c}^{i}\left(
r\right)  $ is the total weight of all letters of $i$ (since $i$ has $r$
letters), whereas $h_{c}^{KP\left(  i\right)  }\left(  r\right)  $ is the
total weight of all letters of $KP\left(  i\right)  $ (for similar reasons).
Since the words $i$ and $KP\left(  i\right)  $ have the same letters (just in
a different order), this shows that $h_{c}^{i}\left(  r\right)  =h_{c}%
^{KP\left(  i\right)  }\left(  r\right)  $.

But $\widetilde{e}_{c}\left(  i\right)  =\infty$, and thus $M_{c}^{i}%
=h_{c}^{i}\left(  r\right)  $ (by the definition of $\widetilde{e}_{c}$ in
(A.3f)). Hence, $M_{c}^{KP\left(  i\right)  }=M_{c}^{i}=h_{c}^{i}\left(
r\right)  =h_{c}^{KP\left(  i\right)  }\left(  r\right)  $. Therefore,
$\widetilde{e}_{c}\left(  KP\left(  i\right)  \right)  =\infty$ (again by the
definition of $\widetilde{e}_{c}$). Thus, (D.7e') is proved in Case 2.
$\blacksquare$
\end{itemize}

\begin{noncompile}
However, there is a way to sidestep this gap and give a nicer proof of (D.7f),
without involving $\mathsf{Q}^{\left(  \lambda\right)  }$. This proof uses the
following simple lemma:

\textbf{(D.7f') Lemma.} Let $P\in\mathcal{Q}\left(  \lambda\right)  $ and
$Q\in\mathcal{Q}\left(  \lambda\right)  $ and $c\in\left\{  1,2,\ldots
,n-1\right\}  $. If the tableau $\widetilde{f}_{c}P$ defined in \S C.5 (that
is, the tableau $\widetilde{P}$ from Theorem (C.5b)) exists, then
$\widetilde{f}_{c}\left(  P:Q\right)  =\left(  \widetilde{f}_{c}P\right)  :Q$.
Otherwise, $\widetilde{f}_{c}\left(  P:Q\right)  =\infty$.

\textit{Proof.} If the tableau $\widetilde{f}_{c}P$ defined in \S C.5 exists,
then it satisfies $K\left(  \widetilde{f}_{c}P\right)  =\widetilde{f}%
_{c}\left(  KP\right)  $ (by Theorem (C.5b) (4)), and thus $Q\left(  K\left(
\widetilde{f}_{c}P\right)  \right)  =Q\left(  \widetilde{f}_{c}\left(
KP\right)  \right)  =Q\left(  KP\right)  $

TODO

Fix $Q,R\in\mathcal{Q}\left(  \lambda\right)  $. Recall that the algebra $L$
is generated by the $\widetilde{f}_{c}$ and the $\widetilde{e}_{c}$ operators.
Thus, it suffices to show that $\gamma_{Q,R}\left(  \widetilde{f}_{c}v\right)
=\widetilde{f}_{c}\left(  \gamma_{Q,R}v\right)  $ and $\gamma_{Q,R}\left(
\widetilde{e}_{c}v\right)  =\widetilde{e}_{c}\left(  \gamma_{Q,R}v\right)  $
for all $i\in I_{\lambda}\left(  Q,\approx\right)  $ and all $c\in\left\{
1,2,\ldots,n-1\right\}  $.

TODO

$\gamma_{Q,R}\left(  \widetilde{f}_{c}v_{i}\right)  =\widetilde{f}_{c}\left(
\gamma_{Q,R}v_{i}\right)  $ and $\gamma_{Q,R}\left(  \widetilde{e}_{c}%
v_{i}\right)  =\widetilde{e}_{c}\left(  \gamma_{Q,R}v_{i}\right)  $ for all
$i\in I_{\lambda}\left(  Q,\approx\right)  $ and all $c\in\left\{
1,2,\ldots,n-1\right\}  $.
\end{noncompile}

\item \textbf{page 136, proof of (D.7h):} Replace \textquotedblleft%
$\operatorname*{Hom}$\textquotedblright\ by \textquotedblleft$\mathsf{Hom}%
$\textquotedblright\ (different font).

\item \textbf{page 141, (D.10b):} Replace \textquotedblleft$t\geq
0$\textquotedblright\ by \textquotedblleft$t\geq1$\textquotedblright.

\item \textbf{page 142:} On the first line, replace \textquotedblleft
tableaux\textquotedblright\ by \textquotedblleft tableau\textquotedblright.

\item \textbf{page 142:} The first displayed equation on this page is saying
\[
\text{\textquotedblleft\ }t^{r_{tt}}\left(  t-1\right)  ^{r_{t-1,t}}%
\ldots1^{r_{1t}}\ \text{\textquotedblright},
\]
but it should be the other way round:
\[
\text{\textquotedblleft\ }1^{r_{1}t}2^{r_{2}t}\cdots t^{r_{tt}}%
\ \text{\textquotedblright}.
\]


\item \textbf{page 142, item (1):} \textquotedblleft the sum of the entries in
row $s$\textquotedblright\ $\rightarrow$ \textquotedblleft the number of the
entries in row $s$\textquotedblright.

\item \textbf{page 142, item (3):} Remove the period at the end of the
displayed inequality.

\item \textbf{page 142, last paragraph:} This is a bit trickier than you make
it sound: Why exactly is the matrix $U$ upper-triangular?

In truth, the triangularity of the matrix $U$ is a red herring, and it would
do no harm to ignore it completely; it would only simplify the proof.

\begin{noncompile}
I believe the whole proof of (D.10b) would profit from taking a more general
and symmetric viewpoint:

Define $\mathbb{N}$ to be the set $\left\{  0,1,2,\ldots\right\}  $. Let
$\mu=\left(  \mu_{1},\mu_{2},\ldots,\mu_{n}\right)  \in\Lambda^{+}\left(
n,r\right)  $ be a dominant weight and $\kappa=\left(  \kappa_{1},\kappa
_{2},\ldots,\kappa_{n}\right)  \in\mathbb{N}^{n}$ be any $n$-tuple of
nonnegative integers. Let $m\in\mathbb{N}$ be arbitrary. Then, we define a
$\mu$\emph{-shifted }$\kappa$\emph{-array} with entries in $\underline{m}$ to
be a map $T$ from the set%
\[
\left[  \mu+\kappa\right]  \setminus\left[  \mu\right]  =\left\{  \left(
i,j\right)  \ \mid\ i\in\underline{n}\text{ and }j\in\mathbb{Z}\text{ and }%
\mu_{i}<j\leq\mu_{i}+\kappa_{i}\right\}
\]
to the set $\underline{m}$ (that is, a way of filling elements of
$\underline{m}$ into the cells $\left(  i,j\right)  $ with $i\in\underline{n}$
and $j\in\mathbb{Z}$ and $\mu_{i}<j\leq\mu_{i}+\kappa_{i}$) in such a way that
the entries (i.e., the values of this map) weakly increase left-to-right along
each row (i.e., we have $T\left(  i,j\right)  \leq T\left(  i,j+1\right)  $
whenever $\left(  i,j\right)  $ and $\left(  i,j+1\right)  $ both belong to
$\left[  \mu+\kappa\right]  \setminus\left[  \mu\right]  $). We say that a
$\mu$-shifted $\kappa$-array $T$ is \emph{standard} if it furthermore has the
property that $\mu+\kappa$ is a dominant weight (i.e., we have $\mu_{i}%
+\kappa_{i}\geq\mu_{i+1}+\kappa_{i+1}$ for all $i<n$) and that the entries of
$T$ strictly increase top-to-bottom along each column (i.e., we have $T\left(
i,j\right)  <T\left(  i^{\prime},j\right)  $ whenever $\left(  i,j\right)  $
and $\left(  i^{\prime},j\right)  $ both belong to $\left[  \mu+\kappa\right]
\setminus\left[  \mu\right]  $ and satisfy $i<i^{\prime}$). For example,%
\begin{equation}
T_{1}=%
\begin{tabular}
[c]{cc|c|cc}\cline{3-5}
&  & $2$ & $2$ & \multicolumn{1}{|c|}{$3$}\\\cline{3-5}
&  & $2$ & $3$ & \multicolumn{1}{|c}{}\\\cline{2-5}\cline{3-4}
& \multicolumn{1}{|c|}{$1$} & $3$ & $3$ & \multicolumn{1}{|c|}{$4$}\\\hline
\multicolumn{1}{|c}{$1$} & \multicolumn{1}{|c|}{$2$} & $2$ &  & \\\cline{1-3}%
\end{tabular}
\label{p142.1}%
\end{equation}
is a $\left(  2,2,1,0\right)  $-shifted $\left(  3,2,4,2\right)  $-array (with
entries in $\underline{m}$ for any $m\geq4$) that is not standard (for many
reasons, e.g, because it fails the $\mu_{2}+\kappa_{3}\geq\mu_{3}+\kappa_{3}$
test), whereas%
\begin{equation}
T_{2}=%
\begin{tabular}
[c]{cc|c|cc}\cline{3-5}
&  & $2$ & $2$ & \multicolumn{1}{|c|}{$3$}\\\cline{3-5}\cline{4-5}
&  & $3$ & $4$ & \multicolumn{1}{|c|}{$4$}\\\cline{2-5}\cline{3-5}
& \multicolumn{1}{|c|}{$1$} & $5$ & $5$ & \multicolumn{1}{|c|}{$7$}\\\hline
\multicolumn{1}{|c}{$1$} & \multicolumn{1}{|c|}{$2$} & $6$ &  & \\\cline{1-3}%
\end{tabular}
\label{p142.2}%
\end{equation}
is a $\left(  2,2,1,0\right)  $-shifted $\left(  3,3,4,3\right)  $-array (with
entries in $\underline{m}$ for any $m\geq7$) that is standard. The following
are two important examples:

\begin{itemize}
\item If $\mu$ is the trivial weight $\mathbf{0}=\left(  0,0,\ldots,0\right)
\in\Lambda^{+}\left(  n,0\right)  $, then a $\mu$-shifted $\kappa$-array is
just a row-standard but not necessarily standard $\kappa$-tableau.

\item Skew tableaux of shape $\nu\setminus\mu$ are $\mu$-shifted $\left(
\nu-\mu\right)  $-arrays (where $\nu-\mu$ denotes the entrywise difference of
$\nu$ and $\mu$).
\end{itemize}

Any $\mu$-shifted $\kappa$-array $T$ has a \emph{Semitic reading word}
$w\left(  T\right)  $, which is obtained by reading each row of $T$ from right
to left and concatenating these readings from top to bottom (i.e., first the
reversed $1$-st row, then the reversed $2$-nd row, etc.). This is the reverse
of the Knuth unwinding $KT$. For instance, the $\mu$-shifted $\kappa$-array
$T_{1}$ in (\ref{p142.1}) has Semitic reading word $w\left(  T_{1}\right)
=322324331221$.

For any $\mu$-shifted $\kappa$-array $T$ with entries in $\underline{m}$, we
define the \emph{row-multiplicity matrix} $U_{T}$ to be the $n\times n$-matrix
$\left(  u_{i,j}\right)  _{i\in\underline{n},\ j\in\underline{m}}$, where
$u_{i,j}$ is the number of letters $j$ in the $i$-th row of $T$. For instance,
the $\mu$-array $T_{1}$ in (\ref{p142.1}) has row-multiplicity matrix
$U_{T_{1}}=\left(
\begin{array}
[c]{cccc}%
0 & 2 & 1 & 0\\
0 & 1 & 1 & 0\\
1 & 0 & 2 & 1\\
1 & 2 & 0 & 0
\end{array}
\right)  $ (for $m=4$).

Clearly, a $\mu$-shifted $\kappa$-array $T$ can be uniquely reconstructed from
its row-multiplicity matrix $U_{T}$ and the dominant weight $\mu$ (since the
matrix $U_{T}$ tells us how many times each number appears in each row, and
the \textquotedblleft weakly increasing along rows\textquotedblright%
\ condition dictates us where to put these numbers). Even better: If we don't
regard $\kappa$ as fixed but rather allow it to vary over all $n$-tuples of
nonnegative integers, then the map%
\begin{align*}
\Phi_{\mu}:\bigsqcup_{\kappa\in\mathbb{N}^{n}}\left\{  \mu\text{-shifted
}\kappa\text{-arrays with entries in }\underline{m}\right\}   &
\rightarrow\mathbb{N}^{n\times m},\\
T  &  \mapsto U_{T}%
\end{align*}
is a bijection. It furthermore has the following properties:

\begin{statement}
\textbf{(D.10c) Proposition.} Let $T$ be any $\mu$-shifted $\kappa$-array with
entries in $\underline{m}$ and with row-multiplicity matrix $U_{T}=\left(
u_{i,j}\right)  _{i\in\underline{n},\ j\in\underline{m}}$. Let $j\in
\underline{m}$. Then, the number of entries $j$ in $T$ is $\sum_{i\in
\underline{n}}u_{i,j}$.
\end{statement}

\textit{Proof.} Obvious. $\blacksquare$

\begin{statement}
\textbf{(D.10d) Proposition.} A $\mu$-shifted $\kappa$-array $T$ is standard
if and only if its row-multiplicity matrix $U_{T}=\left(  u_{i,j}\right)
_{i\in\underline{n},\ j\in\underline{m}}$ satisfies%
\[
\mu_{s}+\sum_{y=1}^{v}u_{y,s}\geq\mu_{s+1}+\sum_{y=1}^{v+1}u_{y,s+1}%
\ \ \ \ \ \ \ \ \ \ \text{for all }s<n\text{ and }v<m
\]
(we allow $v=0$ here).
\end{statement}

\textit{Proof sketch.} We extend $T$ to a $\left(  \mu+\kappa\right)
$-tableau, by filling all cells in $\left[  \mu\right]  $ with the letter $0$.

Let $s<n$ and $v<m$ be arbitrary. Then, the rightmost column in which the
$s$-th row of $T$ has an entry $\leq v$ (this includes the case when this
entry is $0$) is the $\left(  \mu_{s}+\sum_{y=1}^{v}u_{y,s}\right)  $-th
column, whereas the rightmost column in which the $\left(  s+1\right)  $-th
row of $T$ has an entry $\leq v+1$ is the $\left(  \mu_{s+1}+\sum_{y=1}%
^{v+1}u_{y,s+1}\right)  $-th column. Thus, if $\mu_{s}+\sum_{y=1}^{v}%
u_{y,s}<\mu_{s+1}+\sum_{y=1}^{v+1}u_{y,s+1}$, then there is a column of $T$
that has an entry $>v$ in the $s$-th row but an entry $\leq v+1$ in the
$\left(  s+1\right)  $-st row. The existence of such two entries prevents $T$
from being standard (because the entry in the $s$-th row is not smaller than
the one in the $\left(  s+1\right)  $-st row). Conversely, if no two such
entries exist (for all choices of $s<n$ and $v<m$), then $T$ is easily seen to
be standard. Combining this all, we quickly obtain (D.10d). $\blacksquare$

Now, let $\nu=\left(  \nu_{1},\nu_{2},\ldots,\nu_{n}\right)  \in\Lambda
^{+}\left(  n,r\right)  $ be a further dominant weight and $\rho=\left(
\rho_{1},\rho_{2},\ldots,\rho_{n}\right)  \in\mathbb{N}^{n}$ be any further
$n$-tuple of nonnegative integers.

\begin{statement}
\textbf{(D.10e) Proposition.} Let $T$ be a $\mu$-shifted $\kappa$-array with
entries in $\underline{m}$. Then, the concatenation $B\left(  i^{\nu}\right)
\mid w\left(  T\right)  $ is a lattice permutation if and only if the
row-multiplicity matrix $U_{T}=\left(  u_{i,j}\right)  _{i\in\underline{n}%
,\ j\in\underline{m}}$ of $T$ satisfies%
\[
\nu_{c}+\sum_{y=1}^{r}u_{y,c}\geq\nu_{c+1}+\sum_{y=1}^{r+1}u_{y,c+1}%
\ \ \ \ \ \ \ \ \ \ \text{for all }c<m\text{ and }r<n
\]
(we allow $r=0$ here).
\end{statement}

\textit{Proof sketch.} The word $B\left(  i^{\nu}\right)  \mid w\left(
T\right)  $ (over the alphabet $\left\{  1,2,\ldots,m\right\}  $) looks as
follows:%
\[
\underbrace{\left(  1^{\nu_{1}}2^{\nu_{2}}3^{\nu_{3}}\cdots\right)
}_{\text{this is }B\left(  i^{\nu}\right)  }\underbrace{\left(  \cdots
x^{u_{1,x}}\cdots2^{u_{1,2}}1^{u_{1,1}}\right)  \left(  \cdots x^{u_{2,x}%
}\cdots2^{u_{2,2}}1^{u_{2,1}}\right)  \left(  \cdots x^{u_{3,x}}%
\cdots2^{u_{3,2}}1^{u_{1,1}}\right)  \cdots}_{\text{this is }w\left(
T\right)  }.
\]
In order for this word $B\left(  i^{\nu}\right)  \mid w\left(  T\right)  $ to
be a lattice permutation, it needs to have the property that each of its
prefixes (i.e., initial sections) has at least as many $c$'s as it has
$\left(  c+1\right)  $'s, for each $c\in\left\{  1,2,\ldots,m-1\right\}  $. In
other words, it needs to have the property that its height function
$h_{c}^{B\left(  i^{\nu}\right)  \mid w\left(  T\right)  }$ stays always
$\geq0$ for each $c\in\left\{  1,2,\ldots,m-1\right\}  $. This property needs
only to be checked at the local minima of this height function; and it is easy
to see that these local minima can only be attained immediately after a block
of $\left(  c+1\right)  $'s (and at the very beginning of the word, but this
value is clearly $\geq0$). So it suffices to check that the values of the
height function $h_{c}^{B\left(  i^{\nu}\right)  \mid w\left(  T\right)  }$
are $\geq0$

\begin{itemize}
\item at the position immediately after the first $\left(  c+1\right)
^{\nu_{c+1}}$ within the $\left(  1^{\nu_{1}}2^{\nu_{2}}3^{\nu_{3}}%
\cdots\right)  $ block;

\item at the position immediately after an $\left(  c+1\right)  ^{u_{r,c+1}}$
block (in the part of $w\left(  T\right)  $ coming from the $r$-th row of $T$)
for each $r\geq1$.
\end{itemize}

But the value at the former position is $\nu_{c}-\nu_{c+1}$, which is always
$\geq0$ since $\nu\in\Lambda^{+}\left(  n,r\right)  $. At a position of the
latter kind, the value of $h_{c}^{B\left(  i^{\nu}\right)  \mid w\left(
T\right)  }$ is%
\begin{equation}
\left(  \nu_{c}+\sum_{y=1}^{r-1}u_{y,c}\right)  -\left(  \nu_{c+1}+\sum
_{y=1}^{r}u_{y,c+1}\right)  . \label{p142.6}%
\end{equation}
So we need all these values (\ref{p142.6}) to be $\geq0$ in order for the word
$B\left(  i^{\nu}\right)  \mid w\left(  T\right)  $ to be a lattice
permutation. In other words, we need%
\[
\nu_{c}+\sum_{y=1}^{r-1}u_{y,c}\geq\nu_{c+1}+\sum_{y=1}^{r}u_{y,c+1}%
\ \ \ \ \ \ \ \ \ \ \text{for all }c<m\text{ and }r\in\left\{  1,2,\ldots
,n\right\}
\]
in order for the word $B\left(  i^{\nu}\right)  \mid w\left(  T\right)  $ to
be a lattice permutation. Substituting $r+1$ for $r$ in this claim, we
conclude that we need%
\[
\nu_{c}+\sum_{y=1}^{r}u_{y,c}\geq\nu_{c+1}+\sum_{y=1}^{r+1}u_{y,c+1}%
\ \ \ \ \ \ \ \ \ \ \text{for all }c<m\text{ and }r<n
\]
in order for the word $B\left(  i^{\nu}\right)  \mid w\left(  T\right)  $ to
be a lattice permutation. This yields (D.10e). $\blacksquare$

\begin{statement}
\textbf{(D.10f) Proposition.} There is a bijection $\Psi=\Psi_{\mu,\nu}$ from
$\left\{  \mu\text{-shifted }\ast\text{-arrays}\right\}  $ to $\left\{
\nu\text{-shifted }\ast\text{-arrays}\right\}  $ with the following properties:

\textbf{(i)} A given $\mu$-shifted $\ast$-array $T$ is a $\mu$-shifted
$\kappa$-array if and only if its image $\Psi\left(  T\right)  $ has weight
$\kappa$.

\textbf{(ii)} A given $\mu$-shifted $\ast$-array $T$ has weight $\rho$ if and
only if its image $\Psi\left(  T\right)  $ is a $\nu$-shifted $\rho$-array.

\textbf{(iii)} A given $\mu$-shifted $\ast$-array $T$ has the property that
the word $B\left(  i^{\mu}\right)  \mid w\left(  T\right)  $ is a lattice
permutation if and only if its image $\Psi\left(  T\right)  $ is standard.

\textbf{(iv)} ...
\end{statement}
\end{noncompile}

\begin{noncompile}
TODO: FInish this!
\end{noncompile}

\item \textbf{page 143, \S D.11:} \textquotedblleft Chapter 6 of the
collective work\textquotedblright\ should be \textquotedblleft Chapter 5 of
the collective work\textquotedblright. Likewise, all references to LLT in this
section suffer from the same numbering shift: any number starting with
\textquotedblleft6.\textquotedblright\ should be changed to start with
\textquotedblleft5.\textquotedblright.

Also, it is worth warning the reader that LLT use the notation $0$ for what is
called $\infty$ in this appendix (that is, the undefined value of
$\widetilde{f}_{c}$ and $\widetilde{e}_{c}$), and that the definition of
$\sigma_{i}$ in LLT's Section 5.5 suffers from a typo (it says
\textquotedblleft$\sigma_{i}\left(  a_{i}^{r}a_{i+1}^{s}\right)  =a_{i}%
^{s}a_{i+1}^{s}$\textquotedblright\ while meaning \textquotedblleft$\sigma
_{i}\left(  a_{i}^{r}a_{i+1}^{s}\right)  =a_{i}^{s}a_{i+1}^{r}$%
\textquotedblright).

\item \textbf{page 143, (D.11e):} \textquotedblleft\textit{plactid}%
\textquotedblright\ $\rightarrow$ \textquotedblleft\textit{plactic}%
\textquotedblright.

\item \textbf{page 144, (D.11e):} \textquotedblleft implies $uu^{\prime}\sim
vv^{\prime}$\textquotedblright\ should be \textquotedblleft implies $uv\sim
u^{\prime}v^{\prime}$\textquotedblright.

\item \textbf{page 144, (D.11e):} Remove the comma after \textquotedblleft
with the $\sim$-class of $u^{\prime}$\textquotedblright.

\item \textbf{page 144, (D.11f):} It would be worth mentioning that
$\mathbb{Z}\left[  M\right]  $ is known as the \emph{monoid ring} of $M$.

\item \textbf{page 145, (D.11g):} Remove the comma after \textquotedblleft
Then a further generalization\textquotedblright.

\item \textbf{page 145, (D.11g):} The last sentence here (\textquotedblleft
Notice that $d_{\lambda}$ appears in the \textquotedblleft$\lambda
$-rectangle\textquotedblright\ (D.6d)\textquotedblright) looks wrong to me:
The number of plactic classes of \textbf{weight} $\lambda$ is not the same as
the number of plactic classes of \textbf{shape} $\lambda$ (which is the
$d_{\lambda}$ from (D.6d)).

\item \textbf{page 146, (D.11i):} After \textquotedblleft is the last
place\textquotedblright, add \textquotedblleft of $k$\textquotedblright.

\item \textbf{page 146, (D.11i):} After \textquotedblleft is the
place\textquotedblright, add \textquotedblleft of $k$\textquotedblright.
\end{enumerate}

\subsection{Corrections to \S E}

\begin{enumerate}
\item \textbf{page 149:} \textquotedblleft the directed graph $\Gamma$ in
(D.11i)\textquotedblright\ should be \textquotedblleft the directed graph
$\Gamma$ in (D.11h)\textquotedblright.

\item \textbf{page 149:} On the last line, \textquotedblleft%
(C.2c)\textquotedblright\ should be \textquotedblleft(C.2h)\textquotedblright.
\end{enumerate}

\bigskip

\section{\label{sec.LRproof}A shorter path to the Littlewood--Richardson rule}

\textbf{pages 129--140, sections D.4 till D.9:} This is all nice and
interesting, but the Littlewood--Richardson rule (D.10a) can be proved quite
easily without any use of the Littelmann algebra. Here is how this proof goes:

\begin{statement}
\textbf{(D.12a) Lemma.} Let $i\in I\left(  n,r\right)  $ and $j\in I\left(
n,s\right)  $ be two words. Then, any word that is $\approx$-equivalent to the
concatenation $i\mid j$ must itself be a concatenation $i^{\prime}\mid
j^{\prime}$, where $i^{\prime}\in I\left(  n,r\right)  $ and $j^{\prime}\in
I\left(  n,s\right)  $ are two words with $i^{\prime}\approx i$ and
$j^{\prime}\approx j$.
\end{statement}

\textit{Proof.} Let $w$ be a word that is $\approx$-equivalent to the
concatenation $i\mid j$. We must prove that $w$ is itself a concatenation
$i^{\prime}\mid j^{\prime}$, where $i^{\prime}\in I\left(  n,r\right)  $ and
$j^{\prime}\in I\left(  n,s\right)  $ are two words with $i^{\prime}\approx i$
and $j^{\prime}\approx j$.

By Theorem A (see (D.2a)), the word $w$ can be obtained from $i\mid j$ by a
sequence of operations of the form $\widetilde{e}_{c}$ and $\widetilde{f}_{c}%
$. Hence, it suffices to consider the case when $w=\widetilde{e}_{c}\left(
i\mid j\right)  $ or $w=\widetilde{f}_{c}\left(  i\mid j\right)  $ for some
$c\in\left\{  1,2,\ldots,n-1\right\}  $ (the general case will then follow by
induction). Let us thus assume that we are in this case. Fix the $c\in\left\{
1,2,\ldots,n-1\right\}  $ such that $w=\widetilde{e}_{c}\left(  i\mid
j\right)  $ or $w=\widetilde{f}_{c}\left(  i\mid j\right)  $.

Now, (A.3g) (6) shows that $\widetilde{f}_{c}\left(  i\mid j\right)  $ is
either $\widetilde{f}_{c}\left(  i\right)  \mid j$ or $i\mid\widetilde{f}%
_{c}\left(  j\right)  $ (or $\infty$). Thus, if $w=\widetilde{f}_{c}\left(
i\mid j\right)  $, then $w=\widetilde{f}_{c}\left(  i\right)  \mid j$ or
$w=i\mid\widetilde{f}_{c}\left(  j\right)  $. In either case, $w$ thus has the
form $i^{\prime}\mid j^{\prime}$, where $i^{\prime}\in I\left(  n,r\right)  $
and $j^{\prime}\in I\left(  n,s\right)  $ are two words with $i^{\prime
}\approx i$ and $j^{\prime}\approx j$ (because Theorem A shows that
$\widetilde{f}_{c}\left(  i\right)  \approx i$ and $\widetilde{f}_{c}\left(
j\right)  \approx j$). Hence, we are done in the case when $w=\widetilde{f}%
_{c}\left(  i\mid j\right)  $.

The case when $w=\widetilde{e}_{c}\left(  i\mid j\right)  $ is completely
analogous. Hence, we are done in all cases, and Lemma (D.12a) is proved.
$\blacksquare$ \medskip

Now, we let $A^{\ast}$ denote the set of all words with letters in
$\underline{n}$. Thus, $A^{\ast}=\bigcup_{r\geq0}I\left(  n,r\right)  $, and
this is a disjoint union. The set $A^{\ast}$ is a monoid with respect to
concatenation ($ij=i\mid j$ for any two words $i,j\in A^{\ast}$), and thus has
a monoid ring $\mathbb{Z}\left[  A^{\ast}\right]  $. The elements of the
latter ring $\mathbb{Z}\left[  A^{\ast}\right]  $ are formal $\mathbb{Z}%
$-linear combinations of words.

An element $f\in\mathbb{Z}\left[  A^{\ast}\right]  $ is said to be $\approx
$\textit{-invariant} if it has the property that whenever $i,j\in A^{\ast}$
are two words satisfying $i\approx j$, the coefficients of $i$ and $j$ in $f$
are equal. For instance, the element%
\[
2\cdot\left[  133\right]  -3\cdot\left[  331\right]  -3\cdot\left[
313\right]  +5\cdot\left[  1234\right]
\]
(here we put words in square brackets, to avoid confusing them with the scalar
factors in front of them) is $\approx$-invariant, but the element $\left[
331\right]  $ itself is not (since the word $313$ is $\approx$-equivalent to
$331$ but appears with a different coefficient in $\left[  331\right]  $).

Let $\mathbb{Z}\left[  A^{\ast}\right]  _{\approx}$ denote the set of all
$\approx$-invariant elements of $\mathbb{Z}\left[  A^{\ast}\right]  $.

We use the notations $\mathcal{P}\left(  \lambda\right)  $ and $\mathcal{Q}%
\left(  \lambda\right)  $ and $P:Q$ defined in (D.6a) and (D.6b) and (D.6c).
For any $\lambda\in\Lambda^{+}\left(  n,r\right)  $ and any $Q\in
\mathcal{Q}\left(  \lambda\right)  $, we let%
\[
\left(  \ast:Q\right)  :=\sum_{P\in\mathcal{P}\left(  \lambda\right)  }\left(
P:Q\right)  =\sum_{\substack{i\in A^{\ast};\\Q\left(  i\right)  =Q}%
}i=\sum_{i\in I_{\lambda}\left(  Q,\approx\right)  }i\in\mathbb{Z}\left[
A^{\ast}\right]  .
\]
This element $\left(  \ast:Q\right)  $ is the sum of all words in the
$\approx$-equivalence class $I_{\lambda}\left(  Q,\approx\right)  $; hence,
all words in this class appear in it with coefficient $1$, while all other
words appear with coefficient $0$. Therefore, this element $\left(
\ast:Q\right)  $ is $\approx$-invariant, and thus belongs to $\mathbb{Z}%
\left[  A^{\ast}\right]  _{\approx}$. Moreover:

\begin{statement}
\textbf{(D.12b) Lemma.} The set $\mathbb{Z}\left[  A^{\ast}\right]  _{\approx
}$ is the $\mathbb{Z}$-linear span of the elements $\left(  \ast:Q\right)  $
for all $\lambda\in\Lambda^{+}\left(  n,r\right)  $ and $Q\in\mathcal{Q}%
\left(  \lambda\right)  $.
\end{statement}

\textit{Proof.} These elements $\left(  \ast:Q\right)  $ are precisely the
sums of the $\approx$-equivalence classes in $A^{\ast}$. But the span of these
sums is precisely $\mathbb{Z}\left[  A^{\ast}\right]  _{\approx}$, by the
definition of $\mathbb{Z}\left[  A^{\ast}\right]  _{\approx}$. $\blacksquare$

\begin{statement}
\textbf{(D.12c) Lemma.} The set $\mathbb{Z}\left[  A^{\ast}\right]  _{\approx
}$ is a unital $\mathbb{Z}$-subalgebra of $\mathbb{Z}\left[  A^{\ast}\right]
$.
\end{statement}

\textit{Proof.} Clearly, $\mathbb{Z}\left[  A^{\ast}\right]  _{\approx}$ is a
$\mathbb{Z}$-submodule, and contains the unity of $\mathbb{Z}\left[  A^{\ast
}\right]  $ (which is the empty word). It remains to show that it is closed
under multiplication. By Lemma (D.12b), it suffices to show that $\left(
\ast:Q\right)  \left(  \ast:S\right)  \in\mathbb{Z}\left[  A^{\ast}\right]
_{\approx}$ for all $\lambda\in\Lambda^{+}\left(  n,r\right)  $,
$Q\in\mathcal{Q}\left(  \lambda\right)  $, $\mu\in\Lambda^{+}\left(
n,s\right)  $ and $S\in\mathcal{Q}\left(  \mu\right)  $. So let us consider
such $\lambda$, $Q$, $\mu$ and $S$.

Recall that $\left(  \ast:Q\right)  =\sum_{i\in I_{\lambda}\left(
Q,\approx\right)  }i$ and $\left(  \ast:S\right)  =\sum_{j\in I_{\mu}\left(
S,\approx\right)  }j$ (similarly). Hence,
\[
\left(  \ast:Q\right)  \left(  \ast:S\right)  =\left(  \sum_{i\in I_{\lambda
}\left(  Q,\approx\right)  }i\right)  \left(  \sum_{j\in I_{\mu}\left(
S,\approx\right)  }j\right)  =\sum_{i\in I_{\lambda}\left(  Q,\approx\right)
}\ \ \sum_{j\in I_{\mu}\left(  S,\approx\right)  }i\mid j.
\]
Hence, $\left(  \ast:Q\right)  \left(  \ast:S\right)  $ is the sum of all
concatenations $i\mid j$ with $i\in I_{\lambda}\left(  Q,\approx\right)  $ and
$j\in I_{\mu}\left(  S,\approx\right)  $. Note that any two such
concatenations are distinct (because all words in $I_{\lambda}\left(
Q,\approx\right)  $ have length $r$ and all words in $I_{\mu}\left(
S,\approx\right)  $ have length $s$), and thus this sum has no repeated
addends. Therefore,
\begin{equation}
\left(  \ast:Q\right)  \left(  \ast:S\right)  =\sum_{w\in X}w, \label{p199.1}%
\end{equation}
where $X$ is the set of all words $w\in A^{\ast}$ that can be written as
concatenations $i\mid j$ with $i\in I_{\lambda}\left(  Q,\approx\right)  $ and
$j\in I_{\mu}\left(  S,\approx\right)  $.

But Lemma (D.12a) shows that if $i\mid j$ is such a concatenation, then so is
every word $w$ that is $\approx$-equivalent to $i\mid j$ (because Lemma
(D.12a) tells us that any such word $w$ can be written as a concatenation
$i^{\prime}\mid j^{\prime}$ with $i^{\prime}\approx i$ and $j^{\prime}\approx
j$, and of course these $\approx$-relations entail that $i^{\prime}\in
I_{\lambda}\left(  Q,\approx\right)  $ and $j^{\prime}\in I_{\mu}\left(
S,\approx\right)  $ because of $i\in I_{\lambda}\left(  Q,\approx\right)  $
and $j\in I_{\mu}\left(  S,\approx\right)  $). In other words, the set $X$ is
closed under $\approx$ (that is, every element that is $\approx$-equivalent to
an element of $X$ must itself lie in $X$). In other words, $X$ is a union of
$\approx$-equivalence classes. Hence, the sum $\sum_{w\in X}w$ is $\approx
$-invariant, i.e., belongs to $\mathbb{Z}\left[  A^{\ast}\right]  _{\approx}$.
In view of (\ref{p199.1}), we can rewrite this as follows: $\left(
\ast:Q\right)  \left(  \ast:S\right)  $ belongs to $\mathbb{Z}\left[  A^{\ast
}\right]  _{\approx}$. As explained above, this completes the proof of Lemma
(D.12c). $\blacksquare$

\begin{statement}
\textbf{(D.12d) Definition.} Let $\Lambda^{+}\left(  n,\ast\right)
:=\bigsqcup\limits_{r\in\mathbb{N}}\Lambda^{+}\left(  n,r\right)  $ and
$\mathcal{Q}\left(  \ast\right)  :=\bigsqcup\limits_{\lambda\in\Lambda
^{+}\left(  n,\ast\right)  }\mathcal{Q}\left(  \lambda\right)  $.
\end{statement}

\begin{statement}
\textbf{(D.12e) Proposition.} The map%
\begin{align*}
\mathsf{T}  &  \rightarrow\mathcal{Q}\left(  \ast\right)  ,\\
w  &  \mapsto Q\left(  w\right)
\end{align*}
is a bijection.
\end{statement}

\textit{Proof.} Clearly, this map is well-defined. It remains to prove that
each $Q\in\mathcal{Q}\left(  \ast\right)  $ has exactly one preimage under it.
In other words, it remains to prove that for each $Q\in\mathcal{Q}\left(
\ast\right)  $, there is exactly one $w\in\mathsf{T}$ satisfying $Q\left(
w\right)  =Q$. In other words, it remains to prove that for each
$Q\in\mathcal{Q}\left(  \ast\right)  $, there is exactly one $w\in\mathsf{T}$
lying in the set $I\left(  Q,\approx\right)  $. In other words, it remains to
prove that for each $Q\in\mathcal{Q}\left(  \ast\right)  $, there is exactly
one $w\in I\left(  Q,\approx\right)  $ lying in the set $\mathsf{T}$. But this
is part of Proposition (D.2d) (i). $\blacksquare$

\begin{statement}
\textbf{(D.12f) Definition.} For any $\lambda\in\Lambda^{+}\left(  n,r\right)
$, we define the \emph{Schur polynomial} $s_{\lambda}\in\mathbb{Z}\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ by%
\[
s_{\lambda}:=\sum_{P\in\mathcal{P}\left(  \lambda\right)  }X_{P}%
,\ \ \ \ \ \ \ \ \ \ \text{where }X_{P}:=\prod_{c\in\left[  \lambda\right]
}X_{P\left(  c\right)  }.
\]
Splitting this sum by the weight of $P$, we can rewrite this formula as%
\begin{align*}
s_{\lambda}:=  &  \ \sum_{\alpha\in\Lambda\left(  n,r\right)  }\ \ \sum
_{P\in\mathcal{P}\left(  \lambda\right)  \text{ has weight }\alpha}%
X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}\cdots X_{n}^{\alpha_{n}}\\
=\  &  \sum_{\alpha\in\Lambda\left(  n,r\right)  }\left\vert \left\{
P\in\mathcal{P}\left(  \lambda\right)  \ \mid\ P\text{ has weight }%
\alpha\right\}  \right\vert \cdot X_{1}^{\alpha_{1}}X_{2}^{\alpha_{2}}\cdots
X_{n}^{\alpha_{n}}.
\end{align*}

\end{statement}

This polynomial $s_{\lambda}$ is precisely the formal character $\Phi
_{M_{\lambda}}$ from \S D.9 and also is precisely the formal character
$\Phi_{V_{\lambda}}$ from \S 3.4 and \S 5. This is shown in the proof of
(D.9a). As a consequence, the equality%
\[
\Phi_{V_{\lambda}}\cdot\Phi_{V_{\mu}}=\sum_{\nu}c_{\lambda,\mu}^{\nu}%
\Phi_{V_{\nu}}%
\]
(which defines the Littlewood--Richardson coefficients $c_{\lambda,\mu}^{\nu}%
$) can be rewritten as
\begin{equation}
s_{\lambda}s_{\mu}=\sum_{\nu}c_{\lambda,\mu}^{\nu}s_{\nu}. \label{p100.ss=s}%
\end{equation}
We shall now obtain a different formula for $s_{\lambda}s_{\mu}$, which will
then (by comparing coefficients) yield an expression for $c_{\lambda,\mu}%
^{\nu}$.

\begin{statement}
\textbf{(D.12g) Theorem (Littlewood--Richardson rule, general form).} Let
$\lambda\in\Lambda^{+}\left(  n,r\right)  $ and $\mu\in\Lambda^{+}\left(
n,s\right)  $. Let $Q\in\mathcal{Q}\left(  \lambda\right)  $ and
$S\in\mathcal{Q}\left(  \mu\right)  $. Let $\mathcal{W}$ be the set of all
words $k\in I\left(  n,r+s\right)  $ of the form $k=i\mid j$, where

\begin{enumerate}
\item[\textbf{(a1)}] the word $i\in I\left(  n,r\right)  $ satisfies $Q\left(
i\right)  =Q$;

\item[\textbf{(b1)}] the word $j\in I\left(  n,s\right)  $ satisfies $Q\left(
j\right)  =S$;

\item[\textbf{(c1)}] the word $k$ belongs to $\mathsf{T}$.
\end{enumerate}

Then,%
\[
s_{\lambda}s_{\mu}=\sum_{k\in\mathcal{W}}s_{\mathsf{wt}\left(  k\right)  }.
\]
Here, $\mathsf{wt}\left(  k\right)  $ denotes the weight of a word $k$.
\end{statement}

\textit{Proof.} Consider the surjective unital $\mathbb{Z}$-algebra morphism
$\pi:\mathbb{Z}\left[  A^{\ast}\right]  \rightarrow\mathbb{Z}\left[
X_{1},X_{2},\ldots,X_{n}\right]  $ that sends each word $i_{1}i_{2}\cdots
i_{r}$ to the (commutative) monomial $X_{i_{1}}X_{i_{2}}\cdots X_{i_{r}}$.
Then,%
\begin{align*}
\pi\left(  \left(  \ast:Q\right)  \right)   &  =\pi\left(  \sum_{P\in
\mathcal{P}\left(  \lambda\right)  }\left(  P:Q\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  \ast:Q\right)  =\sum
_{P\in\mathcal{P}\left(  \lambda\right)  }\left(  P:Q\right)  \right) \\
&  =\sum_{P\in\mathcal{P}\left(  \lambda\right)  }\underbrace{\pi\left(
\left(  P:Q\right)  \right)  }_{\substack{=X_{P}\\\text{(since the letters of
}P:Q\\\text{are the entries of }P\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }\pi\text{ is }\mathbb{Z}\text{-linear}\right) \\
&  =\sum_{P\in\mathcal{P}\left(  \lambda\right)  }X_{P}=s_{\lambda
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }s_{\lambda}\right)  .
\end{align*}
Similarly,%
\[
\pi\left(  \left(  \ast:S\right)  \right)  =s_{\mu}.
\]
Multiplying these two equalities, we find%
\[
\pi\left(  \left(  \ast:Q\right)  \right)  \pi\left(  \left(  \ast:S\right)
\right)  =s_{\lambda}s_{\mu},
\]
so that
\begin{align}
s_{\lambda}s_{\mu}  &  =\pi\left(  \left(  \ast:Q\right)  \right)  \pi\left(
\left(  \ast:S\right)  \right) \nonumber\\
&  =\pi\left(  \left(  \ast:Q\right)  \left(  \ast:S\right)  \right)
\label{p100.5}%
\end{align}
(since $\pi$ is an algebra morphism).

But the product $\left(  \ast:Q\right)  \left(  \ast:S\right)  $ lies in the
subalgebra $\mathbb{Z}\left[  A^{\ast}\right]  _{\approx}$ (by Lemma (D.12c)).
Thus, it can be written as a linear combination
\begin{equation}
\left(  \ast:Q\right)  \left(  \ast:S\right)  =\sum_{W\in\mathcal{Q}\left(
\ast\right)  }c_{W}\left(  \ast:W\right)  \label{p100.6}%
\end{equation}
with coefficients $c_{W}\in\mathbb{Z}$ (since Lemma (D.12b) says that
$\mathbb{Z}\left[  A^{\ast}\right]  _{\approx}$ is the $\mathbb{Z}$-linear
span of the elements $\left(  \ast:W\right)  $ for all $W\in\mathcal{Q}\left(
\ast\right)  $).

Now, let $\pi_{\mathsf{T}}:\mathbb{Z}\left[  A^{\ast}\right]  \rightarrow
\mathbb{Z}\left[  A^{\ast}\right]  $ be the $\mathbb{Z}$-linear map that
\[
\text{sends each word }w\in A^{\ast}\text{ to }%
\begin{cases}
w, & \text{if }w\in\mathsf{T};\\
0, & \text{else.}%
\end{cases}
\]
Thus, $\pi_{\mathsf{T}}$ is the canonical projection from $\mathbb{Z}\left[
A^{\ast}\right]  $ to the span of the basis vectors $w$ that belong to
$\mathsf{T}$. While $\pi_{\mathsf{T}}$ is not a $\mathbb{Z}$-algebra morphism,
we can notice a few useful properties. Most importantly, applying
$\pi_{\mathsf{T}}$ to a sum of words has the effect of filtering out all the
words that don't belong to $\mathsf{T}$ from this sum, leaving only the words
that do belong to $\mathsf{T}$ in place. Thus,
\begin{equation}
\pi_{\mathsf{T}}\left(  \left(  \ast:Q\right)  \right)  =i^{Q} \label{p100.7}%
\end{equation}
(because among all the words $i\in I_{\lambda}\left(  Q,\approx\right)  $ that
make up the sum $\left(  \ast:Q\right)  =\sum_{i\in I_{\lambda}\left(
Q,\approx\right)  }i$, there is only one that belongs to $\mathsf{T}$, namely
the word $i^{Q}$, as we know from Proposition (D.2d) (i)). The same argument
(applied to $W$ instead of $Q$) shows that for any $W\in\mathcal{Q}\left(
\ast\right)  $, we have%
\begin{equation}
\pi_{\mathsf{T}}\left(  \left(  \ast:W\right)  \right)  =i^{W}.
\label{p100.7W}%
\end{equation}
Hence, applying the map $\pi_{\mathsf{T}}$ to the equality (\ref{p100.6}), we
find%
\begin{align}
\pi_{\mathsf{T}}\left(  \left(  \ast:Q\right)  \left(  \ast:S\right)  \right)
&  =\pi_{\mathsf{T}}\left(  \sum_{W\in\mathcal{Q}\left(  \ast\right)  }%
c_{W}\left(  \ast:W\right)  \right) \nonumber\\
&  =\sum_{W\in\mathcal{Q}\left(  \ast\right)  }c_{W}\underbrace{\pi
_{\mathsf{T}}\left(  \left(  \ast:W\right)  \right)  }_{\substack{=i^{W}%
\\\text{(by (\ref{p100.7W}))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\pi_{\mathsf{T}}\text{ is }\mathbb{Z}\text{-linear}\right) \nonumber\\
&  =\sum_{W\in\mathcal{Q}\left(  \ast\right)  }c_{W}i^{W}. \label{p100.pit1}%
\end{align}


On the other hand,
\[
\left(  \ast:Q\right)  =\sum_{\substack{i\in A^{\ast};\\Q\left(  i\right)
=Q}}i=\sum_{\substack{i\in I\left(  n,r\right)  ;\\Q\left(  i\right)  =Q}}i
\]
(since any word $i\in A^{\ast}$ satisfying $Q\left(  i\right)  =Q$ must have
$\left\vert Q\right\vert =\left\vert \lambda\right\vert =r$ letters and thus
belong to $I\left(  n,r\right)  $) and similarly
\[
\left(  \ast:S\right)  =\sum_{\substack{j\in I\left(  n,s\right)  ;\\Q\left(
j\right)  =S}}j.
\]
Multiplying these two equalities, we find%
\begin{align}
\left(  \ast:Q\right)  \left(  \ast:S\right)   &  =\left(  \sum
_{\substack{i\in I\left(  n,r\right)  ;\\Q\left(  i\right)  =Q}}i\right)
\left(  \sum_{\substack{j\in I\left(  n,s\right)  ;\\Q\left(  j\right)
=S}}j\right) \nonumber\\
&  =\sum_{\substack{i\in I\left(  n,r\right)  ;\\Q\left(  i\right)
=Q}}\ \ \sum_{\substack{j\in I\left(  n,s\right)  ;\\Q\left(  j\right)
=S}}i\mid j. \label{p100.8}%
\end{align}
This sum has no repeated addends (since we can uniquely reconstruct the two
words $i$ and $j$ from the concatenation $i\mid j$ when we know that $i$ lies
in $I\left(  n,r\right)  $), and thus can be rewritten as the sum of all words
$k\in A^{\ast}$ that can be written as $i\mid j$ where the word $i\in I\left(
n,r\right)  $ satisfies $Q\left(  i\right)  =Q$ and where the word $j\in
I\left(  n,s\right)  $ satisfies $Q\left(  j\right)  =S$. Hence,
(\ref{p100.8}) rewrites as follows:%
\[
\left(  \ast:Q\right)  \left(  \ast:S\right)  =\sum_{\substack{k\in A^{\ast
};\\k\text{ can be written as }i\mid j\\\text{with }i\in I\left(  n,r\right)
\text{ and }j\in I\left(  n,s\right)  \\\text{and }Q\left(  i\right)  =Q\text{
and }Q\left(  j\right)  =S}}k=\sum_{\substack{k\in A^{\ast};\\k\text{ can be
written as }i\mid j\\\text{with }Q\left(  i\right)  =Q\text{ and }Q\left(
j\right)  =S}}k
\]
(here we have dropped the requirements \textquotedblleft$i\in I\left(
n,r\right)  $\textquotedblright\ and \textquotedblleft$j\in I\left(
n,s\right)  $\textquotedblright, since they follow automatically from
$Q\left(  i\right)  =Q$ and $Q\left(  j\right)  =S$). Applying the map
$\pi_{\mathsf{T}}$ to this equality, we find%
\[
\pi_{\mathsf{T}}\left(  \left(  \ast:Q\right)  \left(  \ast:S\right)  \right)
=\pi_{\mathsf{T}}\left(  \sum_{\substack{k\in A^{\ast};\\k\text{ can be
written as }i\mid j\\\text{with }Q\left(  i\right)  =Q\text{ and }Q\left(
j\right)  =S}}k\right)  =\sum_{\substack{k\in\mathsf{T};\\k\text{ can be
written as }i\mid j\\\text{with }Q\left(  i\right)  =Q\text{ and }Q\left(
j\right)  =S}}k
\]
(since $\pi_{\mathsf{T}}$ filters the addends for belonging to $\mathsf{T}$).
Comparing this with (\ref{p100.pit1}), we obtain%
\[
\sum_{W\in\mathcal{Q}\left(  \ast\right)  }c_{W}i^{W}=\sum_{\substack{k\in
\mathsf{T};\\k\text{ can be written as }i\mid j\\\text{with }Q\left(
i\right)  =Q\text{ and }Q\left(  j\right)  =S}}k.
\]
Comparing the coefficients of the word $i^{W}$ on both sides of this equality
(for a given $W\in\mathcal{Q}\left(  \ast\right)  $), we obtain%
\begin{equation}
c_{W}=%
\begin{cases}
1, & \text{if }i^{W}\text{ can be written as }i\mid j\text{ with }Q\left(
i\right)  =Q\text{ and }Q\left(  j\right)  =S;\\
0, & \text{else}%
\end{cases}
\label{p100.cW=}%
\end{equation}
for any $W\in\mathcal{Q}\left(  \ast\right)  $ (since Proposition (D.2d) (i)
shows that $i^{W}\in\mathsf{T}$).

For any tableau $W$, we let $\mathsf{shape}\left(  W\right)  $ denote the
shape of $W$. Thus, if $W\in\mathcal{Q}\left(  \nu\right)  $ for some $\nu$,
then $\mathsf{shape}\left(  W\right)  =\nu$.

Now, let us apply the algebra morphism $\pi$ to both sides of (\ref{p100.6}).
We find%
\begin{align*}
\pi\left(  \left(  \ast:Q\right)  \left(  \ast:S\right)  \right)   &
=\pi\left(  \sum_{W\in\mathcal{Q}\left(  \ast\right)  }c_{W}\left(
\ast:W\right)  \right) \\
&  =\sum_{W\in\mathcal{Q}\left(  \ast\right)  }c_{W}\underbrace{\pi\left(
\left(  \ast:W\right)  \right)  }_{\substack{=s_{\mathsf{shape}\left(
W\right)  }\\\text{(this is proved just like}\\\text{we showed }\pi\left(
\left(  \ast:Q\right)  \right)  =s_{\lambda}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi\text{ is linear}\right) \\
&  =\sum_{W\in\mathcal{Q}\left(  \ast\right)  }c_{W}s_{\mathsf{shape}\left(
W\right)  }.
\end{align*}
In view of (\ref{p100.5}), we can rewrite this as%
\begin{align}
s_{\lambda}s_{\mu}  &  =\sum_{W\in\mathcal{Q}\left(  \ast\right)  }%
c_{W}s_{\mathsf{shape}\left(  W\right)  }\nonumber\\
&  =\sum_{w\in\mathsf{T}}c_{Q\left(  w\right)  }s_{\mathsf{shape}\left(
Q\left(  w\right)  \right)  } \label{p100.15}%
\end{align}
(here, we substituted $Q\left(  w\right)  $ for $W$ in the sum, using the
bijection from Proposition (D.12e)).

Now, let $w\in\mathsf{T}$. Then, $i^{Q\left(  w\right)  }=w$ (indeed,
$i^{Q\left(  w\right)  }$ is defined as the unique word in $I\left(  Q\left(
w\right)  ,\approx\right)  $ lying in $\mathsf{T}$; but the word $w$ itself
also lies in $I\left(  Q\left(  w\right)  ,\approx\right)  $ and in
$\mathsf{T}$, and therefore we conclude from the uniqueness that $w$ must be
$i^{Q\left(  w\right)  }$). But (\ref{p100.cW=}) yields%
\begin{align}
c_{Q\left(  w\right)  }  &  =%
\begin{cases}
1, & \text{if }i^{Q\left(  w\right)  }\text{ can be written as }i\mid j\text{
with }Q\left(  i\right)  =Q\text{ and }Q\left(  j\right)  =S;\\
0, & \text{else}%
\end{cases}
\nonumber\\
&  =%
\begin{cases}
1, & \text{if }w\text{ can be written as }i\mid j\text{ with }Q\left(
i\right)  =Q\text{ and }Q\left(  j\right)  =S;\\
0, & \text{else}%
\end{cases}
\label{p100.cQw}%
\end{align}
(since $i^{Q\left(  w\right)  }=w$). Furthermore, Proposition (D.2f) (i)
yields $\mathsf{wt}\left(  i^{Q\left(  w\right)  }\right)  =\mathsf{shape}%
\left(  Q\left(  w\right)  \right)  $, and thus
\begin{equation}
\mathsf{shape}\left(  Q\left(  w\right)  \right)  =\mathsf{wt}\left(
i^{Q\left(  w\right)  }\right)  =\mathsf{wt}\left(  w\right)
\label{p100.shQw}%
\end{equation}
(since $i^{Q\left(  w\right)  }=w$).

Forget that we fixed $w$. We thus have shown that each $w\in\mathsf{T}$
satisfies (\ref{p100.cQw}) and (\ref{p100.shQw}). Hence, we can rewrite
(\ref{p100.15}) as%
\begin{align}
s_{\lambda}s_{\mu}  &  =\sum_{w\in\mathsf{T}}%
\begin{cases}
1, & \text{if }w\text{ can be written as }i\mid j\text{ with }Q\left(
i\right)  =Q\text{ and }Q\left(  j\right)  =S;\\
0, & \text{else}%
\end{cases}
\ \ s_{\mathsf{wt}\left(  w\right)  }\nonumber\\
&  =\sum_{\substack{w\in\mathsf{T};\\w\text{ can be written as }i\mid
j\\\text{with }Q\left(  i\right)  =Q\text{ and }Q\left(  j\right)
=S}}s_{\mathsf{wt}\left(  w\right)  }=\sum_{\substack{k\in\mathsf{T};\\k\text{
can be written as }i\mid j\\\text{with }Q\left(  i\right)  =Q\text{ and
}Q\left(  j\right)  =S}}s_{\mathsf{wt}\left(  k\right)  }\nonumber\\
&  =\sum_{\substack{k\in I\left(  n,r+s\right)  ;\\k\text{ can be written as
}i\mid j\\\text{with }Q\left(  i\right)  =Q\text{ and }Q\left(  j\right)
=S;\\k\in\mathsf{T}}}s_{\mathsf{wt}\left(  k\right)  } \label{p100.slsm=}%
\end{align}
(here, we have restricted the sum to $k\in I\left(  n,r+s\right)  $ only,
because only a word of length $r+s$ has any chance of being written as $i\mid
j$ with $Q\left(  i\right)  =Q$ and $Q\left(  j\right)  =S$). But the
conditions under the summation sign in (\ref{p100.slsm=}) are precisely the
conditions (a1), (b1) and (c1) in the definition of the set $\mathcal{W}$ in
Theorem (D.12g). Hence, the summation sign can be rewritten as
\textquotedblleft$\sum_{k\in\mathcal{W}}$\textquotedblright.

Hence, (\ref{p100.slsm=}) rewrites as
\[
s_{\lambda}s_{\mu}=\sum_{k\in\mathcal{W}}s_{\mathsf{wt}\left(  k\right)  }.
\]
This proves Theorem (D.12g). $\blacksquare$

\begin{statement}
\textbf{(D.12h) Theorem (Littlewood--Richardson rule, restated form).} Let
$\lambda\in\Lambda^{+}\left(  n,r\right)  $ and $\mu\in\Lambda^{+}\left(
n,s\right)  $. Let $Q\in\mathcal{Q}\left(  \lambda\right)  $ and
$S\in\mathcal{Q}\left(  \mu\right)  $. Let $\mathcal{W}$ be the set of all
words $k\in I\left(  n,r+s\right)  $ of the form $k=i\mid j$, where

\begin{enumerate}
\item[\textbf{(a2)}] the word $i\in I\left(  n,r\right)  $ satisfies $Q\left(
i\right)  =Q$;

\item[\textbf{(b2)}] the word $j$ is $i^{S}$;

\item[\textbf{(c2)}] the reverse $B\left(  k\right)  $ of the word $k$ is a
lattice permutation.
\end{enumerate}

Then,%
\[
s_{\lambda}s_{\mu}=\sum_{k\in\mathcal{W}}s_{\mathsf{wt}\left(  k\right)  }.
\]
Here, $\mathsf{wt}\left(  k\right)  $ denotes the weight of a word $k$.
\end{statement}

\textit{Proof.} This will follow from Theorem (D.12g), once we can show that
the set $\mathcal{W}$ defined in Theorem (D.12g) is precisely the set
$\mathcal{W}$ defined in Theorem (D.12h). For this purpose, we must show that
the conditions (a1), (b1) and (c1) from Theorem (D.12g) (taken together) are
equivalent to the conditions (a2), (b2) and (c2) from Theorem (D.12h).

Let us do this. Consider two words $i$ and $j$ and their concatenation
$k=i\mid j\in I\left(  n,r+s\right)  $.

We observed in \S D.1 that a word $i$ belongs to $\mathsf{T}$ if and only if
$B\left(  i\right)  $ is a lattice permutation. Hence, $k\in\mathsf{T}$ if and
only if $B\left(  k\right)  $ is a lattice permutation. In other words,
condition (c1) is equivalent to condition (c2). Thus, we WLOG assume that both
conditions (c1) and (c2) hold (since otherwise, neither of them holds, and we
are done). Hence, $k\in\mathsf{T}$, and the word $B\left(  k\right)  $ is a
lattice permutation.

If a word $w=w_{1}w_{2}\cdots w_{p}$ is a lattice permutation, then any prefix
of $w$ (that is, any word of the form $w_{1}w_{2}\cdots w_{q}$ with $q\leq p$)
is a lattice permutation again (by the definition of \textquotedblleft lattice
permutation\textquotedblright). Hence, any prefix of $B\left(  k\right)  $ is
a lattice permutation (since $B\left(  k\right)  $ is a lattice permutation).
But $B\left(  j\right)  $ is a prefix of $B\left(  k\right)  $ (since $k=i\mid
j$ and thus $B\left(  k\right)  =B\left(  i\mid j\right)  =B\left(  j\right)
\mid B\left(  i\right)  $). Thus, we conclude that $B\left(  j\right)  $ is a
lattice permutation. In other words, $j\in\mathsf{T}$ (since a word $i$
belongs to $\mathsf{T}$ if and only if $B\left(  i\right)  $ is a lattice permutation).

Therefore, if $Q\left(  j\right)  =S$, then $j\in I\left(  S,\approx\right)  $
and thus $j=i^{S}$ (because Proposition (D.2d) (i) tells us that $i^{S}$ is
the only word in $I\left(  S,\approx\right)  $ that belongs to $\mathsf{T}$).
Hence, condition (b1) implies condition (b2).

Conversely, if the word $j$ is $i^{S}$, then $Q\left(  j\right)  =Q\left(
i^{S}\right)  =S$ (by the definition of $i^{S}$) and thus also $j\in I\left(
n,s\right)  $ (since $Q\left(  j\right)  =S$ shows that $j$ has as many
letters as $S$ has entries; but $S\in\mathcal{Q}\left(  \mu\right)  $ shows
that $S$ has $\left\vert \mu\right\vert =s$ many entries). Hence, condition
(b2) implies (b1). Thus, we have shown that the two conditions (b1) and (b2)
imply one another; in other words, they are equivalent.

Finally, conditions (a1) and (a2) are equivalent, since they say the same thing.

Altogether, we now have seen that the conditions (a1), (b1) and (c1) from
Theorem (D.12g) (taken together) are equivalent to the conditions (a2), (b2)
and (c2) from Theorem (D.12h). This completes our proof of (D.12h).
$\blacksquare$

\begin{statement}
\textbf{(D.12i) Theorem (Littlewood--Richardson rule, special form).} Let
$\lambda\in\Lambda^{+}\left(  n,r\right)  $ and $\mu\in\Lambda^{+}\left(
n,s\right)  $. Let $\mathcal{W}$ be the set of all words $k\in I\left(
n,r+s\right)  $ of the form $k=i\mid j$, where

\begin{enumerate}
\item[\textbf{(a3)}] the word $i$ satisfies $KP\left(  i\right)  =i$, and the
tableau $P\left(  i\right)  $ has shape $\lambda$;

\item[\textbf{(b3)}] the word $j$ is $i^{\mu}$;

\item[\textbf{(c3)}] the reverse $B\left(  k\right)  $ of the word $k$ is a
lattice permutation.
\end{enumerate}

Then,%
\[
s_{\lambda}s_{\mu}=\sum_{k\in\mathcal{W}}s_{\mathsf{wt}\left(  k\right)  }.
\]
Here, $\mathsf{wt}\left(  k\right)  $ denotes the weight of a word $k$.
\end{statement}

\textit{Proof.} We set $Q:=\mathsf{Q}^{\left(  \lambda\right)  }$ (see (C.2h)
for the definition of this) and $S:=\mathsf{Q}^{\left(  \mu\right)  }$. Then,
we apply Theorem (D.12h). As a result, we obtain
\[
s_{\lambda}s_{\mu}=\sum_{k\in\mathcal{W}}s_{\mathsf{wt}\left(  k\right)  },
\]
where the set $\mathcal{W}$ is the one defined in Theorem (D.12h). It remains
to show that the set $\mathcal{W}$ defined in Theorem (D.12h) is precisely the
set $\mathcal{W}$ defined in Theorem (D.12i). For this purpose, we must show
that the conditions (a2), (b2) and (c2) from Theorem (D.12h) (taken together)
are equivalent to the conditions (a3), (b3) and (c3) from Theorem (D.12i).

Let us do this. Consider two words $i$ and $j$ and their concatenation
$k=i\mid j\in I\left(  n,r+s\right)  $.

We have $i^{\mu}=i^{\mathsf{Q}^{\left(  \mu\right)  }}$ by Proposition (D.1i).
In other words, $i^{\mu}=i^{S}$ (since $S=\mathsf{Q}^{\left(  \mu\right)  }$).
Hence, condition (b2) is equivalent to (b3). Moreover, condition (c2) is
equivalent to (c3), since they are literally the same.

We know from Exercise (C.2i) that a word $i$ of shape $\lambda$ satisfies
$KP\left(  i\right)  =i$ if and only if $Q\left(  i\right)  =\mathsf{Q}%
^{\left(  \lambda\right)  }$. In other words, a word $i$ of shape $\lambda$
satisfies $KP\left(  i\right)  =i$ if and only if $Q\left(  i\right)  =Q$
(since $Q=\mathsf{Q}^{\left(  \lambda\right)  }$). Hence, if a word $i$ of
shape $\lambda$ satisfies $KP\left(  i\right)  =i$, then $Q\left(  i\right)
=Q$ and therefore also $i\in I\left(  n,r\right)  $ (since $Q\left(  i\right)
=Q$ shows that $i$ has as many letters as $Q$ has entries; but $Q=\mathsf{Q}%
^{\left(  \lambda\right)  }$ shows that $Q$ has $\left\vert \lambda\right\vert
=r$ many entries). In other words, condition (a3) implies (a2). Conversely,
condition (a2) implies (a3), since $Q\left(  i\right)  =Q=\mathsf{Q}^{\left(
\lambda\right)  }\in\mathcal{Q}\left(  \lambda\right)  $ entails that $i$ has
shape $\lambda$ and thus satisfies $KP\left(  i\right)  =i$ (by Exercise
(C.2i), since $Q\left(  i\right)  =\mathsf{Q}^{\left(  \lambda\right)  }$).
Thus, the two conditions (a2) and (a3) mutually imply one another. In other
words, they are equivalent.

Altogether, we now have seen that the conditions (a2), (b2) and (c2) from
Theorem (D.12h) (taken together) are equivalent to the conditions (a3), (b3)
and (c3) from Theorem (D.12i). This completes our proof of (D.12i).
$\blacksquare$

\begin{statement}
\textbf{(D.12j) Theorem (Littlewood--Richardson rule, rewritten special
form).} Let $\lambda\in\Lambda^{+}\left(  n,r\right)  $ and $\mu\in\Lambda
^{+}\left(  n,s\right)  $ and $\nu\in\Lambda^{+}\left(  n,r+s\right)  $. Let
$\mathcal{W}_{\nu}$ be the set of all words $k\in I\left(  n,r+s\right)  $ of
the form $k=i\mid j$, where

\begin{enumerate}
\item[\textbf{(a3)}] the word $i$ satisfies $KP\left(  i\right)  =i$, and the
tableau $P\left(  i\right)  $ has shape $\lambda$;

\item[\textbf{(b3)}] the word $j$ is $i^{\mu}$;

\item[\textbf{(c3')}] the reverse $B\left(  k\right)  $ of the word $k$ is a
lattice permutation of weight $\nu$.
\end{enumerate}

Then,%
\[
c_{\lambda,\mu}^{\nu}=\left\vert \mathcal{W}_{\nu}\right\vert .
\]

\end{statement}

\textit{Proof.} Take the formula $s_{\lambda}s_{\mu}=\sum_{k\in\mathcal{W}%
}s_{\mathsf{wt}\left(  k\right)  }$ from Theorem (D.12i), and compare the
coefficients of $s_{\nu}$ on both of its sides. The coefficient on the left
hand side is $c_{\lambda,\mu}^{\nu}$ (by (\ref{p100.ss=s})), while the
coefficient on the right hand side is $\left\vert \mathcal{W}_{\nu}\right\vert
$. $\blacksquare$ \medskip

\textit{Proof of (D.10a):} The claim of (D.10a) is just Theorem (D.12j), with
the words $k,i,j$ and the set $\mathcal{W}_{\nu}$ renamed as $i,j,k$ and
$\mathcal{W}$. $\blacksquare$


\end{document}