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\ihead{Errata to ``An Introduction to Symmetric Functions...''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{\href{https://www.ericegge.net/cofsf/}{\textbf{An Introduction to
Symmetric Functions and Their Combinatorics}}}

\textit{Eric S. Egge}

AMS 2019

\textbf{Errata and comments}

\bigskip
\end{center}

\setcounter{section}{15}

\section*{Errata and comments}

The following are all comments I have noted down when reading your text
\textquotedblleft\href{https://www.ericegge.net/cofsf/}{\textit{An
Introduction to Symmetric Functions and Their Combinatorics}}%
\textquotedblright\ (Student Mathematical Library \#91, AMS 2019). My
apologies for mixing real corrections with subjective suggestions, pedantic
misreadings and the occasional alternative proof I couldn't help writing up. I
also have never proofread the list below, so it might be in need of its own
error hunter.

\textquotedblleft Line $-n$\textquotedblright\ (for negative $n$) means
\textquotedblleft line $n$ from the bottom\textquotedblright. Lines in
displayed formulas are counted as lines.

\begin{enumerate}
\item \textbf{page 1:} You write: \textquotedblleft The algebra of symmetric
polynomials in $n$ variables is isomorphic to the algebra generated by the
characters of the irreducible representations of the symmetric group $S_{n}%
$\textquotedblright. I am pretty sure this is wrong in any possible
interpretation. The former algebra is a polynomial ring; the latter is
finite-dimensional. The closest thing that is true is that the algebra of
symmetric functions(!) is isomorphic to the direct sum of the character rings
of $S_{0},S_{1},S_{2},\ldots$, equipped with a special multiplication obtained
from induction (not just from multiplying characters pointwise).

\item \textbf{page 2:} I doubt that \textquotedblleft the set of all symmetric
functions with coefficients in $\mathbb{Q}$ is a finite-dimensional vector
space over $\mathbb{Q}$\textquotedblright. I assume you are talking about a
graded component of this space.

\item \textbf{page 3:} When defining $X_{n}$, you should require $n\geq0$
instead of $n\geq1$. Indeed, you use $X_{0}=\left\{  {}\right\}  $ (and the
respective symmetric functions $e_{k}\left(  X_{0}\right)  $ and $h_{k}\left(
X_{0}\right)  $) in Proposition 3.1, for example (since Proposition 3.1 for
$n=1$ mentions $X_{1-1}$).

\item \textbf{page 9, proof of Proposition 1.11:} The expression
\textquotedblleft$\mu_{1},\ldots,\mu_{j+1},\mu_{j},\ldots,\mu_{n}%
$\textquotedblright\ can be interpreted in two ways. The most obvious
interpretation (namely, as the concatenation of the $\left(  j+1\right)
$-tuple $\left(  \mu_{1},\ldots,\mu_{j+1}\right)  $ with the $\left(
n-j+1\right)  $-tuple $\left(  \mu_{j},\ldots,\mu_{n}\right)  $) is the wrong
one; the right interpretation is \textquotedblleft the $n$-tuple $\left(
\mu_{1},\ldots,\mu_{n}\right)  $ with its $j$-th and $\left(  j+1\right)  $-st
entries swapped\textquotedblright. Worth pointing out.

\item \textbf{page 14:} When defining the total degree, replace
\textquotedblleft$a_{1}+a_{2}\cdots$\textquotedblright\ by \textquotedblleft%
$a_{1}+a_{2}+\cdots$\textquotedblright.

\item \textbf{page 17, between Definition 1.19 and Definition 1.20:}
\textquotedblleft we constructed the monomial symmetric
functions\textquotedblright\ $\rightarrow$ \textquotedblleft we constructed
the monomial symmetric polynomials\textquotedblright.

\item \textbf{page 17, Definition 1.19:} This definition of $\Lambda$ has the
consequence that $\prod_{i=1}^{\infty}\left(  1+x_{i}\right)  $ is a symmetric
function. This conflicts with the standard definition of symmetric functions
in the literature (which requires a symmetric function to be of bounded
degree, so that $\prod_{i=1}^{\infty}\left(  1+x_{i}\right)  $ does not fit
the bill). It also conflicts with your own Definition 7.5, because if you
allow $f:=\prod_{i=1}^{\infty}\left(  1+x_{i}\right)  $ to be an element of
$\Lambda$, then the Hall inner product $\left\langle f,f\right\rangle $ is
undefined. It makes sense to study both \textquotedblleft
versions\textquotedblright\ of symmetric functions (i.e., the symmetric power
series of bounded degree, and the merely symmetric power series), but they
need different names and different notations.

\item \textbf{page 17, solution to Exercise 1.21:} Replace \textquotedblleft
sum of the images of $x_{1}^{2}x_{2}$ under all permutations\textquotedblright%
\ by \textquotedblleft sum of all monomials that can be obtained from
$x_{1}^{2}x_{2}$ through permutations\textquotedblright. (There is a subtle
difference: The former wording suggests a sum over permutations, whereas the
latter suggests a sum over monomials. The latter is right, because one and the
same monomial can be obtained using different permutations and thus would be
counted multiple times according to the former wording.)

\item \textbf{page 20, Exercise 1.21:} This exercise relies on a concept of
infinite products of formal power series, which is not completely obvious.
Since it is not easily found in algebra textbooks (usually, textbooks only
define convergence of formal power series in finitely many variables, and not
all of these definitions are easily adapted to the case of infinitely many
variables; for example, $\prod_{i=1}^{\infty}\left(  1+x_{i}\right)  $ does
\textbf{not} converge with respect to total degree, but rather converges with
respect to the product topology), I suggest saying some words about how it is defined.

\item \textbf{page 24, between Example 2.1 and Definition 2.2:}
\textquotedblleft is the sum of all products\textquotedblright\ $\rightarrow$
\textquotedblleft is (up to sign) the sum of all products\textquotedblright.

\item \textbf{page 27, second paragraph:} \textquotedblleft in any linear
combination of monomial symmetric functions\textquotedblright\ $\rightarrow$
\textquotedblleft in any linear combination of monomial symmetric polynomials
(in a given set of variables $X_{n}$)\textquotedblright. After this sentence,
add \textquotedblleft The same applies to monomial symmetric
functions.\textquotedblright.

\item \textbf{page 27, proof of Proposition 2.6:} The \textquotedblleft If
$n\geq\left\vert \lambda\right\vert $\textquotedblright\ here is misplaced:
The sentence in which it appears does not require $n\geq\left\vert
\lambda\right\vert $ (but the next sentence does).

\item \textbf{page 28, solution to Example 2.7:} I would replace
\textquotedblleft with distinct positive integers\textquotedblright\ by
\textquotedblleft with $k$ distinct positive integers\textquotedblright.

\item \textbf{page 30, solution to Example 2.7:} \textquotedblleft case
A4\textquotedblright\ $\rightarrow$ \textquotedblleft case
A3\textquotedblright.

\item \textbf{page 31, proof of Proposition 2.10:} When you say
\textquotedblleft By (2.2) we have\textquotedblright, you mean not (2.2) but
rather the next displayed equality after (2.2).

\item \textbf{page 32, proof of Proposition 2.10:} Before \textquotedblleft
each such $e_{\lambda_{j}}$ must contribute exactly one factor $x_{2}%
$\textquotedblright, add \textquotedblleft$\mu_{1}=\lambda_{1}^{\prime}$ and
$\mu_{2}=\lambda_{2}^{\prime}$ imply that\textquotedblright\ (it is easy to
lose track of the context here).

\item \textbf{page 34, Proposition 2.12 (c):} After \textquotedblleft$l\left(
\lambda\right)  $ urns\textquotedblright, I'd add \textquotedblleft(which are
distinguishable and labelled $1,2,\ldots,l\left(  \lambda\right)
$)\textquotedblright.

\item \textbf{page 37, proof of Theorem 2.17:} I'd mention that different
monomials of $f$ lead to different $e_{\lambda}$'s in $f\left(  e_{1}%
,e_{2},\ldots,e_{n}\right)  $; otherwise the same argument would show the
obviously wrong claim that $e_{1},e_{1},e_{1},\ldots$ are algebraically independent.

\item \textbf{page 38, lines 1--4:} What you are saying here is not a
restatement of Theorem 2.17 alone, but also incorporates the fact that the
$e_{1},e_{2},e_{3},\ldots$ generate the $\mathbb{Q}$-algebra $\Lambda$. (This
follows from Corollary 2.11, too.)

\item \textbf{page 39, Definition 2.19:} \textquotedblleft\textit{complete
homogeneous polynomial}\textquotedblright\ $\rightarrow$ \textquotedblleft%
\textit{complete homogeneous symmetric polynomial}\textquotedblright.

\item \textbf{page 39, Definition 2.19:} Usually your text isn't exactly
suffering from a shortage of examples, but here is one place where an example
would be really helpful: after this definition. The undefined concepts around
multisets (e.g., what is the size of a multisubset? is a sum over
$J\subseteq\left[  \left[  n\right]  \right]  $ itself a sum over a set or
over a multisubset?) might confuse some readers; it would be helpful to dispel
such confusion by showing (say) $h_{1},h_{2},h_{3}$ and $h_{2}\left(
X_{3}\right)  $.

\item \textbf{page 41:} \textquotedblleft In Example 2.20 we
found\textquotedblright\ $\rightarrow$ \textquotedblleft In Example 2.22 we
found\textquotedblright.

\item \textbf{page 43, Combinatorial Proof of Proposition 2.24:}
\textquotedblleft then we just need to give an involution\textquotedblright%
\ might be a bit abrupt for those readers who aren't used to proofs by
sign-reversing involutions. What about including a simple example for this
method before (say, a proof of $\sum_{T\subseteq S}\left(  -1\right)
^{\left\vert T\right\vert }=0$ for any nonempty finite set $S$) in an appendix?

(Strangely enough, you do handhold the reader about sign-reversing involutions
-- but not until later, when you prove Proposition 3.2.)

\item \textbf{page 44, line 1:} \textquotedblleft new way\textquotedblright%
\ $\rightarrow$ \textquotedblleft way\textquotedblright\ (I don't think you
showed this fact in any other way before).

\item \textbf{page 44, proof of Proposition 2.25:} It might be worth pointing
out that the induction step tacitly uses the fact that $h_{\lambda}h_{\mu
}=h_{\nu}$ for some partition $\nu$ (namely, for $\nu=\lambda\cup\mu$ as
defined on page 13). Actually, it may help making this fact into a separate
lemma; you use it on page 180 as well.

\item \textbf{page 45, Proposition 2.28:} The word \textquotedblleft
nonempty\textquotedblright\ isn't needed. (Likewise on the few lines before
Proposition 2.28.)

\item \textbf{page 46, proof of Proposition 2.30 (i):} The fact that
$\mu>_{\operatorname*{lex}}\lambda$ if $\mu$ is obtained from $\lambda$ by
merging parts and rearranging the resulting numbers into weakly decreasing
order is not as obvious as it might appear; I would suggest at least an
example to illustrate the reasons for it.

\item \textbf{page 48:} When you write \textquotedblleft logarithms of formal
power series behave in analogy with logarithms of
polynomials\textquotedblright, do you really mean \textquotedblleft
polynomials\textquotedblright? I think logarithms of polynomials aren't much
simpler than logarithms of arbitrary formal power series.

\item \textbf{page 48, proof of Proposition 2.33:} I'd remove the
\textquotedblleft$P\left(  t\right)  =$\textquotedblright\ at the beginning of
the displayed computation, since it's not clear which side of (2.16) it is
referring to. (Do you even need the notation $P\left(  t\right)  $ anywhere?
If not, I'd remove it from the statement of the proposition as well. It's not
standard; Macdonald's book uses $P\left(  t\right)  $ for what you would call
$\dfrac{d}{dt}P\left(  t\right)  $.)

\item \textbf{page 49, Problem 2.9:} The theorem you are asking the reader to
prove here is the
\href{https://en.wikipedia.org/wiki/Gale-Ryser_theorem}{Gale--Ryser theorem},
which is much harder than most other problems in this section. As far as
symmetric functions are concerned, only the \textquotedblleft only
if\textquotedblright\ part is useful, while the \textquotedblleft
if\textquotedblright\ part is a nice curiosity. (Mark Wildon, in \S 5.2 of
\href{http://www.ma.rhul.ac.uk/~uvah099/Maths/Sym/SymFuncs2017.pdf}{his
\textit{An Involutive Introduction to Symmetric Functions}}, gives a proof of
the \textquotedblleft if\textquotedblright\ part using symmetric functions,
but it uses more tools than at the reader's disposal in \S 2 of your book.)

\item \textbf{page 49, Problem 2.14:} Replace \textquotedblleft$n\geq
0$\textquotedblright\ by \textquotedblleft$n\geq1$\textquotedblright\ twice in
this problem, since there is no $f_{0}$.

\item \textbf{page 50, Problem 2.24:} Replace \textquotedblleft$n-1-k$%
\textquotedblright\ by \textquotedblleft$n-1+k$\textquotedblright. Also worth
pointing out that $k$ can be any integer $\geq1-n$ here (with the standard
convention that $h_{i}\left(  X_{n}\right)  =0$ for all $i<0$).

\item \textbf{page 51, Problem 2.27:} Why are you repeating the equality
(2.15) here?

\item \textbf{page 51, Problem 2.28 (a):} Replace \textquotedblleft$n\times
n$\textquotedblright\ by \textquotedblleft$k\times k$\textquotedblright.

\item \textbf{page 51, Problem 2.28 (a):} \textquotedblleft exactly one entry
of column $j$ is $\lambda_{j}$\textquotedblright\ is literally false if
$\lambda_{j}=0$. There are two ways to fix this: either you make the matrix
$k\times l\left(  \lambda\right)  $ (as opposed to $k\times k$), or you
replace \textquotedblleft exactly one entry of column $j$ is $\lambda_{j}$ and
all other entries in column $j$ are $0$\textquotedblright\ by
\textquotedblleft column $j$ has an entry equal to $\lambda_{j}$ while all
other entries of this column are $0$\textquotedblright.

\item \textbf{page 51, Problem 2.28 (b):} \textquotedblleft for
all\textquotedblright\ $\rightarrow$ \textquotedblleft for
each\textquotedblright. (Otherwise, the \textquotedblleft all of which have
the same type\textquotedblright\ part sounds like all balls in all columns
must have the same type.)

\item \textbf{page 51:} Additional problem suggestions: It is worth adding an
exercise that asks to show that if $\lambda,\mu\vdash k$ satisfy
$M_{\lambda,\mu}\left(  p,m\right)  \neq0$, then $\lambda\trianglelefteq\mu$
in dominance order. (This would strengthen Proposition 2.30 (i).) Another
exercise could ask to show that the relation $M_{\lambda,\mu}\left(
p,m\right)  \neq0$ between $\lambda$ and $\mu$ is in itself a partial order on
the set of all partitions of $k$.

\item \textbf{page 54, proof of Proposition 3.2:} After \textquotedblleft
followed by a choice of one of the variables $x_{s}$ in that
term\textquotedblright, I would add \textquotedblleft(there are $j$ possible
choices, to explain the factor $j$ in (3.4))\textquotedblright.

\item \textbf{page 55, proof of Proposition 3.2:} \textquotedblleft For
example, when $k=7$ and $j=3$\textquotedblright\ $\rightarrow$
\textquotedblleft For example, when $k=7$ and $j=4$\textquotedblright.

\item \textbf{page 57, proof of Proposition 3.3:} \textquotedblleft after the
rightmost $r$\textquotedblright\ $\rightarrow$ \textquotedblleft after the
rightmost number $\leq r$\textquotedblright\ (there might not be an $r$ in the
first title).

\item \textbf{page 57, proof of Proposition 3.3:} I would suggest at least
mentioning the existence of a simple and slick proof of (3.6) that uses
neither bijections nor power series: Expand the $p_{j}$ in $\sum_{j=1}%
^{k}h_{k-j}p_{j}$ using (3.5), then interchange the summation signs, then get
rid of the inner sum using (2.12) to obtain $kh_{k}$. Probably (3.7) can be
proved in the same way.

\item \textbf{page 58, proof of Proposition 3.5:} Here you are tacitly using
the fact that evaluating the symmetric \textbf{polynomial} $e_{j}\left(
X_{n}\right)  $ at $\underbrace{1,1,\ldots,1}_{n\text{ times}}$ gives the same
result as evaluating the symmetric \textbf{function} $e_{j}$ at
$\underbrace{1,1,\ldots,1}_{n\text{ times}},0,0,0,\ldots$. (And the same fact
about $h_{j}$.) This is close to obvious, but worth stating explicitly.

\item \textbf{page 58, between Proposition 3.5 and Proposition 3.6:} The
restatement of (3.10) using multinomial coefficients is broken: The addend for
$j=0$ is undefined (there is a division by zero and a multinomial coefficient
with a negative integer in it).

\item \textbf{page 59, after proof of Proposition 3.6:} The restatement of
(3.11) using multinomial coefficients may involve addends in which the $n-k+j$
is negative. To make sense of them, $\dbinom{a+b+c}{a,b,c}$ should be defined
to be $0$ when one of $a,b,c$ is negative.

\item \textbf{page 60, Definition 3.10:} \textquotedblleft For all $n\geq1$
and all $k\geq1$\textquotedblright\ $\rightarrow$ \textquotedblleft For all
$n\geq0$ and $k\in\mathbb{Z}$\textquotedblright. (You use the $0$-cases later
on, e.g., in Proposition 3.13. The case of negative $k$ is more of a luxury,
but it has the advantage of making (3.17) work for all $k\geq0$, which to me
is (pardon) a plus. It is worth saying that $%
%TCIMACRO{\QDATOPD{[}{]}{n}{k}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n}{k}%
%EndExpansion
=0$ whenever $k<0$.)

\item \textbf{page 60, Definition 3.12:} \textquotedblleft For all $n\geq1$
and all $k\geq1$\textquotedblright\ $\rightarrow$ \textquotedblleft For all
$n\geq0$ and $k\in\mathbb{Z}$\textquotedblright. (I'm not sure if you use the
$0$-cases later on, but there is no reason to exclude them. Once again, the
case of negative $k$ is more of a luxury. It is worth saying that $%
%TCIMACRO{\QDATOPD{\{}{\}}{n}{k}}%
%BeginExpansion
\genfrac{\{}{\}}{0pt}{0}{n}{k}%
%EndExpansion
=0$ whenever $k<0$.)

\item \textbf{page 61, Proposition 3.13:} I'd replace \textquotedblleft for
all $n\geq2$ and $k$ with $1\leq k\leq n-1$\textquotedblright\ by
\textquotedblleft for all $n\geq1$ and $k\in\mathbb{Z}$\textquotedblright.
There is no need to rule out the trivial cases; the proof works equally well
in them.

\item \textbf{page 62, proof of Proposition 3.13:} When you say
\textquotedblleft remove the $n$ from one of these
permutations\textquotedblright, you mean removing it from the cycle notation
rather than (say) from the one-line notation. This should be said -- it's the
natural choice but not the only one.

\item \textbf{page 62, proof of Proposition 3.13:} \textquotedblleft
Therefore, we can construct each permutation of\textquotedblright%
\ $\rightarrow$ \textquotedblleft Therefore, we can construct each second-type
permutation of\textquotedblright.

\item \textbf{page 62, proof of Proposition 3.13:} \textquotedblleft
wth\textquotedblright\ $\rightarrow$ \textquotedblleft with\textquotedblright.

\item \textbf{page 62, Proposition 3.14:} Again, I'd replace \textquotedblleft
for all $n\geq2$ and all $k$ with $2\leq k\leq n-1$\textquotedblright\ by
\textquotedblleft for all $n\geq1$ and $k\in\mathbb{Z}$\textquotedblright.
There is no difference to the proof, but the generality makes it easier to use
the proposition.

\item \textbf{page 62, Proposition 3.15:} You can replace \textquotedblleft%
$n\geq1$\textquotedblright\ by \textquotedblleft$n\geq0$\textquotedblright.

You also don't want to require $k\leq n$ -- certainly not for (3.18), but even
for (3.17) it is not very helpful. (As it stands, it forces you to use $%
%TCIMACRO{\QDATOPD{[}{]}{n+1}{n+1}}%
%BeginExpansion
\genfrac{[}{]}{0pt}{0}{n+1}{n+1}%
%EndExpansion
=n!$ in the proof, which you haven't shown.)

Assuming that you have changed Definitions 3.10 and 3.12 as I suggested above,
you can simply replace \textquotedblleft and all $k$ with $0\leq k\leq
n$\textquotedblright\ here by \textquotedblleft and all $k\geq0$%
\textquotedblright\ here, and the proposition remains valid. The proof also
remains valid, provided that you use induction on $n$ and (in the induction
step) treat the $k=0$ case separately.

\item \textbf{page 63, (3.20):} \textquotedblleft$\left(  -1\right)  ^{j-1}%
$\textquotedblright\ should be \textquotedblleft$\left(  -1\right)  ^{j-1}%
j$\textquotedblright\ on the right hand side.

\item \textbf{page 64, (3.21):} Replace \textquotedblleft$j$\textquotedblright%
\ by \textquotedblleft$k$\textquotedblright\ on the left hand side.

\item \textbf{page 65, Example 3.21:} It is worth saying somewhere that
\textquotedblleft word\textquotedblright\ means \textquotedblleft finite tuple
of positive integers\textquotedblright\ in this book, and that every
permutation $\pi\in S_{n}$ is identified with the word $\left(  \pi\left(
1\right)  ,\pi\left(  2\right)  ,\ldots,\pi\left(  n\right)  \right)  $. (This
becomes particularly important in Chapter 10.)

\item \textbf{page 69, proof of Proposition 3.26:} Your induction step uses
$n\geq2$ and $k\geq1$, but you only have $n\geq2$ \textbf{or} $k\geq1$. It
thus remains to deal with the case $n=1$ and the case $k=0$ separately. (Of
course, both of these cases can be left to the reader, but the structure of
the proof should still be adapted to mention them.)

\item \textbf{page 69, between Proposition 3.26 and Proposition 3.27:}
\textquotedblleft which can appear is $1\cdot q\cdot q^{2}\cdots q^{k}%
$\textquotedblright\ $\rightarrow$ \textquotedblleft which can appear is
$1\cdot q\cdot q^{2}\cdots q^{k-1}$\textquotedblright.

\item \textbf{page 70, (3.28):} Replace \textquotedblleft$k$\textquotedblright%
\ by \textquotedblleft$j$\textquotedblright\ on the right hand side.

\item \textbf{page 71, Problem 3.3:} The \textquotedblleft$n$%
\textquotedblright\ atop the summation sign should be a \textquotedblleft%
$k$\textquotedblright.

\item \textbf{page 75, \S 4.1:} Not sure what you mean by \textquotedblleft In
Chapter 2.16 and 2.17\textquotedblright.

\item \textbf{page 81:} \textquotedblleft seminstandard\textquotedblright%
\ $\rightarrow$ \textquotedblleft semistandard\textquotedblright.

\item \textbf{page 82, Definition 4.11:} Something tells me that you will use
the notion of \textquotedblleft content\textquotedblright\ not just for
semistandard tableaux but for any fillings of a partition shape (and even more
generally, as in the proof of Proposition 5.20), so it may be worth making
this definition a bit more general.

\item \textbf{page 83:} \textquotedblleft Therefore, in $T$ we
have\textquotedblright\ $\rightarrow$ \textquotedblleft Therefore, in $R$ we
have\textquotedblright.

\item \textbf{page 87:} \textquotedblleft our solution to Example 4.12
raises\textquotedblright\ $\rightarrow$ \textquotedblleft our solution to
Example 4.9 raises\textquotedblright.

\item \textbf{page 87, (4.6):} I'd say a couple of words about why this
equality holds. (Namely, it follows from Proposition 4.15 using the same trick
that was used in the proof of Proposition 2.6 and in the solution to Example 2.7.)

\item \textbf{page 90, proof of Proposition 4.23:} This is mis-organized: The
displayed equality relies on $\mu_{j}=\mu_{l}$, so you should say
\textquotedblleft if $\mu_{j}=\mu_{l}$\textquotedblright\ in the first (not in
the second) sentence of this proof. On the other hand, the second sentence
needs an \textquotedblleft if $f\left(  X_{n}\right)  $ is
alternating\textquotedblright.

\item \textbf{page 91, proof of Proposition 4.24:} \textquotedblleft Then by
Proposition 1.2(i),(ii)\textquotedblright\ $\rightarrow$ \textquotedblleft
Then by Proposition 1.2(i),(ii),(iii)\textquotedblright.

\item \textbf{page 91, proof of Proposition 4.24:} In the second displayed
equation of this proof, replace \textquotedblleft$s_{\tau\left(  n\right)
}^{\mu_{n}}$\textquotedblright\ by \textquotedblleft$x_{\tau\left(  n\right)
}^{\mu_{n}}$\textquotedblright.

\item \textbf{page 92, line 1:} \textquotedblleft is the sequence
$n-1,n-2,\ldots,2,1$\textquotedblright\ $\rightarrow$ \textquotedblleft is the
sequence $n-1,n-2,\ldots,2,1,0$\textquotedblright.

\item \textbf{page 92, proof of Proposition 4.25:} It is worth pointing out
that this proof can be skipped, as the result of Proposition 4.25 will not be
used anywhere (or will it? I don't see any direct references).

\item \textbf{page 97, after the proof of Proposition 4.31:} \textquotedblleft
the left side of (4.12)\textquotedblright\ $\rightarrow$ \textquotedblleft the
left side of (4.13)\textquotedblright.

\item \textbf{page 97, Example 4.33:} \textquotedblleft$\pi$%
-tableaux\textquotedblright\ $\rightarrow$ \textquotedblleft semistandard
$\pi$-tableaux\textquotedblright. The same mistake (lack of \textquotedblleft
semistandard\textquotedblright\ even though it is meant) is made a few times
later on, so perhaps it is worth defining \textquotedblleft$\pi$%
-tableau\textquotedblright\ to mean \textquotedblleft semistandard $\pi
$-tableau\textquotedblright.

\item \textbf{page 100, Definition 4.37:} Replace \textquotedblleft which are
equal to $\pi\left(  l_{j}+1\right)  $\textquotedblright\ by \textquotedblleft
which are equal to $\pi\left(  l_{j}-1\right)  $\textquotedblright, and
replace \textquotedblleft such that $t_{j}>r_{j}$\textquotedblright\ by
\textquotedblleft such that $t_{j}<r_{j}$\textquotedblright. (Otherwise, you
are only comparing the numbers of $\pi\left(  l_{j}\right)  $'s and
$\pi\left(  l_{j}+1\right)  $'s in the tail segments that begin with a
$\pi\left(  l_{j}\right)  $; but such segments are biased towards having more
$\pi\left(  l_{j}\right)  $'s, whereas you are interested in them having as
few $\pi\left(  l_{j}\right)  $'s as possible. In particular, if the word has
no $\pi\left(  l_{j}\right)  $'s altogether but some $\pi\left(
l_{j}+1\right)  $'s, then it should not count as a Littlewood--Richardson
$\pi$-word, but your current definition lets it pass as one.)

With my suggested changes, you need to require that $l_{j}>1$ (in
\textquotedblleft for each entry\textquotedblright).

\item \textbf{page 101, line 2:} \textquotedblleft each filling is a $\pi
$-tableau\textquotedblright\ $\rightarrow$ \textquotedblleft each filling is a
semistandard $\pi$-tableau\textquotedblright.

\item \textbf{page 102, line 4:} \textquotedblleft will not start with
$\pi\left(  1\right)  $\textquotedblright\ $\rightarrow$ \textquotedblleft
will not end with $\pi\left(  1\right)  $\textquotedblright.

\item \textbf{page 103, Proposition 4.40:} Replace \textquotedblleft$l\left(
\lambda\right)  \geq n$\textquotedblright\ by \textquotedblleft$l\left(
\lambda\right)  \leq n$\textquotedblright.

\item \textbf{page 103, Proposition 4.40 (ii):} Replace \textquotedblleft
leftmost\textquotedblright\ by \textquotedblleft rightmost\textquotedblright%
\ in \textquotedblleft it is the leftmost entry in $\operatorname*{word}%
\left(  T\right)  $ which is in row $j$ but is not equal to $\pi\left(
j\right)  $\textquotedblright. For example, in the non-Littlewood--Richardson
tableau%
\[%
\begin{tabular}
[c]{|c|c|c|c}\cline{1-3}%
$2$ & $3$ & $4$ & \\\hline
$1$ & $1$ & $1$ & \multicolumn{1}{|c|}{$1$}\\\hline
\end{tabular}
\ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{for }\pi=\operatorname*{id}\right)  ,
\]
the $\pi$-climber is the $4$ in row $2$, which is not the leftmost entry in
row $2$ that doesn't equal $\pi\left(  2\right)  =2$ (there is a $3$ to its left).

\item \textbf{page 104, proof of Lemma 4.41:} Replace \textquotedblleft the
results follow\textquotedblright\ by \textquotedblleft the result
follows\textquotedblright.

\item \textbf{page 105:} \textquotedblleft isn't even a $\kappa\left(
\pi\right)  $-tableau\textquotedblright\ $\rightarrow$ \textquotedblleft isn't
even a semistandard $\kappa\left(  \pi\right)  $-tableau\textquotedblright.

\item \textbf{pages 106--109:} Figures 4.16, 4.19 and 4.20 use the symbol
\textquotedblleft$\ell$\textquotedblright\ for what is denoted by
\textquotedblleft$l$\textquotedblright\ in the main text. These look different
enough to suggest different meanings.

\item \textbf{page 107:} \textquotedblleft middle tableau\textquotedblright%
\ $\rightarrow$ \textquotedblleft second tableau from the
left\textquotedblright.

\item \textbf{page 107:} \textquotedblleft This means $k=1$, $j=2$, and
$l=3$\textquotedblright\ $\rightarrow$ \textquotedblleft This means $k=1$,
$j=3$, and $l=3$\textquotedblright.

\item \textbf{page 108, Lemma 4.43 (ii):} \textquotedblleft not a semistandard
$\pi$-tableau\textquotedblright\ $\rightarrow$ \textquotedblleft not a
Littlewood--Richardson $\pi$-tableau\textquotedblright.

\item \textbf{page 110, proof of Lemma 4.43 (ii):} It is not true that
\textquotedblleft in each row other than the row of the $\pi$-climber, $T$ and
$\kappa\left(  T\right)  $ have the same number of $k$'s and the same number
of $l$'s\textquotedblright. Indeed, this analysis correctly accounts for the
free $k$'s and $l$'s, but ignores the existence of paired $k$'s and $l$'s. The
latter have to be handled separately: In each column other than the column of
the $\pi$-climber, $T$ and $\kappa\left(  T\right)  $ have the same number of
paired $k$'s and the same number of paired $l$'s. Now it remains to add up the
numbers for free and for paired entries.

\item \textbf{page 111, proof of Proposition 4.28:} In \textquotedblleft over
all Littlewood--Richardson $\pi$-tableau\textquotedblright, replace
\textquotedblleft$\pi$-tableau\textquotedblright\ by \textquotedblleft$\pi
$-tableaux\textquotedblright.

\item \textbf{page 111, proof of Proposition 4.28:} In the last sentence of
this proof, replace \textquotedblleft we have\textquotedblright\ by
\textquotedblleft this rewrites as\textquotedblright.

\item \textbf{page 113, Problem 4.20:} Replace \textquotedblleft$n\geq
k$\textquotedblright\ by \textquotedblleft$n>k$\textquotedblright. Indeed,
this doesn't hold for $n=k$ unless you interpret $s_{0,1^{n}}$ to mean $0$.

\item \textbf{page 115, Problem 4.30:} What does \textquotedblleft
sign-reversing\textquotedblright\ mean? Do you mean that
$\operatorname*{weight}\left(  T^{\prime}\right)  =-\operatorname*{weight}%
\left(  T\right)  $ ?

\item \textbf{page 115, Problem 4.31:} Why do you require $n\geq k$ here?

\item \textbf{page 116, Problem 4.36:} After \textquotedblleft Let
$\widetilde{\lambda}$ be the partition with\textquotedblright, add
\textquotedblleft at most $n$ parts and with\textquotedblright.

\item \textbf{page 116, Problem 4.36:} Replace \textquotedblleft%
$\lambda_{r-j+1}$\textquotedblright\ by \textquotedblleft$\lambda_{n-j+1}%
$\textquotedblright.

\item \textbf{page 117:} \textquotedblleft done this\textquotedblright\ is an
overstatement: I believe Billey, Rhoades and Tewari have only showed that
there exist expansions with nonnegative coefficients, but have not found any
formula for these coefficients.

\item \textbf{page 121:} After \textquotedblleft By our definition of the skew
Schur functions\textquotedblright, add \textquotedblleft(and by Proposition
5.4)\textquotedblright.

\item \textbf{page 122, Proposition 5.7:} Here and in the following, you omit
the word \textquotedblleft skew\textquotedblright\ in \textquotedblleft
semistandard skew tableau\textquotedblright. Worth mentioning, I think.

\item \textbf{page 126, proof of Proposition 5.11:} Remove \textquotedblleft
with $\alpha_{1}+\cdots+\alpha_{n}=n$\textquotedblright\ (this requirement is
unnecessary and distracting).

\item \textbf{page 126, proof of Proposition 5.11:} \textquotedblleft a
semistandard tableaux\textquotedblright\ $\rightarrow$ \textquotedblleft a
semistandard tableau\textquotedblright.

\item \textbf{page 127, proof of Proposition 5.12:} \textquotedblleft in the
its\textquotedblright\ $\rightarrow$ \textquotedblleft in
its\textquotedblright.

\item \textbf{page 128:} \textquotedblleft We might hope\textquotedblright%
\ $\rightarrow$ \textquotedblleft We might expect\textquotedblright\ (I don't
see a reason to hope this -- don't we want more identities?).

\item \textbf{page 130, Definition 5.16:} \textquotedblleft over all
set-valued tableaux\textquotedblright\ $\rightarrow$ \textquotedblleft over
all set-valued semistandard tableaux\textquotedblright.

\item \textbf{page 132:} When you say \textquotedblleft we adapt the
Bender--Knuth involutions\textquotedblright, it is worth explaining one little
subtlety: If a cell a set-valued semistandard tableau contains both $j$ and
$j+1$, then these two entries are considered to be free, not paired (even
though the definition of \textquotedblleft paired\textquotedblright\ you give
on page 83 would count them as paired).

\item \textbf{page 134:} After \textquotedblleft Every semistandard tableau is
a set-valued semistandard tableau\textquotedblright, I would add
\textquotedblleft(by regarding each of its entries $i$ as a singleton set
$\left\{  i\right\}  $)\textquotedblright.

\item \textbf{page 134:} When you say \textquotedblleft the symmetric function
$G_{\lambda}$\textquotedblright, you are using a nonstandard concept of
\textquotedblleft symmetric function\textquotedblright\ that does not require
bounded degree. Likewise, on page 135, the words \textquotedblleft linear
combination\textquotedblright\ seem to refer to an infinite linear
combination, which is also nonstandard usage. At the risk of pedantry, I think
this is worth a disclaimer.

\item \textbf{page 136, Definition 5.22:} \textquotedblleft of elegant
tableau\textquotedblright\ $\rightarrow$ \textquotedblleft of elegant
tableaux\textquotedblright.

\item \textbf{page 136, Example 5.24:} I would require $n>0$ here, since the
answer is wrong for $k=n=0$ at least.

\item \textbf{page 136, solution to Example 5.24:} On the last line of page
136, replace \textquotedblleft$n+j$\textquotedblright\ by \textquotedblleft%
$n+j-1$\textquotedblright.

\item \textbf{page 137, solution to Example 5.24:} \textquotedblleft can be at
most $n+j$\textquotedblright\ $\rightarrow$ \textquotedblleft can be at most
$n+j-1$\textquotedblright.

\item \textbf{page 138:} Add \textquotedblleft of shape $\lambda
$\textquotedblright\ before \textquotedblleft with entries in $\left[
n\right]  $\textquotedblright. Likewise, add \textquotedblleft of shape
$\lambda$\textquotedblright\ before \textquotedblleft with entries in
$\mathbb{P}$\textquotedblright.

\item \textbf{page 141, first bullet point:} In \textquotedblleft we replace
each $j$ with $j+1$ in the left column and we replace each $j+1$ with $j$ in
the right column\textquotedblright, the words \textquotedblleft
left\textquotedblright\ and \textquotedblleft right\textquotedblright\ should
be interchanged.

\item \textbf{page 142:} Replace \textquotedblleft be $\sum_{k}jv_{k}%
$\textquotedblright\ by \textquotedblleft be $\sum_{k}kv_{k}$%
\textquotedblright.

\item \textbf{page 143, solution to Example 5.31:} \textquotedblleft Since a
semistandard tableau $T$ of shape $\left(  n\right)  $ cannot have repeated
entries in a column, these tableaux are exactly the reverse plane partitions
of shape $\left(  n\right)  $\textquotedblright\ $\rightarrow$
\textquotedblleft Since a reverse plane partition $T$ of shape $\left(
n\right)  $ cannot have repeated entries in a column, these reverse plane
partitions are exactly the semistandard tableaux of shape $\left(  n\right)
$\textquotedblright.

\item \textbf{page 143, Example 5.32:} You need to require $n\geq1$ for the
solution to work.

\item \textbf{page 143, solution to Example 5.32:} \textquotedblleft with at
least one $j$ for $1\leq j\leq k$\textquotedblright\ $\rightarrow$
\textquotedblleft with at least one $j$ for each $1\leq j\leq k$%
\textquotedblright.

\item \textbf{page 145, line 1:} I would replace \textquotedblleft$n\geq
1$\textquotedblright\ by \textquotedblleft$n\geq0$\textquotedblright. There is
no reason to exclude the trivial case $n=0$; it gives extra information about
the chromatic polynomial (namely, that its constant term is $0$ unless the
graph has no vertices).

\item \textbf{page 146, shortly after the solution to Example 5.35:} The
formulation \textquotedblleft we can remove all of the vertices of a graph $G$
by removing one vertex of degree $1$ at a time\textquotedblright\ is, strictly
speaking, incorrect: What you mean is that by successively removing one vertex
of degree $1$ at a time, you can end up with a $1$-vertex graph. You cannot,
of course, get rid of the final vertex, as it has degree $0$ rather than $1$.

Also, you are using the notion of the degree of a vertex in a graph; this
should be defined. (Namely: If $v$ is a vertex of a graph $G$, then the
\textit{degree} of $v$ means the number of edges of $G$ that contain $v$.)

\item \textbf{page 146, Proposition 5.36:} Once again, you can replace
\textquotedblleft$n\geq1$\textquotedblright\ by \textquotedblleft$n\geq
0$\textquotedblright. The proof works fine for $n=0$: In this case, it is
still true that any proper coloring of $U$ can be extended to a proper
coloring of $T$ in $n-1$ many ways (even though $n-1$ is negative); indeed,
this is vacuously true, since there is no proper coloring of $U$.

\item \textbf{page 146, proof of Proposition 5.36:} \textquotedblleft for the
endpoint\textquotedblright\ $\rightarrow$ \textquotedblleft for the
vertex\textquotedblright.

\item \textbf{page 146, proof of Proposition 5.36:} \textquotedblleft for the
color of the endpoint\textquotedblright\ $\rightarrow$ \textquotedblleft for
the color of the remaining vertex\textquotedblright.

\item \textbf{page 146, Definition 5.37:} Replace \textquotedblleft proper
coloring\textquotedblright\ by \textquotedblleft coloring\textquotedblright%
\ here. Later (on page 147) you refer to the weight of any coloring (not
necessarily proper).

\item \textbf{page 148, proof of Proposition 5.41:} On line 1 of the proof,
replace \textquotedblleft$x_{\lambda_{l\left(  \lambda\right)  }}%
^{\lambda_{l\left(  \lambda\right)  }}$\textquotedblright\ by
\textquotedblleft$x_{l\left(  \lambda\right)  }^{\lambda_{l\left(
\lambda\right)  }}$\textquotedblright.

\item \textbf{page 148, proof of Proposition 5.41:} Replace \textquotedblleft
of $m_{\lambda}$ in $X_{G}$)\textquotedblright\ by \textquotedblleft of
$m_{\lambda}$) in $X_{G}$\textquotedblright.

\item \textbf{page 148, proof of Proposition 5.41:} After \textquotedblleft by
choosing a stable partition $V_{1},V_{2},\ldots$ of $G$\textquotedblright, add
\textquotedblleft satisfying $\left\vert V_{j}\right\vert =\lambda_{j}$ for
all $j$,\textquotedblright.

\item \textbf{page 153, Problem 5.5:} Do you really mean to ask this as
stated? Because the first row of $\lambda$ is longer than that of $\mu$, so
the answer is $0$.

\item \textbf{page 154, Problem 5.8:} I'd add a \textquotedblleft for $n\geq
1$\textquotedblright\ before the comma here.

\item \textbf{page 155, Problem 5.22:} The left hand side of the equality
needs a $t^{n}$ factor (I believe).

\item \textbf{page 156:} \textquotedblleft of graphs\textquotedblright%
\ $\rightarrow$ \textquotedblleft of connected graphs\textquotedblright.

\item \textbf{page 160:} After \textquotedblleft is an infinite sequence of
north $\left(  0,1\right)  $ and east $\left(  1,0\right)  $ steps which
contains exactly $n$ east steps\textquotedblright, I suggest adding
\textquotedblleft and has its first step start at $\left(  a,1\right)  $
whereas each further step starts at the endpoint of the previous
step\textquotedblright. (This is probably clear to everyone who has ever
counted lattice paths, but I'm not sure whether you want to make this
assumption on the reader's experience.)

\item \textbf{page 161:} After the first displayed equation, add
\textquotedblleft(where $a$ is an arbitrary integer, which may even be chosen
differently for each factor in the product)\textquotedblright.

\item \textbf{pages 161, 164, 168, 171, 173, 174, 175, 176, 178:} I would
replace every \textquotedblleft$\left(  -1\right)  ^{\operatorname*{inv}%
\left(  \pi\right)  }$\textquotedblright\ by a \textquotedblleft%
$\operatorname*{sgn}\left(  \pi\right)  $\textquotedblright\ on these pages.
Of course, these are synonyms, but you use the $\operatorname*{sgn}$ notation
in Lemma 6.5 and Proposition C.7, and you speak of a \textquotedblleft
sign-reversing\textquotedblright\ involution (as opposed to an
\textquotedblleft inversion-parity-reversing\textquotedblright\ one), so I
think it makes sense to stick to one choice of notation.

\item \textbf{page 162:} Replace \textquotedblleft moves $\lambda_{j}+\pi
_{j}-j$\textquotedblright\ by \textquotedblleft moves $\lambda_{j}+\pi\left(
j\right)  -j$\textquotedblright.

\item \textbf{page 162:} \textquotedblleft ending $j$ units from the
right\textquotedblright\ $\rightarrow$ \textquotedblleft whose ending point is
the $j$-th one from the right\textquotedblright. (The ending points are
usually not evenly spaced, so I wouldn't speak of \textquotedblleft
units\textquotedblright.)

\item \textbf{page 167, proof of Lemma 6.5 (ii):} \textquotedblleft same set
of east steps\textquotedblright\ $\rightarrow$ \textquotedblleft same multiset
of east steps\textquotedblright.

\item \textbf{page 167, Figure 6.8:} I suggest switching the colors of the red
and blue paths here. After all, the tail-swapping procedure results in each
path keeping its upper end (but possibly changing its lower end), so unless
the colors of the paths do not correspond to the paths, it makes sense for the
upper ends to preserve their colors.

\item \textbf{page 168, proof of Lemma 6.5 (iii):} I suggest replacing
\textquotedblleft decompose as $\alpha_{m}^{-},\alpha_{w}^{+}$ and $\alpha
_{w}^{-},\alpha_{m}^{+}$\textquotedblright\ by \textquotedblleft decompose as
$\alpha_{w}^{-},\alpha_{m}^{+}$ and $\alpha_{m}^{-},\alpha_{w}^{+}%
$\textquotedblright\ in order not to swap the two paths unnecessarily.

\item \textbf{page 168, proof of Lemma 6.6:} The proof could be made clearer
by some reorganization. You aren't cancelling the addends corresponding to the
$\beta$'s with $\pi\neq12\cdots k$; you are cancelling the addends
corresponding to the $\beta$'s that have intersection. The former are a subset
of the latter, but $\operatorname*{tswp}$ does not generally send the former
to the former, so it is misleading to claim that it's the former that are
getting cancelled.

I think the best way to organize the proof is by first invoking
$\operatorname*{tswp}$ to cancel the intersecting $\beta$'s, thus obtaining%
\[
\sum_{\beta\in H_{\lambda,k}}\operatorname*{sgn}\left(  \pi\right)
\operatorname*{wt}\nolimits_{h}\left(  \beta\right)  =\sum_{\beta\in
II_{\lambda,k}}\operatorname*{sgn}\left(  \pi\right)  \operatorname*{wt}%
\nolimits_{h}\left(  \beta\right)  ;
\]
and then arguing that each $\beta\in II_{\lambda,k}$ satisfies $\pi=12\cdots
k$ and therefore $\operatorname*{sgn}\left(  \pi\right)  =1$, so the equality
above simplifies to (6.5).

\item \textbf{page 168, proof of Lemma 6.7:} \textquotedblleft Suppose
$\beta\in H_{\lambda,k}$\textquotedblright\ $\rightarrow$ \textquotedblleft
Suppose $\beta\in II_{\lambda,k}$\textquotedblright.

\item \textbf{page 168, proof of Lemma 6.7:} When you write \textquotedblleft
where $\alpha_{m}\in\Gamma_{-m,\lambda_{m}}$\textquotedblright, you are
implicitly using the fact that the permutation $\pi$ corresponding to any
$\beta\in II_{\lambda,k}$ is the identity $12\cdots k$. Since you only stated
this during the proof of Lemma 6.6, it is worth repeating it here.

\item \textbf{page 173:} \textquotedblleft product of the
weights\textquotedblright\ $\rightarrow$ \textquotedblleft product of the
$e$-weights\textquotedblright.

\item \textbf{page 174:} At the end of the second paragraph of this page, I
would add: \textquotedblleft Thus each path $\alpha_{m}$ ends at $\left(
\lambda_{m}^{\prime}-m,\infty\right)  $, which shows that $\alpha_{m}$ is the
path with the $m$-th rightmost ending point among $\alpha_{1},\alpha
_{2},\ldots,\alpha_{k}$ (and the ending points are
distinct).\textquotedblright.

\item \textbf{page 176, proof of Lemma 6.12:} \textquotedblleft have the same
sets of east steps\textquotedblright\ $\rightarrow$ \textquotedblleft have the
same multisets of east steps\textquotedblright.

\item \textbf{page 176, proof of Lemma 6.12:} \textquotedblleft has leftmost
point $\left(  l,m\right)  $\textquotedblright\ $\rightarrow$
\textquotedblleft has rightmost point $\left(  l,m\right)  $\textquotedblright%
\ (otherwise, you would need to increment the subscripts by $1$).

\item \textbf{page 176, Lemma 6.14:} Replace \textquotedblleft%
$\operatorname*{SST}\left(  \lambda^{\prime}\right)  $\textquotedblright\ by
\textquotedblleft$\operatorname*{SST}\left(  \lambda\right)  $%
\textquotedblright.

\item \textbf{page 176, proof of Lemma 6.14:} \textquotedblleft Ferrers
diagram of $\lambda^{\prime}$\textquotedblright\ $\rightarrow$
\textquotedblleft Ferrers diagram of $\lambda$\textquotedblright.

\item \textbf{page 176, proof of Lemma 6.14:} \textquotedblleft the
weights\textquotedblright\ $\rightarrow$ \textquotedblleft the $e$%
-weights\textquotedblright.

\item \textbf{page 176, proof of Lemma 6.14:} \textquotedblleft column
strict\textquotedblright\ $\rightarrow$ \textquotedblleft
column-strict\textquotedblright.

\item \textbf{page 176, proof of Lemma 6.14:} \textquotedblleft if and only
if\textquotedblright\ $\rightarrow$ \textquotedblleft
because\textquotedblright\ (you assumed that $\beta\in\Xi_{\lambda,k}$, so you
cannot have any intersecting paths in $\beta$).

\item \textbf{page 178, proof of Theorem 6.10:} \textquotedblleft%
$T\in\operatorname*{SST}\left(  \lambda^{\prime}\right)  $\textquotedblright%
\ $\rightarrow$ \textquotedblleft$T\in\operatorname*{SST}\left(
\lambda\right)  $\textquotedblright.

\item \textbf{page 178, proof of Theorem 6.10:} \textquotedblleft%
$s_{\lambda^{\prime}}$\textquotedblright\ $\rightarrow$ \textquotedblleft%
$s_{\lambda}$\textquotedblright.

\item \textbf{page 179, proof of Proposition 6.16:} The equality $\omega
_{h}\left(  \det\left(  h_{\lambda_{i}+j-i}\right)  _{1\leq i,j\leq k}\right)
=\det\left(  e_{\lambda_{i}+j-i}\right)  _{1\leq i,j\leq k}$ is not
immediately obvious at this stage. It relies on three things: (1) Proposition
C.7 (which shows that the determinant of a matrix is a very specific linear
combination of products of its entries), (2) the fact that $\omega_{h}$ is
linear, and (3) the fact that $\omega_{h}\left(  h_{\nu_{1}}h_{\nu_{2}}\cdots
h_{\nu_{p}}\right)  =e_{\nu_{1}}e_{\nu_{2}}\cdots e_{\nu_{p}}$ for any finite
list $\left(  \nu_{1},\nu_{2},\ldots,\nu_{p}\right)  $ of integers. Among
these three things, (3) warrants some justification (in my opinion). Namely:
If all entries of the list $\left(  \nu_{1},\nu_{2},\ldots,\nu_{p}\right)  $
are positive, then the equality in question ($\omega_{h}\left(  h_{\nu_{1}%
}h_{\nu_{2}}\cdots h_{\nu_{p}}\right)  =e_{\nu_{1}}e_{\nu_{2}}\cdots
e_{\nu_{p}}$) can be rewritten as $\omega_{h}\left(  h_{\mu}\right)  =e_{\mu}%
$, where $\mu$ is the partition obtained by sorting the list $\left(  \nu
_{1},\nu_{2},\ldots,\nu_{p}\right)  $ into weakly decreasing order; but the
latter equality follows from the definition of $\omega_{h}$. Thus, it remains
to handle the case when not all entries of the list $\left(  \nu_{1},\nu
_{2},\ldots,\nu_{p}\right)  $ are positive. If at least one entry of the list
$\left(  \nu_{1},\nu_{2},\ldots,\nu_{p}\right)  $ is negative, then the
equality in question ($\omega_{h}\left(  h_{\nu_{1}}h_{\nu_{2}}\cdots
h_{\nu_{p}}\right)  =e_{\nu_{1}}e_{\nu_{2}}\cdots e_{\nu_{p}}$) boils down to
the obvious equality $\omega_{h}\left(  0\right)  =0$ (since $h_{m}=0$ for
every $m<0$), and thus is true. Otherwise, at least one entry of the list
$\left(  \nu_{1},\nu_{2},\ldots,\nu_{p}\right)  $ is $0$; but then we can
remove this entry from the list without changing the equality (since $h_{0}%
=1$), which reduces our problem to a simpler case (thus leading to an
inductive proof over $p$).

\item \textbf{page 189, Problem 6.27:} I'd add \textquotedblleft Let $n$ be a
positive integer.\textquotedblright\ at the beginning of this problem.

\item \textbf{page 191, \S 7.1:} Worth saying that you will sometimes
abbreviate $\delta_{\lambda,\mu}$ as $\delta_{\lambda\mu}$ (for example, in
Proposition 7.4).

\item \textbf{page 193, Definition 7.2:} I don't think you need to define
$\Lambda_{k}\left(  X,Y\right)  $; you can just as well work in the whole
space of formal power series in \newline$x_{1},x_{2},x_{3},\ldots,y_{1}%
,y_{2},y_{3},\ldots$.

\item \textbf{page 194, proof of Proposition 7.3:} The second sentence of this
proof makes no sense as stated. I suggest replacing it by the following argument:

You know that%
\begin{equation}
\sum_{j=1}^{l}\sum_{m=1}^{n}A_{jm}u_{j}\left(  X\right)  v_{m}\left(
Y\right)  =0. \label{p194a.1}%
\end{equation}
You want to prove that each $m$ with $1\leq m\leq n$ satisfies%
\begin{equation}
\sum_{j=1}^{l}A_{jm}u_{j}\left(  X\right)  =0. \label{p194a.2}%
\end{equation}


Fix any monomial $\mathfrak{m}$ in the variables $x_{1},x_{2},x_{3},\ldots$.
Then, if we compare coefficients before $\mathfrak{m}$ on both sides of
(\ref{p194a.1}) (while considering $y_{1},y_{2},y_{3},\ldots$ as scalars), we
obtain%
\[
\sum_{j=1}^{l}\sum_{m=1}^{n}A_{jm}\cdot\left(  \text{the coefficient of
}\mathfrak{m}\text{ in }u_{j}\left(  X\right)  \right)  \cdot v_{m}\left(
Y\right)  =0.
\]
Renaming the variables $y_{1},y_{2},y_{3},\ldots$ as $x_{1},x_{2},x_{3}%
,\ldots$ in this equality, we obtain%
\[
\sum_{j=1}^{l}\sum_{m=1}^{n}A_{jm}\cdot\left(  \text{the coefficient of
}\mathfrak{m}\text{ in }u_{j}\left(  X\right)  \right)  \cdot v_{m}=0.
\]
In other words,%
\[
\sum_{m=1}^{n}\left(  \sum_{j=1}^{l}A_{jm}\cdot\left(  \text{the coefficient
of }\mathfrak{m}\text{ in }u_{j}\left(  X\right)  \right)  \right)  \cdot
v_{m}=0.
\]
Since $\left\{  v_{m}\ \mid\ 1\leq m\leq n\right\}  $ is linearly independent,
we thus conclude that each $m$ with $1\leq m\leq n$ satisfies%
\[
\sum_{j=1}^{l}A_{jm}\cdot\left(  \text{the coefficient of }\mathfrak{m}\text{
in }u_{j}\left(  X\right)  \right)  =0.
\]
Multiplying both sides of this equality by $\mathfrak{m}$, we obtain%
\begin{equation}
\sum_{j=1}^{l}A_{jm}\cdot\left(  \text{the coefficient of }\mathfrak{m}\text{
in }u_{j}\left(  X\right)  \right)  \cdot\mathfrak{m}=0. \label{p194a.6}%
\end{equation}


Now, forget that we fixed $\mathfrak{m}$. We thus have proved the equality
(\ref{p194a.6}) for each monomial $\mathfrak{m}$ in the variables $x_{1}%
,x_{2},x_{3},\ldots$ and each $m$ with $1\leq m\leq n$. Summing these
equalities over all monomials $\mathfrak{m}$ (while $m$ is fixed), we obtain
precisely (\ref{p194a.2}). Thus, (\ref{p194a.2}) is proven.

Note that this argument is completely bulletproof; there are no
\textquotedblleft technical issues\textquotedblright\ that you are leaving
aside here, so the whole first paragraph of page 194 (= last paragraph of page
193) is unnecessary.

\item \textbf{page 194, the paragraph before Proposition 7.4:}
\textquotedblleft inner product\textquotedblright\ $\rightarrow$
\textquotedblleft bilinear map on $\Lambda_{k}\times\Lambda_{k}$%
\textquotedblright.

\item \textbf{page 194, Proposition 7.4:} \textquotedblleft is the
function\textquotedblright\ $\rightarrow$ \textquotedblleft is the bilinear
map\textquotedblright\ (twice).

\item \textbf{page 194, Proposition 7.4:} A reminder about the meaning of
$p\left(  k\right)  $ would be useful (you haven't used that notation for a while).

\item \textbf{page 195, proof of Proposition 7.4:} \textquotedblleft(by
Definition A.15(i)--(iii))\textquotedblright\ $\rightarrow$ \textquotedblleft%
(by Definition A.15(1)--(3))\textquotedblright.

\item \textbf{page 195, proof of Proposition 7.4:} \textquotedblleft
function\textquotedblright\ $\rightarrow$ \textquotedblleft bilinear
function\textquotedblright.

\item \textbf{page 195, proof of Proposition 7.4:} In the first sentence of
the proof of (i) $\Longleftrightarrow$ (iii), replace \textquotedblleft%
$B_{\mu\beta}$\textquotedblright\ by \textquotedblleft$B_{\lambda\beta}%
$\textquotedblright.

\item \textbf{page 195, proof of Proposition 7.4:} After \textquotedblleft
then we can simplify the innermost sum to\textquotedblright, replace
\textquotedblleft get to\textquotedblright\ by \textquotedblleft%
$\delta_{\alpha,\beta}$, whence the whole right hand side simplifies
to\textquotedblright.

\item \textbf{page 197, proof of Proposition 7.6:} The second computation
relies on the fact that $\delta_{\lambda^{\prime},\mu^{\prime}}=\delta
_{\lambda,\mu}$, which is fairly obvious but (I think) still worth mentioning.

\item \textbf{page 198:} An example of a generalized permutation and its two
weights would be useful.

\item \textbf{pages 199--200, proof of Proposition 7.10:} In the first
paragraph of the proof, both \textquotedblleft$\mathbb{P}\times\mathbb{P}$
matrix\textquotedblright\ and \textquotedblleft domino\textquotedblright%
\ should probably be explained. (Also, I'd speak of a \textquotedblleft domino
of $\pi$\textquotedblright\ rather than just of a \textquotedblleft
domino\textquotedblright.)

\item \textbf{page 200, proof of Proposition 7.10:} When you apply Problem
2.17(a), you are assuming $\lambda$ and $\mu$ to be partitions, which is not
generally satisfied when $\lambda$ and $\mu$ come from an arbitrary
generalized permutation.

The easiest way to fix this, I believe, is to recognize that both sides of
(7.6) are symmetric in $x_{1},x_{2},x_{3},\ldots$ and also symmetric in
$y_{1},y_{2},y_{3},\ldots$, and thus it suffices to verify that they have
identical coefficients of $x^{\lambda}y^{\mu}$ whenever $\lambda$ and $\mu$
are partitions. Thus, you can restrict yourself to the case when $\lambda$ and
$\mu$ are partitions. In this case, you can furthermore WLOG assume that the
partitions $\lambda$ and $\mu$ have the same size (because if $\left\vert
\lambda\right\vert \neq\left\vert \mu\right\vert $, then the coefficients on
both sides of (7.6) are easily seen to be $0$). Thus, you can apply Problem 2.17(a).

Restricting yourself to specific $\lambda$ and $\mu$ also allows you to work
with $n\times n$-matrices (for $n=\left\vert \lambda\right\vert =\left\vert
\mu\right\vert $) instead of $\mathbb{P}\times\mathbb{P}$-matrices, which
makes the application of Problem 2.17(a) a lot more direct.

\item \textbf{page 200, proof of Corollary 7.11:} In order to apply
Proposition 7.4, you cannot immediately use the equality%
\[
\sum_{\lambda}m_{\lambda}\left(  X\right)  h_{\lambda}\left(  Y\right)
=\sum_{\lambda}s_{\lambda}\left(  X\right)  s_{\lambda}\left(  Y\right)  ;
\]
instead, you need its \textquotedblleft finite\textquotedblright\ variant%
\[
\sum_{\lambda\vdash n}m_{\lambda}\left(  X\right)  h_{\lambda}\left(
Y\right)  =\sum_{\lambda\vdash n}s_{\lambda}\left(  X\right)  s_{\lambda
}\left(  Y\right)  \ \ \ \ \ \ \ \ \ \ \text{for }n\in\mathbb{N}.
\]
So I would spend a sentence or so explaining how the latter can be derived
from the former.

Likewise, your argument only gives you $\left\langle m_{\lambda},h_{\mu
}\right\rangle =0$ in the case when $\left\vert \lambda\right\vert =\left\vert
\mu\right\vert $ (since the bilinear forms in Proposition 7.4 are defined on
$\Lambda_{k}\times\Lambda_{k}$, not on the whole $\Lambda\times\Lambda$). A
few words should be said about why $\left\langle m_{\lambda},h_{\mu
}\right\rangle =0$ also holds for even simpler reasons when $\left\vert
\lambda\right\vert \neq\left\vert \mu\right\vert $ (maybe a good problem?).

\item \textbf{page 202, solution to Example 7.16:} I'd replace
\textquotedblleft$h_{0}=p_{0}$\textquotedblright\ by \textquotedblleft%
$h_{0}=p_{\varnothing}$\textquotedblright\ in order to clarify that the
\textquotedblleft$0$\textquotedblright\ in \textquotedblleft$p_{0}%
$\textquotedblright\ stands not for the number $0$ (as $p_{0}$ would be
undefined for the number $0$) but for the empty partition $\varnothing$.

\item \textbf{pages 203--204, proof of Proposition 7.17:} It's probably a
correct proof, but my head is spinning from trying to verify the mutual
inverseness of the two maps, so let me suggest a simpler proof. I will use a
tiny bit of algebra -- namely, the orbit-stabilizer formula (in one of the
most down-to-earth settings: the symmetric group $S_{n}$ acting on
$n$-tuples). Feel free to include the following proof (or split it into exercises).

\textit{Alternative proof of Proposition 7.17 (sketched).} Fix $n\geq0$.

\textit{Step 1:} If $j=\left(  j_{1},j_{2},\ldots,j_{n}\right)  \in
\mathbb{P}^{n}$ is an $n$-tuple, then we let $x_{j}$ be the monomial
$x_{j_{1}}x_{j_{2}}\cdots x_{j_{n}}$. An $n$-tuple $\left(  j_{1},j_{2}%
,\ldots,j_{n}\right)  \in\mathbb{P}^{n}$ is said to be \textit{weakly
increasing} if $j_{1}\leq j_{2}\leq\cdots\leq j_{n}$. The definition of
$h_{n}$ yields%
\begin{equation}
h_{n}=\sum_{\substack{j\in\mathbb{P}^{n}\text{ is weakly}\\\text{increasing}%
}}x_{j}. \label{p203.altpf.p.7.17.1}%
\end{equation}


\textit{Step 2:} We let the symmetric group $S_{n}$ act on the set
$\mathbb{P}^{n}$ from the right by permuting the entries: i.e., we set%
\begin{equation}
\left(  k_{1},k_{2},\ldots,k_{n}\right)  \cdot\sigma=\left(  k_{\sigma\left(
1\right)  },k_{\sigma\left(  2\right)  },\ldots,k_{\sigma\left(  n\right)
}\right)  \label{p203.altpf.p.7.17.2}%
\end{equation}
for any $\sigma\in S_{n}$ and $\left(  k_{1},k_{2},\ldots,k_{n}\right)
\in\mathbb{P}^{n}$.

It is clear that if $k\in\mathbb{P}^{n}$ and $\sigma\in S_{n}$, then%
\begin{equation}
x_{k\cdot\sigma}=x_{k}. \label{p203.altpf.p.7.17.3}%
\end{equation}
In other words, if $j\in\mathbb{P}^{n}$ and $k\in\mathbb{P}^{n}$ satisfy $j\in
k\cdot S_{n}$ (where $k\cdot S_{n}$ denotes the orbit $\left\{  k\cdot
\sigma\ \mid\ \sigma\in S_{n}\right\}  $ of $k$), then%
\begin{equation}
x_{j}=x_{k}. \label{p203.altpf.p.7.17.3kj}%
\end{equation}


For each $n$-tuple $k\in\mathbb{P}^{n}$, the orbit $k\cdot S_{n}=\left\{
k\cdot\sigma\ \mid\ \sigma\in S_{n}\right\}  $ contains exactly one weakly
increasing $n$-tuple (namely, the result of sorting the entries of $k$ into
increasing order). In other words, for each $n$-tuple $k\in\mathbb{P}^{n}$,
there is exactly one weakly increasing $n$-tuple $j\in\mathbb{P}^{n}$
satisfying $j\in k\cdot S_{n}$.

The orbit-stabilizer theorem (applied to the action of $S_{n}$ on
$\mathbb{P}^{n}$) yields that each $k\in\mathbb{P}^{n}$ satisfies%
\[
\left\vert S_{n}\right\vert =\left\vert k\cdot S_{n}\right\vert \cdot
\left\vert \left\{  \sigma\in S_{n}\ \mid\ k\cdot\sigma=k\right\}  \right\vert
.
\]
In view of $\left\vert S_{n}\right\vert =n!$, this rewrites as%
\begin{equation}
n!=\left\vert k\cdot S_{n}\right\vert \cdot\left\vert \left\{  \sigma\in
S_{n}\ \mid\ k\cdot\sigma=k\right\}  \right\vert .
\label{p203.altpf.p.7.17.os2}%
\end{equation}


\textit{Step 3:} Let $\operatorname*{type}\sigma$ denote the cycle type of any
permutation $\sigma\in S_{n}$. (This is a partition of $n$.)

For any permutation $\sigma\in S_{n}$, we have%
\begin{equation}
p_{\operatorname*{type}\sigma}=\sum_{\substack{k\in\mathbb{P}^{n}%
;\\k\cdot\sigma=k}}x_{k}. \label{p203.altpf.p.7.17.ptype}%
\end{equation}


[\textit{Proof of (\ref{p203.altpf.p.7.17.ptype}):} Let $\sigma\in S_{n}$ be a
permutation. Let $z_{1},z_{2},\ldots,z_{m}$ be all cycles of $\sigma$
(including the $1$-cycles), listed in the order of decreasing length (with
ties broken in some arbitrary way, without repetitions). Then,
$\operatorname*{type}\sigma=\left(  \left\vert z_{1}\right\vert ,\left\vert
z_{2}\right\vert ,\ldots,\left\vert z_{m}\right\vert \right)  $ (by the
definition of $\operatorname*{type}\sigma$).

Fix some $k=\left(  k_{1},k_{2},\ldots,k_{n}\right)  \in\mathbb{P}^{n}$
satisfying $k\cdot\sigma=k$. Then,%
\begin{align*}
\left(  k_{1},k_{2},\ldots,k_{n}\right)   &  =k=\underbrace{k}_{=\left(
k_{1},k_{2},\ldots,k_{n}\right)  }\cdot\sigma=\left(  k_{1},k_{2},\ldots
,k_{n}\right)  \cdot\sigma\\
&  =\left(  k_{\sigma\left(  1\right)  },k_{\sigma\left(  2\right)  }%
,\ldots,k_{\sigma\left(  n\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{p203.altpf.p.7.17.2})}\right)  .
\end{align*}
In other words, $k_{i}=k_{\sigma\left(  i\right)  }$ for each $i\in\left[
n\right]  $. Thus, if two elements $i$ and $j$ of $\left[  n\right]  $ can be
obtained from one another by applying $\sigma$ some number of times, then
$k_{i}=k_{j}$. Hence, the values of $k_{j}$ for all $j\in z_{1}$ are equal
(since all $j\in z_{1}$ can be obtained from one another by applying $\sigma$
some number of times\footnote{because $z_{1}$ is a cycle of $\sigma$}). In
other words, there exists some $r_{1}\in\mathbb{P}$ such that all $j\in z_{1}$
satisfy $k_{j}=r_{1}$. This $r_{1}$ is unique (since $z_{1}$ is nonempty).
Consider this $r_{1}$. The same logic can be applied to the other cycles
$z_{2},z_{3},\ldots,z_{m}$ of $\sigma$; thus, for each $i\in\left\{
1,2,\ldots,m\right\}  $, we obtain a value $r_{i}\in\mathbb{P}$ such that all
$j\in z_{i}$ satisfy $k_{j}=r_{i}$. It is now easy to see that the $m$-tuple
$\left(  r_{1},r_{2},\ldots,r_{m}\right)  \in\mathbb{P}^{m}$ satisfies
$x_{r_{1}}^{\left\vert z_{1}\right\vert }x_{r_{2}}^{\left\vert z_{2}%
\right\vert }\cdots x_{r_{m}}^{\left\vert z_{m}\right\vert }=x_{k}$.

Forget that we fixed $k$. Thus, for each $n$-tuple $k=\left(  k_{1}%
,k_{2},\ldots,k_{n}\right)  \in\mathbb{P}^{n}$ satisfying $k\cdot\sigma=k$, we
have constructed an $m$-tuple $\left(  r_{1},r_{2},\ldots,r_{m}\right)
\in\mathbb{P}^{m}$ satisfying
\begin{equation}
x_{r_{1}}^{\left\vert z_{1}\right\vert }x_{r_{2}}^{\left\vert z_{2}\right\vert
}\cdots x_{r_{m}}^{\left\vert z_{m}\right\vert }=x_{k}.
\label{p203.altpf.p.7.17.ptype.pf.4}%
\end{equation}
(Namely, $r_{i}$ is the unique positive integer such that all $j\in z_{i}$
satisfy $k_{j}=r_{i}$.) This defines a map%
\begin{align*}
R:\left\{  k\in\mathbb{P}^{n}\ \mid\ k\cdot\sigma=k\right\}   &
\rightarrow\mathbb{P}^{m},\\
k  &  \mapsto\left(  r_{1},r_{2},\ldots,r_{m}\right)  .
\end{align*}
It is easy to see that this map $R$ is injective\footnote{Indeed, any
$k\in\mathbb{P}^{n}$ satisfying $k\cdot\sigma=k$ can be uniquely reconstructed
from its image $R\left(  k\right)  =\left(  r_{1},r_{2},\ldots,r_{m}\right)
$: Indeed, the value $r_{i}$ determines the entries $k_{j}$ for all $j\in
z_{i}$ (where $k=\left(  k_{1},k_{2},\ldots,k_{n}\right)  $), and thus all
entries $k_{j}$ of $k$ are determined (since each $j\in\left[  n\right]  $
belongs to some $z_{i}$).} and surjective\footnote{Indeed, if $\left(
r_{1},r_{2},\ldots,r_{m}\right)  \in\mathbb{P}^{m}$ is given, then we can
easily construct an $n$-tuple $k\in\mathbb{P}^{n}$ (satisfying $k\cdot
\sigma=k$) that gets mapped to $\left(  r_{1},r_{2},\ldots,r_{m}\right)  $
under $R$. Namely, the $j$-th entry of this $n$-tuple $k$ will be $r_{i}$
where $i$ is the unique element of $\left\{  1,2,\ldots,m\right\}  $ that
satisfies $j\in z_{i}$.}; thus, $R$ is bijective. Hence, we can substitute
$R\left(  k\right)  $ for $\left(  r_{1},r_{2},\ldots,r_{m}\right)  $ in the
sum $\sum_{\left(  r_{1},r_{2},\ldots,r_{m}\right)  \in\mathbb{P}^{m}}%
x_{r_{1}}^{\left\vert z_{1}\right\vert }x_{r_{2}}^{\left\vert z_{2}\right\vert
}\cdots x_{r_{m}}^{\left\vert z_{m}\right\vert }$, and thus obtain%
\[
\sum_{\left(  r_{1},r_{2},\ldots,r_{m}\right)  \in\mathbb{P}^{m}}x_{r_{1}%
}^{\left\vert z_{1}\right\vert }x_{r_{2}}^{\left\vert z_{2}\right\vert }\cdots
x_{r_{m}}^{\left\vert z_{m}\right\vert }=\sum_{\substack{k\in\mathbb{P}%
^{n};\\k\cdot\sigma=k}}x_{k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{p203.altpf.p.7.17.ptype.pf.4})}\right)  .
\]
But $\operatorname*{type}\sigma=\left(  \left\vert z_{1}\right\vert
,\left\vert z_{2}\right\vert ,\ldots,\left\vert z_{m}\right\vert \right)  $;
thus,%
\begin{align*}
p_{\operatorname*{type}\sigma}  &  =p_{\left(  \left\vert z_{1}\right\vert
,\left\vert z_{2}\right\vert ,\ldots,\left\vert z_{m}\right\vert \right)
}=p_{\left\vert z_{1}\right\vert }p_{\left\vert z_{2}\right\vert }\cdots
p_{\left\vert z_{m}\right\vert }=\prod_{i=1}^{m}\underbrace{p_{\left\vert
z_{i}\right\vert }}_{=\sum_{r\in\mathbb{P}}x_{r}^{\left\vert z_{i}\right\vert
}}=\prod_{i=1}^{m}\sum_{r\in\mathbb{P}}x_{r}^{\left\vert z_{i}\right\vert }\\
&  =\sum_{\left(  r_{1},r_{2},\ldots,r_{m}\right)  \in\mathbb{P}^{m}}x_{r_{1}%
}^{\left\vert z_{1}\right\vert }x_{r_{2}}^{\left\vert z_{2}\right\vert }\cdots
x_{r_{m}}^{\left\vert z_{m}\right\vert }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
the product rule}\right) \\
&  =\sum_{\substack{k\in\mathbb{P}^{n};\\k\cdot\sigma=k}}x_{k}.
\end{align*}
This proves (\ref{p203.altpf.p.7.17.ptype}).]

\textit{Step 4:} Problem C.4 shows that each $\lambda\vdash n$ satisfies%
\begin{equation}
\left(  \text{the number of all }\sigma\in S_{n}\text{ satisfying
}\operatorname*{type}\sigma=\lambda\right)  \cdot z_{\lambda}=n!.
\label{p203.altpf.p.7.17.cla1}%
\end{equation}
(Indeed, this is precisely the claim $c_{\lambda}z_{\lambda}=n!$ from Problem
C.4, because $c_{\lambda}$ stands for the number of all $\sigma\in S_{n}$
satisfying $\operatorname*{type}\sigma=\lambda$ in Problem C.4.) Now,
$\operatorname*{type}\sigma\vdash n$ for each $\sigma\in S_{n}$; hence, we can
split the sum $\sum_{\sigma\in S_{n}}p_{\operatorname*{type}\sigma}$ according
to the value of $\operatorname*{type}\sigma$ as follows:
\begin{align*}
\sum_{\sigma\in S_{n}}p_{\operatorname*{type}\sigma}  &  =\sum_{\lambda\vdash
n}\sum_{\substack{\sigma\in S_{n};\\\operatorname*{type}\sigma=\lambda
}}\underbrace{p_{\operatorname*{type}\sigma}}_{\substack{=p_{\lambda
}\\\text{(since }\operatorname*{type}\sigma=\lambda\text{)}}}\\
&  =\sum_{\lambda\vdash n}\underbrace{\sum_{\substack{\sigma\in S_{n}%
;\\\operatorname*{type}\sigma=\lambda}}p_{\lambda}}_{=\left(  \text{the number
of all }\sigma\in S_{n}\text{ satisfying }\operatorname*{type}\sigma
=\lambda\right)  \cdot p_{\lambda}}\\
&  =\sum_{\lambda\vdash n}\underbrace{\left(  \text{the number of all }%
\sigma\in S_{n}\text{ satisfying }\operatorname*{type}\sigma=\lambda\right)
}_{\substack{=\dfrac{n!}{z_{\lambda}}\\\text{(by (\ref{p203.altpf.p.7.17.cla1}%
))}}}\cdot p_{\lambda}\\
&  =\sum_{\lambda\vdash n}\dfrac{n!}{z_{\lambda}}p_{\lambda}.
\end{align*}
Thus,%
\begin{align}
\sum_{\lambda\vdash n}\dfrac{n!}{z_{\lambda}}p_{\lambda}  &  =\sum_{\sigma\in
S_{n}}\underbrace{p_{\operatorname*{type}\sigma}}_{\substack{=\sum
_{\substack{k\in\mathbb{P}^{n};\\k\cdot\sigma=k}}x_{k}\\\text{(by
(\ref{p203.altpf.p.7.17.ptype}))}}}=\underbrace{\sum_{\sigma\in S_{n}}%
\ \ \sum_{\substack{k\in\mathbb{P}^{n};\\k\cdot\sigma=k}}}_{=\sum
_{k\in\mathbb{P}^{n}}\ \ \sum_{\substack{\sigma\in S_{n};\\k\cdot\sigma=k}%
}}x_{k}\nonumber\\
&  =\sum_{k\in\mathbb{P}^{n}}\ \ \underbrace{\sum_{\substack{\sigma\in
S_{n};\\k\cdot\sigma=k}}x_{k}}_{=\left(  \text{the number of all }\sigma\in
S_{n}\text{ satisfying }k\cdot\sigma=k\right)  x_{k}}\nonumber\\
&  =\sum_{k\in\mathbb{P}^{n}}\underbrace{\left(  \text{the number of all
}\sigma\in S_{n}\text{ satisfying }k\cdot\sigma=k\right)  }%
_{\substack{=\left\vert \left\{  \sigma\in S_{n}\ \mid\ k\cdot\sigma
=k\right\}  \right\vert \\=\dfrac{n!}{\left\vert k\cdot S_{n}\right\vert
}\\\text{(by (\ref{p203.altpf.p.7.17.os2}))}}}x_{k}\nonumber\\
&  =\sum_{k\in\mathbb{P}^{n}}\dfrac{n!}{\left\vert k\cdot S_{n}\right\vert
}x_{k}. \label{p203.altpf.p.7.17.4b}%
\end{align}


But recall that (as we saw in Step 2) for each $n$-tuple $k\in\mathbb{P}^{n}$,
there is exactly one weakly increasing $n$-tuple $j\in\mathbb{P}^{n}$
satisfying $j\in k\cdot S_{n}$. Thus, we can split the sum $\sum
_{k\in\mathbb{P}^{n}}\dfrac{n!}{\left\vert k\cdot S_{n}\right\vert }x_{k}$
according to the value of this $j$. We thus find%
\begin{align*}
&  \sum_{k\in\mathbb{P}^{n}}\dfrac{n!}{\left\vert k\cdot S_{n}\right\vert
}x_{k}\\
&  =\sum_{\substack{j\in\mathbb{P}^{n}\text{ is weakly}\\\text{increasing}%
}}\ \ \sum_{\substack{k\in\mathbb{P}^{n};\\j\in k\cdot S_{n}}%
}\underbrace{\dfrac{n!}{\left\vert k\cdot S_{n}\right\vert }}%
_{\substack{=\dfrac{n!}{\left\vert j\cdot S_{n}\right\vert }\\\text{(since
}k\cdot S_{n}=j\cdot S_{n}\\\text{(because }j\in k\cdot S_{n}\text{))}%
}}\underbrace{x_{k}}_{\substack{=x_{j}\\\text{(by (\ref{p203.altpf.p.7.17.3kj}%
))}}}\\
&  =\sum_{\substack{j\in\mathbb{P}^{n}\text{ is weakly}\\\text{increasing}%
}}\ \ \underbrace{\sum_{\substack{k\in\mathbb{P}^{n};\\j\in k\cdot S_{n}%
}}\dfrac{n!}{\left\vert j\cdot S_{n}\right\vert }x_{j}}_{=\left(  \text{the
number of all }k\in\mathbb{P}^{n}\text{ satisfying }j\in k\cdot S_{n}\right)
\cdot\dfrac{n!}{\left\vert j\cdot S_{n}\right\vert }x_{j}}\\
&  =\sum_{\substack{j\in\mathbb{P}^{n}\text{ is weakly}\\\text{increasing}%
}}\underbrace{\left(  \text{the number of all }k\in\mathbb{P}^{n}\text{
satisfying }j\in k\cdot S_{n}\right)  }_{\substack{=\left(  \text{the number
of all }k\in\mathbb{P}^{n}\text{ satisfying }k\in j\cdot S_{n}\right)
\\\text{(since \textquotedblleft}j\in k\cdot S_{n}\text{\textquotedblright\ is
equivalent to \textquotedblleft}k\in j\cdot S_{n}\text{\textquotedblright)}%
}}\cdot\dfrac{n!}{\left\vert j\cdot S_{n}\right\vert }x_{j}\\
&  =\sum_{\substack{j\in\mathbb{P}^{n}\text{ is weakly}\\\text{increasing}%
}}\underbrace{\left(  \text{the number of all }k\in\mathbb{P}^{n}\text{
satisfying }k\in j\cdot S_{n}\right)  }_{=\left\vert j\cdot S_{n}\right\vert
}\cdot\dfrac{n!}{\left\vert j\cdot S_{n}\right\vert }x_{j}\\
&  =\sum_{\substack{j\in\mathbb{P}^{n}\text{ is weakly}\\\text{increasing}%
}}\underbrace{\left\vert j\cdot S_{n}\right\vert \cdot\dfrac{n!}{\left\vert
j\cdot S_{n}\right\vert }}_{=n!}x_{j}=n!\underbrace{\sum_{\substack{j\in
\mathbb{P}^{n}\text{ is weakly}\\\text{increasing}}}x_{j}}_{\substack{=h_{n}%
\\\text{(by (\ref{p203.altpf.p.7.17.1}))}}}\\
&  =n!h_{n}.
\end{align*}
Hence, (\ref{p203.altpf.p.7.17.4b}) becomes%
\[
\sum_{\lambda\vdash n}\dfrac{n!}{z_{\lambda}}p_{\lambda}=\sum_{k\in
\mathbb{P}^{n}}\dfrac{n!}{\left\vert k\cdot S_{n}\right\vert }x_{k}=n!h_{n}.
\]
Dividing both sides of this equality by $n!$, we obtain%
\[
\sum_{\lambda\vdash n}\dfrac{1}{z_{\lambda}}p_{\lambda}=h_{n}.
\]
This proves Proposition 7.17 (again). $\blacksquare$

\item \textbf{page 206, proof of Proposition 7.15:} In \textquotedblleft for
any $n\geq0$, we have $p_{n}\left(  XY\right)  $\textquotedblright, replace
\textquotedblleft$n\geq0$\textquotedblright\ by \textquotedblleft$n\geq
1$\textquotedblright.

\item \textbf{page 210, second paragraph:} I'd add a footnote reminding the
reader that $\operatorname*{sh}\left(  T\right)  $ means the shape of a
tableau $T$. (You defined this notation long ago and never used it before.)

\item \textbf{page 213, first paragraph:} I suggest replacing
\textquotedblleft and we replace $b$ with $a$\textquotedblright\ by
\textquotedblleft and we replace this $b$ with $a$\textquotedblright\ to make
it clearer that only one copy of $b$ is getting replaced.

\item \textbf{page 218, Lemma 8.8:} I'd split the first sentence into two, in
order for \textquotedblleft$c\leq d$\textquotedblright\ and \textquotedblleft%
$s_{1},\ldots,s_{n}$\textquotedblright\ to be separated by more than a comma.

\item \textbf{page 218, proof of Lemma 8.8:} The sentence \textquotedblleft
Therefore, $d_{1}$ was to the right of $c_{1}$ in $T$, so $c_{1}\leq d_{1}%
$\textquotedblright\ only makes sense if $c_{1}$ and $d_{1}$ are both defined.
You probably also want to say that if $d_{1}$ exists (i.e., the letter $d$
doesn't just end up as a new last entry in the first row), then so does
$c_{1}$.

\item \textbf{pages 218--220, proof of Proposition 8.9:} This proof is not as
clear as it could be.

First of all, I believe it makes more sense to split it into two: First, show
a lemma that $r_{c}\left(  T\right)  $ is a semistandard tableau whenever $T$
is a semistandard tableau; then use this lemma to quickly obtain Proposition
8.9 by induction. (This is worth already because you are already using this
lemma! On page 251, when you say \textquotedblleft By Proposition
8.9\textquotedblright, you really mean \textquotedblleft By the
lemma\textquotedblright, because you aren't starting with an empty tableau
there. Moreover, you are already stating the analogue of this lemma for
reverse row insertion explicitly (as Proposition 8.11); thus, for symmetry
reasons alone, it is worth giving it the same treatment.)

Second, the last two paragraphs of your proof (which are where the real work
is happening) are somewhat confusing. They refer to Figures 8.14 and 8.15, but
the configurations shown there do not seem representative of all cases: what
if there is no $b$? what if there is no $c_{j-1}$? what if there is no
$c_{j-2}$? Finally, it is not completely obvious that the shape of the tableau
remains a partition (i.e., columns don't grow holes) throughout the process,
and this is somewhat implicit in your argument. If interpreted correctly, the
proof works, but I think you can make it easier for the reader to interpret it
correctly if you reorganize it as follows:

After the sentence \textquotedblleft In fact, we only change one entry of that
column\textquotedblright, add the following: \textquotedblleft\ -- namely, we
insert $c_{j-1}$ either into the position previously occupied by $c_{j}$
(somewhere in the existing $j$-th row) or into a hitherto empty position (thus
extending the $j$-th row). Thus, we need to prove that:

\begin{enumerate}
\item[(A)] no \textquotedblleft hole\textquotedblright\ is gaping below this
position (i.e., there is an entry immediately below it), unless it is in the
first row;

\item[(B)] the number $c_{j-1}$ (in its new position) is larger than the entry
immediately below it, unless it is in the first row; and

\item[(C)] the number $c_{j-1}$ (in its new position) is smaller than the
entry immediately above it (if there is such an entry).
\end{enumerate}

Among these three properties, (C) is trivial: If there is an entry $b$
immediately above the new position, then the position into which we inserted
$c_{j-1}$ was previously occupied by $c_{j}$, and we must have $c_{j}<b$
(since the tableau was semistandard before the insertion), and thus by Lemma
8.6 we have $c_{j-1}<c_{j}<b$, which means precisely that $c_{j-1}$ is smaller
than the entry immediately above the position where it was
inserted.\textquotedblright

Now, of course, it remains to prove (A) and (B). Now you can forget about $b$,
and the existence of $c_{j-1}$ and $c_{j-2}$ is not in question (we are
assuming that we aren't in the first row anymore, so we have $j\geq2$).

Obviously, my writing is neither economical nor very readable, but I think the
organization of my argument is better.

\item \textbf{pages 220--221, first paragraph of page 221 (= last paragraph of
page 220) and second paragraph of page 221:} I find this rather confusing. No
clarity is gained by handling the $n=2$ case separately from the $n\geq3$
case; your description of the latter is actually clearer (at least to me), and
the former is obscured by the fact that \textquotedblleft the intermediate
pair $\left(  S_{int},U_{int}\right)  $ we obtain by applying our bijection to
the reverse plane partition consisting of just $T_{2}$\textquotedblright\ is
simply the pair $\left(  T_{2},\varnothing\right)  $, which makes the reader
wonder whether she is misreading the text (or why otherwise you are
introducing new names for old things).

Also, \textquotedblleft By Lemma 8.8, no two boxes of $U$ are in the same
column\textquotedblright\ is not quite correct: This is true not by Lemma 8.8,
but rather by the very fact that we have only added one new box per column.

\item \textbf{page 221, last paragraph:} \textquotedblleft add
one\textquotedblright\ $\rightarrow$ \textquotedblleft add $1$%
\textquotedblright. (The word \textquotedblleft one\textquotedblright\ can
easily be mistaken for a pronoun, which would even make some sense here but
definitely not reflect your intent.)

\item \textbf{page 222, first paragraph:} You write: \textquotedblleft This
will fill some of the empty boxes, but by Lemma 8.8 we will only need the
empty boxes we have already added\textquotedblright. Be careful: It is true
that the newly filled boxes form a horizontal strip (because of Lemma 8.8);
but this only shows that the newly filled boxes in rows $2,3,4,\ldots$ do not
stick out of $\lambda$. You need to argue separately for why the newly filled
boxes in row $1$ do not stick out of $\lambda$ (after all, a horizontal strip
could have an arbitrarily long first row). This follows from the fact that the
bottom row of $S$ is $T_{1}$; but you need to state this fact here (and prove
it -- frankly, I don't find it obvious enough to just leave as an exercise at
this stage in the treatment of RSK).

In general, I suggest first dealing with $S$ completely (including showing
that $S$ is semistandard and does not stick out of $\lambda$), and only then
defining $U$.

I would also refer to Figures 8.18 and 8.19 earlier on (not just at the end of
the construction), and I would also say in the caption of Figure 8.19 what
step of the construction each filling corresponds to, and what $\widehat{T}%
_{1}$ is (namely, $\widehat{T}_{1}=\left(
\begin{array}
[c]{ccc}%
5 & 5 & 6
\end{array}
\right)  $).

\item \textbf{page 223, first paragraph:} \textquotedblleft subtract
one\textquotedblright\ $\rightarrow$ \textquotedblleft subtract $1$%
\textquotedblright. (Again, the word \textquotedblleft one\textquotedblright%
\ looks too much like a pronoun.)

\item \textbf{page 226, between Proposition 8.11 and Definition 8.12:}
\textquotedblleft to ensure $x^{Q\left(  \pi\right)  }%
=x^{\operatorname*{topwt}\left(  \pi\right)  }$\textquotedblright%
\ $\rightarrow$ \textquotedblleft to ensure $x^{Q\left(  \pi\right)
}=\operatorname*{topwt}\left(  \pi\right)  $\textquotedblright.

\item \textbf{page 227, proof of Proposition 8.14:} After \textquotedblleft
Now by Lemma 8.8\textquotedblright, add \textquotedblleft and Lemma
8.7\textquotedblright\ (and, ideally, elaborate on the details). Indeed, you
want to know that the bumping path that results in $a_{j}$ ends further right
than the bumping path that results in $a_{j-r}$ does. Lemma 8.8 only shows
that the former bumping path stays to the right of the latter in each row; but
since the two paths can end in different rows, this does not mean that the
former path ends further right than the latter. You only get to that
conclusion if you know that bumping paths trend left (Lemma 8.7).

Better yet, this consequence of Lemma 8.8 and Lemma 8.7 could be made an extra
lemma, since you use it several times. (Note that you already state its
analogue for reverse row insertion as a lemma -- Lemma 8.17.)

\item \textbf{page 229, first paragraph:} Again, when you say
\textquotedblleft by Lemma 8.8\textquotedblright, you mean not Lemma 8.8 but
the consequence of Lemma 8.8 and Lemma 8.7 mentioned above (page 227).

\item \textbf{page 229, last paragraph:} \textquotedblleft and building
$E(P,Q)$\textquotedblright\ $\rightarrow$ \textquotedblleft and building
$E(P,Q)=\left[
\begin{array}
[c]{cccc}%
a_{1} & a_{2} & \cdots & a_{n}\\
b_{1} & b_{2} & \cdots & b_{n}%
\end{array}
\right]  $\textquotedblright\ (since you are referring to $a_{j}$ and $b_{j}$
a few sentences further on).

\item \textbf{page 230, proof of Lemma 8.17:} In the first paragraph of this
proof, it is not precise to say that $c_{0}$ and $d_{0}$ are moved to a lower
row; they are rather being expelled from the tableau altogether. (Perhaps
worth making precise in a footnote.)

\item \textbf{page 231, proof of Lemma 8.17:} The last paragraph of this proof
is confusing: When you refer to \textquotedblleft$P$\textquotedblright, it is
not always clear which version of $P$ you mean (as $P$ changes during the
procedure that computes $E\left(  P,Q\right)  $). Moreover, you have $a_{j}$
and $a_{j+1}$ switched a few times. I suggest denoting the versions of $P$ by
$P_{n},P_{n-1},\ldots,P_{0}$, where $P_{i}$ is the version of $P$ obtained
after removing $n-i$ boxes (so that $i$ boxes remain). Also, some
pseudo-formulas like%
\[
P_{j+2}\underset{\text{obtain }c_{0}}{\overset{\text{remove }c_{k}%
}{\longrightarrow}}P_{j+1}\underset{\text{obtain }d_{0}}{\overset{\text{remove
}d_{m}}{\longrightarrow}}P_{j}%
\]
(if properly explained beforehand) might be easier to parse than long sentences.

\item \textbf{page 231, Theorem 8.19:} In the last sentence, replace
\textquotedblleft$x^{\operatorname*{topwt}\left(  \pi\right)  }%
y^{\operatorname*{bottomwt}\left(  \pi\right)  }=x^{Q\left(  \pi\right)
}y^{P\left(  \pi\right)  }$\textquotedblright\ by \textquotedblleft%
$\operatorname*{topwt}\left(  \pi\right)  =x^{Q\left(  \pi\right)  }$ and
$\operatorname*{bottomwt}\left(  \pi\right)  =y^{P\left(  \pi\right)  }%
$\textquotedblright\ (to get the notations right, but also to match the
formulation with requirements on $\mathfrak{R}$ stated on page 210).

\item \textbf{page 232:} You write: \textquotedblleft if $\pi$ is a
permutation in the usual sense\textquotedblright. This assumes that the reader
understands how a permutation can be viewed as a generalized permutation
(viz., by writing it in two-line notation, with the top row being sorted to be
$1,2,\ldots,n$). Have you ever explained that? (I'd say it's worth doing so
already in Chapter 7, when you define the concept of generalized permutations;
it explains the name.)

\item \textbf{Page 233:} Your construction of $\alpha$ does not guarantee that
$\alpha_{j}$ will be positive integers (or even nonnegative integers), nor
does it properly explain when the sequence should end and what the $j$-th
entry of a finite sequence should be understood as when the sequence has fewer
than $j$ entries. I would suggest rephrasing it in terms of infinite sequences
of integers. Then you can show that what if you start with $\varnothing$'s on
the left and bottom borders, then all the sequences you actually get are
partitions (padded with infinitely many zeroes as usually). This can be proved
by induction using a lemma which says that if, in the situation of Figure
8.31, all of $\lambda$, $\mu$ and $\nu$ are partitions such that $\mu/\lambda$
and $\nu/\lambda$ are horizontal strips, then (a) all of $m$, $M$ and $\alpha$
are partitions as well (where $m$, $M$ and $\alpha$ are as defined before -- I
suggest labeling the defining formulas so that you can refer to them cleanly),
and (b) both $\alpha/\mu$ and $\alpha/\nu$ are horizontal strips.

\item \textbf{page 235, Lemma 8.22:} You probably want to restate this lemma
so that it says that $\lambda_{j}\leq\mu_{j}$ whenever $\mu$ is the right or
upper neighbor of $\lambda$ in the growth diagram. Indeed, the way you are
currently stating it, it does not claim that $\lambda_{j}\leq\mu_{j}$ when
$\lambda$ and $\mu$ are on the eastern or northern border of the growth
diagram; but you do use this very case later (in the third paragraph of page
237, when you say \textquotedblleft By Lemma 8.22\textquotedblright).

\item \textbf{page 237, proof of Proposition 8.25:} You define
$\operatorname*{Ins}\left(  a,b\right)  $ thus: \textquotedblleft For each
vertex $\left(  a,b\right)  $, let $\operatorname*{Ins}\left(  a,b\right)  $
be the semistandard tableau we obtain by using the RSK correspondence to
insert the entries of the bottom row of $\pi$ which correspond to the squares
below and to the left of the vertex $\left(  a,b\right)  $ in the growth
diagram.\textquotedblright

I find this confusing. First, the RSK correspondence yields two tableaux, so
\textquotedblleft the semistandard tableau we obtain by using the RSK
correspondence\textquotedblright\ isn't as unambiguous as it might appear.
Second, \textquotedblleft below and to the left\textquotedblright\ may mean
either a whole quadrant or just two half-axes; you mean the quadrant, but this
isn't immediately clear. Finally, there is too much going on in a single sentence.

I suggest fixing this by defining the notation $\pi\left(  a,b\right)  $ as in
the proof of Proposition 8.28 further below (except that the meaning of
\textquotedblleft below and to the left\textquotedblright\ should be
clarified, and ideally illustrated on an example), and then defining
$\operatorname*{Ins}\left(  a,b\right)  $ to be $P\left(  \pi\left(
a,b\right)  \right)  $. (Note that $\pi\left(  a,b\right)  $ can also be
defined more directly as the generalized permutation obtained from $\pi$ by
removing all columns $\left(
\begin{array}
[c]{c}%
a_{i}\\
b_{i}%
\end{array}
\right)  $ with $a_{i}>a$ and removing all columns $\left(
\begin{array}
[c]{c}%
a_{i}\\
b_{i}%
\end{array}
\right)  $ with $b_{i}>b$.)

Such a definition of $\operatorname*{Ins}\left(  a,b\right)  $ would have the
additional advantage of making the similarity between the $P$-filling and the
$Q$-filling more explicit.

\item \textbf{page 239, second paragraph:} After \textquotedblleft Since
$\operatorname*{Ins}\left(  W,H-1\right)  $ is the tableau made by the numbers
less than $H$\textquotedblright, add \textquotedblleft in the tableau
$\operatorname*{Ins}\left(  W,H\right)  $\textquotedblright.

\item \textbf{page 239, second paragraph:} After \textquotedblleft a copy of
$H$ in $\operatorname*{Ins}\left(  W-1,H\right)  $ is bumped up to the next
row if and only if it is in a box in $\mu$ that also appears in $\nu
$\textquotedblright, I would add \textquotedblleft(because it is bumped up to
the next row if and only if its box is filled with a number less than $H$
afterwards)\textquotedblright.

\item \textbf{page 241, proof of Proposition 8.28:} In the last sentence of
this proof, before \textquotedblleft we must have\textquotedblright, add
\textquotedblleft and since $\operatorname*{Fill}\left(  W,H\right)  $ and
$Q\left(  W,H\right)  $ have the same shape (because $\operatorname*{Fill}%
\left(  W,H\right)  $ has the same shape as the tableau $\operatorname*{Ins}%
\left(  W,H\right)  $ from the proof of Proposition 8.25, which in turn is
identical to the tableau $P\left(  W,H\right)  $ from the same proof, which in
turn has the same shape as $Q\left(  W,H\right)  $), \textquotedblright.

\item \textbf{page 242, proof of Theorem 8.29:} \textquotedblleft over a line
from the lower left corner to the upper right corner\textquotedblright%
\ $\rightarrow$ \textquotedblleft over the line of slope $1$ passing through
the lower left corner\textquotedblright. (This line will not pass through the
upper right corner unless $W=H$.)

\item \textbf{page 242, proof of Theorem 8.29:} \textquotedblleft the
$P$-labeling and the $Q$-labeling\textquotedblright\ $\rightarrow$
\textquotedblleft the $P$-filling and the $Q$-filling\textquotedblright.

\item \textbf{page 243, Problem 8.6:} The words \textquotedblleft with finite
support\textquotedblright\ should be explained.

\item \textbf{page 243, Problem 8.9:} The word \textquotedblleft
hook\textquotedblright\ has never been defined.

\item \textbf{page 243, Problem 8.10:} \textquotedblleft whose last entry is
$a_{j}$\textquotedblright\ is ambiguous -- you mean that the last entry is
picked from the $j$-th position, not just that it equals $a_{j}$ (which would
be the literal interpretation).

\item \textbf{page 245, Problem 8.21:} The \textquotedblleft for which
boxes\textquotedblright\ formulation is somewhat ambiguous (the
\textquotedblleft boxes\textquotedblright\ looks like it refers to the
\textquotedblleft which\textquotedblright). I would replace it by
\textquotedblleft whose order relation is defined as follows: two
boxes\textquotedblright.

\item \textbf{page 246:} Replace \textquotedblleft the hook
lengths\textquotedblright\ by \textquotedblleft the
hook-length\textquotedblright\ (in order to match the hyphenation you use in
Problem B.5).

\item \textbf{page 248, \S 9.1:} One line above Example 9.1, replace
\textquotedblleft$s_{11}=h_{2}$, and $s_{2}=e_{2}$\textquotedblright\ by
\textquotedblleft$s_{11}=e_{2}$, and $s_{2}=h_{2}$\textquotedblright.

\item \textbf{page 251, proof of Theorem 9.3:} At the end of the first
paragraph of this proof, replace \textquotedblleft for which $\mu
\subseteq\operatorname*{sh}\left(  P^{\prime}\right)  $, $\operatorname*{sh}%
\left(  P^{\prime}\right)  /\mu$ is a horizontal strip of length $n$, and
$x^{P}\prod_{j\in J}x_{j}=x^{P^{\prime}}$\textquotedblright\ by
\textquotedblleft for which $\mu\subseteq\operatorname*{sh}\left(  P^{\prime
}\right)  $ and $\operatorname*{sh}\left(  P^{\prime}\right)  /\mu$ is a
horizontal strip of length $n$, which bijection has the property that
$x^{P}\prod_{j\in J}x_{j}=x^{P^{\prime}}$\textquotedblright. (Your current
wording belies the fact that \textquotedblleft$\mu\subseteq\operatorname*{sh}%
\left(  P^{\prime}\right)  $\textquotedblright\ and \textquotedblleft%
$\operatorname*{sh}\left(  P^{\prime}\right)  /\mu$ is a horizontal strip of
length $n$\textquotedblright\ are properties of the tableaux $P^{\prime}$ in
the set, whereas \textquotedblleft$x^{P}\prod_{j\in J}x_{j}=x^{P^{\prime}}%
$\textquotedblright\ is a property of the bijection.)

\item \textbf{page 251, proof of Theorem 9.3:} After \textquotedblleft to
obtain a filling $P^{\prime}$ of shape $\lambda$.\textquotedblright, I'd add
\textquotedblleft Thus, $P^{\prime}=r_{j_{n}}\left(  r_{j_{n-1}}\left(
\cdots\left(  r_{j_{1}}\left(  P\right)  \right)  \cdots\right)  \right)
$.\textquotedblright.

\item \textbf{page 251, proof of Theorem 9.3:} When you write
\textquotedblleft combined with Lemma 8.8\textquotedblright, you mean not
Lemma 8.8 itself but rather the corollary of Lemma 8.8 and Lemma 8.7 that says
that (in the notation of Lemma 8.8) the new cell in $r_{d}\left(  r_{c}\left(
T\right)  \right)  $ is in a column further right than the new cell in
$r_{c}\left(  T\right)  $. (As I already said above, I think it makes sense to
state this corollary as a lemma in its own right; this is not the only place
where you say you are using Lemma 8.8, but in truth are using that corollary.)

\item \textbf{page 252, proof of Proposition 9.4:} After \textquotedblleft But
a horizontal strip of length $\lambda_{1}$ has exactly one filling of content
$\lambda_{1}$\textquotedblright, add \textquotedblleft, whereas any other skew
shape has none\textquotedblright. (You want to know that the sum in (9.2) has
no other addends than the ones coming from horizontal strips of length
$\lambda_{1}$.)

\item \textbf{page 252, proof of Proposition 9.4:} \textquotedblleft we can
group fillings of $\nu/\mu$ of content $\lambda$\textquotedblright%
\ $\rightarrow$ \textquotedblleft we can group the semistandard tableaux of
shape $\nu/\mu$ and content $\lambda$\textquotedblright. (For arbitrary
fillings, the boxes which contain $l$ will not generally form a horizontal strip.)

\item \textbf{page 254, first paragraph:} In the first display on page 254,
replace \textquotedblleft$\sum_{\zeta}K_{\nu,\zeta}m_{\nu}$\textquotedblright%
\ by \textquotedblleft$\sum_{\zeta}K_{\nu,\zeta}m_{\zeta}$\textquotedblright.
Besides, you are using the equality (4.6) here, which may be worth mentioning.

\item \textbf{page 254, last paragraph:} Replace \textquotedblleft for which
$\mu\subseteq\operatorname*{sh}\left(  P^{\prime}\right)  $,
$\operatorname*{sh}\left(  P^{\prime}\right)  /\mu$ is a vertical strip of
length $n$, and $x^{P}\prod_{j\in J}x_{j}=x^{P^{\prime}}$\textquotedblright%
\ by \textquotedblleft for which $\mu\subseteq\operatorname*{sh}\left(
P^{\prime}\right)  $ and $\operatorname*{sh}\left(  P^{\prime}\right)  /\mu$
is a vertical strip of length $n$, which bijection has the property that
$x^{P}\prod_{j\in J}x_{j}=x^{P^{\prime}}$\textquotedblright. (This is the same
issue as on page 251 above.)

\item \textbf{page 260, (9.5):} Replace the outer summation sign
\textquotedblleft$\sum_{\mu_{vh}\supseteq\mu}$\textquotedblright\ by
\textquotedblleft$\sum_{\substack{\mu_{vh}\supseteq\mu;\\\left\vert \mu
_{vh}/\mu\right\vert =n}}$\textquotedblright\ (since otherwise, the condition
$\left\vert \mu_{vh}/\mu\right\vert =n$ is lost).

\item \textbf{page 260:} Before \textquotedblleft We suspect many
terms\textquotedblright, I'd add \textquotedblleft Each such choice of
$\mu_{v}$ partitions the skew shape $\mu_{vh}/\mu$ into two subshapes $\mu
_{v}/\mu$ and $\mu_{vh}/\mu_{v}$, which shall be called the \textit{inner
vertical strip} and the \textit{outer horizontal strip},
respectively.\textquotedblright. (This introduces the terminology that is used
later on.)

\item \textbf{page 264, first paragraph:} Here it should also be explained why
the traversal of the component starting at the head actually is a traversal
(i.e., it passes through each box).

\item \textbf{page 264, Lemma 9.16:} In the first sentence of the lemma:
\textquotedblleft Suppose $\mu\subseteq\mu_{vh}$ are partitions and $\mu
_{vh}/\mu$ can be separated into an inner vertical strip and an outer
horizontal strip\textquotedblright\ $\rightarrow$ \textquotedblleft Suppose
$\mu\subseteq\mu_{vh}$ are partitions such that $\mu_{vh}/\mu$ is a border
strip\textquotedblright.

In the last sentence of the lemma, after \textquotedblleft there is a
separation of $\mu_{vh}/\mu$\textquotedblright, add \textquotedblleft into an
inner vertical strip and an outer horizontal strip\textquotedblright.

\item \textbf{page 264, proof of Lemma 9.16:} In the first paragraph of this
proof, you say that \textquotedblleft we can assume without loss of generality
that $\mu_{vh}/\mu$ is connected\textquotedblright. This hinges on some tacit
arguments that is not completely trivial. As you correctly notice, one of
these arguments is saying that two boxes in different connected components
cannot lie in the same row or column. But here is another: If we separate each
connected component of a skew shape into an inner and an outer part, then we
get a separation of the whole skew shape into the union of all inner parts and
the union of all outer parts. (That is, the union of all inner parts is a skew
shape, and the union of all outer parts is a skew shape.) The easiest way to
prove this is by arguing that if we do not get a separation this way, then we
must have some inner-part box that is a right or top neighbor of some
outer-part box; but due to their adjacency, these two boxes have to belong to
the same connected component, and thus cause a contradiction.

\item \textbf{page 264, proof of Lemma 9.16:} In the third paragraph of this
proof, replace \textquotedblleft is to the left (resp., below) the
box\textquotedblright\ by \textquotedblleft is to the left of (resp., below)
the box\textquotedblright.

\item \textbf{page 265, first line:} On the first line of page 265, replace
\textquotedblleft Lemma 9.16 tells\textquotedblright\ by \textquotedblleft
Lemma 9.15 and Lemma 9.16 tell\textquotedblright\ (you need Lemma 9.15 to know
that all $\mu_{vh}/\mu$'s appearing in (9.5) are border strips).

\item \textbf{page 265, (9.6):} Replace the outer summation sign
\textquotedblleft$\sum_{\mu_{vh}\supseteq\mu}$\textquotedblright\ by
\textquotedblleft$\sum_{\substack{\mu_{vh}\supseteq\mu;\\\left\vert \mu
_{vh}/\mu\right\vert =n;\\\mu_{vh}/\mu\text{ is a border strip}}%
}$\textquotedblright.

The same change should be done in (9.7).

\item \textbf{page 265, Theorem 9.17:} Replace \textquotedblleft$n\geq
0$\textquotedblright\ by \textquotedblleft$n\geq1$\textquotedblright\ here.
Even if you did define $p_{0}$ to be $1$, the theorem would still fail for
$n=0$ due to an incorrect sign.

\item \textbf{page 266, Definition 9.18:} Before \textquotedblleft%
$\lambda\left(  j+1\right)  /\lambda\left(  j\right)  $ is a connected border
strip\textquotedblright, I would add \textquotedblleft the set $\lambda\left(
j\right)  $ is (the Ferrers diagram of) a partition and\textquotedblright.
(This might be redundant -- I am not 100\% sure, but I'm pretty sure it makes
everything clearer; you haven't even properly defined what $\alpha/\beta$ is
when $\alpha$ and $\beta$ are not partition shapes. Equivalently, you can
require that the entries of the tableau increase weakly left-to-right and top-to-bottom.)

\item \textbf{page 266, Definition 9.18:} The upper bound of the product in
the displayed formula should be $l\left(  \mu\right)  -1$ rather than
$l\left(  \mu\right)  $.

I would also replace \textquotedblleft for all $j$ with $0\leq j\leq l\left(
\mu\right)  $\textquotedblright\ by \textquotedblleft for all $j$ with $0\leq
j\leq l\left(  \mu\right)  -1$\textquotedblright. (Strictly speaking, this is
unnecessary, at least if you consider an empty skew shape to be a connected
border strip; but I think it is natural to talk of $0\leq j\leq l\left(
\mu\right)  -1$.)

Finally, I would replace \textquotedblleft we set\textquotedblright\ by
\textquotedblleft we define the \textit{sign} $\operatorname*{sgn}\left(
T\right)  $ of $T$ by\textquotedblright\ (so that the word \textquotedblleft
sign\textquotedblright\ is explained).

\item \textbf{page 267, proof of Theorem 9.20:} A better induction base for
this proof would be the case $l\left(  \lambda\right)  =0$ (which has to be
handled either way). There is nothing in your induction step that really
requires $l\left(  \lambda\right)  \geq2$ (as opposed to just $l\left(
\lambda\right)  \geq1$).

\item \textbf{page 268, proof of Theorem 9.20:} After \textquotedblleft When
we combine $T_{\alpha}$ with $\mu/\nu$\textquotedblright, I would add
\textquotedblleft(filling the boxes in $\mu/\nu$ with the number $l$ and
adding them to $T_{\alpha}$)\textquotedblright. I would also replace
\textquotedblleft combine $T_{\alpha}$\textquotedblright\ by \textquotedblleft
combine a $T_{\alpha}\in BST\left(  \nu,\alpha\right)  $\textquotedblright.

\item \textbf{page 269, second paragraph:} Wouldn't it be helpful to actually
cite a proof of this claim about irreducible characters? One place where this
is proved is \S 5.5 of Mark Wildon's nice (albeit much less detailed than your
book) notes \url{http://www.ma.rhul.ac.uk/~uvah099/Maths/Sym/SymFuncs2017.pdf}
. Another is Proposition 5.21.1 in
\href{http://www-math.mit.edu/~etingof/reprbook.pdf}{Pavel Etingof et al.,
\textit{Introduction to representation theory}, AMS 2011 (updated version
2018)}.

\item \textbf{page 272, solution to Example 10.2:} It would be helpful to
clarify whether the rows are $T_{1}$ and the columns are $T_{2}$, or vice versa.

\item \textbf{page 272, (10.3):} You are missing an $s_{411}\left(
X_{3}\right)  $ term on the right hand side.

\item \textbf{page 273:} After \textquotedblleft will have the shapes $\left(
4,2\right)  $,\textquotedblright\ add \textquotedblleft$\left(  4,1,1\right)
$,\textquotedblright.

\item \textbf{page 270, Problem 9.10:} Are you sure you don't want to require
the border strips in the decompositions to be connected?

\item \textbf{page 274, second paragraph:} In the description of jeu de
taquin, you write: \textquotedblleft There is a unique way to slide a box into
the blank space so that the resulting object is column strict and row
nondecreasing\textquotedblright. This is not quite obvious: The uniqueness is
easy, but the existence (specifically, the proof that the sliding entry will
not be smaller than its new left neighbor or smaller-or-equal to its new
bottom neighbor) needs a proof. And this proof is not completely trivial; it
relies on the fact that there is only one empty box in an otherwise filled
skew diagram (otherwise it wouldn't work, which is why the tableau switching
paper of Benkart/Sottile/Stroomer requires the rather technical
\textquotedblleft staircases\textquotedblright\ condition).

\item \textbf{page 274:} \textquotedblleft column strict and row
nondecreasing\textquotedblright\ $\rightarrow$ \textquotedblleft column-strict
and row-nondecreasing\textquotedblright.

\item \textbf{page 278, \S 10.2:} In the second paragraph of \S 10.2, replace
\textquotedblleft suppose $\operatorname*{word}\left(  T_{1}\right)  $ has
just one entry\textquotedblright\ by \textquotedblleft suppose
$\operatorname*{word}\left(  T_{2}\right)  $ has just one
entry\textquotedblright.

\item \textbf{page 279, third paragraph:} Replace \textquotedblleft In one
type of step we have consecutive entries $x$ and $z$\textquotedblright\ by
\textquotedblleft In one type of step we have consecutive entries $z$ and
$x$\textquotedblright. (It is a bit confusing to label consecutive steps
starting with the second one.)

\item \textbf{page 279, third paragraph:} Replace \textquotedblleft for some
$i\geq j$\textquotedblright\ by \textquotedblleft for some $i>j$%
\textquotedblright. (You aren't switching $c$ with $a_{j}$.)

\item \textbf{page 279, third paragraph:} Replace \textquotedblleft as well as
$x=a_{i}$ and $y=a_{i+1}$ for some $i<j$\textquotedblright\ by
\textquotedblleft as well as $x=a_{i}$ and $y=%
\begin{cases}
a_{i+1}, & \text{if }i<j-1;\\
c, & \text{if }i=j-1
\end{cases}
$ for some $i<j$\textquotedblright. (When you first swap $a_{j}$ with
$a_{j-1}$, the catalyzing neighbor on the right is $c$, not $a_{j}$.)

\item \textbf{page 279, third paragraph:} I think an example of the traveler
moving through a row would be helpful here (showing what $x,y,z$ are in each step).

\item \textbf{page 280:} After Definition 10.10, I'd add a sentence along the
lines of \textquotedblleft This relation $\sim_{K}$ is an equivalence
relation, since any elementary Knuth transformation can be undone by another
elementary Knuth transformation.\textquotedblright.

\item \textbf{page 281, first paragraph:} In the sentence \textquotedblleft In
general, our definition of Knuth equivalence guarantees that the Knuth
equivalence classes for a set of words are exactly the connected components of
the corresponding graph on those words\textquotedblright, replace
\textquotedblleft a set of words\textquotedblright\ by \textquotedblleft a set
of words closed under elementary Knuth transformations\textquotedblright%
\footnote{This is not the best condition, but the easiest one to state. In
truth, it suffices to have a \textquotedblleft Knuth-convex\textquotedblright%
\ set of words, i.e., a set of words such that any two Knuth-equivalent words
in the set are connected by a sequence of elementary Knuth transformations
with all intermediate words being in the set as well.}. For example, if $S$ is
the set of all words that correspond to even permutations $\pi\in S_{n}$, then
the graph corresponding to this set $S$ is totally disconnected (i.e., each
vertex is isolated), but the Knuth equivalence classes are nontrivial (for
large $n$).

\item \textbf{page 281, second paragraph:} \textquotedblleft as a product on
words\textquotedblright\ $\rightarrow$ \textquotedblleft as a product on
tableaux\textquotedblright.

\item \textbf{page 282:} One line above Corollary 10.13, replace
\textquotedblleft tableaux\textquotedblright\ by \textquotedblleft
tableau\textquotedblright.

\item \textbf{page 283, proof of Theorem 10.14:} \textquotedblleft using
using\textquotedblright.

\item \textbf{page 283, proof of Theorem 10.14:} When you refer to Figures
10.8 and 10.9, I suggest explaining that the part left of $a$ and the part
right of $d_{k}$ are not shown (i.e., there may be more filled boxes left of
$a$ or right of $d_{k}$, but they do not matter). I would actually remove the
$a$ from the figures as well, and only re-include it when it is needed in the
proof later on (so I would duplicate the figures, removing the $a$'s from the
first copies but leaving them in the second).

\item \textbf{pages 283--285, proof of Theorem 10.14:} I have given up trying
to understand this proof. I think it is too long and yet too terse. I
understand the idea, but I'm not sure I can fill in the details. I am also a
bit skeptical: You start by moving the $c_{j}$'s past the $b_{i}$'s using $x$
as a catalyst, and you finish by claiming that \textquotedblleft we can
reverse our initial steps\textquotedblright\ -- but $x$ is no longer available
as a catalyst at that point. (Something tells me that we need more than just
the single $x$ entry -- I would expect the upper neighbors of $c_{1}%
,c_{2},\ldots,c_{m}$ to all come useful.)

I believe there is a good case for such a proof, but I believe it should be
organized differently. It lends itself ideally to being factored into a
sequence of relatively simple lemmas such as \textquotedblleft If $c_{1}\leq
c_{2}\leq\cdots\leq c_{m}<x\leq b_{1}\leq b_{2}\leq\cdots\leq b_{k}$ is a
sequence of letters, then $xb_{1}\cdots b_{k}c_{1}\cdots c_{m}\sim_{K}%
xc_{1}\cdots c_{m}b_{1}\cdots b_{k}$\textquotedblright\ (I'm not actually if
this one is true, but I think you get the gist), and I see no reason not to
factor it. Moreover, these proofs can preferably be done by induction rather
than by showing the first step and handwaving the rest

However, given the position in which Theorem 10.14 stands in your book, I
would suggest a different way of proof:

\textit{Alternative proof of Theorem 10.14 (sketched).} It suffices to prove
the following lemma:

\textbf{Lemma 10.14a.} Let $a_{1}\leq a_{2}\leq\cdots\leq a_{m}\leq x\leq
b_{1}\leq b_{2}\leq\cdots\leq b_{k}$ and $c_{1}\leq c_{2}\leq\cdots\leq
c_{m}\leq x\leq d_{1}\leq d_{2}\leq\cdots\leq d_{k}$ be two weakly increasing
numbers such that

\begin{itemize}
\item every $i$ satisfies $c_{i}<a_{i}$;

\item every $j$ satisfies $d_{j}<b_{j}$.
\end{itemize}

Then,
\[
a_{1}\cdots a_{m}xb_{1}\cdots b_{k}c_{1}\cdots c_{m}d_{1}\cdots d_{k}\sim
_{K}a_{1}\cdots a_{m}b_{1}\cdots b_{k}c_{1}\cdots c_{m}xd_{1}\cdots d_{k}.
\]


[\textit{Proof of Lemma 10.14a.} Let $T$ be the tableau that consists of just
one row, with entries $a_{1},a_{2},\ldots,a_{m},x,b_{1},\ldots,b_{k}$ (from
left to right). This is clearly a semistandard tableau, and satisfies
\begin{equation}
\operatorname*{word}\left(  T\right)  =a_{1}\cdots a_{m}xb_{1}\cdots b_{k}.
\label{l10.14a.pf.1}%
\end{equation}
Now, let us insert the numbers $c_{1},c_{2},\ldots,c_{m},d_{1},d_{2}%
,\ldots,d_{k}$ into $T$ (using RSK insertion, in this order). Here is what happens:

\begin{itemize}
\item The insertion of $c_{1}$ bumps $a_{1}$ into the $2$nd row (since
$c_{1}<a_{1}$). The first row becomes $c_{1}a_{2}\cdots a_{m}xb_{1}\cdots
b_{k}$.

\item The insertion of $c_{2}$ bumps $a_{2}$ into the $2$nd row (since
$c_{1}\leq c_{2}<a_{2}$). The first row becomes $c_{1}c_{2}a_{3}\cdots
a_{m}xb_{1}\cdots b_{k}$.

\item The insertion of $c_{3}$ bumps $a_{3}$ into the $2$nd row (since
$c_{2}\leq c_{3}<a_{3}$). The first row becomes $c_{1}c_{2}c_{3}a_{4}\cdots
a_{m}xb_{1}\cdots b_{k}$.

\item And so on. After $c_{1},c_{2},\ldots,c_{k}$ have been inserted, we
obtain the tableau%
\[%
\begin{tabular}
[c]{|c|c|c|c|ccccc}\cline{1-4}%
$a_{1}$ & $a_{2}$ & $\cdots$ & $a_{m}$ &  &  &  &  & \\\hline
$c_{1}$ & $c_{2}$ & $\cdots$ & $c_{m}$ & $x$ & \multicolumn{1}{|c}{$b_{1}$} &
\multicolumn{1}{|c}{$b_{2}$} & \multicolumn{1}{|c}{$\cdots$} &
\multicolumn{1}{|c|}{$b_{k}$}\\\hline
\end{tabular}
\ \ .
\]


\item The insertion of $d_{1}$ bumps $b_{1}$ into the $2$nd row (since $x\leq
d_{1}<b_{1}$). The first row becomes $c_{1}\cdots c_{m}xd_{1}b_{2}\cdots
b_{k}$.

\item The insertion of $d_{2}$ bumps $b_{2}$ into the $2$nd row (since
$d_{1}\leq d_{2}<b_{2}$). The first row becomes $c_{1}\cdots c_{m}xd_{1}%
d_{2}b_{3}\cdots b_{k}$.

\item The insertion of $d_{3}$ bumps $b_{3}$ into the $2$nd row (since
$d_{2}\leq d_{3}<b_{3}$). The first row becomes $c_{1}\cdots c_{m}xd_{1}%
d_{2}d_{3}b_{4}\cdots b_{k}$.

\item And so on. After $d_{1},d_{2},\ldots,d_{k}$ have been inserted, we
obtain the tableau%
\[%
\begin{tabular}
[c]{|c|c|c|c|c|c|c|c|c}\cline{1-8}%
$a_{1}$ & $a_{2}$ & $\cdots$ & $a_{m}$ & $b_{1}$ & $b_{2}$ & $\cdots$ &
$b_{k}$ & \\\hline
$c_{1}$ & $c_{2}$ & $\cdots$ & $c_{m}$ & $x$ & $d_{1}$ & $d_{2}$ & $\cdots$ &
\multicolumn{1}{|c|}{$d_{k}$}\\\hline
\end{tabular}
\ \ .
\]

\end{itemize}

Thus,%
\begin{align*}
&  r_{d_{k}}\left(  r_{d_{k-1}}\left(  \cdots\left(  r_{d_{1}}\left(
r_{c_{m}}\left(  r_{c_{m-1}}\left(  \cdots\left(  r_{c_{1}}\left(  T\right)
\right)  \right)  \right)  \right)  \right)  \right)  \right) \\
&  =%
\begin{tabular}
[c]{|c|c|c|c|c|c|c|c|c}\cline{1-8}%
$a_{1}$ & $a_{2}$ & $\cdots$ & $a_{m}$ & $b_{1}$ & $b_{2}$ & $\cdots$ &
$b_{k}$ & \\\hline
$c_{1}$ & $c_{2}$ & $\cdots$ & $c_{m}$ & $x$ & $d_{1}$ & $d_{2}$ & $\cdots$ &
\multicolumn{1}{|c|}{$d_{k}$}\\\hline
\end{tabular}
\ \ \ \ ,
\end{align*}
so that%
\begin{align}
&  \operatorname*{word}\left(  r_{d_{k}}\left(  r_{d_{k-1}}\left(
\cdots\left(  r_{d_{1}}\left(  r_{c_{m}}\left(  r_{c_{m-1}}\left(
\cdots\left(  r_{c_{1}}\left(  T\right)  \right)  \right)  \right)  \right)
\right)  \right)  \right)  \right) \nonumber\\
&  =a_{1}\cdots a_{m}b_{1}\cdots b_{k}c_{1}\cdots c_{m}xd_{1}\cdots d_{k}.
\label{l10.14a.pf.4}%
\end{align}
But Theorem 10.11 (applied to $\left(  c_{1},c_{2},\ldots,c_{m},d_{1}%
,d_{2},\ldots,d_{k}\right)  $ instead of $\left(  c_{1},c_{2},\ldots
,c_{n}\right)  $) yields%
\begin{align*}
&  \operatorname*{word}\left(  T\right)  c_{1}\cdots c_{m}d_{1}\cdots d_{k}\\
&  \sim_{K}\operatorname*{word}\left(  r_{d_{k}}\left(  r_{d_{k-1}}\left(
\cdots\left(  r_{d_{1}}\left(  r_{c_{m}}\left(  r_{c_{m-1}}\left(
\cdots\left(  r_{c_{1}}\left(  T\right)  \right)  \right)  \right)  \right)
\right)  \right)  \right)  \right)  .
\end{align*}
In view of (\ref{l10.14a.pf.1}) and (\ref{l10.14a.pf.4}), this rewrites as%
\[
a_{1}\cdots a_{m}xb_{1}\cdots b_{k}c_{1}\cdots c_{m}d_{1}\cdots d_{k}\sim
_{K}a_{1}\cdots a_{m}b_{1}\cdots b_{k}c_{1}\cdots c_{m}xd_{1}\cdots d_{k}.
\]
This proves Lemma 10.14a. $\blacksquare$ ]

\item \textbf{page 285, \S 10.3:} It is worth explaining that a
\textit{permutation} of a multiset $A$ is defined to be a tuple of elements of
$A$ such that every element of $A$ appears in this tuple as often as it
appears in $A$.

\item \textbf{page 286:} \textquotedblleft identical to the left of $x$ and to
the right of $z$\textquotedblright\ $\rightarrow$ \textquotedblleft identical
to the left and to the right of the respective $yzx$ or $yxz$
blocks\textquotedblright.

\item \textbf{page 286:} I think this argument is somewhat incomplete. In
order to obtain $r_{x}\left(  r_{z}\left(  r_{y}\left(  T\right)  \right)
\right)  =r_{z}\left(  r_{x}\left(  r_{y}\left(  T\right)  \right)  \right)
$, it isn't enough to show that the bumping paths of $z$ and $x$ in
$r_{y}\left(  T\right)  $ have no common boxes; it also needs to be shown that

\begin{itemize}
\item the bumping path of $z$ in $r_{y}\left(  T\right)  $ is identical to the
bumping path of $z$ in $r_{x}\left(  r_{y}\left(  T\right)  \right)  $, and

\item the bumping path of $x$ in $r_{y}\left(  T\right)  $ is identical to the
bumping path of $x$ in $r_{z}\left(  r_{y}\left(  T\right)  \right)  $.
\end{itemize}

The second of these two statements is an automatic consequence of the
disjointness of the bumping paths, but the first is not, since the bumping
procedure is influenced not only by the numbers in the boxes on the bumping
path itself but also by the numbers in their left neighbor boxes. Fortunately,
bumping an entry can only replace it by a smaller entry, so whatever entries
have changed from $r_{y}\left(  T\right)  $ to $r_{x}\left(  r_{y}\left(
T\right)  \right)  $ will not affect the bumping path of $z$. Thus, the first
statement holds as well. But I believe this should be explained.

\item \textbf{pages 287--288:} I think the case distinction in the
\textquotedblleft$r_{y}\left(  r_{z}\left(  r_{x}\left(  T\right)  \right)
\right)  =r_{y}\left(  r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)
$\textquotedblright\ part of the proof can be much improved. While each case
is indeed reasonably straightforward, I would argue that identifying the right
cases to distinguish between is not. Thus, I suggest listing all the cases
with their outcomes.

Here is my version of this list:

We suppose that the entries of the first row of $T$ are $a_{1},\ldots
,a_{j+k+l+n}$, where%
\begin{align*}
&  a_{1}\leq\cdots\leq a_{j}\leq x,\\
x<\  &  a_{j+1}\leq\cdots\leq a_{j+k}\leq y,\\
y<\  &  a_{j+k+1}\leq\cdots\leq a_{j+k+l}\leq z,\\
z<\  &  a_{j+k+l+1}\leq\cdots\leq a_{j+k+l+n}.
\end{align*}
Here, if $k=0$, then the chain of inequalities $x<a_{j+1}\leq\cdots\leq
a_{j+k}\leq y$ should be interpreted as being vacuous; it is thus not saying
that $x<y$. (The same applies to the chain $y<a_{j+k+1}\leq\cdots\leq
a_{j+k+l}\leq z$ when $l=0$, but this does not matter, since we already are
given that $y<z$.) Now, the cases I distinguish are the following:

\begin{itemize}
\item \textit{Case 1:} We have $n=0$.

\begin{itemize}
\item \textit{Subcase 1.1:} We have $k+l=0$. In this case, the first row of
$T$ has the form $a_{1}\cdots a_{j}$. Hence, when we construct $r_{y}\left(
r_{z}\left(  r_{x}\left(  T\right)  \right)  \right)  $, we first have $x$
insert itself at the end of the first row; then $z$ inserts itself at the end
of the first row; then $y$ bumps out $z$. When we construct $r_{y}\left(
r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)  $, we first have $z$
insert itself at the end of the first row; then $x$ bumps out $z$; then $y$
inserts itself at the end of the first row. Thus, the first rows of
$r_{y}\left(  r_{z}\left(  r_{x}\left(  T\right)  \right)  \right)  $ and
$r_{y}\left(  r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)  $ are
identical, and both times the same letter is being bumped into the next row.

\item \textit{Subcase 1.2:} We have $k+l>0$. In this case, when we construct
$r_{z}\left(  r_{x}\left(  T\right)  \right)  $, we first have $x$ bump out
$a_{j+1}$; then $z$ inserts itself at the end of the first row. When we
construct $r_{x}\left(  r_{z}\left(  T\right)  \right)  $, we first have $z$
insert itself at the end of the first row; then $x$ bumps out $a_{j+1}$. Thus,
the first rows of $r_{z}\left(  r_{x}\left(  T\right)  \right)  $ and
$r_{x}\left(  r_{z}\left(  T\right)  \right)  $ are identical, and both times
the same letter is being bumped into the next row. Obviously, this identity
does not change when we apply $r_{y}$ again; thus, the same holds for
$r_{y}\left(  r_{z}\left(  r_{x}\left(  T\right)  \right)  \right)  $ and
$r_{y}\left(  r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)  $.
\end{itemize}

\item \textit{Case 2:} We have $n>0$.

\begin{itemize}
\item \textit{Subcase 2.1:} We have $k+l=0$.

\begin{itemize}
\item \textit{Subsubcase 2.1.1:} We have $n=1$. In this case, $j+k+l=j$.
Hence, the first row of $T$ has the form $a_{1}a_{2}\cdots a_{j}a_{j+1}$, with
$x\leq y<z<a_{j+1}$. Hence, when we construct $r_{y}\left(  r_{z}\left(
r_{x}\left(  T\right)  \right)  \right)  $, we first have $x$ bump out
$a_{j+1}$; then $z$ inserts itself at the end of the first row; then $y$ bumps
out $z$. When we construct $r_{y}\left(  r_{x}\left(  r_{z}\left(  T\right)
\right)  \right)  $, we first have $z$ bump out $a_{j+1}$; then $x$ bumps out
$z$; then $y$ inserts itself at the end of the first row. Thus, the first rows
of $r_{y}\left(  r_{z}\left(  r_{x}\left(  T\right)  \right)  \right)  $ and
$r_{y}\left(  r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)  $ are
identical, and both times the same two letters are being bumped into the next
row (in the same order).

\item \textit{Subsubcase 2.1.2:} We have $n>1$. This is the second case you
consider on page 288.
\end{itemize}

\item \textit{Subcase 2.2:} We have $k+l>0$.

\begin{itemize}
\item \textit{Subsubcase 2.2.1:} We have $l=0$. In this case, $k+l=k$. Hence,
the first row of $T$ has the form $a_{1}a_{2}\cdots a_{j}a_{j+1}\cdots
a_{j+k}a_{j+k+1}\cdots a_{j+k+n}$, with $y<z<a_{j+k+1}\leq\cdots\leq
a_{j+k+n}$. Thus, when we construct $r_{y}\left(  r_{z}\left(  r_{x}\left(
T\right)  \right)  \right)  $, we first have $x$ bump out $a_{j+1}$ (which is
strictly left of $a_{j+k+1}$, because $k=k+l>0$); then $z$ bumps out
$a_{j+k+1}$; then $y$ bumps out $z$. When we construct $r_{y}\left(
r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)  $, we first have $z$
bump out $a_{j+k+1}$; then $x$ bumps out $a_{j+1}$ (which is strictly left of
$a_{j+k+1}$); then $y$ bumps out $z$. Thus, the first rows of $r_{y}\left(
r_{z}\left(  r_{x}\left(  T\right)  \right)  \right)  $ and $r_{y}\left(
r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)  $ are identical, and the
letters that are bumped out from these first row are (respectively)
$a_{j+1},a_{j+k+1},z$ (in this order) and $a_{j+k+1},a_{j+1},z$ (in this
order). Since we know that $a_{j+1}\leq x<z$, we can conclude from the
induction hypothesis that inserting these bumped-out letters into the
remaining rows of $T$ will again have the same result.

\item \textit{Subsubcase 2.2.2:} We have $k=0$. In this case, $k+l=l$. Hence,
the first row of $T$ has the form $a_{1}a_{2}\cdots a_{j}a_{j+1}\cdots
a_{j+l}a_{j+l+1}\cdots a_{j+l+n}$, with $x\leq y<a_{j+1}\leq\cdots\leq
a_{k+l}\leq z$. Thus, when we construct $r_{y}\left(  r_{z}\left(
r_{x}\left(  T\right)  \right)  \right)  $, we first have $x$ bump out
$a_{j+1}$ (which is strictly left of $a_{j+l+1}$, because $l=k+l>0$); then $z$
bumps out $a_{j+l+1}$; then $y$ bumps out $a_{j+2}^{\prime}$, where
$a_{j+2}^{\prime}=%
\begin{cases}
a_{j+2}, & \text{if }l\geq2;\\
z, & \text{if }l=1
\end{cases}
$ is the entry currently occupying position $j+2$ (here, we are using the
facts that $y<a_{j+1}\leq a_{j+2}$ and that $y<z$). When we construct
$r_{y}\left(  r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)  $, we
first have $z$ bump out $a_{j+l+1}$; then $x$ bumps out $a_{j+1}$ (which is
strictly left of $a_{j+l+1}$); then $y$ bumps out $a_{j+2}^{\prime}$, where
$a_{j+2}^{\prime}$ is the same as in the previous sentence. Thus, the first
rows of $r_{y}\left(  r_{z}\left(  r_{x}\left(  T\right)  \right)  \right)  $
and $r_{y}\left(  r_{x}\left(  r_{z}\left(  T\right)  \right)  \right)  $ are
identical, and the letters that are bumped out from these first row are
(respectively) $a_{j+1},a_{j+l+1},a_{j+2}^{\prime}$ (in this order) and
$a_{j+l+1},a_{j+1},a_{j+2}^{\prime}$ (in this order). Since we know that
$a_{j+1}\leq a_{j+2}^{\prime}<a_{j+l+1}$ (indeed, if $l\geq2$, then this
follows from $a_{j+1}\leq a_{j+2}\leq z<a_{j+l+1}$; but if $l=1$, then this
follows from $a_{j+1}\leq z<a_{j+l+1}$), we can conclude from the induction
hypothesis that inserting these bumped-out letters into the remaining rows of
$T$ will again have the same result.

\item \textit{Subsubcase 2.2.3:} We have $k>0$ and $l>0$. This is the
\textquotedblleft most generic\textquotedblright\ case, as all the relevant
intervals $\left[  j+1,j+k\right]  $, $\left[  j+k+1,j+k+l\right]  $ and
$\left[  j+k+l,j+k+n\right]  $ are nonempty. Thus, when we construct
$r_{y}\left(  r_{z}\left(  r_{x}\left(  T\right)  \right)  \right)  $, we
first have $x$ bump out $a_{j+1}$; then $z$ bumps out $a_{j+k+l+1}$; then $y$
bumps out $a_{j+k+1}$. When we construct $r_{y}\left(  r_{x}\left(
r_{z}\left(  T\right)  \right)  \right)  $, we first have $z$ bump out
$a_{j+k+l+1}$; then $x$ bumps out $a_{j+1}$; then $y$ bumps out $a_{j+k+1}$.
Thus, the first rows of $r_{y}\left(  r_{z}\left(  r_{x}\left(  T\right)
\right)  \right)  $ and $r_{y}\left(  r_{x}\left(  r_{z}\left(  T\right)
\right)  \right)  $ are identical, and the letters that are bumped out from
these first row are (respectively) $a_{j+1},a_{j+k+l+1},a_{j+k+1}$ (in this
order) and $a_{j+k+l+1},a_{j+1},a_{j+k+1}$ (in this order). Since we know that
$a_{j+1}\leq a_{j+k+1}<a_{j+k+l+1}$ (indeed, if $a_{j+k+1}<a_{j+k+l+1}$
follows from $a_{j+k+1}\leq z<a_{j+k+l+1}$), we can conclude from the
induction hypothesis that inserting these bumped-out letters into the
remaining rows of $T$ will again have the same result.
\end{itemize}
\end{itemize}
\end{itemize}

Note that I am \textbf{not} distinguishing between the cases $x=y$ and $x<y$,
as I don't see a reason to do so. I believe the same arguments apply to both
$x=y$ and $x<y$, as long as you make sure to interpret the chain of
$x<a_{j+1}\leq\cdots\leq a_{j+k}\leq y$ as vacuous when $k=0$.

\item \textbf{page 289, proof of Theorem 10.17:} \textquotedblleft to its
right\textquotedblright\ $\rightarrow$ \textquotedblleft to its
left\textquotedblright.

\item \textbf{page 289, proof of Theorem 10.17:} Before \textquotedblleft each
entry of $r_{l-1}$ bumps\textquotedblright, add \textquotedblleft when we
insert $r_{l-1}$ into $P\left(  r_{l}\right)  $ using RSK
insertion,\textquotedblright.

\item \textbf{page 290, proof of Corollary 10.22:} Since you haven't mentioned
it before but are using it now, it feels reasonable to mention the monoidal
property of Knuth equivalence here (i.e., the fact that if $\alpha_{1}\sim
_{K}\alpha_{2}$ and $\beta_{1}\sim_{K}\beta_{2}$, then $\alpha_{1}\beta
_{1}\sim_{K}\alpha_{2}\beta_{2}$).

\item \textbf{page 292, line 7:} \textquotedblleft tableaux\textquotedblright%
\ $\rightarrow$ \textquotedblleft tableau\textquotedblright.

\item \textbf{page 292:} Two lines above Definition 10.25, you write
\textquotedblleft into the new boxes of $\lambda/\mu$\textquotedblright. This
presupposes the following two facts:

\begin{itemize}
\item The boxes created by inserting $b_{1},\ldots,b_{\left\vert
\nu\right\vert }$ into $U$ are actually the boxes of $\lambda/\mu$.

\item Inserting $s_{1},\ldots,s_{\left\vert \nu\right\vert }$ into these boxes
(in the order in which they appear) yields a semistandard skew tableau.
\end{itemize}

These facts are not hard to check, but at least something should be said about
the proofs. (The first fact comes from observing that
\begin{align*}
&  r_{b_{\left\vert \nu\right\vert }}\left(  r_{b_{\left\vert \nu\right\vert
-1}}\left(  \cdots r_{b_{1}}\left(  U\right)  \right)  \right) \\
&  =\left(  \text{result of inserting the word }b_{1}\cdots b_{\left\vert
\nu\right\vert }\text{ into }U\right) \\
&  =\left(  \text{result of inserting the word }\operatorname*{word}\left(
V\right)  \text{ into }U\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by an analogue of Theorem 10.16,}\\
\text{since }b_{1}\cdots b_{\left\vert \nu\right\vert }\sim_{K}%
\operatorname*{word}\left(  \underbrace{P\left(  b_{1}\cdots b_{\left\vert
\nu\right\vert }\right)  }_{=P\left(  \sigma\right)  =V}\right)
=\operatorname*{word}\left(  V\right)
\end{array}
\right) \\
&  =U\APLstar_{r}V\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}U\APLstar_{r}V\right) \\
&  =U\APLstar V=T\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  U,V\right)
\in\mathcal{T}\left(  \mu,\nu,T\right)  \right)
\end{align*}
is a tableau of shape $\lambda$. Here, the \textquotedblleft analogue of
Theorem 10.16\textquotedblright\ means the fact that Knuth-equivalent words
not only produce the same $P$-tableau, but also lead to the same result when
inserted in an already existing tableau. While this fact follows easily from
Theorem 10.16 (just write the already existing tableau as the $P$-tableau of
some word), I find this fact sufficiently useful to state in its own right.
The second fact presupposed above is proved in the same way as Proposition 8.14.)

\item \textbf{page 294:} You write: \textquotedblleft Fill in the boxes of
$\mu$ in $S$ with entries which are smaller than all entries of $S$ to create
a semistandard tableau $US$ of shape $\lambda$\textquotedblright. I think this
is only possible if you allow negative integers as entries. Example 10.28
illustrates why: The only reason why you are able to avoid negative (or zero)
entries in Figure 10.17 is that you are not taking the \textquotedblleft
smaller than all entries of $S$\textquotedblright\ requirement seriously (the
$2$ you add in row $2$ is not smaller than the $2$ in $S$).

You should probably say that you are going to allow tableaux (and generalized
permutations) with negative entries when necessary, seeing that you use them
in the proof of Lemma 10.30 as well.

\item \textbf{pages 294--295:} I don't understand your argument for why $U$ is
independent of the choice of entries to fill $\mu$ with. (I think you are
using the letter $U$ for two different tableaux in that argument.) I also am
missing a proof for why $V$ is independent of this choice. Fortunately, I
don't believe these arguments are necessary: You can instead fix a choice of
entries to fill $\mu$ with, and define the map $\Omega$ using this choice;
then Theorem 10.33 entails that $\Omega$ is actually independent of this
choice because it is the inverse of the map $\Psi$ (and inverses are unique).

\item \textbf{page 295, Example 10.28:} As I said above, your $US$ in Figure
10.17 does not quite fit the \textquotedblleft entries smaller than all
entries of $S$\textquotedblright\ bill.

\item \textbf{page 296, proof of Lemma 10.29:} After \textquotedblleft let
$v_{j}^{\prime}$ be $v_{j}$ with $b$ removed\textquotedblright, I'd add
\textquotedblleft(where $b$ refers to the leftmost appearance of $b$ in each
word)\textquotedblright.

\item \textbf{page 297, proof of Lemma 10.30:} You are tacitly using that the
array $\tau$ is actually a generalized permutation. This follows from the fact
that $c_{j}<a_{k}$ for all $j$ and $k$ (which, in turn, follows from the fact
that $c_{1},\ldots,c_{m}$ are the entries of $Q\left(  \sigma\right)  =T^{-}$,
which are nonpositive).

\item \textbf{page 298, proof of Proposition 10.31:} I'd replace
\textquotedblleft$\operatorname*{rect}\left(  \Psi\left(  U,V\right)  \right)
=W$\textquotedblright\ by \textquotedblleft$\operatorname*{rect}\left(
\Psi\left(  U,V\right)  \right)  =Q\left(  \sigma\right)  =W$%
\textquotedblright\ just in order to make the reasoning clearer.

\item \textbf{page 298, proof of Proposition 10.32:} After \textquotedblleft
the entries in $\mu$ must be $u_{1},\ldots,u_{\left\vert \mu\right\vert }%
$\textquotedblright, add \textquotedblleft(since $\pi$ is a generalized
permutation, so that $u_{1}\leq\cdots\leq u_{\left\vert \mu\right\vert }\leq
t_{1}\leq\cdots\leq t_{\left\vert \lambda\right\vert -\left\vert
\mu\right\vert }$, and this entails that $u_{1},\ldots,u_{\left\vert
\mu\right\vert }$ are the smallest $\left\vert \mu\right\vert $ many entries
of $US$)\textquotedblright.

\item \textbf{page 298, proof of Proposition 10.32:} After \textquotedblleft
In particular, the shape of $U$ is $\mu$\textquotedblright, I'd add
\textquotedblleft(since $U=P\left(  a_{1},\ldots,a_{\left\vert \mu\right\vert
}\right)  $ is the intermediate result in the construction of $P\left(
\pi\right)  $ achieved after the first $\left\vert \mu\right\vert $ many
insertions, and the $Q$-tableau achieved at that point is precisely the part
of $Q\left(  \pi\right)  =US$ that contains the smallest $\left\vert
\mu\right\vert $ many entries of $US$; but clearly the $P$-tableau and the
$Q$-tableau always have the same shape)\textquotedblright.

\item \textbf{page 298, proof of Proposition 10.32:} Replace \textquotedblleft
By Lemma 10.30\textquotedblright\ by \textquotedblleft Note that the tableau
$P\left(  \pi\right)  $ can be obtained from $U$ by inserting $c_{1}%
,\ldots,c_{\left\vert \lambda\right\vert -\left\vert \mu\right\vert }$ in this
order; and the $S$ part of $US$ is obtained by filling the boxes of
$\lambda/\mu$ with $t_{1},\ldots,t_{\left\vert \lambda\right\vert -\left\vert
\mu\right\vert }$ in the order in which they are created during this insertion
process. Hence, by Lemma 10.30 (applied to $\tau$, $\left\vert \lambda
\right\vert -\left\vert \mu\right\vert $, $t_{i}$, $c_{i}$, $U$ and $P\left(
\pi\right)  $ instead of $\pi$, $n$, $a_{i}$, $b_{i}$, $T$ and $U$%
)\textquotedblright.

\item \textbf{page 298, proof of Theorem 10.33:} On the second line of this
proof, replace \textquotedblleft, and let $Q\left(  \sigma\right)
=W$\textquotedblright\ by \textquotedblleft and $Q\left(  \sigma\right)
=W$\textquotedblright. (This is part of the definition of $\sigma$, not of $W$.)

\item \textbf{page 299, proof of Theorem 10.33:} You are tacitly using the
fact that the array $\pi^{\prime}$ (which you define as the concatenation
$\sigma^{\prime}\sigma$) is actually a generalized permutation. This follows
from the fact that $v_{1},\ldots,v_{\left\vert \mu\right\vert }$ (which are
the entries of $U_{c}$, since $Q\left(  \sigma^{\prime}\right)  =U_{c}$) are
smaller than $s_{1},\ldots,s_{\left\vert \nu\right\vert }$ (which are the
entries of $S$).

\item \textbf{page 299, proof of Theorem 10.33:} After \textquotedblleft so we
must have $\pi=\pi^{\prime}$.\textquotedblright, I would add \textquotedblleft
Thus, $\left(  d_{1},d_{2},\ldots,d_{\left\vert \mu\right\vert }\right)
=\left(  a_{1},a_{2},\ldots,a_{\left\vert \mu\right\vert }\right)  $ and
$\left(  b_{1},b_{2},\ldots,b_{\left\vert \nu\right\vert }\right)  =\left(
c_{1},c_{2},\ldots,c_{\left\vert \lambda\right\vert -\left\vert \mu\right\vert
}\right)  $.\textquotedblright

\item \textbf{page 299, proof of Theorem 10.33:} Instead of saying
\textquotedblleft by Lemma 10.30 and our construction\textquotedblright, I'd
just say \textquotedblleft as we have seen in the proof of Proposition 10.32
above\textquotedblright. After all, your notations here are exactly the same
as in that proof, so there is no need to redo anything.

\item \textbf{page 299, proof of Theorem 10.33:} In the last displayed
equation on page 299, replace \textquotedblleft$W=\operatorname*{rect}\left(
S_{1}\right)  =Q\left(  \sigma\right)  $\textquotedblright\ by
\textquotedblleft$W=Q\left(  \sigma\right)  $\textquotedblright. Indeed, the
$\operatorname*{rect}\left(  S_{1}\right)  $ term is useless and misleading
(the equality $W=Q\left(  \sigma\right)  $ comes immediately from the
definition of $\sigma$).

\item \textbf{page 299, proof of Theorem 10.33:} After \textquotedblleft
Therefore, $\tau=\sigma$\textquotedblright, I would add \textquotedblleft, so
that $\pi=\left[
\begin{array}
[c]{cccccc}%
u_{1} & \cdots & u_{\left\vert \mu\right\vert } & s_{1} & \cdots &
s_{\left\vert \nu\right\vert }\\
a_{1} & \cdots & a_{\left\vert \mu\right\vert } & b_{1} & \cdots &
b_{\left\vert \nu\right\vert }%
\end{array}
\right]  $\textquotedblright. (This is closer to what you actually use in the
next paragraph.)

\item \textbf{page 301, proof of Theorem 10.39:} I'd replace \textquotedblleft%
$\operatorname*{word}\left(  S\right)  \sim_{K}n\cdots n\cdots1\cdots
1$\textquotedblright\ by \textquotedblleft$\operatorname*{word}\left(
S\right)  \sim_{K}\operatorname*{word}\left(  W\right)  =n\cdots
n\cdots1\cdots1$\textquotedblright\ (since you are talking about
$\operatorname*{word}\left(  W\right)  $ in the next paragraph).

\item \textbf{page 302, second paragraph:} \textquotedblleft is if $xyw_{2}$
is not a Littlewood--Richardson word\textquotedblright\ $\rightarrow$
\textquotedblleft is if the word $xyw_{2}$ contains more copies of $x$ than of
$x-1$ despite $x>1$\textquotedblright.

\item \textbf{page 302, third paragraph:} \textquotedblleft would be if
$zyw_{2}$ were not a Littlewood--Richardson word\textquotedblright%
\ $\rightarrow$ \textquotedblleft would be if the word $zyw_{2}$ contained
more copies of $z$ than of $z-1$\textquotedblright.

\item \textbf{page 302, third paragraph:} I'm skeptical about the claim that
\textquotedblleft$xyw_{2}$ has exactly the same number of $x$'s as
$x+1$'s\textquotedblright. Fortunately, a weaker claim suffices: Since $w_{2}$
is Littlewood--Richardson, and since $y=x$, the word $xyw_{2}$ must have at
least two more $x$'s than it has $x+1$'s. In view of $z=x+1$, this rewrites as
follows: The word $xyw_{2}$ must have at least two more $z-1$'s than it has
$z$'s. Therefore, the word $zyw_{2}$ (which has one more $z$ and one fewer
$x=z-1$ than $xyw_{2}$) must have at least as many $z-1$'s than it has $z$'s.
But this contradicts the fact that the word $zyw_{2}$ contains more copies of
$z$ than of $z-1$.

\item \textbf{page 304, Problem 10.9:} Remove the spurious \textquotedblleft%
$2$\textquotedblright\ at the end of this exercise.

\item \textbf{page 307, Problem 10.12:} The last sentence should not be part
of the last bullet point.

\item \textbf{page 307, Figure 10.20:} Are you sure the labelling of the
middle rhombus is correct? Here $b$ and $c$ are two adjacent vertices, and so
are $a$ and $d$; the inequality $b+c\geq a+d$ then doesn't seem to correspond
to the hive inequality in the works of Knutson and Tao. (But I'm no expert on
those works.)

\item \textbf{page 315, proof of Proposition A.14:} On line $-$3 of the page,
replace \textquotedblleft$F$\textquotedblright\ by \textquotedblleft$F^{n}%
$\textquotedblright.

\item \textbf{page 317, line }$-$\textbf{4:} \textquotedblleft we say
$u_{1},\ldots,u_{n}$\textquotedblright\ $\rightarrow$ \textquotedblleft we say
$\overrightarrow{u}_{1},\ldots,\overrightarrow{u}_{n}$\textquotedblright.

\item \textbf{page 318, Proposition A.18:} It is worth saying that the field
is supposed to be $\mathbb{Q}$ or $\mathbb{R}$ here (but not $\mathbb{C}$,
which would render axiom (4) meaningless and axiom (5) false).

\item \textbf{page 319, proof of Proposition A.18:} On the first line of the
displayed equation, replace \textquotedblleft$u_{k}$\textquotedblright\ by
\textquotedblleft$\overrightarrow{u}_{k}$\textquotedblright.

\item \textbf{page 319, proof of Proposition A.18:} On line 10 of the proof
(and of the page), replace \textquotedblleft and $w=$\textquotedblright\ by
\textquotedblleft and $\overrightarrow{w}=$\textquotedblright.

\item \textbf{page 320, line 2:} \textquotedblleft function\textquotedblright%
\ $\rightarrow$ \textquotedblleft bilinear map\textquotedblright\ (which
should be defined). If it was just an arbitrary function, then the values of
$\left\langle \overrightarrow{u}_{j},\overrightarrow{v}_{k}\right\rangle $
would not uniquely determine it.

\item \textbf{page 321, Problem A.12:} On line 2 of the problem, replace
\textquotedblleft$\overrightarrow{w},\ldots,\overrightarrow{w}_{n}%
$\textquotedblright\ by \textquotedblleft$\overrightarrow{w}_{1}%
,\ldots,\overrightarrow{w}_{n}$\textquotedblright.

\item \textbf{page 321, Problem A.13 (a):} \textquotedblleft%
$\overrightarrow{v}=\left\langle a,b\right\rangle $\textquotedblright%
\ $\rightarrow$ \textquotedblleft$\overrightarrow{v}=\left(  a,b\right)
$\textquotedblright.

\item \textbf{page 324 (ca.):} It is worth pointing out that any partition
$\lambda$ of $k$ has $l\left(  \lambda\right)  \leq k$ and thus can be written
as $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{k}\right)  $. This little
notational trick gets tacitly used in the text many times, but may be somewhat
confusing to readers who aren't used to it (they may wonder whether $\lambda$
actually has $k$ nonzero entries).

\item \textbf{page 333, proof of Proposition C.7:} Replace \textquotedblleft
of the form $1,j$\textquotedblright\ by \textquotedblleft of the form $\left(
1,j\right)  $\textquotedblright.

\item \textbf{page 335:} I'd add an exercise asking to prove that
$\operatorname*{inv}\left(  \pi\sigma\right)  \leq\operatorname*{inv}%
\pi+\operatorname*{inv}\sigma$ for any $\pi,\sigma\in S_{n}$.

\item \textbf{back cover:} \textquotedblleft the involution $\Omega
$\textquotedblright\ $\rightarrow$ \textquotedblleft the involution $\omega
$\textquotedblright.
\end{enumerate}


\end{document}