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\ihead{Errata to ``Notes on Schubert Polynomials''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Notes on Schubert Polynomials}

\textit{I. G. Macdonald}

Publications du LaCIM \textbf{6}, 1991

\href{https://lacim.uqam.ca/les-parutions/LACIM-Publications-Volume-06.pdf}{LaTeX-recompiled
version of 19 February 2025}

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the book \textquotedblleft Notes on
Schubert Polynomials\textquotedblright\ by the numbers under which they appear
in this book. All page numbers are from the
\href{https://lacim.uqam.ca/les-parutions/LACIM-Publications-Volume-06.pdf}{LaTeX-recompiled
version of 19 February 2025}, not from the original 1991 edition.

I have read pages 1--12, 23--45, 70--77 and 103--105 of the book so far.

\begin{noncompile}
... and also pages 49--50 from line 2 of page 49 on.
\end{noncompile}

The list below contains both actual corrections and what I believe to be
clarifications and pertinent comments. I have not tried to separate the former
from the latter, as I suspect that the precise boundary is in the eyes of the beholder.

\appendix \setcounter{section}{1}

\section{Errata}

\begin{enumerate}
\item \textbf{Page 1, line 1 of the main text:} \textquotedblleft of degree
n\textquotedblright\ $\rightarrow$ \textquotedblleft of degree $n$%
\textquotedblright. (That is, the \textquotedblleft n\textquotedblright%
\ should be in mathmode. This typo does not appear in the original printed version.)

\item \textbf{Page 1, second displayed equation from the bottom:} The
expression \textquotedblleft$\left(  w\left(  1\right)  ,\ldots,w\left(
r+1\right)  ,w\left(  r\right)  ,\ldots,w\left(  n\right)  \right)
$\textquotedblright\ could be misconstrued. Its correct interpretation is
$\left(  w\left(  1\right)  ,\ldots,w\left(  r-1\right)  ,w\left(  r+1\right)
,w\left(  r\right)  ,w\left(  r+2\right)  ,\ldots,w\left(  n\right)  \right)
$ (that is, the result of swapping the $r$-th and $\left(  r+1\right)  $-st
entries of the $n$-tuple $\left(  w\left(  1\right)  ,w\left(  2\right)
,\ldots,w\left(  n\right)  \right)  $).

\item \textbf{Page 2, line 1:} The spacing in \textquotedblleft$\ell\left(
w\right)  =\operatorname*{Card}I\left(  w\right)  $\textquotedblright\ is off.
(This typo does not appear in the original printed version.)

\item \textbf{Page 2, proof of (1.4):} A whitespace is missing in
\textquotedblleft$\ell\left(  w\right)  $that\textquotedblright. (This typo
does not appear in the original printed version. I suggest fixing it
throughout the book, possibly by a regex search for whatever caused it. My
guess is that some linebreaks or whitespaces got hurt as the file was moved
across operating systems.)

\item \textbf{Page 2, proof of (1.6) (iv):} It is not true that
\textquotedblleft$I\left(  ww_{0}\right)  $ is the complement of $I\left(
w\right)  $ in $I\left(  w_{0}\right)  $\textquotedblright. Instead, $I\left(
w_{0}w\right)  $ is the complement of $I\left(  w\right)  $ in $I\left(
w_{0}\right)  $. Thus, the roles of $w_{0}w$ and $ww_{0}$ in this proof should
be swapped (except in the equality $ww_{0}=\left(  w\left(  n\right)
,w\left(  n-1\right)  ,\ldots,w\left(  1\right)  \right)  $, which is correct
but not relevant here).

\item \textbf{Page 3, before (1.7):} After \textquotedblleft Such sequences
are called \textsl{reduced words} for $w$\textquotedblright, add
\textquotedblleft, and the corresponding expressions $w=s_{a_{1}}\ldots
s_{a_{p}}$ are called \textsl{reduced expressions} for $w$\textquotedblright.
This notation is used several times below.

\item \textbf{Page 4, (1.9):} A whitespace is missing in \textquotedblleft%
$\left(  p,q\right)  $such that\textquotedblright.

\item \textbf{Page 4, proof of (1.9):} \textquotedblleft it follows that
$v=s_{a_{1}}\cdots\widehat{s}_{a_{p}}\cdots s_{a_{q-1}}$\textquotedblright%
\ should be \textquotedblleft it follows that $v=s_{a_{1}}\cdots
\widehat{s}_{a_{p}}\cdots s_{a_{q-1}}s_{a_{q}}$\textquotedblright.

\item \textbf{Page 4, before (1.11):} \textquotedblleft if and only if
$e_{ij}=0$\textquotedblright\ should be \textquotedblleft if and only if
$e_{ij}\left(  w\right)  =0$\textquotedblright.

\item \textbf{Page 4, proof of (1.11):} \textquotedblleft we have $\left(
i,j\right)  =s_{a_{p}}\cdots s_{a_{r+1}}\left(  a_{r},a_{r+1}\right)
$\textquotedblright\ should be \textquotedblleft we have $\left(  i,j\right)
=s_{a_{p}}\cdots s_{a_{r+1}}\left(  a_{r},a_{r}+1\right)  $\textquotedblright.
(NB: Only the last $a_{r+1}$ should be replaced by $a_{r}+1$.)

\item \textbf{Pages 5--6, proof of (1.15):} \textquotedblleft it follows from
(1.13) that $s_{i}v=s_{i}s_{a_{1}}\cdots\widehat{s}_{a_{r}}\cdots s_{a_{p}}%
$\textquotedblright\ should be \textquotedblleft it follows from (1.13) that
there is a reduced expression $s_{i}v=s_{i}s_{a_{1}}\cdots\widehat{s}_{a_{r}%
}\cdots s_{a_{p}}$\textquotedblright. The reducedness of this expression is
important, since it is (tacitly!) used later in showing that $v=s_{a_{1}%
}\cdots\widehat{s}_{a_{r}}\cdots s_{a_{p}}$ is a reduced expression.

\item \textbf{Page 6, proof of (1.16):} \textquotedblleft should have
$s_{i}v_{j}=v_{j}-1$\textquotedblright\ should be \textquotedblleft should
have $s_{i}v_{j}=v_{j-1}$\textquotedblright.

\item \textbf{Page 6, proof of (1.16):} In equality (1), replace the
\textquotedblleft$\neq$\textquotedblright\ sign by a \textquotedblleft%
$\rightarrow$\textquotedblright\ sign.

\item \textbf{Page 6, proof of (1.16):} \textquotedblleft This shows that (i)
and (iii) are equivalent\textquotedblright\ should be \textquotedblleft This
shows that (i) and (ii) are equivalent\textquotedblright.

\item \textbf{Page 7, proof of (1.16):} \textquotedblleft by $\left(
a\right)  \Longleftrightarrow\left(  b\right)  $\textquotedblright\ should be
\textquotedblleft by (i)$\Longleftrightarrow$(ii)\textquotedblright.

\item \textbf{Page 7, (1.17):} In condition (iii), replace \textquotedblleft
exits\textquotedblright\ by \textquotedblleft exists\textquotedblright.

\item \textbf{Page 7, proof of (1.17):} Add a whitespace after
\textquotedblleft$\left(  b\right)  $\textquotedblright.

\item \textbf{Page 7, proof of (1.17):} In Case $\left(  b\right)  $, replace
\textquotedblleft$p-1+q$\textquotedblright\ by \textquotedblleft%
$q-1+p$\textquotedblright. (Of course, the original \textquotedblleft%
$p-1+q$\textquotedblright\ is equally valid, but less clear in its provenance.)

\item \textbf{Page 7, (1.19):} \textquotedblleft then\textquotedblright%
\ should be \textquotedblleft Then\textquotedblright.

\item \textbf{Page 8, proof of (1.19):} Why is \textquotedblleft$v^{\prime
}\leq w$ by the induction hypothesis\textquotedblright? This would require
showing that $K\left(  v^{\prime}\right)  \leq K\left(  w\right)  $, but this
is not done anywhere in this proof.

What looks like a complete proof of (1.19) (I haven't checked) can be found in
Proposition 3.2 of

\begin{statement}
Yufei Zhao, \textit{On the Bruhat order of the symmetric group and its
shellability}, \url{https://web.mit.edu/yufeiz/www/papers/bruhat.pdf} .
\end{statement}

An even stronger result is proved in Theorem 2.6.3 of

\begin{statement}
Anders Bj\"{o}rner, Francesco Brenti, \textit{Combinatorics of Coxeter
groups}, Springer 2005.
\end{statement}

(To be specific, we only need the direction (ii)$\Longrightarrow$(i) of this
stronger result, and only in the weaker form in which \textquotedblleft for
all $k\in D_{R}\left(  x\right)  $\textquotedblright\ is replaced by
\textquotedblleft for all $k\in\left\{  1,2,\ldots,n\right\}  $%
\textquotedblright.)

\item \textbf{Page 8, after (1.20):} \textquotedblleft either $i\geq w^{-1}$
or\textquotedblright\ should be \textquotedblleft either $i\geq w^{-1}j$
or\textquotedblright.

\item \textbf{Page 9, (1.21) (i):} Add a whitespace in \textquotedblleft%
$D\left(  w^{-1}\right)  $is\textquotedblright.

\item \textbf{Page 9, (1.22):} Remove the extraneous closing parentheses in
\textquotedblleft$\ell\left(  ws_{r}\right)  >\ell\left(  w\right)
)$\textquotedblright\ and \textquotedblleft$\ell\left(  s_{r}w\right)
>\ell\left(  w\right)  )$\textquotedblright.

\item \textbf{Page 10, (1.23) (ii):} Add a comma before \textquotedblleft%
$a_{p}$\textquotedblright\ in \textquotedblleft$\left(  a_{1},\ldots
a_{p}\right)  $\textquotedblright.

\item \textbf{Page 10, proof of (1.23):} Replace \textquotedblleft from
(1.21)\textquotedblright\ by \textquotedblleft from (1.22)\textquotedblright.

\item \textbf{Page 11:} The notation $\lambda^{\prime}$ stands for the
conjugate partition of $\lambda$. This should probably be said somewhere.

\item \textbf{Page 11, (1.27):} You are using the notation $D\left(
\lambda\right)  $ for the Young diagram $\left\{  \left(  i,j\right)
\in\left\{  1,2,3,\ldots\right\}  ^{2}\ \mid\ j\leq\lambda_{i}\right\}  $ of a
partition $\lambda$. This should also be defined.

\item \textbf{Page 11, proof of (1.27):} In the condition \textquotedblleft
there exist $a,\beta,c,\delta\in\left[  1,n\right]  $ such that $a<c,\ \beta
<\delta$ and $\left(  a,\beta\right)  ,\left(  c,\delta\right)  $ belong to
$D\left(  w\right)  $, whilst $\left(  a,\delta\right)  ,\left(
c,\beta\right)  $ do not\textquotedblright, the inequalities \textquotedblleft%
$a<c,\ \beta<\delta$\textquotedblright\ cannot necessarily be guaranteed to
hold at the same time. Instead, you can either have $a\neq c$ and $\beta
\neq\delta$ (in this form, the criterion is symmetric in rows vs. columns,
which yields the equivalence (i)$\Longleftrightarrow$(ii)), or you can ensure
that $a\neq c$ and $\beta<\delta$ (this is used in the proof of
(ii)$\Longleftrightarrow$(iii)).

\item \textbf{Page 11, proof of (1.27):} The proof of (v)$\Longrightarrow$(iv)
here tacitly relies on the following lemma (which ensures that the
\textquotedblleft suitable permutations\textquotedblright\ turn $D\left(
w\right)  $ into the actual diagram $D\left(  \lambda\right)  $ rather than
merely into a diagram that has the same numbers of cells in each row and in
each column):

\begin{statement}
\textbf{Lemma.} Let $\lambda$ be a partition. Let $U$ be a subset of $\left\{
1,2,3,\ldots\right\}  ^{2}$ such that the number of elements in the $i$-th row
of $U$ is $\lambda_{i}$ for each $i$, and such that the number of elements in
the $j$-th column of $U$ is $\lambda_{j}^{\prime}$ for each $j$. Then,
$U=D\left(  \lambda\right)  $.
\end{statement}

This lemma is fairly well-known when restated in the language of symmetric
functions as \textquotedblleft the elementary symmetric function
$e_{\lambda^{\prime}}$ has leading monomial $x^{\lambda}$ and $x^{\lambda}%
$-coefficient $1$\textquotedblright. In this form, it is very easy to prove
(since the leading terms of polynomials multiply when we multiply the
polynomials), but it is less obvious when stated in the above combinatorial form.

\item \textbf{Page 11:} \textquotedblleft it it satisfies\textquotedblright%
\ should be \textquotedblleft if it satisfies\textquotedblright.

\item \textbf{Page 12, (1.28) (iii):} This claim is false in general (and its
proof appears to suffer from a confusion between $I\left(  w\right)  $ and
$D\left(  w\right)  $). A counterexample is $w=\left(  2,1,4,3\right)  \in
S_{4}$, in which case $\overline{w}=w$ but $\lambda\left(  w\right)  $ is not self-conjugate.

However, what is true is that
\[
\lambda\left(  \overline{w}\right)  =\lambda\left(  w^{-1}\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }w\in S_{n}.
\]
This is because each $i\in\left\{  1,2,\ldots,n\right\}  $ satisfies
$c_{n+1-i}\left(  \overline{w}\right)  =c_{w\left(  i\right)  }\left(
w^{-1}\right)  $, as can be easily verified by comparing the definitions of
both sides. This (combined with (1.27)(v)) shows that the original claim
$\lambda\left(  \overline{w}\right)  =\lambda\left(  w\right)  ^{\prime}$ is
true if and only if $w$ is vexillary.

\item \textbf{Page 12, proof of (1.30):} I don't understand why
\textquotedblleft Then we have $w\left(  k\right)  \leq\lambda_{i-1}$ for
$1\leq k\leq i-1$, and $w\left(  k\right)  =\lambda_{i-1}$ for some $k\leq
i-1$\textquotedblright.

However, I see a different way to prove the implication (i)$\Longrightarrow
$(iii). Namely, let us add two more equivalent conditions to (1.30):

\begin{enumerate}
\item[(iv)] there exist no $a\geq1$ and $c>a+1$ such that $w\left(  a\right)
<w\left(  c\right)  <w\left(  a+1\right)  $;

\item[(v)] there exist no $c>b>a\geq1$ such that $w\left(  a\right)  <w\left(
c\right)  <w\left(  b\right)  $ (in modern parlance, this is called
\textquotedblleft$w$ is $132$-avoiding\textquotedblright).
\end{enumerate}

The implication (i)$\Longrightarrow$(iv) follows easily from a closer look at
(1.24) (indeed, for a given $a\geq1$, the existence of a $c>a+1$ satisfying
$w\left(  a\right)  <w\left(  c\right)  <w\left(  a+1\right)  $ is equivalent
to $c_{a}\left(  w\right)  <c_{a+1}\left(  w\right)  $). The implication
(iv)$\Longrightarrow$(v) is fairly easy (fix a $c>1$ such that some $a<b<c$
satisfy $w\left(  a\right)  <w\left(  c\right)  <w\left(  b\right)  $; then
pick such $a,b$ with minimum $b-a$, and argue that $b$ must be $a+1$). The
implication (v)$\Longrightarrow$(i) follows easily from the definition of
$D\left(  w\right)  $. Combining the three implications, (i)$\Longrightarrow
$(iii) follows.

\item \textbf{Page 13, (1.31):} After \textquotedblleft Let $w\in S_{\infty}%
$\textquotedblright, add \textquotedblleft and $r\geq0$\textquotedblright.

\item \textbf{Page 13, proof of (1.31):} \textquotedblleft By
(1.15)\textquotedblright\ should be \textquotedblleft By
(1.24)\textquotedblright.

\item \textbf{Page 13:} \textquotedblleft sequence c\textquotedblright\ should
be \textquotedblleft sequence $c$\textquotedblright\ (mathmode).

\item \textbf{Page 16, (1.38):} Insert whitespace in \textquotedblleft$\left[
1,f_{1}\right]  $;then\textquotedblright.

\item \textbf{Page 18, large picture, last diagram on the second line:} This
has a circle too much. Namely, the rightmost circle on the second line from
the bottom should not be a circle but rather start a fat hook.

\item \textbf{Page 18, large picture, middle diagram on the bottom line:} The
code here should be \textquotedblleft$30221$\textquotedblright, not
\textquotedblleft$302201$\textquotedblright. (The \textquotedblleft%
$302201$\textquotedblright\ on the bottom right, however, is correct.)

\item \textbf{Page 19:} Insert whitespace in \textquotedblleft$\lambda
^{\prime}$,by (1.27)\textquotedblright.

\item \textbf{Page 20, footnote:} Indeed, $N_{8}=24553$. For more terms of the
sequence, or rather for the numbers $V_{n}=n!-N_{n}$ of vexillary
permutations, see the OEIS
(\href{https://oeis.org/A005802}{https://oeis.org/A005802}).

\item \textbf{Page 22, Added in proof:} A proof of (1) can now be found in:
Julian West, \textit{Generating trees and the Catalan and Schr\"{o}der
numbers}, Discrete Mathematics \textbf{146} (1995), pp. 247--262. (More
precisely, combine his Corollary 3.5 with the well-known enumeration of
$1234$-avoiding permutations in terms of standard tableaux. See Exercise 7.22
(e) of Richard Stanley, \textit{Enumerative Combinatorics, volume 2}, 2nd
edition 2024.)

\item \textbf{Page 23:} \textquotedblleft If $f$ is a
function\textquotedblright\ should probably be \textquotedblleft If $f$ is a
rational function\textquotedblright\ (you don't want actual functions here,
lest it have a hole at $x=y$).

\item \textbf{Page 23 onward:} The symbols \textquotedblleft$s_{xy}%
$\textquotedblright\ and \textquotedblleft$t_{xy}$\textquotedblright\ mean the
same thing here. The only difference is that $t_{xy}$ denotes the
transposition in $S_{n}$ (or $S_{\infty}$) while $s_{xy}$ denotes its action
on rational functions; but this distinction is not made for other permutations
(such as $s_{i}$).

\item \textbf{Page 23:} In the displayed equation \textquotedblleft%
$\partial_{xy}f=\left(  x-y\right)  ^{-1}\left(  1-s_{xy}\right)
$\textquotedblright, remove the \textquotedblleft$f$\textquotedblright.

\item \textbf{Page 26, proof of (2.10):} \textquotedblleft in $\partial_{0}$
is\textquotedblright\ should be \textquotedblleft in $\partial_{w_{0}}$
is\textquotedblright.

\item \textbf{Page 26, proof of (2.10):} The claim that \textquotedblleft%
$c_{w_{0}}=\epsilon\left(  w_{0}\right)  a_{\delta}^{-1}$\textquotedblright%
\ behooves a more detailed justification. It is a particular case (for
$w=w_{0}$) of the following lemma:

\begin{statement}
\textbf{Lemma (also mentioned on pages 28--29):} For each $w\in S_{n}$, we can
write $\partial_{w}$ as%
\[
\partial_{w}=\epsilon\left(  w\right)  \prod_{\left(  i,j\right)  \in I\left(
w^{-1}\right)  }\left(  x_{i}-x_{j}\right)  ^{-1}\cdot w+\sum
_{\substack{v<w\\\text{(in Bruhat order)}}}d_{v}v
\]
for some rational functions $d_{v}$.
\end{statement}

This lemma can be proved by induction on $\ell\left(  w\right)  $. (The
induction step proceeds from $ws_{r}$ to $w$, using (1.17) to transform the
sum as well as a fact that is similar to (1.2): If $w\in S_{n}$ and $1\leq
r\leq n-1$, then%
\[
I\left(  s_{r}w\right)  =%
\begin{cases}
I\left(  w\right)  \cup\left\{  w^{-1}\left(  r\right)  ,w^{-1}\left(
r+1\right)  \right\}  , & \text{if }w^{-1}\left(  r\right)  <w^{-1}\left(
r+1\right)  ;\\
I\left(  w\right)  -\left\{  w^{-1}\left(  r+1\right)  ,w^{-1}\left(
r\right)  \right\}  , & \text{if }w^{-1}\left(  r\right)  >w^{-1}\left(
r+1\right)  .
\end{cases}
\]
Note that this proof does not require the specific sequence (2.9); it just
needs the fact that $I\left(  w_{0}\right)  $ is the set of all pairs $\left(
i,j\right)  $ with $1\leq i<j\leq n$.)

\item \textbf{Page 27, between (2.11) and (2.12):} You write:
\textquotedblleft Thus $\partial_{w_{0}}$ is a $\Lambda_{n}$-linear mapping of
$P_{n}$ onto $\Lambda_{n}$\textquotedblright.\ Strictly speaking, the
$\Lambda_{n}$-linearity of $\partial_{w_{0}}$ follows not from (2.11) but from (2.10).

\item \textbf{Page 27, (2.13):} It is worth saying that the $\alpha_{1}%
,\alpha_{2},\ldots$ are supposed to be scalars.

\item \textbf{Page 27, proof of (2.13):} In the first display, I would replace
\textquotedblleft$s_{a_{1}}\cdots\partial_{a_{r}}\cdots s_{a_{p}}%
$\textquotedblright\ by \textquotedblleft$s_{a_{1}}\cdots s_{a_{r-1}}%
\partial_{a_{r}}s_{a_{r+1}}\cdots s_{a_{p}}$\textquotedblright\ for clarity.

\item \textbf{Page 27, proof of (2.13):} Replace \textquotedblleft where
$wt=s_{a_{p}}\cdots\widehat{s}_{a_{r}}\cdots s_{a_{p}}$\textquotedblright\ by
\textquotedblleft where $wt=s_{a_{1}}\cdots\widehat{s}_{a_{r}}\cdots s_{a_{p}%
}$\textquotedblright.

\item \textbf{Page 27, proof of (2.13):} Replace \textquotedblleft$\left(
a_{r},a_{r+1}\right)  $\textquotedblright\ by \textquotedblleft$\left(
a_{r},a_{r}+1\right)  $\textquotedblright.

\item \textbf{Page 27, proof of (2.13):} The last step (rewriting the
$\sum_{r=1}^{p}$ sum as a sum over pairs $\left(  i<j\right)  $) relies on (1.7).

\item \textbf{Page 28, proof of (2.16) and (2.16'):} I think \textquotedblleft
From this and (2.10)\textquotedblright\ should be \textquotedblleft From this
and (2.9) and (2.2')\textquotedblright. More precisely, the preceding equality
(applied many times) shows that%
\begin{align*}
&  \left(  \pi_{1}\cdots\pi_{n-1}\right)  \left(  \pi_{1}\cdots\pi
_{n-2}\right)  \cdots\left(  \pi_{1}\pi_{2}\pi_{3}\right)  \left(  \pi_{1}%
\pi_{2}\right)  \pi_{1}f\\
&  =\left(  \partial_{1}\cdots\partial_{n-1}\right)  \left(  \left(
x_{1}\cdots x_{n-1}\right)  \left(  \partial_{1}\cdots\partial_{n-2}\right)
\left(  \left(  x_{1}\cdots x_{n-2}\right)  \cdots\right.  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \partial_{1}\partial
_{2}\partial_{3}\right)  \left(  \left(  x_{1}x_{2}x_{3}\right)  \left(
\partial_{1}\partial_{2}\right)  \left(  \left(  x_{1}x_{2}\right)
\partial_{1}\left(  x_{1}f\right)  \right)  \right)  ))\\
&  =\underbrace{\left(  \partial_{1}\cdots\partial_{n-1}\right)  \left(
\partial_{1}\cdots\partial_{n-2}\right)  \cdots\left(  \partial_{1}%
\partial_{2}\partial_{3}\right)  \left(  \partial_{1}\partial_{2}\right)
\partial_{1}}_{\substack{=\partial_{w_{0}}\\\text{(by (2.9))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \underbrace{\left(  x_{1}\cdots x_{n-1}\right)
\left(  x_{1}\cdots x_{n-2}\right)  \cdots\left(  x_{1}x_{2}x_{3}\right)
\left(  x_{1}x_{2}\right)  x_{1}}_{=x^{\delta}}f\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we used (2.2') to commute the }x\text{-products past}\\
\text{the }\partial\text{-products, since each }x\text{-product }x_{1}\cdots
x_{k}\\
\text{is symmetric in }x_{1},x_{2},\ldots,x_{k}\text{ and thus commutes
with}\\
\text{any product of }\partial_{i}\text{'s for }i<k\text{ (by (2.2'))}%
\end{array}
\right) \\
&  =\partial_{w_{0}}\left(  x^{\delta}f\right)  .
\end{align*}


\item \textbf{Page 29:} When you say \textquotedblleft It follows that the
$\partial_{w}$ are linearly independent over the field of rational functions
$\mathbf{Q}_{\infty}$\textquotedblright, you are using the fact that the
(actions of the) permutations $w\in S_{\infty}$ are linearly independent over
this field. This fact is a consequence of Dedekind's linear independence of
characters (Theorem 12 in: Emil Artin, \textit{Galois Theory}, Notre Dame
Mathematical Lectures \textbf{2}, second edition, 1971; the theorem must be
applied to $G$ being the multiplicative group of $\mathbf{Q}_{\infty}$ and $F$
being the field $\mathbf{Q}_{\infty}$).

\item \textbf{Page 29:} Add a period at the end of the first formula for
$\partial_{w}\circ\mu$ (the one before \textquotedblleft On
expansion\textquotedblright).

\item \textbf{Page 30, Example 3:} There is a lone period that somehow ended
up on its own line in the align environment.

\item \textbf{Page 30, (2.18):} Replace \textquotedblleft$\partial_{w/u}%
g$\textquotedblright\ by \textquotedblleft$\partial_{v/u}g$\textquotedblright%
\ on the right hand side.

\item \textbf{Page 31, proof of (2.18):} Replace \textquotedblleft%
$v\partial_{w/u}\left(  f\right)  $\textquotedblright\ by \textquotedblleft%
$v\partial_{w/v}\left(  f\right)  $\textquotedblright\ (twice).

\item \textbf{Page 32:} There is no need to \textquotedblleft work in an
arbitrary $\lambda$-ring $R$\textquotedblright. Everything in Chapter III
makes sense in the following more general setting: Let $R$ be a commutative
ring, and let $V$ be a subgroup of the additive group $\left(  R,+,0\right)
$. For each $r\in\mathbb{Z}$, let $e_{r}:V\rightarrow R$ be a map such that
all $X\in V$ satisfy $e_{0}\left(  X\right)  =1$ and $e_{1}\left(  X\right)
=X$. Assume that $e_{r}\left(  X\right)  =0$ for all $r<0$ and $X\in V$.
Assume furthermore that%
\[
e_{r}\left(  X+Y\right)  =\sum_{p+q=r}e_{p}\left(  X\right)  e_{q}\left(
Y\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }X,Y\in V\text{ and }r\geq0.
\]
I shall refer to this setting as the \textquotedblleft
VR-setting\textquotedblright.

The VR-setting is all that is needed in order to make the arguments of Chapter
III work. Having $V=R$ (as in the case of a $\lambda$-ring), or having
additional axioms for $e_{r}\left(  XY\right)  $ and $e_{r}\left(
e_{s}\left(  X\right)  \right)  $ (as in the case of a $\lambda$-ring, at
least according to the definition in Knutson's 1973 LNM book or in Berthelot's
SGA 6 expos\'{e}) is entirely unnecessary here.

Combinatorialists often use the instance of the VR-setting when $R$ is the
ring of formal power series $\mathbf{Z}\left[  \left[  x_{1},x_{2}%
,x_{3},\ldots\right]  \right]  $, and $V$ is the $R$ (as an additive group);
the maps $e_{r}:V\rightarrow R$ are given by the following rule: If $f\in V$
is any formal power series, then we can write $f$ as $f=\left(  \mathfrak{m}%
_{1}+\mathfrak{m}_{2}+\mathfrak{m}_{3}+\cdots\right)  -\left(  \mathfrak{n}%
_{1}+\mathfrak{n}_{2}+\mathfrak{n}_{3}+\cdots\right)  $ for two countable
(possibly finite) sequences $\left(  \mathfrak{m}_{1},\mathfrak{m}%
_{2},\mathfrak{m}_{3},\ldots\right)  $ and $\left(  \mathfrak{n}%
_{1},\mathfrak{n}_{2},\mathfrak{n}_{3},\ldots\right)  $ of monomials in the
$x_{1},x_{2},x_{3},\ldots$, and then we set%
\[
e_{r}\left(  f\right)  =\sum_{p+q=r}e_{p}\left(  \mathfrak{m}_{1}%
,\mathfrak{m}_{2},\mathfrak{m}_{3},\ldots\right)  \cdot\left(  -1\right)
^{q}h_{q}\left(  \mathfrak{n}_{1},\mathfrak{n}_{2},\mathfrak{n}_{3}%
,\ldots\right)  ,
\]
where $e_{p}\left(  \mathfrak{m}_{1},\mathfrak{m}_{2},\mathfrak{m}_{3}%
,\ldots\right)  =\sum_{i_{1}<i_{2}<\cdots<i_{p}}\mathfrak{m}_{i_{1}%
}\mathfrak{m}_{i_{2}}\cdots\mathfrak{m}_{i_{p}}$ denotes the $p$-th elementary
symmetric function in the $\mathfrak{m}_{1},\mathfrak{m}_{2},\mathfrak{m}%
_{3},\ldots$, whereas $h_{q}\left(  \mathfrak{n}_{1},\mathfrak{n}%
_{2},\mathfrak{n}_{3},\ldots\right)  =\sum_{i_{1}\leq i_{2}\leq\cdots\leq
i_{q}}\mathfrak{n}_{i_{1}}\mathfrak{n}_{i_{2}}\cdots\mathfrak{n}_{i_{q}}$
denotes the $q$-th complete homogeneous symmetric function in the
$\mathfrak{n}_{1},\mathfrak{n}_{2},\mathfrak{n}_{3},\ldots$.

Another instance of the VR-setting is obtained when both $V$ and $R$ are the
polynomial ring $\mathbf{Z}\left[  x_{1},x_{2},x_{3},\ldots\right]  $, and the
maps $e_{r}$ are defined as above. The same formula for $e_{r}\left(
f\right)  $ holds here, but now the sequences $\left(  \mathfrak{m}%
_{1},\mathfrak{m}_{2},\mathfrak{m}_{3},\ldots\right)  $ and $\left(
\mathfrak{n}_{1},\mathfrak{n}_{2},\mathfrak{n}_{3},\ldots\right)  $ are always
finite. This instance is used in (3.9).

An even smaller instance of the VR-setting is when $V$ and $R$ are the ring of
symmetric functions in $x_{1},x_{2},x_{3},\ldots$; the maps $e_{r}$ are still
defined in the same way as above (but now the monomials no longer belong to
$V$).

Returning to the general case of an arbitrary VR-setting (with $V$ a group and
$R$ a ring and $e_{r}:V\rightarrow R$ being maps satisfying certain axioms),
we define maps $h_{r}:V\rightarrow R$ for all $r\in\mathbb{Z}$ by setting%
\[
h_{r}\left(  X\right)  :=\left(  -1\right)  ^{r}e_{r}\left(  -X\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }X\in V.
\]
Then, it is very easy to see that%
\[
h_{r}\left(  X+Y\right)  =\sum_{p+q=r}h_{p}\left(  X\right)  h_{q}\left(
Y\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }X,Y\in V\text{ and }r\geq0.
\]
It is also easy to see (by induction) that if $X\in V$ is any element
satisfying $e_{r}\left(  X\right)  =0$ for all $r>1$, then $h_{r}\left(
X\right)  =X^{r}$ for all $r\geq0$. This fact is used on page 35, in the proof
of (3.5).

\item \textbf{Page 33, (3.3):} Add a comma before \textquotedblleft$X_{n}%
$\textquotedblright\ in \textquotedblleft$s_{\lambda/\mu}\left(  X_{1},\ldots
X_{n}\right)  $\textquotedblright.

\item \textbf{Page 34, proof of (3.4):} Replace \textquotedblleft%
$h_{i-j}\left(  -Y_{j}\right)  $\textquotedblright\ by \textquotedblleft%
$h_{j-i}\left(  -Y_{j}\right)  $\textquotedblright.

\item \textbf{Page 34, (3.5):} It is worth saying that \textquotedblleft%
$\left(  i,j\right)  \in\lambda$\textquotedblright\ is shorthand for
\textquotedblleft$\left(  i,j\right)  \in D\left(  \lambda\right)
$\textquotedblright.

\item \textbf{Page 35, proof of (3.5):} In \textquotedblleft$\sum_{r\geq
0}\left(  -1\right)  ^{r}e_{r}\left(  Y_{\alpha_{i}}\right)  x_{i}^{\alpha
_{i}-r}$\textquotedblright, the sum should range from $r=0$ to $\alpha_{i}$
(not to $\infty$).

\item \textbf{Page 35, between (3.5') and (3.5\textquotedblright):}
\textquotedblleft(and $\alpha$ is a partition $\lambda$)\textquotedblright%
\ should better be \textquotedblleft(and $\alpha$ is a partition $\lambda$ of
length $\leq n$)\textquotedblright.

\item \textbf{Page 35, (3.5\textquotedblright):} This equality requires
$\ell\left(  \lambda^{\prime}\right)  \leq n$.

\item \textbf{Page 35, (3.6):} I would replace \textquotedblleft Suppose that
$i<j$\textquotedblright\ by \textquotedblleft Suppose that $i\leq
j$\textquotedblright. This does not add any real generality, but makes the
theorem easier to apply and also simplifies the induction proof.

\item \textbf{Page 35, proof of (3.6):} Replace \textquotedblleft$rk\left(
Y\right)  \leq j-1$\textquotedblright\ by \textquotedblleft$rk\left(
Y\right)  \leq j-i$\textquotedblright.

\item \textbf{Page 36, proof of (3.8):} In the second displayed equality,
\textquotedblleft$\left(  1\leq j\leq n\right)  $\textquotedblright\ should be
\textquotedblleft$\left(  1\leq j\leq m\right)  $\textquotedblright.

\item \textbf{Page 36, proof of (3.8):} \textquotedblleft the interval
$\left[  -m,n-1\right]  $\textquotedblright\ should be \textquotedblleft the
interval $\left[  -n,m-1\right]  $\textquotedblright.

\item \textbf{Page 37, proof of (3.8):} In the first display, replace
\textquotedblleft$-X_{\xi_{i}}$\textquotedblright\ by \textquotedblleft%
$-X_{\xi_{1}}$\textquotedblright.

\item \textbf{Page 37, proof of (3.8):} Replace \textquotedblleft$\left(
h_{j-i}\left(  X_{i+1}\right)  \right)  _{-m\leq i,j\leq n-1}$%
\textquotedblright\ by \textquotedblleft$\left(  h_{j-i}\left(  X_{i+1}%
\right)  \right)  _{-n\leq i,j\leq m-1}$\textquotedblright.

\item \textbf{Page 37, proof of (3.8):} In the last display, replace
\textquotedblleft$\left(  -X_{\xi_{1}^{\prime}},\ldots,-X_{\xi_{m}^{\prime}%
}\right)  $\textquotedblright\ by \textquotedblleft$\left(  X_{-\xi
_{1}^{\prime}},\ldots,X_{-\xi_{m}^{\prime}}\right)  $\textquotedblright.

\item \textbf{Page 37, Remark:} \textquotedblleft Hence (3.8) gives a duality
theorem for the multi-Schur function $s_{\lambda/\mu}\left(  Y_{1}%
,\ldots,Y_{n}\right)  $ provided that $rk\left(  Y_{i+1}-Y_{i}\right)
\leq\lambda_{i}-\lambda_{i+1}+1$ for $1\leq i\leq n-1$\textquotedblright\ is
perhaps an overstatement. It relies on being able to write each $Y_{i+1}%
-Y_{i}$ as a sum of $\lambda_{i}-\lambda_{i+1}+1$ elements $X_{m}-X_{m-1}$ of
rank $\leq1$ each; it is not clear to me how this can be ensured. There might
be a splitting principle that allows for this, but even then, if the right
hand side of (3.8) cannot be reexpressed in terms of the original $Y_{i}$'s,
then the usefulness of such a theorem is questionable.

\item \textbf{Page 37:} \textquotedblleft becuase\textquotedblright\ should be
\textquotedblleft because\textquotedblright.

\item \textbf{Pages 37--38, paragraph above (3.8'):} This becomes clearer when
illustrated by a picture (showing the case $k=4$, which however is pretty
representative of the general case):%
\[%
%TCIMACRO{\TeXButton{tikz young diagram}{\begin{tikzpicture}%
%[scale=0.6, every node/.style={font=\small}]
%\draw[fill=green] (13,12) rectangle (12,13);
%\node(a) at (12.5,12.5) {$s_1$};
%\draw[fill=green] (9,8) rectangle (8,9);
%\node(a) at (8.5,8.5) {$s_2$};
%\draw[fill=green] (6,3) rectangle (5,4);
%\node(a) at (5.5,3.5) {$s_3$};
%\draw[fill=green] (3,0) rectangle (2,1);
%\node(a) at (2.5,0.5) {$s_4$};
%\draw[line width=1.4pt]
%(0,15) -- (13,15) -- (13,12) -- (9,12) -- (9,8) -- (6,8) -- (6,3) -- (3,3) -- (3,0) -- (0,0) -- cycle;
%\draw[<->] (13.4,14.9) -- node[right]{$m_1$} (13.4,12.1);
%\draw[<->] (12.9,11.4) -- node[below]{$n_4$} (9.1,11.4);
%\draw[<->] (9.4,11.9) -- node[right]{$m_2$} (9.4,8.1);
%\draw[<->] (8.9,7.4) -- node[below]{$n_3$} (6.1,7.4);
%\draw[<->] (6.4,7.9) -- node[right]{$m_3$} (6.4,3.1);
%\draw[<->] (5.9,2.4) -- node[below right]{$n_2$} (3.1,2.4);
%\draw[<->] (3.4,2.9) -- node[below right]{$m_4$} (3.4,0.1);
%\draw[<->] (2.9,-0.6) -- node[below]{$n_1$} (0.1,-0.6);
%\draw[<->] (1.5,14.9) -- node[right]{$q_1$} (1.5,0.1);
%\draw[<->] (4.5,14.9) -- node[below right]{$q_2$} (4.5,3.1);
%\draw[<->] (7.5,14.9) -- node[right]{$q_3$} (7.5,8.1);
%\draw[<->] (11,14.9) -- node[right]{$q_4$} (11,12.1);
%\draw[<->] (0.1,13.5) -- node[below]{$p_1$} (12.9,13.5);
%\draw[<->] (0.1,10) -- node[above left]{$p_2\quad$} (8.9,10);
%\draw[<->] (0.1,5.5) -- node[below]{$p_3$} (5.9,5.5);
%\draw[<->] (0.1,1.5) -- node[below]{$p_4$} (2.9,1.5);
%\end{tikzpicture}}}%
%BeginExpansion
\begin{tikzpicture}[scale=0.6, every node/.style={font=\small}]
\draw[fill=green] (13,12) rectangle (12,13);
\node(a) at (12.5,12.5) {$s_1$};
\draw[fill=green] (9,8) rectangle (8,9);
\node(a) at (8.5,8.5) {$s_2$};
\draw[fill=green] (6,3) rectangle (5,4);
\node(a) at (5.5,3.5) {$s_3$};
\draw[fill=green] (3,0) rectangle (2,1);
\node(a) at (2.5,0.5) {$s_4$};
\draw[line width=1.4pt]
(0,15) -- (13,15) -- (13,12) -- (9,12) -- (9,8) -- (6,8) -- (6,3) -- (3,3) -- (3,0) -- (0,0) -- cycle;
\draw[<->] (13.4,14.9) -- node[right]{$m_1$} (13.4,12.1);
\draw[<->] (12.9,11.4) -- node[below]{$n_4$} (9.1,11.4);
\draw[<->] (9.4,11.9) -- node[right]{$m_2$} (9.4,8.1);
\draw[<->] (8.9,7.4) -- node[below]{$n_3$} (6.1,7.4);
\draw[<->] (6.4,7.9) -- node[right]{$m_3$} (6.4,3.1);
\draw[<->] (5.9,2.4) -- node[below right]{$n_2$} (3.1,2.4);
\draw[<->] (3.4,2.9) -- node[below right]{$m_4$} (3.4,0.1);
\draw[<->] (2.9,-0.6) -- node[below]{$n_1$} (0.1,-0.6);
\draw[<->] (1.5,14.9) -- node[right]{$q_1$} (1.5,0.1);
\draw[<->] (4.5,14.9) -- node[below right]{$q_2$} (4.5,3.1);
\draw[<->] (7.5,14.9) -- node[right]{$q_3$} (7.5,8.1);
\draw[<->] (11,14.9) -- node[right]{$q_4$} (11,12.1);
\draw[<->] (0.1,13.5) -- node[below]{$p_1$} (12.9,13.5);
\draw[<->] (0.1,10) -- node[above left]{$p_2\quad$} (8.9,10);
\draw[<->] (0.1,5.5) -- node[below]{$p_3$} (5.9,5.5);
\draw[<->] (0.1,1.5) -- node[below]{$p_4$} (2.9,1.5);
\end{tikzpicture}%
%EndExpansion
\]


\item \textbf{Page 38, (3.8\textquotedblright):} Strictly speaking, this has
only been proved under the assumption that each $Z_{i+1}-Z_{i}$ can be written
as sum of at most $m_{i+1}+n_{k+1-i}$ elements of rank $\leq1$ each. This
assumption is stronger than the given requirement $rk\left(  Z_{i+1}%
-Z_{i}\right)  \leq m_{i+1}+n_{k+1-i}$. From what I see, however, this
stronger assumption is always satisfied when (3.8\textquotedblright) is
applied in this book.

\item \textbf{Page 38, before (3.9):} \textquotedblleft Let $X_{i}%
=x_{1}+\cdots+x_{i}$ for each $i\geq1$\textquotedblright\ should be
\textquotedblleft Let $X_{i}=x_{1}+\cdots+x_{i}$ for each $i\geq
0$\textquotedblright. (Otherwise, the \textquotedblleft$X_{i-1}$%
\textquotedblright\ in (3.9) makes no sense.)

\item \textbf{Page 38:} It is worth warning the reader that the Schur function
$s_{\alpha}\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ used in Chapter II is
not $s_{\alpha}\left(  x_{1},x_{2},\ldots,x_{n}\right)  $ in the notation of
Chapter III (the latter is not even a symmetric polynomial!), but rather is
$s_{\alpha}\left(  X_{n}\right)  =s_{\alpha}\left(  x_{1}+x_{2}+\cdots
+x_{n}\right)  $ in the notation of Chapter III. (This is not obvious, but
follows from the Jacobi--Trudi formula.)

\item \textbf{Page 39, (3.10):} This should also require $r_{i}>0$ (otherwise,
the operators make no sense).

\item \textbf{Page 39, proof of (3.10):} \textquotedblleft By definition, we
have $s_{\alpha}=$\textquotedblright\ should be \textquotedblleft By
definition, we have $s_{\alpha}\left(  X_{r_{1}},\ldots,X_{r_{n}}\right)
=$\textquotedblright.

\item \textbf{Page 39, proof of (3.8\textquotedblright'):} I think
\textquotedblleft equal to $\left(  X_{1}\cdots X_{m}\right)  ^{n}%
$\textquotedblright\ should be \textquotedblleft equal to $\left(  X_{1}\cdots
X_{n}\right)  ^{m}$\textquotedblright.

\item \textbf{Page 39, proof of (3.11):} Replace \textquotedblleft$w_{o}%
$\textquotedblright\ by \textquotedblleft$w_{0}$\textquotedblright\ twice.

\item \textbf{Page 39, proof of (3.11):} The claim that \textquotedblleft By
(3.10), $\pi_{1}\pi_{2}\cdots\pi_{n-1}$ applied to $s_{\alpha}\left(
X_{1}+Z_{1},\ldots,X_{n}+Z_{n}\right)  $ will produce%
\[
s_{\alpha}\left(  X_{2}+Z_{1},X_{3}+Z_{2},\ldots,X_{n}+Z_{n-1},X_{n}%
+Z_{n}\right)
\]
\textquotedblright\ is not as self-evident as it is being sold here. It does
not follow from (3.10) directly, since there are no $Z$-variables in (3.10).
Instead, this claim can be justified as follows: We have%
\[
s_{\alpha}\left(  X_{1}+Z_{1},\ldots,X_{n}+Z_{n}\right)  =\det\left(
h_{\alpha_{i}-i+j}\left(  X_{i}+Z_{i}\right)  \right)  _{1\leq i,j\leq n}.
\]
In the matrix on the right hand side, all entries except for those in the
$\left(  n-1\right)  $-st row are symmetric in $x_{n-1}$ and $x_{n}$, and thus
can \textquotedblleft pass through\textquotedblright\ the map $\pi_{n-1}$
unchanged (this follows from the fact that $\pi_{n-1}\left(  fg\right)
=f\pi_{n-1}\left(  g\right)  $ whenever $f$ is symmetric in $x_{n-1}$ and
$x_{n}$; this follows easily from (2.2')), while the entries $h_{\alpha
_{n-1}-\left(  n-1\right)  +j}\left(  X_{n-1}+Z_{i}\right)  $ in the $\left(
n-1\right)  $-st row are transformed by $\pi_{n-1}$ into%
\begin{align*}
&  \pi_{n-1}\underbrace{h_{\alpha_{n-1}-\left(  n-1\right)  +j}\left(
X_{n-1}+Z_{i}\right)  }_{=\sum_{p+q=\alpha_{n-1}-\left(  n-1\right)  +j}%
h_{p}\left(  X_{n-1}\right)  h_{q}\left(  Z_{i}\right)  }\\
&  =\sum_{p+q=\alpha_{n-1}-\left(  n-1\right)  +j}\underbrace{\pi_{n-1}\left(
h_{p}\left(  X_{n-1}\right)  h_{q}\left(  Z_{i}\right)  \right)
}_{\substack{=\pi_{n-1}\left(  h_{p}\left(  X_{n-1}\right)  \right)  \cdot
h_{q}\left(  Z_{i}\right)  \\=h_{p}\left(  X_{n}\right)  \cdot h_{q}\left(
Z_{i}\right)  \\\text{(since (3.9) yields }\pi_{n-1}\left(  h_{p}\left(
X_{n-1}\right)  \right)  =h_{p}\left(  X_{n}\right)  \text{)}}}\\
&  =\sum_{p+q=\alpha_{n-1}-\left(  n-1\right)  +j}h_{p}\left(  X_{n}\right)
\cdot h_{q}\left(  Z_{i}\right) \\
&  =h_{\alpha_{n-1}-\left(  n-1\right)  +j}\left(  X_{n}+Z_{i}\right)  .
\end{align*}
Hence, applying $\pi_{n-1}$ to the determinant of this matrix results in the
same determinant, except that the $X_{n-1}$'s in the $\left(  n-1\right)  $-st
row are turned into $X_{n}$'s. That is, applying $\pi_{n-1}$ to $s_{\alpha
}\left(  X_{1}+Z_{1},\ldots,X_{n}+Z_{n}\right)  $ turns the $X_{n-1}$ in the
second-to-last argument into an $X_{n}$. Subsequently applying $\pi_{n-2}$ to
the result then turns the $X_{n-2}$ in the third-to-last argument into an
$X_{n-1}$ (for a similar reason). Subsequently applying $\pi_{n-3}$ then turns
the $X_{n-3}$ in the fourth-to-last argument into an $X_{n-2}$. And so on,
until we eventually conclude that $\pi_{1}\pi_{2}\cdots\pi_{n-1}$ applied to
$s_{\alpha}\left(  X_{1}+Z_{1},\ldots,X_{n}+Z_{n}\right)  $ results in each of
the subscripts under $X_{1},X_{2},\ldots,X_{n-1}$ being increased by $1$. That
is, $\pi_{1}\pi_{2}\cdots\pi_{n-1}$ applied to $s_{\alpha}\left(  X_{1}%
+Z_{1},\ldots,X_{n}+Z_{n}\right)  $ produces%
\[
s_{\alpha}\left(  X_{2}+Z_{1},X_{3}+Z_{2},\ldots,X_{n}+Z_{n-1},X_{n}%
+Z_{n}\right)  .
\]
The same reasoning is then applied again to conclude that $\pi_{2}\pi
_{3}\cdots\pi_{n-1}$ applied to $s_{\alpha}\left(  X_{2}+Z_{1},X_{3}%
+Z_{2},\ldots,X_{n}+Z_{n-1},X_{n}+Z_{n}\right)  $ produces%
\[
s_{\alpha}\left(  X_{3}+Z_{1},X_{4}+Z_{2},\ldots,X_{n}+Z_{n-2},X_{n}%
+Z_{n-1},X_{n}+Z_{n}\right)  ,
\]
and so on.

\item \textbf{Page 40, Remark:} This does indeed yield \textquotedblleft
independent proofs of (2.11) and (2.16')\textquotedblright\ provided we know
that the $s_{\alpha}\left(  X_{n}\right)  $ in the terminology of Chapter III
(defined as $\det\left(  h_{\alpha_{i}-i+j}\left(  X_{n}\right)  \right)
_{1\leq i,j\leq n}$) agrees with the Schur polynomials $s_{\alpha}\left(
x_{1},x_{2},\ldots,x_{n}\right)  $ in the notations of Chapter II (defined,
e.g., as the ratio of alternants $a_{\alpha+\delta}/a_{\delta}$, where
$a_{\beta}:=\det\left(  x_{i}^{\beta_{j}}\right)  _{1\leq i,j\leq n}$). The
latter fact is known as the Jacobi--Trudi formula (or, rather, its
generalization to indices in $\mathbf{Z}^{n}$, not just partitions). If we
don't know this formula, then we instead obtain an independent proof of this formula.

\item \textbf{Page 40, Remark:} The words \textquotedblleft(by
linearity)\textquotedblright\ here are somewhat of an overstatement: By
linearity, the formulas (2.11) and (2.16') yield (2.10) and (2.16) for
polynomials only, not for all rational functions. However, the case of
rational functions can easily be reduced to the case of polynomials using the
(easy) fact that every rational function in $x_{1},x_{2},\ldots,x_{n}$ can be
written as a ratio $\dfrac{f}{g}$ where $f$ is a polynomial and $g$ is a
nonzero symmetric polynomial. (This is useful since symmetric polynomials
\textquotedblleft pass through\textquotedblright\ $\partial_{w_{0}}$ and
$\pi_{w_{0}}$ undisturbed.)

\item \textbf{Page 40, (3.12):} The first displayed formula here could use
some clarification. First, any pair $w=\left(  u,v\right)  \in S_{m}\times
S_{n}$ acts on polynomials in the variables $x_{1},\ldots,x_{m},y_{1}%
,\ldots,y_{n}$ in the \textquotedblleft obvious\textquotedblright\ way (i.e.,
by $w\left(  x_{i}\right)  =x_{u\left(  i\right)  }$ for all $i$ and $w\left(
y_{j}\right)  =y_{v\left(  j\right)  }$ for all $j$). Second,
\textquotedblleft$f_{\lambda}\left(  x,y\right)  /D\left(  x\right)  D\left(
y\right)  $\textquotedblright\ is to be understood as \textquotedblleft%
$f_{\lambda}\left(  x,y\right)  /\left(  D\left(  x\right)  D\left(  y\right)
\right)  $\textquotedblright.

\item \textbf{Page 40, proof of (3.12):} Replace \textquotedblleft$w_{o}%
$\textquotedblright\ by \textquotedblleft$w_{0}$\textquotedblright.

\item \textbf{Page 40, proof of (3.12):} \textquotedblleft and in view of
(2.16)\textquotedblright\ could better be explained as \textquotedblleft and
in view of the fact that each $f\in P_{n}$ satisfies%
\begin{align*}
\pi_{w_{0}}\left(  f\right)   &  =a_{\delta}^{-1}\sum_{w\in S_{n}}%
\epsilon\left(  w\right)  w\left(  x^{\delta}f\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (2.16)}\right) \\
&  =\sum_{w\in S_{n}}\underbrace{\epsilon\left(  w\right)  a_{\delta}^{-1}%
}_{=w\left(  a_{\delta}^{-1}\right)  }w\left(  x^{\delta}f\right)  =\sum_{w\in
S_{n}}\underbrace{w\left(  a_{\delta}^{-1}\right)  w\left(  x^{\delta
}f\right)  }_{=w\left(  a_{\delta}^{-1}x^{\delta}f\right)  }\\
&  =\sum_{w\in S_{n}}w\underbrace{\left(  a_{\delta}^{-1}x^{\delta}f\right)
}_{=f/\prod_{1\leq i<j\leq n}\left(  1-x_{i}^{-1}x_{j}\right)  }=\sum_{w\in
S_{n}}w\left(  f/\prod_{1\leq i<j\leq n}\left(  1-x_{i}^{-1}x_{j}\right)
\right)
\end{align*}
\textquotedblright.

\item \textbf{Pages 40--41, proof of (3.12):} This proof appears to have a
subtle flaw: Parts of it (specifically, the part where (3.11) is applied) only
work when $n,m\geq r$.

This flaw can be fixed with a bit of preparation. We define
$\operatorname*{zero}\nolimits_{m+1}:P_{m+1}\rightarrow P_{m}$ to be the
algebra morphism that substitutes $0$ for $x_{m+1}$ in a polynomial (i.e., it
sends each $f\left(  x_{1},x_{2},\ldots,x_{m+1}\right)  \in P_{m+1}$ to
$f\left(  x_{1},x_{2},\ldots,x_{m},0\right)  \in P_{m}$). We need two simple lemmas:

\begin{statement}
\textbf{(3.12\textquotedblright)} Let $\alpha=\left(  \alpha_{1},\alpha
_{2},\ldots,\alpha_{m+1}\right)  \in\mathbf{N}^{m+1}$, and set $\alpha
^{\prime}=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{m}\right)
\in\mathbf{N}^{m}$. Then,%
\[
\operatorname*{zero}\nolimits_{m+1}\left(  s_{\alpha}\left(  x_{1}%
,x_{2},\ldots,x_{m+1}\right)  \right)  =%
\begin{cases}
s_{\alpha^{\prime}}\left(  x_{1},x_{2},\ldots,x_{m}\right)  , & \text{if
}\alpha_{m+1}=0;\\
0, & \text{if }\alpha_{m+1}>0.
\end{cases}
\]

\end{statement}

To prove this, we recall the definition of Schur polynomials using alternants:
For each $k\in\mathbf{N}$ and each $\beta=\left(  \beta_{1},\beta_{2}%
,\ldots,\beta_{k}\right)  \in\mathbf{N}^{k}$, let $a_{\beta}:=\det\left(
x_{i}^{\beta_{j}}\right)  _{1\leq i,j\leq k}$ be the corresponding alternant.
Also, let $\delta^{\left(  k\right)  }$ be the $k$-tuple $\left(
k-1,k-2,\ldots,1,0\right)  $ for each $k\in\mathbf{N}$. Then, the definition
of Schur polynomials yields
\[
s_{\alpha}=a_{\alpha+\delta^{\left(  m+1\right)  }}/a_{\delta^{\left(
m+1\right)  }}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ s_{\alpha
^{\prime}}=a_{\alpha^{\prime}+\delta^{\left(  m\right)  }}/a_{\delta^{\left(
m\right)  }}.
\]


Now, as we apply the map $\operatorname*{zero}\nolimits_{m+1}$ (that is,
substitute $0$ for $x_{m+1}$), the denominator $a_{\delta^{\left(  m+1\right)
}}=\prod_{1\leq i<j\leq m+1}\left(  x_{i}-x_{j}\right)  $ of $s_{\alpha}$
becomes $x_{1}x_{2}\cdots x_{m}\cdot a_{\delta^{\left(  m\right)  }}$, whereas
the numerator $a_{\alpha+\delta^{\left(  m+1\right)  }}=\det\left(
x_{i}^{\alpha_{j}+\left(  m+1-j\right)  }\right)  _{1\leq i,j\leq m+1}$

\begin{itemize}
\item becomes $0$ if $\alpha_{m+1}>0$ (because in this case, the last row of
the matrix $\left(  x_{i}^{\alpha_{j}+\left(  m+1-j\right)  }\right)  _{1\leq
i,j\leq m+1}$ becomes $\left(  0,0,\ldots,0\right)  $),

\item and becomes $x_{1}x_{2}\cdots x_{m}a_{\alpha^{\prime}}$ if $\alpha
_{m+1}=0$ (because in this case, the last row of the matrix $\left(
x_{i}^{\alpha_{j}+\left(  m+1-j\right)  }\right)  _{1\leq i,j\leq m+1}$
becomes $\left(  0,0,\ldots,0,1\right)  $, and so its determinant becomes
\begin{align*}
&  \det\left(  x_{i}^{\alpha_{j}+\left(  m+1-j\right)  }\right)  _{1\leq
i,j\leq m}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Laplace expansion along the
last row}\right) \\
&  =\det\left(  x_{i}\cdot x_{i}^{\alpha_{j}+\left(  m-j\right)  }\right)
_{1\leq i,j\leq m}=x_{1}x_{2}\cdots x_{m}\underbrace{\det\left(  x_{i}%
^{\alpha_{j}+m-j}\right)  _{1\leq i,j\leq m}}_{=a_{\alpha^{\prime}%
+\delta^{\left(  m\right)  }}}\\
&  =x_{1}x_{2}\cdots x_{m}\cdot a_{\alpha^{\prime}+\delta^{\left(  m\right)
}}%
\end{align*}
).
\end{itemize}

Hence, the map $\operatorname*{zero}\nolimits_{m+1}$ takes $s_{\alpha
}=a_{\alpha+\delta^{\left(  m+1\right)  }}/a_{\delta^{\left(  m+1\right)  }}$
to $0$ if $\alpha_{m+1}>0$, and to $\left(  x_{1}x_{2}\cdots x_{m}\cdot
a_{\alpha^{\prime}+\delta^{\left(  m\right)  }}\right)  /\left(  x_{1}%
x_{2}\cdots x_{m}\cdot a_{\delta^{\left(  m\right)  }}\right)  =a_{\alpha
^{\prime}+\delta^{\left(  m\right)  }}/a_{\delta^{\left(  m\right)  }%
}=s_{\alpha^{\prime}}$ if $\alpha_{m+1}=0$. This proves
(3.12\textquotedblright).

\begin{statement}
\textbf{(3.12\textquotedblright')} We have $\operatorname*{zero}%
\nolimits_{m+1}\circ\pi_{w_{0}^{\left(  m+1\right)  }}=\pi_{w_{0}^{\left(
m\right)  }}$ on $P_{m}$.
\end{statement}

By linearity, it suffices to check this for each monomial $x^{\alpha}%
=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}\cdots x_{m}^{\alpha_{m}}$; but in this
case, it follows from (3.12') and (2.16'). This proves
(3.12'\textquotedblright).

Of course, (3.12\textquotedblright') can be generalized to polynomials that
involve several additional independent indeterminates $y_{1},y_{2}%
,\ldots,y_{n}$ as well. The operator $\operatorname*{zero}\nolimits_{m+1}$ is
supposed to leave these extra indeterminates undisturbed (i.e., treat them as constants).

Now, let us fix the proof of (3.12). It suffices to prove (3.12'), since all
the reasoning before has been correct. Now, let us introduce a new variable
$x_{m+1}$ (replacing our convention $x_{m+1}=0$), and write $f_{\lambda
}\left(  x^{+},y\right)  :=\prod_{\left(  i,j\right)  \in\lambda}\left(
x_{i}-y_{j}\right)  $ (this looks the same as $f_{\lambda}\left(  x,y\right)
$, but keep in mind that $x_{m+1}$ was $0$ in the definition of $f_{\lambda
}\left(  x,y\right)  $) and $\pi_{x^{+}}:=\pi_{w_{0}^{\left(  m+1\right)  }}$
and $X_{m+1}^{+}:=x_{1}+x_{2}+\cdots+x_{m+1}$ (this is not $X_{m+1}$ since
$X_{m+1}$ had $x_{m+1}=0$). Clearly, $\operatorname*{zero}\nolimits_{m+1}%
\left(  f_{\lambda}\left(  x^{+},y\right)  \right)  =f_{\lambda}\left(
x,y\right)  $ and $\operatorname*{zero}\nolimits_{m+1}\left(  X_{m+1}%
^{+}\right)  =X_{m+1}=X_{m}$. But (3.12\textquotedblright') says that
$\operatorname*{zero}\nolimits_{m+1}\circ\pi_{x^{+}}=\pi_{x}$ on $P_{m}$.
Hence, $\operatorname*{zero}\nolimits_{m+1}\circ\pi_{x^{+}}=\pi_{x}$ holds
also for polynomials in $x_{1},x_{2},\ldots,x_{m},y_{1},y_{2},\ldots,y_{n}$
(since the extra variables $y_{1},y_{2},\ldots,y_{n}$ \textquotedblleft pass
through\textquotedblright\ the $\pi$ and $\operatorname*{zero}\nolimits_{m+1}$
operators unchanged). Hence,%
\begin{align*}
&  \pi_{y}\underbrace{\pi_{x}}_{=\operatorname*{zero}\nolimits_{m+1}\circ
\pi_{x^{+}}}f_{\lambda}\left(  x,y\right) \\
&  =\pi_{y}\operatorname*{zero}\nolimits_{m+1}\pi_{x^{+}}f_{\lambda}\left(
x,y\right) \\
&  =\operatorname*{zero}\nolimits_{m+1}\pi_{y}\pi_{x^{+}}f_{\lambda}\left(
x,y\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi_{y}\text{ and
}\operatorname*{zero}\nolimits_{m+1}\text{ commute}\right) \\
&  =\operatorname*{zero}\nolimits_{m+1}\left(  \pi_{y}\pi_{x^{+}}f_{\lambda
}\left(  x^{+},y\right)  \right)
\end{align*}
and furthermore%
\[
s_{\lambda}\left(  X_{m}-Y_{n}\right)  =\operatorname*{zero}\nolimits_{m+1}%
\left(  s_{\lambda}\left(  X_{m+1}^{+}-Y_{n}\right)  \right)
\]
(since $\operatorname*{zero}\nolimits_{m+1}$ is an algebra morphism that takes
$X_{m+1}^{+}$ to $X_{m}$ and leaves $Y_{n}$ unchanged). Therefore, in order to
prove (3.12'), it suffices to prove the equality%
\[
s_{\lambda}\left(  X_{m+1}^{+}-Y_{n}\right)  =\pi_{y}\pi_{x^{+}}f_{\lambda
}\left(  x^{+},y\right)  ,
\]
since applying $\operatorname*{zero}\nolimits_{m+1}$ to this latter equality
will then yield (3.12'). But this latter equality is nothing but the equality
(3.12') written down for $m+1$ instead of $m$. Hence, if we can prove (3.12')
for $m+1$ instead of $m$, then (3.12') also holds for the original $m$.

Thus, we can replace $m$ by $m+1$ in our proof of (3.12'). Iterating this
argument, we can therefore replace $m$ by any integer larger than $m$. In
particular, we can WLOG assume that $m\geq\ell\left(  \lambda\right)  $ and
$m\geq\ell\left(  \lambda^{\prime}\right)  $. Likewise, we can WLOG assume
that $n\geq\ell\left(  \lambda\right)  $ and $n\geq\ell\left(  \lambda
^{\prime}\right)  $. With these assumptions made, the proof in the text goes through.

\item \textbf{Page 40, proof of (3.12):} The derivation of (2) still needs a
few more words. As we said above, we are working under the WLOG assumption
that $m\geq\ell\left(  \lambda\right)  $ (among other things). That is, $m\geq
r$. Hence, (1) becomes%
\[
f_{\lambda}\left(  x,y\right)  =s_{\lambda}\left(  X_{1}-Y_{\lambda_{1}%
},\ldots,X_{r}-Y_{\lambda_{r}}\right)  =s_{\lambda}\left(  X_{1}%
-Y_{\lambda_{1}},\ldots,X_{m}-Y_{\lambda_{m}}\right)
\]
by (3.3). Therefore,%
\begin{align}
\pi_{x}f_{\lambda}\left(  x,y\right)   &  =\pi_{x}s_{\lambda}\left(
X_{1}-Y_{\lambda_{1}},\ldots,X_{m}-Y_{\lambda_{m}}\right) \nonumber\\
&  =s_{\lambda}\left(  X_{m}-Y_{\lambda_{1}},\ldots,X_{m}-Y_{\lambda_{m}%
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (3.11)}\right) \tag{2'}\\
&  =s_{\lambda}\left(  X_{m}-Y_{\lambda_{1}},\ldots,X_{m}-Y_{\lambda_{r}%
}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (3.3) again}\right)  .\nonumber
\end{align}
This proves (2).

\item \textbf{Page 41, proof of (3.12):} I would replace \textquotedblleft%
$rk\left(  Y_{p_{i}}-Y_{p_{i+1}}\right)  =p_{i}-p_{i+1}$\textquotedblright\ by
\textquotedblleft$rk\left(  Y_{p_{i}}-Y_{p_{i+1}}\right)  \leq p_{i}-p_{i+1}%
$\textquotedblright.

\item \textbf{Page 41, proof of (3.12):} I don't understand the proof of (4)
given here. But I have a simpler proof. More precisely, I will show not the
equality (4), but rather the equality%
\begin{align}
&  s_{\lambda}\left(  X_{m}-Y_{\lambda_{1}},\ldots,X_{m}-Y_{\lambda_{m}%
}\right) \nonumber\\
&  =\left(  -1\right)  ^{\left\vert \lambda\right\vert }s_{\lambda^{\prime}%
}\left(  Y_{1}-X_{m},Y_{2}-X_{m},\ldots,Y_{n}-X_{m}\right)  , \tag{4'}%
\end{align}
which is equivalent to (4) by way of (3.3) but is even closer to what is
actually needed for proving (3.12).

My proof of (4') relies on the duality theorem (3.8), not
(3.8\textquotedblright). First, we need some combinatorics: For each
$k\in\mathbf{Z}$, let $g_{k}\in\mathbf{N}$ denote the largest $i\geq1$ such
that $\lambda_{i}-i\geq k$ (or $0$ if no such $i$ exists). Visually speaking,
this is the index of the bottommost row of $D\left(  \lambda\right)  $ that
contains a cell in the $k$-th diagonal (i.e., in the main diagonal shifted by
$k$ upwards). It is easy to see that $g_{k-1}-g_{k}\in\left\{  0,1\right\}  $
for each $k\in\mathbf{Z}$ (because if the diagram $D\left(  \lambda\right)  $
contains a cell $\left(  i,j\right)  $ in the $\left(  k-1\right)  $-th
diagonal, then it also contains the cell $\left(  i-1,j\right)  $ in the
$k$-th diagonal, unless $i=1$; and thus we must have $g_{k}\geq g_{k-1}-1$;
and a similar argument shows $g_{k-1}\geq g_{k}$). Now, set%
\[
Z_{k}:=X_{m}-Y_{g_{k}}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\mathbf{Z}%
\text{.}%
\]
Then, $rk\left(  Z_{k}-Z_{k-1}\right)  \leq1$ for all $k\in\mathbf{Z}$ (since
$g_{k-1}-g_{k}\in\left\{  0,1\right\}  $ for each $k\in\mathbf{Z}$).
Therefore, (3.8) can be applied to $Z_{k}$ instead of $X_{k}$. As a
consequence, we obtain%
\begin{equation}
s_{\lambda}\left(  -Z_{\lambda_{1}-1},\ldots,-Z_{\lambda_{n}-n}\right)
=\left(  -1\right)  ^{\left\vert \lambda\right\vert }s_{\lambda^{\prime}%
}\left(  Z_{1-\lambda_{1}^{\prime}},\ldots,Z_{m-\lambda_{m}^{\prime}}\right)
. \tag{4\textquotedblright}%
\end{equation}
But each $1\leq i\leq n$ satisfies $g_{\lambda_{i}-i}=i$ (since the cell
$\left(  i,\lambda_{i}\right)  $ of $D\left(  \lambda\right)  $ lies in the
$\left(  \lambda_{i}-i\right)  $-th diagonal, but the next cell $\left(
i+1,\lambda_{i}+1\right)  $ of this diagonal is already outside of $D\left(
\lambda\right)  $) and therefore%
\begin{equation}
Z_{\lambda_{i}-i}=X_{m}-Y_{g_{\lambda_{i}-i}}=X_{m}-Y_{i}\nonumber
\end{equation}
and thus
\begin{equation}
-Z_{\lambda_{i}-i}=Y_{i}-X_{m}. \tag{5a}%
\end{equation}
Meanwhile, each $1\leq j\leq m$ satisfies $g_{j-\lambda_{j}^{\prime}}%
=\lambda_{j}^{\prime}$ (since the cell $\left(  \lambda_{j}^{\prime},j\right)
$ of $D\left(  \lambda\right)  $ lies in the $\left(  j-\lambda_{j}^{\prime
}\right)  $-th diagonal, but the next cell $\left(  \lambda_{j}^{\prime
}+1,j+1\right)  $ of this diagonal is already outside of $D\left(
\lambda\right)  $) and therefore
\begin{equation}
Z_{j-\lambda_{j}^{\prime}}=X_{m}-Y_{g_{j-\lambda_{j}^{\prime}}}=X_{m}%
-Y_{\lambda_{j}^{\prime}}. \tag{5b}%
\end{equation}
In view of (5a) and (5b), we can rewrite (4\textquotedblright) as
\begin{align*}
&  s_{\lambda}\left(  Y_{1}-X_{m},Y_{2}-X_{m},\ldots,Y_{n}-X_{m}\right) \\
&  =\left(  -1\right)  ^{\left\vert \lambda\right\vert }s_{\lambda^{\prime}%
}\left(  X_{m}-Y_{\lambda_{1}^{\prime}},X_{m}-Y_{\lambda_{2}^{\prime}}%
,\ldots,X_{m}-Y_{\lambda_{m}^{\prime}}\right)  .
\end{align*}
In other words,%
\begin{align*}
&  s_{\lambda^{\prime}}\left(  X_{m}-Y_{\lambda_{1}^{\prime}},X_{m}%
-Y_{\lambda_{2}^{\prime}},\ldots,X_{m}-Y_{\lambda_{m}^{\prime}}\right) \\
&  =\left(  -1\right)  ^{\left\vert \lambda\right\vert }s_{\lambda}\left(
Y_{1}-X_{m},Y_{2}-X_{m},\ldots,Y_{n}-X_{m}\right)  .
\end{align*}
Substituting $\lambda^{\prime}$ for $\lambda$ here (and recalling that
$\left\vert \lambda^{\prime}\right\vert =\left\vert \lambda\right\vert $ and
$\lambda^{\prime\prime}=\lambda$), we obtain precisely the desired equality (4').

\item \textbf{Page 41, proof of (3.12):} The last equality sign might use some
explanation: It is obtained by applying (3.8) to the elements $X_{k}%
:=Y_{n}-X_{m}$ (for all $k\in\mathbf{Z}$).

\item \textbf{Page 42, proof of (4.2):} On the second-to-last line of the
page, after \textquotedblleft if $\ell\left(  w\right)  -\ell\left(  v\right)
$\textquotedblright, add \textquotedblleft$=\ell\left(  wv^{-1}\right)
$\textquotedblright.

\item \textbf{Page 43, proof of (4.3) (ii):} Replace \textquotedblleft%
$\max\left(  \beta_{i},\beta_{i+1}\right)  \leq\max\left(  \alpha_{i}%
,\alpha_{i+1}\right)  -1\leq n-i-1$\textquotedblright\ by \textquotedblleft%
$\max\left(  \beta_{r},\beta_{r+1}\right)  \leq\max\left(  \alpha_{r}%
,\alpha_{r+1}\right)  -1\leq n-r-1$\textquotedblright.

\item \textbf{Page 43, proof of (4.3) (ii):} The claim that $\mathfrak{S}_{w}$
is nonzero is not proved here. (But it follows easily from part (i): Indeed,
$\ell\left(  w_{0}\right)  =\ell\left(  w\right)  +\ell\left(  w^{-1}%
w_{0}\right)  $, thus $\partial_{w_{0}}=\partial_{w}\partial_{w^{-1}w_{0}}$,
and therefore (4.1) yields $\partial_{w}\mathfrak{S}_{w}=\underbrace{\partial
_{w}\partial_{w^{-1}w_{0}}}_{=\partial_{w_{0}}}x^{\delta}=\partial_{w_{0}%
}x^{\delta}=\mathfrak{S}_{1}=1$ (by part (i)), and thus $\mathfrak{S}_{w}%
\neq0$.)

\item \textbf{Page 43, proof of (4.3) (iv):} The claim that $\mathfrak{S}%
_{w}\notin P_{r-1}$ is not proved here. (But again, it is easy to see: If
$\mathfrak{S}_{w}$ beloned to $P_{r-1}$, then $\mathfrak{S}_{w}$ would be
symmetric in $x_{r},x_{r+1}$, and therefore, by part (iii), we would have
$w\left(  r\right)  <w\left(  r+1\right)  $.)

\item \textbf{Page 44, proof of (4.5):} \textquotedblleft hence $\partial
_{2}\partial_{1}\left(  x_{1}^{n}x_{2}^{n-2}\cdots x_{n}\right)
$\textquotedblright\ should be \textquotedblleft hence $\partial_{2}%
\partial_{1}\left(  x_{1}^{n}x_{2}^{n-1}\cdots x_{n}\right)  $%
\textquotedblright.

\item \textbf{Page 45, proof of (4.6):} In the display%
\[
\partial_{i}F\left(  u,v\right)  =%
\begin{cases}
F\left(  u,vs_{i}\right)  & \text{if }\ell\left(  vs_{i}\right)  <\ell\left(
v\right)  \text{,}\\
0 & \text{otherwise,}%
\end{cases}
\]
replace both \textquotedblleft$s_{i}$\textquotedblright s by \textquotedblleft%
$s_{i-m}$\textquotedblright.

\item \textbf{Page 45:} It might be worth mentioning that (4.6) gives a second
proof of (4.5), because the embedding $i:S_{n}\rightarrow S_{m}$ has the
property that $i\left(  w\right)  =w\times1_{m-n}$ for all $w\in S_{n}$ (and
of course, we have $\mathfrak{S}_{1_{n}\times1_{m-n}}=\mathfrak{S}_{1_{n}}=1$).

\item \textbf{Page 45:} \textquotedblleft premutations\textquotedblright%
\ should be \textquotedblleft permutations\textquotedblright.

\item \textbf{Page 49:} \textquotedblleft indentically\textquotedblright%
\ should be \textquotedblleft identically\textquotedblright.

\item \textbf{Page 51, (4.16):} Replace \textquotedblleft$\ell\left(
v\right)  =\ell\left(  v\right)  -1$\textquotedblright\ by \textquotedblleft%
$\ell\left(  v\right)  =\ell\left(  w\right)  -1$\textquotedblright.

\item \textbf{Page 51, (4.16):} After \textquotedblleft and each $w^{\prime}$
in the sum precedes $w$ in the reverse lexicographical
ordering\textquotedblright, add \textquotedblleft and satisfies $\ell\left(
w^{\prime}\right)  =\ell\left(  w\right)  $\textquotedblright.

\item \textbf{Page 51, (4.17):} Replace \textquotedblleft$S_{w}$%
\textquotedblright\ by \textquotedblleft$\mathfrak{S}_{w}$\textquotedblright.

\item \textbf{Page 51:} In the last sentence of this page, \textquotedblleft
By (4.12) we can express\textquotedblright\ should be \textquotedblleft By
(4.13) we can express\textquotedblright.

\item \textbf{Page 51, proof of (4.17):} \textquotedblleft such that
$\ell\left(  v\right)  -\ell\left(  w\right)  $\textquotedblright\ should be
\textquotedblleft such that $\ell\left(  v\right)  <\ell\left(  w\right)
$\textquotedblright.

\item \textbf{Page 52, proof of (4.19):} Replace \textquotedblleft$\rho\left(
x_{i}\right)  $\textquotedblright\ by \textquotedblleft$\rho_{m}\left(
x_{i}\right)  $\textquotedblright\ twice.

\item \textbf{Page 54:} \textquotedblleft let $\rho_{n}:P_{\infty}\mapsto
P_{n}$\textquotedblright\ should be \textquotedblleft let $\rho_{n}:P_{\infty
}\rightarrow P_{n}$\textquotedblright\ (different kind of arrow).

\item \textbf{Page 59, proof of (4.35):} The \textquotedblleft$d_{o}%
$\textquotedblright\ should probably be \textquotedblleft$d_{0}$%
\textquotedblright.

\item \textbf{Page 59, proof of (4.35):} The \textquotedblleft diagram on p.
58\textquotedblright\ is not actually on p. 58 of the reedition. Instead it is
on p. 57, or (to give a less fickle anchor for the reference) between (4.30)
and (4.31).

\item \textbf{Page 61:} \textquotedblleft aribtrarily\textquotedblright%
\ should be \textquotedblleft arbitrarily\textquotedblright.

\item \textbf{Page 70, proof of (5.1):} In the second displayed equation of
the proof, \textquotedblleft$e_{r}\left(  x_{1},\ldots,x_{n}\right)
$\textquotedblright\ should be \textquotedblleft$e_{r}\left(  x_{1}%
,\ldots,x_{n-1}\right)  $\textquotedblright.

\item \textbf{Page 70, proof of (5.1):} \textquotedblleft
follws\textquotedblright\ should be \textquotedblleft
follows\textquotedblright.

\item \textbf{Page 71, (5.3):} The angular brackets here are misaligned with
the equality sign. This is a LaTeX issue; use
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
left%
%TCIMACRO{\TEXTsymbol{<} }%
%BeginExpansion
$<$
%EndExpansion
instead of
%TCIMACRO{\TEXTsymbol{<}}%
%BeginExpansion
$<$%
%EndExpansion
.

\item \textbf{Page 71, proof of (5.3) (i):} \textquotedblleft for $i\leq
i$\textquotedblright\ should be \textquotedblleft for $1\leq i$%
\textquotedblright.

\item \textbf{Page 72, proof of (5.5):} After \textquotedblleft and then it is
equal to $\epsilon\left(  v\right)  \left\langle w_{0}\mathfrak{S}_{uv^{-1}%
},x^{\delta}\right\rangle $\textquotedblright, I would add \textquotedblleft%
$=\epsilon\left(  v\right)  \partial_{w_{0}}\left(  w_{0}\left(
\mathfrak{S}_{uv^{-1}}\right)  x^{\delta}\right)  $\textquotedblright\ to make
it clear what will happen next.

\item \textbf{Page 72, proof of (5.5):} After \textquotedblleft Now
$\mathfrak{S}_{uv^{-1}}$ is a linear combination of monomials $x^{\alpha}%
$\textquotedblright, add \textquotedblleft with $\alpha\in\mathbf{N}^{n}%
$\textquotedblright\ (to avoid the alternative interpretation $\alpha
\in\mathbf{N}^{n-1}$, which would lead to awkwardness further on).

\item \textbf{Page 72, proof of (5.5):} After \textquotedblleft Now
$\partial_{w_{0}}x^{\beta}=0$ unless all components $\beta_{i}$ of $\beta$ are
distinct\textquotedblright, I would add \textquotedblleft(by
(2.11))\textquotedblright.

\item \textbf{Page 72, proof of (5.6):} The expansions (1) and (2) exist
because of (5.1'). This is perhaps worth pointing out.

\item \textbf{Page 73, proof of (5.7):} The set $A_{i}$ is sometimes denoted
by $A$ (without subscript) here.

\item \textbf{Page 74, the line above (5.8):} In \textquotedblleft$x=\left(
x_{1},\ldots x_{n}\right)  $\textquotedblright, add a comma before
\textquotedblleft$x_{n}$\textquotedblright.

\item \textbf{Page 75:} The first displayed equation on this page,%
\[
T_{w}\left(  y\right)  =\left\langle \Delta\left(  x,y\right)  ,\ w_{0}%
\mathfrak{S}_{ww_{0}}\left(  -x\right)  \right\rangle _{x},
\]
could use a bit more justification. Namely, if $\alpha_{u}\in P_{n}\left(
y\right)  $ is a polynomial in $y_{1},y_{2},\ldots,y_{n}$ for each $u\in
S_{n}$, and if $v\in\mathfrak{S}_{n}$ is any permutation, then%
\begin{align*}
\left\langle \sum_{u\in S_{n}}\mathfrak{S}_{u}\left(  x\right)  \alpha
_{u},\ w_{0}\mathfrak{S}_{vw_{0}}\left(  -x\right)  \right\rangle _{x} &
=\sum_{u\in S_{n}}\alpha_{u}\left\langle \mathfrak{S}_{u}\left(  x\right)
,\ w_{0}\underbrace{\mathfrak{S}_{vw_{0}}\left(  -x\right)  }%
_{\substack{=\epsilon\left(  vw_{0}\right)  \mathfrak{S}_{vw_{0}}\left(
x\right)  \\\text{(by homogeneity of }\mathfrak{S}_{vw_{0}}\text{)}%
}}\right\rangle \\
&  =\sum_{u\in S_{n}}\alpha_{u}\underbrace{\left\langle \mathfrak{S}%
_{u}\left(  x\right)  ,\ w_{0}\epsilon\left(  vw_{0}\right)  \mathfrak{S}%
_{vw_{0}}\left(  x\right)  \right\rangle }_{\substack{=\epsilon\left(
w_{0}\right)  \left\langle w_{0}\mathfrak{S}_{u}\left(  x\right)
,\ \epsilon\left(  vw_{0}\right)  \mathfrak{S}_{vw_{0}}\left(  x\right)
\right\rangle \\\text{(by (5.3) (ii), since }w_{0}^{-1}=w_{0}\text{)}}}\\
&  =\sum_{u\in S_{n}}\alpha_{u}\underbrace{\epsilon\left(  w_{0}\right)
\left\langle w_{0}\mathfrak{S}_{u}\left(  x\right)  ,\ \epsilon\left(
vw_{0}\right)  \mathfrak{S}_{vw_{0}}\left(  x\right)  \right\rangle
}_{=\epsilon\left(  v\right)  \left\langle w_{0}\mathfrak{S}_{u}\left(
x\right)  ,\ \mathfrak{S}_{vw_{0}}\left(  x\right)  \right\rangle }\\
&  =\epsilon\left(  v\right)  \sum_{u\in S_{n}}\alpha_{u}%
\underbrace{\left\langle w_{0}\mathfrak{S}_{u}\left(  x\right)
,\ \mathfrak{S}_{vw_{0}}\left(  x\right)  \right\rangle }_{=\epsilon\left(
v\right)  \delta_{u,v}}\\
&  =\sum_{u\in S_{n}}\alpha_{u}\delta_{u,v}=\alpha_{v}.
\end{align*}
Applying this to $\alpha_{u}=T_{u}\left(  y\right)  $ (so that $\sum_{u\in
S_{n}}\mathfrak{S}_{u}\left(  x\right)  \alpha_{u}=\sum_{u\in S_{n}%
}\mathfrak{S}_{u}\left(  x\right)  T_{u}\left(  y\right)  =\Delta\left(
x,y\right)  $) and $v=w$, we conclude that%
\[
\left\langle \Delta\left(  x,y\right)  ,\ w_{0}\mathfrak{S}_{ww_{0}}\left(
-x\right)  \right\rangle _{x}=T_{w}\left(  y\right)  ,
\]
qed.

\item \textbf{Page 75, Remark:} \textquotedblleft if $wu$ is Grassmannian
(with its only descent at $r$)\textquotedblright\ should be \textquotedblleft
if $wu$ is Grassmannian (with its only descent at $r$, or with no descents at
all)\textquotedblright.

\item \textbf{Page 75, Remark:} \textquotedblleft Grassmannian permutations
$v$ with descent at $r$ (i.e. $v\left(  i\right)  <v\left(  i+i\right)  $ if
$i\neq r$)\textquotedblright\ should be \textquotedblleft Grassmannian
permutations $v$ with descent at $r$ or no descent at all (i.e., satisfying
$v\left(  i\right)  <v\left(  i+1\right)  $ for all $i\neq r$%
)\textquotedblright.

\item \textbf{Page 75, Remark:} Let me explain how the \textquotedblleft
easily verified\textquotedblright\ equality%
\[
\partial_{u}\Delta\left(  x,y\right)  =\prod_{i=1}^{r}\ \ \prod_{j=1}%
^{s}\left(  x_{i}-y_{j}\right)
\]
is verified.

We begin by factoring the product $\Delta\left(  x,y\right)  =\prod
\limits_{i+j\leq n}\left(  x_{i}-y_{j}\right)  $ into three subproducts:%
\begin{align*}
\Delta\left(  x,y\right)   &  =ABC,\ \ \ \ \ \ \ \ \ \ \text{where}\\
A &  =\prod\limits_{\substack{i+j\leq n;\\i\leq r\text{ and }j\leq s}}\left(
x_{i}-y_{j}\right)  ,\\
B &  =\prod\limits_{\substack{i+j\leq n;\\i\leq r\text{ and }j>s}}\left(
x_{i}-y_{j}\right)  ,\\
C &  =\prod\limits_{\substack{i+j\leq n;\\i>r\text{ and }j\leq s}}\left(
x_{i}-y_{j}\right)  .
\end{align*}
(The reader might expect a fourth subproduct $\prod\limits_{\substack{i+j\leq
n;\\i>r\text{ and }j>s}}\left(  x_{i}-y_{j}\right)  $, but this subproduct
would be empty, since $i+j>r+s=n$ would contradict $i+j\leq n$.)

Note that%
\[
A=\prod\limits_{\substack{i+j\leq n;\\i\leq r\text{ and }j\leq s}}\left(
x_{i}-y_{j}\right)  =\prod_{i=1}^{r}\ \ \prod_{j=1}^{s}\left(  x_{i}%
-y_{j}\right)
\]
(since the conditions $i\leq r$ and $j\leq s$ entail $i+j\leq r+s=n$ and thus
render the condition $i+j\leq n$ superfluous). This shows that the polynomial
$A$ is symmetric in $x_{1},x_{2},\ldots,x_{r}$ and also symmetric in
$x_{r+1},x_{r+2},x_{r+3},\ldots$ (since the latter variables don't appear in
$A$ at all). But the operator $\partial_{u}$ is linear with respect to such
polynomials (since $u$ has a reduced expression built entirely of $s_{1}%
,s_{2},\ldots,s_{r-1},s_{r+1},s_{r+2},s_{r+3},\ldots$, and thus $\partial_{u}$
is built entirely of the operators $\partial_{1},\partial_{2},\ldots
,\partial_{r-1},\partial_{r+1},\partial_{r+2},\partial_{r+3},\ldots$). Hence,
\[
\partial_{u}\left(  ABC\right)  =A\partial_{u}\left(  BC\right)  .
\]


Next, we observe that $\ell\left(  u\right)  =\ell\left(  w_{0}^{\left(
r\right)  }\times w_{0}^{\left(  s\right)  }\right)  =\ell\left(
w_{0}^{\left(  r\right)  }\right)  +\ell\left(  w_{0}^{\left(  s\right)
}\right)  =\dbinom{r}{2}+\dbinom{s}{2}$. But
\begin{align*}
B &  =\prod\limits_{\substack{i+j\leq n;\\i\leq r\text{ and }j>s}}\left(
x_{i}-y_{j}\right)  \\
&  =\underbrace{\prod\limits_{\substack{i+j\leq n;\\i\leq r\text{ and }%
j>s}}x_{i}}_{=x_{1}^{r-1}x_{2}^{r-2}\cdots x_{r}^{0}}+\left(  \text{terms of
degree }<\dbinom{r}{2}\text{ in the }x\text{-variables}\right)  \\
&  =x_{1}^{r-1}x_{2}^{r-2}\cdots x_{r}^{0}+\left(  \text{terms of degree
}<\dbinom{r}{2}\text{ in the }x\text{-variables}\right)  .
\end{align*}
Similarly,%
\[
C=x_{r+1}^{s-1}x_{r+2}^{s-2}\cdots x_{n}^{0}+\left(  \text{terms of degree
}<\dbinom{s}{2}\text{ in the }x\text{-variables}\right)  .
\]
Multiplying these two equalities, we obtain%
\begin{align*}
BC &  =\left(  x_{1}^{r-1}x_{2}^{r-2}\cdots x_{r}^{0}\right)  \left(
x_{r+1}^{s-1}x_{r+2}^{s-2}\cdots x_{n}^{0}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \text{terms of degree }<\dbinom{r}{2}%
+\dbinom{s}{2}\text{ in the }x\text{-variables}\right)  .
\end{align*}
Applying the map $\partial_{u}$ to this equality, we obtain%
\[
\partial_{u}\left(  BC\right)  =\partial_{u}\left(  \left(  x_{1}^{r-1}%
x_{2}^{r-2}\cdots x_{r}^{0}\right)  \left(  x_{r+1}^{s-1}x_{r+2}^{s-2}\cdots
x_{n}^{0}\right)  \right)  ,
\]
since the operator $\partial_{u}$ lowers degrees by $\ell\left(  u\right)
=\dbinom{r}{2}+\dbinom{s}{2}$ and thus annihilates all terms of degree
$<\dbinom{r}{2}+\dbinom{s}{2}$ in the $x$-variables.

From $u=w_{0}^{\left(  r\right)  }\times w_{0}^{\left(  s\right)  }$, we
easily obtain $\partial_{u}=\partial_{w_{0}^{\left(  r\right)  }}%
\partial_{w_{0}^{\left(  s\right)  }}^{\prime}$, where $\partial
_{w_{0}^{\left(  s\right)  }}^{\prime}$ is like $\partial_{w_{0}^{\left(
s\right)  }}$ but acting on the variables $x_{r+1},x_{r+2},x_{r+3},\ldots$
instead of $x_{1},x_{2},x_{3},\ldots$. Hence,%
\begin{align*}
\partial_{u}\left(  BC\right)   &  =\partial_{w_{0}^{\left(  r\right)  }%
}\partial_{w_{0}^{\left(  s\right)  }}^{\prime}\left(  BC\right)
=\partial_{w_{0}^{\left(  r\right)  }}\left(  B\left(  \partial_{w_{0}%
^{\left(  s\right)  }}^{\prime}C\right)  \right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }B\text{ does not involve}\\
\text{the variables }x_{r+1},x_{r+2},x_{r+3},\ldots\\
\text{and thus can \textquotedblleft pass through\textquotedblright\ }%
\partial_{w_{0}^{\left(  s\right)  }}^{\prime}%
\end{array}
\right)  \\
&  =\left(  \partial_{w_{0}^{\left(  r\right)  }}B\right)  \left(
\partial_{w_{0}^{\left(  s\right)  }}^{\prime}C\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\partial_{w_{0}^{\left(  s\right)  }}^{\prime}C\text{ does not
involve}\\
\text{the variables }x_{1},x_{2},\ldots,x_{r}\\
\text{and thus can \textquotedblleft pass through\textquotedblright\ }%
\partial_{w_{0}^{\left(  r\right)  }}%
\end{array}
\right)  .
\end{align*}


Finally, recall that
\[
B=x_{1}^{r-1}x_{2}^{r-2}\cdots x_{r}^{0}+\left(  \text{terms of degree
}<\dbinom{r}{2}\text{ in the }x\text{-variables}\right)  .
\]
Hence,%
\[
\partial_{w_{0}^{\left(  r\right)  }}B=\partial_{w_{0}^{\left(  r\right)  }%
}\left(  x_{1}^{r-1}x_{2}^{r-2}\cdots x_{r}^{0}\right)
\]
(since the operator $\partial_{w_{0}^{\left(  r\right)  }}$ lowers degrees by
$\ell\left(  w_{0}^{\left(  r\right)  }\right)  =\dbinom{r}{2}$ and thus
annihilates all terms of degree $<\dbinom{r}{2}$ in the $x$-variables). Thus,%
\begin{align*}
\partial_{w_{0}^{\left(  r\right)  }}B &  =\partial_{w_{0}^{\left(  r\right)
}}\left(  x_{1}^{r-1}x_{2}^{r-2}\cdots x_{r}^{0}\right)  \\
&  =\partial_{w_{0}^{\left(  r\right)  }}\left(  x^{\delta^{\left(  r\right)
}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{where }\delta^{\left(  r\right)
}=\left(  r-1,r-2,\ldots,1,0\right)  \right)  \\
&  =s_{\delta^{\left(  r\right)  }-\delta^{\left(  r\right)  }}\left(
x_{1},x_{2},\ldots,x_{r}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(2.11)}\right)  \\
&  =s_{\varnothing}\left(  x_{1},x_{2},\ldots,x_{r}\right)  =1.
\end{align*}
Similarly,%
\[
\partial_{w_{0}^{\left(  s\right)  }}^{\prime}C=1
\]
as well. Altogether, applying the map $\partial_{u}$ to the equality
$\Delta\left(  x,y\right)  =ABC$, we obtain%
\begin{align*}
\partial_{u}\left(  \Delta\left(  x,y\right)  \right)   &  =\partial
_{u}\left(  ABC\right)  =A\underbrace{\partial_{u}\left(  BC\right)
}_{=\left(  \partial_{w_{0}^{\left(  r\right)  }}B\right)  \left(
\partial_{w_{0}^{\left(  s\right)  }}^{\prime}C\right)  }\\
&  =A\underbrace{\left(  \partial_{w_{0}^{\left(  r\right)  }}B\right)  }%
_{=1}\underbrace{\left(  \partial_{w_{0}^{\left(  s\right)  }}^{\prime
}C\right)  }_{=1}=A=\prod_{i=1}^{r}\ \ \prod_{j=1}^{s}\left(  x_{i}%
-y_{j}\right)  .
\end{align*}
This proves the \textquotedblleft easily verified\textquotedblright\ equality.

\item \textbf{Page 76:} The definition of the basis $\left(  \mathfrak{S}%
^{w}\right)  _{w\in S_{n}}$ as the \textquotedblleft$\Lambda_{n}$-basis of
$P_{n}$ dual to the basis $\left(  \mathfrak{S}_{w}\right)  _{w\in S_{n}}$
relative to the scalar product (5.2)\textquotedblright\ tacitly relies on the
fact that the scalar product (5.2) is nondegenerate. Arguably, this is an easy
consequence of (5.5). Still, it would perhaps be easier to turn the argument
around -- i.e., to define the $\mathfrak{S}^{w}$ by (5.12) instead, and then
show that they form a $\Lambda_{n}$-basis of $P_{n}$ dual to the basis
$\left(  \mathfrak{S}_{w}\right)  _{w\in S_{n}}$ relative to the scalar
product (5.2).

\item \textbf{Page 77:} In the displayed equation \textquotedblleft%
$\sum_{\alpha}a_{u\alpha}b_{v\beta}=\delta_{uv}$\textquotedblright, replace
\textquotedblleft$\beta$\textquotedblright\ by \textquotedblleft$\alpha
$\textquotedblright.

\item \textbf{Pages 77--78, proof of (5.15):} I don't understand this proof (I
don't know where the first equality comes from), but here is another:

We have
\begin{equation}
C\left(  x,y\right)  =\prod_{i<j}\left(  y_{i}-x_{j}\right)  =\sum_{w\in
S_{n}}\mathfrak{S}_{w}\left(  y\right)  \mathfrak{S}^{w}\left(  x\right)
\label{eq.77.2}%
\end{equation}
(by (5.13), with $x_{i}$ and $y_{i}$ substituted for $y_{i}$ and $x_{i}$). But
the bases $\left(  \mathfrak{S}^{w}\right)  _{w\in S_{n}}$ and $\left(
\mathfrak{S}_{w}\right)  _{w\in S_{n}}$ of $P_{n}$ are dual; thus, each $f\in
P_{n}$ satisfies%
\begin{equation}
\sum_{w\in S_{n}}\left\langle f,\ \mathfrak{S}^{w}\right\rangle \mathfrak{S}%
_{w}=f.\label{eq.77.3}%
\end{equation}
Furthermore, if $f\in H_{n}$, then all scalar products $\left\langle
f,\ \mathfrak{S}^{w}\right\rangle $ belong to $\mathbf{Z}$ (because (4.11)
shows that $f$ is a linear combination $\sum_{u\in S_{n}}q_{u}\mathfrak{S}%
_{u}$ with $q_{u}\in\mathbf{Z}$, and then we have $\left\langle
f,\ \mathfrak{S}^{w}\right\rangle =q_{w}$ because $\left(  \mathfrak{S}%
^{w}\right)  _{w\in S_{n}}$ and $\left(  \mathfrak{S}_{w}\right)  _{w\in
S_{n}}$ are dual bases). Thus, substituting the variables $y_{i}$ for $x_{i}$
in (\ref{eq.77.3}), we obtain%
\begin{equation}
\sum_{w\in S_{n}}\left\langle f,\ \mathfrak{S}^{w}\right\rangle \mathfrak{S}%
_{w}\left(  y\right)  =f\left(  y\right)  \label{eq.77.4}%
\end{equation}
(for $f\in H_{n}$) However, from (\ref{eq.77.2}), we obtain (again for $f\in
H_{n}$)%
\begin{align*}
\left\langle f\left(  x\right)  ,\ C\left(  x,y\right)  \right\rangle _{x} &
=\left\langle f\left(  x\right)  ,\ \sum_{w\in S_{n}}\mathfrak{S}_{w}\left(
y\right)  \mathfrak{S}^{w}\left(  x\right)  \right\rangle _{x}\\
&  =\sum_{w\in S_{n}}\left\langle f,\ \mathfrak{S}^{w}\right\rangle
\mathfrak{S}_{w}\left(  y\right)  =f\left(  y\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.77.4})}\right)  ,
\end{align*}
which proves (5.15). $\blacksquare$

\item \textbf{Page 78:} I don't understand how the equalities (3) and (4) here
are obtained. To apply (5.15'), we need $\partial_{u}v^{-1}C\left(
x,z\right)  $ to belong to $H_{n}$ (allowing for the $z_{i}$ to be treated as
scalars), but I don't see why this should be true unless $v=w_{0}$.

\item \textbf{Page 79, proof of (5.20):} The first equality (which follows
from (5.18)) is missing an \textquotedblleft$\epsilon\left(  w_{0}\right)
$\textquotedblright\ factor. Or, alternatively, the \textquotedblleft%
$\epsilon\left(  w_{0}\right)  $\textquotedblright\ factor in (5.18) should
not be there (either fix works, but I don't know whether the factor is needed
for further uses of (5.18)).

\item \textbf{Page 92:} \textquotedblleft Recall the decomposition
(4.17)\textquotedblright\ should be \textquotedblleft Recall the decomposition
(4.18)\textquotedblright.

\item \textbf{Page 100, proof of (7.23):} \textquotedblleft precedes
$\lambda\left(  i^{\prime},j^{\prime}\right)  $\textquotedblright\ should be
\textquotedblleft precedes $\left(  i^{\prime},j^{\prime}\right)
$\textquotedblright.

\item \textbf{Page 103:} Replace \textquotedblleft$GL_{n}\left(  k\right)
$\textquotedblright\ by \textquotedblleft$GL_{n}\left(  K\right)
$\textquotedblright\ twice.

\item \textbf{Page 104:} \textquotedblleft it follows that
that\textquotedblright\ $\rightarrow$ \textquotedblleft it follows
that\textquotedblright.

\item \textbf{Page 104, proof of (A.2):} The symbol \textquotedblleft$\supset
$\textquotedblright\ here means \textquotedblleft contains as a proper
subset\textquotedblright. This should be explained, since the symbol is
otherwise used for just \textquotedblleft contains as a
subset\textquotedblright.

\item \textbf{Page 105, proof of (A.2):} On the last line of this proof,
replace \textquotedblleft$\phi\left(  b\mathbf{V}\right)  $\textquotedblright%
\ by \textquotedblleft$\phi\left(  b\mathbf{U}\right)  $\textquotedblright.

\item \textbf{Page 106:} Replace \textquotedblleft$\pi:F\mapsto P\left(
E\right)  $\textquotedblright\ by \textquotedblleft$\pi:F\rightarrow P\left(
E\right)  $\textquotedblright.

\item \textbf{Page 106:} The injectivity of the Pl\"{u}cker embedding $\pi$
follows from the known theorems that:

\begin{enumerate}
\item If two nonzero wedges $v_{1}\wedge v_{2}\wedge\cdots\wedge v_{k}$ and
$w_{1}\wedge w_{2}\wedge\cdots\wedge w_{k}$ in the $k$-th exterior power of a
finite-dimensional vector space are equal, then $\operatorname*{span}\left\{
v_{1},v_{2},\ldots,v_{k}\right\}  =\operatorname*{span}\left\{  w_{1}%
,w_{2},\ldots,w_{k}\right\}  $.

\item If two nonzero pure tensors $v_{1}\otimes v_{2}\otimes\cdots\otimes
v_{k}$ and $w_{1}\otimes w_{2}\otimes\cdots\otimes w_{k}$ in the tensor
product of $k$ finite-dimensional vector spaces are equal, then there exist
scalars $\lambda_{1},\lambda_{2},\ldots,\lambda_{k}$ in the base field such
that $v_{i}=\lambda_{i}w_{i}$ for all $1\leq i\leq k$.
\end{enumerate}

\item \textbf{Page 107, proof of (A.5):} \textquotedblleft Define $\xi
,\ldots,\xi_{n}$\textquotedblright\ should be \textquotedblleft Define
$\xi_{1},\ldots,\xi_{n}$\textquotedblright.

\item \textbf{Page 109:} \textquotedblleft theorm\textquotedblright\ should be
\textquotedblleft theorem\textquotedblright.
\end{enumerate}


\end{document}