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\ihead{Errata to ``Algorithms in Invariant Theory''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Algorithms in Invariant Theory}

\textit{Bernd Sturmfels}

\href{https://www.math.ens.psl.eu/~benoist/refs/Sturmfels.pdf}{Second Edition
2008}

\textbf{Comments and corrections by Darij Grinberg}

\bigskip
\end{center}

\setcounter{section}{6}

The following list contains my comments to Bernd Sturmfels's book
\textit{Algorithms in Invariant Theory}, specifically its 2nd edition
(Springer, 2008). Some of these comments are corrections; others are more
subjective improvements.

I have read the following parts of the book: Chapter 1, the elementary parts
of Chapter 2 (\S 2.1, \S 2.2 until Example 2.2.4, \S 2.4), most of Chapter 3
(\S 3.1, \S 3.2, \S 3.3 until Algorithm 3.3.4, \S 3.6, \S 3.7 sans the second
proof of Proposition 3.7.4 and until Lemma 3.7.6), and the early bits of
Chapter 4 (\S 4.1, \S 4.3).

\section{Errata and comments}

\begin{enumerate}
\item \textbf{page 4, proof of Proposition 1.1.2:} It should be said that
$p_{0}$ is to be understood as $1$. (Otherwise, \textquotedblleft$p_{i_{1}%
}p_{i_{2}}\ldots p_{i_{n}}$\textquotedblright\ might not make sense.)

\item \textbf{page 5, Proposition 1.1.3:} This proposition requires $n>1$.

\item \textbf{page 5, proof of Proposition 1.1.3:} After \textquotedblleft and
therefore $D$ divides $\widetilde{h}$.\textquotedblright, add
\textquotedblleft Furthermore, the polynomial $\widetilde{h}/D$ is symmetric,
since each permutation $\sigma\in S_{n}$ satisfies%
\begin{align*}
\left(  \widetilde{h}/D\right)  \left(  x_{\sigma_{1}},\ldots,x_{\sigma_{n}%
}\right)   &  =\dfrac{\widetilde{h}\left(  x_{\sigma_{1}},\ldots,x_{\sigma
_{n}}\right)  }{D\left(  x_{\sigma_{1}},\ldots,x_{\sigma_{n}}\right)  }%
=\dfrac{\operatorname*{sign}\left(  \sigma\right)  \cdot\widetilde{h}\left(
x_{1},\ldots,x_{n}\right)  }{\operatorname*{sign}\left(  \sigma\right)  \cdot
D\left(  x_{1},\ldots,x_{n}\right)  }\\
&  =\dfrac{\widetilde{h}\left(  x_{1},\ldots,x_{n}\right)  }{D\left(
x_{1},\ldots,x_{n}\right)  }=\widetilde{h}/D.
\end{align*}
\textquotedblright.

\item \textbf{page 6:} After \textquotedblleft The polynomials $a_{\lambda}$
are precisely the nonzero images of monomials under
antisymmetrization\textquotedblright, add \textquotedblleft(up to
sign)\textquotedblright.

\item \textbf{page 6:} The definition of \textquotedblleft
antisymmetrization\textquotedblright\ here uses the notion of
\textquotedblleft canonical projection\textquotedblright, which is not
explained here. Better to say that the \emph{antisymmetrization} of a
polynomial $f\left(  x_{1},\ldots,x_{n}\right)  $ is defined to be
$\sum_{\sigma\in S_{n}}\operatorname*{sign}\left(  \sigma\right)  \cdot
f\left(  x_{\sigma_{1}},\ldots,x_{\sigma_{n}}\right)  $.

\item \textbf{page 6:} The word \textquotedblleft
discriminant\textquotedblright, as it is used here, simply means the
polynomial $D=D\left(  x_{1},\ldots,x_{n}\right)  $. This should be clarified.

\item \textbf{page 6, Corollary 1.1.4:} It should be explained that
\textquotedblleft partitions of $d$ into at most $n$ parts\textquotedblright%
\ means \textquotedblleft partitions of $d$ with $n$ entries (not necessarily
all positive)\textquotedblright.

\item \textbf{page 8:} \textquotedblleft The largest monomial of a
polynomial\textquotedblright\ $\rightarrow$ \textquotedblleft The largest
monomial of a nonzero polynomial\textquotedblright.

\item \textbf{page 8:} \textquotedblleft by the initial monomials of all
polynomials\textquotedblright\ $\rightarrow$ \textquotedblleft by the initial
monomials of all nonzero polynomials\textquotedblright.

\item \textbf{page 8, Example 1.2.1:} \textquotedblleft versus a Gr\"{o}bner
bases\textquotedblright\ $\rightarrow$ \textquotedblleft versus a Gr\"{o}bner
basis\textquotedblright.

\item \textbf{page 11, proof of Theorem 1.2.4:} \textquotedblleft By Corollary
1.2.3, the set of initial monomials $\left\{  \operatorname*{init}\left(
f\right)  :f\in I\setminus\left\{  \mathcal{G}\right\}  \right\}  $ has a
minimal element $\operatorname*{init}\left(  f_{0}\right)  $ with respect to
\textquotedblleft$\prec$\textquotedblright\textquotedblright: In my opinion,
this is not obvious enough to be left unexplained, at least for a reader who
has not already seen this kind of (highly non-constructive) argument before.
The argument being made here is the following: The set of initial monomials
$\left\{  \operatorname*{init}\left(  f\right)  :f\in I\setminus\left\langle
\mathcal{G}\right\rangle \right\}  $ is nonempty (since we assumed that
$I\setminus\left\langle \mathcal{G}\right\rangle $ is nonempty). If this set
had no minimal element (with respect to \textquotedblleft$\prec$%
\textquotedblright), then we could recursively construct an infinite chain
$m_{1}\succ m_{2}\succ m_{3}\succ\cdots$ of monomials in this set by picking
$m_{1}$ arbitrary (this is possible, since the set $\left\{
\operatorname*{init}\left(  f\right)  :f\in I\setminus\left\langle
\mathcal{G}\right\rangle \right\}  $ is nonempty) and then choosing each
further element $m_{i}$ to be smaller than $m_{i-1}$ (this is possible, since
$m_{i-1}$ is not minimal in this set, because this set has no minimal
element). But such a chain would contradict Corollary 1.2.3. Hence, the set
$\left\{  \operatorname*{init}\left(  f\right)  :f\in I\setminus\left\langle
\mathcal{G}\right\rangle \right\}  $ must have a minimal element after all.

\item \textbf{page 12, proof of Theorem 1.2.7:} Here is a proof of the
identity%
\begin{equation}
h_{k}\left(  x_{k},\ldots,x_{n}\right)  +\sum_{i=1}^{k}\left(  -1\right)
^{i}h_{k-i}\left(  x_{k},\ldots,x_{n}\right)  \sigma_{i}\left(  x_{1}%
,\ldots,x_{k-1},x_{k},\ldots,x_{n}\right)  =0. \label{p12.id}%
\end{equation}


\textit{Proof of (\ref{p12.id}):} We work in the polynomial ring
$\mathbf{C}\left[  x_{1},x_{2},\ldots,x_{n}\right]  $ (actually, $\mathbf{C}$
could be replaced by any base ring here). Let $\mathcal{P}\left(  n\right)  $
denote the set of all subsets of $\left\{  1,2,\ldots,n\right\}  $. We define
the complete homogeneous symmetric polynomials%
\begin{equation}
h_{i}\left(  x_{k..n}\right)  :=\left(  \text{sum of all monomials in }%
x_{k},x_{k+1},\ldots,x_{n}\text{ that have degree }i\right)  \label{p12.h=}%
\end{equation}
for all $i\in\mathbf{N}$ (so that $h_{0}\left(  x_{k..n}\right)  =1$ because
the only monomial of degree $0$ is $1$) and the elementary symmetric
polynomials%
\begin{align}
\sigma_{i}\left(  x_{1..n}\right)  :=  &  \sum_{j_{1}<j_{2}<\cdots<j_{i}\text{
in }\left\{  1,2,\ldots,n\right\}  }x_{j_{1}}x_{j_{2}}\cdots x_{j_{i}%
}\nonumber\\
=  &  \sum_{\substack{S\in\mathcal{P}\left(  n\right)  ;\\\left\vert
S\right\vert =i}}\ \ \prod_{s\in S}x_{s} \label{p12.e=}%
\end{align}
for all $i\in\mathbf{N}$ (so that $\sigma_{0}\left(  x_{1..n}\right)  =1$
because the only $0$-element subset of $\left\{  1,2,\ldots,n\right\}  $ is
$\varnothing$). Note that these polynomials $h_{i}\left(  x_{k..n}\right)  $
and $\sigma_{i}\left(  x_{1..n}\right)  $ are denoted by $h_{i}\left(
x_{k},\ldots,x_{n}\right)  $ and $\sigma_{i}\left(  x_{1},\ldots,x_{k-1}%
,x_{k},\ldots,x_{n}\right)  $ in the book. Thus, the identity (\ref{p12.id}),
which we must prove, can be rewritten as follows:%
\begin{equation}
h_{k}\left(  x_{k..n}\right)  +\sum_{i=1}^{k}\left(  -1\right)  ^{i}%
h_{k-i}\left(  x_{k..n}\right)  \sigma_{i}\left(  x_{1..n}\right)  =0.
\label{p12.id2}%
\end{equation}
But we can simplify this identity even further. Namely, we observe that%
\begin{align*}
&  \sum_{i=0}^{k}\left(  -1\right)  ^{i}h_{k-i}\left(  x_{k..n}\right)
\sigma_{i}\left(  x_{1..n}\right) \\
&  =\underbrace{\left(  -1\right)  ^{0}}_{=1}\underbrace{h_{k-0}\left(
x_{k..n}\right)  }_{=h_{k}\left(  x_{k..n}\right)  }\underbrace{\sigma
_{0}\left(  x_{1..n}\right)  }_{=1}+\sum_{i=1}^{k}\left(  -1\right)
^{i}h_{k-i}\left(  x_{k..n}\right)  \sigma_{i}\left(  x_{1..n}\right) \\
&  =h_{k}\left(  x_{k..n}\right)  +\sum_{i=1}^{k}\left(  -1\right)
^{i}h_{k-i}\left(  x_{k..n}\right)  \sigma_{i}\left(  x_{1..n}\right)  .
\end{align*}
Hence, the identity (\ref{p12.id2}), which we must prove, can be rewritten as%
\begin{equation}
\sum_{i=0}^{k}\left(  -1\right)  ^{i}h_{k-i}\left(  x_{k..n}\right)
\sigma_{i}\left(  x_{1..n}\right)  =0. \label{p12.id3}%
\end{equation}
It is this latter identity (\ref{p12.id3}) that we shall now prove; the
original identity (\ref{p12.id}) will then follow.

Let $\mathfrak{M}$ be the set of all monomials in the variables $x_{k}%
,x_{k+1},\ldots,x_{n}$. Thus, we can restate the definition (\ref{p12.h=}) of
$h_{i}\left(  x_{k..n}\right)  $ as follows: For all $i\in\mathbf{N}$, we have%
\begin{equation}
h_{i}\left(  x_{k..n}\right)  =\sum_{\substack{\mathfrak{m}\in\mathfrak{M}%
;\\\deg\mathfrak{m}=i}}\mathfrak{m}. \label{p12.h=2}%
\end{equation}


For any subset $S$ of $\left[  n\right]  $, let $x_{S}$ denote the monomial
$\prod_{s\in S}x_{s}$. Thus, we can restate the definition (\ref{p12.e=}) of
$\sigma_{i}\left(  x_{1..n}\right)  $ as follows: For all $i\in\mathbf{N}$, we
have%
\begin{equation}
\sigma_{i}\left(  x_{1..n}\right)  =\sum_{\substack{S\in\mathcal{P}\left(
n\right)  ;\\\left\vert S\right\vert =i}}x_{S}. \label{p12.e=2}%
\end{equation}


Thus,%
\begin{align}
&  \sum_{i=0}^{k}\left(  -1\right)  ^{i}\underbrace{h_{k-i}\left(
x_{k..n}\right)  }_{\substack{=\sum_{\substack{\mathfrak{m}\in\mathfrak{M}%
;\\\deg\mathfrak{m}=k-i}}\mathfrak{m}\\\text{(by (\ref{p12.h=2}), applied to
}k-i\\\text{instead of }i\text{)}}}\ \ \underbrace{\sigma_{i}\left(
x_{1..n}\right)  }_{\substack{=\sum_{\substack{S\in\mathcal{P}\left(
n\right)  ;\\\left\vert S\right\vert =i}}x_{S}\\\text{(by (\ref{p12.e=2}))}%
}}\nonumber\\
&  =\sum_{i=0}^{k}\left(  -1\right)  ^{i}\left(  \sum_{\substack{\mathfrak{m}%
\in\mathfrak{M};\\\deg\mathfrak{m}=k-i}}\mathfrak{m}\right)  \sum
_{\substack{S\in\mathcal{P}\left(  n\right)  ;\\\left\vert S\right\vert
=i}}x_{S}\nonumber\\
&  =\sum_{i=0}^{k}\ \ \sum_{\substack{\mathfrak{m}\in\mathfrak{M}%
;\\\deg\mathfrak{m}=k-i}}\ \ \sum_{\substack{S\in\mathcal{P}\left(  n\right)
;\\\left\vert S\right\vert =i}}\underbrace{\left(  -1\right)  ^{i}%
}_{\substack{=\left(  -1\right)  ^{\left\vert S\right\vert }\\\text{(since
}i=\left\vert S\right\vert \text{)}}}\mathfrak{m}x_{S}\nonumber\\
&  =\sum_{i=0}^{k}\ \ \sum_{\substack{\mathfrak{m}\in\mathfrak{M}%
;\\\deg\mathfrak{m}=k-i}}\ \ \sum_{\substack{S\in\mathcal{P}\left(  n\right)
;\\\left\vert S\right\vert =i}}\left(  -1\right)  ^{\left\vert S\right\vert
}\mathfrak{m}x_{S}. \label{p12.trisum}%
\end{align}
However, the triple summation sign $\sum_{i=0}^{k}\ \ \sum
_{\substack{\mathfrak{m}\in\mathfrak{M};\\\deg\mathfrak{m}=k-i}}\ \ \sum
_{\substack{S\in\mathcal{P}\left(  n\right)  ;\\\left\vert S\right\vert =i}}$
can be rewritten as a single summation sign $\sum_{\substack{\left(
\mathfrak{m},S\right)  \in\mathfrak{M}\times\mathcal{P}\left(  n\right)
;\\\deg\mathfrak{m}+\left\vert S\right\vert =k}}$, because the conditions
$\deg\mathfrak{m}=k-i$ and $\left\vert S\right\vert =i$ add up to
$\deg\mathfrak{m}+\left\vert S\right\vert =\left(  k-i\right)  +i=k$ (and
conversely, $\deg\mathfrak{m}+\left\vert S\right\vert =k$ implies that
$\deg\mathfrak{m}=k-i$ and $\left\vert S\right\vert =i$ for a unique $i$).
Hence, we can rewrite (\ref{p12.trisum}) as%
\begin{align}
&  \sum_{i=0}^{k}\left(  -1\right)  ^{i}h_{k-i}\left(  x_{k..n}\right)
\sigma_{i}\left(  x_{1..n}\right) \nonumber\\
&  =\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{M}%
\times\mathcal{P}\left(  n\right)  ;\\\deg\mathfrak{m}+\left\vert S\right\vert
=k}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}.
\label{p12.sisum}%
\end{align}


Now, for any monomial $\mathfrak{m}$ in the variables $x_{1},x_{2}%
,\ldots,x_{n}$, let $\operatorname*{Supp}\mathfrak{m}$ denote the set of all
$i\in\left\{  1,2,\ldots,n\right\}  $ such that the variable $x_{i}$ appears
in $\mathfrak{m}$ (that is, such that $\mathfrak{m}$ is divisible by $x_{i}$
as a monomial). Clearly, if $\mathfrak{m}\in\mathfrak{M}$, then
$\operatorname*{Supp}\mathfrak{m}\subseteq\left\{  k,k+1,\ldots,n\right\}  $
(since $\mathfrak{m}\in\mathfrak{M}$ means that $\mathfrak{m}$ is a monomial
in $x_{k},x_{k+1},\ldots,x_{n}$ only). Furthermore, for any $S\in
\mathcal{P}\left(  n\right)  $, we have $\operatorname*{Supp}\left(
x_{S}\right)  =S$. More generally, for any $\mathfrak{m}\in\mathfrak{M}$ and
$S\in\mathcal{P}\left(  n\right)  $, we have%
\begin{equation}
\operatorname*{Supp}\left(  \mathfrak{m}x_{S}\right)  =S\cup
\operatorname*{Supp}\mathfrak{m} \label{p12.Suppmxs}%
\end{equation}
(and more generally, we have $\operatorname*{Supp}\left(  \mathfrak{mn}%
\right)  =\operatorname*{Supp}\mathfrak{m}\cup\operatorname*{Supp}%
\mathfrak{n}$ for any two monomials $\mathfrak{m}$ and $\mathfrak{n}$).

It is easy to see that for every pair $\left(  \mathfrak{m},S\right)
\in\mathfrak{M}\times\mathcal{P}\left(  n\right)  $ satisfying $\deg
\mathfrak{m}+\left\vert S\right\vert =k$, we have
\begin{equation}
\max\left(  \operatorname*{Supp}\left(  \mathfrak{m}x_{S}\right)  \right)
\in\left\{  k,k+1,\ldots,n\right\}  \label{p12.maxSupp}%
\end{equation}
\footnote{\textit{Proof of (\ref{p12.maxSupp}):} Let $\left(  \mathfrak{m}%
,S\right)  \in\mathfrak{M}\times\mathcal{P}\left(  n\right)  $ be a pair
satisfying $\deg\mathfrak{m}+\left\vert S\right\vert =k$. We must prove that
$\max\left(  \operatorname*{Supp}\left(  \mathfrak{m}x_{S}\right)  \right)
\in\left\{  k,k+1,\ldots,n\right\}  $. In other words, we must prove that
$\max\left(  S\cup\operatorname*{Supp}\mathfrak{m}\right)  \in\left\{
k,k+1,\ldots,n\right\}  $ (since (\ref{p12.Suppmxs}) says that
$\operatorname*{Supp}\left(  \mathfrak{m}x_{S}\right)  =S\cup
\operatorname*{Supp}\mathfrak{m}$).
\par
We are in one of the following two cases:
\par
\begin{itemize}
\item \textit{Case 1:} We have $\mathfrak{m}=1$. Then, $\deg\mathfrak{m}=0$
and thus $k=\underbrace{\deg\mathfrak{m}}_{=0}+\left\vert S\right\vert
=\left\vert S\right\vert $. Hence, $S$ is a $k$-element subset of $\left\{
1,2,\ldots,n\right\}  $ and thus contains at least one of the numbers
$k,k+1,\ldots,n$ (since otherwise, $S$ would be a subset of $\left\{
1,2,\ldots,k-1\right\}  $ and therefore would have at most $k-1$ many
elements). Therefore, the union $S\cup\operatorname*{Supp}\mathfrak{m}$ must
also contain at least one of the numbers $k,k+1,\ldots,n$ (since it contains
any element that $S$ contains). Since $S\cup\operatorname*{Supp}\mathfrak{m}$
is a subset of $\left\{  1,2,\ldots,n\right\}  $, we conclude that
$\max\left(  S\cup\operatorname*{Supp}\mathfrak{m}\right)  \in\left\{
k,k+1,\ldots,n\right\}  $.
\par
\item \textit{Case 2:} We have $\mathfrak{m}\neq1$. Hence, the monomial
$\mathfrak{m}$ is not constant, and thus must contain at least one of the
indeterminates $x_{k},x_{k+1},\ldots,x_{n}$ (since it is a monomial in these
indeterminates). In other words, the set $\operatorname*{Supp}\mathfrak{m}$
contains at least one of the numbers $k,k+1,\ldots,n$. Therefore, the union
$S\cup\operatorname*{Supp}\mathfrak{m}$ must also contain at least one of the
numbers $k,k+1,\ldots,n$ (since it contains any element that
$\operatorname*{Supp}\mathfrak{m}$ contains). Since $S\cup\operatorname*{Supp}%
\mathfrak{m}$ is a subset of $\left\{  1,2,\ldots,n\right\}  $, we conclude
that $\max\left(  S\cup\operatorname*{Supp}\mathfrak{m}\right)  \in\left\{
k,k+1,\ldots,n\right\}  $.
\end{itemize}
\par
Hence, in both of these cases, we have shown that $\max\left(  S\cup
\operatorname*{Supp}\mathfrak{m}\right)  \in\left\{  k,k+1,\ldots,n\right\}
$. This completes the proof of (\ref{p12.maxSupp}).}. Hence, we can split up
the sum on the right hand side of (\ref{p12.sisum}) according to the value of
$\max\left(  \operatorname*{Supp}\left(  \mathfrak{m}x_{S}\right)  \right)  $
as follows:%
\begin{align}
&  \sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{M}%
\times\mathcal{P}\left(  n\right)  ;\\\deg\mathfrak{m}+\left\vert S\right\vert
=k}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}\nonumber\\
&  =\sum_{j=k}^{n}\ \ \sum_{\substack{\left(  \mathfrak{m},S\right)
\in\mathfrak{M}\times\mathcal{P}\left(  n\right)  ;\\\deg\mathfrak{m}%
+\left\vert S\right\vert =k;\\\max\left(  \operatorname*{Supp}\left(
\mathfrak{m}x_{S}\right)  \right)  =j}}\left(  -1\right)  ^{\left\vert
S\right\vert }\mathfrak{m}x_{S}. \label{p12.bisum}%
\end{align}


Now, fix $j\in\left\{  k,k+1,\ldots,n\right\}  $. We shall prove that the sum
\begin{equation}
\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{M}\times
\mathcal{P}\left(  n\right)  ;\\\deg\mathfrak{m}+\left\vert S\right\vert
=k;\\\max\left(  \operatorname*{Supp}\left(  \mathfrak{m}x_{S}\right)
\right)  =j}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}
\label{p12.innersum}%
\end{equation}
is $0$ by breaking it up into two mutually cancelling parts. Namely, we define
the set%
\[
\mathfrak{A}:=\left\{  \left(  \mathfrak{m},S\right)  \in\mathfrak{M}%
\times\mathcal{P}\left(  n\right)  \ \mid\ \deg\mathfrak{m}+\left\vert
S\right\vert =k\text{ and }\max\left(  \operatorname*{Supp}\left(
\mathfrak{m}x_{S}\right)  \right)  =j\right\}  ;
\]
this is the indexing set of our sum (\ref{p12.innersum}).

Now, if $\left(  \mathfrak{m},S\right)  \in\mathfrak{A}$ is a pair satisfying
$j\notin S$, then $x_{j}\mid\mathfrak{m}$ (because $\left(  \mathfrak{m}%
,S\right)  \in\mathfrak{A}$ entails $\max\left(  \operatorname*{Supp}\left(
\mathfrak{m}x_{S}\right)  \right)  =j$, so that $j=\max\left(
\operatorname*{Supp}\left(  \mathfrak{m}x_{S}\right)  \right)  \in
\operatorname*{Supp}\left(  \mathfrak{m}x_{S}\right)  =S\cup
\operatorname*{Supp}\mathfrak{m}$ (by (\ref{p12.Suppmxs})), which yields
$j\in\operatorname*{Supp}\mathfrak{m}$ (because $j\notin S$) and therefore
$x_{j}\mid\mathfrak{m}$), and therefore $\mathfrak{m}/x_{j}\in\mathfrak{M}$
and thus $\left(  \mathfrak{m}/x_{j},S\cup\left\{  j\right\}  \right)
\in\mathfrak{A}$\ \ \ \ \footnote{\textit{Proof.} We must show that $\left(
\mathfrak{m}/x_{j},S\cup\left\{  j\right\}  \right)  \in\mathfrak{M}%
\times\mathcal{P}\left(  n\right)  $ and $\deg\left(  \mathfrak{m}%
/x_{j}\right)  +\left\vert S\cup\left\{  j\right\}  \right\vert =k$ and
$\max\left(  \operatorname*{Supp}\left(  \left(  \mathfrak{m}/x_{j}\right)
x_{S\cup\left\{  j\right\}  }\right)  \right)  =j$.
\par
The first of these three statements is clear because $\mathfrak{m}/x_{j}%
\in\mathfrak{M}$ and $S\cup\left\{  j\right\}  \in\mathcal{P}\left(  n\right)
$. The second statement follows from%
\begin{align*}
\underbrace{\deg\left(  \mathfrak{m}/x_{j}\right)  }_{=\deg\mathfrak{m}%
-1}+\underbrace{\left\vert S\cup\left\{  j\right\}  \right\vert }%
_{\substack{=\left\vert S\right\vert +1\\\text{(since }j\notin S\text{)}}}  &
=\left(  \deg\mathfrak{m}-1\right)  +\left(  \left\vert S\right\vert +1\right)
\\
&  =\deg\mathfrak{m}+\left\vert S\right\vert =k\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  \mathfrak{m},S\right)  \in\mathfrak{A}\right)  .
\end{align*}
It remains to prove the third statement. But $x_{S\cup\left\{  j\right\}
}=x_{S}x_{j}$ (since $j\notin S$) and thus $\left(  \mathfrak{m}/x_{j}\right)
x_{S\cup\left\{  j\right\}  }=\left(  \mathfrak{m}/x_{j}\right)  \left(
x_{S}x_{j}\right)  =\mathfrak{m}x_{S}$. Hence, $\max\left(
\operatorname*{Supp}\left(  \left(  \mathfrak{m}/x_{j}\right)  x_{S\cup
\left\{  j\right\}  }\right)  \right)  =\max\left(  \operatorname*{Supp}%
\left(  \mathfrak{m}x_{S}\right)  \right)  =j$ because of $\left(
\mathfrak{m},S\right)  \in\mathfrak{A}$. This proves the third of the three
statements we needed.
\par
Hence, $\left(  \mathfrak{m}/x_{j},S\cup\left\{  j\right\}  \right)
\in\mathfrak{A}$ follows.}. Thus, we obtain a map%
\begin{align}
\left\{  \left(  \mathfrak{m},S\right)  \in\mathfrak{A}\ \mid\ j\notin
S\right\}   &  \rightarrow\left\{  \left(  \mathfrak{m},S\right)
\in\mathfrak{A}\ \mid\ j\in S\right\}  ,\nonumber\\
\left(  \mathfrak{m},S\right)   &  \mapsto\left(  \mathfrak{m}/x_{j}%
,S\cup\left\{  j\right\}  \right)  . \label{p12.map1}%
\end{align}


Conversely, if $\left(  \mathfrak{m},S\right)  \in\mathfrak{A}$ is a pair
satisfying $j\in S$, then $\left(  \mathfrak{m}x_{j},S\setminus\left\{
j\right\}  \right)  \in\mathfrak{A}$\ \ \ \ \footnote{\textit{Proof.} We must
show that $\left(  \mathfrak{m}x_{j},S\setminus\left\{  j\right\}  \right)
\in\mathfrak{M}\times\mathcal{P}\left(  n\right)  $ and $\deg\left(
\mathfrak{m}x_{j}\right)  +\left\vert S\setminus\left\{  j\right\}
\right\vert =k$ and $\max\left(  \operatorname*{Supp}\left(  \left(
\mathfrak{m}x_{j}\right)  x_{S\setminus\left\{  j\right\}  }\right)  \right)
=j$.
\par
The first of these three statements is clear because $\mathfrak{m}x_{j}%
\in\mathfrak{M}$ (since $j\in\left\{  k,k+1,\ldots,n\right\}  $) and
$S\setminus\left\{  j\right\}  \in\mathcal{P}\left(  n\right)  $. The second
statement follows from%
\begin{align*}
\underbrace{\deg\left(  \mathfrak{m}x_{j}\right)  }_{=\deg\mathfrak{m}%
+1}+\underbrace{\left\vert S\setminus\left\{  j\right\}  \right\vert
}_{\substack{=\left\vert S\right\vert -1\\\text{(since }j\in S\text{)}}}  &
=\left(  \deg\mathfrak{m}+1\right)  +\left(  \left\vert S\right\vert -1\right)
\\
&  =\deg\mathfrak{m}+\left\vert S\right\vert =k\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\left(  \mathfrak{m},S\right)  \in\mathfrak{A}\right)  .
\end{align*}
It remains to prove the third statement. But $x_{S}=x_{S\setminus\left\{
j\right\}  }x_{j}$ (since $j\in S$) and thus $x_{S\setminus\left\{  j\right\}
}=x_{S}/x_{j}$, so that $\left(  \mathfrak{m}x_{j}\right)  x_{S\setminus
\left\{  j\right\}  }=\left(  \mathfrak{m}x_{j}\right)  \left(  x_{S}%
/x_{j}\right)  =\mathfrak{m}x_{S}$. Hence, $\max\left(  \operatorname*{Supp}%
\left(  \left(  \mathfrak{m}x_{j}\right)  x_{S\setminus\left\{  j\right\}
}\right)  \right)  =\max\left(  \operatorname*{Supp}\left(  \mathfrak{m}%
x_{S}\right)  \right)  =j$ because of $\left(  \mathfrak{m},S\right)
\in\mathfrak{A}$. This proves the third of the three statements we needed.
\par
Hence, $\left(  \mathfrak{m}x_{j},S\setminus\left\{  j\right\}  \right)
\in\mathfrak{A}$ follows.}. Thus, we obtain a map%
\begin{align}
\left\{  \left(  \mathfrak{m},S\right)  \in\mathfrak{A}\ \mid\ j\in S\right\}
&  \rightarrow\left\{  \left(  \mathfrak{m},S\right)  \in\mathfrak{A}%
\ \mid\ j\notin S\right\}  ,\nonumber\\
\left(  \mathfrak{m},S\right)   &  \mapsto\left(  \mathfrak{m}x_{j}%
,S\setminus\left\{  j\right\}  \right)  . \label{p12.map2}%
\end{align}


These two maps (\ref{p12.map1}) and (\ref{p12.map2}) are mutually inverse.
Hence, they are bijections.

Now,%
\begin{align}
&  \underbrace{\sum_{\substack{\left(  \mathfrak{m},S\right)  \in
\mathfrak{M}\times\mathcal{P}\left(  n\right)  ;\\\deg\mathfrak{m}+\left\vert
S\right\vert =k;\\\max\left(  \operatorname*{Supp}\left(  \mathfrak{m}%
x_{S}\right)  \right)  =j}}}_{\substack{=\sum_{\left(  \mathfrak{m},S\right)
\in\mathfrak{A}}\\\text{(by the definition of }\mathfrak{A}\text{)}}}\left(
-1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}\nonumber\\
&  =\sum_{\left(  \mathfrak{m},S\right)  \in\mathfrak{A}}\left(  -1\right)
^{\left\vert S\right\vert }\mathfrak{m}x_{S}\nonumber\\
&  =\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{A};\\j\in
S}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}%
+\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{A};\\j\notin
S}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }\left(
\mathfrak{m},S\right)  \in\mathfrak{A}\text{ satisfies }j\in S\text{ or
}j\notin S\text{ but never both}\right) \nonumber\\
&  =\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{A};\\j\notin
S}}\underbrace{\left(  -1\right)  ^{\left\vert S\cup\left\{  j\right\}
\right\vert }}_{\substack{=\left(  -1\right)  ^{\left\vert S\right\vert
+1}\\\text{(since }j\notin S\\\text{entails }\left\vert S\cup\left\{
j\right\}  \right\vert =\left\vert S\right\vert +1\text{)}}}\left(
\mathfrak{m}/x_{j}\right)  \underbrace{x_{S\cup\left\{  j\right\}  }%
}_{\substack{=x_{S}x_{j}\\\text{(since }j\notin S\text{)}}}+\sum
_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{A};\\j\notin
S}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left(  \mathfrak{m}/x_{j},S\cup\left\{
j\right\}  \right)  \text{ for }\left(  \mathfrak{m},S\right) \\
\text{in the first sum, since the map (\ref{p12.map1}) is a bijection}%
\end{array}
\right) \nonumber\\
&  =\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{A};\\j\notin
S}}\underbrace{\left(  -1\right)  ^{\left\vert S\right\vert +1}}_{=-\left(
-1\right)  ^{\left\vert S\right\vert }}\underbrace{\left(  \mathfrak{m}%
/x_{j}\right)  x_{S}x_{j}}_{=\mathfrak{m}x_{S}}+\sum_{\substack{\left(
\mathfrak{m},S\right)  \in\mathfrak{A};\\j\notin S}}\left(  -1\right)
^{\left\vert S\right\vert }\mathfrak{m}x_{S}\nonumber\\
&  =-\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{A};\\j\notin
S}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}%
+\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{A};\\j\notin
S}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}=0.
\label{p12.jsum}%
\end{align}


Forget that we fixed $j$. We thus have proved (\ref{p12.jsum}) for each
$j\in\left\{  k,k+1,\ldots,n\right\}  $. Hence, (\ref{p12.bisum}) becomes%
\[
\sum_{\substack{\left(  \mathfrak{m},S\right)  \in\mathfrak{M}\times
\mathcal{P}\left(  n\right)  ;\\\deg\mathfrak{m}+\left\vert S\right\vert
=k}}\left(  -1\right)  ^{\left\vert S\right\vert }\mathfrak{m}x_{S}=\sum
_{j=k}^{n}\ \ \underbrace{\sum_{\substack{\left(  \mathfrak{m},S\right)
\in\mathfrak{M}\times\mathcal{P}\left(  n\right)  ;\\\deg\mathfrak{m}%
+\left\vert S\right\vert =k;\\\max\left(  \operatorname*{Supp}\left(
\mathfrak{m}x_{S}\right)  \right)  =j}}\left(  -1\right)  ^{\left\vert
S\right\vert }\mathfrak{m}x_{S}}_{\substack{=0\\\text{(by (\ref{p12.jsum}))}%
}}=\sum_{j=k}^{n}0=0.
\]
Thus, (\ref{p12.sisum}) rewrites as%
\begin{equation}
\sum_{i=0}^{k}\left(  -1\right)  ^{i}h_{k-i}\left(  x_{k..n}\right)
\sigma_{i}\left(  x_{1..n}\right)  =0.\nonumber
\end{equation}
This proves (\ref{p12.id3}), and thus proves (\ref{p12.id}) as well.
$\blacksquare$

\item \textbf{pages 12--13, proof of Theorem 1.2.7:} The uniqueness of the
reduced Gr\"{o}bner basis of $I$ is not proved here. But it follows from a
general fact saying that any ideal of a polynomial ring over a field has a
unique reduced Gr\"{o}bner basis (at least unique up to scaling; or literally
unique, if we require the elements of a Gr\"{o}bner basis to have leading
coefficients $1$). For a proof of this fact, see, for example, Satz 2.6.4 in
\href{https://annas-archive.org/md5/18e7f4c79d74ce0f6031ddd5a15ea1dd}{Birgit
Reinert, \textit{Gr\"{o}bnerbasen}, Wintersemester 1996/1997}, or Theorem 5 in
\S 2.7 of \href{https://www.idealsvarietiesalgorithms.org/}{David A. Cox, John
Little, Donal O'Shea, \textit{Ideals, Varieties, and Algorithms}, 5th edition,
Springer 2025}.

\item \textbf{page 13, before the Exercises:} \textquotedblleft monomials
$\sigma_{1}^{i_{1}}\sigma_{2}^{i_{2}}\cdots\sigma_{n}^{i_{n}}$ in the
elementary symmetric polynomial\textquotedblright\ $\rightarrow$
\textquotedblleft monomials $\sigma_{1}^{i_{1}}\sigma_{2}^{i_{2}}\cdots
\sigma_{n}^{i_{n}}$ in the elementary symmetric polynomials\textquotedblright.

\item \textbf{page 17, Example 1.3.4:} \textquotedblleft has distance
$7$\textquotedblright\ should be \textquotedblleft has distance $\sqrt{7}%
$\textquotedblright.

\item \textbf{page 19, \S 1.4:} The \textquotedblleft\textit{monoid defined by
}$\mathcal{A}$\textquotedblright\ is a semigroup, not a monoid (since it does
not contain $\left(  0,0,\ldots,0\right)  $ and thus has no neutral element).

\item \textbf{page 20:} \textquotedblleft The invariant monomials are in
bijection with the elements of the monoid $\mathcal{M}_{\mathcal{A}}%
$\textquotedblright\ should be \textquotedblleft The invariant monomials are
in bijection with the elements of the monoid $\mathcal{M}_{\mathcal{A}}%
\cup\left\{  0\right\}  $\textquotedblright\ (since $\mathcal{M}_{\mathcal{A}%
}$ itself is not a monoid).

\item \textbf{page 20, Lemma 1.4.2:} Replace \textquotedblleft$\in
\mathcal{M}_{\mathcal{A}}$\textquotedblright\ by \textquotedblleft%
$\in\mathcal{M}_{\mathcal{A}}\cup\left\{  0\right\}  $\textquotedblright.

\item \textbf{page 20, proof of Lemma 1.4.2:} Replace \textquotedblleft of the
monoid $\mathcal{M}_{\mathcal{A}}$\textquotedblright\ by \textquotedblleft of
the monoid $\mathcal{M}_{\mathcal{A}}\cup\left\{  0\right\}  $%
\textquotedblright.

\item \textbf{page 21, proof of correctness for Algorithms 1.4.3 and 1.4.4:}
Here it is claimed that \textquotedblleft In each step in the reduction of
$\mathbf{x}^{\beta}$ a monomial reduces to another monomial\textquotedblright.
This tacitly uses the fact that the reduced Gr\"{o}bner basis $\mathcal{G}$ of
$I$ consists of \textquotedblleft monomial differences\textquotedblright%
\ (i.e., of polynomials of the form $\mathfrak{m}-\mathfrak{n}$ where
$\mathfrak{m}$ and $\mathfrak{n}$ are two monomials). To prove this fact, it
suffices to show the same about $\mathcal{G}^{\prime}$ (since $\mathcal{G}%
\subseteq\mathcal{G}^{\prime}$). But $\mathcal{G}^{\prime}$ is the reduced
Gr\"{o}bner basis of an ideal of $\mathbf{C}\left[  t_{0},t_{1},\ldots
,t_{d},x_{1},\ldots,x_{n}\right]  $, and the latter ideal is generated by
\textquotedblleft monomial differences\textquotedblright\ (namely, the
differences $t_{0}t_{1}\cdots t_{d}-1$ and $x_{i}-\prod_{j=1}^{d}t_{j}%
^{a_{ij}}$ for $i=1,2,\ldots,n$, where all negative powers $t_{j}^{a_{i,j}}$
in the latter monomials should be replaced by $\left(  t_{0}t_{1}\cdots
t_{j-1}t_{j+1}t_{j+2}\cdots t_{d}\right)  ^{-a_{i,j}}$). It is easy to see
that if a ideal of a polynomial ring is generated by \textquotedblleft
monomial differences\textquotedblright, then its reduced Gr\"{o}bner basis
(with respect to any monomial order) consists of \textquotedblleft monomial
differences\textquotedblright. Thus, $\mathcal{G}^{\prime}$ consists of
\textquotedblleft monomial differences\textquotedblright, and therefore so
does $\mathcal{G}$.

\item \textbf{page 25, proof of Proposition 2.1.1:} The polynomial $P_{i}$ is
not \textquotedblleft monic\textquotedblright, but very close (its leading
coefficient is $\left(  -1\right)  ^{\left\vert \Gamma\right\vert }$, which is
invertible; thus, the polynomial $\left(  -1\right)  ^{\left\vert
\Gamma\right\vert }P_{i}$ is monic).

\item \textbf{page 25, proof of Proposition 2.1.1:} \textquotedblleft Hence
the invariant subring $\mathbf{C}\left[  \mathbf{x}\right]  ^{\Gamma}$ and the
full polynomial ring $\mathbf{C}\left[  \mathbf{x}\right]  $ have the same
transcendence degree $n$ over the ground field $\mathbf{C}$\textquotedblright:
This argument seems to tacitly use the facts that

\begin{enumerate}
\item the polynomial ring $\mathbf{C}\left[  \mathbf{x}\right]  $ has
transcendence degree $n$ over the ground field $\mathbf{C}$, and

\item if $\mathbf{C}\subseteq A\subseteq B$ is an inclusion of integral
domains such that $B$ is integral over $A$, then the transcendence degrees of
$A$ and $B$ are equal.
\end{enumerate}

I know how to prove the first of these facts (indeed, it would follow from
Lemma 4.7.2 if not for the homogeneity requirement in the latter lemma; but
this homogeneity requirement can be easily removed by introducing an extra
slack variable), but I'm less sure about the second.

\begin{noncompile}
Attempt at proof: Both of these facts can be easily derived (using fraction
fields) from the analogous properties of fields ???

Indeed, Theorem 4.8.4 (ii) in the book
\href{https://link.springer.com/book/10.1007/978-3-030-61595-6}{Antoine
Chambert-Loir, \textit{(Mostly) Commutative Algebra}, Springer 2021} (applied
to $C=L$) shows that if $K\subseteq L$ is a field extension, then any subset
of $L$ that is algebraically independent over $K$ is a subset of a
transcendence basis of $L$ over $K$. Hence, the transcendence degree of $L$
over $K$ as defined in that book is precisely the maximum size of a subset of
$L$ that is algebraically independent over $K$; that is, it is precisely the
transcendence degree of $L$ over $K$ as defined in Sturmfels's book.
Furthermore, Example 4.8.10 in Chambert-Loir's book shows that the polynomial
ring $\mathbf{C}\left[  \mathbf{x}\right]  $ has transcendence degree $n$ over
the ground field $\mathbf{C}$. Finally, if $\mathbf{C}\subseteq A\subseteq B$
is an inclusion of fields such that $B$ is integral (= algebraic) over $A$,
then the transcendence degrees of $A$ and $B$ are equal, since any
transcendence basis of $A$ is also a transcendence basis of $B$.

In order to apply this to integral domains rather than fields, we need to show
that if $A$ is an integral domain that contains a field $K$, then its fraction
field $\operatorname*{Frac}A$ has the same transcendence degree over $K$ as
$A$ does. That is, we need to show that if $\operatorname*{Frac}A$ has $m$
algebraically independent elements for some $m\in\mathbf{N}$, then so does $A$
(since the converse inclusion is obvious).
\end{noncompile}

\item \textbf{page 26, proof of Theorem 2.1.3:} This proof tacitly uses the
fact that if $I\in\mathbf{C}\left[  \mathbf{x}\right]  $ is an invariant of
$\Gamma$, then all homogeneous components of $I$ are invariants of $\Gamma$ as
well. (This is easy, since the action of $\Gamma$ does not change the degree
of a homogeneous polynomial.) This fact allows us to restrict ourselves to
homogeneous invariants.

\item \textbf{page 26, proof of Theorem 2.1.3:} \textquotedblleft Hence there
exist finitely many \textit{homogeneous} invariants $I_{1},I_{2},\ldots,I_{m}$
such that $\mathcal{I}_{\Gamma}=\left\langle I_{1},I_{2},\ldots,I_{m}%
\right\rangle $\textquotedblright. Let me explain in more detail why this
holds: Indeed, this is a consequence of the following general fact:

\begin{statement}
\textit{Fact.} If an ideal $J$ of some commutative ring is finitely generated,
and if $G$ is any generating set of $J$ (not necessarily finite), then there
exists a \textbf{finite} subset of $G$ that already generates $J$.
\end{statement}

(To prove this general fact, pick a finite generating set $K$ of $J$, and
expand each $k\in K$ in terms of the generators in $G$. Only finitely many
elements of $G$ altogether are used in these expansions, and so they form a
finite subset of $G$ that already generates $J$.)

\item \textbf{page 26, proof of Theorem 2.1.3:} \textquotedblleft Since
$I\in\mathcal{I}_{\Gamma}$, we have $I=\sum_{j=1}^{s}f_{j}I_{j}$ for some
homogeneous polynomials $f_{j}\in\mathbf{C}\left[  \mathbf{x}\right]  $ of
degree less than $\deg\left(  I\right)  $\textquotedblright: More precisely,
this follows from the fact that $I\in\mathcal{I}_{\Gamma}$ is homogeneous and
that the polynomials $I_{1},I_{2},\ldots,I_{m}$ are also homogeneous of
positive degrees. Namely, the fact that $I\in\mathcal{I}_{\Gamma}=\left\langle
I_{1},I_{2},\ldots,I_{m}\right\rangle $ shows that $I$ can be written as
$I=\sum_{j=1}^{s}g_{j}I_{j}$ for some polynomials $g_{j}\in\mathbf{C}\left[
\mathbf{x}\right]  $ that are not necessarily homogeneous. But now, projecting
this equality onto the $\deg I$-th homogeneous component of $\mathbf{C}\left[
\mathbf{x}\right]  $, we conclude that $I=\sum_{j=1}^{s}f_{j}I_{j}$, where
$f_{j}$ is the $\left(  \deg I-\deg I_{j}\right)  $-th homogeneous component
of $g_{j}$; and this is the expansion we are looking for.

\item \textbf{page 26:} In \textquotedblleft the remarkable statement that
every \textit{ideal basis} $\left\{  I_{1},\ldots,I_{m}\right\}  $ of
$\mathcal{I}_{\Gamma}$ is automatically an \textit{algebra basis} for
$\mathbf{C}\left[  \mathbf{x}\right]  ^{\Gamma}$\textquotedblright, add the
word \textquotedblleft homogeneous\textquotedblright\ in front of
\textquotedblleft\textit{ideal basis}\textquotedblright\ (though arguably, the
whole statement is probably meant informally, since there is no definition of
\textquotedblleft ideal basis\textquotedblright\ anywhere in the book).

It is worth remarking that in some places (e.g., in Proposition 2.1.5),
\textquotedblleft algebra basis\textquotedblright\ means \textquotedblleft
graded generating set of $\mathbf{C}\left[  \mathbf{x}\right]  ^{\Gamma}%
$\textquotedblright\ (that is, \textquotedblleft generating set consisting of
homogeneous elements\textquotedblright).

\item \textbf{page 26:} \textquotedblleft the finiteness of the group $\Gamma$
has not been used until the last paragraph\textquotedblright\ is not literally
true: The Reynolds operator $\ast$ has already made an appearance in the first
paragraph of the proof.

\item \textbf{page 28, Proposition 2.1.5:} Replace \textquotedblleft$n,p\geq
2$\textquotedblright\ by \textquotedblleft$n,p\geq1$\textquotedblright.

\item \textbf{page 29, Theorem 2.2.1:} Add \textquotedblleft Let
$\Gamma\subset GL\left(  \mathbf{C}^{n}\right)  $ be a finite matrix
group.\textquotedblright\ at the beginning of this theorem.

\item \textbf{page 29, proof of Theorem 2.2.1:} The word \textquotedblleft%
$d$-form\textquotedblright\ means \textquotedblleft homogeneous polynomial of
degree $d$\textquotedblright\ here.

\item \textbf{page 30, proof of Theorem 2.2.1:} When applying Lemma 2.2.2
here, one should be careful: As stated, Lemma 2.2.2 would yield a sum over all
the elements of the group $\left\{  \pi^{\left(  d\right)  }\ \mid\ \pi
\in\Gamma\right\}  $, not over all the elements of $\Gamma$. This difference
sometimes matters, since different $\pi$'s in $\Gamma$ might lead to the same
$\pi^{\left(  d\right)  }$'s. The cleanest way to correct this little
discrepancy is to generalize Lemma 2.2.2 by replacing the subgroup
$\Gamma\subset GL\left(  \mathbf{C}^{n}\right)  $ by an arbitrary finite group
$\Gamma$ that acts on $\mathbf{C}^{n}$. (Of course, $\operatorname*{trace}%
\left(  \pi\right)  $ must then be understood as the trace of the action of
$\pi$ on $\mathbf{C}^{n}$.) If we generalize Lemma 2.2.2 this way, then we can
apply it to the group $\Gamma$ acting on $\mathbf{C}\left[  \mathbf{x}\right]
_{d}^{\Gamma}$ via $\pi\mapsto\pi^{\left(  d\right)  }$, and this immediately
yields%
\[
\dim\left(  \mathbf{C}\left[  \mathbf{x}\right]  _{d}^{\Gamma}\right)
=\dfrac{1}{\left\vert \Gamma\right\vert }\sum_{\pi\in\Gamma}%
\operatorname*{trace}\left(  \pi^{\left(  d\right)  }\right)  =\dfrac
{1}{\left\vert \Gamma\right\vert }\sum_{\pi\in\Gamma}\ \ \sum_{d_{1}%
+\cdots+d_{n}=d}\rho_{\pi,1}^{d_{1}}\cdots\rho_{\pi,n}^{d_{n}},
\]
as desired.

\item \textbf{page 30, Lemma 2.2.3:} Replace \textquotedblleft$\sum
_{n=0}^{\infty}$\textquotedblright\ by \textquotedblleft$\sum_{d=0}^{\infty}%
$\textquotedblright.

\item \textbf{page 31, proof of Example 2.2.4:} \textquotedblleft Molien
series\textquotedblright\ just means \textquotedblleft Hilbert
series\textquotedblright\ here (computed using Molien's formula, i.e., Theorem 2.2.1).

\item \textbf{page 37, \S 2.3:} The claims about Krull dimensions and
h.s.o.p.s made here need some further assumptions. Clearly, the algebra $R$
must be commutative, but even this does not seem to be enough; e.g., the
algebra $\mathbf{C}\left[  x_{1},x_{2},x_{3},\ldots\right]  /\left(  x_{1}%
^{2},x_{2}^{2},x_{3}^{2},\ldots\right)  $ has no algebraically independent
elements at all, but is not finitely generated as a module over its subring
$\mathbf{C}$, so it is \textbf{not} free over a subalgebra generated by an
h.s.o.p., at least not if h.s.o.p.s are defined as they are here.

The claim that the maximal number of algebraically independent elements of
$\mathbf{C}$ is the Krull dimension of $R$ is true when $R$ is finitely
generated as a $\mathbf{C}$-algebra, according to Theorem 5.9 in Kemper's book
(\href{https://doi.org/10.1007/978-3-642-03545-6}{Gregor Kemper, \textit{A
Course in Commutative Algebra}, Springer 2011}).

\item \textbf{page 41, Proposition 2.3.6 (ii):} Replace \textquotedblleft%
$\phi_{\Gamma}\left(  z\right)  $\textquotedblright\ by \textquotedblleft%
$\Phi_{\Gamma}\left(  z\right)  $\textquotedblright.

\item \textbf{pages 44--50, \S 2.4:} For most of the book (and, in particular,
for all of the text before \S 2.4), the group $\Gamma$ has always been acting
from the left on $\mathbf{C}^{n}$, and thus from the right on the polynomial
ring $\mathbf{C}\left[  V\right]  $ (since the polynomials in $\mathbf{C}%
\left[  V\right]  $ are viewed as polynomial functions from $V$ to
$\mathbf{C}$). However, in \S 2.4, the group $\Gamma$ suddenly acts from the
left on $\mathbf{C}\left[  V\right]  $ instead (as witnessed, e.g., in the
notation \textquotedblleft$\sigma f$\textquotedblright\ in Lemma 2.4.2). Here
are two ways how this can be reconciled with the rest of the book:

\begin{itemize}
\item One way is to replace all the \textquotedblleft$\sigma f$%
\textquotedblright s (for $\sigma\in\Gamma$ and $f\in\mathbf{C}\left[
V\right]  $) in \S 2.4 by \textquotedblleft$f\circ\sigma$\textquotedblright s.
(Thus, for example, in the proof of Proposition 2.4.3, each \textquotedblleft%
$\sigma h_{1}$\textquotedblright\ should become a \textquotedblleft$h_{1}%
\circ\sigma$\textquotedblright.) Also, in the proof of Proposition 2.4.3, on
page 46, the computation \textquotedblleft%
\begin{align*}
\pi h_{1}-h_{1}  &  =\sum_{i=1}^{l-1}\left(  \sigma_{1}\ldots\sigma_{i}%
\sigma_{i+1}h_{1}-\sigma_{1}\ldots\sigma_{i}h_{1}\right) \\
&  =\sum_{i=1}^{l-1}\left(  \sigma_{1}\ldots\sigma_{i}\left(  \sigma
_{i+1}h_{1}-h_{1}\right)  \right)  \in\mathcal{I}_{\Gamma}%
\end{align*}
\textquotedblright\ should be replaced by \textquotedblleft%
\begin{align*}
h_{1}\circ\pi-h_{1}  &  =\sum_{i=1}^{l-1}\left(  h_{1}\circ\sigma_{i}%
\sigma_{i+1}\ldots\sigma_{l}-h_{1}\circ\sigma_{i+1}\sigma_{i+2}\ldots
\sigma_{l}\right) \\
&  =\sum_{i=1}^{l-1}\left(  \left(  h_{1}\circ\sigma_{i}-h_{1}\right)
\circ\sigma_{i+1}\sigma_{i+2}\ldots\sigma_{l}\right)  \in\mathcal{I}_{\Gamma}%
\end{align*}
\textquotedblright. This is likely the intended way.

\item An alternative way is to define $\sigma f:=f\circ\sigma^{-1}$ for each
$\sigma\in\Gamma$ and $f\in\mathbf{C}\left[  V\right]  $. This way, the right
$\Gamma$-action on $\mathbf{C}\left[  V\right]  $ is \textquotedblleft
translated\textquotedblright\ into a left $\Gamma$-action on $\mathbf{C}%
\left[  V\right]  $ that carries the same information and has the same
invariants (since a polynomial $f\in\mathbf{C}\left[  V\right]  $ and a group
element $\sigma\in\Gamma$ satisfy $\sigma f=f$ if and only if $f\circ
\sigma^{-1}=f$, that is, if and only if $f\circ\sigma=f$).
\end{itemize}

\item \textbf{page 44, \S 2.4:} The definition of a reflection given here (as
a linear transformation $\pi\in GL\left(  \mathbf{C}^{n}\right)  $ such that
\textquotedblleft precisely one eigenvalue of $\pi$ is not equal to
one\textquotedblright) is somewhat eccentric. In the context of a finite
reflection group, it does its job well; however, viewed in isolation, it does
not agree with any of the commonly used definitions. Normally, one defines a
reflection (or pseudo-reflection) to be a linear transformation $\pi\in
GL\left(  \mathbf{C}^{n}\right)  $ such that $\dim\left(  \operatorname*{Ker}%
\left(  \pi-id\right)  \right)  =n-1$. Some authors also require $\pi$ to be
of finite order, but this is automatically satisfies when $\pi$ is an element
of a finite group. The fact that precisely one eigenvalue of $\pi$ is not
equal to one follows automatically from the condition $\dim\left(
\operatorname*{Ker}\left(  \pi-id\right)  \right)  =n-1$, but the converse
implication again requires the assumption that $\pi$ is an element of a finite
group (this ensures that $\pi$ is diagonalizable, so that all the $n-1$
eigenvalues equal to $1$ cause $\operatorname*{Ker}\left(  \pi-id\right)  $ to
have dimension $n-1$). If not for the latter assumption, the matrix $\left(
\begin{array}
[c]{ccc}%
1 & -1 & 0\\
0 & 1 & 0\\
0 & 0 & -1
\end{array}
\right)  $ would be a \textquotedblleft reflection\textquotedblright\ in the
sense of the book (a linear transformation $\pi\in GL\left(  \mathbf{C}%
^{n}\right)  $ such that precisely one eigenvalue of $\pi$ is not equal to
one), but would satisfy rather little of what a reflection is commonly
expected to satisfy.

\item \textbf{page 45, proof of Lemma 2.4.2:} The use of Hilbert's
Nullstellensatz here is overkill. The only thing needed is the following easy fact:

\begin{statement}
\textit{Fact:} Let $L\in\mathbf{C}\left[  \mathbf{x}\right]  $ be a
homogeneous linear polynomial (i.e., a homogeneous polynomial of degree $1$).
Let $g\in\mathbf{C}\left[  \mathbf{x}\right]  $ be a polynomial such that each
$\mathbf{v}\in\mathbf{C}^{n}$ satisfying $L\left(  \mathbf{v}\right)  =0$
satisfies $g\left(  \mathbf{v}\right)  =0$. Then, $L$ is a divisor of $g$ in
$\mathbf{C}\left[  \mathbf{x}\right]  $.
\end{statement}

(\textit{Proof of the fact:} By an appropriate coordinate transformation
(i.e., composition with some invertible matrix $\pi\in GL\left(
\mathbf{C}^{n}\right)  $), we can ensure that the linear polynomial $L$ is
simply $x_{1}$. Thus, WLOG assume that $L=x_{1}$. Then, the condition
\textquotedblleft each $\mathbf{v}\in\mathbf{C}^{n}$ satisfying $L\left(
\mathbf{v}\right)  =0$ satisfies $g\left(  \mathbf{v}\right)  =0$%
\textquotedblright\ becomes \textquotedblleft$g$ vanishes on each vector whose
$1$-st coordinate is $0$\textquotedblright, that is, \textquotedblleft%
$g\left(  0,x_{2},x_{3},\ldots,x_{n}\right)  =0$\textquotedblright. But this
entails that all monomials in $g$ that have degree $0$ with respect to $x_{1}$
have coefficient $0$, and thus all monomials that actually appear in $g$ must
have $x_{1}$ in some nonzero power. Consequently, $g$ is divisible by $x_{1}$,
that is, by $L$. In other words, $L$ is a divisor of $g$. This proves the fact.)

\item \textbf{page 45, proof of Proposition 2.4.3:} The claim that
\textquotedblleft$\widetilde{h}_{1}\cdot L_{\sigma}\in\mathcal{I}_{\Gamma}%
$\textquotedblright\ at the end of page 45 is not entirely obvious: It
requires showing that $\deg\widetilde{h}_{1}>0$ (since $\mathcal{I}_{\Gamma}$
is the ideal generated by all homogeneous invariants \textbf{of positive
degree}). But this is easy: If $\widetilde{h}_{1}=0$, then $\widetilde{h}%
_{1}\cdot L_{\sigma}=0\in\mathcal{I}_{\Gamma}$ is obvious; if $\widetilde{h}%
_{1}$ is a nonzero constant, then $g_{1}\widetilde{h}_{1}+g_{2}\widetilde{h}%
_{2}+\cdots+g_{m}\widetilde{h}_{m}=0$ entails $g_{1}=-\dfrac{1}{\widetilde{h}%
_{1}}\left(  g_{2}\widetilde{h}_{2}+\cdots+g_{m}\widetilde{h}_{m}\right)
\in\left\langle g_{2},\ldots,g_{m}\right\rangle $, contradicting $g_{1}%
\notin\left\langle g_{2},\ldots,g_{m}\right\rangle $.

\item \textbf{page 46, proof of Theorem 2.4.1 (if-part):} The first sentence
of this proof again uses the following fact (already mentioned above):

\begin{statement}
\textit{Fact.} If an ideal $J$ of some commutative ring is finitely generated,
and if $G$ is any generating set of $J$ (not necessarily finite), then there
exists a \textbf{finite} subset of $G$ that already generates $J$.
\end{statement}

Here this fact is being applied to $J=\mathcal{I}_{\Gamma}$ and $G=\left\{
\text{homogeneous invariants of }\Gamma\right\}  $.

\item \textbf{page 46, proof of Theorem 2.4.1 (if-part):} \textquotedblleft We
need to prove that $m=n$, or, equivalently, that the invariants $f_{1}%
,f_{2},\ldots,f_{m}$ are algebraically independent over $\mathbf{C}%
$\textquotedblright: I don't see why these two claims are equivalent. The
proof given here is showing the second claim (i.e., that the invariants
$f_{1},f_{2},\ldots,f_{m}$ are algebraically independent over $\mathbf{C}$).
The first claim $m=n$ can then be derived from this by asymptotically
comparing the Hilbert series (essentially following the proof of Corollary
2.4.5, but without assuming that the number of generators is $n$%
\ \ \ \ \footnote{That is, we need the following generalization of Corollary
2.4.5:
\par
\begin{statement}
\textit{Corollary 2.4.5'.} Let $\Gamma\subset GL\left(  \mathbf{C}^{n}\right)
$ be a finite matrix group whose invariant ring $\mathbf{C}\left[
\mathbf{x}\right]  ^{\Gamma}$ is generated by $m$ algebraically independent
homogeneous invariants $\theta_{1},\ldots,\theta_{m}$ where $d_{i}:=\deg
\theta_{i}$. Let $r$ be the number of reflections in $\Gamma$. Then,%
\[
m=n\qquad\text{and}\qquad\left\vert \Gamma\right\vert =d_{1}d_{2}\cdots
d_{n}\qquad\text{and}\qquad r=d_{1}+d_{2}+\cdots+d_{n}-n.
\]
\end{statement}
\par
The proof of this generalization is just a slight modification of the proof of
the original Corollary 2.4.5.}).

\item \textbf{page 47, proof of Theorem 2.4.1 (if-part):} \textquotedblleft
Euler's formula\textquotedblright\ is the (easily verified) fact that any
homogeneous polynomial $f\in\mathbf{C}\left[  \mathbf{x}\right]  $ satisfies
$\sum_{s=1}^{n}x_{s}\dfrac{\partial f}{\partial x_{s}}=\left(  \deg f\right)
f$.

\item \textbf{page 47, Lemma 2.4.4:} Talking about \textquotedblleft the
Laurent expansion of the Molien series about $z=1$\textquotedblright, it is
important to keep in mind that the Laurent series is a rational function (not
just a formal power series), by Theorem 2.2.1, and therefore can be expanded
into a Laurent series about any complex number.

\item \textbf{page 49, proof of Theorem 2.4.2 (only-if-part):} The claim that
\textquotedblleft the Jacobian determinant $\det\left(  \partial\theta
_{i}/\partial\psi_{j}\right)  $ is nonzero\textquotedblright\ could use a bit
of explanation. It relies on the so-called \textit{Jacobian criterion}, which
says that $n$ polynomials $p_{1},p_{2},\ldots,p_{n}$ in the polynomial ring
$\mathbf{C}\left[  \mathbf{x}\right]  =\mathbf{C}\left[  x_{1},x_{2}%
,\ldots,x_{n}\right]  $ are algebraically independent over $\mathbf{C}$ if and
only if their Jacobian $\det\left(  \partial p_{i}/\partial x_{j}\right)  $ is
nonzero. Here, this criterion is being applied not to the polynomial ring
$\mathbf{C}\left[  \mathbf{x}\right]  =\mathbf{C}\left[  x_{1},x_{2}%
,\ldots,x_{n}\right]  $ but rather to its subring $\mathbf{C}\left[  \psi
_{1},\psi_{2},\ldots,\psi_{n}\right]  $ (which, too, is a polynomial ring,
since the polynomials $\psi_{1},\psi_{2},\ldots,\psi_{n}$ are algebraically
independent) and to the $n$ polynomials $\theta_{1},\theta_{2},\ldots
,\theta_{n}$ therein.

For a proof of the Jacobian criterion, see, e.g., Theorem 2.2 in the paper
\href{https://doi.org/10.1006/eujc.1993.1022}{Richard Ehrenborg, Gian-Carlo
Rota, \textit{Apolarity and Canonical Forms for Homogeneous Polynomials},
European Journal of Combinatorics \textbf{14}, Issue 3, May 1993, pp.
157--181}. (This proof relies on the fact that if $p_{1},p_{2},\ldots,p_{n}$
are $n$ algebraically independent polynomials in $\mathbf{C}\left[
\mathbf{x}\right]  $, then each of the $\left(  n+1\right)  $-tuples $\left(
x_{i},p_{1},p_{2},\ldots,p_{n}\right)  $ is algebraically dependent. This
follows from Lemma 4.7.2 in Sturmfels's book, after some tweaking to make the
polynomials homogeneous.)

\item \textbf{page 52:} The formula \textquotedblleft$f^{\ast}:=\dfrac
{1}{\left\vert \Gamma\right\vert }\sum\limits_{\sigma\in\Gamma}\sigma\left(
f\right)  $\textquotedblright\ should be \textquotedblleft$f^{\ast}:=\dfrac
{1}{\left\vert \Gamma\right\vert }\sum\limits_{\sigma\in\Gamma}f\circ\sigma
$\textquotedblright\ (since the group $\Gamma$ acts on $\mathbf{C}\left[
\mathbf{x}\right]  $ from the right, not from the left).

\item \textbf{page 53, proof of Lemma 2.5.7:} This is unnecessary. After all,
Lemma 2.5.7 simply follows from Proposition 2.6.4 (applied to $M$ instead of
$I$).

\item \textbf{page 55, proof of Lemma 2.5.11:} In the first displayed
equation, replace \textquotedblleft$\prod_{i=1}^{n}\left(  1-z^{d_{j}}\right)
$\textquotedblright\ by either \textquotedblleft$\prod_{i=1}^{n}\left(
1-z^{d_{i}}\right)  $\textquotedblright\ or \textquotedblleft$\prod_{j=1}%
^{n}\left(  1-z^{d_{j}}\right)  $\textquotedblright.

\item \textbf{page 56:} \textquotedblleft where $\alpha\in$
ranges\textquotedblright\ should be \textquotedblleft where $\alpha$
ranges\textquotedblright.

\item \textbf{page 57, Algorithm 2.5.14:} In step 0, replace \textquotedblleft%
$\Phi\left(  z\right)  $\textquotedblright\ by \textquotedblleft$\Phi_{\Gamma
}\left(  z\right)  $\textquotedblright.

\item \textbf{page 57, Algorithm 2.5.14:} In step 2, replace \textquotedblleft%
$\prod_{i=1}^{n}\left(  1-z^{d_{j}}\right)  $\textquotedblright\ by either
\textquotedblleft$\prod_{i=1}^{n}\left(  1-z^{d_{i}}\right)  $%
\textquotedblright\ or \textquotedblleft$\prod_{j=1}^{n}\left(  1-z^{d_{j}%
}\right)  $\textquotedblright.

\item \textbf{page 59, Algorithm 2.6.2:} \textquotedblleft Gr\"{o}bner basis
$\mathcal{G}_{1}$ for $\mathcal{F}\cup\mathcal{G}_{0}$\textquotedblright%
\ should be \textquotedblleft Gr\"{o}bner basis $\mathcal{G}_{1}$ for
$\left\langle \mathcal{F}\cup\mathcal{G}_{0}\right\rangle $\textquotedblright.

\item \textbf{page 61, proof of Proposition 2.6.4:} This proof can be
significantly simplified, removing the use of the Nullstellensatz:

\textit{Simpler proof of Proposition 2.6.4.} Since $I^{\prime}\subseteq I$, we
clearly have $\operatorname*{Rad}\left(  I^{\prime}\right)  \subseteq
\operatorname*{Rad}\left(  I\right)  $. It remains to show that
$\operatorname*{Rad}\left(  I\right)  \subseteq\operatorname*{Rad}\left(
I^{\prime}\right)  $. For this purpose, it suffices to show that
$I\subseteq\operatorname*{Rad}\left(  I^{\prime}\right)  $ (since this would
entail $\operatorname*{Rad}\left(  I\right)  \subseteq\operatorname*{Rad}%
\left(  \operatorname*{Rad}\left(  I^{\prime}\right)  \right)
=\operatorname*{Rad}\left(  I^{\prime}\right)  $).

So let $f\in I$. We must show that $f\in\operatorname*{Rad}\left(  I^{\prime
}\right)  $. We shall show that $f^{\left\vert \Gamma\right\vert }\in
I^{\prime}$; this will clearly do the trick.

Consider the polynomial $\prod_{\sigma\in\Gamma}\left(  z-f\left(
\sigma\mathbf{x}\right)  \right)  \in\left(  \mathbf{C}\left[  \mathbf{x}%
\right]  \right)  \left[  z\right]  $ in the new indeterminate $z$ over
$\mathbf{C}\left[  \mathbf{x}\right]  $. This is clearly a monic polynomial of
degree $\left\vert \Gamma\right\vert $ in $z$, hence can be written as%
\begin{equation}
\prod_{\sigma\in\Gamma}\left(  z-f\left(  \sigma\mathbf{x}\right)  \right)
=z^{\left\vert \Gamma\right\vert }+\sum_{j=0}^{\left\vert \Gamma\right\vert
-1}p_{j}\left(  \mathbf{x}\right)  z^{j}, \label{p61.prod=}%
\end{equation}
where the $p_{j}\left(  \mathbf{x}\right)  \in\mathbf{C}\left[  \mathbf{x}%
\right]  $ are its coefficients (and are themselves polynomials in
$\mathbf{x}$). Consider these $p_{j}\left(  \mathbf{x}\right)  $. Furthermore,
the polynomial $\prod_{\sigma\in\Gamma}\left(  z-f\left(  \sigma
\mathbf{x}\right)  \right)  $ is invariant under the action of $\Gamma$ (since
the action of $\Gamma$ merely permutes the factors of the product
$\prod_{\sigma\in\Gamma}\left(  z-f\left(  \sigma\mathbf{x}\right)  \right)
$). Hence, all its coefficients $p_{j}\left(  \mathbf{x}\right)  $ are
invariant under $\Gamma$ as well.

Now, let $j\in\left\{  0,1,\ldots,\left\vert \Gamma\right\vert -1\right\}  $
be arbitrary. Then, as we just showed, $p_{j}\left(  \mathbf{x}\right)  $ is
invariant under $\Gamma$. Furthermore, (\ref{p61.prod=}) shows that
$p_{j}\left(  \mathbf{x}\right)  $ equals (up to sign) the $\left(  \left\vert
\Gamma\right\vert -j\right)  $-th elementary symmetric polynomial evaluated at
the $\left\vert \Gamma\right\vert $ many inputs $f\left(  \sigma
\mathbf{x}\right)  $ for $\sigma\in\Gamma$; thus, in particular, it is a
polynomial in these $\left\vert \Gamma\right\vert $ many inputs $f\left(
\sigma\mathbf{x}\right)  $ with no constant term (since $j<\left\vert
\Gamma\right\vert $ entails $\left\vert \Gamma\right\vert -j>0$). Hence,
$p_{j}\left(  \mathbf{x}\right)  $ belongs to the ideal $I$ (since all the
$f\left(  \sigma\mathbf{x}\right)  $ belong to $I$ (because $f$ belongs to
$I$, and because $I$ is $\Gamma$-invariant)). Since $p_{j}\left(
\mathbf{x}\right)  $ is furthermore invariant under $\Gamma$, we thus conclude
that $p_{j}\left(  \mathbf{x}\right)  $ is an invariant in $I$. Hence,
$p_{j}\left(  \mathbf{x}\right)  \in I^{\prime}$ (by the definition of
$I^{\prime}$).

Forget that we fixed $j$. We thus have shown that
\begin{equation}
p_{j}\left(  \mathbf{x}\right)  \in I^{\prime}\ \ \ \ \ \ \ \ \ \ \text{for
each }j\in\left\{  0,1,\ldots,\left\vert \Gamma\right\vert -1\right\}  .
\label{p61.inI'}%
\end{equation}


Now, substituting $f=f\left(  \mathbf{x}\right)  $ for $z$ on both sides of
(\ref{p61.prod=}), we obtain%
\[
\prod_{\sigma\in\Gamma}\left(  f-f\left(  \sigma\mathbf{x}\right)  \right)
=f^{\left\vert \Gamma\right\vert }+\sum_{j=0}^{\left\vert \Gamma\right\vert
-1}p_{j}\left(  \mathbf{x}\right)  f^{j}.
\]
Since the product $\prod_{\sigma\in\Gamma}\left(  f-f\left(  \sigma
\mathbf{x}\right)  \right)  $ is $0$ (because one of the factors of this
product is $f-f\left(  \operatorname*{id}\mathbf{x}\right)  =f-f\left(
\mathbf{x}\right)  =f-f=0$), this can be rewritten as
\[
0=f^{\left\vert \Gamma\right\vert }+\sum_{j=0}^{\left\vert \Gamma\right\vert
-1}p_{j}\left(  \mathbf{x}\right)  f^{j}.
\]
Hence,%
\[
f^{\left\vert \Gamma\right\vert }=-\sum_{j=0}^{\left\vert \Gamma\right\vert
-1}\underbrace{p_{j}\left(  \mathbf{x}\right)  }_{\substack{\in I^{\prime
}\\\text{(by (\ref{p61.inI'}))}}}f^{j}\in I^{\prime}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }I^{\prime}\text{ is an ideal}\right)
,
\]
and therefore $f\in\operatorname*{Rad}\left(  I^{\prime}\right)  $, qed.
$\blacksquare$

\item \textbf{page 63, proof of Proposition 2.6.6:} \textquotedblleft By
Hilbert's Nullstellensatz\textquotedblright\ $\rightarrow$ \textquotedblleft
By Hilbert's Nullstellensatz, this shows that $\mathbf{C}\left[
\mathbf{x}\right]  =I+\mathcal{I}\left(  \Gamma\mathbf{a}\right)  $, where
$\Gamma\mathbf{a}$ is the ideal of polynomials vanishing on the finite set
$\Gamma\mathbf{a}$. Hence, by the Chinese Remainder Theorem, $\mathbf{C}%
\left[  \mathbf{x}\right]  /\left(  I\right)  $\textquotedblright.

(Alternatively, we can avoid the use of Hilbert's Nullstellensatz by defining
a maximal ideal
\[
\mathcal{I}_{\mathbf{b}}:=\left\langle x_{1}-b_{1},\ x_{2}-b_{2}%
,\ \ldots,\ x_{n}-b_{n}\right\rangle
\]
of $\mathbf{C}\left[  \mathbf{x}\right]  $ for each point $\mathbf{b}=\left(
b_{1},b_{2},\ldots,b_{n}\right)  \in\Gamma\mathbf{a}$. Then, any two distinct
points $\mathbf{b},\mathbf{c}\in\Gamma\mathbf{a}$ satisfy $\mathbf{C}\left[
\mathbf{x}\right]  =\mathcal{I}_{\mathbf{b}}+\mathcal{I}_{\mathbf{c}}$, and
furthermore $\Gamma\mathbf{a}\cap\mathcal{V}\left(  I\right)  =\varnothing$
shows that all $\mathbf{b}\in\Gamma\mathbf{a}$ satisfy $\mathbf{C}\left[
\mathbf{x}\right]  =I+\mathcal{I}_{\mathbf{b}}$ (since $\mathbf{b}%
\notin\mathcal{V}\left(  I\right)  $ and thus $I\not \subseteq \mathcal{I}%
_{\mathbf{b}}$). Hence, the Chinese Remainder Theorem can be applied to the
ideals $I,\mathcal{I}_{\mathbf{b}_{1}},\mathcal{I}_{\mathbf{b}_{2}}%
,\ldots,\mathcal{I}_{\mathbf{b}_{m}}$ where $\left\{  \mathbf{b}%
_{1},\mathbf{b}_{2},\ldots,\mathbf{b}_{m}\right\}  :=\Gamma\mathbf{a}$.)

\item \textbf{page 69, Algorithm 2.7.3, step 3:} \textquotedblleft solution
monoid\textquotedblright\ $\rightarrow$ \textquotedblleft solution monoid
$\mathcal{F}\subset\mathbf{N}^{n}$\textquotedblright.

\item \textbf{page 71:} \textquotedblleft This means that each factor group
$\Gamma_{i}/\Gamma_{i+1}$ is cyclic of prime order $p_{i}$\textquotedblright:
This is not what a composition series means. Rather, it is an extra
requirement that you want to impose here.

\item \textbf{page 71, algorithm:} \textquotedblleft It follows from Theorem
2.3.5 that $\mathbf{C}\left[  \mathbf{x}\right]  ^{\Gamma_{i+1}}$ is a free
module of rank $p_{i}$ over $\mathbf{C}\left[  \mathbf{x}\right]  ^{\Gamma
_{i}}$\textquotedblright\ is not true in general. For example, in Example
2.7.6, $\mathbf{C}\left[  \mathbf{x}\right]  ^{\left\{  \operatorname*{id}%
\right\}  }$ is not a free module over $\mathbf{C}\left[  \mathbf{x}\right]
^{\Gamma=A_{3}}$.

\begin{noncompile}
\textbf{page 72, Theorem 2.7.6:} This can also be derived from the well-known
fact that as a representation of $S_{n}$ over the ring $\mathbf{C}\left[
\mathbf{x}\right]  ^{S_{n}}$ of symmetric polynomials, $\mathbf{C}\left[
\mathbf{x}\right]  $ is isomorphic to the regular representation (i.e., to the
group algebra $\mathbf{C}\left[  \mathbf{x}\right]  ^{S_{n}}\left[
S_{n}\right]  $ equipped with the action by left multiplication). This fact is
\end{noncompile}

\item \textbf{page 72, Remark 2.7.7:} The meaning of \textquotedblleft cycle
type $\ell\left(  \sigma\right)  =\left(  \ell_{1},\ell_{2},\ldots,\ell
_{n}\right)  $\textquotedblright\ should be explained (since the notation is
quite nonstandard). It means that $\sigma$ has exactly $\ell_{i}$ cycles of
length $i$ for each $i\in\left\{  1,2,\ldots,n\right\}  $.

\item \textbf{page 81, second paragraph:} What is called a \textquotedblleft
standard tableau\textquotedblright\ here is better known as a
\textquotedblleft transpose semistandard tableau of rectangular
shape\textquotedblright\ to combinatorialists (although invariant theorists
often do call it \textquotedblleft standard tableau\textquotedblright).

\item \textbf{page 81, third paragraph:} In the definition of
\textquotedblleft straightening syzygy\textquotedblright, the condition
\textquotedblleft$\beta_{s}<\gamma_{1}$\textquotedblright\ is understood to be
tautologically true if $\gamma_{1}$ does not exist (i.e., if $s=d$).

\item \textbf{page 82, proof of Theorem 3.1.7:} Lemma 3.1.8 is not entirely
obvious. Its proof requires checking three facts:

\begin{enumerate}
\item[Fact 1:] Sorting each column in an arbitrary tableau $T$ yields a
standard tableau $\widetilde{T}$.

\item[Fact 2:] This resulting tableau $\widetilde{T}$ satisfies
$\operatorname*{init}\phi_{n,d}\left(  \widetilde{T}\right)  =\prod_{i=1}%
^{k}x_{\lambda_{1}^{i}1}x_{\lambda_{2}^{i}2}\cdots x_{\lambda_{d}^{i}d}$.

\item[Fact 3:] This tableau $\widetilde{T}$ is the only standard tableau $S$
such that $\operatorname*{init}\phi_{n,d}\left(  S\right)  =\prod_{i=1}%
^{k}x_{\lambda_{1}^{i}1}x_{\lambda_{2}^{i}2}\cdots x_{\lambda_{d}^{i}d}$.
\end{enumerate}

Fact 1 is a version of the \emph{non-messing-up lemma}, saying that if the
entries of an integer matrix are strictly increasing along the rows, then
sorting each column of the matrix will result in a matrix whose entries are
again strictly increasing along the rows (as well as weakly increasing down
the columns, which is clear because we just sorted them). Proving this is a
nice and easy exercise\footnote{The analogous claim with \textquotedblleft
strictly increasing\textquotedblright\ replaced by \textquotedblleft weakly
increasing\textquotedblright\ is Remark 6.42 in
\href{https://arxiv.org/abs/2008.09862v3}{Darij Grinberg, \textit{Notes on the
combinatorial fundamentals of algebra}, arXiv:2008.09862v3}. The version with
\textquotedblleft strictly increasing\textquotedblright\ can be proved in the
exact same way, or reduced to the \textquotedblleft weakly
increasing\textquotedblright\ version by an appropriate tweak of the matrix
(if we subtract $j$ from each entry in the $j$-th column of a matrix, then its
strictly increasing rows become weakly increasing, while the relative order of
entries in a given column does not change).}. Fact 2 is clear, because if we
write $\widetilde{T}$ as $\widetilde{T}=\left[  \widetilde{\lambda}%
^{1}\right]  \left[  \widetilde{\lambda}^{2}\right]  \cdots\left[
\widetilde{\lambda}^{k}\right]  $, then the standardness of $\widetilde{T}$
(see Fact 1) yields%
\begin{align*}
\operatorname*{init}\phi_{n,d}\left(  \widetilde{T}\right)   &  =\prod
_{i=1}^{k}x_{\widetilde{\lambda}_{1}^{i}1}x_{\widetilde{\lambda}_{2}^{i}%
2}\cdots x_{\widetilde{\lambda}_{d}^{i}d}=\prod_{i=1}^{k}\ \ \prod_{j=1}%
^{d}x_{\widetilde{\lambda}_{j}^{i}j}\\
&  =\prod_{j=1}^{d}\ \ \underbrace{\prod_{i=1}^{k}x_{\widetilde{\lambda}%
_{j}^{i}j}}_{\substack{=\prod_{i=1}^{k}x_{\lambda_{j}^{i}j}\\\text{(since the
entries in the }j\text{-th column of }\widetilde{T}\\\text{are the same as
those in the }j\text{-th column of }T\text{)}}}=\prod_{j=1}^{d}\ \ \prod
_{i=1}^{k}x_{\lambda_{j}^{i}j}\\
&  =\prod_{i=1}^{k}\ \ \prod_{j=1}^{d}x_{\lambda_{j}^{i}j}=\prod_{i=1}%
^{k}x_{\lambda_{1}^{i}1}x_{\lambda_{2}^{i}2}\cdots x_{\lambda_{d}^{i}d}.
\end{align*}
Finally, it remains to prove Fact 3. It suffices to show that different
standard tableaux $S$ induce different monomials $\operatorname*{init}%
\phi_{n,d}\left(  S\right)  $. But this is not hard: If we write a standard
tableau $S$ as $S=\left[  \mu^{1}\right]  \left[  \mu^{2}\right]
\cdots\left[  \mu^{k}\right]  $, then the monomial $\operatorname*{init}%
\phi_{n,d}\left(  S\right)  =\prod_{i=1}^{k}x_{\mu_{1}^{i}1}x_{\mu_{2}^{i}%
2}\cdots x_{\mu_{d}^{i}d}$ uniquely determines which entries lie in which
column of $S$ (namely: for each $j\leq d$, the entries in the $j$-th column of
$S$ are exactly the integers $k$ such that $x_{kd}$ appears in this monomial,
and their multiplicities in the $j$-th column are precisely their
multiplicities in this monomial), and thus uniquely determines $S$ itself
(since $S$ is standard, so the entries in each column appear in increasing
order). In other words, different standard tableaux $S$ induce different
monomials $\operatorname*{init}\phi_{n,d}\left(  S\right)  $. This yields Fact 3.

This all said, Lemma 3.1.8 is not actually used in the proof of Theorem 3.1.7
(though it is used later on). Indeed, in the last sentence of the proof of
Theorem 3.1.7 (\textquotedblleft This is a contradiction to Lemma
3.1.8\textquotedblright), the only thing that is actually being used is that
different standard tableaux $S$ induce different monomials
$\operatorname*{init}\phi_{n,d}\left(  S\right)  $. This is a much simpler
claim than Lemma 3.1.8 (and we have proved it above, during our proof of Fact 3).

\item \textbf{page 88, proof of Lemma 3.2.5:} Replace \textquotedblleft%
$\phi_{n+2d,n}$\textquotedblright\ by \textquotedblleft$\phi_{n+2d,d}%
$\textquotedblright\ twice in this proof.

\item \textbf{page 88, proof of Lemma 3.2.5:} The matrix denoted by
$\operatorname*{Adj}\left(  A\right)  $ and called the \textquotedblleft
adjoint matrix of $A$\textquotedblright\ in this proof is not actually the
adjoint matrix of $A$ in the standard meaning of this word, but rather the
transpose of the adjoint matrix of $A$. However, this does not affect the
argument, since its determinant is the same as that of the actual adjoint matrix.

\item \textbf{page 88, proof of Lemma 3.2.5:} The computation (specifically,
the equality \textquotedblleft$\det\left(  A\right)  ^{p\left(  d-1\right)
}\cdot I\left(  x_{ij}\right)  =\det\left(  \operatorname*{Adj}\left(
A\right)  \right)  ^{p}\cdot I\left(  x_{ij}\right)  $\textquotedblright)
relies on the classical result that $\det\left(  \operatorname*{Adj}\left(
A\right)  \right)  =\left(  \det\left(  A\right)  \right)  ^{d-1}$ (see, for
instance, Theorem 5.12 \textbf{(a)} in
\href{https://arxiv.org/abs/2510.20689v1}{Darij Grinberg, \textit{The trace
Cayley-Hamilton theorem}, arXiv:2510.20689v1}).

\item \textbf{page 89, proof of Theorem 3.2.1:} In the first paragraph of this
proof, the straightening algorithm is applied not to the polynomials%
\begin{align*}
&  \left[  b_{1}b_{2}\cdots b_{d}\right]  ^{p\left(  d-1\right)  }\cdot
I\left(  \left[  a_{1}\ldots a_{j-1}x_{i}a_{j+1}\ldots a_{d}\right]  \right)
\ \ \ \ \ \ \ \ \ \ \text{and}\\
&  \left[  a_{1}a_{2}\cdots a_{d}\right]  ^{p\left(  d-1\right)  }\cdot
I\left(  \left[  b_{1}\ldots b_{j-1}xb_{j+1}\ldots b_{d}\right]  \right)  ,
\end{align*}
but rather to the polynomials%
\begin{align*}
&  I\left(  \left[  a_{1}\ldots a_{j-1}x_{i}a_{j+1}\ldots a_{d}\right]
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
&  I\left(  \left[  b_{1}\ldots b_{j-1}xb_{j+1}\ldots b_{d}\right]  \right)  .
\end{align*}
The factors $\left[  b_{1}b_{2}\cdots b_{d}\right]  ^{p\left(  d-1\right)  }$
and $\left[  a_{1}a_{2}\cdots a_{d}\right]  ^{p\left(  d-1\right)  }$ are then
multiplied onto the resulting expansions. This does not destroy the
standardness of the tableaux in these expansions, because

\begin{itemize}
\item the factor $\left[  b_{1}b_{2}\cdots b_{d}\right]  ^{p\left(
d-1\right)  }$ merely inserts $p\left(  d-1\right)  $ many rows of the form
$\left[  b_{1}b_{2}\cdots b_{d}\right]  $ at the bottom of the standard
tableaux appearing in the expansion of $I\left(  \left[  a_{1}\ldots
a_{j-1}x_{i}a_{j+1}\ldots a_{d}\right]  \right)  $ (and these new rows do not
destroy the standardness of the tableaux, since each of the $b_{j}$'s is
larger than any of the letters $a_{1},a_{2},\ldots,a_{d},x_{1},x_{2}%
,\ldots,x_{n}$ that can appear in the expansion of $I\left(  \left[
a_{1}\ldots a_{j-1}x_{i}a_{j+1}\ldots a_{d}\right]  \right)  $);

\item the factor $\left[  a_{1}a_{2}\cdots a_{d}\right]  ^{p\left(
d-1\right)  }$ merely inserts $p\left(  d-1\right)  $ many rows of the form
$\left[  a_{1}a_{2}\cdots a_{d}\right]  $ at the top of the standard tableaux
appearing in the expansion of $I\left(  \left[  b_{1}\ldots b_{j-1}%
xb_{j+1}\ldots b_{d}\right]  \right)  $ (and these new rows do not destroy the
standardness of the tableaux, since each of the $a_{j}$'s is smaller than any
of the letters $b_{1},b_{2},\ldots,b_{d},x_{1},x_{2},\ldots,x_{n}$ that can
appear in the expansion of $I\left(  \left[  b_{1}\ldots b_{j-1}xb_{j+1}\ldots
b_{d}\right]  \right)  $).
\end{itemize}

\item \textbf{page 89, proof of Theorem 3.2.1:} \textquotedblleft both
polynomials in question are equal to\textquotedblright\ should be
\textquotedblleft both polynomials in question evaluate in $\mathbf{C}\left[
x_{ij}\right]  $ to\textquotedblright.

\item \textbf{page 90, third paragraph:} I think that the order
\textquotedblleft$\prec$\textquotedblright\ defined here is \textbf{not} the
\textquotedblleft diagonal order\textquotedblright\ from Sect. 3.1, but a
close relative of it (the variable order is exactly the opposite one);
fortunately all its important properties are analogous.

\item \textbf{page 91:} In the description of the \textquotedblleft subduction
algorithm\textquotedblright, replace \textquotedblleft$f-f_{1}^{\nu_{1}}\cdots
f_{m}^{\nu_{m}}$\textquotedblright\ by \textquotedblleft$f-cf_{1}^{\nu_{1}%
}\cdots f_{m}^{\nu_{m}}$, where $c$ is the coefficient of
$\operatorname*{init}\nolimits_{\prec}f$ in $f$\textquotedblright\ (don't
forget that the leading coefficient is not necessarily $1$).

\item \textbf{page 95, definition of an extensor:} An \textquotedblleft%
\textit{extensor (of step }$k$\textit{)}\textquotedblright\ should be defined
not as an element of the form $A=a_{1}\vee a_{2}\vee\cdots\vee a_{k}$ for some
$a_{1},\ldots,a_{k}\in V$, but rather as an element of the form $A=\lambda
\cdot a_{1}\vee a_{2}\vee\cdots\vee a_{k}$ for some $a_{1},\ldots,a_{k}\in V$
and $\lambda\in\mathbf{C}$. The scalar factor $\lambda$ does not make a real
difference when $k\geq1$ (since it can be simply incorporated into the $a_{1}$
factor), but it becomes important for $k=0$, where it allows any scalar
(rather than just $1$) to be considered as an extensor. And this is necessary
for Theorem 3.3.2 \textbf{(b)} to be true (since the meet of two extensors of
steps $j$ and $k$ with $j+k=d$ can be any scalar).

\item \textbf{page 96, definition of the meet }$A\wedge B$\textbf{:} It is not
obvious that the meet operation $\wedge$ on $\Lambda\left(  V\right)  $ is
well-defined. To prove this, we need to show that the right hand side of
(3.3.6) is multilinear in the inputs $a_{1},a_{2},\ldots,a_{j},b_{1}%
,b_{2},\ldots,b_{k}$ as well as alternating in $a_{1},a_{2},\ldots,a_{j}$ and
alternating in $b_{1},b_{2},\ldots,b_{k}$ (because then, by the universal
property of the exterior powers $\Lambda^{j}\left(  V\right)  $ and
$\Lambda^{k}\left(  V\right)  $, it will follow that this right hand side is a
function of $a_{1}a_{2}\cdots a_{j}$ and $b_{1}b_{2}\cdots b_{k}$).

The multilinearity is obvious (since the determinant form $\left[  v_{1}%
,v_{2},\ldots,v_{d}\right]  $ is multilinear in its inputs $v_{1},v_{2}%
,\ldots,v_{d}$, and since the wedge product $w_{1}w_{2}\cdots w_{p}$ is
multilinear in its factors $w_{1},w_{2},\ldots,w_{p}$). The alternatingness in
$b_{1},b_{2},\ldots,b_{k}$ is also clear (since the determinant form $\left[
v_{1},v_{2},\ldots,v_{d}\right]  $ is alternating in its inputs $v_{1}%
,v_{2},\ldots,v_{d}$). It remains to show that the right hand side of (3.3.6)
is alternating in $a_{1},a_{2},\ldots,a_{j}$. In other words, we must prove
that if two of $a_{1},a_{2},\ldots,a_{j}$ are equal, then the right hand side
of (3.3.6) is $0$. Actually, it suffices to show that if some $p\in\left\{
1,2,\ldots,j-1\right\}  $ satisfies $a_{p}=a_{p+1}$, then the right hand side
of (3.3.6) is $0$ (because it is well-known that this condition, combined with
the multilinearity of the right hand side of (3.3.6), entails that this right
hand side is alternating\footnote{Let me explicitly state the fact that I am
using here:
\par
\begin{statement}
\textit{Fact:} Let $V$ be a vector space over a field $\mathbf{k}$. Let
$f:V^{j}\rightarrow\mathbf{k}$ be any multilinear form on $V$. If we have%
\[
f\left(  v_{1},v_{2},\ldots,v_{j}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{whenever
some }p\in\left\{  1,2,\ldots,j-1\right\}  \text{ satisfies }v_{p}=v_{p+1},
\]
then $f$ is alternating (i.e., satisfies $f\left(  v_{1},v_{2},\ldots
,v_{j}\right)  =0$ whenever two of the $j$ vectors $v_{1},v_{2},\ldots,v_{j}$
are equal).
\end{statement}
\par
For a proof of this fact, see Lemma 1.6 in
\href{https://www.math.lmu.de/~leeb/lehre/texte/ana3/ana3_multilinalg.pdf}{Bernhard
Leeb, \textit{Some multilinear algebra}, January 25, 2020}, or Corollary 2.9
in \href{https://kconrad.math.uconn.edu/blurbs/linmultialg/extmod.pdf}{Keith
Conrad, \textit{Exterior powers}, 2026}, or Proposition 6.4 in
\href{http://www.seas.upenn.edu/~cis515/linalg-I.pdf}{Jean Gallier and Jocelyn
Quaintance, \textit{Linear Algebra for Computer Vision, Robotics, and Machine
Learning}, October 24, 2025}, or Theorem 6 in
\href{https://jordanbell.info/LaTeX/mathematics/alternating/alternating.pdf}{Jordan
Bell, \textit{Alternating multilinear forms}, August 21, 2018} (but beware
that the last two sources define \textquotedblleft
alternating\textquotedblright\ differently from us).}).

Let me prove this now. Assume that some $p\in\left\{  1,2,\ldots,j-1\right\}
$ satisfies $a_{p}=a_{p+1}$. Consider this $p$. Now, the right hand side of
(3.3.6) is%
\[
\sum_{\sigma}\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma
\left(  1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots
,b_{k}\right]  \cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(
j\right)  },
\]
where the sum is over all permutations $\sigma$ of $\left\{  1,2,\ldots
,j\right\}  $ such that $\sigma\left(  1\right)  <\sigma\left(  2\right)
<\cdots<\sigma\left(  d-k\right)  $ and $\sigma\left(  d-k+1\right)
<\sigma\left(  d-k+2\right)  <\cdots<\sigma\left(  j\right)  $. We shall refer
to these permutations $\sigma$ as the \emph{shuffles}. For any permutation
$\sigma\in S_{j}$ (thus, in particular, for any shuffle $\sigma$), we shall
denote the sets $\left\{  \sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  d-k\right)  \right\}  $ and $\left\{  \sigma\left(
d-k+1\right)  ,\sigma\left(  d-k+2\right)  ,\ldots,\sigma\left(  j\right)
\right\}  $ as $L_{\sigma}$ and $R_{\sigma}$, respectively; we call them the
\emph{left half} and the \emph{right half} of $\sigma$ (although they are not
really \textquotedblleft halves\textquotedblright\ as they usually have
different sizes). Clearly, any $\sigma\in S_{j}$ satisfies $L_{\sigma}\cap
R_{\sigma}=\varnothing$ and $L_{\sigma}\cup R_{\sigma}=\left\{  1,2,\ldots
,d\right\}  $, so that each $i\in\left\{  1,2,\ldots,d\right\}  $ lies in
either $L_{\sigma}$ or $R_{\sigma}$ (but not in both). Hence, we can classify
the permutations $\sigma\in S_{j}$ into four classes:

\begin{itemize}
\item \textit{Class LL:} those that satisfy $p,p+1\in L_{\sigma}$.

\item \textit{Class RR:} those that satisfy $p,p+1\in R_{\sigma}$.

\item \textit{Class LR:} those that satisfy $p\in L_{\sigma}$ and $p+1\in
R_{\sigma}$.

\item \textit{Class RL:} those that satisfy $p\in R_{\sigma}$ and $p+1\in
L_{\sigma}$.
\end{itemize}

Each permutation $\sigma\in S_{j}$ is of exactly one of the four Classes LL,
RR, LR and RL.

If $\sigma\in S_{j}$ is a permutation of Class LL, then
\begin{align}
&  \operatorname*{sign}\left(  \sigma\right)  \underbrace{\left[
a_{\sigma\left(  1\right)  },\ldots,a_{\sigma\left(  d-k\right)  }%
,b_{1},\ldots,b_{k}\right]  }_{\substack{=0\\\text{(since this bracket
contains}\\\text{the equal vectors }a_{p}\text{ and }a_{p+1}\\\text{(because
}p,p+1\in L_{\sigma}=\left\{  \sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  d-k\right)  \right\}  \text{))}}}\cdot\,a_{\sigma\left(
d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\nonumber\\
&  =0. \label{p96.0a}%
\end{align}
If $\sigma\in S_{j}$ is a permutation of Class RR, then
\begin{align}
&  \operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot\underbrace{a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(
j\right)  }}_{\substack{=0\\\text{(since this wedge product contains}%
\\\text{the equal vectors }a_{p}\text{ and }a_{p+1}\\\text{(because }p,p+1\in
R_{\sigma}=\left\{  \sigma\left(  d-k+1\right)  ,\sigma\left(  d-k+2\right)
,\ldots,\sigma\left(  j\right)  \right\}  \text{))}}}\nonumber\\
&  =0. \label{p96.0b}%
\end{align}


Hence, in the sum on the right hand side of (3.3.6), all the addends
corresponding to shuffles $\sigma$ of Class LL and of Class RR are $0$ and can
thus be discarded. The remaining addends correspond to shuffles $\sigma$ of
Class LR and of Class RL. These addends are not (usually) $0$, but rather can
be paired up in such a way that any two addends in a pair cancel out; here is
how this pairing works: Let $s_{p}\in S_{j}$ be the transposition that swaps
$p$ and $p+1$. Clearly, $s_{p}^{2}=\operatorname*{id}$ and
$\operatorname*{sign}\left(  s_{p}\right)  =-1$. If $\sigma$ is a shuffle of
Class LR, then $s_{p}\sigma\in S_{j}$ is a shuffle of Class
RL\footnote{\textit{Proof.} Let $\sigma$ be a shuffle of Class LR. Then,
$s_{p}\sigma$ is obtained from $\sigma$ by swapping the entries $p$ and $p+1$
in the list $\left(  \sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  j\right)  \right)  $. Since $\sigma$ is of Class LR, we
know that
\begin{align*}
p  &  \in L_{\sigma}=\left\{  \sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  d-k\right)  \right\}  \ \ \ \ \ \ \ \ \ \ \text{and}\\
p+1  &  \in R_{\sigma}=\left\{  \sigma\left(  d-k+1\right)  ,\sigma\left(
d-k+2\right)  ,\ldots,\sigma\left(  j\right)  \right\}  .
\end{align*}
In other words, $p$ is one of the first $d-k$ entries of the list $\left(
\sigma\left(  1\right)  ,\sigma\left(  2\right)  ,\ldots,\sigma\left(
j\right)  \right)  $, while $p+1$ is one of the last $j-\left(  d-k\right)  $
entries of this list. When we swap the entries $p$ and $p+1$ in the list
$\left(  \sigma\left(  1\right)  ,\sigma\left(  2\right)  ,\ldots
,\sigma\left(  j\right)  \right)  $, this positionality clearly gets reversed;
thus, we have $p\in R_{s_{p}\sigma}$ and $p+1\in L_{s_{p}\sigma}$. Hence, the
permutation $s_{p}\sigma$ is of Class RL.
\par
Moreover, $\sigma$ is a shuffle, so we have $\sigma\left(  1\right)
<\sigma\left(  2\right)  <\cdots<\sigma\left(  d-k\right)  $ and
$\sigma\left(  d-k+1\right)  <\sigma\left(  d-k+2\right)  <\cdots
<\sigma\left(  j\right)  $. The number $p$ appears in the chain of
inequalities $\sigma\left(  1\right)  <\sigma\left(  2\right)  <\cdots
<\sigma\left(  d-k\right)  $ (since $p\in L_{\sigma}=\left\{  \sigma\left(
1\right)  ,\sigma\left(  2\right)  ,\ldots,\sigma\left(  d-k\right)  \right\}
$), but the number $p+1$ does not (since $p+1\in R_{\sigma}$ and thus
$p+1\notin L_{\sigma}=\left\{  \sigma\left(  1\right)  ,\sigma\left(
2\right)  ,\ldots,\sigma\left(  d-k\right)  \right\}  $). When we swap the
entries $p$ and $p+1$ in the list $\left(  \sigma\left(  1\right)
,\sigma\left(  2\right)  ,\ldots,\sigma\left(  j\right)  \right)  $, the
number $p$ is replaced by $p+1$, but the chain of inequalities $\sigma\left(
1\right)  <\sigma\left(  2\right)  <\cdots<\sigma\left(  d-k\right)  $ remains
valid (indeed, the only inequality that would be invalidated when we swap the
entries $p$ and $p+1$ is the inequality $p<p+1$; but this inequality is not
part of the chain, because $p+1$ does not appear in this chain), and so does
the chain of inequalities $\sigma\left(  d-k+1\right)  <\sigma\left(
d-k+2\right)  <\cdots<\sigma\left(  j\right)  $ (for a similar reason: the
number $p+1$ appears in this chain, but $p$ does not). Thus, when we swap the
entries $p$ and $p+1$ in the list $\left(  \sigma\left(  1\right)
,\sigma\left(  2\right)  ,\ldots,\sigma\left(  j\right)  \right)  $, the
permutation $\sigma$ remains a shuffle. In other words, $s_{p}\sigma$ is a
shuffle.
\par
Hence, $s_{p}\sigma$ is a shuffle of Class RL (since we have shown that
$s_{p}\sigma$ is of Class RL).}. Hence, we can define a map%
\begin{align}
\left\{  \text{shuffles of Class LR}\right\}   &  \rightarrow\left\{
\text{shuffles of Class RL}\right\}  ,\nonumber\\
\sigma &  \mapsto s_{p}\sigma. \label{p96.bij1}%
\end{align}
Likewise, we can define a map%
\begin{align*}
\left\{  \text{shuffles of Class RL}\right\}   &  \rightarrow\left\{
\text{shuffles of Class LR}\right\}  ,\\
\sigma &  \mapsto s_{p}\sigma.
\end{align*}
These two maps are mutually inverse (since $s_{p}\left(  s_{p}\sigma\right)
=\underbrace{s_{p}^{2}}_{=\operatorname*{id}}\sigma=\sigma$ for any $\sigma\in
S_{j}$), and thus are bijections.

However, if we replace a given shuffle $\sigma$ by $s_{p}\sigma$, then the
product%
\begin{equation}
\left[  a_{\sigma\left(  1\right)  },\ldots,a_{\sigma\left(  d-k\right)
},b_{1},\ldots,b_{k}\right]  \cdot a_{\sigma\left(  d-k+1\right)  }\cdots
a_{\sigma\left(  j\right)  } \label{p96.exp}%
\end{equation}
does not change (indeed, $s_{p}\sigma$ is obtained from $\sigma$ by swapping
the values $p$ and $p+1$; but our assumption $a_{p}=a_{p+1}$ ensures that this
swap does not change the vectors $a_{1},a_{2},\ldots,a_{j}$ and therefore the
expression (\ref{p96.exp})). In other words, for any shuffle $\sigma$, we have%
\begin{align}
&  \left[  a_{\left(  s_{p}\sigma\right)  \left(  1\right)  },\ldots
,a_{\left(  s_{p}\sigma\right)  \left(  d-k\right)  },b_{1},\ldots
,b_{k}\right]  \cdot a_{\left(  s_{p}\sigma\right)  \left(  d-k+1\right)
}\cdots a_{\left(  s_{p}\sigma\right)  \left(  j\right)  }\nonumber\\
&  =\left[  a_{\sigma\left(  1\right)  },\ldots,a_{\sigma\left(  d-k\right)
},b_{1},\ldots,b_{k}\right]  \cdot a_{\sigma\left(  d-k+1\right)  }\cdots
a_{\sigma\left(  j\right)  }. \label{p96.rep}%
\end{align}
Hence, for any shuffle $\sigma$, we have%
\begin{align}
&  \underbrace{\operatorname*{sign}\left(  s_{p}\sigma\right)  }%
_{\substack{=\operatorname*{sign}\left(  s_{p}\right)  \cdot
\operatorname*{sign}\left(  \sigma\right)  \\=-\operatorname*{sign}\left(
\sigma\right)  \\\text{(since }\operatorname*{sign}\left(  s_{p}\right)
=-1\text{)}}}\underbrace{\left[  a_{\left(  s_{p}\sigma\right)  \left(
1\right)  },\ldots,a_{\left(  s_{p}\sigma\right)  \left(  d-k\right)  }%
,b_{1},\ldots,b_{k}\right]  \cdot a_{\left(  s_{p}\sigma\right)  \left(
d-k+1\right)  }\cdots a_{\left(  s_{p}\sigma\right)  \left(  j\right)  }%
}_{\substack{=\left[  a_{\sigma\left(  1\right)  },\ldots,a_{\sigma\left(
d-k\right)  },b_{1},\ldots,b_{k}\right]  \cdot a_{\sigma\left(  d-k+1\right)
}\cdots a_{\sigma\left(  j\right)  }\\\text{(by (\ref{p96.rep}))}}}\nonumber\\
&  =-\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }.
\label{p96.rep2}%
\end{align}


Now let us summarize: The right hand side of (3.3.6) is
\begin{align*}
&  \sum_{\sigma\text{ is a shuffle}}\operatorname*{sign}\left(  \sigma\right)
\left[  a_{\sigma\left(  1\right)  },\ldots,a_{\sigma\left(  d-k\right)
},b_{1},\ldots,b_{k}\right]  \cdot a_{\sigma\left(  d-k+1\right)  }\cdots
a_{\sigma\left(  j\right)  }\\
&  =\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of Class LL}%
}}\underbrace{\operatorname*{sign}\left(  \sigma\right)  \left[
a_{\sigma\left(  1\right)  },\ldots,a_{\sigma\left(  d-k\right)  }%
,b_{1},\ldots,b_{k}\right]  \cdot a_{\sigma\left(  d-k+1\right)  }\cdots
a_{\sigma\left(  j\right)  }}_{\substack{=0\\\text{(by (\ref{p96.0a}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of
Class RR}}}\underbrace{\operatorname*{sign}\left(  \sigma\right)  \left[
a_{\sigma\left(  1\right)  },\ldots,a_{\sigma\left(  d-k\right)  }%
,b_{1},\ldots,b_{k}\right]  \cdot a_{\sigma\left(  d-k+1\right)  }\cdots
a_{\sigma\left(  j\right)  }}_{\substack{=0\\\text{(by (\ref{p96.0b}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of
Class RL}}}\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of
Class LR}}}\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each shuffle belongs to exactly one of the}\\
\text{four Classes LL, RR, RL and LR}%
\end{array}
\right) \\
&  =\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of Class RL}%
}}\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of
Class LR}}}\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\\
&  =\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of Class LR}%
}}\underbrace{\operatorname*{sign}\left(  s_{p}\sigma\right)  \left[
a_{\left(  s_{p}\sigma\right)  \left(  1\right)  },\ldots,a_{\left(
s_{p}\sigma\right)  \left(  d-k\right)  },b_{1},\ldots,b_{k}\right]  \cdot
a_{\left(  s_{p}\sigma\right)  \left(  d-k+1\right)  }\cdots a_{\left(
s_{p}\sigma\right)  \left(  j\right)  }}_{\substack{=-\operatorname*{sign}%
\left(  \sigma\right)  \left[  a_{\sigma\left(  1\right)  },\ldots
,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]  \cdot
a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\\\text{(by
(\ref{p96.rep2}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of
Class LR}}}\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }s_{p}\sigma\text{ for }\sigma\text{ in the
first sum,}\\
\text{since the map (\ref{p96.bij1}) is a bijection}%
\end{array}
\right) \\
&  =-\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of Class LR}%
}}\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{\sigma\text{ is a shuffle}\\\text{of
Class LR}}}\operatorname*{sign}\left(  \sigma\right)  \left[  a_{\sigma\left(
1\right)  },\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]
\cdot a_{\sigma\left(  d-k+1\right)  }\cdots a_{\sigma\left(  j\right)  }\\
&  =0.
\end{align*}


This completes the proof of the claim that the right hand side of (3.3.6) is
alternating in $a_{1},a_{2},\ldots,a_{j}$, and thus (ultimately) the proof
that the meet operation on $\Lambda\left(  V\right)  $ is well-defined.
$\blacksquare$

\item \textbf{page 97, proof of Theorem 3.3.2:} I am not sure how part
\textbf{(a)} of the theorem is being proved here (e.g., where is the
associativity of $\wedge$ proved?). Let me instead prove Theorem 3.3.2 using
the Hodge star operation. Here is an outline:

\begin{itemize}
\item Let $\left[  d\right]  :=\left\{  1,2,\ldots,d\right\}  $. For any
subset $J=\left\{  j_{1}<j_{2}<\cdots<j_{p}\right\}  $ of $\left[  d\right]
$, let $e_{J}$ be the extensor $e_{j_{1}}\vee e_{j_{2}}\vee\cdots\vee
e_{j_{p}}\in\Lambda^{p}\left(  V\right)  $. It is known that the family
$\left(  e_{J}\right)  _{J\subseteq\left[  d\right]  }$ is a basis of the
$\mathbf{C}$-vector space $\Lambda\left(  V\right)  $.

In particular, $e_{\varnothing}=1\in\Lambda^{0}\left(  V\right)  $ and
$e_{\left[  d\right]  }=e_{1}\vee e_{2}\vee\cdots\vee e_{d}\in\Lambda
^{d}\left(  V\right)  $.

It is well-known (and easy to see) that any two subsets $U$ and $V$ of
$\left[  d\right]  $ satisfy%
\begin{equation}
e_{U}\vee e_{V}=\left(  -1\right)  ^{\left\vert U\right\vert \cdot\left\vert
V\right\vert }e_{V}\vee e_{U}. \label{p97.eUeV}%
\end{equation}
More generally, any $a\in\Lambda^{k}\left(  V\right)  $ and any $b\in
\Lambda^{l}\left(  V\right)  $ satisfy%
\begin{equation}
a\vee b=\left(  -1\right)  ^{kl}b\vee a. \label{p97.grcomm}%
\end{equation}
Moreover, if $U$ and $V$ are two subsets of $\left[  d\right]  $ that are not
disjoint, then%
\begin{equation}
e_{U}\vee e_{V}=0 \label{p97.eUeV=0}%
\end{equation}
(since there is a common element $r\in U\cap V$, and the corresponding basis
vector $e_{r}$ appears as a factor in both $e_{U}$ and $e_{V}$). On the other
hand, if $U$ and $V$ are two disjoint subsets of $\left[  d\right]  $, then%
\begin{equation}
e_{U}\vee e_{V}=\operatorname*{sign}\left(  \eta\right)  \cdot e_{U\cup V},
\label{p97.eUeV=eUV}%
\end{equation}
where $\eta\in S_{\left\vert U\right\vert +\left\vert V\right\vert }$ is the
permutation that transforms the increasing list\footnote{The \emph{increasing
list} of a finite set $K$ of integers means the list of all elements of $K$ in
increasing order.} of $U\cup V$ into the concatenation of the increasing lists
of $U$ and of $V$.

\item For each subset $J$ of $\left[  d\right]  $, we let $\sigma_{J}\in
S_{d}$ be the permutation that sends the numbers $1,2,\ldots,\left\vert
J\right\vert $ to the elements of $J$ listed in increasing order and sends the
numbers $\left\vert J\right\vert +1,\left\vert J\right\vert +2,\ldots,d$ to
the elements of $\left[  d\right]  \setminus J$ listed in increasing order.
For instance:

\begin{itemize}
\item If $n=5$ and $J=\left\{  2,5\right\}  $, then $\sigma_{J}\in S_{5}$ is
the permutation sending $1,2,3,4,5$ to $2,5,1,3,4$.

\item If $n=4$ and $J=\left\{  2\right\}  $, then $\sigma_{J}\in S_{4}$ is the
permutation sending $1,2,3,4$ to $2,1,3,4$.
\end{itemize}

It is easy to see that any subset $J$ of $\left[  d\right]  $ satisfies
\begin{equation}
e_{J}\vee e_{\left[  d\right]  \setminus J}=\operatorname*{sign}\left(
\sigma_{J}\right)  \cdot e_{\left[  d\right]  } \label{p97.eJe-J}%
\end{equation}
(since $e_{J}\vee e_{\left[  d\right]  \setminus J}$ is the extensor obtained
by multiplying the $e_{j}$ for $j\in J$ in increasing order and then the
$e_{j}$ for $j\in\left[  d\right]  \setminus J$ in increasing order; but this
is just the wedge product $e_{1}\vee e_{2}\vee\cdots\vee e_{d}=e_{\left[
d\right]  }$ with its factors permuted by $\sigma_{J}$).

\item Define the \emph{Hodge star} operation to be the $\mathbf{C}$-linear map
$\star:\Lambda\left(  V\right)  \rightarrow\Lambda\left(  V\right)  $ that
sends each element $e_{J}$ (with $J\subseteq\left[  d\right]  $) to
$\operatorname*{sign}\left(  \sigma_{J}\right)  \cdot e_{\left[  d\right]
\setminus J}$. For instance:

\begin{itemize}
\item If $n=5$ and $J=\left\{  2,5\right\}  $, then $\sigma_{J}\in S_{5}$ is
the permutation sending $1,2,3,4,5$ to $2,5,1,3,4$, and thus we have $\star
e_{\left\{  2,5\right\}  }=\underbrace{\operatorname*{sign}\left(
\sigma_{\left\{  2,5\right\}  }\right)  }_{=1}\cdot\underbrace{e_{\left[
5\right]  \setminus\left\{  2,5\right\}  }}_{=e_{\left\{  1,3,4\right\}  }%
}=e_{\left\{  1,3,4\right\}  }$.

\item If $n=4$ and $J=\left\{  2\right\}  $, then $\sigma_{J}\in S_{4}$ is the
permutation sending $1,2,3,4$ to $2,1,3,4$, and thus we have $\star
e_{\left\{  2\right\}  }=\underbrace{\operatorname*{sign}\left(
\sigma_{\left\{  2\right\}  }\right)  }_{=-1}\cdot\underbrace{e_{\left[
4\right]  \setminus\left\{  2\right\}  }}_{=e_{\left\{  1,3,4\right\}  }%
}=-e_{\left\{  1,3,4\right\}  }$.
\end{itemize}

It is easy to see that every $k\in\mathbf{N}$ and $A\in\Lambda^{k}\left(
V\right)  $ satisfy
\begin{equation}
\star A\in\Lambda^{d-k}\left(  V\right)  \label{p97.*deg}%
\end{equation}
and%
\begin{equation}
\star\left(  \star A\right)  =\left(  -1\right)  ^{k\left(  d-k\right)  }A.
\label{p97.**}%
\end{equation}
(\textit{Proof sketch:} By linearity, it suffices to prove both
(\ref{p97.*deg}) and (\ref{p97.**}) in the case when $A=e_{J}$ for some
$k$-element subset $J$ of $\left[  d\right]  $. So let $J$ be a $k$-element
subset of $\left[  d\right]  $, and let $A=e_{J}$. Then, $\left\vert
J\right\vert =k$, so that $\left\vert \left[  d\right]  \setminus J\right\vert
=d-k$. But $A=e_{J}$, and thus the definition of the Hodge star $\star$ yields
$\star A=\operatorname*{sign}\left(  \sigma_{J}\right)  \cdot e_{\left[
d\right]  \setminus J}\in\Lambda^{d-k}\left(  V\right)  $ (since $\left\vert
\left[  d\right]  \setminus J\right\vert =d-k$); this immediately proves
(\ref{p97.*deg}). It remains to prove (\ref{p97.**}).

From $\star A=\operatorname*{sign}\left(  \sigma_{J}\right)  \cdot e_{\left[
d\right]  \setminus J}$, we obtain%
\begin{align*}
\star\left(  \star A\right)   &  =\star\left(  \operatorname*{sign}\left(
\sigma_{J}\right)  \cdot e_{\left[  d\right]  \setminus J}\right)
=\operatorname*{sign}\left(  \sigma_{J}\right)  \cdot\underbrace{\star
\,e_{\left[  d\right]  \setminus J}}_{\substack{=\operatorname*{sign}\left(
\sigma_{\left[  d\right]  \setminus J}\right)  \cdot e_{\left[  d\right]
\setminus\left(  \left[  d\right]  \setminus J\right)  }\\\text{(by the
definition of }\star\text{)}}}\\
&  =\underbrace{\operatorname*{sign}\left(  \sigma_{J}\right)  \cdot
\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
}_{=\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot\operatorname*{sign}\left(  \sigma_{J}\right)  }\cdot
\underbrace{e_{\left[  d\right]  \setminus\left(  \left[  d\right]  \setminus
J\right)  }}_{\substack{=e_{J}\\\text{(since }\left[  d\right]  \setminus
\left(  \left[  d\right]  \setminus J\right)  =J\text{)}}}\\
&  =\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot\operatorname*{sign}\left(  \sigma_{J}\right)  \cdot\underbrace{e_{J}%
}_{=A}\\
&  =\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot\operatorname*{sign}\left(  \sigma_{J}\right)  \cdot A.
\end{align*}
Hence, in order to complete the proof of (\ref{p97.**}), we only need to show
that%
\begin{equation}
\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot\operatorname*{sign}\left(  \sigma_{J}\right)  =\left(  -1\right)
^{k\left(  d-k\right)  }. \label{p97.**.pf.goal}%
\end{equation}
For this purpose, we let $\eta_{k}\in S_{d}$ be the permutation that sends the
numbers $1,2,\ldots,k$ to $d-k+1,d-k+2,\ldots,d$ and sends the numbers
$k+1,k+2,\ldots,d$ to the numbers $1,2,\ldots,d-k$. Then, it is easy to see
that $\operatorname*{sign}\left(  \eta_{k}\right)  =\left(  -1\right)
^{k\left(  d-k\right)  }$ and $\sigma_{J}=\sigma_{\left[  d\right]  \setminus
J}\eta_{k}$\ \ \ \ \footnote{\textit{Proof.} The permutation $\sigma_{\left[
d\right]  \setminus J}\eta_{k}$ sends the numbers $1,2,\ldots,k$ to the
elements of $J$ in increasing order (because the permutation $\eta_{k}$ sends
the numbers $1,2,\ldots,k$ to the numbers $d-k+1,d-k+2,\ldots,d$, and then the
permutation $\sigma_{\left[  d\right]  \setminus J}$ sends the latter numbers
to the elements of $\left[  d\right]  \setminus\left(  \left[  d\right]
\setminus J\right)  =J$ in increasing order), and furthermore sends the
numbers $k+1,k+2,\ldots,d$ to the elements of $\left[  d\right]  \setminus J$
in increasing order (because the permutation $\eta_{k}$ sends the numbers
$k+1,k+2,\ldots,d$ to the numbers $1,2,\ldots,d-k$, and then the permutation
$\sigma_{\left[  d\right]  \setminus J}$ sends the latter numbers to the
elements of $\left[  d\right]  \setminus J$ in increasing order). But the
permutation $\sigma_{J}$ does the exact same things (by its definition).
Hence, the two permutations $\sigma_{\left[  d\right]  \setminus J}\eta_{k}$
and $\sigma_{J}$ agree on all the numbers $1,2,\ldots,k$ and on all the
numbers $k+1,k+2,\ldots,d$. In other words, these two permutations are equal.
That is, $\sigma_{J}=\sigma_{\left[  d\right]  \setminus J}\eta_{k}$.}. The
latter equality yields
\begin{align*}
\operatorname*{sign}\left(  \sigma_{J}\right)   &  =\operatorname*{sign}%
\left(  \sigma_{\left[  d\right]  \setminus J}\eta_{k}\right)
=\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot\underbrace{\operatorname*{sign}\left(  \eta_{k}\right)  }_{=\left(
-1\right)  ^{k\left(  d-k\right)  }}\\
&  =\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot\left(  -1\right)  ^{k\left(  d-k\right)  },
\end{align*}
and therefore we have
\begin{align}
\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot\operatorname*{sign}\left(  \sigma_{J}\right)   &
=\underbrace{\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus
J}\right)  \cdot\operatorname*{sign}\left(  \sigma_{\left[  d\right]
\setminus J}\right)  }_{\substack{=\left(  \operatorname*{sign}\left(
\sigma_{\left[  d\right]  \setminus J}\right)  \right)  ^{2}=1\\\text{(since
}\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\text{ is }1\text{ or }-1\text{)}}}\cdot\left(  -1\right)  ^{k\left(
d-k\right)  }\nonumber\\
&  =\left(  -1\right)  ^{k\left(  d-k\right)  }. \label{p97.**.pf.1}%
\end{align}
This proves (\ref{p97.**.pf.goal}), and so the proof of (\ref{p97.**}) is complete.)

\item The Hodge star $\star:\Lambda\left(  V\right)  \rightarrow\Lambda\left(
V\right)  $ is a vector space isomorphism. (Indeed, (\ref{p97.**}) shows that
the composition $\star\circ\star$ is the linear map that sends each
$A\in\Lambda^{k}\left(  V\right)  $ to $\left(  -1\right)  ^{k\left(
d-k\right)  }A$. But the latter map is clearly a vector space isomorphism.
Thus, $\star$ is an isomorphism as well.)

\item Now we claim that any $a,b\in\Lambda\left(  V\right)  $ satisfy%
\begin{equation}
\star\left(  a\wedge b\right)  =\left(  \star a\right)  \vee\left(  \star
b\right)  . \label{p97.*and}%
\end{equation}


(\textit{Proof sketch:} By linearity, we WLOG assume that $a=e_{I}$ and
$b=e_{J}$, where $I$ and $J$ are two subsets of $\left[  d\right]  $. Then,
the definition of the Hodge star entails that $\star a=\operatorname*{sign}%
\left(  \sigma_{I}\right)  \cdot e_{\left[  d\right]  \setminus I}$ and $\star
b=\operatorname*{sign}\left(  \sigma_{J}\right)  \cdot e_{\left[  d\right]
\setminus J}$. Thus,
\begin{align}
\left(  \star a\right)  \vee\left(  \star b\right)   &  =\left(
\operatorname*{sign}\left(  \sigma_{I}\right)  \cdot e_{\left[  d\right]
\setminus I}\right)  \vee\left(  \operatorname*{sign}\left(  \sigma
_{J}\right)  \cdot e_{\left[  d\right]  \setminus J}\right) \nonumber\\
&  =\operatorname*{sign}\left(  \sigma_{I}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{J}\right)  \cdot e_{\left[  d\right]  \setminus I}\vee
e_{\left[  d\right]  \setminus J}. \label{p97.*and.pf.1}%
\end{align}


Meanwhile, let us write the subsets $I$ and $J$ as $I=\left\{  i_{1}%
<i_{2}<\cdots<i_{p}\right\}  $ and $J=\left\{  j_{1}<j_{2}<\cdots
<j_{q}\right\}  $. Thus, $a=e_{I}=e_{i_{1}}e_{i_{2}}\cdots e_{i_{p}}$ and
$b=e_{J}=e_{j_{1}}e_{j_{2}}\cdots e_{j_{q}}$, so that
\begin{align}
a\wedge b  &  =\left(  e_{i_{1}}e_{i_{2}}\cdots e_{i_{p}}\right)
\wedge\left(  e_{j_{1}}e_{j_{2}}\cdots e_{j_{q}}\right) \nonumber\\
&  =\sum_{\sigma}\operatorname*{sign}\left(  \sigma\right)  \left[
e_{i_{\sigma\left(  1\right)  }},\ldots,e_{i_{\sigma\left(  d-q\right)  }%
},e_{j_{1}},\ldots,e_{j_{q}}\right]  \cdot e_{i_{\sigma\left(  d-q+1\right)
}}\cdots e_{i_{\sigma\left(  p\right)  }}\ \ \ \ \ \ \ \label{p97.*and.pf.4}%
\end{align}
(by the definition of the meet operation), where the sum ranges over certain
permutations $\sigma\in S_{p}$ (namely, those that satisfy $\sigma\left(
1\right)  <\sigma\left(  2\right)  <\cdots<\sigma\left(  d-q\right)  $ and
$\sigma\left(  d-q+1\right)  <\sigma\left(  d-q+2\right)  <\cdots
<\sigma\left(  p\right)  $). We shall denote the latter permutations $\sigma$
as the \emph{shuffles}. Thus, the sum in (\ref{p97.*and.pf.4}) ranges over the
shuffles $\sigma$.

Now we are in one of the following two cases:

\textit{Case 1:} We have $I\cup J\neq\left[  d\right]  $.

\textit{Case 2:} We have $I\cup J=\left[  d\right]  $.

Let us first consider Case 1. In this case, we have $I\cup J\neq\left[
d\right]  $. Hence, there exists some $r\in\left[  d\right]  $ such that
$r\notin I$ and $r\notin J$. Consider this $r$. Now, \textbf{each} permutation
$\sigma\in S_{p}$ satisfies $\left[  e_{i_{\sigma\left(  1\right)  }}%
,\ldots,e_{i_{\sigma\left(  d-q\right)  }},e_{j_{1}},\ldots,e_{j_{q}}\right]
=0$ (since $r\notin I$ and $r\notin J$ shows that the basis vector $e_{r}$
does not appear among $e_{i_{\sigma\left(  1\right)  }},\ldots,e_{i_{\sigma
\left(  d-q\right)  }},e_{j_{1}},\ldots,e_{j_{q}}$, which entails by the
pigeonhole principle that two entries of the list $e_{i_{\sigma\left(
1\right)  }},\ldots,e_{i_{\sigma\left(  d-q\right)  }},e_{j_{1}}%
,\ldots,e_{j_{q}}$ are equal\footnote{Indeed, there are only $d$ basis vectors
$e_{1},e_{2},\ldots,e_{d}$ available. Thus, any list that contains $d$ of
these vectors but does not contain $e_{r}$ must contain two equal entries.}).
Hence, all the addends in the sum on the right hand side of
(\ref{p97.*and.pf.4}) equal $0$. Thus, (\ref{p97.*and.pf.4}) simplifies to
$a\wedge b=0$. Hence $\star\left(  a\wedge b\right)  =\star0=0$. On the other
hand, we have $r\in\left[  d\right]  \setminus I$ (since $r\notin I$) and
$r\in\left[  d\right]  \setminus J$ (since $r\notin J$); thus, the extensor
$e_{\left[  d\right]  \setminus I}\vee e_{\left[  d\right]  \setminus J}$ has
two equal factors (namely, $e_{r}$, which appears as a factor in both
$e_{\left[  d\right]  \setminus I}$ and $e_{\left[  d\right]  \setminus J}$).
Hence, this extensor is $0$. Thus, $e_{\left[  d\right]  \setminus I}\vee
e_{\left[  d\right]  \setminus J}=0$. Therefore, (\ref{p97.*and.pf.1})
simplifies to $\left(  \star a\right)  \vee\left(  \star b\right)  =0$.
Comparing this with $\star\left(  a\wedge b\right)  =0$, we obtain
$\star\left(  a\wedge b\right)  =\left(  \star a\right)  \vee\left(  \star
b\right)  $. Thus, (\ref{p97.*and}) is proved in Case 1.

Now, let us consider Case 2. In this case, we have $I\cup J=\left[  d\right]
$. Hence, $\left[  d\right]  \setminus J\subseteq I$. Thus, $\left[  d\right]
\setminus J$ is a $\left(  d-q\right)  $-element subset of $I$ (since
$\left\vert J\right\vert =q$ and thus $\left\vert \left[  d\right]  \setminus
J\right\vert =d-q$). In other words,
\[
\left[  d\right]  \setminus J=\left\{  i_{r_{1}}<i_{r_{2}}<\cdots<i_{r_{d-q}%
}\right\}
\]
for some $d-q$ elements $r_{1}<r_{2}<\cdots<r_{d-q}$ of $\left\{
1,2,\ldots,p\right\}  $ (since $I=\left\{  i_{1}<i_{2}<\cdots<i_{p}\right\}
$). Consider these elements.

It is easy to see that for any $d-q$ elements $h_{1}<h_{2}<\cdots<h_{d-q}$ of
$\left\{  1,2,\ldots,p\right\}  $, there is a unique shuffle $\sigma_{h}$
whose first $d-q$ values $\sigma_{h}\left(  1\right)  ,\sigma_{h}\left(
2\right)  ,\ldots,\sigma_{h}\left(  d-q\right)  $ are the elements
$h_{1},h_{2},\ldots,h_{d-q}$ (indeed, its remaining $p-\left(  d-q\right)  $
values $\sigma_{h}\left(  d-q+1\right)  ,\sigma_{h}\left(  d-q+2\right)
,\ldots,\sigma_{h}\left(  p\right)  $ must then be the remaining $p-\left(
d-q\right)  $ elements of $\left\{  1,2,\ldots,p\right\}  $ listed in
increasing order). Applying this to $h_{k}=r_{k}$, we see that there is a
unique shuffle $\sigma_{0}$ whose first $d-q$ values $\sigma_{0}\left(
1\right)  ,\sigma_{0}\left(  2\right)  ,\ldots,\sigma_{0}\left(  d-q\right)  $
are the elements $r_{1},r_{2},\ldots,r_{d-q}$ (because $r_{1}<r_{2}%
<\cdots<r_{d-q}$). Consider this shuffle $\sigma_{0}$. It satisfies
\begin{align}
\left[  e_{i_{\sigma_{0}\left(  1\right)  }},\ldots,e_{i_{\sigma_{0}\left(
d-q\right)  }},e_{j_{1}},\ldots,e_{j_{q}}\right]   &  =\left[  e_{i_{r_{1}}%
},\ldots,e_{i_{r_{d-q}}},e_{j_{1}},\ldots,e_{j_{q}}\right] \nonumber\\
&  =\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
, \label{p97.*and.pf.7}%
\end{align}
since the vectors $e_{i_{r_{1}}},\ldots,e_{i_{r_{d-q}}},e_{j_{1}}%
,\ldots,e_{j_{q}}$ inside the bracket are precisely the $d$ basis vectors
$e_{1},e_{2},\ldots,e_{d}$ of $V$ permuted using the permutation
$\sigma_{\left[  d\right]  \setminus J}$ (because we have $\left\{  i_{r_{1}%
}<i_{r_{2}}<\cdots<i_{r_{d-q}}\right\}  =\left[  d\right]  \setminus J$ and
$\left\{  j_{1}<j_{2}<\cdots<j_{q}\right\}  =J$).

Furthermore, since $\sigma_{0}$ is a shuffle, we have $\sigma_{0}\left(
d-q+1\right)  <\sigma_{0}\left(  d-q+2\right)  <\cdots<\sigma_{0}\left(
p\right)  $, so that $i_{\sigma_{0}\left(  d-q+1\right)  }<i_{\sigma
_{0}\left(  d-q+2\right)  }<\cdots<i_{\sigma_{0}\left(  p\right)  }$ (since
$i_{1}<i_{2}<\cdots<i_{p}$). Since we also have%
\begin{align*}
&  \left\{  i_{\sigma_{0}\left(  d-q+1\right)  },i_{\sigma_{0}\left(
d-q+2\right)  },\ldots,i_{\sigma_{0}\left(  p\right)  }\right\} \\
&  =\underbrace{\left\{  i_{1},i_{2},\ldots,i_{p}\right\}  }%
_{\substack{=\left\{  i_{1}<i_{2}<\cdots<i_{p}\right\}  \\=I}}\setminus
\underbrace{\left\{  i_{\sigma_{0}\left(  1\right)  },i_{\sigma_{0}\left(
2\right)  },\ldots,i_{\sigma_{0}\left(  d-q\right)  }\right\}  }%
_{\substack{=\left\{  i_{r_{1}},i_{r_{2}},\ldots,i_{r_{d-q}}\right\}
\\\text{(since the numbers }\sigma_{0}\left(  1\right)  ,\sigma_{0}\left(
2\right)  ,\ldots,\sigma_{0}\left(  d-q\right)  \\\text{are the elements
}r_{1},r_{2},\ldots,r_{d-q}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{because }\sigma
_{0}\text{ is a permutation of }\left\{  1,2,\ldots,p\right\}  \right) \\
&  =I\setminus\underbrace{\left\{  i_{r_{1}},i_{r_{2}},\ldots,i_{r_{d-q}%
}\right\}  }_{\substack{=\left\{  i_{r_{1}}<i_{r_{2}}<\cdots<i_{r_{d-q}%
}\right\}  \\=\left[  d\right]  \setminus J}}=I\setminus\left(  \left[
d\right]  \setminus J\right)  =I\cap J,
\end{align*}
we thus conclude that the numbers $i_{\sigma_{0}\left(  d-q+1\right)
},i_{\sigma_{0}\left(  d-q+2\right)  },\ldots,i_{\sigma_{0}\left(  p\right)
}$ are the elements of $I\cap J$ listed in increasing order. Hence,%
\begin{equation}
e_{i_{\sigma_{0}\left(  d-q+1\right)  }}e_{i_{\sigma_{0}\left(  d-q+2\right)
}}\cdots e_{i_{\sigma_{0}\left(  p\right)  }}=e_{I\cap J}.
\label{p97.*and.pf.8}%
\end{equation}


Recall that the sum on the right hand side of (\ref{p97.*and.pf.4}) ranges
over all shuffles $\sigma$; one of these shuffles $\sigma$ is $\sigma_{0}$.
All other shuffles $\sigma$ are distinct from $\sigma_{0}$, and thus satisfy%
\[
\left[  e_{i_{\sigma\left(  1\right)  }},\ldots,e_{i_{\sigma\left(
d-q\right)  }},e_{j_{1}},\ldots,e_{j_{q}}\right]  =0,
\]
since the bracket $\left[  e_{i_{\sigma\left(  1\right)  }},\ldots
,e_{i_{\sigma\left(  d-q\right)  }},e_{j_{1}},\ldots,e_{j_{q}}\right]  $
contains two equal vectors\footnote{Why? Because if it didn't, then the
indices $i_{\sigma\left(  1\right)  },\ldots,i_{\sigma\left(  d-q\right)
},j_{1},\ldots,j_{q}$ would be distinct, so that we would have
\[
\left\{  i_{\sigma\left(  1\right)  },\ldots,i_{\sigma\left(  d-q\right)
}\right\}  =\left[  d\right]  \setminus\underbrace{\left\{  j_{1},\ldots
,j_{q}\right\}  }_{=J}=\left[  d\right]  \setminus J=\left\{  i_{r_{1}%
}<i_{r_{2}}<\cdots<i_{r_{d-q}}\right\}  ;
\]
but this would entail $\left\{  \sigma\left(  1\right)  ,\ldots,\sigma\left(
d-q\right)  \right\}  =\left\{  r_{1},r_{2},\ldots,r_{d-q}\right\}  $ (since
the $i_{1},i_{2},\ldots,i_{p}$ are distinct), and therefore $\left(
\sigma\left(  1\right)  ,\ldots,\sigma\left(  d-q\right)  \right)  =\left(
r_{1},r_{2},\ldots,r_{d-q}\right)  $ (since $\sigma\left(  1\right)
<\cdots<\sigma\left(  d-q\right)  $ (because $\sigma$ is a shuffle) and
$r_{1}<r_{2}<\cdots<r_{d-q}$), and this would entail $\sigma=\sigma_{0}$
(since $\sigma$ is a shuffle, but $\sigma_{0}$ is the only shuffle such that
the numbers $\sigma_{0}\left(  1\right)  ,\sigma_{0}\left(  2\right)
,\ldots,\sigma_{0}\left(  d-q\right)  $ are the elements $r_{1},r_{2}%
,\ldots,r_{d-q}$).}. Thus, all the addends in the sum on the right hand side
of (\ref{p97.*and.pf.4}) equal $0$, except for the addend for $\sigma
=\sigma_{0}$, which is%
\begin{align*}
&  \operatorname*{sign}\left(  \sigma_{0}\right)  \underbrace{\left[
e_{i_{\sigma_{0}\left(  1\right)  }},\ldots,e_{i_{\sigma_{0}\left(
d-q\right)  }},e_{j_{1}},\ldots,e_{j_{q}}\right]  }%
_{\substack{=\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus
J}\right)  \\\text{(by (\ref{p97.*and.pf.7}))}}}\cdot\underbrace{e_{i_{\sigma
_{0}\left(  d-q+1\right)  }}\cdots e_{i_{\sigma_{0}\left(  p\right)  }}%
}_{\substack{=e_{I\cap J}\\\text{(by (\ref{p97.*and.pf.8}))}}}\\
&  =\operatorname*{sign}\left(  \sigma_{0}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{\left[  d\right]  \setminus J}\right)  \cdot e_{I\cap J}.
\end{align*}
Thus, (\ref{p97.*and.pf.4}) rewrites as
\[
a\wedge b=\operatorname*{sign}\left(  \sigma_{0}\right)  \cdot
\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot e_{I\cap J}.
\]
Thus,%
\begin{align}
\star\left(  a\wedge b\right)   &  =\star\left(  \operatorname*{sign}\left(
\sigma_{0}\right)  \cdot\operatorname*{sign}\left(  \sigma_{\left[  d\right]
\setminus J}\right)  \cdot e_{I\cap J}\right) \nonumber\\
&  =\operatorname*{sign}\left(  \sigma_{0}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{\left[  d\right]  \setminus J}\right)  \cdot\underbrace{\star
\,e_{I\cap J}}_{\substack{=\operatorname*{sign}\left(  \sigma_{I\cap
J}\right)  \cdot e_{\left[  d\right]  \setminus\left(  I\cap J\right)
}\\\text{(by the definition of }\star\text{)}}}\nonumber\\
&  =\operatorname*{sign}\left(  \sigma_{0}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{\left[  d\right]  \setminus J}\right)  \cdot
\operatorname*{sign}\left(  \sigma_{I\cap J}\right)  \cdot e_{\left[
d\right]  \setminus\left(  I\cap J\right)  }%
.\ \ \ \ \ \ \ \label{p97.*and.pf.17}%
\end{align}
On the other hand, $I\cup J=\left[  d\right]  $ entails that the sets $\left[
d\right]  \setminus I$ and $\left[  d\right]  \setminus J$ are disjoint.
Hence, (\ref{p97.eUeV=eUV}) shows that%
\[
e_{\left[  d\right]  \setminus I}\vee e_{\left[  d\right]  \setminus
J}=\operatorname*{sign}\left(  \eta\right)  \cdot e_{\left(  \left[  d\right]
\setminus I\right)  \cup\left(  \left[  d\right]  \setminus J\right)  },
\]
where $\eta\in S_{\left\vert \left[  d\right]  \setminus I\right\vert
+\left\vert \left[  d\right]  \setminus J\right\vert }$ is the permutation
that transforms the increasing list of $\left(  \left[  d\right]  \setminus
I\right)  \cup\left(  \left[  d\right]  \setminus J\right)  $ into the
concatenation of the increasing lists of $\left[  d\right]  \setminus I$ and
of $\left[  d\right]  \setminus J$. In view of $\left(  \left[  d\right]
\setminus I\right)  \cup\left(  \left[  d\right]  \setminus J\right)  =\left[
d\right]  \setminus\left(  I\cap J\right)  $, we can rewrite this as%
\begin{equation}
e_{\left[  d\right]  \setminus I}\vee e_{\left[  d\right]  \setminus
J}=\operatorname*{sign}\left(  \eta\right)  \cdot e_{\left[  d\right]
\setminus\left(  I\cap J\right)  }. \label{p97.*and.pf.18}%
\end{equation}


Hence, (\ref{p97.*and.pf.1}) rewrites as%
\[
\left(  \star a\right)  \vee\left(  \star b\right)  =\operatorname*{sign}%
\left(  \sigma_{I}\right)  \cdot\operatorname*{sign}\left(  \sigma_{J}\right)
\cdot\operatorname*{sign}\left(  \eta\right)  \cdot e_{\left[  d\right]
\setminus\left(  I\cap J\right)  }.
\]
Comparing this with (\ref{p97.*and.pf.17}), we see that the vectors
$\star\left(  a\wedge b\right)  $ and $\left(  \star a\right)  \vee\left(
\star b\right)  $ are equal up to sign. All that remains to be proved now is
that their signs agree as well. In other words, we must prove that%
\begin{align}
&  \operatorname*{sign}\left(  \sigma_{0}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{\left[  d\right]  \setminus J}\right)  \cdot
\operatorname*{sign}\left(  \sigma_{I\cap J}\right) \nonumber\\
&  =\operatorname*{sign}\left(  \sigma_{I}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{J}\right)  \cdot\operatorname*{sign}\left(  \eta\right)  .
\label{pf.p97.*and.sign}%
\end{align}


The easiest way to prove this equality is as follows: First, we observe that
$\left[  d\right]  \setminus I=J\setminus I$ (since $I\cup J=\left[  d\right]
$) and thus $\left\vert I\cap J\right\vert +\left\vert \left[  d\right]
\setminus I\right\vert =\left\vert I\cap J\right\vert +\left\vert J\setminus
I\right\vert =\left\vert J\right\vert =q$. Thus, the extensor $e_{I\cap J}\vee
e_{\left[  d\right]  \setminus I}$ has step $\left\vert I\cap J\right\vert
+\left\vert \left[  d\right]  \setminus I\right\vert =q$. Hence, $e_{I\cap
J}\vee e_{\left[  d\right]  \setminus I}\in\Lambda^{q}\left(  V\right)  $.
Moreover, $\left\vert \left[  d\right]  \setminus J\right\vert =d-q$ (since
$\left\vert J\right\vert =q$) and thus $e_{\left[  d\right]  \setminus J}%
\in\Lambda^{d-q}\left(  V\right)  $. Hence, (\ref{p97.grcomm}) (applied to
$k=q$ and $l=d-q$ and $a=e_{I\cap J}\vee e_{\left[  d\right]  \setminus I}$
and $b=e_{\left[  d\right]  \setminus J}$) yields%
\begin{equation}
e_{I\cap J}\vee e_{\left[  d\right]  \setminus I}\vee e_{\left[  d\right]
\setminus J}=\left(  -1\right)  ^{q\left(  d-q\right)  }e_{\left[  d\right]
\setminus J}\vee e_{I\cap J}\vee e_{\left[  d\right]  \setminus I}.
\label{pf.p97.*and.sign1}%
\end{equation}


But (\ref{p97.eJe-J}) (applied to $I\cap J$ instead of $J$) yields
\[
e_{I\cap J}\vee e_{\left[  d\right]  \setminus\left(  I\cap J\right)
}=\operatorname*{sign}\left(  \sigma_{I\cap J}\right)  \cdot e_{\left[
d\right]  }.
\]
Hence,%
\begin{align}
&  \operatorname*{sign}\left(  \sigma_{I\cap J}\right)  \cdot e_{\left[
d\right]  }\nonumber\\
&  =e_{I\cap J}\vee\underbrace{e_{\left[  d\right]  \setminus\left(  I\cap
J\right)  }}_{\substack{=\dfrac{1}{\operatorname*{sign}\left(  \eta\right)
}e_{\left[  d\right]  \setminus I}\vee e_{\left[  d\right]  \setminus
J}\\\text{(by (\ref{p97.*and.pf.18}))}}}\nonumber\\
&  =\underbrace{\dfrac{1}{\operatorname*{sign}\left(  \eta\right)  }%
}_{\substack{=\operatorname*{sign}\left(  \eta\right)  \\\text{(since
}\operatorname*{sign}\left(  \eta\right)  \\\text{is }1\text{ or }-1\text{)}%
}}\underbrace{e_{I\cap J}\vee e_{\left[  d\right]  \setminus I}\vee e_{\left[
d\right]  \setminus J}}_{\substack{=\left(  -1\right)  ^{q\left(  d-q\right)
}e_{\left[  d\right]  \setminus J}\vee e_{I\cap J}\vee e_{\left[  d\right]
\setminus I}\\\text{(by (\ref{pf.p97.*and.sign1}))}}}\nonumber\\
&  =\operatorname*{sign}\left(  \eta\right)  \cdot\left(  -1\right)
^{q\left(  d-q\right)  }e_{\left[  d\right]  \setminus J}\vee e_{I\cap J}\vee
e_{\left[  d\right]  \setminus I}. \label{pf.p97.*and.sign3}%
\end{align}


Now, recall that the values $\sigma_{0}\left(  1\right)  ,\sigma_{0}\left(
2\right)  ,\ldots,\sigma_{0}\left(  d-q\right)  $ are the elements
$r_{1},r_{2},\ldots,r_{d-q}$ (by the definition of $\sigma_{0}$). Hence,%
\[
e_{i_{\sigma_{0}\left(  1\right)  }}e_{i_{\sigma_{0}\left(  2\right)  }}\cdots
e_{i_{\sigma_{0}\left(  d-q\right)  }}=e_{i_{r_{1}}}e_{i_{r_{2}}}\cdots
e_{i_{r_{d-q}}}=e_{\left[  d\right]  \setminus J}%
\]
(since $\left[  d\right]  \setminus J=\left\{  i_{r_{1}}<i_{r_{2}}%
<\cdots<i_{r_{d-q}}\right\}  $). Multiplying this equality with
(\ref{p97.*and.pf.8}) (using the join operation $\vee$), we obtain%
\begin{align*}
&  \left(  e_{i_{\sigma_{0}\left(  1\right)  }}e_{i_{\sigma_{0}\left(
2\right)  }}\cdots e_{i_{\sigma_{0}\left(  d-q\right)  }}\right)  \vee\left(
e_{i_{\sigma_{0}\left(  d-q+1\right)  }}e_{i_{\sigma_{0}\left(  d-q+2\right)
}}\cdots e_{i_{\sigma_{0}\left(  p\right)  }}\right) \\
&  =e_{\left[  d\right]  \setminus J}\vee e_{I\cap J}.
\end{align*}
Hence,%
\begin{align*}
&  e_{\left[  d\right]  \setminus J}\vee e_{I\cap J}\\
&  =\left(  e_{i_{\sigma_{0}\left(  1\right)  }}e_{i_{\sigma_{0}\left(
2\right)  }}\cdots e_{i_{\sigma_{0}\left(  d-q\right)  }}\right)  \vee\left(
e_{i_{\sigma_{0}\left(  d-q+1\right)  }}e_{i_{\sigma_{0}\left(  d-q+2\right)
}}\cdots e_{i_{\sigma_{0}\left(  p\right)  }}\right) \\
&  =e_{i_{\sigma_{0}\left(  1\right)  }}e_{i_{\sigma_{0}\left(  2\right)  }%
}\cdots e_{i_{\sigma_{0}\left(  p\right)  }}=\operatorname*{sign}\left(
\sigma_{0}\right)  \cdot\underbrace{e_{i_{1}}e_{i_{2}}\cdots e_{i_{p}}%
}_{=e_{I}}=\operatorname*{sign}\left(  \sigma_{0}\right)  \cdot e_{I}.
\end{align*}
Hence, (\ref{pf.p97.*and.sign3}) becomes%
\begin{align*}
&  \operatorname*{sign}\left(  \sigma_{I\cap J}\right)  \cdot e_{\left[
d\right]  }\\
&  =\operatorname*{sign}\left(  \eta\right)  \cdot\left(  -1\right)
^{q\left(  d-q\right)  }\underbrace{e_{\left[  d\right]  \setminus J}\vee
e_{I\cap J}}_{=\operatorname*{sign}\left(  \sigma_{0}\right)  \cdot e_{I}}%
\vee\,e_{\left[  d\right]  \setminus I}\\
&  =\operatorname*{sign}\left(  \eta\right)  \cdot\left(  -1\right)
^{q\left(  d-q\right)  }\cdot\operatorname*{sign}\left(  \sigma_{0}\right)
\cdot\underbrace{e_{I}\vee e_{\left[  d\right]  \setminus I}}%
_{\substack{=\operatorname*{sign}\left(  \sigma_{I}\right)  \cdot e_{\left[
d\right]  }\\\text{(by (\ref{p97.eJe-J}))}}}\\
&  =\operatorname*{sign}\left(  \eta\right)  \cdot\left(  -1\right)
^{q\left(  d-q\right)  }\cdot\operatorname*{sign}\left(  \sigma_{0}\right)
\cdot\operatorname*{sign}\left(  \sigma_{I}\right)  \cdot e_{\left[  d\right]
}.
\end{align*}
Since $e_{\left[  d\right]  }$ is nonzero (and in fact an element of a basis
of $\Lambda\left(  V\right)  $), we can cancel $e_{\left[  d\right]  }$ from
this equality, and obtain%
\begin{align*}
&  \operatorname*{sign}\left(  \sigma_{I\cap J}\right) \\
&  =\operatorname*{sign}\left(  \eta\right)  \cdot\underbrace{\left(
-1\right)  ^{q\left(  d-q\right)  }}_{\substack{=\operatorname*{sign}\left(
\sigma_{\left[  d\right]  \setminus J}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{J}\right)  \\\text{(since }\left\vert J\right\vert =q\text{,
and thus (\ref{p97.**.pf.1}) (applied to }k=q\text{)}\\\text{yields
}\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus J}\right)
\cdot\operatorname*{sign}\left(  \sigma_{J}\right)  =\left(  -1\right)
^{q\left(  d-q\right)  }\text{)}}}\cdot\operatorname*{sign}\left(  \sigma
_{0}\right)  \cdot\operatorname*{sign}\left(  \sigma_{I}\right) \\
&  =\operatorname*{sign}\left(  \eta\right)  \cdot\operatorname*{sign}\left(
\sigma_{\left[  d\right]  \setminus J}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{J}\right)  \cdot\operatorname*{sign}\left(  \sigma_{0}\right)
\cdot\operatorname*{sign}\left(  \sigma_{I}\right)  .
\end{align*}
Multiplying this equality by $\operatorname*{sign}\left(  \sigma_{0}\right)
\cdot\operatorname*{sign}\left(  \sigma_{\left[  d\right]  \setminus
J}\right)  $, we obtain
\begin{align*}
&  \operatorname*{sign}\left(  \sigma_{0}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{\left[  d\right]  \setminus J}\right)  \cdot
\operatorname*{sign}\left(  \sigma_{I\cap J}\right) \\
&  =\operatorname*{sign}\left(  \sigma_{0}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{\left[  d\right]  \setminus J}\right)  \cdot
\operatorname*{sign}\left(  \eta\right)  \cdot\operatorname*{sign}\left(
\sigma_{\left[  d\right]  \setminus J}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{J}\right)  \cdot\operatorname*{sign}\left(  \sigma_{0}\right)
\cdot\operatorname*{sign}\left(  \sigma_{I}\right) \\
&  =\operatorname*{sign}\left(  \sigma_{I}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{J}\right)  \cdot\operatorname*{sign}\left(  \eta\right)
\cdot\underbrace{\left(  \operatorname*{sign}\left(  \sigma_{\left[  d\right]
\setminus J}\right)  \right)  ^{2}}_{=1}\cdot\underbrace{\left(
\operatorname*{sign}\left(  \sigma_{0}\right)  \right)  ^{2}}_{=1}\\
&  =\operatorname*{sign}\left(  \sigma_{I}\right)  \cdot\operatorname*{sign}%
\left(  \sigma_{J}\right)  \cdot\operatorname*{sign}\left(  \eta\right)  .
\end{align*}
This proves (\ref{pf.p97.*and.sign}), and thus completes the proof of
(\ref{p97.*and}) in Case 2.

Thus, (\ref{p97.*and}) is proved in both Cases 1 and 2; this completes the
proof of (\ref{p97.*and}).)

\item Now, it is easy to see that the meet is anticommutative: i.e., any
$a\in\Lambda^{j}\left(  V\right)  $ and $b\in\Lambda^{k}\left(  V\right)  $
satisfy%
\begin{equation}
a\wedge b=\left(  -1\right)  ^{\left(  d-k\right)  \left(  d-j\right)
}b\wedge a. \label{p97.grcomm2}%
\end{equation}


(\textit{Proof sketch:} Let $a\in\Lambda^{j}\left(  V\right)  $ and
$b\in\Lambda^{k}\left(  V\right)  $. Then, (\ref{p97.*deg}) yields $\star
a\in\Lambda^{d-j}\left(  V\right)  $ and $\star b\in\Lambda^{d-k}\left(
V\right)  $. Hence, (\ref{p97.grcomm}) (applied to $d-j$, $d-k$, $\star a$ and
$\star b$ instead of $k$, $j$, $a$ and $b$) yields
\begin{equation}
\left(  \star a\right)  \vee\left(  \star b\right)  =\left(  -1\right)
^{\left(  d-j\right)  \left(  d-k\right)  }\left(  \star b\right)  \vee\left(
\star a\right)  . \label{p97.grcomm2.pf.1}%
\end{equation}


However, (\ref{p97.*and}) yields $\star\left(  a\wedge b\right)  =\left(
\star a\right)  \vee\left(  \star b\right)  $ and $\star\left(  b\wedge
a\right)  =\left(  \star b\right)  \vee\left(  \star a\right)  $. In light of
these two equalities, we can rewrite (\ref{p97.grcomm2.pf.1}) as%
\[
\star\left(  a\wedge b\right)  =\left(  -1\right)  ^{\left(  d-j\right)
\left(  d-k\right)  }\star\left(  b\wedge a\right)  =\star\left(  \left(
-1\right)  ^{\left(  d-j\right)  \left(  d-k\right)  }\left(  b\wedge
a\right)  \right)  .
\]
Since the map $\star$ is injective (because $\star$ is a vector space
isomorphism), this entails $a\wedge b=\left(  -1\right)  ^{\left(  d-j\right)
\left(  d-k\right)  }\left(  b\wedge a\right)  =\left(  -1\right)  ^{\left(
d-k\right)  \left(  d-j\right)  }b\wedge a$. This proves (\ref{p97.grcomm2}).)

\item Now, it is easy to see that the meet is associative: i.e., any
$a,b,c\in\Lambda\left(  V\right)  $ satisfy%
\begin{equation}
a\wedge\left(  b\wedge c\right)  =\left(  a\wedge b\right)  \wedge c.
\label{p97.ass}%
\end{equation}


(\textit{Proof sketch:} This is similar to the proof of (\ref{p97.grcomm2}):
Again, we use (\ref{p97.*and}) to reduce the claim $a\wedge\left(  b\wedge
c\right)  =\left(  a\wedge b\right)  \wedge c$ to the equality $\left(  \star
a\right)  \vee\left(  \left(  \star b\right)  \vee\left(  \star c\right)
\right)  =\left(  \left(  \star a\right)  \vee\left(  \star b\right)  \right)
\vee\left(  \star c\right)  $, which is true because the join is associative.)
\end{itemize}

Thus, Theorem 3.3.2 \textbf{(a)} is proved. $\blacksquare$

\item \textbf{page 97, proof of Theorem 3.3.2:} \textquotedblleft In view of
the assumption $\overline{A}+\overline{B}=V$\textquotedblright: This is
somewhat inappropriate, since no such assumption has been made. Instead, the
case $\overline{A}+\overline{B}\neq V$ must be considered separately.
Fortunately, this case is very easy: In this case, we can easily see that
$A\wedge B=0$, because all the brackets $\left[  a_{\sigma\left(  1\right)
},\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]  $ on the
right hand side of (3.3.6) are $0$ (indeed, $\overline{A}+\overline{B}\neq V$
shows that $\overline{A}+\overline{B}$ is a proper subspace of $V$, and
obviously all the $d$ vectors $a_{\sigma\left(  1\right)  },\ldots
,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}$ belong to this proper
subspace; but this entails that these $d$ vectors cannot be linearly
independent, and thus must satisfy $\left[  a_{\sigma\left(  1\right)
},\ldots,a_{\sigma\left(  d-k\right)  },b_{1},\ldots,b_{k}\right]  =0$).

\item \textbf{page 97, proof of Theorem 3.3.2:} Let me add a few remarks about
Theorem 3.3.2 \textbf{(b)}.

\begin{enumerate}
\item Somewhat surprisingly, Theorem 3.3.2 \textbf{(b)} is not true if we
replace the base field $\mathbf{C}$ by a commutative ring. For a specific
counterexample, let $\mathbf{k}$ be the polynomial ring $\mathbf{R}\left[
x,y,z\right]  $, and let $d=4$ and $k=3$ and $j=3$. In the exterior algebra
$\Lambda_{\mathbf{k}}\left(  \mathbf{k}^{4}\right)  $, let $A=e_{1}e_{2}e_{3}$
and $B=b_{1}b_{2}b_{3}$, where $\left(  e_{1},e_{2},e_{3},e_{4}\right)  $ is
the standard basis of $\mathbf{k}^{4}$, and where $b_{1},b_{2},b_{3}$ are the
vectors $\left(  1,0,0,x\right)  ^{T},\left(  0,1,0,y\right)  ^{T},\left(
0,0,1,z\right)  ^{T}$. Then, $A$ and $B$ are extensors, but%
\begin{align*}
A\wedge B  &  =\left[  e_{1},b_{1},b_{2},b_{3}\right]  e_{2}e_{3}-\left[
e_{2},b_{1},b_{2},b_{3}\right]  e_{1}e_{3}+\left[  e_{3},b_{1},b_{2}%
,b_{3}\right]  e_{1}e_{2}\\
&  =xe_{2}e_{3}-ye_{1}e_{3}+ze_{1}e_{2}%
\end{align*}
is not. (Indeed, if $xe_{2}e_{3}-ye_{1}e_{3}+ze_{1}e_{2}$ was an extensor
$v\vee w$, then we could write $v$ as $v=\left(  \alpha,\beta,\gamma
,\delta\right)  ^{T}$, and obtain
\begin{align*}
0  &  =\underbrace{v}_{\substack{=\left(  \alpha,\beta,\gamma,\delta\right)
^{T}\\=\alpha e_{1}+\beta e_{2}+\gamma e_{3}+\delta e_{4}}}\vee
\underbrace{v\vee w}_{=xe_{2}e_{3}-ye_{1}e_{3}+ze_{1}e_{2}}\\
&  =\left(  \alpha e_{1}+\beta e_{2}+\gamma e_{3}+\delta e_{4}\right)
\vee\left(  xe_{2}e_{3}-ye_{1}e_{3}+ze_{1}e_{2}\right) \\
&  =\left(  \alpha x+\beta y+\gamma z\right)  e_{1}e_{2}e_{3}+\delta
e_{4}\left(  xe_{2}e_{3}-ye_{1}e_{3}+ze_{1}e_{2}\right)  ,
\end{align*}
which would entail $\alpha x+\beta y+\gamma z=0$ and $\delta=0$, and thus the
vector $\left(  \alpha,\beta,\gamma\right)  ^{T}$ would be a nonvanishing
tangent vector field on the $2$-sphere $\mathbf{R}^{3}$ (nonvanishing because
$v\vee w=xe_{2}e_{3}-ye_{1}e_{3}+ze_{1}e_{2}$ is nonvanishing and thus $v$ is
nonvanishing); but this would contradict
\href{https://en.wikipedia.org/wiki/Hairy_ball_theorem}{the hairy ball
theorem} (which says that no such tangent vector fields exist, even if we
replace polynomials by continuous functions). Thus, $xe_{2}e_{3}-ye_{1}%
e_{3}+ze_{1}e_{2}$ is not an extensor.)

This explains why the proof of Theorem 3.3.2 \textbf{(b)} must use linear
algebra that is specific to vector spaces (over fields).

\item The Hodge star $\star$ has a similar property to Theorem 3.3.2
\textbf{(b)}: If $A\in\Lambda\left(  V\right)  $ is any extensor, then $\star
A$ is an extensor again. This, too, would fail if we replaced $\mathbf{C}$ by
a commutative ring such as $\mathbf{R}\left[  x,y,z\right]  $ (and again, a
counterexample can be built based on the hairy ball theorem).
\end{enumerate}

\item \textbf{page 103, Theorem:} This theorem is very easy to prove without
any algebra.

\textit{Proof sketch:} WLOG assume that $\left(  \ell_{3},\ell_{4}\right)  $
is the pair of lines whose intersection we must prove (while all the other
five pairs are known to intersect). The two lines $\ell_{1}$ and $\ell_{2}$
are distinct (otherwise, $\ell_{1},\ell_{2},\ell_{3}$ would lie in a plane),
and thus span a plane $H$ and intersect at a point $P$. The line $\ell_{3}$
intersects both $\ell_{1}$ and $\ell_{2}$ but cannot lie on the plane $H$
(otherwise, $\ell_{1},\ell_{2},\ell_{3}$ would lie in a plane); thus,
$\ell_{3}$ must pass through $P$. Similarly, $\ell_{4}$ must pass through $P$.
Hence, all four lines $\ell_{1},\ell_{2},\ell_{3},\ell_{4}$ pass through $P$;
in particular, the pair $\left(  \ell_{3},\ell_{4}\right)  $ intersects.
$\blacksquare$

\item \textbf{page 111:} \textquotedblleft the \textit{weight} of a
tableaux\textquotedblright\ should be \textquotedblleft the \textit{weight} of
a tableau\textquotedblright.

\item \textbf{page 111:} It is not true that \textquotedblleft the property of
being homogeneous depends only on the image in $\mathcal{B}_{n,d}%
$\textquotedblright. Indeed, an inhomogeneous bracket polyomial can lie in
$I_{n,d}$ (an example is a linear combination of syzygies of different
weights) and thus turn into the homogeneous polynomial $0$ in $\mathcal{B}%
_{n,d}$.

\item \textbf{page 118, Proposition 3.6.1:} It is worth saying that the proof
of Proposition 3.6.1 is entirely straightforward:

\textit{Proof of Proposition 3.6.1.} The equality (3.6.3) becomes\footnote{Any
sum with no upper or lower limits is understood to range over all integers.}%
\begin{align*}
&  \overline{f}\left(  \overline{x},\overline{y}\right) \\
&  =\sum_{k=0}^{n}\dbinom{n}{k}a_{k}\underbrace{\left(  c_{11}\overline
{x}+c_{12}\overline{y}\right)  ^{k}}_{\substack{=\sum_{j}\dbinom{k}{j}\left(
c_{11}\overline{x}\right)  ^{j}\left(  c_{12}\overline{y}\right)
^{k-j}\\\text{(by the binomial theorem)}}}\underbrace{\left(  c_{21}%
\overline{x}+c_{22}\overline{y}\right)  ^{n-k}}_{\substack{=\sum_{l}%
\dbinom{n-k}{l}\left(  c_{21}\overline{x}\right)  ^{l}\left(  c_{22}%
\overline{y}\right)  ^{n-k-l}\\\text{(by the binomial theorem)}}}\\
&  =\sum_{k=0}^{n}\dbinom{n}{k}a_{k}\sum_{j}\dbinom{k}{j}\left(
c_{11}\overline{x}\right)  ^{j}\left(  c_{12}\overline{y}\right)  ^{k-j}%
\sum_{l}\dbinom{n-k}{l}\left(  c_{21}\overline{x}\right)  ^{l}\left(
c_{22}\overline{y}\right)  ^{n-k-l}\\
&  =\sum_{k=0}^{n}\ \ \sum_{j}\ \ \sum_{l}\underbrace{\dbinom{n}{k}%
a_{k}\dbinom{k}{j}\left(  c_{11}\overline{x}\right)  ^{j}\left(
c_{12}\overline{y}\right)  ^{k-j}\dbinom{n-k}{l}\left(  c_{21}\overline
{x}\right)  ^{l}\left(  c_{22}\overline{y}\right)  ^{n-k-l}}_{=\dbinom{n}%
{k}\dbinom{k}{j}\dbinom{n-k}{l}a_{k}c_{11}^{j}c_{12}^{k-j}c_{21}^{l}%
c_{22}^{n-k-l}\overline{x}^{j+l}\overline{y}^{n-j-l}}\\
&  =\sum_{k=0}^{n}\ \ \sum_{j}\ \ \sum_{l}\dbinom{n}{k}\dbinom{k}{j}%
\dbinom{n-k}{l}a_{k}c_{11}^{j}c_{12}^{k-j}c_{21}^{l}c_{22}^{n-k-l}\overline
{x}^{j+l}\overline{y}^{n-j-l}\\
&  =\underbrace{\sum_{k=0}^{n}\ \ \sum_{j}\ \ \sum_{i}}_{=\sum_{i}%
\ \ \sum_{k=0}^{n}\ \ \sum_{j}}\underbrace{\dbinom{n}{k}\dbinom{k}{j}%
\dbinom{n-k}{i-j}}_{\substack{=\dbinom{n}{i}\dbinom{i}{j}\dbinom{n-i}%
{k-j}\\\text{(this is easy to check using the}\\\text{definition of binomial
coefficients)}}}a_{k}c_{11}^{j}c_{12}^{k-j}c_{21}^{i-j}\underbrace{c_{22}%
^{n-k-\left(  i-j\right)  }}_{=c_{22}^{n-i-k+j}}\underbrace{\overline
{x}^{j+\left(  i-j\right)  }}_{=\overline{x}^{i}}\underbrace{\overline
{y}^{n-j-\left(  i-j\right)  }}_{=\overline{y}^{n-i}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted
}i-j\text{ for }l\text{ in the third sum}\right) \\
&  =\sum_{i}\ \ \sum_{k=0}^{n}\ \ \sum_{j}\dbinom{n}{i}\dbinom{i}{j}%
\dbinom{n-i}{k-j}a_{k}c_{11}^{j}c_{12}^{k-j}c_{21}^{i-j}c_{22}^{n-i-k+j}%
\overline{x}^{i}\overline{y}^{n-i}\\
&  =\sum_{k}\ \ \sum_{i=0}^{n}\ \ \sum_{j}\dbinom{n}{k}\dbinom{k}{j}%
\dbinom{n-k}{i-j}a_{i}c_{11}^{j}c_{12}^{i-j}c_{21}^{k-j}c_{22}^{n-k-i+j}%
\overline{x}^{k}\overline{y}^{n-k}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed the
summation indices }i\text{ and }k\text{ as }k\text{ and }i\right) \\
&  =\sum_{k}\dbinom{n}{k}\left(  \sum_{i=0}^{n}\left(  \sum_{j}\dbinom{k}%
{j}\dbinom{n-k}{i-j}c_{11}^{j}c_{12}^{i-j}c_{21}^{k-j}c_{22}^{n-k-i+j}\right)
a_{i}\right)  \overline{x}^{k}\overline{y}^{n-k}.
\end{align*}
In other words,
\[
\overline{f}\left(  \overline{x},\overline{y}\right)  =\sum_{k}\dbinom{n}%
{k}\overline{a}_{k}\overline{x}^{k}\overline{y}^{n-k},
\]
where%
\begin{equation}
\overline{a}_{k}=\sum_{i=0}^{n}\left(  \sum_{j}\dbinom{k}{j}\dbinom{n-k}%
{i-j}c_{11}^{j}c_{12}^{i-j}c_{21}^{k-j}c_{22}^{n-k-i+j}\right)  a_{i}.
\label{p118.ak=}%
\end{equation}
The latter equality (\ref{p118.ak=}) is precisely the equality (3.6.5), except
that the second sum ranges over all $j\in\mathbf{Z}$ instead of only ranging
from $\max\left(  0,i-n+k\right)  $ to $\min\left(  i,k\right)  $. But this
little difference in summation ranges does not affect the value of the sum,
since all addends with $j<\max\left(  0,i-n+k\right)  $ or $j>\min\left(
i,k\right)  $ are $0$ anyway (indeed, $\dbinom{k}{j}=0$ if $j<0$ or $j>k$,
whereas $\dbinom{n-k}{i-j}=0$ if $j<i-n+k$ or $j>i$). Thus, by proving
(\ref{p118.ak=}), we have proved (3.6.5) and therefore Proposition 3.6.1.
$\blacksquare$

\item \textbf{page 122:} \textquotedblleft defines a $\mathbf{C}$-algebra
homomorphism\textquotedblright\ should be \textquotedblleft defines a
$\mathbf{C}\left[  x,y\right]  $-algebra homomorphism\textquotedblright\ (or
you should say separately that $\Psi$ is supposed to send $x\mapsto x$ and
$y\mapsto y$).

\item \textbf{page 122, (3.6.7):} In \textquotedblleft$\dfrac{\left(
-1\right)  ^{k}}{n!}\mu_{1}\cdots\mu_{n}\cdot\sigma_{k}\left(  \dfrac{\nu_{1}%
}{\mu_{1}},\ldots,\dfrac{\nu_{n}}{\mu_{n}}\right)  $\textquotedblright, the
\textquotedblleft$n!$\textquotedblright\ should be \textquotedblleft%
$\dbinom{n}{k}$\textquotedblright. (But the \textquotedblleft$n!$%
\textquotedblright\ one line above is correct.)

\item \textbf{page 123, proof of Lemma 3.6.3:} Here is an alternative proof:

\textit{Second proof of Lemma 3.6.3.} Equip the polynomial ring $\mathbf{C}%
\left[  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right]  $ with a monomial
order that is lexicographic with $\mu_{1}>\mu_{2}>\cdots>\mu_{n}>\nu_{1}%
>\nu_{2}>\cdots>\nu_{n}>x>y$. Then, for each $k\in\left\{  0,1,\ldots
,n\right\}  $, we have%
\[
\Psi\left(  a_{k}\right)  =\dfrac{\left(  -1\right)  ^{n-k}}{n!}\sum_{\pi\in
S_{n}}\nu_{\pi\left(  1\right)  }\nu_{\pi\left(  2\right)  }\cdots\nu
_{\pi\left(  n-k\right)  }\mu_{\pi\left(  n-k+1\right)  }\mu_{\pi\left(
n-k+2\right)  }\cdots\mu_{\pi\left(  n\right)  }%
\]
and therefore
\begin{equation}
\operatorname*{init}\left(  \Psi\left(  a_{k}\right)  \right)  =\mu_{1}\mu
_{2}\cdots\mu_{k}\nu_{k+1}\nu_{k+2}\cdots\nu_{n} \label{p123.inita}%
\end{equation}
(with coefficient $\dfrac{\left(  -1\right)  ^{n-k}}{n!}$, which we ignore).
Also, of course, $\operatorname*{init}\left(  \Psi\left(  x\right)  \right)
=x$ (since $\Psi\left(  x\right)  =x$) and $\operatorname*{init}\left(
\Psi\left(  y\right)  \right)  =y$ (likewise). Hence, for any product
$a_{0}^{u_{0}}a_{1}^{u_{1}}\cdots a_{n}^{u_{n}}x^{v}y^{w}$ (with $u_{0}%
,u_{1},\ldots,u_{n},v,w\in\mathbf{N}$), we have%
\begin{align*}
&  \operatorname*{init}\left(  \Psi\left(  a_{0}^{u_{0}}a_{1}^{u_{1}}\cdots
a_{n}^{u_{n}}x^{v}y^{w}\right)  \right) \\
&  =\operatorname*{init}\left(  \Psi\left(  x^{v}y^{w}a_{0}^{u_{0}}%
a_{1}^{u_{1}}\cdots a_{n}^{u_{n}}\right)  \right) \\
&  =\operatorname*{init}\left(  \Psi\left(  x\right)  ^{v}\Psi\left(
y\right)  ^{w}\Psi\left(  a_{0}\right)  ^{u_{0}}\Psi\left(  a_{1}\right)
^{u_{1}}\cdots\Psi\left(  a_{n}\right)  ^{u_{n}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Psi\text{ is
an algebra morphism}\right) \\
&  =\left(  \underbrace{\operatorname*{init}\left(  \Psi\left(  x\right)
\right)  }_{=x}\right)  ^{v}\left(  \underbrace{\operatorname*{init}\left(
\Psi\left(  y\right)  \right)  }_{=y}\right)  ^{w}\prod_{k=0}^{n}\left(
\underbrace{\operatorname*{init}\left(  \Psi\left(  a_{k}\right)  \right)
}_{\substack{=\mu_{1}\mu_{2}\cdots\mu_{k}\nu_{k+1}\nu_{k+2}\cdots\nu
_{n}\\\text{(by (\ref{p123.inita}))}}}\right)  ^{u_{k}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the initial monomial of a product of several polynomials}\\
\text{is the product of the initial monomials of the factors}%
\end{array}
\right) \\
&  =x^{v}y^{w}\underbrace{\prod_{k=0}^{n}\left(  \mu_{1}\mu_{2}\cdots\mu
_{k}\nu_{k+1}\nu_{k+2}\cdots\nu_{n}\right)  ^{u_{k}}}_{=\prod_{j=1}^{n}\mu
_{j}^{u_{j}+u_{j+1}+\cdots+u_{n}}\nu_{j}^{u_{0}+u_{1}+\cdots+u_{j-1}}}\\
&  =x^{v}y^{w}\prod_{j=1}^{n}\mu_{j}^{u_{j}+u_{j+1}+\cdots+u_{n}}\nu
_{j}^{u_{0}+u_{1}+\cdots+u_{j-1}}.
\end{align*}
We can easily recover the original exponents $u_{0},u_{1},\ldots,u_{n},v,w$
from this monomial (indeed, $v$ and $w$ are simply the exponents of $x$ and
$y$, whereas the exponents $u_{j}+u_{j+1}+\cdots+u_{n}$ on the $\mu_{j}$'s
allow us to find $u_{1},u_{2},\ldots,u_{n}$ by taking differences, and the
exponents $u_{0}+u_{1}+\cdots+u_{j-1}$ on the $\nu_{j}$'s allow us to recover
$u_{0},u_{1},\ldots,u_{n-1}$ by taking differences). Thus, the initial
monomials $\operatorname*{init}\left(  \Psi\left(  a_{0}^{u_{0}}a_{1}^{u_{1}%
}\cdots a_{n}^{u_{n}}x^{v}y^{w}\right)  \right)  $ for all $u_{0},u_{1}%
,\ldots,u_{n},v,w\in\mathbf{N}$ are pairwise distinct. Hence, the polynomials
$\Psi\left(  a_{0}^{u_{0}}a_{1}^{u_{1}}\cdots a_{n}^{u_{n}}x^{v}y^{w}\right)
$ are $\mathbf{C}$-linearly independent (since a family of polynomials that
have pairwise distinct initial monomials must always be $\mathbf{C}$-linearly independent).

But the family $\left(  a_{0}^{u_{0}}a_{1}^{u_{1}}\cdots a_{n}^{u_{n}}%
x^{v}y^{w}\right)  _{u_{0},u_{1},\ldots,u_{n},v,w\in\mathbf{N}}$ of monomials
is a basis of the $\mathbf{C}$-vector space $\mathbf{C}\left[  a_{0}%
,a_{1},\ldots,a_{n},x,y\right]  $. We have just proved that the map $\Psi$
sends this basis to a linearly independent family (since the polynomials
$\Psi\left(  a_{0}^{u_{0}}a_{1}^{u_{1}}\cdots a_{n}^{u_{n}}x^{v}y^{w}\right)
$ are $\mathbf{C}$-linearly independent). Thus, the map $\Psi$ is injective
(since a linear map that sends a basis of its domain to a linearly independent
family in its target must always be injective). This proves Lemma 3.6.3 again.
$\blacksquare$

\item \textbf{page 124, proof of Proposition 3.6.4:} \textquotedblleft Then
$R$ can be rewritten as%
\[
R\left(  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right)  =\left(  \mu
_{1}\ldots\mu_{n}\right)  ^{d}\cdot\widehat{R}\left(  \dfrac{\nu_{1}}{\mu_{1}%
},\ldots,\dfrac{\nu_{n}}{\mu_{n}},x,y\right)  ,
\]
where $\widehat{R}$ is a symmetric function (in the usual sense) in the ratios
$\dfrac{\nu_{i}}{\mu_{i}}$\textquotedblright: Let me explain why this is the
case. We define the polynomial $\widehat{R}=\widehat{R}\left(  \alpha
_{1},\ldots,\alpha_{n},x,y\right)  \in\mathbf{C}\left[  \alpha_{1}%
,\ldots,\alpha_{n},x,y\right]  $ by%
\[
\widehat{R}\left(  \alpha_{1},\ldots,\alpha_{n},x,y\right)  :=R\left(
1,\alpha_{1},1,\alpha_{2},\ldots,1,\alpha_{n},x,y\right)
\]
(that is, $\widehat{R}$ is the evaluation of $R$ at $\mu_{i}:=1$ and $\nu
_{i}:=\alpha_{i}$). This polynomial $\widehat{R}$ is symmetric in $\alpha
_{1},\ldots,\alpha_{n}$, since $R$ is symmetric. Furthermore, since $R$ is
regular of degree $d$, we know that if we multiply the inputs $\mu_{1},\nu
_{1},\mu_{2},\nu_{2},\ldots,\mu_{n},\nu_{n}$ by some scalars (or
indeterminates) $\omega_{1},\omega_{1},\omega_{2},\omega_{2},\ldots,\omega
_{n},\omega_{n}$, respectively (that is, if we multiply both $\mu_{i}$ and
$\nu_{i}$ by $\omega_{i}$ for each $i\in\left\{  1,2,\ldots,n\right\}  $),
then the value $R\left(  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right)  $
gets multiplied by $\omega_{1}^{d}\omega_{2}^{d}\cdots\omega_{n}^{d}$. In
other words, for any $\omega_{1},\omega_{2},\ldots,\omega_{n}$, we have%
\begin{align*}
&  R\left(  \omega_{1}\mu_{1},\omega_{1}\nu_{1},\ldots,\omega_{n}\mu
_{n},\omega_{n}\nu_{n},x,y\right) \\
&  =\left(  \omega_{1}^{d}\omega_{2}^{d}\cdots\omega_{n}^{d}\right)  R\left(
\mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right)  .
\end{align*}
Applying this to $\omega_{i}=\dfrac{1}{\mu_{i}}$, we obtain%
\begin{align*}
&  R\left(  \dfrac{1}{\mu_{1}}\mu_{1},\dfrac{1}{\mu_{1}}\nu_{1},\ldots
,\dfrac{1}{\mu_{n}}\mu_{n},\dfrac{1}{\mu_{n}}\nu_{n},x,y\right) \\
&  =\underbrace{\left(  \left(  \dfrac{1}{\mu_{1}}\right)  ^{d}\left(
\dfrac{1}{\mu_{2}}\right)  ^{d}\cdots\left(  \dfrac{1}{\mu_{n}}\right)
^{d}\right)  }_{=\dfrac{1}{\left(  \mu_{1}\ldots\mu_{n}\right)  ^{d}}}R\left(
\mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right) \\
&  =\dfrac{1}{\left(  \mu_{1}\ldots\mu_{n}\right)  ^{d}}R\left(  \mu_{1}%
,\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right)  .
\end{align*}
Solving this for $R\left(  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right)
$, we obtain%
\begin{align*}
&  R\left(  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right) \\
&  =\left(  \mu_{1}\ldots\mu_{n}\right)  ^{d}\cdot R\left(  \underbrace{\dfrac
{1}{\mu_{1}}\mu_{1}}_{=1},\underbrace{\dfrac{1}{\mu_{1}}\nu_{1}}_{=\dfrac
{\nu_{1}}{\mu_{1}}},\ldots,\underbrace{\dfrac{1}{\mu_{n}}\mu_{n}}%
_{=1},\underbrace{\dfrac{1}{\mu_{n}}\nu_{n}}_{=\dfrac{\nu_{n}}{\mu_{n}}%
},x,y\right) \\
&  =\left(  \mu_{1}\ldots\mu_{n}\right)  ^{d}\cdot\underbrace{R\left(
1,\dfrac{\nu_{1}}{\mu_{1}},\ldots,1,\dfrac{\nu_{n}}{\mu_{n}},x,y\right)
}_{\substack{=\widehat{R}\left(  \dfrac{\nu_{1}}{\mu_{1}},\ldots,\dfrac
{\nu_{n}}{\mu_{n}},x,y\right)  \\\text{(by the definition of }\widehat{R}%
\text{)}}}\\
&  =\left(  \mu_{1}\ldots\mu_{n}\right)  ^{d}\cdot\widehat{R}\left(
\dfrac{\nu_{1}}{\mu_{1}},\ldots,\dfrac{\nu_{n}}{\mu_{n}},x,y\right)  .
\end{align*}
Thus, we have rewritten $R$ as desired.

\item \textbf{page 124, proof of Proposition 3.6.4:} \textquotedblleft
Multiplying $Q$ by $\left(  \mu_{1}\ldots\mu_{n}\right)  ^{d}$ and
distributing factors of $\mu_{1}\ldots\mu_{n}$, we obtain a representation of
$R$ as a polynomial function in the magnitudes $\mu_{1}\cdots\mu_{n}\sigma
_{k}\left(  \dfrac{\nu_{1}}{\mu_{1}},\ldots,\dfrac{\nu_{n}}{\mu_{n}}\right)
$\textquotedblright: This argument is slightly incomplete. If we do what is
suggested here, then we obtain a representation of $R$ as a polynomial
function in the magnitudes $\mu_{1}\cdots\mu_{n}$ and $\sigma_{k}\left(
\dfrac{\nu_{1}}{\mu_{1}},\ldots,\dfrac{\nu_{n}}{\mu_{n}}\right)  $. In order
to rewrite this as a polynomial function in the products $\mu_{1}\cdots\mu
_{n}\sigma_{k}\left(  \dfrac{\nu_{1}}{\mu_{1}},\ldots,\dfrac{\nu_{n}}{\mu_{n}%
}\right)  $, we must make sure that each of the monomials contains at least as
many $\mu_{1}\cdots\mu_{n}$'s as it contains $\sigma_{k}\left(  \dfrac{\nu
_{1}}{\mu_{1}},\ldots,\dfrac{\nu_{n}}{\mu_{n}}\right)  $'s; otherwise, we get
only a Laurent polynomial.

There are several ways to fix this. In my opinion, the easiest way to prove
the \textquotedblleft if\textquotedblright\ direction of Proposition 3.6.4
correctly is to proceed as in the proof of Theorem 1.1.1 (that is, in essence,
by showing that the $\Psi\left(  a_{0}\right)  ,\Psi\left(  a_{1}\right)
,\ldots,\Psi\left(  a_{n}\right)  $ form a Sagbi basis of the ring of regular
symmetric polynomials $R\in\mathbf{C}\left[  \mu_{1},\nu_{1},\ldots,\mu
_{n},\nu_{n},x,y\right]  $ as a $\mathbf{C}\left[  x,y\right]  $-algebra, even
though we don't ever have to use this language). Here are the details:

\textit{Proof of the \textquotedblleft if\textquotedblright\ direction of
Proposition 3.6.4:} We give an algorithm to express each regular symmetric
polynomial $R\in\mathbf{C}\left[  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu
_{n},x,y\right]  $ as a polynomial function of the images $\Psi\left(
a_{0}\right)  ,\Psi\left(  a_{1}\right)  ,\ldots,\Psi\left(  a_{n}\right)  $
over $\mathbf{C}\left[  x,y\right]  $ (which will, of course, show that $R$
lies in the image of $\Psi$). To do so, we view $x$ and $y$ as constants, and
we equip the polynomial ring $\mathbf{C}\left[  \mu_{1},\nu_{1},\ldots,\mu
_{n},\nu_{n},x,y\right]  =\left(  \mathbf{C}\left[  x,y\right]  \right)
\left[  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n}\right]  $ with a monomial order
that is lexicographic with $\mu_{1}>\mu_{2}>\cdots>\mu_{n}>\nu_{1}>\nu
_{2}>\cdots>\nu_{n}$. Then, for a nonzero regular symmetric polynomial
$R\in\mathbf{C}\left[  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right]  $,
the initial monomial $\operatorname*{init}R$ must have the form $\mu
_{1}^{u_{1}}\mu_{2}^{u_{2}}\cdots\mu_{n}^{u_{n}}\nu_{1}^{d-u_{1}}\nu
_{2}^{d-u_{2}}\cdots\nu_{n}^{d-u_{n}}$ (since $R$ is regular) with $u_{1}\geq
u_{2}\geq\cdots\geq u_{n}$ (since otherwise, by the symmetry of $R$, we could
obtain a larger monomial of $R$ by swapping two of the $\mu_{i}$'s along with
the corresponding $\nu_{i}$'s). But this is precisely the initial monomial
$\operatorname*{init}\left(  R^{\prime}\right)  $ of the polynomial
$R^{\prime}:=\prod_{k=0}^{n}\left(  \Psi\left(  a_{k}\right)  \right)
^{u_{k}-u_{k+1}}$, where we set $u_{0}:=d$ and $u_{n+1}:=0$ (this follows
easily from (\ref{p123.inita})). Thus, subtracting an appropriate scalar
multiple of $R^{\prime}$ from $R$, we cancel the initial term of $R$ and
obtain a smaller polynomial in $\mathbf{C}\left[  \mu_{1},\nu_{1},\ldots
,\mu_{n},\nu_{n},x,y\right]  $ that is still regular and symmetric (since $R$
and $R^{\prime}$ are regular and symmetric). Repeating this procedure again
and again, we eventually end up with the polynomial $0$ (since Corollary 1.2.3
shows that this process cannot go on forever). Thus, $R$ equals the sum of all
the polynomials we have subtracted, hence a polynomial function of the
$\Psi\left(  a_{0}\right)  ,\Psi\left(  a_{1}\right)  ,\ldots,\Psi\left(
a_{n}\right)  $ over $\mathbf{C}\left[  x,y\right]  $. Therefore, $R$ lies in
the image of $\Psi$. This finishes the proof. $\blacksquare$

\item \textbf{page 125, definition of the bracket }$\left[  i\ \mathbf{u}%
\right]  $\textbf{:} Replace \textquotedblleft$\left[  i\ \mathbf{u}\right]
:=\mu_{i}y-\nu_{i}x$\textquotedblright\ by \textquotedblleft$\left[
i\ \mathbf{u}\right]  :=\mu_{i}x-\nu_{i}y$\textquotedblright.

\item \textbf{page 125, proof of Lemma 3.6.5:} Add commas before
\textquotedblleft$\left[  1\ n\right]  $\textquotedblright, before
\textquotedblleft$\left[  n-1\ n\right]  $\textquotedblright, and before
\textquotedblleft$\left[  n\ \mathbf{u}\right]  $\textquotedblright.

\item \textbf{page 125, Theorem 3.6.6:} It is worth pointing out that
\textquotedblleft bracket polynomial\textquotedblright\ here means a
polynomial in $\mathbf{C}\left[  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu
_{n},x,y\right]  $ that lies in the bracket ring. The notions of
\textquotedblleft symmetric\textquotedblright\ and \textquotedblleft
regular\textquotedblright\ are inherited from $\mathbf{C}\left[  \mu_{1}%
,\nu_{1},\ldots,\mu_{n},\nu_{n},x,y\right]  $. (Hence, in particular, a
\textquotedblleft symmetric bracket polynomial\textquotedblright\ means a
symmetric polynomial that lies in the bracket ring. Such a polynomial is
symmetric under all permutations of the subscripts $1,2,\ldots,n$ when written
as a polynomial in $\mu_{1},\nu_{1},\ldots,\mu_{n},\nu_{n},x,y$, and
furthermore can be expressed as a polynomial in the brackets $\left[
i\ j\right]  $ and $\left[  i\ \mathbf{u}\right]  $; but the latter expression
might not be invariant under all permutations of the symbols $1,2,\ldots,n$.)

\item \textbf{page 126, proof of Theorem 3.6.6:} \textquotedblleft The
expansion map $\Psi$ commutes with\textquotedblright\ should better be
\textquotedblleft The expansion map $\Psi$ is injective (by Lemma 3.6.3) and
commutes with\textquotedblright.

\item \textbf{page 127, definition of the bracket }$\left[  i^{\left(
k\right)  }\ \mathbf{u}\right]  $\textbf{:} Replace \textquotedblleft$\left[
i^{\left(  k\right)  }\ \mathbf{u}\right]  :=\mu_{i}^{\left(  k\right)  }%
x-\mu_{j}^{\left(  l\right)  }y$\textquotedblright\ by \textquotedblleft%
$\left[  i^{\left(  k\right)  }\ \mathbf{u}\right]  :=\mu_{i}^{\left(
k\right)  }x-\nu_{i}^{\left(  k\right)  }y$\textquotedblright.

\item \textbf{page 127, middle of the page:} In \textquotedblleft$f\left(
x,y\right)  =a_{2}x^{2}+2a_{1}xy+a_{0}^{2}y$\textquotedblright, replace
\textquotedblleft$a_{0}^{2}y$\textquotedblright\ by \textquotedblleft%
$a_{0}y^{2}$\textquotedblright. Likewise, on the next line, replace
\textquotedblleft$b_{0}^{2}y$\textquotedblright\ by \textquotedblleft%
$b_{0}y^{2}$\textquotedblright.

\item \textbf{page 129, proof of Lemma 3.7.2:} I find this proof confusing
(what does \textquotedblleft canonical preimage\textquotedblright\ mean), so
let me give my own:

\textit{Proof of Lemma 3.7.2.} \textbf{(a)} For any polynomial $f\in
\mathbf{C}\left[  x_{1},\ldots,x_{r}\right]  $, we let $\overline{f}$ denote
its residue class $f+I$ (that is, the projection of $f$ onto the quotient ring
$\mathbf{C}\left[  x_{1},\ldots,x_{r}\right]  /I$).

Clearly, the residue classes $\overline{p_{1}},\ldots,\overline{p_{m}}$ are
invariant under $\Gamma$ (since the polynomials $p_{1},\ldots,p_{m}$ are
invariant under $\Gamma$).

Let $\overline{p}\in\left(  \mathbf{C}\left[  x_{1},\ldots,x_{r}\right]
/I\right)  ^{\Gamma}$ be any invariant residue class in $\mathbf{C}\left[
x_{1},\ldots,x_{r}\right]  /I$. We must show that $\overline{p}$ can be
written as a polynomial function of $\overline{p_{1}},\ldots,\overline{p_{m}}$.

The polynomial $p\in\mathbf{C}\left[  x_{1},\ldots,x_{r}\right]  $ itself
might not be $\Gamma$-invariant, but its image $p^{\ast}$ under the Reynolds
operator is. Furthermore, $\overline{p^{\ast}}=\overline{p}^{\ast}$, because
the Reynolds operator (being defined as the average of the actions of all
elements of $\Gamma$) commutes with the projection onto $\mathbf{C}\left[
x_{1},\ldots,x_{r}\right]  /I$. But $\overline{p}^{\ast}=\overline{p}$ because
$\overline{p}$ is $\Gamma$-invariant. Hence, $\overline{p}=\overline{p}^{\ast
}=\overline{p^{\ast}}$.

But $p^{\ast}$ belongs to $\mathbf{C}\left[  x_{1},\ldots,x_{r}\right]
^{\Gamma}$ and thus can be written as a polynomial function $u\left(
p_{1},\ldots,p_{m}\right)  $ of $p_{1},\ldots,p_{m}$. Hence, $\overline{p}$
can be written as the same polynomial function $u\left(  \overline{p_{1}%
},\ldots,\overline{p_{m}}\right)  $ of $\overline{p_{1}},\ldots,\overline
{p_{m}}$ (since $p^{\ast}=u\left(  p_{1},\ldots,p_{m}\right)  $ entails
$\overline{p^{\ast}}=\overline{u\left(  p_{1},\ldots,p_{m}\right)  }=u\left(
\overline{p_{1}},\ldots,\overline{p_{m}}\right)  $ and thus $\overline
{p}=\overline{p^{\ast}}=u\left(  \overline{p_{1}},\ldots,\overline{p_{m}%
}\right)  $). This completes the proof of Lemma 3.7.2 \textbf{(a)}.

\textbf{(b)} In the proof of Noether's degree bound (Theorem 2.1.4), it was
shown that the invariant ring $\mathbf{C}\left[  x_{1},\ldots,x_{r}\right]
^{\Gamma}$ is generated by the Reynolds images $\left(  x_{1}^{i_{1}}%
x_{2}^{i_{2}}\cdots x_{r}^{i_{r}}\right)  ^{\ast}$ of all monomials
$x_{1}^{i_{1}}x_{2}^{i_{2}}\cdots x_{r}^{i_{r}}$ with degree $i_{1}%
+i_{2}+\cdots+i_{r}\leq\left\vert \Gamma\right\vert $. Hence, applying part
\textbf{(a)} (with the $p_{1},\ldots,p_{m}$ being these Reynolds images), we
conclude that the ring $\left(  \mathbf{C}\left[  x_{1},\ldots,x_{r}\right]
/I\right)  ^{\Gamma}$ is generated by the images of these Reynolds images
under the canonical surjection $\mathbf{C}\left[  x_{1},\ldots,x_{r}\right]
\rightarrow\mathbf{C}\left[  x_{1},\ldots,x_{r}\right]  /I$. This proves Lemma
3.7.2 \textbf{(b)}. $\blacksquare$

\item \textbf{page 130, proof of Theorem 3.7.1:} In the first sentence of this
proof, add commas before \textquotedblleft$\left[  1\ n\right]  $%
\textquotedblright, before \textquotedblleft$\left[  n-1\ n\right]
$\textquotedblright, and before \textquotedblleft$\left[  n\ \mathbf{u}%
\right]  $\textquotedblright.

\item \textbf{page 130, proof of Theorem 3.7.1:} \textquotedblleft Since
minimally regular monomials remain minimally regular after permuting
letters\textquotedblright: This is not literally true, since (e.g.) the
transposition $t_{1,2}$ sends the bracket $\left[  1\ 2\right]  $ to $-\left[
1\ 2\right]  $. The action of a permutation $\sigma\in S_{n}$ sends a bracket
monomial to $\pm$ a bracket monomial.

The easiest way to fix this is by introducing additional brackets $\left[
i\ j\right]  $ for $i>j$. These new brackets are defined in the same way as
the old brackets (that is, $\left[  i\ j\right]  :=\mu_{i}\nu_{j}-\nu_{i}%
\mu_{j}$). Of course, they are redundant as generators, since $\left[
i\ j\right]  =-\left[  j\ i\right]  $ for all $i\neq j$; but they serve to
make the set of generators more symmetric. After we introduce these new
brackets, the set of bracket monomials (i.e., products of brackets) becomes
fixed under the $S_{n}$-action that permutes the letters $1,2,\ldots,n$
(fixing $\mathbf{u}$), and so is the subset of regular bracket monomials;
hence, the minimially regular bracket monomials also form an $S_{n}$-set. With
these changes made, the proof goes through as stated.

\item \textbf{page 131, discussion of the monomial order }$\prec
_{\operatorname*{circ}}$\textbf{:} \textquotedblleft Establishing the
existence of such a monomial order is a nontrivial exercise\textquotedblright:
But not a hard exercise at any rate. For instance, we can introduce a (rather
exotic) grading on the polynomial ring $\mathbf{C}\left[  \Lambda\left(
n,2\right)  \right]  $ by assigning to each indeterminate $\left[
i\ j\right]  $ the degree $\left(  2-\dfrac{1}{j-i}\right)  n!$ (this is a
positive integer, because $j-i\in\left\{  1,2,\ldots,n\right\}  $ is a
positive divisor of $n!$). Then, we totally order the bracket monomials as
follows: If two monomials $\mathfrak{m}$ and $\mathfrak{n}$ have different
degrees, then we set $\mathfrak{m}\succ\mathfrak{n}$ if and only if
$\mathfrak{m}$ has larger degree than $\mathfrak{n}$; otherwise, we let
$\mathfrak{m}\succ\mathfrak{n}$ if $\mathfrak{m}$ is lexicographically larger
than $\mathfrak{n}$. This is easily seen to be a monomial order (indeed, this
holds for any grading on a polynomial ring in which each indeterminate is
homogeneous and has its degree equal to a positive integer). Moreover, for any
$1\leq i_{1}<i_{2}<i_{3}<i_{4}\leq n$, the initial monomial of the syzygy%
\[
P_{i_{1}i_{2}i_{3}i_{4}}=\left[  i_{1}\ i_{3}\right]  \left[  i_{2}%
\ i_{4}\right]  -\left[  i_{1}\ i_{2}\right]  \left[  i_{3}\ i_{4}\right]
-\left[  i_{1}\ i_{4}\right]  \left[  i_{2}\ i_{3}\right]
\]
is $\left[  i_{1}\ i_{3}\right]  \left[  i_{2}\ i_{4}\right]  $. (Indeed, the
three monomials $\left[  i_{1}\ i_{3}\right]  \left[  i_{2}\ i_{4}\right]  $,
$\left[  i_{1}\ i_{2}\right]  \left[  i_{3}\ i_{4}\right]  $ and $\left[
i_{1}\ i_{4}\right]  \left[  i_{2}\ i_{3}\right]  $ have respective degrees
\begin{align*}
&  \left(  2-\dfrac{1}{i_{3}-i_{1}}\right)  n!+\left(  2-\dfrac{1}{i_{4}%
-i_{2}}\right)  n!,\\
&  \left(  2-\dfrac{1}{i_{2}-i_{1}}\right)  n!+\left(  2-\dfrac{1}{i_{4}%
-i_{3}}\right)  n!\ \ \ \ \ \ \ \ \ \ \text{and}\\
&  \left(  2-\dfrac{1}{i_{4}-i_{1}}\right)  n!+\left(  2-\dfrac{1}{i_{3}%
-i_{2}}\right)  n!,
\end{align*}
but the first of these three degrees is larger than the other two\footnote{Let
us show this. We must prove that
\begin{align*}
\left(  2-\dfrac{1}{i_{3}-i_{1}}\right)  n!+\left(  2-\dfrac{1}{i_{4}-i_{2}%
}\right)  n!  &  >\left(  2-\dfrac{1}{i_{2}-i_{1}}\right)  n!+\left(
2-\dfrac{1}{i_{4}-i_{3}}\right)  n!\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(  2-\dfrac{1}{i_{3}-i_{1}}\right)  n!+\left(  2-\dfrac{1}{i_{4}-i_{2}%
}\right)  n!  &  >\left(  2-\dfrac{1}{i_{4}-i_{1}}\right)  n!+\left(
2-\dfrac{1}{i_{3}-i_{2}}\right)  n!.
\end{align*}
Upon cancelling the $2$'s and $n!$'s, these two inequalities rewrite as%
\begin{align*}
\dfrac{1}{i_{3}-i_{1}}+\dfrac{1}{i_{4}-i_{2}}  &  <\dfrac{1}{i_{2}-i_{1}%
}+\dfrac{1}{i_{4}-i_{3}}\ \ \ \ \ \ \ \ \ \ \text{and}\\
\dfrac{1}{i_{3}-i_{1}}+\dfrac{1}{i_{4}-i_{2}}  &  <\dfrac{1}{i_{4}-i_{1}%
}+\dfrac{1}{i_{3}-i_{2}}.
\end{align*}
Upon setting $p:=i_{2}-i_{1}$ and $q:=i_{3}-i_{2}$ and $r:=i_{4}-i_{2}$ (so
that $i_{3}-i_{1}=p+q$ and $i_{4}-i_{2}=q+r$ and $i_{4}-i_{1}=p+q+r$), these
two inequalities rewrite as%
\begin{align*}
\dfrac{1}{p+q}+\dfrac{1}{q+r}  &  <\dfrac{1}{p}+\dfrac{1}{r}%
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\dfrac{1}{p+q}+\dfrac{1}{q+r}  &  <\dfrac{1}{p+q+r}+\dfrac{1}{q}.
\end{align*}
Since $p,q,r$ are positive (because $i_{1}<i_{2}<i_{3}<i_{4}$), the first of
these two inequalities is obvious (since $\dfrac{1}{p+q}<\dfrac{1}{p}$ and
$\dfrac{1}{q+r}<\dfrac{1}{r}$), whereas the second boils down (upon
cross-multiplying and cancelling the numerator $p+2q+r$) to $\left(
p+q+r\right)  q<\left(  p+q\right)  \left(  q+r\right)  $, which is again
clear (since $\left(  p+q\right)  \left(  q+r\right)  -\left(  p+q+r\right)
q=pr>0$). Thus, both inequalities are true.}.)

Thus, the monomial order we have just introduced is (one option for) the
monomial order $\prec_{\operatorname*{circ}}$ that we need.

\item \textbf{page 131, first proof of Proposition 3.7.4:} This is somewhat
confused. The set $\mathcal{S}_{n,2}$ is a subset of the bracket polynomial
ring $\mathbf{C}\left[  \Lambda\left(  n,2\right)  \right]  $, not of the
polynomial ring $\mathbf{C}\left[  \mu_{1},\nu_{1},\ldots,\mu_{n},\nu
_{n},x,y\right]  $; hence, the brackets $\left[  i\ j\right]  $ are the
indeterminates here. Thus, if the initial monomials of $P_{i_{1}i_{2}%
i_{3}i_{4}}$ and $P_{j_{1}j_{2}j_{3}j_{4}}$ are not relatively prime, then the
sets $\left\{  \left[  i_{1}\ i_{3}\right]  ,\ \left[  i_{2}\ i_{4}\right]
\right\}  $ and $\left\{  \left[  j_{1}\ j_{3}\right]  ,\ \left[  j_{2}%
\ j_{4}\right]  \right\}  $ have nonempty intersection, which is a stronger
claim than \textquotedblleft the set of indices $\left\{  i_{1},i_{2}%
,i_{3},i_{4},j_{1},j_{2},j_{3},j_{4}\right\}  $ has cardinality at most
seven\textquotedblright\ (in fact, it shows that the set of indices $\left\{
i_{1},i_{2},i_{3},i_{4},j_{1},j_{2},j_{3},j_{4}\right\}  $ has cardinality at
most $6$, but even this is unnecessarily weak a statement). As a consequence,
we only need to verify the Gr\"{o}bner basis property in the cases $n=5,6$,
not in the cases $n=5,6,7$.

This is not as painful as it sounds, even without the use of a computer. All
we have to do is show that for any two distinct $4$-tuples $\left(
i_{1}<i_{2}<i_{3}<i_{4}\right)  $ and $\left(  j_{1}<j_{2}<j_{3}<j_{4}\right)
$ such that the initial monomials of $P_{i_{1}i_{2}i_{3}i_{4}}$ and
$P_{j_{1}j_{2}j_{3}j_{4}}$ are not relatively prime (i.e., the the sets
$\left\{  \left[  i_{1}\ i_{3}\right]  ,\ \left[  i_{2}\ i_{4}\right]
\right\}  $ and $\left\{  \left[  j_{1}\ j_{3}\right]  ,\ \left[  j_{2}%
\ j_{4}\right]  \right\}  $ have nonempty intersection), the $\mathcal{S}%
$-polynomial $\mathcal{S}\left(  P_{i_{1}i_{2}i_{3}i_{4}},P_{j_{1}j_{2}%
j_{3}j_{4}}\right)  $ can be reduced to $0$ modulo $\mathcal{S}_{n,2}$. For
$n=5$, the only $\mathcal{S}$-polynomials we need to check are%
\begin{align*}
&  \mathcal{S}\left(  P_{1,2,3,4},P_{1,2,3,5}\right)
,\ \ \ \ \ \ \ \ \ \ \mathcal{S}\left(  P_{1,2,4,5},P_{1,3,4,5}\right)  ,\\
&  \mathcal{S}\left(  P_{1,2,3,4},P_{2,3,4,5}\right)
,\ \ \ \ \ \ \ \ \ \ \mathcal{S}\left(  P_{1,2,3,5},P_{1,2,4,5}\right)  ,\\
&  \mathcal{S}\left(  P_{1,3,4,5},P_{2,3,4,5}\right)
\end{align*}
\footnote{I have omitted the \textquotedblleft mirror
versions\textquotedblright\ of these $\mathcal{S}$-polynomials (where the
\emph{mirror version} of an $\mathcal{S}$-polynomial $\mathcal{S}\left(
f,g\right)  $ means the $\mathcal{S}$-polynomial\ $\mathcal{S}\left(
g,f\right)  $), because these need not be checked separately (in fact, we
always have $\mathcal{S}\left(  g,f\right)  =-\mathcal{S}\left(  f,g\right)
$).
\par
There is an additional symmetry in the definitions of $\Lambda\left(
n,2\right)  $ and $\mathcal{S}_{n,2}$ that we could use to simplify our life
(namely, we can replace $\left[  u\ v\right]  $ by $\left[  v^{\prime
}\ u^{\prime}\right]  $, where $u^{\prime}=n+1-u$ and $v^{\prime}=n+1-v$). We
leave it to the reader to check how it can be exploited.}. For $n=6$, the only
additional $\mathcal{S}$-polynomials we need to check (aside from the above
ones and their variants with relabelled variables) are%
\begin{align*}
&  \mathcal{S}\left(  P_{1,2,3,5},P_{2,4,5,6}\right)
,\ \ \ \ \ \ \ \ \ \ \mathcal{S}\left(  P_{1,2,4,5},P_{2,3,5,6}\right)  ,\\
&  \mathcal{S}\left(  P_{1,3,4,6},\mathcal{S}_{2,3,5,6}\right)
,\ \ \ \ \ \ \ \ \ \ \mathcal{S}\left(  P_{1,3,5,6},P_{2,3,4,6}\right)  ,\\
&  \mathcal{S}\left(  P_{1,2,4,5},P_{1,3,4,6}\right)
,\ \ \ \ \ \ \ \ \ \ \mathcal{S}\left(  P_{1,2,4,6},P_{1,3,4,5}\right)
\end{align*}
\footnote{Again, I have omitted the \textquotedblleft mirror
versions\textquotedblright\ of these $\mathcal{S}$-polynomials.}. In each
case, we must check that the respective $\mathcal{S}$-polynomial can be
reduced to $0$ modulo $\mathcal{S}_{n,2}$. According to a known
criterion\footnote{See, e.g., Lemma 3.6 in
\href{https://arxiv.org/abs/1704.00839v6}{Darij Grinberg, \textit{t-unique
reductions for M\'{e}sz\'{a}ros's subdivision algebra [detailed version]},
ancillary file to arXiv:1704.00839v6}. (This lemma appears as Lemma 4.6 in the
published version, but without the proof.)}, it suffices to express these
$\mathcal{S}$-polynomials in the form $\sum_{s\in S}a_{s}\mathfrak{s}_{s}%
g_{s}$, where $S$ is a finite set, where $a_{s}\in\mathbf{C}$ are scalars,
where $\mathfrak{s}_{s}$ are monomials and where $g_{s}$ are (not necessarily
distinct) elements of $\mathcal{S}_{n,2}$ such that all the monomials
$\mathfrak{s}_{s}\cdot\operatorname*{init}\left(  g_{s}\right)  $ are distinct
(where $\operatorname*{init}$ means the initial monomial with respect to the
monomial order $\prec_{\operatorname*{circ}}$). But such expressions can be
explicitly provided:%
\begin{align*}
\mathcal{S}\left(  P_{1,2,3,4},P_{1,2,3,5}\right)   &  =\left[  2\ 5\right]
\ P_{1,2,3,4}-\left[  2\ 4\right]  \ P_{1,2,3,5}\\
&  =\left[  1\ 2\right]  \ P_{2,3,4,5}-\left[  2\ 3\right]  \ P_{1,2,4,5}%
\end{align*}
(with $\left[  1\ 2\right]  $ and $\left[  2\ 3\right]  $ playing the role of
the monomials $\mathfrak{s}_{s}$ and with $P_{2,3,4,5}$ and $P_{1,2,4,5}$
acting as the $g_{s}$; it is easy to see that the monomials $\left[
1\ 2\right]  \cdot\operatorname*{init}P_{2,3,4,5}=\left[  1\ 2\right]
\cdot\left[  2\ 4\right]  \cdot\left[  3\ 5\right]  $ and $\left[
2\ 3\right]  \cdot\operatorname*{init}P_{1,2,4,5}=\left[  2\ 3\right]
\cdot\left[  1\ 4\right]  \cdot\left[  2\ 5\right]  $ are distinct) and
similarly%
\begin{align*}
\mathcal{S}\left(  P_{1,2,4,5},P_{1,3,4,5}\right)   &  =\left[  4\ 5\right]
\ P_{1,2,3,5}-\left[  1\ 5\right]  \ P_{2,3,4,5};\\
\mathcal{S}\left(  P_{1,2,3,4},P_{2,3,4,5}\right)   &  =\left[  3\ 4\right]
\ P_{1,2,3,5}-\left[  2\ 3\right]  \ P_{1,3,4,5};\\
\mathcal{S}\left(  P_{1,2,3,5},P_{1,2,4,5}\right)   &  =\left[  1\ 5\right]
\ P_{1,2,3,4}-\left[  1\ 2\right]  \ P_{1,3,4,5};\\
\mathcal{S}\left(  P_{1,3,4,5},P_{2,3,4,5}\right)   &  =\left[  3\ 4\right]
\ P_{1,2,4,5}-\left[  4\ 5\right]  \ P_{1,2,3,4};\\
\mathcal{S}\left(  P_{1,2,3,5},P_{2,4,5,6}\right)   &  =\left[  5\ 6\right]
\ P_{1,2,3,4}+\left[  4\ 5\right]  \ P_{1,2,3,6}-\left[  2\ 3\right]
\ P_{1,4,5,6}-\left[  1\ 2\right]  \ P_{3,4,5,6};\\
\mathcal{S}\left(  P_{1,2,4,5},P_{2,3,5,6}\right)   &  =\left[  3\ 5\right]
\ P_{1,2,4,6}-\left[  2\ 4\right]  \ P_{1,3,5,6}-\left[  5\ 6\right]
\ P_{1,2,3,4}+\left[  1\ 2\right]  \ P_{3,4,5,6};\\
\mathcal{S}\left(  P_{1,3,4,6},P_{2,3,5,6}\right)   &  =\left[  1\ 6\right]
\ P_{2,3,4,5}-\left[  2\ 3\right]  \ P_{1,4,5,6}-\left[  4\ 6\right]
\ P_{1,2,3,5}+\left[  3\ 5\right]  \ P_{1,2,4,6};\\
\mathcal{S}\left(  P_{1,3,5,6},P_{2,3,4,6}\right)   &  =-\left[  1\ 6\right]
\ P_{2,3,4,5}+\left[  2\ 3\right]  \ P_{1,4,5,6}-\left[  5\ 6\right]
\ P_{1,2,3,4}+\left[  3\ 4\right]  \ P_{1,2,5,6};\\
\mathcal{S}\left(  P_{1,2,4,5},P_{1,3,4,6}\right)   &  =-\left[  1\ 6\right]
\ P_{2,3,4,5}+\left[  4\ 5\right]  \ P_{1,2,3,6}+\left[  1\ 3\right]
\ P_{2,4,5,6}-\left[  2\ 4\right]  \ P_{1,3,5,6};\\
\mathcal{S}\left(  P_{1,2,4,6},P_{1,3,4,5}\right)   &  =-\left[  1\ 6\right]
\ P_{2,3,4,5}+\left[  4\ 5\right]  \ P_{1,2,3,6}-\left[  1\ 2\right]
\ P_{3,4,5,6}+\left[  3\ 4\right]  \ P_{1,2,5,6}.
\end{align*}


\item \textbf{page 132, Lemma 3.7.5:} This lemma is false. A counterexample is
the bracket monomial%
\[
\left[  1\ 2\right]  \left[  1\ 3\right]  \left[  2\ 3\right]  ^{2}%
\cdot\left[  4\ 5\right]  \left[  4\ 6\right]  \left[  5\ 6\right]  ^{2}%
\cdot\left[  7\ 8\right]  \left[  7\ 9\right]  \left[  8\ 9\right]  ^{2}%
\cdot\left[  1\ \mathbf{x}\right]  \left[  4\ \mathbf{x}\right]  \left[
7\ \mathbf{x}\right]
\]
in $\mathbf{C}\left[  \Lambda\left(  10,2\right)  \right]  $, which is regular
of degree $3$ and elemental.

However, there is an alternative way of proving Theorem 3.7.3, which bypasses
Lemma 3.7.5. It goes back to the paper
\href{https://math.stanford.edu/~vakil/files/hmsv030708.pdf}{B. Howard, J.
Millson, A. Snowden, R. Vakil, \textit{The equations for the moduli space of n
points on the line}, Duke Math. J. \textbf{146} (2009), no. 2, pp. 175--226}
(see also the proof of Theorem 31 in
\href{https://arxiv.org/abs/1305.2749v1}{Giorgio Ottaviani, \textit{Five
Lectures on Projective Invariants}, arXiv:1305.2749v1}). Here is an outline:

\textit{Correct proof of Theorem 3.7.3 (sketched).} We rename the letter
$\mathbf{u}$ as $n+1$ and rename the indeterminates $y$ and $x$ as $\mu_{n+1}$
and $\nu_{n+1}$ (so that our brackets have the form $\left[  i\ j\right]  $
for $1\leq i<j\leq n+1$). Furthermore, we rename $n$ as $n-1$ (so that our
brackets now have the form $\left[  i\ j\right]  $ for $1\leq i<j\leq n$). We
also introduce the shorthand $\left[  i\ j\right]  :=\left[  j\ i\right]  $
for all $i>j$, so that the bracket $\left[  i\ j\right]  $ is defined for all
$i\neq j$ (not only for $i<j$). (Note that this differs from the convention I
suggested on page 130.)

It is easy to see (indeed, it follows from Example 3.1.6 upon renaming the
variables) that
\begin{equation}
\left[  i_{1}\ i_{3}\right]  \left[  i_{2}\ i_{4}\right]  =\left[
i_{1}\ i_{2}\right]  \left[  i_{3}\ i_{4}\right]  +\left[  i_{1}%
\ i_{4}\right]  \left[  i_{2}\ i_{3}\right]  \label{p133.ptolemy}%
\end{equation}
for any $1\leq i_{1}<i_{2}<i_{3}<i_{4}\leq n$. Hence, for any four distinct
elements $\alpha,\beta,\gamma,\delta$ of $\left\{  1,2,\ldots,n\right\}  $, we
have%
\begin{equation}
\left[  \alpha\ \gamma\right]  \left[  \beta\ \delta\right]  =\pm\left[
\alpha\ \beta\right]  \left[  \gamma\ \delta\right]  \pm\left[  \alpha
\ \delta\right]  \left[  \beta\ \gamma\right]  \label{p133.ptolemy2}%
\end{equation}
for an appropriate choice of $\pm$ signs (the two $\pm$ signs need not be the
same sign). (Indeed, (\ref{p133.ptolemy2}) follows by applying
(\ref{p133.ptolemy}) with $i_{1},i_{2},i_{3},i_{4}$ being the numbers
$\alpha,\beta,\gamma,\delta$ listed in increasing order, and then solving for
$\left[  \alpha\ \gamma\right]  \left[  \beta\ \delta\right]  $.)

In the following, a \textit{lowlie} will mean a bracket monomial that is
regular of degree $1$ or $2$. For example, $\left[  1\ 2\right]  \left[
2\ 3\right]  \cdots\left[  n-1\ n\right]  \left[  n\ 1\right]  $ is a lowlie
that is regular of degree $2$. Lowlies that are regular of degree $1$ exist
only when $n$ is even; an example of such a lowlie is $\left[  1\ 2\right]
\left[  3\ 4\right]  \cdots\left[  n-1\ n\right]  $.

We view any bracket monomial $\left[  i_{1}\ j_{1}\right]  \left[
i_{2}\ j_{2}\right]  \cdots\left[  i_{k}\ j_{k}\right]  $ as a multigraph with
vertex set $\left\{  1,2,\ldots,n\right\}  $ and edges $\left\{  i_{1}%
,j_{1}\right\}  ,\ \left\{  i_{2},j_{2}\right\}  ,\ \ldots,\ \left\{
i_{k},j_{k}\right\}  $. Such multigraphs can have parallel edges, but cannot
have loops. A bracket monomial is regular of degree $d$ if and only if it is a
$d$-regular multigraph (i.e., each vertex has degree $d$).

Now we want to prove Theorem 3.7.3. This theorem claims that the ring
$\mathcal{B}_{\operatorname*{reg}}$ of regular bracket polynomials is
generated by the lowlies. In other words, it claims that every bracket
polynomial that is regular of degree $d$ (for some $d\in\mathbf{N}$) is a
polynomial function of the lowlies. We shall prove this by distinguishing
between two cases, depending on the parity of $d$:

\begin{itemize}
\item \textit{Case 1:} The number $d$ is even. Hence,
\href{https://en.wikipedia.org/wiki/2-factor_theorem}{Petersen's 2-factor
theorem} shows that every $d$-regular multigraph (that is, every bracket
monomial that is regular of degree $d$) can be partitioned into $d/2$
edge-disjoint $2$-factors, i.e., (regarded as a bracket monomial), can be
factored as a product of $d/2$ bracket monomials that are regular of degree
$2$. Of course, the latter monomials are lowlies. Hence, we have shown that
every bracket monomial that is regular of degree $d$ is a product of lowlies.
Thus, every bracket polynomial that is regular of degree $d$ is a polynomial
function of the lowlies. So our proof is complete in Case 1.

\item \textit{Case 2:} The number $d$ is odd. We must prove that every bracket
polynomial that is regular of degree $d$ is a polynomial function of the
lowlies. Clearly, it is enough to prove this for bracket monomials. Thus, we
let $\mathfrak{m}$ be a bracket monomial that is regular of degree $d$. As a
multigraph, $\mathfrak{m}$ is thus a $d$-regular multigraph with $n$ vertices.
Since the sum of the degrees of all vertices in a multigraph is even, we thus
conclude that $nd$ is even, so that $n$ must be even (since $d$ is odd).

If the multigraph $\mathfrak{m}$ is bipartite, then
\href{https://math.stackexchange.com/questions/1805181/prove-that-a-k-regular-bipartite-graph-has-a-perfect-matching}{Frobenius's
matching theorem} shows that $\mathfrak{m}$ has a perfect matching (since
$\mathfrak{m}$ is $d$-regular), i.e., a spanning $1$-regular subgraph.
Rewritten in terms of brackets, this is saying that if $\mathfrak{m}$ is
bipartite, then $\mathfrak{m}$ is divisible by a bracket monomial
$\mathfrak{n}$ that is regular of degree $1$. The quotient $\mathfrak{m}%
/\mathfrak{n}$ of this division must then be a bracket monomial that is
regular of degree $d-1$, and hence is a polynomial function of the lowlies (by
Case 1, since $d-1$ is even). Multiplying this polynomial function by
$\mathfrak{n}$ (which itself is a lowlie), we obtain an expression for
$\mathfrak{m}$ as a polynomial function of the lowlies. Thus, our proof is
finished if the multigraph $\mathfrak{m}$ is bipartite.

What can we do if $\mathfrak{m}$ is not bipartite? In this case,
$\mathfrak{m}$ contains at least one edge $\left\{  e_{1},e_{2}\right\}  $
whose both endpoints are even\footnote{Recall that $n$ is even, so the even
vertices are $2,4,6,\ldots,n$, while the odd vertices are $1,3,5,\ldots,n-1$%
.}, as well as at least one edge $\left\{  o_{1},o_{2}\right\}  $ whose both
endpoints are odd\footnote{Indeed:
\par
\begin{itemize}
\item If neither $\left\{  e_{1},e_{2}\right\}  $ nor $\left\{  o_{1}%
,o_{2}\right\}  $ existed, then $\mathfrak{m}$ would be bipartite (with all
the even vertices being left vertices, and all the odd vertices being right
vertices); but we have assumed that $\mathfrak{m}$ is not.
\par
\item If $\left\{  e_{1},e_{2}\right\}  $ existed but $\left\{  o_{1}%
,o_{2}\right\}  $ did not, then the sum of the degrees of all odd vertices
would equal the number of even-odd edges (i.e., edges that have an even and an
odd endpoint), whereas the sum of the degrees of all even vertices would be
larger than the number of even-odd edges (since it would count the edge
$\left\{  e_{1},e_{2}\right\}  $ as well); but this is impossible, since both
of these sums are $\underbrace{d+d+\cdots+d}_{n/2\text{ times}}=dn/2$.
\par
\item If $\left\{  o_{1},o_{2}\right\}  $ existed but $\left\{  e_{1}%
,e_{2}\right\}  $ did not, then we would obtain a similar contradiction.
\end{itemize}
\par
So the only possibility is that both $\left\{  e_{1},e_{2}\right\}  $ and
$\left\{  o_{1},o_{2}\right\}  $ exist.}. Consider these two edges. Thus, the
bracket monomial $\mathfrak{m}$ is divisible by $\left[  e_{1}\ e_{2}\right]
\left[  o_{1}\ o_{2}\right]  $. In other words,
\begin{equation}
\mathfrak{m}=\left[  e_{1}\ e_{2}\right]  \left[  o_{1}\ o_{2}\right]
\ \mathfrak{p} \label{p133.mp}%
\end{equation}
for some bracket monomial $\mathfrak{p}$. Consider this $\mathfrak{p}$. But
(\ref{p133.ptolemy2}) (applied to $\alpha=e_{1}$, $\beta=o_{1}$, $\gamma
=e_{2}$ and $\delta=o_{2}$) shows that%
\[
\left[  e_{1}\ e_{2}\right]  \left[  o_{1}\ o_{2}\right]  =\pm\left[
e_{1}\ o_{1}\right]  \left[  e_{2}\ o_{2}\right]  \pm\left[  e_{1}%
\ o_{2}\right]  \left[  o_{1}\ e_{2}\right]  .
\]
Hence, we can rewrite (\ref{p133.mp}) as
\begin{align}
\mathfrak{m}  &  =\left(  \pm\left[  e_{1}\ o_{1}\right]  \left[  e_{2}%
\ o_{2}\right]  \pm\left[  e_{1}\ o_{2}\right]  \left[  o_{1}\ e_{2}\right]
\right)  \ \mathfrak{p}\nonumber\\
&  =\pm\left[  e_{1}\ o_{1}\right]  \left[  e_{2}\ o_{2}\right]
\ \mathfrak{p}\pm\left[  e_{1}\ o_{2}\right]  \left[  o_{1}\ e_{2}\right]
\ \mathfrak{p}. \label{p133.m2}%
\end{align}
However, both bracket monomials $\left[  e_{1}\ o_{1}\right]  \left[
e_{2}\ o_{2}\right]  \ \mathfrak{p}$ and $\left[  e_{1}\ o_{2}\right]  \left[
o_{1}\ e_{2}\right]  \ \mathfrak{p}$ are regular of degree $d$ again (since
they are obtained from $\left[  e_{1}\ e_{2}\right]  \left[  o_{1}%
\ o_{2}\right]  \ \mathfrak{p}=\mathfrak{m}$ by replacing the $\left[
e_{1}\ e_{2}\right]  \left[  o_{1}\ o_{2}\right]  $ factor with $\left[
e_{1}\ o_{1}\right]  \left[  e_{2}\ o_{2}\right]  $ or $\left[  e_{1}%
\ o_{2}\right]  \left[  o_{1}\ e_{2}\right]  $, respectively; but this
replacement takes away equally many edges from each vertex as it adds to that
vertex), but have a higher number of even-odd edges\footnote{An \emph{even-odd
edge} means an edge that has an even and an odd endpoint.} than $\mathfrak{m}$
does (since all the four new edges $\left\{  e_{1},o_{1}\right\}  ,\ \left\{
e_{2},\ o_{2}\right\}  ,\ \left\{  e_{1},\ o_{2}\right\}  ,\ \left\{
o_{1},\ e_{2}\right\}  $ are even-odd edges, while the original two edges
$\left\{  e_{1},e_{2}\right\}  $ and $\left\{  o_{1},o_{2}\right\}  $ are
not). By descending induction on the number of even-odd edges, we can thus
assume that $\left[  e_{1}\ o_{1}\right]  \left[  e_{2}\ o_{2}\right]
\ \mathfrak{p}$ and $\left[  e_{1}\ o_{2}\right]  \left[  o_{1}\ e_{2}\right]
\ \mathfrak{p}$ are polynomial functions of the lowlies (the base case is the
case when $\mathfrak{m}$ is bipartite; this case has already been handled).
Then, the equality (\ref{p133.m2}) shows that $\mathfrak{m}$ is a polynomial
function of the lowlies as well. This completes our proof in Case 2.
\end{itemize}

Hence, the proof of Theorem 3.7.3 is complete in both cases. $\blacksquare$

(Note that Case 1 and Case 2 in the above proof are not as different as they
look, since the proof of
\href{https://en.wikipedia.org/wiki/2-factor_theorem}{Petersen's 2-factor
theorem} also relies on
\href{https://math.stackexchange.com/questions/1805181/prove-that-a-k-regular-bipartite-graph-has-a-perfect-matching}{Frobenius's
matching theorem}.)

(Note also that the proof of Theorem 3.7.3 given in \S 6.3 of
\href{https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-10/issue-1/The-invariant-theory-of-binary-forms/bams/1183551414.full}{Joseph
P. S. Kung, Gian-Carlo Rota, \textit{The invariant theory of binary forms},
Bull. Amer. Math. Soc. (N.S.) \textbf{10}(1) (1984), pp. 27--85}. appears
fairly similar to the above.)

\item \textbf{page 135, exercise (4):} \textquotedblleft set
for\textquotedblright\ should be \textquotedblleft sets for\textquotedblright.

\item \textbf{page 137:} \textquotedblleft A comprehensive introduction with
many geometric applications can be found in Fulton and Harris
(1991)\textquotedblright: Here are a few other references (which I found more
readable than Fulton and Harris):

\begin{itemize}
\item \href{https://doi.org/10.1017/CBO9780511626241}{William Fulton,
\textit{Young Tableaux, With Applications to Representation Theory and
Geometry}, Cambridge University Press 1999}, Chapter 8.
(\href{https://mathoverflow.net/questions/456463/errata-for-fultons-young-tableaux}{Errata}%
.)

\item \href{https://math.mit.edu/~etingof/books.html}{Pavel Etingof, Oleg
Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob,
Elena Yudovina, \textit{Introduction to Representation Theory}, Student
Mathematical Library \textbf{59}, AMS 2011}, Sections \S 5.19--5.23.

\item \href{https://www.math.uni-bielefeld.de/~wcrawley/repinv.pdf}{William
Crawley-Boevey, \textit{Lectures on representation theory and invariant
theory}, 1999.}
(\href{https://www.math.uni-bielefeld.de/~wcrawley/repinv-errata-by-darij-grinberg-2010.pdf}{Errata}%
.)

\item
\href{http://web.archive.org/web/20190619201616/http://www.math.iitb.ac.in/~shripad/Wilberd/KP-Primer.pdf}{Hanspeter
Kraft, Claudio Procesi, \textit{Classical Invariant Theory: A Primer}, July
1996}.
(\href{https://www.cip.ifi.lmu.de/~grinberg/algebra/KP-errata-web.pdf}{Errata}.)
\end{itemize}

\item \textbf{page 138:} \textquotedblleft Note that $\det\left(  A\right)
^{n-1}$ is a common denominator\textquotedblright\ should be \textquotedblleft
Note that $\det\left(  A\right)  $ is a common denominator\textquotedblright.

\item \textbf{page 139, Examples 4.1.5 (e):} Replace \textquotedblleft%
$V=S_{2}\mathbf{C}^{3}$\textquotedblright\ by \textquotedblleft$V=S_{3}%
\mathbf{C}^{2}$\textquotedblright.

\item \textbf{page 140, definition of irreducibility:} \textquotedblleft We
say that $\left(  V,\rho\right)  $ is \textit{irreducible}
if\textquotedblright\ should be \textquotedblleft We say that a nonzero
$\Gamma$-representation $\left(  V,\rho\right)  $ is \textit{irreducible}
if\textquotedblright.

\item \textbf{page 140:} After \textquotedblleft For instance, the
representations $S_{d}\mathbf{C}^{n}$ and $\wedge_{d}\mathbf{C}^{n}$ are
irreducible\textquotedblright, add \textquotedblleft(except that $\wedge
_{d}\mathbf{C}^{n}=0$ for $d>n$)\textquotedblright.

\item \textbf{page 140, Theorem 4.1.7:} \textquotedblleft of irreducible
representation\textquotedblright\ should be \textquotedblleft of irreducible
representations\textquotedblright.

\item \textbf{page 141, definition of standard Young tableaux:} It is worth
pointing out that the notion of a \textquotedblleft standard Young
tableaux\textquotedblright\ as defined here is different from than the notion
of a \textquotedblleft standard tableau\textquotedblright\ defined in \S 3.1.
(The latter tableaux can have equal entries, while the former can have
non-rectangular shapes.)

\item \textbf{page 141, definition of the Young symmetrizer:} The Young
symmetrizer $c_{T}$ is not \textquotedblleft an idempotent linear
map\textquotedblright, but only a quasi-idempotent linear map -- i.e., we have
$c_{T}^{2}=kc_{T}$ for some nonzero rational number $k$. Specifically, this
factor $k$ is the positive integer $n!/f^{\lambda}$, where $f^{\lambda}$ is
the number of \textsc{syt}$\lambda$. Thus, the scalar multiple $\dfrac
{f^{\lambda}}{n!}c_{T}$ of $c_{T}$ is an actual idempotent. Maybe it is this
multiple that you want to call the \textquotedblleft Young
symmetrizer\textquotedblright.

(It is worth noting that many authors define the Young symmetrizer to be not
the linear operator $c_{T}$, but rather the element
\[
\sum_{\sigma\in\operatorname*{colstb}\left(  T\right)  }\ \ \sum
_{T\in\operatorname*{rowstb}\left(  T\right)  }\left(  \operatorname*{sign}%
\sigma\right)  \cdot\sigma\tau\ \ \ \ \ \ \ \ \ \ \text{in the group algebra
}\mathbf{C}\left[  S_{d}\right]  ,
\]
whose action on $\otimes_{d}\mathbf{C}^{n}$ (from the right, by permuting
tensorands) is your linear operator $c_{T}$. Some also define it as the
element
\[
\sum_{\sigma\in\operatorname*{colstb}\left(  T\right)  }\ \ \sum
_{T\in\operatorname*{rowstb}\left(  T\right)  }\left(  \operatorname*{sign}%
\sigma\right)  \cdot\tau\sigma\ \ \ \ \ \ \ \ \ \ \text{in the group algebra
}\mathbf{C}\left[  S_{d}\right]  ,
\]
whose action on $\otimes_{d}\mathbf{C}^{n}$ (now from the left, by permuting
tensorands) is your linear operator $c_{T}$.)

\item \textbf{page 141:} \textquotedblleft Thus $W_{T}\mathbf{C}^{n}$ is the
subspace of all tensors in $\otimes_{d}\mathbf{C}^{n}$ which are symmetric
with respect to the rows of $T$ and antisymmetric with respect to the columns
of $T$\textquotedblright: This is false. For instance, for $n=2$ and $d=3$ and
$\lambda=\left(  2,1\right)  $ and $T=%
\begin{array}
[c]{cc}%
1 & 2\\
3 &
\end{array}
$, the only tensor in $\otimes_{d}\mathbf{C}^{n}$ that is symmetric in its
first two tensor factors and antisymmetric in its first and third tensor
factors is $0$, but $W_{T}\mathbf{C}^{n}$ is not $0$.

\item \textbf{page 141, Theorem 4.1.11:} Let me outline a proof of this
theorem, as it is not easily found in the literature.

\textit{Proof of Theorem 4.1.11 (sketched).} We shall show that%
\begin{equation}
\otimes_{d}\mathbf{C}^{n}=\bigoplus_{\lambda\vdash d}\ \ \bigoplus_{T\text{
\textsc{syt}}\lambda}W_{T}\mathbf{C}^{n} \label{p141.main}%
\end{equation}
(an internal direct sum, not just an isomorphism). For this purpose, we define
some notations. If $T$ is any standard Young tableau with $d$ entries, then
$r\left(  T\right)  $ shall denote the sequence $\left(  r_{d},r_{d-1}%
,\ldots,r_{1}\right)  \in\mathbf{N}^{d}$, where $r_{i}$ is the number of the
row of $T$ that contains the entry $i$ (that is, entry $i$ appears in the
$r_{i}$-th row of $T$). For example, $r\left(
\begin{array}
[c]{ccc}%
1 & 3 & 4\\
2 & 5 & \\
6 & 7 &
\end{array}
\right)  =\left(  3,3,2,1,1,2,1\right)  $. Clearly, a standard Young tableau
$T$ is uniquely determined by this sequence $r\left(  T\right)  $ (since
$r\left(  T\right)  $ tells us which row contains which entries, and the order
of the entries in a row must be increasing in order for $T$ to be standard).

Now, consider the set
\[
\operatorname*{SYT}\left(  d\right)  :=\bigcup_{\lambda\vdash d}\left\{
T\text{ \textsc{syt}}\lambda\right\}
\]
of all standard Young tableaux of shape $\lambda$ for all partitions $\lambda$
of $d$. We equip this set $\operatorname*{SYT}\left(  d\right)  $ with the
following total order: For two standard Young tableaux $S$ and $T$ of shapes
$\lambda$ and $\mu$, we say that $S>T$ if and only if

\begin{itemize}
\item either $\lambda>\mu$ in the lexicographic order on $\mathbf{N}^{n}$,

\item or $\lambda=\mu$ and $r\left(  S\right)  >r\left(  T\right)  $ in the
lexicographic order on $\mathbf{N}^{d}$.
\end{itemize}

It is clear that this gives a well-defined total order on $\operatorname*{SYT}%
\left(  d\right)  $ (in fact, it is simply the lexicographic order for the
concatenations $\lambda\oplus r\left(  T\right)  $).

Next, for each $T\in\operatorname*{SYT}\left(  d\right)  $, we let
$\mathbf{E}_{T}$ denote the element%
\begin{align}
&  \left(  \sum_{\sigma\in\operatorname*{colstb}\left(  T\right)  }\left(
\operatorname*{sign}\sigma\right)  \cdot\sigma\right)  \left(  \sum
_{T\in\operatorname*{rowstb}\left(  T\right)  }\tau\right) \nonumber\\
&  =\sum_{\sigma\in\operatorname*{colstb}\left(  T\right)  }\ \ \sum
_{T\in\operatorname*{rowstb}\left(  T\right)  }\left(  \operatorname*{sign}%
\sigma\right)  \cdot\sigma\tau\label{p141.ET}%
\end{align}
in the group algebra $\mathbf{C}\left[  S_{d}\right]  $. This element
$\mathbf{E}_{T}$ is called the \emph{Young symmetrizer} of $T$ (and is denoted
by $\mathbf{E}_{T}$) in my notes
\href{https://arxiv.org/abs/2507.20706v1}{Darij Grinberg, \textit{An
introduction to the symmetric group algebra [Math 701, Spring 2024 lecture
notes]}, arXiv:2507.20706v1} (which I shall henceforth cite as [sga]). Note
that this element $\mathbf{E}_{T}$ is quasi-idempotent, and specifically, its
square is
\begin{equation}
\mathbf{E}_{T}^{2}=\dfrac{n!}{f^{\lambda}}\mathbf{E}_{T}, \label{p141.ETET}%
\end{equation}
where $f^{\lambda}$ is the number of \textsc{syt}$\lambda$ (by Theorem 5.11.3
in [sga]). The Young symmetrizer $c_{T}$ as you define it is the action of
this element $\mathbf{E}_{T}$ on $\otimes_{d}\mathbf{C}^{n}$ from the right
(by permuting tensorands). Thus,
\begin{equation}
W_{T}\mathbf{C}^{n}=c_{T}\cdot\otimes_{d}\mathbf{C}^{n}=\left(  \otimes
_{d}\mathbf{C}^{n}\right)  \mathbf{E}_{T} \label{p141.WTET}%
\end{equation}
for each $T\in\operatorname*{SYT}\left(  d\right)  $.

Now, it can be shown that any two standard Young tableaux $S,T\in
\operatorname*{SYT}\left(  d\right)  $ satisfy%
\begin{equation}
\mathbf{E}_{S}\mathbf{E}_{T}=0\ \ \ \ \ \ \ \ \ \ \text{if}%
\ \ \ \ \ \ \ \ \ \ S>T \label{p141.ESET}%
\end{equation}
(with respect to the above-defined total order on $\operatorname*{SYT}\left(
d\right)  $). Indeed, letting $S$ and $T$ have shapes $\lambda$ and $\mu$,
respectively, we can argue as follows:

\begin{itemize}
\item If $\lambda>\mu$ in the lexicographic order on $\mathbf{N}^{n}$, then
$\lambda\neq\mu$, and thus $\mathbf{E}_{S}\mathbf{E}_{T}=0$ follows from
Proposition 5.11.15 in [sga] (which makes the even stronger claim that
$\mathbf{E}_{S}\mathbf{aE}_{T}=0$ for each $\mathbf{a}\in\mathbf{C}\left[
S_{d}\right]  $).

\item If $\lambda=\mu$ and $r\left(  S\right)  >r\left(  T\right)  $ in the
lexicographic order on $\mathbf{N}^{d}$, then $\mathbf{E}_{S}\mathbf{E}_{T}=0$
follows from Lemma 5.15.21 in [sga], since $r\left(  S\right)  >r\left(
T\right)  $ is saying that $\overline{S}>\overline{T}$ with respect to the
Young last letter order on $n$-tabloids (see [sga] for details).
\end{itemize}

In either case, we get $\mathbf{E}_{S}\mathbf{E}_{T}=0$, so that
(\ref{p141.ESET}) is proved.

Now, we claim the following:

\begin{statement}
\textit{Claim 1: }The sum
\begin{equation}
\sum_{T\in\operatorname*{SYT}\left(  d\right)  }W_{T}\mathbf{C}^{n}
\label{p141.sum}%
\end{equation}
of subspaces of $\otimes_{d}\mathbf{C}^{n}$ is a direct sum.
\end{statement}

\begin{statement}
\textit{Claim 2:} We have%
\[
\otimes_{d}\mathbf{C}^{n}=\sum_{T\in\operatorname*{SYT}\left(  d\right)
}W_{T}\mathbf{C}^{n}.
\]

\end{statement}

\begin{proof}
[Proof of Claim 1.]Assume the contrary. Thus, there exists a family $\left(
w_{T}\right)  _{T\in\operatorname*{SYT}\left(  d\right)  }$ of vectors
$w_{T}\in W_{T}\mathbf{C}^{n}$ such that not all these vectors $w_{T}$ are
zero, but the sum $\sum_{T\in\operatorname*{SYT}\left(  d\right)  }w_{T}$ is
$0$. Consider such a family. Thus, there exists some $Q\in\operatorname*{SYT}%
\left(  d\right)  $ such that $w_{Q}\neq0$. Pick the smallest such $Q$ (with
respect to the above-defined total order). Thus,
\begin{equation}
w_{P}=0\ \ \ \ \ \ \ \ \ \ \text{for all }P<Q. \label{p141.c1.pf.1}%
\end{equation}


We have $w_{Q}\in W_{Q}\mathbf{C}^{n}=\left(  \otimes_{d}\mathbf{C}%
^{n}\right)  \mathbf{E}_{Q}$ (by (\ref{p141.WTET})). But (\ref{p141.ETET})
shows that $\mathbf{E}_{Q}^{2}=\dfrac{n!}{f^{\lambda}}\mathbf{E}_{Q}$. Hence,
each $a\in W_{Q}\mathbf{C}^{n}$ satisfies $a\mathbf{E}_{Q}=\dfrac
{n!}{f^{\lambda}}a$ (because $a\in W_{Q}\mathbf{C}^{n}=\left(  \otimes
_{d}\mathbf{C}^{n}\right)  \mathbf{E}_{Q}$ allows us to write $a$ as
$a=b\mathbf{E}_{Q}$ for some $b\in\otimes_{d}\mathbf{C}^{n}$, and therefore we
have $a\mathbf{E}_{Q}=b\underbrace{\mathbf{E}_{Q}\mathbf{E}_{Q}}%
_{=\mathbf{E}_{Q}^{2}=\dfrac{n!}{f^{\lambda}}\mathbf{E}_{Q}}=\dfrac
{n!}{f^{\lambda}}\underbrace{b\mathbf{E}_{Q}}_{=a}=\dfrac{n!}{f^{\lambda}}a$).
Applying this to $a=w_{Q}$, we obtain $w_{Q}\mathbf{E}_{Q}=\dfrac
{n!}{f^{\lambda}}w_{Q}\neq0$ (since $\dfrac{n!}{f^{\lambda}}\neq0$ and
$w_{Q}\neq0$).

On the other hand, if $P\in\operatorname*{SYT}\left(  d\right)  $ satisfies
$P>Q$, then $\mathbf{E}_{P}\mathbf{E}_{Q}=0$ (by (\ref{p141.ESET})) and thus
\[
\underbrace{w_{P}}_{\substack{\in W_{P}\mathbf{C}^{n}\\=\left(  \otimes
_{d}\mathbf{C}^{n}\right)  \mathbf{E}_{P}}}\mathbf{E}_{Q}\in\left(
\otimes_{d}\mathbf{C}^{n}\right)  \underbrace{\mathbf{E}_{P}\mathbf{E}_{Q}%
}_{=0}=0,
\]
so that
\begin{equation}
w_{P}\mathbf{E}_{Q}=0. \label{p141.c1.pf.3}%
\end{equation}


Now, we assumed that the sum $\sum_{T\in\operatorname*{SYT}\left(  d\right)
}w_{T}$ is $0$. Thus, $0=\sum_{T\in\operatorname*{SYT}\left(  d\right)  }%
w_{T}=\sum_{P\in\operatorname*{SYT}\left(  d\right)  }w_{P}$. Multiplying this
equality by $\mathbf{E}_{Q}$ from the right, we obtain%
\begin{align*}
0  &  =\left(  \sum_{P\in\operatorname*{SYT}\left(  d\right)  }w_{P}\right)
\mathbf{E}_{Q}=\sum_{P\in\operatorname*{SYT}\left(  d\right)  }w_{P}%
\mathbf{E}_{Q}\\
&  =\sum_{\substack{P\in\operatorname*{SYT}\left(  d\right)  ;\\P<Q}%
}\underbrace{w_{P}}_{\substack{=0\\\text{(by (\ref{p141.c1.pf.1}))}%
}}\mathbf{E}_{Q}+w_{Q}\mathbf{E}_{Q}+\sum_{\substack{P\in\operatorname*{SYT}%
\left(  d\right)  ;\\P>Q}}\underbrace{w_{P}\mathbf{E}_{Q}}%
_{\substack{=0\\\text{(by (\ref{p141.c1.pf.3}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }P\in\operatorname*{SYT}\left(  d\right)  \text{ satisfies}\\
\text{either }P<Q\text{ or }P=Q\text{ or }P>Q
\end{array}
\right) \\
&  =w_{Q}\mathbf{E}_{Q}\neq0.
\end{align*}
This obvious contradiction shows that our assumption was false; hence, Claim 1
is proved.
\end{proof}

\begin{proof}
[Proof of Claim 2.]Let $\mathcal{A}=\mathbf{C}\left[  S_{d}\right]  $. From
Corollary 5.15.25 in [sga], we know that%
\[
\mathcal{A}=\underbrace{\bigoplus_{\lambda\vdash d}\ \ \bigoplus_{T\text{
\textsc{syt}}\lambda}}_{=\bigoplus\limits_{T\in\operatorname*{SYT}\left(
d\right)  }}\mathcal{A}\mathbf{E}_{T}=\bigoplus\limits_{T\in
\operatorname*{SYT}\left(  d\right)  }\mathcal{A}\mathbf{E}_{T}=\sum
_{T\in\operatorname*{SYT}\left(  d\right)  }\mathcal{A}\mathbf{E}_{T}.
\]
Hence, in particular, $1_{\mathcal{A}}\in\sum_{T\in\operatorname*{SYT}\left(
d\right)  }\mathcal{A}\mathbf{E}_{T}$, so that we can write $1_{\mathcal{A}}$
in the form%
\begin{equation}
1_{\mathcal{A}}=\sum_{T\in\operatorname*{SYT}\left(  d\right)  }%
a_{T}\mathbf{E}_{T}\ \ \ \ \ \ \ \ \ \ \text{for some }a_{T}\in\mathcal{A}.
\label{p141.c2.pf.2}%
\end{equation}
Consider these $a_{T}$. Now, for each $w\in\otimes_{d}\mathbf{C}^{n}$, we have%
\begin{align*}
w  &  =w1_{\mathcal{A}}=w\sum_{T\in\operatorname*{SYT}\left(  d\right)  }%
a_{T}\mathbf{E}_{T}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{p141.c2.pf.2}%
)}\right) \\
&  =\sum_{T\in\operatorname*{SYT}\left(  d\right)  }\underbrace{wa_{T}%
\mathbf{E}_{T}}_{\substack{\in\left(  \otimes_{d}\mathbf{C}^{n}\right)
\mathbf{E}_{T}\\=W_{T}\mathbf{C}^{n}}}\in\sum_{T\in\operatorname*{SYT}\left(
d\right)  }W_{T}\mathbf{C}^{n}.
\end{align*}
In other words, $\otimes_{d}\mathbf{C}^{n}\subseteq\sum_{T\in
\operatorname*{SYT}\left(  d\right)  }W_{T}\mathbf{C}^{n}$. Hence,%
\[
\otimes_{d}\mathbf{C}^{n}=\sum_{T\in\operatorname*{SYT}\left(  d\right)
}W_{T}\mathbf{C}^{n}.
\]
This proves Claim 2.
\end{proof}

Combining Claim 1 with Claim 2, we obtain
\begin{align*}
\otimes_{d}\mathbf{C}^{n}  &  =\bigoplus_{T\in\operatorname*{SYT}\left(
d\right)  }W_{T}\mathbf{C}^{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{internal
direct sum}\right) \\
&  =\bigoplus_{\lambda\vdash d}\ \ \bigoplus_{T\text{ \textsc{syt}}\lambda
}W_{T}\mathbf{C}^{n}.
\end{align*}
This proves (\ref{p141.main}) and therefore also Theorem 4.1.11.
$\blacksquare$

\item \textbf{page 142, (4.1.5):} Replace \textquotedblleft$c_{\substack{1\\2}%
}\oplus c_{1\ 2}$\textquotedblright\ by \textquotedblleft$c_{1\ 2}\oplus
c_{\substack{1\\2}}$\textquotedblright\ in order to match the order of the
addends to the right of the arrow.

Also, it would be better to multiply \textquotedblleft$\left(  v_{1}\otimes
v_{2}+v_{2}\otimes v_{1}\right)  +\left(  v_{1}\otimes v_{2}-v_{2}\otimes
v_{1}\right)  $\textquotedblright\ by $\dfrac{1}{2}$, so that the map becomes
the identity map.

\item \textbf{page 142, Theorem 4.1.12:} It is worth saying that this theorem
is a direct consequence of Corollary 4.1.14 (1) (which I prove below) and of
the direct sum decomposition (\ref{p141.main}).

\item \textbf{page 143, Corollary 4.1.14:} Let me give a reference for the
proof of this corollary, since it is not easy to locate in the literature.

I will use the book \href{https://doi.org/10.1017/CBO9780511626241}{William
Fulton, \textit{Young Tableaux, With Applications to Representation Theory and
Geometry}, Cambridge University Press 1999}, (see also the
\href{https://mathoverflow.net/questions/456463/errata-for-fultons-young-tableaux}{errata}%
). In the following, this book will be cited as [Fulton].

\textit{Proof of Corollary 4.1.14.} \textbf{(1)} Set $E:=\mathbf{C}^{n}$. Let
$\mathbf{E}_{T}\in\mathbf{C}\left[  S_{d}\right]  $ be the Young symmetrizer
defined in (\ref{p141.ET}). By the definition of $W_{T}\mathbf{C}^{n}$, we
have%
\begin{align*}
W_{T}\mathbf{C}^{n}  &  =c_{T}\left(  \otimes_{d}\mathbf{C}^{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }W_{T}\mathbf{C}%
^{n}\right) \\
&  =c_{T}\left(  \otimes_{d}E\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathbf{C}^{n}=E\right) \\
&  =c_{T}\left(  E^{\otimes d}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we switched from the notation }\otimes_{d}E\text{ to the}\\
\text{more convenient notation }E^{\otimes d}\text{ for the same thing}%
\end{array}
\right) \\
&  =E^{\otimes d}\mathbf{E}_{T}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
c_{T}\text{ is right multiplication by }\mathbf{E}_{T}\right)  .
\end{align*}
Note that [Fulton] denotes $\mathbf{E}_{T}$ as $c_{T}$.

Now, [Fulton] considers the $\Gamma$-module $E^{\lambda}$, about which he
claims (on page 119 of [Fulton], in the paragraph below Exercise 11 in \S 8.3)
that (in his notations, which are different from ours!) \textquotedblleft%
$E^{\lambda}$ is isomorphic to the image of the map $E^{\otimes n}\rightarrow
E^{\otimes n}$ that is right multiplication by $c_{U}$\textquotedblright.
Translated into our notations (noting that our $d,n,E,T,U,\mathbf{E}_{T}$
correspond to [Fulton]'s $n,d,\mathbf{C}^{n},U,T,c_{U}$), this is saying that
[Fulton]'s $E^{\lambda}$ is isomorphic to the image of the map $E^{\otimes
d}\rightarrow E^{\otimes d}$ that is right multiplication by $\mathbf{E}_{T}$.
In other words, [Fulton]'s $E^{\lambda}$ is isomorphic to $E^{\otimes
d}\mathbf{E}_{T}=W_{T}\mathbf{C}^{n}$. This isomorphism is constructed as
follows:%
\begin{align}
E^{\lambda}  &  \cong E\left(  S^{\lambda}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by Proposition 1 in [Fulton]'s \S 8.3}\right) \nonumber\\
&  =E^{\otimes d}\otimes_{\mathbf{C}\left[  S_{d}\right]  }%
\underbrace{S^{\lambda}}_{\cong\mathbf{C}\left[  S_{d}\right]  \cdot
\mathbf{E}_{T}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the
}E\text{ functor}\right) \nonumber\\
&  \cong E^{\otimes d}\otimes_{\mathbf{C}\left[  S_{d}\right]  }\left(
\mathbf{C}\left[  S_{d}\right]  \cdot\mathbf{E}_{T}\right) \nonumber\\
&  \cong E^{\otimes d}\mathbf{E}_{T} \label{p143.iso}%
\end{align}
(where the last $\cong$ sign is a particular case of the isomorphism
$M\otimes_{A}\left(  Ae\right)  \cong Me$ for each ring $A$, each right
$A$-module $M$ and each idempotent $e\in A$).

Using the above isomorphism $E^{\lambda}\rightarrow W_{T}\mathbf{C}^{n}$, the
basis of $E^{\lambda}$ constructed in Theorem 1 of [Fulton]'s \S 8.1 can be
translated into a basis of $W_{T}\mathbf{C}^{n}$. Let us see what comes out of
this translation. Theorem 1 of [Fulton]'s \S 8.1 claims that the vectors
$e_{U}$, where $U$ ranges over all semistandard tableaux of shape $\lambda$,
form a basis of the vector space $E^{\lambda}$ (recall that our $T$ is
[Fulton]'s $U$, and vice versa!). Consider a semistandard tableau $U$ of shape
$\lambda$, and the corresponding basis vector $e_{U}$. The isomorphism
$E^{\lambda}\rightarrow E\left(  S^{\lambda}\right)  $ from Proposition 1 in
[Fulton]'s \S 8.3 maps this basis vector $e_{U}\in E^{\lambda}$ to the tensor
$\mathbf{v}\left(  T\right)  \otimes v_{T}$ (again, our $T$ is [Fulton]'s $U$,
so this looks like \textquotedblleft$\mathbf{v}\left(  U\right)  \otimes
v_{U}$\textquotedblright\ in [Fulton]), where $\mathbf{v}\left(  T\right)  \in
E^{\otimes d}$ is the pure tensor that we call $e_{\left(  T,U\right)  }$
(defined in the exact same way), while $v_{T}$ is the polytabloid in the
Specht module $S^{\lambda}$ corresponding to the standard tableau $T$. The
isomorphism $S^{\lambda}\rightarrow\mathbf{C}\left[  S_{d}\right]
\cdot\mathbf{E}_{T}$ sends this latter polytabloid $v_{T}$ to $\mathbf{E}_{T}%
$, and thus the induced isomorphism $E^{\otimes d}\otimes_{\mathbf{C}\left[
S_{d}\right]  }S^{\lambda}\rightarrow E^{\otimes d}\otimes_{\mathbf{C}\left[
S_{d}\right]  }\left(  \mathbf{C}\left[  S_{d}\right]  \cdot\mathbf{E}%
_{T}\right)  $ sends $\mathbf{v}\left(  T\right)  \otimes v_{T}$ to
$\mathbf{v}\left(  T\right)  \otimes\mathbf{E}_{T}$. Finally, the isomorphism
$E^{\otimes d}\otimes_{\mathbf{C}\left[  S_{d}\right]  }\left(  \mathbf{C}%
\left[  S_{d}\right]  \cdot\mathbf{E}_{T}\right)  \rightarrow E^{\otimes
d}\mathbf{E}_{T}$ sends $\mathbf{v}\left(  T\right)  \otimes\mathbf{E}_{T}$ to
$\mathbf{v}\left(  T\right)  \mathbf{E}_{T}=e_{\left(  T,U\right)  }%
\mathbf{E}_{T}$ (since $\mathbf{v}\left(  T\right)  =e_{\left(  T,U\right)  }%
$). Altogether, we thus see that our isomorphism $E^{\lambda}\rightarrow
W_{T}\mathbf{C}^{n}$ constructed in (\ref{p143.iso}) sends $e_{U}$ to
$e_{\left(  T,U\right)  }\mathbf{E}_{T}$ (with the intermediate steps being
$e_{U}\mapsto\mathbf{v}\left(  T\right)  \otimes v_{T}\mapsto\mathbf{v}\left(
T\right)  \otimes\mathbf{E}_{T}\mapsto e_{\left(  T,U\right)  }\mathbf{E}_{T}%
$). Hence, it sends the whole basis $\left(  e_{U}\right)  _{U\text{
\textsc{ssyt}}\lambda}$ of $E^{\lambda}$ to a basis $\left(  e_{\left(
T,U\right)  }\mathbf{E}_{T}\right)  _{U\text{ \textsc{ssyt}}\lambda}$ of
$W_{T}\mathbf{C}^{n}$. We can rewrite this latter basis further as $\left(
c_{T}\left(  e_{\left(  T,U\right)  }\right)  \right)  _{U\text{
\textsc{ssyt}}\lambda}$ (since $c_{T}$ is right multiplication by
$\mathbf{E}_{T}$, so that we have $c_{T}\left(  e_{\left(  T,U\right)
}\right)  =e_{\left(  T,U\right)  }\mathbf{E}_{T}$ for each $U$ \textsc{ssyt}%
$\lambda$). Thus, we have shown that $\left(  c_{T}\left(  e_{\left(
T,U\right)  }\right)  \right)  _{U\text{ \textsc{ssyt}}\lambda}$ is a basis of
$W_{T}\mathbf{C}^{n}$. This proves Corollary 4.1.14 \textbf{(1)}.

\textbf{(2)} Consider a diagonal matrix $\operatorname*{diag}\left(
t_{1},t_{2},\ldots,t_{n}\right)  \in\Gamma$. We compute the trace
$\operatorname*{trace}\left(  \rho\left(  \operatorname*{diag}\left(
t_{1},t_{2},\ldots,t_{n}\right)  \right)  \right)  $ using the basis of
$W_{T}\mathbf{C}^{n}$ given in Corollary 4.1.14 \textbf{(1)}. The map
$\rho\left(  \operatorname*{diag}\left(  t_{1},t_{2},\ldots,t_{n}\right)
\right)  $ acts on a basis vector $c_{T}\left(  e_{\left(  T,U\right)
}\right)  $ by scaling it by the factor%
\[
\rho\left(  \operatorname*{diag}\left(  t_{1},t_{2},\ldots,t_{n}\right)
\right)  \cdot c_{T}\left(  e_{\left(  T,U\right)  }\right)  =\prod_{i=1}%
^{n}t_{i}^{\#\text{ }i\text{'s in }U}%
\]
(since $c_{T}\left(  e_{\left(  T,U\right)  }\right)  $ is a linear
combination of permutations of the pure tensor $e_{\left(  T,U\right)  }$, and
the latter pure tensor contains each $e_{i}$ as many times as there are $i$'s
in $U$). Thus, $\operatorname*{trace}\left(  \rho\left(  \operatorname*{diag}%
\left(  t_{1},t_{2},\ldots,t_{n}\right)  \right)  \right)  $ is the sum of
these factors $\prod_{i=1}^{n}t_{i}^{\#\text{ }i\text{'s in }U}$ over all $U$
\textsc{ssyt}$\lambda$. That is,%
\[
\operatorname*{trace}\left(  \rho\left(  \operatorname*{diag}\left(
t_{1},t_{2},\ldots,t_{n}\right)  \right)  \right)  =\sum_{U\text{
\textsc{ssyt}}\lambda}\ \ \prod_{i=1}^{n}t_{i}^{\#\text{ }i\text{'s in }%
U}=s_{\lambda}\left(  t_{1},t_{2},\ldots,t_{n}\right)  .
\]
This proves Corollary 4.1.14 \textbf{(2)}. $\blacksquare$

\item \textbf{page 144:} \textquotedblleft Any partition $\lambda\vdash d$ can
be encoded into a monomial $\omega\left(  \lambda\right)  :=t_{1}^{\nu_{1}%
}t_{2}^{\nu_{2}}\cdots t_{n}^{\nu_{n}}$ as follows: the exponent $\nu_{i}$ is
the cardinality of the $i$-th column in the Ferrers diagram of $\lambda$.
Equivalently, $\nu_{i}=\#\left\{  j:\lambda_{j}\geq i\right\}  $. It is easy
to see that $\omega\left(  \lambda\right)  $ is the lexicographically leading
monomial of the Schur polynomial $s_{\lambda}\left(  t_{1},t_{2},\ldots
,t_{n}\right)  $\textquotedblright\ should be \textquotedblleft Any partition
$\lambda\vdash d$ can be encoded into a monomial $\omega\left(  \lambda
\right)  :=t_{1}^{\lambda_{1}}t_{2}^{\lambda_{2}}\cdots t_{n}^{\lambda_{n}}$
as follows. It is easy to see that $\omega\left(  \lambda\right)  $ is the
lexicographically leading monomial of the Schur polynomial $s_{\lambda}\left(
t_{1},t_{2},\ldots,t_{n}\right)  $ (where the variables are ordered by
$t_{1}>t_{2}>\cdots>t_{n}$)\textquotedblright.

\item \textbf{page 144, (4.1.8):} The \textquotedblleft$c_{\lambda
}\ W_{\lambda}\mathbf{C}^{n}$\textquotedblright\ here means a direct sum of
$c_{\lambda}$ many copies of $W_{\lambda}\mathbf{C}^{n}$.

\item \textbf{page 144:} I am not sure I would call Algorithm 4.1.16 a
\textquotedblleft subduction algorithm (cf. Algorithm 3.2.8)\textquotedblright%
; the Schur polynomials form a $\mathbf{Z}$-module basis of $\mathbf{Z}\left[
t_{1},\ldots,t_{n}\right]  ^{S_{n}}$, not a sagbi basis.

\item \textbf{page 145, Algorithm 4.1.16:} In Step 2, replace
\textquotedblleft$t_{1}^{\nu_{1}}t_{2}^{\nu_{2}}\cdots t_{n}^{\nu_{n}}%
$\textquotedblright\ by \textquotedblleft$t_{1}^{\lambda_{1}}t_{2}%
^{\lambda_{2}}\cdots t_{n}^{\lambda_{n}}$\textquotedblright.

\item \textbf{page 145, Example 4.1.17:} \textquotedblleft the space of
polynomial functions\textquotedblright\ should be \textquotedblleft the space
of homogeneous polynomial functions\textquotedblright.

\item \textbf{page 145, Example 4.1.17:} In (4.1.12), replace
\textquotedblleft$s_{\left(  3,3,3\right)  }$\textquotedblright\ by
\textquotedblleft$s_{\left(  2,2,2\right)  }$\textquotedblright.

\item \textbf{page 146:} \textquotedblleft the space of homogeneous
polynomials of degree $d$ in the coefficients of a generic homogeneous
polynomials of degree $m$\textquotedblright\ should be \textquotedblleft the
space of homogeneous polynomials of degree $m$ in the coefficients of a
generic homogeneous polynomial of degree $d$\textquotedblright.

\item \textbf{page 147, \S 4.2:} Replace \textquotedblleft$A\circ
f$\textquotedblright\ by \textquotedblleft$f\circ A$\textquotedblright%
\ everywhere in the third paragraph of \S 4.2. (The group $\Gamma$ acts on
$\mathbf{C}\left[  V\right]  $ is a right action, not a left action.)

\item \textbf{page 147, \S 4.2:} \textquotedblleft The symmetric power
$S_{k}\left(  V\right)  $ is a vector space of dimension $\dbinom{m+k-1}{k}$.
We identify it with the space of homogeneous polynomial functions of degree
$k$ on $V$\textquotedblright: This identification is unnatural, and does not
respect the $\Gamma$-action in general. The vector space $\mathbf{C}\left[
V\right]  _{k}^{\Gamma}$ of all homogeneous polynomial functions of degree $k$
on $V$ can be naturally identified with the symmetric power $S_{k}\left(
V^{\ast}\right)  $, where $\Gamma$ acts on the dual space $V^{\ast}$ of $V$ by
the transpose matrix (i.e., where the action of $\Gamma$ on $V^{\ast}$ is
given by $\left(  f\circ A\right)  \left(  v\right)  =f\left(  Av\right)  $
for all $A\in\Gamma$, $f\in V^{\ast}$ and $v\in V$). On both of these spaces,
$\Gamma$ acts \textbf{from the right}. Meanwhile, on the symmetric power
$S_{k}\left(  V\right)  $, the group $\Gamma$ acts \textbf{from the left}.
Even if we transform the left action into a right action by taking inverses
(i.e., we define $Af$ to be $f\circ A^{-1}$), and even if we assume that
$V=\mathbf{C}^{n}$ (in which case there is a \textquotedblleft
natural\textquotedblright\ vector space isomorphism $\mathbf{C}\left[
V\right]  _{k}^{\Gamma}\cong S_{k}\left(  V^{\ast}\right)  \cong S_{k}\left(
V\right)  $ by way of the standard basis), the actions do not become the same
(unless you restrict them to the orthogonal group $\operatorname*{O}\left(
\mathbf{C}^{n}\right)  $).

This discrepancy prevents much of what is being done in Chapter 4 of the book
from being true \textquotedblleft on the nose\textquotedblright. However, the
approach to invariants by means of symmetric powers is nevertheless a valid
one, and much of what is done in Chapter 4 can be adapted once proper fixes
are made. In many cases, it suffices to read $S_{k}\left(  V^{\ast}\right)  $
for $S_{k}\left(  V\right)  $ and make sure to adapt the representation theory
of $\Gamma$ to right (as opposed to left) $\Gamma$-actions. Even without such
an adaptation, the invariant-theoretical insights can often be salvaged, since
it can be shown that for any $k\in\mathbf{N}$ and any representation $\left(
V,\rho\right)  $ of $\Gamma$, we have%
\begin{equation}
\dim\left(  \left(  S_{k}\left(  V^{\ast}\right)  \right)  ^{\Gamma}\right)
=\dim\left(  \left(  S_{k}\left(  V\right)  \right)  ^{\Gamma}\right)
\label{p147.dim=dim}%
\end{equation}
(despite the lack of a $\Gamma$-equivariant isomorphism $S_{k}\left(  V^{\ast
}\right)  \cong S_{k}\left(  V\right)  $).

Let me outline a proof of (\ref{p147.dim=dim}). It relies on the following two facts:

\begin{statement}
\textit{Fact 1:} Let $\left(  W,\tau\right)  $ be any $\Gamma$-representation.
Then, $\dim\left(  \left(  W^{\ast}\right)  ^{\Gamma}\right)  =\dim\left(
W^{\Gamma}\right)  $.
\end{statement}

\begin{statement}
\textit{Fact 2:} Let $\left(  V,\rho\right)  $ be any $\Gamma$-representation.
Let $k\in\mathbf{N}$. Then, $S_{k}\left(  V^{\ast}\right)  \cong\left(
S_{k}\left(  V\right)  \right)  ^{\ast}$ as $\Gamma$-representations.
\end{statement}

\begin{proof}
[Proof of Fact 1 (sketched).]Let's say that $W$ is a left $\Gamma
$-representation. Decompose $W$ into a direct sum $I_{1}\oplus I_{2}%
\oplus\cdots\oplus I_{k}$ of irreducible left $\Gamma$-representations. Then,
$\dim\left(  W^{\Gamma}\right)  $ is the number of $i\in\left\{
1,2,\ldots,k\right\}  $ for which $I_{i}$ is isomorphic to the trivial
$1$-dimensional representation $\mathbf{C}$. But $W\cong I_{1}\oplus
I_{2}\oplus\cdots\oplus I_{k}$ as left $\Gamma$-representations yields
\[
W^{\ast}\cong\left(  I_{1}\oplus I_{2}\oplus\cdots\oplus I_{k}\right)  ^{\ast
}\cong I_{1}^{\ast}\oplus I_{2}^{\ast}\oplus\cdots\oplus I_{k}^{\ast}%
\]
as right $\Gamma$-representations, and moreover, the addends $I_{j}^{\ast}$
here are again irreducible (since the dual of an irreducible left $\Gamma
$-representation is an irreducible right $\Gamma$-representation). Hence,
$\dim\left(  \left(  W^{\ast}\right)  ^{\Gamma}\right)  $ is the number of
$i\in\left\{  1,2,\ldots,k\right\}  $ for which $I_{i}^{\ast}$ is isomorphic
to the trivial $1$-dimensional representation $\mathbf{C}$.

But the $i\in\left\{  1,2,\ldots,k\right\}  $ that satisfy $I_{i}%
\cong\mathbf{C}$ are precisely those $i\in\left\{  1,2,\ldots,k\right\}  $
that satisfy $I_{i}^{\ast}\cong\mathbf{C}$ (since $\mathbf{C}^{\ast}%
\cong\mathbf{C}$). Thus, the above descriptions of $\dim\left(  \left(
W^{\ast}\right)  ^{\Gamma}\right)  $ and of $\dim\left(  W^{\Gamma}\right)  $
boil down to the same thing, and we conclude that $\dim\left(  \left(
W^{\ast}\right)  ^{\Gamma}\right)  =\dim\left(  W^{\Gamma}\right)  $. This
proves Fact 1.
\end{proof}

\begin{proof}
[Proof of Fact 2 (sketched).]There is an obvious bilinear form%
\begin{align*}
\beta:S_{k}\left(  V^{\ast}\right)  \times S_{k}\left(  V\right)   &
\rightarrow\mathbf{C},\\
\left(  f_{1}f_{2}\cdots f_{k},\ v_{1}v_{2}\cdots v_{k}\right)   &
\mapsto\sum_{\sigma\in S_{k}}f_{1}\left(  v_{\sigma\left(  1\right)  }\right)
f_{2}\left(  v_{\sigma\left(  2\right)  }\right)  \cdots f_{k}\left(
v_{\sigma\left(  k\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ =\sum_{\sigma\in S_{k}}f_{\sigma\left(  1\right)
}\left(  v_{1}\right)  f_{\sigma\left(  2\right)  }\left(  v_{2}\right)
\cdots f_{\sigma\left(  k\right)  }\left(  v_{k}\right)  ,
\end{align*}
which is easily seen to be $\Gamma$-tensorial (i.e., any $g\in S_{k}\left(
V^{\ast}\right)  $ and $w\in S_{k}\left(  V\right)  $ and $A\in\Gamma$ satisfy
$\beta\left(  g\circ A,\ w\right)  =\beta\left(  g,\ Aw\right)  $) and
nondegenerate (i.e., it produces an isomorphism $S_{k}\left(  V^{\ast}\right)
\rightarrow\left(  S_{k}\left(  V\right)  \right)  ^{\ast}$%
)\ \ \ \ \footnote{For the nondegeneracy, we have to thank the facts that $V$
is finite-dimensional and that our base field $\mathbf{C}$ has characteristic
$0$.}. Thus, it gives rise to an isomorphism $S_{k}\left(  V^{\ast}\right)
\cong\left(  S_{k}\left(  V\right)  \right)  ^{\ast}$ of right $\Gamma
$-representations. This proves Fact 2.
\end{proof}

Now (\ref{p147.dim=dim}) follows from
\begin{align*}
\dim\left(  \left(  S_{k}\left(  V^{\ast}\right)  \right)  ^{\Gamma}\right)
&  =\dim\left(  \left(  \left(  S_{k}\left(  V\right)  \right)  ^{\ast
}\right)  ^{\Gamma}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Fact
2}\right) \\
&  =\dim\left(  \left(  S_{k}\left(  V\right)  \right)  ^{\Gamma}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Fact 1, applied to }W=S_{k}\left(
V\right)  \right)  .
\end{align*}


\item \textbf{page 147, \S 4.2:} After \textquotedblleft This implies that $f$
is a homogeneous polynomial of degree $gn/d$\textquotedblright, I would add
\textquotedblleft(in fact, we can WLOG assume that $f$ is homogeneous; then,
the assumption $f\circ A=\left(  \det A\right)  ^{g}\cdot f$ rewrites as
$f\left(  \rho\left(  A\right)  \cdot\mathbf{v}\right)  =\left(  \det
A\right)  ^{g}\cdot f\left(  \mathbf{v}\right)  $, but the left hand side of
the latter equality is homogeneous of degree $d\cdot\deg f$ in the entries of
$A$, while the right hand side is homogeneous of degree $gn$ in the entries of
$A$; thus, $d\cdot\deg f=gn$ and therefore $\deg f=gn/d$)\textquotedblright.

\item \textbf{page 148, Proposition 4.2.1:} As remarked above, the polynomial
ring $\mathbf{C}\left[  V\right]  $ is not isomorphic to the symmetric algebra
$S\left(  V\right)  $ of $V$ as a $\Gamma$-module. Consequently,
\textquotedblleft$\mathbf{C}\left[  V\right]  $\textquotedblright\ should be
replaced by \textquotedblleft$S\left(  V\right)  $\textquotedblright%
\ throughout this proposition and its proof.

\item \textbf{page 150, proof of Lemma 4.2.4:} The \textquotedblleft typical
such \textsc{ssyt}$\lambda$\textquotedblright\ looks not like
\[%
\begin{tabular}
[c]{ccccccccc}%
$1$ & $\cdots$ & $1$ & $1$ & $\cdots$ & $1$ & $1$ & $\cdots$ & $1$\\
$1$ & $\cdots$ & $1$ & $2$ & $\cdots$ & $2$ &  &  &
\end{tabular}
\ \ \
\]
but like%
\[%
\begin{tabular}
[c]{ccccccccc}%
$1$ & $\cdots$ & $1$ & $1$ & $\cdots$ & $1$ & $2$ & $\cdots$ & $2$\\
$2$ & $\cdots$ & $2$ &  &  &  &  &  &
\end{tabular}
\ \ \ \ \ .
\]


\item \textbf{page 150, proof of Lemma 4.2.4:} \textquotedblleft By Corollary
4.1.14 (b)\textquotedblright\ should be \textquotedblleft By Corollary 4.1.14
(2)\textquotedblright.

\item \textbf{page 157:} \textquotedblleft which we denote with $\Omega
_{\mathbf{s}}$, $\Omega_{\mathbf{t}}$ and $\Omega_{\mathbf{u}}$
respectively\textquotedblright\ should be \textquotedblleft which we denote
with $\Omega_{\mathbf{t}}$, $\Omega_{\mathbf{s}}$ and $\Omega_{\mathbf{u}}$
respectively\textquotedblright.

\item \textbf{page 157:} \textquotedblleft Given a matrix-valued polynomial
function $\phi$\textquotedblright\ should be \textquotedblleft Given a
polynomial function $\phi\in\mathbf{C}\left[  \mathbf{t}\right]
$\textquotedblright.

\item \textbf{page 157, proof of Theorem 4.3.4:} Replace each
\textquotedblleft$k$\textquotedblright\ in this proof by an \textquotedblleft%
$n$\textquotedblright.

\item \textbf{page 157, proof of Theorem 4.3.4:} In the equality (4.3.5) and
further on, the notation%
\[
\dfrac{\partial^{n}\phi\left(  \mathbf{u}=\mathbf{st}\right)  }{\partial
u_{\sigma_{1},\pi_{1}}\ \partial u_{\sigma_{2},\pi_{2}}\ \cdots\ \partial
u_{\sigma_{n},\pi_{n}}}%
\]
should be understood as
\[
\dfrac{\partial^{n}\phi\left(  \mathbf{u}\right)  }{\partial u_{\sigma_{1}%
,\pi_{1}}\ \partial u_{\sigma_{2},\pi_{2}}\ \cdots\ \partial u_{\sigma_{n}%
,\pi_{n}}}\mid_{\mathbf{u}=\mathbf{st}}%
\]
(that is, we first compute the $n$-th partial derivative $\dfrac{\partial
^{n}\phi\left(  \mathbf{u}\right)  }{\partial u_{\sigma_{1},\pi_{1}}\ \partial
u_{\sigma_{2},\pi_{2}}\ \cdots\ \partial u_{\sigma_{n},\pi_{n}}}$ in the ring
$\mathbf{C}\left[  \mathbf{u}\right]  $, and then substitute the entries of
the matrix $\mathbf{st}$ for the respective entries of $\mathbf{u}$ in the
result we obtained). An alternative (and more standard) way to express the
same thing is%
\[
\dfrac{\partial^{n}\phi\left(  \mathbf{u}\right)  }{\partial u_{\sigma_{1}%
,\pi_{1}}\ \partial u_{\sigma_{2},\pi_{2}}\ \cdots\ \partial u_{\sigma_{n}%
,\pi_{n}}}\left(  \mathbf{st}\right)  .
\]


\item \textbf{page 157, proof of Theorem 4.3.4:} Here is a quick outline of
how (4.3.5) can be proved. Namely, (4.3.5) follows by $n$-times repeated
application of the formula%
\begin{equation}
\dfrac{\partial\phi\left(  \mathbf{st}\right)  }{\partial t_{p,q}}%
=\sum_{\sigma=1}^{n}\left(  \dfrac{\partial\phi\left(  \mathbf{u}\right)
}{\partial u_{\sigma,q}}\right)  \left(  \mathbf{st}\right)  \cdot
s_{\sigma,p}, \label{p157.1diff}%
\end{equation}
which holds for all $p,q\in\left\{  1,2,\ldots,n\right\}  $. This formula
(\ref{p157.1diff}), in turn, is a particular case of the multivariate chain
rule (since $\phi\left(  \mathbf{st}\right)  =\phi\left(  \psi\left(
\mathbf{t}\right)  \right)  $, where $\psi$ is the function sending each entry
$t_{i,j}$ of the matrix $\mathbf{t}$ to the respective entry $\sum_{h=1}%
^{n}s_{i,h}t_{h,j}$ of $\mathbf{st}$). Alternatively, (\ref{p157.1diff}) can
be proved by induction on $\deg\phi$, using the fact that both sides of
(\ref{p157.1diff}) are functions $\omega\left(  \phi\right)  $ of $\phi$ that
satisfy the recurrence%
\[
\omega\left(  \phi\psi\right)  =\phi\left(  \mathbf{st}\right)  \cdot
\omega\left(  \psi\right)  +\omega\left(  \phi\right)  \cdot\omega\left(
\mathbf{st}\right)
\]
(by the Leibniz rule) and send each indeterminate $t_{i,j}$ to $%
\begin{cases}
0, & \text{if }q\neq j;\\
s_{i,p}, & \text{if }q=j
\end{cases}
$ (this is easy to check directly).

\item \textbf{page 157, proof of Theorem 4.3.4:} \textquotedblleft
antisymmetrize the expression (4.3.5) with respect to $\pi\in S_{n}%
$\textquotedblright\ simply means \textquotedblleft multiply (4.3.5) by
$\operatorname*{sign}\left(  \pi\right)  $ and sum the result over all $\pi\in
S_{n}$\textquotedblright. (No further substitution of variables is required,
unlike in a usual antisymmetrization of polynomials.)

\item \textbf{page 158, proof of Theorem 4.3.4:} \textquotedblleft The proof
of the second identity in (4.3.4) is analogous\textquotedblright: This needs
some caveats. Instead of applying (4.3.3) directly with each $t_{i,j}$
replaced by $s_{i,j}$, we have to apply the identity%
\[
\Omega\left(  \phi\right)  =\sum_{\pi\in S_{n}}\operatorname*{sign}\left(
\pi\right)  \cdot\dfrac{\partial^{n}\phi}{\partial t_{\pi_{1},1}\ \partial
t_{\pi_{2},2}\ \cdots\ \partial t_{\pi_{n},n}}%
\]
(which follows from (4.3.3) by first reindexing the sum using the bijection
$\pi\mapsto\pi^{-1}$ and then rewriting the expression $\dfrac{\partial
^{n}\phi}{\partial t_{1,\pi_{1}^{-1}}\ \partial t_{2,\pi_{2}^{-1}}%
\ \cdots\ \partial t_{n,\pi_{n}^{-1}}}$ as $\dfrac{\partial^{n}\phi}{\partial
t_{\pi_{1},1}\ \partial t_{\pi_{2},2}\ \cdots\ \partial t_{\pi_{n},n}}$).

Instead of (\ref{p157.1diff}), we need the analogous formula%
\begin{equation}
\dfrac{\partial\phi\left(  \mathbf{st}\right)  }{\partial s_{p,q}}%
=\sum_{\sigma=1}^{n}\left(  \dfrac{\partial\phi\left(  \mathbf{u}\right)
}{\partial u_{p,\sigma}}\right)  \left(  \mathbf{st}\right)  \cdot
t_{q,\sigma}. \label{p157.1diff2}%
\end{equation}
Instead of (4.3.5), we need the formula%
\begin{align*}
&  \dfrac{\partial^{n}\phi\left(  \mathbf{st}\right)  }{\partial s_{\pi_{1}%
,1}\ \partial s_{\pi_{2},2}\ \cdots\ \partial s_{\pi_{n},n}}\\
&  =\sum_{\sigma_{1},\sigma_{2},\ldots,\sigma_{n}=1}^{n}\dfrac{\partial
^{n}\phi\left(  \mathbf{u}=\mathbf{st}\right)  }{\partial u_{\pi_{1}%
,\sigma_{1}}\ \partial u_{\pi_{2},\sigma_{2}}\ \cdots\ \partial u_{\pi
_{n},\sigma_{n}}}t_{1,\sigma_{1}}t_{2,\sigma_{2}}\cdots t_{n,\sigma_{n}}.
\end{align*}


\item \textbf{page 158, Corollary 4.3.6:} It can be shown that this constant
$c_{p}$ equals $p\left(  p+1\right)  \left(  p+2\right)  \cdots\left(
p+n-1\right)  $. Indeed, Corollary 4.3.6 with this particular formula for
$c_{p}$ is known as the \emph{Cayley identity}, and appears (e.g.) as
Corollary A.4 in \href{https://arxiv.org/abs/0809.3516v2}{Sergio Caracciolo,
Andrea Sportiello, Alan D. Sokal, \textit{Noncommutative determinants,
Cauchy-Binet formulae, and Capelli-type identities. I. Generalizations of the
Capelli and Turnbull identities}, arXiv:0809.3516v2} (see
\href{https://www.cip.ifi.lmu.de/~grinberg/algebra/ncd-errata.pdf}{my errata}).

\item \textbf{page 159:} \textquotedblleft its image $\mathbf{t}\circ
f=f\left(  \mathbf{tv}\right)  $\textquotedblright\ should be
\textquotedblleft its image $f\circ\mathbf{t}=f\left(  \mathbf{tv}\right)
$\textquotedblright.

\item \textbf{page 163, proof of Proposition 4.4.2:} \textquotedblleft as
polynomial functions\textquotedblright\ should be \textquotedblleft as
rational functions\textquotedblright.

\item \textbf{page 163, proof of Proposition 4.4.2:} In (4.4.5), the
\textquotedblleft$\partial x_{i}\ \partial x_{j}$\textquotedblright\ on the
right hand side should be \textquotedblleft$\partial x_{k}\ \partial x_{l}%
$\textquotedblright.

\item \textbf{page 163, definition of the discriminant of a quadratic form:}
You probably want to replace \textquotedblleft$\sum_{i=1}^{n}\sum_{j=i}%
^{n}a_{ij}x_{i}x_{j}$\textquotedblright\ by \textquotedblleft$\sum_{i=1}%
^{n}\sum_{j=1}^{n}a_{ij}x_{i}x_{j}$ (with $a_{ij}=a_{ji}$)\textquotedblright%
\ in order to recover the factors $2$ in (4.4.7).

\item \textbf{page 164, Theorem 4.4.3:} \textquotedblleft ring a
quadratic\textquotedblright\ should be \textquotedblleft ring of a
quadratic\textquotedblright.

\item \textbf{page 169, proof of Lemma 4.5.1:} The \textquotedblleft
if\textquotedblright\ direction of the lemma can also be proved without any
reference to connectedness and Lie groups. Here is an outline of this proof:

\textit{Proof of the \textquotedblleft if\textquotedblright\ direction of
Lemma 4.5.1.} Let $v\in V$ be a vector. Assume that%
\[
\rho^{\ast}\left(  E_{i,j}\right)  \cdot v=0\ \ \ \ \ \ \ \ \ \ \text{for all
}i,j\in\left\{  1,2,\ldots,n\right\}  \text{ with }i\neq j,
\]
and that%
\[
\rho^{\ast}\left(  E_{i,i}\right)  \cdot v=g\cdot
v\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left\{  1,2,\ldots,n\right\}  .
\]
We can combine these two assumptions into one: namely, for all $i,j\in\left\{
1,2,\ldots,n\right\}  $, we have%
\begin{equation}
\rho^{\ast}\left(  E_{i,j}\right)  \cdot v=\delta_{i,j}g\cdot v,
\label{p169.ass2}%
\end{equation}
where $\delta_{i,j}$ is the Kronecker delta of $i$ and $j$. (This covers both
the case $i=j$ and the case $i\neq j$.)

Let us set%
\[
\gamma\left(  T\right)  :=\left(  \det T\right)  ^{-g}\cdot\rho\left(
T\right)  \in\operatorname*{GL}\left(  V\right)  \ \ \ \ \ \ \ \ \ \ \text{for
each }T\in\Gamma.
\]
Thus, $\gamma$ is a rational map from $\Gamma$ to $\operatorname*{GL}\left(
V\right)  $ (that is, if we choose a basis of $V$ and coordinatize
$\operatorname*{GL}\left(  V\right)  \subseteq\operatorname*{End}\left(
V\right)  $ in the obvious way using this basis, then all the coordinates of
$\gamma\left(  T\right)  $ are rational functions in the coordinates of $T$).
By the definition of $\gamma$, we have%
\[
\gamma\left(  \mathbf{1}\right)  =\underbrace{\left(  \det\mathbf{1}\right)
^{-g}}_{=1^{-g}=1}\cdot\underbrace{\rho\left(  \mathbf{1}\right)
}_{=\operatorname*{id}\nolimits_{V}}=\operatorname*{id}\nolimits_{V}.
\]
Moreover, for each $T\in\Gamma$, we have%
\begin{equation}
\rho\left(  T\right)  =\left(  \det T\right)  ^{g}\cdot\gamma\left(  T\right)
\label{p169.rhogam}%
\end{equation}
(since $\gamma\left(  T\right)  =\left(  \det T\right)  ^{-g}\cdot\rho\left(
T\right)  $). Thus, for each $i,j\in\left\{  1,2,\ldots,n\right\}  $, we have%
\begin{align*}
&  \left(  \dfrac{\partial}{\partial t_{i,j}}\rho\left(  T\right)  \right)
_{T=\mathbf{1}}\\
&  =\left(  \dfrac{\partial}{\partial t_{i,j}}\left(  \left(  \det T\right)
^{g}\cdot\gamma\left(  T\right)  \right)  \right)  _{T=\mathbf{1}}\\
&  =\underbrace{\left(  \dfrac{\partial}{\partial t_{i,j}}\left(  \det
T\right)  ^{g}\right)  _{T=\mathbf{1}}}_{\substack{=\delta_{i,j}g\\\text{(this
is easy to check)}}}\cdot\underbrace{\gamma\left(  \mathbf{1}\right)
}_{=\operatorname*{id}\nolimits_{V}}+\underbrace{\left(  \det\mathbf{1}%
\right)  ^{g}}_{=1^{g}=1}\cdot\left(  \dfrac{\partial}{\partial t_{i,j}}%
\gamma\left(  T\right)  \right)  _{T=\mathbf{1}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by the Leibniz rule
for noncommutative products}\right) \\
&  =\delta_{i,j}g\operatorname*{id}\nolimits_{V}+\left(  \dfrac{\partial
}{\partial t_{i,j}}\gamma\left(  T\right)  \right)  _{T=\mathbf{1}},
\end{align*}
so that%
\[
\left(  \dfrac{\partial}{\partial t_{i,j}}\gamma\left(  T\right)  \right)
_{T=\mathbf{1}}=\underbrace{\left(  \dfrac{\partial}{\partial t_{i,j}}%
\rho\left(  T\right)  \right)  _{T=\mathbf{1}}}_{\substack{=\rho^{\ast}\left(
E_{i,j}\right)  \\\text{(by (4.5.2))}}}-\,\delta_{i,j}g\operatorname*{id}%
\nolimits_{V}=\rho^{\ast}\left(  E_{i,j}\right)  -\delta_{i,j}%
g\operatorname*{id}\nolimits_{V}%
\]
and therefore%
\begin{align*}
\left(  \dfrac{\partial}{\partial t_{i,j}}\gamma\left(  T\right)  \right)
_{T=\mathbf{1}}\cdot v  &  =\left(  \rho^{\ast}\left(  E_{i,j}\right)
-\delta_{i,j}g\operatorname*{id}\nolimits_{V}\right)  \cdot v\\
&  =\rho^{\ast}\left(  E_{i,j}\right)  \cdot v-\delta_{i,j}g\cdot
v=0\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{p169.ass2})}\right)  .
\end{align*}
Thus, we have shown that%
\begin{equation}
\left(  \dfrac{\partial}{\partial t_{i,j}}\gamma\left(  T\right)  \right)
_{T=\mathbf{1}}\cdot v=0\ \ \ \ \ \ \ \ \ \ \text{for each }i,j\in\left\{
1,2,\ldots,n\right\}  . \label{p169.ass4}%
\end{equation}


Now, fix $S\in\Gamma$. Let $i,j\in\left\{  1,2,\ldots,n\right\}  $. For each
$T\in\Gamma$, we have%
\[
\rho\left(  ST\right)  =\rho\left(  S\right)  \cdot\rho\left(  T\right)
\]
(since $\rho$ is a group action) and thus%
\begin{align*}
\gamma\left(  ST\right)   &  =\left(  \underbrace{\det\left(  ST\right)
}_{=\det S\cdot\det T}\right)  ^{-g}\cdot\underbrace{\rho\left(  ST\right)
}_{=\rho\left(  S\right)  \cdot\rho\left(  T\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\gamma\right) \\
&  =\underbrace{\left(  \det S\cdot\det T\right)  ^{-g}}_{=\left(  \det
S\right)  ^{-g}\cdot\left(  \det T\right)  ^{-g}}\cdot\rho\left(  S\right)
\cdot\rho\left(  T\right) \\
&  =\underbrace{\left(  \det S\right)  ^{-g}\cdot\rho\left(  S\right)
}_{\substack{=\gamma\left(  S\right)  \\\text{(by the definition of }%
\gamma\text{)}}}\cdot\underbrace{\left(  \det T\right)  ^{-g}\cdot\rho\left(
T\right)  }_{\substack{=\gamma\left(  T\right)  \\\text{(by the definition of
}\gamma\text{)}}}\\
&  =\gamma\left(  S\right)  \cdot\gamma\left(  T\right)
\end{align*}
and therefore%
\[
\gamma\left(  ST\right)  \cdot v=\gamma\left(  S\right)  \cdot\gamma\left(
T\right)  \cdot v.
\]
Applying the operator $\dfrac{\partial}{\partial t_{i,j}}\mid_{T=\mathbf{1}}$
to this equality (i.e., taking the derivative with respect to $t_{i,j}$ at the
point $T=\mathbf{1}$), we obtain%
\begin{align}
\dfrac{\partial}{\partial t_{i,j}}\left(  \gamma\left(  ST\right)  \cdot
v\right)  _{T=\mathbf{1}}  &  =\dfrac{\partial}{\partial t_{i,j}}\left(
\gamma\left(  S\right)  \cdot\gamma\left(  T\right)  \cdot v\right)
_{T=\mathbf{1}}=\gamma\left(  S\right)  \cdot\underbrace{\left(
\dfrac{\partial}{\partial t_{i,j}}\gamma\left(  T\right)  \right)
_{T=\mathbf{1}}\cdot v}_{\substack{=0\\\text{(by (\ref{p169.ass4}))}%
}}\nonumber\\
&  =0. \label{p169.3}%
\end{align}
On the other hand, the map $\Gamma\rightarrow V,\ T\mapsto\gamma\left(
ST\right)  \cdot v$ can be written as the composition of the maps
$\Gamma\rightarrow\Gamma,\ T\mapsto ST$ and $\Gamma\rightarrow V,\ U\mapsto
\gamma\left(  U\right)  \cdot v$. Thus, the chain rule for multivariate
functions yields%
\begin{align*}
&  \dfrac{\partial}{\partial t_{i,j}}\left(  \gamma\left(  ST\right)  \cdot
v\right)  _{T=\mathbf{1}}\\
&  =\sum_{k,l=1}^{n}\underbrace{\left(  \dfrac{\partial}{\partial t_{i,j}%
}\left(  \left(  ST\right)  _{k,l}\right)  \right)  _{T=\mathbf{1}}%
}_{\substack{=%
\begin{cases}
s_{k,i}, & \text{if }l=j;\\
0, & \text{if }l\neq j
\end{cases}
\\\text{(since }\left(  ST\right)  _{k,l}=\sum_{h=1}^{n}s_{k,h}t_{h,l}%
\text{)}}}\cdot\underbrace{\left(  \dfrac{\partial}{\partial u_{k,l}}\left(
\gamma\left(  U\right)  \cdot v\right)  \right)  _{U=S\mathbf{1}}%
}_{\substack{=\left(  \dfrac{\partial}{\partial u_{k,l}}\gamma\left(
U\right)  \right)  _{U=S\mathbf{1}}\cdot v\\=\left(  \dfrac{\partial}{\partial
u_{k,l}}\gamma\left(  U\right)  \right)  _{U=S}\cdot v\\\text{(since
}S\mathbf{1}=S\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{where }u_{k,l}\text{ denotes the }\left(  k,l\right)  \text{-th entry of
the matrix }U\in\Gamma\text{,}\\
\text{and where }\left(  ST\right)  _{k,l}\text{ denotes the }\left(
k,l\right)  \text{-th entry of the matrix }ST
\end{array}
\right) \\
&  =\sum_{k,l=1}^{n}%
\begin{cases}
s_{k,i}, & \text{if }l=j;\\
0, & \text{if }l\neq j
\end{cases}
\ \ \cdot\left(  \dfrac{\partial}{\partial u_{k,l}}\gamma\left(  U\right)
\right)  _{U=S}\cdot v\\
&  =\sum_{k=1}^{n}s_{k,i}\cdot\left(  \dfrac{\partial}{\partial u_{k,j}}%
\gamma\left(  U\right)  \right)  _{U=S}\cdot v\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have removed all addends with }l\neq j\\
\text{from the sum, since these addends are }0
\end{array}
\right)  .
\end{align*}
Comparing this with (\ref{p169.3}), we obtain%
\begin{equation}
\sum_{k=1}^{n}s_{k,i}\cdot\left(  \dfrac{\partial}{\partial u_{k,j}}%
\gamma\left(  U\right)  \right)  _{U=S}\cdot v=0. \label{p169.6}%
\end{equation}


Now forget that we fixed $i$. We thus have proved the equality (\ref{p169.6})
for all $i\in\left\{  1,2,\ldots,n\right\}  $.

But the matrix $S$ is invertible (since $S\in\Gamma=\operatorname*{GL}\left(
\mathbf{C}^{n}\right)  $). Let $S^{-1}=\left(  s_{p,q}^{\prime}\right)
_{p,q\in\left\{  1,2,\ldots,n\right\}  }$ be its inverse matrix. Then,
$SS^{-1}=\mathbf{1}$, so that all $k,q\in\left\{  1,2,\ldots,n\right\}  $
satisfy%
\begin{equation}
\sum_{i=1}^{n}s_{k,i}s_{i,q}^{\prime}=\delta_{k,q} \label{p169.10}%
\end{equation}
(by comparing the $\left(  k,q\right)  $-th entries of the matrices $SS^{-1}$
and $\mathbf{1}$). Now, for any $q\in\left\{  1,2,\ldots,n\right\}  $, we have%
\[
\sum_{i=1}^{n}s_{i,q}^{\prime}\cdot\underbrace{\sum_{k=1}^{n}s_{k,i}%
\cdot\left(  \dfrac{\partial}{\partial u_{k,j}}\gamma\left(  U\right)
\right)  _{U=S}\cdot v}_{\substack{=0\\\text{(by (\ref{p169.6}))}}}=0
\]
and thus%
\begin{align*}
0  &  =\sum_{i=1}^{n}s_{i,q}^{\prime}\cdot\sum_{k=1}^{n}s_{k,i}\cdot\left(
\dfrac{\partial}{\partial u_{k,j}}\gamma\left(  U\right)  \right)  _{U=S}\cdot
v\\
&  =\sum_{k=1}^{n}\ \ \underbrace{\sum_{i=1}^{n}s_{k,i}s_{i,q}^{\prime}%
}_{\substack{=\delta_{k,q}\\\text{(by (\ref{p169.10}))}}}\cdot
\underbrace{\left(  \dfrac{\partial}{\partial u_{k,j}}\gamma\left(  U\right)
\right)  _{U=S}\cdot v}_{=\left(  \dfrac{\partial}{\partial u_{k,j}}\left(
\gamma\left(  U\right)  \cdot v\right)  \right)  _{U=S}}\\
&  =\sum_{k=1}^{n}\delta_{k,q}\cdot\left(  \dfrac{\partial}{\partial u_{k,j}%
}\left(  \gamma\left(  U\right)  \cdot v\right)  \right)  _{U=S}=\left(
\dfrac{\partial}{\partial u_{q,j}}\left(  \gamma\left(  U\right)  \cdot
v\right)  \right)  _{U=S}%
\end{align*}
(since all the addends of the sum are $0$ except for the addend for $k=q$). In
other words,%
\[
\left(  \dfrac{\partial}{\partial u_{q,j}}\left(  \gamma\left(  U\right)
\cdot v\right)  \right)  _{U=S}=0\ \ \ \ \ \ \ \ \ \ \text{for all }%
q\in\left\{  1,2,\ldots,n\right\}  .
\]


Now, forget that we fixed $j$. We thus have shown that
\[
\left(  \dfrac{\partial}{\partial u_{q,j}}\left(  \gamma\left(  U\right)
\cdot v\right)  \right)  _{U=S}=0\ \ \ \ \ \ \ \ \ \ \text{for all }%
q,j\in\left\{  1,2,\ldots,n\right\}  .
\]
In other words, all the partial derivatives of $\gamma\left(  U\right)  \cdot
v$ (as a function in $U\in\Gamma$) at the point $U=S$ are $0$. Since
$S\in\Gamma$ was also chosen arbitrary, this shows that all the partial
derivatives of $\gamma\left(  U\right)  \cdot v$ (as a function in $U\in
\Gamma$) at every point are $0$. Therefore, this function $\gamma\left(
U\right)  \cdot v$ must be constant (since a rational function whose all
partial derivatives at every point are $0$ must be constant\footnote{For those
who want to prove this without the use of analysis: First, reduce the result
to the case of a univariate rational function with values in $\mathbf{C}$ (by
focussing on a single coordinate of the output and on a change in a single
coordinate of the input); then express this rational function as a ratio $f/g$
of two coprime polynomials $f$ and $g$, and argue that $\left(  f/g\right)
^{\prime}=0$ entails $f^{\prime}g=fg^{\prime}$, which can be combined with the
coprimality of $f$ and $g$ to yield $f\mid f^{\prime}$ and $g\mid g^{\prime}$,
which is absurd unless $f$ and $g$ are constant.}). In particular, every
$T\in\Gamma$ thus satisfies
\[
\gamma\left(  T\right)  \cdot v=\underbrace{\gamma\left(  \mathbf{1}\right)
}_{=\operatorname*{id}\nolimits_{V}}\cdot\,v=v
\]
and therefore%
\begin{align*}
\rho\left(  T\right)  \cdot v  &  =\left(  \det T\right)  ^{g}\cdot
\underbrace{\gamma\left(  T\right)  \cdot v}_{=v}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{p169.rhogam})}\right) \\
&  =\left(  \det T\right)  ^{g}\cdot v.
\end{align*}
In other words, $v$ is a $\Gamma$-invariant of index $g$. This proves the
\textquotedblleft if\textquotedblright\ direction of Lemma 4.5.1.

\item \textbf{page 170, Theorem 4.5.2:} In the first equality in the display
(4.5.7), replace \textquotedblleft for $i=j,2,\ldots,n$\textquotedblright\ by
\textquotedblleft for $j=1,2,\ldots,n$\textquotedblright.

\item \textbf{page 174, proof of Lemma 4.5.4:} On the right hand side of the
display, replace \textquotedblleft$x_{\sigma\left(  i\right)  ,j}^{\nu
_{\sigma\left(  i\right)  ,j}}$\textquotedblright\ by \textquotedblleft%
$x_{\sigma\left(  i\right)  ,j}^{\nu_{i,j}}$\textquotedblright\ (otherwise, we
don't get an $S_{m}$-invariant sum).

\item \textbf{page 185, Lemma 4.7.2:} Remark: If $t\geq1$, then the bound
\textquotedblleft$\leq s\left(  t+1\right)  ^{s-1}$\textquotedblright\ can be
replaced by the (slightly less horrible) bound \textquotedblleft$\leq s\left(
t^{s-1}-1\right)  $\textquotedblright\ (since $\dfrac{r+k}{rt+k}\geq\dfrac
{1}{t}$ for each $k\geq0$).
\end{enumerate}


\end{document}