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\ihead{Errata to ``Invariant Theory with Applications''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Invariant Theory with Applications}

\textit{Jan Draisma and Dion Gijswijt}

\url{http://www.win.tue.nl/~jdraisma/teaching/invtheory0910/lecturenotes12.pdf}

version of 7 December 2009

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

The following is a haphazard list of errors I found in \textquotedblleft
Invariant Theory with Applications\textquotedblright\ by Jan Draisma and Dion Gijswijt.

\begin{verlong}
All page numbers given below are to be understood as page numbers in these notes.
\end{verlong}

\setcounter{section}{15}

\section{Errata}

\begin{itemize}
\item \textbf{Page 5, \S 1.1:} Replace \textquotedblleft Clearly, the elements
of $V^{\ast}$ are regular of degree\textquotedblright\ by \textquotedblleft
Clearly, the elements of $V^{\ast}$ are regular functions and are homogeneous
of degree\textquotedblright.

\item \textbf{Page 7, \S 1.3:} \textquotedblleft discribed\textquotedblright%
\ $\rightarrow$ \textquotedblleft described\textquotedblright.

\item \textbf{Page 8, Example 1.3.2:} \textquotedblleft
althought\textquotedblright\ $\rightarrow$ \textquotedblleft
although\textquotedblright.

\item \textbf{Page 8, Example 1.3.3:} \textquotedblleft with the same
exponent\textquotedblright\ $\rightarrow$ \textquotedblleft with the same
coefficient\textquotedblright.

\item \textbf{Page 9, proof of Proposition 1.4.1:} Replace \textquotedblleft
To each $c=(c_{1},\ldots,c_{n}\in\mathbb{C}^{n}$\textquotedblright\ by
\textquotedblleft To each $c=\left(  c_{1},\ldots,c_{n}\right)  \in
\mathbb{C}^{n}$\textquotedblright.

\item \textbf{Page 9, proof of Proposition 1.4.1:} In (1.8), the entries in
the last column should be $-c_{n},-c_{n-1},\ldots,-c_{2},-c_{1}$ (not
$-c_{n},-c_{n-1},\ldots,c_{2},c_{1}$).

\item \textbf{Page 9, proof of Proposition 1.4.1:} Replace \textquotedblleft
shows that $\chi_{A_{c}}\left(  t\right)  =t^{n}+c_{n-1}t^{n-1}+\cdots
+c_{1}t+c_{0}$\textquotedblright\ by \textquotedblleft shows that $\chi
_{A_{c}}\left(  t\right)  =t^{n}+c_{1}t^{n-1}+\cdots+c_{n-1}t+c_{n}%
$\textquotedblright.

\item \textbf{Page 9, Exercise 1.4.2:} Replace \textquotedblleft that
$\chi_{A_{c}}\left(  t\right)  =t^{n}+c_{n-1}t^{n-1}+\cdots+c_{1}t+c_{0}%
$\textquotedblright\ by \textquotedblleft that $\chi_{A_{c}}\left(  t\right)
=t^{n}+c_{1}t^{n-1}+\cdots+c_{n-1}t+c_{n}$\textquotedblright.

\item \textbf{Page 9, proof of Proposition 1.4.1:} In (1.9), replace
\textquotedblleft$(s_{1}\left(  A_{c}\right)  ,s_{2}\left(  A_{c}\right)
,\ldots,s_{n}\left(  A_{c}\right)  $\textquotedblright\ by \textquotedblleft%
$\left(  s_{1}\left(  A_{c}\right)  ,s_{2}\left(  A_{c}\right)  ,\ldots
,s_{n}\left(  A_{c}\right)  \right)  $\textquotedblright.

\item \textbf{Page 9, proof of Proposition 1.4.1:} Replace \textquotedblleft
dense in $\mathcal{O}\left(  \operatorname*{Mat}\nolimits_{n}\left(
\mathbb{C}\right)  \right)  $\textquotedblright\ by \textquotedblleft dense in
$\operatorname*{Mat}\nolimits_{n}\left(  \mathbb{C}\right)  $%
\textquotedblright. (This mistake appears twice.)

\item \textbf{Page 9, Exercise 1.4.3:} Replace \textquotedblleft nonzero
eigenvalues\textquotedblright\ by \textquotedblleft
eigenvalues\textquotedblright.

\item \textbf{Page 10, Exercise 1.4.3:} Replace \textquotedblleft distinct and
nonzero\textquotedblright\ by \textquotedblleft nonzero\textquotedblright.

\item \textbf{Page 10, Exercise 1.4.3:} It might be worth noticing that
\textquotedblleft the fact\textquotedblright\ you are mentioning about the
Vandermonde determinant is a consequence of Lemma 2.2.4 below (using the
well-known fact that the determinant of a square matrix equals the determinant
of its transpose).

\item \textbf{Page 15, Theorem 2.2.9:} You misspell \textquotedblleft
Sylvester\textquotedblright\ as \textquotedblleft Sylverster\textquotedblright.

\item \textbf{Page 15, proof of Theorem 2.2.9:} Remove the comma in
\textquotedblleft Since, $\widetilde{A}$ contains\textquotedblright.

\item \textbf{Page 15, proof of Theorem 2.2.9:} You write: \textquotedblleft
it follows that $\operatorname*{Bez}\left(  f\right)  $ has rank
$2k+r$\textquotedblright. How does this follow? I only see that
$\operatorname*{Bez}\left(  f\right)  $ has rank $\leq2k+r$.

\item \textbf{Page 20:} Replace \textquotedblleft every element $T\in U\otimes
V$\textquotedblright\ by \textquotedblleft every element $t\in U\otimes
V$\textquotedblright.

\item \textbf{Page 20:} Replace \textquotedblleft for $T$ to
zero\textquotedblright\ by \textquotedblleft for $t$ to zero\textquotedblright.

\item \textbf{Page 21:} \textquotedblleft with of $k$%
-tensors\textquotedblright\ $\rightarrow$ \textquotedblleft with
$k$-tensors\textquotedblright.

\item \textbf{Page 23:} \textquotedblleft so that the $v^{\alpha},\ \left\vert
\alpha\right\vert =k$ a basis of $V$\textquotedblright\ should be
\textquotedblleft so that the $v^{\alpha}$ with $\left\vert \alpha\right\vert
=k$ form a basis of $S^{k}V$\textquotedblright.

\item \textbf{Page 23:} Replace \textquotedblleft$\pi\left(  v_{1}%
\otimes\cdots v_{k}\right)  $\textquotedblright\ by \textquotedblleft%
$\pi\left(  v_{1}\otimes\cdots\otimes v_{k}\right)  $\textquotedblright.

\item \textbf{Page 23, Exercise 3.0.13:} Replace \textquotedblleft$v\otimes
v\cdots\otimes v$\textquotedblright\ by \textquotedblleft$v\otimes
v\otimes\cdots\otimes v$\textquotedblright.

\item \textbf{Page 24, Exercise 3.1.4:} You should require that at least one
of $U$ and $V$ is finite-dimensional.

\item \textbf{Page 24, Exercise 3.1.4:} Replace \textquotedblleft
isomorhism\textquotedblright\ by \textquotedblleft
isomorphism\textquotedblright.

\item \textbf{Page 25:} Replace \textquotedblleft so that $g\left(  hf\right)
=\left(  hg\right)  f$\textquotedblright\ by \textquotedblleft so that
$g\left(  hf\right)  =\left(  gh\right)  f$\textquotedblright.

\item \textbf{Page 25, Example 4.0.8:} Replace \textquotedblleft$G$
module\textquotedblright\ by \textquotedblleft$G$-module\textquotedblright.

\item \textbf{Page 26, Example 4.0.9:} Replace \textquotedblleft$G$
module\textquotedblright\ by \textquotedblleft$G$-module\textquotedblright.

\item \textbf{Page 27, proof of Proposition 4.0.7:} \textquotedblleft$\left(
v\mid v\right)  =\sum_{g\in G}\left(  gv\mid gv\right)  $\textquotedblright%
\ should be \textquotedblleft$\left(  v\mid v\right)  =\sum_{g\in G}\left(
gv\mid gv\right)  ^{\prime}$\textquotedblright.

\item \textbf{Page 28, Lemma 4.1.1:} Replace \textquotedblleft$G$
modules\textquotedblright\ by \textquotedblleft$G$-modules\textquotedblright.

\item \textbf{Page 28, \S 4.1:} \textquotedblleft of the isomorphism classes
of $G$-modules\textquotedblright\ should be \textquotedblleft of the
isomorphism classes of irreducible $G$-modules\textquotedblright.

\item \textbf{Page 29, Exercise 4.1.2:} Remove the superscript
\textquotedblleft$^{G}$\textquotedblright.

\item \textbf{Page 31, Lemma 5.0.9:} \textquotedblleft
Dixon's\textquotedblright\ $\rightarrow$ \textquotedblleft
Dickson's\textquotedblright.

\item \textbf{Page 32, proof of Hilbert's Basis Theorem:} \textquotedblleft
Dixon's\textquotedblright\ $\rightarrow$ \textquotedblleft
Dickson's\textquotedblright.

\item \textbf{Page 32:} In (5.2), add a whitespace before \textquotedblleft
for all $f\in V_{1}$\textquotedblright.

\item \textbf{Page 32:} \textquotedblleft This a $G$-module
morphism\textquotedblright\ $\rightarrow$ \textquotedblleft This is a
$G$-module morphism\textquotedblright.

\item \textbf{Page 33, Exercise 5.0.13:} \textquotedblleft with zero
coefficient\textquotedblright\ should be \textquotedblleft with constant
coefficient equal to $0$\textquotedblright.

\item \textbf{Page 33, Exercise 5.0.13:} I am wondering whether you really
mean \textquotedblleft subalgebra\textquotedblright\ here and not
\textquotedblleft graded subalgebra\textquotedblright.

\item \textbf{Page 34, proof of Theorem 5.1.1:} Replace \textquotedblleft%
$\left\vert \beta\right\vert <=\left\vert G\right\vert $\textquotedblright\ by
\textquotedblleft$\left\vert \beta\right\vert \leq\left\vert G\right\vert
$\textquotedblright.

\item \textbf{Page 34, proof of Theorem 5.1.1:} Replace \textquotedblleft%
$p_{j}=\sum_{\left\vert \alpha\right\vert =j}f_{\alpha}z_{1}^{\alpha_{1}%
}\cdots x_{n}^{\alpha_{n}}$\textquotedblright\ by \textquotedblleft$p_{j}%
=\sum_{\left\vert \alpha\right\vert =j}f_{\alpha}z_{1}^{\alpha_{1}}\cdots
z_{n}^{\alpha_{n}}$\textquotedblright.

\item \textbf{Page 34, proof of Theorem 5.1.1:} You write: \textquotedblleft
Recall that $p_{j}$ is a polynomial in $p_{1},\ldots,p_{\left\vert
G\right\vert }$\textquotedblright. Did you actually prove this anywhere? (This
is a particular case of the following fact: In the polynomial ring
$\mathbb{C}\left[  x_{1},x_{2},\ldots,x_{n}\right]  $, each $S_{n}$-invariant
polynomial $f\in\mathbb{C}\left[  x_{1},x_{2},\ldots,x_{n}\right]  ^{S_{n}}$
can be written as a polynomial in the Newton polynomials $p_{1},p_{2}%
,\ldots,p_{n}$.\ \ \ \ \footnote{The \textit{proof} of this fact is easy: By
Theorem 2.1.1, it suffices to show that the $s_{1},s_{2},\ldots,s_{n}$ are
polynomials in $p_{1},p_{2},\ldots,p_{n}$. In other words, it suffices to show
that $s_{k}$ is a polynomial in $p_{1},p_{2},\ldots,p_{n}$ for each
$k\in\left\{  1,2,\ldots,n\right\}  $. But this easily follows by strong
induction over $k$ (indeed, (2.18) gives a way to write each $s_{k}$ for
$k\in\left\{  1,2,\ldots,n\right\}  $ as a polynomial in $p_{1},p_{2}%
,\ldots,p_{n}$, provided that $s_{1},s_{2},\ldots,s_{k-1}$ have already been
written in this form).} This is probably worth stating as an exercise in
Chapter 2.

\item \textbf{Page 37, proof of the weak Nullstellensatz:} Replace
\textquotedblleft$f_{k,\xi}:=\left(  x_{1},\ldots,x_{n-1},\xi\right)
$\textquotedblright\ by \textquotedblleft$f_{k,\xi}:=f_{k}\left(  x_{1}%
,\ldots,x_{n-1},\xi\right)  $\textquotedblright.

\item \textbf{Page 37, proof of the weak Nullstellensatz:} Replace all three
\textquotedblleft$\sum_{i=1}^{k}$\textquotedblright\ signs by
\textquotedblleft$\sum_{j=1}^{k}$\textquotedblright\ signs.

\item \textbf{Page 37:} \textquotedblleft Nulstellensatz\textquotedblright%
\ $\rightarrow$ \textquotedblleft Nullstellensatz\textquotedblright.

\item \textbf{Page 39, proof of Theorem 6.1.10:} Replace the \textquotedblleft%
$\sum_{i=1}^{k}$\textquotedblright\ sign by a \textquotedblleft$\sum_{j=1}%
^{k}$\textquotedblright\ sign.

\item \textbf{Page 41, Lemma 6.2.6:} Replace \textquotedblleft from
$\mathbb{C}\left[  Y\right]  $ $\mathbb{C}\left[  X\right]  $%
\textquotedblright\ by \textquotedblleft from $\mathbb{C}\left[  Y\right]  $
to $\mathbb{C}\left[  X\right]  $\textquotedblright.

\item \textbf{Page 41, proof of Lemma 6.2.8:} \textquotedblleft are a regular
maps\textquotedblright\ $\rightarrow$ \textquotedblleft are regular
maps\textquotedblright.

\item \textbf{Page 42, Example 6.3.3:} Replace \textquotedblleft act on the
$W$\textquotedblright\ by \textquotedblleft act on the vector space
$W$\textquotedblright.

\item \textbf{Page 43, Theorem 6.3.4:} In property 4, replace
\textquotedblleft$\phi:Z\mapsto\mathbb{C}^{m}$\textquotedblright\ by
\textquotedblleft$\phi:Z\rightarrow\mathbb{C}^{m}$\textquotedblright.

\item \textbf{Page 43, proof of Theorem 6.3.4:} In the proof of property 3,
replace \textquotedblleft$\phi:Z\mapsto U$\textquotedblright\ by
\textquotedblleft$\phi:Z\rightarrow\mathbb{C}^{m}$\textquotedblright.

\item \textbf{Page 47, proof of Theorem 7.0.14:} Replace \textquotedblleft
Hence $w$ is in the null-cone $N_{V}$\textquotedblright\ by \textquotedblleft
Hence $w$ is in the null-cone $N_{W}$\textquotedblright.

\item \textbf{Page 49:} Replace \textquotedblleft Let $W\bigoplus
_{d=0}^{\infty}W_{d}$ be a direct sum\textquotedblright\ by \textquotedblleft
Let $W=\bigoplus_{d=0}^{\infty}W_{d}$ be a direct sum\textquotedblright.

\item \textbf{Page 49:} In (8.1), replace \textquotedblleft$V$%
\textquotedblright\ and \textquotedblleft$V_{d}$\textquotedblright\ by
\textquotedblleft$W$\textquotedblright\ and \textquotedblleft$W_{d}%
$\textquotedblright, respectively.

\item \textbf{Page 49, Example 8.0.18:} Replace \textquotedblleft$H\left(
\mathbb{C}\left[  x_{1},\ldots,x_{n}\right]  \right)  $\textquotedblright\ by
\textquotedblleft$H\left(  \mathbb{C}\left[  x_{1},\ldots,x_{n}\right]
,t\right)  $\textquotedblright.

\item \textbf{Page 50, Theorem 8.1.1:} Replace \textquotedblleft of a finite
group\textquotedblright\ by \textquotedblleft of a finite group $G$%
\textquotedblright.

\item \textbf{Page 50, proof of Theorem 8.1.1:} In (8.5), replace
\textquotedblleft$\operatorname*{tr}\left(  L_{d}\left(  g\right)  \right)
$\textquotedblright\ by \textquotedblleft$t^{d}\operatorname*{tr}\left(
L_{d}\left(  g\right)  \right)  $\textquotedblright.

\item \textbf{Page 50, proof of Theorem 8.1.1:} Replace \textquotedblleft lets
fix\textquotedblright\ by \textquotedblleft let's fix\textquotedblright.

\item \textbf{Page 50, proof of Theorem 8.1.1:} Replace \textquotedblleft the
inner sum $\sum_{d=0}^{\infty}\operatorname*{tr}\left(  L_{d}\left(  g\right)
\right)  $\textquotedblright\ by \textquotedblleft the inner sum $\sum
_{d=0}^{\infty}t^{d}\operatorname*{tr}\left(  L_{d}\left(  g\right)  \right)
$\textquotedblright.

\item \textbf{Page 50, proof of Theorem 8.1.1:} You write: \textquotedblleft
Pick a basis $x_{1},\ldots,x_{n}$ of $V^{\ast}$ that is a system of
eigenvectors for $L_{1}\left(  g\right)  $\textquotedblright. It is worth
justifying why such a basis exists. (Namely, you are using the apocryphal
theorem from linear algebra that says that if $U$ is a finite-dimensional
$\mathbb{C}$-vector space, and if $\alpha$ is an element of
$\operatorname*{GL}\left(  U\right)  $ having finite order, then $\alpha$ is
diagonalizable. You are applying this theorem to $U=V^{\ast}$ and
$\alpha=L_{1}\left(  g\right)  $, which is allowed because the element
$L_{1}\left(  g\right)  $ of $\operatorname*{GL}\left(  V^{\ast}\right)  $ has
finite order (since the element $g$ of $G$ has finite order). This is not a
difficult argument, but I don't think it is obvious enough to be entirely left
to the reader.)

\item \textbf{Page 50, proof of Theorem 8.1.1:} Replace \textquotedblleft for
a system\textquotedblright\ by \textquotedblleft form a
system\textquotedblright.

\item \textbf{Page 50, proof of Theorem 8.1.1:} On the first line of the
computation (8.7), replace \textquotedblleft$\left(  1+\lambda_{n}%
t+\lambda_{n}t^{2}+\cdots\right)  $\textquotedblright\ by \textquotedblleft%
$\left(  1+\lambda_{n}t+\lambda_{n}^{2}t^{2}+\cdots\right)  $%
\textquotedblright.

\item \textbf{Page 50, proof of Theorem 8.1.1:} On the first line of the
computation (8.8), replace \textquotedblleft$\operatorname*{tr}\left(
L_{d}\left(  g\right)  \right)  $\textquotedblright\ by \textquotedblleft%
$t^{d}\operatorname*{tr}\left(  L_{d}\left(  g\right)  \right)  $%
\textquotedblright.

\item \textbf{Page 50, proof of Theorem 8.1.1:} On the third line of the
computation (8.8), replace \textquotedblleft$\det(I-\rho\left(  g\right)
t$\textquotedblright\ by \textquotedblleft$\det\left(  I-\rho\left(  g\right)
t\right)  $\textquotedblright.

\item \textbf{Page 51, \S 8.2:} Replace \textquotedblleft which $u$ and $v$,
differ\textquotedblright\ by \textquotedblleft which $u$ and $v$
differ\textquotedblright.

\item \textbf{Page 51, \S 8.1:} It is worth pointing out that you use the word
\textquotedblleft code\textquotedblright\ to mean \textquotedblleft linear
code\textquotedblright.

\item \textbf{Page 52:} \textquotedblleft Furhermore\textquotedblright%
\ $\rightarrow$ \textquotedblleft Furthermore\textquotedblright.

\item \textbf{Page 53, Theorem 8.2.6:} Replace \textquotedblleft$\left(
x^{4}-y^{4}\right)  $\textquotedblright\ by \textquotedblleft$\left(
x^{4}-y^{4}\right)  ^{4}$\textquotedblright.

\item \textbf{Page 57, Example 9.1.8:} Replace \textquotedblleft$\prod
_{k}\prod_{l}$\textquotedblright\ by \textquotedblleft$\sum_{k}\sum_{l}%
$\textquotedblright.

\item \textbf{Page 59, \S 9.2:} Replace \textquotedblleft Consider the map
$\lambda:G\rightarrow\operatorname*{GL}[\mathbb{C}\left[  x_{ij},1/\det\left(
x\right)  \right]  )$\textquotedblright\ by \textquotedblleft Consider the map
$\lambda:G\rightarrow\operatorname*{GL}\left(  \mathbb{C}\left[  x_{ij}%
,1/\det\left(  x\right)  \right]  \right)  $\textquotedblright.

\item \textbf{Page 65, Exercise 10.0.12:} Replace \textquotedblleft larger
enough\textquotedblright\ by \textquotedblleft large enough\textquotedblright.

\item \textbf{Page 65, proof of Proposition 10.0.13:} Replace
\textquotedblleft standard basis $\mathbb{C}^{2}$\textquotedblright\ by
\textquotedblleft standard basis of $\mathbb{C}^{2}$\textquotedblright.

\item \textbf{Page 65, proof of Proposition 10.0.13:} Replace
\textquotedblleft induced basis of $S^{d}\left(  V\right)  $\textquotedblright%
\ by \textquotedblleft induced basis of $S^{k}\left(  V\right)  $%
\textquotedblright.

\item \textbf{Page 65, proof of Proposition 10.0.13:} Replace
\textquotedblleft$\sum_{i}d\left(  \lambda\right)  x^{i}y^{k-i}$%
\textquotedblright\ by \textquotedblleft$\sum_{i}d_{i}\left(  \lambda\right)
x^{i}y^{k-i}$\textquotedblright.

\item \textbf{Page 65, proof of Proposition 10.0.13:} You claim that
\textquotedblleft$d_{0}$ and every $d_{i}$ with $c_{i}\neq0$ are nonzero
polynomials with $\lambda$\textquotedblright. I would suggest explaining why
they are nonzero. (Namely, the polynomial $d_{0}$ is nonzero because
$d_{0}=\sum_{i}c_{i}\lambda^{i}y^{k}$ (and because not all $c_{i}$ are $0$);
meanwhile, the polynomials $d_{i}$ with $c_{i}\neq0$ are nonzero because they
satisfy $d_{i}\left(  0\right)  =c_{i}\neq0$.)

\item \textbf{Page 66, proof of Proposition 10.0.13:} Replace
\textquotedblleft Then for every $i$ the vector $\mu^{k}\left(
\begin{array}
[c]{cc}%
\mu & 0\\
0 & \mu^{-1}%
\end{array}
\right)  u=\sum_{i}\lambda^{i}c_{i}x^{i}y^{k-i}$ belongs to $U$%
\textquotedblright\ by \textquotedblleft Then for every $\mu\in\left\{
\mu_{0},\mu_{1},\ldots,\mu_{k}\right\}  $ the vector $\mu^{k}\left(
\begin{array}
[c]{cc}%
\mu & 0\\
0 & \mu^{-1}%
\end{array}
\right)  u=\sum_{i}\lambda^{i}c_{i}x^{i}y^{k-i}$ (with $\lambda=\mu^{2}$)
belongs to $U$\textquotedblright.

\item \textbf{Page 66, proof of Proposition 10.0.13:} In (10.4), replace
\textquotedblleft$S^{d}\left(  \operatorname*{End}\left(  S^{2}\left(
V\right)  \oplus\mathbb{C}\right)  \right)  $\textquotedblright\ by
\textquotedblleft$S^{d}\left(  S^{2}\left(  V\right)  \oplus\mathbb{C}\right)
$\textquotedblright.

\item \textbf{Page 66, Exercise 10.0.14:} Replace \textquotedblleft%
$\operatorname*{SL}\nolimits_{2}\left(  \mathbb{C}\right)  $
module\textquotedblright\ by \textquotedblleft$\operatorname*{SL}%
\nolimits_{2}\left(  \mathbb{C}\right)  $-module\textquotedblright.

\item \textbf{Page 68, \S 11.1:} Replace \textquotedblleft it identifies the
space $\operatorname*{End}\left(  V^{\otimes k}\right)  ^{S_{k}}$ with
$\left(  \operatorname*{End}\left(  V\right)  ^{\otimes k}\right)  ^{S_{k}}$
of symmetric tensors\textquotedblright\ by \textquotedblleft it identifies the
space $\operatorname*{End}\left(  V^{\otimes k}\right)  ^{S_{k}}$ with the
space $\left(  \operatorname*{End}\left(  V\right)  ^{\otimes k}\right)
^{S_{k}}$ of symmetric tensors\textquotedblright.

\item \textbf{Page 68, \S 11.1:} Replace \textquotedblleft Applying the
following theorem to $H=S_{n}$\textquotedblright\ by \textquotedblleft
Applying the following theorem to $H=S_{k}$\textquotedblright.

\item \textbf{Page 69, proof of Theorem 11.2.1:} \textquotedblleft
represations\textquotedblright\ $\rightarrow$ \textquotedblleft
representations\textquotedblright.

\item \textbf{Page 69, proof of Theorem 11.2.1:} You write: \textquotedblleft
By complete reducibility, the map $\left(  \left(  U^{\ast}\right)  ^{\otimes
d}\right)  ^{G}\rightarrow\left(  S^{d}U^{\ast}\right)  ^{G}$ is
surjective\textquotedblright. Actually, you don't need to use complete
reducibility here: The projection map
\[
\pi:\left(  U^{\ast}\right)  ^{\otimes d}\rightarrow S^{d}U^{\ast
},\ \ \ \ \ \ \ \ \ \ u_{1}\otimes u_{2}\otimes\cdots\otimes u_{d}\mapsto
u_{1}u_{2}\cdots u_{d}%
\]
has a $G$-equivariant section -- namely, the linear map%
\[
\psi:S^{d}U^{\ast}\rightarrow\left(  U^{\ast}\right)  ^{\otimes d}%
,\ \ \ \ \ \ \ \ \ \ u_{1}u_{2}\cdots u_{d}\mapsto\dfrac{1}{d!}\sum_{\sigma\in
S_{d}}u_{\sigma\left(  1\right)  }\otimes u_{\sigma\left(  2\right)  }%
\otimes\cdots\otimes u_{\sigma\left(  d\right)  }.
\]
Hence, the restriction $\left(  \left(  U^{\ast}\right)  ^{\otimes d}\right)
^{G}\rightarrow\left(  S^{d}U^{\ast}\right)  ^{G}$ of the map $\pi$ to the
$G$-invariants has a section as well (namely, the restriction of the section
$\psi:S^{d}U^{\ast}\rightarrow\left(  U^{\ast}\right)  ^{\otimes d}$ to the
$G$-invariants). Therefore, this restriction is surjective.

I like this argument more not just because it avoids the use of complete
reducibility, but also because it is more general (it works for any subgroup
$G$ of $\operatorname*{GL}\nolimits_{n}$, including those for which the
representations involved fail to be completely reducible).

\item \textbf{Page 70, proof of Theorem 11.2.1:} \textquotedblleft If
$d=k$\textquotedblright\ should be \textquotedblleft If $k=d-k$%
\textquotedblright.

\item \textbf{Page 74:} \textquotedblleft a fix a stochastic\textquotedblright%
\ $\rightarrow$ \textquotedblleft we fix a stochastic\textquotedblright.

\item \textbf{Page 75, \S 12.3:} Replace \textquotedblleft formal linear
combinations of the alphabet $V$\textquotedblright\ by \textquotedblleft
formal linear combinations of the alphabet $B$\textquotedblright.

\item \textbf{Page 75, \S 12.3:} Replace \textquotedblleft Next we define a
polynomial map $\psi_{T}:\operatorname*{rep}\left(  T\right)  \rightarrow
\bigotimes_{p\in\operatorname*{leaf}\left(  T\right)  }$\textquotedblright\ by
\textquotedblleft Next we define a polynomial map $\Psi_{T}%
:\operatorname*{rep}\left(  T\right)  \rightarrow\bigotimes_{p\in
\operatorname*{leaf}\left(  T\right)  }V_{p}$\textquotedblright. (There were
two typos here: \textquotedblleft$\psi_{T}$\textquotedblright\ should be
\textquotedblleft$\Psi_{T}$\textquotedblright, and the \textquotedblleft%
$V_{p}$\textquotedblright\ was missing.)

\item \textbf{Page 76, \S 12.3:} I suppose that \textquotedblleft%
$\bigotimes_{p\in\operatorname*{leaf}\left(  T\right)  f\left(  p\right)  }%
$\textquotedblright\ should be \textquotedblleft$\bigotimes_{p\in
\operatorname*{leaf}\left(  T\right)  }f\left(  p\right)  $\textquotedblright.
\end{itemize}


\end{document}