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\ihead{Errata to ``Hyperplane arrangements and descent algebras''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Hyperplane arrangements and descent algebras}

\textit{Franco V Saliola}

\texttt{\href{http://thales.math.uqam.ca/~saliola/maths/publications/LectureNotes/DesAlgLectureNotes.pdf}{\texttt{saliola
- DesAlgLectureNotes.pdf}}}

version of 10 January 2006

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

I will refer to the results appearing in the article \textquotedblleft A
Hyperplane arrangements and descent algebras\textquotedblright\ by the numbers
under which they appear in this article.

\setcounter{section}{5}

\section{Errata}

\begin{itemize}
\item \textbf{Various places (for example, \S 2.1):} You use the notations
$\subseteq$ and $\subset$ synonymously. It might be better if you consistently
keep to one of them, as the appearance of both of them in your notes suggests
that $\subset$ means proper inclusion (but it does not).

\item \textbf{Page 1, \S 1.1:} Replace \textquotedblleft$v_{1}+v_{2}+v_{3}%
=0$\textquotedblright\ by \textquotedblleft$v_{1}+v_{2}+\cdots+v_{n}%
=0$\textquotedblright.

\item \textbf{Page 3:} Replace \textquotedblleft the nonempty intersections of
the open half spaces\textquotedblright\ by \textquotedblleft a nonempty
intersection of open half spaces\textquotedblright. (Maybe also add
\textquotedblleft(one for each hyperplane)\textquotedblright\ at the end of
the sentence.)

\item \textbf{Page 4, Figure 3:} I think the \textquotedblleft$\left(
+0-\right)  $\textquotedblright\ label is wrong, and should be a
\textquotedblleft$\left(  -0-\right)  $\textquotedblright\ label instead.

\item \textbf{Page 4:} Replace \textquotedblleft and that the
closure\textquotedblright\ by \textquotedblleft and that the
closures\textquotedblright.

\item \textbf{Page 5, \S 1.3:} Your claim that \textquotedblleft the
\textit{join} $X\vee Y$ of $X$ and $Y$ is $X+Y$\textquotedblright\ is
generally false (even when $\mathcal{A}$ is the braid
arrangement)\footnote{For a counterexample, set $n=4$, $X=\left\{  \left(
x_{1},x_{2},x_{3},x_{4}\right)  \in\mathbb{R}^{4}\ \mid\ x_{1}=x_{2}\text{ and
}x_{3}=x_{4}\right\}  $ and $Y=\left\{  \left(  x_{1},x_{2},x_{3}%
,x_{4}\right)  \in\mathbb{R}^{4}\ \mid\ x_{1}=x_{3}\text{ and }x_{2}%
=x_{4}\right\}  $. Then, the join $X\vee Y$ is the whole space, while the sum
$X+Y$ is the hyperplane $\left\{  \left(  x_{1},x_{2},x_{3},x_{4}\right)
\in\mathbb{R}^{4}\ \mid\ x_{1}-x_{2}-x_{3}+x_{4}=0\right\}  $ (which is not an
element of $\mathcal{L}$).}. I don't think the join can be characterized this
easily. (Of course, the existence of a join follows from the existence of the
meet using the fact that any finite meet-semilattice having a greatest element
is a lattice.)

\item \textbf{Page 5, \S 1.3:} I don't think your claim that \textquotedblleft
The rank of $X\in\mathcal{L}$ is the dimension of the subspace $X\subset
\mathbb{R}^{d}$\textquotedblright\ is true.

\item \textbf{Page 8, Exercise 2:} I think it would be useful to add the
following claim between (2) and (3): \textquotedblleft$x\leq xy$%
\textquotedblright.

\item \textbf{Page 9:} In the formula for $\sigma_{H_{ij}}\left(  BC\right)
$, why do you write \textquotedblleft$C\left(  j\right)  <C\left(  i\right)
$\textquotedblright\ instead of \textquotedblleft$C\left(  i\right)  >C\left(
j\right)  $\textquotedblright? Of course, this is equivalent, but it looks out
of place.

\item \textbf{Page 10, Example 4:} Replace \textquotedblleft subset of
the\textquotedblright\ by \textquotedblleft subset of\textquotedblright.

\item \textbf{Page 11, \S 1.6:} You write: \textquotedblleft and with $\left(
s,t\right)  $-entry the probability of moving to the state $s$ from the state
$t$\textquotedblright. I think you want to interchange the words
\textquotedblleft from\textquotedblright\ and \textquotedblleft
to\textquotedblright\ here, since otherwise (I believe) this definition does
not match the equations further below.

\item \textbf{Page 12, Theorem 1.4:} Replace \textquotedblleft
left\textquotedblright\ by \textquotedblleft let\textquotedblright.

\item \textbf{Page 12:} Replace \textquotedblleft defined in the next
section\textquotedblright\ by \textquotedblleft defined in the previous
section\textquotedblright.

\item \textbf{Page 12:} Replace \textquotedblleft$\sum_{c\in\mathcal{F}}%
$\textquotedblright\ by \textquotedblleft$\sum_{c\in\mathcal{C}}%
$\textquotedblright.

\item \textbf{Page 12:} Replace \textquotedblleft Since $k\mathcal{C}%
$\textquotedblright\ by \textquotedblleft Since $\mathbb{R}\mathcal{C}%
$\textquotedblright.

\item \textbf{Page 13, \S 2.1:} It would be good to explain how you define the
descent algebra\footnote{along the lines of: \textquotedblleft We define
$\mathcal{D}\left(  S_{n}\right)  $ to be the vector subspace of $kS_{n}$
spanned by the $x_{J}$ with $J\subseteq\left[  n-1\right]  $. We call
$\mathcal{D}\left(  S_{n}\right)  $ the \textit{descent algebra} of $S_{n}$,
although we do not yet know that it is an algebra (we shall see this in the
proof of Theorem 2.1).\textquotedblright}. You use this notion in the proof of
Theorem 2.1, yet before you show that the $x_{J}$ span an algebra, and it is
not immediately clear whether you mean the span of the $x_{J}$ or the
subalgebra they generate or something else.

\item \textbf{Page 13, \S 2.2:} Replace \textquotedblleft
endomorphisms\textquotedblright\ by \textquotedblleft
automorphisms\textquotedblright.

\item \textbf{Page 13, \S 2.2:} Replace \textquotedblleft
endomorphism\textquotedblright\ by \textquotedblleft
automorphism\textquotedblright.

\item \textbf{Page 13, \S 2.2:} Replace \textquotedblleft if $a,a^{\prime}\in
A$\textquotedblright\ by \textquotedblleft if $a,a^{\prime}\in A^{G}%
$\textquotedblright.

\item \textbf{Page 14, \S 2.3:} Since you have recalled the definition of an
automorphism previously, it would seem reasonable to also give the definition
of an anti-isomorphism and what it means for two algebras to be anti-isomorphic.

\item \textbf{Page 14, proof of Theorem 2.1:} Replace \textquotedblleft that
$w\left(  a\right)  =a$\textquotedblright\ by \textquotedblleft that
$\omega\left(  a\right)  =a$\textquotedblright.

\item \textbf{Page 14, proof of Theorem 2.1:} Replace \textquotedblleft This
gives an algebra homomorphism\textquotedblright\ by \textquotedblleft Thus,
$a\mapsto f_{a}$ gives an algebra homomorphism\textquotedblright.

\item \textbf{Page 15:} Replace both \textquotedblleft$\operatorname*{End}%
\nolimits_{S_{n}}$\textquotedblright's by \textquotedblleft%
$\operatorname*{End}\nolimits_{kS_{n}}$\textquotedblright's.

\item \textbf{Page 15:} Replace each \textquotedblleft$\psi$\textquotedblright%
\ and each \textquotedblleft$\psi^{-1}$\textquotedblright\ appearing on page
15 by \textquotedblleft$\psi^{-1}$\textquotedblright\ and \textquotedblleft%
$\psi$\textquotedblright, respectively.

\item \textbf{Page 15:} When you introduce $a_{B}$, it would be useful to
point out that $a_{B}$ is the sum of all set compositions of $\left[
n\right]  $ having the form $\left(  C_{1},C_{2},\ldots,C_{m}\right)  $ which
satisfy $\left\vert C_{i}\right\vert =\left\vert B_{i}\right\vert $ for each
$i\in\left\{  1,2,\ldots,m\right\}  $. This alternative description of $a_{B}$
is what is used on page 16 to find $a_{B}\left(  1,2,\ldots,n\right)  $.

\item \textbf{Page 16, proof of Theorem 2.1:} You need to WLOG apply $n\geq1$
in the last paragraph (otherwise, the descent algebra is not of dimension
$2^{n-1}$).

\item \textbf{Page 16, proof of Theorem 2.1:} Replace \textquotedblleft and
the sum\textquotedblright\ by \textquotedblleft and the sums\textquotedblright.

\item \textbf{Page 16, \S 2.3:} I know this flies in the face of the
underlying philosophy of your article, but methinks it wouldn't hurt to point
out that the geometric language you are using (i.e., the language of
hyperplane arrangements, faces and chambers) wasn't necessary for the proof of
Theorem 2.1. In fact, Theorem 2.1 (and, with it, the fact that $\mathcal{D}%
\left(  S_{n}\right)  $ is a subalgebra of $kS_{n}$) becomes a purely
elementary combinatorial statement if we just define $\mathcal{F}$ as the set
of all set compositions of $\left[  n\right]  $ (and we define the action of
$S_{n}$ on $\mathcal{F}$ by setting%
\begin{align}
\omega\left(  \left(  B_{1},B_{2},\ldots,B_{m}\right)  \right)   &  =\left(
\omega\left(  B_{1}\right)  ,\omega\left(  B_{2}\right)  ,\ldots,\omega\left(
B_{m}\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }\omega\in S_{n}\text{ and }\left(
B_{1},B_{2},\ldots,B_{m}\right)  \in\mathcal{F} \label{p16.a.1}%
\end{align}
). Moreover, your proof of Theorem 2.1 becomes a purely combinatorial proof of
this combinatorial statement if we make the following changes:

\begin{itemize}
\item We define $\mathcal{F}$ as the set of all set compositions of $\left[
n\right]  $.

\item We define the action of $S_{n}$ on $\mathcal{F}$ by setting
(\ref{p16.a.1}).

\item For any two set compositions $\left(  B_{1},B_{2},\ldots,B_{l}\right)  $
and $\left(  C_{1},C_{2},\ldots,C_{m}\right)  $ in $\mathcal{F}$, we define
the product $\left(  B_{1},B_{2},\ldots,B_{l}\right)  \left(  C_{1}%
,C_{2},\ldots,C_{m}\right)  $ by%
\begin{align*}
&  \left(  B_{1},B_{2},\ldots,B_{l}\right)  \left(  C_{1},C_{2},\ldots
,C_{m}\right) \\
&  =\left(  B_{1}\cap C_{1},B_{1}\cap C_{2},\ldots,B_{1}\cap C_{m},\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  B_{2}\cap C_{1},B_{2}\cap C_{2},\ldots
,B_{2}\cap C_{m},\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \ldots,\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  B_{l}\cap C_{1},B_{l}\cap C_{2},\ldots
,B_{l}\cap C_{m}\right)  ^{\text{\ScissorHollowRight}},
\end{align*}
where $\text{\ScissorHollowRight}$ means \textquotedblleft delete empty
intersections from the list\textquotedblright.

\item We define $\mathcal{C}$ as the set of all set compositions of $\left[
n\right]  $ into singleton blocks. This is a subset of $\mathcal{F}$.

\item We replace \textquotedblleft faces of the chamber $\left(
1,2,\ldots,n\right)  $\textquotedblright\ by \textquotedblleft set
compositions of $\left[  n\right]  $ having the form%
\begin{align*}
&  \left(  \left\{  1,2,\ldots,i_{1}\right\}  ,\left\{  i_{1}+1,i_{1}%
+2,\ldots,i_{2}\right\}  ,\left\{  i_{2}+1,i_{2}+2,\ldots,i_{3}\right\}
,\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \ldots,\left\{  i_{k}+1,i_{k}+2,\ldots
,n\right\}  \right)
\end{align*}
where $i_{1},i_{2},\ldots,i_{k}$ are elements of $\left[  n-1\right]  $
satisfying $i_{1}<i_{2}<\cdots<i_{k}$\textquotedblright. (This includes both
the set composition $\left(  \left\{  1,2,\ldots,n\right\}  \right)  $, which
is obtained for $k=0$, and the set composition $\left(  1,2,\ldots,n\right)
=\left(  \left\{  1\right\}  ,\left\{  2\right\}  ,\ldots,\left\{  n\right\}
\right)  $, which is obtained for $k=n-1$ and $i_{j}=j$.)
\end{itemize}

\item \textbf{Page 17:} When you speak of cosets, it might be helpful to
explain whether they are left or right cosets.

\item \textbf{Page 17, Theorem 2.2:} The definition of $\mathcal{C}$ should be
moved from the proof of Theorem 2.2 into the statement of Theorem 2.2.

\item \textbf{Page 17, proof of Theorem 2.2:} Replace \textquotedblleft any
$kW$-endomorphism commuting with the action of $W$\textquotedblright\ by
\textquotedblleft any $kW$-endomorphism of $kW$\textquotedblright\ (or
\textquotedblleft any endomorphism of $kW$ commuting with the action of
$W$\textquotedblright).

\item \textbf{Page 17, proof of Theorem 2.2:} Replace \textquotedblleft be
reversing\textquotedblright\ by \textquotedblleft by
reversing\textquotedblright.

\item \textbf{Page 18, \S 2.5:} At the beginning of \S 2.5, it would be good
to explicitly say that you are returning to the setting in which $\mathcal{A}$
is the braid arrangement.

\item \textbf{Page 18, \S 2.5:} You say that the elements $x_{\lambda}$ (with
$\lambda$ ranging over the integer compositions of $n$) form a basis of
$\left(  k\mathcal{F}\right)  ^{S_{n}}$. At this point it is natural to
observe that we have already seen these elements $x_{\lambda}$: namely, the
elements $a_{B}$ defined in the proof of Theorem 2.1 satisfy $a_{B}%
=x_{\lambda\left(  B\right)  }$ for all set compositions $B$ of $\left[
n\right]  $.

\item \textbf{Page 18, proof of Proposition 2.3:} Replace \textquotedblleft
The coefficient of $C$ is exactly\textquotedblright\ by \textquotedblleft The
coefficient of $C$ in the left hand side is exactly\textquotedblright.

\item \textbf{Page 19, proof of Corollary 2.4:} Replace \textquotedblleft%
$\sum_{i=1}^{k-1}\gamma_{i}$\textquotedblright\ by \textquotedblleft%
$\sum_{i=1}^{k-1}\gamma_{i}+1$\textquotedblright.

\item \textbf{Page 19, proof of Corollary 2.4:} After \textquotedblleft give
the integer composition $\gamma$\textquotedblright, add \textquotedblleft(when
$AB=C$)\textquotedblright.

\item \textbf{Page 19, Exercise 5:} It would be good to clarify that
$\lambda\left(  J\right)  $ is defined to be $\left(  n\right)  $ when $J$ is
the empty set. (Your definition of $\lambda\left(  J\right)  $ is slightly
unclear in this case.)

\item \textbf{Page 20, Exercise 5:} Replace \textquotedblleft on
$n$\textquotedblright\ by \textquotedblleft of $n$\textquotedblright.

\item \textbf{Page 20, Exercise 5:} I think this exercise is wrong as stated.
For instance (it would be good if you could double-check me), if
$K=\varnothing$, then $c_{\lambda\left(  K\right)  ,\lambda\left(  J\right)
,\lambda\left(  L\right)  }=\delta_{J,L}$, whereas
\begin{align*}
&  \left\vert \left\{  \omega\in X_{J}^{-1}\cap\underbrace{X_{K}}_{=\left\{
\operatorname*{id}\right\}  }\ :\ L=\underbrace{K}_{=\varnothing}\cap
\,\omega^{-1}\left(  J\right)  \right\}  \right\vert \\
&  =\left\vert \left\{  \omega\in X_{J}^{-1}\cap\left\{  \operatorname*{id}%
\right\}  \ :\ L=\varnothing\cap\omega^{-1}\left(  J\right)  \right\}
\right\vert \\
&  =\left\vert \left\{  \omega\in\left\{  \operatorname*{id}\right\}
\ :\ L=\varnothing\right\}  \right\vert =\delta_{L,\varnothing}.
\end{align*}
Is Theorem 1 in
\texttt{\href{http://arxiv.org/abs/0706.2714v1}{arXiv:0706.2714v1}} what you
are trying to get at?

\item \textbf{Page 20, \S 3.1:} Replace \textquotedblleft if $e$ is
idempotent\textquotedblright\ by \textquotedblleft if $e$ is a nonzero
idempotent\textquotedblright.

\item \textbf{Page 20, \S 3.1:} The claim \textquotedblleft Moreover, if $M$
is any $A$-module, then there is an $A$-module decomposition of $M$ given by
the idempotents: $M\cong\bigoplus_{i\in I}Me_{i}$\textquotedblright\ is false
as stated. In truth, if $M$ is a left $A$-module, then $M=\bigoplus_{i\in
I}e_{i}M$ (not $Me_{i}$, which makes no sense), but this is only a
decomposition of $k$-vector spaces, not of $A$-modules (unless $M$ is an
$\left(  A,A\right)  $-bimodule, in which case this is a decomposition of
right $A$-modules).

An example for this is when $A$ is a matrix algebra $k^{n\times n}$ and $M$ is
the vector space $k^{n}$ (with the obvious left $A$-action by matrix-vector
multiplication). The diagonal unit matrices $e_{i}:=E_{i,i}\in A$ for
$i\in\left\{  1,2,\ldots,n\right\}  $ form a complete system of primitive
orthogonal idempotents for $A$, but $M$ is indecomposable as an $A$-module,
and its subspaces $e_{i}M$ are only $k$-vector subspaces, not $A$-submodules.

\item \textbf{Page 20, \S 3.1:} It would be good to give references for all
claims about primitive idempotents made here (in the first two paragraphs of
\S 3.1). One such reference is \S I.4 and Corollary I.5.17 in: Ibrahim Assem,
Daniel Simson, Andrzej Skowr\'{o}nski, \textit{Elements of the Representation
Theory of Associative Algebras, Volume 1: Techniques of Representation
Theory}, Cambridge University Press 2006.

\item \textbf{Page 21, Example 5:} Two of the minus signs in the computation
of $\left(  e_{\left\{  13,2\right\}  }\right)  ^{2}$ should be plus signs
(namely, the last minus signs on the second and on the third line of the computation).

\item \textbf{Page 22, proof of Lemma 3.1:} Replace \textquotedblleft
Proposition 2 (2)\textquotedblright\ by \textquotedblleft Exercise 2
(2)\textquotedblright.

\item \textbf{Page 22, proof of Lemma 3.1:} Replace \textquotedblleft$W\vee
X$\textquotedblright\ by \textquotedblleft$X\vee W$\textquotedblright\ twice.
(Of course, \textquotedblleft$W\vee X$\textquotedblright\ is correct, too, but
it helps to keep notations consistent.)

\item \textbf{Page 22, proof of Lemma 3.1:} In your place, I would explain how
you get $z=\sum_{Y\geq X\vee W}ze_{Y}$. (Namely, you have $ze_{X\vee
W}=e_{X\vee W}=z-\sum_{Y>X\vee W}ze_{Y}$, so that $z=ze_{X\vee W}%
+\sum_{Y>X\vee W}ze_{Y}=\sum_{Y\geq X\vee W}ze_{Y}$).

\item \textbf{Page 22, proof of Lemma 3.1:} Replace \textquotedblleft
Proposition 2 (5)\textquotedblright\ by \textquotedblleft Exercise 2
(5)\textquotedblright.

\item \textbf{Page 22, proof of Theorem 3.2:} Replace \textquotedblleft
Proposition 2 (5)\textquotedblright\ by \textquotedblleft Exercise 2
(5)\textquotedblright.

\item \textbf{Page 22, proof of Theorem 3.2:} In the \textquotedblleft
Idempotent\textquotedblright\ part of the proof, please define $x$ and $y$.
(Namely, $x$ is the element of support $X$ that was chosen while defining
$e_{X}$, and $y$ is the element of support $Y$ that was chosen while defining
$e_{Y}$.)

\item \textbf{Page 22, proof of Theorem 3.2:} After you observe that
\textquotedblleft$e_{Y}z=e_{Y}$ for any $z$ with $\operatorname*{supp}\left(
z\right)  \leq Y$\textquotedblright, it would be helpful to point out that
this, in particular, shows that $e_{Y}y=e_{Y}$. (You use this equality a few
lines later.)

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Idempotent\textquotedblright\ part of the proof, it would help to clarify why
$\sum_{Y>X}xe_{Y}\left(  ye_{X}\right)  $. (Indeed, this is because every
$Y>X$ satisfies $ye_{X}=0$ (by Lemma 3.1, applied to $w=y$)).

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Orthogonal\textquotedblright\ part of the proof, please define $x$. (Namely,
$x$ is the element of support $X$ that was chosen while defining $e_{X}$.)

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Orthogonal\textquotedblright\ part of the proof, you are writing
\textquotedblleft$e_{X}e_{Y}=e_{X}xe_{Y}$\textquotedblright. It would be good
to explain why this holds. (Namely, it follows from the equality $e_{X}%
=e_{X}x$. This equality can be proven just as the equality $e_{Y}=e_{Y}y$ in
the \textquotedblleft Idempotent\textquotedblright\ part of the proof. In my
opinion it wouldn't hurt to explicitly state both equalities $x=e_{X}x$ and
$x=xe_{X}$ as a lemma, given that you are applying them several times.)

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Primitive\textquotedblright\ part of the proof, you are using some notations
which, in my opinion, you should define:

\begin{itemize}
\item You extend the map $\operatorname*{supp}:\mathcal{F}\rightarrow
\mathcal{L}$ to a $k$-linear map $k\mathcal{F}\rightarrow k\mathcal{L}$, and
denote the latter map again by $\operatorname*{supp}$. This allows you to
speak of $\operatorname*{supp}w$ for arbitrary $w\in k\mathcal{F}$, not only
for $w\in\mathcal{F}$.

\item You extend the binary operation $\vee:\mathcal{L}\times\mathcal{L}%
\rightarrow\mathcal{L}$ to a $k$-bilinear binary operation $k\mathcal{L}\times
k\mathcal{L}\rightarrow k\mathcal{L}$, and denote this latter operation again
by $\vee$. This operation $\vee$ turns $k\mathcal{L}$ into a commutative
$k$-algebra. (This allows you to speak of the $E_{X}$ as being idempotents.)
The map $\operatorname*{supp}:k\mathcal{F}\rightarrow k\mathcal{L}$ becomes a
surjective $k$-algebra homomorphism.
\end{itemize}

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Primitive\textquotedblright\ part of the proof, you write: \textquotedblleft
Then the above arguments show that the elements $E_{X}$ are orthogonal
idempotents in $k\mathcal{L}$ summing to $1$\textquotedblright. This looks a
bit like a non-sequitur (although I understand what you apparently want to
say). In my opinion, it would be easier to first prove that the elements
$e_{X}$ of $k\mathcal{F}$ lift the elements $E_{X}$ of $k\mathcal{L}$ (that
is, $\operatorname*{supp}\left(  e_{X}\right)  =E_{X}$ for every
$X\in\mathcal{L}$), and then conclude that the $E_{X}$ are orthogonal
idempotents in $k\mathcal{L}$ summing up to $1$ (since the $e_{X}$ are
orthogonal idempotents in $k\mathcal{F}$ summing up to $1$). The fact that the
elements $e_{X}$ of $k\mathcal{F}$ lift the elements $E_{X}$ of $k\mathcal{L}$
is proven in the next paragraph of your proof of Theorem 3.2.

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Primitive\textquotedblright\ part of the proof, replace \textquotedblleft
orthognal\textquotedblright\ by \textquotedblleft orthogonal\textquotedblright.

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Primitive\textquotedblright\ part of the proof, please define $x$. (Namely,
$x$ is the element of support $X$ that was chosen while defining $e_{X}$.)

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Primitive\textquotedblright\ part of the proof, replace \textquotedblleft%
$\sum_{Y\geq X}E_{X}$\textquotedblright\ by \textquotedblleft$\sum_{Y\geq
X}E_{Y}$\textquotedblright.

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Primitive\textquotedblright\ part of the proof, replace \textquotedblleft%
$X-\sum_{Y>X}E_{Y}=E_{Y}$\textquotedblright\ by \textquotedblleft$X-\sum
_{Y>X}E_{Y}=E_{X}$\textquotedblright.

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Primitive\textquotedblright\ part of the proof, you claim that
\textquotedblleft This kernel is nilpotent\textquotedblright\ (speaking of the
kernel of $\operatorname*{supp}$). This is correct, but I don't see any
previous statement from which this would follow easily. Let me outline my
proof of this nilpotency.

\textbf{Theorem 3.1a.} The kernel of the $k$-algebra homomorphism
$\operatorname*{supp}:k\mathcal{F}\rightarrow k\mathcal{L}$ (which is defined
by extending the map $\operatorname*{supp}:\mathcal{F}\rightarrow\mathcal{L}$,
as above) is nilpotent.

\textit{Proof sketch.} Let us denote this kernel by $P$. Let us furthermore
define a few more notations:

For every $X\in\mathcal{L}$, we define the \textit{corank} of $X$ to be the
largest $\ell\in\mathbb{N}$ such that there exist elements $X_{0},X_{1}%
,\ldots,X_{\ell}\in\mathcal{L}$ with $X_{0}=X$ and $X_{0}<X_{1}<\cdots
<X_{\ell}$. Notice that this is well-defined (because such an $\ell$ exists
(namely, $\ell=0$ fits the bill), but the finiteness of $X$ forces any such
$\ell$ to be $<\left\vert X\right\vert $). We denote the corank of $X$ by
$\operatorname*{corank}X$. The following property of coranks is obvious: If
$X$ and $Y$ are two elements of $\mathcal{L}$ such that $X<Y$, then%
\begin{equation}
\operatorname*{corank}X>\operatorname*{corank}Y. \label{pf.thm.3.1a.triv}%
\end{equation}


For every $N\in\mathbb{Z}$, we define a subset $\mathcal{L}_{N}$ of
$\mathcal{L}$ by
\[
\mathcal{L}_{N}=\left\{  X\in\mathcal{L}\ \mid\ \operatorname*{corank}%
X<N\right\}  .
\]
Then, $\mathcal{L}_{0}=\varnothing$ (since no $X\in\mathcal{L}$ has corank
$<0$), while $\mathcal{L}_{\left\vert X\right\vert }=\mathcal{L}$ (since each
$X\in\mathcal{L}$ has corank $<\left\vert X\right\vert $).

For every $N\in\mathbb{Z}$, we define a subset $\mathcal{F}_{N}$ of
$\mathcal{F}$ by $\mathcal{F}_{N}=\operatorname*{supp}\nolimits^{-1}%
\mathcal{L}_{N}$ (where $\operatorname*{supp}$ here means the map
$\operatorname*{supp}:\mathcal{F}\rightarrow\mathcal{L}$, not the map
$\operatorname*{supp}:k\mathcal{F}\rightarrow k\mathcal{L}$). Clearly,
$\mathcal{F}_{0}=\operatorname*{supp}\nolimits^{-1}\underbrace{\mathcal{L}%
_{0}}_{=\varnothing}=\operatorname*{supp}\nolimits^{-1}\varnothing
=\varnothing$ and $\mathcal{F}_{\left\vert X\right\vert }=\operatorname*{supp}%
\nolimits^{-1}\underbrace{\mathcal{L}_{\left\vert X\right\vert }%
}_{=\mathcal{L}}=\operatorname*{supp}\nolimits^{-1}\mathcal{L}=\mathcal{F}$.

For every $N\in\mathbb{N}$, the set $\mathcal{F}_{N}$ is a subset of
$\mathcal{F}$, and thus $k\mathcal{F}_{N}$ becomes a $k$-vector subspace of
$k\mathcal{F}$. We are now going to show that%
\begin{equation}
\left(  k\mathcal{F}_{N}\right)  \cdot P\subseteq k\mathcal{F}_{N-1}%
\ \ \ \ \ \ \ \ \ \ \text{for every }N\in\mathbb{Z}. \label{pf.thm.3.1a.rec}%
\end{equation}


\textit{Proof of (\ref{pf.thm.3.1a.rec}):} Let $N\in\mathbb{Z}$. We need to
prove (\ref{pf.thm.3.1a.rec}). It is clearly enough to show that $xp\in
k\mathcal{F}_{N-1}$ for each $x\in k\mathcal{F}_{N}$ and $p\in P$. So fix
$x\in k\mathcal{F}_{N}$ and $p\in P$. We need to show that $xp\in
k\mathcal{F}_{N-1}$. Since this relation is $k$-linear in $x$, we can WLOG
assume that $x\in\mathcal{F}_{N}$. Assume this.

Every element $u\in\mathcal{F}_{N}$ satisfying $\operatorname*{supp}%
u\not \leq \operatorname*{supp}x$ satisfies%
\begin{equation}
xu\in\mathcal{F}_{N-1}. \label{pf.thm.3.1a.rec.pf.0}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.3.1a.rec.pf.0}):} Let $u$ be an
element of $\mathcal{F}_{N}$ satisfying $\operatorname*{supp}u\not \leq
\operatorname*{supp}x$. Then, $\operatorname*{supp}u\vee\operatorname*{supp}%
x\neq\operatorname*{supp}x$ (because otherwise, we would have
$\operatorname*{supp}u\leq\operatorname*{supp}u\vee\operatorname*{supp}%
x=\operatorname*{supp}x$, which would contradict $\operatorname*{supp}%
u\not \leq \operatorname*{supp}x$). Combined with $\operatorname*{supp}%
x\leq\operatorname*{supp}u\vee\operatorname*{supp}x$, this yields
$\operatorname*{supp}x<\operatorname*{supp}u\vee\operatorname*{supp}x$. Hence,
(\ref{pf.thm.3.1a.triv}) (applied to $X=\operatorname*{supp}x$ and
$Y=\operatorname*{supp}u\vee\operatorname*{supp}x$) yields
$\operatorname*{corank}\left(  \operatorname*{supp}x\right)
>\operatorname*{corank}\left(  \operatorname*{supp}u\vee\operatorname*{supp}%
x\right)  $. Since $\operatorname*{supp}u\vee\operatorname*{supp}%
x=\operatorname*{supp}x\vee\operatorname*{supp}u=\operatorname*{supp}\left(
xu\right)  $, this rewrites as $\operatorname*{corank}\left(
\operatorname*{supp}x\right)  >\operatorname*{corank}\left(
\operatorname*{supp}\left(  xu\right)  \right)  $. Hence,
$\operatorname*{corank}\left(  \operatorname*{supp}\left(  xu\right)  \right)
<\operatorname*{corank}\left(  \operatorname*{supp}x\right)  $ and thus%
\[
\operatorname*{corank}\left(  \operatorname*{supp}\left(  xu\right)  \right)
\leq\underbrace{\operatorname*{corank}\left(  \operatorname*{supp}x\right)
}_{\substack{<N\\\text{(since }\operatorname*{supp}x\in\mathcal{L}%
_{N}\\\text{(since }x\in\mathcal{F}_{N}=\operatorname*{supp}\nolimits^{-1}%
\mathcal{L}_{N}\text{))}}}-\,1<N-1.
\]
In other words, $\operatorname*{supp}\left(  xu\right)  \in\mathcal{L}_{N-1}$,
so that $xu\in\operatorname*{supp}\nolimits^{-1}\mathcal{L}_{N-1}%
=\mathcal{F}_{N-1}$. This proves (\ref{pf.thm.3.1a.rec.pf.0}).}

We can write $p$ in the form $p=\sum_{u\in\mathcal{F}}\lambda_{u}u$ for some
family $\left(  \lambda_{u}\right)  _{u\in\mathcal{F}}$ of elements of $k$.
Consider this family $\left(  \lambda_{u}\right)  _{u\in\mathcal{F}}$. We have%
\begin{equation}
\sum_{\substack{u\in\mathcal{F};\\\operatorname*{supp}u=U}}\lambda
_{u}=0\ \ \ \ \ \ \ \ \ \ \text{for every }U\in\mathcal{L}
\label{pf.thm.3.1a.rec.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.3.1a.rec.pf.1}):} We have $p\in P$. In
other words, $p$ belongs to the kernel of the $k$-algebra homomorphism
$\operatorname*{supp}:k\mathcal{F}\rightarrow k\mathcal{L}$ (since $P$ is the
kernel of the $k$-algebra homomorphism $\operatorname*{supp}:k\mathcal{F}%
\rightarrow k\mathcal{L}$). In other words, $\operatorname*{supp}p=0$. Hence,%
\begin{align*}
0  &  =\operatorname*{supp}\underbrace{p}_{=\sum_{u\in\mathcal{F}}\lambda
_{u}u}=\operatorname*{supp}\left(  \sum_{u\in\mathcal{F}}\lambda_{u}u\right)
\\
&  =\sum_{u\in\mathcal{F}}\lambda_{u}\operatorname*{supp}u=\sum_{U\in
\mathcal{L}}\sum_{\substack{u\in\mathcal{F};\\\operatorname*{supp}u=U}%
}\lambda_{u}U=\sum_{U\in\mathcal{L}}\left(  \sum_{\substack{u\in
\mathcal{F};\\\operatorname*{supp}u=U}}\lambda_{u}\right)  U.
\end{align*}
Thus, the element $\sum_{U\in\mathcal{L}}\left(  \sum_{\substack{u\in
\mathcal{F};\\\operatorname*{supp}u=U}}\lambda_{u}\right)  U$ of
$k\mathcal{L}$ is $0$. Since the elements $U$ of $\mathcal{L}$ are
$k$-linearly independent in $k\mathcal{L}$, this shows that $\sum
_{\substack{u\in\mathcal{F};\\\operatorname*{supp}u=U}}\lambda_{u}=0$ for
every $U\in\mathcal{L}$. This proves (\ref{pf.thm.3.1a.rec.pf.1}).}. Hence,%
\begin{equation}
\sum_{\substack{u\in\mathcal{F};\\\operatorname*{supp}u\leq
\operatorname*{supp}x}}\lambda_{u}=\sum_{\substack{U\in\mathcal{L}%
;\\U\leq\operatorname*{supp}x}}\ \ \underbrace{\sum_{\substack{u\in
\mathcal{F};\\\operatorname*{supp}u=U}}\lambda_{u}}_{\substack{=0\\\text{(by
(\ref{pf.thm.3.1a.rec.pf.1}))}}}=0. \label{pf.thm.3.1a.rec.pf.2}%
\end{equation}


Now, multiplying both sides of the equality $p=\sum_{u\in\mathcal{F}}%
\lambda_{u}u$ by $x$ from the left, we obtain%
\begin{align*}
xp  &  =x\sum_{u\in\mathcal{F}}\lambda_{u}u=\sum_{u\in\mathcal{F}}\lambda
_{u}xu\\
&  =\sum_{\substack{u\in\mathcal{F};\\\operatorname*{supp}u\leq
\operatorname*{supp}x}}\lambda_{u}\underbrace{xu}_{\substack{=x\\\text{(by
Exercise 2 (5),}\\\text{since }\operatorname*{supp}u\leq\operatorname*{supp}%
x\text{)}}}+\sum_{\substack{u\in\mathcal{F};\\\operatorname*{supp}%
u\not \leq \operatorname*{supp}x}}\lambda_{u}\underbrace{xu}_{\substack{\in
\mathcal{F}_{N-1}\\\text{(by (\ref{pf.thm.3.1a.rec.pf.0}))}}}\\
&  \in\underbrace{\sum_{\substack{u\in\mathcal{F};\\\operatorname*{supp}%
u\leq\operatorname*{supp}x}}\lambda_{u}}_{\substack{=0\\\text{(by
(\ref{pf.thm.3.1a.rec.pf.2}))}}}x+\underbrace{\sum_{\substack{u\in
\mathcal{F};\\\operatorname*{supp}u\not \leq \operatorname*{supp}x}%
}\lambda_{u}\mathcal{F}_{N-1}}_{\subseteq k\mathcal{F}_{N-1}}\subseteq
0x+k\mathcal{F}_{N-1}=k\mathcal{F}_{N-1}.
\end{align*}
This finishes the proof of (\ref{pf.thm.3.1a.rec}).

Now, for every $M\in\mathbb{N}$ and every $N\in\mathbb{Z}$, we have%
\begin{equation}
\left(  k\mathcal{F}_{N}\right)  \cdot P^{M}\subseteq k\mathcal{F}_{N-M}.
\label{pf.thm.3.1a.recM}%
\end{equation}
(In fact, this can be proven by straightforward induction over $M$ using
(\ref{pf.thm.3.1a.rec}).) Applying (\ref{pf.thm.3.1a.recM}) to $N=\left\vert
X\right\vert $ and $M=\left\vert X\right\vert $, we obtain%
\[
\left(  k\mathcal{F}_{\left\vert X\right\vert }\right)  \cdot P^{\left\vert
X\right\vert }\subseteq k\underbrace{\mathcal{F}_{\left\vert X\right\vert
-\left\vert X\right\vert }}_{=\mathcal{F}_{0}=\varnothing}=k\varnothing=0.
\]
Since $\mathcal{F}_{\left\vert X\right\vert }=\mathcal{F}$, this rewrites as
$\left(  k\mathcal{F}\right)  \cdot P^{\left\vert X\right\vert }\subseteq0$.
Now, $P^{\left\vert X\right\vert }=\underbrace{1}_{\in k\mathcal{F}}%
\cdot\,P^{\left\vert X\right\vert }\subseteq\left(  k\mathcal{F}\right)  \cdot
P^{\left\vert X\right\vert }\subseteq0$, so that $P^{\left\vert X\right\vert
}=0$. Thus, $P$ is nilpotent. This proves Theorem 3.1a.

\item \textbf{Page 23, proof of Theorem 3.2:} In the \textquotedblleft
Primitive\textquotedblright\ part of the proof, replace \textquotedblleft for
some $n\geq0$\textquotedblright\ by \textquotedblleft for some $n\geq
1$\textquotedblright\ (since you end up using $e_{1}=e_{1}^{n}$ on the next line).

\item \textbf{Page 24, Remark 3.3:} Replace \textquotedblleft in
for\textquotedblright\ by \textquotedblleft for\textquotedblright.

\item \textbf{Page 24, Remark 3.3:} One further observation needs to be
checked to ensure that the proofs still hold if $x$ is replaced by
$\widetilde{x}$: Namely, it should be checked that $\widetilde{x}%
y\widetilde{x}=\widetilde{x}y$ for every $y\in\mathcal{F}$. This, fortunately,
is easy.\footnote{\textit{Proof.} Let $y\in\mathcal{F}$. Then,%
\begin{align*}
\widetilde{x}y\widetilde{x}  &  =\left(  \sum_{\substack{x\in\mathcal{F}%
;\\\operatorname*{supp}x=X}}\lambda_{x}x\right)  y\left(  \sum_{\substack{x\in
\mathcal{F};\\\operatorname*{supp}x=X}}\lambda_{x}x\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widetilde{x}=\sum_{\substack{x\in
\mathcal{F};\\\operatorname*{supp}x=X}}\lambda_{x}x\right) \\
&  =\left(  \sum_{\substack{x\in\mathcal{F};\\\operatorname*{supp}x=X}%
}\lambda_{x}x\right)  y\left(  \sum_{\substack{x^{\prime}\in\mathcal{F}%
;\\\operatorname*{supp}\left(  x^{\prime}\right)  =X}}\lambda_{x^{\prime}%
}x^{\prime}\right) \\
&  =\sum_{\substack{x\in\mathcal{F};\\\operatorname*{supp}x=X}}\lambda_{x}%
\sum_{\substack{x^{\prime}\in\mathcal{F};\\\operatorname*{supp}\left(
x^{\prime}\right)  =X}}\lambda_{x^{\prime}}\underbrace{xyx^{\prime}%
}_{\substack{=xy\\\text{(by Exercise 2 (5) (applied to }xy\text{ and
}x^{\prime}\text{ instead of }x\text{ and }y\text{),}\\\text{since
}\operatorname*{supp}\left(  x^{\prime}\right)  =X=\operatorname*{supp}%
x\leq\operatorname*{supp}x\vee\operatorname*{supp}y=\operatorname*{supp}%
\left(  xy\right)  \text{)}}}\\
&  =\sum_{\substack{x\in\mathcal{F};\\\operatorname*{supp}x=X}}\lambda
_{x}\underbrace{\sum_{\substack{x^{\prime}\in\mathcal{F}%
;\\\operatorname*{supp}\left(  x^{\prime}\right)  =X}}\lambda_{x^{\prime}}%
}_{=1}xy=\underbrace{\sum_{\substack{x\in\mathcal{F};\\\operatorname*{supp}%
x=X}}\lambda_{x}x}_{=\widetilde{x}}y=\widetilde{x}y,
\end{align*}
qed.}

\item \textbf{Page 24, Remark 3.4:} I think \textquotedblleft left regular
band\textquotedblright\ should be replaced by \textquotedblleft finite left
regular band\textquotedblright\ in the first sentence; or at least my proof of
the existence of $L$ and $\operatorname*{supp}$ only works in the case when
$S$ is finite.\footnote{Here is a rough sketch of my proof (but I expect it is
the same as yours):
\par
We define a binary relation $\equiv$ on the set $S$ by setting%
\[
\left(  x\equiv y\right)  \Longleftrightarrow\left(  xy=x\text{ and
}yx=y\right)
\]
for any $x\in S$ and $y\in S$. It is easy to see that this relation $\equiv$
is transitive, reflexive and symmetric (the transitivity follows from arguing
that if $x\equiv y$ and $y\equiv z$, then $\underbrace{x}_{=xy}%
z=x\underbrace{yz}_{=y}=xy=x$ and $\underbrace{z}_{=zy}x=z\underbrace{yx}%
_{=y}=zy=z$). Thus, $\equiv$ is an equivalence relation. Also, if $x$, $y$,
$x^{\prime}$ and $y^{\prime}$ are four elements of $S$ satisfying $x\equiv y$
and $x^{\prime}\equiv y^{\prime}$, then $xx^{\prime}\equiv yy^{\prime}$
(because $\underbrace{xy}_{=xyx}x^{\prime}y^{\prime}=xy\underbrace{xx^{\prime
}}_{=x}y^{\prime}=\underbrace{xyx}_{=xy}y^{\prime}=x\underbrace{yy^{\prime}%
}_{=y}=xy$ and similarly $x^{\prime}y^{\prime}xy\equiv x^{\prime}y^{\prime}$).
In other words, the relation $\equiv$ respects the multiplication of $S$.
Hence, the quotient set $S/\equiv$ (which is well-defined since $\equiv$ is an
equivalence relation) is a semigroup. We denote this semigroup by $L$, and we
denote the canonical projection $S\rightarrow L$ by $\operatorname*{supp}$.
The semigroup $L$ is clearly a left regular band and therefore, in particular,
idempotent. Moreover, any $x\in S$ and $y\in S$ satisfy $xy\equiv yx$ (because
$x\underbrace{yy}_{=y}x=xyx=xy$ and similarly $yxxy=yx$). Thus, the semigroup
$L=S/\equiv$ is commutative. We regard this semigroup $L$ as a
join-semilattice, with join operation $\vee$ defined to be a product. (This is
a well-defined semilattice because $S$ is a commutative idempotent semigroup.)
Then, $L$ is a lattice (since every finite join-semilattice is a lattice).
Moreover, it is clear that $\operatorname*{supp}\left(  xy\right)
=\operatorname*{supp}x\vee\operatorname*{supp}y$ for all $x\in S$ and $y\in S$
(since $\operatorname*{supp}$ is the canonical projection $S\rightarrow L$,
and since $\vee$ is the multiplication in $L$). Finally, for all $x,y\in S$,
we have $xy=x$ if and only if $\operatorname*{supp}\left(  y\right)
\leq\operatorname*{supp}\left(  x\right)  $ (this is easy to check).}

\item \textbf{Page 24, proof of Corollary 3.5:} The letter \textquotedblleft%
$x$\textquotedblright\ is used in two different meanings here: Up until
\textquotedblleft$\operatorname*{supp}\left(  yx\right)  =\operatorname*{supp}%
\left(  y\right)  \vee\operatorname*{supp}\left(  x\right)  =X$%
\textquotedblright, it means the element of support $X$ that was chosen while
defining $e_{X}$; but from \textquotedblleft the face $y$\textquotedblright%
\ on, it means an arbitrary element of $\mathcal{F}$.

\item \textbf{Page 24, proof of Corollary 3.5:} You write: \textquotedblleft%
$xe_{X}$ lifts the primitive idempotent $E_{X}$\textquotedblright. It might be
helpful to justify why this is the case. (In fact, it is because
\begin{align*}
\operatorname*{supp}\left(  xe_{X}\right)   &
=\underbrace{\operatorname*{supp}x}_{=X}\vee\underbrace{\operatorname*{supp}%
\left(  e_{X}\right)  }_{=E_{X}}=X\vee E_{X}=X\vee\sum_{Y\geq X}\mu\left(
X,Y\right)  Y\\
&  =\sum_{Y\geq X}\mu\left(  X,Y\right)  \underbrace{X\vee Y}%
_{\substack{=Y\\\text{(since }Y\geq X\text{)}}}=\sum_{Y\geq X}\mu\left(
X,Y\right)  Y=E_{X}.
\end{align*}
)

\item \textbf{Page 24, proof of Corollary 3.5:} Replace \textquotedblleft
Corollary 3.2\textquotedblright\ by \textquotedblleft Theorem
3.2\textquotedblright.

\item \textbf{Page 24, \S 3.2:} At some point here you should require that the
field $k$ have characteristic $0$.

\item \textbf{Page 25:} Once you have defined the new family $\left(
e_{X}\right)  _{X\in\mathcal{L}}$ of idempotents (the one that relies on the
normalized sums $\widehat{X}$), it would be good to show an example. Here is
one possible example: For the braid arrangement for $n=3$, we have%
\begin{align*}
e_{\left\{  1,2,3\right\}  }  &  =\widehat{\left\{  1,2,3\right\}  }\\
&  =\dfrac{1}{6}\left(  \left(  1,2,3\right)  +\left(  1,3,2\right)  +\left(
2,1,3\right)  +\left(  2,3,1\right)  +\left(  3,1,2\right)  +\left(
3,2,1\right)  \right)  ;\\
e_{\left\{  12,3\right\}  }  &  =\dfrac{1}{2}\left(  \left(  12,3\right)
+\left(  3,12\right)  \right)  -\dfrac{1}{2}\left(  \left(  12,3\right)
+\left(  3,12\right)  \right)  e_{\left\{  1,2,3\right\}  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widehat{\left\{  12,3\right\}
}=\dfrac{1}{2}\left(  \left(  12,3\right)  +\left(  3,12\right)  \right)
\right) \\
&  =\dfrac{1}{2}\left(  \left(  12,3\right)  +\left(  3,12\right)  -\dfrac
{1}{2}\left(  \left(  1,2,3\right)  +\left(  2,1,3\right)  +\left(
3,1,2\right)  +\left(  3,2,1\right)  \right)  \right)  ;\\
e_{\left\{  13,2\right\}  }  &  =\dfrac{1}{2}\left(  \left(  13,2\right)
+\left(  2,13\right)  -\dfrac{1}{2}\left(  \left(  1,3,2\right)  +\left(
3,1,2\right)  +\left(  2,1,3\right)  +\left(  2,3,1\right)  \right)  \right)
;\\
e_{\left\{  1,23\right\}  }  &  =\dfrac{1}{2}\left(  \left(  1,23\right)
+\left(  23,1\right)  -\dfrac{1}{2}\left(  \left(  1,2,3\right)  +\left(
1,3,2\right)  +\left(  2,3,1\right)  +\left(  3,2,1\right)  \right)  \right)
;\\
e_{\left\{  123\right\}  }  &  =\left(  123\right)  -e_{\left\{  12,3\right\}
}-e_{\left\{  13,2\right\}  }-e_{\left\{  1,23\right\}  }-e_{\left\{
1,2,3\right\}  }\\
&  =\left(  123\right)  -\dfrac{1}{2}\left(  \left(  12,3\right)  +\left(
3,12\right)  +\left(  13,2\right)  +\left(  2,13\right)  +\left(  1,23\right)
+\left(  23,1\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\dfrac{1}{3}\left(  \left(  1,2,3\right)  +\left(
1,3,2\right)  +\left(  2,1,3\right)  +\left(  2,3,1\right)  +\left(
3,1,2\right)  +\left(  3,2,1\right)  \right)  .
\end{align*}


\item \textbf{Page 25:} You say that \textquotedblleft it is not difficult to
show that these are also primitive\textquotedblright. I didn't find it too
easy either; let me sketch my argument:

The surjective $k$-algebra homomorphism $\operatorname*{supp}:k\mathcal{F}%
\rightarrow k\mathcal{L}$ restricts to a $k$-algebra homomorphism
$\operatorname*{supp}:\left(  k\mathcal{F}\right)  ^{W}\rightarrow\left(
k\mathcal{L}\right)  ^{W}$ (which is also surjective, even though we will not
use this). With respect to this latter homomorphism, the idempotent
$\varepsilon_{X}=\sum_{Y\in\left[  X\right]  }e_{Y}$ of $\left(
k\mathcal{F}\right)  ^{W}$ lifts the idempotent $\sum_{Y\in\left[  X\right]
}E_{Y}$ of $\left(  k\mathcal{L}\right)  ^{W}$ (since each $e_{Y}$ lifts the
corresponding $E_{Y}$ with respect to $\operatorname*{supp}:k\mathcal{F}%
\rightarrow k\mathcal{L}$). Hence, in order to prove that the idempotent
$\varepsilon_{X}$ of $\left(  k\mathcal{F}\right)  ^{W}$ is primitive, it is
enough to show that the idempotent $\sum_{Y\in\left[  X\right]  }E_{Y}$ of
$\left(  k\mathcal{L}\right)  ^{W}$ is primitive (since, at the end of the
proof of Theorem 3.2, we argued that a lift of a primitive idempotent with
respect to a $k$-algebra homomorphism with nilpotent kernel must always be
primitive). But the latter can be proven similarly to how we showed the
primitivity of the $E_{Y}$: The elements $\sum_{Y\in\left[  X\right]  }E_{Y}$
of $\left(  k\mathcal{L}\right)  ^{W}$ of $k\mathcal{L}$ (with $\left[
X\right]  $ ranging over all the $W$-orbits on $\mathcal{L}$, with each orbit
only counting once) are orthogonal idempotents and sum to $1$ (because they
are the images of the elements $\varepsilon_{X}$ under $\operatorname*{supp}$,
and we know that this holds for the elements $\varepsilon_{X}$). But their
number is the number of all $W$-orbits on $\mathcal{L}$; this is the same as
the dimension of $\left(  k\mathcal{L}\right)  ^{W}$. Hence, these elements
$\sum_{Y\in\left[  X\right]  }E_{Y}$ form a basis of the $k$-module $\left(
k\mathcal{L}\right)  ^{W}$, and each $W$-orbit $\left[  X\right]  $ on
$\mathcal{L}$ satisfies $\left(  k\mathcal{L}\right)  ^{W}\left(  \sum
_{Y\in\left[  X\right]  }E_{Y}\right)  =\operatorname*{span}\nolimits_{k}%
\left(  \sum_{Y\in\left[  X\right]  }E_{Y}\right)  \cong k$, which is an
indecomposable $\left(  k\mathcal{L}\right)  ^{W}$-module. Hence, each of the
idempotents $\sum_{Y\in\left[  X\right]  }E_{Y}$ is primitive, and this
completes our proof.

\item \textbf{Page 26:} Replace \textquotedblleft elements $\varepsilon_{X}$
sum to $1$\textquotedblright\ by \textquotedblleft elements $\varepsilon_{X}$
(with $\left[  X\right]  $ ranging over all the $W$-orbits on $\mathcal{L}$,
with each orbit only counting once)\textquotedblright.

\item \textbf{Page 26, Corollary 3.8:} This is not quite correct: The elements
mentioned here are not idempotent, but rather quasi-idempotent\footnote{For
example, for $x=1=\widehat{0}$, the element $\sum_{w\in W}w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  $ equals $\left\vert
W\right\vert =n!$.}. Also, in order for them to form a basis of $\left(
k\mathcal{F}\right)  ^{W}$, you need to remove repetitions (that is, not the
whole family $\left(  \sum_{w\in W}w\left(  xe_{\operatorname*{supp}\left(
x\right)  }\right)  \right)  _{x\in\mathcal{F}}$, but only its subfamily
$\left(  \sum_{w\in W}w\left(  xe_{\operatorname*{supp}\left(  x\right)
}\right)  \right)  _{x\in T}$ forms a basis of $\left(  k\mathcal{F}\right)
^{W}$, where $T$ is any system of distinct representatives for the $W$-orbits
on $\mathcal{F}$)\ \ \ \ \footnote{Indeed, the element $\sum_{w\in W}w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  $ only depends on the
$W$-orbit on $x$, not on $x$ itself, and so should only be picked once per
$W$-orbit.}.

Let me sketch a proof of the quasi-idempotency of the elements described in
Corollary 3.8:

\textbf{Corollary 3.8a.} \textbf{(a)} We have $\varepsilon_{X}\in\left(
k\mathcal{F}\right)  ^{W}$ for every $X\in\mathcal{L}$.

\textbf{(b)} We have $\sum_{w\in W}w\left(  xe_{\operatorname*{supp}\left(
x\right)  }\right)  =\left(  \sum_{w\in W}w\left(  x\right)  \right)
\varepsilon_{\left[  \operatorname*{supp}\left(  x\right)  \right]  }$ for
every $x\in\mathcal{F}$.

\textbf{(c)} For every $X\in\mathcal{L}$, let $\operatorname*{Stab}%
\nolimits_{W}X$ denote the stabilizer of $X$ with respect to the $W$-action on
$\mathcal{L}$. (This is the subset $\left\{  w\in W\ \mid\ w\left(  X\right)
=X\right\}  $ of $W$.) Let $x\in\mathcal{F}$. Let $\mathfrak{E}_{x}$ denote
the element $\sum_{w\in W}w\left(  xe_{\operatorname*{supp}\left(  x\right)
}\right)  $ of $k\mathcal{F}$. Then,%
\begin{equation}
\mathfrak{E}_{x}xe_{\operatorname*{supp}\left(  x\right)  }=\sum
_{w\in\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}\left(
x\right)  \right)  }w\left(  xe_{\operatorname*{supp}\left(  x\right)
}\right)  \label{eq.cor.3.8a.c.1}%
\end{equation}
and
\begin{equation}
\mathfrak{E}_{x}^{2}=\left\vert \operatorname*{Stab}\nolimits_{W}\left(
\operatorname*{supp}\left(  x\right)  \right)  \right\vert \mathfrak{E}_{x}.
\label{eq.cor.3.8a.c.2}%
\end{equation}


Before we prove Corollary 3.8a, let us first show three very simple lemmas:

\textbf{Lemma 3.8b.} Let $w\in\mathcal{F}$, $p\in k\mathcal{F}$ and
$X\in\mathcal{L}$ be such that $\operatorname*{supp}\left(  w\right)
\not \leq X$. Then, $wpe_{X}=0$.

\textit{Proof of Lemma 3.8b.} The equality that we want to prove (namely,
$wpe_{X}=0$) is $k$-linear in $p$. Thus, we can WLOG assume that
$p\in\mathcal{F}$ (since $\mathcal{F}$ is a basis of the $k$-vector space
$k\mathcal{F}$). Assume this. Exercise 2 (6) (applied to $w$ and $p$ instead
of $x$ and $y$) yields that $\operatorname*{supp}\left(  wp\right)
=\operatorname*{supp}\left(  w\right)  \vee\operatorname*{supp}\left(
p\right)  \geq\operatorname*{supp}\left(  w\right)  $, so that
$\operatorname*{supp}\left(  w\right)  \leq\operatorname*{supp}\left(
wp\right)  $. Hence, if we had $\operatorname*{supp}\left(  wp\right)  \leq
X$, then we would have $\operatorname*{supp}\left(  w\right)  \leq
\operatorname*{supp}\left(  wp\right)  \leq X$, which would contradict
$\operatorname*{supp}\left(  w\right)  \not \leq X$. Thus, we cannot have
$\operatorname*{supp}\left(  wp\right)  \leq X$. In other words, we have
$\operatorname*{supp}\left(  wp\right)  \not \leq X$. Hence, Lemma 3.1
(applied to $wp$ instead of $w$) yields $wpe_{X}=0$. This proves Lemma 3.8b.

\textbf{Lemma 3.8c.} Let $X$ and $Y$ be two distinct elements of $\mathcal{L}$
which belong to one and the same $W$-orbit. Then, $Y\not \leq X$.

\textit{Proof of Lemma 3.8c.} Assume the contrary. Thus, $Y\leq X$. In other
words, the set partition $Y$ is obtained from $X$ by repeated merging of
blocks. Therefore, either the number of blocks of $Y$ is smaller than the
number of blocks of $X$, or we have $Y=X$. Since we cannot have $Y=X$ (because
$X$ and $Y$ are distinct), this shows that the number of blocks of $Y$ is
smaller than the number of blocks of $X$.

But $X$ and $Y$ belong to one and the same $W$-orbit. Thus, $Y=w\left(
X\right)  $ for some $w\in W$. Consequently, the blocks of $Y$ are obtained by
applying $w$ (pointwise) to the blocks of $X$. Hence, the number of blocks of
$Y$ equals the number of blocks of $X$. This contradicts the fact that the
number of blocks of $Y$ is smaller than the number of blocks of $X$. This
contradiction shows that our assumption was wrong. Lemma 3.8c is thus proven.

\textbf{Lemma 3.8d.} Let $w\in\mathcal{F}$, $p\in k\mathcal{F}$ and
$t\in\mathcal{F}$ be such that $\operatorname*{supp}\left(  t\right)
\leq\operatorname*{supp}\left(  w\right)  $. Then, $wpt=wp$.

\textit{Proof of Lemma 3.8d.} The equality that we want to prove (namely,
$wpt=wp$) is $k$-linear in $p$. Thus, we can WLOG assume that $p\in
\mathcal{F}$ (since $\mathcal{F}$ is a basis of the $k$-vector space
$k\mathcal{F}$). Assume this. Exercise 2 (6) (applied to $w$ and $p$ instead
of $x$ and $y$) yields that $\operatorname*{supp}\left(  wp\right)
=\operatorname*{supp}\left(  w\right)  \vee\operatorname*{supp}\left(
p\right)  \geq\operatorname*{supp}\left(  w\right)  $, so that
$\operatorname*{supp}\left(  w\right)  \leq\operatorname*{supp}\left(
wp\right)  $. Thus, $\operatorname*{supp}\left(  t\right)  \leq
\operatorname*{supp}\left(  w\right)  \leq\operatorname*{supp}\left(
wp\right)  $. But Exercise 2 (5) (applied to $wp$ and $t$ instead of $x$ and
$y$) yields that $wpt=wp$ if and only if $\operatorname*{supp}\left(
t\right)  \leq\operatorname*{supp}\left(  wp\right)  $. Hence, $wpt=wp$ (since
$\operatorname*{supp}\left(  t\right)  \leq\operatorname*{supp}\left(
wp\right)  $). This proves Lemma 3.8d.

\textit{Proof of Corollary 3.8a.} \textbf{(a)} Let $X\in\mathcal{L}$. Let
$w\in W$. The set $\left[  X\right]  $ is a $W$-orbit. Hence, the action of
$w$ on $\mathcal{L}$ restricts to a permutation of this set $\left[  X\right]
$. Consequently, we can substitute $w\left(  Y\right)  $ for $Y$ in the sum
$\sum_{Y\in\left[  X\right]  }e_{Y}$. As a result, we obtain%
\[
\sum_{Y\in\left[  X\right]  }e_{Y}=\sum_{Y\in\left[  X\right]  }%
\underbrace{e_{w\left(  Y\right)  }}_{\substack{=w\left(  e_{Y}\right)
\\\text{(since Lemma 3.6}\\\text{(applied to }Y\text{ instead of}\\X\text{)
yields }w\left(  e_{Y}\right)  =e_{w\left(  Y\right)  }\text{)}}}=\sum
_{Y\in\left[  X\right]  }w\left(  e_{Y}\right)  =w\left(  \sum_{Y\in\left[
X\right]  }e_{Y}\right)  .
\]
In other words, $\varepsilon_{\left[  X\right]  }=w\left(  \varepsilon
_{\left[  X\right]  }\right)  $ (since $\varepsilon_{\left[  X\right]  }%
=\sum_{Y\in\left[  X\right]  }e_{Y}$).

Let us now forget that we fixed $w$. We thus have shown that $\varepsilon
_{\left[  X\right]  }=w\left(  \varepsilon_{\left[  X\right]  }\right)  $ for
every $w\in W$. In other words, $\varepsilon_{\left[  X\right]  }\in\left(
k\mathcal{F}\right)  ^{W}$. This proves Corollary 3.8a \textbf{(a)}.

\textbf{(b)} We have $x\varepsilon_{\left[  \operatorname*{supp}\left(
x\right)  \right]  }=xe_{\operatorname*{supp}\left(  x\right)  }%
$\ \ \ \ \footnote{\textit{Proof.} Let $Y\in\left[  \operatorname*{supp}%
\left(  x\right)  \right]  $ be such that $Y\neq\operatorname*{supp}\left(
x\right)  $. Then, the elements $Y$ and $\operatorname*{supp}\left(  x\right)
$ of $\mathcal{L}$ are distinct (since $Y\neq\operatorname*{supp}\left(
x\right)  $) and belong to one and the same $W$-orbit (namely, to the
$W$-orbit $\left[  \operatorname*{supp}\left(  x\right)  \right]  $). Hence,
$\operatorname*{supp}\left(  x\right)  \not \leq Y$ (by Lemma 3.8c, applied to
$Y$ and $\operatorname*{supp}\left(  x\right)  $ instead of $X$ and $Y$).
Therefore, $xe_{Y}=0$ (by Lemma 3.1, applied to $x$ and $Y$ instead of $w$ and
$X$).
\par
Let us now forget that we fixed $Y$. We thus have shown that every
$Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]  $ satisfying
$Y\neq\operatorname*{supp}\left(  x\right)  $ satisfies $xe_{Y}=0$. Thus,
$\sum_{\substack{Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]
;\\Y\neq\operatorname*{supp}\left(  x\right)  }}\underbrace{xe_{Y}}_{=0}%
=\sum_{\substack{Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]
;\\Y\neq\operatorname*{supp}\left(  x\right)  }}0=0$.
\par
Now, the definition of $\varepsilon_{\left[  \operatorname*{supp}\left(
x\right)  \right]  }$ yields $\varepsilon_{\left[  \operatorname*{supp}\left(
x\right)  \right]  }=\sum_{Y\in\left[  \operatorname*{supp}\left(  x\right)
\right]  }e_{Y}$. Hence,%
\begin{align*}
x\underbrace{\varepsilon_{\left[  \operatorname*{supp}\left(  x\right)
\right]  }}_{=\sum_{Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]
}e_{Y}}  &  =x\sum_{Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]
}e_{Y}=\sum_{Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]
}xe_{Y}\\
&  =\sum_{\substack{Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]
;\\Y=\operatorname*{supp}\left(  x\right)  }}x\underbrace{e_{Y}}%
_{\substack{=e_{\operatorname*{supp}\left(  x\right)  }\\\text{(since
}Y=\operatorname*{supp}\left(  x\right)  \text{)}}}+\sum_{\substack{Y\in
\left[  \operatorname*{supp}\left(  x\right)  \right]  ;\\Y\neq
\operatorname*{supp}\left(  x\right)  }}\underbrace{xe_{Y}}_{=0}%
=\sum_{\substack{Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]
;\\Y=\operatorname*{supp}\left(  x\right)  }}xe_{\operatorname*{supp}\left(
x\right)  }+\underbrace{\sum_{\substack{Y\in\left[  \operatorname*{supp}%
\left(  x\right)  \right]  ;\\Y\neq\operatorname*{supp}\left(  x\right)  }%
}0}_{=0}\\
&  =\sum_{\substack{Y\in\left[  \operatorname*{supp}\left(  x\right)  \right]
;\\Y=\operatorname*{supp}\left(  x\right)  }}xe_{\operatorname*{supp}\left(
x\right)  }=xe_{\operatorname*{supp}\left(  x\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{supp}\left(  x\right)
\in\left[  \operatorname*{supp}\left(  x\right)  \right]  \right)  ,
\end{align*}
qed.}. Hence, every $w\in W$ satisfies $w\left(  x\right)  \varepsilon
_{\left[  \operatorname*{supp}\left(  x\right)  \right]  }=w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $w\in W$. Then, $w\left(  \varepsilon
_{\left[  \operatorname*{supp}\left(  x\right)  \right]  }\right)
=\varepsilon_{\left[  \operatorname*{supp}\left(  x\right)  \right]  }$ (since
$\varepsilon_{\left[  \operatorname*{supp}\left(  x\right)  \right]  }%
\in\left(  k\mathcal{F}\right)  ^{W}$ (by Corollary 3.8a \textbf{(a)}, applied
to $X=\operatorname*{supp}\left(  x\right)  $)). But $w\in W$ acts on
$k\mathcal{F}$ by an algebra homomorphism, and therefore we have%
\[
w\left(  x\varepsilon_{\left[  \operatorname*{supp}\left(  x\right)  \right]
}\right)  =w\left(  x\right)  \underbrace{w\left(  \varepsilon_{\left[
\operatorname*{supp}\left(  x\right)  \right]  }\right)  }_{=\varepsilon
_{\left[  \operatorname*{supp}\left(  x\right)  \right]  }}=w\left(  x\right)
\varepsilon_{\left[  \operatorname*{supp}\left(  x\right)  \right]  }.
\]
Hence, $w\left(  x\right)  \varepsilon_{\left[  \operatorname*{supp}\left(
x\right)  \right]  }=w\left(  \underbrace{x\varepsilon_{\left[
\operatorname*{supp}\left(  x\right)  \right]  }}_{=xe_{\operatorname*{supp}%
\left(  x\right)  }}\right)  =w\left(  xe_{\operatorname*{supp}\left(
x\right)  }\right)  $, qed.}. Now,%
\[
\left(  \sum_{w\in W}w\left(  x\right)  \right)  \varepsilon_{\left[
\operatorname*{supp}\left(  x\right)  \right]  }=\sum_{w\in W}%
\underbrace{w\left(  x\right)  \varepsilon_{\left[  \operatorname*{supp}%
\left(  x\right)  \right]  }}_{=w\left(  xe_{\operatorname*{supp}\left(
x\right)  }\right)  }=\sum_{w\in W}w\left(  xe_{\operatorname*{supp}\left(
x\right)  }\right)  .
\]
This proves Corollary 3.8a \textbf{(b)}.

\textbf{(c)} Let us first fix some $w\in W$ such that $w\left(
\operatorname*{supp}\left(  x\right)  \right)  =\operatorname*{supp}\left(
x\right)  $. Now, $\operatorname*{supp}\left(  wx\right)  =w\left(
\operatorname*{supp}\left(  x\right)  \right)  $. Hence, $\operatorname*{supp}%
\left(  wx\right)  =w\left(  \operatorname*{supp}\left(  x\right)  \right)
=\operatorname*{supp}\left(  x\right)  $, so that $\operatorname*{supp}\left(
x\right)  =\operatorname*{supp}\left(  wx\right)  \leq\operatorname*{supp}%
\left(  wx\right)  $. Therefore, Lemma 3.8d (applied to $wx$,
$we_{\operatorname*{supp}\left(  x\right)  }$ and $x$ instead of $w$, $p$ and
$t$) yields $\left(  wx\right)  \left(  we_{\operatorname*{supp}\left(
x\right)  }\right)  x=\left(  wx\right)  \left(  we_{\operatorname*{supp}%
\left(  x\right)  }\right)  $.

But Lemma 3.6 (applied to $X=\operatorname*{supp}\left(  x\right)  $) yields
$we_{\operatorname*{supp}\left(  x\right)  }=e_{w\left(  \operatorname*{supp}%
\left(  x\right)  \right)  }=e_{\operatorname*{supp}\left(  x\right)  }$
(since $w\left(  \operatorname*{supp}\left(  x\right)  \right)
=\operatorname*{supp}\left(  x\right)  $). Thus,%
\[
\left(  wx\right)  \left(  we_{\operatorname*{supp}\left(  x\right)  }\right)
x=\left(  wx\right)  \underbrace{\left(  we_{\operatorname*{supp}\left(
x\right)  }\right)  }_{=e_{\operatorname*{supp}\left(  x\right)  }}=\left(
wx\right)  e_{\operatorname*{supp}\left(  x\right)  }.
\]
Since $\left(  wx\right)  \left(  we_{\operatorname*{supp}\left(  x\right)
}\right)  =w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)  $
(because $w\in W$ acts on $k\mathcal{F}$ by an algebra homomorphism), this
rewrites as $w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
x=\left(  wx\right)  e_{\operatorname*{supp}\left(  x\right)  }$. Thus,%
\begin{align*}
\underbrace{w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
x}_{=\left(  wx\right)  e_{\operatorname*{supp}\left(  x\right)  }%
}e_{\operatorname*{supp}\left(  x\right)  }  &  =\left(  wx\right)
\underbrace{e_{\operatorname*{supp}\left(  x\right)  }e_{\operatorname*{supp}%
\left(  x\right)  }}_{\substack{=e_{\operatorname*{supp}\left(  x\right)
}\\\text{(since }e_{\operatorname*{supp}\left(  x\right)  }\text{ is an
idempotent)}}}=wx\underbrace{e_{\operatorname*{supp}\left(  x\right)  }%
}_{\substack{=we_{\operatorname*{supp}\left(  x\right)  }\\\text{(since
}we_{\operatorname*{supp}\left(  x\right)  }=e_{\operatorname*{supp}\left(
x\right)  }\text{)}}}\\
&  =\left(  wx\right)  \left(  we_{\operatorname*{supp}\left(  x\right)
}\right)  =w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)  .
\end{align*}


Let us now forget that we fixed $w$. We thus have proven that%
\begin{equation}
w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
xe_{\operatorname*{supp}\left(  x\right)  }=w\left(  xe_{\operatorname*{supp}%
\left(  x\right)  }\right)  \label{pf.cor.3.8a.c.1.1}%
\end{equation}
for every $w\in W$ such that $w\left(  \operatorname*{supp}\left(  x\right)
\right)  =\operatorname*{supp}\left(  x\right)  $.

On the other hand, let us fix some $w\in W$ such that $w\left(
\operatorname*{supp}\left(  x\right)  \right)  \neq\operatorname*{supp}\left(
x\right)  $. Thus, $\operatorname*{supp}\left(  x\right)  $ and $w\left(
\operatorname*{supp}\left(  x\right)  \right)  $ are two distinct elements of
$\mathcal{L}$ which belong to one and the same $W$-orbit. Thus, Lemma 3.8c
(applied to $X=\operatorname*{supp}\left(  x\right)  $ and $Y=w\left(
\operatorname*{supp}\left(  x\right)  \right)  $) yields $w\left(
\operatorname*{supp}\left(  x\right)  \right)  \not \leq \operatorname*{supp}%
\left(  x\right)  $. Thus, $\operatorname*{supp}\left(  wx\right)  =w\left(
\operatorname*{supp}\left(  x\right)  \right)  \not \leq \operatorname*{supp}%
\left(  x\right)  $. Hence, Lemma 3.8b (applied to $wx$, $\left(
we_{\operatorname*{supp}\left(  x\right)  }\right)  x$ and
$\operatorname*{supp}\left(  x\right)  $ instead of $w$, $p$ and $X$) yields
$\left(  wx\right)  \left(  we_{\operatorname*{supp}\left(  x\right)
}\right)  xe_{\operatorname*{supp}\left(  x\right)  }=0$. Since $\left(
wx\right)  \left(  we_{\operatorname*{supp}\left(  x\right)  }\right)
=w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)  $ (because $w\in
W$ acts on $k\mathcal{F}$ by an algebra homomorphism), this rewrites as
$w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
xe_{\operatorname*{supp}\left(  x\right)  }=0$.

Let us now forget that we fixed $w$. We thus have proven that%
\begin{equation}
w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
xe_{\operatorname*{supp}\left(  x\right)  }=0 \label{pf.cor.3.8a.c.1.2}%
\end{equation}
for every $w\in W$ such that $w\left(  \operatorname*{supp}\left(  x\right)
\right)  \neq\operatorname*{supp}\left(  x\right)  $.

But recall that $\mathfrak{E}_{x}=\sum_{w\in W}w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  $. Multiplying both sides
of this equality with $xe_{\operatorname*{supp}\left(  x\right)  }$ from the
right, we obtain%
\begin{align*}
\mathfrak{E}_{x}xe_{\operatorname*{supp}\left(  x\right)  }  &  =\left(
\sum_{w\in W}w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
\right)  xe_{\operatorname*{supp}\left(  x\right)  }=\sum_{w\in W}w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  xe_{\operatorname*{supp}%
\left(  x\right)  }\\
&  =\underbrace{\sum_{\substack{w\in W;\\w\left(  \operatorname*{supp}\left(
x\right)  \right)  =\operatorname*{supp}\left(  x\right)  }}}_{\substack{=\sum
_{w\in\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}\left(
x\right)  \right)  }\\\text{(since }\operatorname*{Stab}\nolimits_{W}\left(
\operatorname*{supp}\left(  x\right)  \right)  \\=\left\{  w\in W\ \mid
\ w\left(  \operatorname*{supp}\left(  x\right)  \right)
=\operatorname*{supp}\left(  x\right)  \right\}  \\\text{(by the definition of
}\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}\left(
x\right)  \right)  \text{))}}}\underbrace{w\left(  xe_{\operatorname*{supp}%
\left(  x\right)  }\right)  xe_{\operatorname*{supp}\left(  x\right)  }%
}_{\substack{=w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
\\\text{(by (\ref{pf.cor.3.8a.c.1.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{w\in W;\\w\left(  \operatorname*{supp}%
\left(  x\right)  \right)  \neq\operatorname*{supp}\left(  x\right)
}}\underbrace{w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
xe_{\operatorname*{supp}\left(  x\right)  }}_{\substack{=0\\\text{(by
(\ref{pf.cor.3.8a.c.1.2}))}}}\\
&  =\sum_{w\in\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}%
\left(  x\right)  \right)  }w\left(  xe_{\operatorname*{supp}\left(  x\right)
}\right)  +\underbrace{\sum_{\substack{w\in W;\\w\left(  \operatorname*{supp}%
\left(  x\right)  \right)  \neq\operatorname*{supp}\left(  x\right)  }}0}%
_{=0}\\
&  =\sum_{w\in\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}%
\left(  x\right)  \right)  }w\left(  xe_{\operatorname*{supp}\left(  x\right)
}\right)  .
\end{align*}
This proves (\ref{eq.cor.3.8a.c.1}).

Now, recall that $\mathfrak{E}_{x}=\sum_{w\in W}w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  =\sum_{p\in W}p\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  $ (here, we renamed the
summation index $w$ as $p$). But every $p\in W$ satisfies%
\begin{align}
p\left(  \underbrace{\mathfrak{E}_{x}}_{=\sum_{w\in W}w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  }\right)   &  =p\left(
\sum_{w\in W}w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
\right)  =\sum_{w\in W}\underbrace{p\left(  w\left(  xe_{\operatorname*{supp}%
\left(  x\right)  }\right)  \right)  }_{=\left(  pw\right)  \left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  }\nonumber\\
&  =\sum_{w\in W}\left(  pw\right)  \left(  xe_{\operatorname*{supp}\left(
x\right)  }\right)  =\sum_{w\in W}w\left(  xe_{\operatorname*{supp}\left(
x\right)  }\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }w\text{ for }pw\text{ in the sum,}\\
\text{since the map }W\rightarrow W,\ w\mapsto pw\text{ is a bijection}\\
\text{(because }W\text{ is a group)}%
\end{array}
\right) \nonumber\\
&  =\mathfrak{E}_{x}. \label{pf.cor.3.8a.c.2.2}%
\end{align}
Now,%
\begin{align*}
\mathfrak{E}_{x}^{2}  &  =\mathfrak{E}_{x}\underbrace{\mathfrak{E}_{x}}%
_{=\sum_{p\in W}p\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)
}=\mathfrak{E}_{x}\sum_{p\in W}p\left(  xe_{\operatorname*{supp}\left(
x\right)  }\right) \\
&  =\sum_{p\in W}\underbrace{\mathfrak{E}_{x}}_{\substack{=p\left(
\mathfrak{E}_{x}\right)  \\\text{(by (\ref{pf.cor.3.8a.c.2.2}))}}}p\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  =\sum_{p\in W}%
\underbrace{p\left(  \mathfrak{E}_{x}\right)  p\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  }_{\substack{=p\left(
\mathfrak{E}_{x}xe_{\operatorname*{supp}\left(  x\right)  }\right)
\\\text{(since }p\in W\text{ acts on }k\mathcal{F}\text{ by}\\\text{a
}k\text{-algebra homomorphism)}}}\\
&  =\sum_{p\in W}p\left(  \underbrace{\mathfrak{E}_{x}xe_{\operatorname*{supp}%
\left(  x\right)  }}_{\substack{=\sum_{w\in\operatorname*{Stab}\nolimits_{W}%
\left(  \operatorname*{supp}\left(  x\right)  \right)  }w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  \\\text{(by
(\ref{eq.cor.3.8a.c.1}))}}}\right)  =\sum_{p\in W}p\left(  \sum_{w\in
\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}\left(  x\right)
\right)  }w\left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)  \right)
\\
&  =\underbrace{\sum_{p\in W}\sum_{w\in\operatorname*{Stab}\nolimits_{W}%
\left(  \operatorname*{supp}\left(  x\right)  \right)  }}_{=\sum
_{w\in\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}\left(
x\right)  \right)  }\sum_{p\in W}}\underbrace{p\left(  w\left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  \right)  }_{=\left(
pw\right)  \left(  xe_{\operatorname*{supp}\left(  x\right)  }\right)  }\\
&  =\sum_{w\in\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}%
\left(  x\right)  \right)  }\sum_{p\in W}\left(  pw\right)  \left(
xe_{\operatorname*{supp}\left(  x\right)  }\right)  =\sum_{w\in
\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}\left(  x\right)
\right)  }\underbrace{\sum_{p\in W}p\left(  xe_{\operatorname*{supp}\left(
x\right)  }\right)  }_{=\mathfrak{E}_{x}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }p\text{ for }pw\text{ in the second sum,}\\
\text{since the map }W\rightarrow W,\ p\mapsto pw\text{ is a bijection}\\
\text{(because }W\text{ is a group)}%
\end{array}
\right) \\
&  =\sum_{w\in\operatorname*{Stab}\nolimits_{W}\left(  \operatorname*{supp}%
\left(  x\right)  \right)  }\mathfrak{E}_{x}=\left\vert \operatorname*{Stab}%
\nolimits_{W}\left(  \operatorname*{supp}\left(  x\right)  \right)
\right\vert \mathfrak{E}_{x}.
\end{align*}
This proves (\ref{eq.cor.3.8a.c.2}).

\item \textbf{Page 26, Exercise 6:} Replace \textquotedblleft an equivalence
class\textquotedblright\ by \textquotedblleft an equivalence
relation\textquotedblright.

\item \textbf{Page 26, Exercise 6:} I haven't been reading the things
pertaining to the general Coxeter-group case anywhere near carefully, but
shouldn't $x_{J}$ be $w_{J}$ to match your notations in \S 2.4?

\item \textbf{Page 26, Exercise 6:} Replace \textquotedblleft as
recursively\textquotedblright\ by \textquotedblleft
recursively\textquotedblright.
\end{itemize}


\end{document}