\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}% \usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage} \usepackage[all,cmtip]{xy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{framed} \usepackage{amsmath} \usepackage{comment} \usepackage{bbding} \usepackage{color} \usepackage{hyperref} \usepackage[sc]{mathpazo} \usepackage[T1]{fontenc} \usepackage{amsthm} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=5.50.0.2960} %TCIDATA{LastRevised=Wednesday, March 25, 2015 00:37:28} %TCIDATA{SuppressPackageManagement} %TCIDATA{} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %BeginMSIPreambleData \providecommand{\U}{\protect\rule{.1in}{.1in}} %EndMSIPreambleData \theoremstyle{definition} \newtheorem{theo}{Theorem}[section] \newenvironment{theorem}[] {\begin{theo}[#1]\begin{leftbar}} {\end{leftbar}\end{theo}} \newtheorem{lem}[theo]{Lemma} \newenvironment{lemma}[] {\begin{lem}[#1]\begin{leftbar}} {\end{leftbar}\end{lem}} \newtheorem{prop}[theo]{Proposition} \newenvironment{proposition}[] {\begin{prop}[#1]\begin{leftbar}} {\end{leftbar}\end{prop}} \newtheorem{defi}[theo]{Definition} \newenvironment{definition}[] {\begin{defi}[#1]\begin{leftbar}} {\end{leftbar}\end{defi}} \newtheorem{remk}[theo]{Remark} \newenvironment{remark}[] {\begin{remk}[#1]\begin{leftbar}} {\end{leftbar}\end{remk}} \newtheorem{coro}[theo]{Corollary} \newenvironment{corollary}[] {\begin{coro}[#1]\begin{leftbar}} {\end{leftbar}\end{coro}} \newtheorem{conv}[theo]{Convention} \newenvironment{condition}[] {\begin{conv}[#1]\begin{leftbar}} {\end{leftbar}\end{conv}} \newtheorem{quest}[theo]{Question} \newenvironment{algorithm}[] {\begin{quest}[#1]\begin{leftbar}} {\end{leftbar}\end{quest}} \newtheorem{warn}[theo]{Warning} \newenvironment{conclusion}[] {\begin{warn}[#1]\begin{leftbar}} {\end{leftbar}\end{warn}} \newtheorem{conj}[theo]{Conjecture} \newenvironment{conjecture}[] {\begin{conj}[#1]\begin{leftbar}} {\end{leftbar}\end{conj}} \newtheorem{exmp}[theo]{Example} \newenvironment{example}[] {\begin{exmp}[#1]\begin{leftbar}} {\end{leftbar}\end{exmp}} \newenvironment{proof2}[] {\begin{proof}[#1]}{\end{proof}} \newenvironment{verlong}{}{} \newenvironment{vershort}{}{} \newenvironment{noncompile}{}{} \excludecomment{verlong} \includecomment{vershort} \excludecomment{noncompile} \newcommand{\kk}{\mathbf{k}} \newcommand{\id}{\operatorname{id}} \newcommand{\ev}{\operatorname{ev}} \newcommand{\Comp}{\operatorname{Comp}} \newcommand{\bk}{\mathbf{k}} \newcommand{\Nplus}{\mathbb{N}_{+}} \newcommand{\NN}{\mathbb{N}} \newcommand{\scissors}{\text{\ScissorHollowRight}} \let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \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$C\left( j\right) C\left( j\right)$\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 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 $i_{1}X\vee W}ze_{Y}$, 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 $X_{0}\operatorname*{corank}Y. \label{pf.thm.3.1a.triv}% \end{equation} For every$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\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{