\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{color} \usepackage{hyperref} \usepackage[sc]{mathpazo} \usepackage[T1]{fontenc} \usepackage{amsthm} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=5.50.0.2960} %TCIDATA{LastRevised=Friday, June 12, 2015 22:40:01} %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}} \let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \ihead{Errata to An Introduction to Hyperplane Arrangements''} \ohead{12 June 2015} \begin{document} \begin{center} \textbf{An Introduction to Hyperplane Arrangements} \textit{Richard P. Stanley} IAS/Park City Mathematics Series Volume 14, 2004 version of February 26, 2006 \url{http://www.cis.upenn.edu/~cis610/sp06stanley.pdf} --------------------------------------------------------------------------------------- \textbf{List of additional errata and questions - I} (version 2) Darij Grinberg, 11 June 2015 \bigskip \end{center} Page numbers refer to the page numbers at the top of the pages, not to the page count of the PDF file. \begin{itemize} \item \textbf{page 2:} You speak of the \textquotedblleft usual dot product\textquotedblright, but there is no \textquotedblleft usual dot product\textquotedblright\ on a general finite-dimensional $K$-vector space. If you want to work in this generality, you should use the dual space. (Or else just identify $V$ with $K^{n}$ for this definition, or require the choice of a nondegenerate symmetric bilinear form $V\times V\rightarrow K$ which is to serve as a dot product. I personally find it easier to invoke the dual space, because as soon as one introduces additional structures like a basis or a bilinear form, it starts clouding further definitions. Besides, you use linear forms in the next paragraph, even though in the next paragraph you actually use a basis!) Similarly, in the next paragraph, \textquotedblleft$x=\left( x_{1}% ,\ldots,x_{n}\right)$\textquotedblright\ assumes an isomorphism $V\rightarrow K^{n}$ to be given. And after that, the notion of \textquotedblleft normals\textquotedblright\ assumes a dot product. \item \textbf{pages 2-3:} You write: \textquotedblleft Let $Y$ be a complementary space in $K^{n}$ to the subspace $X$ spanned by the normals to hyperplanes in $\mathcal{A}$. Define% $W=\left\{ v\in V\ :\ v\cdot y=0\ \forall y\in Y\right\} .$ If $\operatorname*{char}\left( K\right) =0$ then we can simply take $W=X$.\textquotedblright I understand what you mean here, but it is not correctly explained. First, even if $\operatorname*{char}\left( K\right) =0$, then we cannot take $W=X$ unless $Y$ is \textit{the orthogonal complement} to $X$ (instead of just a random complementary space in $K^{n}$ to the subspace $X$); I think you should say \textquotedblleft we can simply take $Y=X^{\perp}$ and $W=X$% \textquotedblright\ instead of \textquotedblleft we can simply take $W=X$\textquotedblright\ because otherwise your wording suggests that $Y$ can be taken arbitrary here. Second, $\operatorname*{char}\left( K\right) =0$ does not guarantee that the orthogonal complement to $X$ is indeed a complementary subspace to $X$ (for example, the orthogonal complement to the subspace $\left\langle \left( 1,i\right) \right\rangle$ of $\mathbb{C}^{2}$ is \textit{not} a complementary subspace to $\left\langle \left( 1,i\right) \right\rangle$, even though $\operatorname*{char}\mathbb{C}=0$). So to make sure that we can actually take $W=X$, we need to require that $K$ is a formally real field (which is a stronger assertion than $\operatorname*{char}% \left( K\right) =0$). Once again, the whole situation simplifies if you use the dual space. Now, the \textquotedblleft normals to hyperplanes in $\mathcal{A}$\textquotedblright% \ become the \textquotedblleft linear forms defining the hyperplanes in $\mathcal{A}$\textquotedblright, and they form a subspace $\widetilde{X}$ of the dual space $V^{\ast}$. The orthogonal space of this $\widetilde{X}$ (= the joint kernel of the linear forms defining the hyperplanes in $\mathcal{A}$) is a subspace of $V$; we call it $\widetilde{Y}$. Then, the images of the hyperplanes in $\mathcal{A}$ under the projection map $V\rightarrow V/\widetilde{Y}$ are hyperplanes in $V/\widetilde{Y}$ (this is easy to prove\footnote{All that needs to be checked is that every hyperplane $H\in\mathcal{A}$ satisfies $\widetilde{Y}\subseteq H^{\prime}$, where $H^{\prime}$ is the translate of $H$ that passes through the origin. The proof is easy: From $H^{\perp}\subseteq\widetilde{X}$, we obtain $\widetilde{X}% ^{\perp}\subseteq\left( H^{\perp}\right) ^{\perp}=H$, thus $Y=\widetilde{X}% ^{\perp}\subseteq H$. Here, the $\perp$ sign is defined with reference to the canonical pairing $V^{\ast}\times V\rightarrow K$, not to any non-canonical bilinear form on $V$.} -- a lot easier than checking your equality (1)). The arrangement they form in $V/\widetilde{Y}$ is isomorphic to your $\mathcal{A}_{W}$, but defined canonically (thus eliminating the necessity of checking that your $\mathcal{A}_{W}$ is independent on the choice of $W$ up to isomorphism). \item \textbf{page 3:} When you write \textquotedblleft$H^{\prime}% \in\mathcal{A}_{W}$ if and only if $H^{\prime}\oplus W^{\perp}\in\mathcal{A}%$\textquotedblright, it would be good to point out that $H^{\prime}$ is supposed to be a subspace of $W$. I would also replace the $\oplus$ sign by a $+$ sign, since $\oplus$ has not been defined for \textit{affine} subspaces (and its standard meaning that involves the intersection being $0$ is not correct for affine subspaces). \item \textbf{page 3:} You write: \textquotedblleft in characteristic $p$ this type of reasoning fails\textquotedblright. Yes, but not only in characteristic $p$. Also for $K=\mathbb{C}$, as I explained above. \item \textbf{page 3:} You write: \textquotedblleft then $R\in\mathcal{R}% \left( \mathcal{A}\right)$ if and only if $R\cap W\in\mathcal{R}\left( \mathcal{A}_{W}\right)$\textquotedblright. This makes little sense ($R$ cannot be any subset of $\mathbb{R}^{n}$, but what should it be?). I assume you mean that the map% \begin{align*} \mathcal{R}\left( \mathcal{A}\right) & \rightarrow\mathcal{R}\left( \mathcal{A}_{W}\right) ,\\ R & \mapsto R\cap W \end{align*} is well-defined and bijective, with inverse% \begin{align*} \mathcal{R}\left( \mathcal{A}_{W}\right) & \rightarrow\mathcal{R}\left( \mathcal{A}\right) ,\\ R & \mapsto R+W^{\perp}. \end{align*} \item \textbf{page 4:} Trivial nitpick: In the definition of \textquotedblleft general position\textquotedblright, the $H_{1},\ldots,H_{p}$ should be assumed distinct in the formulas. \item \textbf{page 4, second line of Example 1.2:} \textquotedblleft$L$ line\textquotedblright\ should be \textquotedblleft line $L$\textquotedblright. \item \textbf{page 4, second line of Example 1.2:} \textquotedblleft% $\mathcal{A}_{K}$\textquotedblright\ should be \textquotedblleft% $\mathcal{A}_{k}$\textquotedblright. \item \textbf{page 8:} You define saturated chains, but you do not define maximal chains. The fact that you use the word \textquotedblleft maximal\textquotedblright\ in the next sentence (\textquotedblleft if every maximal chain of $P$ has length $n$\textquotedblright) creates the incorrect impression that \textquotedblleft maximal\textquotedblright\ is a synonym for \textquotedblleft saturated\textquotedblright. \item \textbf{page 8:} You write: \textquotedblleft If $x\widehat{0}$. In the M\"{o}bius algebra $A\left( L\right)$ (defined in \S 2), we have $\sigma_{\widehat{0}}=\underbrace{\sum_{y\geq\widehat{0}}% }_{=\sum_{y\in L}}\underbrace{\mu\left( \widehat{0},y\right) }_{=\mu\left( y\right) }y=\sum_{y\in L}\mu\left( y\right) y=\sum_{x\in L}\mu\left( x\right) x$ and thus% \begin{align*} \sum_{x\in L}\mu\left( x\right) \underbrace{x\vee a}_{=xa} & =\sum_{x\in L}\mu\left( x\right) xa=\underbrace{\left( \sum_{x\in L}\mu\left( x\right) x\right) }_{=\sigma_{\widehat{0}}}\underbrace{a}_{\substack{=\sum _{y\geq a}\sigma_{y}\\\text{(by (9))}}}=\sigma_{\widehat{0}}\left( \sum_{y\geq a}\sigma_{y}\right) \\ & =\sum_{y\geq a}\underbrace{\sigma_{\widehat{0}}\sigma_{y}}% _{\substack{=\delta_{\widehat{0},y}\sigma_{\widehat{0}}\\\text{(by the second sentence}\\\text{of Theorem 2.3)}}}=\sum_{y\geq a}\underbrace{\delta _{\widehat{0},y}}_{\substack{=0\\\text{(since }y\geq a>\widehat{0}\\\text{and thus }y\neq\widehat{0}\text{)}}}\sigma_{\widehat{0}}=0. \end{align*} Comparing coefficients before $\widehat{1}$ in this equality yields $\sum_{\substack{x\in L;\\x\vee a=\widehat{1}}}\mu\left( x\right) =0$. Theorem 3.9 is proven. \item \textbf{page 38, proof of Theorem 3.10:} It took me a while to understand why \textquotedblleft The sum on the right is nonempty\textquotedblright. The simplest proof of this that I can find is the following: Let $A$ be the set of all atoms of $M$. The map $\Psi$ defined by (\ref{p36.Psi}) is injective (since we have found an inverse to it). Thus, $\bigvee A\neq\bigvee\left( A\setminus\left\{ a\right\} \right)$. Hence, $\bigvee\left( A\setminus\left\{ a\right\} \right) <\bigvee A$. Semimodularity of $L$ easily shows that $\bigvee\left( A\setminus\left\{ a\right\} \right) \lessdot\bigvee A$. But $\bigvee A=\widehat{1}$ (which is easy to prove using atomicity of $L$: the element $\widehat{1}$ must be a join of \textit{some} set of atoms, and thus also of the set $A$ of all atoms), so this becomes $\bigvee\left( A\setminus\left\{ a\right\} \right) \lessdot 1$. But $a\not \leq \bigvee\left( A\setminus\left\{ a\right\} \right)$ (since otherwise, we would have $\bigvee\left( A\setminus\left\{ a\right\} \right) =\bigvee A$, contradicting $\bigvee A\neq\bigvee\left( A\setminus\left\{ a\right\} \right)$). Thus, there exists an $x\in L$ satisfying $a\not \leq x\lessdot1$ (namely, $x=\bigvee\left( A\setminus \left\{ a\right\} \right)$). \item \textbf{page 38, (26):} Replace \textquotedblleft$M$\textquotedblright% \ by \textquotedblleft$M_{\mathcal{A}}$\textquotedblright. \item \textbf{page 41, \S 4.1:} Throughout this section, whenever you work with $\operatorname*{BC}\left( M\right)$, you need to require $M$ to have no loops. Otherwise, $\operatorname*{BC}\left( M\right)$ is the empty set (since the empty set is a broken circuit), and thus not a simplicial complex. \item \textbf{page 42, Lemma 4.4:} Replace \textquotedblleft$-c_{1}% +c_{2}-c_{3}+\cdots$\textquotedblright\ by \textquotedblleft$c_{0}-c_{1}% +c_{2}-c_{3}+\cdots$\textquotedblright, in order for the lemma to still be valid when $P$ is the one-element poset. \item \textbf{page 42, }\textsc{Note}\textbf{ after Lemma 4.4:} The equality \textquotedblleft$\mu\left( \widehat{0},\widehat{1}\right) =\widetilde{\chi }\left( \Delta\left( P^{\prime}\right) \right)$\textquotedblright% \ requires that $P$ contain more than one element. \item \textbf{page 43:} This is absolutely not an erratum, and not even a suggestion, but I just felt like sharing another proof of Theorem 4.11 (though the probability that it is not new to you is high). \textit{Proof sketch for Theorem 4.11.} Let $K=\mathbb{Q}$. For every $i\in\mathbb{P}$, define a function $\zeta_{i}:\operatorname*{Int}P\rightarrow K$ by% $\zeta_{i}\left[ x,y\right] =\left[ x\lessdot y\text{ and }\lambda\left( x,y\right) =i\right] .$ Here, we are using the Iverson bracket notation (that is, $\left[ \mathcal{A}\right] =\left\{ \begin{array} [c]{c}% 1,\text{ if }\mathcal{A}\text{ is true;}\\ 0,\text{ if }\mathcal{A}\text{ is false}% \end{array} \right.$ for any logical statement $\mathcal{A}$). Recall that the functions $\operatorname*{Int}P\rightarrow K$ form a $K$-algebra $\mathcal{I}\left( P\right)$ (defined in \S 1.3, and called the incidence algebra of $P$). So all of the $\zeta_{i}$ are elements of this $K$-algebra $\mathcal{I}\left( P\right)$. These elements $\zeta_{i}$ are locally nilpotent (since they send one-element intervals $\left[ x,x\right]$ to $0$), and the infinite products $\cdots\left( 1-\zeta_{3}\right) \left( 1-\zeta_{2}\right) \left( 1-\zeta_{1}\right)$ and $\left( 1-\zeta_{1}\right) ^{-1}\left( 1-\zeta_{2}\right) ^{-1}\left( 1-\zeta _{3}\right) ^{-1}\cdots$ is well-defined. Here, $1$ stands for the unity of the $K$-algebra $\mathcal{I}\left( P\right)$; this is its element $\delta$. We have% \begin{align*} & \underbrace{\left( 1-\zeta_{1}\right) ^{-1}}_{=\sum_{m\in\mathbb{N}}% \zeta_{1}^{m}}\underbrace{\left( 1-\zeta_{2}\right) ^{-1}}_{=\sum _{m\in\mathbb{N}}\zeta_{2}^{m}}\underbrace{\left( 1-\zeta_{3}\right) ^{-1}% }_{=\sum_{m\in\mathbb{N}}\zeta_{3}^{m}}\cdots\\ & =\left( \sum_{m\in\mathbb{N}}\zeta_{1}^{m}\right) \left( \sum _{m\in\mathbb{N}}\zeta_{2}^{m}\right) \left( \sum_{m\in\mathbb{N}}\zeta _{3}^{m}\right) \cdots\\ & =\sum_{\substack{\left( m_{1},m_{2},m_{3},\ldots\right) \text{ is a}\\\text{weak composition}}}\zeta_{1}^{m_{1}}\zeta_{2}^{m_{2}}\zeta _{3}^{m_{3}}\cdots=\sum_{1\leq a_{1}\leq a_{2}\leq\cdots\leq a_{k}}% \zeta_{a_{1}}\zeta_{a_{2}}\cdots\zeta_{a_{k}}. \end{align*} Hence, every $x\leq y$ in $P$ satisfy% \begin{align} & \left( \left( 1-\zeta_{1}\right) ^{-1}\left( 1-\zeta_{2}\right) ^{-1}\left( 1-\zeta_{3}\right) ^{-1}\cdots\right) \left[ x,y\right] \nonumber\\ & =\left( \sum_{1\leq a_{1}\leq a_{2}\leq\cdots\leq a_{k}}\zeta_{a_{1}}% \zeta_{a_{2}}\cdots\zeta_{a_{k}}\right) \left[ x,y\right] \label{p43.5a}\\ & =\sum_{1\leq a_{1}\leq a_{2}\leq\cdots\leq a_{k}}\underbrace{\left( \zeta_{a_{1}}\zeta_{a_{2}}\cdots\zeta_{a_{k}}\right) \left[ x,y\right] }_{=\sum_{x=x_{0}\leq x_{1}\leq x_{2}\leq\cdots\leq x_{k}\leq y}\prod _{i=1}^{k}\zeta_{a_{i}}\left[ x_{i-1},x_{i}\right] }\nonumber\\ & =\sum_{1\leq a_{1}\leq a_{2}\leq\cdots\leq a_{k}}\sum_{x=x_{0}\leq x_{1}\leq x_{2}\leq\cdots\leq x_{k}\leq y}\prod_{i=1}^{k}\underbrace{\zeta _{a_{i}}\left[ x_{i-1},x_{i}\right] }_{\substack{=\left[ x_{i-1}\lessdot x_{i}\text{ and }\lambda\left( x_{i-1},x_{i}\right) =a_{i}\right] \\\text{(by the definition of }\zeta_{a_{i}}\text{)}}}\nonumber\\ & =\sum_{1\leq a_{1}\leq a_{2}\leq\cdots\leq a_{k}}\sum_{x=x_{0}\leq x_{1}\leq x_{2}\leq\cdots\leq x_{k}\leq y}\underbrace{\prod_{i=1}^{k}\left[ x_{i-1}\lessdot x_{i}\text{ and }\lambda\left( x_{i-1},x_{i}\right) =a_{i}\right] }_{=\left[ x_{0}\lessdot x_{1}\lessdot\cdots\lessdot x_{k}\text{ and each }i\text{ satisfies }\lambda\left( x_{i-1},x_{i}\right) =i\right] }\nonumber\\ & =\sum_{1\leq a_{1}\leq a_{2}\leq\cdots\leq a_{k}}\underbrace{\sum _{x=x_{0}\leq x_{1}\leq x_{2}\leq\cdots\leq x_{k}\leq y}\left[ x_{0}\lessdot x_{1}\lessdot\cdots\lessdot x_{k}\text{ and each }i\text{ satisfies }% \lambda\left( x_{i-1},x_{i}\right) =i\right] }_{=\sharp\left\{ x=x_{0}\lessdot x_{1}\lessdot\cdots\lessdot x_{k}=y\ :\ \text{each }i\text{ satisfies }\lambda\left( x_{i-1},x_{i}\right) =i\right\} }\nonumber\\ & =\sum_{1\leq a_{1}\leq a_{2}\leq\cdots\leq a_{k}}\sharp\left\{ x=x_{0}\lessdot x_{1}\lessdot\cdots\lessdot x_{k}=y\ :\ \text{each }i\text{ satisfies }\lambda\left( x_{i-1},x_{i}\right) =i\right\} \nonumber\\ & =\sharp\left\{ x=x_{0}\lessdot x_{1}\lessdot\cdots\lessdot x_{k}% =y\ :\ \lambda\left( x_{0},x_{1}\right) \leq\lambda\left( x_{1}% ,x_{2}\right) \leq\cdots\leq\lambda\left( x_{k-1},x_{k}\right) \right\} \label{p43.5b}\\ & =1\ \ \ \ \ \ \ \ \ \ \left( \text{by Definition 4.11}\right) \nonumber\\ & =\zeta\left[ x,y\right] .\nonumber \end{align} Hence, $\left( 1-\zeta_{1}\right) ^{-1}\left( 1-\zeta_{2}\right) ^{-1}\left( 1-\zeta_{3}\right) ^{-1}\cdots=\zeta$. Inverting both sides of this equality, we obtain $\cdots\left( 1-\zeta_{3}\right) \left( 1-\zeta_{2}\right) \left( 1-\zeta_{1}\right) =\zeta^{-1}=\mu$. Thus,% $\mu=\cdots\left( 1-\zeta_{3}\right) \left( 1-\zeta_{2}\right) \left( 1-\zeta_{1}\right) =\sum_{a_{1}>a_{2}>\cdots>a_{k}\geq1}\left( -1\right) ^{k}\zeta_{a_{1}}\zeta_{a_{2}}\cdots\zeta_{a_{k}},$ Hence, every $x\leq y$ satisfy% \begin{align*} & \mu\left[ x,y\right] \\ & =\left( \sum_{a_{1}>a_{2}>\cdots>a_{k}\geq1}\left( -1\right) ^{k}% \zeta_{a_{1}}\zeta_{a_{2}}\cdots\zeta_{a_{k}}\right) \left[ x,y\right] \\ & =\sum_{k\in\mathbb{N}}\left( -1\right) ^{k}\sharp\left\{ x=x_{0}\lessdot x_{1}\lessdot\cdots\lessdot x_{k}=y\ :\ \lambda\left( x_{0},x_{1}\right) >\lambda\left( x_{1},x_{2}\right) >\cdots>\lambda\left( x_{k-1}% ,x_{k}\right) \right\} \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{similarly to how we got from (\ref{p43.5a}) to (\ref{p43.5b})}\right) . \end{align*} The sum on the right hand side has only one (or, rather, at most one) nonzero term, namely the one for $k=\operatorname*{rk}\left( x,y\right)$ (since $P$ is graded). Hence, this equality rewrites as% \begin{align*} & \mu\left[ x,y\right] \\ & =\left( -1\right) ^{\operatorname*{rk}\left( x,y\right) }\sharp\left\{ x=x_{0}\lessdot x_{1}\lessdot\cdots\lessdot x_{k}=y\ :\ \lambda\left( x_{0},x_{1}\right) >\lambda\left( x_{1},x_{2}\right) >\cdots>\lambda\left( x_{k-1},x_{k}\right) \right\} , \end{align*} and we are done. Now that I have written up this proof, I guess I understand why you didn't want to do it... \item \textbf{page 44, proof of Theorem 4.11:} In (27), replace \textquotedblleft$n$\textquotedblright\ by \textquotedblleft$n-1$% \textquotedblright. \item \textbf{page 44, proof of Theorem 4.11:} Replace \textquotedblleft% $\lambda\left( x_{i}\right) >\lambda\left( x_{i+1}\right)$% \textquotedblright\ by \textquotedblleft$\lambda\left( x_{i-1},x_{i}\right) >\lambda\left( x_{i},x_{i+1}\right)$\textquotedblright. \item \textbf{page 44, proof of Theorem 4.11:} The case of $n=0$ should be ruled out somewhere near the beginning of the proof, as there are arguments that tacitly use $n>0$ throughout the proof. (Compare what I wrote about Lemma 4.4.) \item \textbf{page 45, proof of Theorem 4.12:} Again, I think that assuming that $M$ is simple is not worth the hassle. We already have the hypothesis that $M$ has no loops (else, $\operatorname*{BC}\left( M\right)$ is not a simplicial complex). Without the assumption that $M$ be simple, we can no longer identify the atoms of $L\left( M\right)$ with the points of $M$. But we still have a surjective map% \begin{align*} M & \rightarrow\left\{ \text{atoms of }L\left( M\right) \right\} ,\\ x_{i} & \mapsto\overline{\left\{ x_{i}\right\} }, \end{align*} and your proof goes through if some of the $x_{i}$'s appearing in it are replaced by the corresponding $\overline{\left\{ x_{i}\right\} }$'s. \item \textbf{page 45, proof of Theorem 4.12:} In \textquotedblleft Figure 3 shows the lattice of flats of the matroid $M$ of Figure 1 with the edge labeling (30)\textquotedblright, add \textquotedblleft the ordering $\mathcal{O}$ and\textquotedblright\ after the \textquotedblleft with\textquotedblright. \item \textbf{page 45, proof of Theorem 4.12:} In \textquotedblleft Moreover, there is a unique $y_{1}$ satisfying $x=x_{0}\lessdot y_{1}\leq y$ and $\widetilde{\lambda}\left( x_{0},y_{1}\right) =j$, viz., $y_{1}=x_{0}\vee x_{j}$. (Note that $y_{1}\gtrdot x_{0}$ by semimodularity.)\textquotedblright, replace every of the four occurrences of \textquotedblleft$x_{0}%$\textquotedblright\ by \textquotedblleft$x$\textquotedblright. Then, define $y_{0}$ (not $x_{0}$) to mean $x$ (this notation is used in the next sentence). \item \textbf{page 46, proof of Theorem 4.12:} In \textquotedblleft% $\widetilde{\lambda}\left( y_{0},y_{1}\right) =j>\widetilde{\lambda}\left( y_{1},y_{2}\right)$\textquotedblright, replace the \textquotedblleft% $>$\textquotedblright\ sign by a \textquotedblleft$\geq$\textquotedblright\ sign. \item \textbf{page 46, proof of Theorem 4.12:} In Claim 2, replace both appearances of \textquotedblleft$\lambda\left( C\right)$\textquotedblright% \ by \textquotedblleft$\widetilde{\lambda}\left( C\right)$% \textquotedblright. Also, it would be good to define what $\widetilde{\lambda}\left( C\right)$ means, and explain the abuse of notation. As far as I understand, you define $\widetilde{\lambda}\left( C\right)$ as follows: If $C$ is a chain $0=y_{0}\lessdot y_{1}\lessdot\cdots\lessdot y_{k}$, then you define $\widetilde{\lambda}\left( C\right)$ to be the sequence $\left( \widetilde{\lambda}\left( y_{0},y_{1}\right) ,\widetilde{\lambda}\left( y_{1},y_{2}\right) ,\ldots,\widetilde{\lambda}\left( y_{k-1},y_{k}\right) \right)$. Sometimes you denote the set of the entries of this sequence (rather than this sequence itself) as $\widetilde{\lambda}\left( C\right)$. Moreover, you identify this set with the set $\left\{ x_{\widetilde{\lambda }\left( y_{0},y_{1}\right) },x_{\widetilde{\lambda}\left( y_{1}% ,y_{2}\right) },\ldots,x_{\widetilde{\lambda}\left( y_{k-1},y_{k}\right) }\right\} \subseteq S$. \item \textbf{page 46, proof of Theorem 4.12:} In Claim 2, replace \textquotedblleft increasing chain\textquotedblright\ by \textquotedblleft strictly increasing chain\textquotedblright. \item \textbf{page 46, proof of Theorem 4.12:} In the proof of Claim 2, replace \textquotedblleft$\lambda\left( C\right)$\textquotedblright\ by \textquotedblleft$\widetilde{\lambda}\left( C\right)$\textquotedblright% \ (in \textquotedblleft To prove the distinctness of the labels $\lambda \left( C\right)$\textquotedblright). \item \textbf{page 46, proof of Theorem 4.12:} In the proof of Claim 2, replace \textquotedblleft$\widehat{0}:=y_{0}\lessdot y_{1}\lessdot \cdots\lessdot y_{k}$\textquotedblright\ by \textquotedblleft$\widehat{0}% =y_{0}\lessdot y_{1}\lessdot\cdots\lessdot y_{k}$\textquotedblright\ (no colon, since you are not defining anything). \item \textbf{page 46, proof of Theorem 4.12:} In the proof of Claim 2, you write: \textquotedblleft Note that $C$ is saturated by semimodularity\textquotedblright. This is only half of the story, because it has to be checked that no $y_{i}$ equals $y_{i-1}$. This latter statement follows from% \begin{align*} \operatorname*{rk}\left( y_{i}\right) & =\operatorname*{rk}\left( \overline{\left\{ x_{a_{1}}\right\} }\vee\overline{\left\{ x_{a_{2}% }\right\} }\vee\cdots\vee\overline{\left\{ x_{a_{i}}\right\} }\right) =\operatorname*{rk}\overline{\left\{ x_{a_{1}},x_{a_{2}},\ldots,x_{a_{i}% }\right\} }\\ & =\operatorname*{rk}\left\{ x_{a_{1}},x_{a_{2}},\ldots,x_{a_{i}}\right\} =i\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\left\{ x_{a_{1}},x_{a_{2}},\ldots,x_{a_{i}}\right\} \text{ is a subset of the independent set }T\text{,}\\ \text{and thus itself independent}% \end{array} \right) , \end{align*} where we are tacitly using that the lattice $L\left( M\right)$ is graded by the rank of flats in the matroid $M$. \item \textbf{page 46, proof of Theorem 4.12:} In the proof of Claim 2, you write: \textquotedblleft Thus% $\operatorname*{rk}\left( T\right) =\operatorname*{rk}\left( T\cup\left\{ x_{j}\right\} \right) =i.$ Since $T$ is independent, $T\cup\left\{ x_{j}\right\}$ contains a circuit $Q$ satisfying $x_{j}\in Q$, so $T$ contains a broken circuit.\textquotedblright\ This is wrong in two places: first, $\operatorname*{rk}\left( T\right)$ is not $i$, and second, $x_{j}$ might not be larger than $\max T$. Let me suggest the following corrected argument: \textquotedblleft Let $T_{i}$ be the subset $\left\{ x_{a_{1}},\ldots ,x_{a_{i}}\right\}$ of $T$; then, $T_{i}$ is independent (since $T$ is independent). Moreover, $y_{i}=\bigvee_{t\in T_{i}}\overline{\left\{ t\right\} }=\overline{T_{i}}$. Hence, from $y_{i-1}\vee\overline{\left\{ x_{j}\right\} }=y_{i}$, we obtain $\overline{\left\{ x_{j}\right\} }\leq y_{i}$, so $\overline{\left\{ x_{j}\right\} }\subseteq y_{i}=\overline {T_{i}}$. Thus, the set $T_{i}\cup\left\{ x_{j}\right\}$ is dependent, and thus contains a circuit $Q$ satisfying $x_{j}\in Q$ (since $T_{i}$ is independent). Therefore, $T_{i}$ contains a broken circuit (namely, $Q\setminus\left\{ x_{j}\right\}$, since $j>a_{i}>a_{i-1}>\cdots>a_{1}$). Thus, $T$ contains a broken circuit (since $T_{i}\subseteq T$), which is absurd.\textquotedblright \item \textbf{page 47, Example 4.9 (c):} Replace \textquotedblleft and $\operatorname*{rk}\left( y\right) =2$\textquotedblright\ by \textquotedblleft with $\operatorname*{rk}\left( y\right) =2$% \textquotedblright. \item \textbf{page 47, Example 4.9 (e):} Replace \textquotedblleft% $\mathbb{F}_{n}\left( q\right)$\textquotedblright\ by \textquotedblleft% $\mathbb{F}_{q}^{n}$\textquotedblright. \item \textbf{page 48, Example 4.9 (e):} Replace \textquotedblleft$L$ is a modular geometric lattice\textquotedblright\ by \textquotedblleft$B_{n}\left( q\right)$ is a modular geometric lattice\textquotedblright. \item \textbf{page 48, Example 4.9 (e):} Replace \textquotedblleft every $x\in L$ is modular\textquotedblright\ by \textquotedblleft every $x\in B_{n}\left( q\right)$ is modular\textquotedblright. \item \textbf{page 48, Example 4.9 (e), }\textsc{Note}\textbf{:} Replace \textquotedblleft every two points\textquotedblright\ by \textquotedblleft every two distinct points\textquotedblright. Similarly, replace \textquotedblleft every two lines\textquotedblright\ by \textquotedblleft every two distinct lines\textquotedblright. \item \textbf{page 49, Example 4.9 (f):} In \textquotedblleft$\left\{ a,b,B_{1}-a,B_{2}-b,\ldots,B_{3},\ldots,B_{k}\right\}$\textquotedblright, remove the first \textquotedblleft$\ldots$\textquotedblright. \item \textbf{page 49, Theorem 4.13:} The \textquotedblleft of rank $n$\textquotedblright\ is slightly ambiguous: does it refer to the lattice or to $z$ ? (It is meant to refer to $L$, of course, rather unsurprisingly, but I'd still split such a sentence into two if I were to write it.) \item \textbf{page 49, Theorem 4.13:} If I am not mistaken, $\chi_{L}$ and $\chi_{\left[ \widehat{0},z\right] }$ have never been defined: You defined $\chi_{M}$ for matroids, but not $\chi_{L}$ for lattices. I guess it wouldn't be wrong to address this on a more general level and define $\chi_{P}$ for every finite graded poset $P$ which has a $\widehat{0}$ and a $\widehat{1}$, by setting% $\chi_{P}\left( t\right) =\sum_{x\in P}\mu\left( \widehat{0},x\right) t^{\operatorname*{rk}\widehat{1}-\operatorname*{rk}x}.$ \item \textbf{page 50:} In the first paragraph of this page, \textquotedblleft begins $x^{n}-ax^{n-1}+\cdots$\textquotedblright\ should be \textquotedblleft begins $t^{n}-at^{n-1}+\cdots$\textquotedblright. \item \textbf{page 52, proof of Theorem 4.13:} It took me a while to understand what you mean by \textquotedblleft the product will be preserved\textquotedblright. Your argument, set up more algebraically, seems to be this: We define a $K$-module homomorphism $\alpha:K\left[ \widehat{0},z\right] \rightarrow K\left[ t\right]$ by $\alpha\left( v\right) =t^{\operatorname*{rk}z-\operatorname*{rk}v}$ for every $v\in\left[ \widehat{0},z\right]$. We define a $K$-module homomorphism $\beta:K\left\{ w\in L\ \mid\ w\wedge z=\widehat{0}\right\} \rightarrow K\left[ t\right]$ by $\beta\left( y\right) =t^{n-\operatorname*{rk}y-\operatorname*{rk}z}$ for every $y\in L$ satisfying $y\wedge z=\widehat{0}$. We define a $K$-module homomorphism $\gamma:KL\rightarrow K\left[ t\right]$ by $\gamma\left( x\right) =t^{n-\operatorname*{rk}x}$ for every $x\in L$. Then, you show (using Claim 2) the equality% \begin{equation} \alpha\left( v\right) \beta\left( y\right) =\gamma\left( v\vee y\right) \label{p52.3}% \end{equation} for every $v\in L$ and $y\in L$ satisfying $v\leq z$ and $y\wedge z=\widehat{0}$. By linearity, the same equality thus holds for every $v\in K\left[ \widehat{0},z\right]$ and $y\in K\left\{ y\in L\ \mid\ y\wedge z=\widehat{0}\right\}$. Now, you apply the map $\gamma$ to both sides of (33), and simplify the right hand side using (\ref{p52.3}). \item \textbf{page 53, Definition 4.13:} Replace \textquotedblleft% $L_{\mathcal{A}}$\textquotedblright\ by \textquotedblleft$L\left( \mathcal{A}\right)$\textquotedblright. \item \textbf{page 54, Example 4.11 (c):} In \textquotedblleft$B_{1}\subset B_{2}\cdots\subset B_{n-1}$\textquotedblright, you forgot a \textquotedblleft% $\subset$\textquotedblright\ sign. \item \textbf{page 54, Example 4.11 (c):} \textquotedblleft The atoms covered by $\pi_{i}$\textquotedblright\ should be \textquotedblleft The atoms $\leq \pi_{i}$\textquotedblright. \item \textbf{page 54, Example 4.11 (c):} On the last line of the page, replace \textquotedblleft$\mathcal{B}_{n}\left( t\right)$\textquotedblright% \ by \textquotedblleft$\mathcal{B}_{n}$\textquotedblright. \item \textbf{page 55:} Again, \textquotedblleft$L_{\mathcal{A}}%$\textquotedblright\ should be \textquotedblleft$L\left( \mathcal{A}\right)$\textquotedblright\ (two lines above Theorem 4.14). \item \textbf{page 61, proof of Proposition 5.13:} On line 2 of the proof, replace \textquotedblleft$v_{i},a_{i}\in\mathbb{Z}^{n}$\textquotedblright\ by \textquotedblleft$v_{i}\in\mathbb{Z}^{n}$ and $a_{i}\in\mathbb{Z}%$\textquotedblright. \item \textbf{page 62, proof of Proposition 5.13:} On the second line of the page, you write \textquotedblleft if and only if at least one\textquotedblright. I understand the \textquotedblleft only if\textquotedblright. The \textquotedblleft if\textquotedblright\ might be true, but is probably not easy to prove (the point is to rule out accidental isomorphisms $L\left( \mathcal{A}\right) \cong L\left( \mathcal{A}% _{p}\right)$ that could happen if hyperplanes becoming parallel \textquotedblleft undo\textquotedblright\ the damage done by hyperplanes becoming concurrent); either way it is a distraction from the proof. \item \textbf{page 62, proof of Theorem 5.15:} Replace \textquotedblleft% $F_{q}$\textquotedblright\ by \textquotedblleft$\mathbb{F}_{q}$% \textquotedblright. \end{itemize} \end{document}