\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{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\theoremstyle{definition}
\newtheorem{theo}{Theorem}[section]
\newenvironment{theorem}[1][]
{\begin{theo}[#1]\begin{leftbar}}
{\end{leftbar}\end{theo}}
\newtheorem{lem}[theo]{Lemma}
\newenvironment{lemma}[1][]
{\begin{lem}[#1]\begin{leftbar}}
{\end{leftbar}\end{lem}}
\newtheorem{prop}[theo]{Proposition}
\newenvironment{proposition}[1][]
{\begin{prop}[#1]\begin{leftbar}}
{\end{leftbar}\end{prop}}
\newtheorem{defi}[theo]{Definition}
\newenvironment{definition}[1][]
{\begin{defi}[#1]\begin{leftbar}}
{\end{leftbar}\end{defi}}
\newtheorem{remk}[theo]{Remark}
\newenvironment{remark}[1][]
{\begin{remk}[#1]\begin{leftbar}}
{\end{leftbar}\end{remk}}
\newtheorem{coro}[theo]{Corollary}
\newenvironment{corollary}[1][]
{\begin{coro}[#1]\begin{leftbar}}
{\end{leftbar}\end{coro}}
\newtheorem{conv}[theo]{Convention}
\newenvironment{condition}[1][]
{\begin{conv}[#1]\begin{leftbar}}
{\end{leftbar}\end{conv}}
\newtheorem{quest}[theo]{Question}
\newenvironment{algorithm}[1][]
{\begin{quest}[#1]\begin{leftbar}}
{\end{leftbar}\end{quest}}
\newtheorem{warn}[theo]{Warning}
\newenvironment{conclusion}[1][]
{\begin{warn}[#1]\begin{leftbar}}
{\end{leftbar}\end{warn}}
\newtheorem{conj}[theo]{Conjecture}
\newenvironment{conjecture}[1][]
{\begin{conj}[#1]\begin{leftbar}}
{\end{leftbar}\end{conj}}
\newtheorem{exmp}[theo]{Example}
\newenvironment{example}[1][]
{\begin{exmp}[#1]\begin{leftbar}}
{\end{leftbar}\end{exmp}}
\newenvironment{proof2}[1][]
{\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 $x<y$ in a graded poset
$P$ then we write $\operatorname*{rk}\left(  x,y\right)  =\operatorname*{rk}%
\left(  y\right)  -\operatorname*{rk}\left(  x\right)  $\textquotedblright. It
would be better to replace \textquotedblleft$x<y$\textquotedblright\ by
\textquotedblleft$x\leq y$\textquotedblright\ here, since you later use the
notation in this mildly greater generality.

\item \textbf{page 10:} You never seem to explicitly say that Theorem 1.1 is
the \textquotedblleft M\"{o}bius inversion formula\textquotedblright. When
that name appears in the proof of Theorem 1.1, the reader has to guess.

\item \textbf{page 12, Exercise (7):} The word \textquotedblleft
face\textquotedblright\ has not been defined.

\item \textbf{page 14, Lemma 2.2:} Do you really want the word
\textquotedblleft real\textquotedblright\ here? I haven't read the proof in
much detail, but I don't see how you are using this condition (actually, you
even use the letter $K$ for the ground field in the proof).

\item \textbf{page 16, Theorem 2.3:} I find it interesting to observe that
this theorem holds whenever $L$ is a finite join-semilattice, not necessarily
a lattice. (Of course, the same proof applies, although the algebras now
become nonunital algebras.) In this generality, $L$ as a semigroup needs not
have a neutral element, but the algebra $A\left(  L\right)  $ has a unity
nevertheless (because $A^{\prime}\left(  L\right)  $ has a unity, namely
$\sum_{x\in L}\sigma_{x}^{\prime}$).

\item \textbf{page 24:} In the last paragraph of this page, replace
\textquotedblleft by first choosing the size $i=\sharp\kappa\left(  \left[
n\right]  \right)  $ of its image in $\dbinom{q}{i}$ ways\textquotedblright%
\ by \textquotedblleft by first choosing the size $i=\sharp\kappa\left(
\left[  n\right]  \right)  $ of its image, then choosing its image
$\kappa\left(  \left[  n\right]  \right)  $ itself in $\dbinom{q}{i}$
ways\textquotedblright.

\item \textbf{page 28:} In the \textquotedblleft\textsc{Note}%
\textquotedblright, I think \textquotedblleft linear extension of
$\mathfrak{o}$\textquotedblright\ should be \textquotedblleft linear extension
of $\overline{\mathfrak{o}}$\textquotedblright. (At least I usually have seen
the notion of a \textquotedblleft linear extension\textquotedblright\ defined
for posets, sometimes for preorders, but not for acyclic orientations themselves.)

\item \textbf{page 32:} \textquotedblleft diagam\textquotedblright%
\ $\rightarrow$ \textquotedblleft diagram\textquotedblright.

\item \textbf{page 33:} After \textquotedblleft$M$ is isomorphic to the set of
nonzero vectors in the vector space $\mathbb{F}_{2}^{3}$\textquotedblright,
add \textquotedblleft(with independence defined as linear
independence)\textquotedblright. (This is to say, not affine independence.)

\item \textbf{page 35, proof of Proposition 3.6:} Replace \textquotedblleft%
$\overline{B}=\overline{B}^{\prime}$\textquotedblright\ by \textquotedblleft%
$\overline{\mathcal{B}}=\overline{\mathcal{B}^{\prime}}$\textquotedblright%
\ (notice the $^{\prime}$ being under the overline).

\item \textbf{page 35, proof of Proposition 3.6:} You show that $X=X^{\prime}$
if and only if $\overline{\mathcal{B}}=\overline{\mathcal{B}^{\prime}}$. But
in order to check that $L\left(  M\right)  \cong L\left(  \mathcal{A}\right)
$ as lattices, it is also important to prove that $X\subseteq X^{\prime}$ if
and only if $\overline{\mathcal{B}}\supseteq\overline{\mathcal{B}^{\prime}}$.
(Not difficult but worth making explicit.)

\item \textbf{page 36, Definition 3.9:} In the first sentence of this
definition, replace \textquotedblleft A finite lattice\textquotedblright\ by
\textquotedblleft A finite graded lattice\textquotedblright. Similarly, in the
last sentence, replace \textquotedblleft a finite lattice\textquotedblright%
\ by \textquotedblleft a finite graded lattice\textquotedblright.

\item \textbf{page 36, proof of Theorem 3.8:} In the first displayed equation
of the proof, replace \textquotedblleft$\operatorname*{rk}\left(  \vee
I\right)  $\textquotedblright\ by \textquotedblleft$\operatorname*{rk}\left(
\bigvee I\right)  $\textquotedblright.

\item \textbf{page 36, proof of Theorem 3.8:} You write: \textquotedblleft
Since $L$ is atomic, there exists $y\in S$ such that $y\not \leq \bigvee
T$\textquotedblright. (Here I have applied the correction that already appears
in your errata.) In my opinion, the existence of such a $y$ is not a
consequence of $L$ being atomic. Instead, it is the consequence of the
argument that if no such $y$ existed, then every $z\in S$ would satisfy
$z\leq\bigvee T$, and thus we would have $\bigvee S\leq\bigvee T<\bigvee
T^{\prime}\leq\bigvee S$ (since $T^{\prime}\subseteq S$), which is absurd.

\item \textbf{page 36, proof of Theorem 3.8:} You claim that \textquotedblleft%
$L\cong L\left(  M\right)  $\textquotedblright. I think this claim is roughly
the difficulty of your [2+] exercises (certainly harder than the
\textquotedblleft\lbrack why?]\textquotedblright\ in the same paragraph) and
should get at least a hint to its proof.

\textit{Proof sketch.} The idea is to show that the maps%
\begin{align*}
\Phi:L  &  \rightarrow L\left(  M\right)  ,\\
x  &  \mapsto\left\{  a\in A\ \mid\ a\leq x\right\}
\end{align*}
and%
\begin{align}
\Psi:L\left(  M\right)   &  \rightarrow L,\label{p36.Psi}\\
F  &  \mapsto\bigvee F\nonumber
\end{align}
are well-defined, mutually inverse and lattice homomorphisms. The
well-definedness of $\Psi$ is obvious. The well-definedness of $\Phi$ uses the
neat but not completely trivial observation that every $B\subseteq A$
satisfies
\begin{equation}
\operatorname*{rk}B=\operatorname*{rk}\left(  \bigvee B\right)  \label{p36.5}%
\end{equation}
(where the first $\operatorname*{rk}$ means rank in the matroid $M$, while the
second $\operatorname*{rk}$ means rank in the lattice $L$). The proof of
$\Psi\circ\Phi=\operatorname*{id}$ uses the fact that every $x\in L$ satisfies
$x=\bigvee\left(  \Phi\left(  x\right)  \right)  $, which has a neat indirect
proof (the lattice $L$ is atomic, so $x$ is the join $\bigvee C$ of
\textit{some} subset $C$ of $A$, and clearly this $C$ satisfies $C\subseteq
\Phi\left(  x\right)  $, so that $x=\bigvee\underbrace{C}_{\subseteq
\Phi\left(  x\right)  }\leq\bigvee\left(  \Phi\left(  x\right)  \right)  $,
but on the other hand $\bigvee\left(  \Phi\left(  x\right)  \right)  \leq x$
for obvious reasons, and thus $x=\bigvee\left(  \Phi\left(  x\right)  \right)
=\Psi\left(  \Phi\left(  x\right)  \right)  =\left(  \Psi\circ\Phi\right)
\left(  x\right)  $). The proof of $\Phi\circ\Psi=\operatorname*{id}$ proceeds
by observing that if $a\in A$ and $F\in L\left(  M\right)  $ satisfy
$a\leq\bigvee F$, then $\bigvee\left(  F\cup\left\{  a\right\}  \right)
=\bigvee F$, and thus (\ref{p36.5}) yields $\operatorname*{rk}\left(
F\cup\left\{  a\right\}  \right)  =\operatorname*{rk}\left(
\underbrace{\bigvee\left(  F\cup\left\{  a\right\}  \right)  }_{=\bigvee
F}\right)  =\operatorname*{rk}\left(  \bigvee F\right)  =\operatorname*{rk}F$
(by (\ref{p36.5}) again), which shows that $a\in F$ (since $F$ is a flat). So
the maps $\Phi$ and $\Psi$ are mutually inverse. In order to prove that they
are lattice homomorphisms, I proceed as follows: The maps $\Psi$ and $\Phi$
are poset homomorphisms and form a monotone Galois connection between the
posets $L\left(  M\right)  $ and $L$; thus, $\Psi$ preserves joins and $\Phi$
preserves meets. Since $\Psi$ and $\Phi$ are mutually inverse, this yields
that both of these maps preserve both joins and meets, and we are done. This
would be more noticeably painful to show without knowing the Galois-connection trick.

Am I missing something obvious?

\item \textbf{page 36, proof of Theorem 3.8:} The assumption that $M$ is
simple (in the proof of (2) $\Longrightarrow$ (1)) seems like overkill to me.
It can be lifted easily (you just have to replace \textquotedblleft every flat
is the join of its elements\textquotedblright\ by \textquotedblleft every flat
$F$ satisfies $F=\bigvee_{x\in F}\overline{\left\{  x\right\}  }%
$\textquotedblright), whereas justifying this assumption requires the use of
the $L\left(  M\right)  \cong L\left(  \widehat{M}\right)  $ statement on page
34, which does not look easy to prove at all.

\item \textbf{page 37:} In the paragraph directly after the proof of Theorem
3.8, replace \textquotedblleft the intersection lattice $L_{\mathcal{A}}%
$\textquotedblright\ by \textquotedblleft the intersection lattice $L\left(
\mathcal{A}\right)  $\textquotedblright.

\item \textbf{page 37:} I am surprised that you don't prove Theorem 3.9 here.
In my opinion, the proof is very short (possibly even shorter than justifying
why it really is dual to [31, Cor. 3.9.3] -- there is an asymmetry in the
definition of the M\"{o}bius function which \textquotedblleft
favors\textquotedblright\ the lower end of the interval\footnote{The formula
(2) says $\mu\left(  x,y\right)  =-\sum_{x\leq z<y}\mu\left(  x,z\right)  $,
rather than $\mu\left(  x,y\right)  =-\sum_{x<z\leq y}\mu\left(  z,y\right)
$. Maybe it is worth explaining that the latter equality also holds (because
$\mu=\zeta^{-1}$ is a two-sided inverse) and therefore the M\"{o}bius function
is invariant under \textquotedblleft turning the poset upside
down\textquotedblright.
\par
(While at that, it's also worth pointing out that $\mu\left(  x,y\right)  $
depends only on the poset $\left[  x,y\right]  $, not on the whole poset $P$.
You use this a lot, but you leave it tacit, defining $\mu$ as a function on
the whole $\operatorname*{Int}P$ instead.)}), and fits perfectly with what you
did in \S 2:

\textit{Proof of Theorem 3.9.} We have $a\neq\widehat{0}$ and thus
$a>\widehat{0}$. 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}