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\ihead{Errata to ``The Geometry and Topology of Coxeter Groups''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{The Geometry and Topology of Coxeter Groups}

\textit{Michael W. Davis}

\href{https://people.math.osu.edu/davis.12/second_edition.pdf}{Second Edition
2025}

\textbf{Comments and corrections by Darij Grinberg}

\bigskip
\end{center}

\appendix\setcounter{section}{13}

The following list contains my comments to Michael W. Davis's book \textit{The
Geometry and Topology of Coxeter Groups}, specifically its 2nd edition
(Springer, 2025). Among these comments are a few corrections (most
importantly, the missing condition in Definition 3.2.1), several minor
nitpicks, a few reformulations that I made for my own understanding, and a
number of alternative proofs that I invented as I was unable to follow some of
the existing proofs. In the latter case, I genuinely don't know whether the
fault lies with me or the book, as I am new to the subject and come from a
combinatorics background. 

I have read Chapter 3 (minus Example 3.1.7, the Remark after Theorem 3.3.4,
the paragraph after Lemma 3.3.5, and \S 3.5) and Chapter 4 until Remark 4.8.1,
as well as \S 4.10 until Definition 4.10.3.

\section{Errata and comments}

\begin{enumerate}
\item \textbf{Page 23, Example 3.1.4:} In \textquotedblleft of all pairs
$\left(  x,\varepsilon\right)  \in\mathbf{C}_{m}\times C_{2}$%
\textquotedblright, replace \textquotedblleft$C_{2}$\textquotedblright\ by
\textquotedblleft$\mathbf{C}_{2}$\textquotedblright.

\item \textbf{Page 23, Example 3.1.4:} The equality \textquotedblleft$\left(
x,\varepsilon\right)  \cdot\left(  x^{\prime},\varepsilon^{\prime}\right)
=\left(  x^{\varepsilon^{\prime}}x,\varepsilon\varepsilon^{\prime}\right)
$\textquotedblright\ should be \textquotedblleft$\left(  x,\varepsilon\right)
\cdot\left(  x^{\prime},\varepsilon^{\prime}\right)  =\left(  x^{\varepsilon
^{\prime}}x^{\prime},\varepsilon\varepsilon^{\prime}\right)  $%
\textquotedblright.

\item \textbf{Page 23, Lemma 3.1.5:} After \textquotedblleft by distinct
elements $s$ and $t$\textquotedblright, add \textquotedblleft of order
$2$\textquotedblright.

\item \textbf{Page 23, Lemma 3.1.5 (i):} Remove the comma in \textquotedblleft
the semidirect product,\textquotedblright.

\item \textbf{Page 26, proof of Lemma 3.1.5 (ii):} In the last sentence, it is
worth pointing out that $sps^{-1}=p^{-1}$ is being used to show that $\varphi$
really is an isomorphism.

\item \textbf{Page 26, Example 3.1.7:} \textquotedblleft the the real
line\textquotedblright\ $\rightarrow$ \textquotedblleft the real
line\textquotedblright.

\item \textbf{Page 26, proof of Lemma 3.1.8:} \textquotedblleft Extend this to
a function\textquotedblright\ $\rightarrow$ \textquotedblleft Extend this to a
multiplicative function\textquotedblright.

\item \textbf{Page 27, Definition 3.2.1:} This needs one more condition:

\begin{enumerate}
\item[\textbf{(e)}] each $r\in R$ flips at least one edge of $\Omega$.
\end{enumerate}

This is important (among other things) in order to make Lemma 3.2.5 hold, and
even earlier than that, it is used in arguing that $R$ is the set of all
conjugates of $S$. And indeed, without the requirement \textbf{(e)}, Lemma
3.2.5 would be false\footnote{A counterexample: Let $\Omega$ be a $6$-cycle,
visualized as a regular hexagon. Let $W$ be the dihedral group $D_{6}$, acting
on $\Omega$ in the obvious way. Let $R$ be the subset of $W$ that consists of
the three reflections across the perpendicular bisectors of the sides of the
hexagon (there are $6$ sides but only $3$ perpendicular bisectors) as well as
the $180^{\circ}$-rotation around its center. Then, all four conditions
\textbf{(a)}, \textbf{(b)}, \textbf{(c)} and \textbf{(d)} of Definition 3.2.1
are satisfied, but \textbf{(e)} is not, and Lemma 3.2.5 fails as well (the set
$S$ contains only the three perpendicular bisector reflections from $R$, not
the $180^{\circ}$-rotation, and thus only generates a $D_{3}$ inside the
$D_{6}$).}.

\item \textbf{Page 27:} \textquotedblleft the involution $wsw^{-1}$ flips $e$
and since $r$ is the unique such involution\textquotedblright\ $\rightarrow$
\textquotedblleft the pre-reflection $wsw^{-1}$ flips $e$ and since $r$ is the
unique such pre-reflection\textquotedblright.

\item \textbf{Page 27, (3.1):} I would extend this equality as follows:%
\[
r_{i}:=w_{i-1}s_{i}w_{i-1}^{-1}=w_{i}w_{i-1}^{-1}=\left(  s_{1}s_{2}\cdots
s_{i}\right)  \left(  s_{1}s_{2}\cdots s_{i-1}\right)  ^{-1}.
\]
These equalities are easy to check, but they are used several times (tacitly)
in the text that follows.

\item \textbf{Page 28:} \textquotedblleft edge path\textquotedblright\ in a
graph means what graph theorists commonly call a walk, not a path.
(\textquotedblleft Path\textquotedblright, to a graph theorist, requires the
vertices to be distinct.)

\item \textbf{Page 28, after the proof of Lemma 3.2.5:} Here it would be
helpful to add \textquotedblleft Thus, if $\left(  \Omega,v_{0}\right)  $ is a
pre-reflection system for $W$ and if $S=S\left(  v_{0}\right)  $, then
$\left(  W,S\right)  $ is a pre-Coxeter system.\textquotedblright.

\item \textbf{Page 29:} The combinatorially minded reader who defines graphs
as collections of vertices and edges (rather than as topological spaces)
should read \textquotedblleft midpoints of those edges\textquotedblright\ as
\textquotedblleft those edges\textquotedblright\ here. Similarly, he should
read \textquotedblleft fixed point set of $r$\textquotedblright\ as
\textquotedblleft fixed point set of $r$ acting on the set of vertices
\textbf{and} edges of $\Omega$\textquotedblright. (Once you get to reflection
systems, a pre-reflection $r$ cannot fix any vertices, since Lemma 3.2.12
shows that $W$ acts freely on $\operatorname*{Vert}\left(  \Omega\right)  $.
Thus, $\Omega^{r}$ consists entirely of edges when $\left(  W,S\right)  $ is a
reflection system. But when we are just dealing with a pre-reflection system,
both vertices and edges can be fixed.\footnote{An example is obtained when the
graph $\Omega$ is a $3$-cycle $C_{3}$, and $W$ is its symmetry group $S_{3}$,
with $R$ being the set of all three involutions in $S_{3}$. This defines a
pre-reflection system (choosing $v_{0}\in\operatorname*{Vert}\left(
\Omega\right)  $ arbitrarily), but not a reflection system, since each
$\Omega^{r}$ contains a fixed vertex \textbf{and} a fixed edge.})

\item \textbf{Page 29:} \textquotedblleft if and only $r$
occurs\textquotedblright\ $\rightarrow$ \textquotedblleft if and only if $r$
occurs\textquotedblright.

\item \textbf{Page 30, Lemma 3.2.14:} I would replace \textquotedblleft%
$\operatorname*{Cay}\left(  W,S\right)  $\textquotedblright\ by
\textquotedblleft$\Omega$\textquotedblright\ here for the sake of familiarity
(of course, this makes no real difference, since $\Omega$ is isomorphic to
$\operatorname*{Cay}\left(  W,S\right)  $).

\item \textbf{Page 31, condition (D):} The notion of an \textquotedblleft
expression\textquotedblright\ should be defined somewhere (not just
\textquotedblleft reduced expression\textquotedblright): An \emph{expression}
for an element $w\in W$ means a word $\left(  t_{1},\ldots,t_{p}\right)  $ in
$S$ such that $w=t_{1}\cdots t_{p}$.

\item \textbf{Page 33, proof of (E) }$\Longrightarrow$\textbf{ (F):} After
\textquotedblleft Applying Condition (E) to the word $\left(  s_{1}%
,\ldots,s_{k},t\right)  $\textquotedblright, add \textquotedblleft(which is a
reduced expression of $wt$)\textquotedblright.

\item \textbf{Page 33, Definition 3.3.1:} The display%
\[
m_{st}=%
\begin{cases}
1, & \text{if }s=t;\\
\geq2, & \text{otherwise}%
\end{cases}
\]
looks strange (it reads \textquotedblleft$m_{st}=\geq2$\textquotedblright\ in
the \textquotedblleft otherwise\textquotedblright\ case). A better way to
write it would be%
\[
m_{s,t}%
\begin{cases}
=1, & \text{if }s=t;\\
\geq2, & \text{otherwise}%
\end{cases}
\]
(or just split it into two separate displays).

\item \textbf{Page 34:} Remove the comma in \textquotedblleft by the formula,
$m_{st}=m\left(  s,t\right)  $\textquotedblright.

\item \textbf{Page 34:} In the middle of the page, before \textquotedblleft
The next theorem is the main result\textquotedblright, I would add
\textquotedblleft Recall from Example 3.2.4 that any pre-Coxeter system
$\left(  W,S\right)  $ becomes a pre-reflection system $\Omega
=\operatorname*{Cay}\left(  W,S\right)  $ where we define $v_{0}=1$ and let
$R$ be the set of all elements of $W$ that are conjugate to an element of
$S$.\textquotedblright.

\item \textbf{Page 34, Remark after Theorem 3.3.4:} \textquotedblleft denote
dihedral subgroup\textquotedblright\ $\rightarrow$ \textquotedblleft denote
the dihedral subgroup\textquotedblright.

\item \textbf{Page 35, Lemma 3.3.5 (i):} After \textquotedblleft depends only
on $w$\textquotedblright, add \textquotedblleft and $r$\textquotedblright.

\item \textbf{Page 35, Lemma 3.3.5 (ii):} Remove the comma after
\textquotedblleft There is a homomorphism\textquotedblright.

\item \textbf{Page 35, proof of Lemma 3.3.5:} After \textquotedblleft Let
$\mathbf{s}=\left(  s_{1},\ldots,s_{k}\right)  $\textquotedblright, add
\textquotedblleft$\in S^{k}$\textquotedblright.

\item \textbf{Pages 35--36, proof of Lemma 3.3.5:} Every \textquotedblleft%
$n\left(  \mathbf{s},r\right)  $\textquotedblright\ here should actually be an
\textquotedblleft$n\left(  r,\mathbf{s}\right)  $\textquotedblright.

\item \textbf{Page 35, proof of Lemma 3.3.5:} \textquotedblleft The case $k=1$
is trivial\textquotedblright\ $\rightarrow$ \textquotedblleft The case $k=0$
is trivial\textquotedblright.

\item \textbf{Page 35, proof of Lemma 3.3.5:} \textquotedblleft Suppose
$k>1$\textquotedblright\ $\rightarrow$ \textquotedblleft Suppose
$k>0$\textquotedblright.

\item \textbf{Page 35, proof of Lemma 3.3.5:} In \textquotedblleft$\Phi\left(
\mathbf{s}\right)  =\left(  \Phi\left(  \mathbf{s}^{\prime}\right)
,u^{-1}s_{k}u\right)  $\textquotedblright, there is a little abuse of
notation: The right hand side is not to be understood as a pair consisting of
$\Phi\left(  \mathbf{s}^{\prime}\right)  $ and $u^{-1}s_{k}u$, but rather as
the $\left(  k-1\right)  $-tuple $\Phi\left(  \mathbf{s}^{\prime}\right)  $
with an extra entry $u^{-1}s_{k}u$ inserted into it at its end.

\item \textbf{Page 36, proof of Lemma 3.3.5:} After \textquotedblleft the map
$\mathbf{s}\rightarrow\phi_{\mathbf{s}}$ descends to a homomorphism from $W$
to the group of permutations of $R\times\left\{  \pm1\right\}  $%
\textquotedblright, add \textquotedblleft defined by%
\begin{align*}
W  &  \rightarrow\left\{  \text{permutations of }R\times\left\{  \pm1\right\}
\right\}  ,\\
w  &  \mapsto\phi_{\mathbf{s}}\text{ where }\mathbf{s}=\left(  s_{1}%
,s_{2},\ldots,s_{k}\right)  \text{ is a word in }S\text{ such that }%
w=s_{k}s_{k-1}\cdots s_{1}%
\end{align*}
\textquotedblright. (The important part here is that we want $w=s_{k}%
s_{k-1}\cdots s_{1}$, not $w=s_{1}s_{2}\cdots s_{k}$.)

\item \textbf{Page 36, proof of Proposition 3.3.6:} I think this proof could
be worded more clearly as follows: \textquotedblleft Lemma 3.3.5 (i) shows
that the condition (ii) of Lemma 3.2.9 is satisfied for each $r\in R$. Thus,
condition (i) is also satisfied for each $r\in R$; but this means that
$\left(  \Omega,v_{0}\right)  $ is a reflection system. Finally, the claim
about $\widehat{R}_{w}$ follows from the parenthetical statement in Lemma
3.2.9 condition (ii).\textquotedblright

\item \textbf{Page 36, Definition 3.4.1:} The word \textquotedblleft
subword\textquotedblright\ means a contiguous subword here, unlike in Lemma
3.2.6. It would probably better to call it a \textquotedblleft
factor\textquotedblright\ (as combinatorialists usually do).

\item \textbf{Page 37, proof of Theorem 3.4.2:} \textquotedblleft($\mathbf{s}$
and $\mathbf{u}$ begin with the same letter as do $\mathbf{u}^{\prime}$ and
$\mathbf{t}$)\textquotedblright\ could be reworded less ambiguously as
follows: \textquotedblleft($\mathbf{s}$ and $\mathbf{u}$ begin with the same
letter, and so do $\mathbf{u}^{\prime}$ and $\mathbf{t}$)\textquotedblright.

\item \textbf{Page 37, proof of Theorem 3.4.2:} Remove the comma before
\textquotedblleft for a $t$ in front\textquotedblright.

\item \textbf{Page 38, proof of Theorem 3.3.4:} It would be helpful to explain
why \textquotedblleft the Coxeter system associated to the Coxeter matrix of
$\left(  W,S\right)  $\textquotedblright\ exists in the first place. Its
construction has been given in \S 3.3 (in the last paragraph of page 33), but
it has not been shown there that it is a Coxeter system. So let us make up for
this now: The generators $\widetilde{s}\in\widetilde{S}$ are involutions in
$\widetilde{W}$, since the set $\mathcal{R}$ of relations imposed on them
include the relations $\left(  \widetilde{s}\widetilde{s}\right)  ^{m_{ss}%
}=\left(  \widetilde{s}\widetilde{s}\right)  ^{1}=\widetilde{s}\widetilde{s}$.
Moreover, these generators $\widetilde{s}$ differ from each other and from the
identity of $\widetilde{W}$, because the same holds for their images under the
canonical projection $\widetilde{W}\rightarrow W$. Thus, these generators
$\widetilde{s}$ have order $2$ in $\widetilde{W}$. Finally, these generators
$\widetilde{s}$ generate $\widetilde{W}$ by the definition of $\widetilde{W}$.
Hence, $\left(  \widetilde{W},\widetilde{S}\right)  $ is a pre-Coxeter system.
Since the canonical epimorphism $\widetilde{\widetilde{W}}\rightarrow
\widetilde{W},\ \widetilde{\widetilde{s}}\mapsto\widetilde{s}$ is an
isomorphism (indeed, $\widetilde{\widetilde{W}}$ is just an isomorphic copy of
$\widetilde{W}$ with the generators $s$ renamed as $\widetilde{s}$, because if
$s,t\in S$ are two elements, then the order $m_{\widetilde{s}\widetilde{t}}$
of $\widetilde{s}\widetilde{t}$ in $\widetilde{W}$ always equals the order
$m_{st}$ of $st$ in $W$\ \ \ \ \footnote{\textit{Proof:} Let $s,t\in S$. We
must show that $m_{\widetilde{s}\widetilde{t}}=m_{st}$. Recall that the
canonical projection $\widetilde{W}\rightarrow W$ sends $\widetilde{s}%
\widetilde{t}$ to $st$. Thus, if $m_{st}=\infty$, then $m_{\widetilde{s}%
\widetilde{t}}=\infty$ as well, and thus $m_{\widetilde{s}\widetilde{t}%
}=m_{st}$ is obvious in this case. Hence, we can WLOG assume that $m_{st}%
\neq\infty$ from now on.
\par
Then, $m_{\widetilde{s}\widetilde{t}}\mid m_{st}$ because the relation
$\left(  \widetilde{s}\widetilde{t}\right)  ^{m_{st}}\in\mathcal{R}$ forces
$\left(  \widetilde{s}\widetilde{t}\right)  ^{m_{st}}$ to be $1$ in
$\widetilde{W}$. Conversely, $m_{st}\mid m_{\widetilde{s}\widetilde{t}}$
because the canonical projection $\widetilde{W}\rightarrow W$ sends
$\widetilde{s}\widetilde{t}$ to $st$. Combining these two divisibilities, we
obtain $m_{\widetilde{s}\widetilde{t}}=m_{st}$, qed.}), this entails that
$\left(  \widetilde{W},\widetilde{S}\right)  $ is a Coxeter system, qed.

\item \textbf{Page 41, proof of Proposition 4.1.1:} I would replace
\textquotedblleft from Theorem 3.4.2\textquotedblright\ by \textquotedblleft
from Theorem 3.4.2 (ii)\textquotedblright\ here just to be a bit more specific.

\item \textbf{Page 42, proof of Corollary 4.1.5:} I would replace
\textquotedblleft By Proposition 4.1.1\textquotedblright\ by \textquotedblleft
By Corollary 4.1.2, we have $S\left(  w\right)  \subset T$, so
that\textquotedblright\ here.

\item \textbf{Page 42, proof of Theorem 4.1.6:} I would replace
\textquotedblleft$l_{T}\left(  tw\right)  <l_{T}\left(  w\right)
$\textquotedblright\ and \textquotedblleft$l_{S}\left(  tw\right)
<l_{S}\left(  w\right)  $\textquotedblright\ by \textquotedblleft$l_{T}\left(
tw\right)  \neq l_{T}\left(  w\right)  +1$\textquotedblright\ and
\textquotedblleft$l_{S}\left(  tw\right)  \neq l_{T}\left(  w\right)
+1$\textquotedblright, respectively, in order to match the wording of the
Exchange Condition more closely.

\item \textbf{Page 43, proof of Lemma 4.2.1:} \textquotedblleft(iv) follows
from (ii) and (iii)\textquotedblright\ should probably be \textquotedblleft%
(iv) follows from (i) and (iii)\textquotedblright.

\item \textbf{Page 43, proof of Lemma 4.2.2:} \textquotedblleft this to new
path\textquotedblright\ $\rightarrow$ \textquotedblleft this new
path\textquotedblright.

\item \textbf{Page 44, proof of Lemma 4.2.3:} \textquotedblleft
gallery\textquotedblright\ should be \textquotedblleft edge
path\textquotedblright\ (galleries have not been defined yet).

\item \textbf{Page 44, end of \S 4.2:} For later use, let me add another lemma here:

\begin{statement}
\textbf{Lemma 4.2.4.} Let $x,y\in W$ and $t\in R$ satisfy $l\left(  xt\right)
>l\left(  x\right)  $ and $l\left(  ty\right)  >l\left(  y\right)  $. Then,
$l\left(  xty\right)  >l\left(  xy\right)  $.
\end{statement}

\textit{Proof of Lemma 4.2.4.} Assume the contrary. Thus, $l\left(
xty\right)  \leq l\left(  xy\right)  $, so that $l\left(  xty\right)
<l\left(  xy\right)  $ (since Remark 3.2.18 ensures that $l\left(  xty\right)
$ and $l\left(  xy\right)  $ have different parities (because $l\left(
t\right)  $ is odd), and thus cannot be equal).

Set $r:=xtx^{-1}\in R$. Thus, $rx=xt$, so that $l\left(  rxy\right)  =l\left(
xty\right)  <l\left(  xy\right)  $. Hence, Lemma 4.2.2 (applied to $w=xy$)
yields that
\[
r\in R\left(  1,xy\right)  \subseteq R\left(  1,x\right)  \cup R\left(
x,xy\right)
\]
(since Lemma 4.2.1 (ii) shows that $R\left(  1,xy\right)  $ is the symmetric
difference of $R\left(  1,x\right)  $ and $R\left(  x,xy\right)  $). But
$rx=xt$ also leads to $l\left(  rx\right)  =l\left(  xt\right)  >l\left(
x\right)  $, so that we don't have $l\left(  x\right)  >l\left(  rx\right)  $.
Thus, $r\notin R\left(  1,x\right)  $ (since Lemma 4.2.2 shows that $r\in
R\left(  1,x\right)  $ if and only if $l\left(  x\right)  >l\left(  rx\right)
$). Combining this with $r\in R\left(  1,x\right)  \cup R\left(  x,xy\right)
$, we obtain $r\in R\left(  x,xy\right)  =xR\left(  1,y\right)  x^{-1}$ (by
Lemma 4.2.1 (i)). In view of $r=xtx^{-1}$, we can rewrite this as $xtx^{-1}\in
xR\left(  1,y\right)  x^{-1}$. In other words, $t\in R\left(  1,y\right)  $.
By Lemma 4.2.2 (applied to $t$ and $y$ instead of $r$ and $w$), this means
that $l\left(  y\right)  >l\left(  ty\right)  $. But this contradicts
$l\left(  ty\right)  >l\left(  y\right)  $. This contradiction completes our
proof of Lemma 4.2.4.

\item \textbf{Page 44, Lemma 4.3.1:} Remove the comma after \textquotedblleft
in the double coset\textquotedblright\ and the comma after \textquotedblleft
written in the form\textquotedblright.

\item \textbf{Page 44, proof of Lemma 4.3.1:} \textquotedblleft Hence, one of
the deleted letters must occur in $\mathbf{s}$ and the other in $\mathbf{s}%
^{\prime}$.\textquotedblright: This is true for the first deletion step, but
might not be literally true for the following ones, since the sequences
$\mathbf{s}$ and $\mathbf{s}^{\prime}$ may lose their reducedness after the
first step (or at least I don't see why they necessarily remain reduced).
However, fortunately, you don't need this claim in its full force; all you
need is that no letters are ever deleted from $\mathbf{u}$.

\item \textbf{Page 44, proof of Lemma 4.3.1:} Replace \textquotedblleft%
$W_{T}^{\prime}$\textquotedblright\ by \textquotedblleft$W_{T^{\prime}}%
$\textquotedblright\ (on the third-to-last line of the proof).

\item \textbf{Page 44, Definition 4.3.2:} Replace \textquotedblleft%
$W_{T}^{\prime}$\textquotedblright\ by \textquotedblleft$W_{T^{\prime}}%
$\textquotedblright.

\item \textbf{Page 44, after Definition 4.3.2:} It might be worth reminding
the reader that both left and right cosets are instance of double cosets:
$W_{T}a=W_{T}aW_{\varnothing}$ and $aW_{T}=W_{\varnothing}aW_{T}$ for any
$a\in W$.

\item \textbf{Page 44, proof of Lemma 4.3.3:} The equalities \textquotedblleft%
$l\left(  tw\right)  =l\left(  w\right)  +l\left(  t\right)  $%
\textquotedblright\ and \textquotedblleft$l\left(  tw\right)  =l\left(
ta\right)  +l\left(  u\right)  $\textquotedblright\ are somewhat unjustified,
seeing that the equality $l\left(  w^{\prime}\right)  =l\left(  a\right)
+l\left(  w\right)  +l\left(  a^{\prime}\right)  $ in Lemma 4.3.1 is stated
only for \textbf{one} factorization $w^{\prime}=awa^{\prime}$ of $w^{\prime}$,
not for \textbf{every} such factorization.

However, this can be fixed: At the end of Lemma 4.3.3 (i), insert the
additional claim \textquotedblleft Moreover, $l\left(  w^{\prime}\right)
=l\left(  a\right)  +l\left(  w\right)  $.\textquotedblright. This additional
claim follows from Lemma 4.3.1. But now, the equalities \textquotedblleft%
$l\left(  tw\right)  =l\left(  w\right)  +l\left(  t\right)  $%
\textquotedblright\ and \textquotedblleft$l\left(  tw\right)  =l\left(
ta\right)  +l\left(  u\right)  $\textquotedblright\ in the proof of Lemma
4.3.3 (ii) follow from this additional claim (because the factorization
$w^{\prime}=aw$ in Lemma 4.3.3 (i) is unique, and thus the additional claim
holds for every such factorization).

\item \textbf{Page 45, proof of Lemma 4.3.3:} \textquotedblleft an element
$T$\textquotedblright\ should be \textquotedblleft an element of $T$
satisfying $l\left(  at\right)  <l\left(  a\right)  $\textquotedblright.

\item \textbf{Page 45, after the proof of Lemma 4.3.3:} For later use, let me
state two corollaries of Lemma 4.3.3 and Lemma 4.3.1 here:

\begin{statement}
\textbf{Corollary 4.3.4.} Suppose $T\subset S$.

\textbf{(a)} If $m\in W$ is $\left(  T,\varnothing\right)  $-reduced, then
$l\left(  um\right)  =l\left(  u\right)  +l\left(  m\right)  $ for each $u\in
W_{T}$.

\textbf{(b)} If $m\in W$ is $\left(  \varnothing,T\right)  $-reduced, then
$l\left(  mu\right)  =l\left(  u\right)  +l\left(  m\right)  $ for each $u\in
W_{T}$.

\textbf{(c)} Suppose $U\subset S$. If $m\in W$ is simultaneously $\left(
U,\varnothing\right)  $-reduced and $\left(  \varnothing,T\right)  $-reduced,
then $l\left(  m^{-1}um\right)  =l\left(  u\right)  $ for each $u\in W_{U}\cap
mW_{T}m^{-1}$.
\end{statement}

\textit{Proof of Corollary 4.3.4.} \textbf{(a)} Let $m\in W$ be $\left(
T,\varnothing\right)  $-reduced. Thus, $m$ is the unique shortest element in
the double coset $W_{T}mW_{\varnothing}=W_{T}m$. Hence, $m$ is the only
$\left(  T,\varnothing\right)  $-reduced element in the coset $W_{T}m$.

Let $u\in W_{T}$. Lemma 4.3.3 (i) (applied to $w^{\prime}=um$) shows that $um$
can be written uniquely in the form $um=aw$, where $a\in W_{T}$ and $w$ is
$\left(  T,\varnothing\right)  $-reduced, and moreover, that we have $l\left(
um\right)  =l\left(  a\right)  +l\left(  w\right)  $ (this follows from the
\textquotedblleft Moreover\textquotedblright\ claim that I have inserted into
Lemma 4.3.3 (i) in one of my comments above). But this factorization $um=aw$
must satisfy $w=m$ (since%
\[
w=\underbrace{a^{-1}}_{\substack{\in W_{T}\\\text{(since }a\in W_{T}\text{)}%
}}\underbrace{aw}_{=um}\in\underbrace{W_{T}u}_{\substack{=W_{T}\\\text{(since
}u\in W_{T}\text{)}}}m=W_{T}m
\]
shows that $w$ is a $\left(  T,\varnothing\right)  $-reduced element in the
coset $W_{T}m$; but we know that $m$ is the only $\left(  T,\varnothing
\right)  $-reduced element in the coset $W_{T}m$) and thus $a=u$ (since
$um=a\underbrace{w}_{=m}=am$ and thus $u=a$). Hence, the equality $l\left(
um\right)  =l\left(  a\right)  +l\left(  w\right)  $ shown above can be
rewritten as $l\left(  um\right)  =l\left(  u\right)  +l\left(  m\right)  $.
This proves part (a).

\textbf{(b)} This is analogous to part (a) using left instead of right cosets.
(Or apply part (a) to $m^{-1}$ and $u^{-1}$ instead of $m$ and $u$, and recall
that taking inverses in $W$ turns left cosets into right cosets and vice versa.)

\textbf{(c)} Let $m\in W$ be simultaneously $\left(  U,\varnothing\right)
$-reduced and $\left(  \varnothing,T\right)  $-reduced. Let $u\in W_{U}\cap
mW_{T}m^{-1}$. Thus, $u\in W_{U}$ and $u\in mW_{T}m^{-1}$, so that
$m^{-1}um\in W_{T}$. Thus, part (b) (applied to $m^{-1}um$ instead of $u$)
shows that $l\left(  mm^{-1}um\right)  =l\left(  m^{-1}um\right)  +l\left(
m\right)  $, so that
\[
l\left(  m^{-1}um\right)  +l\left(  m\right)  =l\left(  \underbrace{mm^{-1}%
um}_{=um}\right)  =l\left(  um\right)  =l\left(  u\right)  +l\left(  m\right)
\]
(by part (a), applied to $U$ instead of $T$). Subtracting $l\left(  m\right)
$ from this, we obtain $l\left(  m^{-1}um\right)  =l\left(  u\right)  $. This
proves part (c).

\begin{statement}
\textbf{Corollary 4.3.5.} Let $T\subset S$ and $T^{\prime}\subset S$ and $w\in
S$.

\textbf{(a)} The element $w$ is $\left(  T,T^{\prime}\right)  $-reduced if and
only if each $s\in T$ satisfies $l\left(  sw\right)  >l\left(  w\right)  $ and
each $s\in T^{\prime}$ satisfies $l\left(  ws\right)  >l\left(  w\right)  $.

\textbf{(b)} The element $w$ is $\left(  T,\varnothing\right)  $-reduced if
and only if each $s\in T$ satisfies $l\left(  sw\right)  >l\left(  w\right)  $.

\textbf{(c)} The element $w$ is $\left(  \varnothing,T^{\prime}\right)
$-reduced if and only if each $s\in T^{\prime}$ satisfies $l\left(  ws\right)
>l\left(  w\right)  $.

\textbf{(d)} The element $w$ is $\left(  T,T^{\prime}\right)  $-reduced if and
only if it is both $\left(  T,\varnothing\right)  $-reduced and $\left(
\varnothing,T^{\prime}\right)  $-reduced.
\end{statement}

\textit{Proof of Corollary 4.3.5.} \textbf{(a)} The \textquotedblleft only
if\textquotedblright\ direction is clear (since both $sw$ and $ws$ belong to
the double coset $W_{T}wW_{T^{\prime}}$ and are distinct from $w$). It remains
to prove the \textquotedblleft if\textquotedblright\ direction. So let us
assume that each $s\in T$ satisfies $l\left(  sw\right)  >l\left(  w\right)  $
and each $s\in T^{\prime}$ satisfies $l\left(  ws\right)  >l\left(  w\right)
$. We must show that $w$ is $\left(  T,T^{\prime}\right)  $-reduced.

Assume the contrary. Let $m$ be the unique element of minimum length in the
double coset $W_{T}wW_{T^{\prime}}$ (this is unique by Lemma 4.3.1). Then, $m$
is $\left(  T,T^{\prime}\right)  $-reduced. Thus, $w\neq m$ (since we assumed
that $w$ is not $\left(  T,T^{\prime}\right)  $-reduced). But we have $m\in
W_{T}wW_{T^{\prime}}$ and thus $W_{T}mW_{T^{\prime}}=W_{T}wW_{T^{\prime}}$, so
that $w\in W_{T}mW_{T^{\prime}}$. Hence, Lemma 4.3.1 (applied to $m$ and $w$
instead of $w$ and $w^{\prime}$) yields that $w$ can be written in the form
$w=ama^{\prime}$ with $a\in W_{T}$ and $a^{\prime}\in W_{T^{\prime}}$ and
$l\left(  w\right)  =l\left(  a\right)  +l\left(  m\right)  +l\left(
a^{\prime}\right)  $. Consider these $a$ and $a^{\prime}$.

We must have $a\neq1$ or $a^{\prime}\neq1$ (since otherwise, $w=ama^{\prime}$
would rewrite as $w=m$, which would contradict $w\neq m$). We WLOG assume that
$a\neq1$ (since the case $a^{\prime}\neq1$ is analogous). Thus, $a$ has a
reduced expression $\left(  s_{1},s_{2},\ldots,s_{k}\right)  $ with $k>0$ and
$s_{1},s_{2},\ldots,s_{k}\in T$ (since $a\in W_{T}$). Consider this reduced
expression. Then, $a=s_{1}s_{2}\cdots s_{k}$, so that $s_{1}a=s_{2}s_{3}\cdots
s_{k}$ and therefore $l\left(  s_{1}a\right)  \leq k-1=l\left(  a\right)  -1$.
Furthermore, from $w=ama^{\prime}$, we obtain%
\begin{align*}
&  l\left(  s_{1}w\right)  \\
&  =l\left(  s_{1}ama^{\prime}\right)  \\
&  \leq\underbrace{l\left(  s_{1}a\right)  }_{\leq l\left(  a\right)
-1}+\,l\left(  m\right)  +l\left(  a^{\prime}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }l\left(  uv\right)  \leq l\left(
u\right)  +l\left(  v\right)  \text{ for all }u,v\in W\right)  \\
&  \leq\underbrace{l\left(  a\right)  +l\left(  m\right)  +l\left(  a^{\prime
}\right)  }_{=l\left(  w\right)  }-\,1=l\left(  w\right)  -1<l\left(
w\right)  .
\end{align*}
But we have assumed that each $s\in T$ satisfies $l\left(  sw\right)
>l\left(  w\right)  $. Hence, $l\left(  s_{1}w\right)  >l\left(  w\right)  $
(since $s_{1}\in T$), contradicting $l\left(  s_{1}w\right)  <l\left(
w\right)  $. This contradiction shows that our assumption is wrong, at least
in the case when $a\neq1$. As we said, the case $a^{\prime}\neq1$ is analogous
(but now we must fix a reduced expression $\left(  s_{1},s_{2},\ldots
,s_{k}\right)  $ of $a^{\prime}$ and argue that $l\left(  a^{\prime}%
s_{k}\right)  <l\left(  a^{\prime}\right)  -1$ and $l\left(  ws_{k}\right)
<l\left(  w\right)  $, which contradicts the assumption that each $s\in
T^{\prime}$ satisfies $l\left(  ws\right)  >l\left(  w\right)  $). Hence, we
always have a contradiction, and this completes the proof of Corollary 4.3.6 (a).

\textbf{(b)} This follows from part (a), applied to $T^{\prime}=\varnothing$.

\textbf{(c)} This follows from part (a), applied to $T=\varnothing$.

\textbf{(d)} This follows from parts (a), (b) and (c).

\item \textbf{Page 45, (4.3):} A \textquotedblleft$\}$\textquotedblright%
\ bracket is missing after \textquotedblleft$l\left(  sw\right)  >l\left(
w\right)  $\textquotedblright.

\item \textbf{Page 45, Remark 4.4.1:} After \textquotedblleft Then
$r=wsw^{-1}$ for some $s\in S$\textquotedblright, add \textquotedblleft and
$w\in W$\textquotedblright.

\item \textbf{Page 45, Remark 4.4.1:} It would be helpful to refer to Lemma
4.2.2 and Proposition 3.3.6 for the reasons why these claims are true.

\item \textbf{Page 45:} \textquotedblleft the \textit{initial} and
\textit{final} elements of the $\gamma$\textquotedblright\ should be
\textquotedblleft the \textit{initial} and \textit{final} elements of the
gallery $\gamma$\textquotedblright.

\item \textbf{Page 46, proof of Proposition 4.4.2:} In case (a), replace the
equality \textquotedblleft$sw=w_{1}\cdots w_{i-1}w_{i+1}\cdots w_{k}%
$\textquotedblright\ by \textquotedblleft$sw=s_{1}\cdots s_{i-1}s_{i+1}\cdots
s_{k}$\textquotedblright.

\item \textbf{Page 46, proof of Proposition 4.4.2:} In case (b), after
\textquotedblleft$l\left(  sw\right)  =l\left(  w^{\prime}\right)  >l\left(
sw^{\prime}\right)  =l\left(  w\right)  $\textquotedblright, I would add
\textquotedblleft and thus $l\left(  sw\right)  =l\left(  w\right)
+1$\textquotedblright.

\item \textbf{Page 46, proof of Lemma 4.5.3:} This is correct, but it's
missing the most important part: What is the relation between these properties
of $R$'s and minimal galleries? What you are tacitly using here is the fact
that a gallery is minimal if and only if it intersects each wall $\Omega^{r}$
at most once. (This is easy to show: If the gallery starts at the vertex $1$,
then this is a restatement of the \textquotedblleft if and only
if\textquotedblright\ part of Lemma 3.2.14. In the general case, we can apply
$w_{0}^{-1}$ to this gallery to reduce it to the previous case, where $w_{0}$
is the first vertex of the gallery.) This fact is used so often that it should
probably be stated explicitly.

\item \textbf{Page 47, proof of Lemma 4.5.7:} \textquotedblleft chamber in
$U$\textquotedblright\ should be \textquotedblleft element of $U$%
\textquotedblright. (Chambers have not been defined yet.)

\item \textbf{Page 47, proof of Lemma 4.5.7:} I don't understand why
\textquotedblleft The first wall $\Omega^{r}$ that it crosses must separate
$v$ from $U$\textquotedblright.

Here is how I would prove $D=U$ instead (sadly my proof is much more involved):

Assume the contrary. Since $U\subset D$ is clear (by definition of $D$), we
thus must have $D\not \subset U$. Hence, there exists some $d\in D-U$. Pick
such a $d$. There clearly exists a gallery from $d$ to some vertex in $U$
(here, we need $U\neq\varnothing$; but we can easily assume this to be WLOG
this case, since the case $U=\varnothing$ is easy to handle\footnote{To be
very pedantic, Lemma 4.5.7 does not hold in the very degenerate case when $W$
is the trivial group and the subset is empty, since there are no half-spaces
to intersect. But we leave this silly case aside.}). Let $\gamma$ be such a
gallery \textbf{of minimum length}. This gallery has at least two vertices
(since it starts at $d\notin U$ but ends at a vertex in $U$). Let $u$ be the
last vertex and $v$ the second-to-last vertex of this gallery. Then, $v\notin
U$ (otherwise, we could get an even shorter gallery from $d$ to some vertex in
$U$ by removing the last vertex from $\gamma$) but $u\in U$ (by definition of
$\gamma$). The graph $\Omega$ has an edge $\left\{  u,v\right\}  $ (since $v$
and $u$ are the last two vertices of the gallery $\gamma$). Also, $u\in
U\subset D$.

Note that $\gamma$ is a minimal gallery (since it has minimum length). Hence,
any convex set that contains its initial element $d$ and its final element $u$
must also contain all its intermediate elements. In particular, any such set
must contain $v$. But $D$ is such a set (indeed, $D$ is an intersection of
half-spaces, hence convex, and furthermore contains $d$ and $u$). Hence, we
conclude that $v\in D$.

Let $r\in R$ be the reflection that flips the edge $\left\{  u,v\right\}  $ of
$\Omega$. Then, the vertices $u$ and $v$ belong to different half-spaces of
the wall $\Omega^{r}$. Let us call these half-spaces $H_{u}$ and $H_{v}$, so
that $u\in H_{u}$ and $v\in H_{v}$ and $W=H_{u}\sqcup H_{v}$. If the halfspace
$H_{u}$ contained $U$ as a subset, then we would have $D\subset H_{u}$ (since
$D$ is the intersection of all half-spaces that contain $U$ as a subset) and
thus $v\in D\subset H_{u}$, which would contradict $v\in H_{v}$ (since $H_{u}$
and $H_{v}$ are disjoint). Hence, $H_{u}$ does not contain $U$ as a subset. In
other words, there exists some $w\in U-H_{u}$. Consider this $w$. Then,
$w\notin H_{u}$. Combined with $u\in H_{u}$, this shows that the vertices $w$
and $u$ lie on different sides of the wall $\Omega^{r}$.

Now, let $l\left(  \delta\right)  $ denote the number of edges in a gallery
$\delta$; we call this the \emph{length} of $\delta$. The graph $\Omega$ is
bipartite (since Remark 3.2.18 constructs a function $\varepsilon
:W\rightarrow\left\{  \pm1\right\}  $ such that any two adjacent vertices
$p,q$ of $\Omega$ satisfy $\varepsilon\left(  p\right)  =-\varepsilon\left(
q\right)  $). Hence, every closed gallery (i.e., closed edge path of $\Omega$)
has even length.

Let $\alpha$ be a minimal gallery from $u$ to $w$, and let $\beta$ be a
minimal gallery from $w$ to $v$. Concatenating these two galleries $\alpha$
and $\beta$ as well as the edge $\left\{  u,v\right\}  $, we obtain a closed
gallery from $u$ to $u$ having length $l\left(  \alpha\right)  +l\left(
\beta\right)  +1$. Since every closed gallery has even length, this entails
that $l\left(  \alpha\right)  +l\left(  \beta\right)  +1$ is even. Hence,
$l\left(  \alpha\right)  \not \equiv l\left(  \beta\right)  \operatorname{mod}%
2$.

However, we can obtain an edge path $\alpha^{\prime}$ from $u$ to $w$ by
concatenating the edge $\left\{  u,v\right\}  $ with the gallery $\beta$ (from
$w$ to $v$) walked in reverse. This edge path $\alpha^{\prime}$ has its
initial element $u$ belong to $U$ and its final element $w$ belong to $U$ as
well (since $w\in U-H_{u}$), but its intermediate element $v$ does not belong
to $U$ (since $v\notin U$). Since $U$ is a convex subset, this entails that
this edge path $\alpha^{\prime}$ cannot be a minimal gallery (because a
minimal gallery starting and ending inside the convex subset $U$ would force
all its intermediate elements to belong to $U$ as well). Thus, its length
$l\left(  \alpha^{\prime}\right)  =1+l\left(  \beta\right)  $ must be
\textbf{larger} than the length of a minimal gallery from $u$ to $w$, which is
$l\left(  \alpha\right)  $. That is, $1+l\left(  \beta\right)  >l\left(
\alpha\right)  $. In other words, $l\left(  \beta\right)  \geq l\left(
\alpha\right)  $.

On the other hand, we can obtain an edge path $\beta^{\prime}$ from $w$ to $v$
by concatenating the gallery $\alpha$ (from $u$ to $w$) walked in reverse with
the edge $\left\{  u,v\right\}  $. This edge path $\beta^{\prime}$ crosses the
wall $\Omega^{r}$ twice (since the gallery $\alpha$ crosses this wall (because
the vertices $w$ and $u$ lie on different sides of the wall $\Omega^{r}$), but
the edge $\left\{  u,v\right\}  $ also crosses this wall); hence it cannot be
a minimal gallery (since a minimal gallery cannot cross any wall more than
once). Thus, its length $l\left(  \beta^{\prime}\right)  =l\left(
\alpha\right)  +1$ must be \textbf{larger} than the length of a minimal
gallery from $w$ to $v$, which is $l\left(  \beta\right)  $. That is,
$l\left(  \alpha\right)  +1>l\left(  \beta\right)  $. In other words,
$l\left(  \alpha\right)  \geq l\left(  \beta\right)  $.

Combining the inequalities $l\left(  \alpha\right)  \geq l\left(
\beta\right)  $ and $l\left(  \beta\right)  \geq l\left(  \alpha\right)  $, we
obtain $l\left(  \alpha\right)  =l\left(  \beta\right)  $. But this
contradicts $l\left(  \alpha\right)  \not \equiv l\left(  \beta\right)
\operatorname{mod}2$. This contradiction completes our proof.

\item \textbf{Page 47, Lemma 4.5.9:} I don't understand how this follows from
Example 4.5.8. However, this is a particular case of the claim that
\textquotedblleft$D$ is the intersection of the $A_{r}\left(  D\right)
$\textquotedblright\ made in \S 4.9 (on page 53), obtained by letting $X=T$
(so that $R^{\prime}=R\cap W_{T}$ and $Z\left(  X\right)  =\Omega\left(
T\right)  $) and letting $D$ be the chamber (i.e., connected component of
$Z\left(  X\right)  =\Omega\left(  T\right)  $) that contains the identity
vertex. (Indeed, the $A_{r}\left(  D\right)  $ in this case are precisely the
$A_{s}$ for $s\in T$, and thus their intersection is the sector $A_{T}$.)

\item \textbf{Page 47, proof of Proposition 4.5.10:} In \textquotedblleft By
Lemma 4.2.3, conjugation by $w$ maps $R\cap W_{T}$ to $R\cap W_{U}%
$\textquotedblright, I don't see any need for Lemma 4.2.3.

\item \textbf{Page 47, proof of Proposition 4.5.10:} \textquotedblleft hence,
it takes\textquotedblright\ should be \textquotedblleft hence, $w$
takes\textquotedblright\ (the action of $w$ on $\Omega$ is not by conjugation).

\item \textbf{Pages 47--48, proof of Proposition 4.5.10:} Here is a simpler
proof of Proposition 4.5.10:

\textit{Proof of Proposition 4.5.10.} By assumption, $W_{U}=wW_{T}w^{-1}$ for
some $w\in W$. Consider this $w$. Thus, $W_{U}w=wW_{T}$. Hence, the set
$W_{U}w=wW_{T}$ is simultaneously a left coset, a right coset and a double
coset. Thus, by Lemma 4.3.1, it has a unique element $m$ of minimum length.
This element $m$ is both $\left(  U,\varnothing\right)  $-reduced (since it
has minimum length in $W_{U}w=W_{U}wW_{\varnothing}$) and $\left(
\varnothing,T\right)  $-reduced (since it has minimum length in $wW_{T}%
=W_{\varnothing}wW_{T}$).

From $m\in wW_{T}$, we obtain $mW_{T}=wW_{T}=W_{U}w=W_{U}m$ (since $m\in
W_{U}w$), so that $mW_{T}m^{-1}=W_{U}$. Thus, $W_{U}\cap mW_{T}m^{-1}%
=W_{U}\cap W_{U}=W_{U}$.

Now, let $s\in T$. Then, $s\in T\subset W_{T}$, so that $ms\in mW_{T}=W_{U}m$.
In other words, $ms=um$ for some $u\in W_{U}$. Consider this $u$. From
$ms=um$, we obtain $m^{-1}um=s$ and $u=msm^{-1}$.

Since $m$ is simultaneously $\left(  U,\varnothing\right)  $-reduced and
$\left(  \varnothing,T\right)  $-reduced, and since $u\in W_{U}=W_{U}\cap
mW_{T}m^{-1}$, we obtain $l\left(  m^{-1}um\right)  =l\left(  u\right)  $ by
Corollary 4.3.4 (c). In view of $m^{-1}um=s$, we can rewrite this as $l\left(
s\right)  =l\left(  u\right)  $. Hence, $l\left(  u\right)  =l\left(
s\right)  =1$ (since $s\in T\subset S$). Therefore, $u\in S$, so that $u\in
W_{U}\cap S=U$ (by Corollary 4.1.3). In view of $u=msm^{-1}$, we can rewrite
this as $msm^{-1}\in U$.

Forget that we fixed $s$. We thus have shown that $msm^{-1}\in U$ for each
$s\in T$. In other words, $mTm^{-1}\subset U$. Similarly, we can see that
$m^{-1}Um\subset T$, so that $U\subset mTm^{-1}$. Combining this with
$mTm^{-1}\subset U$, we find $mTm^{-1}=U$. This proves Proposition 4.5.10.

\item \textbf{Page 48, \S 4.6:} Why do you call $w_{0}$ the \textquotedblleft%
\textit{element of longest length}\textquotedblright\ and not just the
\textquotedblleft\textit{longest element}\textquotedblright? After all, you
use the latter notation for maximum-length elements of cosets.

\item \textbf{Page 49, Lemma 4.6.1:} This contains all the basic properties of
$w_{0}$.\ Should it really be just a \textquotedblleft Lemma\textquotedblright?

\item \textbf{Page 49, proof of Lemma 4.6.1:} Remove the comma after
\textquotedblleft Rewriting (a) as\textquotedblright.

\item \textbf{Page 49, proof of Lemma 4.6.2:} Remove the comma after
\textquotedblleft to obtain\textquotedblright.

\item \textbf{Page 49, proof of Lemma 4.6.2:} When you say \textquotedblleft
Hence, condition (a) of Lemma 4.6.1 holds\textquotedblright, you are applying
your (previously proved) claim to a sequence $\mathbf{s}=\left(  s_{1}%
,s_{2},\ldots,s_{l\left(  u\right)  }\right)  $ that is a reduced expression
of $u\in W$ (this claim is applicable here, since any initial subsequence of a
reduced expression is again a reduced expression, and this property is
preserved if you write it in reverse order). The claim then shows that $w_{0}$
has a reduced expression beginning with $\mathbf{s}$ (and thus ending with a
reduced expression of $u^{-1}w_{0}$); this entails $l\left(  w_{0}\right)
=l\left(  u\right)  +l\left(  u^{-1}w_{0}\right)  $, hence showing that
condition (a) of Lemma 4.6.1 holds indeed.

\item \textbf{Page 50, proofs of Lemma 4.7.2 and of Lemma 4.7.3:} I struggle
to understand the proofs of Lemma 4.7.2 and of Lemma 4.7.3 (i), so let me
rewrite them for my own clarity.

Let $w\in W$. We shall show that

\begin{enumerate}
\item[\textbf{(i)}] the subgroup $W_{\operatorname*{In}\left(  w\right)  }$ is finite;

\item[\textbf{(ii)}] we have $l\left(  wu\right)  =l\left(  w\right)
-l\left(  u\right)  $ for each $u\in W_{\operatorname*{In}\left(  w\right)  }$;

\item[\textbf{(iii)}] the element $w$ is the longest element of
$wW_{\operatorname*{In}\left(  w\right)  }$.
\end{enumerate}

We begin by setting $T:=\operatorname*{In}\left(  w\right)  \subset S$. Let
$m$ be the shortest element of the left coset $wW_{T}$ (this is unique by
Definition 4.3.2, since $wW_{T}$ is the double coset $W_{\varnothing}wW_{T}$).
In other words, $m$ is the unique $\left(  \varnothing,T\right)  $-reduced
element in this left coset $wW_{T}$. Of course, $m^{-1}w\in W_{T}$ because
$m\in wW_{T}$. Hence, Corollary 4.3.4 (b)\footnote{This corollary was stated
in one of my comments above.} (applied to $u=m^{-1}w$) yields%
\[
l\left(  mm^{-1}w\right)  =l\left(  m^{-1}w\right)  +l\left(  m\right)  .
\]
In other words,
\begin{equation}
l\left(  w\right)  =l\left(  m^{-1}w\right)  +l\left(  m\right)
\label{eq.50.2}%
\end{equation}
(since $mm^{-1}w=w$). However, if $s\in T$ is arbitrary, then we have
$m^{-1}ws\in W_{T}$ (since $m^{-1}w\in W_{T}$ and $s\in T\subset W_{T}$) and%
\[
l\left(  \underbrace{ws}_{=mm^{-1}ws}\right)  =l\left(  mm^{-1}ws\right)
=l\left(  m^{-1}ws\right)  +l\left(  m\right)
\]
(by Corollary 4.3.4 (b), applied to $u=m^{-1}ws$ because $m^{-1}ws\in W_{T}$)
and thus%
\[
l\left(  m^{-1}ws\right)  =\underbrace{l\left(  ws\right)  }%
_{\substack{<l\left(  w\right)  \\\text{(since }s\in T=\operatorname*{In}%
\left(  w\right)  \text{)}}}-\,l\left(  m\right)  <l\left(  w\right)
-l\left(  m\right)  =l\left(  m^{-1}w\right)
\]
(by (\ref{eq.50.2})). Therefore, Lemma 4.6.2 (applied to $T$, $W_{T}$ and
$m^{-1}w$ instead of $S$, $W$ and $w_{0}$) shows that $W_{T}$ is finite and
that $m^{-1}w$ is its element of longest length (since $m^{-1}w\in W_{T}$). In
particular, $W_{T}$ is finite; this proves claim \textbf{(i)} (since
$T=\operatorname*{In}\left(  w\right)  $).

But we have also shown that $m^{-1}w$ is the element of longest length in
$W_{T}$. Thus, we can apply Lemma 4.6.1 to $T$, $W_{T}$ and $m^{-1}w$ instead
of $S$, $W$ and $w_{0}$. As a consequence, we see that $m^{-1}w$ is an
involution (by Lemma 4.6.1 (iii)), and that each $u\in W_{T}$ satisfies%
\begin{equation}
l\left(  m^{-1}w\right)  =l\left(  u\right)  +l\left(  u^{-1}m^{-1}w\right)
\label{eq.50.3}%
\end{equation}
(by Lemma 4.6.1 (a)). Note that $\left(  m^{-1}w\right)  ^{-1}=m^{-1}w$ (since
$m^{-1}w$ is an involution).

Now, for each $u\in W_{T}$, we have%
\begin{align*}
l\left(  wu\right)   &  =l\left(  mm^{-1}wu\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }wu=mm^{-1}wu\right) \\
&  =\underbrace{l\left(  m^{-1}wu\right)  }_{\substack{=l\left(  \left(
m^{-1}wu\right)  ^{-1}\right)  \\\text{(since }l\left(  p\right)  =l\left(
p^{-1}\right)  \\\text{for any }p\in W\text{)}}}+\,l\left(  m\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Corollary 4.3.4 (b), applied to }u=m^{-1}wu\\
\text{(since }m^{-1}w\in W_{T}\text{ and }u\in W_{T}\text{ entail }m^{-1}wu\in
W_{T}\text{)}%
\end{array}
\right) \\
&  =l\left(  \left(  m^{-1}wu\right)  ^{-1}\right)  +l\left(  m\right)
=\underbrace{l\left(  u^{-1}m^{-1}w\right)  }_{\substack{=l\left(
m^{-1}w\right)  -l\left(  u\right)  \\\text{(by (\ref{eq.50.3}))}}}+\,l\left(
m\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
m^{-1}wu\right)  ^{-1}=u^{-1}\underbrace{\left(  m^{-1}w\right)  ^{-1}%
}_{\substack{=m^{-1}w}}=u^{-1}m^{-1}w\right) \\
&  =l\left(  m^{-1}w\right)  -l\left(  u\right)  +l\left(  m\right)
=\underbrace{l\left(  m^{-1}w\right)  +l\left(  m\right)  }%
_{\substack{=l\left(  w\right)  \\\text{(by (\ref{eq.50.2}))}}}-\,l\left(
u\right)  =l\left(  w\right)  -l\left(  u\right)  .
\end{align*}
This proves our claim \textbf{(ii)} (since $T=\operatorname*{In}\left(
w\right)  $).

Finally, claim \textbf{(iii)} follows from claim \textbf{(ii)}, because the
latter claim shows that each $u\in W_{\operatorname*{In}\left(  w\right)  }$
satisfies $l\left(  wu\right)  =l\left(  w\right)  -l\left(  u\right)  \leq
l\left(  w\right)  $.

\item \textbf{Page 50, Lemma 4.7.6:} The comma in \textquotedblleft Let $s\in
S-T$,\textquotedblright\ should be a period.

\item \textbf{Page 51, proof of Lemma 4.7.6:} \textquotedblleft By Lemma 4.6.1
(i)\textquotedblright\ should be \textquotedblleft By Lemma 4.7.3
(i)\textquotedblright.

\item \textbf{Page 51, proof of Lemma 4.7.5:} \textquotedblleft By Lemma 4.6.1
(ii)\textquotedblright\ should probably be \textquotedblleft By Lemma 4.6.1
(b)\textquotedblright.

\item \textbf{Page 51, Remark 4.8.1:} Why is $wB\not \subset sB_{s}$ ?

\item \textbf{Page 53, \S 4.9:} The claim that \textquotedblleft$D$ is the
intersection of the $A_{r}\left(  D\right)  $\textquotedblright\ is not
completely obvious, so let me prove it:

Clearly, $D\subset A_{r}\left(  D\right)  $ for each $r\in R^{\prime}$, since
$A_{r}\left(  D\right)  $ is the vertex set of a component of $\Omega
-\Omega^{r}$ that contains $D$. Thus, $D$ is a subset of the intersection of
the $A_{r}\left(  D\right)  $ over all $r\in R^{\prime}$. It remains to prove
the converse inclusion -- i.e., that the latter intersection is a subset of
$D$.

So let $u$ be an element of this intersection. We must show that $u\in D$.
Choose any $v\in D$ (this exists since $D$ is a path component and thus is
nonempty). Note that both $u$ and $v$ belong to the intersection of the
$A_{r}\left(  D\right)  $ over all $r\in R^{\prime}$ (since $D$ is a subset of
this intersection). In other words, both $u$ and $v$ belong to $A_{r}\left(
D\right)  $ for each $r\in R^{\prime}$.

Pick a minimal gallery $\gamma$ from $u$ to $v$. This minimal gallery $\gamma$
cannot cross any of the walls $\Omega^{r}$ for $r\in R^{\prime}$ (because if
it crossed such a wall, then it would put $u$ and $v$ on different sides of
this wall, i.e., in two different components of $\Omega-\Omega^{r}$; but this
would contradict the fact that both $u$ and $v$ lie in the same component
$A_{r}\left(  D\right)  $ of $\Omega-\Omega^{r}$, as we saw in the previous
paragraph). Hence, $\gamma$ is an edge path of the graph $\Omega-\bigcup_{r\in
R^{\prime}}\Omega^{r}=Z\left(  X\right)  $. Thus, $u$ and $v$ are connected by
a path of $Z\left(  X\right)  $. In other words, $u$ and $v$ belong to the
same chamber (since the chambers are the path components of $Z\left(
X\right)  $). Since $v$ belongs to the chamber $D$, this shows that $u$ must
belong to the chamber $D$ as well. In other words, $u\in D$, qed.

\item \textbf{Page 54, Theorem 4.9.2:} A closing parenthesis should be
inserted at the end of \textquotedblleft for $(W^{\prime},S^{\prime}\left(
D\right)  ,\left\{  A_{s}\left(  D\right)  \right\}  _{s\in S^{\prime}\left(
D\right)  }$\textquotedblright.

\item \textbf{Page 55:} \textquotedblleft Here [D] denotes the
vertex\textquotedblright\ $\rightarrow$ \textquotedblleft Here $\left[
D\right]  $ denotes the vertex\textquotedblright.

\item \textbf{Page 56, proof of Lemma 4.10.1:} Once again, I struggle to
understand this proof (why is it that \textquotedblleft For any $w\in W$,
$wA_{T}$ is a sector for $wW_{T}w^{-1}$\textquotedblright?), so let me write
down my own.

\textit{Proof of Lemma 4.10.1.} We begin with part (ii):

(ii) Let $w\in G_{T}$. Then, $wA_{T}=A_{T}$ by the definition of $G_{T}$.
Hence, $w=w\underbrace{1}_{\in A_{T}}\in wA_{T}=A_{T}$, so that $w$ is
$\left(  T,\varnothing\right)  $-reduced (since $A_{T}$ is the set of all
$\left(  T,\varnothing\right)  $-reduced elements of $W$). Furthermore,
$ww^{-1}=1\in A_{T}=wA_{T}$, thus $w^{-1}\in A_{T}$, so that $w^{-1}$ is
$\left(  T,\varnothing\right)  $-reduced. In other words, $w$ is $\left(
\varnothing,T\right)  $-reduced (since taking inverses of elements of $W$
preserves length but turns left cosets into right cosets and vice versa).
Since $w$ is both $\left(  T,\varnothing\right)  $-reduced and $\left(
\varnothing,T\right)  $-reduced, we conclude by Corollary 4.3.5 (d) (see one
of my other comments above) that $w$ is $\left(  T,T\right)  $-reduced. This
proves part (ii).

(iii) As you point out, this follows from (ii), since the only $\left(
T,T\right)  $-reduced (or even just $\left(  T,\varnothing\right)  $-reduced)
element of $W_{T}$ is $1$.

(i) We must prove that $mW_{T}m^{-1}=W_{T}$ for each $m\in G_{T}$. It suffices
to prove the inclusion $mW_{T}m^{-1}\subset W_{T}$ for each $m\in G_{T}$
(since $G_{T}$ is a subgroup, so that we can then apply this inclusion to
$m^{-1}$ instead of $m$ and obtain $m^{-1}W_{T}m\subset W_{T}$, that is,
$W_{T}\subset mW_{T}m^{-1}$, which we can then combine with $mW_{T}%
m^{-1}\subset W_{T}$ to obtain the desired equality $mW_{T}m^{-1}=W_{T}$).

So let $m\in G_{T}$. We must prove that $mW_{T}m^{-1}\subset W_{T}$. Since
conjugation by $m$ is a group automorphism, it will suffice to show that
$mtm^{-1}\in W_{T}$ for each $t\in T$. So let us do this.

Fix $t\in T$. By part (ii), the element $m\in G_{T}$ is $\left(  T,T\right)
$-reduced, thus both $\left(  T,\varnothing\right)  $-reduced and $\left(
\varnothing,T\right)  $-reduced.

We have $m\in G_{T}$, so that $mA_{T}=A_{T}$ (by the definition of $G_{T}$).
But $t\in T$ entails $W_{T}t=W_{T}$ and thus $t\notin A_{T}$ (since the coset
$W_{T}t=W_{T}$ contains the shorter element $1$). Thus, $mt\notin mA_{T}%
=A_{T}$. Hence, $mt$ is not $\left(  T,\varnothing\right)  $-reduced. Thus,
there exists some $s\in T$ such that $l\left(  smt\right)  \leq l\left(
mt\right)  $ (by Corollary 4.3.5 (b), applied to $w=mt$). Consider this $s$.
We have $l\left(  sm\right)  =l\left(  m\right)  +1$ (by Corollary 4.3.4 (a),
since $m$ is $\left(  T,\varnothing\right)  $-reduced and $s\in T\subset
W_{T}$) and $l\left(  mt\right)  =l\left(  m\right)  +1$ (by Corollary 4.3.4
(b), since $m$ is $\left(  \varnothing,T\right)  $-reduced and $t\in T\subset
W_{T}$). Hence, the Folding Condition (F) entails that either $l\left(
smt\right)  =l\left(  m\right)  +2$ or $smt=m$. Since $l\left(  smt\right)
\leq l\left(  mt\right)  =l\left(  m\right)  +1$, we cannot have $l\left(
smt\right)  =l\left(  m\right)  +2$, and thus we must have $smt=m$. Thus,
$sm=mt$, so that $mtm^{-1}=s\in T\subset W_{T}$, just as we wanted to prove.
Thus the proof of part (i) is complete.

(iv) Let $w\in N\left(  W_{T}\right)  $. Thus, $wW_{T}=W_{T}w$.

Let me set $U:=T$. Thus, the equality $wW_{T}=W_{T}w$ can be rewritten as
$wW_{T}=W_{U}w$. Hence, $W_{U}=wW_{T}w^{-1}$. This shows that we are in the
exact same situation as in my above proof of Proposition 4.5.10. As in this
latter proof, let $m$ be the unique element of minimum length in the coset
$W_{U}w=wW_{T}$. As was shown in the proof of Proposition 4.5.10, this element
$m$ is both $\left(  U,\varnothing\right)  $-reduced and $\left(
\varnothing,T\right)  $-reduced and satisfies $mTm^{-1}=U$. In particular, $m$
is $\left(  U,\varnothing\right)  $-reduced; in other words, $m\in A_{U}$.

Now we shall show that $mA_{T}\subset A_{U}$. For this purpose, let $a\in
A_{T}$. We want to show that $ma\in A_{U}$. In other words, we want to show
that $ma$ is $\left(  U,\varnothing\right)  $-reduced (since $A_{U}$ is the
set of all $\left(  U,\varnothing\right)  $-reduced elements of $W$). By
Corollary 4.3.5 (b), this only requires us to show that each $u\in U$
satisfies $l\left(  uma\right)  >l\left(  ma\right)  $. So let $u\in U$. Then,
$m^{-1}um\in m^{-1}Um=T$ (since $mTm^{-1}=U$). Set $t:=m^{-1}um$; thus,
$mt=um$ and $t=m^{-1}um\in T$. But $a$ is $\left(  T,\varnothing\right)
$-reduced (since $a\in A_{T}$), and thus we have $l\left(  ta\right)
>l\left(  a\right)  $ (by Corollary 4.3.5 (b), since $t\in T$). Furthermore,
$m$ is $\left(  \varnothing,T\right)  $-reduced; thus Corollary 4.3.5 (a)
yields $l\left(  mt\right)  >l\left(  m\right)  $ (since $t\in T$).

Thus, we know that $t\in T\subset S\subset R$ and $l\left(  mt\right)
>l\left(  m\right)  $ and $l\left(  ta\right)  >l\left(  a\right)  $. Hence,
Lemma 4.2.4 (applied to $x=m$ and $y=a$) yields $l\left(  mta\right)
>l\left(  ma\right)  $. In view of $mt=um$, this rewrites as $l\left(
uma\right)  >l\left(  ma\right)  $.

Forget that we fixed $u$. We thus have shown that each $u\in U$ satisfies
$l\left(  uma\right)  >l\left(  ma\right)  $. By Corollary 4.3.5 (b), this
means that $ma$ is $\left(  U,\varnothing\right)  $-reduced. In other words,
$ma\in A_{U}$.

Forget that we fixed $a$. We thus have shown that $ma\in A_{U}$ for each $a\in
A_{T}$. In other words, $mA_{T}\subset A_{U}$.

Recall that taking the inverse of an element of $W$ preserves its length, but
changes left cosets into right cosets. Thus, since $m$ is $\left(
U,\varnothing\right)  $-reduced, its inverse $m^{-1}$ is $\left(
\varnothing,U\right)  $-reduced. Moreover, from $mTm^{-1}=U$, we obtain
$m^{-1}U\left(  m^{-1}\right)  ^{-1}=T$. Thus, just as we have shown that
$mA_{T}\subset A_{U}$, we can prove that $m^{-1}A_{U}\subset A_{T}$ (we just
need to apply the same arguments to $m^{-1}$, $U$ and $T$ instead of $m$, $T$
and $U$). In other words, $A_{U}\subset mA_{T}$. Combining this with
$mA_{T}=A_{U}$.

But $m\in wW_{T}=W_{T}w$. Hence, there exists some $u\in W_{T}$ that satisfies
$w=um$ and therefore also satisfies $wA_{T}=u\underbrace{mA_{T}}_{=A_{U}%
}=uA_{U}=uA_{T}$ (since $U=T$). This proves the existence part of Lemma 4.10.1
(iv). It remains to prove the uniqueness part. But it is easy: If $u\in W_{T}$
satisfies $wA_{T}=uA_{T}$, then
\begin{align*}
w  &  =w1\in wA_{T}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }1\in A_{T}\right)
\\
&  =uA_{T},
\end{align*}
so that $w$ can be factored as $w=uq$ for some $q\in A_{T}$; but Lemma 4.3.3
(i) shows that such a factorization of $w$ is unique (since $q\in A_{T}$ is
just saying that $q$ is $\left(  T,\varnothing\right)  $-reduced), and thus
$u$ is unique as well. This proves the uniqueness of $u$, and thus finishes
the proof of Lemma 4.10.1 (iv).
\end{enumerate}


\end{document}