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\ihead{Errata to ``The Steinberg module and the Hecke algebra''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{The Steinberg module and the Hecke algebra}

\textit{Neil P. Strickland}

\url{https://neil-strickland.staff.shef.ac.uk/research/jordan.pdf}

version of 2 May 2012

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

The list below contains corrections and comments to the preprint
\textquotedblleft The Steinberg module and the Hecke algebra\textquotedblright%
\ by Neil P. Strickland. The comments include alternative proofs and
additional details (indeed, most of the comments below are of the latter kind,
and they are the main reason why this list is so large). I have written this
list while I was reading the preprint (over the course of several
months\footnote{The preprint packs a whole lot of material into just 15 pages.
Partly, I wish it would proceed more slowly and leave less work to the reader;
the below comments fill in lots of details that are omitted.}); since I am not
myself an expert in the subject, my comments are not always particularly
learned (I suspect that many of the proofs I am giving below can be
drastically simplified), and they are probably full of mistakes of their own.
(I have tried to be detailed, partly in order to avoid mistakes.)

I will refer to the results appearing in Strickland's preprint by the numbers
under which they appear in it (specifically, in its version of 2 May 2012,
available from \url{https://neil-strickland.staff.shef.ac.uk/research/jordan.pdf}).

\setcounter{section}{10}

\section*{Errata and addenda}

\begin{itemize}
\item \textbf{\S 2:} I think it would be better if you spent a bit of time
defining some of your notations:

\begin{itemize}
\item For any nonnegative integer $n$, you let $\Sigma_{n}$ denote the
symmetric group of the set $\left\{  1,2,\ldots,n\right\}  $. (This is not a
notation I have seen very often. Most combinatorialists call it either $S_{n}$
or $\mathfrak{S}_{n}$ or $\mathcal{S}_{n}$.)

\item The composition $\alpha\beta$ of two maps $\alpha:Y\rightarrow Z$ and
$\beta:X\rightarrow Y$ is defined as the map $X\rightarrow Z$ that sends each
$x\in X$ to $\alpha\left(  \beta\left(  x\right)  \right)  $. (This might
sound obvious, but irritatingly, a lot of people use the opposite convention
for the order of multiplication, particularly when permutations are concerned.)

\item If $\alpha:X\rightarrow Y$ is a map, then $\alpha_{\ast}$ means the map
$\mathcal{P}\left(  X\right)  \rightarrow\mathcal{P}\left(  Y\right)  $
canonically induced by $\alpha$ (where $\mathcal{P}\left(  Z\right)  $ denotes
the powerset of a set $Z$). This is the map that sends every subset $T$ of $X$
to the subset $\alpha\left(  T\right)  $ of $Y$.
\end{itemize}

\item \textbf{\S 2:} Do you ever use the notation $L^{+}\left(  \sigma\right)
$ that you define in the beginning of \S 2? (I don't know for sure; just asking.)

\item \textbf{Lemma 2.6:} Replace \textquotedblleft$\Sigma$\textquotedblright%
\ by \textquotedblleft$\Sigma_{n}$\textquotedblright.

\item \textbf{Lemma 2.6:} The period at the end of the sentence should be
outside of the parentheses.

\item \textbf{Proof of Proposition 2.11:} After \textquotedblleft are
disjoint, and\textquotedblright, add \textquotedblleft we have $\sigma
=t_{m_{n}}^{n}\tau$; thus, Lemma 2.6 (applied to $t_{m_{n}}^{n}$ instead of
$\sigma$) yields%
\[
\overline{L}\left(  \sigma\right)  =\overline{L}\left(  \tau\right)
\Delta\tau_{\ast}^{-1}\overline{L}\left(  t_{m_{n}}^{n}\right)  =\overline
{L}\left(  \tau\right)  \sqcup\tau_{\ast}^{-1}\overline{L}\left(  t_{m_{n}%
}^{n}\right)
\]
and therefore\textquotedblright.

\item \textbf{Proof of Proposition 2.11:} You have not proven the uniqueness
that is claimed in Proposition 2.11. This is not a large gap to fill, and
becomes obvious later on\footnote{Namely: In the proof of Proposition 2.13,
you show that the map $\epsilon:X_{n}\rightarrow\Sigma_{n}$ is surjective.
Since $\left\vert X_{n}\right\vert \leq n!=\left\vert \Sigma_{n}\right\vert $,
this entails that the map $\epsilon$ also is injective. But this means
precisely that no two distinct sequences $\left(  m_{1},m_{2},\ldots
,m_{n}\right)  $ with $1\leq m_{k}\leq k$ give rise to one and the same
permutation $t_{m_{n}}^{n}t_{m_{n-1}}^{n-1}\cdots t_{m_{2}}^{2}t_{m_{1}}^{1}$.
And this is exactly the uniqueness claim of Proposition 2.11.}; but I think it
is worth at least briefly mentioning how it is proven.

\item \textbf{Definition 2.12:} Replace \textquotedblleft there is a canonical
map\textquotedblright\ by \textquotedblleft there is a canonical
homomorphism\textquotedblright.

\item \textbf{Proof of Proposition 2.13:} You are slightly abusing notation
here: When you write \textquotedblleft$X_{n}=\bigcup_{m=1}^{n}t_{m}^{n}%
X_{n-1}$\textquotedblright, you are implicitly suggesting that
$\widetilde{\Sigma}_{n-1}$ can be embedded into $\widetilde{\Sigma}_{n}$. This
is correct, but is not obvious until Proposition 2.13 is already proven (at
which point it is not useful anymore). A-priori, it is plausible that some
nontrivial elements of $\widetilde{\Sigma}_{n-1}$ would collapse to the
identity upon adding the extra generator $s_{n-1}$ of $\widetilde{\Sigma}_{n}$
and the extra relations that come with it.

Fortunately, the proof is easy to fix, by introducing a group homomorphism
$\widetilde{\Sigma}_{n-1}\rightarrow\widetilde{\Sigma}_{n}$: Namely, observe
that all the generators and the relations appearing in the definition of
$\widetilde{\Sigma}_{n-1}$ also appear in the definition of $\widetilde{\Sigma
}_{n}$ (along with one new generator $s_{n-1}$ and some new relation). Thus,
there is a group homomorphism $\eta:\widetilde{\Sigma}_{n-1}\rightarrow
\widetilde{\Sigma}_{n}$ sending $s_{i}\mapsto s_{i}$ for each $i\in\left\{
1,2,\ldots,n-2\right\}  $. Consider this $\eta$. Regard $\widetilde{\Sigma
}_{n}$ as a right $\widetilde{\Sigma}_{n-1}$-set by having $\widetilde{\Sigma
}_{n-1}$ act through $\eta$ (that is, set $xy=x\eta\left(  y\right)  $ for all
$x\in\widetilde{\Sigma}_{n}$ and $y\in\widetilde{\Sigma}_{n-1}$). Then,
$X_{n}=\bigcup_{m=1}^{n}t_{m}^{n}X_{n-1}$ is still correct (where the implied
multiplication in $t_{m}^{n}X_{n-1}=\left\{  t_{m}^{n}x\ \mid\ x\in
X_{n-1}\right\}  $ is now to be understood as the $\widetilde{\Sigma}_{n-1}%
$-action on $\widetilde{\Sigma}_{n}$). All the rest of the proof goes through
unchanged, except for one simple modification (namely, \textquotedblleft%
$\widetilde{\Sigma}_{n}$ is generated by $\widetilde{\Sigma}_{n-1}$ and
$s_{n-1}$\textquotedblright\ must become \textquotedblleft$\widetilde{\Sigma
}_{n}$ is generated by $\eta\left(  \widetilde{\Sigma}_{n-1}\right)  $ and
$s_{n-1}$\textquotedblright).

\item \textbf{Proof of Lemma 2.14:} Replace \textquotedblleft identity
permutation\textquotedblright\ by \textquotedblleft identity of
$\widetilde{\Sigma}_{n}$\textquotedblright. (The identification between
permutations and elements $\widetilde{\Sigma}_{n}$ cannot yet be used at this point.)

\item \textbf{Definition 2.15:} Replace \textquotedblleft$s_{1},\ldots,s_{n}%
$\textquotedblright\ by \textquotedblleft$s_{1},\ldots,s_{n-1}$%
\textquotedblright\ twice in this definition.

\item \textbf{Definition 2.15:} Replace \textquotedblleft$us_{i}s_{j}%
s_{i}v=us_{j}s_{i}s_{j}v$\textquotedblright\ by \textquotedblleft$us_{i}%
s_{j}s_{i}v\sim us_{j}s_{i}s_{j}v$\textquotedblright.

\item \textbf{Definition 2.15:} Replace \textquotedblleft$\Sigma
$\textquotedblright\ by \textquotedblleft$\Sigma_{n}$\textquotedblright\ twice
in this definition (the second time is inside the commutative diagram). Or
just define $\Sigma$ to be an abbreviation for $\Sigma_{n}$ ?

\item \textbf{Definition 2.15:} Please explain that $\sim$ is defined to be
the disjoint union of the relations $\sim_{r}$ over all $r\in\mathbb{N}$.
(This is a relation on $\coprod_{r}W_{r}=W$.)

\item \textbf{Definition 2.17:} At the end of condition (c), add
\textquotedblleft and the word $uv$ is reduced\textquotedblright. Otherwise,
condition (c) would always hold!

\item \textbf{Definition 2.17:} I think the justification for the equivalence
of the four conditions would be clearer if you replaced \textquotedblleft and
it follows from Lemma 2.6 that (a) is equivalent to (d)\textquotedblright\ by
\textquotedblleft and it follows from Lemma 2.6 (applied to $\sigma\tau^{-1}$
instead of $\sigma$) that (b) is equivalent to (d)\textquotedblright.

\item \textbf{Proof of Lemma 2.18:} Replace \textquotedblleft the $3$-cycle
$\left(  i,i+1,i+2\right)  $\textquotedblright\ by \textquotedblleft the
transposition $\left(  i,i+2\right)  $\textquotedblright.

\item \textbf{Proof of Theorem 2.16:} I suspect the LaTeX here is slightly
broken: You want to start the proof by \textquotedblleft\textit{Proof of
Theorem 2.16.}\textquotedblright\ and not by \textquotedblleft\textit{Proof.}
Proof of Theorem 2.16.\textquotedblright.

\item \textbf{Proof of Theorem 2.16:} Replace \textquotedblleft so
$u=v$\textquotedblright\ by \textquotedblleft so $u\sim v$\textquotedblright.

\item \textbf{\S 2:} I suggest adding the following fact to \S 2 (which is
used later, in \S 9):

\textbf{Corollary 2.19.} \textbf{(a)} We have $l\left(  \sigma^{-1}%
\rho\right)  =n\left(  n-1\right)  /2-l\left(  \sigma\right)  $ for each
$\sigma\in\Sigma_{n}$.

\textbf{(b)} We have $l\left(  \sigma\rho\right)  =n\left(  n-1\right)
/2-l\left(  \sigma\right)  $ for each $\sigma\in\Sigma_{n}$.

\textbf{(c)} We have $l\left(  \rho\sigma^{-1}\right)  =n\left(  n-1\right)
/2-l\left(  \sigma\right)  $ for each $\sigma\in\Sigma_{n}$.

\textbf{(d)} We have $l\left(  \rho\sigma\right)  =n\left(  n-1\right)
/2-l\left(  \sigma\right)  $ for each $\sigma\in\Sigma_{n}$.

[\textit{Proof of Corollary 2.19.} \textbf{(c)} Recall the four equivalent
conditions (a), (b), (c) and (d) in Definition 2.17. In particular, the two
conditions (b) and (d) are equivalent for each $\sigma\in\Sigma_{n}$ and
$\tau\in\Sigma_{n}$. In other words, for each $\sigma\in\Sigma_{n}$ and
$\tau\in\Sigma_{n}$, we have the following equivalence:%
\begin{equation}
\left(  l\left(  \sigma\tau^{-1}\right)  =l\left(  \sigma\right)  -l\left(
\tau\right)  \right)  \ \Longleftrightarrow\ \left(  \overline{L}\left(
\tau\right)  \subseteq\overline{L}\left(  \sigma\right)  \right)  .
\label{pf.c2.19.1}%
\end{equation}


Now, fix $\sigma\in\Sigma_{n}$. Then, $\overline{L}\left(  \sigma\right)
\subseteq\overline{L}\left(  \rho\right)  $\ \ \ \ \footnote{\textit{Proof.}
Let $U\in\overline{L}\left(  \sigma\right)  $. Then, $U\in\overline{L}\left(
\sigma\right)  =\left\{  \left\{  i,j\right\}  \ \mid\ \left(  i,j\right)  \in
L\left(  \sigma\right)  \right\}  =\left\{  \left\{  u,v\right\}
\ \mid\ \left(  u,v\right)  \in L\left(  \sigma\right)  \right\}  $. In other
words, $U=\left\{  u,v\right\}  $ for some $\left(  u,v\right)  \in L\left(
\sigma\right)  $. Fix this $\left(  u,v\right)  $. We have $0<u<v\leq n$
(since $\left(  u,v\right)  \in L\left(  \sigma\right)  $). The definition of
$\rho$ yields $\rho\left(  u\right)  =n+1-u$ and $\rho\left(  v\right)
=n+1-v$. Hence, $\rho\left(  u\right)  =n+1-\underbrace{u}_{<v}>n+1-v=\rho
\left(  v\right)  $. Thus, $0<u<v\leq n$ and $\rho\left(  u\right)
>\rho\left(  v\right)  $. In other words, $\left(  u,v\right)  $ is an element
$\left(  i,j\right)  $ such that $0<i<j\leq n$ and $\rho\left(  i\right)
>\rho\left(  j\right)  $. In other words, $\left(  u,v\right)  \in\left\{
\left(  i,j\right)  \ \mid\ 0<i<j\leq n\text{ and }\rho\left(  u\right)
>\rho\left(  v\right)  \right\}  $. This rewrites as $\left(  u,v\right)  \in
L\left(  \rho\right)  $ (since $L\left(  \rho\right)  =\left\{  \left(
i,j\right)  \ \mid\ 0<i<j\leq n\text{ and }\rho\left(  u\right)  >\rho\left(
v\right)  \right\}  $ (by the definition of $L\left(  \rho\right)  $)). Now,
recall that $U=\left\{  u,v\right\}  $. Hence, $U=\left\{  i,j\right\}  $ for
some $\left(  i,j\right)  \in L\left(  \rho\right)  $ (namely, for $\left(
i,j\right)  =\left(  u,v\right)  $). In other words, $U\in\left\{  \left\{
i,j\right\}  \ \mid\ \left(  i,j\right)  \in L\left(  \rho\right)  \right\}
$. This rewrites as $U\in\overline{L}\left(  \rho\right)  $ (since
$\overline{L}\left(  \rho\right)  =\left\{  \left\{  i,j\right\}
\ \mid\ \left(  i,j\right)  \in L\left(  \rho\right)  \right\}  $ (by the
definition of $\overline{L}\left(  \rho\right)  $)).
\par
Now, forget that we fixed $U$. We thus have shown that $U\in\overline
{L}\left(  \rho\right)  $ for each $U\in\overline{L}\left(  \sigma\right)  $.
In other words, $\overline{L}\left(  \sigma\right)  \subseteq\overline
{L}\left(  \rho\right)  $, qed.}. But the equivalence (\ref{pf.c2.19.1})
(applied to $\rho$ and $\sigma$ instead of $\sigma$ and $\rho$) shows that%
\[
\left(  l\left(  \rho\sigma^{-1}\right)  =l\left(  \rho\right)  -l\left(
\sigma\right)  \right)  \ \Longleftrightarrow\ \left(  \overline{L}\left(
\sigma\right)  \subseteq\overline{L}\left(  \rho\right)  \right)  .
\]
Therefore, we have $l\left(  \rho\sigma^{-1}\right)  =l\left(  \rho\right)
-l\left(  \sigma\right)  $ (since we have $\overline{L}\left(  \sigma\right)
\subseteq\overline{L}\left(  \rho\right)  $). Thus, $l\left(  \rho\sigma
^{-1}\right)  =\underbrace{l\left(  \rho\right)  }_{=n\left(  n-1\right)
/2}-l\left(  \sigma\right)  =n\left(  n-1\right)  /2-l\left(  \sigma\right)
$. This proves Corollary 2.19 \textbf{(c)}.

\textbf{(d)} Let $\sigma\in\Sigma_{n}$. Corollary 2.19 \textbf{(c)} (applied
to $\sigma^{-1}$ instead of $\sigma$) shows that $l\left(  \rho\left(
\sigma^{-1}\right)  ^{-1}\right)  =n\left(  n-1\right)  /2-l\left(
\sigma^{-1}\right)  $. Since $\left(  \sigma^{-1}\right)  ^{-1}=\sigma$, this
rewrites as%
\[
l\left(  \rho\sigma\right)  =n\left(  n-1\right)  /2-\underbrace{l\left(
\sigma^{-1}\right)  }_{\substack{=l\left(  \sigma\right)  \\\text{(by Lemma
2.3)}}}=n\left(  n-1\right)  /2-l\left(  \sigma\right)  .
\]
This proves Corollary 2.19 \textbf{(d)}.

\textbf{(a)} Let $\sigma\in\Sigma_{n}$. Then, Lemma 2.3 (applied to
$\rho\sigma$ instead of $\sigma$) shows that $l\left(  \left(  \rho
\sigma\right)  ^{-1}\right)  =l\left(  \rho\sigma\right)  =n\left(
n-1\right)  /2-l\left(  \sigma\right)  $ (by Corollary 2.19 \textbf{(d)}).
Since $\left(  \rho\sigma\right)  ^{-1}=\sigma^{-1}\underbrace{\rho^{-1}%
}_{=\rho}=\sigma^{-1}\rho$, this rewrites as $l\left(  \sigma^{-1}\rho\right)
=n\left(  n-1\right)  /2-l\left(  \sigma\right)  $. This proves Corollary 2.19
\textbf{(a)}.

\textbf{(b)} Let $\sigma\in\Sigma_{n}$. Then, Lemma 2.3 (applied to
$\sigma\rho$ instead of $\sigma$) shows that $l\left(  \left(  \sigma
\rho\right)  ^{-1}\right)  =l\left(  \sigma\rho\right)  $. Since $\left(
\sigma\rho\right)  ^{-1}=\underbrace{\rho^{-1}}_{=\rho}\sigma^{-1}=\rho
\sigma^{-1}$, this rewrites as $l\left(  \rho\sigma^{-1}\right)  =l\left(
\sigma\rho\right)  $. Hence, $l\left(  \sigma\rho\right)  =l\left(  \rho
\sigma^{-1}\right)  =n\left(  n-1\right)  /2-l\left(  \sigma\right)  $ (by
Corollary 2.19 \textbf{(c)}). This proves Corollary 2.19 \textbf{(b)}.
$\square$ ]

\item \textbf{\S 3:} In the definition of $T$, a whitespace is missing between
\textquotedblleft$ge_{i}\in\mathbb{F}_{p}e_{i}$\textquotedblright\ and
\textquotedblleft for all $i$\textquotedblright.

\item \textbf{\S 3:} Add a period before \textquotedblleft With this
convention\textquotedblright.

\item \textbf{\S 3:} After \textquotedblleft and $\left(  g.x\right)
_{i}=\sum_{j}g_{ij}x_{j}$\textquotedblright, add \textquotedblleft for
$x=\left(  x_{1},x_{2},\ldots,x_{n}\right)  ^{T}$\textquotedblright.

\item \textbf{\S 4:} Before Example 4.1, it would be good to say the
following: \textquotedblleft We shall call $\delta\left(  \underline{U}%
,\underline{V}\right)  $ the \textit{Jordan permutation} of the flags
$\underline{U}$ and $\underline{V}$.\textquotedblright. This way, the words
\textquotedblleft Jordan permutation\textquotedblright\ (which are used in
Definition 8.2) are actually defined.

\item \textbf{\S 4:} It would also be useful to point out explicitly that
$\delta\left(  \underline{U},\underline{U}\right)  =1$ for each $\underline{U}%
\in\operatorname*{Flag}\left(  W\right)  $. This is very easy to prove (it is
a corollary of Lemma 4.5, but it is also pretty easy to check using just the
definition); but I think it's worth explicitly stating.

\item \textbf{Example 4.1:} At the very beginning of this example, add the
following sentence: \textquotedblleft Set $\underline{E}=\left(  E_{0}%
<E_{1}<\cdots<E_{n}=\mathbb{F}_{p}^{n}\right)  \in\operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  $.\textquotedblright.

\item \textbf{Example 4.1:} At the end of the last formula on page 5 (the
formula that defines $Q_{ij}$), add a period.

\item \textbf{Example 4.1:} At the end of the first displayed formula on page
6 (the formula that defines $C_{ij}$), a closing parenthesis is missing.

\item \textbf{Example 4.2:} Before \textquotedblleft Then $Q_{15}%
$\textquotedblright, add: \textquotedblleft Set $U_{i}=\operatorname*{span}%
\left\{  u_{1},\ldots,u_{i}\right\}  $ and $V_{i}=E_{i}$ for each
$i$.\textquotedblright. (Otherwise it is not clear how the $Q_{ij}$ are defined.)

\item \textbf{Lemma 4.5:} It might be better to split the second sentence of
this proposition as follows: \textquotedblleft Assume that $U_{i-1}=V_{i-1}$.
Then, $\sigma\left(  i\right)  =i$ iff $U_{i}=V_{i}$.\textquotedblright%
\ (Otherwise, the order of precedence between the \textquotedblleft
then\textquotedblright\ and the \textquotedblleft iff\textquotedblright\ is ambiguous.)

\item \textbf{Proof of Proposition 4.7:} Replace \textquotedblleft%
$U_{i}<V_{i+1}$\textquotedblright\ by \textquotedblleft$U_{i}\leq V_{i+1}%
$\textquotedblright\ (unless you really mean to use the symbol $<$ for
not-necessarily-proper submodules; but in that case, you would need to replace
several $<$'s by other symbols).

\item \textbf{Proof of Proposition 4.7:} Remove the word \textquotedblleft
we\textquotedblright\ in \textquotedblleft Of course we also\textquotedblright.

\item \textbf{Proof of Proposition 4.7:} You replace $\dim\left(  U_{i}\cap
V_{i}\right)  $ by $i-1$ in the long displayed equation. I would suggest
explaining why this is allowed: Namely, you argue that $U_{i}\cap
V_{i}=A=U_{i-1}$, and thus $\dim\left(  U_{i}\cap V_{i}\right)  =\dim\left(
U_{i-1}\right)  =i-1$.

\item \textbf{Proof of Proposition 4.7:} After \textquotedblleft
Symmetrically, we have $U_{i+1}=U_{i}+V_{i}$.\textquotedblright, I would add
\textquotedblleft Hence, $U_{i+1}=V_{i+1}$.\textquotedblright.

\item \textbf{Proof of Proposition 4.7:} Replace \textquotedblleft$U_{j}%
=V_{j}$ for all such $j$\textquotedblright\ by \textquotedblleft$U_{j}=V_{j}$
for all $j\geq i+1$\textquotedblright\ (since the induction starts at $j=i+1$,
not at $j=i+2$).

\item \textbf{Proof of Lemma 5.1:} After the first sentence, add:
\textquotedblleft Thus, assume that $g_{ii}=1$ from now on.\textquotedblright

\item \textbf{Proof of Lemma 5.1:} Replace \textquotedblleft iff $a_{ij}%
=0$\textquotedblright\ by \textquotedblleft iff $g_{ij}=0$\textquotedblright.

\item \textbf{Proposition 5.2:} You write: \textquotedblleft and thus
$\left\vert Y\left(  \sigma,\underline{V}\right)  \right\vert =p^{l\left(
\sigma\right)  }$ for any $\underline{V}$\textquotedblright. This is correct,
but I find it a bit too nontrivial to just state without further explanation.
Maybe it's even worth moving this claim into a separate corollary:

\textbf{Corollary 5.2a.} Let $\sigma\in\Sigma_{n}$ and $\underline{V}%
\in\operatorname{Flag}\left(  W\right)  $, where $W$ is an $n$-dimensional
$\mathbb{F}_{p}$-vector space. Then, $\left\vert Y\left(  \sigma
,\underline{V}\right)  \right\vert =p^{l\left(  \sigma\right)  }$.

[\textit{Proof of Corollary 5.2a.} Pick a basis $\left(  f_{1},f_{2}%
,\ldots,f_{n}\right)  $ of $W$ such that each $0\leq i\leq n$ satisfies
$V_{i}=\operatorname{span}\left\{  f_{1},f_{2},\ldots,f_{i}\right\}  $. (Such
a basis exists, because we can start with the empty basis of $V_{0}$, then
extend it to a basis of $V_{1}$, then extend it to a basis of $V_{2}$, etc..)
Let $\alpha$ be the $\mathbb{F}_{p}$-linear map $\mathbb{F}_{p}^{n}\rightarrow
W$ that sends each $e_{k}$ to $f_{k}$. Then, $\alpha$ is a vector space
isomorphism (since it sends the basis $\left(  e_{1},e_{2},\ldots
,e_{n}\right)  $ of $\mathbb{F}_{p}^{n}$ to the basis $\left(  f_{1}%
,f_{2},\ldots,f_{n}\right)  $ of $W$). Moreover, it satisfies $\alpha
E_{i}=V_{i}$ for each $0\leq i\leq n$. In other words, $\alpha\underline{E}%
=\underline{V}$.

But the naturality of the definition of $\delta\left(  \underline{U}%
,\underline{V}\right)  $ yields an important fact (which is a slight
generalization of your Remark 4.3): If $W_{1}$ and $W_{2}$ are two
$\mathbb{F}_{p}$-vector spaces, and if $\beta:W_{1}\rightarrow W_{2}$ is an
isomorphism, then:

(a) any $\underline{P},\underline{Q}\in\operatorname{Flag}\left(
W_{1}\right)  $ satisfy $\delta\left(  \beta\underline{P},\beta\underline{Q}%
\right)  =\delta\left(  \underline{P},\underline{Q}\right)  $.

(b) for any $\sigma\in\Sigma_{n}$ and any $\underline{Q}\in\operatorname{Flag}%
\left(  W_{1}\right)  $, the map $\operatorname{Flag}\left(  W_{1}\right)
\rightarrow\operatorname{Flag}\left(  W_{2}\right)  ,\ \underline{P}%
\mapsto\beta\underline{P}$ maps the subset $Y\left(  \sigma,\underline{Q}%
\right)  $ bijectively onto $Y\left(  \sigma,\beta\underline{Q}\right)  $.

Applying part (b) of this fact to $W_{1}=\mathbb{F}_{p}^{n}$, $W_{2}=W$,
$\beta=\alpha$ and $\underline{Q}=\underline{E}$, we see that the map
$\operatorname{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \rightarrow
\operatorname{Flag}\left(  W\right)  ,\ \underline{P}\mapsto\alpha
\underline{P}$ maps the subset $Y\left(  \sigma,\underline{E}\right)  $
bijectively onto $Y\left(  \sigma,\alpha\underline{E}\right)  $. Thus,
$\left\vert Y\left(  \sigma,\alpha\underline{E}\right)  \right\vert
=\left\vert Y\left(  \sigma,\underline{E}\right)  \right\vert $. Since
$\alpha\underline{E}=\underline{V}$ and $Y\left(  \sigma,\underline{E}\right)
=Y\left(  \sigma\right)  $, this rewrites as $\left\vert Y\left(
\sigma,\underline{V}\right)  \right\vert =\left\vert Y\left(  \sigma\right)
\right\vert $.

But the first sentence of Proposition 5.2 entails $\left\vert Y\left(
\sigma\right)  \right\vert =\left\vert X\left(  \sigma\right)  \right\vert
=p^{l\left(  \sigma\right)  }$ (by Lemma 5.1). Hence, $\left\vert Y\left(
\sigma,\underline{V}\right)  \right\vert =\left\vert Y\left(  \sigma\right)
\right\vert =p^{l\left(  \sigma\right)  }$. This proves Corollary 5.2a.
$\square$ ]

\item \textbf{Proof of Proposition 5.2:} Replace \textquotedblleft that
$X\left(  \sigma\right)  $ acts freely on $\sigma\underline{E}$%
\textquotedblright\ by \textquotedblleft that $X\left(  \sigma\right)  $ acts
freely on the $X\left(  \sigma\right)  $-orbit of $\sigma\underline{E}%
$\textquotedblright\ (just to use more standard terminology).

\item \textbf{Proof of Proposition 5.2:} Replace \textquotedblleft We claim
that there is a unique element $v_{i}\in V_{i}\cap T_{i}$ such that
$\epsilon_{\sigma\left(  i\right)  }\left(  v_{i}\right)  =1$, and moreover
that $v_{1},\ldots,v_{i}$ is a basis for $V_{i}$ over $\mathbb{F}_{p}%
$\textquotedblright\ by \textquotedblleft We claim that, for each
$i\in\left\{  1,2,\ldots,n\right\}  $, there is a unique element $v_{i}\in
V_{i}\cap T_{i}$ such that $\epsilon_{\sigma\left(  i\right)  }\left(
v_{i}\right)  =1$, and moreover that $v_{1},\ldots,v_{i}$ is a basis for
$V_{i}$ over $\mathbb{F}_{p}$ (if the elements $v_{j}$ for all $j<i$ are
defined similarly)\textquotedblright.

\item \textbf{Proof of Proposition 5.2:} Add a period at the end of the
displayed equation that defines $S_{i}$.

\item \textbf{Proof of Proposition 5.2:} The word \textquotedblleft leading
term\textquotedblright\ might use a definition. Actually, it is probably best
to introduce a number of related notions \textbf{before} the proof of
Proposition 5.2:

\begin{itemize}
\item For each $i\in\left\{  1,2,\ldots,n\right\}  $, let $\epsilon
_{i}:\mathbb{F}_{p}^{n}\rightarrow\mathbb{F}_{p}$ be the $i$-th coordinate
projection. Thus, for any $x\in\mathbb{F}_{p}^{n}$ and $i\in\left\{
1,2,\ldots,n\right\}  $, the element $\epsilon_{i}\left(  x\right)
\in\mathbb{F}_{p}$ is the $i$-th coordinate of the vector $x$.

\item The \textit{leading index} of a nonzero vector $x\in\mathbb{F}_{p}^{n}$
is defined to be the highest $k\in\left\{  1,2,\ldots,n\right\}  $ satisfying
$\epsilon_{k}\left(  x\right)  \neq0$. This leading index is denoted by
$\operatorname*{lind}\left(  x\right)  $. The definition of the leading index
can be rewritten as follows: The \textit{leading index} of a nonzero vector
$x=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)  \in
\mathbb{F}_{p}^{n}$ is the highest $k\in\left\{  1,2,\ldots,n\right\}  $
satisfying $\lambda_{k}\neq0$.

\item The \textit{leading term} of a nonzero vector $x\in\mathbb{F}_{p}^{n}$
is defined to be $\epsilon_{k}\left(  x\right)  e_{k}$, where
$k=\operatorname*{lind}\left(  x\right)  $. In other words, the
\textit{leading term} of a nonzero vector $x=\left(  \lambda_{1},\lambda
_{2},\ldots,\lambda_{n}\right)  \in\mathbb{F}_{p}^{n}$ is $\lambda_{k}e_{k}$,
where $k$ is the highest element of $\left\{  1,2,\ldots,n\right\}  $
satisfying $\lambda_{k}\neq0$.
\end{itemize}

Here are some basic properties of leading indices:

\begin{itemize}
\item \textbf{Lemma 5.2b.} Let $x\in\mathbb{F}_{p}^{n}$ is a nonzero vector.

\textbf{(a)} If $i\in\left\{  1,2,\ldots,n\right\}  $ is such that
$i>\operatorname*{lind}\left(  x\right)  $, then $\epsilon_{i}\left(
x\right)  =0$.

\textbf{(b)} We have $\epsilon_{\operatorname*{lind}\left(  x\right)  }\left(
x\right)  \neq0$.

[\textit{Proof of Lemma 5.2b:} \textbf{(a)} Let $i\in\left\{  1,2,\ldots
,n\right\}  $ be such that $i>\operatorname*{lind}\left(  x\right)  $. But
$\operatorname*{lind}\left(  x\right)  $ is the highest $k\in\left\{
1,2,\ldots,n\right\}  $ satisfying $\epsilon_{k}\left(  x\right)  \neq0$ (by
the definition of $\operatorname*{lind}\left(  x\right)  $). Hence, every
$k\in\left\{  1,2,\ldots,n\right\}  $ satisfying $\epsilon_{k}\left(
x\right)  \neq0$ must satisfy $k\leq\operatorname*{lind}\left(  x\right)  $.
Applying this to $k=i$, we conclude that if $\epsilon_{i}\left(  x\right)
\neq0$, then $i\leq\operatorname*{lind}\left(  x\right)  $. Hence, we cannot
have $\epsilon_{i}\left(  x\right)  \neq0$ (since we cannot have
$i\leq\operatorname*{lind}\left(  x\right)  $ (since we have
$i>\operatorname*{lind}\left(  x\right)  $)). In other words, we have
$\epsilon_{i}\left(  x\right)  =0$. This proves Lemma 5.2b \textbf{(a)}.

\textbf{(b)} We know that $\operatorname*{lind}\left(  x\right)  $ is the
highest $k\in\left\{  1,2,\ldots,n\right\}  $ satisfying $\epsilon_{k}\left(
x\right)  \neq0$ (by the definition of $\operatorname*{lind}\left(  x\right)
$). Hence, $\operatorname*{lind}\left(  x\right)  $ is a $k\in\left\{
1,2,\ldots,n\right\}  $ satisfying $\epsilon_{k}\left(  x\right)  \neq0$.
Thus, $\epsilon_{\operatorname*{lind}\left(  x\right)  }\left(  x\right)
\neq0$. This proves Lemma 5.2b \textbf{(b)}. $\square$]

\item \textbf{Lemma 5.2c.} Let $\left(  w_{s}\right)  _{s\in S}$ be a finite
family of nonzero vectors in $\mathbb{F}_{p}^{n}$. Assume that the leading
indices of the $w_{s}$ (for $s\in S$) are pairwise distinct.

\textbf{(a)} Each nonzero vector $x\in\mathbb{F}_{p}\left\{  w_{s}\ \mid\ s\in
S\right\}  $ satisfies
\[
\operatorname*{lind}\left(  x\right)  \in\left\{  \operatorname*{lind}\left(
w_{s}\right)  \ \mid\ s\in S\right\}  .
\]


\textbf{(b)} The family $\left(  w_{s}\right)  _{s\in S}$ is $\mathbb{F}_{p}%
$-linearly independent.

\textbf{(c)} Let $\left(  \lambda_{s}\right)  _{s\in S}\in\mathbb{F}_{p}^{S}$
be a family of elements of $\mathbb{F}_{p}$. Assume that there exists at least
one $s\in S$ satisfying $\lambda_{s}\neq0$. Set $x=\sum_{s\in S}\lambda
_{s}w_{s}$. Then, $x\neq0$ and $\operatorname*{lind}\left(  x\right)
\in\left\{  \operatorname*{lind}\left(  w_{s}\right)  \ \mid\ s\in S\right\}
$.

[\textit{Proof of Lemma 5.2c:} \textbf{(c)} There exists at least one $s\in S$
satisfying $\lambda_{s}\neq0$. Among all these $s\in S$ satisfying
$\lambda_{s}\neq0$, pick one for which $\operatorname*{lind}\left(
w_{s}\right)  $ is maximum, and denote this $s$ by $t$. Thus, $\lambda_{t}%
\neq0$, and $\operatorname*{lind}\left(  w_{t}\right)  $ is the highest among
the $\operatorname*{lind}\left(  w_{s}\right)  $ for all $s\in S$ satisfying
$\lambda_{s}\neq0$. As a consequence,%
\begin{equation}
\text{every }s\in S\text{ satisfying }\lambda_{s}\neq0\text{ satisfies
}\operatorname*{lind}\left(  w_{s}\right)  \leq\operatorname*{lind}\left(
w_{t}\right)  \label{pf.prop.5.2.lt-in-set.pf.leq}%
\end{equation}
(since $\operatorname*{lind}\left(  w_{t}\right)  $ is the highest among the
$\operatorname*{lind}\left(  w_{s}\right)  $ for all $s\in S$ satisfying
$\lambda_{s}\neq0$).

Moreover, recall that the leading indices of the $w_{s}$ (for $s\in S$) are
pairwise distinct. In other words,%
\begin{equation}
\text{every two distinct elements }p\text{ and }q\text{ of }S\text{ satisfy
}\operatorname*{lind}\left(  w_{p}\right)  \neq\operatorname*{lind}\left(
w_{q}\right)  . \label{pf.prop.5.2.lt-in-set.pf.neq}%
\end{equation}


Now,%
\begin{equation}
\epsilon_{\operatorname*{lind}\left(  w_{t}\right)  }\left(  \lambda_{s}%
w_{s}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }s\in S\text{ satisfying
}s\neq t \label{pf.prop.5.2.lt-in-set.pf.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.5.2.lt-in-set.pf.1}):} Let $s\in S$
be such that $s\neq t$. Then, (\ref{pf.prop.5.2.lt-in-set.pf.neq}) (applied to
$p=s$ and $q=t$) yields $\operatorname*{lind}\left(  w_{s}\right)
\neq\operatorname*{lind}\left(  w_{t}\right)  $.
\par
If $\lambda_{s}=0$, then $\epsilon_{\operatorname*{lind}\left(  w_{t}\right)
}\left(  \underbrace{\lambda_{s}}_{=0}w_{s}\right)  =\epsilon
_{\operatorname*{lind}\left(  w_{t}\right)  }\left(  0w_{s}\right)  =0$.
Hence, if $\lambda_{s}=0$, then (\ref{pf.prop.5.2.lt-in-set.pf.1}) holds.
Thus, for the rest of this proof of (\ref{pf.prop.5.2.lt-in-set.pf.1}), we
WLOG asume that $\lambda_{s}\neq0$. Hence, (\ref{pf.prop.5.2.lt-in-set.pf.leq}%
) shows that $\operatorname*{lind}\left(  w_{s}\right)  \leq
\operatorname*{lind}\left(  w_{t}\right)  $. Combining this with
$\operatorname*{lind}\left(  w_{s}\right)  \neq\operatorname*{lind}\left(
w_{t}\right)  $, we obtain $\operatorname*{lind}\left(  w_{s}\right)
<\operatorname*{lind}\left(  w_{t}\right)  $. Thus, $\operatorname*{lind}%
\left(  w_{t}\right)  >\operatorname*{lind}\left(  w_{s}\right)  $. Hence,
Lemma 5.2b \textbf{(a)} (applied to $x=w_{s}$ and $i=\operatorname*{lind}%
\left(  w_{t}\right)  $) yields $\epsilon_{\operatorname*{lind}\left(
w_{t}\right)  }\left(  w_{s}\right)  =0$. Thus, $\epsilon
_{\operatorname*{lind}\left(  w_{t}\right)  }\left(  \lambda_{s}w_{s}\right)
=\lambda_{s}\underbrace{\epsilon_{\operatorname*{lind}\left(  w_{t}\right)
}\left(  w_{s}\right)  }_{=0}=0$. This proves
(\ref{pf.prop.5.2.lt-in-set.pf.1}).}. Furthermore, if $i\in\left\{
1,2,\ldots,n\right\}  $ is such that $i>\operatorname*{lind}\left(
w_{t}\right)  $, then
\begin{equation}
\epsilon_{i}\left(  \lambda_{s}w_{s}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for
every }s\in S \label{pf.prop.5.2.lt-in-set.pf.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.5.2.lt-in-set.pf.2}):} Let $s\in S$.
If $\lambda_{s}=0$, then $\epsilon_{i}\left(  \underbrace{\lambda_{s}}%
_{=0}w_{s}\right)  =\epsilon_{i}\left(  0w_{s}\right)  =0$. Hence, if
$\lambda_{s}=0$, then (\ref{pf.prop.5.2.lt-in-set.pf.2}) holds. Thus, for the
rest of this proof of (\ref{pf.prop.5.2.lt-in-set.pf.2}), we WLOG asume that
$\lambda_{s}\neq0$. Hence, (\ref{pf.prop.5.2.lt-in-set.pf.leq}) shows that
$\operatorname*{lind}\left(  w_{s}\right)  \leq\operatorname*{lind}\left(
w_{t}\right)  <i$ (since $i>\operatorname*{lind}\left(  w_{t}\right)  $).
Thus, $i>\operatorname*{lind}\left(  w_{s}\right)  $. Hence, Lemma 5.2b
\textbf{(a)} (applied to $x=w_{s}$) yields $\epsilon_{i}\left(  w_{s}\right)
=0$. Thus, $\epsilon_{i}\left(  \lambda_{s}w_{s}\right)  =\lambda
_{s}\underbrace{\epsilon_{i}\left(  w_{s}\right)  }_{=0}=0$. This proves
(\ref{pf.prop.5.2.lt-in-set.pf.2}).}. If $i\in\left\{  1,2,\ldots,n\right\}  $
is such that $i>\operatorname*{lind}\left(  w_{t}\right)  $, then%
\begin{align}
\epsilon_{i}\left(  \underbrace{x}_{=\sum_{s\in S}\lambda_{s}w_{s}}\right)
&  =\epsilon_{i}\left(  \sum_{s\in S}\lambda_{s}w_{s}\right)  =\sum_{s\in
S}\underbrace{\epsilon_{i}\left(  \lambda_{s}w_{s}\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.prop.5.2.lt-in-set.pf.2}))}}}\nonumber\\
&  =\sum_{s\in S}0=0. \label{pf.prop.5.2.lt-in-set.pf.4}%
\end{align}
But%
\begin{align*}
&  \epsilon_{\operatorname*{lind}\left(  w_{t}\right)  }\left(  \underbrace{x}%
_{=\sum_{s\in S}\lambda_{s}w_{s}}\right) \\
&  =\epsilon_{\operatorname*{lind}\left(  w_{t}\right)  }\left(  \sum_{s\in
S}\lambda_{s}w_{s}\right)  =\sum_{s\in S}\epsilon_{\operatorname*{lind}\left(
w_{t}\right)  }\left(  \lambda_{s}w_{s}\right) \\
&  =\underbrace{\epsilon_{\operatorname*{lind}\left(  w_{t}\right)  }\left(
\lambda_{t}w_{t}\right)  }_{=\lambda_{t}\epsilon_{\operatorname*{lind}\left(
w_{t}\right)  }\left(  w_{t}\right)  }+\sum_{\substack{s\in S;\\s\neq
t}}\underbrace{\epsilon_{\operatorname*{lind}\left(  w_{t}\right)  }\left(
\lambda_{s}w_{s}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.prop.5.2.lt-in-set.pf.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}s=t\text{ from the sum}\right) \\
&  =\lambda_{t}\epsilon_{\operatorname*{lind}\left(  w_{t}\right)  }\left(
w_{t}\right)  +\underbrace{\sum_{\substack{s\in S;\\s\neq t}}0}_{=0}%
=\underbrace{\lambda_{t}}_{\neq0}\underbrace{\epsilon_{\operatorname*{lind}%
\left(  w_{t}\right)  }\left(  w_{t}\right)  }_{\substack{\neq0\\\text{(by
Lemma 5.2b \textbf{(b)}, applied}\\\text{to }x=w_{t}\text{)}}}\neq0.
\end{align*}
Hence, $x\neq0$. It remains to show that $\operatorname*{lind}\left(
x\right)  \in\left\{  \operatorname*{lind}\left(  w_{s}\right)  \ \mid\ s\in
S\right\}  $.

But $\operatorname*{lind}\left(  w_{t}\right)  $ is a $k\in\left\{
1,2,\ldots,n\right\}  $ satisfying $\epsilon_{k}\left(  x\right)  \neq0$
(since $\epsilon_{\operatorname*{lind}\left(  w_{t}\right)  }\left(  x\right)
\neq0$). Moreover, $\operatorname*{lind}\left(  w_{t}\right)  $ is the
\textbf{highest} such $k$ (because any $i\in\left\{  1,2,\ldots,n\right\}  $
satisfying $i>\operatorname*{lind}\left(  w_{t}\right)  $ satisfies
$\epsilon_{i}\left(  x\right)  =0$ (by (\ref{pf.prop.5.2.lt-in-set.pf.4}))).
Thus, $\operatorname*{lind}\left(  w_{t}\right)  $ is the highest
$k\in\left\{  1,2,\ldots,n\right\}  $ satisfying $\epsilon_{k}\left(
x\right)  \neq0$. In other words, $\operatorname*{lind}\left(  w_{t}\right)  $
is the leading index of $x$ (by the definition of the leading index). In other
words, $\operatorname*{lind}\left(  w_{t}\right)  =\operatorname*{lind}\left(
x\right)  $. Thus,%
\[
\operatorname*{lind}\left(  x\right)  =\operatorname*{lind}\left(
w_{t}\right)  \in\left\{  \operatorname*{lind}\left(  w_{s}\right)
\ \mid\ s\in S\right\}  .
\]
This completes the proof of Lemma 5.2c \textbf{(c)}.

\textbf{(b)} Let $\left(  \lambda_{s}\right)  _{s\in S}\in\mathbb{F}_{p}^{S}$
be a family of elements of $\mathbb{F}_{p}$ satisfying $\sum_{s\in S}%
\lambda_{s}w_{s}=0$. Thus, $0=\sum_{s\in S}\lambda_{s}w_{s}$. Then, no $s\in
S$ satisfies $\lambda_{s}\neq0$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Thus, there exists at least one $s\in S$ satisfying $\lambda_{s}%
\neq0$. Hence, Lemma 5.2c \textbf{(c)} (applied to $x=0$) yields $0\neq0$ and
$\operatorname*{lind}\left(  0\right)  \in\left\{  \operatorname*{lind}\left(
w_{s}\right)  \ \mid\ s\in S\right\}  $. But $0\neq0$ is clearly absurd.
Hence, we have obtained a contradiction. This shows that our assumption was
wrong. Qed.}. In other words, every $s\in S$ satisfies $\lambda_{s}=0$.

Now, forget that we fixed $\left(  \lambda_{s}\right)  _{s\in S}$. We thus
have shown that if $\left(  \lambda_{s}\right)  _{s\in S}\in\mathbb{F}_{p}%
^{S}$ is a family of elements of $\mathbb{F}_{p}$ satisfying $\sum_{s\in
S}\lambda_{s}w_{s}=0$, then every $s\in S$ satisfies $\lambda_{s}=0$. In other
words, the family $\left(  w_{s}\right)  _{s\in S}$ is $\mathbb{F}_{p}%
$-linearly independent. This proves Lemma 5.2c \textbf{(b)}.

\textbf{(a)} Let $x\in\mathbb{F}_{p}\left\{  w_{s}\ \mid\ s\in S\right\}  $ be
a nonzero vector. We must prove that $\operatorname*{lind}\left(  x\right)
\in\left\{  \operatorname*{lind}\left(  w_{s}\right)  \ \mid\ s\in S\right\}
$.

We have $x\in\mathbb{F}_{p}\left\{  w_{s}\ \mid\ s\in S\right\}  $. Hence, we
can write $x$ in the form $x=\sum_{s\in S}\lambda_{s}w_{s}$ for some elements
$\lambda_{s}$ of $\mathbb{F}_{p}$. Consider these $\lambda_{s}$. There exists
at least one $s\in S$ satisfying $\lambda_{s}\neq0$%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $\lambda_{s}=0$
for all $s\in S$. Now, $x=\sum_{s\in S}\underbrace{\lambda_{s}}_{=0}w_{s}%
=\sum_{s\in S}0w_{s}=0$. This contradicts the fact that $x$ is nonzero. This
contradiction shows that our assumption was wrong, qed.}. Thus, Lemma 5.2c
\textbf{(c)} yields $x\neq0$ and $\operatorname*{lind}\left(  x\right)
\in\left\{  \operatorname*{lind}\left(  w_{s}\right)  \ \mid\ s\in S\right\}
$. This proves Lemma 5.2c \textbf{(a)}. $\square$]

\item \textbf{Lemma 5.2d.} If $\mathfrak{W}$ is a vector subspace of
$\mathbb{F}_{p}^{n}$, then
\[
\dim\mathfrak{W}=\left\vert \left\{  \operatorname*{lind}\left(  x\right)
\ \mid\ x\in\mathfrak{W}\setminus\left\{  0\right\}  \right\}  \right\vert .
\]


[\textit{Proof of Lemma 5.2d (sketched).} Lemma 5.2d is well-known and not
hard to prove. We shall only use it on one occasion, which is not central to
our argument; thus, I shall only outline the proof.

Define the \textit{energy} of a basis $\left(  w_{1},w_{2},\ldots
,w_{k}\right)  $ of $\mathfrak{W}$ to be the nonnegative integer
$\operatorname*{lind}\left(  w_{1}\right)  +\operatorname*{lind}\left(
w_{2}\right)  +\cdots+\operatorname*{lind}\left(  w_{k}\right)  $. Then, there
clearly exists a basis $\left(  w_{1},w_{2},\ldots,w_{k}\right)  $ having
minimum energy. Fix such a basis\footnote{This argument is not constructive,
but we could easily replace it by a constructive argument by induction.}.
Then, no two among the elements $w_{1},w_{2},\ldots,w_{k}$ can have equal
leading indices (because if $w_{i}$ and $w_{j}$ had equal leading indices for
some $i$ and $j$, then we could replace $w_{j}$ by some linear combination
$\alpha w_{i}+w_{j}$ with $\alpha\in\mathbb{F}_{p}$, and by choosing $\alpha$
and $\beta$ appropriately we would ensure that $\operatorname*{lind}\left(
\alpha w_{i}+w_{j}\right)  <\operatorname*{lind}\left(  w_{j}\right)  $, so
that the resulting basis would have a smaller energy than $\left(  w_{1}%
,w_{2},\ldots,w_{k}\right)  $; but this would contradict our choice of
$\left(  w_{1},w_{2},\ldots,w_{k}\right)  $ as the basis with minimum energy).
Hence, the indices of the elements $w_{1},w_{2},\ldots,w_{k}$ are distinct.
Thus, $\left\vert \left\{  \operatorname*{lind}\left(  x\right)  \ \mid
\ x\in\mathfrak{W}\setminus\left\{  0\right\}  \right\}  \right\vert \geq
k=\dim\mathfrak{W}$. It remains to prove that $\dim\mathfrak{W}\geq\left\vert
\left\{  \operatorname*{lind}\left(  x\right)  \ \mid\ x\in\mathfrak{W}%
\setminus\left\{  0\right\}  \right\}  \right\vert $. In order to do so, we
assume the contrary. Thus, $\left\vert \left\{  \operatorname*{lind}\left(
x\right)  \ \mid\ x\in\mathfrak{W}\setminus\left\{  0\right\}  \right\}
\right\vert >\dim\mathfrak{W}=k$. Hence, there exists some $x\in
\mathfrak{W}\setminus\left\{  0\right\}  $ such that $\operatorname*{lind}%
\left(  x\right)  $ equals none of $\operatorname*{lind}\left(  w_{1}\right)
,\operatorname*{lind}\left(  w_{2}\right)  ,\ldots,\operatorname*{lind}\left(
w_{k}\right)  $. Consider the $x$. The $k+1$ nonzero vectors $w_{1}%
,w_{2},\ldots,w_{k},x$ in $\mathbb{F}_{p}^{n}$ have the property that their
leading indices are pairwise distinct. Thus, Lemma 5.2c \textbf{(b)} shows
that they are $\mathbb{F}_{p}$-linearly independent. Since these $k+1$ vectors
all belong to $\mathfrak{W}$, we thus have found $k+1$ linearly independent
vectors in $\mathfrak{W}$. But this contradicts the fact that $\dim
\mathfrak{W}=k<k+1$. This contradiction completes our proof. $\square$]
\end{itemize}

\item \textbf{Proof of Proposition 5.2:} Before the sentence that begins with
\textquotedblleft The leading terms\textquotedblright, I would suggest adding
the following text: \textquotedblleft For each $j<i$, the leading term of the
vector $v_{j}$ is $e_{\sigma\left(  j\right)  }$ (since $e_{j}\in V_{j}\cap
T_{j}\subseteq T_{j}\subseteq E_{\sigma\left(  j\right)  }$ and $\epsilon
_{\sigma\left(  j\right)  }\left(  v_{j}\right)  =1$). Hence, the leading
terms of the vectors $v_{j}$ in $S_{i}$ are precisely the vectors
$e_{\sigma\left(  j\right)  }$ with $j<i$ and $\sigma\left(  j\right)
<\sigma\left(  i\right)  $. In other words,\textquotedblright.

\item \textbf{Proof of Proposition 5.2:} Before the sentence that begins with
\textquotedblleft Using this, we see that $E_{\sigma\left(  i\right)  }%
=S_{i}\oplus T_{i}$\textquotedblright, I would add the following:
\textquotedblleft Now, consider the vectors $v_{j}$ spanning $S_{i}$ and the
vectors $e_{m}$ spanning $T_{i}$. Altogether, these are $\sigma\left(
i\right)  $ vectors lying in the $\sigma\left(  i\right)  $-dimensional space
$E_{\sigma\left(  i\right)  }$. Each of the vectors $e_{m}$ with $m\leq
\sigma\left(  i\right)  $ is the leading term of exactly one of these
$\sigma\left(  i\right)  $ vectors (as we have just shown). Thus, each of the
numbers $1,2,\ldots,\sigma\left(  i\right)  $ is the leading index of exactly
one of these $\sigma\left(  i\right)  $ vectors. Consequently, the leading
indices of these $\sigma\left(  i\right)  $ vectors are pairwise distinct.
Lemma 5.2c \textbf{(b)} (applied to the family of these $\sigma\left(
i\right)  $ vectors) therefore shows that these $\sigma\left(  i\right)  $
vectors are $\mathbb{F}_{p}$-.linearly independent. Hence, these
$\sigma\left(  i\right)  $ vectors form a basis of $E_{\sigma\left(  i\right)
}$ (because they are $\sigma\left(  i\right)  $ linearly independent vectors
lying in the $\sigma\left(  i\right)  $-dimensional space $E_{\sigma\left(
i\right)  }$).\textquotedblright.

\item \textbf{Proof of Proposition 5.2:} You claim that \textquotedblleft and
thus that $V_{i}\cap E_{\sigma\left(  i\right)  }=S_{i}\oplus L_{i}$ for some
(unique) subspace $L_{i}\leq T_{i}$\textquotedblright. It would be friendlier
to the reader to explain why this follows.

Namely, you are using the following (easy) fact from linear algebra: If
$\mathfrak{A}$ and $\mathfrak{B}$ are two subspaces of a vector space
$\mathfrak{V}$, and if $\mathfrak{C}$ is a subspace of $\mathfrak{V}$
satisfying $\mathfrak{A}\leq\mathfrak{C}\leq\mathfrak{A}\oplus\mathfrak{B}$,
then there is a unique subspace $\mathfrak{D}\leq\mathfrak{B}$ satisfying
$\mathfrak{C}=\mathfrak{A}\oplus\mathfrak{D}$. (Namely, this $\mathfrak{D}$
can be constructed as $\mathfrak{D}=\mathfrak{C}\cap\mathfrak{B}$.)

Applying this fact to $\mathfrak{V}=\mathbb{F}_{p}^{n}$, $\mathfrak{A}=S_{i}$,
$\mathfrak{B}=T_{i}$ and $\mathfrak{C}=V_{i}\cap E_{\sigma\left(  i\right)  }$
(and renaming $\mathfrak{D}$ as $L_{i}$) shows that there is a unique subspace
$L_{i}\leq T_{i}$ satisfying $V_{i}\cap E_{\sigma\left(  i\right)  }%
=S_{i}\oplus L_{i}$ (since $S_{i}\leq V_{i}\cap E_{\sigma\left(  i\right)
}\leq E_{\sigma\left(  i\right)  }=S_{i}\oplus T_{i}$). This is exactly your
claim. $\square$

\item \textbf{Proof of Proposition 5.2:} You claim that \textquotedblleft%
$S_{i}=V_{i-1}\cap E_{\sigma\left(  i\right)  }$\textquotedblright\ and write
that \textquotedblleft This is straightforward\textquotedblright. Again, I do
not agree that this is straightforward enough to be left to the reader. Let me
flesh out this proof; more precisely, let me give one proof of the fact that
$S_{i}=V_{i-1}\cap E_{\sigma\left(  i\right)  }$, and one alternative proof of
the existence of $v_{i}$ that sidesteps this and other confusing points in
your proof.

\textit{Proof of }$S_{i}=V_{i-1}\cap E_{\sigma\left(  i\right)  }$\textit{:}
Let $x\in\left(  V_{i-1}\cap E_{\sigma\left(  i\right)  }\right)
\setminus\left\{  0\right\}  $. Thus, $x$ is nonzero. Also, $x\in\left(
V_{i-1}\cap E_{\sigma\left(  i\right)  }\right)  \setminus\left\{  0\right\}
\subseteq V_{i-1}\cap E_{\sigma\left(  i\right)  }\subseteq E_{\sigma\left(
i\right)  }$, so that $\operatorname*{lind}\left(  x\right)  \in\left\{
1,2,\ldots,\sigma\left(  i\right)  \right\}  $. But we also have
\begin{align*}
x  &  \in\left(  V_{i-1}\cap E_{\sigma\left(  i\right)  }\right)
\setminus\left\{  0\right\}  \subseteq V_{i-1}\cap E_{\sigma\left(  i\right)
}\\
&  \subseteq V_{i-1}=\mathbb{F}_{p}\left\{  v_{1},v_{2},\ldots,v_{i-1}%
\right\}  =\mathbb{F}_{p}\left\{  v_{j}\ \mid\ j\in\left\{  1,2,\ldots
,i-1\right\}  \right\}  .
\end{align*}
For each $j<i$, the leading term of $v_{j}$ is $e_{\sigma\left(  j\right)  }$
(since $e_{j}\in V_{j}\cap T_{j}\leq T_{j}\leq E_{\sigma\left(  j\right)  }$
and $\epsilon_{\sigma\left(  j\right)  }\left(  v_{j}\right)  =1$). Thus, for
each $j<i$, the leading index of $v_{j}$ is $\sigma\left(  j\right)  $. Thus,
the leading indices of $v_{1},v_{2},\ldots,v_{i-1}$ are $\sigma\left(
1\right)  ,\sigma\left(  2\right)  ,\ldots,\sigma\left(  i-1\right)  $.
Therefore, these leading indices are pairwise distinct (since $\sigma$ is
injective). In other words, the family $\left(  v_{j}\right)  _{j\in\left\{
1,2,\ldots,i-1\right\}  }$ of nonzero vectors in $\mathbb{F}_{p}^{n}$ has the
property that the leading indices of the $v_{j}$ (for $j\in\left\{
1,2,\ldots,i-1\right\}  $) are pairwise distinct. Thus, Lemma 5.2c
\textbf{(c)} (applied to $\left\{  1,2,\ldots,i-1\right\}  $ and $\left(
v_{j}\right)  _{j\in\left\{  1,2,\ldots,i-1\right\}  }$ instead of $S$ and
$\left(  w_{s}\right)  _{s\in S}$) shows that%
\begin{align*}
\operatorname*{lind}\left(  x\right)   &  \in\left\{  \operatorname*{lind}%
\left(  v_{j}\right)  \ \mid\ j\in\left\{  1,2,\ldots,i-1\right\}  \right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }x\in\mathbb{F}_{p}\left\{
v_{j}\ \mid\ j\in\left\{  1,2,\ldots,i-1\right\}  \right\}  \text{ is a
nonzero vector}\right) \\
&  =\left\{  \underbrace{\operatorname*{lind}\left(  v_{j}\right)
}_{\substack{=\sigma\left(  j\right)  \\\text{(since the leading index of
}v_{j}\text{ is }\sigma\left(  j\right)  \text{)}}}\ \mid\ j<i\right\}
=\left\{  \sigma\left(  j\right)  \ \mid\ j<i\right\}  .
\end{align*}
Combining this with $\operatorname*{lind}\left(  x\right)  \in\left\{
1,2,\ldots,\sigma\left(  i\right)  \right\}  $, we find%
\begin{align*}
\operatorname*{lind}\left(  x\right)   &  \in\left\{  1,2,\ldots,\sigma\left(
i\right)  \right\}  \cap\left\{  \sigma\left(  j\right)  \ \mid\ j<i\right\}
\\
&  =\left\{  \sigma\left(  j\right)  \ \mid\ j<i\text{ and }\sigma\left(
j\right)  <\sigma\left(  i\right)  \right\}  .
\end{align*}


Now, forget that we fixed $x$. We thus have proven that \newline%
$\operatorname*{lind}\left(  x\right)  \in\left\{  \sigma\left(  j\right)
\ \mid\ j<i\text{ and }\sigma\left(  j\right)  <\sigma\left(  i\right)
\right\}  $ for each $x\in\left(  V_{i-1}\cap E_{\sigma\left(  i\right)
}\right)  \setminus\left\{  0\right\}  $. In other words,%
\begin{align}
&  \left\{  \operatorname*{lind}\left(  x\right)  \ \mid\ x\in\left(
V_{i-1}\cap E_{\sigma\left(  i\right)  }\right)  \setminus\left\{  0\right\}
\right\} \nonumber\\
&  \subseteq\left\{  \sigma\left(  j\right)  \ \mid\ j<i\text{ and }%
\sigma\left(  j\right)  <\sigma\left(  i\right)  \right\}  .
\label{pf.prop.5.2.subst}%
\end{align}


But Lemma 5.2d (applied to $\mathfrak{W}=V_{i-1}\cap E_{\sigma\left(
i\right)  }$) yields%
\begin{align}
\dim\left(  V_{i-1}\cap E_{\sigma\left(  i\right)  }\right)   &  =\left\vert
\left\{  \operatorname*{lind}\left(  x\right)  \ \mid\ x\in\left(  V_{i-1}\cap
E_{\sigma\left(  i\right)  }\right)  \setminus\left\{  0\right\}  \right\}
\right\vert \nonumber\\
&  \leq\left\vert \left\{  \sigma\left(  j\right)  \ \mid\ j<i\text{ and
}\sigma\left(  j\right)  <\sigma\left(  i\right)  \right\}  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{because of (\ref{pf.prop.5.2.subst})}\right)
\nonumber\\
&  =\left\vert \left\{  j\ \mid\ j<i\text{ and }\sigma\left(  j\right)
<\sigma\left(  i\right)  \right\}  \right\vert \label{pf.prop.5.2.leq}%
\end{align}
(since the map $\sigma$ is injective).

But $\left(  v_{1},v_{2},\ldots,v_{i-1}\right)  $ is a basis for $V_{j-1}$ (by
the induction hypothesis). Hence, the vectors $v_{1},v_{2},\ldots,v_{i-1}$ are
linearly independent. Thus, the vectors $v_{j}$ for all $j<i$ satisfying
$\sigma\left(  j\right)  <\sigma\left(  i\right)  $ are also linearly
independent (since these vectors form a subfamily of the vectors $v_{1}%
,v_{2},\ldots,v_{i-1}$), and therefore distinct. The definition of $S_{i}$
yields that the vector space $S_{i}$ is spanned by these vectors $v_{j}$ for
all $j<i$ satisfying $\sigma\left(  j\right)  <\sigma\left(  i\right)  $.
Therefore, these vectors $v_{j}$ for all $j<i$ satisfying $\sigma\left(
j\right)  <\sigma\left(  i\right)  $ form a basis of $S_{i}$ (because they are
linearly independent). Hence, the dimension of $S_{i}$ equals the number of
these vectors. In other words,%
\begin{align}
\dim\left(  S_{i}\right)   &  =\left\vert \left\{  v_{j}\ \mid\ j<i\text{ and
}\sigma\left(  j\right)  <\sigma\left(  i\right)  \right\}  \right\vert
\nonumber\\
&  =\left\vert \left\{  j\ \mid\ j<i\text{ and }\sigma\left(  j\right)
<\sigma\left(  i\right)  \right\}  \right\vert \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the vectors }v_{j}\text{ for all
}j<i\text{ satisfying }\sigma\left(  j\right)  <\sigma\left(  i\right)  \text{
are distinct}\right) \nonumber\\
&  \geq\dim\left(  V_{i-1}\cap E_{\sigma\left(  i\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.5.2.leq})}\right)  .
\label{pf.prop.5.2.geq}%
\end{align}


On the other hand, $S_{i}\leq V_{i-1}\cap E_{\sigma\left(  i\right)  }$ (this
follows by combining%
\[
S_{i}=\mathbb{F}_{p}\underbrace{\left\{  v_{j}\ \mid\ j<i\text{ and }%
\sigma\left(  j\right)  <\sigma\left(  i\right)  \right\}  }_{\subseteq
\left\{  v_{j}\ \mid\ j<i\right\}  }\subseteq\mathbb{F}_{p}\left\{
v_{j}\ \mid\ j<i\right\}  =V_{i-1}%
\]
with%
\begin{align*}
S_{i}  &  =\mathbb{F}_{p}\left\{  v_{j}\ \mid\ j<i\text{ and }\sigma\left(
j\right)  <\sigma\left(  i\right)  \right\}  \subseteq E_{\sigma\left(
i\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since each }j<i\text{ satisfying }\sigma\left(  j\right)  <\sigma\left(
i\right)  \text{ satisfies}\\
\operatorname*{lind}\left(  v_{j}\right)  =\sigma\left(  j\right)  \in\left\{
1,2,\ldots,\sigma\left(  i\right)  \right\}  \text{ and thus}\\
v_{j}\in\mathbb{F}_{p}\left\{  e_{1},e_{2},\ldots,e_{\sigma\left(  i\right)
}\right\}  =E_{\sigma\left(  i\right)  }%
\end{array}
\right)
\end{align*}
). Thus, $S_{i}$ is a vector subspace of the finite-dimensional vector space
$V_{i-1}\cap E_{\sigma\left(  i\right)  }$. Since the dimension of this
subspace $S_{i}$ is at least as high as the dimension of $V_{i-1}\cap
E_{\sigma\left(  i\right)  }$ (indeed, this is what (\ref{pf.prop.5.2.geq})
says), we conclude that this subspace $S_{i}$ is the whole $V_{i-1}\cap
E_{\sigma\left(  i\right)  }$. In other words, $S_{i}=V_{i-1}\cap
E_{\sigma\left(  i\right)  }$, qed. $\square$

\textit{Alternative proof of the existence of an }$v_{i}\in V_{i}\cap T_{i}$
\textit{such that }$\epsilon_{\sigma\left(  i\right)  }\left(  v_{i}\right)
=1$\textit{ and moreover that }$v_{1},v_{2},\ldots,v_{i}$ \textit{is a basis
for }$V_{i}$\textit{ over }$\mathbb{F}_{p}$\textit{:} Let me now show another
way to prove that there is an element $v_{i}\in V_{i}\cap T_{i}$ such that
$\epsilon_{\sigma\left(  i\right)  }\left(  v_{i}\right)  =1$ and moreover
that $v_{1},v_{2},\ldots,v_{i}$ is a basis for $V_{i}$ over $\mathbb{F}_{p}$.
This argument will not show the uniqueness of this $v_{i}$ (but you don't ever
use this uniqueness anyway).

I proceed by induction over $i$ (as you do). As in your proof, I define
$S_{i}$, and show that $S_{i}\leq V_{i}\cap E_{\sigma\left(  i\right)  }$ and
that $E_{\sigma\left(  i\right)  }=S_{i}\oplus T_{i}$.

But $\underline{V}\in Y\left(  \sigma\right)  =Y\left(  \sigma,\underline{E}%
\right)  $. In other words, $\delta\left(  \underline{V},\underline{E}\right)
=\sigma$ (by the definition of $Y\left(  \sigma,\underline{E}\right)  $). By
the definition of $\delta$, this yields that $Q_{i,\sigma\left(  i\right)
}\neq0$, where $Q_{i,j}=\dfrac{V_{i}\cap E_{j}}{V_{i-1}\cap E_{j}+V_{i}\cap
E_{j-1}}$. In other words,%
\[
\dfrac{V_{i}\cap E_{\sigma\left(  i\right)  }}{V_{i-1}\cap E_{\sigma\left(
i\right)  }+V_{i}\cap E_{\sigma\left(  i\right)  -1}}\neq0.
\]
Hence,%
\begin{equation}
V_{i}\cap E_{\sigma\left(  i\right)  }>V_{i-1}\cap E_{\sigma\left(  i\right)
}+V_{i}\cap E_{\sigma\left(  i\right)  -1}\geq V_{i}\cap E_{\sigma\left(
i\right)  -1}. \label{pf.prop.5.2.alt}%
\end{equation}


Now, $\epsilon_{\sigma\left(  i\right)  }\mid_{V_{i}\cap E_{\sigma\left(
i\right)  }}\neq0$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus,
$\epsilon_{\sigma\left(  i\right)  }\mid_{V_{i}\cap E_{\sigma\left(  i\right)
}}=0$.
\par
Fix $x\in V_{i}\cap E_{\sigma\left(  i\right)  }$. Then, $x\in V_{i}\cap
E_{\sigma\left(  i\right)  }\subseteq E_{\sigma\left(  i\right)  }%
=\mathbb{F}_{p}\left\{  e_{1},e_{2},\ldots,e_{\sigma\left(  i\right)
}\right\}  $. Also, $x\in V_{i}\cap E_{\sigma\left(  i\right)  }$, so that
$\epsilon_{\sigma\left(  i\right)  }\left(  x\right)  =\underbrace{\left(
\epsilon_{\sigma\left(  i\right)  }\mid_{V_{i}\cap E_{\sigma\left(  i\right)
}}\right)  }_{=0}\left(  x\right)  =0$. In other words, the $\sigma\left(
i\right)  $-th coordinate of the vector $x$ is $0$. Combining this with
$x\in\mathbb{F}_{p}\left\{  e_{1},e_{2},\ldots,e_{\sigma\left(  i\right)
}\right\}  $, we conclude $x\in\mathbb{F}_{p}\left\{  e_{1},e_{2}%
,\ldots,e_{\sigma\left(  i\right)  -1}\right\}  =E_{\sigma\left(  i\right)
-1}$. Combining $x\in V_{i}\cap E_{\sigma\left(  i\right)  }\subseteq V_{i}$
with $x\in E_{\sigma\left(  i\right)  -1}$, we find $x\in V_{i}\cap
E_{\sigma\left(  i\right)  -1}$.
\par
Now, forget that we fixed $x$. We thus have proven that $x\in V_{i}\cap
E_{\sigma\left(  i\right)  -1}$ for each $x\in V_{i}\cap E_{\sigma\left(
i\right)  }$. In other words, $V_{i}\cap E_{\sigma\left(  i\right)  }\subseteq
V_{i}\cap E_{\sigma\left(  i\right)  -1}$. Hence, $V_{i}\cap E_{\sigma\left(
i\right)  -1}$ is not a proper subset of $V_{i}\cap E_{\sigma\left(  i\right)
}$. This contradicts (\ref{pf.prop.5.2.alt}). This contradiction shows that
our assumption was wrong, qed.}. Hence, the $\mathbb{F}_{p}$-linear map
$\epsilon_{\sigma\left(  i\right)  }\mid_{V_{i}\cap E_{\sigma\left(  i\right)
}}:V_{i}\cap E_{\sigma\left(  i\right)  }\rightarrow\mathbb{F}_{p}$ has rank
$\geq1$, and therefore must be surjective (since its target is the
$1$-dimensional $\mathbb{F}_{p}$-vector space $\mathbb{F}_{p}$). Therefore,
there exists some $x\in V_{i}\cap E_{\sigma\left(  i\right)  }$ satisfying
$\left(  \epsilon_{\sigma\left(  i\right)  }\mid_{V_{i}\cap E_{\sigma\left(
i\right)  }}\right)  \left(  x\right)  =1$. Consider this $x$.

We have $\epsilon_{\sigma\left(  i\right)  }\left(  x\right)  =\left(
\epsilon_{\sigma\left(  i\right)  }\mid_{V_{i}\cap E_{\sigma\left(  i\right)
}}\right)  \left(  x\right)  =1$. Furthermore, $x\in V_{i}\cap E_{\sigma
\left(  i\right)  }\subseteq E_{\sigma\left(  i\right)  }=S_{i}\oplus T_{i}$.
In other words, there exist $y\in S_{i}$ and $z\in T_{i}$ such that $x=y+z$.
Consider these $y$ and $z$. We have $x\in V_{i}\cap E_{\sigma\left(  i\right)
}\leq V_{i}$ and $y\in S_{i}\leq V_{i}\cap E_{\sigma\left(  i\right)  }\leq
V_{i}$. Now, $x=y+z$, so that $z=x-y\in V_{i}$ (since $x\in V_{i}$ and $y\in
V_{i}$, and since $V_{i}$ is an $\mathbb{F}_{p}$-vector space). Combining this
with $z\in T_{i}$, we obtain $z\in V_{i}\cap T_{i}$.

For each $j<i$, the leading term of $v_{j}$ is $e_{\sigma\left(  j\right)  }$
(since $e_{j}\in V_{j}\cap T_{j}\leq T_{j}\leq E_{\sigma\left(  j\right)  }$
and $\epsilon_{\sigma\left(  j\right)  }\left(  v_{j}\right)  =1$). Thus, for
each $j<i$, the leading index of $v_{j}$ is $\sigma\left(  j\right)  $. In
other words, for each $j<i$, we have $\operatorname*{lind}\left(
v_{j}\right)  =\sigma\left(  j\right)  $. Thus, every $j<i$ satisfying
$\sigma\left(  j\right)  <\sigma\left(  i\right)  $ must satisfy%
\[
\operatorname*{lind}\left(  v_{j}\right)  =\sigma\left(  j\right)  \in\left\{
1,2,\ldots,\sigma\left(  i\right)  -1\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sigma\left(  j\right)  <\sigma\left(  i\right)  \right)
\]
and therefore
\begin{equation}
v_{j}\in\mathbb{F}_{p}\left\{  e_{1},e_{2},\ldots,e_{\sigma\left(  i\right)
-1}\right\}  =E_{\sigma\left(  i\right)  -1}. \label{pf.prop.5.2.alt.lt}%
\end{equation}
Now,%
\begin{align*}
y  &  \in S_{i}=\mathbb{F}_{p}\left\{  v_{j}\ \mid\ j<i\text{ and }%
\sigma\left(  j\right)  <\sigma\left(  i\right)  \right\}  \subseteq
E_{\sigma\left(  i\right)  -1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since each }j<i\text{ satisfying }%
\sigma\left(  j\right)  <\sigma\left(  i\right)  \text{ satisfies
(\ref{pf.prop.5.2.alt.lt})}\right)
\end{align*}
and thus $\epsilon_{\sigma\left(  i\right)  }\left(  y\right)  =0$. Now,
\[
\epsilon_{\sigma\left(  i\right)  }\left(  \underbrace{x}_{=y+z}\right)
=\epsilon_{\sigma\left(  i\right)  }\left(  y+z\right)  =\underbrace{\epsilon
_{\sigma\left(  i\right)  }\left(  y\right)  }_{=0}+\epsilon_{\sigma\left(
i\right)  }\left(  z\right)  =\epsilon_{\sigma\left(  i\right)  }\left(
z\right)  .
\]
Comparing this with $\epsilon_{\sigma\left(  i\right)  }\left(  x\right)  =1$,
we find $\epsilon_{\sigma\left(  i\right)  }\left(  z\right)  =1$.

Furthermore, $z\notin V_{i-1}$\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Then, $z\in V_{i-1}=\mathbb{F}_{p}\left\{  v_{1},v_{2}%
,\ldots,v_{i-1}\right\}  =\mathbb{F}_{p}\left\{  v_{j}\ \mid\ j\in\left\{
1,2,\ldots,i-1\right\}  \right\}  $. Also, $z$ is nonzero (since
$\epsilon_{\sigma\left(  i\right)  }\left(  z\right)  =1\neq0$).
\par
For each $j<i$, the leading term of $v_{j}$ is $e_{\sigma\left(  j\right)  }$
(as we have already seen). Thus, for each $j<i$, the leading index of $v_{j}$
is $\sigma\left(  j\right)  $. Thus, the leading indices of $v_{1}%
,v_{2},\ldots,v_{i-1}$ are $\sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  i-1\right)  $. Therefore, these leading indices are
pairwise distinct (since $\sigma$ is injective). In other words, the family
$\left(  v_{j}\right)  _{j\in\left\{  1,2,\ldots,i-1\right\}  }$ of nonzero
vectors in $\mathbb{F}_{p}^{n}$ has the property that the leading indices of
the $v_{j}$ (for $j\in\left\{  1,2,\ldots,i-1\right\}  $) are pairwise
distinct. Thus, Lemma 5.2c \textbf{(c)} (applied to $\left\{  1,2,\ldots
,i-1\right\}  $, $\left(  v_{j}\right)  _{j\in\left\{  1,2,\ldots,i-1\right\}
}$ and $z$ instead of $S$, $\left(  w_{s}\right)  _{s\in S}$ and $x$) shows
that%
\begin{align*}
\operatorname*{lind}\left(  z\right)   &  \in\left\{  \operatorname*{lind}%
\left(  v_{j}\right)  \ \mid\ j\in\left\{  1,2,\ldots,i-1\right\}  \right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }z\in\mathbb{F}_{p}\left\{
v_{j}\ \mid\ j\in\left\{  1,2,\ldots,i-1\right\}  \right\}  \text{ is a
nonzero vector}\right) \\
&  =\left\{  \underbrace{\operatorname*{lind}\left(  v_{j}\right)
}_{\substack{=\sigma\left(  j\right)  \\\text{(since the leading index of
}v_{j}\text{ is }\sigma\left(  j\right)  \text{)}}}\ \mid\ j<i\right\}
=\left\{  \sigma\left(  j\right)  \ \mid\ j<i\right\}  .
\end{align*}
But $z\in T_{i}\leq E_{\sigma\left(  i\right)  }=\mathbb{F}_{p}\left\{
e_{1},e_{2},\ldots,e_{\sigma\left(  i\right)  }\right\}  $. Hence, the $k$-th
coordinate of $z$ is $0$ for all $k>\sigma\left(  i\right)  $. But the
$\sigma\left(  i\right)  $-th coordinate of $z$ is $\epsilon_{\sigma\left(
i\right)  }\left(  z\right)  =1\neq0$. Combining the preceding two sentences,
we conclude that the leading index of $z$ is $\sigma\left(  i\right)  $. In
other words, $\operatorname*{lind}\left(  z\right)  =\sigma\left(  i\right)
$. Hence, $\sigma\left(  i\right)  =\operatorname*{lind}\left(  z\right)
\in\left\{  \sigma\left(  j\right)  \ \mid\ j<i\right\}  $, so that
$\sigma\left(  i\right)  =\sigma\left(  j\right)  $ for some $j<i$. This is
absurd, since $\sigma$ is injective. Thus, we have obtained a contradiction.
This completes our proof of the fact that $z\notin V_{i-1}$.}.

Finally, the list $v_{1},v_{2},\ldots,v_{i-1},z$ is a basis for $V_{i}$ over
$\mathbb{F}_{p}$\ \ \ \ \footnote{\textit{Proof.} We have $V_{i-1}%
+\mathbb{F}_{p}z>V_{i-1}$ (since $z\in\mathbb{F}_{p}z\subseteq V_{i-1}%
+\mathbb{F}_{p}z$ but $z\notin V_{i-1}$). Hence, $\dim\left(  V_{i-1}%
+\mathbb{F}_{p}z\right)  >\dim\left(  V_{i-1}\right)  =i-1$. Since
$\dim\left(  V_{i-1}+\mathbb{F}_{p}z\right)  $ and $i-1$ are integers, this
entails that $\dim\left(  V_{i-1}+\mathbb{F}_{p}z\right)  \geq\left(
i-1\right)  +1=i=\dim\left(  V_{i}\right)  $. Furthermore,
$\underbrace{V_{i-1}}_{\leq V_{i}}+\underbrace{\mathbb{F}_{p}z}%
_{\substack{\leq V_{i}\\\text{(since }z\in V_{i}\text{)}}}\leq V_{i}%
+V_{i}=V_{i}$.
\par
Now, it is well-known that if $\mathfrak{U}$ is a subspace of a
finite-dimensional vector space $\mathfrak{V}$, and if $\dim\mathfrak{U}%
\geq\dim\mathfrak{V}$, then $\mathfrak{U}=\mathfrak{V}$. Applying this to
$\mathfrak{U}=V_{i-1}+\mathbb{F}_{p}z$ and $\mathfrak{V}=V_{i}$, we obtain
$V_{i-1}+\mathbb{F}_{p}z=V_{i}$ (since $V_{i-1}+\mathbb{F}_{p}z$ is a subspace
of $V_{i}$ and satisfies $\dim\left(  V_{i-1}+\mathbb{F}_{p}z\right)  \geq
\dim\left(  V_{i}\right)  $). Now,%
\[
\mathbb{F}_{p}\left\{  v_{1},v_{2},\ldots,v_{i-1},z\right\}
=\underbrace{\mathbb{F}_{p}\left\{  v_{1},v_{2},\ldots,v_{i-1}\right\}
}_{\substack{=V_{i-1}\\\text{(since }\left(  v_{1},v_{2},\ldots,v_{i-1}%
\right)  \\\text{is a basis for }V_{i-1}\text{)}}}+\mathbb{F}_{p}%
z=V_{i-1}+\mathbb{F}_{p}z=V_{i}.
\]
Hence, the list $\left(  v_{1},v_{2},\ldots,v_{i-1},z\right)  $ spans the
$\mathbb{F}_{p}$-vector space $V_{i}$. Since the size $i$ of this list equals
the dimension of $V_{i}$ (because $\dim\left(  V_{i}\right)  =i$), this shows
that the list $\left(  v_{1},v_{2},\ldots,v_{i-1},z\right)  $ is a basis for
$V_{i}$. Qed.}.

Thus, we have shown that $z$ is an element of $V_{i}\cap T_{i}$ such that
$\epsilon_{\sigma\left(  i\right)  }\left(  z\right)  =1$ and moreover that
$v_{1},v_{2},\ldots,v_{i-1},z$ is a basis for $V_{i}$ over $\mathbb{F}_{p}$.
Hence, there is an element $v_{i}\in V_{i}\cap T_{i}$ such that $\epsilon
_{\sigma\left(  i\right)  }\left(  v_{i}\right)  =1$ and moreover that
$v_{1},v_{2},\ldots,v_{i}$ is a basis for $V_{i}$ over $\mathbb{F}_{p}$
(namely, $v_{i}=z$). This completes our proof. (As I have said, the uniqueness
of this $v_{i}$ is not proven here, but it is not needed in your argument
either.) $\square$

\item \textbf{Proof of Proposition 5.2:} Before the words \textquotedblleft
Now define $g$\textquotedblright, add the following sentences:
\textquotedblleft Notice that, for each $i\in\left\{  1,2,\ldots,n\right\}  $,
the leading term of $v_{i}$ is $e_{\sigma\left(  i\right)  }$ (because
$v_{i}\in T_{i}\leq E_{\sigma\left(  i\right)  }=\mathbb{F}_{p}\left\{
e_{1},e_{2},\ldots,e_{\sigma\left(  i\right)  }\right\}  $ has its
$\sigma\left(  i\right)  $-th coordinate equal to $\epsilon_{\sigma\left(
i\right)  }\left(  v_{i}\right)  =1$). Hence, for each $i\in\left\{
1,2,\ldots,n\right\}  $, the leading term of $v_{\sigma^{-1}\left(  i\right)
}$ is $e_{i}$.\textquotedblright

\item \textbf{Proof of Proposition 5.2:} Replace \textquotedblleft Now define
$g$\textquotedblright\ by \textquotedblleft Now define an $\mathbb{F}_{p}%
$-linear map $g$\textquotedblright.

\item \textbf{Proof of Proposition 5.2:} Replace \textquotedblleft%
$\mathbb{F}_{p}\left\{  e_{\sigma\left(  k\right)  },e_{\sigma\left(
k+1\right)  },\ldots,e_{\sigma\left(  m\right)  }\right\}  $\textquotedblright%
\ by \newline\textquotedblleft$\mathbb{F}_{p}\left\{  e_{\sigma\left(
k\right)  },e_{\sigma\left(  k+1\right)  },\ldots,e_{\sigma\left(  n\right)
}\right\}  $\textquotedblright.

\item \textbf{Proof of Proposition 5.2:} Replace \textquotedblleft so
$\sigma^{-1}g\sigma$ is a lower-triangular matrix\textquotedblright\ by:
\textquotedblleft so $\sigma^{-1}g\sigma\left(  e_{k}\right)  \in
\mathbb{F}_{p}\left\{  e_{k},e_{k+1},\ldots,e_{n}\right\}  $. Hence,
$\sigma^{-1}g\sigma$ is a lower-triangular matrix\textquotedblright.

\item \textbf{Proof of Proposition 5.2:} Replace \textquotedblleft so $g\in
U^{\rho\sigma^{-1}}$\textquotedblright\ by \textquotedblleft so $g\in
U^{\rho\sigma^{-1}}=U^{\left(  \sigma\rho\right)  ^{-1}}$\textquotedblright.

\item \textbf{Proof of Proposition 5.2:} Replace \textquotedblleft and
$\phi\left(  g\right)  =\underline{V}$\textquotedblright\ by \textquotedblleft
and $\phi\left(  g\right)  =g\sigma\left(  \underline{E}\right)
=\underline{V}$\textquotedblright.

\item \textbf{Example 5.3:} Replace \textquotedblleft$g=\left[
\begin{array}
[c]{ccccc}%
1 & 0 & 0 & b & a\\
0 & 1 & 0 & d & c\\
0 & 0 & 1 & f & e\\
0 & 0 & 0 & 1 & g\\
0 & 0 & 0 & 0 & 1
\end{array}
\right]  $\textquotedblright\ by \textquotedblleft$g=\left[
\begin{array}
[c]{ccccc}%
1 & 0 & 0 & e & a\\
0 & 1 & 0 & f & b\\
0 & 0 & 1 & g & c\\
0 & 0 & 0 & 1 & d\\
0 & 0 & 0 & 0 & 1
\end{array}
\right]  $\textquotedblright. (Otherwise, the equation after it wouldn't be true.)

\item \textbf{Example 5.3:} Rename \textquotedblleft$g$\textquotedblright\ as
\textquotedblleft$h$\textquotedblright\ in the contexts \textquotedblleft%
$g=$\textquotedblright, \textquotedblleft For such $g$\textquotedblright,
\textquotedblleft$g\sigma=$\textquotedblright\ and \textquotedblleft%
$\phi\left(  g\right)  $\textquotedblright. (In fact, the notation
\textquotedblleft$g$\textquotedblright\ here clashes with the notation
\textquotedblleft$g$\textquotedblright\ for the $\left(  4,5\right)  $-th
entry of the matrix $g$.)

\item \textbf{Corollary 5.4:} I think you should define what you mean by
\textquotedblleft isomorphism\textquotedblright\ here. Namely, an
\textit{isomorphism} from a triple $\left(  V,\underline{U},\underline{W}%
\right)  $ (where $V$ is an $n$-dimensional $\mathbb{F}_{p}$-vector space, and
$\underline{U}$ and $\underline{W}$ are two complete flags in $V$) to a triple
$\left(  V^{\prime},\underline{U^{\prime}},\underline{W^{\prime}}\right)  $
(where $V^{\prime}$ is an $n$-dimensional $\mathbb{F}_{p}$-vector space, and
$\underline{U^{\prime}}$ and $\underline{W^{\prime}}$ are two complete flags
in $V^{\prime}$) means an isomorphism $\phi:V\rightarrow V^{\prime}$ of
$\mathbb{F}_{p}$-vector spaces satisfying $\phi U=U^{\prime}$ and $\phi
W=W^{\prime}$.

\item \textbf{Corollary 5.4:} Replace \textquotedblleft$\Sigma$%
\textquotedblright\ by \textquotedblleft$\Sigma_{n}$\textquotedblright.

\item \textbf{Corollary 5.4:} Replace \textquotedblleft if and only
iff\textquotedblright\ by \textquotedblleft if and only if\textquotedblright%
\ (or by \textquotedblleft iff\textquotedblright).

\item \textbf{Proof of Corollary 5.4:} Replace \textquotedblleft a pair as
above\textquotedblright\ by \textquotedblleft a triple as
above\textquotedblright.

\item \textbf{Proof of Corollary 5.4:} Replace \textquotedblleft by $f\left(
a\right)  =\sum_{i}a_{i}w_{i}$\textquotedblright\ by \textquotedblleft by
$f\left(  a_{1},a_{2},\ldots,a_{n}\right)  =\sum_{i}a_{i}w_{i}$%
\textquotedblright.

\item \textbf{Proof of Corollary 5.4:} Replace \textquotedblleft so
$\underline{F}=x\sigma\underline{E}$\textquotedblright\ by: \textquotedblleft.
Since the map $X\left(  \sigma\right)  \rightarrow Y\left(  \sigma\right)
,\ g\mapsto g\sigma\underline{E}$ is a bijection (by Proposition 5.2), we thus
see that $\underline{F}=x\sigma\underline{E}$\textquotedblright.

\item \textbf{Proof of Corollary 5.4:} At the end of this proof, add the
following sentence: \textquotedblleft Hence, $\left(  \mathbb{F}_{p}%
^{n},\sigma\underline{E},\underline{E}\right)  \simeq\left(  \mathbb{F}%
_{p}^{n},\underline{F},\underline{E}\right)  \simeq\left(  V,\underline{U}%
,\underline{W}\right)  $ (since the map $f:\mathbb{F}_{p}^{n}\rightarrow V$ is
an isomorphism $\left(  \mathbb{F}_{p}^{n},\underline{F},\underline{E}\right)
\rightarrow\left(  V,\underline{U},\underline{W}\right)  $).\textquotedblright

\item \textbf{Proof of Corollary 5.5:} After \textquotedblleft such bases
exist iff $\delta\left(  \underline{U},\underline{W}\right)  =\sigma
$\textquotedblright, add \textquotedblleft(because the first claim of
Corollary 5.4 shows that $\left(  \mathbb{F}_{p}^{n},\sigma\underline{E}%
,\underline{E}\right)  \cong\left(  V,\underline{U},\underline{W}\right)  $
holds if and only if $\delta\left(  \underline{U},\underline{W}\right)
=\delta\left(  \sigma\underline{E},\underline{E}\right)  $; but in light of
Example 4.1 this condition rewrites as $\delta\left(  \underline{U}%
,\underline{W}\right)  =\sigma$)\textquotedblright.

\item \textbf{Proof of Corollary 5.5:} I think the last sentence of this proof
would be better off taken out into a separate result:

\textbf{Lemma 5.5a.} Let $\sigma\in\Sigma_{n}$. Then, $\left\vert B^{\sigma
}\cap B\right\vert =\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}%
\rho\right)  }$.

[\textit{Proof of Lemma 5.5a.} The definition of $X\left(  \sigma^{-1}%
\rho\right)  $ yields%
\begin{align*}
X\left(  \sigma^{-1}\rho\right)   &  =U\cap U^{\left(  \sigma^{-1}\rho
\rho\right)  ^{-1}}\\
&  =U\cap U^{\sigma}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
\sigma^{-1}\underbrace{\rho\rho}_{=1}\right)  ^{-1}=\left(  \sigma
^{-1}\right)  ^{-1}=\sigma\right) \\
&  =U^{\sigma}\cap U.
\end{align*}
But the last sentence of Lemma 5.1 (applied to $\sigma^{-1}\rho$ instead of
$\sigma$) yields $\left\vert X\left(  \sigma^{-1}\rho\right)  \right\vert
=p^{l\left(  \sigma^{-1}\rho\right)  }$.

Consider the short exact sequence%
\[
1\longrightarrow U^{\sigma}\cap U\longrightarrow B^{\sigma}\cap
B\longrightarrow T\longrightarrow1,
\]
where the arrow $U^{\sigma}\cap U\longrightarrow B^{\sigma}\cap B$ is the
canonical inclusion, and where the arrow $B^{\sigma}\cap B\longrightarrow T$
is the map that replaces all off-diagonal entries of a matrix $g\in B^{\sigma
}\cap B$ by $0$.\ \ \ \ \footnote{This is indeed a short exact sequence,
because the matrices $g\in B^{\sigma}\cap B$ whose diagonal entries all equal
$1$ are exactly the elements of $U^{\sigma}\cap U$. It is actually a split
extension, since the arrow $B^{\sigma}\cap B\longrightarrow T$ is split by the
canonical inclusion $T\longrightarrow B^{\sigma}\cap B$.} This short exact
sequence shows that%
\begin{align}
\left\vert B^{\sigma}\cap B\right\vert  &  =\left\vert T\right\vert
\cdot\left\vert \underbrace{U^{\sigma}\cap U}_{=X\left(  \sigma^{-1}%
\rho\right)  }\right\vert =\left\vert T\right\vert \cdot\underbrace{\left\vert
X\left(  \sigma^{-1}\rho\right)  \right\vert }_{=p^{l\left(  \sigma^{-1}%
\rho\right)  }}=\underbrace{\left\vert T\right\vert }_{=\left(  p-1\right)
^{n}}\cdot p^{l\left(  \sigma^{-1}\rho\right)  }\label{l5.5a.pf.1}\\
&  =\left(  p-1\right)  ^{n}\cdot p^{l\left(  \sigma^{-1}\rho\right)
}.\nonumber
\end{align}
Thus, Lemma 5.5a is proven. $\square$]

\item \textbf{Proof of Proposition 6.1:} Before \textquotedblleft Moreover,
if\textquotedblright, add the following sentence: \textquotedblleft Thus,
$\pi\left(  BgB\right)  =\left\{  \pi\left(  g\right)  \right\}  $ for each
$g\in G$.\textquotedblright.

\item \textbf{Proof of Proposition 6.1:} Replace \textquotedblleft from which
it follows directly that $\pi\left(  \sigma\right)  =\sigma$\textquotedblright%
\ by \textquotedblleft and the definition of $\pi$ yields $\pi\left(
\sigma\right)  =\delta\left(  \sigma\underline{E},\underline{E}\right)
=\sigma$ (by Example 4.1)\textquotedblright.

\item \textbf{Proof of Proposition 6.1:} After \textquotedblleft This means
that $\pi\left(  B\sigma B\right)  =\left\{  \sigma\right\}  $%
.\textquotedblright, add \textquotedblleft Hence, $B\sigma B\subseteq\pi
^{-1}\left\{  \sigma\right\}  $.\textquotedblright.

\item \textbf{Proof of Proposition 6.1:} Replace \textquotedblleft Conversely,
suppose that $\pi\left(  h\right)  =\sigma$.\textquotedblright\ by
\textquotedblleft Conversely, let $h\in\pi^{-1}\left\{  \sigma\right\}  $.
Thus, $h\in G$ and $\pi\left(  h\right)  =\sigma$. Hence, $\sigma=\pi\left(
h\right)  =\delta\left(  h\underline{E},\underline{E}\right)  $, so that
$h\underline{E}\in Y\left(  \sigma,\underline{E}\right)  =Y\left(
\sigma\right)  $. Hence,\textquotedblright.

\item \textbf{Proof of Proposition 6.1:} Replace \textquotedblleft
Propsition\textquotedblright\ by \textquotedblleft
Proposition\textquotedblright.

\item \textbf{Proof of Proposition 6.1:} After \textquotedblleft we find that
$b\in B$\textquotedblright, add \textquotedblleft(since $b\underline{E}%
=\left(  g\sigma\right)  ^{-1}\underbrace{h\underline{E}}_{=g\sigma
\underline{E}}=\left(  g\sigma\right)  ^{-1}g\sigma\underline{E}%
=\underline{E}$)\textquotedblright.

\item \textbf{Proof of Proposition 6.1:} After \textquotedblleft shows that
$h\in B\sigma B$\textquotedblright, add \textquotedblleft(since $g\in X\left(
\sigma\right)  \leq U\leq B$ and $b\in B$). Hence, we have shown that
$\pi^{-1}\left\{  \sigma\right\}  \subseteq B\sigma B$ (since $h$ was assumed
to be any element of $G$ satisfying $\pi\left(  h\right)  =\sigma
$)\textquotedblright.

\item \textbf{Proof of Proposition 6.1:} Replace \textquotedblleft we see that
$g\sigma\underline{E}=g^{\prime}\sigma\underline{E}$\textquotedblright\ by
\textquotedblleft and $g\sigma b=h=g^{\prime}\sigma b^{\prime}$, we see that
$g\sigma\underbrace{\underline{E}}_{=b\underline{E}}=\underbrace{g\sigma
b}_{=g^{\prime}\sigma b^{\prime}}\underline{E}=g^{\prime}\sigma
\underbrace{b^{\prime}\underline{E}}_{=\underline{E}}=g^{\prime}%
\sigma\underline{E}$\textquotedblright.

\item \textbf{Proof of Corollary 6.2:} The claim \textquotedblleft$B=T\times
U$\textquotedblright\ is wrong, or at least seriously misleading (the group
$B$ is not a direct product of $T$ and $U$). I would replace the whole
sentence containing it by \textquotedblleft Applying (\ref{l5.5a.pf.1}) to
$\rho\sigma^{-1}$ instead of $\sigma$, we obtain $\left\vert B^{\rho
\sigma^{-1}}\cap B\right\vert =\left\vert T\right\vert \cdot p^{l\left(
\left(  \rho\sigma^{-1}\right)  ^{-1}\rho\right)  }$. Since $B^{\rho
\sigma^{-1}}\cap B=B\cap B^{\rho\sigma^{-1}}$ and $\left(  \rho\sigma
^{-1}\right)  ^{-1}\rho=\sigma$, this rewrites as $\left\vert B\cap
B^{\rho\sigma^{-1}}\right\vert =\left\vert T\right\vert \cdot p^{l\left(
\sigma\right)  }$.

\item \textbf{Proof of Corollary 6.2:} After the displayed equation
\textquotedblleft$\left\vert B\sigma B\right\vert =p^{l\left(  \sigma\right)
}\left\vert B\right\vert =\left\vert U\right\vert \left\vert T\right\vert
p^{l\left(  \sigma\right)  }$\textquotedblright, add \textquotedblleft(since
the short exact sequence $1\longrightarrow U\longrightarrow B\longrightarrow
T\longrightarrow1$ yields $\left\vert B\right\vert =\left\vert U\right\vert
\left\vert T\right\vert $)\textquotedblright.

\item \textbf{Proof of Corollary 6.3:} Before \textquotedblleft We have a
bijection\textquotedblright, add \textquotedblleft Proposition 6.1 (applied to
$\sigma^{-1}$ instead of $\sigma$) shows that\textquotedblright.

\item \textbf{Proof of Corollary 6.3:} Replace \textquotedblleft given by
$\phi\left(  g,b\right)  =g\sigma b$\textquotedblright\ by \textquotedblleft
given by $\phi\left(  g,b\right)  =g\sigma^{-1}b$\textquotedblright.

\item \textbf{Proof of Proposition 6.4:} The statement in the first sentence
of your proof is worth stating as a separate lemma:

\textbf{Lemma 2.6a.} Let $\sigma,\tau\in\Sigma_{n}$. Then, the following two
conditions are equivalent:

\textit{Condition }$\mathcal{C}_{1}$\textit{:} We have $l\left(  \sigma
\tau\right)  =l\left(  \sigma\right)  +l\left(  \tau\right)  $.

\textit{Condition $\mathcal{C}$}$_{2}$\textit{:} For any $T\subseteq\left\{
1,2,\ldots,n\right\}  $ with $\left\vert T\right\vert =2$, at most one of the
two maps $T\overset{\tau}{\longrightarrow}\tau\left(  T\right)
\overset{\sigma}{\longrightarrow}\sigma\tau\left(  T\right)  $ is order-reversing.

[\textit{Proof of Lemma 2.6a.} Let us prove the implications $\mathcal{C}%
_{1}\Longrightarrow\mathcal{C}_{2}$ and $\mathcal{C}_{2}\Longrightarrow
\mathcal{C}_{1}$ separately.

\textit{Proof of the implication }$\mathcal{C}_{1}\Longrightarrow
\mathcal{C}_{2}$\textit{:} Assume that Condition $\mathcal{C}_{1}$ holds. In
other words, we have $l\left(  \sigma\tau\right)  =l\left(  \sigma\right)
+l\left(  \tau\right)  $.

Lemma 2.6 shows that $\overline{L}\left(  \sigma\tau\right)  =\overline
{L}\left(  \tau\right)  \Delta\tau_{\ast}^{-1}\overline{L}\left(
\sigma\right)  $. Also, $l\left(  \sigma\right)  =\left\vert L\left(
\sigma\right)  \right\vert =\left\vert \overline{L}\left(  \sigma\right)
\right\vert $ and similarly $l\left(  \tau\right)  =\left\vert \overline
{L}\left(  \tau\right)  \right\vert $ and $l\left(  \sigma\tau\right)
=\left\vert \overline{L}\left(  \sigma\tau\right)  \right\vert $. Now,%
\begin{align*}
\left\vert \overline{L}\left(  \tau\right)  \right\vert
+\underbrace{\left\vert \tau_{\ast}^{-1}\overline{L}\left(  \sigma\right)
\right\vert }_{\substack{=\left\vert \overline{L}\left(  \sigma\right)
\right\vert \\\text{(since }\tau_{\ast}\text{ is a bijection)}}}  &
=\underbrace{\left\vert \overline{L}\left(  \tau\right)  \right\vert
}_{=l\left(  \tau\right)  }+\underbrace{\left\vert \overline{L}\left(
\sigma\right)  \right\vert }_{=l\left(  \sigma\right)  }=l\left(  \tau\right)
+l\left(  \sigma\right)  =l\left(  \sigma\right)  +l\left(  \tau\right) \\
&  =l\left(  \sigma\tau\right)  =\left\vert \underbrace{\overline{L}\left(
\sigma\tau\right)  }_{=\overline{L}\left(  \tau\right)  \Delta\tau_{\ast}%
^{-1}\overline{L}\left(  \sigma\right)  }\right\vert =\left\vert \overline
{L}\left(  \tau\right)  \Delta\tau_{\ast}^{-1}\overline{L}\left(
\sigma\right)  \right\vert .
\end{align*}
But a simple and fundamental fact states that if $A$ and $B$ are two finite
sets satisfying $\left\vert A\right\vert +\left\vert B\right\vert =\left\vert
A\Delta B\right\vert $, then $A\cap B=\varnothing$. Applying this to
$A=\overline{L}\left(  \tau\right)  $ and $B=\tau_{\ast}^{-1}\overline
{L}\left(  \sigma\right)  $, we find that $\overline{L}\left(  \tau\right)
\cap\tau_{\ast}^{-1}\overline{L}\left(  \sigma\right)  =\varnothing$ (since
$\left\vert \overline{L}\left(  \tau\right)  \right\vert +\left\vert
\tau_{\ast}^{-1}\overline{L}\left(  \sigma\right)  \right\vert =\left\vert
\overline{L}\left(  \tau\right)  \Delta\tau_{\ast}^{-1}\overline{L}\left(
\sigma\right)  \right\vert $).

Now, let $T\subseteq\left\{  1,2,\ldots,n\right\}  $ with $\left\vert
T\right\vert =2$. We shall show that at most one of the two maps
$T\overset{\tau}{\longrightarrow}\tau\left(  T\right)  \overset{\sigma
}{\longrightarrow}\sigma\tau\left(  T\right)  $ is order-reversing.

Indeed, assume the contrary. Thus, both maps $T\overset{\tau}{\longrightarrow
}\tau\left(  T\right)  \overset{\sigma}{\longrightarrow}\sigma\tau\left(
T\right)  $ are order-reversing.

One of the characterizations of $\overline{L}\left(  \sigma\right)  $ shows
that the two-element subset $\tau\left(  T\right)  $ of $\left\{
1,2,\ldots,n\right\}  $ belongs to $\overline{L}\left(  \sigma\right)  $ if
and only if the map $\sigma:\tau\left(  T\right)  \rightarrow\sigma\tau\left(
T\right)  $ is order-reversing. Hence, the two-element subset $\tau\left(
T\right)  $ of $\left\{  1,2,\ldots,n\right\}  $ belongs to $\overline
{L}\left(  \sigma\right)  $ (since the map $\sigma:\tau\left(  T\right)
\rightarrow\sigma\tau\left(  T\right)  $ is order-reversing). Thus,
$\tau\left(  T\right)  \in\overline{L}\left(  \sigma\right)  $. But
$\tau_{\ast}\left(  T\right)  =\tau\left(  T\right)  \in\overline{L}\left(
\sigma\right)  $, so that $T\in\tau_{\ast}^{-1}\overline{L}\left(
\sigma\right)  $.

One of the characterizations of $\overline{L}\left(  \tau\right)  $ shows that
the two-element subset $T$ of $\left\{  1,2,\ldots,n\right\}  $ belongs to
$\overline{L}\left(  \tau\right)  $ if and only if the map $\tau
:T\rightarrow\tau\left(  T\right)  $ is order-reversing. Hence, the
two-element subset $T$ of $\left\{  1,2,\ldots,n\right\}  $ belongs to
$\overline{L}\left(  \tau\right)  $ (since the map $\tau:T\rightarrow
\tau\left(  T\right)  $ is order-reversing). In other words, $T\in\overline
{L}\left(  \tau\right)  $. Combining this with $T\in\tau_{\ast}^{-1}%
\overline{L}\left(  \sigma\right)  $, we obtain $T\in\overline{L}\left(
\tau\right)  \cap\tau_{\ast}^{-1}\overline{L}\left(  \sigma\right)
=\varnothing$. In other words, $T$ belongs to the empty set. This is clearly absurd.

Thus, we have obtained a contradiction. Hence, our assumption was wrong. We
thus have proven that at most one of the two maps $T\overset{\tau
}{\longrightarrow}\tau\left(  T\right)  \overset{\sigma}{\longrightarrow
}\sigma\tau\left(  T\right)  $ is order-reversing.

Now, forget that we fixed $T$. We thus have shown that for any $T\subseteq
\left\{  1,2,\ldots,n\right\}  $ with $\left\vert T\right\vert =2$, at most
one of the two maps $T\overset{\tau}{\longrightarrow}\tau\left(  T\right)
\overset{\sigma}{\longrightarrow}\sigma\tau\left(  T\right)  $ is
order-reversing. In other words, Condition $\mathcal{C}_{2}$ holds.

Thus, we have derived Condition $\mathcal{C}_{2}$ from Condition
$\mathcal{C}_{1}$. In other words, we have proven the implication
$\mathcal{C}_{1}\Longrightarrow\mathcal{C}_{2}$.

We omit the proof of the implication $\mathcal{C}_{2}\Longrightarrow
\mathcal{C}_{1}$ (since you don't actually use this implication in your
arguments, and since this proof is rather easy to obtained by
\textquotedblleft walking backwards\textquotedblright\ our above proof of the
implication $\mathcal{C}_{1}\Longrightarrow\mathcal{C}_{2}$).

Combining the implications $\mathcal{C}_{1}\Longrightarrow\mathcal{C}_{2}$ and
$\mathcal{C}_{2}\Longrightarrow\mathcal{C}_{1}$, we obtain the equivalence
$\mathcal{C}_{1}\Longleftrightarrow\mathcal{C}_{2}$. Thus, Lemma 2.6a is proven.]

\item \textbf{\S 6:} Let me suggest an alternative way of proving Proposition
6.4 and Proposition 6.5. This alternative way has the advantage that it does
not use the finiteness of the field $\mathbb{F}_{p}$, and so can be directly
generalized to an arbitrary field.\footnote{A few comments on your proofs:
\par
-- In the proof of Proposition 6.4, before \textquotedblleft We now show that
$\phi\left(  g,h\right)  \in X\left(  \sigma\tau\right)  $\textquotedblright,
I would suggest adding \textquotedblleft Let $g\in X\left(  \sigma\right)  $
and $h\in X\left(  \tau\right)  $.\textquotedblright.
\par
-- In the proof of Proposition 6.4, \textquotedblleft apply Corollary
6.1\textquotedblright\ should be replaced by \textquotedblleft apply
Proposition 6.1\textquotedblright.}

First, let me show a few useful lemmas:

\textbf{Lemma 6.4a.} Let $\sigma\in\Sigma_{n}$ and $g=\left(  g_{i,j}\right)
_{i,j=1}^{n}\in G$.

\textbf{(a)} If $g\in X\left(  \sigma\right)  $, then%
\begin{equation}
\left(  g_{i,i}=1\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right)
\label{eq.l6.4a.1}%
\end{equation}
and%
\begin{equation}
\left(  g_{i,j}=0\text{ for any }i,j\in\left\{  1,2,\ldots,n\right\}  \text{
satisfying }i\neq j\text{ and }\left(  i,j\right)  \notin L\left(  \sigma
^{-1}\right)  \right)  . \label{eq.l6.4a.2}%
\end{equation}


\textbf{(b)} If (\ref{eq.l6.4a.1}) and (\ref{eq.l6.4a.2}) hold, then $g\in
X\left(  \sigma\right)  $.

[\textit{Proof of Lemma 6.4a.} From the first sentence of Lemma 5.1, we see
that $g\in X\left(  \sigma\right)  $ holds if and only if we have%
\[
g_{i,j}=%
\begin{cases}
1, & \text{if }i=j;\\
\text{arbitrary,} & \text{if }\left(  i,j\right)  \in L\left(  \sigma
^{-1}\right)  ;\\
0, & \text{otherwise}%
\end{cases}
.
\]
In other words, $g\in X\left(  \sigma\right)  $ holds if and only if
(\ref{eq.l6.4a.1}) and (\ref{eq.l6.4a.2}) hold. This proves both parts
\textbf{(a)} and \textbf{(b)} of Lemma 6.4a. $\square$]

\textbf{Lemma 6.4b.} Let $g=\left(  g_{i,j}\right)  _{i,j=1}^{n}\in G$.

\textbf{(a)} If $g\in U$, then%
\begin{equation}
\left(  g_{i,i}=1\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right)
\label{eq.l6.4b.1}%
\end{equation}
and%
\begin{equation}
\left(  g_{i,j}=0\text{ for any }i,j\in\left\{  1,2,\ldots,n\right\}  \text{
satisfying }i>j\right)  . \label{eq.l6.4b.2}%
\end{equation}


\textbf{(b)} If (\ref{eq.l6.4b.1}) and (\ref{eq.l6.4b.2}) hold, then $g\in U$.

[\textit{Proof of Lemma 6.4b.} Recall that $U$ is the set of all
upper-unitriangular $n\times n$-matrices. Hence, $g\in U$ holds if and only if
$g$ is upper-unitriangular. By the definition of upper-triangular, this
rewrites as follows: $g\in U$ holds if and only if (\ref{eq.l6.4b.1}) and
(\ref{eq.l6.4b.2}) hold. This proves both parts \textbf{(a)} and \textbf{(b)}
of Lemma 6.4b. $\square$]

\textbf{Lemma 6.4c.} Let $\sigma\in\Sigma_{n}$. Then, the map $X\left(
\sigma\right)  \times B\rightarrow B\sigma B,\ \left(  g,b\right)  \mapsto
g\sigma b$ is well-defined and is a bijection.

[\textit{Proof of Lemma 6.4c.} Lemma 6.4c is the first claim of Proposition
6.1, and thus we already have proven it. $\square$]

\textbf{Lemma 6.4d.} Let $\sigma\in\Sigma_{n}$ and $\tau\in\Sigma_{n}$ be such
that $l\left(  \sigma\tau\right)  =l\left(  \sigma\right)  +l\left(
\tau\right)  $. Then:

\textbf{(a)} We have $\overline{L}\left(  \tau\right)  \cap\tau_{\ast}%
^{-1}\overline{L}\left(  \sigma\right)  =\varnothing$.

\textbf{(b)} We have $X\left(  \sigma\right)  \cap\left(  X\left(
\tau\right)  \right)  ^{\sigma^{-1}}=1$.

\textbf{(c)} We have $X\left(  \sigma\right)  \subseteq U^{\rho\tau^{-1}%
\sigma^{-1}}$.

\textbf{(d)} We have $\left(  X\left(  \tau\right)  \right)  ^{\sigma^{-1}%
}\subseteq U$.

\textbf{(e)} We have $X\left(  \sigma\right)  \cdot\left(  X\left(
\tau\right)  \right)  ^{\sigma^{-1}}\subseteq X\left(  \sigma\tau\right)  $.

\textbf{(f)} We have $B\sigma B\tau B=B\sigma\tau B$.

\textbf{(g)} The map
\[
X\left(  \sigma\right)  \times X\left(  \tau\right)  \rightarrow X\left(
\sigma\tau\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  g,h\right)  \mapsto
gh^{\sigma^{-1}}%
\]
is well-defined and bijective.

[\textit{Proof of Lemma 6.4d.} \textbf{(a)} This was shown in the proof of
Claim 1 during the proof of Lemma 2.6a given above.

\textbf{(b)} Let $g\in X\left(  \sigma\right)  \cap U^{\sigma^{-1}}$. Thus,
$g\in X\left(  \sigma\right)  $ and $g\in U^{\sigma^{-1}}$. Write the matrix
$g$ in the form $g=\left(  g_{i,j}\right)  _{i,j=1}^{n}$. Then, $g^{\sigma
}=\left(  g_{\sigma\left(  i\right)  ,\sigma\left(  j\right)  }\right)
_{i,j=1}^{n}$. But $g\in U^{\sigma^{-1}}$, so that%
\[
g^{\sigma}\in\left(  U^{\sigma^{-1}}\right)  ^{\sigma}=U^{\sigma^{-1}\sigma
}=U^{1}=U.
\]


Hence, Lemma 6.4a \textbf{(a)} (applied to $g^{\sigma}$ and $g_{\sigma\left(
i\right)  ,\sigma\left(  j\right)  }$ instead of $g$ and $g_{i,j}$) yields
that%
\begin{equation}
\left(  g_{\sigma\left(  i\right)  ,\sigma\left(  i\right)  }=1\text{ for all
}i\in\left\{  1,2,\ldots,n\right\}  \right)  \label{pf.l6.4b.b.1}%
\end{equation}
and%
\begin{equation}
\left(  g_{\sigma\left(  i\right)  ,\sigma\left(  j\right)  }=0\text{ for any
}i,j\in\left\{  1,2,\ldots,n\right\}  \text{ satisfying }i>j\right)  .
\label{pf.l6.4b.b.2}%
\end{equation}


On the other hand, $g\in X\left(  \sigma\right)  $. Hence, Lemma 6.4a
\textbf{(a)} yields that%
\begin{equation}
\left(  g_{i,i}=1\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right)
\label{pf.l6.4b.b.3}%
\end{equation}
and%
\begin{equation}
\left(  g_{i,j}=0\text{ for any }i,j\in\left\{  1,2,\ldots,n\right\}  \text{
satisfying }i\neq j\text{ and }\left(  i,j\right)  \notin L\left(  \sigma
^{-1}\right)  \right)  . \label{pf.l6.4b.b.4}%
\end{equation}


Now, let $i,j\in\left\{  1,2,\ldots,n\right\}  $. We shall show that
$g_{i,j}=\delta_{i,j}$.

Indeed, assume the contrary. Thus, $g_{i,j}\neq\delta_{i,j}$. Hence, $i\neq
j$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $i=j$. Hence,
$j=i$, so that $g_{i,j}=g_{i,i}=1$ (by (\ref{pf.l6.4b.b.3})). But from $i=j$,
we also obtain $\delta_{i,j}=1$. Comparing this with $g_{i,j}=1$, we obtain
$g_{i,j}=\delta_{i,j}$. This contradicts $g_{i,j}\neq\delta_{i,j}$. This
contradiction shows that our assumption was wrong. Qed.}, so that
$\delta_{i,j}=0$ and thus $g_{i,j}\neq\delta_{i,j}=0$.

If we had $\left(  i,j\right)  \notin L\left(  \sigma^{-1}\right)  $, then we
would have $g_{i,j}=0$ (by (\ref{pf.l6.4b.b.4})), which would contradict
$g_{i,j}\neq0$. Thus, we cannot have $\left(  i,j\right)  \notin L\left(
\sigma^{-1}\right)  $. Hence, we have $\left(  i,j\right)  \in L\left(
\sigma^{-1}\right)  $. In other words, $i<j$ and $\sigma^{-1}\left(  i\right)
>\sigma^{-1}\left(  j\right)  $ (by the definition of $L\left(  \sigma
^{-1}\right)  $). Thus, (\ref{pf.l6.4b.b.2}) (applied to $\left(  \sigma
^{-1}\left(  i\right)  ,\sigma^{-1}\left(  j\right)  \right)  $ instead of
$\left(  i,j\right)  $) yields

$g_{\sigma\left(  \sigma^{-1}\left(  i\right)  \right)  ,\sigma\left(
\sigma^{-1}\left(  j\right)  \right)  }=0$. This contradicts $g_{\sigma\left(
\sigma^{-1}\left(  i\right)  \right)  ,\sigma\left(  \sigma^{-1}\left(
j\right)  \right)  }=g_{i,j}\neq0$. This contradiction shows that our
assumption was false. Hence, $g_{i,j}=\delta_{i,j}$ is proven.

Now, forget that we fixed $i,j$. We thus have shown that $g_{i,j}=\delta
_{i,j}$ for all $i,j\in\left\{  1,2,\ldots,n\right\}  $. In other words,
$\left(  g_{i,j}\right)  _{i,j=1}^{n}=\left(  \delta_{i,j}\right)
_{i,j=1}^{n}=1$. Hence, $g=\left(  g_{i,j}\right)  _{i,j=1}^{n}=1$.

Now, forget that we fixed $g$. Thus we have proven that $g=1$ for each $g\in
X\left(  \sigma\right)  \cap U^{\sigma^{-1}}$. In other words, $X\left(
\sigma\right)  \cap U^{\sigma^{-1}}=1$.

But the definition of $X\left(  \tau\right)  $ yields $X\left(  \tau\right)
=U\cap U^{\left(  \tau\rho\right)  ^{-1}}\subseteq U$. Thus, $\left(  X\left(
\tau\right)  \right)  ^{\sigma^{-1}}\subseteq U^{\sigma^{-1}}$, so that
$X\left(  \sigma\right)  \cap\underbrace{\left(  X\left(  \tau\right)
\right)  ^{\sigma^{-1}}}_{\subseteq U^{\sigma^{-1}}}\subseteq X\left(
\sigma\right)  \cap U^{\sigma^{-1}}=1$. Hence, $X\left(  \sigma\right)
\cap\left(  X\left(  \tau\right)  \right)  ^{\sigma^{-1}}=1$. This proves
Lemma 6.4d \textbf{(b)}.

\textbf{(c)} Let $g\in X\left(  \sigma\right)  $. We shall prove that $g\in
U^{\rho\tau^{-1}\sigma^{-1}}$.

Write the matrix $g$ in the form $g=\left(  g_{i,j}\right)  _{i,j=1}^{n}$.

We have $g\in X\left(  \sigma\right)  $. Hence, Lemma 6.4a \textbf{(a)} yields
that%
\begin{equation}
\left(  g_{i,i}=1\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right)
\label{pf.l6.4b.c.1}%
\end{equation}
and%
\begin{equation}
\left(  g_{i,j}=0\text{ for any }i,j\in\left\{  1,2,\ldots,n\right\}  \text{
satisfying }i\neq j\text{ and }\left(  i,j\right)  \notin L\left(  \sigma
^{-1}\right)  \right)  . \label{pf.l6.4b.c.2}%
\end{equation}


Let $\eta=\sigma\tau\rho$. Thus, $\eta\in\Sigma_{n}$, so that $\eta$ is an
injective map. From $\eta=\sigma\tau\rho$, we obtain $\eta^{-1}=\left(
\sigma\tau\rho\right)  ^{-1}=\underbrace{\rho^{-1}}_{=\rho}\tau^{-1}%
\sigma^{-1}=\rho\tau^{-1}\sigma^{-1}$.

Let $i,j\in\left\{  1,2,\ldots,n\right\}  $ be such that $i>j$. We shall show
that $g_{\eta\left(  i\right)  ,\eta\left(  j\right)  }=0$.

Indeed, assume the contrary. Thus, $g_{\eta\left(  i\right)  ,\eta\left(
j\right)  }\neq0$. From $i>j$, we obtain $i\neq j$ and therefore $\eta\left(
i\right)  \neq\eta\left(  j\right)  $ (since $\eta$ is injective). If we had
$\left(  \eta\left(  i\right)  ,\eta\left(  j\right)  \right)  \notin L\left(
\sigma^{-1}\right)  $, then we would have $g_{\eta\left(  i\right)
,\eta\left(  j\right)  }=0$ (by (\ref{pf.l6.4b.c.2}), applied to $\eta\left(
i\right)  $ and $\eta\left(  j\right)  $ instead of $i$ and $j$), which would
contradict $g_{\eta\left(  i\right)  ,\eta\left(  j\right)  }\neq0$. Hence, we
cannot have $\left(  \eta\left(  i\right)  ,\eta\left(  j\right)  \right)
\notin L\left(  \sigma^{-1}\right)  $. Thus, we must have $\left(  \eta\left(
i\right)  ,\eta\left(  j\right)  \right)  \in L\left(  \sigma^{-1}\right)  $.
In other words, $\eta\left(  i\right)  <\eta\left(  j\right)  $ and
$\sigma^{-1}\left(  \eta\left(  i\right)  \right)  >\sigma^{-1}\left(
\eta\left(  j\right)  \right)  $ (by the definition of $L\left(  \sigma
^{-1}\right)  $). From $\eta\left(  i\right)  <\eta\left(  j\right)  $, we
obtain $\eta\left(  j\right)  >\eta\left(  i\right)  $.

Now, the definition of $\rho$ yields $\rho\left(  i\right)
=n+1-\underbrace{i}_{>j}<n+1-j=\rho\left(  j\right)  $ (since $\rho\left(
j\right)  =n+1-j$ (by the definition of $\rho$)). But $\eta=\sigma\tau\rho$,
so that $\sigma^{-1}\eta=\tau\rho$. Now, $\sigma^{-1}\left(  \eta\left(
i\right)  \right)  =\underbrace{\left(  \sigma^{-1}\eta\right)  }_{=\tau\rho
}\left(  i\right)  =\left(  \tau\rho\right)  \left(  i\right)  =\tau\left(
\rho\left(  i\right)  \right)  $ and similarly $\sigma^{-1}\left(  \eta\left(
j\right)  \right)  =\tau\left(  \rho\left(  j\right)  \right)  $. Thus,
$\tau\left(  \rho\left(  i\right)  \right)  =\sigma^{-1}\left(  \eta\left(
i\right)  \right)  >\sigma^{-1}\left(  \eta\left(  j\right)  \right)
=\tau\left(  \rho\left(  j\right)  \right)  $.

Combining $\rho\left(  i\right)  <\rho\left(  j\right)  $ with $\tau\left(
\rho\left(  i\right)  \right)  >\tau\left(  \rho\left(  j\right)  \right)  $,
we obtain $\left(  \rho\left(  i\right)  ,\rho\left(  j\right)  \right)  \in
L\left(  \tau\right)  $ (by the definition of $L\left(  \tau\right)  $). Thus,
$\left\{  \rho\left(  i\right)  ,\rho\left(  j\right)  \right\}  \in
\overline{L}\left(  \tau\right)  $ (by the definition of $\overline{L}\left(
\tau\right)  $).

On the other hand, $\tau\left(  \rho\left(  j\right)  \right)  <\tau\left(
\rho\left(  i\right)  \right)  $ (since $\tau\left(  \rho\left(  i\right)
\right)  >\tau\left(  \rho\left(  j\right)  \right)  $) and $\sigma\left(
\tau\left(  \rho\left(  j\right)  \right)  \right)  =\underbrace{\left(
\sigma\tau\rho\right)  }_{=\eta}\left(  j\right)  =\eta\left(  j\right)
>\underbrace{\eta}_{=\sigma\tau\rho}\left(  i\right)  =\left(  \sigma\tau
\rho\right)  \left(  i\right)  =\sigma\left(  \tau\left(  \rho\left(
i\right)  \right)  \right)  $. Combining these two inequalities, we obtain
$\left(  \tau\left(  \rho\left(  j\right)  \right)  ,\tau\left(  \rho\left(
i\right)  \right)  \right)  \in L\left(  \sigma\right)  $ (by the definition
of $L\left(  \sigma\right)  $). Hence, $\left\{  \tau\left(  \rho\left(
j\right)  \right)  ,\tau\left(  \rho\left(  i\right)  \right)  \right\}
\in\overline{L}\left(  \sigma\right)  $ (by the definition of $\overline
{L}\left(  \sigma\right)  $).

Now,%
\begin{align*}
\tau_{\ast}\left(  \left\{  \rho\left(  i\right)  ,\rho\left(  j\right)
\right\}  \right)   &  =\tau\left(  \left\{  \rho\left(  i\right)
,\rho\left(  j\right)  \right\}  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
the definition of }\tau_{\ast}\right) \\
&  =\left\{  \tau\left(  \rho\left(  i\right)  \right)  ,\tau\left(
\rho\left(  j\right)  \right)  \right\}  =\left\{  \tau\left(  \rho\left(
j\right)  \right)  ,\tau\left(  \rho\left(  i\right)  \right)  \right\}
\in\overline{L}\left(  \sigma\right)  ,
\end{align*}
so that $\left\{  \rho\left(  i\right)  ,\rho\left(  j\right)  \right\}
\in\tau_{\ast}^{-1}\overline{L}\left(  \sigma\right)  $. Combining this with
$\left\{  \rho\left(  i\right)  ,\rho\left(  j\right)  \right\}  \in
\overline{L}\left(  \tau\right)  $, we obtain $\left\{  \rho\left(  i\right)
,\rho\left(  j\right)  \right\}  \in\overline{L}\left(  \tau\right)  \cap
\tau_{\ast}^{-1}\overline{L}\left(  \sigma\right)  =\varnothing$. Thus,
$\left\{  \rho\left(  i\right)  ,\rho\left(  j\right)  \right\}  $ belongs to
the empty set. This is clearly absurd. Thus, we have obtained a contradiction.
This shows that our assumption was false. Hence, $g_{\eta\left(  i\right)
,\eta\left(  j\right)  }=0$ is proven.

Now, forget that we fixed $i,j$. We thus have shown that
\begin{equation}
\left(  g_{\eta\left(  i\right)  ,\eta\left(  j\right)  }=0\text{ for any
}i,j\in\left\{  1,2,\ldots,n\right\}  \text{ satisfying }i>j\right)  .
\label{pf.l6.4b.c.7}%
\end{equation}


Moreover, if $i\in\left\{  1,2,\ldots,n\right\}  $, then $g_{\eta\left(
i\right)  ,\eta\left(  i\right)  }=1$ (by (\ref{pf.l6.4b.c.1}), applied to
$\eta\left(  i\right)  $ instead of $i$). Thus, we have shown that%
\begin{equation}
\left(  g_{\eta\left(  i\right)  ,\eta\left(  i\right)  }=1\text{ for all
}i\in\left\{  1,2,\ldots,n\right\}  \right)  . \label{pf.l6.4b.c.8}%
\end{equation}


Now, $g^{\eta}=\left(  g_{\eta\left(  i\right)  ,\eta\left(  j\right)
}\right)  _{i,j=1}^{n}$ (since $g=\left(  g_{i,j}\right)  _{i,j=1}^{n}$).
Hence, Lemma 6.4b \textbf{(b)} (applied to $g^{\eta}$ and $g_{\eta\left(
i\right)  ,\eta\left(  j\right)  }$ instead of $g$ and $g_{i,j}$) shows that
$g^{\eta}\in U$ (since (\ref{pf.l6.4b.c.8}) and (\ref{pf.l6.4b.c.7}) hold).
Hence, $\left(  g^{\eta}\right)  ^{\eta^{-1}}\in U^{\eta^{-1}}$. Since
$\left(  g^{\eta}\right)  ^{\eta^{-1}}=g^{\eta\eta^{-1}}=g^{1}=g$, this
rewrites as $g\in U^{\eta^{-1}}$. Since $\eta^{-1}=\rho\tau^{-1}\sigma^{-1}$,
this rewrites as $g\in U^{\rho\tau^{-1}\sigma^{-1}}$.

Now, forget that we fixed $g$. We thus have shown that $g\in U^{\rho\tau
^{-1}\sigma^{-1}}$ for each $g\in X\left(  \sigma\right)  $. In other words,
$X\left(  \sigma\right)  \subseteq U^{\rho\tau^{-1}\sigma^{-1}}$. This proves
Lemma 6.4d \textbf{(c)}.

\textbf{(d)} Let $g\in X\left(  \tau\right)  $. We shall prove that $g\in
U^{\sigma}$.

Write the matrix $g$ in the form $g=\left(  g_{i,j}\right)  _{i,j=1}^{n}$.

We have $g\in X\left(  \tau\right)  $. Hence, Lemma 6.4a \textbf{(a)} (applied
to $\tau$ instead of $\sigma$) yields that%
\begin{equation}
\left(  g_{i,i}=1\text{ for all }i\in\left\{  1,2,\ldots,n\right\}  \right)
\label{pf.l6.4b.d.1}%
\end{equation}
and%
\begin{equation}
\left(  g_{i,j}=0\text{ for any }i,j\in\left\{  1,2,\ldots,n\right\}  \text{
satisfying }i\neq j\text{ and }\left(  i,j\right)  \notin L\left(  \tau
^{-1}\right)  \right)  . \label{pf.l6.4b.d.2}%
\end{equation}


Let $\eta=\sigma^{-1}$. Thus, $\eta\in\Sigma_{n}$. Hence, $\eta$ is an
injective map.

Let $i,j\in\left\{  1,2,\ldots,n\right\}  $ be such that $i>j$. We shall show
that $g_{\eta\left(  i\right)  ,\eta\left(  j\right)  }=0$.

Indeed, assume the contrary. Thus, $g_{\eta\left(  i\right)  ,\eta\left(
j\right)  }\neq0$. From $i>j$, we obtain $i\neq j$ and therefore $\eta\left(
i\right)  \neq\eta\left(  j\right)  $ (since $\eta$ is injective). If we had
$\left(  \eta\left(  i\right)  ,\eta\left(  j\right)  \right)  \notin L\left(
\tau^{-1}\right)  $, then we would have $g_{\eta\left(  i\right)  ,\eta\left(
j\right)  }=0$ (by (\ref{pf.l6.4b.d.2}), applied to $\eta\left(  i\right)  $
and $\eta\left(  j\right)  $ instead of $i$ and $j$), which would contradict
$g_{\eta\left(  i\right)  ,\eta\left(  j\right)  }\neq0$. Hence, we cannot
have $\left(  \eta\left(  i\right)  ,\eta\left(  j\right)  \right)  \notin
L\left(  \tau^{-1}\right)  $. Thus, we must have $\left(  \eta\left(
i\right)  ,\eta\left(  j\right)  \right)  \in L\left(  \tau^{-1}\right)  $. In
other words, $\eta\left(  i\right)  <\eta\left(  j\right)  $ and $\tau
^{-1}\left(  \eta\left(  i\right)  \right)  >\tau^{-1}\left(  \eta\left(
j\right)  \right)  $ (by the definition of $L\left(  \tau^{-1}\right)  $).
From $\eta\left(  i\right)  <\eta\left(  j\right)  $, we obtain $\eta\left(
j\right)  >\eta\left(  i\right)  $.

Set $x=\tau^{-1}\left(  \eta\left(  j\right)  \right)  $ and $y=\tau
^{-1}\left(  \eta\left(  i\right)  \right)  $. Then, $x,y$ are elements of
$\left\{  1,2,\ldots,n\right\}  $. Furthermore, $y=\tau^{-1}\left(
\eta\left(  i\right)  \right)  >\tau^{-1}\left(  \eta\left(  j\right)
\right)  =x$, so that $x<y$. Besides, from $x=\tau^{-1}\left(  \eta\left(
j\right)  \right)  $, we obtain $\tau\left(  x\right)  =\eta\left(  j\right)
$. From $y=\tau^{-1}\left(  \eta\left(  i\right)  \right)  $, we obtain
$\tau\left(  y\right)  =\eta\left(  i\right)  $. Thus, $\tau\left(  y\right)
=\eta\left(  i\right)  <\eta\left(  j\right)  =\tau\left(  x\right)  $, so
that $\tau\left(  x\right)  >\tau\left(  y\right)  $.

From $x<y$ and $\tau\left(  x\right)  >\tau\left(  y\right)  $, we obtain
$\left(  x,y\right)  \in L\left(  \tau\right)  $ (by the definition of
$L\left(  \tau\right)  $). Thus, $\left\{  x,y\right\}  \in\overline{L}\left(
\tau\right)  $ (by the definition of $\overline{L}\left(  \tau\right)  $).

On the other hand, the elements $\eta\left(  i\right)  ,\eta\left(  j\right)
\in\left\{  1,2,\ldots,n\right\}  $ satisfy $\eta\left(  i\right)
<\eta\left(  j\right)  $ and $\sigma\left(  \eta\left(  i\right)  \right)
>\sigma\left(  \eta\left(  j\right)  \right)  $ (since $\sigma\left(
\underbrace{\eta}_{=\sigma^{-1}}\left(  i\right)  \right)  =\sigma\left(
\sigma^{-1}\left(  i\right)  \right)  =i>j=\sigma\left(  \underbrace{\sigma
^{-1}}_{=\eta}\left(  j\right)  \right)  =\sigma\left(  \eta\left(  j\right)
\right)  $). In other words, $\left(  \eta\left(  i\right)  ,\eta\left(
j\right)  \right)  \in L\left(  \sigma\right)  $ (by the definition of
$L\left(  \sigma\right)  $). Hence, $\left\{  \eta\left(  i\right)
,\eta\left(  j\right)  \right\}  \in\overline{L}\left(  \sigma\right)  $ (by
the definition of $\overline{L}\left(  \sigma\right)  $).

But%
\begin{align*}
\tau_{\ast}\left(  \left\{  x,y\right\}  \right)   &  =\tau\left(  \left\{
x,y\right\}  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of
}\tau_{\ast}\right) \\
&  =\left\{  \underbrace{\tau\left(  x\right)  }_{=\eta\left(  j\right)
},\underbrace{\tau\left(  y\right)  }_{=\eta\left(  i\right)  }\right\}
=\left\{  \eta\left(  j\right)  ,\eta\left(  i\right)  \right\}  =\left\{
\eta\left(  i\right)  ,\eta\left(  j\right)  \right\}  \in\overline{L}\left(
\sigma\right)  ,
\end{align*}
so that $\left\{  x,y\right\}  \in\tau_{\ast}^{-1}\overline{L}\left(
\sigma\right)  $. Combining this with $\left\{  x,y\right\}  \in\overline
{L}\left(  \tau\right)  $, we obtain $\left\{  x,y\right\}  \in\overline
{L}\left(  \tau\right)  \cap\tau_{\ast}^{-1}\overline{L}\left(  \sigma\right)
=\varnothing$. Thus, $\left\{  x,y\right\}  $ belongs to the empty set. This
is clearly absurd. Thus, we have obtained a contradiction. This shows that our
assumption was false. Hence, $g_{\eta\left(  i\right)  ,\eta\left(  j\right)
}=0$ is proven.

Now, forget that we fixed $i,j$. We thus have shown that
\begin{equation}
\left(  g_{\eta\left(  i\right)  ,\eta\left(  j\right)  }=0\text{ for any
}i,j\in\left\{  1,2,\ldots,n\right\}  \text{ satisfying }i>j\right)  .
\label{pf.l6.4b.d.7}%
\end{equation}


Moreover, if $i\in\left\{  1,2,\ldots,n\right\}  $, then $g_{\eta\left(
i\right)  ,\eta\left(  i\right)  }=1$ (by (\ref{pf.l6.4b.d.1}), applied to
$\eta\left(  i\right)  $ instead of $i$). Thus, we have shown that%
\begin{equation}
\left(  g_{\eta\left(  i\right)  ,\eta\left(  i\right)  }=1\text{ for all
}i\in\left\{  1,2,\ldots,n\right\}  \right)  . \label{pf.l6.4b.d.8}%
\end{equation}


Now, $g^{\eta}=\left(  g_{\eta\left(  i\right)  ,\eta\left(  j\right)
}\right)  _{i,j=1}^{n}$ (since $g=\left(  g_{i,j}\right)  _{i,j=1}^{n}$).
Hence, Lemma 6.4b \textbf{(b)} (applied to $g^{\eta}$ and $g_{\eta\left(
i\right)  ,\eta\left(  j\right)  }$ instead of $g$ and $g_{i,j}$) shows that
$g^{\eta}\in U$ (since (\ref{pf.l6.4b.d.8}) and (\ref{pf.l6.4b.d.7}) hold).
Hence, $\left(  g^{\eta}\right)  ^{\sigma}\in U^{\sigma}$. Since
\begin{align*}
\left(  g^{\eta}\right)  ^{\sigma}  &  =g^{\eta\sigma}=g^{1}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\underbrace{\eta}_{=\sigma^{-1}%
}\sigma=\sigma^{-1}\sigma=1\right) \\
&  =g,
\end{align*}
this rewrites as $g\in U^{\sigma}$.

Now, forget that we fixed $g$. We thus have shown that $g\in U^{\sigma}$ for
each $g\in X\left(  \tau\right)  $. In other words, $X\left(  \tau\right)
\subseteq U^{\sigma}$. Hence, $\left(  X\left(  \tau\right)  \right)
^{\sigma^{-1}}\subseteq\left(  U^{\sigma}\right)  ^{\sigma^{-1}}%
=U^{\sigma\sigma^{-1}}=U^{1}=U$. This proves Lemma 6.4d \textbf{(d)}.

\textbf{(e)} The definition of $X\left(  \sigma\tau\right)  $ yields $X\left(
\sigma\tau\right)  =U\cap U^{\left(  \sigma\tau\rho\right)  ^{-1}}$. Hence,
$X\left(  \sigma\tau\right)  $ is the intersection of two subgroups of $G$
(namely, of the subgroup $U$ and of the subgroup $U^{\left(  \sigma\tau
\rho\right)  ^{-1}}$). Thus, $X\left(  \sigma\tau\right)  $ is itself a
subgroup of $G$. Therefore, $X\left(  \sigma\tau\right)  \cdot X\left(
\sigma\tau\right)  \subseteq X\left(  \sigma\tau\right)  $.

Now, $\left(  \sigma\tau\rho\right)  ^{-1}=\underbrace{\rho^{-1}}_{=\rho}%
\tau^{-1}\sigma^{-1}=\rho\tau^{-1}\sigma^{-1}$, so that $U^{\left(  \sigma
\tau\rho\right)  ^{-1}}=U^{\rho\tau^{-1}\sigma^{-1}}$.

The definition of $X\left(  \sigma\right)  $ yields $X\left(  \sigma\right)
=U\cap U^{\left(  \sigma\rho\right)  ^{-1}}\subseteq U$. But Lemma 6.4d
\textbf{(c)} yields $X\left(  \sigma\right)  \subseteq U^{\rho\tau^{-1}%
\sigma^{-1}}=U^{\left(  \sigma\tau\rho\right)  ^{-1}}$. Combining $X\left(
\sigma\right)  \subseteq U$ with $X\left(  \sigma\right)  \subseteq U^{\left(
\sigma\tau\rho\right)  ^{-1}}$, we obtain $X\left(  \sigma\right)  \subseteq
U\cap U^{\left(  \sigma\tau\rho\right)  ^{-1}}=X\left(  \sigma\tau\right)  $.

On the other hand, the definition of $X\left(  \tau\right)  $ yields $X\left(
\tau\right)  =U\cap U^{\left(  \tau\rho\right)  ^{-1}}\subseteq U^{\left(
\tau\rho\right)  ^{-1}}$. Hence, $\left(  X\left(  \tau\right)  \right)
^{\sigma^{-1}}\subseteq\left(  U^{\left(  \tau\rho\right)  ^{-1}}\right)
^{\sigma^{-1}}=U^{\left(  \tau\rho\right)  ^{-1}\sigma^{-1}}=U^{\left(
\sigma\tau\rho\right)  ^{-1}}$ (since $\left(  \tau\rho\right)  ^{-1}%
\sigma^{-1}=\left(  \sigma\tau\rho\right)  ^{-1}$). Combining $\left(
X\left(  \tau\right)  \right)  ^{\sigma^{-1}}\subseteq U$ (which follows from
Lemma 6.4d \textbf{(d)}) with $\left(  X\left(  \tau\right)  \right)
^{\sigma^{-1}}\subseteq U^{\left(  \sigma\tau\rho\right)  ^{-1}}$, we obtain
$\left(  X\left(  \tau\right)  \right)  ^{\sigma^{-1}}\subseteq U\cap
U^{\left(  \sigma\tau\rho\right)  ^{-1}}=X\left(  \sigma\tau\right)  $.

Now,%
\[
\underbrace{X\left(  \sigma\right)  }_{\subseteq X\left(  \sigma\tau\right)
}\cdot\underbrace{\left(  X\left(  \tau\right)  \right)  ^{\sigma^{-1}}%
}_{\subseteq X\left(  \sigma\tau\right)  }\subseteq X\left(  \sigma
\tau\right)  \cdot X\left(  \sigma\tau\right)  \subseteq X\left(  \sigma
\tau\right)  .
\]
This proves Lemma 6.4d \textbf{(e)}.

\textbf{(f)} Let $r\in B\sigma B\tau B$. Then, $r\in B\sigma B\tau
B=B\sigma\left(  B\tau B\right)  $. In other words, there exist $c\in B$ and
$p\in B\tau B$ such that $r=c\sigma p$. Consider these $c$ and $p$.

Lemma 6.4c (applied to $\tau$ instead of $\sigma$) yields that the map
$X\left(  \tau\right)  \times B\rightarrow B\tau B,\ \left(  g,b\right)
\mapsto g\tau b$ is well-defined and is a bijection. Hence, the element $p\in
B\tau B$ is an image under this map. In other words, there exists some
$\left(  g,b\right)  \in X\left(  \tau\right)  \times B$ such that $p=g\tau
b$. Consider this $\left(  g,b\right)  $.

From $\left(  g,b\right)  \in X\left(  \tau\right)  \times B$, we obtain $g\in
X\left(  \tau\right)  $ and $b\in B$. From $g\in X\left(  \tau\right)  $, we
obtain $g^{\sigma^{-1}}\in\left(  X\left(  \tau\right)  \right)  ^{\sigma
^{-1}}\subseteq U$ (by Lemma 6.4c \textbf{(d)}), so that $g^{\sigma^{-1}}\in
U\subseteq B$. Since $g^{\sigma^{-1}}=\underbrace{\left(  \sigma^{-1}\right)
^{-1}}_{=\sigma}g\sigma^{-1}=\sigma g\sigma^{-1}$, this rewrites as $\sigma
g\sigma^{-1}\in B$.

Now, $\sigma g\underbrace{\sigma^{-1}\sigma}_{=1}\tau b=\sigma
\underbrace{g\tau b}_{=p}=\sigma p$, so that $\sigma p=\underbrace{\sigma
g\sigma^{-1}}_{\in B}\sigma\tau\underbrace{b}_{\in B}\in B\sigma\tau B$. Now,%
\[
r=\underbrace{c}_{\in B}\underbrace{\sigma p}_{\in B\sigma\tau B}%
\in\underbrace{BB}_{\substack{\subseteq B\\\text{(since }B\text{ is a group)}%
}}\sigma\tau B\subseteq B\sigma\tau B.
\]


Now, forget that we fixed $r$. We thus have proven that $r\in B\sigma\tau B$
for each $r\in B\sigma B\tau B$. In other words, $B\sigma B\tau B\subseteq
B\sigma\tau B$. Combining this with the inclusion%
\[
B\underbrace{\sigma}_{=\sigma1}\tau B=B\sigma\underbrace{1}_{\in B}\tau
B\subseteq B\sigma B\tau B,
\]
we obtain $B\sigma B\tau B=B\sigma\tau B$. This proves Lemma 6.4d \textbf{(f)}.

\textbf{(g)} For every $\left(  g,h\right)  \in X\left(  \sigma\right)  \times
X\left(  \tau\right)  $, we have%
\[
\underbrace{g}_{\substack{\in X\left(  \sigma\right)  \\\text{(since }\left(
g,h\right)  \in X\left(  \sigma\right)  \times X\left(  \tau\right)  \text{)}%
}}\underbrace{h^{\sigma^{-1}}}_{\substack{\in\left(  X\left(  \tau\right)
\right)  ^{\sigma^{-1}}\\\text{(since }h\in X\left(  \tau\right)
\\\text{(since }\left(  g,h\right)  \in X\left(  \sigma\right)  \times
X\left(  \tau\right)  \text{))}}}\in X\left(  \sigma\right)  \cdot\left(
X\left(  \tau\right)  \right)  ^{\sigma^{-1}}\subseteq X\left(  \sigma
\tau\right)
\]
(by Lemma 6.4d \textbf{(e)}). Thus, the map
\[
X\left(  \sigma\right)  \times X\left(  \tau\right)  \rightarrow X\left(
\sigma\tau\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  g,h\right)  \mapsto
gh^{\sigma^{-1}}%
\]
is well-defined. It remains to prove that this map is bijective. In order to
do so, we denote this map by $\Phi$. Thus, $\Phi\left(  g,h\right)
=gh^{\sigma^{-1}}$ for each $\left(  g,h\right)  \in X\left(  \sigma\right)
\times X\left(  \tau\right)  $. Our goal is to prove that $\Phi$ is bijective.

Let us first prove that $\Phi$ is surjective. Indeed, let $k\in X\left(
\sigma\tau\right)  $. Then, $k\in X\left(  \sigma\tau\right)  =U\cap
U^{\left(  \sigma\tau\rho\right)  ^{-1}}$ (by the definition of $X\left(
\sigma\tau\right)  $), so that $k\in U\cap U^{\left(  \sigma\tau\rho\right)
^{-1}}\subseteq U\subseteq B$. Hence, $k\sigma=\underbrace{k}_{\in B}%
\sigma\underbrace{1}_{\in B}\in B\sigma B$.

Lemma 6.4c yields that the map $X\left(  \sigma\right)  \times B\rightarrow
B\sigma B,\ \left(  g,b\right)  \mapsto g\sigma b$ is well-defined and is a
bijection. Hence, the element $k\sigma\in B\sigma B$ is an image under this
map. In other words, there exists some $\left(  u,d\right)  \in X\left(
\sigma\right)  \times B$ such that $k\sigma=u\sigma d$. Consider this $\left(
u,d\right)  $. From $\left(  u,d\right)  \in X\left(  \sigma\right)  \times
B$, we obtain $u\in X\left(  \sigma\right)  $ and $d\in B$.

We have $d\tau=\underbrace{d}_{\in B}\tau\underbrace{1}_{\in B}\in B\tau B$.

Lemma 6.4c (applied to $\tau$ instead of $\sigma$) yields that the map
$X\left(  \tau\right)  \times B\rightarrow B\tau B,\ \left(  g,b\right)
\mapsto g\tau b$ is well-defined and is a bijection. Hence, the element
$d\tau\in B\tau B$ is an image under this map. In other words, there exists
some $\left(  h,c\right)  \in X\left(  \tau\right)  \times B$ such that
$d\tau=h\tau c$. Consider this $\left(  h,c\right)  $. From $\left(
h,c\right)  \in X\left(  \tau\right)  \times B$, we obtain $h\in X\left(
\tau\right)  $ and $c\in B$.

We have $\left(  u,h\right)  \in X\left(  \sigma\right)  \times X\left(
\tau\right)  $ (since $u\in X\left(  \sigma\right)  $ and $h\in X\left(
\tau\right)  $). Thus, the definition of $\Phi$ yields $\Phi\left(
u,h\right)  =u\underbrace{h^{\sigma^{-1}}}_{=\left(  \sigma^{-1}\right)
^{-1}h\sigma^{-1}}=u\underbrace{\left(  \sigma^{-1}\right)  ^{-1}}_{=\sigma
}h\sigma^{-1}=u\sigma h\sigma^{-1}$, so that%
\begin{equation}
\Phi\left(  u,h\right)  \sigma=u\sigma h. \label{pf.l6.4b.g.surj.1}%
\end{equation}


We have%
\[
\underbrace{k\sigma}_{=u\sigma d}\underbrace{\tau1}_{=\tau}=u\sigma
\underbrace{d\tau}_{=h\tau c}=\underbrace{u\sigma h}_{\substack{=\Phi\left(
u,h\right)  \sigma\\\text{(by (\ref{pf.l6.4b.g.surj.1}))}}}\tau c=\Phi\left(
u,h\right)  \sigma\tau c.
\]
Notice that $\left(  k,1\right)  \in X\left(  \sigma\tau\right)  \times B$
(since $k\in X\left(  \sigma\tau\right)  $ and $1\in B$) and $\left(
\Phi\left(  u,h\right)  ,c\right)  \in X\left(  \sigma\tau\right)  \times B$
(since $\Phi\left(  u,h\right)  \in X\left(  \sigma\tau\right)  $ and $c\in B$).

But Lemma 6.4c (applied to $\sigma\tau$ instead of $\sigma$) yields that the
map $X\left(  \sigma\tau\right)  \times B\rightarrow B\sigma\tau B,\ \left(
g,b\right)  \mapsto g\sigma\tau b$ is well-defined and is a bijection. In
particular, this map is bijective, thus injective. In other words, if $\left(
g_{1},b_{1}\right)  $ and $\left(  g_{2},b_{2}\right)  $ are two elements of
$X\left(  \sigma\tau\right)  \times B$ satisfying $g_{1}\sigma\tau b_{1}%
=g_{2}\sigma\tau b_{2}$, then we have $\left(  g_{1},b_{1}\right)  =\left(
g_{2},b_{2}\right)  $. Applying this to $\left(  g_{1},b_{1}\right)  =\left(
k,1\right)  $ and $\left(  g_{2},b_{2}\right)  =\left(  \Phi\left(
u,h\right)  ,c\right)  $, we obtain $\left(  k,1\right)  =\left(  \Phi\left(
u,h\right)  ,c\right)  $ (since $\left(  k,1\right)  \in X\left(  \sigma
\tau\right)  \times B$ and $\left(  \Phi\left(  u,h\right)  ,c\right)  \in
X\left(  \sigma\tau\right)  \times B$ and $k\sigma\tau1=\Phi\left(
u,h\right)  \sigma\tau c$). In other words, $k=\Phi\left(  u,h\right)  $ and
$1=c$. Hence, $k=\Phi\left(  u,h\right)  \in\Phi\left(  X\left(
\sigma\right)  \times X\left(  \tau\right)  \right)  $.

Now, forget that we fixed $k$. We thus have shown that every $k\in X\left(
\sigma\tau\right)  $ satisfies $k\in\Phi\left(  X\left(  \sigma\right)  \times
X\left(  \tau\right)  \right)  $. In other words, $X\left(  \sigma\tau\right)
\subseteq\Phi\left(  X\left(  \sigma\right)  \times X\left(  \tau\right)
\right)  $. In other words, the map $\Phi$ is surjective. (This proof was a
more detailed paraphrase of an argument that you included in your proof of
Proposition 6.4.)

Let us now show that the map $\Phi$ is injective. Indeed, let $\left(
g_{1},h_{1}\right)  $ and $\left(  g_{2},h_{2}\right)  $ be two elements of
$X\left(  \sigma\right)  \times X\left(  \tau\right)  $ satisfying
$\Phi\left(  g_{1},h_{1}\right)  =\Phi\left(  g_{2},h_{2}\right)  $. We shall
show that $\left(  g_{1},h_{1}\right)  =\left(  g_{2},h_{2}\right)  $.

We have $\left(  g_{1},h_{1}\right)  \in X\left(  \sigma\right)  \times
X\left(  \tau\right)  $. In other words, $g_{1}\in X\left(  \sigma\right)  $
and $h_{1}\in X\left(  \tau\right)  $.

We have $\left(  g_{2},h_{2}\right)  \in X\left(  \sigma\right)  \times
X\left(  \tau\right)  $. In other words, $g_{2}\in X\left(  \sigma\right)  $
and $h_{2}\in X\left(  \tau\right)  $.

The definition of $\Phi$ yields $\Phi\left(  g_{1},h_{1}\right)  =g_{1}\left(
h_{1}\right)  ^{\sigma^{-1}}$. The definition of $\Phi$ yields $\Phi\left(
g_{2},h_{2}\right)  =g_{2}\left(  h_{2}\right)  ^{\sigma^{-1}}$. Now,%
\[
g_{1}\left(  h_{1}\right)  ^{\sigma^{-1}}=\Phi\left(  g_{1},h_{1}\right)
=\Phi\left(  g_{2},h_{2}\right)  =g_{2}\left(  h_{2}\right)  ^{\sigma^{-1}}.
\]
Multiplying both sides of this equality by $g_{2}^{-1}$ from the left and by
$\left(  \left(  h_{1}\right)  ^{\sigma^{-1}}\right)  ^{-1}$ from the right,
we obtain%
\[
g_{2}^{-1}g_{1}=\left(  h_{2}\right)  ^{\sigma^{-1}}\left(  \left(
h_{1}\right)  ^{\sigma^{-1}}\right)  ^{-1}=\left(  h_{2}h_{1}^{-1}\right)
^{\sigma^{-1}}%
\]
(since the map $G\rightarrow G,\ x\mapsto x^{\sigma^{-1}}$ is a group automorphism).

But the definition of $X\left(  \sigma\right)  $ yields $X\left(
\sigma\right)  =U\cap U^{\left(  \sigma\rho\right)  ^{-1}}$. Hence, $X\left(
\sigma\right)  $ is the intersection of two subgroups of $G$ (namely, of the
subgroup $U$ and of the subgroup $U^{\left(  \sigma\rho\right)  ^{-1}}$).
Thus, $X\left(  \sigma\right)  $ is itself a subgroup of $G$. The same
argument (applied to $\tau$ instead of $\sigma$) shows that $X\left(
\tau\right)  $ is a subgroup of $G$.

From $g_{2}\in X\left(  \sigma\right)  $ and $g_{1}\in X\left(  \sigma\right)
$, we obtain $g_{2}^{-1}g_{1}\in X\left(  \sigma\right)  $ (since $X\left(
\sigma\right)  $ is a subgroup of $G$).

From $h_{2}\in X\left(  \tau\right)  $ and $h_{1}\in X\left(  \tau\right)  $,
we obtain $h_{2}h_{1}^{-1}\in X\left(  \tau\right)  $ (since $X\left(
\tau\right)  $ is a subgroup of $G$), so that $\left(  h_{2}h_{1}^{-1}\right)
^{\sigma^{-1}}\in\left(  X\left(  \tau\right)  \right)  ^{\sigma^{-1}}$.

Combining $g_{2}^{-1}g_{1}\in X\left(  \sigma\right)  $ with $g_{2}^{-1}%
g_{1}=\left(  h_{2}h_{1}^{-1}\right)  ^{\sigma^{-1}}\in\left(  X\left(
\tau\right)  \right)  ^{\sigma^{-1}}$, we obtain $g_{2}^{-1}g_{1}\in X\left(
\sigma\right)  \cap\left(  X\left(  \tau\right)  \right)  ^{\sigma^{-1}}=1$
(by Lemma 6.4d \textbf{(b)}). In other words, $g_{2}^{-1}g_{1}=1$. Thus,
$g_{1}=g_{2}$.

Comparing $g_{2}^{-1}g_{1}=1$ with $g_{2}^{-1}g_{1}=\left(  h_{2}h_{1}%
^{-1}\right)  ^{\sigma^{-1}}$, we obtain $1=\left(  h_{2}h_{1}^{-1}\right)
^{\sigma^{-1}}=\underbrace{\left(  \sigma^{-1}\right)  ^{-1}}_{=\sigma}%
h_{2}h_{1}^{-1}\sigma^{-1}=\sigma h_{2}h_{1}^{-1}\sigma^{-1}$. Multiplying
both sides of this equality by $\sigma$ from the right, we obtain
$\sigma=\sigma h_{2}h_{1}^{-1}$. Cancelling $\sigma$ from this equality, we
find $1=h_{2}h_{1}^{-1}$. Thus, $h_{1}=h_{2}$.

Now, $\left(  \underbrace{g_{1}}_{=g_{2}},\underbrace{h_{1}}_{=h_{2}}\right)
=\left(  g_{2},h_{2}\right)  $.

Let us now forget that we fixed $\left(  g_{1},h_{1}\right)  $ and $\left(
g_{2},h_{2}\right)  $. We thus have shown that if $\left(  g_{1},h_{1}\right)
$ and $\left(  g_{2},h_{2}\right)  $ are two elements of $X\left(
\sigma\right)  \times X\left(  \tau\right)  $ satisfying $\Phi\left(
g_{1},h_{1}\right)  =\Phi\left(  g_{2},h_{2}\right)  $, then $\left(
g_{1},h_{1}\right)  =\left(  g_{2},h_{2}\right)  $. In other words, the map
$\Phi$ is injective. Since we also know that $\Phi$ is surjective, we
therefore conclude that the map $\Phi$ is bijective. In other words, the map
\[
X\left(  \sigma\right)  \times X\left(  \tau\right)  \rightarrow X\left(
\sigma\tau\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  g,h\right)  \mapsto
gh^{\sigma^{-1}}%
\]
is bijective (since this map is $\Phi$). This proves Lemma 6.4d \textbf{(g)}.
$\square$]

Now, your Proposition 6.4 is precisely Lemma 6.4d \textbf{(g)}, whereas your
Proposition 6.5 is exactly Lemma 6.4d \textbf{(f)}. Hence, both Proposition
6.4 and Proposition 6.5 are proven.

\item \textbf{Definition 7.1:} Remove the comma in \textquotedblleft preserves
the sets,\textquotedblright.

\item \textbf{Definition 7.1:} After \textquotedblleft it is the largest
subgroup of $\Sigma_{n}$ that preserves these sets\textquotedblright, I would
add \textquotedblleft(actually, it is the set of all permutations $\sigma
\in\Sigma_{n}$ that preserve these sets)\textquotedblright. (This is a more
concrete description of $\Sigma_{I}$, and you use it in the proof of
Proposition 7.3 below.)

I would also suggest replacing the word \textquotedblleft
sets\textquotedblright\ by \textquotedblleft intervals\textquotedblright%
\ whenever you are talking about these sets.

\item \textbf{Definition 7.2:} At the beginning of this definition, I would
add the following sentences: \textquotedblleft Let $I$ be a subset of
$\left\{  1,2,\ldots,n-1\right\}  $. Set $I^{c}=\left\{  0,1,\ldots,n\right\}
\setminus I$. Again, write this set $I^{c}$ as $\left\{  i_{0},i_{1}%
,\ldots,i_{r}\right\}  $ with $0=i_{0}<i_{1}<\cdots<i_{r}=n$%
.\textquotedblright.

\item \textbf{Definition 7.2:} Replace \textquotedblleft write $P_{i}$ for
$P_{\{}i\}$ and $P_{ij}$ for $P_{\{}i,j\}$\textquotedblright\ by
\textquotedblleft write $P_{i}$ for $P_{\left\{  i\right\}  }$ and $P_{ij}$
for $P_{\left\{  i,j\right\}  }$\textquotedblright.

\item \textbf{Definition 7.2:} Replace \textquotedblleft This gives a functor
from $n$-dimensional vector spaces to sets\textquotedblright\ by
\textquotedblleft This gives a functor from the category \newline$\left(
n\text{-dimensional vector spaces, isomorphisms}\right)  $ to the category
$\left(  \text{sets, bijections}\right)  $\textquotedblright. (If you try to
apply a linear map that is not an isomorphism to a flag, then the resulting
flag might have different dimensions.)

\item \textbf{Definition 7.2:} After \textquotedblleft We let $P_{I}$ denote
the stabilizer of this flag\textquotedblright, add \textquotedblleft in
$G$\textquotedblright.

\item \textbf{Proof of Proposition 7.3:} For the sake of clarity, I would
replace the two sentences

\textquotedblleft Let $\sigma$ be the permutation such that $g\in B\sigma B$.
Recall from Section 6 that this is characterised by characterised by
$Q_{i,\sigma\left(  i\right)  }\neq0$, where%
\[
Q_{ij}=\left(  U_{i}\cap E_{j}\right)  /\left(  \left(  U_{i-1}\cap
E_{j}\right)  +\left(  U_{i}\cap E_{j-1}\right)  \right)  .
\]
\textquotedblright

by

\textquotedblleft Let $\sigma=\delta\left(  g\underline{E},\underline{E}%
\right)  $. Then, $\sigma\in\Sigma_{n}$. If we define the map $\pi
:G\rightarrow\Sigma_{n}$ as in the proof of Proposition 6.1, then $\pi\left(
g\right)  =\delta\left(  g\underline{E},\underline{E}\right)  =\sigma$, so
that $g\in\pi^{-1}\left\{  \sigma\right\}  =B\sigma B$ (as was proven in the
proof of Proposition 6.1). Hence, it suffices to show that $\sigma\in
\Sigma_{I}$ (because then, it will follow that $g\in B\underbrace{\sigma}%
_{\in\Sigma_{I}}B\subseteq B\Sigma_{I}B$, so that $P_{I}\subseteq B\Sigma
_{I}B$ and therefore $P_{I}=B\Sigma_{I}B$). We have $\sigma=\delta\left(
g\underline{E},\underline{E}\right)  $; in other words, each $i\in\left\{
1,2,\ldots,n\right\}  $ satisfies $Q_{i,\sigma\left(  i\right)  }\neq0$, where%
\[
Q_{ij}=\left(  U_{i}\cap E_{j}\right)  /\left(  \left(  U_{i-1}\cap
E_{j}\right)  +\left(  U_{i}\cap E_{j-1}\right)  \right)  .
\]
\textquotedblright.

\item \textbf{Proof of Proposition 7.3:} After \textquotedblleft$U_{a}\leq
U_{i-1}$\textquotedblright, add \textquotedblleft(since $a\leq i-1$%
)\textquotedblright.

\item \textbf{Proof of Lemma 7.4:} Replace \textquotedblleft the element
$g=\sigma g\sigma^{-1}$\textquotedblright\ by \textquotedblleft the element
$b=\sigma g\sigma^{-1}$\textquotedblright.

\item \textbf{Proof of Lemma 7.4:} Replace \textquotedblleft$b\left(
e_{\sigma\left(  k\right)  }\right)  =e_{\sigma\left(  k+1\right)  }%
$\textquotedblright\ by \textquotedblleft$b\left(  e_{\sigma\left(  k\right)
}\right)  =\underbrace{b}_{=\sigma g\sigma^{-1}}\sigma\left(  e_{k}\right)
=\sigma g\underbrace{\sigma^{-1}\sigma}_{=1}\left(  e_{k}\right)
=\sigma\underbrace{g\left(  e_{k}\right)  }_{=e_{k}+e_{k+1}}=\sigma\left(
e_{k}+e_{k+1}\right)  =e_{\sigma\left(  k\right)  }+e_{\sigma\left(
k+1\right)  }$\textquotedblright.

\item \textbf{Proof of Proposition 7.5:} After \textquotedblleft Put
$I=\left\{  i\ \mid\ s_{i}\in P\right\}  $\textquotedblright, add
\textquotedblleft. Hence, $\Sigma_{I}\leq P$ and thus $P_{I}=B\Sigma_{I}B\leq
P$, \textquotedblright.

\item \textbf{Proof of Proposition 7.5:} Replace the sentence
\textquotedblleft Suppose that $g\in P$, and put $\sigma=\pi\left(  g\right)
$.\textquotedblright\ by the following: \textquotedblleft Thus, it remains to
show that $P\leq P_{I}$. Let $g\in P$; our goal is to prove that $g\in P_{I}$.
Define the map $\pi:G\rightarrow\Sigma_{n}$ as in the proof of Proposition
6.1. Set $\sigma=\pi\left(  g\right)  $. We showed in the proof of Proposition
6.1 that $\pi^{-1}\left\{  \sigma\right\}  =B\sigma B$. Now, from $\sigma
=\pi\left(  g\right)  $, we obtain $g\in\pi^{-1}\left\{  \sigma\right\}
=B\sigma B$, so that $BgB=B\sigma B$.\textquotedblright.

\item \textbf{Proof of Proposition 7.5:} Before \textquotedblleft If $l\left(
\sigma\right)  =0$\textquotedblright, add the following sentences:
\textquotedblleft Thus, it suffices to prove that $\sigma\in P_{I}$. Now, let
us forget how $\sigma$ was defined. Our goal is to show that $\sigma\in P_{I}$
for every $\sigma\in\Sigma_{n}$ satisfying $\sigma\in P$. We shall do this by
induction over $l\left(  \sigma\right)  $:\textquotedblright.

\item \textbf{Proof of Proposition 7.5:} Replace \textquotedblleft Now suppose
that $l\left(  \sigma\right)  >1$\textquotedblright\ by \textquotedblleft Now
suppose that $l\left(  \sigma\right)  >0$\textquotedblright. Also, remove the
preceding sentence (\textquotedblleft If $l\left(  \sigma\right)  =1$ then
$\sigma=s_{i}$ for some $i$ and $\sigma\in P$ so $i\in I$ so $\sigma\in P_{I}%
$\textquotedblright) completely (it is unnecessary and complicates the
structure of the proof).

\item \textbf{Proof of Proposition 7.5:} Replace \textquotedblleft so
$s_{k}\in P_{I}$\textquotedblright\ by \textquotedblleft so $s_{k}\in
\Sigma_{I}\subseteq B\Sigma_{I}B=P_{I}$\textquotedblright.

\item \textbf{Proof of Proposition 7.5:} Replace \textquotedblleft so we can
assume by induction that\textquotedblright\ by \textquotedblleft so the
induction hypothesis yields\textquotedblright.

\item \textbf{Proof of Proposition 7.5:} Remove \textquotedblleft and thus
that $g\in P_{I}$\textquotedblright.

\item \textbf{Proof of Proposition 7.5:} I would notice that your proof of
Proposition 7.5 proves a slightly slonger claim:

\textbf{Proposition 7.5a.} Let $P$ be a subgroup of $G$ such that $P\geq B$.
Let $I=\left\{  i\ \mid\ s_{i}\in P\right\}  $. Then, $P=P_{I}$.

Furthermore, let me state another fact (that will be used later):

\textbf{Lemma 7.5b.} Let $I\subseteq\left\{  1,2,\ldots,n-1\right\}  $. Then,
$I=\left\{  i\ \mid\ s_{i}\in P_{I}\right\}  $.

[\textit{Proof of Lemma 7.5b.} Define a subset $J$ of $\left\{  1,2,\ldots
,n-1\right\}  $ by $J=\left\{  i\ \mid\ s_{i}\in P_{I}\right\}  $. Then,
$J=\left\{  i\ \mid\ s_{i}\in P_{I}\right\}  \supseteq I$ (since every $i\in
I$ satisfies $s_{i}\in\Sigma_{I}\subseteq B\Sigma_{I}B=P_{I}$).

Now, let $j\in J$. We are going to prove that $j\in I$.

Indeed, assume the contrary. Hence, $j\notin I$, so that $j\in I^{c}$. Hence,
$E_{j}$ is one of the entries of the obvious flag $\underline{E}%
\in\operatorname*{Flag}\nolimits_{I}\left(  \mathbb{F}_{p}^{n}\right)  $.
Therefore, the group $P_{I}$ fixes the subspace $E_{j}$ (since the group
$P_{I}$ fixes the obvious flag $\underline{E}\in\operatorname*{Flag}%
\nolimits_{I}\left(  \mathbb{F}_{p}^{n}\right)  $). In other words,
$pE_{j}\subseteq E_{j}$ for each $p\in P_{I}$.

But $j\in J=\left\{  i\ \mid\ s_{i}\in P_{I}\right\}  $. In other words,
$s_{j}\in P_{I}$. But we have $pE_{j}\subseteq E_{j}$ for each $p\in P_{I}$.
Applying this to $p=s_{j}$, we conclude that $s_{j}E_{j}\subseteq E_{j}$
(since $s_{j}\in P_{I}$). Now, $s_{j}e_{j}=e_{j+1}$, so that $e_{j+1}%
=s_{j}\underbrace{e_{j}}_{\in E_{j}}\in s_{j}E_{j}\subseteq E_{j}$. This
contradicts the (obvious) fact that $e_{j+1}\notin E_{j}$. This contradiction
proves that our assumption was false. Hence, we have $j\in I$.

Now, forget that we fixed $j$. We thus have shown that $j\in I$ for each $j\in
J$. In other words, $J\subseteq I$. Combining this with $J\supseteq I$, we
obtain $J=I$. Hence, $I=J=\left\{  i\ \mid\ s_{i}\in P_{I}\right\}  $. This
proves Lemma 7.5b. $\square$]

\item \textbf{Proposition 7.7:} Add \textquotedblleft Let $V=\mathbb{F}%
_{p}^{n}$.\textquotedblright\ at the beginning of the theorem. (Otherwise,
$\operatorname*{Flag}\nolimits_{I}\left(  V\right)  $ and
$\operatorname*{Flag}\nolimits_{J}\left(  V\right)  $ wouldn't canonically be
$G$-sets.)

\item \textbf{Proof of Proposition 7.7:} Again, I'd prefer some more details:

\textbf{1.} You claim that \textquotedblleft the orbits of $B$ in $V$ are
precisely the sets $E_{k}\setminus E_{k-1}$\textquotedblright. In order for
this claim to be fully correct, you should set $E_{-1}=\varnothing$, and allow
$k$ to range over $\left\{  0,1,\ldots,n\right\}  $ (rather than $\left\{
1,2,\ldots,n\right\}  $ only). (Otherwise, you are missing the orbit $\left\{
0\right\}  =E_{0}\setminus E_{-1}$.)

Let me also prove this claim:

[\textit{Proof of the fact that the orbits of }$B$ \textit{in }$V$ \textit{are
precisely the sets }$E_{k}\setminus E_{k-1}$\textit{:} For every $k\in\left\{
1,2,\ldots,n\right\}  $, we have%
\begin{equation}
E_{k}\setminus E_{k-1}=Be_{k} \label{pf.p7.7.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.p7.7.1}):} Let $k\in\left\{
1,2,\ldots,n\right\}  $. Hence, the vector $e_{k}$ is well-defined.
\par
Let $b\in B$. Then, $bE_{k}=E_{k}$ and $bE_{k-1}=E_{k-1}$ (by the definition
of $B$). But the element $b$ of $G$ is invertible (since $G$ is a group), thus
a bijection. Hence, $b\left(  E_{k}\setminus E_{k-1}\right)
=\underbrace{bE_{k}}_{=E_{k}}\setminus\underbrace{bE_{k-1}}_{=E_{k-1}}%
=E_{k}\setminus E_{k-1}$.
\par
Now, $e_{k}\in E_{k}\setminus E_{k-1}$ (since $e_{k}\in E_{k}$ and
$e_{k}\notin E_{k-1}$), and thus $b\underbrace{e_{k}}_{\in E_{k}\setminus
E_{k-1}}\in b\left(  E_{k}\setminus E_{k-1}\right)  =E_{k}\setminus E_{k-1}$.
\par
Now, forget that we fixed $b$. We thus have shown that $be_{k}\in
E_{k}\setminus E_{k-1}$ for each $b\in B$. In other words, $Be_{k}\subseteq
E_{k}\setminus E_{k-1}$.
\par
On the other hand, fix $\zeta\in E_{k}\setminus E_{k-1}$. Thus, $\zeta\in
E_{k}$ and $\zeta\notin E_{k-1}$. The last $n-k$ coordinates of $\zeta$ are
zero (since $\zeta\in E_{k}$), but the last $n-k+1$ coordinates of $\zeta$ are
not all zero (since $\zeta\notin E_{k-1}$). Hence, the $k$-th coordinate of
$\zeta$ must be nonzero.
\par
Let $c\in\mathbb{F}_{p}^{n\times n}$ be the $n\times n$-matrix whose $k$-th
column is $\zeta$ whereas all its other columns are the corresponding columns
of the identity matrix (i.e., for each $i\neq k$, the $i$-th column of $c$
shall be $e_{i}$). Then, the $k$-th column of $c$ is the vector $\zeta$, whose
last $n-k$ coordinates are zero. Thus, the last $n-k$ entries of the $k$-th
column of $c$ are zero. Moreover, the $k$-th column of $c$ is the vector
$\zeta$, whose $k$-th coordinate is nonzero. Hence, the $k$-th entry of the
$k$-th column of $c$ is nonzero. Now, the matrix $c$ is upper-triangular
(since the last $n-k$ entries of the $k$-th column of $c$ are zero, while all
other columns are the corresponding columns of the identity matrix) and its
diagonal entries are nonzero (since the $k$-th entry of the $k$-th column of
$c$ is nonzero, while all other columns are the corresponding columns of the
identity matrix). Thus, the matrix $c$ is an invertible upper-triangular
matrix. In other words, $c\in B$. Now, $ce_{k}=\left(  \text{the }k\text{-th
column of }c\right)  =\zeta$, so that $\zeta=\underbrace{c}_{\in B}e_{k}\in
Be_{k}$.
\par
Now, forget that we fixed $\zeta$. We thus have proven that $\zeta\in Be_{k}$
for each $\zeta\in E_{k}\setminus E_{k-1}$. In other words, $E_{k}\setminus
E_{k-1}\subseteq Be_{k}$. Combining this with $Be_{k}\subseteq E_{k}\setminus
E_{k-1}$, we obtain $E_{k}\setminus E_{k-1}=Be_{k}$. This proves
(\ref{pf.p7.7.1}).}.

Every set of the form $E_{k}\setminus E_{k-1}$ (with $k\in\left\{
0,1,\ldots,n\right\}  $) is an orbit of $B$ in $V$%
\ \ \ \ \footnote{\textit{Proof.} Let $k\in\left\{  0,1,\ldots,n\right\}  $.
We must show that the set $E_{k}\setminus E_{k-1}$ is an orbit of $B$ in $V$.
\par
If $k=0$, then this is fairly clear (indeed, if $k=0$, then $\underbrace{E_{k}%
}_{=E_{0}=\left\{  0\right\}  }\setminus\underbrace{E_{k-1}}_{=E_{-1}%
=\varnothing}=\left\{  0\right\}  \setminus\varnothing=\left\{  0\right\}
=B0$, which is clearly an orbit of $B$ in $V$). Thus, we WLOG assume that
$k\neq0$. Hence, $k\in\left\{  1,2,\ldots,n\right\}  $ (since $k\in\left\{
1,2,\ldots,n\right\}  $). Hence, (\ref{pf.p7.7.1}) shows that $E_{k}\setminus
E_{k-1}=Be_{k}$. Thus, $E_{k}\setminus E_{k-1}$ is an orbit of $B$ in $V$
(since $Be_{k}$ is an orbit of $B$ in $V$). Qed.}. The union of these orbits
$E_{k}\setminus E_{k-1}$ is the whole space $\mathbb{F}_{p}^{n}$ (because this
union is $\bigcup_{k=0}^{n}\left(  E_{k}\setminus E_{k-1}\right)
=\underbrace{E_{n}}_{=\mathbb{F}_{p}^{n}}\setminus\underbrace{E_{-1}%
}_{=\varnothing}=\mathbb{F}_{p}^{n}\setminus\varnothing=\mathbb{F}_{p}^{n}$).
Hence, these orbits $E_{k}\setminus E_{k-1}$ are \textbf{all} the orbits of
$B$ in $V$. This is exactly what we wanted to prove. $\square$]

\textbf{2.} You claim that \textquotedblleft the spaces $E_{k}$ are the only
$B$-invariant subspace of $V$\textquotedblright. This claim has a little typo
in it (\textquotedblleft subspace\textquotedblright\ should be
\textquotedblleft subspaces\textquotedblright), and again needs a proof.

[\textit{Proof of the fact that the subspaces }$E_{k}$ \textit{(for }%
$k\in\left\{  0,1,\ldots,n\right\}  $\textit{) are the only }$B$%
\textit{-invariant subspaces of }$V$\textit{:} For each $k\in\left\{
0,1,\ldots,n\right\}  $, the subspace $E_{k}$ is a $B$-invariant subspace of
$V$ (because every $b\in B$ satisfies $bE_{k}=E_{k}$). Conversely, every
$B$-invariant subspace of $V$ has the form $E_{k}$ for some $k\in\left\{
0,1,\ldots,n\right\}  $\ \ \ \ \footnote{\textit{Proof.} Let $Q$ be a
$B$-invariant subspace of $V$. We must show that $Q$ has the form $E_{k}$ for
some $k\in\left\{  0,1,\ldots,n\right\}  $.
\par
We have $Q\subseteq V=\mathbb{F}_{p}^{n}=E_{n}$. Hence, there exists some
$k\in\left\{  -1,0,\ldots,n\right\}  $ such that $Q\subseteq E_{k}$ (namely,
$k=n$). Let $\ell$ be the \textbf{largest} such $k$. Thus, $Q\subseteq
E_{\ell}$.
\par
We have $0\in Q$ (since $Q$ is a subspace of $V$) but $0\notin\varnothing$. If
we had $E_{\ell}=\varnothing$, then we would have $0\in Q\subseteq E_{\ell
}=\varnothing$, which would contradict $0\notin\varnothing$. Hence, we cannot
have $E_{\ell}=\varnothing$. Thus, we have $E_{\ell}\neq\varnothing=E_{-1}$.
Consequently, $\ell\neq-1$. Hence, $\ell\geq0$, so that $\ell\in\left\{
0,1,\ldots,n\right\}  $ and therefore $\ell-1\in\left\{  -1,0,\ldots
,n\right\}  $.
\par
But $\ell$ is the \textbf{largest} $k\in\left\{  -1,0,\ldots,n\right\}  $ such
that $Q\subseteq E_{k}$ (by the definition of $\ell$). Thus, every
$k\in\left\{  -1,0,\ldots,n\right\}  $ satisfying $k<\ell$ satisfies
$Q\not \subseteq E_{k}$. Applying this to $k=\ell-1$, we obtain
$Q\not \subseteq E_{\ell-1}$ (since $\ell-1<\ell$). Thus, there exists some
$q\in Q$ such that $q\notin E_{\ell-1}$. Consider this $q$. Combining $q\in
Q\subseteq E_{\ell}$ with $q\notin E_{\ell-1}$, we obtain $q\in E_{\ell
}\setminus E_{\ell-1}=Be_{\ell}$ (by (\ref{pf.p7.7.1})). In other words, there
exists some $b\in B$ such that $q=be_{\ell}$. Consider this $b$. Since $B$ is
a group, we have $Bb=B$ (since $b\in B$). Now, $B\underbrace{q}_{=be_{\ell}%
}=\underbrace{Bb}_{=B}e_{\ell}=Be_{\ell}=E_{\ell}\setminus E_{\ell-1}$. Thus,
$E_{\ell}\setminus E_{\ell-1}=B\underbrace{q}_{\in Q}\subseteq BQ\subseteq Q$
(since the subspace $Q$ is $B$-invariant).
\par
Now, let $r\in E_{\ell}$. We will show that $r\in Q$. In fact, if $r\in
E_{\ell}\setminus E_{\ell-1}$, then this is obvious (because if $r\in
E_{\ell-1}$, then $r\in E_{\ell}\setminus E_{\ell-1}\subseteq Q$). Thus, we
WLOG assume that we don't have $r\in E_{\ell}\setminus E_{\ell-1}$. In other
words, we have $r\notin E_{\ell}\setminus E_{\ell-1}$. Combining $r\in
E_{\ell}$ with $r\notin E_{\ell}\setminus E_{\ell-1}$, we obtain $r\in
E_{\ell}\setminus\left(  E_{\ell}\setminus E_{\ell-1}\right)  \subseteq
E_{\ell-1}$.
\par
If we had $r-q\in E_{\ell-1}$, then we would have $q=\underbrace{r}_{\in
E_{\ell-1}}-\underbrace{\left(  r-q\right)  }_{\in E_{\ell-1}}\in E_{\ell
-1}-E_{\ell-1}\subseteq E_{\ell-1}$ (since $E_{\ell-1}$ is a vector space),
which would contradict $q\notin E_{\ell-1}$. Hence, we do not have $r-q\in
E_{\ell-1}$. In other words, we have $r-q\notin E_{\ell-1}$. But
$\underbrace{r}_{\in E_{\ell}\setminus E_{\ell-1}\subseteq E_{\ell}%
}-\underbrace{q}_{\in E_{\ell}}\in E_{\ell}-E_{\ell}\subseteq E_{\ell}$ (since
$E_{\ell}$ is a vector space). Combining this with $r-q\notin E_{\ell-1}$, we
obtain $r-q\in E_{\ell}\setminus E_{\ell-1}\subseteq Q$. Now,
$r=\underbrace{q}_{\in Q}+\underbrace{\left(  r-q\right)  }_{\in Q}\in
Q+Q\subseteq Q$ (since $Q$ is a vector space). Hence, we have proven that
$r\in Q$.
\par
Now, forget that we fixed $r$. We thus have shown that $r\in Q$ for each $r\in
E_{\ell}$. In other words, $E_{\ell}\subseteq Q$. Combining this with
$Q\subseteq E_{\ell}$, we obtain $Q=E_{\ell}$. Thus, $Q=E_{k}$ for some
$k\in\left\{  0,1,\ldots,n\right\}  $ (namely, $k=\ell$). In other words, $Q$
has the form $E_{k}$ for some $k\in\left\{  0,1,\ldots,n\right\}  $. Qed.}.
Hence, the subspaces $E_{k}$ (for $k\in\left\{  0,1,\ldots,n\right\}  $) are
the only $B$-invariant subspaces of $V$. This completes the proof. $\square$]

\item \textbf{Proof of Proposition 7.7:} Replace \textquotedblleft the point
$\underline{E}_{I}\in\operatorname*{Flag}\nolimits_{I}\left(  V\right)
$\textquotedblright\ by \textquotedblleft the point $\underline{E}%
\in\operatorname*{Flag}\nolimits_{I}\left(  V\right)  $\textquotedblright.

\item \textbf{Proof of Proposition 7.7:} Replace \textquotedblleft any
map\textquotedblright\ by \textquotedblleft any $G$-equivariant
map\textquotedblright.

\item \textbf{Proof of Proposition 7.7:} You write that \textquotedblleft It
is also clear that $P_{I}\leq P_{J}$ iff $I\subseteq J$\textquotedblright.
Maybe it is worth giving a proof of this:

[\textit{Proof of the fact that }$P_{I}\leq P_{J}$ \textit{iff }$I\subseteq
J$\textit{:} We want to show that $P_{I}\leq P_{J}$ iff $I\subseteq J$. One
direction of this equivalence is clear (namely: if $I\subseteq J$, then
$P_{I}\leq P_{J}$). It remains to prove the other. In other words, it remains
to prove that if $P_{I}\leq P_{J}$, then $I\subseteq J$. So let us assume that
$P_{I}\leq P_{J}$. We must show that $I\subseteq J$.

Lemma 7.5b yields $I=\left\{  i\ \mid\ s_{i}\in P_{I}\right\}  \subseteq
\left\{  i\ \mid\ s_{i}\in P_{J}\right\}  $ (since $P_{I}\leq P_{J}$).

But Lemma 7.5b (applied to $J$ instead of $I$) yields $J=\left\{
i\ \mid\ s_{i}\in P_{J}\right\}  $. Hence, $I\subseteq\left\{  i\ \mid
\ s_{i}\in P_{J}\right\}  =J$. Thus, $I\subseteq J$ is proven. This completes
our proof. $\square$]

\item \textbf{\S 7:} I think it is worthwhile stating three additional facts
as consequences of the proof of Proposition 7.7:

\textbf{Proposition 7.7a.} Let $V=\mathbb{F}_{p}^{n}$. Let $I\subseteq\left\{
1,2,\ldots,n-1\right\}  $. Then, $\operatorname*{Flag}\nolimits_{I}\left(
V\right)  \cong G/P_{I}$ as $G$-sets.

[\textit{Proof of Proposition 7.7a.} In the proof of Proposition 7.7, we have
shown that $G$ acts transitivitely on $\operatorname*{Flag}\nolimits_{I}%
\left(  V\right)  $. Thus, for any $\underline{X}\in\operatorname*{Flag}%
\nolimits_{I}\left(  V\right)  $, we have $\operatorname*{Flag}\nolimits_{I}%
\left(  V\right)  \cong G/G_{\underline{X}}$ as $G$-sets, where
$G_{\underline{X}}$ denotes the stabilizer of $\underline{X}$. Applying this
to $\underline{X}=\underline{E}$ (where $\underline{E}$ is the
\textquotedblleft obvious flag\textquotedblright\ defined in Definition 7.2),
we conclude that $\operatorname*{Flag}\nolimits_{I}\left(  V\right)  \cong
G/G_{\underline{E}}$ as $G$-sets. But the stabilizer of $\underline{E}$ is
$P_{I}$ (by the definition of $P_{I}$). In other words, $G_{\underline{E}%
}=P_{I}$. Hence, $\operatorname*{Flag}\nolimits_{I}\left(  V\right)  \cong
G/\underbrace{G_{\underline{E}}}_{=P_{I}}=G/P_{I}$ as $G$-sets. This proves
Proposition 7.7a. $\square$]

\textbf{Proposition 7.7b.} Let $V=\mathbb{F}_{p}^{n}$. Let $I\subseteq\left\{
1,2,\ldots,n-1\right\}  $. For any $g\in G$, we shall use the notation
$\overline{g}$ for the coset $gP_{I}$ of $g$ in $G/P_{I}$.

\textbf{(a)} There is precisely one $B$-fixed point in $G/P_{I}$. This
$B$-fixed point is $\overline{1}$, and will be called the \textit{basepoint}
of $G/P_{I}$. We have%
\begin{equation}
\left(  G/P_{I}\right)  ^{B}=\left\{  \overline{1}\right\}  .
\label{eq.p7.7b.1}%
\end{equation}


\textbf{(b)} Let $i\in\left\{  1,2,\ldots,n-1\right\}  $. Then, $\left(
G/P_{I}\right)  ^{P_{i}}=%
\begin{cases}
\left\{  \overline{1}\right\}  , & \text{if }i\in I\text{;}\\
\varnothing, & \text{otherwise}%
\end{cases}
$.

[\textit{Proof of Proposition 7.7b.} Proposition 7.7a yields
$\operatorname*{Flag}\nolimits_{I}\left(  V\right)  \cong G/P_{I}$ as $G$-sets.

We have the following general fact about group actions: If $\mathfrak{A}$ is a
subgroup of a group $\mathfrak{G}$, and if $\mathfrak{X}$ is a $\mathfrak{G}%
$-set, then
\begin{equation}
\mathfrak{X}^{\mathfrak{A}}\cong\operatorname*{Map}\nolimits_{\mathfrak{G}%
}\left(  \mathfrak{G}/\mathfrak{A},\mathfrak{X}\right)  \label{pf.p7.7b.1}%
\end{equation}
as sets\footnote{This is easy to prove. (In fact, for each $g\in\mathfrak{G}$,
let $\overline{g}$ denote the coset $g\mathfrak{A}$ of $g$ in $\mathfrak{G}%
/\mathfrak{A}$. Then, the map $\operatorname*{Map}\nolimits_{\mathfrak{G}%
}\left(  \mathfrak{G}/\mathfrak{A},\mathfrak{X}\right)  \rightarrow
\mathfrak{X}^{\mathfrak{A}}$ sending each $f\in\operatorname*{Map}%
\nolimits_{\mathfrak{G}}\left(  \mathfrak{G}/\mathfrak{A},\mathfrak{X}\right)
$ to $f\left(  \overline{1}\right)  \in\mathfrak{X}^{\mathfrak{A}}$ is a
bijection. Indeed, its inverse map sends each $u\in\mathfrak{X}^{\mathfrak{A}%
}$ to the $G$-map $\mathfrak{G}/\mathfrak{A}\rightarrow\mathfrak{X}%
,\ \overline{g}\mapsto gu$.)}.

\textbf{(a)} In the proof of Proposition 7.7, we have seen that the point
$\underline{E}\in\operatorname*{Flag}\nolimits_{I}\left(  V\right)  $ is the
unique $B$-fixed point in $\operatorname*{Flag}\nolimits_{I}\left(  V\right)
$. In other words, $\left(  \operatorname*{Flag}\nolimits_{I}\left(  V\right)
\right)  ^{B}=\left\{  \underline{E}\right\}  $. But $\operatorname*{Flag}%
\nolimits_{I}\left(  V\right)  \cong G/P_{I}$ as $G$-sets. Hence, $\left(
\operatorname*{Flag}\nolimits_{I}\left(  V\right)  \right)  ^{B}\cong\left(
G/P_{I}\right)  ^{B}$ as sets. Thus, $\left(  G/P_{I}\right)  ^{B}\cong\left(
\operatorname*{Flag}\nolimits_{I}\left(  V\right)  \right)  ^{B}=\left\{
\underline{E}\right\}  $ as sets. Hence, $\left(  G/P_{I}\right)  ^{B}$ is a
$1$-element set (since $\left\{  \underline{E}\right\}  $ is a $1$-element set).

For every $b\in B$, we have $b\overline{1}=\overline{b1}=\overline
{b}=\overline{1}$ in $G/P_{I}$ (since $b\in B\leq P_{I}$). Therefore,
$\overline{1}\in\left(  G/P_{I}\right)  ^{B}$. Therefore, $\left(
G/P_{I}\right)  ^{B}=\left\{  \overline{1}\right\}  $ (since $\left(
G/P_{I}\right)  ^{B}$ is a $1$-element set). In other words, the set of all
$B$-fixed points in $G/P_{I}$ is $\left\{  \overline{1}\right\}  $. In other
words, there is precisely one $B$-fixed point in $G/P_{I}$, and this $B$-fixed
point is $\overline{1}$. This completes the proof of Proposition 7.7b
\textbf{(a)}.

\textbf{(b)} Applying (\ref{pf.p7.7b.1}) to $\mathfrak{G}=G$, $\mathfrak{A}%
=P_{i}$ and $\mathfrak{X}=G/P_{I}$, we conclude that $\left(  G/P_{I}\right)
^{P_{i}}\cong\operatorname*{Map}\nolimits_{G}\left(  G/P_{i},G/P_{I}\right)  $
as sets. But recall that $\operatorname*{Flag}\nolimits_{I}\left(  V\right)
\cong G/P_{I}$ as $G$-sets. Also, Proposition 7.7a (applied to $\left\{
i\right\}  $ instead of $I$) yields $\operatorname*{Flag}\nolimits_{\left\{
i\right\}  }\left(  V\right)  \cong G/P_{\left\{  i\right\}  }=G/P_{i}$ as
$G$-sets. Hence,%
\begin{align*}
\left(  G/P_{I}\right)  ^{P_{i}}  &  \cong\operatorname*{Map}\nolimits_{G}%
\left(  \underbrace{G/P_{i}}_{\substack{\cong\operatorname*{Flag}%
\nolimits_{\left\{  i\right\}  }\left(  V\right)  \\=\operatorname*{Flag}%
\nolimits_{i}\left(  V\right)  }},\underbrace{G/P_{I}}_{\cong%
\operatorname*{Flag}\nolimits_{I}\left(  V\right)  }\right)  \cong%
\operatorname*{Map}\nolimits_{G}\left(  \operatorname*{Flag}\nolimits_{i}%
\left(  V\right)  ,\operatorname*{Flag}\nolimits_{I}\left(  V\right)  \right)
\\
&  =%
\begin{cases}
\left(  \text{a singleton}\right)  , & \text{if }\left\{  i\right\}  \subseteq
I\text{;}\\
\varnothing, & \text{otherwise}%
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition 7.7 (applied to }\left\{  i\right\}  \text{ and }I\\
\text{instead of }I\text{ and }J\text{)}%
\end{array}
\right) \\
&  =%
\begin{cases}
\left(  \text{a singleton}\right)  , & \text{if }i\in I\text{;}\\
\varnothing, & \text{otherwise}%
\end{cases}
.
\end{align*}
Therefore, if $i\notin I$, then $\left(  G/P_{I}\right)  ^{P_{i}}%
\cong\varnothing$ and thus $\left(  G/P_{I}\right)  ^{P_{i}}=\varnothing$.
Hence, Proposition 7.7b \textbf{(b)} is proven in the case when $i\notin I$.
We thus WLOG assume that we don't have $i\notin I$. Hence, we have $i\in I$.
We must show that $\left(  G/P_{I}\right)  ^{P_{i}}=\left\{  \overline
{1}\right\}  $.

Now, $\left(  G/P_{I}\right)  ^{P_{i}}\cong%
\begin{cases}
\left(  \text{a singleton}\right)  , & \text{if }i\in I\text{;}\\
\varnothing, & \text{otherwise}%
\end{cases}
=\left(  \text{a singleton}\right)  $ (since $i\in I$). Hence, $\left(
G/P_{I}\right)  ^{P_{i}}$ is a $1$-element set.

But $i\in I$, so that $\left\{  i\right\}  \subseteq I$ and thus $P_{\left\{
i\right\}  }\subseteq P_{I}$. Therefore, every $p\in P_{i}$ satisfies
$p\overline{1}=\overline{p1}=\overline{p}=\overline{1}$ in $G/P_{I}$ (since
$p\in P_{i}=P_{\left\{  i\right\}  }\subseteq P_{I}$). In other words,
$\overline{1}\in\left(  G/P_{I}\right)  ^{P_{i}}$. Since $\left(
G/P_{I}\right)  ^{P_{i}}$ is a $1$-element set, we can therefore conclude that
$\left(  G/P_{I}\right)  ^{P_{i}}=\left\{  \overline{1}\right\}  $. This
completes the proof of Proposition 7.7b \textbf{(b)}.]

\textbf{Proposition 7.7c.} Let $V=\mathbb{F}_{p}^{n}$. Let $X$ be a parabolic
$G$-set. For each $y\in X^{B}$, set $I_{y}=\left\{  i\in\left\{
1,2,\ldots,n-1\right\}  \ \mid\ y\in X^{P_{i}}\right\}  $. For each $y\in X$,
let $G_{y}$ denote the stabilizer of $y$ in $G$.

\textbf{(a)} If $y\in X^{B}$, then $P_{I_{y}}=G_{y}$.

\textbf{(b)} Let $y_{1}\in X^{B}$, $y_{2}\in X^{B}$, $q_{1}\in G$ and
$q_{2}\in G$ be such that $q_{1}y_{1}=q_{2}y_{2}$. Then, $y_{1}=y_{2}$ and
$q_{1}G_{y_{1}}=q_{2}G_{y_{1}}$.

[\textit{Proof of Proposition 7.7c.} \textbf{(a)} Let $y\in X^{B}$. Then,
$B\subseteq G_{y}$ (since $y\in X^{B}$), so that $G_{y}\geq B$. Thus, $G_{y}$
is a subgroup of $G$ such that $G_{y}\geq B$.

Let $I=\left\{  i\ \mid\ s_{i}\in G_{y}\right\}  $. Hence, Proposition 7.5a
(applied to $P=G_{y}$) shows that $G_{y}=P_{I}$.

Now, let $j\in I_{y}$. Then, $j\in I_{y}=\left\{  i\in\left\{  1,2,\ldots
,n-1\right\}  \ \mid\ y\in X^{P_{i}}\right\}  $. In other words, $j$ is an
element of $\left\{  1,2,\ldots,n-1\right\}  $ and satisfies $y\in X^{P_{j}}$.
But $P_{j}=P_{\left\{  j\right\}  }=B\Sigma_{\left\{  j\right\}  }B$ (by
Proposition 7.3, applied to $\left\{  j\right\}  $ instead of $I$). Now,
$s_{j}=\underbrace{1}_{\in B}\underbrace{s_{j}}_{\in\Sigma_{\left\{
j\right\}  }}\underbrace{1}_{\in B}\in B\Sigma_{\left\{  j\right\}  }B=P_{j}$.

From $y\in X^{P_{j}}$, we conclude that $py=y$ for each $p\in P_{j}$. Applying
this to $p=s_{j}$, we obtain $s_{j}y=y$ (since $s_{j}\in P_{j}$). In other
words, $s_{j}\in G_{y}$. In other words, $j\in\left\{  i\ \mid\ s_{i}\in
G_{y}\right\}  $. This rewrites as $j\in I$ (since $I=\left\{  i\ \mid
\ s_{i}\in G_{y}\right\}  $).

Now, forget that we fixed $j$. We thus have proven that $j\in I$ for each
$j\in I_{y}$. In other words, $I_{y}\subseteq I$.

On the other hand, let $k\in I$. Thus, $k\in I=\left\{  i\ \mid\ s_{i}\in
G_{y}\right\}  $. In other words, $k$ is an element of $\left\{
1,2,\ldots,n-1\right\}  $ and satisfies $s_{k}\in G_{y}$. In other words,
$s_{k}y=y$.

But $P_{k}=P_{\left\{  k\right\}  }=B\Sigma_{\left\{  k\right\}  }B$ (by
Proposition 7.3, applied to $\left\{  k\right\}  $ instead of $I$). The
definition of $\Sigma_{\left\{  k\right\}  }$ yields $\Sigma_{\left\{
k\right\}  }=\left\langle s_{k}\right\rangle =\left\{  1,s_{k}\right\}  $.
Hence, $gy=y$ for each $g\in\Sigma_{\left\{  k\right\}  }$ (since $1y=y$ and
$s_{k}y=y$). In other words, $y\in X^{\Sigma_{\left\{  k\right\}  }}$.

Now, let $p\in P_{k}$. Then, $p\in P_{k}=B\Sigma_{\left\{  k\right\}  }B$. In
other words, there exist $b_{1}\in B$, $g\in\Sigma_{\left\{  k\right\}  }$ and
$b_{2}\in B$ such that $p=b_{1}gb_{2}$. Consider these $b_{1}$, $g$ and
$b_{2}$. Now,%
\[
\underbrace{p}_{=b_{1}gb_{2}}y=b_{1}g\underbrace{b_{2}y}%
_{\substack{=y\\\text{(since }y\in X^{B}\text{)}}}=b_{1}\underbrace{gy}%
_{\substack{=y\\\text{(since }y\in X^{\Sigma_{\left\{  k\right\}  }}\text{)}%
}}=b_{1}y=y
\]
(since $y\in X^{B}$).

Now, forget that we fixed $p$. We thus have proven that $py=y$ for each $p\in
P_{k}$. In other words, $y\in X^{P_{k}}$. Hence, $k$ is an element of
$\left\{  1,2,\ldots,n-1\right\}  $ and satisfies $y\in X^{P_{k}}$. In other
words, $k\in\left\{  i\in\left\{  1,2,\ldots,n-1\right\}  \ \mid\ y\in
X^{P_{i}}\right\}  $. In other words, $k\in I_{y}$ (since $I_{y}=\left\{
i\in\left\{  1,2,\ldots,n-1\right\}  \ \mid\ y\in X^{P_{i}}\right\}  $).

Now, forget that we fixed $k$. We thus have proven that $k\in I_{y}$ for each
$k\in I$. In other words, $I\subseteq I_{y}$. Combining this with
$I_{y}\subseteq I$, we obtain $I_{y}=I$. Hence, $P_{I_{y}}=P_{I}=G_{y}$. This
proves Proposition 7.7c \textbf{(a)}.

\textbf{(b)} Let $Y=Gy_{1}$ be the $G$-orbit of $y_{1}$. Then, $Y$ is a
$G$-subset of $X^{B}$. Moreover, $Y\cong G/G_{y_{1}}$ as $G$-sets (by the
orbit-stabilizer theorem). Proposition 7.7c \textbf{(a)} (applied to $y=y_{1}%
$) yields $P_{I_{y_{1}}}=G_{y_{1}}$. But Proposition 7.7b \textbf{(a)}
(applied to $I=I_{y_{1}}$) yields $\left(  G/P_{I_{y_{1}}}\right)
^{B}=\left\{  \overline{1}\right\}  $. Hence, $\left\vert \left(
G/P_{I_{y_{1}}}\right)  ^{B}\right\vert =\left\vert \left\{  \overline
{1}\right\}  \right\vert =1$.

We have $Y\cong G/\underbrace{G_{y_{1}}}_{=P_{I_{y_{1}}}}=G/P_{I_{y_{1}}}$ as
$G$-sets, and thus $Y^{B}\cong\left(  G/P_{I_{y_{1}}}\right)  ^{B}$ as sets.
Hence, $\left\vert Y^{B}\right\vert =\left\vert \left(  G/P_{I_{y_{1}}%
}\right)  ^{B}\right\vert =1$.

Both $y_{1}$ and $y_{2}$ are $B$-fixed points (since $y_{1}\in X^{B}$ and
$y_{2}\in X^{B}$). We have $q_{1}y_{1}=q_{2}y_{2}$. Multiplying both sides of
this equality by $q_{2}^{-1}$, we obtain $q_{2}^{-1}q_{1}y_{1}%
=\underbrace{q_{2}^{-1}q_{2}}_{=1}y_{2}=y_{2}$, so that $y_{2}%
=\underbrace{q_{2}^{-1}q_{1}}_{\in G}y_{1}\in Gy_{1}=Y$.

We have $y_{1}\in Y$ (since $Y$ is the $G$-orbit of $y_{1}$). Thus, $y_{1}\in
Y^{B}$ (since $y_{1}$ is a $B$-fixed point). We also have $y_{2}\in Y$. Thus,
$y_{2}\in Y^{B}$ (since $y_{2}$ is a $B$-fixed point).

But $Y^{B}$ is a $1$-element set (since $\left\vert Y^{B}\right\vert =1$).
Thus, any two elements of $Y^{B}$ are identical. Applying this to the two
elements $y_{1}$ and $y_{2}$ of $Y^{B}$, we conclude that $y_{1}$ and $y_{2}$
are identical (since $y_{1}\in Y^{B}$ and $y_{2}\in Y^{B}$). In other words,
$y_{1}=y_{2}$.

Now, $q_{2}^{-1}q_{1}y_{1}=y_{2}=y_{1}$. In other words, $q_{2}^{-1}q_{1}\in
G_{y_{1}}$. In other words, $q_{1}G_{y_{1}}=q_{2}G_{y_{1}}$. This completes
the proof of Proposition 7.7c \textbf{(b)}. $\square$]

\item \textbf{Definition 7.8:} Replace \textquotedblleft the category of
finite sets $Y$ equipped with a list $\left(  Y_{1},\ldots,Y_{n-1}\right)  $
of subsets.\textquotedblright\ by \textquotedblleft the category whose objects
are finite sets $Y$ equipped with a list $\left(  Y_{1},\ldots,Y_{n-1}\right)
$ of subsets. Such an object will be denoted $\left(  Y;Y_{1},\ldots
,Y_{n-1}\right)  $. Morphisms $\left(  Y;Y_{1},\ldots,Y_{n-1}\right)
\rightarrow\left(  Z;Z_{1},\ldots,Z_{n-1}\right)  $ in $\mathcal{P}^{\prime}$
shall be maps $Y\rightarrow Z$ mapping each $Y_{i}$ into $Z_{i}$%
.\textquotedblright

\item \textbf{Proof of Proposition 7.9:} Replace \textquotedblleft Consider an
object $Y\in\mathcal{P}^{\prime}$.\textquotedblright\ by \textquotedblleft
Consider an object $\left(  Y;Y_{1},\ldots,Y_{n-1}\right)  \in\mathcal{P}%
^{\prime}$ (abbreviated as $Y$).\textquotedblright.

\item \textbf{Proof of Proposition 7.9:} Replace \textquotedblleft Now
consider a morphism $f:Y\rightarrow Z$ in $\mathcal{P}^{\prime}$%
\textquotedblright\ by \textquotedblleft Now consider a morphism $f:\left(
Y;Y_{1},\ldots,Y_{n-1}\right)  \rightarrow\left(  Z;Z_{1},\ldots
,Z_{n-1}\right)  $ in $\mathcal{P}^{\prime}$\textquotedblright.

\item \textbf{Proof of Proposition 7.9:} You write: \textquotedblleft so there
is a unique $G$-map $G/P_{I_{y}}\rightarrow G/P_{I_{f\left(  y\right)  }}%
$\textquotedblright. The uniqueness of this $G$-map might need a
proof\footnote{\textit{Proof.} We have $I_{y}\subseteq I_{f\left(  y\right)
}$. Thus, $P_{I_{y}}\subseteq P_{I_{f\left(  y\right)  }}$ (because if two
subsets $I$ and $J$ of $\left\{  1,2,\ldots,n-1\right\}  $ satisfy $I\subseteq
J$, then they also satisfy $P_{I}\subseteq P_{J}$). Hence, there clearly
exists a $G$-map $G/P_{I_{y}}\rightarrow G/P_{I_{f\left(  y\right)  }}$
(namely, the map that sends any coset $gP_{I_{y}}$ of $P_{I_{y}}$ to the coset
$gP_{I_{f\left(  y\right)  }}$ of $P_{I_{f\left(  y\right)  }}$). It remains
to prove that there exists \textbf{at most one} $G$-map $G/P_{I_{y}%
}\rightarrow G/P_{I_{f\left(  y\right)  }}$.
\par
Let $V=\mathbb{F}_{p}^{n}$. If $I$ is any subset of $\left\{  1,2,\ldots
,n-1\right\}  $, then we have $\operatorname*{Flag}\nolimits_{I}\left(
V\right)  \cong G/P_{I}$ as $G$-sets (because $G$ acts transitively on the
$G$-set $\operatorname*{Flag}\nolimits_{I}\left(  V\right)  $, and the
stabilizer of the element $\underline{E}\in\operatorname*{Flag}\nolimits_{I}%
\left(  V\right)  $ is $P_{I}$). In other words, if $I$ is a subset of
$\left\{  1,2,\ldots,n-1\right\}  $, then $G/P_{I}\cong\operatorname*{Flag}%
\nolimits_{I}\left(  V\right)  $ as $G$-sets. Thus, if $I$ and $J$ are two
subsets of $\left\{  1,2,\ldots,n-1\right\}  $, then%
\[
\operatorname*{Map}\nolimits_{G}\left(  \underbrace{G/P_{I}}_{\cong%
\operatorname*{Flag}\nolimits_{I}\left(  V\right)  },\underbrace{G/P_{J}%
}_{\cong\operatorname*{Flag}\nolimits_{J}\left(  V\right)  }\right)
\cong\operatorname*{Map}\nolimits_{G}\left(  \operatorname*{Flag}%
\nolimits_{I}\left(  V\right)  ,\operatorname*{Flag}\nolimits_{J}\left(
V\right)  \right)  =%
\begin{cases}
\text{a singleton}, & \text{if }I\subseteq J\\
\varnothing, & \text{otherwise}%
\end{cases}
\]
(by Proposition 7.7). Hence, if $I$ and $J$ are two subsets of $\left\{
1,2,\ldots,n-1\right\}  $, then the set $\operatorname*{Map}\nolimits_{G}%
\left(  G/P_{I},G/P_{J}\right)  $ has at most one element. In other words, if
$I$ and $J$ are two subsets of $\left\{  1,2,\ldots,n-1\right\}  $, then there
exists \textbf{at most one} $G$-map $G/P_{I}\rightarrow G/P_{J}$. Applying
this to $I=I_{y}$ and $J=I_{f\left(  y\right)  }$, we conclude that there
exists \textbf{at most one} $G$-map $G/P_{I_{y}}\rightarrow G/P_{I_{f\left(
y\right)  }}$. This concludes our proof.}.

\item \textbf{Proof of Proposition 7.9:} Replace \textquotedblleft this gives
us a functor $\mathcal{P}^{\prime}\rightarrow\mathcal{P}$\textquotedblright%
\ by \textquotedblleft this gives us a functor $F^{\prime}:\mathcal{P}%
^{\prime}\rightarrow\mathcal{P}$\textquotedblright.

\item \textbf{Proof of Proposition 7.9:} You write: \textquotedblleft Note
that%
\[
\left(  G/P_{I}\right)  ^{P_{i}}=\operatorname*{Map}\nolimits_{G}\left(
\operatorname*{Flag}\nolimits_{i}\left(  V\right)  ,\operatorname*{Flag}%
\nolimits_{I}\left(  V\right)  \right)  =%
\begin{cases}
1 & \text{if }i\in I\\
\varnothing & \text{otherwise.}%
\end{cases}
\]
Using this we see that $FF^{\prime}=1_{\mathcal{P}^{\prime}}$%
.\textquotedblright.

I would suggest replacing this by the following (more detailed) argument:

\textquotedblleft Let $Y\in\mathcal{P}^{\prime}$ be an object. Then,
$F^{\prime}Y=\coprod_{y\in Y}G/P_{I_{y}}$, so that
\[
\left(  F^{\prime}Y\right)  ^{B}=\left(  \coprod_{y\in Y}G/P_{I_{y}}\right)
^{B}\cong\coprod_{y\in Y}\underbrace{\left(  G/P_{I_{y}}\right)  ^{B}%
}_{\substack{=\left\{  \overline{1}\right\}  \\\text{(by (\ref{eq.p7.7b.1}),
applied to }I=I_{y}\text{)}}}=\coprod_{y\in Y}\left\{  \overline{1}\right\}
.
\]
Hence, there exists a bijection $Y\rightarrow\left(  F^{\prime}Y\right)  ^{B}$
that sends each $y\in Y$ to the element $\overline{1}$ of $G/P_{I_{y}}$.
Denote this bijection by $\eta_{Y}$.

We have $FF^{\prime}Y=\left(  F^{\prime}Y\right)  ^{B}$ as sets (by the
definition of the functor $F$). Thus, the bijection $\eta_{Y}:Y\rightarrow
\left(  F^{\prime}Y\right)  ^{B}$ is a bijection $Y\rightarrow FF^{\prime}Y$.

This bijection $\eta_{Y}$ is an isomorphism in the category $\mathcal{P}%
^{\prime}$\ \ \ \ \footnote{\textit{Proof.} Fix $j\in\left\{  1,2,\ldots
,n-1\right\}  $.
\par
For each $y\in Y$, we have $I_{y}=\left\{  i\in\left\{  1,2,\ldots
,n-1\right\}  \ \mid\ y\in Y_{i}\right\}  $ (by the definition of $I_{y}$).
Hence, for each $y\in Y$ and $i\in\left\{  1,2,\ldots,n-1\right\}  $, we have
the following logical equivalence:%
\[
\left(  i\in I_{y}\right)  \ \Longleftrightarrow\ \left(  y\in Y_{i}\right)
.
\]
Applying this to $i=j$, we conclude that for each $y\in Y$, we have the
following logical equivalence:%
\begin{equation}
\left(  j\in I_{y}\right)  \ \Longleftrightarrow\ \left(  y\in Y_{j}\right)  .
\label{pf.p7.9.u.fn1.2}%
\end{equation}
\par
For every $y\in Y$, we have%
\begin{align}
\left(  G/P_{I_{y}}\right)  ^{P_{j}}  &  =%
\begin{cases}
\left\{  \overline{1}\right\}  , & \text{if }j\in I_{y}\text{;}\\
\varnothing, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition 7.7b \textbf{(b)}, applied to
}j\text{ instead of }i\right) \nonumber\\
&  =%
\begin{cases}
\left\{  \overline{1}\right\}  , & \text{if }y\in Y_{j}\text{;}\\
\varnothing, & \text{otherwise}%
\end{cases}
\label{pf.p7.9.u.fn1.3}%
\end{align}
(because of the equivalence (\ref{pf.p7.9.u.fn1.2})).
\par
From $F^{\prime}Y=\coprod_{y\in Y}G/P_{I_{y}}$, we obtain%
\begin{align*}
\left(  F^{\prime}Y\right)  ^{P_{j}}  &  =\left(  \coprod_{y\in Y}G/P_{I_{y}%
}\right)  ^{P_{j}}\cong\coprod_{y\in Y}\underbrace{\left(  G/P_{I_{y}}\right)
^{P_{j}}}_{\substack{=%
\begin{cases}
\left\{  \overline{1}\right\}  , & \text{if }y\in Y_{j}\text{;}\\
\varnothing, & \text{otherwise}%
\end{cases}
\\\text{(by (\ref{pf.p7.9.u.fn1.3}))}}}=\coprod_{y\in Y}%
\begin{cases}
\left\{  \overline{1}\right\}  , & \text{if }y\in Y_{j}\text{;}\\
\varnothing, & \text{otherwise}%
\end{cases}
\\
&  =\coprod_{y\in Y_{j}}\left\{  \overline{1}\right\}  =\eta_{Y}\left(
Y_{j}\right)
\end{align*}
(because the definition of $\eta_{Y}$ yields $\eta_{Y}\left(  Y_{j}\right)
=\coprod_{y\in Y_{j}}\left\{  \overline{1}\right\}  $).
\par
Now, forget that we fixed $j$. We thus have proven that
\[
\left(  F^{\prime}Y\right)  ^{P_{j}}=\eta_{Y}\left(  Y_{j}\right)
\]
for each $j\in\left\{  1,2,\ldots,n-1\right\}  $. Hence, we have $\eta
_{Y}\left(  Y_{j}\right)  \subseteq\left(  F^{\prime}Y\right)  ^{P_{j}}$ and
$\left(  \eta_{Y}\right)  ^{-1}\left(  \left(  F^{\prime}Y\right)  ^{P_{j}%
}\right)  \subseteq Y_{j}$ for each $j\in\left\{  1,2,\ldots,n-1\right\}  $.
\par
Recall that $Y=\left(  Y;Y_{1},\ldots,Y_{n-1}\right)  $ and $FF^{\prime
}Y=\left(  \left(  F^{\prime}Y\right)  ^{B};\left(  F^{\prime}Y\right)
^{P_{1}},\ldots,\left(  F^{\prime}Y\right)  ^{P_{n-1}}\right)  $ (by the
definition of $F$). Hence, the bijection $\eta_{Y}:Y\rightarrow FF^{\prime}Y$
is a morphism in $\mathcal{P}^{\prime}$ (since $\eta_{Y}\left(  Y_{j}\right)
\subseteq\left(  F^{\prime}Y\right)  ^{P_{j}}$ for each $j\in\left\{
1,2,\ldots,n-1\right\}  $), and its inverse $\left(  \eta_{Y}\right)
^{-1}:FF^{\prime}Y\rightarrow Y$ is also a morphism in $\mathcal{P}^{\prime}$
(since $\left(  \eta_{Y}\right)  ^{-1}\left(  \left(  F^{\prime}Y\right)
^{P_{j}}\right)  \subseteq Y_{j}$ for each $j\in\left\{  1,2,\ldots
,n-1\right\}  $). Thus, $\eta_{Y}$ is an isomorphism in the category
$\mathcal{P}^{\prime}$. Qed.}.

Now, forget that we fixed $Y$. Thus, for each object $Y\in\mathcal{P}^{\prime
}$, we have constructed an isomorphism $\eta_{Y}:Y\rightarrow FF^{\prime}Y$ in
the category $\mathcal{P}^{\prime}$. It is straightforward to see that this
isomorphism $\eta_{Y}$ is functorial in $Y$. Thus, we have defined a natural
isomorphism $\eta:1_{\mathcal{P}^{\prime}}\rightarrow FF^{\prime}$. Therefore,
$FF^{\prime}\cong1_{\mathcal{P}^{\prime}}$ as functors.\textquotedblright.

\item \textbf{Proof of Proposition 7.9:} After \textquotedblleft The only
$B$-fixed point in $G/P_{I}$ is the basepoint\textquotedblright, I would add
\textquotedblleft(by Proposition 7.7b \textbf{(a)})\textquotedblright.

\item \textbf{Proof of Proposition 7.9:} After \textquotedblleft and the
basepoint is fixed by $P_{i}$ iff $i\in I$\textquotedblright, I would add
\textquotedblleft(by Proposition 7.7b \textbf{(b)})\textquotedblright.

\item \textbf{Proof of Proposition 7.9:} Replace \textquotedblleft%
$X=F^{\prime}FX$\textquotedblright\ by \textquotedblleft$X\cong F^{\prime}FX$
by a functorial isomorphism (i.e., we have $1_{\mathcal{P}}\cong F^{\prime}%
F$)\textquotedblright.

More importantly, I believe that this claim should be proven. Here is my proof:

[\textit{Proof of the functorial isomorphism }$1_{\mathcal{P}}\cong F^{\prime
}F$\textit{:} Let $X\in\mathcal{P}$ be an object. Then, $FX=\left(
X^{B};X^{P_{1}},\ldots,X^{P_{n-1}}\right)  $ (by the definition of the functor
$F$). Hence, the definition of the functor $F^{\prime}$ shows that $F^{\prime
}FX=\coprod_{y\in X^{B}}G/P_{I_{y}}$, where we set $I_{y}=\left\{
i\in\left\{  1,2,\ldots,n-1\right\}  \ \mid\ y\in X^{P_{i}}\right\}  $ for
each $y\in X^{B}$. We shall now define a map $\varepsilon_{X}:F^{\prime
}FX\rightarrow X$ as follows:

Let $p\in F^{\prime}FX$. Then, $p\in F^{\prime}FX=\coprod_{y\in X^{B}%
}G/P_{I_{y}}$. In other words, $p\in G/P_{I_{y}}$ for some $y\in X^{B}$.
Consider this $y$. Write $p$ in the form $p=\overline{q}$ for some $q\in G$
(where $\overline{q}$ denotes the coset $qP_{I_{y}}$ of $q$ in $G/P_{I_{y}}$).
Then, the element $qy$ of $X$ does not depend on the choice of $q$%
\ \ \ \ \footnote{\textit{Proof.} Let $q_{1}$ and $q_{2}$ be two elements
$q\in G$ satisfying $p=\overline{q}$. We must prove that $q_{1}y=q_{2}y$.
\par
We know that $q_{1}$ is an element $q\in G$ satisfying $p=\overline{q}$. In
other words, $q_{1}$ is an element of $G$ and satisfies $p=\overline{q_{1}}$.
Similarly, $q_{2}$ is an element of $G$ and satisfies $p=\overline{q_{2}}$.
From $\overline{q_{1}}=p=\overline{q_{2}}$, we conclude that $q_{1}P_{I_{y}%
}=q_{2}P_{I_{y}}$. In other words, $q_{1}=q_{2}h$ for some $h\in P_{I_{y}}$.
\par
Let $G_{y}$ denote the stabilizer of $y$ in $G$. Then, Proposition 7.7c
\textbf{(a)} yields $P_{I_{y}}=G_{y}$. Therefore, $h\in P_{I_{y}}=G_{y}$. In
other words, $hy=y$.
\par
Now, $\underbrace{q_{1}}_{=q_{2}h}y=q_{2}\underbrace{hy}_{=y}=q_{2}y$. This
completes our proof.}. Hence, we can define $\varepsilon_{X}\left(  p\right)
$ to be the element $qy$ of $X$. Thus, a map $\varepsilon_{X}:F^{\prime
}FX\rightarrow X$ is defined.

This map $\varepsilon_{X}:F^{\prime}FX\rightarrow X$ is $G$%
-equivariant\footnote{\textit{Proof.} Let $p\in FF^{\prime}X$ and $g\in G$. We
must show that $\varepsilon_{X}\left(  gp\right)  =g\varepsilon_{X}\left(
p\right)  $.
\par
Indeed, we have $p\in F^{\prime}FX=\coprod_{y\in X^{B}}G/P_{I_{y}}$. In other
words, $p\in G/P_{I_{y}}$ for some $y\in X^{B}$. Consider this $y$. Write $p$
in the form $p=\overline{q}$ for some $q\in G$ (where $\overline{q}$ denotes
the coset $qP_{I_{y}}$ of $q$ in $G/P_{I_{y}}$). Then, $\varepsilon_{X}\left(
p\right)  =qy$ (by the definition of $\varepsilon_{X}$). On the other hand,
$gp\in G/P_{I_{y}}$ (since $p\in G/P_{I_{y}}$) and $g\underbrace{p}%
_{=\overline{q}}=g\overline{q}=\overline{gq}$ with $gq\in G$. Hence, the
definition of $\varepsilon_{X}$ yields $\varepsilon_{X}\left(  gp\right)
=g\underbrace{qy}_{=\varepsilon_{X}\left(  p\right)  }=g\varepsilon_{X}\left(
p\right)  $.
\par
Now, forgot that we fixed $p$ and $g$. We thus have shown that $\varepsilon
_{X}\left(  gp\right)  =g\varepsilon_{X}\left(  p\right)  $ for each $p\in
FF^{\prime}X$ and $g\in G$. In other words, the map $\varepsilon_{X}%
:F^{\prime}FX\rightarrow X$ is $G$-equivariant. Qed.}. Moreover, this map
$\varepsilon_{X}$ is injective\footnote{\textit{Proof.} Let $p_{1}\in
F^{\prime}FX$ and $p_{2}\in F^{\prime}FX$ be such that $\varepsilon_{X}\left(
p_{1}\right)  =\varepsilon_{X}\left(  p_{2}\right)  $. We shall show that
$p_{1}=p_{2}$.
\par
For each $y\in X$, let $G_{y}$ denote the stabilizer of $y$ in $G$.
\par
We have $p_{1}\in F^{\prime}FX=\coprod_{y\in X^{B}}G/P_{I_{y}}$. In other
words, $p_{1}\in G/P_{I_{y}}$ for some $y\in X^{B}$. Denote this $y$ by
$y_{1}$. Thus, $y_{1}\in X^{B}$ and $p_{1}\in G/P_{I_{y_{1}}}$. Write $p_{1}$
in the form $p_{1}=\overline{q_{1}}^{/I_{y_{1}}}$ for some $q_{1}\in G$ (where
$\overline{q_{1}}^{/I_{y_{1}}}$ denotes the coset $q_{1}P_{I_{y_{1}}}$ of
$q_{1}$ in $G/P_{I_{y_{1}}}$). Then, $\varepsilon_{X}\left(  p_{1}\right)
=q_{1}y_{1}$ (by the definition of $\varepsilon_{X}$).
\par
We have $p_{2}\in F^{\prime}FX=\coprod_{y\in X^{B}}G/P_{I_{y}}$. In other
words, $p_{2}\in G/P_{I_{y}}$ for some $y\in X^{B}$. Denote this $y$ by
$y_{2}$. Thus, $y_{2}\in X^{B}$ and $p_{2}\in G/P_{I_{y_{2}}}$. Write $p_{2}$
in the form $p_{2}=\overline{q_{2}}^{/I_{y_{2}}}$ for some $q_{2}\in G$ (where
$\overline{q_{2}}^{/I_{y_{2}}}$ denotes the coset $q_{2}P_{I_{y_{2}}}$ of
$q_{2}$ in $G/P_{I_{y_{2}}}$). Then, $\varepsilon_{X}\left(  p_{2}\right)
=q_{2}y_{2}$ (by the definition of $\varepsilon_{X}$).
\par
Now, $q_{1}y_{1}=\varepsilon_{X}\left(  p_{1}\right)  =\varepsilon_{X}\left(
p_{2}\right)  =q_{2}y_{2}$. Hence, Proposition 7.7c \textbf{(b)} shows that
$y_{1}=y_{2}$ and $q_{1}G_{y_{1}}=q_{2}G_{y_{1}}$.
\par
On the other hand, Proposition 7.7c \textbf{(a)} (applied to $y=y_{1}$) shows
that $P_{I_{y_{1}}}=G_{y_{1}}$. Thus, the equality $q_{1}G_{y_{1}}%
=q_{2}G_{y_{1}}$ rewrites as $q_{1}P_{I_{y_{1}}}=q_{2}P_{I_{y_{1}}}$.
\par
The definition of $\overline{q_{1}}^{/I_{y_{1}}}$ yields $\overline{q_{1}%
}^{/I_{y_{1}}}=q_{1}P_{I_{y_{1}}}=q_{2}P_{I_{y_{1}}}=q_{2}P_{I_{y_{2}}}$
(since $y_{1}=y_{2}$). Hence, $p_{1}=\overline{q_{1}}^{/I_{y_{1}}}%
=q_{2}P_{I_{y_{2}}}$.
\par
The definition of $\overline{q_{2}}^{/I_{y_{2}}}$ yields $\overline{q_{2}%
}^{/I_{y_{2}}}=q_{2}P_{I_{y_{2}}}$. Hence, $p_{2}=\overline{q_{2}}^{/I_{y_{2}%
}}=q_{2}P_{I_{y_{2}}}$. Comparing this with $p_{1}=q_{2}P_{I_{y_{2}}}$, we
obtain $p_{1}=p_{2}$.
\par
Now, let us forget that we fixed $p_{1}$ and $p_{2}$. We thus have proven that
if $p_{1}\in F^{\prime}FX$ and $p_{2}\in F^{\prime}FX$ are such that
$\varepsilon_{X}\left(  p_{1}\right)  =\varepsilon_{X}\left(  p_{2}\right)  $,
then $p_{1}=p_{2}$. In other words, the map $\varepsilon_{X}$ is injective.
Qed.} and surjective\footnote{\textit{Proof.} Let $x\in X$.
\par
Let $Y=Gx$ be the $G$-orbit of $x$. Then, $Y\cong G/G_{x}$ (by the
orbit-stabilizer theorem). However, $X$ is a parabolic $G$-set; thus, the
stabilizer of every element of $X$ is parabolic. In other words, for every
$\xi\in X$, the subgroup $G_{\xi}$ of $G$ is parabolic. Applying this to
$\xi=x$, we conclude that the subgroup $G_{x}$ of $G$ is parabolic. In other
words, $G_{x}$ contains a conjugate of $B$ (by the definition of
\textquotedblleft parabolic\textquotedblright). In other words, there exists
some $q\in G$ such that $G_{x}\supseteq qBq^{-1}$. Consider this $q$.
\par
Set $z=q^{-1}x$. Clearly, $z=\underbrace{q^{-1}}_{\in G}x\in Gx=Y$.
\par
Recall that $G_{rx}=rG_{x}r^{-1}$ for each $r\in G$. Applying this to
$r=q^{-1}$, we obtain $G_{q^{-1}x}=q^{-1}\underbrace{G_{x}}_{\supseteq
qBq^{-1}}\underbrace{\left(  q^{-1}\right)  ^{-1}}_{=q}\supseteq
\underbrace{q^{-1}q}_{=1}B\underbrace{q^{-1}q}_{=1}=B$. Since $z=q^{-1}x$, we
obtain $G_{z}=G_{q^{-1}x}\supseteq B$. Hence, $B\subseteq G_{z}$. In other
words, $z\in Y^{B}$ (since $z\in Y$). Thus, $G/P_{I_{z}}$ is a component of
the disjoint union $\coprod_{y\in X^{B}}G/P_{I_{y}}$.
\par
Let $\overline{q}$ denote the coset $qP_{I_{z}}$ of $q$ in $G/P_{I_{z}}$. We
have $\overline{q}\in G/P_{I_{z}}\subseteq\coprod_{y\in X^{B}}G/P_{I_{y}}$
(since $G/P_{I_{z}}$ is a component of the disjoint union $\coprod_{y\in
X^{B}}G/P_{I_{y}}$). Thus, $\overline{q}\in\coprod_{y\in X^{B}}G/P_{I_{y}%
}=FF^{\prime}X$. Therefore, $\varepsilon_{X}\left(  \overline{q}\right)  $ is
well-defined. Moreover, the definition of $\varepsilon_{X}$ shows that
$\varepsilon_{X}\left(  \overline{q}\right)  =qz$ (since $\overline{q}\in
G/P_{I_{z}}$ and since $\overline{q}=\overline{q}$). Thus, $\varepsilon
_{X}\left(  \overline{q}\right)  =q\underbrace{z}_{=q^{-1}x}%
=\underbrace{qq^{-1}}_{=1}x=x$, so taht $x=\varepsilon_{X}\left(
\underbrace{\overline{q}}_{\in F^{\prime}FX}\right)  \in\varepsilon_{X}\left(
F^{\prime}FX\right)  $.
\par
Now, forget that we fixed $x$. We thus have shown that $x\in\varepsilon
_{X}\left(  F^{\prime}FX\right)  $ for each $x\in X$. In other words,
$X\subseteq\varepsilon_{X}\left(  F^{\prime}FX\right)  $. In other words, the
map $\varepsilon_{X}$ is surjective. Qed.}. Hence, the map $\varepsilon_{X}$
is bijective, and thus is a $G$-set isomorphism (since it is $G$-equivariant).

Now, forget that we fixed $X$. Thus, for each object $X\in\mathcal{P}$, we
have constructed a $G$-set isomorphism $\varepsilon_{X}:F^{\prime
}FX\rightarrow X$. Moreover, this isomorphism $\varepsilon_{X}$ is functorial
in $X$\ \ \ \ \footnote{\textit{Proof.} Let $Y$ and $Z$ be two objects of
$\mathcal{P}$, and let $f:Y\rightarrow Z$ be a $G$-equivariant map. We must
prove that the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%F^\prime F Y \ar[r]^-{\varepsilon_Y} \ar[d]_{F^\prime F f} & Y \ar[d]^{f} \\
%F^\prime F Z \ar[r]_-{\varepsilon_Z} & Z
%}} }%
%BeginExpansion
\xymatrix{
F^\prime F Y \ar[r]^-{\varepsilon_Y} \ar[d]_{F^\prime F f} & Y \ar[d]^{f} \\
F^\prime F Z \ar[r]_-{\varepsilon_Z} & Z
}
%EndExpansion
\label{pf.p7.9.epsilon-is-functorial.diagram}%
\end{equation}
is commutative.
\par
Let $p\in F^{\prime}FY$. Then, $FY=\left(  Y^{B};Y^{P_{1}},\ldots,Y^{P_{n-1}%
}\right)  $ (by the definition of the functor $F$). Hence, the definition of
the functor $F^{\prime}$ shows that $F^{\prime}FY=\coprod_{y\in Y^{B}%
}G/P_{I_{y}}$, where we set $I_{y}=\left\{  i\in\left\{  1,2,\ldots
,n-1\right\}  \ \mid\ y\in Y^{P_{i}}\right\}  $ for each $y\in Y^{B}$.
\par
We have $p\in F^{\prime}FY=\coprod_{y\in Y^{B}}G/P_{I_{y}}$. In other words,
$p\in G/P_{I_{y}}$ for some $y\in Y^{B}$. Consider this $y$. Write $p$ in the
form $p=\overline{q}$ for some $q\in G$ (where $\overline{q}$ denotes the
coset $qP_{I_{y}}$ of $q$ in $G/P_{I_{y}}$). Then, $\varepsilon_{Y}\left(
p\right)  =qy$ (by the definition of $\varepsilon_{Y}$).
\par
The definition of the action of the functor $F$ on the morphism
$f:Y\rightarrow Z$ shows that $Ff:Y^{B}\rightarrow Z^{B}$ is the restriction
of the map $f:Y\rightarrow Z$ to the $B$-fixed points. Thus, $\left(
Ff\right)  \left(  y\right)  =f\left(  y\right)  $.
\par
On the other hand, the definition of the action of the functor $F^{\prime}$ on
the morphism $Ff:FY\rightarrow FZ$ yields $\left(  F^{\prime}Ff\right)
\left(  \overline{q}\right)  =\overline{q}\in G/P_{I_{\left(  Ff\right)
\left(  y\right)  }}=G/P_{I_{f\left(  y\right)  }}$ (since $\left(  Ff\right)
\left(  y\right)  =f\left(  y\right)  $). Hence, the definition of
$\varepsilon_{Z}$ yields $\varepsilon_{Z}\left(  \left(  F^{\prime}Ff\right)
\left(  \overline{q}\right)  \right)  =qf\left(  y\right)  $. Hence,%
\[
\left(  \varepsilon_{Z}\circ\left(  F^{\prime}Ff\right)  \right)  \left(
\underbrace{p}_{=\overline{q}}\right)  =\left(  \varepsilon_{Z}\circ\left(
F^{\prime}Ff\right)  \right)  \left(  \overline{q}\right)  =\varepsilon
_{Z}\left(  \left(  F^{\prime}Ff\right)  \left(  \overline{q}\right)  \right)
=qf\left(  y\right)  .
\]
Comparing this with%
\[
\left(  f\circ\varepsilon_{Y}\right)  \left(  p\right)  =f\left(
\underbrace{\varepsilon_{Y}\left(  p\right)  }_{=qy}\right)  =f\left(
qy\right)  =qf\left(  y\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the
map }f\text{ is }G\text{-equivariant}\right)  ,
\]
we obtain $\left(  \varepsilon_{Z}\circ\left(  F^{\prime}Ff\right)  \right)
\left(  p\right)  =\left(  f\circ\varepsilon_{Y}\right)  \left(  p\right)  $.
\par
Now, let us forget that we fixed $p$. We thus have proven that $\left(
\varepsilon_{Z}\circ\left(  F^{\prime}Ff\right)  \right)  \left(  p\right)
=\left(  f\circ\varepsilon_{Y}\right)  \left(  p\right)  $ for each $p\in
F^{\prime}FY$. In other words, $\varepsilon_{Z}\circ\left(  F^{\prime
}Ff\right)  =f\circ\varepsilon_{Y}$. In other words, the diagram
(\ref{pf.p7.9.epsilon-is-functorial.diagram}) is commutative. This completes
the proof.}. Hence, we have defined a natural isomorphism $\varepsilon
:F^{\prime}F\rightarrow1_{\mathcal{P}}$. Therefore, $1_{\mathcal{P}}\cong
F^{\prime}F$ as functors.]

\item \textbf{\S 8:} I would begin this section with the following introduction:

\textquotedblleft We consider the localization $\mathbb{Z}_{\left(  p\right)
}$ of the ring $\mathbb{Z}$ at its prime ideal $\left(  p\right)
=p\mathbb{Z}$. Explicitly, $\mathbb{Z}_{\left(  p\right)  }$ is the subring%
\[
\left\{  \dfrac{a}{b}\ \mid\ \left(  a,b\right)  \in\mathbb{Z}\times
\mathbb{Z}\text{ and }\gcd\left(  b,p\right)  =1\right\}
\]
of $\mathbb{Q}$.

\textbf{Lemma 8.0a.} Let $V$ be an $n$-dimensional $\mathbb{F}_{p}$-vector
space. Then:

\textbf{(a)} We have $\left\vert \operatorname*{Flag}\left(  V\right)
\right\vert =\sum_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma\right)  }$.

\textbf{(b)} We have $\left\vert \operatorname*{Flag}\left(  V\right)
\right\vert \equiv1\operatorname{mod}p$ and $\left\vert \operatorname*{Flag}%
\left(  V\right)  \right\vert ^{-1}\in\mathbb{Z}_{\left(  p\right)  }$.

[\textit{Proof of Lemma 8.0a.} Fix some $\underline{W}\in\operatorname*{Flag}%
\left(  V\right)  $. (Such a $\underline{W}$ clearly exists.) For each
$\sigma\in\Sigma_{n}$, there is a subset $Y\left(  \sigma,\underline{W}%
\right)  \subset\operatorname*{Flag}\left(  V\right)  $ defined by $Y\left(
\sigma,\underline{W}\right)  =\left\{  \underline{U}\in\operatorname*{Flag}%
\left(  V\right)  \ \mid\ \delta\left(  \underline{U},\underline{W}\right)
=\sigma\right\}  $. (Here, $\delta\left(  \underline{U},\underline{W}\right)
$ is the Jordan permutation, defined as in \S 4.)

Clearly, $\operatorname*{Flag}\left(  V\right)  $ is the union of its disjoint
subsets $\left\{  \underline{U}\in\operatorname*{Flag}\left(  V\right)
\ \mid\ \delta\left(  \underline{U},\underline{W}\right)  =\sigma\right\}  $
for all $\sigma\in\Sigma_{n}$ (because for each $\underline{U}\in
\operatorname*{Flag}\left(  V\right)  $, there is exactly one $\sigma\in
\Sigma_{n}$ satisfying $\delta\left(  \underline{U},\underline{W}\right)
=\sigma$). Hence,%
\begin{align*}
\left\vert \operatorname*{Flag}\left(  V\right)  \right\vert  &  =\sum
_{\sigma\in\Sigma_{n}}\left\vert \underbrace{\left\{  \underline{U}%
\in\operatorname*{Flag}\left(  V\right)  \ \mid\ \delta\left(  \underline{U}%
,\underline{W}\right)  =\sigma\right\}  }_{=Y\left(  \sigma,\underline{W}%
\right)  }\right\vert =\sum_{\sigma\in\Sigma_{n}}\underbrace{\left\vert
Y\left(  \sigma,\underline{W}\right)  \right\vert }_{\substack{=p^{l\left(
\sigma\right)  }\\\text{(by Corollary 5.2a, applied}\\\text{to }V\text{ and
}\underline{W}\text{ instead of }W\text{ and }\underline{V}\text{)}}}\\
&  =\sum_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma\right)  }.
\end{align*}
This proves Lemma 8.0a \textbf{(a)}.

\textbf{(b)} Lemma 8.0a \textbf{(a)} yields%
\begin{align*}
\left\vert \operatorname*{Flag}\left(  V\right)  \right\vert  &  =\sum
_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma\right)  }=\underbrace{p^{l\left(
\operatorname*{id}\right)  }}_{=p^{0}=1}+\sum_{\substack{\sigma\in\Sigma
_{n};\\\sigma\neq\operatorname*{id}}}\underbrace{p^{l\left(  \sigma\right)  }%
}_{\substack{\equiv0\operatorname{mod}p\\\text{(since }l\left(  \sigma\right)
\geq1\\\text{(since }\sigma\neq\operatorname*{id}\text{))}}}\\
&  \equiv1+\underbrace{\sum_{\substack{\sigma\in\Sigma_{n};\\\sigma
\neq\operatorname*{id}}}0}_{=0}=1\operatorname{mod}p.
\end{align*}
Hence, $\left\vert \operatorname*{Flag}\left(  V\right)  \right\vert $ is
coprime to $p$. Thus, $\left\vert \operatorname*{Flag}\left(  V\right)
\right\vert ^{-1}\in\mathbb{Z}_{\left(  p\right)  }$. This proves Lemma 8.0a
\textbf{(b)}. $\square$ ]\textquotedblright

You use Lemma 8.0a \textbf{(b)} implicitly in Definition 8.5.

\item \textbf{Definition 8.1:} Replace \textquotedblleft
ring\textquotedblright\ by \textquotedblleft$\mathbb{Z}_{\left(  p\right)  }%
$-algebra\textquotedblright.

\item \textbf{Definition 8.1:} Replace \textquotedblleft$\mathbb{Z}\left[
\operatorname*{Flag}\right]  $\textquotedblright\ by \textquotedblleft%
$\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]
$\textquotedblright.

\item \textbf{\S 8, between Definition 8.1 and Definition 8.2:} You write:
\textquotedblleft this construction gives an equivalence $\left[
\mathcal{V},\left\{  \operatorname*{sets}\right\}  \right]  =\left\{
G-\text{sets}\right\}  $\textquotedblright.

This argument is non-constructive\footnote{Namely, it seems to use
\par
\begin{itemize}
\item either the fact that every category is equivalent to its skeleton,
\par
\item or the fact that any functor that is essentially surjective, full and
faithful must be an equivalence of categories.
\end{itemize}
\par
As far as I know, none of these two facts has a constructive proof.} and (in
my opinion) overkill. It appears to me that you are not actually using the
full power of this equivalence either; instead, you seem to only use the
natural $\mathbb{Z}_{\left(  p\right)  }$-algebra isomorphism%
\begin{align*}
\mathcal{H}  &  =\operatorname*{End}\nolimits_{\mathcal{VA}}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]  \right)
\cong\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  }\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right]  \right) \\
&  \cong\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  }\left(  \mathbb{Z}_{\left(  p\right)  }\left[  G/B\right]  \right)
,
\end{align*}
which has an elementary and constructive proof. Namely, this isomorphism
follows from Proposition 8.1b further below.

Before I state this proposition, let me state a simple fact from category theory:

\textbf{Proposition 8.1a.} Let $\mathcal{C}$ and $\mathcal{D}$ be two
categories. Let $C\in\mathcal{C}$ be an object. Let $F:\mathcal{C}%
\rightarrow\mathcal{D}$ is a functor. Let $\mathcal{E}_{F}\left(  C\right)  $
denote the subset%
\[
\left\{  f\in\operatorname*{End}\nolimits_{\mathcal{D}}\left(  F\left(
C\right)  \right)  \ \mid\ f\circ F\left(  k\right)  =F\left(  k\right)  \circ
f\text{ for each }k\in\operatorname*{End}\nolimits_{\mathcal{C}}C\right\}
\]
of $\operatorname*{End}\nolimits_{\mathcal{D}}\left(  F\left(  C\right)
\right)  $.

\textbf{(a)} The subset $\mathcal{E}_{F}\left(  C\right)  $ is a submonoid of
$\operatorname*{End}\nolimits_{\mathcal{D}}\left(  F\left(  C\right)  \right)
$. Moreover, there is a canonical monoid homomorphism $\varepsilon
:\operatorname*{End}\nolimits_{\left[  \mathcal{C},\mathcal{D}\right]
}F\rightarrow\mathcal{E}_{F}\left(  C\right)  $ that sends each natural
transformation $\alpha:F\Longrightarrow F$ to its component $\alpha
_{C}:F\left(  C\right)  \rightarrow F\left(  C\right)  $.

\textbf{(b)} Consider this $\varepsilon$. Assume that each two objects of
$\mathcal{C}$ are isomorphic. Then, $\varepsilon$ is a monoid isomorphism.

[\textit{Proof of Proposition 8.1a.} \textbf{(a)} This is a simple exercise in
category theory.

\textbf{(b)} The map $\varepsilon$ is injective\footnote{\textit{Proof.} Let
$\alpha$ and $\beta$ be two elements of $\operatorname*{End}\nolimits_{\left[
\mathcal{C},\mathcal{D}\right]  }F$ such that $\varepsilon\left(
\alpha\right)  =\varepsilon\left(  \beta\right)  $. We shall show that
$\alpha=\beta$.
\par
Let $A\in\mathcal{C}$ be any object. Then, the objects $A$ and $C$ of
$\mathcal{C}$ are isomorphic (since each two objects of $\mathcal{C}$ are
isomorphic). In other words, there exists an isomorphism $j:A\rightarrow C$ in
$\mathcal{C}$. Consider this $j$. Thus, the morphism $j^{-1}$ exists (since
$j$ is an isomorphism).
\par
Recall that $\alpha\in\operatorname*{End}\nolimits_{\left[  \mathcal{C}%
,\mathcal{D}\right]  }F$. In other words, $\alpha$ is a natural transformation
from $F$ to $F$. Thus, the diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%F\left(A\right) \ar[r]^{\alpha_A} \ar[d]_{F\left(j\right)} & F\left
%(A\right) \ar[d]^{F\left(j\right)} \\
%F\left(C\right) \ar[r]_{\alpha_C} & F\left(C\right)
%}}}%
%BeginExpansion
\xymatrix{
F\left(A\right) \ar[r]^{\alpha_A} \ar[d]_{F\left(j\right)} & F\left
(A\right) \ar[d]^{F\left(j\right)} \\
F\left(C\right) \ar[r]_{\alpha_C} & F\left(C\right)
}%
%EndExpansion
\]
is commutative. In other words, we have $\alpha_{A}\circ F\left(  j\right)
=F\left(  j\right)  \circ\alpha_{C}$. But $F$ is a functor; thus, $F\left(
j^{-1}\right)  =\left(  F\left(  j\right)  \right)  ^{-1}$. The definition of
$\varepsilon$ yields $\varepsilon\left(  \alpha\right)  =\alpha_{C}$. Thus,%
\[
F\left(  j\right)  \circ\underbrace{\varepsilon\left(  \alpha\right)
}_{=\alpha_{C}}\circ\underbrace{F\left(  j^{-1}\right)  }_{=\left(  F\left(
j\right)  \right)  ^{-1}}=\underbrace{F\left(  j\right)  \circ\alpha_{C}%
}_{=\alpha_{A}\circ F\left(  j\right)  }\circ\left(  F\left(  j\right)
\right)  ^{-1}=\alpha_{A}\circ\underbrace{F\left(  j\right)  \circ\left(
F\left(  j\right)  \right)  ^{-1}}_{=\operatorname*{id}}=\alpha_{A}.
\]
Hence, $\alpha_{A}=F\left(  j\right)  \circ\varepsilon\left(  \alpha\right)
\circ F\left(  j^{-1}\right)  $. The same argument (applied to $\beta$ instead
of $\alpha$) shows that $\beta_{A}=F\left(  j\right)  \circ\varepsilon\left(
\beta\right)  \circ F\left(  j^{-1}\right)  $.
\par
Now,%
\[
\alpha_{A}=F\left(  j\right)  \circ\underbrace{\varepsilon\left(
\alpha\right)  }_{=\varepsilon\left(  \beta\right)  }\circ F\left(
j^{-1}\right)  =F\left(  j\right)  \circ\varepsilon\left(  \beta\right)  \circ
F\left(  j^{-1}\right)  =\beta_{A}.
\]
\par
Now, forget that we fixed $A$. We thus have proven that $\alpha_{A}=\beta_{A}$
for each object $A\in\mathcal{C}$. In other words, $\alpha=\beta$.
\par
Now, forget that we fixed $\alpha$ and $\beta$. We thus have shown that if
$\alpha$ and $\beta$ are two elements of $\operatorname*{End}%
\nolimits_{\left[  \mathcal{C},\mathcal{D}\right]  }F$ such that
$\varepsilon\left(  \alpha\right)  =\varepsilon\left(  \beta\right)  $, then
$\alpha=\beta$. In other words, the map $\varepsilon$ is injective. Qed.}. We
shall now show that $\varepsilon$ is surjective.

Indeed, fix any $\rho\in\mathcal{E}_{F}\left(  C\right)  $. Let $A\in
\mathcal{C}$ be any object. We are going to construct a morphism $\alpha
_{A}:F\left(  C\right)  \rightarrow F\left(  C\right)  $ in $\mathcal{D}$.

We have
\[
\rho\in\mathcal{E}_{F}\left(  C\right)  =\left\{  f\in\operatorname*{End}%
\nolimits_{\mathcal{D}}\left(  F\left(  C\right)  \right)  \ \mid\ f\circ
F\left(  k\right)  =F\left(  k\right)  \circ f\text{ for each }k\in
\operatorname*{End}\nolimits_{\mathcal{C}}C\right\}
\]
(by the definition of $\mathcal{E}_{F}\left(  C\right)  $). In other words,
$\rho$ is an element of $\operatorname*{End}\nolimits_{\mathcal{D}}\left(
F\left(  C\right)  \right)  $ and satisfies%
\begin{equation}
\left(  \rho\circ F\left(  k\right)  =F\left(  k\right)  \circ\rho\text{ for
each }k\in\operatorname*{End}\nolimits_{\mathcal{C}}C\right)  .
\label{pf.p8.1a.b.rhoF}%
\end{equation}


The objects $A$ and $C$ of $\mathcal{C}$ are isomorphic (since each two
objects of $\mathcal{C}$ are isomorphic). In other words, there exists an
isomorphism $j:C\rightarrow A$ in $\mathcal{C}$. Consider this $j$. Thus, the
morphism $j^{-1}$ exists (since $j$ is an isomorphism).

Now, define a morphism $\alpha_{A}:F\left(  C\right)  \rightarrow F\left(
C\right)  $ in $\mathcal{D}$ by $\alpha_{A}=F\left(  j\right)  \circ\rho\circ
F\left(  j^{-1}\right)  $. This morphism $\alpha_{A}$ is independent on the
choice of $j$\ \ \ \ \footnote{\textit{Proof.} Let $j_{1}$ and $j_{2}$ be two
isomorphisms $j:C\rightarrow A$ in $\mathcal{C}$. We will prove that $F\left(
j_{1}\right)  \circ\rho\circ F\left(  j_{1}^{-1}\right)  =F\left(
j_{2}\right)  \circ\rho\circ F\left(  j_{2}^{-1}\right)  $.
\par
We recall that $j_{1}$ and $j_{2}$ are two isomorphisms $C\rightarrow A$ in
$\mathcal{C}$. Thus, $j_{1}^{-1}\circ j_{2}:C\rightarrow C$ is an isomorphism
in $\mathcal{C}$ as well. In particular, $j_{1}^{-1}\circ j_{2}\in
\operatorname*{End}\nolimits_{\mathcal{C}}C$. Thus, (\ref{pf.p8.1a.b.rhoF})
(applied to $k=j_{1}^{-1}\circ j_{2}$) yields $\rho\circ F\left(  j_{1}%
^{-1}\circ j_{2}\right)  =F\left(  j_{1}^{-1}\circ j_{2}\right)  \circ\rho$.
But $F$ is a functor; thus, $F\left(  j_{1}^{-1}\circ j_{2}\right)  =\left(
F\left(  j_{1}\right)  \right)  ^{-1}\circ F\left(  j_{2}\right)  $. But%
\begin{align*}
&  F\left(  j_{1}\right)  \circ\rho\circ\underbrace{F\left(  j_{1}%
^{-1}\right)  }_{\substack{=\left(  F\left(  j_{1}\right)  \right)
^{-1}\\\text{(since }F\text{ is a functor)}}}\\
&  =F\left(  j_{1}\right)  \circ\rho\circ\left(  F\left(  j_{1}\right)
\right)  ^{-1}\\
&  =F\left(  j_{1}\right)  \circ\rho\circ\underbrace{\left(  F\left(
j_{1}\right)  \right)  ^{-1}\circ F\left(  j_{2}\right)  }_{=F\left(
j_{1}^{-1}\circ j_{2}\right)  }\circ\underbrace{\left(  F\left(  j_{2}\right)
\right)  ^{-1}}_{\substack{=F\left(  j_{2}^{-1}\right)  \\\text{(since
}F\text{ is a functor)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\left(  j_{1}\right)  \circ
\rho\circ\left(  F\left(  j_{1}\right)  \right)  ^{-1}\circ
\underbrace{F\left(  j_{2}\right)  \circ\left(  F\left(  j_{2}\right)
\right)  ^{-1}}_{=\operatorname*{id}}=F\left(  j_{1}\right)  \circ\rho
\circ\left(  F\left(  j_{1}\right)  \right)  ^{-1}\right) \\
&  =F\left(  j_{1}\right)  \circ\underbrace{\rho\circ F\left(  j_{1}^{-1}\circ
j_{2}\right)  }_{=F\left(  j_{1}^{-1}\circ j_{2}\right)  \circ\rho}\circ
F\left(  j_{2}^{-1}\right) \\
&  =\underbrace{F\left(  j_{1}\right)  \circ F\left(  j_{1}^{-1}\circ
j_{2}\right)  }_{\substack{=F\left(  j_{1}\circ j_{1}^{-1}\circ j_{2}\right)
\\\text{(since }F\text{ is a functor)}}}\circ\rho\circ F\left(  j_{2}%
^{-1}\right) \\
&  =F\left(  \underbrace{j_{1}\circ j_{1}^{-1}}_{=\operatorname*{id}}\circ
j_{2}\right)  \circ\rho\circ F\left(  j_{2}^{-1}\right)  =F\left(
j_{2}\right)  \circ\rho\circ F\left(  j_{2}^{-1}\right)  .
\end{align*}
\par
Now, forget that we fixed $j_{1}$ and $j_{2}$. We thus have shown that if
$j_{1}$ and $j_{2}$ are two isomorphisms $j:C\rightarrow A$ in $\mathcal{C}$,
then $F\left(  j_{1}\right)  \circ\rho\circ F\left(  j_{1}^{-1}\right)
=F\left(  j_{2}\right)  \circ\rho\circ F\left(  j_{2}^{-1}\right)  $. In other
words, the morphism $F\left(  j\right)  \circ\rho\circ F\left(  j^{-1}\right)
$ is independent on the choice on $j$. In other words, the morphism
$\alpha_{A}$ is independent on the choice of $j$ (since $\alpha_{A}=F\left(
j\right)  \circ\rho\circ F\left(  j^{-1}\right)  $). Qed.}.

Now, forget that we fixed $A$. Thus, for each $A\in\mathcal{C}$, we have
defined a morphism $\alpha_{A}:F\left(  C\right)  \rightarrow F\left(
C\right)  $ in $\mathcal{D}$. This morphism $\alpha_{A}$ satisfies%
\begin{equation}
\alpha_{A}=F\left(  j\right)  \circ\rho\circ F\left(  j^{-1}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every isomorphism }j:C\rightarrow A\text{ in
}\mathcal{C} \label{pf.p8.1a.b.alphaA}%
\end{equation}
(by the definition of $\alpha_{A}$).

If $A$ and $B$ are two objects in $\mathcal{C}$, and if $f:A\rightarrow B$ is
a morphism in $\mathcal{C}$, then the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%F\left(A\right) \ar[r]^{\alpha_A} \ar[d]_{F\left(f\right)} & F\left
%(A\right) \ar[d]^{F\left(f\right)} \\
%F\left(B\right) \ar[r]_{\alpha_B} & F\left(B\right)
%}} }%
%BeginExpansion
\xymatrix{
F\left(A\right) \ar[r]^{\alpha_A} \ar[d]_{F\left(f\right)} & F\left
(A\right) \ar[d]^{F\left(f\right)} \\
F\left(B\right) \ar[r]_{\alpha_B} & F\left(B\right)
}
%EndExpansion
\label{pf.p8.1a.b.naturality}%
\end{equation}
is commutative\footnote{\textit{Proof.} Let $A$ and $B$ be two objects in
$\mathcal{C}$. Let $f:A\rightarrow B$ be a morphism in $\mathcal{C}$.
\par
The objects $A$ and $C$ of $\mathcal{C}$ are isomorphic (since each two
objects of $\mathcal{C}$ are isomorphic). In other words, there exists an
isomorphism $j:C\rightarrow A$ in $\mathcal{C}$. Consider this $j$. Thus,
(\ref{pf.p8.1a.b.alphaA}) yields $\alpha_{A}=F\left(  j\right)  \circ\rho\circ
F\left(  j^{-1}\right)  $.
\par
The objects $B$ and $C$ of $\mathcal{C}$ are isomorphic (since each two
objects of $\mathcal{C}$ are isomorphic). In other words, there exists an
isomorphism $i:C\rightarrow B$ in $\mathcal{C}$. Consider this $i$. Thus,
(\ref{pf.p8.1a.b.alphaA}) (applied to $B$ and $i$ instead of $A$ and $j$)
yields $\alpha_{B}=F\left(  i\right)  \circ\rho\circ F\left(  i^{-1}\right)
$.
\par
Since $i:C\rightarrow B$ is an isomorphism, its inverse $i^{-1}:B\rightarrow
C$ is well-defined and an isomorphism as well. The composition $i^{-1}\circ
f\circ j:C\rightarrow C$ is thus an endomorphism of $C$ in $\mathcal{C}$. In
other words, $i^{-1}\circ f\circ j\in\operatorname*{End}\nolimits_{\mathcal{C}%
}C$. Hence, (\ref{pf.p8.1a.b.rhoF}) (applied to $k=i^{-1}\circ f\circ j$)
yields $\rho\circ F\left(  i^{-1}\circ f\circ j\right)  =F\left(  i^{-1}\circ
f\circ j\right)  \circ\rho$. But $F$ is a functor; thus, $F\left(  i^{-1}\circ
f\circ j\right)  =\left(  F\left(  i\right)  \right)  ^{-1}\circ F\left(
f\right)  \circ F\left(  j\right)  $. Now,%
\begin{align*}
\underbrace{\alpha_{B}}_{=F\left(  i\right)  \circ\rho\circ F\left(
i^{-1}\right)  }\circ F\left(  f\right)  \circ F\left(  j\right)   &
=F\left(  i\right)  \circ\rho\circ\underbrace{F\left(  i^{-1}\right)  \circ
F\left(  f\right)  \circ F\left(  j\right)  }_{=F\left(  i^{-1}\circ f\circ
j\right)  }=F\left(  i\right)  \circ\underbrace{\rho\circ F\left(  i^{-1}\circ
f\circ j\right)  }_{=F\left(  i^{-1}\circ f\circ j\right)  \circ\rho}\\
&  =F\left(  i\right)  \circ\underbrace{F\left(  i^{-1}\circ f\circ j\right)
}_{=\left(  F\left(  i\right)  \right)  ^{-1}\circ F\left(  f\right)  \circ
F\left(  j\right)  }\circ\rho=\underbrace{F\left(  i\right)  \circ\left(
F\left(  i\right)  \right)  ^{-1}}_{=\operatorname*{id}}\circ F\left(
f\right)  \circ F\left(  j\right)  \circ\rho\\
&  =F\left(  f\right)  \circ F\left(  j\right)  \circ\rho,
\end{align*}
so that%
\[
\underbrace{\alpha_{B}\circ F\left(  f\right)  \circ F\left(  j\right)
}_{=F\left(  f\right)  \circ F\left(  j\right)  \circ\rho}\circ F\left(
j^{-1}\right)  =F\left(  f\right)  \circ\underbrace{F\left(  j\right)
\circ\rho\circ F\left(  j^{-1}\right)  }_{=\alpha_{A}}=F\left(  f\right)
\circ\alpha_{A}.
\]
Comparing this with%
\[
\alpha_{B}\circ F\left(  f\right)  \circ F\left(  j\right)  \circ
\underbrace{F\left(  j^{-1}\right)  }_{\substack{=\left(  F\left(  j\right)
\right)  ^{-1}\\\text{(since }F\text{ is a functor)}}}=\alpha_{B}\circ
F\left(  f\right)  \circ\underbrace{F\left(  j\right)  \circ\left(  F\left(
j\right)  \right)  ^{-1}}_{=\operatorname*{id}}=\alpha_{B}\circ F\left(
f\right)  ,
\]
we obtain $F\left(  f\right)  \circ\alpha_{A}=\alpha_{B}\circ F\left(
f\right)  $. In other words, the diagram (\ref{pf.p8.1a.b.naturality}) is
commutative. Qed.}. Therefore, the morphisms $\alpha_{A}$ (defined for all
objects $A\in\mathcal{C}$) can be assembled to a natural transformation
$\alpha:F\Longrightarrow F$. Consider this $\alpha$. We have $\alpha
\in\operatorname*{End}\nolimits_{\left[  \mathcal{C},\mathcal{D}\right]  }F$
(since $\alpha$ is a natural transformation $F\Longrightarrow F$). Moreover,
the definition of $\varepsilon$ shows that $\varepsilon\left(  \alpha\right)
=\alpha_{C}$.

But $\operatorname*{id}:C\rightarrow C$ is an isomorphism in $\mathcal{C}$.
Hence, (\ref{pf.p8.1a.b.alphaA}) (applied to $A=C$ and $j=\operatorname*{id}$)
yields%
\[
\alpha_{C}=F\left(  \operatorname*{id}\right)  \circ\rho\circ F\left(
\underbrace{\operatorname*{id}\nolimits^{-1}}_{=\operatorname*{id}}\right)
=\underbrace{F\left(  \operatorname*{id}\right)  }%
_{\substack{=\operatorname*{id}\\\text{(since }F\text{ is a functor)}}%
}\circ\rho\circ\underbrace{F\left(  \operatorname*{id}\right)  }%
_{\substack{=\operatorname*{id}\\\text{(since }F\text{ is a functor)}}}=\rho.
\]
Comparing this with $\varepsilon\left(  \alpha\right)  =\alpha_{C}$, we obtain
$\rho=\varepsilon\left(  \underbrace{\alpha}_{\in\operatorname*{End}%
\nolimits_{\left[  \mathcal{C},\mathcal{D}\right]  }F}\right)  \in
\varepsilon\left(  \operatorname*{End}\nolimits_{\left[  \mathcal{C}%
,\mathcal{D}\right]  }F\right)  $.

Now, forget that we fixed $\rho$. We thus have shown that $\rho\in
\varepsilon\left(  \operatorname*{End}\nolimits_{\left[  \mathcal{C}%
,\mathcal{D}\right]  }F\right)  $ for each $\rho\in\mathcal{E}_{F}\left(
C\right)  $. In other words, $\mathcal{E}_{F}\left(  C\right)  \subseteq
\varepsilon\left(  \operatorname*{End}\nolimits_{\left[  \mathcal{C}%
,\mathcal{D}\right]  }F\right)  $. In other words, the map $\varepsilon$ is surjective.

So we know that the map $\varepsilon$ is both injective and surjective. Thus,
$\varepsilon$ is bijective. Furthermore, $\varepsilon$ is a monoid
homomorphism. Thus, $\varepsilon$ is a bijective monoid homomorphism.
Therefore, $\varepsilon$ is a monoid isomorphism. This proves Proposition 8.1a
\textbf{(b)}. $\square$ ]

\textbf{Proposition 8.1b.} We have%
\begin{align*}
\mathcal{H}  &  =\operatorname*{End}\nolimits_{\mathcal{VA}}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]  \right)
\cong\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  }\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right]  \right) \\
&  \cong\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  }\left(  \mathbb{Z}_{\left(  p\right)  }\left[  G/B\right]  \right)
\end{align*}
as $\mathbb{Z}_{\left(  p\right)  }$-algebras. More precisely, the following holds:

\textbf{(a)} Recall the definition of $\mathcal{E}_{F}\left(  C\right)  $ in
Proposition 8.1a (where $\mathcal{C}$ and $\mathcal{D}$ are two categories,
$C\in\mathcal{C}$ is an object, and $F:\mathcal{C}\rightarrow\mathcal{D}$ is a
functor). Applying this to $\mathcal{C}=\mathcal{V}$, $\mathcal{D}%
=\mathcal{A}$, $C=\mathbb{F}_{p}^{n}$ and $F=\mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Flag}\right]  $, we obtain a set $\mathcal{E}%
_{\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]  }\left(
\mathbb{F}_{p}^{n}\right)  $. Proposition 8.1a \textbf{(a)} (applied to
$\mathcal{C}=\mathcal{V}$, $\mathcal{D}=\mathcal{A}$, $C=\mathbb{F}_{p}^{n}$
and $F=\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]  $)
shows that this set $\mathcal{E}_{\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  }\left(  \mathbb{F}_{p}^{n}\right)  $ is a
submonoid of $\operatorname*{End}\nolimits_{\mathcal{A}}\left(  \left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]  \right)
\left(  \mathbb{F}_{p}^{n}\right)  \right)  $, and that there is a monoid
homomorphism $\varepsilon:\operatorname*{End}\nolimits_{\left[  \mathcal{V}%
,\mathcal{A}\right]  }\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  \right)  \rightarrow\mathcal{E}_{\mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]  }\left(
\mathbb{F}_{p}^{n}\right)  $. Consider this $\varepsilon$. Then, $\varepsilon$
is a $\mathbb{Z}_{\left(  p\right)  }$-algebra isomorphism
$\operatorname*{End}\nolimits_{\mathcal{VA}}\left(  \mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Flag}\right]  \right)  \rightarrow
\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]
}\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  $. Thus,%
\[
\operatorname*{End}\nolimits_{\mathcal{VA}}\left(  \mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Flag}\right]  \right)  \cong%
\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]
}\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)
\]
as $\mathbb{Z}_{\left(  p\right)  }$-algebras.

\textbf{(b)} Consider the complete flag $\underline{E}=\left(  E_{0}%
<E_{1}<\cdots<E_{n}=\mathbb{F}_{p}^{n}\right)  \in\operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  $ defined in \S 4.

There is a natural isomorphism $G/B\rightarrow\operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  $ of $G$-sets, which sends each coset $hB\in G/B$
of $B$ to the complete flag $h\underline{E}\in\operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  $. The inverse of this isomorphism is an
isomorphism $\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)
\rightarrow G/B$ of $G$-sets. This isomorphism gives rise to a $\mathbb{Z}%
_{\left(  p\right)  }$-module isomorphism $\mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right]
\rightarrow\mathbb{Z}_{\left(  p\right)  }\left[  G/B\right]  $ and thus to a
$\mathbb{Z}_{\left(  p\right)  }$-algebra isomorphism $\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  }\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  \rightarrow\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  }\left(
\mathbb{Z}_{\left(  p\right)  }\left[  G/B\right]  \right)  $. Thus,%
\[
\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]
}\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  \cong\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  }\left(
\mathbb{Z}_{\left(  p\right)  }\left[  G/B\right]  \right)
\]
as $\mathbb{Z}_{\left(  p\right)  }$-algebras.

[\textit{Proof of Proposition 8.1b.} \textbf{(a)} Recall that $\mathcal{V}$ is
the category whose objects are $n$-dimensional vector spaces over
$\mathbb{F}_{p}$, and whose morphisms are the isomorphisms between these
vector spaces. In particular, the endomorphisms of $\mathbb{F}_{p}^{n}$ in
$\mathcal{V}$ are the vector space isomorphisms $\mathbb{F}_{p}^{n}%
\rightarrow\mathbb{F}_{p}^{n}$. In other words,%
\begin{align*}
\operatorname*{End}\nolimits_{\mathcal{V}}\left(  \mathbb{F}_{p}^{n}\right)
&  =\left(  \text{the set of the vector space isomorphisms }\mathbb{F}_{p}%
^{n}\rightarrow\mathbb{F}_{p}^{n}\right) \\
&  =\left(  \text{the set of the vector space automorphisms of }\mathbb{F}%
_{p}^{n}\right) \\
&  =\operatorname*{GL}\nolimits_{n}\left(  \mathbb{F}_{p}\right)  =G.
\end{align*}


Let $F$ be the functor $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  :\mathcal{V}\rightarrow\mathcal{A}$. Hence,
$F\left(  \mathbb{F}_{p}^{n}\right)  =\left(  \mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Flag}\right]  \right)  \left(  \mathbb{F}_{p}%
^{n}\right)  =\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right]  $. This $\mathbb{Z}_{\left(
p\right)  }$-module $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right]  $ is a
$\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  $-module, because the set
$\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $ is a $G$-set. The
action of $G$ on $\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right]  $ has the property that%
\begin{equation}
F\left(  k\right)  =\left(  \text{the action of }k\text{ on }\mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(  \mathbb{F}_{p}%
^{n}\right)  \right]  \right)  \label{pf.p8.1b.a.action}%
\end{equation}
for each $k\in G$\ \ \ \ \footnote{\textit{Proof of (\ref{pf.p8.1b.a.action}%
):} Let $k\in G$. Then, $k\in G=\operatorname*{End}\nolimits_{\mathcal{V}%
}\left(  \mathbb{F}_{p}^{n}\right)  $. Hence, $F\left(  k\right)  $ is a
well-defined endomorphism of $F\left(  \mathbb{F}_{p}^{n}\right)
=\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  $. Moreover,
\[
\underbrace{F}_{=\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\right]  }\left(  k\right)  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  \right)  \left(  k\right)  =\mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Flag}\left(  k\right)  \right]  .
\]
\par
Now, fix $\underline{X}\in\operatorname*{Flag}\left(  \mathbb{F}_{p}%
^{n}\right)  $. Then, $\left(  \operatorname*{Flag}\left(  k\right)  \right)
\left(  \underline{X}\right)  =k\underline{X}$ (where the $k\underline{X}$ on
the right hand side means the image of $\underline{X}$ under the action of
$k\in G$ on $\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $). Now,%
\[
\underbrace{\left(  F\left(  k\right)  \right)  }_{=\mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Flag}\left(  k\right)  \right]  }\left(
\underline{X}\right)  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\left(  k\right)  \right]  \right)  \left(  \underline{X}%
\right)  =\left(  \operatorname*{Flag}\left(  k\right)  \right)  \left(
\underline{X}\right)  =k\underline{X}.
\]
\par
Now, forget that we fixed $\underline{X}$. We thus have shown that $\left(
F\left(  k\right)  \right)  \left(  \underline{X}\right)  =k\underline{X}$ for
each $\underline{X}\in\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)
$. In other words,
\begin{align*}
F\left(  k\right)   &  =\left(  \text{the endomorphism of }\mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)
\right]  \text{ that sends each }\underline{X}\in\operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \text{ to }k\underline{X}\right) \\
&  =\left(  \text{the action of }k\text{ on }\mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right]
\right)
\end{align*}
(since the action of $k$ on $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right]  $ is defined
as the endomorphism of $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right]  $ that sends
each $\underline{X}\in\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $
to $k\underline{X}$). This proves (\ref{pf.p8.1b.a.action}).}.

Now, it is easy to see that
\begin{equation}
\mathcal{E}_{F}\left(  \mathbb{F}_{p}^{n}\right)  =\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  }\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  \label{pf.p8.1.b.a.EF}%
\end{equation}
\ \ \ \ \footnote{\textit{Proof of (\ref{pf.p8.1.b.a.EF}):} Recall that
$F\left(  \mathbb{F}_{p}^{n}\right)  =\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right]  $. Also,
$\operatorname*{End}\nolimits_{\mathcal{A}}\left(  F\left(  \mathbb{F}_{p}%
^{n}\right)  \right)  =\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(
p\right)  }}\left(  F\left(  \mathbb{F}_{p}^{n}\right)  \right)  $ (because
the morphisms in the category $\mathcal{A}$ are just the $\mathbb{Z}_{\left(
p\right)  }$-linear maps). Hence,%
\[
\operatorname*{End}\nolimits_{\mathcal{A}}\left(  F\left(  \mathbb{F}_{p}%
^{n}\right)  \right)  =\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(
p\right)  }}\left(  \underbrace{F\left(  \mathbb{F}_{p}^{n}\right)
}_{=\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  }\right)  =\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }}\left(  \mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)
\right]  \right)  .
\]
\par
For each $f\in\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }%
}\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  $, we have the following chain of
equivalences:%
\begin{align}
&  \ \left(  f\circ F\left(  k\right)  =F\left(  k\right)  \circ f\text{ for
each }k\in\underbrace{\operatorname*{End}\nolimits_{\mathcal{V}}\left(
\mathbb{F}_{p}^{n}\right)  }_{=G}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  f\circ F\left(  k\right)  =F\left(  k\right)
\circ f\text{ for each }k\in G\right) \nonumber\\
&  \Longleftrightarrow\ \left(  f\text{ commutes with }\underbrace{F\left(
k\right)  }_{\substack{=\left(  \text{the action of }k\text{ on }%
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  \\\text{(by
(\ref{pf.p8.1b.a.action}))}}}\text{ for each }k\in G\right) \nonumber\\
&  \Longleftrightarrow\ \left(  f\text{ commutes with the action of }k\text{
on }\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \text{ for each }k\in G\right) \nonumber\\
&  \Longleftrightarrow\ \left(  f\text{ is }G\text{-equivariant}\right)
\nonumber\\
&  \Longleftrightarrow\ \left(  f\text{ is a }\mathbb{Z}_{\left(  p\right)
}\left[  G\right]  \text{-linear map}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }f\text{ is a }\mathbb{Z}_{\left(  p\right)  }\text{-linear
map}\right) \nonumber\\
&  \Longleftrightarrow\ \left(  f\in\operatorname*{End}\nolimits_{\mathbb{Z}%
_{\left(  p\right)  }\left[  G\right]  }\left(  \mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right]
\right)  \right)  . \label{pf.p8.1b.a.EF.pf.1}%
\end{align}
Now, the definition of $\mathcal{E}_{F}\left(  \mathbb{F}_{p}^{n}\right)  $
yields%
\begin{align*}
\mathcal{E}_{F}\left(  \mathbb{F}_{p}^{n}\right)   &  =\left\{  f\in
\underbrace{\operatorname*{End}\nolimits_{\mathcal{A}}\left(  F\left(
\mathbb{F}_{p}^{n}\right)  \right)  }_{=\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }}\left(  \mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)
\right]  \right)  }\ \mid\ f\circ F\left(  k\right)  =F\left(  k\right)  \circ
f\text{ for each }k\in\operatorname*{End}\nolimits_{\mathcal{V}}\left(
\mathbb{F}_{p}^{n}\right)  \right\} \\
&  =\left\{  f\in\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)
}}\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  \ \mid\ \underbrace{f\circ
F\left(  k\right)  =F\left(  k\right)  \circ f\text{ for each }k\in
\operatorname*{End}\nolimits_{\mathcal{V}}\left(  \mathbb{F}_{p}^{n}\right)
}_{\substack{\Longleftrightarrow\ \left(  f\in\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  }\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  \right)  \\\text{(by
(\ref{pf.p8.1b.a.EF.pf.1}))}}}\right\} \\
&  =\left\{  f\in\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)
}}\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  \ \mid\ f\in\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  }\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  \right\} \\
&  =\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  }\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right]  \right)
\end{align*}
(since $\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  }\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right]  \right)  \subseteq
\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  $). This proves
(\ref{pf.p8.1.b.a.EF}).}.

Each two objects of $\mathcal{V}$ are isomorphic (since the objects of
$\mathcal{V}$ are $n$-dimensional $\mathbb{F}_{p}$-vector spaces, and since
any two such vector spaces are isomorphic). Hence, Proposition 8.1a
\textbf{(b)} (applied to $\mathcal{C}=\mathcal{V}$, $\mathcal{D}=\mathcal{A}$,
$C=\mathbb{F}_{p}^{n}$ and $F=\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  $) shows that $\varepsilon$ is a monoid
isomorphism. Therefore, the map $\varepsilon$ is bijective. Moreover,
$\varepsilon$ is a $\mathbb{Z}_{\left(  p\right)  }$-algebra homomorphism
(this follows directly from the definition of the $\mathbb{Z}_{\left(
p\right)  }$-algebra structure on $\operatorname*{End}\nolimits_{\left[
\mathcal{V},\mathcal{A}\right]  }\left(  \mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Flag}\right]  \right)  $). Hence, the map
$\varepsilon$ is a $\mathbb{Z}_{\left(  p\right)  }$-algebra isomorphism
$\operatorname*{End}\nolimits_{\left[  \mathcal{V},\mathcal{A}\right]
}\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]
\right)  \rightarrow\mathcal{E}_{\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  }\left(  \mathbb{F}_{p}^{n}\right)  $ (since
$\varepsilon$ is a $\mathbb{Z}_{\left(  p\right)  }$-algebra homomorphism and
is bijective). Since $\left[  \mathcal{V},\mathcal{A}\right]  =\mathcal{VA}$
and
\begin{align*}
\mathcal{E}_{\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\right]  }\left(  \mathbb{F}_{p}^{n}\right)   &  =\mathcal{E}_{F}\left(
\mathbb{F}_{p}^{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]  =F\right)
\\
&  =\operatorname*{End}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  }\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right]  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.p8.1.b.a.EF})}\right)  ,
\end{align*}
this rewrites as follows: The map $\varepsilon$ is a $\mathbb{Z}_{\left(
p\right)  }$-algebra isomorphism \newline$\operatorname*{End}%
\nolimits_{\mathcal{VA}}\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  \right)  \rightarrow\operatorname*{End}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  }\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right]  \right)  $. This proves Proposition 8.1b
\textbf{(a)}.

\textbf{(b)} The orbit $G\underline{E}$ of $\underline{E}\in
\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $ is the whole $G$-set
$\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $ (since the $G$-set
$\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $ is transitive).
Thus, $\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  =G\underline{E}%
$. But the orbit-stabilizer theorem shows that $G\underline{E}\cong
G/G_{\underline{E}}$ as $G$-sets, where $G_{\underline{E}}$ denotes the
stabilizer of $\underline{E}$ in $G$. However, the stabilizer of
$\underline{E}$ in $G$ is $B$ (this is essentially the definition of $B$). In
other words, $G_{\underline{E}}=B$. Altogether, we thus have
$\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  =G\underline{E}\cong
G/\underbrace{G_{\underline{E}}}_{=B}=G/B$ as $G$-sets. This isomorphism is
natural. Hence, there is a natural isomorphism $G/B\rightarrow
\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $ of $G$-sets. This
isomorphism sends each coset $hB\in G/B$ of $B$ to the complete flag
$h\underline{E}\in\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $
(because of how it is constructed). The remaining statements of Proposition
8.1b \textbf{(b)} follow from this immediately.]

\item \textbf{\S 8:} Replace \textquotedblleft Note that $\mathbb{Z}_{\left(
p\right)  }\left[  X\right]  $ has an obvious inner product\textquotedblright%
\ by \textquotedblleft Note that $\mathbb{Z}_{\left(  p\right)  }\left[
X\right]  $ (for any set $X$) has an obvious inner product\textquotedblright.
(This is to disambiguate the meaning of $X$; you have previously used $X$ for
functors as well.)

\item \textbf{\S 8:} Replace \textquotedblleft Given $f:\mathbb{Z}_{\left(
p\right)  }\left[  X\right]  \rightarrow\mathbb{Z}_{\left(  p\right)  }\left[
Y\right]  $ we let $f^{t}:\mathbb{Z}_{\left(  p\right)  }\left[  Y\right]
\rightarrow\mathbb{Z}_{\left(  p\right)  }\left[  X\right]  $%
\textquotedblright\ by \textquotedblleft Given a $\mathbb{Z}_{\left(
p\right)  }$-linear map $f:\mathbb{Z}_{\left(  p\right)  }\left[  X\right]
\rightarrow\mathbb{Z}_{\left(  p\right)  }\left[  Y\right]  $ we let
$f^{t}:\mathbb{Z}_{\left(  p\right)  }\left[  Y\right]  \rightarrow
\mathbb{Z}_{\left(  p\right)  }\left[  X\right]  $\textquotedblright.

\item \textbf{\S 8:} When you write \textquotedblleft if $f$ comes from a map
$f:X\rightarrow Y$\textquotedblright, you are slightly abusing notation (you
are using the same notation for the map $f:X\rightarrow Y$ and the
$\mathbb{Z}_{\left(  p\right)  }$-linear map $\mathbb{Z}_{\left(  p\right)
}\left[  X\right]  \rightarrow\mathbb{Z}_{\left(  p\right)  }\left[  Y\right]
$ induced by it); it might be good to explicitly point this out. Better yet, I
suggest replacing the whole sentence (\textquotedblleft In particular, if $f$
comes from a map $f:X\rightarrow Y$ then $f^{t}\left[  y\right]
=\sum_{f\left(  x\right)  =y}\left[  x\right]  $.\textquotedblright)\ by the
following paragraphs:

\textquotedblleft Notice that if $f:\mathbb{Z}_{\left(  p\right)  }\left[
X\right]  \rightarrow\mathbb{Z}_{\left(  p\right)  }\left[  Y\right]  $ and
$g:\mathbb{Z}_{\left(  p\right)  }\left[  Y\right]  \rightarrow\mathbb{Z}%
_{\left(  p\right)  }\left[  Z\right]  $ are two $\mathbb{Z}_{\left(
p\right)  }$-linear maps, then%
\begin{equation}
\left(  g\circ f\right)  ^{t}=f^{t}\circ g^{t}. \label{par8.Zpf.adjgf}%
\end{equation}


To each map $f:X\rightarrow Y$ between two sets $X$ and $Y$ corresponds a
$\mathbb{Z}_{\left(  p\right)  }$-linear map $\mathbb{Z}_{\left(  p\right)
}\left[  f\right]  :\mathbb{Z}_{\left(  p\right)  }\left[  X\right]
\rightarrow\mathbb{Z}_{\left(  p\right)  }\left[  Y\right]  $ defined by
\[
\left(  \left(  \mathbb{Z}_{\left(  p\right)  }\left[  f\right]  \right)
\left[  x\right]  =\left[  f\left(  x\right)  \right]
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in X\right)  .
\]
The adjoint $\left(  \mathbb{Z}_{\left(  p\right)  }\left[  f\right]  \right)
^{t}$ of this map $\mathbb{Z}_{\left(  p\right)  }\left[  f\right]  $ is given
by%
\begin{equation}
\left(  \mathbb{Z}_{\left(  p\right)  }\left[  f\right]  \right)  ^{t}\left[
y\right]  =\sum_{\substack{x\in X;\\f\left(  x\right)  =y}}\left[  x\right]
\ \ \ \ \ \ \ \ \ \ \text{for each }y\in Y. \label{par8.Zpf.adj.1}%
\end{equation}
We shall often (by abuse of notation) denote the $\mathbb{Z}_{\left(
p\right)  }$-linear map $\mathbb{Z}_{\left(  p\right)  }\left[  f\right]  $ by
$f$ again. (This is a natural thing to do, because if we regard $X$ and $Y$ as
subsets of $\mathbb{Z}_{\left(  p\right)  }\left[  X\right]  $ and
$\mathbb{Z}_{\left(  p\right)  }\left[  Y\right]  $ in the obvious way, then
the original map $f:X\rightarrow Y$ becomes a restriction of the map
$\mathbb{Z}_{\left(  p\right)  }\left[  f\right]  :\mathbb{Z}_{\left(
p\right)  }\left[  X\right]  \rightarrow\mathbb{Z}_{\left(  p\right)  }\left[
Y\right]  $.) Thus, the equality (\ref{par8.Zpf.adj.1}) rewrites as%
\begin{equation}
f^{t}\left[  y\right]  =\sum_{\substack{x\in X;\\f\left(  x\right)
=y}}\left[  x\right]  \ \ \ \ \ \ \ \ \ \ \text{for each }y\in Y.
\label{par8.Zpf.adj.1-abused}%
\end{equation}


Next, let us state two basic facts about adjoints:

\textbf{Lemma 8.1d.} Let $X$ and $Y$ be two sets. Let $f:X\rightarrow Y$ be a
bijection. Then, $\left(  \mathbb{Z}_{\left(  p\right)  }\left[  f\right]
\right)  ^{t}=\mathbb{Z}_{\left(  p\right)  }\left[  f^{-1}\right]  $.
(Relying on our abuse of notation, we can rewrite this as $f^{t}=f^{-1}$.)

[\textit{Proof of Lemma 8.1d.} Let $y\in Y$. Recall that $f$ is a bijection.
Hence, there exists exactly one $x\in X$ satisfying $f\left(  x\right)  =y$
(namely, $x=f^{-1}\left(  y\right)  $). Thus, $\sum_{\substack{x\in
X;\\f\left(  x\right)  =y}}=\sum_{x\in\left\{  f^{-1}\left(  y\right)
\right\}  }$ (an equality of summation signs). Now, (\ref{par8.Zpf.adj.1})
yields
\[
\left(  \mathbb{Z}_{\left(  p\right)  }\left[  f\right]  \right)  ^{t}\left[
y\right]  =\underbrace{\sum_{\substack{x\in X;\\f\left(  x\right)  =y}%
}}_{=\sum_{x\in\left\{  f^{-1}\left(  y\right)  \right\}  }}\left[  x\right]
=\sum_{x\in\left\{  f^{-1}\left(  y\right)  \right\}  }\left[  x\right]
=\left[  f^{-1}\left(  y\right)  \right]  .
\]
Comparing this with $\left(  \mathbb{Z}_{\left(  p\right)  }\left[
f^{-1}\right]  \right)  \left[  y\right]  =\left[  f^{-1}\left(  y\right)
\right]  $, we obtain $\left(  \mathbb{Z}_{\left(  p\right)  }\left[
f\right]  \right)  ^{t}\left[  y\right]  =\left(  \mathbb{Z}_{\left(
p\right)  }\left[  f^{-1}\right]  \right)  \left[  y\right]  $.

Now, forget that we fixed $y$. We thus have proven that $\left(
\mathbb{Z}_{\left(  p\right)  }\left[  f\right]  \right)  ^{t}\left[
y\right]  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[  f^{-1}\right]
\right)  \left[  y\right]  $ for each $y\in Y$. In other words, the two maps
$\left(  \mathbb{Z}_{\left(  p\right)  }\left[  f\right]  \right)  ^{t}$ and
$\mathbb{Z}_{\left(  p\right)  }\left[  f^{-1}\right]  $ are equal to each
other on each of the elements of the basis $\left(  \left[  y\right]  \right)
_{y\in Y}$ of the $\mathbb{Z}_{\left(  p\right)  }$-module $\mathbb{Z}%
_{\left(  p\right)  }\left[  Y\right]  $. Since these two maps are
$\mathbb{Z}_{\left(  p\right)  }$-linear, we can therefore conclude that they
are equal. In other words, $\left(  \mathbb{Z}_{\left(  p\right)  }\left[
f\right]  \right)  ^{t}=\mathbb{Z}_{\left(  p\right)  }\left[  f^{-1}\right]
$. This proves Lemma 8.1d. $\square$ ]

\textbf{Lemma 8.1e.} Let $P$ and $Q$ be two functors from $\mathcal{V}$ to
$\mathcal{F}$. Let $\alpha:P\Longrightarrow Q$ be a natural transformation.

\textbf{(a)} Define a family $\beta=\left(  \beta_{V}\right)  _{V\in
\mathcal{V}}$ of morphisms $\beta_{V}:\mathbb{Z}_{\left(  p\right)  }\left[
P\left(  V\right)  \right]  \rightarrow\mathbb{Z}_{\left(  p\right)  }\left[
Q\left(  V\right)  \right]  $ by
\[
\left(  \beta_{V}=\mathbb{Z}_{\left(  p\right)  }\left[  \alpha_{V}\right]
\ \ \ \ \ \ \ \ \ \ \text{for each }V\in\mathcal{V}\right)  .
\]
Then, $\beta$ is a natural transformation $\mathbb{Z}_{\left(  p\right)
}\left[  P\right]  \Longrightarrow\mathbb{Z}_{\left(  p\right)  }\left[
Q\right]  $.

\textbf{(b)} Define a family $\gamma=\left(  \gamma_{V}\right)  _{V\in
\mathcal{V}}$ of morphisms $\gamma_{V}:\mathbb{Z}_{\left(  p\right)  }\left[
Q\left(  V\right)  \right]  \rightarrow\mathbb{Z}_{\left(  p\right)  }\left[
P\left(  V\right)  \right]  $ by
\[
\left(  \gamma_{V}=\left(  \beta_{V}\right)  ^{t}\ \ \ \ \ \ \ \ \ \ \text{for
each }V\in\mathcal{V}\right)  .
\]
Then, $\gamma$ is a natural transformation $\mathbb{Z}_{\left(  p\right)
}\left[  Q\right]  \Longrightarrow\mathbb{Z}_{\left(  p\right)  }\left[
P\right]  $.

[\textit{Proof of Lemma 8.1e.} \textbf{(a)} This is straightforward to prove.

\textbf{(b)} We need to prove that if $V$ and $W$ are two objects of
$\mathcal{V}$, and if $g:V\rightarrow W$ is a morphism of $\mathcal{V}$, then
the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\Znp\left[Q\left(V\right)\right] \ar[r]^{\gamma_V} \ar[d]_{\Znp\left
%[Q\left(g\right)\right]} & \Znp\left[P\left(V\right)\right] \ar[d]^{\Znp
%\left[P\left(g\right)\right]} \\
%\Znp\left[Q\left(W\right)\right] \ar[r]_{\gamma_W} & \Znp\left[P\left
%(W\right)\right]
%}} }%
%BeginExpansion
\xymatrix{
\Znp\left[Q\left(V\right)\right] \ar[r]^{\gamma_V} \ar[d]_{\Znp\left
[Q\left(g\right)\right]} & \Znp\left[P\left(V\right)\right] \ar[d]^{\Znp
\left[P\left(g\right)\right]} \\
\Znp\left[Q\left(W\right)\right] \ar[r]_{\gamma_W} & \Znp\left[P\left
(W\right)\right]
}
%EndExpansion
\label{pf.l8.1e.b.diagram}%
\end{equation}
is commutative.

So let $V$ and $W$ be two objects of $\mathcal{V}$, and let $g:V\rightarrow W$
be a morphism of $\mathcal{V}$. We must prove that the diagram
(\ref{pf.l8.1e.b.diagram}) is commutative.

The morphisms of $\mathcal{V}$ are isomorphisms of $\mathbb{F}_{p}$ -vector
spaces (by the definition of $\mathcal{V}$). Thus, each morphism of
$\mathcal{V}$ is an isomorphism. In particular, $g$ is an isomorphism (since
$g$ is a morphism of $\mathcal{V}$). Since $P$ is a functor, this shows that
$P\left(  g\right)  $ is an isomorphism and that $\left(  P\left(  g\right)
\right)  ^{-1}=P\left(  g^{-1}\right)  $.

The map $P\left(  g\right)  $ is an isomorphism in $\mathcal{F}$. In other
words, $P\left(  g\right)  $ is a bijection. Hence, Lemma 8.1d (applied to
$X=P\left(  V\right)  $, $Y=P\left(  W\right)  $ and $f=g$) shows that
\[
\left(  \mathbb{Z}_{\left(  p\right)  }\left[  P\left(  g\right)  \right]
\right)  ^{t}=\mathbb{Z}_{\left(  p\right)  }\left[  \underbrace{\left(
P\left(  g\right)  \right)  ^{-1}}_{=P\left(  g^{-1}\right)  }\right]
=\mathbb{Z}_{\left(  p\right)  }\left[  P\left(  g^{-1}\right)  \right]  .
\]
Hence,%
\[
\left(  \underbrace{\mathbb{Z}_{\left(  p\right)  }\left[  P\left(
g^{-1}\right)  \right]  }_{=\left(  \mathbb{Z}_{\left(  p\right)  }\left[
P\left(  g\right)  \right]  \right)  ^{t}}\right)  ^{t}=\left(  \left(
\mathbb{Z}_{\left(  p\right)  }\left[  P\left(  g\right)  \right]  \right)
^{t}\right)  ^{t}=\mathbb{Z}_{\left(  p\right)  }\left[  P\left(  g\right)
\right]
\]
(since the adjoint of the adjoint of a linear map is the original map). The
same argument (applied to $Q$ instead of $P$) shows that $\left(
\mathbb{Z}_{\left(  p\right)  }\left[  Q\left(  g^{-1}\right)  \right]
\right)  ^{t}=\mathbb{Z}_{\left(  p\right)  }\left[  Q\left(  g\right)
\right]  $.

But the definition of $\gamma_{V}$ shows that
\[
\gamma_{V}=\left(  \underbrace{\beta_{V}}_{=\mathbb{Z}_{\left(  p\right)
}\left[  \alpha_{V}\right]  }\right)  ^{t}=\left(  \mathbb{Z}_{\left(
p\right)  }\left[  \alpha_{V}\right]  \right)  ^{t}.
\]
The same argument (applied to $W$ instead of $V$) shows that $\gamma
_{W}=\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \alpha_{W}\right]
\right)  ^{t}$.

But $\alpha$ is a natural transformation. Hence, the diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%P\left(W\right) \ar[r]^{\alpha_W} \ar[d]_{P\left(g^{-1}\right)} & Q\left
%(W\right) \ar[d]^{Q\left(g^{-1}\right)} \\
%P\left(V\right) \ar[r]_{\alpha_V} & Q\left(V\right)
%}}}%
%BeginExpansion
\xymatrix{
P\left(W\right) \ar[r]^{\alpha_W} \ar[d]_{P\left(g^{-1}\right)} & Q\left
(W\right) \ar[d]^{Q\left(g^{-1}\right)} \\
P\left(V\right) \ar[r]_{\alpha_V} & Q\left(V\right)
}%
%EndExpansion
\]
is commutative. In other words, we have $Q\left(  g^{-1}\right)  \circ
\alpha_{W}=\alpha_{V}\circ P\left(  g^{-1}\right)  $.

But%
\begin{align*}
&  \left(  \underbrace{\mathbb{Z}_{\left(  p\right)  }\left[  Q\left(
g^{-1}\right)  \right]  \circ\mathbb{Z}_{\left(  p\right)  }\left[  \alpha
_{W}\right]  }_{=\mathbb{Z}_{\left(  p\right)  }\left[  Q\left(
g^{-1}\right)  \circ\alpha_{W}\right]  }\right)  ^{t}\\
&  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \underbrace{Q\left(
g^{-1}\right)  \circ\alpha_{W}}_{=\alpha_{V}\circ P\left(  g^{-1}\right)
}\right]  \right)  ^{t}\\
&  =\left(  \underbrace{\mathbb{Z}_{\left(  p\right)  }\left[  \alpha_{V}\circ
P\left(  g^{-1}\right)  \right]  }_{=\mathbb{Z}_{\left(  p\right)  }\left[
\alpha_{V}\right]  \circ\mathbb{Z}_{\left(  p\right)  }\left[  P\left(
g^{-1}\right)  \right]  }\right)  ^{t}\\
&  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \alpha_{V}\right]
\circ\mathbb{Z}_{\left(  p\right)  }\left[  P\left(  g^{-1}\right)  \right]
\right)  ^{t}=\underbrace{\left(  \mathbb{Z}_{\left(  p\right)  }\left[
P\left(  g^{-1}\right)  \right]  \right)  ^{t}}_{=\mathbb{Z}_{\left(
p\right)  }\left[  P\left(  g\right)  \right]  }\circ\underbrace{\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \alpha_{V}\right]  \right)  ^{t}%
}_{=\gamma_{V}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{par8.Zpf.adjgf}) (applied to }P\left(  W\right)  \text{,
}P\left(  V\right)  \text{, }Q\left(  V\right)  \text{, }\mathbb{Z}_{\left(
p\right)  }\left[  P\left(  g^{-1}\right)  \right] \\
\text{and }\mathbb{Z}_{\left(  p\right)  }\left[  \alpha_{V}\right]  \text{
instead of }X\text{, }Y\text{, }Z\text{, }f\text{ and }g\text{)}%
\end{array}
\right) \\
&  =\mathbb{Z}_{\left(  p\right)  }\left[  P\left(  g\right)  \right]
\circ\gamma_{V}.
\end{align*}
Hence,%
\begin{align*}
&  \mathbb{Z}_{\left(  p\right)  }\left[  P\left(  g\right)  \right]
\circ\gamma_{V}\\
&  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[  Q\left(  g^{-1}\right)
\right]  \circ\mathbb{Z}_{\left(  p\right)  }\left[  \alpha_{W}\right]
\right)  ^{t}=\underbrace{\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\alpha_{W}\right]  \right)  ^{t}}_{=\gamma_{W}}\circ\underbrace{\left(
\mathbb{Z}_{\left(  p\right)  }\left[  Q\left(  g^{-1}\right)  \right]
\right)  ^{t}}_{=\mathbb{Z}_{\left(  p\right)  }\left[  Q\left(  g\right)
\right]  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{par8.Zpf.adjgf}) (applied to }P\left(  W\right)  \text{,
}Q\left(  W\right)  \text{, }Q\left(  V\right)  \text{, }\mathbb{Z}_{\left(
p\right)  }\left[  \alpha_{W}\right] \\
\text{and }\mathbb{Z}_{\left(  p\right)  }\left[  Q\left(  g^{-1}\right)
\right]  \text{ instead of }X\text{, }Y\text{, }Z\text{, }f\text{ and
}g\text{)}%
\end{array}
\right) \\
&  =\gamma_{W}\circ\mathbb{Z}_{\left(  p\right)  }\left[  Q\left(  g\right)
\right]  .
\end{align*}
In other words, the diagram (\ref{pf.l8.1e.b.diagram}) is commutative. This is
precisely what we wanted to prove. Hence, Lemma 8.1e \textbf{(b)} is proven.
$\square$ ]

The natural transformations $\beta$ and $\gamma$ defined in Lemma 8.1e will be
denoted by $\mathbb{Z}_{\left(  p\right)  }\left[  \alpha\right]  $ and
$\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \alpha\right]  \right)  ^{t}$,
respectively. By abuse of notation, we shall often denote these natural
transformations $\beta$ and $\gamma$ by $\alpha$ and $\alpha^{t}$,
respectively. Notice that this abuse of notation is compatible with
composition of functors, because of the following remark:

\textbf{Remark 8.1f.} \textbf{(a)} For every functor $P:\mathcal{V}%
\rightarrow\mathcal{F}$, we have $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{id}\nolimits_{P}\right]  =\operatorname*{id}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  P\right]  }$.

\textbf{(b)} Let $P$, $Q$ and $R$ be three functors from $\mathcal{V}$ to
$\mathcal{F}$. Let $\alpha_{1}:P\Longrightarrow Q$ and $\alpha_{2}%
:Q\Longrightarrow R$ be two natural transformations. Then, $\mathbb{Z}%
_{\left(  p\right)  }\left[  \alpha_{2}\right]  \circ\mathbb{Z}_{\left(
p\right)  }\left[  \alpha_{1}\right]  =\mathbb{Z}_{\left(  p\right)  }\left[
\alpha_{2}\circ\alpha_{1}\right]  $.

[\textit{Proof of Remark 8.1f.} These are straightforward computations.
$\square$ ]\textquotedblright

The purpose of Lemma 8.1e is to explain the meaning of \textquotedblleft%
$\pi_{1}\pi_{0}^{t}$\textquotedblright\ in Definition 8.2.

\item \textbf{Definition 8.2:} I suggest adding the sentence \textquotedblleft
Let $\sigma\in\Sigma_{n}$.\textquotedblright\ at the beginning of this definition.

\item \textbf{Definition 8.2:} For the sake of completeness, I suggest
actually defining the projection maps $\pi_{0}$ and $\pi_{1}$: Namely, the
natural transformation $\pi_{0}:Z\left(  \sigma\right)  \rightarrow
\operatorname*{Flag}$ (or, rather, its component $\left(  \pi_{0}\right)
_{V}:\left(  Z\left(  \sigma\right)  \right)  \left(  V\right)  \rightarrow
\operatorname*{Flag}\left(  V\right)  $ for a given $V\in\mathcal{V}$) sends
each $\left(  \underline{U},\underline{W}\right)  \in\left(  Z\left(
\sigma\right)  \right)  \left(  V\right)  $ to $\underline{U}$, whereas the
natural transformation $\pi_{1}:Z\left(  \sigma\right)  \rightarrow
\operatorname*{Flag}$ (or, rather, its component $\left(  \pi_{1}\right)
_{V}:\left(  Z\left(  \sigma\right)  \right)  \left(  V\right)  \rightarrow
\operatorname*{Flag}\left(  V\right)  $ for a given $V\in\mathcal{V}$) sends
each $\left(  \underline{U},\underline{W}\right)  \in\left(  Z\left(
\sigma\right)  \right)  \left(  V\right)  $ to $\underline{W}$.

\item \textbf{Definition 8.2:} After \textquotedblleft that $T_{\sigma}%
^{t}=T_{\sigma^{-1}}$\textquotedblright, add \textquotedblleft(since
$\delta\left(  \underline{U},\underline{W}\right)  =\delta\left(
\underline{W},\underline{U}\right)  ^{-1}$ for all $\underline{U}%
,\underline{W}\in\operatorname*{Flag}\left(  V\right)  $).

\item \textbf{Proof of Proposition 8.3:} After \textquotedblleft Consider an
element $f:\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]
\rightarrow\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]
$\textquotedblright, add \textquotedblleft of $\mathcal{H}$\textquotedblright.

\item \textbf{Proof of Proposition 8.3:} I would replace \textquotedblleft
well-defined numbers $n_{\sigma}$\textquotedblright\ by \textquotedblleft
well-defined numbers $n_{\sigma}\in\mathbb{Z}_{\left(  p\right)  }%
$\textquotedblright\ (in order to avoid creating the false impression that
they must necessarily be integers).

\item \textbf{Proposition 8.4:} Replace \textquotedblleft Let $\underline{U}$
be a flag, let $F$ be the set of flags $\underline{W}$ such that $W_{j}=U_{j}$
for all $j\neq i$, and put $a=\sum_{\underline{W}\in F}\left[  \underline{W}%
\right]  $.\textquotedblright\ by: \textquotedblleft Let $V\in\mathcal{V}$.
Let $\underline{U}\in\operatorname*{Flag}\left(  V\right)  $. Let $F$ be the
set of all $\underline{W}\in\operatorname*{Flag}\left(  V\right)  $ satisfying
$W_{j}=U_{j}$ for all $j\neq i$. Set $a=\sum_{\underline{W}\in F}\left[
\underline{W}\right]  \in\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\left(  V\right)  \right]  $.\textquotedblright

\item \textbf{Proof of Proposition 8.4:} I would suggest replacing the
\textquotedblleft$\sum_{\underline{W}}$\textquotedblright\ sign by an
\textquotedblleft$\sum_{\underline{W}\in F}$\textquotedblright\ sign (seeing
that you otherwise always use \textquotedblleft$\sum_{\underline{W}\in F}%
$\textquotedblright\ signs).

\item \textbf{Proof of Proposition 8.4:} Replace \textquotedblleft so
$T_{i}\left(  a\right)  =pa$\textquotedblright\ by \textquotedblleft so
$T_{i}\left(  a\right)  =\left(  \underbrace{\left\vert F\right\vert }%
_{=p+1}-1\right)  a=pa$, thus $\left(  T_{i}-p\right)  a=0$\textquotedblright.

\item \textbf{Proof of Proposition 8.4:} Replace \textquotedblleft It follows
that $\left(  T_{i}-p\right)  \left(  T_{i}+1\right)  \left[  \underline{W}%
\right]  =\left(  T_{i}-p\right)  \left(  a\right)  =0$ as
claimed\textquotedblright\ by \textquotedblleft From $\left(  T_{i}+1\right)
\left[  \underline{U}\right]  =\underbrace{T_{i}\left[  \underline{U}\right]
}_{=a-\left[  \underline{U}\right]  }+\left[  \underline{U}\right]  =a$, it
follows that $\left(  T_{i}-p\right)  \left(  T_{i}+1\right)  \left[
\underline{U}\right]  =\left(  T_{i}-p\right)  \left(  a\right)  =0$ as
claimed\textquotedblright. (The claim of Proposition 8.4 does not involve
$\underline{W}$ !)

\item \textbf{Definition 8.5:} Replace \textquotedblleft We
define\textquotedblright\ by \textquotedblleft For each $V\in\mathcal{V}$, we
define a $\mathbb{Z}_{\left(  p\right)  }$-linear map\textquotedblright.

\item \textbf{Definition 8.5:} Replace \textquotedblleft We also define a
map\textquotedblright\ by \textquotedblleft We also define a $\mathbb{Z}%
_{\left(  p\right)  }$-linear map\textquotedblright.

\item \textbf{Definition 8.5:} Replace \textquotedblleft$\sum_{\sigma
}p^{l\left(  \sigma\right)  }$\textquotedblright\ by \textquotedblleft%
$\sum_{\sigma}n_{\sigma}p^{l\left(  \sigma\right)  }$ whenever $n_{\sigma}%
\in\mathbb{Z}_{\left(  p\right)  }$\textquotedblright.

\item \textbf{Proof of Proposition 8.6:} Replace \textquotedblleft so
$\widehat{e}\left[  \underline{U}\right]  =x$ for all $\underline{W}%
$\textquotedblright\ by \textquotedblleft so $\widehat{e}\left[
\underline{U}\right]  =x$ for all $\underline{U}$\textquotedblright.

\item \textbf{Proof of Proposition 8.6:} Replace \textquotedblleft$\left\vert
\operatorname*{Flag}\right\vert ^{-1}$\textquotedblright\ by \textquotedblleft%
$\left\vert \operatorname*{Flag}\left(  V\right)  \right\vert ^{-1}%
$\textquotedblright.

\item \textbf{Proof of Proposition 8.6:} I'd suggest more detail once again:
Replace \textquotedblleft Next, we have%
\[
\widehat{e}T_{\sigma}\left[  \underline{U}\right]  =\sum_{\delta\left(
\underline{U},\underline{W}\right)  =\sigma}\widehat{e}\left[  \underline{W}%
\right]  =\left\vert \left\{  \underline{W}\mid\delta\left(  \underline{W}%
,\underline{U}\right)  =\sigma^{-1}\right\}  \right\vert x=p^{l\left(
\sigma^{-1}\right)  }x=p^{l\left(  \sigma\right)  }\widehat{e}\left[
\underline{W}\right]  .
\]
(using Proposition 5.2 and the fact that $l\left(  \sigma^{-1}\right)
=l\left(  \sigma\right)  $)\textquotedblright\ by \textquotedblleft Next,
observe that each $\sigma\in\Sigma_{n}$ satisfies%
\begin{align*}
&  \left\{  \underline{W}\mid\underbrace{\delta\left(  \underline{U}%
,\underline{W}\right)  }_{=\delta\left(  \underline{W},\underline{U}\right)
^{-1}}=\sigma\right\} \\
&  =\left\{  \underline{W}\mid\underbrace{\delta\left(  \underline{W}%
,\underline{U}\right)  ^{-1}=\sigma}_{\Longleftrightarrow\ \left(
\delta\left(  \underline{W},\underline{U}\right)  =\sigma^{-1}\right)
}\right\}  =\left\{  \underline{W}\mid\delta\left(  \underline{W}%
,\underline{U}\right)  =\sigma^{-1}\right\} \\
&  =Y\left(  \sigma^{-1},\underline{U}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{using the notation from \S 5}\right)
\end{align*}
and therefore
\begin{align*}
\left\vert \underbrace{\left\{  \underline{W}\mid\delta\left(  \underline{U}%
,\underline{W}\right)  =\sigma\right\}  }_{=Y\left(  \sigma^{-1}%
,\underline{U}\right)  }\right\vert  &  =\left\vert Y\left(  \sigma
^{-1},\underline{U}\right)  \right\vert =p^{l\left(  \sigma^{-1}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Corollary 5.2a, applied to }\sigma^{-1}\text{, }V\text{ and
}\underline{U}\\
\text{instead of }\sigma\text{, }W\text{ and }\underline{V}%
\end{array}
\right) \\
&  =p^{l\left(  \sigma\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}l\left(  \sigma^{-1}\right)  =l\left(  \sigma\right)  \right)
\end{align*}
and thus%
\begin{align*}
\widehat{e}T_{\sigma}\left[  \underline{U}\right]   &  =\sum_{\delta\left(
\underline{U},\underline{W}\right)  =\sigma}\underbrace{\widehat{e}\left[
\underline{W}\right]  }_{=x}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
T_{\sigma}\left[  \underline{U}\right]  =\sum_{\delta\left(  \underline{U}%
,\underline{W}\right)  =\sigma}\left[  \underline{W}\right]  \right) \\
&  =\sum_{\delta\left(  \underline{U},\underline{W}\right)  =\sigma
}x=\underbrace{\left\vert \left\{  \underline{W}\mid\delta\left(
\underline{U},\underline{W}\right)  =\sigma\right\}  \right\vert
}_{=p^{l\left(  \sigma\right)  }}x=p^{l\left(  \sigma\right)  }\underbrace{x}%
_{=\widehat{e}\left[  \underline{U}\right]  }=p^{l\left(  \sigma\right)
}\widehat{e}\left[  \underline{U}\right]
\end{align*}
\textquotedblright. (Notice that I replaced the \textquotedblleft%
$\widehat{e}\left[  \underline{W}\right]  .$\textquotedblright\ at the end of
the long equation by an \textquotedblleft$\widehat{e}\left[  \underline{U}%
\right]  $\textquotedblright, as I think this is what you meant.)

\item \textbf{Proof of Proposition 8.6:} After \textquotedblleft we deduce
that $\widehat{\xi}$ is a ring map\textquotedblright, I suggest adding
\textquotedblleft(in fact, we compare $\widehat{e}ab=\widehat{e}\left(
ab\right)  =\widehat{\xi}\left(  ab\right)  \widehat{e}$ with
$\underbrace{\widehat{e}a}_{=\widehat{\xi}\left(  a\right)  \widehat{e}%
}b=\widehat{\xi}\left(  a\right)  \underbrace{\widehat{e}b}_{=\widehat{\xi
}\left(  b\right)  \widehat{e}}=\widehat{\xi}\left(  a\right)  \widehat{\xi
}\left(  b\right)  \widehat{e}$, and thus we find $\widehat{\xi}\left(
ab\right)  \widehat{e}=\widehat{\xi}\left(  a\right)  \widehat{\xi}\left(
b\right)  \widehat{e}$, which leads to $\widehat{\xi}\left(  ab\right)
=\widehat{\xi}\left(  a\right)  \widehat{\xi}\left(  b\right)  $ because
$\widehat{e}$ is a nonzero vector in a free $\mathbb{Z}_{\left(  p\right)  }%
$-module)\textquotedblright.

\item \textbf{Proof of Proposition 8.6:} Replace \textquotedblleft%
$a\widehat{e}=\xi\left(  a^{t}\right)  \widehat{e}$\textquotedblright\ by
\textquotedblleft$a\widehat{e}=\widehat{\xi}\left(  a^{t}\right)  \widehat{e}%
$\textquotedblright.

\item \textbf{Proof of Lemma 8.7:} I would replace the last sentence of this
proof by the following (more detailed) argument:

\textquotedblleft Since every $\underline{U}\in\operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  $ satisfies the chain of equivalences%
\begin{align*}
\left(  \delta\left(  \underline{E},\underline{U}\right)  =\sigma^{-1}\right)
\  &  \Longleftrightarrow\ \left(  \underbrace{\delta\left(  \underline{E}%
,\underline{U}\right)  ^{-1}}_{=\delta\left(  \underline{U},\underline{E}%
\right)  }=\sigma\right)  \ \Longleftrightarrow\ \left(  \delta\left(
\underline{U},\underline{E}\right)  =\sigma\right) \\
&  \Longleftrightarrow\ \left(  \underline{U}\in\underbrace{Y\left(
\sigma,\underline{E}\right)  }_{=Y\left(  \sigma\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }Y\left(
\sigma,\underline{E}\right)  \right) \\
&  \Longleftrightarrow\ \left(  \underline{U}\in Y\left(  \sigma\right)
\right)  ,
\end{align*}
this rewrites as $T_{\sigma^{-1}}\left[  \underline{E}\right]  =\sum
_{\underline{U}\in Y\left(  \sigma\right)  }\left[  \underline{U}\right]  $.
But Proposition 5.2 shows that the map $g\mapsto g\sigma\underline{E}$ gives a
bijection $X\left(  \sigma\right)  \rightarrow Y\left(  \sigma\right)  $.
Hence, $\sum_{\underline{U}\in Y\left(  \sigma\right)  }\left[  \underline{U}%
\right]  =\sum_{g\in X\left(  \sigma\right)  }\left[  g\sigma\underline{E}%
\right]  $. Hence,
\begin{equation}
T_{\sigma^{-1}}\left[  \underline{E}\right]  =\sum_{\underline{U}\in Y\left(
\sigma\right)  }\left[  \underline{U}\right]  =\sum_{g\in X\left(
\sigma\right)  }\left[  g\sigma\underline{E}\right]  =\sum_{x\in X\left(
\sigma\right)  }\left[  x\sigma\underline{E}\right]  ,
\label{pf.l8.7.intermed}%
\end{equation}
as required.\textquotedblright.

\item \textbf{\S 8:} Between Lemma 8.7 and Corollary 8.8, you write:
\textquotedblleft note that $\operatorname*{Base}\left(  V\right)  $ is
canonically the same as $\operatorname*{Iso}\left(  \mathbb{F}_{p}%
^{n},V\right)  $\textquotedblright. I suggest adding \textquotedblleft(in
fact, there is a canonical bijection $\operatorname*{Iso}\left(
\mathbb{F}_{p}^{n},V\right)  \rightarrow\operatorname*{Base}\left(  V\right)
$ sending each $\phi\in\operatorname*{Iso}\left(  \mathbb{F}_{p}^{n},V\right)
$ to $\left(  \phi e_{1},\phi e_{2},\ldots,\phi e_{n}\right)  \in
\operatorname*{Base}\left(  V\right)  $)\textquotedblright\ after this.

\item \textbf{\S 8:} Between Lemma 8.7 and Corollary 8.8, replace
\textquotedblleft gives a map $g^{\ast}:\operatorname*{Base}\rightarrow
\operatorname*{Base}$\textquotedblright\ by \textquotedblleft gives a natural
transformation $g^{\ast}:\operatorname*{Base}\rightarrow\operatorname*{Base}$
(whose $V$-component for any given $V\in\mathcal{V}$ is the map
\[
\operatorname*{Iso}\left(  \mathbb{F}_{p}^{n},V\right)  \rightarrow
\operatorname*{Iso}\left(  \mathbb{F}_{p}^{n},V\right)
,\ \ \ \ \ \ \ \ \ \ \phi\mapsto\phi\circ g,
\]
interpreted as a map $\operatorname*{Base}\left(  V\right)  \rightarrow
\operatorname*{Base}\left(  V\right)  $ via the canonical bijection
\newline$\operatorname*{Iso}\left(  \mathbb{F}_{p}^{n},V\right)
\rightarrow\operatorname*{Base}\left(  V\right)  $)\textquotedblright.

\item \textbf{\S 8:} Just before Corollary 8.8, I would add the following two lemmas:

\textbf{Lemma 8.7a.} Let $\mathbf{e}\in\operatorname*{Base}\left(
\mathbb{F}_{p}^{n}\right)  $ be the basis $\left(  e_{1},e_{2},\ldots
,e_{n}\right)  $ of $\mathbb{F}_{p}^{n}$. Let $g\in G$.

\textbf{(a)} We have $g^{\ast}\mathbf{e}=\left(  ge_{1},ge_{2},\ldots
,ge_{n}\right)  $.

\textbf{(b)} We have $\pi\left(  g^{\ast}\mathbf{e}\right)  =g\underline{E}$.

[\textit{Proof of Lemma 8.7a.} \textbf{(a)} Set $V=\mathbb{F}_{p}^{n}$. Recall
that the $V$-component of the natural transformation $g^{\ast}%
:\operatorname*{Base}\rightarrow\operatorname*{Base}$ is the map%
\[
\operatorname*{Iso}\left(  \mathbb{F}_{p}^{n},V\right)  \rightarrow
\operatorname*{Iso}\left(  \mathbb{F}_{p}^{n},V\right)
,\ \ \ \ \ \ \ \ \ \ \phi\mapsto\phi\circ g,
\]
interpreted as a map $\operatorname*{Base}\left(  V\right)  \rightarrow
\operatorname*{Base}\left(  V\right)  $ via the canonical bijection
\newline$\operatorname*{Iso}\left(  \mathbb{F}_{p}^{n},V\right)
\rightarrow\operatorname*{Base}\left(  V\right)  $ (because this is how
$g^{\ast}$ was defined). In other words, every $\mathbf{b}\in
\operatorname*{Base}\left(  V\right)  $ satisfies $g^{\ast}\mathbf{b}%
=\alpha\left(  \alpha^{-1}\left(  \mathbf{b}\right)  \circ g\right)  $, where
$\alpha$ is the canonical bijection $\operatorname*{Iso}\left(  \mathbb{F}%
_{p}^{n},V\right)  \rightarrow\operatorname*{Base}\left(  V\right)  $.
Consider this $\alpha$. Notice that $\operatorname*{id}\nolimits_{\mathbb{F}%
_{p}^{n}}\in\operatorname*{Iso}\left(  \mathbb{F}_{p}^{n}%
,\underbrace{\mathbb{F}_{p}^{n}}_{=V}\right)  =\operatorname*{Iso}\left(
\mathbb{F}_{p}^{n},V\right)  $. The definition of $\alpha$ shows that
\[
\alpha\left(  \operatorname*{id}\nolimits_{\mathbb{F}_{p}^{n}}\right)
=\left(  \operatorname*{id}e_{1},\operatorname*{id}e_{2},\ldots
,\operatorname*{id}e_{n}\right)  =\left(  e_{1},e_{2},\ldots,e_{n}\right)
=\mathbf{e}.
\]
Hence, $\alpha^{-1}\left(  \mathbf{e}\right)  =\operatorname*{id}%
\nolimits_{\mathbb{F}_{p}^{n}}$. Now, recall that every $\mathbf{b}%
\in\operatorname*{Base}\left(  V\right)  $ satisfies $g^{\ast}\mathbf{b}%
=\alpha\left(  \alpha^{-1}\left(  \mathbf{b}\right)  \circ g\right)  $.
Applying this to $\mathbf{b}=\mathbf{e}$, we obtain
\[
g^{\ast}\mathbf{e}=\alpha\left(  \underbrace{\alpha^{-1}\left(  \mathbf{e}%
\right)  }_{=\operatorname*{id}\nolimits_{\mathbb{F}_{p}^{n}}}\circ g\right)
=\alpha\left(  g\right)  =\left(  ge_{1},ge_{2},\ldots,ge_{n}\right)  .
\]
This proves Lemma 8.7a \textbf{(a)}.

\textbf{(b)} We have $g^{\ast}\mathbf{e}=\left(  ge_{1},ge_{2},\ldots
,ge_{n}\right)  $ (by Lemma 8.7a \textbf{(a)}). Applying the natural
transformation $\pi$ to both sides of this equality, we obtain%
\[
\pi\left(  g^{\ast}\mathbf{e}\right)  =\pi\left(  ge_{1},ge_{2},\ldots
,ge_{n}\right)  =g\underbrace{\pi\left(  e_{1},e_{2},\ldots,e_{n}\right)
}_{=\underline{E}}=g\underline{E}.
\]
This proves Lemma 8.7a \textbf{(b)}. $\square$ ]

\textbf{Lemma 8.7b.} Let $Y\in\mathcal{VA}$ be a functor. Let $X$ be the
functor $\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]
\in\mathcal{VA}$. Let $\alpha:X\Longrightarrow Y$ and $\beta:X\Longrightarrow
Y$ be two natural transformations. Let $\mathbf{e}\in\operatorname*{Base}%
\left(  \mathbb{F}_{p}^{n}\right)  $ be the basis $\left(  e_{1},e_{2}%
,\ldots,e_{n}\right)  $ of $\mathbb{F}_{p}^{n}$. Assume that $\alpha\left[
\mathbf{e}\right]  =\beta\left[  \mathbf{e}\right]  $ in $Y\left(
\mathbb{F}_{p}^{n}\right)  $. Then, $\alpha=\beta$.

[\textit{Proof of Lemma 8.7b.} Fix $V\in\mathcal{V}$. Let $u\in X\left(
V\right)  $. We are going to show that $\alpha\left(  u\right)  =\beta\left(
u\right)  $. Notice that $\left(  \left[  \mathbf{u}\right]  \right)
_{\mathbf{u}\in\operatorname*{Base}\left(  V\right)  }$ is a basis of the
$\mathbb{Z}_{\left(  p\right)  }$-module $\mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Base}\left(  V\right)  \right]  =\underbrace{\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)
}_{=X}\left(  V\right)  =X\left(  V\right)  $.

The $V$-components of $\alpha$ and $\beta$ are morphisms in the category
$\mathcal{A}$. In other words, the $V$-components of $\alpha$ and $\beta$ are
$\mathbb{Z}_{\left(  p\right)  }$-linear maps. Hence, both $\alpha\left(
u\right)  $ and $\beta\left(  u\right)  $ depend $\mathbb{Z}_{\left(
p\right)  }$-linearly on $u$. Hence, the equation $\alpha\left(  u\right)
=\beta\left(  u\right)  $ (which we want to prove) is linear in $u$. Thus, for
the proof of this equation, we can WLOG assume that $u$ belongs to the basis
$\left(  \left[  \mathbf{u}\right]  \right)  _{\mathbf{u}\in
\operatorname*{Base}\left(  V\right)  }$ of the $\mathbb{Z}_{\left(  p\right)
}$-module $X\left(  V\right)  $. Assume this.

We have assumed that $u$ belongs to the basis $\left(  \left[  \mathbf{u}%
\right]  \right)  _{\mathbf{u}\in\operatorname*{Base}\left(  V\right)  }$ of
the $\mathbb{Z}_{\left(  p\right)  }$-module $X\left(  V\right)  $. In other
words, $u=\left[  \mathbf{u}\right]  $ for some $\mathbf{u}\in
\operatorname*{Base}\left(  V\right)  $. Consider this $\mathbf{u}$.

We have $\mathbf{u}\in\operatorname*{Base}\left(  V\right)  $. Thus,
$\mathbf{u}$ is a basis of the $n$-dimensional $\mathbb{F}_{p}$-vector space
$V$. Write $\mathbf{u}$ in the form $\left(  u_{1},u_{2},\ldots,u_{n}\right)
$.

Let $\phi$ be the $\mathbb{F}_{p}$-linear map $\mathbb{F}_{p}^{n}\rightarrow
V$ that sends the basis vectors $e_{1},e_{2},\ldots,e_{n}$ of $\mathbb{F}%
_{p}^{n}$ to $u_{1},u_{2},\ldots,u_{n}$, respectively. Then, the
$\mathbb{F}_{p}$-linear map $\phi$ sends the basis $\left(  e_{1},e_{2}%
,\ldots,e_{n}\right)  $ of $\mathbb{F}_{p}^{n}$ to the basis $\left(
u_{1},u_{2},\ldots,u_{n}\right)  $ of $V$. Hence, $\phi$ is an isomorphism of
vector spaces. In other words, $\phi$ is a morphism in the category
$\mathcal{V}$.

Now, the morphism $\operatorname*{Base}\phi:\operatorname*{Base}\left(
\mathbb{F}_{p}^{n}\right)  \rightarrow\operatorname*{Base}\left(  V\right)  $
(induced by the morphism $\phi:\mathbb{F}_{p}^{n}\rightarrow V$) satisfies%
\begin{align*}
\left(  \operatorname*{Base}\phi\right)  \left(  \mathbf{e}\right)   &
=\left(  \operatorname*{Base}\phi\right)  \left(  e_{1},e_{2},\ldots
,e_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{e}=\left(
e_{1},e_{2},\ldots,e_{n}\right)  \right) \\
&  =\left(  u_{1},u_{2},\ldots,u_{n}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\phi\text{ sends the basis vectors}\\
e_{1},e_{2},\ldots,e_{n}\text{ to }u_{1},u_{2},\ldots,u_{n}%
\end{array}
\right) \\
&  =\mathbf{u}.
\end{align*}
But the morphism $X\left(  \phi\right)  :X\left(  \mathbb{F}_{p}^{n}\right)
\rightarrow X\left(  V\right)  $ satisfies%
\begin{equation}
\left(  \underbrace{X}_{=\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  }\left(  \phi\right)  \right)  \left[
\mathbf{e}\right]  =\left(  \left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  \left(  \phi\right)  \right)  \left[
\mathbf{e}\right]  =\left[  \underbrace{\left(  \operatorname*{Base}%
\phi\right)  \left(  \mathbf{e}\right)  }_{=\mathbf{u}}\right]  =\left[
\mathbf{u}\right]  =u. \label{pf.l8.7b.u-from-e}%
\end{equation}


But $\alpha:X\Longrightarrow Y$ is a natural transformation. Hence, the
diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%X\left(\Fpn\right) \ar[r]^{\alpha_{\Fpn}} \ar[d]_{X\left(\phi\right)}
%& Y\left(\Fpn\right) \ar[d]^{Y\left(\phi\right)} \\
%X\left(V\right) \ar[r]_{\alpha_V} & Y\left(V\right)
%}}}%
%BeginExpansion
\xymatrix{
X\left(\Fpn\right) \ar[r]^{\alpha_{\Fpn}} \ar[d]_{X\left(\phi\right)}
& Y\left(\Fpn\right) \ar[d]^{Y\left(\phi\right)} \\
X\left(V\right) \ar[r]_{\alpha_V} & Y\left(V\right)
}%
%EndExpansion
\]
is commutative (since $\phi:\mathbb{F}_{p}^{n}\rightarrow V$ is a morphism in
the category $\mathcal{V}$). In other words, we have $Y\left(  \phi\right)
\circ\alpha_{\mathbb{F}_{p}^{n}}=\alpha_{V}\circ X\left(  \phi\right)  $.
Hence,%
\[
\underbrace{\left(  Y\left(  \phi\right)  \circ\alpha_{\mathbb{F}_{p}^{n}%
}\right)  }_{=\alpha_{V}\circ X\left(  \phi\right)  }\left[  \mathbf{e}%
\right]  =\left(  \alpha_{V}\circ X\left(  \phi\right)  \right)  \left[
\mathbf{e}\right]  =\alpha_{V}\left(  \underbrace{\left(  X\left(
\phi\right)  \right)  \left[  \mathbf{e}\right]  }_{=u}\right)  =\alpha
_{V}\left(  u\right)  =\alpha\left(  u\right)  .
\]
Hence,%
\[
\alpha\left(  u\right)  =\left(  Y\left(  \phi\right)  \circ\alpha
_{\mathbb{F}_{p}^{n}}\right)  \left[  \mathbf{e}\right]  =\left(  Y\left(
\phi\right)  \right)  \left(  \underbrace{\alpha_{\mathbb{F}_{p}^{n}}\left[
\mathbf{e}\right]  }_{=\alpha\left[  \mathbf{e}\right]  }\right)  =\left(
Y\left(  \phi\right)  \right)  \left(  \alpha\left[  \mathbf{e}\right]
\right)  .
\]
The same argument (applied to $\beta$ instead of $\alpha$) shows that
$\beta\left(  u\right)  =\left(  Y\left(  \phi\right)  \right)  \left(
\beta\left[  \mathbf{e}\right]  \right)  $. Comparing this with $\alpha\left(
u\right)  =\left(  Y\left(  \phi\right)  \right)  \left(  \underbrace{\alpha
\left[  \mathbf{e}\right]  }_{=\beta\left[  \mathbf{e}\right]  }\right)
=\left(  Y\left(  \phi\right)  \right)  \left(  \beta\left[  \mathbf{e}%
\right]  \right)  $, we obtain $\alpha\left(  u\right)  =\beta\left(
u\right)  $. Thus, $\alpha\left(  u\right)  =\beta\left(  u\right)  $ is proven.

Now, forget that we fixed $u$. We thus have shown that $\alpha\left(
u\right)  =\beta\left(  u\right)  $ for each $u\in X\left(  V\right)  $. In
other words, the $V$-component of $\alpha$ equals the $V$-component of $\beta$.

Now, forget that we fixed $V$. We thus have proven that the $V$-component of
$\alpha$ equals the $V$-component of $\beta$ for each $V\in\mathcal{V}$. In
other words, $\alpha=\beta$. This proves Lemma 8.7b. $\square$ ]

\item \textbf{Corollary 8.8:} The map $\pi$ should be defined! I guess you
want to define it as follows: \textquotedblleft Let $\pi:\operatorname*{Base}%
\rightarrow\operatorname*{Flag}$ be the natural transformation whose
$V$-component (for any given $V\in\mathcal{V}$) is the map
$\operatorname*{Base}\left(  V\right)  \rightarrow\operatorname*{Flag}\left(
V\right)  $ sending each basis $\left(  v_{1},v_{2},\ldots,v_{n}\right)
\in\operatorname*{Base}\left(  V\right)  $ to the flag
\begin{align*}
&  \left(  0=\operatorname*{span}\left\{  {}\right\}  <\operatorname*{span}%
\left\{  v_{1}\right\}  <\operatorname*{span}\left\{  v_{1},v_{2}\right\}
<\cdots<\operatorname*{span}\left\{  v_{1},v_{2},\ldots,v_{n}\right\}
=V\right) \\
&  \in\operatorname*{Flag}\left(  V\right)  .
\end{align*}
\textquotedblright\ (At least, this is the definition you give later, in
Definition 9.1.)

\item \textbf{Proof of Corollary 8.8:} I would replace the proof by the
following (clearer) argument:

[\textit{Proof of Corollary 8.8.} First of all, the diagram makes sense, since
all its arrows are well-defined natural transformations. It thus remains to
prove that it commutes. In other words, it remains to prove that $T_{\sigma
}^{t}\circ\pi=\pi\circ\left(  \sum_{x\in X\left(  \sigma\right)  }\left(
x\sigma\right)  ^{\ast}\right)  $.

Let $\mathbf{e}\in\operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)  $ be
the basis $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ of $\mathbb{F}_{p}^{n}$.
Then, $\pi\left(  \mathbf{e}\right)  =\underline{E}$.

Now, the natural transformation $\pi:\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \rightarrow\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  $ satisfies $\pi\left[  \mathbf{e}\right]
=\left[  \underbrace{\pi\left(  \mathbf{e}\right)  }_{=\underline{E}}\right]
=\left[  \underline{E}\right]  $. Hence,%
\[
\left(  T_{\sigma}^{t}\circ\pi\right)  \left[  \mathbf{e}\right]
=\underbrace{T_{\sigma}^{t}}_{=T_{\sigma^{-1}}}\left(  \underbrace{\pi\left[
\mathbf{e}\right]  }_{=\left[  \underline{E}\right]  }\right)  =T_{\sigma
^{-1}}\left[  \underline{E}\right]  =\sum_{x\in X\left(  \sigma\right)
}\left[  x\sigma\underline{E}\right]
\]
(by (\ref{pf.l8.7.intermed})). Comparing this with%
\begin{align*}
\left(  \pi\circ\left(  \sum_{x\in X\left(  \sigma\right)  }\left(
x\sigma\right)  ^{\ast}\right)  \right)  \left[  \mathbf{e}\right]   &
=\pi\underbrace{\left(  \left(  \sum_{x\in X\left(  \sigma\right)  }\left(
x\sigma\right)  ^{\ast}\right)  \left[  \mathbf{e}\right]  \right)  }%
_{=\sum_{x\in X\left(  \sigma\right)  }\left(  x\sigma\right)  ^{\ast}\left[
\mathbf{e}\right]  }=\pi\left(  \sum_{x\in X\left(  \sigma\right)  }\left(
x\sigma\right)  ^{\ast}\left[  \mathbf{e}\right]  \right) \\
&  =\pi\left(  \sum_{x\in X\left(  \sigma\right)  }\left(  x\sigma\right)
^{\ast}\left[  \mathbf{e}\right]  \right)  =\sum_{x\in X\left(  \sigma\right)
}\underbrace{\pi\left(  \left(  x\sigma\right)  ^{\ast}\left[  \mathbf{e}%
\right]  \right)  }_{=\left[  \pi\left(  \left(  x\sigma\right)  ^{\ast
}\mathbf{e}\right)  \right]  }\\
&  =\sum_{x\in X\left(  \sigma\right)  }\left[  \underbrace{\pi\left(  \left(
x\sigma\right)  ^{\ast}\mathbf{e}\right)  }_{\substack{=x\sigma\underline{E}%
\\\text{(by Lemma 8.7a \textbf{(b)}}\\\text{(applied to }g=x\sigma\text{))}%
}}\right]  =\sum_{x\in X\left(  \sigma\right)  }\left[  x\sigma\underline{E}%
\right]  ,
\end{align*}
we obtain%
\[
\left(  T_{\sigma}^{t}\circ\pi\right)  \left[  \mathbf{e}\right]  =\left(
\pi\circ\left(  \sum_{x\in X\left(  \sigma\right)  }\left(  x\sigma\right)
^{\ast}\right)  \right)  \left[  \mathbf{e}\right]  .
\]
Hence, Lemma 8.7b (applied to $Y=\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\right]  $, $X=\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  $, $\alpha=T_{\sigma}^{t}\circ\pi$ and $\beta
=\pi\circ\left(  \sum_{x\in X\left(  \sigma\right)  }\left(  x\sigma\right)
^{\ast}\right)  $) yields that $T_{\sigma}^{t}\circ\pi=\pi\circ\left(
\sum_{x\in X\left(  \sigma\right)  }\left(  x\sigma\right)  ^{\ast}\right)  $.
This completes the proof of Corollary 8.8. $\square$ ]

\item \textbf{Proof of Corollary 8.9:} After \textquotedblleft The right hand
side is $\sum_{z\in X\left(  \sigma\tau\right)  }\left[  gz\sigma\tau
B\right]  $\textquotedblright, add \textquotedblleft(by Lemma 8.7, applied to
$\sigma\tau$ instead of $\sigma$)\textquotedblright.

\item \textbf{Corollary 8.10:} The notion of a \textquotedblleft reduced word
for $\sigma$\textquotedblright\ should be explained. (It means a reduced word
$w\in W$ such that $\pi\left(  w\right)  =\sigma$, where $\pi$ is the map
$W\rightarrow\Sigma$ introduced in \S 2.)

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft Now
consider $T_{i}^{2}$\textquotedblright\ by \textquotedblleft Now fix
$i\in\left\{  1,2,\ldots,n-1\right\}  $ and consider $T_{i}^{2}$%
\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} After \textquotedblleft so
$\left\vert A\right\vert =p+1$\textquotedblright, add \textquotedblleft%
(because the elements of $A$ are in bijection with the nonzero proper
subspaces of the $2$-dimensional $\mathbb{F}_{p}$-vector space $U_{i+1}%
/U_{i-1}$, and because the number of the latter subspaces is $p+1$%
)\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft Put%
\[
\phi\left(  W\right)  =\left(  0=U_{0}<\cdots<U_{i-1}<W<U_{i+1}<\cdots
<U_{n}=V\right)
\]
\textquotedblright\ by \textquotedblleft For each $W\in A$, put%
\[
\phi\left(  W\right)  =\left(  0=U_{0}<\cdots<U_{i-1}<W<U_{i+1}<\cdots
<U_{n}=V\right)  \in\operatorname*{Flag}\left(  V\right)  .
\]
\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft One checks
that the flags with $\delta\left(  \underline{W},\underline{U}\right)  =s_{i}$
are\textquotedblright\ by \textquotedblleft Proposition 4.7 shows that the
flags $\underline{W}$ with $\delta\left(  \underline{U},\underline{W}\right)
=s_{i}$ are\textquotedblright. (I have made three changes here. The
replacement of \textquotedblleft$\delta\left(  \underline{W},\underline{U}%
\right)  $\textquotedblright\ by \textquotedblleft$\delta\left(
\underline{U},\underline{W}\right)  $\textquotedblright\ is due to the fact
that the explicit formula for $T_{i}$ shows $T_{i}\left[  \underline{U}%
\right]  =\sum_{\delta\left(  \underline{U},\underline{W}\right)  =s_{i}%
}\left[  \underline{W}\right]  $ rather than $T_{i}\left[  \underline{U}%
\right]  =\sum_{\delta\left(  \underline{W},\underline{U}\right)  =s_{i}%
}\left[  \underline{W}\right]  $, even though both formulas are equivalent
upon a closer look.

\item \textbf{Proof of Proposition 8.11:} I would replace \textquotedblleft It
follows that $T_{i}\left[  \underline{U}\right]  =\sum_{W\neq U_{i}}\left[
\phi\left(  W\right)  \right]  $\textquotedblright\ by \textquotedblleft It
follows that $\sum_{\delta\left(  \underline{U},\underline{W}\right)  =s_{i}%
}\left[  \underline{W}\right]  =\sum_{W\neq U_{i}}\left[  \phi\left(
W\right)  \right]  $. Hence, $T_{i}\left[  \underline{U}\right]  =T_{s_{i}%
}\left[  \underline{U}\right]  =\sum_{\delta\left(  \underline{U}%
,\underline{W}\right)  =s_{i}}\left[  \underline{W}\right]  =\sum_{W\neq
U_{i}}\left[  \phi\left(  W\right)  \right]  $\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft as
claimed.\textquotedblright\ by \textquotedblleft and thus the first relation
$T_{i}^{2}=p+\left(  p-1\right)  T_{i}$ holds. (Alternatively, this also
follows from Proposition 8.4.)\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft Now let
$\mathcal{H}^{\prime}$ be generated\textquotedblright\ by \textquotedblleft
Now let $\mathcal{H}^{\prime}$ be the $\mathbb{Z}_{\left(  p\right)  }%
$-algebra generated\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} After \textquotedblleft subject only
to the relations in the statement of the proposition\textquotedblright, add
\textquotedblleft(with the $T_{k}$ replaced by $T_{k}^{\prime}$%
)\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft and the
$T_{i}$ generate\textquotedblright\ by \textquotedblleft and the $T_{i}$
generate $\mathcal{H}$\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft ring map
$\theta$\textquotedblright\ by \textquotedblleft$\mathbb{Z}_{\left(  p\right)
}$-algebra map $\theta$\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} You write: \textquotedblleft any
reduced word $u=s_{i_{1}}\cdots s_{i_{r}}$ such that $\pi\left(  u\right)
=\sigma$\textquotedblright. Here, $\pi$ denotes the map $\pi:W\rightarrow
\Sigma$ from \S 2, not the natural transformation $\pi:\operatorname*{Base}%
\rightarrow\operatorname*{Flag}$ from Corollary 8.8. I think this should be explained.

\item \textbf{Proof of Proposition 8.11:} After \textquotedblleft This is
well-defined\textquotedblright, add \textquotedblleft(i.e., independent of the
choice of $u$)\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft Define a
map $\phi$\textquotedblright\ by \textquotedblleft Define a $\mathbb{Z}%
_{\left(  p\right)  }$-linear map $\phi$\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft To see
this, consider an element $T_{\sigma}^{\prime}\in A$\textquotedblright\ by
\textquotedblleft To see this, it suffices to check that $T_{\sigma}^{\prime
}T_{i}^{\prime}\in A$ for each $\sigma\in\Sigma_{n}$ (since the $\mathbb{Z}%
_{\left(  p\right)  }$-module $A$ is spanned by the $\phi\left(  T_{\sigma
}\right)  =T_{\sigma}^{\prime}$ for $\sigma\in\Sigma_{n}$). Choose any
$\sigma\in\Sigma_{n}$\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft If
$\sigma\left(  i\right)  >\sigma\left(  i+1\right)  $\textquotedblright\ by
\textquotedblleft If $\sigma\left(  i\right)  <\sigma\left(  i+1\right)
$\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft We choose
any reduced word\textquotedblright\ by \textquotedblleft In this case, we
choose any reduced word\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Before the long equation that begins
with \textquotedblleft$T_{\sigma}^{\prime}T_{i}^{\prime}=T_{\tau}^{\prime
}\left(  T_{i}^{\prime}\right)  ^{2}$\textquotedblright, I would add
\textquotedblleft$T_{\sigma}^{\prime}=T_{\tau}^{\prime}T_{i}^{\prime}$ and
thus\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} I would replace \textquotedblleft%
$pT_{\tau}^{\prime}+\left(  p-1\right)  T_{\tau s_{i}}^{\prime}\in
A$\textquotedblright\ by \textquotedblleft$pT_{\tau}^{\prime}+\left(
p-1\right)  \underbrace{T_{\tau}^{\prime}T_{i}^{\prime}}_{=T_{\sigma}^{\prime
}}=pT_{\tau}^{\prime}+\left(  p-1\right)  T_{\sigma}^{\prime}\in
A$\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} I would replace \textquotedblleft%
$1\in A$\textquotedblright\ by \textquotedblleft$1=T_{\operatorname*{id}%
}^{\prime}\in A$\textquotedblright.

\item \textbf{Proof of Proposition 8.11:} Replace \textquotedblleft
injectve\textquotedblright\ by \textquotedblleft injective\textquotedblright.

\item \textbf{\S 9:} I suggest using the LaTeX syntax \texttt{%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
operatorname\{St\}} instead of \texttt{%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
text\{St\}} in order to achieve the \textquotedblleft$\operatorname*{St}%
$\textquotedblright\ subscripts. Otherwise, these subscripts are italicized
whenever they appear inside propositions (because text in propositions is italicized).

\item \textbf{\S 9:} At the very beginning of \S 9, I would add the following
lemma (which is tacitly used in the definition of $\omega$):

\textbf{Lemma 9.0a.} We have $\left\vert G/U\right\vert \equiv\left(
-1\right)  ^{n}\operatorname{mod}p$ and thus $\left\vert G/U\right\vert
^{-1}\in\mathbb{Z}_{\left(  p\right)  }$.

[\textit{Proof of Lemma 9.0a.} Lemma 8.0a \textbf{(b)} (applied to
$V=\mathbb{F}_{p}^{n}$) yields $\left\vert \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right\vert \equiv1\operatorname{mod}p$ and
$\left\vert \operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right\vert
^{-1}\in\mathbb{Z}_{\left(  p\right)  }$.

Proposition 8.1b \textbf{(b)} shows that there is a natural isomorphism
$G/B\rightarrow\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $ of
$G$-sets. Hence, $\left\vert G/B\right\vert =\left\vert \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right\vert \equiv1\operatorname{mod}p$.

But from $\left\vert G/U\right\vert =\left\vert G\right\vert /\left\vert
U\right\vert $ and $\left\vert G/B\right\vert =\left\vert G\right\vert
/\left\vert B\right\vert $, we obtain%
\begin{align}
\dfrac{\left\vert G/U\right\vert }{\left\vert G/B\right\vert }  &
=\dfrac{\left\vert G\right\vert /\left\vert U\right\vert }{\left\vert
G\right\vert /\left\vert B\right\vert }=\dfrac{\left\vert B\right\vert
}{\left\vert U\right\vert }=\dfrac{\left(  p-1\right)  ^{n}p^{n\left(
n-1\right)  /2}}{p^{n\left(  n-1\right)  /2}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\vert B\right\vert =\left(
p-1\right)  ^{n}p^{n\left(  n-1\right)  /2}\text{ and }\left\vert U\right\vert
=p^{n\left(  n-1\right)  /2}\right) \nonumber\\
&  =\left(  p-1\right)  ^{n}. \label{pf.l9.0a.1}%
\end{align}
Thus,%
\[
\left\vert G/U\right\vert =\left(  \underbrace{p-1}_{\equiv
-1\operatorname{mod}p}\right)  ^{n}\cdot\underbrace{\left\vert G/B\right\vert
}_{\equiv1\operatorname{mod}p}\equiv\left(  -1\right)  ^{n}\operatorname{mod}%
p.
\]
Thus, $\left\vert G/U\right\vert $ is coprime to $p$ (since $\left(
-1\right)  ^{n}$ is coprime to $p$). Hence, $\left\vert G/U\right\vert
^{-1}\in\mathbb{Z}_{\left(  p\right)  }$. This proves Lemma 9.0a. $\square$ ]

\item \textbf{\S 9:} I believe some more work is needed in order to justify
the claim that \textquotedblleft$\operatorname*{End}\left(  \mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)
=\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  ^{\operatorname*{op}}%
$\textquotedblright\ (again, an isomorphism, not a literal equality). Here is
how I would prove this claim:

\textbf{Lemma 9.0b.} Let $\tau:G\rightarrow\operatorname*{Base}\left(
\mathbb{F}_{p}^{n}\right)  $ be the map sending each $g\in G$ to the basis
$\left(  ge_{1},ge_{2},\ldots,ge_{n}\right)  \in\operatorname*{Base}\left(
\mathbb{F}_{p}^{n}\right)  $. This map $\tau$ is well-defined and bijective.

[\textit{Proof of Lemma 9.0b.} If $g\in G$, then $\left(  ge_{1},ge_{2}%
,\ldots,ge_{n}\right)  \in\operatorname*{Base}\left(  \mathbb{F}_{p}%
^{n}\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $g\in G$. Hence, $g$ is an
automorphism of the $\mathbb{F}_{p}$-vector space $\mathbb{F}_{p}^{n}$. But
$\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ is a basis of the $\mathbb{F}_{p}%
$-vector space $\mathbb{F}_{p}^{n}$. Thus, the image of this basis $\left(
e_{1},e_{2},\ldots,e_{n}\right)  $ under $g$ must also be a basis of the
$\mathbb{F}_{p}$-vector space $\mathbb{F}_{p}^{n}$ (since $g$ is an
automorphism of the $\mathbb{F}_{p}$-vector space $\mathbb{F}_{p}^{n}$). In
other words, $\left(  ge_{1},ge_{2},\ldots,ge_{n}\right)  $ must be a basis of
the $\mathbb{F}_{p}$-vector space $\mathbb{F}_{p}^{n}$ (since the image of the
basis $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ under $g$ is $\left(
ge_{1},ge_{2},\ldots,ge_{n}\right)  $). In other words, $\left(  ge_{1}%
,ge_{2},\ldots,ge_{n}\right)  \in\operatorname*{Base}\left(  \mathbb{F}%
_{p}^{n}\right)  $. Qed.}. Hence, the map $\tau$ is well-defined. It remains
to prove that this map $\tau$ is bijective.

The map $\tau$ is injective\footnote{\textit{Proof.} Let $g\in G$ and $h\in G$
be such that $\tau\left(  g\right)  =\tau\left(  h\right)  $. We shall show
that $g=h$.
\par
The definition of $\tau$ yields $\tau\left(  g\right)  =\left(  ge_{1}%
,ge_{2},\ldots,ge_{n}\right)  $ and $\tau\left(  h\right)  =\left(
he_{1},he_{2},\ldots,he_{n}\right)  $. Thus, $\left(  ge_{1},ge_{2}%
,\ldots,ge_{n}\right)  =\tau\left(  g\right)  =\tau\left(  h\right)  =\left(
he_{1},he_{2},\ldots,he_{n}\right)  $. In other words, $ge_{i}=he_{i}$ for
each $i\in\left\{  1,2,\ldots,n\right\}  $. But $g$ and $h$ are elements of
$G=\operatorname*{GL}\nolimits_{n}\left(  \mathbb{F}_{p}\right)  $. Thus, $g$
and $h$ are $\mathbb{F}_{p}$-linear maps. These two $\mathbb{F}_{p}$-linear
maps are equal to each other on each entry of the basis $\left(  e_{1}%
,e_{2},\ldots,e_{n}\right)  $ of the $\mathbb{F}_{p}$-vector space
$\mathbb{F}_{p}^{n}$ (since $ge_{i}=he_{i}$ for each $i\in\left\{
1,2,\ldots,n\right\}  $). Hence, these two maps must be identical. In other
words, $g=h$.
\par
Now, forget that we fixed $g$ and $h$. We thus have shown that if $g\in G$ and
$h\in G$ are such that $\tau\left(  g\right)  =\tau\left(  h\right)  $, then
$g=h$. In other words, the map $\tau$ is injective. Qed.} and
surjective\footnote{\textit{Proof.} Let $b\in\operatorname*{Base}\left(
\mathbb{F}_{p}^{n}\right)  $. Thus, $b$ is a basis of the $\mathbb{F}_{p}%
$-vector space $\mathbb{F}_{p}^{n}$. Hence, $b$ is a list of $\dim\left(
\mathbb{F}_{p}^{n}\right)  =n$ elements of $\mathbb{F}_{p}^{n}$. Write $b$ in
the form $\left(  b_{1},b_{2},\ldots,b_{n}\right)  $. (This is possible, since
$b$ is a list of $n$ elements of $\mathbb{F}_{p}^{n}$). Thus, $\left(
b_{1},b_{2},\ldots,b_{n}\right)  $ is a basis of the $\mathbb{F}_{p}$-vector
space $\mathbb{F}_{p}^{n}$ (since $\left(  b_{1},b_{2},\ldots,b_{n}\right)
=b\in\operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)  $).
\par
Let $g:\mathbb{F}_{p}^{n}\rightarrow\mathbb{F}_{p}^{n}$ be the unique
$\mathbb{F}_{p}$-linear map that sends each $e_{i}$ (with $i\in\left\{
1,2,\ldots,n\right\}  $) to $b_{i}$. (This is well-defined, since $\left(
e_{1},e_{2},\ldots,e_{n}\right)  $ is a basis of the $\mathbb{F}_{p}$-vector
space $\mathbb{F}_{p}^{n}$.) Then, $ge_{i}=b_{i}$ for each $i\in\left\{
1,2,\ldots,n\right\}  $. Hence, the map $g$ sends the basis $\left(
e_{1},e_{2},\ldots,e_{n}\right)  $ of $\mathbb{F}_{p}^{n}$ to the list
$\left(  b_{1},b_{2},\ldots,b_{n}\right)  $. Therefore, the map $g$ sends a
basis of $\mathbb{F}_{p}^{n}$ to a basis of $\mathbb{F}_{p}^{n}$ (since both
lists $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ and $\left(  b_{1}%
,b_{2},\ldots,b_{n}\right)  $ are bases of $\mathbb{F}_{p}^{n}$). Thus, $g$ is
an isomorphism of $\mathbb{F}_{p}$-vector spaces between $\mathbb{F}_{p}^{n}$
and $\mathbb{F}_{p}^{n}$. In other words, $g$ is an automorphism of the
$\mathbb{F}_{p}$-vector space $\mathbb{F}_{p}^{n}$. Thus, $g\in
\operatorname*{Aut}\left(  \mathbb{F}_{p}^{n}\right)  =\operatorname*{GL}%
\nolimits_{n}\left(  \mathbb{F}_{p}\right)  =G$. The definition of $\tau$
yields $\tau\left(  g\right)  =\left(  ge_{1},ge_{2},\ldots,ge_{n}\right)
=\left(  b_{1},b_{2},\ldots,b_{n}\right)  $ (since $ge_{i}=b_{i}$ for each
$i\in\left\{  1,2,\ldots,n\right\}  $). Thus, $\tau\left(  g\right)  =\left(
b_{1},b_{2},\ldots,b_{n}\right)  =b$. Hence, $b=\tau\left(  \underbrace{g}%
_{\in G}\right)  \in\tau\left(  G\right)  $.
\par
Now, forget that we fixed $b$. We thus have proven that $b\in\tau\left(
G\right)  $ for each $b\in\operatorname*{Base}\left(  \mathbb{F}_{p}%
^{n}\right)  $. In other words, $\operatorname*{Base}\left(  \mathbb{F}%
_{p}^{n}\right)  \subseteq\tau\left(  G\right)  $. In other words, the map
$\tau$ is surjective. Qed.}. Hence, the map $\tau$ is bijective. This
completes the proof of Lemma 9.0b. $\square$ ]

\textbf{Lemma 9.0c.} Recall that for every $g\in G$, we have defined a natural
transformation $g^{\ast}:\operatorname*{Base}\Longrightarrow
\operatorname*{Base}$, and thus we also obtain a natural transformation
$\mathbb{Z}_{\left(  p\right)  }\left[  g^{\ast}\right]  :\mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Base}\right]  \Longrightarrow\mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  $. The latter
natural transformation is an element of the ring $\operatorname*{End}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)
$. (By abuse of notation, we can denote this natural transformation
$\mathbb{Z}_{\left(  p\right)  }\left[  g^{\ast}\right]  :\mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Base}\right]  \Longrightarrow\mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  $ by $g^{\ast}$
again; but we shall not do so in this lemma, because this would risk confusing
it with the natural transformation $g^{\ast}:\operatorname*{Base}%
\Longrightarrow\operatorname*{Base}$.)

Let $\gamma:\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  \rightarrow
\operatorname*{End}\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  $ be the $\mathbb{Z}_{\left(  p\right)
}$-linear map that sends each each $\left[  g\right]  \in G$ to $\mathbb{Z}%
_{\left(  p\right)  }\left[  g^{\ast}\right]  \in\operatorname*{End}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)
$. (This is well-defined, since the family $\left(  \left[  g\right]  \right)
_{g\in G}$ is a basis of the $\mathbb{Z}_{\left(  p\right)  }$-module
$\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  $.)

This map $\gamma$ is a $\mathbb{Z}_{\left(  p\right)  }$-algebra isomorphism
$\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  ^{\operatorname*{op}%
}\rightarrow\operatorname*{End}\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  $. Thus, $\operatorname*{End}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)
\cong\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  ^{\operatorname*{op}}$
as $\mathbb{Z}_{\left(  p\right)  }$-algebras.

[\textit{Proof of Lemma 9.0c.} Consider the map $\tau:G\rightarrow
\operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)  $ defined in Lemma
9.0b. Lemma 9.0b shows that this map $\tau$ is well-defined and bijective.
Hence, $\tau$ is an isomorphism of sets. Thus, it induces an isomorphism
$\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  :\mathbb{Z}_{\left(
p\right)  }\left[  G\right]  \rightarrow\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)  \right]  $ of
$\mathbb{Z}_{\left(  p\right)  }$-modules.

Let $\mathbf{e}\in\operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)  $ be
the basis $\left(  e_{1},e_{2},\ldots,e_{n}\right)  $ of $\mathbb{F}_{p}^{n}$.
Then,%
\begin{equation}
g^{\ast}\mathbf{e}=\tau\left(  g\right)  \ \ \ \ \ \ \ \ \ \ \text{for every
}g\in G \label{pf.l9.0c.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.l9.0c.1}):} Let $g\in G$. Then,
$\tau\left(  g\right)  =\left(  ge_{1},ge_{2},\ldots,ge_{n}\right)  $ (by the
definition of $\tau$). Comparing this with $g^{\ast}\mathbf{e}=\left(
ge_{1},ge_{2},\ldots,ge_{n}\right)  $ (by Lemma 8.7a \textbf{(a)}), we obtain
$g^{\ast}\mathbf{e}=\tau\left(  g\right)  $. This proves (\ref{pf.l9.0c.1}).}.

For every $u\in\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  $, the element
$\gamma\left(  u\right)  \in\operatorname*{End}\left(  \mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Base}\right]  \right)  $ can be applied to
the element $\left[  \mathbf{e}\right]  \in\mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)  \right]
=\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]
\right)  \left(  \mathbb{F}_{p}^{n}\right)  $, and the result is a new element
$\gamma\left(  u\right)  \cdot\left[  \mathbf{e}\right]  $ of $\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)
\left(  \mathbb{F}_{p}^{n}\right)  =\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)  \right]  $. We have
\begin{equation}
\gamma\left(  u\right)  \cdot\left[  \mathbf{e}\right]  =\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  \right)  \left(  u\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }u\in\mathbb{Z}_{\left(  p\right)
}\left[  G\right]  \label{pf.l9.0c.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.l9.0c.2}):} Let $u\in\mathbb{Z}_{\left(
p\right)  }\left[  G\right]  $. We must prove the equality (\ref{pf.l9.0c.2}).
\par
Both $\gamma\left(  u\right)  $ and $\left(  \mathbb{Z}_{\left(  p\right)
}\left[  \tau\right]  \right)  \left(  u\right)  $ depend $\mathbb{Z}_{\left(
p\right)  }$-linearly on $u$. Hence, the equality (\ref{pf.l9.0c.2}) is
$\mathbb{Z}_{\left(  p\right)  }$-linear in $u$. Thus, for the proof of this
equality, we can WLOG assume that $u$ belongs to the basis $\left(  \left[
g\right]  \right)  _{g\in G}$ of the $\mathbb{Z}_{\left(  p\right)  }$-module
$\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  $. Assume this. Hence,
$u=\left[  g\right]  $ for some $g\in G$. Consider this $g$.
\par
Now, $\gamma\left(  \underbrace{u}_{=\left[  g\right]  }\right)
=\gamma\left(  \left[  g\right]  \right)  =\mathbb{Z}_{\left(  p\right)
}\left[  g^{\ast}\right]  $ (by the definition of $\gamma$). Hence,
$\underbrace{\gamma\left(  u\right)  }_{=\mathbb{Z}_{\left(  p\right)
}\left[  g^{\ast}\right]  }\cdot\left[  \mathbf{e}\right]  =\left(
\mathbb{Z}_{\left(  p\right)  }\left[  g^{\ast}\right]  \right)  \left[
\mathbf{e}\right]  =\left[  \underbrace{g^{\ast}\mathbf{e}}_{\substack{=\tau
\left(  g\right)  \\\text{(by (\ref{pf.l9.0c.1}))}}}\right]  =\left[
\tau\left(  g\right)  \right]  $.
\par
On the other hand, $\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\tau\right]  \right)  \left(  \underbrace{u}_{=\left[  g\right]  }\right)
=\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  \right)  \left(
\left[  g\right]  \right)  =\left[  \tau\left(  g\right)  \right]  $ (by the
definition of $\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  $).
Comparing this with $\gamma\left(  u\right)  \cdot\left[  \mathbf{e}\right]
=\left[  \tau\left(  g\right)  \right]  $, we obtain $\gamma\left(  u\right)
\cdot\left[  \mathbf{e}\right]  =\left(  \mathbb{Z}_{\left(  p\right)
}\left[  \tau\right]  \right)  \left(  u\right)  $. This proves
(\ref{pf.l9.0c.2}).}.

The map $\gamma$ is injective\footnote{\textit{Proof.} Let $u\in
\operatorname*{Ker}\gamma$. Thus, $u\in\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  $ and $\gamma\left(  u\right)  =0$. But (\ref{pf.l9.0c.2}) yields
$\gamma\left(  u\right)  \cdot\left[  \mathbf{e}\right]  =\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  \right)  \left(  u\right)
$. Hence, $\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]
\right)  \left(  u\right)  =\underbrace{\gamma\left(  u\right)  }_{=0}%
\cdot\left[  \mathbf{e}\right]  =0$, so that $u\in\operatorname*{Ker}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  \right)  $.
\par
But the map $\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  $ is an
isomorphism. Thus, $\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  $ is
injective, so that $\operatorname*{Ker}\left(  \mathbb{Z}_{\left(  p\right)
}\left[  \tau\right]  \right)  =0$. Hence, $u\in\operatorname*{Ker}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  \right)  =0$, so that
$u=0$.
\par
Now, forget that we fixed $u$. We thus have shown that $u=0$ for each
$u\in\operatorname*{Ker}\gamma$. In other words, $\operatorname*{Ker}\gamma
=0$. Hence, the map $\gamma$ is injective (since $\gamma$ is $\mathbb{Z}%
_{\left(  p\right)  }$-linear). Qed.} and surjective\footnote{\textit{Proof.}
Let $X$ be the functor $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \in\mathcal{VA}$. Thus, $X\left(  \mathbb{F}%
_{p}^{n}\right)  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  \left(  \mathbb{F}_{p}^{n}\right)
=\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\left(
\mathbb{F}_{p}^{n}\right)  \right]  $. Hence, $\left[  \mathbf{e}\right]
\in\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\left(
\mathbb{F}_{p}^{n}\right)  \right]  =X\left(  \mathbb{F}_{p}^{n}\right)  $.
\par
Let $\alpha\in\operatorname*{End}\left(  X\right)  $. Then, $\alpha$ can be
applied to the element $\left[  \mathbf{e}\right]  $ of $X\left(
\mathbb{F}_{p}^{n}\right)  $. The result is an element $\alpha\left[
\mathbf{e}\right]  \in X\left(  \mathbb{F}_{p}^{n}\right)  =\mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Base}\left(  \mathbb{F}_{p}%
^{n}\right)  \right]  $.
\par
The map $\mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  :\mathbb{Z}%
_{\left(  p\right)  }\left[  G\right]  \rightarrow\mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)
\right]  $ is an isomorphism, and thus is surjective. Hence, $\mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Base}\left(  \mathbb{F}_{p}%
^{n}\right)  \right]  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\tau\right]  \right)  \left(  \mathbb{Z}_{\left(  p\right)  }\left[  G\right]
\right)  $. Thus, $\alpha\left[  \mathbf{e}\right]  \in\mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Base}\left(  \mathbb{F}_{p}^{n}\right)
\right]  =\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \tau\right]  \right)
\left(  \mathbb{Z}_{\left(  p\right)  }\left[  G\right]  \right)  $. In other
words, there exists an $u\in\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  $
such that $\alpha\left[  \mathbf{e}\right]  =\left(  \mathbb{Z}_{\left(
p\right)  }\left[  \tau\right]  \right)  \left(  u\right)  $. Consider this
$u$.
\par
We have $\alpha\in\operatorname*{End}X$ and $\gamma\left(  u\right)
\in\operatorname*{End}\left(  \underbrace{\mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Base}\right]  }_{=X}\right)  =\operatorname*{End}X$.
Thus, both $\alpha$ and $\gamma\left(  u\right)  $ are elements of
$\operatorname*{End}X$. In other words, both $\alpha$ and $\gamma\left(
u\right)  $ are natural transformations $X\Longrightarrow X$. Also,
$\alpha\left[  \mathbf{e}\right]  =\left(  \mathbb{Z}_{\left(  p\right)
}\left[  \tau\right]  \right)  \left(  u\right)  =\gamma\left(  u\right)
\cdot\left[  \mathbf{e}\right]  $ (by (\ref{pf.l9.0c.2})). Hence, Lemma 8.7b
(applied to $Y=X$ and $\beta=\gamma\left(  u\right)  $) yields $\alpha
=\gamma\left(  \underbrace{u}_{\in\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  }\right)  \in\gamma\left(  \mathbb{Z}_{\left(  p\right)  }\left[
G\right]  \right)  $.
\par
Now, forget that we fixed $\alpha$. We thus have proven that $\alpha\in
\gamma\left(  \mathbb{Z}_{\left(  p\right)  }\left[  G\right]  \right)  $ for
each $\alpha\in\operatorname*{End}\left(  X\right)  $. In other words,
$\operatorname*{End}\left(  X\right)  \subseteq\gamma\left(  \mathbb{Z}%
_{\left(  p\right)  }\left[  G\right]  \right)  $. Since $X=\mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  $, this rewrites as
$\operatorname*{End}\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  \subseteq\gamma\left(  \mathbb{Z}%
_{\left(  p\right)  }\left[  G\right]  \right)  $. In other words, the map
$\gamma$ is surjective. Qed.}. Hence, the map $\gamma$ is bijective. Also,
$\gamma$ is a $\mathbb{Z}_{\left(  p\right)  }$-linear map $\mathbb{Z}%
_{\left(  p\right)  }\left[  G\right]  \rightarrow\operatorname*{End}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)
$. In other words, $\gamma$ is a $\mathbb{Z}_{\left(  p\right)  }$-linear map
$\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  ^{\operatorname*{op}%
}\rightarrow\operatorname*{End}\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  $ (since $\mathbb{Z}_{\left(  p\right)
}\left[  G\right]  ^{\operatorname*{op}}=\mathbb{Z}_{\left(  p\right)
}\left[  G\right]  $ as a $\mathbb{Z}_{\left(  p\right)  }$-module).

We have $\gamma\left(  1\right)  =\operatorname*{id}\nolimits_{\mathbb{Z}%
_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  }$%
\ \ \ \ \footnote{\textit{Proof.} The definition of $\gamma$ yields
$\gamma\left(  1\right)  =\mathbb{Z}_{\left(  p\right)  }\left[
\underbrace{1^{\ast}}_{=\operatorname*{id}\nolimits_{\operatorname*{Base}}%
}\right]  =\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{id}%
\nolimits_{\operatorname*{Base}}\right]  =\operatorname*{id}%
\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]
}$ (by Remark 8.1f \textbf{(a)}, applied to $P=\operatorname*{Base}$). Qed.}.
Also, recall that any $g\in G$ and $h\in G$ satisfy
\begin{equation}
\left(  gh\right)  ^{\ast}=h^{\ast}g^{\ast}. \label{pf.l9.0c.6}%
\end{equation}
Now, any $v\in\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  $ and
$u\in\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  $ satisfy%
\begin{equation}
\gamma\left(  uv\right)  =\gamma\left(  v\right)  \circ\gamma\left(  u\right)
\label{pf.l9.0c.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.l9.0c.7}):} Let $v\in\mathbb{Z}_{\left(
p\right)  }\left[  G\right]  $ and $u\in\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  $. We must prove the equality (\ref{pf.l9.0c.7}). This equality is
$\mathbb{Z}_{\left(  p\right)  }$-linear in each of $u$ and $v$ (since
$\gamma$ is a $\mathbb{Z}_{\left(  p\right)  }$-linear map). Hence, for the
proof of this equality, we can WLOG assume that both $u$ and $v$ belong to the
basis $\left(  \left[  g\right]  \right)  _{g\in G}$ of the $\mathbb{Z}%
_{\left(  p\right)  }$-module $\mathbb{Z}_{\left(  p\right)  }\left[
G\right]  $. Assume this. Thus, $u=\left[  g\right]  $ and $v=\left[
h\right]  $ for some elements $g\in G$ and $h\in G$. Consider these $g$ and
$h$. We have%
\begin{align*}
\gamma\left(  \underbrace{u}_{=\left[  g\right]  }\underbrace{v}_{=\left[
h\right]  }\right)   &  =\gamma\left(  \underbrace{\left[  g\right]  \left[
h\right]  }_{=\left[  gh\right]  }\right)  =\gamma\left(  \left[  gh\right]
\right) \\
&  =\mathbb{Z}_{\left(  p\right)  }\left[  \underbrace{\left(  gh\right)
^{\ast}}_{\substack{=h^{\ast}g^{\ast}\\\text{(by (\ref{pf.l9.0c.6}))}%
}}\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }%
\gamma\right) \\
&  =\mathbb{Z}_{\left(  p\right)  }\left[  h^{\ast}g^{\ast}\right]  .
\end{align*}
Comparing this with%
\begin{align*}
\gamma\left(  \underbrace{v}_{=\left[  h\right]  }\right)  \circ\gamma\left(
\underbrace{u}_{=\left[  g\right]  }\right)   &  =\underbrace{\gamma\left(
\left[  h\right]  \right)  }_{\substack{=\mathbb{Z}_{\left(  p\right)
}\left[  h^{\ast}\right]  \\\text{(by the definition of }\gamma\text{)}}%
}\circ\underbrace{\gamma\left(  \left[  g\right]  \right)  }%
_{\substack{=\mathbb{Z}_{\left(  p\right)  }\left[  g^{\ast}\right]
\\\text{(by the definition of }\gamma\text{)}}}=\mathbb{Z}_{\left(  p\right)
}\left[  h^{\ast}\right]  \circ\mathbb{Z}_{\left(  p\right)  }\left[  g^{\ast
}\right] \\
&  =\mathbb{Z}_{\left(  p\right)  }\left[  h^{\ast}\circ g^{\ast}\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Remark 8.1f \textbf{(b)}, applied to }P=\operatorname*{Base}\text{,
}Q=\operatorname*{Base}\text{, }R=\operatorname*{Base}\text{,}\\
\alpha_{1}=g^{\ast}\text{ and }\alpha_{2}=h^{\ast}%
\end{array}
\right) \\
&  =\mathbb{Z}_{\left(  p\right)  }\left[  h^{\ast}g^{\ast}\right]  ,
\end{align*}
we obtain $\gamma\left(  uv\right)  =\gamma\left(  v\right)  \circ
\gamma\left(  u\right)  $. This proves (\ref{pf.l9.0c.7}).}. In other words,
$\gamma$ is multiplicative when viewed as a map $\mathbb{Z}_{\left(  p\right)
}\left[  G\right]  ^{\operatorname*{op}}\rightarrow\operatorname*{End}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)  $
(because $uv$ is the product of $v$ and $u$ in the $\mathbb{Z}_{\left(
p\right)  }$-algebra $\mathbb{Z}_{\left(  p\right)  }\left[  G\right]
^{\operatorname*{op}}$). Combining this with $\gamma\left(  1\right)
=\operatorname*{id}\nolimits_{\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  }$, we conclude that $\gamma$ is a $\mathbb{Z}%
_{\left(  p\right)  }$-algebra homomorphism $\mathbb{Z}_{\left(  p\right)
}\left[  G\right]  ^{\operatorname*{op}}\rightarrow\operatorname*{End}\left(
\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]  \right)  $
(because $\gamma$ is a $\mathbb{Z}_{\left(  p\right)  }$-linear map
$\mathbb{Z}_{\left(  p\right)  }\left[  G\right]  ^{\operatorname*{op}%
}\rightarrow\operatorname*{End}\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  $). Since $\gamma$ is bijective, we thus
conclude that $\gamma$ is a $\mathbb{Z}_{\left(  p\right)  }$-algebra
isomorphism $\mathbb{Z}_{\left(  p\right)  }\left[  G\right]
^{\operatorname*{op}}\rightarrow\operatorname*{End}\left(  \mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Base}\right]  \right)  $. Thus,
$\operatorname*{End}\left(  \mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  \right)  \cong\mathbb{Z}_{\left(  p\right)
}\left[  G\right]  ^{\operatorname*{op}}$ as $\mathbb{Z}_{\left(  p\right)  }%
$-algebras. This proves Lemma 9.0c. $\square$ ]

\item \textbf{Proposition 9.2:} Replace \textquotedblleft The
map\textquotedblright\ by \textquotedblleft The $\mathbb{Z}_{\left(  p\right)
}$-linear map\textquotedblright.

\item \textbf{Proof of Proposition 9.2:} After \textquotedblleft in the $i$'th
space\textquotedblright, add \textquotedblleft(whenever $\underline{v}%
\in\operatorname*{Base}\left(  V\right)  $)\textquotedblright.

\item \textbf{Proof of Proposition 9.2:} Replace \textquotedblleft a reduced
word $s_{i_{1}}\cdots s_{i_{r}}$ for $\Sigma$\textquotedblright\ by
\textquotedblleft a reduced word $s_{i_{1}}\cdots s_{i_{r}}$ for $\sigma
$\textquotedblright.

\item \textbf{Proof of Proposition 9.2:} Replace every appearance of
\textquotedblleft$\widehat{\xi}$\textquotedblright\ in this proof by
\textquotedblleft$\xi$\textquotedblright.

\item \textbf{Proof of Proposition 9.2:} After \textquotedblleft$\left(
-1\right)  ^{r}\mu=\widehat{\xi}\left(  T_{\sigma}\right)  \mu$%
\textquotedblright, add \textquotedblleft(since $\left(  -1\right)
^{r}=\operatorname*{sgn}\left(  \sigma\right)  =\xi\left(  T_{\sigma}\right)
$)\textquotedblright.

\item \textbf{Proof of Proposition 9.2:} After \textquotedblleft It follows
that $\xi\left(  ab\right)  \mu=\xi\left(  a\right)  \xi\left(  b\right)  \mu
$\textquotedblright, add \textquotedblleft(since $ab\mu=\xi\left(  ab\right)
\mu$ and thus $\xi\left(  ab\right)  \mu=a\underbrace{b\mu}_{=\xi\left(
b\right)  \mu}=a\xi\left(  b\right)  \mu=\xi\left(  b\right)  \underbrace{a\mu
}_{=\xi\left(  a\right)  \mu}=\xi\left(  b\right)  \xi\left(  a\right)
\mu=\xi\left(  a\right)  \xi\left(  b\right)  \mu$)\textquotedblright.

\item \textbf{Proof of Proposition 9.2:} You write: \textquotedblleft(This
could also have been deduced from Proposition 8.11.)\textquotedblright. A few
details about this deduction would be useful. Namely, here is how it works:

[\textit{Proof of the fact that }$\xi$ \textit{is a ring map:} Clearly, we
have%
\begin{align*}
\left(  -1\right)  ^{2}  &  =p+\left(  p-1\right)  \left(  -1\right)  ,\\
\left(  -1\right)  \left(  -1\right)  \left(  -1\right)   &  =\left(
-1\right)  \left(  -1\right)  \left(  -1\right)  ,\\
\left(  -1\right)  \left(  -1\right)   &  =\left(  -1\right)  \left(
-1\right)  .
\end{align*}
Thus, the relations in Proposition 8.11 remain valid if each $T_{k}$ in them
is replaced by $-1$. Hence, Proposition 8.11 shows that there exists a unique
$\mathbb{Z}_{\left(  p\right)  }$-algebra homomorphism $\eta:\mathcal{H}%
\rightarrow\mathbb{Z}_{\left(  p\right)  }$ that sends each $T_{k}$ to $-1$.
Consider this $\eta$. Now, if $\sigma\in\Sigma_{n}$, then we can fix any
reduced word $s_{i_{1}}s_{i_{2}}\cdots s_{i_{r}}$ for $\sigma$, and then we
find%
\begin{align*}
\eta\left(  \underbrace{T_{\sigma}}_{\substack{=T_{i_{1}}T_{i_{2}}\cdots
T_{i_{r}}\\\text{(by Corollary 8.10)}}}\right)   &  =\eta\left(  T_{i_{1}%
}T_{i_{2}}\cdots T_{i_{r}}\right)  =\eta\left(  T_{i_{1}}\right)  \eta\left(
T_{i_{2}}\right)  \cdots\eta\left(  T_{i_{r}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta\text{ is a }\mathbb{Z}%
_{\left(  p\right)  }\text{-algebra homomorphism}\right) \\
&  =\underbrace{\left(  -1\right)  \left(  -1\right)  \cdots\left(  -1\right)
}_{r\text{ times}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta\left(  T_{i_{p}}\right)
=-1\text{ for each }p\text{ (by the definition of }\eta\text{)}\right) \\
&  =\left(  -1\right)  ^{r}=\operatorname*{sgn}\left(  \sigma\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\sigma=s_{i_{1}}s_{i_{2}}\cdots
s_{i_{r}}\text{ and thus }\operatorname*{sgn}\left(  \sigma\right)  =\left(
-1\right)  ^{r}\right) \\
&  =\xi\left(  T_{\sigma}\right)  .
\end{align*}
Thus, we have found that $\eta\left(  T_{\sigma}\right)  =\xi\left(
T_{\sigma}\right)  $ for each $\sigma\in\Sigma_{n}$. In other words, the maps
$\eta$ and $\xi$ are equal to each other on the basis $\left(  T_{\sigma
}\right)  _{\sigma\in\Sigma_{n}}$ of the $\mathbb{Z}_{\left(  p\right)  }%
$-module $\mathcal{H}$. Hence, these two maps $\eta$ and $\xi$ must be
identical (since they are both $\mathbb{Z}_{\left(  p\right)  }$-linear). In
other words, $\xi=\eta$. Thus, $\xi$ is a ring homomorphism (since $\eta$ is a
ring homomorphism). $\square$ ]

\item \textbf{Proof of Proposition 9.3:} This proof has several flaws. In
particular, the expression \textquotedblleft$\left\langle \sigma^{\ast}\pi
^{t}\left[  \underline{U}\right]  ,\left[  \underline{W}\right]  \right\rangle
$\textquotedblright\ makes no sense, and the formula $\pi\sigma^{\ast}\pi
^{t}=\left\vert B/U\right\vert p^{l\left(  \sigma^{-1}\rho\right)  }T_{\sigma
}$ is false. Let me show a correct (and more detailed) proof:

[\textit{Proof of Proposition 9.3.} Fix $\sigma\in\Sigma_{n}$. Let
$V\in\mathcal{V}$, and let $\underline{W}\in\operatorname*{Flag}\left(
V\right)  $. For each $\underline{U}\in\operatorname*{Flag}\left(  V\right)
$, we have%
\begin{align*}
&  \left\langle \pi\sigma^{\ast}\pi^{t}\left[  \underline{W}\right]  ,\left[
\underline{U}\right]  \right\rangle \\
&  =\left\langle \pi\sigma^{\ast}\sum_{\substack{\mathbf{w}\in
\operatorname*{Base}\left(  V\right)  ;\\\pi\left(  \mathbf{w}\right)
=\underline{W}}}\left[  \mathbf{w}\right]  ,\left[  \underline{U}\right]
\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi^{t}\left[
\underline{W}\right]  =\sum_{\substack{\mathbf{w}\in\operatorname*{Base}%
\left(  V\right)  ;\\\pi\left(  \mathbf{w}\right)  =\underline{W}}}\left[
\mathbf{w}\right]  \right) \\
&  =\sum_{\substack{\mathbf{w}\in\operatorname*{Base}\left(  V\right)
;\\\pi\left(  \mathbf{w}\right)  =\underline{W}}}\left\langle \underbrace{\pi
\sigma^{\ast}\left[  \mathbf{w}\right]  }_{=\left[  \pi\left(  \sigma^{\ast
}\mathbf{w}\right)  \right]  },\left[  \underline{U}\right]  \right\rangle
=\sum_{\substack{\mathbf{w}\in\operatorname*{Base}\left(  V\right)
;\\\pi\left(  \mathbf{w}\right)  =\underline{W}}}\underbrace{\left\langle
\left[  \pi\left(  \sigma^{\ast}\mathbf{w}\right)  \right]  ,\left[
\underline{U}\right]  \right\rangle }_{=\delta_{\pi\left(  \sigma^{\ast
}\mathbf{w}\right)  ,\underline{U}}}=\sum_{\substack{\mathbf{w}\in
\operatorname*{Base}\left(  V\right)  ;\\\pi\left(  \mathbf{w}\right)
=\underline{W}}}\delta_{\pi\left(  \sigma^{\ast}\mathbf{w}\right)
,\underline{U}}\\
&  =\left(  \text{the number of all }\mathbf{w}\in\operatorname*{Base}\left(
V\right)  \text{ such that }\pi\left(  \mathbf{w}\right)  =\underline{W}\text{
and }\pi\left(  \sigma^{\ast}\mathbf{w}\right)  =\underline{U}\right) \\
&  =\left(  \text{the number of all }\left(  v_{1},v_{2},\ldots,v_{n}\right)
\in\operatorname*{Base}\left(  V\right)  \text{ such }\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{that }\pi\left(  v_{1},v_{2},\ldots
,v_{n}\right)  =\underline{W}\text{ and }\pi\left(  \underbrace{\sigma^{\ast
}\left(  v_{1},v_{2},\ldots,v_{n}\right)  }_{=\left(  v_{\sigma\left(
1\right)  },v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  n\right)
}\right)  }\right)  =\underline{U}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left(  v_{1},v_{2},\ldots,v_{n}\right)
\text{ for the index }\mathbf{w}\text{,}\\
\text{since each element of }\operatorname*{Base}\left(  V\right)  \text{ is
an }n\text{-tuple}%
\end{array}
\right) \\
&  =\left(  \text{the number of all }\left(  v_{1},v_{2},\ldots,v_{n}\right)
\in\operatorname*{Base}\left(  V\right)  \text{ such }\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{that }\underbrace{\pi\left(  v_{1}%
,v_{2},\ldots,v_{n}\right)  =\underline{W}}_{\Longleftrightarrow\ \left(
W_{i}=\operatorname*{span}\left\{  v_{1},v_{2},\ldots,v_{i}\right\}  \text{
for all }i\right)  }\text{ and }\underbrace{\pi\left(  v_{\sigma\left(
1\right)  },v_{\sigma\left(  2\right)  },\ldots,v_{\sigma\left(  n\right)
}\right)  =\underline{U}}_{\Longleftrightarrow\ \left(  U_{i}%
=\operatorname*{span}\left\{  v_{\sigma\left(  1\right)  },v_{\sigma\left(
2\right)  },\ldots,v_{\sigma\left(  i\right)  }\right\}  \text{ for all
}i\right)  }\right) \\
&  =\left(  \text{the number of all }\left(  v_{1},v_{2},\ldots,v_{n}\right)
\in\operatorname*{Base}\left(  V\right)  \text{ such }\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{that }\left(  W_{i}=\operatorname*{span}%
\left\{  v_{1},v_{2},\ldots,v_{i}\right\}  \text{ for all }i\right)  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{and }\left(  U_{i}=\operatorname*{span}%
\left\{  v_{\sigma\left(  1\right)  },v_{\sigma\left(  2\right)  }%
,\ldots,v_{\sigma\left(  i\right)  }\right\}  \text{ for all }i\right)
\right) \\
&  =\left(  \text{the number of all bases }\left(  v_{1},v_{2},\ldots
,v_{n}\right)  \text{ of }V\text{ such that for all }i\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{we have }U_{i}=\operatorname*{span}%
\left\{  v_{\sigma\left(  1\right)  },v_{\sigma\left(  2\right)  }%
,\ldots,v_{\sigma\left(  i\right)  }\right\}  \text{ and }W_{i}%
=\operatorname*{span}\left\{  v_{1},v_{2},\ldots,v_{i}\right\}  \right) \\
&  =%
\begin{cases}
\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)  }, & \text{if
}\delta\left(  \left[  \underline{U}\right]  ,\left[  \underline{W}\right]
\right)  =\sigma;\\
0, & \text{otherwise}%
\end{cases}
\end{align*}
(by Corollary 5.5). Hence,%
\begin{align*}
\pi\sigma^{\ast}\pi^{t}\left[  \underline{W}\right]   &  =\sum_{\underline{U}%
\in\operatorname*{Flag}\left(  V\right)  }%
\begin{cases}
\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)  }, & \text{if
}\delta\left(  \left[  \underline{U}\right]  ,\left[  \underline{W}\right]
\right)  =\sigma;\\
0, & \text{otherwise}%
\end{cases}
\left[  \underline{U}\right] \\
&  =\sum_{\substack{\underline{U}\in\operatorname*{Flag}\left(  V\right)
;\\\delta\left(  \left[  \underline{U}\right]  ,\left[  \underline{W}\right]
\right)  =\sigma}}\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)
}\left[  \underline{U}\right] \\
&  =\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)  }%
\underbrace{\sum_{\substack{\underline{U}\in\operatorname*{Flag}\left(
V\right)  ;\\\delta\left(  \left[  \underline{U}\right]  ,\left[
\underline{W}\right]  \right)  =\sigma}}}_{\substack{=\sum
_{\substack{\underline{U}\in\operatorname*{Flag}\left(  V\right)
;\\\delta\left(  \left[  \underline{W}\right]  ,\left[  \underline{U}\right]
\right)  =\sigma^{-1}}}\\\text{(because for each }\underline{U}\in
\operatorname*{Flag}\left(  V\right)  \text{,}\\\text{the condition }\left(
\delta\left(  \left[  \underline{U}\right]  ,\left[  \underline{W}\right]
\right)  =\sigma\right)  \text{ is}\\\text{equivalent to }\left(
\delta\left(  \left[  \underline{W}\right]  ,\left[  \underline{U}\right]
\right)  =\sigma^{-1}\right)  \\\text{(since }\delta\left(  \left[
\underline{W}\right]  ,\left[  \underline{U}\right]  \right)  =\delta\left(
\left[  \underline{U}\right]  ,\left[  \underline{W}\right]  \right)
^{-1}\text{))}}}\left[  \underline{U}\right] \\
&  =\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)  }%
\sum_{\substack{\underline{U}\in\operatorname*{Flag}\left(  V\right)
;\\\delta\left(  \left[  \underline{W}\right]  ,\left[  \underline{U}\right]
\right)  =\sigma^{-1}}}\left[  \underline{U}\right]  .
\end{align*}
Comparing this with%
\[
\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)  }%
\underbrace{T_{\sigma^{-1}}\left[  \underline{W}\right]  }_{=\sum
_{\substack{\underline{U}\in\operatorname*{Flag}\left(  V\right)
;\\\delta\left(  \left[  \underline{W}\right]  ,\left[  \underline{U}\right]
\right)  =\sigma^{-1}}}\left[  \underline{U}\right]  }=\left(  p-1\right)
^{n}p^{l\left(  \sigma^{-1}\rho\right)  }\sum_{\substack{\underline{U}%
\in\operatorname*{Flag}\left(  V\right)  ;\\\delta\left(  \left[
\underline{W}\right]  ,\left[  \underline{U}\right]  \right)  =\sigma^{-1}%
}}\left[  \underline{U}\right]  ,
\]
we obtain $\pi\sigma^{\ast}\pi^{t}\left[  \underline{W}\right]  =\left(
p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)  }T_{\sigma^{-1}}\left[
\underline{W}\right]  $.

Now, forget that we fixed $\underline{W}$. We thus have shown that $\pi
\sigma^{\ast}\pi^{t}\left[  \underline{W}\right]  =\left(  p-1\right)
^{n}p^{l\left(  \sigma^{-1}\rho\right)  }T_{\sigma^{-1}}\left[  \underline{W}%
\right]  $ for each $\underline{W}\in\operatorname*{Flag}\left(  V\right)  $.
In other words, the two maps $\pi\sigma^{\ast}\pi^{t}$ and $\left(
p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)  }T_{\sigma^{-1}}$ are
equal to each other on the basis $\left(  \left[  \underline{W}\right]
\right)  _{W\in\operatorname*{Flag}\left(  V\right)  }$ of the $\mathbb{Z}%
_{\left(  p\right)  }$-module $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Flag}\left(  V\right)  \right]  $. Since these two maps are
$\mathbb{Z}_{\left(  p\right)  }$-linear, we can thus conclude that they are
identical. In other words, $\pi\sigma^{\ast}\pi^{t}=\left(  p-1\right)
^{n}p^{l\left(  \sigma^{-1}\rho\right)  }T_{\sigma^{-1}}$.

Now, forget that we fixed $\sigma$. We thus have shown that
\begin{equation}
\pi\sigma^{\ast}\pi^{t}=\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}%
\rho\right)  }T_{\sigma^{-1}}\ \ \ \ \ \ \ \ \ \ \text{for each }\sigma
\in\Sigma_{n}. \label{pf.p9.2.1}%
\end{equation}


On the other hand, every $\sigma\in\Sigma_{n}$ satisfies%
\begin{equation}
l\left(  \sigma\rho\right)  =l\left(  \sigma^{-1}\rho\right)
\label{pf.p9.2.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.p9.2.2}):} Let $\sigma\in\Sigma_{n}$.
Then, Corollary 2.19 \textbf{(a)} yields $l\left(  \sigma^{-1}\rho\right)
=n\left(  n-1\right)  /2-l\left(  \sigma\right)  $. But Corollary 2.19
\textbf{(b)} yields $l\left(  \sigma\rho\right)  =n\left(  n-1\right)
/2-l\left(  \sigma\right)  $. Thus, $l\left(  \sigma^{-1}\rho\right)
=n\left(  n-1\right)  /2-l\left(  \sigma\right)  =l\left(  \sigma\rho\right)
$. This proves (\ref{pf.p9.2.2}).}.

Now,
\begin{align*}
e  &  =\pi\omega\pi^{t}\\
&  =\pi\left\vert G/U\right\vert ^{-1}\sum_{\sigma\in\Sigma_{n}}%
\operatorname*{sgn}\left(  \sigma\right)  \sigma^{\ast}\pi^{t}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega=\left\vert G/U\right\vert
^{-1}\sum_{\sigma\in\Sigma_{n}}\operatorname*{sgn}\left(  \sigma\right)
\sigma^{\ast}\right) \\
&  =\left\vert G/U\right\vert ^{-1}\sum_{\sigma\in\Sigma_{n}}%
\underbrace{\operatorname*{sgn}\left(  \sigma\right)  }_{=\operatorname*{sgn}%
\left(  \sigma^{-1}\right)  }\underbrace{\pi\sigma^{\ast}\pi^{t}%
}_{\substack{=\left(  p-1\right)  ^{n}p^{l\left(  \sigma^{-1}\rho\right)
}T_{\sigma^{-1}}\\\text{(by (\ref{pf.p9.2.1}))}}}\\
&  =\left\vert G/U\right\vert ^{-1}\sum_{\sigma\in\Sigma_{n}}%
\operatorname*{sgn}\left(  \sigma^{-1}\right)  \left(  p-1\right)
^{n}p^{l\left(  \sigma^{-1}\rho\right)  }T_{\sigma^{-1}}\\
&  =\left\vert G/U\right\vert ^{-1}\underbrace{\left(  p-1\right)  ^{n}%
}_{\substack{=\dfrac{\left\vert G/U\right\vert }{\left\vert G/B\right\vert
}\\\text{(by (\ref{pf.l9.0a.1}))}}}\underbrace{\sum_{\sigma\in\Sigma_{n}%
}\operatorname*{sgn}\left(  \sigma^{-1}\right)  p^{l\left(  \sigma^{-1}%
\rho\right)  }T_{\sigma^{-1}}}_{\substack{=\sum_{\sigma\in\Sigma_{n}%
}\operatorname*{sgn}\left(  \sigma\right)  p^{l\left(  \sigma\rho\right)
}T_{\sigma}\\\text{(here, we have substituted }\sigma\text{ for }\sigma
^{-1}\\\text{in the sum, since the map }\Sigma_{n}\rightarrow\Sigma
_{n},\ \sigma\mapsto\sigma^{-1}\\\text{is a bijection)}}}\\
&  =\underbrace{\left\vert G/U\right\vert ^{-1}\dfrac{\left\vert
G/U\right\vert }{\left\vert G/B\right\vert }}_{=\left\vert G/B\right\vert
^{-1}}\sum_{\sigma\in\Sigma_{n}}\operatorname*{sgn}\left(  \sigma\right)
\underbrace{p^{l\left(  \sigma\rho\right)  }}_{\substack{=p^{l\left(
\sigma^{-1}\rho\right)  }\\\text{(since }l\left(  \sigma\rho\right)  =l\left(
\sigma^{-1}\rho\right)  \\\text{(by (\ref{pf.p9.2.2})))}}}T_{\sigma}\\
&  =\left\vert G/B\right\vert ^{-1}\sum_{\sigma\in\Sigma_{n}}%
\operatorname*{sgn}\left(  \sigma\right)  p^{l\left(  \sigma^{-1}\rho\right)
}T_{\sigma}.
\end{align*}
This proves Proposition 9.3. $\square$ ]

\item \textbf{Proof of Proposition 9.4:} Replace \textquotedblleft First, we
have\textquotedblright\ by \textquotedblleft Proposition 9.3
yields\textquotedblright.

\item \textbf{Proof of Proposition 9.4:} The first chain of equalities in the
proof needs some justification (e.g., why do we have $\sum_{\sigma}p^{l\left(
\sigma^{-1}\rho\right)  }=\left\vert \coprod_{\sigma}X\left(  \sigma^{-1}%
\rho\right)  \right\vert $, and why is $\left\vert G/B\right\vert
^{-1}\left\vert \coprod_{\tau}X\left(  \tau\right)  \right\vert =1$ ?). I
would actually suggest the following alternative argument:

Proposition 8.1b \textbf{(b)} shows that there is a natural isomorphism
$G/B\rightarrow\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  $ of
$G$-sets. Hence, $\left\vert G/B\right\vert =\left\vert \operatorname*{Flag}%
\left(  \mathbb{F}_{p}^{n}\right)  \right\vert $. But every $V\in\mathcal{V}$
satisfies $\left\vert \operatorname*{Flag}\left(  V\right)  \right\vert
=\sum_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma\right)  }$ (by Lemma 8.0a
\textbf{(b)}). Applying this to $V=\mathbb{F}_{p}^{n}$, we obtain $\left\vert
\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right\vert
=\sum_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma\right)  }$. The map $\Sigma
_{n}\rightarrow\Sigma_{n},\ \sigma\mapsto\sigma^{-1}\rho$ is a bijection
(since $\Sigma_{n}$ is a group). Hence, we can substitute $\sigma^{-1}\rho$
for $\sigma$ in the sum $\sum_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma\right)
}$. We thus obtain $\sum_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma\right)
}=\sum_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma^{-1}\rho\right)  }$. Hence,%
\begin{equation}
\left\vert G/B\right\vert =\left\vert \operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \right\vert =\sum_{\sigma\in\Sigma_{n}}p^{l\left(
\sigma\right)  }=\sum_{\sigma\in\Sigma_{n}}p^{l\left(  \sigma^{-1}\rho\right)
}. \label{pf.p9.4.1}%
\end{equation}
Now, Proposition 9.3 yields%
\[
e=\left\vert G/B\right\vert ^{-1}\sum_{\sigma\in\Sigma_{n}}\operatorname*{sgn}%
\left(  \sigma\right)  p^{l\left(  \sigma^{-1}\rho\right)  }T_{\sigma}.
\]
Applying the map $\xi$ to both sides of this equality, we find%
\begin{align*}
\xi\left(  e\right)   &  =\xi\left(  \left\vert G/B\right\vert ^{-1}%
\sum_{\sigma\in\Sigma_{n}}\operatorname*{sgn}\left(  \sigma\right)
p^{l\left(  \sigma^{-1}\rho\right)  }T_{\sigma}\right) \\
&  =\left\vert G/B\right\vert ^{-1}\sum_{\sigma\in\Sigma_{n}}%
\operatorname*{sgn}\left(  \sigma\right)  p^{l\left(  \sigma^{-1}\rho\right)
}\underbrace{\xi\left(  T_{\sigma}\right)  }_{\substack{=\operatorname*{sgn}%
\left(  \sigma\right)  \\\text{(by the definition of }\xi\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\xi\text{ is }%
\mathbb{Z}_{\left(  p\right)  }\text{-linear}\right) \\
&  =\left\vert G/B\right\vert ^{-1}\sum_{\sigma\in\Sigma_{n}}%
\operatorname*{sgn}\left(  \sigma\right)  p^{l\left(  \sigma^{-1}\rho\right)
}\operatorname*{sgn}\left(  \sigma\right)  =\left\vert G/B\right\vert
^{-1}\sum_{\sigma\in\Sigma_{n}}\underbrace{\left(  \operatorname*{sgn}\left(
\sigma\right)  \right)  ^{2}}_{=1}p^{l\left(  \sigma^{-1}\rho\right)  }\\
&  =\left\vert G/B\right\vert ^{-1}\sum_{\sigma\in\Sigma_{n}}%
\underbrace{\left(  \operatorname*{sgn}\left(  \sigma\right)  \right)  ^{2}%
}_{=1}p^{l\left(  \sigma^{-1}\rho\right)  }\\
&  =\left\vert G/B\right\vert ^{-1}\underbrace{\sum_{\sigma\in\Sigma_{n}%
}p^{l\left(  \sigma^{-1}\rho\right)  }}_{\substack{=\left\vert G/B\right\vert
\\\text{(by (\ref{pf.p9.4.1}))}}}=\left\vert G/B\right\vert ^{-1}\left\vert
G/B\right\vert =1.
\end{align*}


\item \textbf{Proof of Proposition 9.4:} Replace \textquotedblleft%
$\widehat{\xi}\left(  e\right)  \mu$\textquotedblright\ by \textquotedblleft%
$\xi\left(  e\right)  \mu$\textquotedblright.

\item \textbf{Proof of Proposition 9.4:} I would replace \textquotedblleft As
$e_{\operatorname*{St}}=\pi^{t}\mu$ we see that $\pi^{t}\left(
M_{\operatorname*{St}}\right)  =\operatorname{image}\left(  \pi^{t}\mu\right)
=M_{\operatorname*{St}}^{\prime}$\textquotedblright\ by the more detailed
argument \textquotedblleft As $M_{\operatorname*{St}}=\operatorname{image}%
\left(  \mu\right)  $, we see that $\pi^{t}\left(  M_{\operatorname*{St}%
}\right)  =\operatorname{image}\left(  \underbrace{\pi^{t}\mu}%
_{=e_{\operatorname*{St}}}\right)  =\operatorname{image}\left(
e_{\operatorname*{St}}\right)  =M_{\operatorname*{St}}^{\prime}$%
\textquotedblright.

\item \textbf{Proof of Proposition 9.4:} I would replace \textquotedblleft
Thus, if we let $\beta$ be the restriction of $\pi^{t}$ to
$M_{\operatorname*{St}}$, we see that $\beta$ gives an epimorphism
$M_{\operatorname*{St}}\rightarrow M_{\operatorname*{St}}^{\prime}%
$\textquotedblright\ by \textquotedblleft Thus, $\pi^{t}$ restricts to an
epimorphism $\beta:M_{\operatorname*{St}}\rightarrow M_{\operatorname*{St}%
}^{\prime}$\textquotedblright. (This is both shorter and also explains
unambiguously what the codomain of $\beta$ is.)

\item \textbf{Proof of Proposition 9.4:} After \textquotedblleft As the map
$\mu\pi^{t}=e$ restricts to $1$ on $M_{\operatorname*{St}}$\textquotedblright,
I would add \textquotedblleft(because $e$ is idempotent, and
$M_{\operatorname*{St}}$ is its image)\textquotedblright.

\item \textbf{Proof of Proposition 9.5:} I don't understand the first sentence
of this proof: Why does the Yoneda isomorphism exist? (I only know Yoneda
isomorphisms for functors to $\operatorname*{Set}$, not for functors to
$\mathcal{A}$; but even if I were to write down the obvious generalization,
there remains the question why $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  $ is a Hom-functor.) And supposing that the
Yoneda isomorphism exists, why does it yield that $\mathbb{Z}_{\left(
p\right)  }\left[  \operatorname*{Base}\right]  $ is projective?

What I do see is that the functor $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  $ is \textquotedblleft pointwise
projective\textquotedblright, in the sense that the image of each object of
$\mathcal{V}$ under this functor is a projective $\mathbb{Z}_{\left(
p\right)  }$-module. (This is obvious, because the image of an object
$V\in\mathcal{V}$ under the functor $\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\right]  $ is the $\mathbb{Z}_{\left(  p\right)  }$-module
$\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}\right]
\right)  \left(  V\right)  =\mathbb{Z}_{\left(  p\right)  }\left[
\operatorname*{Base}\left(  V\right)  \right]  $, which is free and therefore
projective.) Therefore, the functor $M_{\operatorname*{St}}$ is
\textquotedblleft pointwise projective\textquotedblright\ as well (since, as
youpoint out, $M_{\operatorname*{St}}\left(  V\right)  $ is a direct summand
in $\left(  \mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Base}%
\right]  \right)  \left(  V\right)  $).

\item \textbf{Proof of Proposition 9.5:} After \textquotedblleft the image of
$e_{\operatorname*{St}}$, which is a summand in $\mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Base}\right]  $\textquotedblright, I suggest adding
\textquotedblleft(since $e_{\operatorname*{St}}$ is
idempotent)\textquotedblright.

\item \textbf{Proof of Proposition 9.5:} After \textquotedblleft%
$M_{\operatorname*{St}}$ is also the image of a self-adjoint idempotent on
$\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}\right]
$\textquotedblright, I suggest adding \textquotedblleft(namely, of
$e$)\textquotedblright.

\item \textbf{Proof of Proposition 9.5:} After \textquotedblleft the rank of
$M_{\operatorname*{St}}$ is the trace of $e$\textquotedblright, I suggest
adding \textquotedblleft(since $M_{\operatorname*{St}}$ is the image of the
idempotent endomorphism $e$)\textquotedblright.

\item \textbf{Proof of Proposition 9.5:} After \textquotedblleft the map
$T_{\sigma}$ is the identity\textquotedblright, I would add \textquotedblleft%
(by Corollary 4.6)\textquotedblright.

\item \textbf{Proof of Proposition 9.5:} I would suggest replacing
\textquotedblleft with trace $\left\vert \operatorname*{Flag}\right\vert
=\left\vert G/B\right\vert $\textquotedblright\ by \textquotedblleft with
trace%
\begin{align*}
\left\vert \operatorname*{Flag}\left(  V\right)  \right\vert  &  =\left\vert
\operatorname*{Flag}\left(  \mathbb{F}_{p}^{n}\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }V\cong\mathbb{F}_{p}^{n}\text{ as
}\mathbb{F}_{p}\text{-vector spaces}\right) \\
&  =\left\vert G/B\right\vert \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Proposition 8.1b \textbf{(b)} shows that there is a}\\
\text{natural isomorphism }G/B\rightarrow\operatorname*{Flag}\left(
\mathbb{F}_{p}^{n}\right)  \text{ of }G\text{-sets}%
\end{array}
\right)
\end{align*}
\textquotedblright.

\item \textbf{Proof of Proposition 9.5:} I suggest replacing \textquotedblleft
Next, note that $\mathbb{Z}_{\left(  p\right)  }\left[  \operatorname*{Flag}%
\right]  =M_{\operatorname*{St}}\oplus N$\textquotedblright\ by
\textquotedblleft But $e$ is idempotent; thus, $\mathbb{Z}_{\left(  p\right)
}\left[  \operatorname*{Flag}\right]  =\underbrace{\operatorname{image}\left(
e\right)  }_{=M_{\operatorname*{St}}}\oplus\underbrace{\operatorname{image}%
\left(  1-e\right)  }_{=N}=M_{\operatorname*{St}}\oplus N$\textquotedblright.
\end{itemize}


\end{document}