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\ihead{Errata to ``Algebraic combinatorics related to the free Lie algebra''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Algebraic combinatorics related to the free Lie algebra}

\textit{D. Blessenohl and H. Laue}

S\'eminaire Lotharingien de Combinatoire \textbf{29}, 1992, B29e.

\url{https://eudml.org/doc/121542}

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

These errata mostly concern the proof of Solomon's Mackey formula (=
Proposition 4.3) and the lemma (Lemma 4.4) used in it.

I will refer to the results appearing in the article \textquotedblleft
Algebraic combinatorics related to the free Lie algebra\textquotedblright\ by
the numbers under which they appear in this article.

\setcounter{section}{6}

\section{Errata and addenda}

\begin{enumerate}
\item \textbf{page 2, Proposition 1.1:} Replace \textquotedblleft$\left\{
x_{1}\ldots,x_{n}\right\}  $\textquotedblright\ by \textquotedblleft$\left\{
x_{1},\ldots,x_{n}\right\}  $\textquotedblright\ (a comma was missing).

\item \textbf{page 3:} It is worth mentioning that the permutations in
$\mathcal{X}_{m}$ are also known as the \emph{V-permutations} (since their
first-decreasing-then-increasing plot resembles the letter \textquotedblleft
V\textquotedblright) or as the \emph{valley permutations}.

\item \textbf{page 5:} It is worth mentioning that the \textquotedblleft
defect set\textquotedblright\ $\operatorname*{D}\left(  \sigma\right)  $ is
more commonly known as the \emph{descent set} of $\sigma$, and is usually
denoted by $\operatorname*{Des}\sigma$.

\item \textbf{page 6, (10):} The equivalence of the three statements in (10)
is not obvious enough to be left unexplained. Of course, the first two
statements ($\pi\in\mathcal{X}_{n}$ and $D\left(  \pi\right)  =\left\{
1,2,\ldots,r\right\}  $) are easily seen to be equivalent, but the tricky part
is to see that the third statement is also equivalent to them. This relies on
the following fact:

\begin{statement}
\textit{Fact:} Let $j_{1},j_{2},\ldots,j_{r}$ be elements of $\left\{
1,2,\ldots,n\right\}  $ satisfying $j_{1}>j_{2}>\cdots>j_{r}$. Then, the
product%
\[
\left(  j_{1}\ldots1\right)  \left(  j_{2}\ldots1\right)  \cdots\left(
j_{r}\ldots1\right)
\]
is the permutation in $\mathrm{S}_{n}$ whose one-line notation (i.e., list of
values at $1,2,\ldots,n$) begins with the numbers $j_{1},j_{2},\ldots,j_{r}$
in this order and ends with the remaining $n-r$ elements of $\left\{
1,2,\ldots,n\right\}  $ in increasing order. (In particular, this product
belongs to $\mathcal{X}_{n}$.)
\end{statement}

This fact is (essentially) Exercise 4 in
\href{https://www.cip.ifi.lmu.de/~grinberg/t/17f/hw7os.pdf}{\textit{Math 4990
Fall 2017 (Darij Grinberg) homework set 7}}, where I give a detailed proof.
(Note that the order in which permutations are multiplied differs between my
writeup and the Blessenohl/Laue paper. Thus, even though the cycles appear in
the order of decreasing length in the former and increasing length in the
latter, the products are the same.)

\item \textbf{page 16, proof of Proposition 4.2:} It is helpful to restate the
definition of $M_{\ell}$ as follows: The set $M_{\ell}$ consists of all
permutations $\mu\in\mathrm{S}_{\ell}$ that satisfy $\left(  i+1\right)
\mu\geq i\mu-1$ for each $i<\ell$.

\item \textbf{page 17, proof of Proposition 4.2:} Replace \textquotedblleft
put \newline$k:=min\left\{  j\mid\left(  i+1\right)  \rho\leq j\leq
i\rho,\ j\rho^{-1}\geq i\right\}  -1$\textquotedblright\ (which is just a
complicated way to say \textquotedblleft put $k:=i$\textquotedblright, clearly
against the authors' intent) by \textquotedblleft let $k$ be the largest $j$
satisfying $i\rho\geq j\geq\left(  i+1\right)  \rho$ and $j\rho^{-1}\geq
i+1$\textquotedblright. To see why this choice of $k$ works, we first note
that it satisfies $k\neq i\rho$ (since $k\rho^{-1}\geq i+1$, but $\left(
i\rho\right)  \rho^{-1}=i$ is not $\geq i+1$), hence $k<i\rho$ (since
$i\rho\geq k$), and thus $\left(  k+1\right)  \rho^{-1}<i+1$ (since otherwise,
$k$ would not be the \textbf{largest} $j$ with $j\rho^{-1}\geq i+1$). Hence,
in order to prove that $k\rho^{-1}-\left(  k+1\right)  \rho^{-1}>1$, it
suffices to show that we cannot have the situation where $k\rho^{-1}=i+1$ and
$\left(  k+1\right)  \rho^{-1}=i$ (since in all other cases, $k\rho
^{-1}-\left(  k+1\right)  \rho^{-1}>1$ follows from $k\rho^{-1}\geq i+1$ and
$\left(  k+1\right)  \rho^{-1}<i+1$). But this situation is indeed impossible,
since it entails $i\rho=k+1$ and $\left(  i+1\right)  \rho=k$ in contradiction
to $i\rho-\left(  i+1\right)  \rho\geq2$.

\item \textbf{page 18:} What is called a \textquotedblleft decomposition of
$n$\textquotedblright\ here is more usually called a \emph{composition} of $n$.

\item \textbf{page 18, last line:} It is also worth pointing out that the
standard partition $P^{q}=\left(  P_{1}^{q},\ldots,P_{\ell}^{q}\right)  $ of a
composition $q=\left(  q_{1},\ldots,q_{\ell}\right)  \models n$ can be defined
as follows: the set $P_{1}^{q}$ consists of the smallest $q_{1}$ elements of
$\left\{  1,2,\ldots,n\right\}  $; the set $P_{2}^{q}$ consists of the
next-smallest $q_{2}$ elements of $\left\{  1,2,\ldots,n\right\}  $; and so on.

\item \textbf{page 20, proof of Lemma 4.4:} \textquotedblleft This implies, by
(40), that $x_{1}<x_{2}$\textquotedblright: This could use a bit more
explanation. Namely, $\rho\in\mathcal{S}^{r}\left(  M\right)  \subseteq
\mathcal{S}^{r}$. Thus, (40) shows that $\rho\mid_{P_{i}^{r}}$ is increasing.
Hence, if we had $x_{1}\geq x_{2}$, then we would have $x_{1}\rho\geq
x_{2}\rho$ (since $x_{1}$ and $x_{2}$ both belong to $P_{i}^{r}$), which would
contradict $x_{1}\rho<x_{2}\rho$. Thus, we cannot have $x_{1}\geq x_{2}$, so
we must have $x_{1}<x_{2}$.

\item \textbf{page 20, proof of Lemma 4.4:} Here is a bit more detail on the
derivation of (44): Applying the bijection $\rho$ to both sides of (43), we
see that%
\begin{equation}
P_{i}^{r}\rho\cap P_{j}^{q}=R_{i,j}\rho\ \ \ \ \ \ \ \ \ \ \text{for all
}i\leq k\text{, }j\leq\ell\text{ and }\rho\in\mathcal{S}^{r}\left(  M\right)
. \tag{43'}%
\end{equation}
Now, taking the union over all $i$ (and recalling that $\bigcup_{i}\left(
P_{i}^{r}\rho\right)  =\left\{  1,2,\ldots,n\right\}  $), we obtain (44).

\item \textbf{page 20, proof of Lemma 4.4:} Let me explain why
\textquotedblleft$\rho\sigma\mid_{R_{ij}}$ is the composition of two
increasing functions\textquotedblright. Indeed, the function $\rho\mid
_{R_{ij}}$ is increasing, because (40) shows that $\rho\mid_{P_{i}^{r}}$ is
increasing (since $\rho\in\mathcal{S}^{r}\left(  M\right)  \subseteq
\mathcal{S}^{r}$) and because $R_{ij}\subseteq P_{i}^{r}$. Furthermore, the
function $\sigma\mid_{R_{ij}\rho}$ is increasing, because (40) shows that
$\sigma\mid_{P_{j}^{q}}$ is increasing (since $\sigma\in\mathcal{S}^{q}$) and
because (43') shows that $R_{ij}\rho=P_{i}^{r}\rho\cap P_{j}^{q}\subseteq
P_{j}^{q}$. Now, $\rho\sigma\mid_{R_{ij}}$ is the composition of these two
increasing functions $\rho\mid_{R_{ij}}$ and $\sigma\mid_{R_{ij}\rho}$.

\item \textbf{pages 20--21, proof of Lemma 4.4:} The explanation for why (48)
holds in the case $j_{1}=j_{2}$ is a bit laconic. Let me give a more detailed one:

Assume that $j_{1}=j_{2}$. Then, $x_{1}$ and $x_{2}$ belong to the same
$R_{ij}$ (namely, $R_{ij_{1}}=R_{ij_{2}}$), and thus we have $x_{1}\tau
<x_{2}\tau$ (as $\tau\in\mathcal{S}^{w\left(  M\right)  }$ and $x_{1}<x_{2}$).
Moreover, each $h\in\left\{  1,2\right\}  $ satisfies $\underbrace{x_{h}}_{\in
R_{i,j_{h}}}\underbrace{\rho}_{=\tau\sigma^{-1}}\in\underbrace{R_{i,j_{h}}%
\tau}_{\subseteq P_{j_{h}}^{q}\sigma}\sigma^{-1}\subseteq P_{j_{h}}^{q}$. In
view of $j_{1}=j_{2}$, this means that both $x_{2}\rho$ and $x_{1}\rho$ belong
to the same $P_{j}^{q}$. Hence, if we had $x_{2}\rho\leq x_{1}\rho$, then
(from $\sigma\in\mathcal{S}^{q}$) we would obtain $x_{2}\rho\sigma\leq
x_{1}\rho\sigma$, that is, $x_{2}\tau\leq x_{1}\tau$ (since $\rho\sigma=\tau
$), which would contradict $x_{1}\tau<x_{2}\tau$. Hence, we cannot have
$x_{2}\rho\leq x_{1}\rho$. Thus, we must have $x_{1}\rho<x_{2}\rho$. This
proves (48).
\end{enumerate}


\end{document}