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\ihead{Errata to Bidigare3}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{On Bidigare's proof of Solomon's theorem}

\textit{Mark Wildon}

\url{http://www.ma.rhul.ac.uk/~uvah099/Maths/Bidigare3.pdf}

version of 15 January 2015

\textbf{Errata and addenda by Darij Grinberg}

\bigskip
\end{center}

%\setcounter{section}{}


\section{Errata}

\begin{itemize}
\item \textbf{page 1:} As usual, I think it's worth explaining that your
functions (or permutations, at least) act on the right of their values and are
multiplied accordingly (so that $\alpha\beta$ sends $i$ to $\left(
i\alpha\right)  \beta$).

\item \textbf{page 1:} Also, I think \textquotedblleft natural
number\textquotedblright\ should be defined (to warn the reader that $0$
doesn't count as a natural number).

\item \textbf{page 1:} Also, the Young subgroup $S_{p}$ should be defined.

\item \textbf{page 1 and later:} You occasionally use \textquotedblleft%
$\Xi_{p}$\textquotedblright\ as a synonym for \textquotedblleft$\Xi^{p}%
$\textquotedblright. (Probably, a search for \textquotedblleft\texttt{%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
Xi\_}\textquotedblright\ will quickly locate all the instances of this.)

\item \textbf{page 1:} In the definition of $\operatorname*{Des}\left(
g\right)  $, replace \textquotedblleft$xg<\left(  x+1\right)  g$%
\textquotedblright\ by \textquotedblleft$xg>\left(  x+1\right)  g$%
\textquotedblright\ (or does gravity, too, work the other way round in Britain?).

\item \textbf{page 1:} \textquotedblleft Given compositions $p$, $q$ and $r$
of $\mathbf{N}$ such that $p$ has $k$ parts and $q$ has $\ell$
parts\textquotedblright\ $\rightarrow$ \textquotedblleft Given compositions
$p=\left(  p_{1},p_{2},\ldots,p_{k}\right)  $, $q=\left(  q_{1},q_{2}%
,\ldots,q_{\ell}\right)  $ and $r$ of $n\in\mathbf{N}$\textquotedblright. (Two
things corrected here: \textquotedblleft of $\mathbf{N}$\textquotedblright%
\ became \textquotedblleft of $n\in\mathbf{N}$\textquotedblright, and the
notations $p_{i}$ and $q_{j}$ have been defined explicitly since you refer to
them later.)

\item \textbf{page 1, Theorem 1:} \textquotedblleft If $p$, $q$ and
$r$\textquotedblright\ $\rightarrow$ \textquotedblleft If $p$ and
$q$\textquotedblright.

\item \textbf{page 1, \S 2:} \textquotedblleft the sets $P_{1},\ldots,P_{n}$
are disjoint\textquotedblright\ $\rightarrow$ \textquotedblleft the sets
$P_{1},\ldots,P_{k}$ are disjoint\textquotedblright.

\item \textbf{page 1, \S 2:} In the first displayed equation of \S 2, add a
comma after \textquotedblleft$P_{1}g$\textquotedblright\ in \textquotedblleft%
$\left(  P_{1}g\ldots,P_{k}g\right)  $\textquotedblright.

\item \textbf{page 1, \S 2:} It might be worth saying a few words about why
this product $\wedge$ is associative. To me, this becomes really clear when I
identify each set composition $P=\left(  P_{1},\ldots,P_{k}\right)  $ of $n$
with a total pre-order on the set $\left\{  1,2,\ldots,n\right\}  $ (namely,
the pre-order under which two elements $i$ and $j$ satisfy $i\leq j$ if and
only if $i\in P_{u}$ and $j\in P_{v}$ for some $u\leq v$), and then the
product $\wedge$ becomes a \textquotedblleft lexicographic
order\textquotedblright\ product (i.e., two elements $i$ and $j$ of $\left\{
1,2,\ldots,n\right\}  $ satisfy $i\leq j$ in $P\wedge Q$ if and only if they
satisfy $i\leq j$ in $P$ and $\left(  \text{if }i\sim j\text{ in }P\text{,
then }i\leq j\text{ in }Q\right)  $). Thus, $P\wedge Q$ means
\textquotedblleft order the elements according to $P$, and use $Q$ to break
ties\textquotedblright.

\item \textbf{page 2, basic property (2):} Here, \textquotedblleft$\left\{
1,\ldots,n\right\}  $\textquotedblright\ should be replaced by
\textquotedblleft$\left(  \left\{  1,\ldots,n\right\}  \right)  $%
\textquotedblright.

\item \textbf{page 2, basic property (3):} This is not correct, at least not
if you use the usual concept of \textquotedblleft refinement\textquotedblright%
\ for set compositions. For example, the type of%
\[
\left(  \left\{  1,2,3\right\}  ,\left\{  4\right\}  \right)  \wedge\left(
\left\{  1,4\right\}  ,\left\{  2,3\right\}  \right)  =\left(  \left\{
1\right\}  ,\left\{  2,3\right\}  ,\left\{  4\right\}  \right)
\]
is $\left(  1,2,1\right)  $, which is a refinement of $\left(  3,1\right)  $
but not a refinement of $\left(  2,2\right)  $.

My suggestion is to replace \textbf{(3)} by \textquotedblleft If $Q\in\Pi_{n}$
has type $\left(  1^{n}\right)  $, then $P\wedge Q$ has type $\left(
1^{n}\right)  $ for any $P\in\Pi_{n}$.\textquotedblright. This is all you need
in the following, and it has the advantage of being obviously true. (Plus, you
haven't defined \textquotedblleft refinement\textquotedblright.)

\item \textbf{page 2, basic property (C):} \textquotedblleft By (3)
above\textquotedblright\ $\rightarrow$ \textquotedblleft By (3) and (4)
above\textquotedblright\ (at least if you follow my suggestion in nerfing (3)).

\item \textbf{page 2, basic property (D):} \textquotedblleft If $q$ has $\ell$
parts\textquotedblright\ $\rightarrow$ \textquotedblleft If $q=\left(
q_{1},\ldots,q_{\ell}\right)  $\textquotedblright.

\item \textbf{page 2, basic property (D):} In the displayed equation that
defines $T^{q}$, replace \textquotedblleft$\left\{  1\ldots q_{1}\right\}
$\textquotedblright\ by \textquotedblleft$\left\{  1,\ldots,q_{1}\right\}
,\left\{  q_{1}+1,\ldots,q_{1}+q_{2}\right\}  $\textquotedblright\ (I've added
missing commas and also added a second set to make the construction clearer).

\item \textbf{page 2, proof of Proposition 3:} \textquotedblleft Let $p$ be a
composition with $k$ parts\textquotedblright\ $\rightarrow$ \textquotedblleft
Let $p=\left(  p_{1},\ldots,p_{k}\right)  $ be a composition\textquotedblright.

\item \textbf{page 3, proof of Proposition 3:} Remove the \textquotedblleft
where $q$ has $\ell$ parts\textquotedblright\ (you never use $\ell$).

\item \textbf{page 3, proof of Proposition 3:} \textquotedblleft for $1\leq
i<k$\textquotedblright\ $\rightarrow$ \textquotedblleft for $1\leq i\leq
k$\textquotedblright.

\item \textbf{page 3, proof of Proposition 3:} Replace \textquotedblleft%
$T_{q}$\textquotedblright\ by \textquotedblleft$T^{q}$\textquotedblright.
This, too, appears several times, so it's worth searching for it.

\item \textbf{page 3:} You write \textquotedblleft is a subalgebra of
$\mathbf{Z}S_{n}$ isomorphic to $\left(  \mathbf{Z}\Pi_{n}\right)  ^{S_{n}}%
$\textquotedblright. You seem to be going a tad too fast here; the isomorphism
only follows once you realize that the $\Xi^{p}$ are linearly independent,
which follows from the distinctness of their \textquotedblleft leading
terms\textquotedblright\ with respect to some order on the permutations; but
this isn't really so obvious that it isn't worth further mention, if you ask me.

\item \textbf{page 3:} \textquotedblleft say that $T\in\Pi_{n}$ is
\textit{increasing}\textquotedblright\ $\rightarrow$ \textquotedblleft say
that $T=\left(  T_{1},\ldots,T_{\ell}\right)  \in\Pi_{n}$ is
\textit{increasing}\textquotedblright.

\item \textbf{page 3:} \textquotedblleft for $1\leq i<i^{\prime}\leq\ell
$\textquotedblright\ $\rightarrow$ \textquotedblleft for $1\leq i<j\leq\ell
$\textquotedblright\ (or rename $j$ as $i^{\prime}$ later).

\item \textbf{page 3, proof of Proposition 4:} \textquotedblleft Suppose that
$p$ has $k$ parts, $q$ has $\ell$ parts and that $r$ has $m$
parts\textquotedblright\ $\rightarrow$ \textquotedblleft Suppose that
$p=\left(  p_{1},\ldots,p_{k}\right)  $ and $q=\left(  q_{1},\ldots,q_{\ell
}\right)  $\textquotedblright. (You don't need $m$.)
\end{itemize}


\end{document}