\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}% \usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage} \usepackage[all,cmtip]{xy} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{framed} \usepackage{amsmath} \usepackage{comment} \usepackage{color} \usepackage{hyperref} \usepackage[sc]{mathpazo} \usepackage[T1]{fontenc} \usepackage{amsthm} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=5.50.0.2960} %TCIDATA{LastRevised=Wednesday, June 12, 2019 02:18:09} %TCIDATA{SuppressPackageManagement} %TCIDATA{} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %BeginMSIPreambleData \providecommand{\U}{\protect\rule{.1in}{.1in}} %EndMSIPreambleData \theoremstyle{definition} \newtheorem{theo}{Theorem}[section] \newenvironment{theorem}[] {\begin{theo}[#1]\begin{leftbar}} {\end{leftbar}\end{theo}} \newtheorem{lem}[theo]{Lemma} \newenvironment{lemma}[] {\begin{lem}[#1]\begin{leftbar}} {\end{leftbar}\end{lem}} \newtheorem{prop}[theo]{Proposition} \newenvironment{proposition}[] {\begin{prop}[#1]\begin{leftbar}} {\end{leftbar}\end{prop}} \newtheorem{defi}[theo]{Definition} \newenvironment{definition}[] {\begin{defi}[#1]\begin{leftbar}} {\end{leftbar}\end{defi}} \newtheorem{remk}[theo]{Remark} \newenvironment{remark}[] {\begin{remk}[#1]\begin{leftbar}} {\end{leftbar}\end{remk}} \newtheorem{coro}[theo]{Corollary} \newenvironment{corollary}[] {\begin{coro}[#1]\begin{leftbar}} {\end{leftbar}\end{coro}} \newtheorem{conv}[theo]{Convention} \newenvironment{condition}[] {\begin{conv}[#1]\begin{leftbar}} {\end{leftbar}\end{conv}} \newtheorem{quest}[theo]{Question} \newenvironment{algorithm}[] {\begin{quest}[#1]\begin{leftbar}} {\end{leftbar}\end{quest}} \newtheorem{warn}[theo]{Warning} \newenvironment{conclusion}[] {\begin{warn}[#1]\begin{leftbar}} {\end{leftbar}\end{warn}} \newtheorem{conj}[theo]{Conjecture} \newenvironment{conjecture}[] {\begin{conj}[#1]\begin{leftbar}} {\end{leftbar}\end{conj}} \newtheorem{exmp}[theo]{Example} \newenvironment{example}[] {\begin{exmp}[#1]\begin{leftbar}} {\end{leftbar}\end{exmp}} \iffalse \newenvironment{proof}[Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \fi \newenvironment{verlong}{}{} \newenvironment{vershort}{}{} \newenvironment{noncompile}{}{} \excludecomment{verlong} \includecomment{vershort} \excludecomment{noncompile} \newcommand{\kk}{\mathbf{k}} \newcommand{\id}{\operatorname{id}} \newcommand{\ev}{\operatorname{ev}} \newcommand{\Comp}{\operatorname{Comp}} \newcommand{\bk}{\mathbf{k}} \newcommand{\Nplus}{\mathbb{N}_{+}} \newcommand{\NN}{\mathbb{N}} \let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \DeclareSymbolFont{bbold}{U}{bbold}{m}{n} \DeclareSymbolFontAlphabet{\mathbbold}{bbold} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \ihead{Errata to Bidigare4} \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/Bidigare4.pdf} version of 6 February 2019 \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 statement relies on a somewhat unusual concept of \textquotedblleft refinement\textquotedblright, as you explain a few paragraphs below; with the normal concept of refinement for compositions, it is false\footnote{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 $\left( 2,2\right)$ in the usual sense of this word.}. Let me, however, suggest to replace \textbf{(3)} by the weaker claim that \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. \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 3, 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