\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 07, 2017 02:48:20} %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} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \ihead{Errata to Approche Duale des repr\'esentations...''} \ohead{\today} \begin{document} \begin{center} \textbf{\href{http://user.math.uzh.ch/feray/Peccot/ApprocheDualeVFeray.pdf}{\textbf{Approche Duale des repr\'{e}sentations du groupe sym\'{e}trique}}} \textit{Valentin F\'{e}ray} \texttt{feray - approche duale des Rep Sn - version 29 aug 2016.pdf} version of 29 August 2016 \textbf{Errata and addenda by Darij Grinberg} \bigskip \end{center} I will refer to the results appearing in the book \textquotedblleft Approche Duale des repr\'{e}sentations du groupe sym\'{e}trique\textquotedblright\ by the numbers under which they appear in this book (specifically, in its version of 29 August 2016). Most of my corrections are written in English, as I don't speak French well enough. \setcounter{section}{8} \section{Errata} \begin{itemize} \item \textbf{Page 2, Exemple 1.5:} The words \textquotedblleft permutation de $\mathcal{S}_{n}$\textquotedblright\ are ambiguous: do you mean a permutation \textbf{in} the set $\mathcal{S}_{n}$ (of which there are $n!$), or a permutation \textbf{of} the set $\mathcal{S}_{n}$ (of which there are $\left( n!\right) !$) ? I know you mean the former, but maybe not every reader does. \item \textbf{Pages 1--13:} Somewhere here you should introduce a few more notations that you are using. Namely: \begin{itemize} \item If $\left( V,\rho\right)$ is a representation of a group $G$, and if $u$ is any element of the group ring $\mathbb{C}\left[ G\right]$, then the endomorphism $\rho\left( u\right)$ of $V$ is defined as follows: Write $u$ in the form $u=\sum_{g\in G}\alpha_{g}e_{g}$ (with $\alpha_{g}\in\mathbb{C}$), and set $\rho\left( u\right) =\sum_{g\in G}\alpha_{g}\rho\left( g\right)$. (You are using this notation, e.g., in the formulas (a) and (b) on page 13.) Notice that $\rho\left( u\right)$ depends $\mathbb{C}$-linearly on $u$, and that we have $\rho\left( e_{g}\right) =\rho\left( g\right)$ for each $g\in G$. Thus, the values $\rho\left( u\right)$ for $u\in \mathbb{C}\left[ G\right]$ encode exactly the same information as the original homomorphism $\rho:G\rightarrow\operatorname*{GL}\left( V\right)$. \item If $G$ is a group, if $\chi:G\rightarrow\mathbb{C}$ is any map, and if $u$ is any element of the group ring $\mathbb{C}\left[ G\right]$, then the complex number $\chi\left( u\right)$ is defined as follows: Write $u$ in the form $u=\sum_{g\in G}\alpha_{g}e_{g}$ (with $\alpha_{g}\in\mathbb{C}$), and set $\chi\left( u\right) =\sum_{g\in G}\alpha_{g}\chi\left( g\right)$. (You are using this notation, e.g., in Lemme 3.4 on page 39.) Notice that $\chi\left( u\right)$ depends $\mathbb{C}$-linearly on $u$, and that we have $\chi\left( e_{g}\right) =\chi\left( g\right)$ for each $g\in G$. Thus, the values $\chi\left( u\right)$ for $u\in\mathbb{C}\left[ G\right]$ encode exactly the same information as the original map $\chi:G\rightarrow \mathbb{C}$. \end{itemize} \item \textbf{Page 3, D\'{e}finition 1.7:} Replace \textquotedblleft Une repr\'{e}sentation est dite \textbf{irr\'{e}ductible}\textquotedblright\ by \textquotedblleft Une repr\'{e}sentation $\left( V,\rho\right)$ est dite \textbf{irr\'{e}ductible}\textquotedblright\ (since you later refer to both $V$ and $\rho$ in this sentence). \item \textbf{Page 3, D\'{e}finition 1.7:} Both in the definition of \textquotedblleft ind\'{e}composable\textquotedblright\ and in the definition of \textquotedblleft irr\'{e}ductible\textquotedblright, a requirement that $V\neq0$ should be added. \item \textbf{Page 3, Remarque 1.8:} You probably want to replace the word \textquotedblleft irr\'{e}ductibles\textquotedblright\ by \textquotedblleft ind\'{e}composables\textquotedblright. (They are, of course, equivalent, but only the version with \textquotedblleft ind\'{e}composables\textquotedblright% \ really follows directly from the definitions). \item \textbf{Page 3, Th\'{e}or\{e}me 1.9:} When you say \textquotedblleft un nombre fini\textquotedblright, it might be useful to point out that you are counting irreducible representations \textbf{up to isomorphism}. \item \textbf{Page 4, D\'{e}monstration de Corollaire 1.15:} Replace \textquotedblleft$\chi_{V_{N}}$\textquotedblright\ by \textquotedblleft% $\chi^{V_{N}}$\textquotedblright. \item \textbf{Page 5, \S 1.2:} When you define the \textquotedblleft% \textbf{type cyclique}\textquotedblright, it would be worthwhile pointing out that cycles of size $1$ (that is, cycles corresponding to fixed points of $\sigma$) are included in the decomposition. (Many authors, particularly those of introductory algebra texts, tend to omit them.) \item \textbf{Page 6:} In \textquotedblleft et donc de repr\'{e}sentations irr\'{e}ductibles de $S_{n}$\textquotedblright, replace \textquotedblleft% $S_{n}$\textquotedblright\ by \textquotedblleft$\mathcal{S}_{n}$% \textquotedblright. \item \textbf{Page 8, D\'{e}monstration de Lemme 1.23:} In \textquotedblleft d'un seul param\{e}tre $\alpha_{\operatorname*{id}}$\textquotedblright, replace \textquotedblleft$\operatorname*{id}$\textquotedblright\ by \textquotedblleft$\operatorname*{Id}$\textquotedblright. \item \textbf{Page 8, D\'{e}monstration de Lemme 1.23:} You write: \textquotedblleft Il n'est pas difficile de v\'{e}rifier la CNS suivante\textquotedblright. From a sufficiently experienced point of view, this CNS is really not hard to verify. However, I don't think that omitting this proof is appropriate when you are targeting an undergraduate readership. The proof is somewhat similar to the proof of Lemme 1.31, and more difficult if anything; it is strange that you present the latter in all its detail while leaving the former entirely to the reader. At the very least, it's reasonable to give some references for the proof of your CNS (more precisely: of the fact that $\pi\in\mathcal{S}_{n}$ can be written in the form $\sigma\tau$ with $\sigma\in\operatorname*{RS}% \left( T\right)$ and $\tau\in\operatorname*{CS}\left( T\right)$ if and only if the condition (P) is satisfied): \begin{itemize} \item The fact that you are claiming (or, more precisely, the sufficiency of your condition (P)) is implicit in the proof of Lemma 4.40 in: \href{https://arxiv.org/abs/0901.0827v5}{Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, Elena Yudovina, \textit{Introduction to representation theory}, arXiv:0901.0827v5}. \item The same fact is a particular case of Lemma 3 (b) in: \href{http://www.math.toronto.edu/murnaghan/courses/mat445/Symmetric.pdf}{Graham Gill, \textit{Representation theory of the symmetric group: basic elements}, 16 January 2006}. \item It is also a consequence of the Lemma in \S 2.3 of: \href{http://www1.mat.uniroma1.it/people/procesi/tensor.pdf}{Claudio Procesi, \textit{Chapter 9: Tensor symmetry}, 28 November 2005}. \item It also follows from Lemma (2.6) in: \href{http://math.cmu.edu/~cnewstea/notes/reptheory.pdf}{Stuart Martin, \textit{Part III Representation Theory, Lent Term 2013 (notes by Clive Newstead)}, 26 April 2013} (applied to $\mu=\lambda$, $t_{\lambda}=\pi^{-1}\left( T\right)$ and $t_{\mu}=T$). \item It also follows from Lemma 2 in \S 2 of: \href{https://www1.maths.leeds.ac.uk/~pmtwc/repinv.pdf}{William Crawley-Boevey, \textit{Lectures on representation theory and invariant theory}, 1999} (or, rather, is proven by a similar argument). \end{itemize} (I have only listed references freely available online; of course, there are many others as well.) \item \textbf{Page 9, D\'{e}monstration de Lemme 1.23:} In \textquotedblleft ne v\'{e}rifiant pas $\left( P\right)$\textquotedblright, replace the \textquotedblleft$\left( P\right)$\textquotedblright\ by a textmode \textquotedblleft(P)\textquotedblright. \item \textbf{Page 10, Exemple 1.26:} On the last line of this example, replace \newline\textquotedblleft$\operatorname*{Vect}\left( \left( u_{1}+\cdots+u_{n}\right) \right)$\textquotedblright\ by \textquotedblleft% $\operatorname*{Vect}\left( \left( e_{1}+\cdots+e_{n}\right) \right)$\textquotedblright. \item \textbf{Page 11, D\'{e}monstration de Proposition 1.27:} Let me suggest a different proof of Proposition 1.27 -- one that does not apply the (somewhat obscure) Lemme 1.19, but instead makes use of Proposition 1.28. (This is not circular reasoning, because the proof of Proposition 1.28 does not rely on Proposition 1.27 anywhere.) \textit{Second proof of Proposition 1.27.} We want to prove that $V_{\lambda}$ is irreducible. Since \textquotedblleft irreducible\textquotedblright\ is equivalent to \textquotedblleft indecomposable\textquotedblright\ (for a representation of a finite group), it suffices to show that $V_{\lambda}$ is indecomposable. Thus, let us prove this. Since $C_{\lambda}\neq0$, we have $V_{\lambda}\neq\left\{ 0\right\}$. Thus, it remains to show that $V_{\lambda}$ cannot be written in the form $\left( V_{1}\oplus V_{2}% ,\rho_{1}\oplus\rho_{2}\right)$ for two nonzero subspaces $V_{1}$ and $V_{2}$. Assume the contrary. Thus, $V_{\lambda}=\left( V_{1}\oplus V_{2},\rho _{1}\oplus\rho_{2}\right)$ for two nonzero subspaces $V_{1}$ and $V_{2}$. Thus, both subspaces $V_{1}$ and $V_{2}$ are stable under the action of $\mathcal{S}_{n}$. We have $V_{\lambda}=V_{1}\oplus V_{2}$ and thus $\dim\left( V_{\lambda}\right) =\dim\left( V_{1}\oplus V_{2}\right) =\dim\left( V_{1}\right) +\underbrace{\dim\left( V_{2}\right) }_{\substack{>0\\\text{(since }V_{2}\text{ is nonzero)}}}>\dim\left( V_{1}\right)$. Hence, $V_{\lambda}\not \subseteq V_{1}$. Let $n_{1}$ and $n_{2}$ be as in Proposition 1.28. Then, Proposition 1.28 yields $n_{1}n_{2}\mid n!$, so that $n_{1}n_{2}\neq0$ and thus $n_{2}\neq0$. Now, $C_{\lambda}^{2}=n_{2}C_{\lambda}\neq0$ (since $n_{2}\neq0$ and $C_{\lambda}\neq0$). We have $C_{\lambda}\in\mathbb{C}\left[ \mathcal{S}_{n}\right] C_{\lambda }=V_{\lambda}=V_{1}\oplus V_{2}$. Hence, there exist some $v_{1}\in V_{1}$ and $v_{2}\in V_{2}$ such that $C_{\lambda}=v_{1}+v_{2}$. Consider these $v_{1}$ and $v_{2}$. If we had $C_{\lambda}\in V_{1}$, then we would have $V_{\lambda}% =\mathbb{C}\left[ \mathcal{S}_{n}\right] \underbrace{C_{\lambda}}_{\in V_{1}}\subseteq\mathbb{C}\left[ \mathcal{S}_{n}\right] V_{1}\subseteq V_{1}$ (since $V_{\lambda}$ is stable under the action of $\mathcal{S}_{n}$), which would contradict $V_{\lambda}\not \subseteq V_{1}$. Hence, we cannot have $C_{\lambda}\in V_{1}$. In other words, we have $C_{\lambda}\notin V_{1}$. We have $v_{1}\in V_{1}\subseteq V_{\lambda}=\mathbb{C}\left[ \mathcal{S}% _{n}\right] C_{\lambda}$. In other words, there exists some $w_{1}% \in\mathbb{C}\left[ \mathcal{S}_{n}\right]$ satisfying $v_{1}% =w_{1}C_{\lambda}$. Consider this $w_{1}$. The element $x=C_{\lambda}w_{1}C_{\lambda}$ satisfies the equalities (1) and (2) from the proof of Lemme 1.23 (because of (1.3) and (1.4)). Hence, $x$ is a scalar multiple of $C_{\lambda}$ (as follows from Lemme 1.23). In other words, $C_{\lambda}w_{1}C_{\lambda}=\alpha C_{\lambda}$ for some $\alpha\in \mathbb{C}$. Consider this $\alpha$. We have $\alpha C_{\lambda}=C_{\lambda }\underbrace{w_{1}C_{\lambda}}_{=v_{1}\in V_{1}}\in C_{\lambda}V_{1}\subseteq V_{1}$ (since $V_{\lambda}$ is stable under the action of $\mathcal{S}_{n}$). If $\alpha$ was nonzero, then we would have $C_{\lambda}=\dfrac{1}{\alpha }\underbrace{C_{\lambda}}_{\in V_{1}}\in\dfrac{1}{\alpha}V_{1}\subseteq V_{1}%$, which would contradict $C_{\lambda}\notin V_{1}$. Hence, $\alpha$ cannot be nonzero. Thus, $\alpha=0$. Hence, $C_{\lambda}w_{1}C_{\lambda}% =\underbrace{\alpha}_{=0}C_{\lambda}=0$. Thus, $C_{\lambda}\underbrace{v_{1}% }_{=w_{1}C_{\lambda}}=C_{\lambda}w_{1}C_{\lambda}=0$. Similarly, $C_{\lambda }v_{2}=0$. Now, $C_{\lambda}^{2}=C_{\lambda}\underbrace{C_{\lambda}}_{=v_{1}+v_{2}}=C_{\lambda }\left( v_{1}+v_{2}\right) =\underbrace{C_{\lambda}v_{1}}_{=0}% +\underbrace{C_{\lambda}v_{2}}_{=0}=0.$ This contradicts $C_{\lambda}^{2}\neq0$. This contradiction shows that our assumption was false. Hence, $V_{\lambda}$ cannot be written in the form $\left( V_{1}\oplus V_{2},\rho_{1}\oplus\rho_{2}\right)$ for two nonzero subspaces $V_{1}$ and $V_{2}$. Thus, the representation $V_{\lambda}$ is indecomposable (since $V_{\lambda}\neq\left\{ 0\right\}$), and therefore irreducible (since Maschke's theorem tells us that \textquotedblleft irreducible\textquotedblright\ and \textquotedblleft indecomposable\textquotedblright\ is the same thing). This proves Proposition 1.27 again. $\square$ \item \textbf{Page 12, D\'{e}monstration de Proposition 1.28:} Replace \textquotedblleft$\operatorname*{id}$\textquotedblright\ by \textquotedblleft% $\operatorname*{Id}$\textquotedblright\ twice in this proof. (Maybe it is easier to just root out these errors by a find-replace? Or do you use $\operatorname*{Id}$ and $\operatorname*{id}$ for two different things at some point?) \item \textbf{Page 18, D\'{e}monstration de Lemma 2.2:} On the second line of this proof, the words \textquotedblleft dans ce cas\textquotedblright\ sound somewhat inappropriate to me (you aren't really focussing on a case). Maybe something like \textquotedblleft for this reason\textquotedblright\ would be better. \item \textbf{Page 18:} The last equality sign in the equality% \begin{align*} \chi^{\lambda}\left( \pi\right) & :=\operatorname*{Tr}\left( \rho\left( \pi\right) \right) =\operatorname*{Tr}\left( \varphi_{\lambda,\pi}\right) \\ & =\dfrac{\dim\left( V_{\lambda}\right) }{n!}\sum_{\substack{\sigma \in\mathcal{S}_{n}\\\sigma\in\operatorname*{RS}\left( T\right) }% }\sum_{\substack{\tau\in\mathcal{S}_{n}\\\tau\in\operatorname*{CS}\left( T\right) }}\varepsilon\left( \tau\right) \operatorname*{Tr}\left( x\mapsto e_{\pi}\cdot x\cdot e_{\sigma}\cdot e_{\tau}\right) \end{align*} requires some work to verify; maybe it's better to add some explanation such as \textquotedblleft because the definition of $\varphi_{\lambda,\pi}$ becomes \begin{align*} \varphi_{\lambda,\pi} & =\dfrac{\dim\left( V_{\lambda}\right) }{n!}\left( x\mapsto e_{\pi}\cdot x\cdot C_{\lambda}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }p_{\lambda}\right) \\ & =\dfrac{\dim\left( V_{\lambda}\right) }{n!}\sum_{\substack{\sigma \in\mathcal{S}_{n}\\\sigma\in\operatorname*{RS}\left( T\right) }% }\sum_{\substack{\tau\in\mathcal{S}_{n}\\\tau\in\operatorname*{CS}\left( T\right) }}\varepsilon\left( \tau\right) \left( x\mapsto e_{\pi}\cdot x\cdot e_{\sigma}\cdot e_{\tau}\right) \\ & \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }C_{\lambda}\right) . \end{align*} \textquotedblright. \item \textbf{Page 18:} Before \textquotedblleft Nous allons manipuler\textquotedblright, I'd suggest adding something like \textquotedblleft Therefore,% \begin{align*} \dfrac{n!\chi^{\lambda}\left( \pi\right) }{\dim\left( V_{\lambda}\right) } & =\sum_{\substack{\sigma\in\mathcal{S}_{n}\\\sigma\in\operatorname*{RS}% \left( T\right) }}\sum_{\substack{\tau\in\mathcal{S}_{n}\\\tau \in\operatorname*{CS}\left( T\right) }}\varepsilon\left( \tau\right) \operatorname*{Card}\left\{ g\in\mathcal{S}_{n}\ \mid\ g=\pi\cdot g\cdot\sigma\cdot\tau\right\} \\ & =\sum_{\substack{\sigma\in\mathcal{S}_{n}\\\sigma\in\operatorname*{RS}% \left( T\right) }}\sum_{\substack{\tau\in\mathcal{S}_{n}\\\tau \in\operatorname*{CS}\left( T\right) }}\varepsilon\left( \tau\right) \sum_{g\in\mathcal{S}_{n}}\delta_{g,\pi\cdot g\cdot\sigma\cdot\tau}=\sum _{g\in\mathcal{S}_{n}}\sum_{\substack{\sigma\in\operatorname*{RS}\left( T\right) \\\tau\in\operatorname*{CS}\left( T\right) }}\varepsilon\left( \tau\right) \delta_{g,\pi\cdot g\cdot\sigma\cdot\tau}. \end{align*} \textquotedblright.\ This would help bridge the gap to the computations on page 19 (which otherwise seem to come out of the blue). \item \textbf{Page 19:} Replace \textquotedblleft$\delta_{\pi=\tau^{\prime }\sigma^{\prime}}$\textquotedblright\ by \textquotedblleft$\delta_{\pi ,\tau^{\prime}\sigma^{\prime}}$\textquotedblright. \item \textbf{Page 19:} Replace \textquotedblleft$\delta_{\pi=\tau\sigma}%$\textquotedblright\ by \textquotedblleft$\delta_{\pi,\tau\sigma}%$\textquotedblright. \item \textbf{Page 19:} Replace \textquotedblleft$CS\left( T\right)$\textquotedblright\ by \textquotedblleft$\operatorname*{CS}\left( T\right)$\textquotedblright\ (twice on this page). Also, replace \textquotedblleft$RS\left( T\right)$\textquotedblright\ by \textquotedblleft$\operatorname*{RS}\left( T\right)$\textquotedblright. (Again, this kind of mistake might be susceptible to an automated search.) \item \textbf{Page 20, Remarque 2.6:} Replace \textquotedblleft$CS\left( T\right)$\textquotedblright\ by \textquotedblleft$\operatorname*{CS}\left( T\right)$\textquotedblright. Also, replace \textquotedblleft$RS\left( T\right)$\textquotedblright\ by \textquotedblleft$\operatorname*{RS}\left( T\right)$\textquotedblright. \item \textbf{Page 21, D\'{e}monstration de Proposition 2.7:} Replace \textquotedblleft de la d\'{e}finition (2.5)\textquotedblright\ by \textquotedblleft de la D\'{e}finition 2.5\textquotedblright. (You are citing a definition, not a formula.) \item \textbf{Page 22, D\'{e}monstration de Proposition 2.7:} Replace \textquotedblleft sont dans la m\^{e}me case\textquotedblright\ by \textquotedblleft sont la m\^{e}me case\textquotedblright. \item \textbf{Page 22, D\'{e}monstration de Proposition 2.7:} In the last equation in this proof, replace the \textquotedblleft$\varepsilon\left( \tau\right)$\textquotedblright\ (on the right hand side) by an \textquotedblleft$\varepsilon\left( \check{\tau}\right)$\textquotedblright% \ (and maybe add a justification, such as \textquotedblleft because $\varepsilon\left( \tau_{\mid\left\{ 1,\ldots,k\right\} }\right) =\varepsilon\left( \tau\right)$\textquotedblright). \item \textbf{Page 22, Exemple 2.9:} Replace \textquotedblleft$\in S_{2}%$\textquotedblright\ by \textquotedblleft$\in\mathcal{S}_{2}$% \textquotedblright. \item \textbf{Page 23, Exemple 2.9:} Replace \textquotedblleft Le th\'{e}or\{e}me nous donne\textquotedblright\ by \textquotedblleft Proposition 2.7 nous donne\textquotedblright. \item \textbf{Page 24, D\'{e}monstration de Th\'{e}or\{e}me 2.13:} In this proof, you are using the notation $\delta_{\mathcal{A}}$ for $% \begin{cases} 1, & \text{if }\mathcal{A}\text{ is true;}\\ 0, & \text{if }\mathcal{A}\text{ is false}% \end{cases}$ when $\mathcal{A}$ is any logical statement. It might be worth defining this notation. \item \textbf{Page 26, Corollaire 2.16:} Replace \textquotedblleft$\dim\left( V^{p\times q}\right)$\textquotedblright\ by \textquotedblleft$\dim\left( V_{p\times q}\right)$\textquotedblright. \item \textbf{Page 26, Remarque 2.17:} Replace \textquotedblleft$\dim\left( V^{p\times q}\right)$\textquotedblright\ by \textquotedblleft$\dim\left( V_{p\times q}\right)$\textquotedblright. \item \textbf{Page 27, Corollaire 2.22:} I'd clarify here that $\lambda$ is considered fixed. \item \textbf{Page 27, Exemple 2.24:} These aren't all the possible graphs; you are missing the ones where one of $V_{\circ}\left( G\right)$ and $V_{\bullet}\left( G\right)$ is empty :) \item \textbf{Page 27, Exemple 2.24:} I don't think you have ever defined the notation $\ell\left( \lambda\right)$ for the length of a partition $\lambda$. \item \textbf{Page 27, Remarque 2.25:} I know it looks totally obvious, but it wouldn't hurt to explain what the \textquotedblleft sommets blancs\textquotedblright\ of a bipartite graph $G$ are (namely, the elements of $V_{\circ}\left( G\right)$), and what the \textquotedblleft sommets noirs\textquotedblright\ of a bipartite graph $G$ are (namely, the elements of $V_{\bullet}\left( G\right)$). \item \textbf{Page 29, Proposition 2.31:} The sum on the right hand side has many undefined addends. You should either replace \textquotedblleft% $\varphi:V_{G}\rightarrow\mathbb{N}^{\ast}$\textquotedblright\ by \textquotedblleft$\varphi:V_{G}\rightarrow\left\{ 1,2,\ldots,m\right\}$\textquotedblright\ under the summation sign (saying that $m$ is the size of $\mathbf{p}$), or define the values $p_{i}$ and $q_{i}$ to be $0$ for $i>m$. \item \textbf{Page 29, D\'{e}monstration de Proposition 2.31:} You write: \textquotedblleft On part du lemme 2.19\textquotedblright, but the setting of Proposition 2.31 is more general than the setting of Lemme 2.19 (for example, the bipartite graph $G$ in Proposition 2.31 may have isolated vertices, which can never happen for a bipartite graph of the form $G_{\sigma,\tau}$). You probably want to refer to D\'{e}finition 2.23 instead. Furthermore, whenever you refer to \textquotedblleft$\left( \star\right) _{h}$\textquotedblright, you mean the third condition in D\'{e}finition 2.23. \item \textbf{Page 29, D\'{e}monstration de Proposition 2.31:} Replace \textquotedblleft$\varphi\left( v\right) =j$ si $q_{1}+\cdots+q_{j-1}% i \end{cases} \ \ \ \ \ \ \ \ \ \ \text{for all }v\in V_{\circ}\sqcup V_{\bullet}% \] where$i$is some number and$\mathcal{C}$is some predicate; \item under some other conditions (the \textquotedblleft merging case\textquotedblright), the involution maps a$k$-packed map$\varphi$to a$\left( k-1\right) $-packed map$\varphi^{\prime}$given by $\varphi^{\prime}\left( v\right) =% \begin{cases} \varphi\left( v\right) , & \text{if }\varphi\left( v\right) i+1 \end{cases} \ \ \ \ \ \ \ \ \ \ \text{for all }v\in V_{\circ}\sqcup V_{\bullet}%$ where$i$is some number; \item in the remaining (rare) cases, the involution sends$\varphi$to$\varphi\$. \end{itemize} If we are lucky, then this involution might generalize to an answer to Question 1 as well. \end{itemize} \end{document}