\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, October 16, 2019 01:55:28} %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}} \let\sumnonlimits\sum \let\prodnonlimits\prod \let\bigcapnonlimits\bigcap \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \renewcommand{\bigcap}{\bigcapnonlimits\limits} \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \ihead{Errata to Lecture Notes Combinatorics''} \ohead{\today} \begin{document} \begin{center} \textbf{Lecture Notes Combinatorics} \textit{Torsten Ueckerdt} version of 30 May 2017, downloadable from \url{http://www.math.kit.edu/iag6/lehre/combinatorics2017s/media/script.pdf} \textbf{Errata and addenda by Darij Grinberg} \bigskip \end{center} I will refer to the results appearing in the notes \textquotedblleft Lecture Notes Combinatorics\textquotedblright\ by the numbers under which they appear in these notes (specifically, in its version of 30 May 2017). \setcounter{section}{7} \section{Errata} \begin{itemize} \item \textbf{general:} Let me clarify the meaning of some notation that is not standardized across the literature: \begin{itemize} \item The symbol $\mathbb{N}$ stands for the set $\left\{ 0,1,2,\ldots \right\}$. \item The symbol $\subset$ means \textquotedblleft proper subset\textquotedblright, while the symbol $\subseteq$ stands for \textquotedblleft subset\textquotedblright. \end{itemize} \item \textbf{page 6:} In the proof of the \textquotedblleft Claim\textquotedblright\ at the top (about the tileability of the $m\times n$ board), replace the equations% \begin{align*} \#\text{ squares colored with }1 & =ks_{1}s_{2}+s_{1}r_{2}+s_{2}r_{1}% +r_{2}\\ \#\text{ squares colored with }2 & =ks_{1}s_{2}+s_{1}r_{2}+s_{2}r_{1}% +r_{2}-1 \end{align*} by% \begin{align*} \#\text{ squares colored with }1 & =ks_{1}s_{2}+s_{1}r_{2}+s_{2}r_{1}% +r_{1}\\ \#\text{ squares colored with }2 & =ks_{1}s_{2}+s_{1}r_{2}+s_{2}r_{1}% +r_{1}-1. \end{align*} (After all, the main diagonal in the $r_{1}\times r_{2}$ rectangle crosses as many squares as the shorter side, and the latter is $r_{1}$, not $r_{2}$.) \item \textbf{page 7:} \textquotedblleft Temperly\textquotedblright% \ $\rightarrow$ \textquotedblleft Temperley\textquotedblright. \item \textbf{page 9, \S 1.1:} \textquotedblleft where\textquotedblright% \ $\rightarrow$ \textquotedblleft were\textquotedblright. \item \textbf{page 10, \S 1.1.1:} I would replace \textquotedblleft the parts are disjoint\textquotedblright\ by \textquotedblleft the parts $S_{1}% ,\ldots,S_{k}$ are disjoint sets\textquotedblright\ in order to clarify what is meant by \textquotedblleft parts\textquotedblright. \item \textbf{page 10, \S 1.1.2:} Pedantic remark: The $S_{i}$ need to be required to be finite. Otherwise, it could happen that one of the $S_{i}$ is infinite but their product is nevertheless $\varnothing$ (since any set times $\varnothing$ is $\varnothing$). \item \textbf{page 10, \S 1.1.2:} In the formula \textquotedblleft$\left\vert S\right\vert =\left\vert S_{1}\right\vert \times\left\vert S_{2}\right\vert \times\cdots\times\left\vert S_{m}\right\vert$\textquotedblright, replace \textquotedblleft$m$\textquotedblright\ by \textquotedblleft$k$% \textquotedblright. \item \textbf{page 11, \S 1.1.3:} In the first sentence of the Example, replace \textquotedblleft If $T$ is\textquotedblright\ by \textquotedblleft Let $T$ be\textquotedblright. \item \textbf{page 11, \S 1.1.3:} \textquotedblleft\textit{substraction}% \textquotedblright\ $\rightarrow$ \textquotedblleft\textit{subtraction}% \textquotedblright \item \textbf{page 11, \S 1.1.4:} \textquotedblleft is a a\textquotedblright% \ $\rightarrow$ \textquotedblleft is a\textquotedblright. \item \textbf{page 11, \S 1.1.5:} The sets $S_{1},\ldots,S_{m}$ (and their disjointness) are irrelevant here, since the statement only uses cardinalities, which can just as well be arbitrary integers. But is this really worth calling the \textquotedblleft pigeonhole principle\textquotedblright? To me, the \textquotedblleft pigeonhole principle\textquotedblright\ is one of the following two facts: \begin{itemize} \item An injective map $f:A\rightarrow B$ between two finite sets $A$ and $B$ is automatically bijective if $\left\vert A\right\vert \geq\left\vert B\right\vert$. \item A surjective map $f:A\rightarrow B$ between two finite sets $A$ and $B$ is automatically bijective if $\left\vert A\right\vert \leq\left\vert B\right\vert$. \end{itemize} Note that the first of these two facts is tacitly used on page 13 (in \textquotedblleft we have $S_{n}=P\left( n,n\right)$\textquotedblright), so it is worth stating these two facts here. In contrast, I know your \textquotedblleft pigeonhole principle\textquotedblright\ as the \textquotedblleft Lake Wobegon principle\textquotedblright\ (\textquotedblleft where all the children are above average\textquotedblright). \item \textbf{page 12, \S 1.1.6:} \textquotedblleft how two count\textquotedblright\ $\rightarrow$ \textquotedblleft how to count\textquotedblright. \item \textbf{page 12, \S 1.2:} \textquotedblleft the first $n$ natural number\textquotedblright\ $\rightarrow$ \textquotedblleft the first $n$ natural numbers\textquotedblright. \item \textbf{page 12, \S 1.2:} I would not require $X$ to be finite in the first sentence here, but only in Definition 1.1. I am sure you end up using tuples of elements from infinite sets at some point... \item \textbf{page 13, \S 1.2:} In the \textquotedblleft string\textquotedblright\ section, I would replace \textquotedblleft is denoted by\textquotedblright\ by \textquotedblleft is defined by\textquotedblright. After all, we are not defining $s\left( i\right)$ here, but rather the notions of \textquotedblleft position\textquotedblright% \ and \textquotedblleft character\textquotedblright. That said, I think the word \textquotedblleft position\textquotedblright\ is out of place here. When I hear \textquotedblleft position\textquotedblright, I rather think of a preimage (i.e., a \textquotedblleft position of $x$ in the string $s$\textquotedblright\ is an $i$ such that $s\left( i\right) =x$), not an image under $s$. (And that's how you use this word several times further down.) It is probably best to replace \textquotedblleft The $i$-th position (or \textit{character})\textquotedblright\ by \textquotedblleft The $i$-th \textit{character} (or the \textit{character in position }$i$% )\textquotedblright. \item \textbf{page 13, \S 1.2:} The same problem with the meaning of \textquotedblleft position\textquotedblright\ appears in the \textquotedblleft sequence\textquotedblright\ section. \item \textbf{page 13, \S 1.2:} After \textquotedblleft but could have, for instance,\textquotedblright, add \textquotedblleft if\textquotedblright. \item \textbf{page 13, \S 1.2.1:} I think it is worth explaining that the definition of a permutation you state here is only one of two (competing) standard definitions of \textquotedblleft permutation\textquotedblright. (The other definition -- used, e.g., in Bourbaki -- defines a permutation of a set $X$ to be a bijection from $X$ to $X$. The two definitions coincide when $X=\left[ n\right]$, but otherwise are not equivalent.) \item \textbf{page 13, \S 1.2.1:} In the definition of \textquotedblleft% $k$-permutation\textquotedblright, I would allow arbitrary $k$; in particular, $k$ can be $0$. This is very likely to come up useful in the base of some induction proof or otherwise... \item \textbf{page 14, Theorem 1.2:} In part (i), I would allow the case $k=0$. (Of course, in part (ii) I would not.) \item \textbf{page 14, proof of Theorem 1.2:} The word \textquotedblleft permutation\textquotedblright\ is being abused for \textquotedblleft% $k$-permutation\textquotedblright\ here. I would suggest correcting this imprecision here and elsewhere where it appears (e.g., on the next page: \textquotedblleft$k$ permutations\textquotedblright\ $\rightarrow$ \textquotedblleft$k$ distinct $k$-permutations\textquotedblright), since it conflicts with your own definition of \textquotedblleft permutation\textquotedblright\ a page ago. \item \textbf{page 15, \S 1.3:} Again, the finiteness of $X$ is unnecessary here (except in Definition 1.3, but you already require it there). \item \textbf{page 15, Definition 1.3:} In the definition of \textquotedblleft% $k$-Combination of a Multiset\textquotedblright, the indexing of the $r_{i}$ and $s_{i}$ is somewhat unclear (it presumes that the elements of $X$ are numbered, and this numbering is inherited by $M$, but this is not given a priori). Wouldn't it be clearer to instead define \textquotedblleft A $k$\textit{-combination} of $M$ is a multiset of size $k$ with types in $X$ in which each element occurs at most as often as in $X$\textquotedblright\ ? \item \textbf{page 16, Definition 1.3:} The definition of \textquotedblleft% $k$-Permutation of a Multiset\textquotedblright\ confuses me: What is an \textquotedblleft ordered multiset\textquotedblright? What are \textquotedblleft different orderings of elements of the same type\textquotedblright? For me, a $k$\textit{-permutation} of a multiset $M$ is simply defined as a $k$-tuple in $X^{k}$ (where $X$ is the set of types of $M$) in which each element of $X$ appears at most as often as in $M$. This definition seems to match the examples you give below. \item \textbf{page 17:} \textquotedblleft There $n-k_{1}-\cdots-k_{j-1}$ indices left to choose from\textquotedblright\ $\rightarrow$ \textquotedblleft There are $n-k_{1}-\cdots-k_{j-1}$ indices left to choose from\textquotedblright. \item \textbf{page 19:} The comma in front of the \textquotedblleft however\textquotedblright\ should be a period or a semicolon; otherwise, it is unclear whether it refers to the part of the sentence before or after the \textquotedblleft however\textquotedblright. \item \textbf{page 19, Example:} The list after \textquotedblleft Trying to list them all\textquotedblright\ has some misplaced commas and parentheses. \item \textbf{page 19, Theorem 1.8:} In \textquotedblleft numbers $r_{1}% ,r_{2},\ldots r_{k}$\textquotedblright, add a comma before \textquotedblleft% $r_{k}$\textquotedblright. \item \textbf{page 19, last line:} \textquotedblleft(i.e. number of subsets of $\left[ n\right]$)\textquotedblright\ $\rightarrow$ \textquotedblleft(i.e. number of $k$-subsets of $\left[ n\right]$)\textquotedblright. \item \textbf{page 21:} \textquotedblleft number of $2$ permutations\textquotedblright\ $\rightarrow$ \textquotedblleft number of $2$-permutations\textquotedblright. \item \textbf{page 21:} \textquotedblleft only of one\textquotedblright% \ $\rightarrow$ \textquotedblleft only one\textquotedblright. \item \textbf{page 21:} \textquotedblleft of $n$ element multisets\textquotedblright\ $\rightarrow$ \textquotedblleft of $n$-element multisets\textquotedblright. \item \textbf{page 24:} In the \textquotedblleft arbitrary number of labels per box\textquotedblright\ section, specifically in its item 1, you should assume $k>0$ (otherwise, it won't be $\left( k-1\right) \cdot\mid$ but rather $0\cdot\mid$ in the multiset). \item \textbf{page 24:} In the \textquotedblleft arbitrary number of labels per box\textquotedblright\ section, specifically in its item 3, replace \textquotedblleft to the remaining boxes\textquotedblright\ by \textquotedblleft to the remaining\textquotedblright, and replace \textquotedblleft non of\textquotedblright\ by \textquotedblleft none of\textquotedblright. \item \textbf{page 25:} In the \textquotedblleft$\geq1$ ball per box\textquotedblright\ section, the comma after \textquotedblleft This amounts to $2^{n}-2$ possibilities\textquotedblright\ should be a period; otherwise, it is ambiguous whether the \textquotedblleft however\textquotedblright% \ affects the part of the sentence before or after it. \item \textbf{page 26:} \textquotedblleft to an arrangements\textquotedblright% \ $\rightarrow$ \textquotedblleft to an arrangement\textquotedblright. \item \textbf{page 27, \S 1.5.2:} \textquotedblleft non of\textquotedblright% \ $\rightarrow$ \textquotedblleft none of\textquotedblright. \item \textbf{page 28:} \textquotedblleft flee\textquotedblright% \ $\rightarrow$ \textquotedblleft flea\textquotedblright. \item \textbf{page 29:} \textquotedblleft(for $0\leq i\leq k$% )\textquotedblright\ $\rightarrow$ \textquotedblleft(for $1\leq i\leq k$)\textquotedblright\ (in the last case of the proof of $p_{k}\left( n\right) =p_{k}\left( n-k\right) +p_{k-1}\left( n-1\right)$). \item \textbf{page 29:} In the \textquotedblleft$\geq1$ ball per box\textquotedblright\ section, replace add a plus sign before the last \textquotedblleft$1$\textquotedblright\ in \textquotedblleft% $n=\underbrace{1+1+\cdots1}_{n\text{times}}$\textquotedblright, and add a whitespace between \textquotedblleft$n$\textquotedblright\ and \textquotedblleft times\textquotedblright. \item \textbf{page 29:} In the \textquotedblleft arbitrary number of balls per box\textquotedblright\ section, replace \textquotedblleft how many boxes $i$ should be non-empty\textquotedblright\ by \textquotedblleft how many boxes should be non-empty (call this number $i$)\textquotedblright. \item \textbf{page 29:} The summation sign in the \textquotedblleft arbitrary number of balls per box\textquotedblright\ section should start at $i=0$ instead of at $i=1$ (unless you want to require $n\geq1$, but why would you?). The same applies to the southeasternmost cell in Table 1 on page 30. \item \textbf{page 30, \S 1.5.5:} In the table, replace \textquotedblleft non-negati ve\textquotedblright\ by \textquotedblleft non-negative\textquotedblright. \item \textbf{page 30, Example 1.14:} \textquotedblleft A monotone lattice paths\textquotedblright\ $\rightarrow$ \textquotedblleft A \textit{monotone lattice path}\textquotedblright. \item \textbf{page 30, Example 1.14:} Add a semicolon before \textquotedblleft see Figure 12\textquotedblright. \item \textbf{page 31, footnote }$^{2}$\textbf{:} \textquotedblleft week\textquotedblright\ $\rightarrow$ \textquotedblleft weak\textquotedblright% . (This is not the first time I am seeing this typo; I remember spotting a \textquotedblleft weekly increasing sequence\textquotedblright\ in the literature.) \item \textbf{page 32, proof of Theorem 1.15:} \textquotedblleft tiles of with\textquotedblright\ $\rightarrow$ \textquotedblleft tiles with\textquotedblright. Better yet, I would replace \textquotedblleft tiles of with sizes of the first $n$ odd numbers\textquotedblright\ by \textquotedblleft tiles with sizes $1,3,5,\ldots,2n-1$ (the first $n$ odd positive integers)\textquotedblright. \item \textbf{page 33, proof of Theorem 1.16 (i):} \textquotedblleft$n$ permutations\textquotedblright\ $\rightarrow$ \textquotedblleft$n$% -permutations\textquotedblright\ (what a difference a little hyphen can make). \item \textbf{page 35, proof of Theorem 1.18:} Remove the colon after \textquotedblleft For general\textquotedblright. \item \textbf{page 35, proof of Theorem 1.18:} \textquotedblleft placed all number\textquotedblright\ $\rightarrow$ \textquotedblleft placed all numbers\textquotedblright. \item \textbf{page 37, \S 1.7.2:} \textquotedblleft to connect i with $\pi\left( i\right)$\textquotedblright\ $\rightarrow$ \textquotedblleft to connect $i$ with $\pi\left( i\right)$\textquotedblright\ (the \textquotedblleft$i$\textquotedblright\ should be in mathmode). Also, I would suggest reformulating this sentence in a less telegraphic way, such as: \textquotedblleft to draw a diagram in which the numbers $1,2,\ldots,n$ are arranged (in order) along a \textquotedblleft top\textquotedblright% \ horizontal line and the same numbers are also arranged (in order) along a \textquotedblleft bottom\textquotedblright\ horizontal line (with each \textquotedblleft top\textquotedblright\ number $i$ vertically facing the corresponding \textquotedblleft bottom\textquotedblright\ $i$), and each top number $i$ is connected with the bottom $\pi\left( i\right)$% \textquotedblright. \item \textbf{page 38:} \textquotedblleft are group by\textquotedblright% \ $\rightarrow$ \textquotedblleft are grouped by\textquotedblright. \item \textbf{page 38:} \textquotedblleft this composition\textquotedblright% \ $\rightarrow$ \textquotedblleft this decomposition\textquotedblright. \item \textbf{page 39, between Theorem 1.20 and Theorem 1.21:} \textquotedblleft smallest $k$\textquotedblright\ $\rightarrow$ \textquotedblleft smallest positive integer $k$\textquotedblright. \item \textbf{page 39, proof of Theorem 1.21:} In \textquotedblleft parsing $i$ through $i$\textquotedblright, remove the second \textquotedblleft% $i$\textquotedblright. \item \textbf{page 39, \S 1.7.3:} After \textquotedblleft\textit{discriminant} of $\pi$\textquotedblright, add \textquotedblleft$\in S_{n}$\textquotedblright% \ in order to clarify what the symbol $n$ later refers to. \item \textbf{page 40, Theorem 1.22:} \textquotedblleft the product\textquotedblright\ $\rightarrow$ \textquotedblleft a product\textquotedblright. (This appears twice in the theorem and twice in the proof.) \item \textbf{page 43, proof of Recursion (i):} In \textquotedblleft(case 2)\textquotedblright, replace every \textquotedblleft$\phi$\textquotedblright% \ by \textquotedblleft$\pi$\textquotedblright\ starting with \textquotedblleft If $ir\geq0$. This makes the proof somewhat easier to grasp, in my opinion. Of course, when you use the Lemma later on (on page 51), you apply it to $m=2n$, but you can just say that. \item \textbf{page 50, proof of Lemma:} When you say \textquotedblleft the remaining positions\textquotedblright\ in Case 1, you mean \textquotedblleft the positions $2,3,\ldots,2n-1$\textquotedblright\ (not \textquotedblleft the positions $2,3,\ldots,2n$\textquotedblright\ as one would normally expect in this context). \item \textbf{page 52, \S 2.1.1:} For the clarity, I'd add a comma before \textquotedblleft the total number\textquotedblright. \item \textbf{page 52:} \textquotedblleft we proof\textquotedblright% \ $\rightarrow$ \textquotedblleft we prove\textquotedblright. \item \textbf{page 53:} \textquotedblleft We now proof\textquotedblright% \ $\rightarrow$ \textquotedblleft We now prove\textquotedblright. \item \textbf{page 53, proof of Theorem 2.8:} I would replace \textquotedblleft for appropriate $c_{T}$ (to be determined!)\textquotedblright\ by \textquotedblleft for $c_{T}=\sum \limits_{T\subseteq S\subseteq A}\left( -1\right) ^{\left\vert A\right\vert -\left\vert S\right\vert }$\textquotedblright; this would be clearer and more concrete. \item \textbf{page 54:} Remove the comma before \textquotedblleft is the number $k=\gcd\left( m,n\right)$\textquotedblright. \item \textbf{page 54:} In the definition of the $\phi$-function, replace \textquotedblleft$n\geq2$\textquotedblright\ by \textquotedblleft$n\geq 1$\textquotedblright. Otherwise, one of the addends in the sum in Theorem 2.10 is undefined. \item \textbf{page 54, Theorem 2.9:} Here, you want to require that $a_{i}>0$ for all $i$, and that the $p_{i}$ are distinct primes. \item \textbf{page 54(?):} It would be good to explain that sums of the form $\sum\limits_{d\mid n}$ are understood to be sums over all \textbf{positive} divisors $d$ of $n$ (rather than all divisors). \item \textbf{page 55:} \textquotedblleft they do not have a square as a factor\textquotedblright\ $\rightarrow$ \textquotedblleft they do not have a square $>1$ as a factor\textquotedblright. \item \textbf{page 55, proof of Corollary 2.12:} \textquotedblleft% $-1^{\left\vert S\right\vert }$\textquotedblright\ $\rightarrow$ \textquotedblleft$\left( -1\right) ^{\left\vert S\right\vert }%$\textquotedblright. \item \textbf{page 56, proof of Theorem 2.13:} I would replace \textquotedblleft Where $c_{d^{\prime}}$ are numbers (to be determined!) that count how often $f\left( d^{\prime}\right)$ occurs as a summand\textquotedblright\ by the shorter and clearer \textquotedblleft where $c_{d^{\prime}}=\sum\limits_{d^{\prime}\mid d\mid n}\mu\left( \dfrac{n}% {d}\right)$\textquotedblright. \item \textbf{page 59, Example 3.1 (i):} \textquotedblleft is one\textquotedblright\ $\rightarrow$ \textquotedblleft is $1$% \textquotedblright. Otherwise, it is too easily misunderstood as \textquotedblleft is an infinite geometric series\textquotedblright. \item \textbf{page 63, proof of Lemma 3.9:} The last line of this proof has a \textquotedblleft$\left( n-1\right) !\left( n-1\right) \cdot n\cdot n$\textquotedblright\ in the denominator; this should be a \textquotedblleft% $\left( n-1\right) !\left( n-1\right) !\cdot n\cdot n$\textquotedblright. \item \textbf{page 64:} You want to replace \textquotedblleft$\mathbb{R}% \left[ x\right]$\textquotedblright\ by \textquotedblleft$\mathbb{R}\left[ \left[ x\right] \right]$\textquotedblright; after all, you are getting a power series, not a polynomial. For the same reason, \textquotedblleft linear combination of the basis\textquotedblright\ should be replaced by \textquotedblleft infinite linear combination of the topological basis\textquotedblright\ (though it is probably more elementary to avoid talking about bases altogether, and just say instead that each $f\in \mathbb{R}\left[ \left[ x\right] \right]$ has a unique representation of the form $\sum_{n=0}^{\infty}a_{n}x^{n}\qquad\text{with }a_{n}\in\mathbb{R},$ a unique representation of the form $\sum_{n=0}^{\infty}a_{n}p\left( x,n\right) \qquad\text{with }a_{n}% \in\mathbb{R},$ a unique representation of the form $\sum_{n=0}^{\infty}a_{n}\dfrac{x^{n}}{n!}\qquad\text{with }a_{n}\in \mathbb{R},$ and a unique representation of the form $\sum_{n=0}^{\infty}a_{n}e^{-x}\dfrac{x^{n}}{n!}\qquad\text{with }a_{n}% \in\mathbb{R}.$ As to the \textquotedblleft basis\textquotedblright\ $\left\{ \dfrac{1}% {n^{x}}\right\} _{n\in\mathbb{N}}$, I would leave it out completely; it is not a basis of $\mathbb{R}\left[ \left[ x\right] \right]$ in any meaning of this word, but rather a topological basis of a much different ring, which looks a bit like formal power series but uses multiplication instead of addition for exponents. (It is the ring of formal Dirichlet series.) Since you seem to only be mentioning Dirichlet series as an aside, I am not sure if they are worth mentioning at all. \item \textbf{page 64, Theorem 3.11 and before:} I am pretty sure all the three sums \textquotedblleft$\sum_{n=0}^{\infty}\dfrac{a_{n}}{n^{s}}%$\textquotedblright, \textquotedblleft$\sum_{n=0}^{\infty}\dfrac{\mu\left( n\right) }{n^{s}}$\textquotedblright\ and \textquotedblleft$\sum _{n=0}^{\infty}\dfrac{1}{n^{s}}$\textquotedblright\ should start at $n=1$ rather than at $n=0$. \item \textbf{page 85:} \textquotedblleft A partition of $\left[ n\right]$ into\textquotedblright\ $\rightarrow$ \textquotedblleft A partition of $\left[ n\right]$\textquotedblright. \item \textbf{page 85, proof of Theorem 4.2:} It is worth pointing out that this proof relies on the concept of power series from complex analysis rather than the formal kind of power series (\textquotedblleft power series = sequence of coefficients\textquotedblright). If you just used the formal kind of power series, then $e^{e^{x}}$ would be undefined. \item \textbf{page 86, \S 4.1.1:} The definition of \textquotedblleft crossing\textquotedblright\ here is not symmetric in $A$ and $B$. You probably want to add \textquotedblleft or $\left\{ i,k\right\} \subseteq B,\left\{ j,l\right\} \subseteq A$\textquotedblright\ at the end of the sentence. \item \textbf{page 87, proof of Theorem 4.3:} \textquotedblleft every part contains either only numbers that are bigger than $k$ or only numbers that are at most $k$\textquotedblright\ is wrong (strictly speaking), since the part $S$ satisfies both. Apparently, you are talking not about the partition $\mathcal{P}$ but about the partition obtained from $\mathcal{P}$ upon removing the part $S$. \item \textbf{page 87, \S 4.2:} \textquotedblleft as the sum\textquotedblright% \ $\rightarrow$ \textquotedblleft as a sum\textquotedblright. \item \textbf{page 87, \S 4.2:} I suggest including a rigorous definition of a partition. (Namely: A \textit{partition} of a nonnegative integer $n$ is defined as a weakly decreasing finite tuple of positive integers whose sum is $n$.) This notion is a bit too important for a \textquotedblleft definition by example\textquotedblright. Also, the notation \textquotedblleft$n=\lambda$\textquotedblright\ for \textquotedblleft$\lambda$ is a partition of $n$\textquotedblright\ is really non-standard and, in my opinion, bad (it abuses the equality sign, tempting to jump to nonsensical conclusions like \textquotedblleft if $n=\lambda$ and $n=\mu$ then $\lambda=\mu$\textquotedblright). \item \textbf{pages 87--88:} The notion of the \textquotedblleft size\textquotedblright\ of a partition should be defined. (Namely: The \textit{size} of a partition $\lambda$ is defined to be the nonnegative integer $n$ such that $\lambda$ is a partition of $n$.) \item \textbf{page 88:} In the recursive formula for $p_{k}\left( n\right)$, the equality sign is missing. \item \textbf{page 88 and further on:} \textquotedblleft Ferrer\textquotedblright\ $\rightarrow$ \textquotedblleft Ferrers\textquotedblright. \item \textbf{page 89, proof of Theorem 4.5:} \textquotedblleft$\sum_{i\geq0}%$\textquotedblright\ $\rightarrow$ \textquotedblleft$\sum_{i\geq1}%$\textquotedblright. \item \textbf{page 89:} In the long formula (after \textquotedblleft This is no coincidence\textquotedblright), the product $\prod_{k=0}^{\infty}\left( 1+x^{k}\right)$ should start at $k=1$, not at $k=0$. \item \textbf{page 91, proof of Lemma 4.7:} The notion of a \textquotedblleft staircase\textquotedblright\ should be defined -- for example, as a sequence of squares of the form% $\left( \left( i,j\right) ,\left( i+1,j-1\right) ,\left( i+2,j-2\right) ,\ldots,\left( i+g,j-g\right) \right) \qquad\text{for some }g\geq0.$ Since the parts of the partition are distinct, it is clear that if a staircase starts in the top-right square of the diagram, then each square of this staircase must be the rightmost square of its row. But this is worth mentioning. \item \textbf{page 91, proof of Lemma 4.7:} \textquotedblleft define the bottom\textquotedblright\ $\rightarrow$ \textquotedblleft define the \textit{bottom}\textquotedblright\ (to italicize the word \textquotedblleft bottom\textquotedblright). \item \textbf{page 91, proof of Lemma 4.7:} \textquotedblleft three types partitions\textquotedblright\ $\rightarrow$ \textquotedblleft three types of partitions\textquotedblright. \item \textbf{page 92, proof of (ii):} \textquotedblleft We can iterate through all of them by alternatingly adding a column and a row\textquotedblright\ is not quite correct; I suggest replacing it by \textquotedblleft We can iterate through all of them by alternatingly adding a column (if $S=B$) or a row and a column (if $S=B-1$)\textquotedblright. \item \textbf{page 92, Claim (iii):} \textquotedblleft partitions\textquotedblright\ $\rightarrow$ \textquotedblleft partitions of $n$\textquotedblright\ (both times). \item \textbf{page 93, \S 4.2:} In the formula at the end of \S 4.2, replace \textquotedblleft$3i+1$\textquotedblright\ by \textquotedblleft$\left( 3i+1\right)$\textquotedblright, since the sum otherwise does not encompass the $1$. \item \textbf{page 94, Theorem 4.10:} After \textquotedblleft of the same shape\textquotedblright, add \textquotedblleft of size $n$\textquotedblright. \item \textbf{page 94, proof of Theorem 4.10:} In the definition of $\min\left( X\right)$, replace \textquotedblleft$X\setminus\left\{ x,y\right\}$\textquotedblright\ by \textquotedblleft$X\setminus\left\{ \left( x,y\right) \right\}$\textquotedblright. \item \textbf{page 94, proof of Theorem 4.10:} It would be great to have a more precise definition of $S\left( X\right)$. Here is my suggestion: \textquotedblleft Number the points $\left( x,y\right) \in\min\left( X\right)$ as $\left( x_{1},y_{1}\right) ,\left( x_{2},y_{2}\right) ,\ldots,\left( x_{p},y_{p}\right)$ in the order of increasing $x$-coordinates. Then, we easily see that $x_{1}y_{2}>\cdots>y_{p}$. The shadowline $S(X)$ is then defined as the broken line \begin{align*} \left( x_{1},+\infty\right) & \rightarrow\left( x_{1},y_{1}\right) \rightarrow\left( x_{2},y_{1}\right) \rightarrow\left( x_{2},y_{2}\right) \rightarrow\left( x_{3},y_{2}\right) \rightarrow\left( x_{3},y_{3}\right) \\ & \rightarrow\cdots\rightarrow\left( x_{p},y_{p-1}\right) \rightarrow \left( x_{p},y_{p}\right) \rightarrow\left( +\infty,y_{p}\right) , \end{align*} which enters each point in $\min\left( X\right)$ going down and leaves it going right.\textquotedblright \item \textbf{page 95, Algorithm 1:} \textquotedblleft convex\textquotedblright\ $\rightarrow$ \textquotedblleft concave\textquotedblright. More precisely, by \textquotedblleft convex bends\textquotedblright, you mean the set of all points at which concave bends occur. \item \textbf{page 96, Example:} \textquotedblleft There are still convex bends\textquotedblright\ $\rightarrow$ \textquotedblleft There are still concave bends\textquotedblright. \item \textbf{page 97:} After defining \textquotedblleft increasing\textquotedblright, it is worth defining \textquotedblleft decreasing\textquotedblright\ as well: Namely, a subset $Y$ of a point set $X$ is said to be \textit{decreasing} if it can be written as $Y=\left\{ \left( x_{1},y_{1}\right) ,\left( x_{2},y_{2}\right) ,\ldots,\left( x_{k}% ,y_{k}\right) \right\}$ with $y_{1}\geq y_{2}\geq\cdots\geq y_{k}$ and $x_{1}\leq x_{2}\leq\cdots\leq x_{k}$. \item \textbf{page 97:} Before the first Claim here, I would add a \textquotedblleft zero-th\textquotedblright\ claim: If $i$ is arbitrary, then no two points in $X_{i}$ have the same $x$-coordinate, and no two points in $X_{i}$ have the same $y$-coordinate. (This is proven by induction on $i$, since the $x$-coordinate of each point in $X_{i+1}$ equals the $x$-coordinate of precisely one point in $X_{i}$ (because the concave bends of a shadowline steal their coordinates from the original convex bends), and the same holds for $y$-coordinates. But this needs to be said, since you keep using this claim tacitly; in particular, it is the reason why you can restrict yourself to considering increasing rather than \textquotedblleft weakly increasing\textquotedblright\ chains.) \item \textbf{page 97, first Claim:} \textquotedblleft shadows lines\textquotedblright\ $\rightarrow$ \textquotedblleft shadowlines\textquotedblright. \item \textbf{page 97, proof of the first Claim:} The \textquotedblleft% $X_{1+\left\vert Y\right\vert }$\textquotedblright\ at the end is false. You probably mean the $X_{i}^{\prime}$ for $j=1+\left\vert Y\right\vert$ or something like that. (I would find an inductive proof clearer here anyway. You could even factor such a proof out of the long proof of Theorem 4.10. Its main tool is the following general statement: If $U$ is a nonempty finite set of points in the plane, no two of which have the same $x$-coordinate and no two of which have the same $y$-coordinate, then each longest chain of $U$ has exactly one point in common with $\min\left( U\right)$, and furthermore, if we remove this common point from the longest chain, then we are left with a longest chain of $U\setminus\min\left( U\right)$. This is easy to prove, and once it is proven, your first Claim follows by an easy induction.) \item \textbf{page 97, proof of the fourth Claim:} \textquotedblleft where constructed\textquotedblright\ $\rightarrow$ \textquotedblleft were constructed\textquotedblright. \item \textbf{page 97, proof of the fourth Claim:} It is worth explaining in a footnote why \textquotedblleft the shadow lines of the same phase are non-crossing\textquotedblright. The proof is not hard (in fact, each point on $S_{i}^{p+1}$ lies northeast of at least one point of $S_{i}^{p}$, because each convex bend of $S_{i}^{p+1}$ lies northeast of at least one convex bend of $S_{i}^{p}$), but I don't consider it trivial. \item \textbf{page 97, proof of the fourth Claim:} \textquotedblleft is therefore at least the $x$-coordinate of the leftmost concave bend of $S_{i}^{j}$\textquotedblright\ $\rightarrow$ \textquotedblleft is therefore at least the $x$-coordinate of the leftmost concave bend of $S_{i}^{j}% ,\ldots,S_{i}^{n_{i}}$\textquotedblright. Indeed, I believe it may happen that one of the paths $S_{i}^{k}$ with $k>j$ has its leftmost concave bend further left than $S_{i}^{j}$ (or at least I am not convinced that this can be ruled out). The argument, however, still works because the shadowlines are non-crossing. \item \textbf{page 98:} \textquotedblleft a corresponding shadowlines\textquotedblright\ $\rightarrow$ \textquotedblleft a corresponding shadowline\textquotedblright. \item \textbf{page 98:} \textquotedblleft do not have convex bends\textquotedblright\ $\rightarrow$ \textquotedblleft do not have concave bends\textquotedblright. \item \textbf{page 98:} Add a semicolon before \textquotedblleft if it didn't\textquotedblright. \item \textbf{page 98:} \textquotedblleft that where not visited\textquotedblright\ $\rightarrow$ \textquotedblleft that were not visited\textquotedblright. \item \textbf{page 98:} I find it far from obvious that the inverse map is well-defined. For example, why are there no unused points left after the \textquotedblleft reconstruction\textquotedblright\ of the shadowlines? \item \textbf{page 100, proof of Lemma 4.11 (iii):} Add a \textquotedblleft% $<$\textquotedblright\ sign before \textquotedblleft$i_{k}$\textquotedblright% \ and a \textquotedblleft$<$\textquotedblright\ sign before \textquotedblleft% $\pi\left( i_{k}\right)$\textquotedblright. \item \textbf{page 102:} In \textquotedblleft which is replaced by $\left( x_{i-1}-x_{i}\right) =-\left( x_{i}-x_{i-1}\right)$\textquotedblright, replace both \textquotedblleft$i-1$\textquotedblright s by \textquotedblleft% $i+1$\textquotedblright s. \item \textbf{page 102:} \textquotedblleft must be fall behind\textquotedblright\ $\rightarrow$ \textquotedblleft must fall behind\textquotedblright. \item \textbf{page 102, Lemma 4.13:} FYI: There is an elementary proof of Lemma 4.13 without any polynomial tricks. I have outlined it in\newline% \texttt{ \href{https://math.stackexchange.com/a/1969211/}{https://math.stackexchange.com/a/1969211/}% } (and exposed it in detail in the reference given therein, which does not however appear very useful). \item \textbf{pages 102--103:} \textquotedblleft then $g$ is a polynomial of degree $n$ over $x_{1}$\textquotedblright\ $\rightarrow$ \textquotedblleft then $g$ is a polynomial of degree at most $m$ in $x_{1}$\textquotedblright. (The word \textquotedblleft over\textquotedblright\ is very ambiguous here -- after all, one usually says \textquotedblleft over the ring of constants\textquotedblright.) \item \textbf{page 103:} I would not use the notation $p^{\prime}$ here, as it (falsely) suggests the derivative of $p$. \item \textbf{page 103:} Remove the period at the end of ($\star$). \item \textbf{page 104, proof of Theorem 4.14:} In the definition of $t$, you should add that $t\left( n_{1},\ldots,n_{m}\right)$ is defined to be $0$ when when $x_{1}\geq\cdots\geq x_{m}\geq0$ is not satisfied. Thus, $t$ is really defined on all $m$-tuples of integers. You use this for the uniqueness argument. \item \textbf{page 105, proof of Theorem 4.14:} In the first computation on page 105 (in the proof of (2)), you have an \textquotedblleft$\left( x_{m}-1\right) !$\textquotedblright\ in the denominator. This should be an \textquotedblleft$\left( x_{m-1}-1\right) !$\textquotedblright. \item \textbf{page 105:} In the definition of a \textquotedblleft hook\textquotedblright, I would point out that the square $\left( i,j\right)$ itself is counted as part of the hook. Also, I would replace \textquotedblleft union\textquotedblright\ by \textquotedblleft set\textquotedblright. \item \textbf{page 106, proof of Theorem 4.15:} You should WLOG assume $n_{m}>0$ here; otherwise, the hook $h_{i,1}$ does not start in $\left( m,1\right)$. \item \textbf{page 108, Table 3:} Strictly speaking, the \textquotedblleft completely left of\textquotedblright\ relation should be defined only on nonempty intervals, not on all intervals (because defining it on all intervals would lead to \$I<\varnothing