\documentclass[ALCO, Unicode]{cedram} \usepackage{color} \usepackage{pythonhighlight} \usepackage{tabls} \equalenv{corollary}{coro} \equalenv{proposition}{prop} \equalenv{remark}{rema} \equalenv{example}{exam} \equalenv{definition}{defi} \theoremstyle{definition} \newtheorem{convention}{Convention} % The following four lines ensure that summation/product limits % always appear below and above the summation/product signs % rather than to their right. That's how I like it. I don't % insist on it, but please make sure to check for bleeding and % overfulls if you change this. \let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} \title{Petrie symmetric functions} \author[\initial{D.} Grinberg]{\firstname{Darij} \lastname{Grinberg}} \address{Drexel University\\ Korman Center\\ 15 S 33rd Street\\ Office \#263\\ Philadelphia, PA 19104 (USA)} \email{darijgrinberg@gmail.com} \thanks{This material is based upon work supported by the Swedish Research Council under grant no. 2016-06596 while the author was in residence at Institut Mittag--Leffler in Djursholm, Sweden during Spring 2020.} \keywords{symmetric functions, Schur functions, Schur polynomials, combinatorial Hopf algebras, Petrie matrices, Pieri rules, Murnaghan--Nakayama rule} \subjclass{05E05} \datereceived{2020-05-08} \daterevised{2021-12-31} \dateaccepted{2022-02-14} \begin{document} \begin{abstract} For any positive integer $k$ and nonnegative integer $m$, we consider the symmetric function $G\left(k,m\right) $ defined as the sum of all monomials of degree $m$ that involve only exponents smaller than $k$. We call $G\left( k,m\right) $ a \textit{Petrie symmetric function} in honor of Flinders Petrie, as the coefficients in its expansion in the Schur basis are determinants of Petrie matrices (and thus belong to $\left\{ 0,1,-1\right\} $ by a classical result of Gordon and Wilkinson). More generally, we prove a Pieri-like rule for expanding a product of the form $G\left( k,m\right) \cdot s_{\mu}$ in the Schur basis whenever $\mu$ is a partition; all coefficients in this expansion belong to $\left\{ 0,1,-1\right\} $. We also show that $G\left( k,1\right) ,G\left( k,2\right) ,G\left( k,3\right) ,\ldots$ form an algebraically independent generating set for the symmetric functions when $1-k$ is invertible in the base ring, and we prove a conjecture of Liu and Polo about the expansion of $G\left( k,2k-1\right) $ in the Schur basis. \end{abstract} \maketitle \noindent Considered as a ring, the \textit{symmetric functions} (which is short for \textquotedblleft formal power series in countably many indeterminates $x_{1},x_{2},x_{3},\ldots$ that are of bounded degree and fixed under permutations of the indeterminates\textquotedblright) are hardly a remarkable object: By a classical result essentially known to Gauss, they form a polynomial ring in countably many indeterminates. The true theory of symmetric functions is rather the study of specific \textit{families} of symmetric functions, often defined by combinatorial formulas (\eg as multivariate generating functions) but interacting deeply with many other fields of mathematics. Classical families are, for example, the \textit{monomial symmetric functions} $m_{\lambda}$, the \textit{complete homogeneous symmetric functions }$h_{n}$, the \textit{power-sum symmetric functions} $p_{n}$, and the \textit{Schur functions }$s_{\lambda}$. Some of these families -- such as the monomial symmetric functions $m_{\lambda}$ and the Schur functions $s_{\lambda}$ -- form bases of the ring of symmetric functions (as a module over the base ring). In this paper, we introduce a new family $\left( G\left( k,m\right) \right) _{k\geq1;\ m\geq0}$ of symmetric functions, which we call the \textit{Petrie symmetric functions} in honor of Flinders Petrie. For any integers $k\geq1$ and $m\geq0$, we define $G\left( k,m\right) $ as the sum of all monomials of degree $m$ (in $x_{1},x_{2},x_{3},\ldots$) that involve only exponents smaller than $k$. When $G\left( k,m\right) $ is expanded in the Schur basis (\ie as a linear combination of Schur functions $s_{\lambda}% $), all coefficients belong to $\left\{ 0,1,-1\right\} $ by a classical result of Gordon and Wilkinson, as they are determinants of so-called \textit{Petrie matrices} (whence our name for $G\left( k,m\right) $). We give an explicit combinatorial description for the coefficients as well. More generally, we prove a Pieri-like rule for expanding a product of the form $G\left( k,m\right) \cdot s_{\mu}$ in the Schur basis whenever $\mu$ is a partition; all coefficients in this expansion again belong to $\left\{ 0,1,-1\right\} $ (although we have no explicit combinatorial rule for them). We show some further properties of $G\left( k,m\right) $ and prove that if $k$ is a fixed positive integer such that $1-k$ is invertible in the base ring, then $G\left( k,1\right) ,G\left( k,2\right) ,G\left( k,3\right) ,\ldots$ form an algebraically independent generating set for the symmetric functions. We prove a conjecture of Liu and Polo in \cite[Remark 1.4.5]{LiuPol19} about the expansion of $G\left( k,2k-1\right) $ in the Schur basis. This paper begins with Section \ref{sect.not}, in which we introduce the notions and notations that the paper will rely on. (Further notations will occasionally be introduced as the need arises.) The rest of the paper consists of two essentially independent parts. The first part comprises Section \ref{sect.thms}, in which we define the Petrie symmetric functions $G\left( k,m\right) $ (and the related power series $G\left( k\right) $) and state several of their properties, and Section \ref{sect.proofs}, in which we prove said properties. The second part is Section \ref{sect.liu}, which is devoted to proving the conjecture of Liu and Polo.\footnote{This proof is independent of the first part of the paper, except that it uses the very simple Proposition \ref{prop.G.basics} \textbf{(c)}.} A final Section \ref{sect.fin} adds comments, formulates two conjectures, and (in its last subsection) explores a more general family of symmetric functions that still shares some of the properties of the Petrie functions $G\left( k,m\right) $. (As a byproduct of the latter generalization, a formula for the antipode of $G\left( k,m\right) $ -- Corollary \ref{cor.Gkm.antipode} -- emerges.) \longthanks{I thank Moussa Ahmia, Per Alexandersson, Fran\c{c}ois Bergeron, Steve Doty, Ira Gessel, Jim Haglund, Linyuan Liu, Patrick Polo, Sasha Postnikov, Christopher Ryba, Richard Stanley, Ole Warnaar and Mark Wildon for interesting and helpful conversations, and two referees for helpful suggestions. Special thanks are due to Sasha Postnikov for his permission to include his generalization of the Petrie symmetric functions. This paper was started at the Mathematisches Forschungsinstitut Oberwolfach, where I was staying as a Leibniz fellow in Summer 2019, and finished during a semester program at the Institut Mittag--Leffler in 2020. I thank both institutes for their hospitality. The SageMath computer algebra system \cite{SageMath} has been used in discovering some of the results.} \subsubsection*{Remarks} \textbf{1.} A short exposition of the main results of this paper (without proofs), along with an additional question motivated by it, can be found in \cite{Grinbe20}. \textbf{2.} While finishing this work, I have become aware of three independent discoveries of the Petrie symmetric functions $G\left( k,m\right) $: \begin{enumerate} \item[\textbf{(a)}] In \cite[\S 3.3]{DotWal92}, Stephen Doty and Grant Walker define a \emph{modular complete symmetric function }$h_{d}^{\prime}$, which is precisely our Petrie symmetric function $G\left( k,m\right) $ up to a renaming of variables (namely, their $m$ and $d$ correspond to our $k$ and $m$). Some of our results appear in their work: Our Theorem \ref{thm.Gkm-genset} is (a slight generalization of) \cite[Corollary 3.9]{DotWal92}; our Theorem \ref{thm.Uk.main} is (part of) \cite[Proposition 3.15]{DotWal92} restated in the language of Hopf algebras. The $h_{d}^{\prime }$ are studied further in Walker's follow-up paper \cite{Walker94}, some of whose results mirror ours again (in particular, the maps $\psi^{p}$ and $\psi_{p}$ from \cite{Walker94} are our $\mathbf{f}_{p}$ and $\mathbf{v}_{p}$). \item[\textbf{(b)}] The preprint \cite{FuMei20} by Houshan Fu and Zhousheng Mei introduces the Petrie symmetric functions $G\left( k,m\right) $ and refers to them as \emph{truncated homogeneous symmetric functions} $h_{m}^{\left[ k-1\right] }$. Some results below are also independently obtained in \cite{FuMei20}. In particular, Theorem \ref{thm.G.main} is a formula in \cite[\S 2]{FuMei20}, and Theorem \ref{thm.petk.explicit} is equivalent to \cite[Proposition 2.9]{FuMei20}. The particular case of Theorem \ref{thm.Gkm-genset} when $\mathbf{k}=\mathbb{Q}$ is part of \cite[Theorem 2.7]{FuMei20}. \item[\textbf{(c)}] The paper \cite{BaAhBe18} by Bazeniar, Ahmia and Belbachir introduces the symmetric functions $G\left( k,m\right) $ as well, or rather their evaluations $\left( G\left( k,m\right) \right) \left( x_{1}% ,x_{2},\ldots,x_{n}\right) $ at finitely many variables $x_1, x_2, \ldots, x_n$; it denotes them by $E_{m}^{\left( k-1\right) }\left( n\right) =E_{m}^{\left( k-1\right) }\left( x_{1},x_{2},\ldots,x_{n}\right) $. Ahmia and Merca continue the study of these $E_{m}^{\left( k-1\right) }\left( x_{1},x_{2},\ldots ,x_{n}\right) $ in \cite{AhmMer20}. Our Theorem \ref{thm.G.frob} is equivalent to the second formula in \cite[Theorem 3.3]{AhmMer20} (although we are using infinitely many variables). \item[\textbf{(d)}] The formal power series $G\left( k\right) $ also appears in \cite[Chapter I, \S 6]{FulLan85}, under the guise of \emph{Bott's cannibalistic class} $\theta^{j}\left( e\right) $ (for $j=k$ and rewritten in the language of $\lambda$-ring operations\footnote{See \cite[\S 16.74]% {Hazewi08} for the connection between symmetric functions (over $\mathbb{Z}$) and universal operations on $\lambda$-rings. To be specific: If $a$ is an element of a $\lambda$-ring $A$, then the canonical $\lambda$-ring morphism $\Lambda_{\mathbb{Z}}\rightarrow A$ (where $\Lambda_{\mathbb{Z}}$ is the ring of symmetric functions over $\mathbb{Z}$) that sends $e_{1}=x_{1}+x_{2}% +x_{3}+\cdots\in\Lambda_{\mathbb{Z}}$ to $a\in A$ will send the Petrie symmetric function $G\left( k,m\right) $ to the \textquotedblleft$m$-th graded component\textquotedblright\ of Bott's cannibalistic class $\theta ^{k}\left( a\right) $. (Bott's cannibalistic class $\theta^{k}\left( a\right) $ itself is defined only if $a$ is a \textquotedblleft positive element\textquotedblright\ in the sense of \cite{FulLan85} (or can only be defined in an appropriate closure of $A$). When it is defined, it is the image of the series $G\left( k\right) $. Otherwise, its \textquotedblleft graded components\textquotedblright\ are the right object to consider.)}); it is used there to prove an abstract Riemann--Roch theorem. An application to group representations appears in \cite{AtiTal69}. \end{enumerate} \textbf{3.} The Petrie symmetric functions have been added to Per Alexandersson's collection of symmetric functions at \url{https://www.math.upenn.edu/~peal/polynomials/petrie.htm} . \subsection*{Remark on alternative versions} This paper also has an arXiv version \cite{vershort}, which includes some proofs that are here omitted for brevity. It also has a detailed version \cite{verlong} with many more details. \section{\label{sect.not}Notations} We will use the following notations (most of which are also used in \cite[\S 2.1]{GriRei}): \begin{itemize} \item We let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. \item The words \textquotedblleft positive\textquotedblright, \textquotedblleft larger\textquotedblright, \etc\ will be used in their Anglophone meaning (so that $0$ is neither positive nor larger than itself). \item We fix a commutative ring $\mathbf{k}$; we will use this $\mathbf{k}$ as the base ring in what follows. \item A \emph{weak composition} means an infinite sequence of nonnegative integers that contains only finitely many nonzero entries (\ie a sequence $\left( \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right) \in\mathbb{N}% ^{\infty}$ such that all but finitely many $i\in\left\{ 1,2,3,\ldots\right\} $ satisfy $\alpha_{i}=0$). \item We let $\operatorname*{WC}$ denote the set of all weak compositions. \item For any weak composition $\alpha$ and any positive integer $i$, we let $\alpha_{i}$ denote the $i$-th entry of $\alpha$ (so that $\alpha=\left( \alpha_{1},\alpha_{2},\alpha_{3},\ldots\right) $). More generally, we use this notation whenever $\alpha$ is an infinite sequence of any kind of objects. \item The \emph{size} $\left\vert \alpha\right\vert $ of a weak composition $\alpha$ is defined to be $\alpha_{1}+\alpha_{2}+\alpha_{3}+\cdots \in\mathbb{N}$. \item A \emph{partition} means a weak composition whose entries weakly decrease (\ie a weak composition $\alpha$ satisfying $\alpha_{1}\geq \alpha_{2}\geq\alpha_{3}\geq\cdots$). \item If $n\in\mathbb{Z}$, then a \emph{partition of} $n$ means a partition $\alpha$ having size $n$ (that is, satisfying $\left\vert \alpha\right\vert =n$). \item We let $\operatorname*{Par}$ denote the set of all partitions. For each $n\in\mathbb{Z}$, we let $\operatorname*{Par}\nolimits_{n}$ denote the set of all partitions of $n$. \item We will sometimes omit trailing zeroes from partitions: \ie a partition $\lambda=\left( \lambda_{1},\lambda_{2},\lambda_{3},\ldots\right) $ will be identified with the $k$-tuple $\left( \lambda_{1},\lambda_{2},\ldots ,\lambda_{k}\right) $ whenever $k\in\mathbb{N}$ satisfies $\lambda _{k+1}=\lambda_{k+2}=\lambda_{k+3}=\cdots=0$. For example, $\left( 3,2,1,0,0,0,\ldots\right) =\left( 3,2,1\right) =\left( 3,2,1,0\right) $. \item The partition $\left( 0,0,0,\ldots\right) =\left( {}\right) $ is called the \emph{empty partition} and denoted by $\varnothing$. \item A \emph{part} of a partition $\lambda$ means a nonzero entry of $\lambda$. For example, the parts of the partition $\left( 3,1,1\right) =\left( 3,1,1,0,0,0,\ldots\right) $ are $3,1,1$. \item We will use the notation $1^{k}$ for \textquotedblleft% $\underbrace{1,1,\ldots,1}_{k\text{ times}}$\textquotedblright\ in partitions. (For example, $\left( 2,1^{4}\right) =\left( 2,1,1,1,1\right) $. This notation is a particular case of the more general notation $m^{k}$ for \textquotedblleft$\underbrace{m,m,\ldots,m}_{k\text{ times}}$% \textquotedblright\ in partitions, used, \eg in \cite[Definition 2.2.1]{GriRei}.) \item We let $\Lambda$ denote the ring of symmetric functions in infinitely many variables $x_{1},x_{2},x_{3},\ldots$ over $\mathbf{k}$. This is a subring of the ring $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ of formal power series. To be more specific, $\Lambda$ consists of all power series in $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3}% ,\ldots\right] \right] $ that are symmetric (\ie invariant under permutations of the variables) and of bounded degree (see \cite[\S 2.1]% {GriRei} for the precise meaning of this). \item A \emph{monomial} shall mean a formal expression of the form $x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha_{3}}\cdots$ with $\alpha \in\operatorname*{WC}$. Formal power series are formal infinite $\mathbf{k}% $-linear combinations of such monomials. \item For any weak composition $\alpha$, we let $\mathbf{x}^{\alpha}$ denote the monomial $x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}x_{3}^{\alpha_{3}}\cdots$. \item The \emph{degree} of a monomial $\mathbf{x}^{\alpha}$ is defined to be $\left\vert \alpha\right\vert $. \item A formal power series is said to be \emph{homogeneous of degree }$n$ (for some $n\in\mathbb{N}$) if all monomials appearing in it (with nonzero coefficient) have degree $n$. In particular, the power series $0$ is homogeneous of any degree. \item If $f\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ is a power series, then there is a unique family $\left( f_{i}\right) _{i\in\mathbb{N}}=\left( f_{0},f_{1},f_{2},\ldots\right) $ of formal power series $f_{i}\in\mathbf{k}\left[ \left[ x_{1},x_{2}% ,x_{3},\ldots\right] \right] $ such that each $f_{i}$ is homogeneous of degree $i$ and such that $f=\sum_{i\in\mathbb{N}}f_{i}$. This family $\left( f_{i}\right) _{i\in\mathbb{N}}$ is called the \emph{homogeneous decomposition} of $f$, and its entry $f_{i}$ (for any given $i\in\mathbb{N}$) is called the $i$\emph{-th degree homogeneous component} of $f$. \item The $\mathbf{k}$-algebra $\Lambda$ is graded: \ie any symmetric function $f$ can be uniquely written as a sum $\sum_{i\in\mathbb{N}}f_{i}$, where each $f_{i}$ is a homogeneous symmetric function of degree $i$, and where all but finitely many $i\in\mathbb{N}$ satisfy $f_{i}=0$. \end{itemize} We shall use the symmetric functions $m_{\lambda},h_{n},e_{n},p_{n}% ,s_{\lambda}$ in $\Lambda$ as defined in \cite[Sections 2.1 and 2.2]{GriRei}. Let us briefly recall how they are defined: \begin{itemize} \item For any partition $\lambda$, we define the \emph{monomial symmetric function }$m_{\lambda}\in\Lambda$ by\footnote{This definition of $m_{\lambda}$ is not the same as the one given in \cite[Definition 2.1.3]{GriRei}; but it is easily seen to be equivalent to the latter (\ie it defines the same $m_{\lambda}$). See \cite{verlong} for the details of the proof.} \[ m_{\lambda}=\sum\mathbf{x}^{\alpha}, \] where the sum ranges over all weak compositions $\alpha\in\operatorname*{WC}$ that can be obtained from $\lambda$ by permuting entries\footnote{Here, we understand $\lambda$ to be an infinite sequence, not a finite tuple, so the entries being permuted include infinitely many $0$'s.}. For example, \[ m_{\left( 2,2,1\right) }=\sum_{im$, then $G\left( k,m\right) =h_{m}$. \label{it:2.3d} \item If $m\in\mathbb{N}$ and $k=2$, then $G\left( k,m\right) =e_{m}$.\label{it:2.3e} \item If $m=k$, then $G\left( k,m\right) =h_{m}-p_{m}$.\label{it:2.3f} \end{enumerate} \end{proposition} Proving Proposition \ref{prop.G.basics} makes good practice in understanding the definitions of $m_{\lambda}$, $h_{n}$, $e_{n}$, $p_{n}$, $G\left( k\right) $ and $G\left( k,n\right) $. We omit the proof here; it can be found in full (hardly necessary) detail in \cite{verlong}. Parts \ref{it:2.3d} and \ref{it:2.3e} of Proposition \ref{prop.G.basics} suggest to regard the Petrie symmetric functions $G\left( k,m\right) $ as an interpolation between the $h_{m}$ and the $e_{m}$. \subsection{\label{subsect.petk.schur}The Schur expansion} The solution to \cite[Exercise 7.3]{Stanley-EC2} gives an expansion of $G\left( 3\right) $ in terms of the elementary symmetric functions (due to I. M. Gessel); this expansion can be rewritten as% \[ G\left( 3\right) =\sum_{n\in\mathbb{N}}e_{n}^{2}+\sum_{m0$ and $m\in\mathbb{N}$) is a linear combination of Schur functions, with all coefficients belonging to $\left\{ -1,0,1\right\} $. It is natural to expect the more general symmetric functions \[ \widetilde{G}\left( k,k^{\prime},m\right) =\sum_{\substack{\alpha \in\operatorname*{WC};\\\left\vert \alpha\right\vert =m;\\k^{\prime}\leq \alpha_{i}0$). Moreover, the empty partition $\varnothing$ can be written as $\varnothing=\left( \varnothing_{1},\varnothing_{2}% ,\ldots,\varnothing_{\ell}\right) $ (since $\varnothing_{i}=0$ for each integer $i>\ell$). Thus, we have $\lambda=\left( \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell }\right) $ and $\varnothing=\left( \varnothing_{1},\varnothing_{2}% ,\ldots,\varnothing_{\ell}\right) $. Hence, the definition of $\operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing\right) $ yields \begin{align} \operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing\right) & =\det\left( \left( \left[ 0\leq\lambda_{i}-\underbrace{\varnothing_{j}% }_{=0}-i+j0. \] \textup{(}This is well-defined, since the sequence $\left( h_{1},h_{2},h_{3}% ,\ldots\right) $ is an algebraically independent generating set of the commutative $\mathbf{k}$-algebra $\Lambda$.\textup{)} This map $\mathbf{v}_{n}$ is called the $n$\emph{-th Verschiebung endomorphism} of $\Lambda$. \end{definition} Again, it is known (\cite[Exercise 2.9.10(e)]{GriRei}) that this map $\mathbf{v}_{n}:\Lambda\rightarrow\Lambda$ is a Hopf algebra endomorphism of $\Lambda$. Moreover, the following holds (\cite[Exercise 2.9.10(f)]{GriRei}): \begin{proposition} \label{prop.f-v-dual}Let $n\in\left\{ 1,2,3,\ldots\right\} $. Then, the maps $\mathbf{f}_{n}:\Lambda\rightarrow\Lambda$ and $\mathbf{v}_{n}:\Lambda \rightarrow\Lambda$ are adjoint with respect to the Hall inner product on $\Lambda$. That is, we have \[ \left\langle a,\mathbf{f}_{n}\left( b\right) \right\rangle =\left\langle \mathbf{v}_{n}\left( a\right) ,b\right\rangle \ \ \ \ \ \ \ \ \ \ \text{for any }a\in\Lambda\text{ and }b\in\Lambda. \] \end{proposition} Furthermore, it is known (see, \eg \cite[Exercise 2.9.10(a)]{GriRei}) that any positive integers $n$ and $m$ satisfy% \begin{equation} \mathbf{v}_{n}\left( p_{m}\right) =% \begin{cases} np_{m/n}, & \text{if }n\mid m;\\ 0, & \text{if }n\nmid m. \end{cases} \label{eq.vnpm}% \end{equation} \subsection{The Hopf endomorphisms $U_{k}$ and $V_{k}$} In this final subsection, we shall show another way to obtain the Petrie symmetric functions $G\left( k,m\right) $ using the machinery of Hopf algebras. We refer, \eg to \cite[Chapters 1 and 2]{GriRei} for everything we will use about Hopf algebras. \begin{convention} As already mentioned, $\Lambda$ is a connected graded Hopf algebra. We let $S$ denote its antipode. \end{convention} \begin{definition} \label{def.convolution}If $C$ is a $\mathbf{k}$-coalgebra and $A$ is a $\mathbf{k}$-algebra, and if $f,g:C\rightarrow A$ are two $\mathbf{k}$-linear maps, then the \emph{convolution} $f\star g$ of $f$ and $g$ is defined to be the $\mathbf{k}$-linear map $m_{A}\circ\left( f\otimes g\right) \circ \Delta_{C}:C\rightarrow A$, where $\Delta_{C}:C\rightarrow C\otimes C$ is the comultiplication of the $\mathbf{k}$-coalgebra $C$, and where $m_{A}:A\otimes A\rightarrow A$ is the $\mathbf{k}$-linear map sending each pure tensor $a\otimes b\in A\otimes A$ to $ab\in A$. \end{definition} We also recall Definition \ref{def.vn} and Definition \ref{def.fn}. We now claim the following. \begin{theorem} \label{thm.Uk.main}Fix a positive integer $k$. Let $U_{k}$ be the map $\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}:\Lambda\rightarrow\Lambda$. Let $V_{k}$ be the map $\operatorname*{id}\nolimits_{\Lambda}\star\, U_{k}% :\Lambda\rightarrow\Lambda$. (This is well-defined by Definition \ref{def.convolution}, since $\Lambda$ is both a $\mathbf{k}$-coalgebra and a $\mathbf{k}$-algebra.) Then: \begin{enumerate}[label=\textbf{(\alph*)}] \item \label{226a} The map $U_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism. \item\label{226b} The map $V_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism. \item\label{226c}We have $V_{k}\left( h_{m}\right) =G\left( k,m\right) $ for each $m\in\mathbb{N}$. \item\label{226d} We have $V_{k}\left( p_{n}\right) =\left( 1-\left[ k\mid n\right] k\right) p_{n}$ for each positive integer $n$. \end{enumerate} \end{theorem} See Subsection \ref{subsect.proofs.Uk.main} for a proof of this theorem. Note that Theorem \ref{thm.Uk.main} can be used to give a second proof of Theorem \ref{thm.DeltaGkm}; see \cite{verlong} for this. We also obtain the following corollary from Theorem \ref{thm.DeltaGkm}: \begin{corollary} \label{cor.p-via-G}Let $k$ and $n$ be two positive integers. Then, there exists a polynomial $f\in\mathbf{k}\left[ x_{1},x_{2},x_{3},\ldots\right] $ such that% \begin{equation} \left( 1-\left[ k\!\!\mid\!\! n\right] k\right) p_{n}=f\left( G\left( k,1\right) ,G\left( k,2\right) ,G\left( k,3\right) ,\ldots\right) . \label{eq.cor.p-via-G.eq}% \end{equation} \end{corollary} This corollary will be proved in Subsection \ref{subsect.proofs.p-via-G}. \section{\label{sect.proofs}Proofs} \subsection{The symmetric functions $h_{\lambda}$} We shall now approach the proofs of the claims made above. First, let us introduce a family of symmetric functions, obtained by multiplying several $h_{n}$'s: \begin{definition} \label{def.hlam}Let $\lambda$ be a partition. Write $\lambda$ in the form $\lambda=\left( \lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}\right) $, where $\lambda_{1},\lambda_{2},\ldots,\lambda_{\ell}$ are positive integers. Then, we define a symmetric function $h_{\lambda}\in\Lambda$ by% \[ h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}}\cdots h_{\lambda_{\ell}}. \] \end{definition} The symmetric function $h_{\lambda}$ is called the \emph{complete homogeneous symmetric function} corresponding to the partition $\lambda$. From \cite[Corollary 2.5.17(a)]{GriRei}, we know that the families $\left( h_{\lambda}\right) _{\lambda\in\operatorname*{Par}}$ and $\left( m_{\lambda }\right) _{\lambda\in\operatorname*{Par}}$ are dual bases with respect to the Hall inner product. Thus,% \begin{equation} \left\langle h_{\lambda},m_{\mu}\right\rangle =\delta_{\lambda,\mu }\ \ \ \ \ \ \ \ \ \ \text{for any }\lambda\in\operatorname*{Par}\text{ and }\mu\in\operatorname*{Par}. \label{eq.hlam-mlam-dual}% \end{equation} We note that $h_{0}=1$; therefore, Definition \ref{def.hlam} can be restated as follows: For any partition $\lambda$, we have% \[ h_{\lambda}=h_{\lambda_{1}}h_{\lambda_{2}}h_{\lambda_{3}}\cdots \] (this infinite product $h_{\lambda_{1}}h_{\lambda_{2}}h_{\lambda_{3}}\cdots$ is well-defined, since every sufficiently high positive integer $i$ satisfies $\lambda_{i}=0$ and thus $h_{\lambda_{i}}=h_{0}=1$). This is how $h_{\lambda}$ is defined in \cite[Section I.2]{Macdon95}. \subsection{\label{subsect.proofs.skew}Skew Schur functions} Let us define a classical partial order on $\operatorname*{Par}$ (see, \eg \cite[Definition 2.3.1]{GriRei}): \begin{definition} Let $\lambda$ and $\mu$ be two partitions. We say that $\mu\subseteq\lambda$ if each $i\in\left\{ 1,2,3,\ldots\right\} $ satisfies $\mu_{i}\leq\lambda_{i}$. We say that $\mu\not \subseteq \lambda$ if we don't have $\mu\subseteq\lambda$. \end{definition} For example, $\left( 3,2\right) \subseteq\left( 4,2,1\right) $, but $\left( 3,2,1\right) \not \subseteq \left( 4,2\right) $ (since $\left( 3,2,1\right) _{3}=1$ is not $\leq$ to $\left( 4,2\right) _{3}=0$). For any two partitions $\lambda$ and $\mu$, a symmetric function $s_{\lambda/\mu}$ called a \textit{skew Schur function} is defined in \cite[Definition 2.3.1]{GriRei} and in \cite[\S I.5]{Macdon95} (see also \cite[Definition 7.10.1]{Stanley-EC2} for the case when $\mu\subseteq\lambda $). We shall not recall its standard definition here, but rather state a few properties. The first property (which can in fact be used as an alternative definition of $s_{\lambda/\mu}$) is the \textit{first Jacobi--Trudi formula} for skew shapes; it states the following: \begin{theorem} \label{thm.JTs.h}Let $\lambda=\left( \lambda_{1},\lambda_{2},\ldots ,\lambda_{\ell}\right) $ and $\mu=\left( \mu_{1},\mu_{2},\ldots,\mu_{\ell }\right) $ be two partitions. Then,% \begin{equation} s_{\lambda/\mu}=\det\left( \left( h_{\lambda_{i}-\mu_{j}-i+j}\right) _{1\leq i\leq\ell,\ 1\leq j\leq\ell}\right) . \label{eq.schur.JT.sh-lammu}% \end{equation} \end{theorem} Theorem \ref{thm.JTs.h} appears (with proof) in \cite[(2.4.16)]{GriRei} and in \cite[Chapter I, (5.4)]{Macdon95}. The following properties of skew Schur functions are easy to see: \begin{itemize} \item If $\lambda$ is any partition, then $s_{\lambda/\varnothing}=s_{\lambda }$. (Recall that $\varnothing$ denotes the empty partition.) \item If $\lambda$ and $\mu$ are two partitions satisfying $\mu\not \subseteq \lambda$, then $s_{\lambda/\mu}=0$. \end{itemize} \subsection{\label{subsect.proofs.cauchy}A Cauchy-like identity} We shall use the following identity, which connects the skew Schur functions $s_{\lambda/\mu}$, the symmetric functions $h_{\lambda}$ from Definition \ref{def.hlam} and the monomial symmetric functions $m_{\lambda}$: \begin{theorem} \label{thm.our-cauchy}Recall the symmetric functions $h_{\lambda}$ defined in Definition \ref{def.hlam}. Let $\mu$ be any partition. Then, in the ring $\mathbf{k}\left[ \left[ \mathbf{x},\mathbf{y}\right] \right] $, we have% \[ \sum_{\lambda\in\operatorname*{Par}}s_{\lambda/\mu}\left( \mathbf{x}\right) s_{\lambda}\left( \mathbf{y}\right) =s_{\mu}\left( \mathbf{y}\right) \cdot\sum_{\lambda\in\operatorname*{Par}}h_{\lambda}\left( \mathbf{x}\right) m_{\lambda}\left( \mathbf{y}\right) . \] Here, we are using the notations introduced in Subsection \ref{subsect.thms.coprod}. \end{theorem} Theorem \ref{thm.our-cauchy} appears in \cite[fourth display on page 70]{Macdon95}, so we omit its proof.\footnote{See also \cite{verlong} for a detailed proof. (The main ingredients are \cite[proof of Proposition 2.5.15]{GriRei} and \cite[Exercise 2.5.11(a)]{GriRei}.)} \subsection{\label{subsect.proofs.petk.alphak}The $\mathbf{k}$-algebra homomorphism $\alpha_{k}:\Lambda\rightarrow\mathbf{k}$} Recall that the family $\left( h_{n}\right) _{n\geq1}=\left( h_{1}% ,h_{2},h_{3},\ldots\right) $ is algebraically independent and generates $\Lambda$ as a $\mathbf{k}$-algebra. Thus, $\Lambda$ can be viewed as a polynomial ring in the (infinitely many) indeterminates $h_{1},h_{2}% ,h_{3},\ldots$. The universal property of a polynomial ring thus shows that if $A$ is any commutative $\mathbf{k}$-algebra, and if $\left( a_{1},a_{2}% ,a_{3},\ldots\right) $ is any sequence of elements of $A$, then there is a unique $\mathbf{k}$-algebra homomorphism from $\Lambda$ to $A$ that sends $h_{i}$ to $a_{i}$ for all positive integers $i$. We shall refer to this as the \textit{h-universal property of }$\Lambda$. It lets us make the following definition:\footnote{We are using the Iverson bracket notation (see Convention \ref{conv.iverson}) here.} \begin{definition} \label{def.alphak}Let $k$ be a positive integer. The h-universal property of $\Lambda$ shows that there is a unique $\mathbf{k}$-algebra homomorphism $\alpha_{k}:\Lambda\rightarrow\mathbf{k}$ that sends $h_{i}$ to $\left[ i\lambda_{2}-2>\cdots>\lambda_{\ell}-\ell$. Hence, the set of all $i\in\left\{ 1,2,\ldots,\ell\right\} $ satisfying $\mu_{j}-j\leq\lambda_{i}-i<\mu_{j}-j+k$ is a (possibly empty) integer interval\footnote{An \emph{integer interval} means a subset of $\mathbb{Z}$ that has the form $\left\{ a,a+1,\ldots,b\right\} $ for some $a\in \mathbb{Z}$ and $b\in\mathbb{Z}$. (If $a>b$, then this is the empty set.)}. Thus, the $j$-th column of the matrix $B$ has a contiguous (but possibly empty) block of $1$'s (in the rows corresponding to all these $i$'s), while all other entries of this column are $0$. In other words, this column has the form (\ref{pf.prop.petk.-101.petcol}) for some nonnegative integers $a,b,c$. Since $j$ was arbitrary, we thus have shown that $B$ is a Petrie matrix. Therefore, (\ref{pf.prop.petk.-101.1}) (applied to $A=B$) yields $\det B\in\left\{ -1,0,1\right\} $. In view of (\ref{pf.prop.petkrel.-101.=detB}), this proves Proposition \ref{prop.petkrel.-101}. \end{proof} \subsection{\label{subsect.proofs.G.pieri.1st}Proof of Theorem \ref{thm.G.pieri}} We are now ready to prove Theorem \ref{thm.G.pieri}: \begin{proof} [Proof of Theorem \ref{thm.G.pieri}.]We shall use the notations $\mathbf{k}% \left[ \left[ \mathbf{x}\right] \right] $, $\mathbf{k}\left[ \left[ \mathbf{x},\mathbf{y}\right] \right] $, $\mathbf{x}$, $\mathbf{y}$, $f\left( \mathbf{x}\right) $ and $f\left( \mathbf{y}\right) $ introduced in Subsection \ref{subsect.thms.coprod}. If $R$ is any commutative ring, then $R\left[ \left[ \mathbf{y}\right] \right] $ shall denote the ring $R\left[ \left[ y_{1},y_{2},y_{3},\ldots\right] \right] $ of formal power series in the indeterminates $y_{1},y_{2},y_{3},\ldots$ over the ring $R$. We will identify the ring $\mathbf{k}\left[ \left[ \mathbf{x},\mathbf{y}% \right] \right] $ with the ring $\left( \mathbf{k}\left[ \left[ \mathbf{x}\right] \right] \right) \left[ \left[ \mathbf{y}\right] \right] =\left( \mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] \right) \left[ \left[ y_{1},y_{2},y_{3},\ldots\right] \right] $. Note that $\Lambda\subseteq\mathbf{k}\left[ \left[ \mathbf{x}\right] \right] $ and thus $\Lambda\left[ \left[ \mathbf{y}\right] \right] \subseteq\left( \mathbf{k}\left[ \left[ \mathbf{x}\right] \right] \right) \left[ \left[ \mathbf{y}\right] \right] =\mathbf{k}\left[ \left[ \mathbf{x},\mathbf{y}\right] \right] $. We equip the rings $\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $, $\Lambda\left[ \left[ \mathbf{y}\right] \right] $ and $\mathbf{k}\left[ \left[ \mathbf{x},\mathbf{y}\right] \right] $ with the usual topologies that are defined on rings of power series, where $\Lambda$ itself is equipped with the discrete topology. This has the somewhat confusing consequence that $\Lambda\left[ \left[ \mathbf{y}\right] \right] \subseteq\mathbf{k}\left[ \left[ \mathbf{x},\mathbf{y}\right] \right] $ is an inclusion of rings but not of topological spaces; however, this will not cause us any trouble, since all infinite sums in $\Lambda\left[ \left[ \mathbf{y}\right] \right] $ we will consider (such as $\sum_{\lambda\in\operatorname*{Par}}s_{\lambda/\mu }\left( \mathbf{x}\right) s_{\lambda}\left( \mathbf{y}\right) $ and $\sum_{\lambda\in\operatorname*{Par}}h_{\lambda}\left( \mathbf{x}\right) m_{\lambda}\left( \mathbf{y}\right) $) will converge to the same value in either topology. We consider both $\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $ and $\Lambda$ as subrings of $\Lambda\left[ \left[ \mathbf{y}\right] \right] $ (indeed, $\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $ embeds into $\Lambda\left[ \left[ \mathbf{y}\right] \right] $ because $\mathbf{k}$ is a subring of $\Lambda$, whereas $\Lambda$ embeds into $\Lambda\left[ \left[ \mathbf{y}\right] \right] $ because $\Lambda\left[ \left[ \mathbf{y}% \right] \right] $ is a ring of power series over $\Lambda$). In this proof, the word \textquotedblleft monomial\textquotedblright\ may refer to a monomial in any set of variables (not necessarily in $x_{1}% ,x_{2},x_{3},\ldots$). Recall the $\mathbf{k}$-algebra homomorphism $\alpha_{k}:\Lambda \rightarrow\mathbf{k}$ from Definition \ref{def.alphak}. This $\mathbf{k}% $-algebra homomorphism $\alpha_{k}:\Lambda\rightarrow\mathbf{k}$ induces a $\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $-algebra homomorphism $\alpha_{k}\left[ \left[ \mathbf{y}\right] \right] :\Lambda\left[ \left[ \mathbf{y}\right] \right] \rightarrow\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $, which is given by the formula% \[ \left( \alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \right) \left( \sum_{\substack{\mathfrak{n}\text{ is a monomial}\\\text{in }y_{1},y_{2}% ,y_{3},\ldots}}f_{\mathfrak{n}}\mathfrak{n}\right) =\sum _{\substack{\mathfrak{n}\text{ is a monomial}\\\text{in }y_{1},y_{2}% ,y_{3},\ldots}}\alpha_{k}\left( f_{\mathfrak{n}}\right) \mathfrak{n}% \] for any family $\left( f_{\mathfrak{n}}\right) _{\mathfrak{n}\text{ is a monomial in }y_{1},y_{2},y_{3},\ldots}$ of elements of $\Lambda$. This induced $\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $-algebra homomorphism $\alpha_{k}\left[ \left[ \mathbf{y}\right] \right] $ is $\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $-linear and continuous (with respect to the usual topologies on the power series rings $\Lambda\left[ \left[ \mathbf{y}\right] \right] $ and $\mathbf{k}\left[ \left[ \mathbf{y}% \right] \right] $), and thus preserves infinite $\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $-linear combinations. Moreover, it extends $\alpha_{k}$ (that is, for any $f\in\Lambda$, we have $\left( \alpha _{k}\left[ \left[ \mathbf{y}\right] \right] \right) \left( f\right) =\alpha_{k}\left( f\right) $). Recall the skew Schur functions $s_{\lambda/\mu}$ defined in Subsection \ref{subsect.proofs.skew}. Also, recall the symmetric functions $h_{\lambda}$ defined in Definition \ref{def.hlam}. Theorem \ref{thm.our-cauchy} yields% \begin{align*} \sum_{\lambda\in\operatorname*{Par}}s_{\lambda/\mu}\left( \mathbf{x}\right) s_{\lambda}\left( \mathbf{y}\right) & =s_{\mu}\left( \mathbf{y}\right) \cdot\sum_{\lambda\in\operatorname*{Par}}h_{\lambda}\left( \mathbf{x}\right) m_{\lambda}\left( \mathbf{y}\right) =\sum_{\lambda\in\operatorname*{Par}% }s_{\mu}\left( \mathbf{y}\right) h_{\lambda}\left( \mathbf{x}\right) m_{\lambda}\left( \mathbf{y}\right)\\ & =\sum_{\lambda\in\operatorname*{Par}}s_{\mu}\left( \mathbf{y}\right) m_{\lambda}\left( \mathbf{y}\right) \underbrace{h_{\lambda}\left( \mathbf{x}\right) }_{=h_{\lambda}}=\sum_{\lambda\in\operatorname*{Par}}% s_{\mu}\left( \mathbf{y}\right) m_{\lambda}\left( \mathbf{y}\right) h_{\lambda}. \end{align*} Comparing this with% \[ \sum_{\lambda\in\operatorname*{Par}}s_{\lambda/\mu}\left( \mathbf{x}\right) s_{\lambda}\left( \mathbf{y}\right) =\sum_{\lambda\in\operatorname*{Par}% }s_{\lambda}\left( \mathbf{y}\right) \underbrace{s_{\lambda/\mu}\left( \mathbf{x}\right) }_{=s_{\lambda/\mu}}=\sum_{\lambda\in\operatorname*{Par}% }s_{\lambda}\left( \mathbf{y}\right) s_{\lambda/\mu}, \] we obtain% \[ \sum_{\lambda\in\operatorname*{Par}}s_{\lambda}\left( \mathbf{y}\right) s_{\lambda/\mu}=\sum_{\lambda\in\operatorname*{Par}}s_{\mu}\left( \mathbf{y}\right) m_{\lambda}\left( \mathbf{y}\right) h_{\lambda}. \] Consider this as an equality in the ring $\Lambda\left[ \left[ \mathbf{y}\right] \right] =\Lambda\left[ \left[ y_{1},y_{2},y_{3}% ,\ldots\right] \right] $. Apply the map $\alpha_{k}\left[ \left[ \mathbf{y}\right] \right] :\Lambda\left[ \left[ \mathbf{y}\right] \right] \rightarrow\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] $ to both sides of this equality. We obtain% \[ \left( \alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \right) \left( \sum_{\lambda\in\operatorname*{Par}}s_{\lambda}\left( \mathbf{y}\right) s_{\lambda/\mu}\right) =\left( \alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \right) \left( \sum_{\lambda\in\operatorname*{Par}}s_{\mu}\left( \mathbf{y}\right) m_{\lambda}\left( \mathbf{y}\right) h_{\lambda}\right) . \] Comparing this with% \begin{align*} & \left( \alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \right) \left( \sum_{\lambda\in\operatorname*{Par}}s_{\lambda}\left( \mathbf{y}% \right) s_{\lambda/\mu}\right)\\ & =\sum_{\lambda\in\operatorname*{Par}}s_{\lambda}\left( \mathbf{y}\right) \cdot\underbrace{\left( \alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \right) \left( s_{\lambda/\mu}\right) }_{\substack{=\alpha_{k}\left( s_{\lambda/\mu}\right) \\\text{(since }\alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \text{ extends }\alpha_{k}\text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since the map }\alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \text{ preserves infinite }\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] \text{-linear combinations}\right)\\ & =\sum_{\lambda\in\operatorname*{Par}}s_{\lambda}\left( \mathbf{y}\right) \cdot\underbrace{\alpha_{k}\left( s_{\lambda/\mu}\right) }% _{\substack{=\operatorname*{pet}\nolimits_{k}\left( \lambda,\mu\right) \\\text{(by (\ref{pf.thm.G.pieri.alks}))}}}=\sum_{\lambda\in \operatorname*{Par}}\operatorname*{pet}\nolimits_{k}\left( \lambda ,\mu\right) \cdot s_{\lambda}\left( \mathbf{y}\right) , \end{align*} we obtain% \begin{align*} & \sum_{\lambda\in\operatorname*{Par}}\operatorname*{pet}\nolimits_{k}\left( \lambda,\mu\right) \cdot s_{\lambda}\left( \mathbf{y}\right)\\ & =\left( \alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \right) \left( \sum_{\lambda\in\operatorname*{Par}}s_{\mu}\left( \mathbf{y}\right) m_{\lambda}\left( \mathbf{y}\right) h_{\lambda}\right) =\sum_{\lambda \in\operatorname*{Par}}s_{\mu}\left( \mathbf{y}\right) m_{\lambda}\left( \mathbf{y}\right) \underbrace{\left( \alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \right) \left( h_{\lambda}\right) }_{\substack{=\alpha_{k}\left( h_{\lambda}\right) \\\text{(since }\alpha _{k}\left[ \left[ \mathbf{y}\right] \right] \text{ extends }\alpha _{k}\text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{since the map }\alpha_{k}\left[ \left[ \mathbf{y}\right] \right] \text{ preserves infinite }\mathbf{k}\left[ \left[ \mathbf{y}\right] \right] \text{-linear combinations}\right)\\ & =\sum_{\lambda\in\operatorname*{Par}}s_{\mu}\left( \mathbf{y}\right) m_{\lambda}\left( \mathbf{y}\right) \underbrace{\alpha_{k}\left( h_{\lambda}\right) }_{\substack{=\left[ \lambda_{i}\gamma_{2}>\cdots>\gamma_{k-1}$, then% \[ \det G=\left( -1\right) ^{\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma _{k-1}\right) -\left( 1+2+\cdots+\left( k-1\right) \right) }. \] \item\label{39c} Assume that the $k-1$ numbers $\gamma_{1},\gamma _{2},\ldots,\gamma_{k-1}$ are distinct. Let% \[ g=\left\vert \left\{ \left( i,j\right) \in\left\{ 1,2,\ldots,k-1\right\} ^{2}\ \mid\ i\gamma_{2}>\cdots>\gamma_{k-1}$. Thus, $\left\{ \gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}\right\} $ is a $\left( k-1\right) $-element subset of $\left\{ 1,2,\ldots,k\right\} $. Hence, $\left\{ \gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}\right\} =\left\{ 1,2,\ldots,k\right\} \setminus\left\{ u\right\} $ for some $u\in\left\{ 1,2,\ldots,k\right\} $ (since any $\left( k-1\right) $-element subset of $\left\{ 1,2,\ldots,k\right\} $ has such a form). Consider this $u$. From $\left\{ \gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}\right\} =\left\{ 1,2,\ldots,k\right\} \setminus\left\{ u\right\} $, we conclude that $\gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}$ are the $k-1$ elements of the set $\left\{ 1,2,\ldots,k\right\} \setminus\left\{ u\right\} $, listed in decreasing order (since $\gamma_{1}>\gamma_{2}>\cdots>\gamma_{k-1}$). In other words,% \begin{equation} \left( \gamma_{1},\gamma_{2},\ldots,\gamma_{k-1}\right) =\left( k,k-1,\ldots,\widehat{u},\ldots,2,1\right) , \label{pf.lem.petk.explicit.2.b.short.2}% \end{equation} where the \textquotedblleft hat\textquotedblright\ over the $u$ signifies that $u$ is omitted from the list (\ie the expression \textquotedblleft$\left( k,k-1,\ldots,\widehat{u},\ldots,2,1\right) $\textquotedblright\ is understood to mean the $\left( k-1\right) $-element list $\left( k,k-1,\ldots ,u+1,u-1,\ldots,2,1\right) $, which contains all $k$ integers from $1$ to $k$ in decreasing order except for $u$). Now, we claim that% \begin{align} & \left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \nonumber\\ & =\left( -1\right) ^{\gamma_{i}+j-k}\left[ \gamma_{i}+j\in\left\{ k,k+1\right\} \right] \label{pf.lem.petk.explicit.2.b.5}% \end{align} for any $i\in\left\{ 1,2,\ldots,k-1\right\} $ and $j\in\left\{ 1,2,\ldots,k-1\right\} $. [\textit{Proof of (\ref{pf.lem.petk.explicit.2.b.5}):} Let $i\in\left\{ 1,2,\ldots,k-1\right\} $ and $j\in\left\{ 1,2,\ldots,k-1\right\} $. We must prove the equality (\ref{pf.lem.petk.explicit.2.b.5}). From $i\in\left\{ 1,2,\ldots,k-1\right\} $, we obtain $1\leq i\leq k-1$ and thus $k-1\geq1$. Thus, $k>k-1\geq1$. Hence, $k+1<2k$, so that $\left( k+1\right) \%k=1$. If we don't have $\left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} $, then both truth values \newline$\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] $ and $\left[ \gamma_{i}+j\in\left\{ k,k+1\right\} \right] $ are $0$ (indeed, the statement \textquotedblleft% $\gamma_{i}+j\in\left\{ k,k+1\right\} $\textquotedblright\ is false, since otherwise it would imply $\left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} $), and therefore the equality (\ref{pf.lem.petk.explicit.2.b.5}) simplifies to $\left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}0=\left( -1\right) ^{\gamma_{i}+j-k}0$ in this case, which is obviously true. Hence, for the rest of this proof, we WLOG assume that we do have $\left( \gamma _{i}+j\right) \%k\in\left\{ 0,1\right\} $. Adding $\gamma_{i}\in\left\{ 1,2,\ldots,k\right\} $ with $j\in\left\{ 1,2,\ldots,k-1\right\} $, we find that $\gamma_{i}+j\in\left\{ 2,3,\ldots ,2k-1\right\} $. Thus, $\gamma_{i}+j$ is an integer in the set $\left\{ 2,3,\ldots,2k-1\right\} $ that leaves the remainder $0$ or $1$ upon division by $k$ (since $\left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} $). But the only such integers are $k$ and $k+1$. Hence, we obtain $\gamma _{i}+j\in\left\{ k,k+1\right\} \subseteq\left\{ k,k+1,\ldots,2k-1\right\} $ (since $k+1<2k$), and therefore $\left( \gamma_{i}+j\right) \%k=\gamma _{i}+j-k$. Thus,% \[ \underbrace{\left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}% }_{\substack{=\left( -1\right) ^{\gamma_{i}+j-k}\\\text{(since }\left( \gamma_{i}+j\right) \%k=\gamma_{i}+j-k\text{)}}}\underbrace{\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] }% _{\substack{=1\\\text{(since }\left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \text{)}}}=\left( -1\right) ^{\gamma_{i}+j-k}. \] \noindent Comparing this with% \[ \left( -1\right) ^{\gamma_{i}+j-k}\underbrace{\left[ \gamma_{i}% +j\in\left\{ k,k+1\right\} \right] }_{\substack{=1\\\text{(since }% \gamma_{i}+j\in\left\{ k,k+1\right\} \text{)}}}=\left( -1\right) ^{\gamma_{i}+j-k}, \] we obtain% \[ \left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] =\left( -1\right) ^{\gamma_{i}+j-k}\left[ \gamma_{i}+j\in\left\{ k,k+1\right\} \right] . \] This proves (\ref{pf.lem.petk.explicit.2.b.5}).] Now, $G$ is a $\left( k-1\right) \times\left( k-1\right) $-matrix. For each $i\in\left\{ 1,2,\ldots,k-1\right\} $ and $j\in\left\{ 1,2,\ldots ,k-1\right\} $, we have% \begin{align*} & \left( \text{the }\left( i,j\right) \text{-th entry of }G\right)\\ & =\left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }G\right)\\ & =\left( -1\right) ^{\gamma_{i}+j-k}\left[ \gamma_{i}+j\in\left\{ k,k+1\right\} \right] \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.lem.petk.explicit.2.b.5})}\right)\\ & =% \begin{cases} 1, & \text{if }\gamma_{i}+j=k;\\ -1, & \text{if }\gamma_{i}+j=k+1;\\ 0, & \text{otherwise}% \end{cases} \quad=% \begin{cases} 1, & \text{if }j=k-\gamma_{i};\\ -1, & \text{if }j=k-\gamma_{i}+1;\\ 0, & \text{otherwise}. \end{cases} \end{align*} \noindent Thus, we can explicitly describe the matrix $G$ as follows: For each $i\in\left\{ 1,2,\ldots,k-1\right\} $, the $i$-th row of $G$ has an entry equal to $1$ in position $k-\gamma_{i}$ if $k-\gamma_{i}>0$, and an entry equal to $-1$ in position $k-\gamma_{i}+1$ if $k-\gamma_{i}+1\gamma_{\sigma\left( 2\right) }>\cdots >\gamma_{\sigma\left( k-1\right) }$. Consider this $\sigma$. Let $\tau$ denote the permutation $\sigma^{-1}$. Let $\delta_{1},\delta_{2},\ldots,\delta_{k-1}$ denote the $k-1$ elements $\gamma_{\sigma\left( 1\right) },\gamma_{\sigma\left( 2\right) }% ,\ldots,\gamma_{\sigma\left( k-1\right) }$ of $\left\{ 1,2,\ldots ,k\right\} $. Thus, for each $j\in\left\{ 1,2,\ldots,k-1\right\} $, we have% \begin{equation} \delta_{j}=\gamma_{\sigma\left( j\right) }. \label{pf.lem.petk.explicit.2.c.1}% \end{equation} Hence, the chain of inequalities $\gamma_{\sigma\left( 1\right) }% >\gamma_{\sigma\left( 2\right) }>\cdots>\gamma_{\sigma\left( k-1\right) }$ (which is true) can be rewritten as $\delta_{1}>\delta_{2}>\cdots>\delta _{k-1}$. Moreover, from (\ref{pf.lem.petk.explicit.2.c.1}), we obtain% \begin{align} \delta_{1}+\delta_{2}+\cdots+\delta_{k-1} & =\gamma_{\sigma\left( 1\right) }+\gamma_{\sigma\left( 2\right) }+\cdots+\gamma_{\sigma\left( k-1\right) }\nonumber\\ & =\gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1} \label{pf.lem.petk.explicit.2.c.sumdel=sumgam}% \end{align} (since $\sigma$ is a permutation of $\left\{ 1,2,\ldots,k-1\right\} $). Moreover, for each $i\in\left\{ 1,2,\ldots,k-1\right\} $, we have% \begin{align} \delta_{\tau\left( i\right) } & =\gamma_{\sigma\left( \tau\left( i\right) \right) }\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.lem.petk.explicit.2.c.1}), applied to }j=\tau\left( i\right) \right) \nonumber\\ & =\gamma_{i}\ \ \ \ \ \ \ \ \ \ \left( \text{since }\tau=\sigma^{-1}\text{ entails }\sigma\left( \tau\left( i\right) \right) =i\right) . \label{pf.lem.petk.explicit.2.c.2}% \end{align} Recall that an \emph{inversion} of the permutation $\tau$ is defined to be a pair $\left( i,j\right) $ of elements of $\left\{ 1,2,\ldots,k-1\right\} $ satisfying $i\tau\left( j\right) $. Hence, the inversions of $\tau$ are precisely the pairs $\left( i,j\right) $ of elements of $\left\{ 1,2,\ldots,k-1\right\} $ satisfying $i\delta_{2}>\cdots>\delta_{k-1}$ shows that the statement \textquotedblleft$\tau\left( i\right) >\tau\left( j\right) $% \textquotedblright\ is equivalent to \textquotedblleft$\delta_{\tau\left( i\right) }<\delta_{\tau\left( j\right) }$\textquotedblright). In other words, the inversions of $\tau$ are precisely the pairs $\left( i,j\right) $ of elements of $\left\{ 1,2,\ldots,k-1\right\} $ satisfying $i\delta_{2}>\cdots>\delta_{k-1}$). We thus obtain% \[ \det H=\left( -1\right) ^{\left( \delta_{1}+\delta_{2}+\cdots+\delta _{k-1}\right) -\left( 1+2+\cdots+\left( k-1\right) \right) }=\left( -1\right) ^{\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right) -\left( 1+2+\cdots+\left( k-1\right) \right) }% \] (by (\ref{pf.lem.petk.explicit.2.c.sumdel=sumgam})). But the definition of $G$ yields% \begin{align*} G & =\left( \left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\\ & =\left( \left( -1\right) ^{\left( \delta_{\tau\left( i\right) }+j\right) \%k}\left[ \left( \delta_{\tau\left( i\right) }+j\right) \%k\in\left\{ 0,1\right\} \right] \right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}% \end{align*} (since (\ref{pf.lem.petk.explicit.2.c.2}) yields $\gamma_{i}=\delta _{\tau\left( i\right) }$ for each $i\in\left\{ 1,2,\ldots,k-1\right\} $). Hence,% \begin{align*} \det G & =\det\left( \left( \left( -1\right) ^{\left( \delta _{\tau\left( i\right) }+j\right) \%k}\left[ \left( \delta_{\tau\left( i\right) }+j\right) \%k\in\left\{ 0,1\right\} \right] \right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\right)\\ & =\underbrace{\left( -1\right) ^{\tau}}_{=\left( -1\right) ^{g}}% \det\left( \underbrace{\left( \left( -1\right) ^{\left( \delta _{i}+j\right) \%k}\left[ \left( \delta_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}% }_{\substack{=H\\\text{(by the definition of }H\text{)}}}\right)\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{by Lemma \ref{lem.petk.explicit.1} \ref{38a}, applied to }m=k-1\text{ and }R=\mathbf{k}\\ \text{and }a_{i,j}=\left( -1\right) ^{\left( \delta_{i}+j\right) \%k}\left[ \left( \delta_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \end{array} \right)\\ & =\left( -1\right) ^{g}\underbrace{\det H}_{=\left( -1\right) ^{\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right) -\left( 1+2+\cdots+\left( k-1\right) \right) }}\\ & =\left( -1\right) ^{g}\left( -1\right) ^{\left( \gamma_{1}+\gamma _{2}+\cdots+\gamma_{k-1}\right) -\left( 1+2+\cdots+\left( k-1\right) \right) }\\ & =\left( -1\right) ^{g+\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma _{k-1}\right) -\left( 1+2+\cdots+\left( k-1\right) \right) }. \end{align*} This proves Lemma \ref{lem.petk.explicit.2} \ref{39c}. \end{proof} Next, we recall a well-known property of symmetric functions (see, \eg \cite[proof of Theorem 7.6.1]{Stanley-EC2} or \cite[(2.4.3)]{GriRei}): \begin{lemma} \label{lem.h-e-reciprocity-t}Consider the ring $\Lambda\left[ \left[ t\right] \right] $ of formal power series in one indeterminate $t$ over $\Lambda$. In this ring, we have% \begin{equation} 1=\left( \sum_{n\geq0}\left( -1\right) ^{n}e_{n}t^{n}\right) \left( \sum_{n\geq0}h_{n}t^{n}\right) . \label{pf.thm.petk.explicit.aker.pf.1}% \end{equation} \end{lemma} Next, we shall prove yet another evaluation of the homomorphism $\alpha_{k}$: \begin{lemma} \label{lem.alphak.e}Let $k$ be a positive integer such that $k>1$. Consider the $\mathbf{k}$-algebra homomorphism $\alpha_{k}:\Lambda\rightarrow \mathbf{k}$ from Definition \ref{def.alphak}. Also, recall Convention \ref{conv.iverson}. Let $r$ be an integer such that $r>-k+1$. Then,% \begin{equation} \alpha_{k}\left( e_{r}\right) =\left( -1\right) ^{r+r\%k}\left[ r\%k\in\left\{ 0,1\right\} \right] . \label{pf.thm.petk.explicit.aker}% \end{equation} \end{lemma} \begin{proof} [Proof of Lemma \ref{lem.alphak.e}.]Consider the ring $\Lambda\left[ \left[ t\right] \right] $ of formal power series in one indeterminate $t$ over $\Lambda$. Consider also the analogous ring $\mathbf{k}\left[ \left[ t\right] \right] $ over $\mathbf{k}$. The map $\alpha_{k}:\Lambda\rightarrow\mathbf{k}$ is a $\mathbf{k}$-algebra homomorphism. Hence, it induces a $\mathbf{k}\left[ \left[ t\right] \right] $-algebra homomorphism% \[ \alpha_{k}\left[ \left[ t\right] \right] :\Lambda\left[ \left[ t\right] \right] \rightarrow\mathbf{k}\left[ \left[ t\right] \right] \] that sends each formal power series $\sum_{n\geq0}a_{n}t^{n}\in\Lambda\left[ \left[ t\right] \right] $ (with $a_{n}\in\Lambda$) to $\sum_{n\geq0}% \alpha_{k}\left( a_{n}\right) t^{n}$. Consider this $\alpha_{k}\left[ \left[ t\right] \right] $. Applying the map $\alpha_{k}\left[ \left[ t\right] \right] $ to both sides of the equality (\ref{pf.thm.petk.explicit.aker.pf.1}), we obtain% \begin{align*} \left( \alpha_{k}\left[ \left[ t\right] \right] \right) \left( 1\right) & =\left( \alpha_{k}\left[ \left[ t\right] \right] \right) \left( \left( \sum_{n\geq0}\left( -1\right) ^{n}e_{n}t^{n}\right) \left( \sum_{n\geq0}h_{n}t^{n}\right) \right)\\ & =\underbrace{\left( \alpha_{k}\left[ \left[ t\right] \right] \right) \left( \sum_{n\geq0}\left( -1\right) ^{n}e_{n}t^{n}\right) }% _{\substack{=\sum_{n\geq0}\alpha_{k}\left( \left( -1\right) ^{n}% e_{n}\right) t^{n}\\\text{(by the definition of }\alpha_{k}\left[ \left[ t\right] \right] \text{)}}}\cdot\underbrace{\left( \alpha_{k}\left[ \left[ t\right] \right] \right) \left( \sum_{n\geq0}h_{n}t^{n}\right) }_{\substack{=\sum_{n\geq0}\alpha_{k}\left( h_{n}\right) t^{n}\\\text{(by the definition of }\alpha_{k}\left[ \left[ t\right] \right] \text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{since }\alpha _{k}\left[ \left[ t\right] \right] \text{ is a }\mathbf{k}\left[ \left[ t\right] \right] \text{-algebra homomorphism}\right)\\ & =\left( \sum_{n\geq0}\underbrace{\alpha_{k}\left( \left( -1\right) ^{n}e_{n}\right) }_{\substack{=\left( -1\right) ^{n}\alpha_{k}\left( e_{n}\right) \\\text{(since }\alpha_{k}\text{ is }\mathbf{k}\text{-linear)}% }}t^{n}\right) \cdot\left( \sum_{n\geq0}\underbrace{\alpha_{k}\left( h_{n}\right) }_{\substack{=\left[ n1$, since for $k=1$ there would be cancellations in the sum $1-t+t^{k}-t^{k+1}+t^{2k}-t^{2k+1}+t^{3k}-t^{3k+1}\pm\cdots$). Comparing coefficients before $t^{m}$ on both sides of this equality, we obtain% \begin{equation} \left( -1\right) ^{m}\alpha_{k}\left( e_{m}\right) =\left( -1\right) ^{m\%k}\left[ m\%k\in\left\{ 0,1\right\} \right] \label{pf.thm.petk.explicit.aker.pf.4}% \end{equation} for each $m\in\mathbb{N}$. Multiplying both sides of this equality by $\left( -1\right) ^{m}$, we obtain% \begin{equation} \alpha_{k}\left( e_{m}\right) =\left( -1\right) ^{m+m\%k}\left[ m\%k\in\left\{ 0,1\right\} \right] . \label{pf.thm.petk.explicit.aker.pf.short.4b}% \end{equation} We must prove that \[ \alpha_{k}\left( e_{r}\right) =\left( -1\right) ^{r+r\%k}\left[ r\%k\in\left\{ 0,1\right\} \right] . \] If $r\in\mathbb{N}$, then this follows by applying (\ref{pf.thm.petk.explicit.aker.pf.short.4b}) to $m=r$. Hence, for the rest of this proof, we WLOG assume that $r\notin\mathbb{N}$. Thus, $r$ is negative, so that $r\in\left\{ -k+2,-k+3,\ldots,-1\right\} $ (since $r>-k+1$). Hence, $r\%k\in\left\{ 2,3,\ldots,k-1\right\} $, so that $r\%k\notin\left\{ 0,1\right\} $. Consequently, $\left[ r\%k\in\left\{ 0,1\right\} \right] =0$. Also, $e_{r}=0$ (since $r$ is negative) and thus $\alpha_{k}\left( e_{r}\right) =\alpha_{k}\left( 0\right) =0$. Comparing this with $\left( -1\right) ^{r+r\%k}\underbrace{\left[ r\%k\in\left\{ 0,1\right\} \right] }_{=0}=0$, we obtain $\alpha_{k}\left( e_{r}\right) =\left( -1\right) ^{r+r\%k}\left[ r\%k\in\left\{ 0,1\right\} \right] $. This concludes the proof of Lemma \ref{lem.alphak.e}. \end{proof} \begin{proof} [Proof of Theorem \ref{thm.petk.explicit}.]\ref{213a} Assume that $\mu _{k}\neq0$. Thus, the partition $\mu$ has at least $k$ entries. But we have $\mu=\lambda^{t}$, and thus (by a known and easy fact) the partition $\mu$ has precisely $\lambda_{1}$ entries. Reconciling the previous two sentences, we see that $\lambda_{1}\geq k$. Thus, Proposition \ref{prop.petk.0} yields $\operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing\right) =0$. This proves Theorem \ref{thm.petk.explicit} \ref{213a}. Now, let us prepare for the proof of parts~\ref{213b} and \ref{213c}. Consider the $\mathbf{k}$-algebra homomorphism $\alpha_{k}:\Lambda \rightarrow\mathbf{k}$ from Definition \ref{def.alphak}. We have $s_{\lambda}=s_{\lambda/\varnothing}$, so that \begin{equation} \alpha_{k}\left( s_{\lambda}\right) =\alpha_{k}\left( s_{\lambda /\varnothing}\right) =\operatorname*{pet}\nolimits_{k}\left( \lambda ,\varnothing\right) \label{pf.thm.petk.explicit.as-det-0}% \end{equation} (by (\ref{pf.thm.G.pieri.alks}), applied to $\varnothing$ instead of $\mu$). For each $i\in\left\{ 1,2,\ldots,k-1\right\} $, we have $\gamma_{i}% \in\left\{ 1,2,\ldots,k\right\} $ (by (\ref{eq.thm.petk.explicit.gammi=}), since the remainder $\left( \beta_{i}-1\right) \%k$ clearly belongs to $\left\{ 0,1,\ldots,k-1\right\} $). In other words, $\gamma_{1},\gamma _{2},\ldots,\gamma_{k-1}$ are $k-1$ elements of the set $\left\{ 1,2,\ldots,k\right\} $. Assume that $\mu_{k}=0$. Thus, $\mu=\left( \mu_{1},\mu_{2},\ldots,\mu _{k-1}\right) $ (since $\mu\in\operatorname*{Par}$). It is known that $\left( \lambda^{t}\right) ^{t}=\lambda$. Hence, $\lambda=\left( \lambda^{t}\right) ^{t}=\mu^{t}$ (since $\lambda^{t}=\mu$). Therefore,% \[ s_{\lambda}=s_{\mu^{t}}=\det\left( \left( e_{\mu_{i}-i+j}\right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\right) \] (by (\ref{eq.schur.JTsh-e}), applied to $\mu$ and $k-1$ instead of $\lambda$ and $\ell$), because $\mu=\left( \mu_{1},\mu_{2},\ldots,\mu_{k-1}\right) $. This rewrites as% \[ s_{\lambda}=\det\left( \left( e_{\beta_{i}+j}\right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\right) \] (since (\ref{eq.thm.petk.explicit.beti=}) yields $\mu_{i}-i=\beta_{i}$ for each $i\in\left\{ 1,2,\ldots,k-1\right\} $). Applying the map $\alpha_{k}$ to both sides of this equality, we find% \[ \alpha_{k}\left( s_{\lambda}\right) =\alpha_{k}\left( \det\left( \left( e_{\beta_{i}+j}\right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\right) \right) =\det\left( \left( \alpha_{k}\left( e_{\beta_{i}+j}\right) \right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\right) \] (since $\alpha_{k}$ is a $\mathbf{k}$-algebra homomorphism, and thus commutes with taking determinants of matrices). Comparing this with (\ref{pf.thm.petk.explicit.as-det-0}), we obtain% \begin{equation} \operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing\right) =\det\left( \left( \alpha_{k}\left( e_{\beta_{i}+j}\right) \right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\right) . \label{pf.thm.petk.explicit.as-det-1}% \end{equation} But each $i\in\left\{ 1,2,\ldots,k-1\right\} $ and $j\in\left\{ 1,2,\ldots,k-1\right\} $ satisfy $k>1$\ \ \ \ \footnote{Indeed, if $i\in\left\{ 1,2,\ldots,k-1\right\} $, then $1\leq i\leq k-1$ and thus $k-1\geq1>0$, so that $k>1$.} and% \[ \underbrace{\beta_{i}}_{\substack{=\mu_{i}-i\\\text{(by (\ref{eq.thm.petk.explicit.beti=}))}}}+j=\underbrace{\mu_{i}}_{\geq 0}-\underbrace{i}_{\leq k-1}+\underbrace{j}_{>0}>0-\left( k-1\right) +0=-k+1 \] and thus% \begin{equation} \alpha_{k}\left( e_{\beta_{i}+j}\right) =\left( -1\right) ^{\left( \beta_{i}+j\right) +\left( \beta_{i}+j\right) \%k}\left[ \left( \beta _{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \label{pf.thm.petk.explicit.akem1a}% \end{equation} (by (\ref{pf.thm.petk.explicit.aker}), applied to $r=\beta_{i}+j$). Furthermore, each $i\in\left\{ 1,2,\ldots,k-1\right\} $ and $j\in\left\{ 1,2,\ldots,k-1\right\} $ satisfy% \begin{align} & \left( -1\right) ^{\left( \beta_{i}+j\right) +\left( \beta _{i}+j\right) \%k}\left[ \left( \beta_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \nonumber\\ & =\left( -1\right) ^{\beta_{i}}\left( -1\right) ^{j}\left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] . \label{pf.thm.petk.explicit.akem2}% \end{align} [\textit{Proof of (\ref{pf.thm.petk.explicit.akem2}):} Let $i\in\left\{ 1,2,\ldots,k-1\right\} $ and $j\in\left\{ 1,2,\ldots,k-1\right\} $. We have $\left( \beta_{i}-1\right) \%k\equiv\beta_{i}-1\operatorname{mod}k$ (since $u\%k\equiv u\operatorname{mod}k$ for any $u\in\mathbb{Z}$), so that $1+\left( \beta_{i}-1\right) \%k\equiv\beta_{i}\operatorname{mod}k$. But the definition of $\gamma_{i}$ yields $\gamma_{i}=1+\left( \beta_{i}-1\right) \%k\equiv\beta_{i}\operatorname{mod}k$. Hence, $\gamma_{i}+j\equiv\beta _{i}+j\operatorname{mod}k$, and therefore $\left( \gamma_{i}+j\right) \%k=\left( \beta_{i}+j\right) \%k$ (because if two integers are congruent modulo $k$, then their remainders upon division by $k$ are equal). Hence,% \begin{align*} & \left( -1\right) ^{\beta_{i}}\left( -1\right) ^{j}\left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right]\\ & =\left( -1\right) ^{\beta_{i}}\left( -1\right) ^{j}\left( -1\right) ^{\left( \beta_{i}+j\right) \%k}\left[ \left( \beta_{i}+j\right) \%k\in\left\{ 0,1\right\} \right]\\ & =\left( -1\right) ^{\left( \beta_{i}+j\right) +\left( \beta _{i}+j\right) \%k}\left[ \left( \beta_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] . \end{align*} This proves (\ref{pf.thm.petk.explicit.akem2}).] Now, for each $i\in\left\{ 1,2,\ldots,k-1\right\} $ and $j\in\left\{ 1,2,\ldots,k-1\right\} $, we have% \begin{align*} \alpha_{k}\left( e_{\beta_{i}+j}\right) & =\left( -1\right) ^{\left( \beta_{i}+j\right) +\left( \beta_{i}+j\right) \%k}\left[ \left( \beta _{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.thm.petk.explicit.akem1a}% )}\right)\\ & =\left( -1\right) ^{\beta_{i}}\left( -1\right) ^{j}\left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.thm.petk.explicit.akem2})}\right) . \end{align*} Thus, (\ref{pf.thm.petk.explicit.as-det-1}) can be rewritten as follows:% \begin{align*} & \operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing\right)\\ & =\det\left( \left( \left( -1\right) ^{\beta_{i}}\left( -1\right) ^{j}\left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\right)\\ & =\left( \prod_{i=1}^{k-1}\left( \left( -1\right) ^{\beta_{i}}\left( -1\right) ^{i}\right) \right)\\ & \ \ \ \ \ \ \ \ \ \ \cdot\det\left( \left( \left( -1\right) ^{\left( \gamma_{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] \right) _{1\leq i\leq k-1,\ 1\leq j\leq k-1}\right) \end{align*} (by Lemma \ref{lem.petk.explicit.1} \ref{38b}, applied to $m=k-1$ and $R=\mathbf{k}$ and \newline$a_{i,j}=\left( -1\right) ^{\left( \gamma _{i}+j\right) \%k}\left[ \left( \gamma_{i}+j\right) \%k\in\left\{ 0,1\right\} \right] $ and $u_{i}=\left( -1\right) ^{\beta_{i}}$ and $v_{j}=\left( -1\right) ^{j}$). Define a $\left( k-1\right) \times\left( k-1\right) $-matrix $G$ as in Lemma \ref{lem.petk.explicit.2}. Then, the matrix whose determinant appears on the right hand side of this equality is precisely $G$; thus, this equality rewrites as% \begin{equation} \operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing\right) =\left( \prod_{i=1}^{k-1}\left( \left( -1\right) ^{\beta_{i}}\left( -1\right) ^{i}\right) \right) \cdot\det G. \label{pf.thm.petk.explicit.petk-via-G}% \end{equation} Now, we can readily prove parts~\ref{213b} and \ref{213c} of Theorem \ref{thm.petk.explicit}: \ref{213b} Assume that the $k-1$ numbers $\gamma_{1},\gamma_{2}% ,\ldots,\gamma_{k-1}$ are not distinct. Then, Lemma \ref{lem.petk.explicit.2} \ref{39a} yields $\det G=0$. Hence, (\ref{pf.thm.petk.explicit.petk-via-G}) simplifies to $\operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing \right) =0$. This proves Theorem \ref{thm.petk.explicit} \ref{213b}. \ref{213c} The equality (\ref{pf.thm.petk.explicit.petk-via-G}) becomes% \begin{align*} & \operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing\right)\\ & =\underbrace{\left( \prod_{i=1}^{k-1}\left( \left( -1\right) ^{\beta_{i}}\left( -1\right) ^{i}\right) \right) }_{=\left( -1\right) ^{\left( \beta_{1}+\beta_{2}+\cdots+\beta_{k-1}\right) +\left( 1+2+\cdots+\left( k-1\right) \right) }}\cdot\underbrace{\det G}% _{\substack{=\left( -1\right) ^{g+\left( \gamma_{1}+\gamma_{2}% +\cdots+\gamma_{k-1}\right) -\left( 1+2+\cdots+\left( k-1\right) \right) }\\\text{(by Lemma \ref{lem.petk.explicit.2} \textbf{(c)})}}}\\ & =\left( -1\right) ^{\left( \beta_{1}+\beta_{2}+\cdots+\beta _{k-1}\right) +\left( 1+2+\cdots+\left( k-1\right) \right) }\cdot\left( -1\right) ^{g+\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right) -\left( 1+2+\cdots+\left( k-1\right) \right) }\\ & =\left( -1\right) ^{\left( \beta_{1}+\beta_{2}+\cdots+\beta _{k-1}\right) +g+\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma_{k-1}\right) }. \end{align*} This proves Theorem \ref{thm.petk.explicit} \ref{213c}. \end{proof} The proof of Proposition \ref{prop.petk.explicit-old} relies on the following known fact: \begin{proposition} \label{prop.partition.transpose.djun}Let $\lambda\in\operatorname*{Par}$. Let $\mu=\lambda^{t}$. Then: \begin{enumerate}[label=\textbf{(\alph*)}] \item\label{312a} If $i$ and $j$ are two positive integers satisfying $\lambda_{i}\geq j$, then $\mu_{j}\geq i$. \item\label{312b} If $i$ and $j$ are two positive integers satisfying $\lambda_{i}\beta_{2}>\cdots>\beta_{k-1}$ and $\lambda_{1}-1>\lambda _{2}-2>\lambda_{3}-3>\cdots$. From $\lambda_{1}\beta_{2}>\cdots>\beta_{k-1}$, so that $\eta_{1}<\eta_{2}<\cdots<\eta_{k-1}$. Hence, $\eta_{1},\eta_{2},\ldots ,\eta_{k-1}$ are the elements of the set $\left\{ \eta_{1},\eta_{2}% ,\ldots,\eta_{k-1}\right\} $ listed in increasing order (with no repetition). Let us define a $\left( k-1\right) $-tuple $\left( \gamma_{1},\gamma _{2},\ldots,\gamma_{k-1}\right) \in\left\{ 1,2,\ldots,k\right\} ^{k-1}$ as in Theorem \ref{thm.petk.explicit}. Now, we have the following chain of logical equivalences:% \begin{align*} & \ \left( \operatorname*{pet}\nolimits_{k}\left( \lambda,\varnothing \right) \neq0\right)\\ & \Longleftrightarrow\ \left( \text{the }k-1\text{ numbers }\gamma _{1},\gamma_{2},\ldots,\gamma_{k-1}\text{ are distinct}\right)\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{by parts \ref{213b} and \ref{213c} of Theorem \ref{thm.petk.explicit}}\right)\\ & \Longleftrightarrow\ \left( \text{the }k-1\text{ numbers }\left( \beta_{1}-1\right) \%k,\left( \beta_{2}-1\right) \%k,\ldots,\left( \beta_{k-1}-1\right) \%k\text{ are distinct}\right)\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{since }\gamma _{i}=1+\left( \beta_{i}-1\right) \%k\text{ for each }i\right)\\ & \Longleftrightarrow\ \left( \text{no two of the }k-1\text{ numbers }% \beta_{1}-1,\beta_{2}-1,\ldots,\beta_{k-1}-1\right.\\ & \ \ \ \ \ \ \ \ \ \ \left. \text{are congruent modulo }k\right)\\ & \Longleftrightarrow\ \left( \text{no two of the }k-1\text{ numbers }% \beta_{1},\beta_{2},\ldots,\beta_{k-1}\text{ are congruent modulo }k\right)\\ & \Longleftrightarrow\ \left( \text{no two of the }k-1\text{ numbers }-1-\beta_{1},-1-\beta_{2},\ldots,-1-\beta_{k-1}\text{ }\right.\\ & \ \ \ \ \ \ \ \ \ \ \left. \text{are congruent modulo }k\right)\\ & \Longleftrightarrow\ \left( \text{no two of the }k-1\text{ numbers }% \eta_{1},\eta_{2},\ldots,\eta_{k-1}\text{ are congruent modulo }k\right)\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{since }\eta _{j}=-1-\beta_{j}\text{ for each }j\right)\\ & \Longleftrightarrow\ \left( \text{no two of the }k-1\text{ elements of }\left\{ \eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\} \text{ are congruent modulo }k\right)\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\eta_{1},\eta_{2},\ldots,\eta_{k-1}\text{ are the elements of the set }\left\{ \eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\}\\ \text{listed in increasing order (with no repetition)}% \end{array} \right)\\ & \Longleftrightarrow\ \left( \text{no two of the }k-1\text{ elements of }W\setminus B\text{ are congruent modulo }k\right)\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left\{ \eta_{1},\eta_{2},\ldots,\eta_{k-1}\right\} =W\setminus B\right)\\ & \Longleftrightarrow\ \left( \text{each congruence class }\overline {i}\text{ has at most }1\text{ element in common with }W\setminus B\right)\\ & \Longleftrightarrow\ \left( \text{each }i\in\left\{ 0,1,\ldots ,k-1\right\} \text{ satisfies }\left\vert \overline{i}\cap\left( W\setminus B\right) \right\vert \leq1\right)\\ & \Longleftrightarrow\ \left( \text{each }i\in\left\{ 0,1,\ldots ,k-1\right\} \text{ satisfies }\left\vert \left( \overline{i}\cap W\right) \setminus B\right\vert \leq1\right) \end{align*} (since $\overline{i}\cap\left( W\setminus B\right) =\left( \overline{i}\cap W\right) \setminus B$ for each $i$). This proves Proposition \ref{prop.petk.explicit-old}. \end{proof} \subsection{\label{subsect.proofs.G.frob}Proof of Theorem \ref{thm.G.frob}} \begin{proof} [Proof of Theorem \ref{thm.G.frob}.]Consider the ring $\left( \mathbf{k}% \left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] \right) \left[ \left[ t\right] \right] $ of formal power series in one indeterminate $t$ over $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $. We equip this ring with the topology that is obtained by identifying it with $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots,t\right] \right] $ (or, equivalently, which is obtained by considering $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ itself as equipped with the standard topology on a ring of formal power series, and then adjoining the extra indeterminate~$t$). Now, consider the map% \begin{align*} \mathbf{F}_{k}:\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] & \rightarrow\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3}% ,\ldots\right] \right] ,\\ a & \mapsto a\left( x_{1}^{k},x_{2}^{k},x_{3}^{k},\ldots\right) . \end{align*} This map $\mathbf{F}_{k}$ is a continuous $\mathbf{k}$-algebra homomorphism (since it is an evaluation homomorphism).\footnote{It is well-defined, since $k$ is positive.} Hence, it induces a continuous\footnote{Continuity is defined with respect to the topology that we defined on $\left( \mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] \right) \left[ \left[ t\right] \right] $.} $\mathbf{k}\left[ \left[ t\right] \right] $-algebra homomorphism% \[ \mathbf{F}_{k}\left[ \left[ t\right] \right] :\left( \mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] \right) \left[ \left[ t\right] \right] \rightarrow\left( \mathbf{k}\left[ \left[ x_{1}% ,x_{2},x_{3},\ldots\right] \right] \right) \left[ \left[ t\right] \right] \] that sends each formal power series $\sum_{n\geq0}a_{n}t^{n}\in\left( \mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] \right) \left[ \left[ t\right] \right] $ (with $a_{n}\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $) to $\sum_{n\geq0}\mathbf{F}% _{k}\left( a_{n}\right) t^{n}$. Consider this $\mathbf{k}\left[ \left[ t\right] \right] $-algebra homomorphism $\mathbf{F}_{k}\left[ \left[ t\right] \right] $. The definition of $\mathbf{F}_{k}$ yields% \begin{equation} \mathbf{F}_{k}\left( x_{i}\right) =x_{i}^{k}\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{ 1,2,3,\ldots\right\} . \label{pf.thm.G.frob.Fkxi}% \end{equation} Also, for each $a\in\Lambda$, we have% \begin{align} \mathbf{F}_{k}\left( a\right) & =a\left( x_{1}^{k},x_{2}^{k},x_{3}% ^{k},\ldots\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\mathbf{F}_{k}\right) \nonumber\\ & =\mathbf{f}_{k}\left( a\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\mathbf{f}_{k}\right) . \label{pf.thm.G.frob.Fka}% \end{align} It is known (\eg \cite[(2.2.18) and (2.2.19)]{GriRei}) that% \begin{equation} \prod_{i=1}^{\infty}\left( 1-x_{i}t\right) ^{-1}=\sum_{n\geq0}h_{n}t^{n} \label{pf.lem.h-e-reciprocity-t.h}% \end{equation} and \[ \prod_{i=1}^{\infty}\left( 1+x_{i}t\right) =\sum_{n\geq0}e_{n}t^{n}. \] Applying the map $\mathbf{F}_{k}\left[ \left[ t\right] \right] $ to both sides of the latter equality, we obtain% \[ \left( \mathbf{F}_{k}\left[ \left[ t\right] \right] \right) \left( \prod_{i=1}^{\infty}\left( 1+x_{i}t\right) \right) =\left( \mathbf{F}% _{k}\left[ \left[ t\right] \right] \right) \left( \sum_{n\geq0}% e_{n}t^{n}\right) =\sum_{n\geq0}\mathbf{F}_{k}\left( e_{n}\right) t^{n}% \] (by the definition of $\mathbf{F}_{k}\left[ \left[ t\right] \right] $). Hence,% \begin{align*} \sum_{n\geq0}\mathbf{F}_{k}\left( e_{n}\right) t^{n} & =\left( \mathbf{F}_{k}\left[ \left[ t\right] \right] \right) \left( \prod _{i=1}^{\infty}\left( 1+x_{i}t\right) \right) =\prod_{i=1}^{\infty }\underbrace{\left( \mathbf{F}_{k}\left[ \left[ t\right] \right] \right) \left( 1+x_{i}t\right) }_{\substack{=1+\mathbf{F}_{k}\left( x_{i}\right) t\\\text{(by the definition of }\mathbf{F}_{k}\left[ \left[ t\right] \right] \text{)}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \begin{array} [c]{c}% \text{since }\mathbf{F}_{k}\left[ \left[ t\right] \right] \text{ is a continuous }\mathbf{k}\left[ \left[ t\right] \right] \text{-algebra homomorphism,}\\ \text{and thus respects infinite products}% \end{array} \right)\\ & =\prod_{i=1}^{\infty}\left( 1+\underbrace{\mathbf{F}_{k}\left( x_{i}\right) }_{\substack{=x_{i}^{k}\\\text{(by (\ref{pf.thm.G.frob.Fkxi}))}% }}t\right) =\prod_{i=1}^{\infty}\left( 1+x_{i}^{k}t\right) . \end{align*} Substituting $-t^{k}$ for $t$ in this equality, we find% \[ \sum_{n\geq0}\mathbf{F}_{k}\left( e_{n}\right) \left( -t^{k}\right) ^{n}=\prod_{i=1}^{\infty}\underbrace{\left( 1+x_{i}^{k}\left( -t^{k}\right) \right) }_{=1-\left( x_{i}t\right) ^{k}}=\prod_{i=1}^{\infty}\left( 1-\left( x_{i}t\right) ^{k}\right) . \] We can divide both sides of this equality by $\prod_{i=1}^{\infty}\left( 1-x_{i}t\right) $ (since the formal power series $\prod_{i=1}^{\infty}\left( 1-x_{i}t\right) $ has constant term $1$ and thus is invertible), and thus obtain% \begin{align*} & \dfrac{\sum_{n\geq0}\mathbf{F}_{k}\left( e_{n}\right) \left( -t^{k}\right) ^{n}}{\prod_{i=1}^{\infty}\left( 1-x_{i}t\right) }\\ & =\dfrac{\prod_{i=1}^{\infty}\left( 1-\left( x_{i}t\right) ^{k}\right) }{\prod_{i=1}^{\infty}\left( 1-x_{i}t\right) }=\prod_{i=1}^{\infty }\underbrace{\dfrac{1-\left( x_{i}t\right) ^{k}}{1-x_{i}t}}% _{\substack{=\left( x_{i}t\right) ^{0}+\left( x_{i}t\right) ^{1}% +\cdots+\left( x_{i}t\right) ^{k-1}\\=\sum_{u=0}^{k-1}\left( x_{i}t\right) ^{u}}}=\prod_{i=1}^{\infty}\ \ \sum_{u=0}^{k-1}\left( x_{i}t\right) ^{u}\\ & =\underbrace{\sum_{\substack{\alpha=\left( \alpha_{1},\alpha_{2}% ,\alpha_{3},\ldots\right) \in\left\{ 0,1,\ldots,k-1\right\} ^{\infty };\\\alpha_{i}=0\text{ for all but finitely many }i}}}_{=\sum _{\substack{\alpha\in\operatorname*{WC};\\\alpha_{i}0$. Thus, $m-ki$ is a positive integer. Also, $i$ is a positive integer (since $i\in\mathbb{N}$ and $i\neq0$), and thus $ki$ is a positive integer (since $k$ is a positive integer). The map $\mathbf{f}_{k}:\Lambda\rightarrow\Lambda$ operates by replacing each $x_{i}$ by $x_{i}^{k}$ in a symmetric function (by the definition of $\mathbf{f}_{k}$). Thus, if $g\in\Lambda$ is any homogeneous symmetric function of some degree $\gamma$, then $\mathbf{f}_{k}\left( g\right) $ is a homogeneous symmetric function of degree $k\gamma$. Therefore, $\mathbf{f}% _{k}\left( e_{i}\right) $ is a homogeneous symmetric function of degree $ki$ (since $e_{i}$ is a homogeneous symmetric function of degree $i$). Also, $h_{m-ki}$ is a homogeneous symmetric function of degree $m-ki$. Hence, Lemma \ref{lem.hall-pmprod} (applied to $\alpha=m-ki$, $a=h_{m-ki}$, $\beta=ki$ and $b=\mathbf{f}_{k}\left( e_{i}\right) $) yields $\left\langle p_{m},h_{m-ki}\cdot\mathbf{f}_{k}\left( e_{i}\right) \right\rangle =0$. This proves (\ref{pf.prop.hall-pmGkm.zeroes}).] Note that $e_{0}=1$ and thus $\mathbf{f}_{k}\left( e_{0}\right) =\mathbf{f}_{k}\left( 1\right) =1$ (by the definition of $\mathbf{f}_{k}$). Note that $m/k>0$ (since $m$ and $k$ are positive). Hence, $m/k\neq0$. Now, we are in one of the following two cases: \textit{Case 1:} We have $k\!\!\mid \!\!m$. \textit{Case 2:} We have $k\!\nmid\! m$. Let us consider Case 1 first. In this case, we have $k\!\!\mid\!\! m$. Hence, $m/k$ is a positive integer (since $m/k>0$). Thus, $0$ and $m/k$ are two distinct elements of $\mathbb{N}$. Now, consider the sum on the right hand side of (\ref{pf.prop.hall-pmGkm.1}). All addends of this sum are $0$, except for the one for $i=0$ and the one for $i=m/k$ (because (\ref{pf.prop.hall-pmGkm.zeroes}) shows that the Hall inner products $\left\langle p_{m},h_{m-ki}\cdot\mathbf{f}_{k}\left( e_{i}\right) \right\rangle $ that appear in these addends vanish whenever $i\notin\left\{ 0,m/k\right\} $). Thus, (\ref{pf.prop.hall-pmGkm.1}) simplifies to\footnote{We note that the addends for $i=0$ and for $i=m/k$ really do exist and are two different addends (since $0$ and $m/k$ are two distinct elements of $\mathbb{N}$).}% \begin{align*} & \left\langle p_{m},G\left( k,m\right) \right\rangle\\ & =\underbrace{\left( -1\right) ^{0}}_{=1}\left\langle p_{m}% ,\underbrace{h_{m-k\cdot0}}_{=h_{m}}\cdot\underbrace{\mathbf{f}_{k}\left( e_{0}\right) }_{=1}\right\rangle +\left( -1\right) ^{m/k}\left\langle p_{m},\underbrace{h_{m-k\cdot m/k}}_{=h_{m-m}=h_{0}=1}\cdot\mathbf{f}% _{k}\left( e_{m/k}\right) \right\rangle\\ & =\underbrace{\left\langle p_{m},h_{m}\right\rangle }% _{\substack{=\left\langle h_{m},p_{m}\right\rangle\\\text{(since the Hall inner}\\\text{product is symmetric)}}}+\left( -1\right) ^{m/k}% \underbrace{\left\langle p_{m},\mathbf{f}_{k}\left( e_{m/k}\right) \right\rangle }_{\substack{=\left( -1\right) ^{m/k-1}\left[ m=k\left( m/k\right) \right] k\\\text{(by Lemma \ref{lem.hall-hfep}, applied to }j=m/k\text{)}}}\\ & =\underbrace{\left\langle h_{m},p_{m}\right\rangle }% _{\substack{=1\\\text{(by Proposition \ref{prop.hjpj})}}}+\underbrace{\left( -1\right) ^{m/k}\left( -1\right) ^{m/k-1}}_{=-1}\underbrace{\left[ m=k\left( m/k\right) \right] }_{\substack{=1\\\text{(since }m=k\left( m/k\right) \text{)}}}k=1-k. \end{align*} Comparing this with% \[ 1-\underbrace{\left[ k\!\!\mid \!\! m\right] }_{\substack{=1\\\text{(since }k\mid m\!\text{)}}}k=1-k, \] we obtain $\left\langle p_{m},G\left( k,m\right) \right\rangle =1-\left[ k\!\!\mid \!\! m\right] k$. Hence, Proposition \ref{prop.hall-pmGkm} is proven in Case 1. Case 2 is similar to Case 1, but simpler because the addend for $i=m/k$ does not exist (since $m/k\notin\mathbb{N}$ in this case). We leave it to the reader. We have now proven Proposition \ref{prop.hall-pmGkm} in both possible cases. \end{proof} Theorem \ref{thm.Gkm-genset} will follow from Proposition \ref{prop.hall-pmGkm} using the following general criterion for generating sets of $\Lambda$: \begin{proposition} \label{prop.genset-crit}For each positive integer $m$, let $v_{m}\in\Lambda$ be a homogeneous symmetric function of degree $m$. Assume that $\left\langle p_{m},v_{m}\right\rangle $ is an invertible element of $\mathbf{k}$ for each positive integer $m$. Then, the family $\left( v_{m}\right) _{m\geq1}=\left( v_{1},v_{2}% ,v_{3},\ldots\right) $ is an algebraically independent generating set of the commutative $\mathbf{k}$-algebra $\Lambda$. \end{proposition} \begin{proof} [Proof of Proposition \ref{prop.genset-crit}.]Proposition \ref{prop.genset-crit} is \cite[Exercise 2.5.24]{GriRei}. \end{proof} \begin{proof} [Proof of Theorem \ref{thm.Gkm-genset}.]Let $m$ be a positive integer. Proposition \ref{prop.hall-pmGkm} yields that% \[ \left\langle p_{m},G\left( k,m\right) \right\rangle =1-\left[ k\!\!\mid\!\! m\right] k=% \begin{cases} 1-k, & \text{if }k\!\!\mid \!\! m;\\ 1, & \text{if }k\!\nmid \!m. \end{cases} \] Hence, $\left\langle p_{m},G\left( k,m\right) \right\rangle $ is an invertible element of $\mathbf{k}$ (because both $1-k$ and $1$ are invertible elements of $\mathbf{k}$). Forget that we fixed $m$. We thus have showed that $\left\langle p_{m},G\left( k,m\right) \right\rangle $ is an invertible element of $\mathbf{k}$ for each positive integer $m$. Also, clearly, for each positive integer $m$, the element $G\left( k,m\right) \in\Lambda$ is a homogeneous symmetric function of degree $m$. Thus, Proposition \ref{prop.genset-crit} (applied to $v_{m}=G\left( k,m\right) $) shows that the family $\left( G\left( k,m\right) \right) _{m\geq1}=\left( G\left( k,1\right) ,G\left( k,2\right) ,G\left( k,3\right) ,\ldots\right) $ is an algebraically independent generating set of the commutative $\mathbf{k}$-algebra $\Lambda$. This proves Theorem \ref{thm.Gkm-genset}. \end{proof} \subsection{Proof of Theorem \ref{thm.Uk.main}}\label{subsect.proofs.Uk.main} \begin{proof} [Proof of Theorem \ref{thm.Uk.main}.]The $\mathbf{k}$-Hopf algebra $\Lambda$ is both commutative and cocommutative (by \cite[Exercise 2.3.7(a)]{GriRei}). Its antipode $S$ is a $\mathbf{k}$-Hopf algebra homomorphism (by \cite[Proposition 2.4.3(g)]{GriRei}). Let $\Delta=\Delta_{\Lambda}:\Lambda\rightarrow\Lambda\otimes\Lambda$ be the comultiplication of the $\mathbf{k}$-coalgebra $\Lambda$. Let $m_{\Lambda }:\Lambda\otimes\Lambda\rightarrow\Lambda$ be the $\mathbf{k}$-linear map sending each pure tensor $a\otimes b\in\Lambda\otimes\Lambda$ to $ab\in \Lambda$. Definition \ref{def.convolution} then yields $\operatorname*{id}% \nolimits_{\Lambda}\star\, U_{k}=m_{\Lambda}\circ\left( \operatorname*{id}% \nolimits_{\Lambda}\otimes U_{k}\right) \circ\Delta$. Thus,% \begin{equation} V_{k}=\operatorname*{id}\nolimits_{\Lambda}\star\, U_{k}=m_{\Lambda}\circ\left( \operatorname*{id}\nolimits_{\Lambda}\otimes U_{k}\right) \circ\Delta. \label{pf.thm.Uk.main.c.conv-def}% \end{equation} \ref{226a} The map $\mathbf{f}_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism (by \cite[Exercise 2.9.9(d)]{GriRei}, applied to $n=k$). The map $\mathbf{v}_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism (by \cite[Exercise 2.9.10(e)]{GriRei}, applied to $n=k$). Thus, we have shown that all three maps $\mathbf{f}_{k}$, $S$ and $\mathbf{v}_{k}$ are $\mathbf{k}% $-Hopf algebra homomorphisms. Hence, their composition $\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism as well. In other words, $U_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism (since $U_{k}=\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}$). This proves Theorem \ref{thm.Uk.main} \ref{226a}. \ref{226b} Recall (from \cite[Exercise 1.5.11(a)]{GriRei}) the following fact: \begin{quote} \textit{Claim 1:} If $H$ is a $\mathbf{k}$-bialgebra and $A$ is a commutative $\mathbf{k}$-algebra, then the convolution $f\star g$ of any two $\mathbf{k}% $-algebra homomorphisms $f,g:H\rightarrow A$ is again a $\mathbf{k}$-algebra homomorphism. \end{quote} The following fact is dual to Claim 1: \begin{quote} \textit{Claim 2:} If $H$ is a $\mathbf{k}$-bialgebra and $C$ is a cocommutative $\mathbf{k}$-coalgebra, then the convolution $f\star g$ of any two $\mathbf{k}$-coalgebra homomorphisms $f,g:C\rightarrow H$ is again a $\mathbf{k}$-coalgebra homomorphism. \end{quote} (See \cite[solution to Exercise 1.5.11(h)]{GriRei} for why exactly Claim 2 is dual to Claim 1, and how it can be proved.) Theorem \ref{thm.Uk.main} \ref{226a} yields that the map $U_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism. Hence, $U_{k}$ is both a $\mathbf{k}% $-algebra homomorphism and a $\mathbf{k}$-coalgebra homomorphism. Now, recall that $\Lambda$ is commutative, and that $\operatorname*{id}% \nolimits_{\Lambda}$ and $U_{k}$ are two $\mathbf{k}$-algebra homomorphisms from $\Lambda$ to $\Lambda$. Hence, Claim 1 (applied to $H=\Lambda$, $A=\Lambda$, $f=\operatorname*{id}\nolimits_{\Lambda}$ and $g=U_{k}$) shows that the convolution $\operatorname*{id}\nolimits_{\Lambda}\star\, U_{k}$ is a $\mathbf{k}$-algebra homomorphism. In other words, $V_{k}$ is a $\mathbf{k}% $-algebra homomorphism (since $V_{k}=\operatorname*{id}\nolimits_{\Lambda }\star\, U_{k}$). Next, recall that $\Lambda$ is cocommutative, and that $\operatorname*{id}% \nolimits_{\Lambda}$ and $U_{k}$ are two $\mathbf{k}$-coalgebra homomorphisms from $\Lambda$ to $\Lambda$. Hence, Claim 2 (applied to $H=\Lambda$, $C=\Lambda$, $f=\operatorname*{id}\nolimits_{\Lambda}$ and $g=U_{k}$) shows that the convolution $\operatorname*{id}\nolimits_{\Lambda}\star\, U_{k}$ is a $\mathbf{k}$-coalgebra homomorphism. In other words, $V_{k}$ is a $\mathbf{k}% $-coalgebra homomorphism (since $V_{k}=\operatorname*{id}\nolimits_{\Lambda }\star\, U_{k}$). So we know that the map $V_{k}$ is both a $\mathbf{k}$-algebra homomorphism and a $\mathbf{k}$-coalgebra homomorphism. Thus, $V_{k}$ is a $\mathbf{k}% $-bialgebra homomorphism, therefore a $\mathbf{k}$-Hopf algebra homomorphism.\footnote{This follows since any $\mathbf{k}$-bialgebra homomorphism between two $\mathbf{k}$-Hopf algebras is automatically a $\mathbf{k}$-Hopf algebra homomorphism.} This proves Theorem \ref{thm.Uk.main} \ref{226b}. \ref{226c} The map $\mathbf{v}_{k}$ is a $\mathbf{k}$-algebra homomorphism; thus, $\mathbf{v}_{k}\left( 1\right) =1$. Now, we have% \begin{equation} \mathbf{v}_{k}\left( h_{m}\right) =% \begin{cases} h_{m/k}, & \text{if }k\!\!\mid\!\! m;\\ 0, & \text{if }k\!\nmid\! m \end{cases} \label{pf.thm.Uk.main.c.vh}% \end{equation} for each $m\in\mathbb{N}$. (Indeed, if $m>0$, then this follows from the definition of $\mathbf{v}_{k}$. But if $m=0$, then this follows from $\mathbf{v}_{k}\left( 1\right) =1$, since $h_{0}=1$.) We have% \begin{equation} S\left( h_{n}\right) =\left( -1\right) ^{n}e_{n}% \ \ \ \ \ \ \ \ \ \ \text{for each }n\in\mathbb{N}. \label{pf.thm.Uk.main.c.Sh}% \end{equation} (This follows from \cite[Proposition 2.4.1(iii)]{GriRei}.) Each $i\in\mathbb{N}$ satisfies% \begin{align} \mathbf{v}_{k}\left( h_{ki}\right) & =% \begin{cases} h_{ki/k}, & \text{if }k\!\!\mid\!\! ki;\\ 0, & \text{if }k\!\nmid\! ki \end{cases} \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.thm.Uk.main.c.vh}), applied to }m=ki\right) \nonumber\\ & =h_{ki/k}\ \ \ \ \ \ \ \ \ \ \left( \text{since }k\!\!\mid\!\! ki\right) \nonumber\\ & =h_{i} \label{pf.thm.Uk.main.c.vkhj1}% \end{align} and \begin{align} U_{k}\left( h_{ki}\right) & =\left( \mathbf{f}_{k}\circ S\circ \mathbf{v}_{k}\right) \left( h_{ki}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }U_{k}=\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}\right) \nonumber\\ & =\mathbf{f}_{k}\left( S\left( \underbrace{\mathbf{v}_{k}\left( h_{ki}\right) }_{\substack{=h_{i}\\\text{(by (\ref{pf.thm.Uk.main.c.vkhj1}% ))}}}\right) \right) =\mathbf{f}_{k}\left( \underbrace{S\left( h_{i}\right) }_{\substack{=\left( -1\right) ^{i}e_{i}\\\text{(by (\ref{pf.thm.Uk.main.c.Sh}))}}}\right) =\mathbf{f}_{k}\left( \left( -1\right) ^{i}e_{i}\right) \nonumber\\ & =\left( -1\right) ^{i}\mathbf{f}_{k}\left( e_{i}\right) . \label{pf.thm.Uk.main.c.Ukhj1}% \end{align} On the other hand, if $j\in\mathbb{N}$ satisfies $k\nmid j$, then% \begin{equation} \mathbf{v}_{k}\left( h_{j}\right) =0\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.thm.Uk.main.c.vh}), applied to }m=j\right) \label{pf.thm.Uk.main.c.vkhj2}% \end{equation} and \begin{align} U_{k}\left( h_{j}\right) & =\left( \mathbf{f}_{k}\circ S\circ \mathbf{v}_{k}\right) \left( h_{j}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }U_{k}=\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}\right) \nonumber\\ & =\left( \mathbf{f}_{k}\circ S\right) \left( \mathbf{v}_{k}\left( h_{j}\right) \right) =\left( \mathbf{f}_{k}\circ S\right) \left( 0\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.thm.Uk.main.c.vkhj2}% )}\right) \nonumber\\ & =0. \label{pf.thm.Uk.main.c.Ukhj2}% \end{align} Let $m\in\mathbb{N}$ (not to be mistaken for the map $m_{\Lambda}$). Then, \cite[Proposition 2.3.6(iii)]{GriRei} (applied to $n=m$) yields \begin{align*} \Delta\left( h_{m}\right) & =\sum_{i+j=m}h_{i}\otimes h_{j}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{where the sum ranges over all pairs }\left( i,j\right) \in\mathbb{N}\times\mathbb{N}\text{ with }i+j=m\right)\\ & =\sum_{j=0}^{m}h_{m-j}\otimes h_{j}% \end{align*} (here, we have substituted $\left( m-j,j\right) $ for $\left( i,j\right) $ in the sum). Applying the map $\operatorname*{id}\nolimits_{\Lambda}\otimes U_{k}$ to both sides of this equality, we obtain% \[ \left( \operatorname*{id}\nolimits_{\Lambda}\otimes U_{k}\right) \left( \Delta\left( h_{m}\right) \right) =\left( \operatorname*{id}% \nolimits_{\Lambda}\otimes U_{k}\right) \left( \sum_{j=0}^{m}h_{m-j}\otimes h_{j}\right) =\sum_{j=0}^{m}h_{m-j}\otimes U_{k}\left( h_{j}\right) . \] Applying the map $m_{\Lambda}$ to both sides of this equality, we find% \begin{align*} & m_{\Lambda}\left( \left( \operatorname*{id}\nolimits_{\Lambda}\otimes U_{k}\right) \left( \Delta\left( h_{m}\right) \right) \right)\\ & =m_{\Lambda}\left( \sum_{j=0}^{m}h_{m-j}\otimes U_{k}\left( h_{j}\right) \right) =\sum_{j=0}^{m}h_{m-j}U_{k}\left( h_{j}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }m_{\Lambda}\right)\\ & =\underbrace{\sum_{j=0}^{\infty}}_{=\sum_{j\in\mathbb{N}}}h_{m-j}% U_{k}\left( h_{j}\right) -\sum_{j=m+1}^{\infty}\underbrace{h_{m-j}% }_{\substack{=0\\\text{(since }m-j<0\text{)}}}U_{k}\left( h_{j}\right)\\ & =\sum_{j\in\mathbb{N}}h_{m-j}U_{k}\left( h_{j}\right) -\underbrace{\sum _{j=m+1}^{\infty}0U_{k}\left( h_{j}\right) }_{=0}\\ & =\sum_{j\in\mathbb{N}}h_{m-j}U_{k}\left( h_{j}\right) =\sum _{\substack{j\in\mathbb{N};\\k\mid j}}h_{m-j}U_{k}\left( h_{j}\right) +\sum_{\substack{j\in\mathbb{N};\\k\nmid j}}h_{m-j}\underbrace{U_{k}\left( h_{j}\right) }_{\substack{=0\\\text{(by (\ref{pf.thm.Uk.main.c.Ukhj2}))}}}\\ & =\sum_{\substack{j\in\mathbb{N};\\k\mid j}}h_{m-j}U_{k}\left( h_{j}\right) +\underbrace{\sum_{\substack{j\in\mathbb{N};\\k\nmid j}% }h_{m-j}0}_{=0}=\sum_{\substack{j\in\mathbb{N};\\k\mid j}}h_{m-j}U_{k}\left( h_{j}\right) =\sum_{i\in\mathbb{N}}h_{m-ki}\underbrace{U_{k}\left( h_{ki}\right) }_{\substack{=\left( -1\right) ^{i}\mathbf{f}_{k}\left( e_{i}\right) \\\text{(by (\ref{pf.thm.Uk.main.c.Ukhj1}))}}}\\ & \ \ \ \ \ \ \ \ \ \ \left( \text{here, we have substituted }ki\text{ for }j\text{ in the sum}\right)\\ & =\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m-ki}\cdot\mathbf{f}% _{k}\left( e_{i}\right) . \end{align*} Comparing this with% \[ G\left( k,m\right) =\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}% h_{m-ki}\cdot\mathbf{f}_{k}\left( e_{i}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by Theorem \ref{thm.G.frob}}\right) , \] we obtain% \[ G\left( k,m\right) =m_{\Lambda}\left( \left( \operatorname*{id}% \nolimits_{\Lambda}\otimes U_{k}\right) \left( \Delta\left( h_{m}\right) \right) \right) =\underbrace{\left( m_{\Lambda}\circ\left( \operatorname*{id}\nolimits_{\Lambda}\otimes U_{k}\right) \circ\Delta\right) }_{\substack{=V_{k}\\\text{(by (\ref{pf.thm.Uk.main.c.conv-def}))}}}\left( h_{m}\right) =V_{k}\left( h_{m}\right) . \] This proves Theorem \ref{thm.Uk.main} \ref{226c}. \ref{226d} From \cite[Exercise 2.9.10(a)]{GriRei}, we know that every positive integers $n$ and $m$ satisfy% \begin{equation} \mathbf{v}_{n}\left( p_{m}\right) =% \begin{cases} np_{m/n}, & \text{if }n\!\!\mid\!\! m;\\ 0, & \text{if }n\!\nmid\! m. \end{cases} \label{pf.thm.Uk.main.vp}% \end{equation} On the other hand, it is easy to see (directly using the definition of $\mathbf{f}_{n}$) that every positive integers $n$ and $m$ satisfy% \begin{equation} \mathbf{f}_{n}\left( p_{m}\right) =p_{nm}. \label{pf.thm.Uk.main.short.fp}% \end{equation} Finally, \cite[Proposition 2.4.1(i)]{GriRei} yields that every positive integer $n$ satisfies% \begin{equation} S\left( p_{n}\right) =-p_{n}. \label{pf.thm.Uk.main.Sp}% \end{equation} Now, let $n$ be a positive integer. We first claim the following: \begin{quote} \textit{Claim 1:} We have $U_{k}\left( p_{n}\right) =-\left[ k\!\!\mid\!\! n\right] kp_{n}$. \end{quote} [\textit{Proof of Claim 1:} We are in one of the following two cases: \textit{Case 1:} We have $k\!\!\mid\!\! n$. \textit{Case 2:} We have $k\!\nmid\! n$. Let us first consider Case 1. In this case, we have $k\!\!\mid\!\! n$. Hence, $n/k$ is a positive integer. Now, (\ref{pf.thm.Uk.main.vp}) (applied to $k$ and $n$ instead of $n$ and $m$) yields% \[ \mathbf{v}_{k}\left( p_{n}\right) =kp_{n/k}\ \ \ \ \ \ \ \ \ \ \left( \text{since }k\!\!\mid\!\! n\right) . \] Applying the map $S$ to both sides of this equality, we find% \[ S\left( \mathbf{v}_{k}\left( p_{n}\right) \right) =S\left( kp_{n/k}% \right) =k\underbrace{S\left( p_{n/k}\right) }_{\substack{=-p_{n/k}% \\\text{(by (\ref{pf.thm.Uk.main.Sp}),}\\\text{applied to }n/k\text{ instead of }n\text{)}}}=k\left( -p_{n/k}\right) =-kp_{n/k}. \] \noindent Applying the map $\mathbf{f}_{k}$ to both sides of this equality, we find \[ \mathbf{f}_{k}\left( S\left( \mathbf{v}_{k}\left( p_{n}\right) \right) \right) =\mathbf{f}_{k}\left( -kp_{n/k}\right) =-k\underbrace{\mathbf{f}% _{k}\left( p_{n/k}\right) }_{\substack{=p_{k\left( n/k\right) }\\\text{(by (\ref{pf.thm.Uk.main.short.fp}),}\\\text{applied to }k\text{ and }n/k\\\text{instead of }n\text{ and }m\text{)}}}=-kp_{k\left( n/k\right) }=-kp_{n}. \] \noindent Now, the definition of $U_{k}$ yields $U_{k}=\mathbf{f}_{k}\circ S\circ\mathbf{v}_{k}$. Hence,% \[ U_{k}\left( p_{n}\right) =\left( \mathbf{f}_{k}\circ S\circ\mathbf{v}% _{k}\right) \left( p_{n}\right) =\mathbf{f}_{k}\left( S\left( \mathbf{v}_{k}\left( p_{n}\right) \right) \right) =-kp_{n}. \] Comparing this with% \[ -\underbrace{\left[ k\!\!\mid\!\! n\right] }_{\substack{=1\\\text{(since }k\mid n\text{)}}}kp_{n}=-kp_{n}, \] we obtain $U_{k}\left( p_{n}\right) =-\left[ k\!\!\mid\!\! n\right] kp_{n}$. Hence, Claim 1 is proved in Case 1. Let us now consider Case 2. In this case, we have $k\nmid n$. Therefore, (\ref{pf.thm.Uk.main.vp}) (applied to $k$ and $n$ instead of $n$ and $m$) yields% \[ \mathbf{v}_{k}\left( p_{n}\right) =0. \] But the definition of $U_{k}$ yields $U_{k}=\mathbf{f}_{k}\circ S\circ \mathbf{v}_{k}$. Hence,% \[ U_{k}\left( p_{n}\right) =\left( \mathbf{f}_{k}\circ S\circ\mathbf{v}% _{k}\right) \left( p_{n}\right) =\left( \mathbf{f}_{k}\circ S\right) \left( \underbrace{\mathbf{v}_{k}\left( p_{n}\right) }_{=0}\right) =\left( \mathbf{f}_{k}\circ S\right) \left( 0\right) =0. \] Comparing this with% \[ -\underbrace{\left[ k\!\!\mid\!\! n\right] }_{\substack{=0\\\text{(since }k\nmid n\text{)}}}kp_{n}=0, \] we obtain $U_{k}\left( p_{n}\right) =-\left[ k\!\!\mid\!\! n\right] kp_{n}$. Hence, Claim 1 is proved in Case 2. We have now proved Claim 1 in both Cases 1 and 2. Thus, Claim 1 always holds.] Theorem \ref{thm.Uk.main} \ref{226a} shows that the map $U_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism. Hence, $U_{k}\left( 1\right) =1$. But \cite[Proposition 2.3.6(i)]{GriRei} yields $\Delta\left( p_{n}\right) =1\otimes p_{n}+p_{n}\otimes1$. Now,% \begin{align*} & \underbrace{V_{k}}_{\substack{=m_{\Lambda}\circ\left( \operatorname*{id}% \nolimits_{\Lambda}\otimes U_{k}\right) \circ\Delta\\\text{(by (\ref{pf.thm.Uk.main.c.conv-def}))}}}\left( p_{n}\right)\\ & =\left( m_{\Lambda}\circ\left( \operatorname*{id}\nolimits_{\Lambda }\otimes U_{k}\right) \circ\Delta\right) \left( p_{n}\right) =m_{\Lambda }\left( \left( \operatorname*{id}\nolimits_{\Lambda}\otimes U_{k}\right) \left( \underbrace{\Delta\left( p_{n}\right) }_{=1\otimes p_{n}% +p_{n}\otimes1}\right) \right)\\ & =m_{\Lambda}\left( \underbrace{\left( \operatorname*{id}% \nolimits_{\Lambda}\otimes U_{k}\right) \left( 1\otimes p_{n}+p_{n}% \otimes1\right) }_{=\operatorname*{id}\nolimits_{\Lambda}\left( 1\right) \otimes U_{k}\left( p_{n}\right) +\operatorname*{id}\nolimits_{\Lambda }\left( p_{n}\right) \otimes U_{k}\left( 1\right) }\right)\\ & =m_{\Lambda}\left( \operatorname*{id}\nolimits_{\Lambda}\left( 1\right) \otimes U_{k}\left( p_{n}\right) +\operatorname*{id}\nolimits_{\Lambda }\left( p_{n}\right) \otimes U_{k}\left( 1\right) \right)\\ & =\underbrace{\operatorname*{id}\nolimits_{\Lambda}\left( 1\right) }% _{=1}\cdot\underbrace{U_{k}\left( p_{n}\right) }_{\substack{=-\left[ k\mid n\right] kp_{n}\\\text{(by Claim 1)}}}+\underbrace{\operatorname*{id}% \nolimits_{\Lambda}\left( p_{n}\right) }_{=p_{n}}\cdot\underbrace{U_{k}% \left( 1\right) }_{=1}\ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }m_{\Lambda}\right)\\ & =-\left[ k\!\!\mid\!\! n\right] kp_{n}+p_{n}=\left( 1-\left[ k\!\!\mid \!\! n\right] k\right) p_{n}. \end{align*} This proves Theorem \ref{thm.Uk.main} \ref{226d}. \end{proof} \subsection{\label{subsect.proofs.p-via-G}Proof of Corollary \ref{cor.p-via-G}% } \begin{proof} [Proof of Corollary \ref{cor.p-via-G}.]Recall that the family $\left( h_{n}\right) _{n\geq1}=\left( h_{1},h_{2},h_{3},\ldots\right) $ generates $\Lambda$ as a $\mathbf{k}$-algebra. Hence, each $g\in\Lambda$ can be written as a polynomial in $h_{1},h_{2},h_{3},\ldots$. Applying this to $g=p_{n}$, we conclude that $p_{n}$ can be written as a polynomial in $h_{1},h_{2}% ,h_{3},\ldots$. In other words, there exists a polynomial $f\in\mathbf{k}% \left[ x_{1},x_{2},x_{3},\ldots\right] $ such that% \begin{equation} p_{n}=f\left( h_{1},h_{2},h_{3},\ldots\right) . \label{pf.cor.p-via-G.0}% \end{equation} Consider this $f$. We shall show that this $f$ satisfies (\ref{eq.cor.p-via-G.eq}). This will clearly prove Corollary \ref{cor.p-via-G}. Consider the map $V_{k}$ defined in Theorem \ref{thm.Uk.main}. Theorem \ref{thm.Uk.main} \ref{226c} yields that $V_{k}\left( h_{m}\right) =G\left( k,m\right) $ for each positive integer $m$. In other words,% \begin{equation} \left( V_{k}\left( h_{1}\right) ,V_{k}\left( h_{2}\right) ,V_{k}\left( h_{3}\right) ,\ldots\right) =\left( G\left( k,1\right) ,G\left( k,2\right) ,G\left( k,3\right) ,\ldots\right) . \label{pf.cor.p-via-G.1}% \end{equation} The map $V_{k}$ is a $\mathbf{k}$-Hopf algebra homomorphism (by Theorem \ref{thm.Uk.main} \ref{226b}, and thus is a $\mathbf{k}$-algebra homomorphism. Hence, it commutes with polynomials over $\mathbf{k}$. Thus,% \begin{align*} V_{k}\left( f\left( h_{1},h_{2},h_{3},\ldots\right) \right) & =f\left( V_{k}\left( h_{1}\right) ,V_{k}\left( h_{2}\right) ,V_{k}\left( h_{3}\right) ,\ldots\right)\\ & =f\left( G\left( k,1\right) ,G\left( k,2\right) ,G\left( k,3\right) ,\ldots\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.cor.p-via-G.1}% )}\right) . \end{align*} Now, applying the map $V_{k}$ to both sides of the equality (\ref{pf.cor.p-via-G.0}), we obtain% \[ V_{k}\left( p_{n}\right) =V_{k}\left( f\left( h_{1},h_{2},h_{3}% ,\ldots\right) \right) =f\left( G\left( k,1\right) ,G\left( k,2\right) ,G\left( k,3\right) ,\ldots\right) . \] Comparing this with% \[ V_{k}\left( p_{n}\right) =\left( 1-\left[ k\!\!\mid\!\! n\right] k\right) p_{n}\ \ \ \ \ \ \ \ \ \ \left( \text{by Theorem \ref{thm.Uk.main} \ref{226d}}\right) , \] we obtain% \[ \left( 1-\left[ k\!\!\mid\!\! n\right] k\right) p_{n}=f\left( G\left( k,1\right) ,G\left( k,2\right) ,G\left( k,3\right) ,\ldots\right) . \] Thus, we have shown that our $f$ satisfies (\ref{eq.cor.p-via-G.eq}). As we said, this proves Corollary \ref{cor.p-via-G}. \end{proof} \section{\label{sect.liu}Proof of the Liu--Polo conjecture} Let us recall a well-known partial order on the set of partitions of a given $n\in\mathbb{N}$: \begin{definition} \label{def.dominance}Let $n\in\mathbb{N}$. We define a binary relation $\triangleright$ on the set $\operatorname*{Par}\nolimits_{n}$ as follows: Two partitions $\lambda,\mu\in\operatorname*{Par}\nolimits_{n}$ shall satisfy $\lambda\triangleright\mu$ if and only if we have% \[ \lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}\geq\mu_{1}+\mu_{2}+\cdots+\mu _{k}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{ 1,2,\ldots,n\right\} . \] This relation $\triangleright$ is the greater-or-equal relation of a partial order on $\operatorname*{Par}\nolimits_{n}$, which is known as the \emph{dominance order} (or the \emph{majorization order}). \end{definition} This definition is precisely \cite[Definition 2.2.7]{GriRei}. Note that if we replace \textquotedblleft for each $k\in\left\{ 1,2,\ldots,n\right\} $\textquotedblright\ by \textquotedblleft for each $k\in\left\{ 1,2,3,\ldots\right\} $\textquotedblright\ in this definition, then the relation $\triangleright$ does not change. Our goal in this section is to prove the conjecture made in \cite[Remark 1.4.5]{LiuPol19}. We state this conjecture as follows:\footnote{Note that $\left( n-1,n-1,1\right) $ is a partition whenever $n>1$ is an integer.} \begin{theorem} \label{thm.claim}Let $n$ be an integer such that $n>1$. Then: \begin{enumerate}[label=\textbf{(\alph*)}] \item\label{42a} We have% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{n};\\\left( n-1,1\right) \triangleright\lambda}}m_{\lambda}=\sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1-i,1^{i+1}\right) }. \] \item\label{42b} We have% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\left( n-1,n-1,1\right) \triangleright\lambda}}m_{\lambda}=\sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1,n-1-i,1^{i+1}\right) }. \] \end{enumerate} \end{theorem} \begin{example} For this example, let $n=3$. Then, $n-1=2$ and $2n-1=5$. Hence, the left hand side of the equality in Theorem \ref{thm.claim} \ref{42b} is% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\left( n-1,n-1,1\right) \triangleright\lambda}}m_{\lambda}=\sum_{\substack{\lambda \in\operatorname*{Par}\nolimits_{5};\\\left( 2,2,1\right) \triangleright \lambda}}m_{\lambda}=m_{\left( 2,2,1\right) }+m_{\left( 2,1,1,1\right) }+m_{\left( 1,1,1,1,1\right) }. \] Meanwhile, the right hand side of the equality in Theorem \ref{thm.claim} \ref{42b} is% \[ \sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1,n-1-i,1^{i+1}\right) }=\sum_{i=0}^{1}\left( -1\right) ^{i}s_{\left( 2,2-i,1^{i+1}\right) }=s_{\left( 2,2,1\right) }-s_{\left( 2,1,1,1\right) }. \] Thus, Theorem \ref{thm.claim} \ref{42b} claims that $m_{\left( 2,2,1\right) }+m_{\left( 2,1,1,1\right) }+m_{\left( 1,1,1,1,1\right) }=s_{\left( 2,2,1\right) }-s_{\left( 2,1,1,1\right) }$ in this case. \end{example} We will pave our way to the proof of Theorem \ref{thm.claim} by several lemmas. We begin with a particularly simple one: \begin{lemma} \label{lem.dominnn-1}Let $n$ be an integer such that $n>1$. Let $\lambda \in\operatorname*{Par}\nolimits_{2n-1}$. Then, $\left( n-1,n-1,1\right) \triangleright\lambda$ if and only if all positive integers $i$ satisfy $\lambda_{i}1$. Let $\lambda \in\operatorname*{Par}\nolimits_{n}$. Then, $\left( n-1,1\right) \triangleright\lambda$ if and only if all positive integers $i$ satisfy $\lambda_{i}1$. Then: \begin{enumerate}[label=\textbf{(\alph*)}] \item \label{it:4.6a} We have% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{n};\\\left( n-1,1\right) \triangleright\lambda}}m_{\lambda}=G\left( n,n\right) . \] \item \label{it:4.6b} We have% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\left( n-1,n-1,1\right) \triangleright\lambda}}m_{\lambda}=G\left( n,2n-1\right) . \] \end{enumerate} \end{corollary} \begin{proof} \ref{it:4.6b} Proposition \ref{prop.G.basics} \ref{it:2.3c} (applied to $k=n$ and $m=2n-1$) yields \begin{equation} G\left( n,2n-1\right) =\sum_{\substack{\alpha\in\operatorname*{WC}% ;\\\left\vert \alpha\right\vert =2n-1;\\\alpha_{i}n+k$). \end{quote} Let $\mathfrak{M}_{k}$ be the set of all monomials of degree $k$. The definition of $h_{k}$ shows that \begin{equation} h_{k}=\sum_{\mathfrak{m}\in\mathfrak{M}_{k}}\mathfrak{m}. \label{pf.prop.h2n-1.hk=}% \end{equation} Let $\mathfrak{M}_{n+k}$ be the set of all monomials of degree $n+k$. The definition of $h_{n+k}$ shows that% \begin{equation} h_{n+k}=\sum_{\mathfrak{n}\in\mathfrak{M}_{n+k}}\mathfrak{n}. \label{pf.prop.h2n-1.hn+k=}% \end{equation} Let $\mathfrak{N}$ be the set of all monomials of degree $n+k$ in which all exponents are $1$. Then: \begin{enumerate}[label=\textbf{(\alph*)}] \item \label{it:4.8a} We have% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{n};\\\left( n-1,1\right) \triangleright\lambda}}m_{\lambda}=h_{n}-p_{n}. \] \item \label{it:4.8b} We have% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\left( n-1,n-1,1\right) \triangleright\lambda}}m_{\lambda}=h_{2n-1}-h_{n-1}p_{n}. \] \end{enumerate} \end{corollary} \begin{proof} \ref{it:4.8b} Corollary \ref{cor.LHS1} \ref{it:4.6b} yields% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\left( n-1,n-1,1\right) \triangleright\lambda}}m_{\lambda}=G\left( n,2n-1\right) =h_{2n-1}-h_{n-1}p_{n}% \] (by Proposition \ref{prop.h2n-1}, applied to $k=n-1$). This proves Corollary \ref{cor.LHS2} \ref{it:4.8b}. \ref{it:4.8a} Corollary \ref{cor.LHS1} \ref{it:4.6a} yields% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{n};\\\left( n-1,1\right) \triangleright\lambda}}m_{\lambda}=G\left( n,n\right) =h_{n}-h_{0}p_{n}% \] (by Proposition \ref{prop.h2n-1}, applied to $k=0$). Since $h_{0}=1$, this proves Corollary \ref{cor.LHS2} \ref{it:4.8a}. \end{proof} Our next claim is an easy consequence of Proposition \ref{prop.p-as-sum}: \begin{corollary} \label{cor.p-as-sum-2}Let $n$ be a positive integer. Then, \[ h_{n}-p_{n}=\sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1-i,1^{i+1}% \right) }. \] \end{corollary} \begin{proof} Proposition \ref{prop.p-as-sum} yields% \begin{align*} p_{n} & =\sum_{i=0}^{n-1}\left( -1\right) ^{i}s_{\left( n-i,1^{i}\right) }=\underbrace{\left( -1\right) ^{0}}_{=1}\underbrace{s_{\left( n-0,1^{0}\right) }}_{=s_{\left( n-0\right) }=s_{\left( n\right) }=h_{n}% }+\sum_{i=1}^{n-1}\left( -1\right) ^{i}s_{\left( n-i,1^{i}\right) }\\ & =h_{n}+\sum_{i=1}^{n-1}\left( -1\right) ^{i}s_{\left( n-i,1^{i}\right) }, \end{align*} so that% \begin{align*} h_{n}-p_{n} & =-\sum_{i=1}^{n-1}\left( -1\right) ^{i}s_{\left( n-i,1^{i}\right) }=\sum_{i=1}^{n-1}\underbrace{\left( -\left( -1\right) ^{i}\right) }_{=\left( -1\right) ^{i-1}}s_{\left( n-i,1^{i}\right) }% =\sum_{i=1}^{n-1}\left( -1\right) ^{i-1}s_{\left( n-i,1^{i}\right) }\\ & =\sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1-i,1^{i+1}\right) }% \end{align*} (here, we have substituted $i+1$ for $i$ in the sum). \end{proof} We can now immediately prove Theorem \ref{thm.claim} \textbf{(a)}: \begin{proof} [Proof of Theorem \ref{thm.claim} \ref{42a}.]Corollary \ref{cor.LHS2} \ref{it:4.8a} yields% \[ \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{n};\\\left( n-1,1\right) \triangleright\lambda}}m_{\lambda}=h_{n}-p_{n}=\sum_{i=0}% ^{n-2}\left( -1\right) ^{i}s_{\left( n-1-i,1^{i+1}\right) }% \ \ \ \ \ \ \ \ \ \ \left( \text{by Corollary \ref{cor.p-as-sum-2}}\right) . \] This proves Theorem \ref{thm.claim} \ref{42a}. \end{proof} We shall use the \emph{skewing operators} $f^{\perp}:\Lambda\rightarrow \Lambda$ for all $f\in\Lambda$ as defined in \cite[\S 2.8]{GriRei} or in \cite[Chapter I,\ Section 5, Example 3]{Macdon95}. The easiest way to define them (following \cite[Chapter I,\ Section 5, Example 3]{Macdon95}) is as follows: For each $f\in\Lambda$, we let $f^{\perp}:\Lambda\rightarrow\Lambda$ be the $\mathbf{k}$-linear map adjoint to the map $L_{f}:\Lambda \rightarrow\Lambda,\ g\mapsto fg$ (that is, to the map that multiplies every element of $\Lambda$ by $f$) with respect to the Hall inner product. That is, $f^{\perp}$ is the $\mathbf{k}$-linear map from $\Lambda$ to $\Lambda$ that satisfies \[ \left\langle g,f^{\perp}\left( a\right) \right\rangle =\left\langle fg,a\right\rangle \ \ \ \ \ \ \ \ \ \ \text{for all }a\in\Lambda\text{ and }g\in\Lambda. \] It is not hard to show that such an operator $f^{\perp}$ exists.\footnote{This is not completely automatic: Not every $\mathbf{k}$-linear map from $\Lambda$ to $\Lambda$ has an adjoint with respect to the Hall inner product! (For example, the $\mathbf{k}$-linear map $\Lambda\rightarrow\Lambda$ that sends each Schur function $s_{\lambda}$ to $1$ has none.) The reason why the map $L_{f}:\Lambda\rightarrow\Lambda,\ g\mapsto fg$ has an adjoint is that when $f$ is homogeneous of degree $k$, this map $L_{f}$ sends each graded component $\Lambda_{m}$ of $\Lambda$ to $\Lambda_{m+k}$, and both of these graded components $\Lambda_{m}$ and $\Lambda_{m+k}$ are $\mathbf{k}$-modules with \textbf{finite} bases. (The case when $f$ is not homogeneous can be reduced to the case when $f$ is homogeneous, since each $f\in\Lambda$ is a sum of finitely many homogeneous elements.)} The definition of $f^{\perp}$ in \cite[\S 2.8]{GriRei} is different but equivalent (because of \cite[Proposition 2.8.2(i)]{GriRei}). One of the most important properties of skewing operators is the following fact (\cite[(2.8.2)]{GriRei}): \begin{lemma} \label{lem.skewing.schur}Let $\lambda$ and $\mu$ be any two partitions. Then,% \begin{equation} s_{\mu}^{\perp}\left( s_{\lambda}\right) =s_{\lambda/\mu}. \label{pf.lem.Bmhn.skew}% \end{equation} \textup{(}Here, $s_{\lambda/\mu}$ is a skew Schur function, defined in Subsection \ref{subsect.proofs.skew}.\textup{)} \end{lemma} Using skewing operators, we can define another helpful family of operators on $\Lambda$: \begin{definition} For any $m\in\mathbb{Z}$, we define a map $\mathbf{B}_{m}:\Lambda \rightarrow\Lambda$ by setting% \[ \mathbf{B}_{m}\left( f\right) =\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}e_{i}^{\perp}f\ \ \ \ \ \ \ \ \ \ \text{for all }f\in \Lambda\text{.}% \] It is known (\cite[Exercise 2.9.1(a)]{GriRei}) that this map $\mathbf{B}_{m}$ is well-defined and $\mathbf{k}$-linear. \end{definition} (Actually, the well-definedness of $\mathbf{B}_{m}$ is easy to check: If $f\in\Lambda$ has degree $d$, then all integers $i>d$ satisfy $e_{i}^{\perp }f=0$ for degree reasons, and thus the sum $\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}e_{i}^{\perp}f$ has only finitely many nonzero addends. The $\mathbf{k}$-linearity of $\mathbf{B}_{m}$ is even clearer.) The operators $\mathbf{B}_{m}$ for $m\in\mathbb{Z}$ have first appeared in Zelevinsky's \cite[\S 4.20]{Zelevi81} (in the different-looking but secretly equivalent setting of a PSH-algebra), where they are credited to J. N. Bernstein. They have since been dubbed the \emph{Bernstein creation operators} and proved useful in various contexts (\eg the definition of the \textquotedblleft dual immaculate functions\textquotedblright\ in \cite{BBSSZ13} takes them for inspiration). One of their most fundamental properties is the following fact (which originates in \cite[4.20, ($\ast\ast $)]{Zelevi81} and appears implicitly in \cite[Chapter I, Section 5, Example 29]{Macdon95}): \begin{proposition} \label{prop.Bmslam}Let $\lambda$ be any partition. Let $m\in\mathbb{Z}$ satisfy $m\geq\lambda_{1}$. Then,% \begin{equation} \sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}e_{i}^{\perp}s_{\lambda }=s_{\left( m,\lambda_{1},\lambda_{2},\lambda_{3},\ldots\right) }. \label{eq.schur-row-adder-1}% \end{equation} \end{proposition} See \cite[Exercise 2.9.1(b)]{GriRei} for a proof of Proposition \ref{prop.Bmslam}. Thus, if $\lambda$ is any partition, and if $m\in \mathbb{Z}$ satisfies $m\geq\lambda_{1}$, then% \begin{align} \mathbf{B}_{m}\left( s_{\lambda}\right) & =\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}e_{i}^{\perp}s_{\lambda}\ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\mathbf{B}_{m}\right) \nonumber\\ & =s_{\left( m,\lambda_{1},\lambda_{2},\lambda_{3},\ldots\right) }\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{eq.schur-row-adder-1})}\right) . \label{eq.schur-row-adder-2}% \end{align} \begin{lemma} \label{lem.Bmhn}Let $n$ be a positive integer. Let $m\in\mathbb{N}$. Then, $\mathbf{B}_{m}\left( h_{n}\right) =h_{m}h_{n}-h_{m+1}h_{n-1}$. \end{lemma} \begin{proof} [Proof of Lemma \ref{lem.Bmhn}.]We have $e_{0}=1$ and thus $e_{0}^{\perp }=1^{\perp}=\operatorname*{id}$. Hence, $e_{0}^{\perp}\left( h_{n}\right) =h_{n}$. We shall use the notion of skew Schur functions $s_{\lambda/\mu}$ (as in Subsection \ref{subsect.proofs.skew}). Recall that $s_{\lambda/\mu}=0$ when $\mu\not \subseteq \lambda$. From $e_{1}=s_{\left( 1\right) }$ and $h_{n}=s_{\left( n\right) }$, we obtain% \[ e_{1}^{\perp}\left( h_{n}\right) =s_{\left( 1\right) }^{\perp}\left( s_{\left( n\right) }\right) =s_{\left( n\right) /\left( 1\right) }\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.lem.Bmhn.skew})}\right) . \] But it is easy to see that $s_{\left( n\right) /\left( 1\right) }=s_{\left( n-1\right) }$. (Indeed, this follows from the combinatorial definition of skew Schur functions, since the skew Ferrers diagram of $\left( n\right) /\left( 1\right) $ can be obtained from the Ferrers diagram of $\left( n-1\right) $ by parallel shift\footnote{See \cite[\S 2.3]{GriRei} for the notions we are using here.}. Alternatively, this follows easily from Theorem \ref{thm.JTs.h}, because $s_{\left( n-1\right) }=h_{n-1}$.) Thus, we obtain% \[ e_{1}^{\perp}\left( h_{n}\right) =s_{\left( n\right) /\left( 1\right) }=s_{\left( n-1\right) }=h_{n-1}. \] For each integer $i>1$, we have% \begin{align} e_{i}^{\perp}\left( h_{n}\right) & =s_{\left( 1^{i}\right) }^{\perp }\left( s_{\left( n\right) }\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }e_{i}=s_{\left( 1^{i}\right) }\text{ and }h_{n}=s_{\left( n\right) }\right) \nonumber\\ & =s_{\left( n\right) /\left( 1^{i}\right) }\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.lem.Bmhn.skew})}\right) \nonumber\\ & =0\ \ \ \ \ \ \ \ \ \ \left( \text{since }\left( 1^{i}\right) \not \subseteq \left( n\right) \text{ (because }i>1\text{)}\right) . \label{pf.lem.Bmhn.3}% \end{align} Now, the definition of $\mathbf{B}_{m}$ yields% \begin{align*} \mathbf{B}_{m}\left( h_{n}\right) & =\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}e_{i}^{\perp}\left( h_{n}\right)\\ & =\underbrace{\left( -1\right) ^{0}}_{=1}\underbrace{h_{m+0}}_{=h_{m}% }\underbrace{e_{0}^{\perp}\left( h_{n}\right) }_{=h_{n}}+\underbrace{\left( -1\right) ^{1}}_{=-1}h_{m+1}\underbrace{e_{1}^{\perp}\left( h_{n}\right) }_{=h_{n-1}}+\sum_{i\geq2}\left( -1\right) ^{i}h_{m+i}\underbrace{e_{i}% ^{\perp}\left( h_{n}\right) }_{\substack{=0\\\text{(by (\ref{pf.lem.Bmhn.3}% ))}}}\\ & =h_{m}h_{n}-h_{m+1}h_{n-1}. \qedhere \end{align*} \end{proof} \begin{corollary} \label{cor.Bn-1Hn}Let $n$ be a positive integer. Then, $\mathbf{B}% _{n-1}\left( h_{n}\right) =0$. \end{corollary} \begin{proof} Apply Lemma \ref{lem.Bmhn} to $m=n-1$ and simplify. \end{proof} \begin{lemma} \label{lem.Bmpn}Let $m\in\mathbb{N}$. Let $n$ be a positive integer. Then, $\mathbf{B}_{m}\left( p_{n}\right) =h_{m}p_{n}-h_{m+n}$. \end{lemma} \begin{proof} This is \cite[Exercise 2.9.1(f)]{GriRei}. But here is a more direct proof: We will use the comultiplication $\Delta:\Lambda\rightarrow\Lambda\otimes\Lambda$ of the Hopf algebra $\Lambda$ (see \cite[\S 2.3]{GriRei}). Here and in the following, the \textquotedblleft$\otimes$\textquotedblright\ sign denotes $\otimes_{\mathbf{k}}$. The power-sum symmetric function $p_{n}$ is primitive\footnote{Recall that an element $x$ of a Hopf algebra $H$ is said to be \textit{primitive} if the comultiplication $\Delta_{H}$ of $H$ satisfies $\Delta_{H}\left( x\right) =1\otimes x+x\otimes1$.} (see \cite[Proposition 2.3.6(i)]{GriRei}); thus, \[ \Delta\left( p_{n}\right) =1\otimes p_{n}+p_{n}\otimes1. \] Hence, for each $i\in\mathbb{N}$, the definition of $e_{i}^{\perp}$ given in \cite[Definition 2.8.1]{GriRei} (not the equivalent definition we gave above) yields% \begin{equation} e_{i}^{\perp}\left( p_{n}\right) =\left\langle e_{i},1\right\rangle p_{n}+\left\langle e_{i},p_{n}\right\rangle 1. \label{pf.lem.Bmpn.eipn}% \end{equation} Now, the definition of $\mathbf{B}_{m}$ yields% \begin{align*} \mathbf{B}_{m}\left( p_{n}\right) & =\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}\underbrace{e_{i}^{\perp}\left( p_{n}\right) }_{\substack{=\left\langle e_{i},1\right\rangle p_{n}+\left\langle e_{i}% ,p_{n}\right\rangle 1\\\text{(by (\ref{pf.lem.Bmpn.eipn}))}}}\\ & =\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}\left( \left\langle e_{i},1\right\rangle p_{n}+\left\langle e_{i},p_{n}\right\rangle 1\right)\\ & =\underbrace{\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}% \cdot\left\langle e_{i},1\right\rangle p_{n}}_{\substack{=\left( -1\right) ^{0}h_{m+0}\cdot\left\langle e_{0},1\right\rangle p_{n}\\\text{(because the Hall inner product }\left\langle e_{i},1\right\rangle\\\text{equals }0\text{ whenever }i\neq0\text{ (by (\ref{eq.Hall.graded})),}\\\text{and thus the only nonzero addend of this}\\\text{sum is the addend for }i=0\text{)}% }}+\underbrace{\sum_{i\in\mathbb{N}}\left( -1\right) ^{i}h_{m+i}% \cdot\left\langle e_{i},p_{n}\right\rangle 1}_{\substack{=\left( -1\right) ^{n}h_{m+n}\cdot\left\langle e_{n},p_{n}\right\rangle 1\\\text{(because the Hall inner product }\left\langle e_{i},p_{n}\right\rangle\\\text{equals }0\text{ whenever }i\neq n\text{ (by (\ref{eq.Hall.graded})),}\\\text{and thus the only nonzero addend of this}\\\text{sum is the addend for }i=n\text{)}}}\\ & =\underbrace{\left( -1\right) ^{0}}_{=1}\underbrace{h_{m+0}}_{=h_{m}% }\cdot\underbrace{\left\langle e_{0},1\right\rangle }_{=\left\langle 1,1\right\rangle =1}p_{n}+\left( -1\right) ^{n}h_{m+n}\cdot \underbrace{\left\langle e_{n},p_{n}\right\rangle }_{\substack{=\left( -1\right) ^{n-1}\\\text{(by Proposition \ref{prop.ejpj})}}}1\\ & =h_{m}p_{n}+\underbrace{\left( -1\right) ^{n}h_{m+n}\cdot\left( -1\right) ^{n-1}1}_{=-h_{m+n}}=h_{m}p_{n}-h_{m+n}. \qedhere \end{align*} \end{proof} \begin{lemma} \label{lem.Bm1}Let $n$ be a positive integer. Then,% \[ \mathbf{B}_{n-1}\left( h_{n}-p_{n}\right) =h_{2n-1}-h_{n-1}p_{n}. \] \end{lemma} \begin{proof} The map $\mathbf{B}_{n-1}$ is $\mathbf{k}$-linear. Thus,% \begin{align*} \mathbf{B}_{n-1}\left( h_{n}-p_{n}\right) & =\underbrace{\mathbf{B}% _{n-1}\left( h_{n}\right) }_{\substack{=0\\\text{(by Corollary \ref{cor.Bn-1Hn})}}}-\underbrace{\mathbf{B}_{n-1}\left( p_{n}\right) }_{\substack{=h_{n-1}p_{n}-h_{\left( n-1\right) +n}\\\text{(by Lemma \ref{lem.Bmpn},}\\\text{applied to }m=n-1\text{)}}}\\ & =-\left( h_{n-1}p_{n}-h_{\left( n-1\right) +n}\right) =\underbrace{h_{\left( n-1\right) +n}}_{=h_{2n-1}}-h_{n-1}p_{n}% =h_{2n-1}-h_{n-1}p_{n}. \end{align*} \end{proof} \begin{lemma} \label{lem.Bm2}Let $n$ be a positive integer. Then,% \[ \mathbf{B}_{n-1}\left( h_{n}-p_{n}\right) =\sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1,n-1-i,1^{i+1}\right) }. \] \end{lemma} \begin{proof} [Proof of Lemma \ref{lem.Bm2}.]We have \begin{align*} \mathbf{B}_{n-1}\left( h_{n}-p_{n}\right) & =\mathbf{B}_{n-1}\left( \sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1-i,1^{i+1}\right) }\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by Corollary \ref{cor.p-as-sum-2}% }\right)\\ & =\sum_{i=0}^{n-2}\left( -1\right) ^{i}\underbrace{\mathbf{B}_{n-1}\left( s_{\left( n-1-i,1^{i+1}\right) }\right) }_{\substack{=s_{\left( n-1,n-1-i,1^{i+1}\right) }\\\text{(by (\ref{eq.schur-row-adder-2}), applied to }m=n-1\\\text{and }\lambda=\left( n-1-i,1^{i+1}\right) \\\text{(since }n-1\geq n-1-i\text{))}}}\ \ \ \ \ \ \ \ \ \ \left( \text{since }% \mathbf{B}_{n-1}\text{ is }\mathbf{k}\text{-linear}\right)\\ & =\sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1,n-1-i,1^{i+1}% \right) }. \qedhere \end{align*} \end{proof} Now the proof of Theorem \ref{thm.claim} \ref{42b} is a trivial concatenation of equalities: \begin{proof} [Proof of Theorem \ref{thm.claim} \ref{42b}.]Corollary \ref{cor.LHS2} \ref{it:4.8b} yields% \begin{align*} \sum_{\substack{\lambda\in\operatorname*{Par}\nolimits_{2n-1};\\\left( n-1,n-1,1\right) \triangleright\lambda}}m_{\lambda} & =h_{2n-1}% -h_{n-1}p_{n}=\mathbf{B}_{n-1}\left( h_{n}-p_{n}\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by Lemma \ref{lem.Bm1}}\right)\\ & =\sum_{i=0}^{n-2}\left( -1\right) ^{i}s_{\left( n-1,n-1-i,1^{i+1}% \right) }\ \ \ \ \ \ \ \ \ \ \left( \text{by Lemma \ref{lem.Bm2}}\right) .\qedhere \end{align*} \end{proof} \section{\label{sect.fin}Final remarks} \subsection{SageMath code} The SageMath computer algebra system \cite{SageMath} does not (yet) natively know the Petrie symmetric functions $G\left( k,m\right) $; but they can be easily constructed in it. For example, the code that follows computes $G\left( k, n\right) $ expanded in the Schur basis: \begin{python} Sym = SymmetricFunctions(QQ) # Replace QQ by your favorite base ring. m = Sym.m() # monomial symmetric functions s = Sym.s() # Schur functions def G(k, n): # a Petrie function return s(m.sum(m[lam] for lam in Partitions(n, max_part=k-1))) \end{python} \subsection{Understanding the Petrie numbers} Combining Corollary \ref{cor.Gkm.main} with Theorem \ref{thm.petk.explicit} yields an explicit expression of all coefficients in the expansion of a Petrie symmetric function $G\left( k,m\right) $ in the Schur basis. It would stand to reason if the identity in Theorem \ref{thm.claim} \ref{42b} (whose left hand side is $G\left( n,2n-1\right) $) could be obtained from this expression. Surprisingly, we have been unable to do so, which suggests that the description of $\operatorname*{pet}\nolimits_{k}\left( \lambda ,\varnothing\right) $ in Theorem \ref{thm.petk.explicit} might not be optimal. As to $\operatorname*{pet}\nolimits_{k}\left( \lambda,\mu\right) $, we do not have an explicit description at all, unless we count the recursive one that can be extracted from the proof in \cite{GorWil74}. \subsection{MNable symmetric functions} Combining Theorem \ref{thm.G.pieri} with Proposition \ref{prop.petkrel.-101}, we conclude that for any $k>0$ and $m\in\mathbb{N}$, the symmetric function $G\left( k,m\right) \in\Lambda$ has the following property: For any $\mu \in\operatorname*{Par}$, its product $G\left( k,m\right) \cdot s_{\mu}$ with $s_{\mu}$ can be written in the form $\sum_{\lambda\in\operatorname*{Par}% }u_{\lambda}s_{\lambda}$ with $u_{\lambda}\in\left\{ -1,0,1\right\} $ for all $\lambda\in\operatorname*{Par}$. It has this property in common with the symmetric functions $h_{m}$ and $e_{m}$ (according to the Pieri rules) and $p_{m}$ (according to the Murnaghan--Nakayama rule) as well as several others. The study of symmetric functions having this property -- which we call \textit{MNable symmetric functions} (in honor of Murnaghan and Nakayama) -- has been initiated in \cite[\S 8]{Grinbe20}, but there is much to be done. \subsection{A conjecture of Per Alexandersson} In February 2020, Per Alexandersson suggested the following conjecture: \begin{conj} \label{conj.Gkmp2}Let $k$ be a positive integer, and $m\in\mathbb{N}$. Then, $G\left( k,m\right) \cdot p_{2}\in\Lambda$ can be written in the form $\sum_{\lambda\in\operatorname*{Par}}u_{\lambda}s_{\lambda}$ with $u_{\lambda }\in\left\{ -1,0,1\right\} $ for all $\lambda\in\operatorname*{Par}$. \end{conj} For example,% \[ G\left( 3,4\right) \cdot p_{2}=s_{\left( 1,1,1,1,1,1\right) }+s_{\left( 2,2,2\right) }-s_{\left( 3,1,1,1\right) }-s_{\left( 3,3\right) }+s_{\left( 4,2\right) }. \] Conjecture \ref{conj.Gkmp2} has been verified for all $k$ and $m$ satisfying $k+m\leq30$. Note that Conjecture \ref{conj.Gkmp2} becomes false if $p_{2}$ is replaced by $p_{3}$. For example,% \[ G\left( 3,4\right) \cdot p_{3}=-s_{\left( 1,1,1,1,1,1,1\right) }+s_{\left( 2,2,1,1,1\right) }-2s_{\left( 2,2,2,1\right) }+s_{\left( 3,2,1,1\right) }-s_{\left( 4,1,1,1\right) }-s_{\left( 4,3\right) }+s_{\left( 5,2\right) }. \] \subsection{A conjecture of Fran\c{c}ois Bergeron} An even more mysterious conjecture was suggested by Fran\c{c}ois Bergeron in April 2020: \begin{conj} \label{conj.nabla-fb}Let $k$ and $n$ be positive integers, and $m\in \mathbb{N}$. Let $\nabla$ be the \emph{nabla operator} as defined (\eg) in \cite[\S 3.2.1]{Berger19}. Then, there exists a sign $\sigma_{n,k,m}% \in\left\{ 1,-1\right\} $ such that $\sigma_{n,k,m}\nabla^{n}\left( G\left( k,m\right) \right) $ is an $\mathbb{N}\left[ q,t\right] $-linear combination of Schur functions. \end{conj} Using SageMath, this conjecture has been checked for $n=1$ and all $k,m\in\left\{ 0,1,\ldots,9\right\} $; the signs $\sigma_{1,k,m}$ are given by the following table:% \[% \begin{tabular} [c]{||c||ccccccccc}\hline\hline & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$\\\hline\hline $2$ & $+$ & $+$ & $+$ & $+$ & $+$ & $+$ & $+$ & $+$ & $+$\\ $3$ & $+$ & $-$ & $-$ & $-$ & $+$ & $+$ & $+$ & $-$ & $-$\\ $4$ & $+$ & $-$ & $+$ & $+$ & $+$ & $-$ & $+$ & $+$ & $+$\\ $5$ & $+$ & $-$ & $+$ & $-$ & $-$ & $-$ & $+$ & $-$ & $+$\\ $6$ & $+$ & $-$ & $+$ & $-$ & $+$ & $+$ & $+$ & $-$ & $+$\\ $7$ & $+$ & $-$ & $+$ & $-$ & $+$ & $-$ & $-$ & $-$ & $+$\\ $8$ & $+$ & $-$ & $+$ & $-$ & $+$ & $-$ & $+$ & $+$ & $+$\\ $9$ & $+$ & $-$ & $+$ & $-$ & $+$ & $-$ & $+$ & $-$ & $-$% \end{tabular} \] (where the entry in the row indexed $k$ and the column indexed $m$ is the sign $\sigma_{1,k,m}$, represented by a \textquotedblleft$+$\textquotedblright% \ sign if it is $1$ and by a \textquotedblleft$-$\textquotedblright\ sign if it is $-1$). I am not aware of a pattern in these signs, apart from the fact that $\sigma_{1,2,m}=1$ for all $m\in\mathbb{N}$ (a consequence of Haiman's famous interpretation of $\nabla\left( e_{m}\right) $ as a character), and that $\sigma_{1,k,m}$ appears to be $\left( -1\right) ^{m-1}$ whenever $1\leq m1$. A weak composition $\alpha$ will be called $k$\emph{-friendly} if each $i\in\left\{ 1,2,3,\ldots\right\} $ satisfies $\alpha_{i}\equiv0\operatorname{mod}k$ or $\alpha_{i}\equiv1\operatorname{mod}k$. If $\alpha$ is a weak composition, then $w\left( \alpha\right) $ shall denote the number of all $i\in\left\{ 1,2,3,\ldots\right\} $ satisfying $\alpha_{i}\equiv1\operatorname{mod}k$. Let $S$ be the antipode of the Hopf algebra $\Lambda$. Then, for each $n\in\mathbb{N}$, we have% \[ S\left( G\left( k,n\right) \right) =\sum_{\substack{\alpha\in \operatorname*{WC};\\\left\vert \alpha\right\vert =n;\\\alpha\text{ is }k\text{-friendly}}}\left( -1\right) ^{w\left( \alpha\right) }% \mathbf{x}^{\alpha}=\sum_{\substack{\lambda\in\operatorname*{Par};\\\left\vert \lambda\right\vert =n;\\\lambda\text{ is }k\text{-friendly}}}\left( -1\right) ^{w\left( \lambda\right) }m_{\lambda}. \] \end{corollary} \begin{proof} [Proof of Corollary \ref{cor.Gkm.antipode}.]Let $F=1+t+t^{2}+\cdots+t^{k-1}% \in\mathbf{k}\left[ \left[ t\right] \right] $. Then, $F$ is a power series whose constant term is $1$. Hence, its reciprocal $F^{-1}$ is well-defined and again has constant term $1$. Let us denote this reciprocal $F^{-1}$ by $Q$; thus, $Q=F^{-1}$. Let $q_{0},q_{1},q_{2},\ldots$ be the coefficients of the formal power series $Q$, so that $Q=\sum_{n\in\mathbb{N}}q_{n}t^{n}$. Thus, $q_{0}=1$ (since the constant term of $Q$ is $1$). On the other hand,% \begin{align*} Q & =F^{-1}=\left( \dfrac{1-t^{k}}{1-t}\right) ^{-1}% \ \ \ \ \ \ \ \ \ \ \left( \text{since }F=1+t+t^{2}+\cdots+t^{k-1}% =\dfrac{1-t^{k}}{1-t}\right)\\ & =\dfrac{1-t}{1-t^{k}}=\left( 1-t\right) \cdot\underbrace{\left( 1-t^{k}\right) ^{-1}}_{=t^{0}+t^{k}+t^{2k}+t^{3k}+\cdots}=\left( 1-t\right) \cdot\left( t^{0}+t^{k}+t^{2k}+t^{3k}+\cdots\right)\\ & =t^{0}-t^{1}+t^{k}-t^{k+1}+t^{2k}-t^{2k+1}\pm\cdots\\ & =\sum_{n\in\mathbb{N}}\left( \left[ n\equiv0\operatorname{mod}k\right] -\left[ n\equiv1\operatorname{mod}k\right] \right) t^{n}. \end{align*} Comparing coefficients on both sides of this equality, we find% \begin{equation} q_{n}=\left[ n\equiv0\operatorname{mod}k\right] -\left[ n\equiv 1\operatorname{mod}k\right] \ \ \ \ \ \ \ \ \ \ \text{for each }% n\in\mathbb{N}. \label{pf.cor.Gkm.antipode.qn=}% \end{equation} For any weak composition $\alpha$, we define an element $q_{\alpha}% \in\mathbf{k}$ by% \[ q_{\alpha}=q_{\alpha_{1}}q_{\alpha_{2}}q_{\alpha_{3}}\cdots. \] (This infinite product is well-defined, since every sufficiently high positive integer $i$ satisfies $\alpha_{i}=0$ and thus $q_{\alpha_{i}}=q_{0}=1$.) It is now easy to see (using (\ref{pf.cor.Gkm.antipode.qn=})) that% \begin{equation} q_{\alpha}=\left[ \alpha\text{ is }k\text{-friendly}\right] \cdot\left( -1\right) ^{w\left( \alpha\right) } \label{pf.cor.Gkm.antipode.qal=}% \end{equation} for any weak composition $\alpha$. Now, let $n\in\mathbb{N}$. Recall that our scalars $q_{i}$ and $q_{\alpha}$ were defined in the exact same way as the scalars $f_{i}$ and $f_{\alpha}$ were defined in Definition \ref{def.GF.GF}, but using the power series $Q$ instead of $F$. Hence, Proposition \ref{prop.GF.basics} \textbf{(c)} (applied to $Q$, $q_{i}$, $q_{\alpha}$ and $n$ instead of $F$, $f_{i}$, $f_{\alpha}$ and $m$) yields that% \[ G_{Q,n}=\sum_{\substack{\alpha\in\operatorname*{WC};\\\left\vert \alpha\right\vert =n}}q_{\alpha}\mathbf{x}^{\alpha}=\sum_{\substack{\lambda \in\operatorname*{Par};\\\left\vert \lambda\right\vert =n}}q_{\lambda }m_{\lambda}\in\Lambda. \] Hence,% \begin{align*} G_{Q,n} & =\sum_{\substack{\alpha\in\operatorname*{WC};\\\left\vert \alpha\right\vert =n}}\ \ \underbrace{q_{\alpha}}_{\substack{=\left[ \alpha\text{ is }k\text{-friendly}\right] \cdot\left( -1\right) ^{w\left( \alpha\right) }\\\text{(by (\ref{pf.cor.Gkm.antipode.qal=}))}}}\mathbf{x}% ^{\alpha}=\sum_{\substack{\alpha\in\operatorname*{WC};\\\left\vert \alpha\right\vert =n}}\left[ \alpha\text{ is }k\text{-friendly}\right] \cdot\left( -1\right) ^{w\left( \alpha\right) }\mathbf{x}^{\alpha}\\ & =\sum_{\substack{\alpha\in\operatorname*{WC};\\\left\vert \alpha\right\vert =n;\\\alpha\text{ is }k\text{-friendly}}}\left( -1\right) ^{w\left( \alpha\right) }\mathbf{x}^{\alpha}% \end{align*} and% \begin{align*} G_{Q,n} & =\sum_{\substack{\lambda\in\operatorname*{Par};\\\left\vert \lambda\right\vert =n}}\ \ \underbrace{q_{\lambda}}_{\substack{=\left[ \lambda\text{ is }k\text{-friendly}\right] \cdot\left( -1\right) ^{w\left( \lambda\right) }\\\text{(by (\ref{pf.cor.Gkm.antipode.qal=}),}\\\text{applied to }\lambda\text{ instead of }\alpha\text{)}}}m_{\lambda}=\sum _{\substack{\lambda\in\operatorname*{Par};\\\left\vert \lambda\right\vert =n}}\left[ \lambda\text{ is }k\text{-friendly}\right] \cdot\left( -1\right) ^{w\left( \lambda\right) }m_{\lambda}\\ & =\sum_{\substack{\lambda\in\operatorname*{Par};\\\left\vert \lambda \right\vert =n;\\\lambda\text{ is }k\text{-friendly}}}\left( -1\right) ^{w\left( \lambda\right) }m_{\lambda}. \end{align*} Combining these two equalities, we find% \begin{equation} G_{Q,n}=\sum_{\substack{\alpha\in\operatorname*{WC};\\\left\vert \alpha\right\vert =n;\\\alpha\text{ is }k\text{-friendly}}}\left( -1\right) ^{w\left( \alpha\right) }\mathbf{x}^{\alpha}=\sum_{\substack{\lambda \in\operatorname*{Par};\\\left\vert \lambda\right\vert =n;\\\lambda\text{ is }k\text{-friendly}}}\left( -1\right) ^{w\left( \lambda\right) }m_{\lambda }. \label{pf.cor.Gkm.antipode.GQn=2}% \end{equation} This will immediately yield the claim of Corollary \ref{cor.Gkm.antipode} if we can show that $S\left( G\left( k,n\right) \right) =G_{Q,n}$. But this is easy: In Example \ref{exa.GF.GF123} \textbf{(c)}, we have seen that $G_{F,m}=G\left( k,m\right) $ for each $m\in\mathbb{N}$. Applying this to $m=n$, we obtain $G_{F,n}=G\left( k,n\right) $. On the other hand, Theorem \ref{thm.GF.antipode} yields $S\left( G_{F,n}\right) =G_{F^{-1},n}=G_{Q,n}$ (since $F^{-1}=Q$). In view of $G_{F,n}=G\left( k,n\right) $, this rewrites as $S\left( G\left( k,n\right) \right) =G_{Q,n}$. This completes our proof of Corollary \ref{cor.Gkm.antipode}. \end{proof} One last property shall be noted in passing: \begin{proposition} \label{prop.VF.VAB-etc}For any power series $F\in\mathbf{k}\left[ \left[ t\right] \right] $ whose constant term is $1$, we define a $\mathbf{k}% $-algebra homomorphism $V_{F}:\Lambda\rightarrow\Lambda$ as in Theorem \ref{thm.VF.main}. Then: \begin{enumerate} \item[\textbf{(a)}] If $A$ and $B$ are two power series in $\mathbf{k}\left[ \left[ t\right] \right] $ whose constant terms are $1$, then $V_{AB}% =V_{A}\star V_{B}$. \item[\textbf{(b)}] We have $V_{1}=\eta\circ\varepsilon$. \item[\textbf{(c)}] For any power series $F\in\mathbf{k}\left[ \left[ t\right] \right] $ whose constant term is $1$, we have $V_{F^{-1}}% =V_{F}\circ S$, where $S$ is the antipode of $\Lambda$. \end{enumerate} \end{proposition} This follows easily from Proposition \ref{prop.GF.GAB1}; we leave the details to the reader. %\nocite{*} \bibliographystyle{mersenne-plain} \bibliography{ALCO_Grinberg_447} \end{document}