0$) and thus $i_{p}\in\operatorname*{Supp}\mathfrak{m}%
\subseteq\operatorname*{Supp}\underbrace{\left( \mathfrak{mn}\right)
}_{=\mathfrak{q}}=\operatorname*{Supp}\mathfrak{q}=\left\{ k_{1},k_{2}%
,\ldots,k_{n}\right\} $.}
\item For every $q\in\left\{ 1,2,\ldots,m\right\} $, we let $f\left(
1,q\right) $ be the unique $r\in\left\{ 1,2,\ldots,n\right\} $ such that
$j_{q}=k_{r}$.\ \ \ \ \footnote{This is again well-defined, for similar
reasons as the $r$ in the definition of $f\left( 0,p\right) $.}
\end{itemize}
It is now straightforward to show that $f$ is a $\gamma$%
-smap.\footnote{Indeed:
\par
\begin{itemize}
\item The first defining property of a $\gamma$-smap holds. (\textit{Proof:}
Let us show that $f\circ\operatorname*{inc}\nolimits_{0}$ is strictly
increasing (the proof for $f\circ\operatorname*{inc}\nolimits_{1}$ is
similar). Assume it is not. Then there exist some $p,p^{\prime}\in\left\{
1,2,\ldots,\ell\right\} $ satisfying $pk
\end{cases}
\label{pf.thm.beldend.monom}%
\end{equation}
\footnote{\textit{Proof.} Let $\mathfrak{m}$ be a monomial. Then,%
\begin{align*}
\mathfrak{B}_{k}\left( \mathfrak{m}\right) & =\mathfrak{m}\left(
x_{1},x_{2},\ldots,x_{k},0,0,0,\ldots\right) \ \ \ \ \ \ \ \ \ \ \left(
\text{by the definition of }\mathfrak{B}_{k}\right) \\
& =\left( \text{the result of replacing the indeterminates }x_{k+1}%
,x_{k+2},x_{k+3},\ldots\text{ by }0\text{ in }\mathfrak{m}\right) \\
& =
\begin{cases}
\mathfrak{m}, & \text{if none of the indeterminates }x_{k+1},x_{k+2}%
,x_{k+3},\ldots\text{ appears in }\mathfrak{m}\text{;}\\
0, & \text{if some of the indeterminates }x_{k+1},x_{k+2},x_{k+3},\ldots\text{
appear in }\mathfrak{m}%
\end{cases}
\\
& =
\begin{cases}
\mathfrak{m}, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right)
\leq k;\\
0, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right) >k
\end{cases}
\end{align*}
(because none of the indeterminates $x_{k+1},x_{k+2},x_{k+3},\ldots$ appears
in $\mathfrak{m}$ if and only if $\max\left( \operatorname*{Supp}%
\mathfrak{m}\right) \leq k$). This proves (\ref{pf.thm.beldend.monom}).}. Any
$p\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $
satisfies%
\begin{equation}
p \bel a=a\cdot\mathfrak{B}_{k}\left( p\right) \label{pf.thm.beldend.B}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.beldend.B}):} Fix $p\in\mathbf{k}%
\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $. Since the
equality (\ref{pf.thm.beldend.B}) is $\mathbf{k}$-linear and continuous in
$p$, we can WLOG assume that $p$ is a monomial. Assume this. Hence,
$p=\mathfrak{m}$ for some monomial $\mathfrak{m}$. Consider this
$\mathfrak{m}$. We have%
\begin{equation}
\mathfrak{B}_{k}\left( \underbrace{p}_{=\mathfrak{m}}\right) =\mathfrak{B}%
_{k}\left( \mathfrak{m}\right) =
\begin{cases}
\mathfrak{m}, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right)
\leq k;\\
0, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right) >k
\end{cases}
\label{pf.thm.beldend.B.pf.2}%
\end{equation}
(by (\ref{pf.thm.beldend.monom})). Now,%
\begin{align*}
\underbrace{p}_{=\mathfrak{m}} \bel \underbrace{a}_{=\mathfrak{n}} &
=\mathfrak{m} \bel \mathfrak{n}=
\begin{cases}
\mathfrak{m}\cdot\mathfrak{n}, & \text{if }\max\left( \operatorname*{Supp}%
\mathfrak{m}\right) \leq\min\left( \operatorname*{Supp}\mathfrak{n}\right)
;\\
0, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right)
>\min\left( \operatorname*{Supp}\mathfrak{n}\right)
\end{cases}
\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of } \bel \right) \\
& =
\begin{cases}
\mathfrak{m}\cdot\mathfrak{n}, & \text{if }\max\left( \operatorname*{Supp}%
\mathfrak{m}\right) \leq k;\\
0, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right) >k
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left( \text{since }\min\left( \operatorname*{Supp}%
\mathfrak{n}\right) =k\right) \\
& =\underbrace{\mathfrak{n}}_{=a}\cdot\underbrace{
\begin{cases}
\mathfrak{m}, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right)
\leq k;\\
0, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right) >k
\end{cases}
}_{\substack{=\mathfrak{B}_{k}\left( p\right) \\\text{(by
(\ref{pf.thm.beldend.B.pf.2}))}}}\\
& =a\cdot\mathfrak{B}_{k}\left( p\right) .
\end{align*}
This proves (\ref{pf.thm.beldend.B}).}. Also, every composition $\alpha
=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) $ satisfies%
\begin{equation}
\mathfrak{B}_{k}\left( M_{\alpha}\right) =\sum_{1\leq i_{1}k
\end{cases}
\\\text{(by (\ref{pf.thm.beldend.monom}), applied to }\mathfrak{m}=x_{i_{1}%
}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}%
}\text{)}}}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since }\mathfrak{B}_{k}\text{ is
}\mathbf{k}\text{-linear and continuous}\right) \\
& =\sum_{1\leq i_{1}k
\end{cases}
}_{\substack{=
\begin{cases}
x_{i_{1}}^{\alpha_{1}}x_{i_{2}}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell
}}, & \text{if }\max\left\{ i_{1},i_{2},\ldots,i_{\ell}\right\} \leq k;\\
0, & \text{if }\max\left\{ i_{1},i_{2},\ldots,i_{\ell}\right\} >k
\end{cases}
\\\text{(since }\operatorname*{Supp}\left( x_{i_{1}}^{\alpha_{1}}x_{i_{2}%
}^{\alpha_{2}}\cdots x_{i_{\ell}}^{\alpha_{\ell}}\right) =\left\{
i_{1},i_{2},\ldots,i_{\ell}\right\} \text{)}}}\\
& =\sum_{1\leq i_{1}k
\end{cases}
\\
& =\underbrace{\sum_{\substack{1\leq i_{1}k
\end{cases}
\quad. \label{pf.thm.beldend.short.monom}%
\end{equation}
Using this (and the definition of $\bel $), we see that any $p\in
\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $
satisfies%
\begin{equation}
p \bel a=a\cdot\mathfrak{B}_{k}\left( p\right)
\label{pf.thm.beldend.short.B}%
\end{equation}
(indeed, this is trivial to check for $p$ being a monomial, and thus follows
by linearity for all $p$). Also, every composition $\alpha=\left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) $ satisfies%
\begin{equation}
\mathfrak{B}_{k}\left( M_{\alpha}\right) =\sum_{1\leq i_{1} 0$ and $q > 0$. Then,
$\left[ \alpha,\beta\right] $ is a composition of $p+q$ satisfying $D\left(
\left[ \alpha,\beta\right] \right) =D\left( \alpha\right) \cup\left\{
p\right\} \cup\left( D\left( \beta\right) +p\right) $.
\end{lemma}
(Actually, part \textbf{(b)} of this lemma will not be used until much later,
but part \textbf{(a)} will be used soon.)
\begin{proof}
[Proof of Lemma \ref{lem.D(a(.)b)}.]Write $\alpha$ in the form $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) $. Thus, $\left\vert
\alpha\right\vert =\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}$, so that
$\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}=\left\vert \alpha\right\vert =p$
(since $\alpha$ is a composition of $p$).
Write $\beta$ in the form $\beta=\left( \beta_{1},\beta_{2},\ldots,\beta
_{m}\right) $. Thus, $\left\vert \beta\right\vert =\beta_{1}+\beta_{2}%
+\cdots+\beta_{m}$, so that $\beta_{1}+\beta_{2}+\cdots+\beta_{m}=\left\vert
\beta\right\vert =q$ (since $\beta$ is a composition of $q$).
We have $\beta=\left( \beta_{1},\beta_{2},\ldots,\beta_{m}\right) $, and
thus%
\begin{align*}
D\left( \beta\right) & =\left\{ \beta_{1},\beta_{1}+\beta_{2},\beta
_{1}+\beta_{2}+\beta_{3},\ldots,\beta_{1}+\beta_{2}+\cdots+\beta_{m-1}\right\}
\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }D\left(
\beta\right) \right) \\
& =\left\{ \beta_{1}+\beta_{2}+\cdots+\beta_{j}\ \mid\ j\in\left\{
1,2,\ldots,m-1\right\} \right\} .
\end{align*}
Also, $\alpha=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) $,
and thus%
\begin{align*}
D\left( \alpha\right) & =\left\{ \alpha_{1},\alpha_{1}+\alpha_{2}%
,\alpha_{1}+\alpha_{2}+\alpha_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots
+\alpha_{\ell-1}\right\} \\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }D\left(
\alpha\right) \right) \\
& =\left\{ \alpha_{1}+\alpha_{2}+\cdots+\alpha_{i}\ \mid\ i\in\left\{
1,2,\ldots,\ell-1\right\} \right\} .
\end{align*}
\textbf{(a)} If $\alpha$ or $\beta$ is empty, then Lemma \ref{lem.D(a(.)b)}
\textbf{(a)} holds for obvious reasons (because of the definition of
$\alpha\odot\beta$ in this case). Thus, we WLOG assume that neither $\alpha$
nor $\beta$ is empty.
We have $\alpha\odot\beta=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1},\alpha_{\ell}+\beta_{1},\beta_{2},\beta_{3},\ldots,\beta_{m}\right) $ (by
the definition of $\alpha\odot\beta$) and thus%
\begin{align*}
\left\vert \alpha\odot\beta\right\vert & =\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell-1}+\left( \alpha_{\ell}+\beta_{1}\right) +\beta
_{2}+\beta_{3}+\cdots+\beta_{m}\\
& =\underbrace{\left( \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}\right)
}_{=p}+\underbrace{\left( \beta_{1}+\beta_{2}+\cdots+\beta_{m}\right) }%
_{=q}=p+q.
\end{align*}
Thus, $\alpha\odot\beta$ is a composition of $p+q$. Hence, it remains to show
that $D\left( \alpha\odot\beta\right) =D\left( \alpha\right) \cup\left(
D\left( \beta\right) +p\right) $.
Now, $\alpha\odot\beta=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1},\alpha_{\ell}+\beta_{1},\beta_{2},\beta_{3},\ldots,\beta_{m}\right) $, so
that%
\begin{align*}
& D\left( \alpha\odot\beta\right) \\
& =\left\{ \alpha_{1},\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha_{2}+\alpha
_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1},\right. \\
& \ \ \ \ \ \ \ \ \ \ \left. \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1}+\left( \alpha_{\ell}+\beta_{1}\right) ,\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell-1}+\left( \alpha_{\ell}+\beta_{1}\right) +\beta
_{2},\right. \\
& \ \ \ \ \ \ \ \ \ \ \left. \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1}+\left( \alpha_{\ell}+\beta_{1}\right) +\beta_{2}+\beta_{3}%
,\ldots,\right. \\
& \ \ \ \ \ \ \ \ \ \ \left. \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1}+\left( \alpha_{\ell}+\beta_{1}\right) +\beta_{2}+\beta_{3}+\cdots
+\beta_{m-1}\right\} \\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }D\left(
\alpha\odot\beta\right) \right) \\
& =\underbrace{\left\{ \alpha_{1}+\alpha_{2}+\cdots+\alpha_{i}\ \mid
\ i\in\left\{ 1,2,\ldots,\ell-1\right\} \right\} }_{=D\left(
\alpha\right) }\\
& \ \ \ \ \ \ \ \ \ \ \cup\left\{ \underbrace{\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell-1}+\left( \alpha_{\ell}+\beta_{1}\right) +\beta
_{2}+\beta_{3}+\cdots+\beta_{j}}_{\substack{=\left( \alpha_{1}+\alpha
_{2}+\cdots+\alpha_{\ell}\right) +\left( \beta_{1}+\beta_{2}+\cdots
+\beta_{j}\right) \\=\left( \beta_{1}+\beta_{2}+\cdots+\beta_{j}\right)
+\left( \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}\right) }}\ \mid
\ j\in\left\{ 1,2,\ldots,m-1\right\} \right\} \\
& =D\left( \alpha\right) \cup\left\{ \left( \beta_{1}+\beta_{2}%
+\cdots+\beta_{j}\right) +\underbrace{\left( \alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell}\right) }_{=p}\ \mid\ j\in\left\{ 1,2,\ldots
,m-1\right\} \right\} \\
& =D\left( \alpha\right) \cup\underbrace{\left\{ \left( \beta_{1}%
+\beta_{2}+\cdots+\beta_{j}\right) +p\ \mid\ j\in\left\{ 1,2,\ldots
,m-1\right\} \right\} }_{=\left\{ \beta_{1}+\beta_{2}+\cdots+\beta
_{j}\ \mid\ j\in\left\{ 1,2,\ldots,m-1\right\} \right\} +p}\\
& =D\left( \alpha\right) \cup\left( \underbrace{\left\{ \beta_{1}%
+\beta_{2}+\cdots+\beta_{j}\ \mid\ j\in\left\{ 1,2,\ldots,m-1\right\}
\right\} }_{=D\left( \beta\right) }+p\right) \\
& =D\left( \alpha\right) \cup\left( D\left( \beta\right) +p\right) .
\end{align*}
This completes the proof of Lemma \ref{lem.D(a(.)b)} \textbf{(a)}.
\textbf{(b)} We have $p > 0$. Thus, the composition $\alpha$ is nonempty
(since $\alpha$ is a composition of $p$). In other words, the composition
$\left( \alpha_{1}, \alpha_{2}, \ldots, \alpha_{\ell}\right) $ is nonempty
(since $\alpha= \left( \alpha_{1}, \alpha_{2}, \ldots, \alpha_{\ell}\right)
$). Hence, $\ell> 0$.
We have $q > 0$. Thus, the composition $\beta$ is nonempty (since $\beta$ is a
composition of $q$). In other words, the composition $\left( \beta_{1},
\beta_{2}, \ldots, \beta_{m}\right) $ is nonempty (since $\beta= \left(
\beta_{1}, \beta_{2}, \ldots, \beta_{m}\right) $). Hence, $m > 0$.
We have $\left[ \alpha,\beta\right] =\left( \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell},\beta_{1},\beta_{2},\ldots,\beta_{m}\right) $ (by the
definition of $\left[ \alpha,\beta\right] $) and thus%
\begin{align*}
\left\vert \alpha\odot\beta\right\vert & =\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell}+\beta_{1}+\beta_{2}+\cdots+\beta_{m}\\
& =\underbrace{\left( \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}\right)
}_{=p}+\underbrace{\left( \beta_{1}+\beta_{2}+\cdots+\beta_{m}\right) }%
_{=q}=p+q.
\end{align*}
Thus, $\left[ \alpha,\beta\right] $ is a composition of $p+q$. Hence, it
remains to show that $D\left( \left[ \alpha,\beta\right] \right) =D\left(
\alpha\right) \cup\left( D\left( \beta\right) +p\right) $.
Now, $\left[ \alpha,\beta\right] =\left( \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell},\beta_{1},\beta_{2},\ldots,\beta_{m}\right) $, so that%
\begin{align*}
& D\left( \left[ \alpha,\beta\right] \right) \\
& =\left\{ \alpha_{1},\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha_{2}+\alpha
_{3},\ldots,\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1},\right. \\
& \ \ \ \ \ \ \ \ \ \ \left. \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1}+\alpha_{\ell},\alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell-1}+\alpha_{\ell
}+\beta_{1},\right. \\
& \ \ \ \ \ \ \ \ \ \ \left. \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1}+\alpha_{\ell}+\beta_{1}+\beta_{2},\ldots,\right. \\
& \ \ \ \ \ \ \ \ \ \ \left. \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell
-1}+\alpha_{\ell}+\beta_{1}+\beta_{2}+\cdots+\beta_{m-1}\right\} \\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }D\left( \left[
\alpha,\beta\right] \right) \right) \\
& =\underbrace{\left\{ \alpha_{1}+\alpha_{2}+\cdots+\alpha_{i}\ \mid
\ i\in\left\{ 1,2,\ldots,\ell-1\right\} \right\} }_{=D\left(
\alpha\right) }\\
& \ \ \ \ \ \ \ \ \ \ \cup\left\{ \underbrace{\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell}}_{=p}\right\} \\
& \ \ \ \ \ \ \ \ \ \ \cup\left\{ \underbrace{\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell-1}+\alpha_{\ell}+\beta_{1}+\beta_{2}+\cdots+\beta_{j}%
}_{\substack{=\left( \alpha_{1}+\alpha_{2}+\cdots+\alpha_{\ell}\right)
+\left( \beta_{1}+\beta_{2}+\cdots+\beta_{j}\right) \\=\left( \beta
_{1}+\beta_{2}+\cdots+\beta_{j}\right) +\left( \alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{\ell}\right) }}\ \mid\ j\in\left\{ 1,2,\ldots,m-1\right\}
\right\} \\
& =D\left( \alpha\right) \cup\left\{ p\right\} \cup\left\{ \left(
\beta_{1}+\beta_{2}+\cdots+\beta_{j}\right) +\underbrace{\left( \alpha
_{1}+\alpha_{2}+\cdots+\alpha_{\ell}\right) }_{=p}\ \mid\ j\in\left\{
1,2,\ldots,m-1\right\} \right\} \\
& =D\left( \alpha\right) \cup\left\{ p\right\} \cup\underbrace{\left\{
\left( \beta_{1}+\beta_{2}+\cdots+\beta_{j}\right) +p\ \mid\ j\in\left\{
1,2,\ldots,m-1\right\} \right\} }_{=\left\{ \beta_{1}+\beta_{2}%
+\cdots+\beta_{j}\ \mid\ j\in\left\{ 1,2,\ldots,m-1\right\} \right\} +p}\\
& =D\left( \alpha\right) \cup\left\{ p\right\} \cup\left(
\underbrace{\left\{ \beta_{1}+\beta_{2}+\cdots+\beta_{j}\ \mid\ j\in\left\{
1,2,\ldots,m-1\right\} \right\} }_{=D\left( \beta\right) }+p\right) \\
& =D\left( \alpha\right) \cup\left\{ p\right\} \cup\left( D\left(
\beta\right) +p\right) .
\end{align*}
This completes the proof of Lemma \ref{lem.D(a(.)b)} \textbf{(b)}.
\end{proof}
\end{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.bel.F}.]If either $\alpha$ or $\beta$ is
empty, then this is obvious (since $\bel $ is unital with $1$ as its unity,
and since $F_{\varnothing}=1$). So let us WLOG assume that neither is. Write
$\alpha$ as $\alpha=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
$, and write $\beta$ as $\beta=\left( \beta_{1},\beta_{2},\ldots,\beta
_{m}\right) $. Thus, $\ell$ and $m$ are positive (since $\alpha$ and $\beta$
are nonempty).
Let $p=\left\vert \alpha\right\vert $ and $q=\left\vert \beta\right\vert $.
Thus, $p$ and $q$ are positive (since $\alpha$ and $\beta$ are nonempty).
Recall that we use the notation $D\left( \alpha\right) $ for the set of
partial sums of a composition $\alpha$. If $G$ is a set of integers and $r$ is
an integer, then we let $G+r$ denote the set $\left\{ g+r\ \mid\ g\in
G\right\} $ of integers.
\begin{verlong}
Lemma \ref{lem.D(a(.)b)} \textbf{(a)} shows that $\alpha\odot\beta$ is a
composition of $p+q$ satisfying $D\left( \alpha\odot\beta\right) =D\left(
\alpha\right) \cup\left( D\left( \beta\right) +p\right) $.
\end{verlong}
Applying (\ref{eq.F.def}) to $p$ instead of $n$, we obtain%
\begin{equation}
F_{\alpha}=\sum_{\substack{i_{1}\leq i_{2}\leq\cdots\leq i_{p};\\i_{j}%
i_{p+1}%
\end{cases}
\\\text{(by the definition of } \bel \text{ on monomials)}}}\nonumber\\
& =\sum_{\substack{i_{1}\leq i_{2}\leq\cdots\leq i_{p};\\i_{j}i_{p+1}%
\end{cases}
\nonumber\\
& =\underbrace{\sum_{\substack{i_{1}\leq i_{2}\leq\cdots\leq i_{p}%
;\\i_{j}$\textquotedblright\ in the definition of $\left. \prec\right. $.)
\end{remark}
\end{vershort}
\begin{verlong}
For the sake of completeness (or, rather, in order not to lose old writing),
let me write down the definitions of some more operations on $\mathbf{k}%
\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $.
\begin{definition}
We define a binary operation $\left. \preceq\right. :\mathbf{k}\left[
\left[ x_{1},x_{2},x_{3},\ldots\right] \right] \times\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] $ (written in infix
notation) by the requirements that it be $\mathbf{k}$-bilinear and continuous
with respect to the topology on $\mathbf{k}\left[ \left[ x_{1},x_{2}%
,x_{3},\ldots\right] \right] $ and that it satisfy%
\[
\mathfrak{m}\left. \preceq\right. \mathfrak{n}=
\begin{cases}
\mathfrak{m}\cdot\mathfrak{n}, & \text{if }\min\left( \operatorname*{Supp}%
\mathfrak{m}\right) \leq\min\left( \operatorname*{Supp}\mathfrak{n}\right)
;\\
0, & \text{if }\min\left( \operatorname*{Supp}\mathfrak{m}\right)
>\min\left( \operatorname*{Supp}\mathfrak{n}\right)
\end{cases}
\]
for any two monomials $\mathfrak{m}$ and $\mathfrak{n}$.
\end{definition}
Here, all the remarks we made after Definition \ref{def.Dless} apply. In
particular, $\min\varnothing=\infty$, and we are using $\left. \preceq
\right. $ as an operation symbol.
We have $\mathfrak{m}\left. \preceq\right. 1=\mathfrak{m}$ for every
monomial $\mathfrak{m}$, and $1\left. \preceq\right. \mathfrak{m}=0$ for
every nonconstant monomial $\mathfrak{m}$.
\begin{remark}
The operation $\left. \preceq\right. $ is part of a dendriform algebra
structure on $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right]
\right] $ (and on $\operatorname*{QSym}$). More precisely, if we define
another binary operation $\left. \succ\right. $ on $\mathbf{k}\left[
\left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ similarly to $\left.
\preceq\right. $ except that we set%
\[
\mathfrak{m}\left. \succ\right. \mathfrak{n}=
\begin{cases}
\mathfrak{m}\cdot\mathfrak{n}, & \text{if }\min\left( \operatorname*{Supp}%
\mathfrak{m}\right) >\min\left( \operatorname*{Supp}\mathfrak{n}\right) ;\\
0, & \text{if }\min\left( \operatorname*{Supp}\mathfrak{m}\right) \leq
\min\left( \operatorname*{Supp}\mathfrak{n}\right)
\end{cases}
\quad,
\]
then the structure $\left( \mathbf{k}\left[ \left[ x_{1},x_{2},x_{3}%
,\ldots\right] \right] ,\left. \preceq\right. ,\left. \succ\right.
\right) $ is a dendriform algebra augmented to satisfy \cite[(15)]{EbrFar08}.
In particular, any three elements $a$, $b$ and $c$ of $\mathbf{k}\left[
\left[ x_{1},x_{2},x_{3},\ldots\right] \right] $ satisfy%
\begin{align*}
a\left. \preceq\right. b+a\left. \succ\right. b & =ab;\\
\left( a\left. \preceq\right. b\right) \left. \preceq\right. c &
=a\left. \preceq\right. \left( bc\right) ;\\
\left( a\left. \succ\right. b\right) \left. \preceq\right. c &
=a\left. \succ\right. \left( b\left. \preceq\right. c\right) ;\\
a\left. \succ\right. \left( b\left. \succ\right. c\right) & =\left(
ab\right) \left. \succ\right. c.
\end{align*}
\end{remark}
And here is an analogue of $\bel $ which has \textit{mostly} similar properties:
\begin{definition}
We define a binary operation $\tvi :\mathbf{k}\left[ \left[ x_{1}%
,x_{2},x_{3},\ldots\right] \right] \times\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] $ (written in infix notation) by the
requirements that it be $\mathbf{k}$-bilinear and continuous with respect to
the topology on $\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right]
\right] $ and that it satisfy%
\[
\mathfrak{m} \tvi \mathfrak{n}=
\begin{cases}
\mathfrak{m}\cdot\mathfrak{n}, & \text{if }\max\left( \operatorname*{Supp}%
\mathfrak{m}\right) <\min\left( \operatorname*{Supp}\mathfrak{n}\right) ;\\
0, & \text{if }\max\left( \operatorname*{Supp}\mathfrak{m}\right) \geq
\min\left( \operatorname*{Supp}\mathfrak{n}\right)
\end{cases}
\]
for any two monomials $\mathfrak{m}$ and $\mathfrak{n}$.
\end{definition}
Here, again, $\max\varnothing$ is understood as $0$. The binary operation
$\tvi$ is associative. It is also unital (with $1$ serving as the unity).
\begin{proposition}
Every $a\in\operatorname*{QSym}$ and $b\in\operatorname*{QSym}$ satisfy
$a\left. \preceq\right. b\in\operatorname*{QSym}$ and $a \tvi b\in
\operatorname*{QSym}$.
\end{proposition}
For example, any two compositions $\alpha$ and $\beta$ satisfy $M_{\alpha}
\tvi M_{\beta}=M_{\left[ \alpha,\beta\right] }$, where $\left[ \alpha
,\beta\right] $ denotes the concatenation of $\alpha$ and $\beta$ (defined by
$\left[ \left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) ,\left(
\beta_{1},\beta_{2},\ldots,\beta_{m}\right) \right] =\left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell},\beta_{1},\beta_{2},\ldots,\beta
_{m}\right) $). (Recall that $\left( M_{\gamma}\right) _{\gamma
\in\operatorname*{Comp}}$ is the monomial basis of $\operatorname*{QSym}$.)
\begin{theorem}
\label{thm.tvidend}Let $S$ denote the antipode of the Hopf algebra
$\operatorname*{QSym}$. Let us use Sweedler's notation $\sum_{\left(
b\right) }b_{\left( 1\right) }\otimes b_{\left( 2\right) }$ for
$\Delta\left( b\right) $, where $b$ is any element of $\operatorname*{QSym}%
$. Then,%
\[
\sum_{\left( b\right) }\left( S\left( b_{\left( 1\right) }\right)
\tvi a\right) b_{\left( 2\right) }=a\left. \preceq\right. b
\]
for any $a\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right]
\right] $ and $b\in\operatorname*{QSym}$.
\end{theorem}
\begin{proof}
[Proof of Theorem \ref{thm.tvidend}.]The following proof is mostly analogous
to the proof of Theorem \ref{thm.beldend}.
Let $a\in\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right]
$. We can WLOG assume that $a$ is a monomial (because all operations in sight
are $\mathbf{k}$-linear and continuous). So assume this. That is,
$a=\mathfrak{n}$ for some monomial $\mathfrak{n}$. Consider this
$\mathfrak{n}$. Let $k=\min\left( \operatorname*{Supp}\mathfrak{n}\right) $.
Notice that $k\in\left\{ 1,2,3,\ldots\right\} \cup\left\{ \infty\right\} $.
(Some remarks about $\infty$ are in order. We use $\infty$ as an object which
is greater than every integer. We will use summation signs like $\sum_{1\leq
i_{1}\min\left\{ i_{1},i_{2},\ldots,i_{\ell}\right\}
\end{cases}
\\\text{(by the definition of }\left. \preceq\right. \text{ on monomials)}%
}}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since }\left. \preceq\right. \text{ is
}\mathbf{k}\text{-bilinear and continuous}\right) \\
& =\sum_{1\leq i_{1}\min\left\{ i_{1},i_{2},\ldots,i_{\ell}\right\}
\end{cases}
\\
& =\underbrace{\sum_{\substack{1\leq i_{1}k
\end{cases}
\quad.
\]
Notice that not every immaculate tableau has a content (with this definition),
because we only allow compositions as contents. More precisely, if $T$ is an
immaculate tableau of shape $\alpha$, then there exists a composition $\beta$
such that $T$ has content $\beta$ if and only if there exists a $k\in
\mathbb{N}$ such that $T\left( Y\left( \alpha\right) \right) =\left\{
1,2,\ldots,k\right\} $.
\textbf{(d)} Let $\beta$ be a composition of $\left\vert \alpha\right\vert $.
Then, $K_{\alpha,\beta}$ denotes the number of immaculate tableaux of shape
$\alpha$ and content $\beta$.
\end{definition}
\begin{noncompile}
If $\alpha$ is a composition and $T$ is an immaculate tableau of shape
$\alpha$, then we refer to the image of $\left( i,j\right) \in Y\left(
\alpha\right) $ under $T$ as the \textit{entry}\ of the cell $\left(
i,j\right) $ in the tableau $T$.
\end{noncompile}
\begin{vershort}
For future reference, let us notice that if $\alpha$ is a composition, if $T$
is an immaculate tableau of shape $\alpha$, and if $\left( i,j\right) \in
Y\left( \alpha\right) $ is such that $i>1$, then%
\begin{equation}
T\left( 1,1\right) 1$, then%
\begin{equation}
T\left( 1,1\right) \ell
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for every positive integer }k
\label{pf.dualImm.fn.ct}%
\end{equation}
(since $Q$ has content $\beta$). Hence, $Q\left( Y\left( \alpha\right)
\right) =\left\{ 1,2,\ldots,\ell\right\} $. As a consequence, the maps
$T:Y\left( \alpha\right) \rightarrow\left\{ 1,2,3,\ldots\right\} $
satisfying $r_{T\left( Y\left( \alpha\right) \right) }^{-1}\circ T=Q$ are
in 1-to-1 correspondence with the $\ell$-element subsets of $\left\{
1,2,3,\ldots\right\} $ (the correspondence sends a map $T$ to the $\ell
$-element subset $T\left( Y\left( \alpha\right) \right) $, and the inverse
correspondence sends an $\ell$-element subset $I$ to the map $r_{I}\circ Q$).
But these latter subsets, in turn, are in 1-to-1 correspondence with the
strictly increasing length-$\ell$ sequences $\left( i_{1}0$. Let
$\overline{\alpha}$ denote the composition $\left( \alpha_{2},\alpha
_{3},\ldots,\alpha_{\ell}\right) $ of $\left\vert \alpha\right\vert
-\alpha_{1}$. Then,%
\[
\mathfrak{S}_{\alpha}^{\ast}=h_{\alpha_{1}}\left. \prec\right.
\mathfrak{S}_{\overline{\alpha}}^{\ast}.
\]
Here, $h_{n}$ denotes the $n$-th complete homogeneous symmetric function for
every $n\in\mathbb{N}$.
\end{corollary}
\begin{proof}
[Proof of Corollary \ref{cor.dualImm.dend}.]Proposition \ref{prop.dualImm}
shows that
\begin{equation}
\mathfrak{S}_{\alpha}^{\ast}=\sum_{\substack{T\text{ is an immaculate}%
\\\text{tableau of shape }\alpha}}\mathbf{x}_{T}=\sum_{\substack{Q\text{ is an
immaculate}\\\text{tableau of shape }\alpha}}\mathbf{x}_{Q}
\label{pf.cor.dualImm.dend.LHS}%
\end{equation}
(here, we have renamed the summation index $T$ as $Q$).
Let $n=\alpha_{1}$. If $i_{1},i_{2},\ldots,i_{n}$ are positive integers
satisfying $i_{1}\leq i_{2}\leq\cdots\leq i_{n}$, and if $T$ is an immaculate
tableau of shape $\overline{\alpha}$, then%
\begin{align}
& \left( x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right) \left. \prec\right.
\mathbf{x}_{T}\nonumber\\
& =
\begin{cases}
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}, & \text{if } \min\left(
\operatorname*{Supp}\left( x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right)
\right) <\min\left( \operatorname*{Supp}\left( \mathbf{x}_{T}\right)
\right) ;\\
0, & \text{if }\min\left( \operatorname*{Supp}\left( x_{i_{1}}x_{i_{2}%
}\cdots x_{i_{n}}\right) \right) \geq\min\left( \operatorname*{Supp}\left(
\mathbf{x}_{T}\right) \right)
\end{cases}
\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\left.
\prec\right. \text{ on monomials}\right) \nonumber\\
& =
\begin{cases}
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}, & \text{if } i_{1}%
<\min\left( T\left( Y\left( \overline{\alpha}\right) \right) \right) ;\\
0, & \text{if }i_{1}\geq\min\left( T\left( Y\left( \overline{\alpha
}\right) \right) \right)
\end{cases}
\label{pf.cor.dualImm.dend.1}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since }\min\left( \operatorname*{Supp}%
\left( x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right) \right) =i_{1}\text{ and
}\operatorname*{Supp}\left( \mathbf{x}_{T}\right) =T\left( Y\left(
\overline{\alpha}\right) \right) \right) .\nonumber
\end{align}
\begin{vershort}
But from $n=\alpha_{1}$, we obtain $h_{n}=h_{\alpha_{1}}$, so that
$h_{\alpha_{1}}=h_{n}=\sum\limits_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}%
}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ and $\mathfrak{S}_{\overline{\alpha}%
}^{\ast}=\sum_{\substack{T\text{ is an immaculate}\\\text{tableau of shape
}\overline{\alpha}}}\mathbf{x}_{T}$ (by Proposition \ref{prop.dualImm}).
Hence,%
\begin{align}
& h_{\alpha_{1}}\left. \prec\right. \mathfrak{S}_{\overline{\alpha}}^{\ast
}\nonumber\\
& =\left( \sum\limits_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}x_{i_{1}}%
x_{i_{2}}\cdots x_{i_{n}}\right) \left. \prec\right. \left( \sum
_{\substack{T\text{ is an immaculate}\\\text{tableau of shape }\overline
{\alpha}}}\mathbf{x}_{T}\right) \nonumber\\
& =\sum\limits_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}\sum_{\substack{T\text{
is an immaculate}\\\text{tableau of shape }\overline{\alpha}}}\left(
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right) \left. \prec\right.
\mathbf{x}_{T}\nonumber\\
& =\sum\limits_{\substack{i_{1}\leq i_{2}\leq\cdots\leq i_{n};\\T\text{ is an
immaculate}\\\text{tableau of shape }\overline{\alpha};\\i_{1}<\min\left(
T\left( Y\left( \overline{\alpha}\right) \right) \right) }}x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}\ \ \ \ \ \ \ \ \ \ \left( \text{by
(\ref{pf.cor.dualImm.dend.1})}\right) . \label{pf.cor.dualImm.dend.short.RHS}%
\end{align}
We need to check that this equals $\mathfrak{S}_{\alpha}^{\ast}=\sum
_{\substack{Q\text{ is an immaculate}\\\text{tableau of shape }\alpha
}}\mathbf{x}_{Q}$.
\end{vershort}
\begin{verlong}
But from $n=\alpha_{1}$, we obtain $h_{n}=h_{\alpha_{1}}$, so that
$h_{\alpha_{1}}=h_{n}=\sum\limits_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}%
}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}$ and $\mathfrak{S}_{\overline{\alpha}%
}^{\ast}=\sum_{\substack{T\text{ is an immaculate}\\\text{tableau of shape
}\overline{\alpha}}}\mathbf{x}_{T}$ (by Proposition \ref{prop.dualImm}).
Hence,%
\begin{align}
& h_{\alpha_{1}}\left. \prec\right. \mathfrak{S}_{\overline{\alpha}}^{\ast
}\nonumber\\
& =\left( \sum\limits_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}x_{i_{1}}%
x_{i_{2}}\cdots x_{i_{n}}\right) \left. \prec\right. \left( \sum
_{\substack{T\text{ is an immaculate}\\\text{tableau of shape }\overline
{\alpha}}}\mathbf{x}_{T}\right) \nonumber\\
& =\sum\limits_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}\sum_{\substack{T\text{
is an immaculate}\\\text{tableau of shape }\overline{\alpha}}%
}\underbrace{\left( x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\right) \left.
\prec\right. \mathbf{x}_{T}}_{\substack{=
\begin{cases}
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}, & \text{if }i_{1}%
<\min\left( T\left( Y\left( \overline{\alpha}\right) \right) \right) ;\\
0, & \text{if }i_{1}\geq\min\left( T\left( Y\left( \overline{\alpha
}\right) \right) \right)
\end{cases}
\\\text{(by (\ref{pf.cor.dualImm.dend.1}))}}}\nonumber\\
& =\sum\limits_{i_{1}\leq i_{2}\leq\cdots\leq i_{n}}\sum_{\substack{T\text{
is an immaculate}\\\text{tableau of shape }\overline{\alpha}}}
\begin{cases}
x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}, & \text{if }i_{1}%
<\min\left( T\left( Y\left( \overline{\alpha}\right) \right) \right) ;\\
0, & \text{if }i_{1}\geq\min\left( T\left( Y\left( \overline{\alpha
}\right) \right) \right)
\end{cases}
\nonumber\\
& =\sum\limits_{\substack{i_{1}\leq i_{2}\leq\cdots\leq i_{n};\\T\text{ is an
immaculate}\\\text{tableau of shape }\overline{\alpha};\\i_{1}<\min\left(
T\left( Y\left( \overline{\alpha}\right) \right) \right) }}x_{i_{1}%
}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}. \label{pf.cor.dualImm.dend.RHS}%
\end{align}
We need to check that this equals $\mathfrak{S}_{\alpha}^{\ast}=\sum
_{\substack{Q\text{ is an immaculate}\\\text{tableau of shape }\alpha
}}\mathbf{x}_{Q}$.
\end{verlong}
Now, let us define a map $\Phi$ from:
\begin{itemize}
\item the set of all pairs $\left( \left( i_{1},i_{2},\ldots,i_{n}\right)
,T\right) $, where $i_{1}$, $i_{2}$, $\ldots$, $i_{n}$ are positive integers
satisfying $i_{1}\leq i_{2}\leq\cdots\leq i_{n}$, and where $T$ is an
immaculate tableau of shape $\overline{\alpha}$ satisfying $i_{1}<\min\left(
T\left( Y\left( \overline{\alpha}\right) \right) \right) $
\end{itemize}
to:
\begin{itemize}
\item the set of all immaculate tableaux of shape $\alpha$.
\end{itemize}
Namely, we define the image of a pair $\left( \left( i_{1},i_{2}%
,\ldots,i_{n}\right) ,T\right) $ under $\Phi$ to be the immaculate tableau
obtained by adding a new row, filled with the entries $i_{1},i_{2}%
,\ldots,i_{n}$ (from left to right), to the top\footnote{Here, we are using
the graphical representation of immaculate tableaux introduced in Definition
\ref{def.immactab}.} of the tableau $T$ \ \ \ \ \footnote{Formally speaking,
this means that the image of $\left( \left( i_{1},i_{2},\ldots,i_{n}\right)
,T\right) $ is the map $Y\left( \alpha\right) \rightarrow\left\{
1,2,3,\ldots\right\} $ which sends every $\left( u,v\right) \in Y\left(
\alpha\right) $ to $%
\begin{cases}
i_{v}, & \text{if }u=1;\\
T\left( u-1,v\right) , & \text{if }u\neq1
\end{cases}
\quad$. Proving that this map is an immaculate tableau is easy.}.
\begin{vershort}
This map $\Phi$ is a bijection\footnote{\textit{Proof.} The injectivity of the
map $\Phi$ is obvious. Its surjectivity follows from the observation that if
$Q$ is an immaculate tableau of shape $\alpha$, then the first entry of its
top row is smaller than the smallest entry of the immaculate tableau formed by
all other rows of $Q$. (This is a consequence of
(\ref{eq.immactab.min.2.short}), applied to $Q$ instead of $T$.)}, and has the
property that if $Q$ denotes the image of a pair $\left( \left( i_{1}%
,i_{2},\ldots,i_{n}\right) ,T\right) $ under the bijection $\Phi$, then
$\mathbf{x}_{Q}=x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}$. Hence,%
\[
\sum_{\substack{Q\text{ is an immaculate}\\\text{tableau of shape }\alpha
}}\mathbf{x}_{Q}=\sum\limits_{\substack{i_{1}\leq i_{2}\leq\cdots\leq
i_{n};\\T\text{ is an immaculate}\\\text{tableau of shape }\overline{\alpha
};\\i_{1}<\min\left( T\left( Y\left( \overline{\alpha}\right) \right)
\right) }}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}.
\]
In light of (\ref{pf.cor.dualImm.dend.LHS}) and
(\ref{pf.cor.dualImm.dend.short.RHS}), this rewrites as $\mathfrak{S}_{\alpha
}^{\ast}=h_{\alpha_{1}}\left. \prec\right. \mathfrak{S}_{\overline{\alpha}%
}^{\ast}$.
\end{vershort}
\begin{verlong}
This map $\Phi$ is a bijection\footnote{\textit{Proof.} The injectivity of the
map $\Phi$ is obvious. Its surjectivity follows from the observation that if
$Q$ is an immaculate tableau of shape $\alpha$, then the first entry of its
top row is smaller than the smallest entry of the immaculate tableau formed by
all other rows of $Q$. (This is a consequence of (\ref{eq.immactab.min.2}),
applied to $Q$ instead of $T$.)}, and has the property that if $Q$ denotes the
image of a pair $\left( \left( i_{1},i_{2},\ldots,i_{n}\right) ,T\right) $
under the bijection $\Phi$, then $\mathbf{x}_{Q}=x_{i_{1}}x_{i_{2}}\cdots
x_{i_{n}}\mathbf{x}_{T}$. Hence,%
\[
\sum_{\substack{Q\text{ is an immaculate}\\\text{tableau of shape }\alpha
}}\mathbf{x}_{Q}=\sum\limits_{\substack{i_{1}\leq i_{2}\leq\cdots\leq
i_{n};\\T\text{ is an immaculate}\\\text{tableau of shape }\overline{\alpha
};\\i_{1}<\min\left( T\left( Y\left( \overline{\alpha}\right) \right)
\right) }}x_{i_{1}}x_{i_{2}}\cdots x_{i_{n}}\mathbf{x}_{T}.
\]
In light of (\ref{pf.cor.dualImm.dend.LHS}) and (\ref{pf.cor.dualImm.dend.RHS}%
), this rewrites as $\mathfrak{S}_{\alpha}^{\ast}=h_{\alpha_{1}}\left.
\prec\right. \mathfrak{S}_{\overline{\alpha}}^{\ast}$. So Corollary
\ref{cor.dualImm.dend} is proven.
\end{verlong}
\end{proof}
\begin{corollary}
\label{cor.dualImm.dend.explicit}Let $\alpha=\left( \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right) $ be a composition. Then,%
\[
\mathfrak{S}_{\alpha}^{\ast}=h_{\alpha_{1}}\left. \prec\right. \left(
h_{\alpha_{2}}\left. \prec\right. \left( \cdots\left. \prec\right.
\left( h_{\alpha_{\ell}}\left. \prec\right. 1\right) \cdots\right)
\right) .
\]
\end{corollary}
\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.dualImm.dend.explicit}.]We prove Corollary
\ref{cor.dualImm.dend.explicit} by induction over $\ell$:
\textit{Induction base:} If $\ell=0$, then $\alpha=\varnothing$ and thus
$\mathfrak{S}_{\alpha}^{\ast}=\mathfrak{S}_{\varnothing}^{\ast}=1$. But if
$\ell=0$, then we also have $h_{\alpha_{1}}\left. \prec\right. \left(
h_{\alpha_{2}}\left. \prec\right. \left( \cdots\left. \prec\right.
\left( h_{\alpha_{\ell}}\left. \prec\right. 1\right) \cdots\right)
\right) =1$. Hence, if $\ell=0$, then $\mathfrak{S}_{\alpha}^{\ast
}=1=h_{\alpha_{1}}\left. \prec\right. \left( h_{\alpha_{2}}\left.
\prec\right. \left( \cdots\left. \prec\right. \left( h_{\alpha_{\ell}%
}\left. \prec\right. 1\right) \cdots\right) \right) $. Thus, Corollary
\ref{cor.dualImm.dend.explicit} is proven when $\ell=0$. The induction base is complete.
\textit{Induction step:} Let $L$ be a positive integer. Assume that Corollary
\ref{cor.dualImm.dend.explicit} holds for $\ell=L-1$. We now need to prove
that Corollary \ref{cor.dualImm.dend.explicit} by holds for $\ell=L$.
So let $\alpha=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) $ be
a composition with $\ell=L$. Then, $\ell=L>0$. Now, let $\overline{\alpha}$
denote the composition $\left( \alpha_{2},\alpha_{3},\ldots,\alpha_{\ell
}\right) $ of $\left\vert \alpha\right\vert -\alpha_{1}$. Then, Corollary
\ref{cor.dualImm.dend} yields $\mathfrak{S}_{\alpha}^{\ast}=h_{\alpha_{1}%
}\left. \prec\right. \mathfrak{S}_{\overline{\alpha}}^{\ast}$. But by our
induction hypothesis, we can apply Corollary \ref{cor.dualImm.dend.explicit}
to $\overline{\alpha}=\left( \alpha_{2},\alpha_{3},\ldots,\alpha_{\ell
}\right) $ instead of $\alpha=\left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right) $ (since $\ell-1=L-1$). As a result, we obtain
$\mathfrak{S}_{\overline{\alpha}}^{\ast}=h_{\alpha_{2}}\left. \prec\right.
\left( h_{\alpha_{3}}\left. \prec\right. \left( \cdots\left.
\prec\right. \left( h_{\alpha_{\ell}}\left. \prec\right. 1\right)
\cdots\right) \right) $. Hence,%
\begin{align*}
\mathfrak{S}_{\alpha}^{\ast} & =h_{\alpha_{1}}\left. \prec\right.
\underbrace{\mathfrak{S}_{\overline{\alpha}}^{\ast}}_{=h_{\alpha_{2}}\left.
\prec\right. \left( h_{\alpha_{3}}\left. \prec\right. \left(
\cdots\left. \prec\right. \left( h_{\alpha_{\ell}}\left. \prec\right.
1\right) \cdots\right) \right) }=h_{\alpha_{1}}\left. \prec\right.
\left( h_{\alpha_{2}}\left. \prec\right. \left( h_{\alpha_{3}}\left.
\prec\right. \left( \cdots\left. \prec\right. \left( h_{\alpha_{\ell}%
}\left. \prec\right. 1\right) \cdots\right) \right) \right) \\
& =h_{\alpha_{1}}\left. \prec\right. \left( h_{\alpha_{2}}\left.
\prec\right. \left( \cdots\left. \prec\right. \left( h_{\alpha_{\ell}%
}\left. \prec\right. 1\right) \cdots\right) \right) .
\end{align*}
Now, let us forget that we fixed $\alpha$. We thus have shown that
\newline$\mathfrak{S}_{\alpha}^{\ast}=h_{\alpha_{1}}\left. \prec\right.
\left( h_{\alpha_{2}}\left. \prec\right. \left( \cdots\left.
\prec\right. \left( h_{\alpha_{\ell}}\left. \prec\right. 1\right)
\cdots\right) \right) $ for every composition $\alpha=\left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) $ which satisfies $\ell=L$. In
other words, Corollary \ref{cor.dualImm.dend.explicit} holds for $\ell=L$.
This completes the induction step. The induction proof of Corollary
\ref{cor.dualImm.dend.explicit} is thus complete.
\end{proof}
\end{verlong}
\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.dualImm.dend.explicit}.]This follows by induction
from Corollary \ref{cor.dualImm.dend} (since $\mathfrak{S}_{\varnothing}%
^{\ast}=1$).
\end{proof}
\end{vershort}
\section{\label{sect.zabrocki}An alternative description of $h_{m}\left.
\prec\right. $}
In this section, we shall also use the Hopf algebra of \textit{noncommutative
symmetric functions}. This Hopf algebra (a noncommutative one, for a change)
is denoted by $\operatorname*{NSym}$ and has been discussed in \cite[Section
5.4]{Reiner} and \cite[Chapter 6]{HGK}; all we need to know about it are the
following properties:
\begin{itemize}
\item There is a nondegenerate pairing between $\operatorname*{NSym}$ and
$\operatorname*{QSym}$, that is, a nondegenerate $\mathbf{k}$-bilinear form
$\operatorname*{NSym}\times\operatorname*{QSym}\rightarrow\mathbf{k}$. We
shall denote this bilinear form by $\left( \cdot,\cdot\right) $. This
$\mathbf{k}$-bilinear form is a Hopf algebra pairing, i.e., it satisfies%
\begin{align}
\left( ab,c\right) & =\sum_{\left( c\right) }\left( a,c_{\left(
1\right) }\right) \left( b,c_{\left( 2\right) }\right) \label{eq.(ab,c)}%
\\
& \ \ \ \ \ \ \ \ \ \ \text{for all }a\in\operatorname*{NSym}\text{, }%
b\in\operatorname*{NSym}\text{ and }c\in\operatorname*{QSym};\nonumber
\end{align}%
\[
\left( 1,c\right) =\varepsilon\left( c\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }c\in\operatorname*{QSym};
\]%
\begin{align*}
\sum_{\left( a\right) }\left( a_{\left( 1\right) },b\right) \left(
a_{\left( 2\right) },c\right) & =\left( a,bc\right) \\
\ \ \ \ \ \ \ \ \ \ \text{for all }a & \in\operatorname*{NSym}\text{, }%
b\in\operatorname*{QSym}\text{ and }c\in\operatorname*{QSym};
\end{align*}%
\[
\left( a,1\right) =\varepsilon\left( a\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\operatorname*{NSym};
\]%
\[
\left( S\left( a\right) ,b\right) =\left( a,S\left( b\right) \right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\operatorname*{NSym}\text{ and }%
b\in\operatorname*{QSym}%
\]
(where we use Sweedler's notation).
\item There is a basis of the $\mathbf{k}$-module $\operatorname*{NSym}$ which
is dual to the fundamental basis $\left( F_{\alpha}\right) _{\alpha
\in\operatorname*{Comp}}$ of $\operatorname*{QSym}$ with respect to the
bilinear form $\left( \cdot,\cdot\right) $. This basis is called the
\textit{ribbon basis} and will be denoted by $\left( R_{\alpha}\right)
_{\alpha\in\operatorname*{Comp}}$.
\end{itemize}
Both of these properties are immediate consequences of the definitions of
$\operatorname*{NSym}$ and of $\left( R_{\alpha}\right) _{\alpha
\in\operatorname*{Comp}}$ given in \cite[Section 5.5]{Reiner} (although other
sources define these objects differently, and then the properties no longer
are immediate). The notations we are using here are the same as the ones used
in \cite[Section 5.5]{Reiner} (except that \cite[Section 5.5]{Reiner} calls
$L_{\alpha}$ what we denote by $F_{\alpha}$), and only slightly differ from
those in \cite{BBSSZ} (namely, \cite{BBSSZ} denotes the pairing $\left(
\cdot,\cdot\right) $ by $\left\langle \cdot,\cdot\right\rangle $ instead).
We need some more definitions. For any $g\in\operatorname*{NSym}$, let
$\operatorname{L}_{g}:\operatorname*{NSym}\rightarrow\operatorname*{NSym}$
denote the left multiplication by $g$ on $\operatorname*{NSym}$ (that is, the
$\mathbf{k}$-linear map $\operatorname{NSym}\rightarrow\operatorname{NSym}%
,\ f\mapsto gf$). For any $g\in\operatorname{NSym}$, let $g^{\perp
}:\operatorname*{QSym}\rightarrow\operatorname*{QSym}$ be the $\mathbf{k}%
$-linear map adjoint to $\operatorname{L}_{g}:\operatorname*{NSym}%
\rightarrow\operatorname*{NSym}$ with respect to the pairing $\left(
\cdot,\cdot\right) $ between $\operatorname*{NSym}$ and $\operatorname*{QSym}%
$. Thus, for any $g\in\operatorname*{NSym}$, $a\in\operatorname*{NSym}$ and
$c\in\operatorname*{QSym}$, we have%
\begin{equation}
\left( a,g^{\perp}c\right) =\left( \underbrace{\operatorname*{L}%
\nolimits_{g}a}_{=ga},c\right) =\left( ga,c\right) .
\label{pf.lem.adjoint-g.pre}%
\end{equation}
The following fact is well-known (and also is an easy formal consequence of
the definition of $g^{\perp}$ and of (\ref{eq.(ab,c)})):
\begin{lemma}
\label{lem.adjoint-g}Every $g\in\operatorname*{NSym}$ and $f\in
\operatorname*{QSym}$ satisfy%
\begin{equation}
g^{\perp}f=\sum_{\left( f\right) }\left( g,f_{\left( 1\right) }\right)
f_{\left( 2\right) }. \label{pf.hmDless.1}%
\end{equation}
\end{lemma}
\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.adjoint-g}.]See the detailed version of this note.
\end{proof}
\end{vershort}
\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.adjoint-g}.]Let $g\in\operatorname*{NSym}$ and
$f\in\operatorname*{QSym}$. For every $a\in\operatorname*{NSym}$, we have%
\begin{align*}
\left( a,g^{\perp}f\right) & =\left( \underbrace{\operatorname*{L}%
\nolimits_{g}a}_{\substack{=ga\\\text{(by the definition of }\operatorname*{L}%
\nolimits_{g}\text{)}}},f\right) \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }g^{\perp}\text{ is adjoint to }\operatorname*{L}%
\nolimits_{g}\\
\text{with respect to the pairing }\left( \cdot,\cdot\right)
\end{array}
\right) \\
& =\left( ga,f\right) =\sum_{\left( f\right) }\left( g,f_{\left(
1\right) }\right) \left( a,f_{\left( 2\right) }\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{eq.(ab,c)}), applied to }g\text{, }a\text{ and }f\\
\text{instead of }a\text{, }b\text{ and }c
\end{array}
\right) \\
& =\left( a,\sum_{\left( f\right) }\left( g,f_{\left( 1\right)
}\right) f_{\left( 2\right) }\right) \ \ \ \ \ \ \ \ \ \ \left(
\text{since the pairing }\left( \cdot,\cdot\right) \text{ is }%
\mathbf{k}\text{-bilinear}\right) .
\end{align*}
Since the pairing $\left( \cdot,\cdot\right) $ is nondegenerate, this
entails that $g^{\perp}f=\sum_{\left( f\right) }\left( g,f_{\left(
1\right) }\right) f_{\left( 2\right) }$. This proves Lemma
\ref{lem.adjoint-g}.
\end{proof}
\end{verlong}
For any composition $\alpha$, we define a composition $\omega\left(
\alpha\right) $ as follows: Let $n=\left\vert \alpha\right\vert $, and write
$\alpha$ as $\alpha=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
$. Let $\operatorname*{rev}\alpha$ denote the composition $\left(
\alpha_{\ell},\alpha_{\ell-1},\ldots,\alpha_{1}\right) $ of $n$. Then,
$\omega\left( \alpha\right) $ shall be the unique composition $\beta$ of $n$
which satisfies $D\left( \beta\right) =\left\{ 1,2,\ldots,n-1\right\}
\setminus D\left( \operatorname*{rev}\alpha\right) $. (This definition is
identical with that in \cite[Definition 5.2.14]{Reiner}. Some authors denote
$\omega\left( \alpha\right) $ by $\alpha^{\prime}$ instead.) We notice that
$\omega\left( \omega\left( \alpha\right) \right) =\alpha$ for any
composition $\alpha$.
\begin{verlong}
Here is a simple property of the composition $\omega\left( \alpha\right) $
that will later be used:
\begin{proposition}
\label{prop.omega.odot}\textbf{(a)} We have $\omega\left( \left[
\alpha,\beta\right] \right) =\omega\left( \beta\right) \odot\omega\left(
\alpha\right) $ for any two compositions $\alpha$ and $\beta$.
\textbf{(b)} We have $\omega\left( \alpha\odot\beta\right) =\left[
\omega\left( \beta\right) ,\omega\left( \alpha\right) \right] $ for any
two compositions $\alpha$ and $\beta$.
\textbf{(c)} We have $\omega\left( \omega\left( \gamma\right) \right)
=\gamma$ for every composition $\gamma$.
\end{proposition}
\begin{proof}
[Proof of Proposition \ref{prop.omega.odot}.]For any composition $\alpha$, we
define a composition $\operatorname*{rev}\alpha$ as follows: Let $n=\left\vert
\alpha\right\vert $, and write $\alpha$ as $\alpha=\left( \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right) $. Let $\operatorname*{rev}\alpha$
denote the composition $\left( \alpha_{\ell},\alpha_{\ell-1},\ldots
,\alpha_{1}\right) $ of $n$. (This definition of $\operatorname*{rev}\alpha$
is the same as the one we gave above during the definition of $\omega\left(
\alpha\right) $.) Clearly,%
\begin{equation}
\left\vert \operatorname*{rev}\gamma\right\vert =\left\vert \gamma\right\vert
\ \ \ \ \ \ \ \ \ \ \text{for any composition }\gamma.
\label{pf.prop.omega.odot.rev0}%
\end{equation}
It is easy to see that%
\begin{align}
\operatorname*{rev}\left( \left[ \alpha,\beta\right] \right) & =\left[
\operatorname*{rev}\beta,\operatorname*{rev}\alpha\right]
\ \ \ \ \ \ \ \ \ \ \text{and}\label{pf.prop.omega.odot.rev1}\\
\operatorname*{rev}\left( \alpha\odot\beta\right) & =\left(
\operatorname*{rev}\beta\right) \odot\left( \operatorname*{rev}%
\alpha\right) \label{pf.prop.omega.odot.rev2}%
\end{align}
for any two compositions $\alpha$ and $\beta$.
Recall that a composition $\gamma$ of a nonnegative integer $n$ is uniquely
determined by the set $D\left( \gamma\right) $ and the number $n$. Thus, if
$\gamma_{1}$ and $\gamma_{2}$ are two compositions of one and the same
nonnegative integer $n$ satisfying $D\left( \gamma_{1}\right) =D\left(
\gamma_{2}\right) $, then%
\begin{equation}
\gamma_{1}=\gamma_{2}. \label{pf.prop.omega.odot.gamma1=2}%
\end{equation}
For every composition $\gamma$, we define a composition $\rho\left(
\gamma\right) $ as follows: Let $n=\left\vert \gamma\right\vert $. Let
$\rho\left( \gamma\right) $ be the unique composition $\beta$ of $n$ which
satisfies $D\left( \beta\right) =\left\{ 1,2,\ldots,n-1\right\} \setminus
D\left( \gamma\right) $. (This is well-defined, because for every subset $T$
of $\left\{ 1,2,\ldots,n-1\right\} $, there exists a unique composition
$\tau$ of $n$ which satisfies $D\left( \tau\right) =T$.) Notice that%
\begin{equation}
\left\vert \rho\left( \gamma\right) \right\vert =\left\vert \gamma
\right\vert \ \ \ \ \ \ \ \ \ \ \text{for any composition }\gamma.
\label{pf.prop.omega.odot.rho1}%
\end{equation}
Also, if $n\in\mathbb{N}$, and if $\gamma$ is a composition of $n$, then%
\begin{equation}
D\left( \rho\left( \gamma\right) \right) =\left\{ 1,2,\ldots,n-1\right\}
\setminus D\left( \gamma\right) \label{pf.prop.omega.odot.c.Drho}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.omega.odot.c.Drho}):} Let
$n\in\mathbb{N}$. Let $\gamma$ be a composition of $n$. Then, $\rho\left(
\gamma\right) $ is the unique composition $\beta$ of $n$ which satisfies
$D\left( \beta\right) =\left\{ 1,2,\ldots,n-1\right\} \setminus D\left(
\gamma\right) $ (because this is how $\rho\left( \gamma\right) $ was
defined). Thus, $\rho\left( \gamma\right) $ is a composition of $n$ which
satisfies $D\left( \rho\left( \gamma\right) \right) =\left\{
1,2,\ldots,n-1\right\} \setminus D\left( \gamma\right) $. This proves
(\ref{pf.prop.omega.odot.c.Drho}).}.
Notice also that%
\begin{equation}
\omega\left( \alpha\right) =\rho\left( \operatorname*{rev}\alpha\right)
\ \ \ \ \ \ \ \ \ \ \text{for any composition }\alpha
\label{pf.prop.omega.odot.omega1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.omega.odot.omega1}):} Let $\alpha$ be
a composition. Let $n=\left\vert \alpha\right\vert $. Thus, $\alpha$ is a
composition of $n$. Hence, $\omega\left( \alpha\right) $ is a composition of
$n$ as well. Also, $\operatorname*{rev}\alpha$ is a composition of $n$. Now,
the definition of $\rho\left( \operatorname*{rev}\alpha\right) $ shows that
$\rho\left( \operatorname*{rev}\alpha\right) $ is the unique composition
$\beta$ of $n$ which satisfies $D\left( \beta\right) =\left\{
1,2,\ldots,n-1\right\} \setminus D\left( \operatorname*{rev}\alpha\right)
$. Hence, $\rho\left( \operatorname*{rev}\alpha\right) $ is a composition of
$n$ and satisfies $D\left( \rho\left( \operatorname*{rev}\alpha\right)
\right) =\left\{ 1,2,\ldots,n-1\right\} \setminus D\left(
\operatorname*{rev}\alpha\right) $.
\par
On the other hand, $\omega\left( \alpha\right) $ is the unique composition
$\beta$ of $n$ which satisfies $D\left( \beta\right) =\left\{
1,2,\ldots,n-1\right\} \setminus D\left( \operatorname*{rev}\alpha\right) $
(by the definition of $\omega\left( \alpha\right) $). Thus, $\omega\left(
\alpha\right) $ is a composition of $n$ and satisfies $D\left( \omega\left(
\alpha\right) \right) =\left\{ 1,2,\ldots,n-1\right\} \setminus D\left(
\operatorname*{rev}\alpha\right) $.
\par
Hence,%
\[
D\left( \rho\left( \operatorname*{rev}\alpha\right) \right) =\left\{
1,2,\ldots,n-1\right\} \setminus D\left( \operatorname*{rev}\alpha\right)
=D\left( \omega\left( \alpha\right) \right) .
\]
Applying (\ref{pf.prop.omega.odot.gamma1=2}) to $\gamma_{1}=\rho\left(
\operatorname*{rev}\alpha\right) $ and $\gamma_{2}=\omega\left(
\alpha\right) $, we therefore obtain $\rho\left( \operatorname*{rev}%
\alpha\right) =\omega\left( \alpha\right) $. Qed.}.
Now, we shall prove that%
\begin{equation}
\rho\left( \left[ \alpha,\beta\right] \right) =\rho\left( \alpha\right)
\odot\rho\left( \beta\right) \label{pf.prop.omega.odot.1}%
\end{equation}
for any two compositions $\alpha$ and $\beta$.
\textit{Proof of (\ref{pf.prop.omega.odot.1}):} Let $\alpha$ and $\beta$ be
two compositions. Let $p=\left\vert \alpha\right\vert $ and $q=\left\vert
\beta\right\vert $; thus, $\alpha$ and $\beta$ are compositions of $p$ and
$q$, respectively. We WLOG assume that both compositions $\alpha$ and $\beta$
are nonempty (since otherwise, (\ref{pf.prop.omega.odot.1}) is fairly
obvious). The composition $\alpha$ is a composition of $p$. Thus, $p > 0$
(since $\alpha$ is nonempty). Similarly, $q > 0$.
Hence, $\left[ \alpha,\beta\right] $ is a composition of $p+q$ satisfying
$D\left( \left[ \alpha,\beta\right] \right) =D\left( \alpha\right)
\cup\left\{ p\right\} \cup\left( D\left( \beta\right) +p\right) $ (by
Lemma \ref{lem.D(a(.)b)} \textbf{(b)}). The definition of $\rho\left( \left[
\alpha,\beta\right] \right) $ thus yields%
\begin{align}
D\left( \rho\left( \left[ \alpha,\beta\right] \right) \right) &
=\left\{ 1,2,\ldots,p+q-1\right\} \setminus\underbrace{D\left( \left[
\alpha,\beta\right] \right) }_{=D\left( \alpha\right) \cup\left\{
p\right\} \cup\left( D\left( \beta\right) +p\right) }\nonumber\\
& =\left\{ 1,2,\ldots,p+q-1\right\} \setminus\left( \left\{ p\right\}
\cup D\left( \alpha\right) \cup\left( D\left( \beta\right) +p\right)
\right) . \label{pf.prop.omega.odot.1.pf.1}%
\end{align}
Applying (\ref{pf.prop.omega.odot.rho1}) to $\gamma=\alpha$, we obtain
$\left\vert \rho\left( \alpha\right) \right\vert =\left\vert \alpha
\right\vert =p$. Thus, $\rho\left( \alpha\right) $ is a composition of $p$.
Similarly, $\rho\left( \beta\right) $ is a composition of $q$. Thus, Lemma
\ref{lem.D(a(.)b)} \textbf{(a)} (applied to $\rho\left( \alpha\right) $ and
$\rho\left( \beta\right) $ instead of $\alpha$ and $\beta$) shows that
$\rho\left( \alpha\right) \odot\rho\left( \beta\right) $ is a composition
of $p+q$ satisfying $D\left( \rho\left( \alpha\right) \odot\rho\left(
\beta\right) \right) =D\left( \rho\left( \alpha\right) \right)
\cup\left( D\left( \rho\left( \beta\right) \right) +p\right) $. Also,
applying (\ref{pf.prop.omega.odot.rho1}) to $\gamma=\left[ \alpha
,\beta\right] $, we obtain $\left| \rho\left( \left[ \alpha,\beta\right]
\right) \right| = \left| \left[ \alpha,\beta\right] \right| = p+q$
(since $\left[ \alpha,\beta\right] $ is a composition of $p+q$). In other
words, $\rho\left( \left[ \alpha,\beta\right] \right) $ is a composition
of $p+q$.
But the definition of $\rho\left( \alpha\right) $ shows that $D\left(
\rho\left( \alpha\right) \right) =\left\{ 1,2,\ldots,p-1\right\}
\setminus D\left( \alpha\right) $. Also, the definition of $\rho\left(
\beta\right) $ shows that $D\left( \rho\left( \beta\right) \right)
=\left\{ 1,2,\ldots,q-1\right\} \setminus D\left( \beta\right) $. Hence,%
\begin{align*}
& \underbrace{D\left( \rho\left( \beta\right) \right) }_{=\left\{
1,2,\ldots,q-1\right\} \setminus D\left( \beta\right) }+p\\
& =\left( \left\{ 1,2,\ldots,q-1\right\} \setminus D\left( \beta\right)
\right) +p\\
& =\underbrace{\left( \left\{ 1,2,\ldots,q-1\right\} +p\right)
}_{=\left\{ p+1,p+2,\ldots,p+q-1\right\} }\setminus\left( D\left(
\beta\right) +p\right) \\
& =\left\{ p+1,p+2,\ldots,p+q-1\right\} \setminus\left( D\left(
\beta\right) +p\right) .
\end{align*}
Also, $D\left( \beta\right) \subseteq\left\{ 1,2,\ldots,q-1\right\} $, so
that $D\left( \beta\right) +p\subseteq\left\{ 1,2,\ldots,q-1\right\}
+p=\left\{ p+1,p+2,\ldots,p+q-1\right\} $.
Now, it is well-known that if $X$, $Y$, $X^{\prime}$ and $Y^{\prime}$ are four
sets such that $X^{\prime}\subseteq X$, $Y^{\prime}\subseteq Y$ and $X\cap
Y=\varnothing$, then
\begin{equation}
\left( X\setminus X^{\prime}\right) \cup\left( Y\setminus Y^{\prime
}\right) =\left( X\cup Y\right) \setminus\left( X^{\prime}\cup Y^{\prime
}\right) . \label{pf.prop.omega.odot.setth}%
\end{equation}
Now,%
\begin{align*}
& D\left( \rho\left( \alpha\right) \odot\rho\left( \beta\right) \right)
\\
& =\underbrace{D\left( \rho\left( \alpha\right) \right) }_{=\left\{
1,2,\ldots,p-1\right\} \setminus D\left( \alpha\right) }\cup
\underbrace{\left( D\left( \rho\left( \beta\right) \right) +p\right)
}_{=\left\{ p+1,p+2,\ldots,p+q-1\right\} \setminus\left( D\left(
\beta\right) +p\right) }\\
& =\left( \left\{ 1,2,\ldots,p-1\right\} \setminus D\left( \alpha\right)
\right) \cup\left( \left\{ p+1,p+2,\ldots,p+q-1\right\} \setminus\left(
D\left( \beta\right) +p\right) \right) \\
& =\underbrace{\left( \left\{ 1,2,\ldots,p-1\right\} \cup\left\{
p+1,p+2,\ldots,p+q-1\right\} \right) }_{=\left\{ 1,2,\ldots,p+q-1\right\}
\setminus\left\{ p\right\} }\\
& \ \ \ \ \ \ \ \ \ \ \setminus\left( D\left( \alpha\right) \cup\left(
D\left( \beta\right) +p\right) \right) \\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{pf.prop.omega.odot.setth}), applied to }X=\left\{
1,2,\ldots,p-1\right\} \text{,}\\
Y=\left\{ p+1,p+2,\ldots,p+q-1\right\} \text{, }X^{\prime}=D\left(
\alpha\right) \text{ and }Y^{\prime}=D\left( \beta\right) +p
\end{array}
\right) \\
& =\left( \left\{ 1,2,\ldots,p+q-1\right\} \setminus\left\{ p\right\}
\right) \setminus\left( D\left( \alpha\right) \cup\left( D\left(
\beta\right) +p\right) \right) \\
& =\left\{ 1,2,\ldots,p+q-1\right\} \setminus\underbrace{\left( \left\{
p\right\} \cup D\left( \alpha\right) \cup\left( D\left( \beta\right)
+p\right) \right) }_{=D\left( \alpha\right) \cup\left\{ p\right\}
\cup\left( D\left( \beta\right) +p\right) }\\
& =\left\{ 1,2,\ldots,p+q-1\right\} \setminus\left( \left\{ p\right\}
\cup D\left( \alpha\right) \cup\left( D\left( \beta\right) +p\right)
\right) \\
& =D\left( \rho\left( \left[ \alpha,\beta\right] \right) \right)
\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.prop.omega.odot.1.pf.1}%
)}\right) .
\end{align*}
Thus, (\ref{pf.prop.omega.odot.gamma1=2}) (applied to $n=p+q$, $\gamma
_{1}=\rho\left( \alpha\right) \odot\rho\left( \beta\right) $ and
$\gamma_{2}=\rho\left( \left[ \alpha,\beta\right] \right) $) shows that
$\rho\left( \alpha\right) \odot\rho\left( \beta\right) =\rho\left(
\left[ \alpha,\beta\right] \right) $. This proves
(\ref{pf.prop.omega.odot.1}).
\textbf{(a)} Let $\alpha$ and $\beta$ be two compositions. Then,
(\ref{pf.prop.omega.odot.omega1}) yields $\omega\left( \alpha\right)
=\rho\left( \operatorname*{rev}\alpha\right) $. Also,
(\ref{pf.prop.omega.odot.omega1}) (applied to $\beta$ instead of $\alpha$)
yields $\omega\left( \beta\right) =\rho\left( \operatorname*{rev}%
\beta\right) $.
From (\ref{pf.prop.omega.odot.omega1}) (applied to $\left[ \alpha
,\beta\right] $ instead of $\alpha$), we obtain%
\begin{align*}
\omega\left( \left[ \alpha,\beta\right] \right) & =\rho\left(
\underbrace{\operatorname*{rev}\left( \left[ \alpha,\beta\right] \right)
}_{\substack{=\left[ \operatorname*{rev}\beta,\operatorname*{rev}%
\alpha\right] \\\text{(by (\ref{pf.prop.omega.odot.rev1}))}}}\right)
=\rho\left( \left[ \operatorname*{rev}\beta,\operatorname*{rev}%
\alpha\right] \right) \\
& =\underbrace{\rho\left( \operatorname*{rev}\beta\right) }_{=\omega\left(
\beta\right) }\odot\underbrace{\rho\left( \operatorname*{rev}\alpha\right)
}_{=\omega\left( \alpha\right) }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{pf.prop.omega.odot.1}), applied to }\operatorname*{rev}\beta\\
\text{and }\operatorname*{rev}\alpha\text{ instead of }\alpha\text{ and }\beta
\end{array}
\right) \\
& =\omega\left( \beta\right) \odot\omega\left( \alpha\right) .
\end{align*}
This proves Proposition \ref{prop.omega.odot} \textbf{(a)}.
\textbf{(c)} First of all, it is clear that%
\begin{equation}
\operatorname*{rev}\left( \operatorname*{rev}\gamma\right) =\gamma
\ \ \ \ \ \ \ \ \ \ \text{for every composition }\gamma.
\label{pf.prop.omega.odot.c.rev}%
\end{equation}
Furthermore,%
\begin{equation}
\rho\left( \rho\left( \gamma\right) \right) =\gamma
\ \ \ \ \ \ \ \ \ \ \text{for every composition }\gamma
\label{pf.prop.omega.odot.c.rho}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.omega.odot.c.rho}):} Let $\gamma$ be
a composition. Let $n=\left\vert \gamma\right\vert $. Thus, $\gamma$ is a
composition of $n$. The definition of $\rho\left( \gamma\right) $ shows that
$\rho\left( \gamma\right) $ is the unique composition $\beta$ of $n$ which
satisfies $D\left( \beta\right) =\left\{ 1,2,\ldots,n-1\right\} \setminus
D\left( \gamma\right) $. Thus, $\rho\left( \gamma\right) $ is a
composition of $n$ and satisfies $D\left( \rho\left( \gamma\right) \right)
=\left\{ 1,2,\ldots,n-1\right\} \setminus D\left( \gamma\right) $.
\par
Therefore, the definition of $\rho\left( \rho\left( \gamma\right) \right)
$ shows that $\rho\left( \rho\left( \gamma\right) \right) $ is the unique
composition $\beta$ of $n$ which satisfies $D\left( \beta\right) =\left\{
1,2,\ldots,n-1\right\} \setminus D\left( \rho\left( \gamma\right) \right)
$. Thus, $\rho\left( \rho\left( \gamma\right) \right) $ is a composition
of $n$ and satisfies $D\left( \rho\left( \rho\left( \gamma\right) \right)
\right) =\left\{ 1,2,\ldots,n-1\right\} \setminus D\left( \rho\left(
\gamma\right) \right) $. Hence,%
\begin{align*}
D\left( \rho\left( \rho\left( \gamma\right) \right) \right) &
=\left\{ 1,2,\ldots,n-1\right\} \setminus\underbrace{D\left( \rho\left(
\gamma\right) \right) }_{=\left\{ 1,2,\ldots,n-1\right\} \setminus
D\left( \gamma\right) }\\
& =\left\{ 1,2,\ldots,n-1\right\} \setminus\left( \left\{ 1,2,\ldots
,n-1\right\} \setminus D\left( \gamma\right) \right) \\
& =D\left( \gamma\right) \ \ \ \ \ \ \ \ \ \ \left( \text{since }D\left(
\gamma\right) \subseteq\left\{ 1,2,\ldots,n-1\right\} \right) .
\end{align*}
Hence, (\ref{pf.prop.omega.odot.gamma1=2}) (applied to $\gamma_{1}=\rho\left(
\rho\left( \gamma\right) \right) $ and $\gamma_{2}=\gamma$) shows that
$\rho\left( \rho\left( \gamma\right) \right) =\gamma$. This proves
(\ref{pf.prop.omega.odot.c.rho}).}.
On the other hand, if $G$ is a set of integers and $r$ is an integer, then we
let $r-G$ denote the set $\left\{ r-g\ \mid\ g\in G\right\} $ of integers.
Then, for any $n\in\mathbb{N}$ and any composition $\gamma$ of $n$, we have%
\begin{equation}
D\left( \operatorname*{rev}\gamma\right) =n-D\left( \gamma\right)
\label{pf.prop.omega.odot.c.Drev}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.omega.odot.c.Drev}):} Let
$n\in\mathbb{N}$. Let $\gamma$ be a composition of $n$. Thus, $\gamma$ is a
composition satisfying $\left\vert \gamma\right\vert =n$.
\par
Write $\gamma$ in the form $\gamma=\left( \gamma_{1},\gamma_{2},\ldots
,\gamma_{\ell}\right) $. Then, $\operatorname*{rev}\gamma=\left(
\gamma_{\ell},\gamma_{\ell-1},\ldots,\gamma_{1}\right) $ (by the definition
of $\operatorname*{rev}\gamma$). Also, from $\gamma=\left( \gamma_{1}%
,\gamma_{2},\ldots,\gamma_{\ell}\right) $, we obtain $\left\vert
\gamma\right\vert =\gamma_{1}+\gamma_{2}+\cdots+\gamma_{\ell}$, whence
$\gamma_{1}+\gamma_{2}+\cdots+\gamma_{\ell}=\left\vert \gamma\right\vert =n$.
Hence, every $i\in\left\{ 1,2,\ldots,\ell-1\right\} $ satisfies%
\begin{align}
n & =\gamma_{1}+\gamma_{2}+\cdots+\gamma_{\ell}=\left( \gamma_{1}%
+\gamma_{2}+\cdots+\gamma_{i}\right) +\underbrace{\left( \gamma_{i+1}%
+\gamma_{i+2}+\cdots+\gamma_{\ell}\right) }_{=\gamma_{\ell}+\gamma_{\ell
-1}+\cdots+\gamma_{i+1}}\nonumber\\
& =\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma_{i}\right) +\left(
\gamma_{\ell}+\gamma_{\ell-1}+\cdots+\gamma_{i+1}\right) .
\label{pf.prop.omega.odot.c.Drev.pf.1}%
\end{align}
Also, $\gamma=\left( \gamma_{1},\gamma_{2},\ldots,\gamma_{\ell}\right) $, so
that the definition of $D\left( \gamma\right) $ yields%
\begin{align}
D\left( \gamma\right) & =\left\{ \gamma_{1},\gamma_{1}+\gamma_{2}%
,\gamma_{1}+\gamma_{2}+\gamma_{3},\ldots,\gamma_{1}+\gamma_{2}+\cdots
+\gamma_{\ell-1}\right\} \nonumber\\
& =\left\{ \gamma_{1}+\gamma_{2}+\cdots+\gamma_{i}\ \mid\ i\in\left\{
1,2,\ldots,\ell-1\right\} \right\} . \label{pf.prop.omega.odot.c.Drev.pf.2}%
\end{align}
\par
But $\operatorname*{rev}\gamma=\left( \gamma_{\ell},\gamma_{\ell-1}%
,\ldots,\gamma_{1}\right) $. Hence, the definition of $D\left(
\operatorname*{rev}\gamma\right) $ yields%
\begin{align*}
D\left( \operatorname*{rev}\gamma\right) & =\left\{ \gamma_{\ell}%
,\gamma_{\ell}+\gamma_{\ell-1},\gamma_{\ell}+\gamma_{\ell-1}+\gamma_{\ell
-2},\ldots,\gamma_{\ell}+\gamma_{\ell-1}+\gamma_{\ell-2}+\cdots+\gamma
_{2}\right\} \\
& =\left\{ \underbrace{\gamma_{\ell}+\gamma_{\ell-1}+\cdots+\gamma_{i+1}%
}_{\substack{=n-\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma_{i}\right)
\\\text{(by (\ref{pf.prop.omega.odot.c.Drev.pf.1}))}}}\ \mid\ i\in\left\{
1,2,\ldots,\ell-1\right\} \right\} \\
& =\left\{ n-\left( \gamma_{1}+\gamma_{2}+\cdots+\gamma_{i}\right)
\ \mid\ i\in\left\{ 1,2,\ldots,\ell-1\right\} \right\} \\
& =n-\underbrace{\left\{ \gamma_{1}+\gamma_{2}+\cdots+\gamma_{i}\ \mid
\ i\in\left\{ 1,2,\ldots,\ell-1\right\} \right\} }_{\substack{=D\left(
\gamma\right) \\\text{(by (\ref{pf.prop.omega.odot.c.Drev.pf.2}))}}}\\
& =n-D\left( \gamma\right) .
\end{align*}
This proves (\ref{pf.prop.omega.odot.c.Drev}).}.
Now,%
\begin{equation}
\rho\left( \operatorname*{rev}\gamma\right) =\operatorname*{rev}\left(
\rho\left( \gamma\right) \right) \ \ \ \ \ \ \ \ \ \ \text{for every
composition }\gamma\label{pf.prop.omega.odot.c.rhorev}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.omega.odot.c.rhorev}):} Let $\gamma$
be a composition. Let $n=\left\vert \gamma\right\vert $. Thus, $\gamma$ is a
composition of $n$.
\par
Now, (\ref{pf.prop.omega.odot.rho1}) (applied to $\operatorname*{rev}\gamma$
instead of $\gamma$) yields $\left\vert \rho\left( \operatorname*{rev}%
\gamma\right) \right\vert =\left\vert \operatorname*{rev}\gamma\right\vert
=\left\vert \gamma\right\vert $ (by (\ref{pf.prop.omega.odot.rev0})). Also,
(\ref{pf.prop.omega.odot.rev0}) (applied to $\rho\left( \gamma\right) $
instead of $\gamma$) yields $\left\vert \operatorname*{rev}\left( \rho\left(
\gamma\right) \right) \right\vert =\left\vert \rho\left( \gamma\right)
\right\vert =\left\vert \gamma\right\vert $ (by (\ref{pf.prop.omega.odot.rho1}%
)). Now, $\left\vert \rho\left( \operatorname*{rev}\gamma\right) \right\vert
=\left\vert \gamma\right\vert =n$, $\left\vert \operatorname*{rev}%
\gamma\right\vert =\left\vert \gamma\right\vert =n$, $\left\vert \rho\left(
\gamma\right) \right\vert =\left\vert \gamma\right\vert =n$ and $\left\vert
\operatorname*{rev}\left( \rho\left( \gamma\right) \right) \right\vert
=\left\vert \gamma\right\vert =n$. Hence, all of $\rho\left(
\operatorname*{rev}\gamma\right) $, $\operatorname*{rev}\gamma$, $\rho\left(
\gamma\right) $ and $\operatorname*{rev}\left( \rho\left( \gamma\right)
\right) $ are compositions of $n$.
\par
Applying (\ref{pf.prop.omega.odot.c.Drho}) to $\operatorname*{rev}\gamma$
instead of $\gamma$, we obtain%
\begin{align*}
D\left( \rho\left( \operatorname*{rev}\gamma\right) \right) &
=\underbrace{\left\{ 1,2,\ldots,n-1\right\} }_{=n-\left\{ 1,2,\ldots
,n-1\right\} }\setminus\underbrace{D\left( \operatorname*{rev}\gamma\right)
}_{\substack{=n-D\left( \gamma\right) \\\text{(by
(\ref{pf.prop.omega.odot.c.Drev}))}}}\\
& =\left( n-\left\{ 1,2,\ldots,n-1\right\} \right) \setminus\left(
n-D\left( \gamma\right) \right) \\
& =n-\underbrace{\left( \left\{ 1,2,\ldots,n-1\right\} \setminus D\left(
\gamma\right) \right) }_{\substack{=D\left( \rho\left( \gamma\right)
\right) \\\text{(by (\ref{pf.prop.omega.odot.c.Drho}))}}}\\
& =n-D\left( \rho\left( \gamma\right) \right) .
\end{align*}
Comparing this with%
\[
D\left( \operatorname*{rev}\left( \rho\left( \gamma\right) \right)
\right) =n-D\left( \rho\left( \gamma\right) \right)
\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.prop.omega.odot.c.Drev}),
applied to }\rho\left( \gamma\right) \text{ instead of }\gamma\right) ,
\]
we obtain $D\left( \rho\left( \operatorname*{rev}\gamma\right) \right)
=D\left( \operatorname*{rev}\left( \rho\left( \gamma\right) \right)
\right) $. Hence, (\ref{pf.prop.omega.odot.gamma1=2}) (applied to $\gamma
_{1}=\rho\left( \operatorname*{rev}\gamma\right) $ and $\gamma
_{2}=\operatorname*{rev}\left( \rho\left( \gamma\right) \right) $) yields
$\rho\left( \operatorname*{rev}\gamma\right) =\operatorname*{rev}\left(
\rho\left( \gamma\right) \right) $. This proves
(\ref{pf.prop.omega.odot.c.rhorev}).}.
Now, let $\gamma$ be a composition. Then, (\ref{pf.prop.omega.odot.omega1})
(applied to $\alpha=\gamma$) yields $\omega\left( \gamma\right) =\rho\left(
\operatorname*{rev}\gamma\right) =\operatorname*{rev}\left( \rho\left(
\gamma\right) \right) $ (by (\ref{pf.prop.omega.odot.c.rhorev})). But
(\ref{pf.prop.omega.odot.omega1}) (applied to $\alpha=\omega\left(
\gamma\right) $) yields%
\begin{align*}
\omega\left( \omega\left( \gamma\right) \right) & =\rho\left(
\operatorname*{rev}\left( \underbrace{\omega\left( \gamma\right)
}_{=\operatorname*{rev}\left( \rho\left( \gamma\right) \right) }\right)
\right) =\rho\left( \underbrace{\operatorname*{rev}\left(
\operatorname*{rev}\left( \rho\left( \gamma\right) \right) \right)
}_{\substack{=\rho\left( \gamma\right) \\\text{(by
(\ref{pf.prop.omega.odot.c.rev}), applied to}\\\rho\left( \gamma\right)
\text{ instead of }\gamma\text{)}}}\right) \\
& =\rho\left( \rho\left( \gamma\right) \right) =\gamma
\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.prop.omega.odot.c.rho})}\right)
.
\end{align*}
This proves Proposition \ref{prop.omega.odot} \textbf{(c)}.
\textbf{(b)} Let $\alpha$ and $\beta$ be two compositions. Then, Proposition
\ref{prop.omega.odot} \textbf{(a)} (applied to $\omega\left( \beta\right) $
and $\omega\left( \alpha\right) $ instead of $\alpha$ and $\beta$) yields%
\begin{align*}
\omega\left( \left[ \omega\left( \beta\right) ,\omega\left(
\alpha\right) \right] \right) & =\underbrace{\omega\left( \omega\left(
\alpha\right) \right) }_{\substack{=\alpha\\\text{(by Proposition
\ref{prop.omega.odot} \textbf{(c)},}\\\text{applied to }\gamma=\alpha\text{)}%
}}\odot\underbrace{\omega\left( \omega\left( \beta\right) \right)
}_{\substack{=\beta\\\text{(by Proposition \ref{prop.omega.odot}
\textbf{(c)},}\\\text{applied to }\gamma=\beta\text{)}}}\\
& =\alpha\odot\beta.
\end{align*}
Hence, $\alpha\odot\beta=\omega\left( \left[ \omega\left( \beta\right)
,\omega\left( \alpha\right) \right] \right) $. Applying the map $\omega$
to both sides of this equality, we conclude that%
\[
\omega\left( \alpha\odot\beta\right) =\omega\left( \omega\left( \left[
\omega\left( \beta\right) ,\omega\left( \alpha\right) \right] \right)
\right) =\left[ \omega\left( \beta\right) ,\omega\left( \alpha\right)
\right]
\]
(by Proposition \ref{prop.omega.odot} \textbf{(c)}, applied to $\gamma=\left[
\omega\left( \beta\right) ,\omega\left( \alpha\right) \right] $). This
proves Proposition \ref{prop.omega.odot} \textbf{(b)}.
\end{proof}
\end{verlong}
The notion of $\omega\left( \alpha\right) $ gives rise to a simple formula
for the antipode $S$ of the Hopf algebra $\operatorname*{QSym}$ in terms of
its fundamental basis:
\begin{proposition}
\label{prop.S.F}Let $\alpha$ be a composition. Then, $S\left( F_{\alpha
}\right) =\left( -1\right) ^{\left\vert \alpha\right\vert }F_{\omega\left(
\alpha\right) }$.
\end{proposition}
This is proven in \cite[Proposition 5.2.15]{Reiner}.
We now state the main result of this note:
\begin{theorem}
\label{thm.hmDless}Let $f\in\operatorname*{QSym}$ and let $m$ be a positive
integer. For any two compositions $\alpha$ and $\beta$, define a composition
$\alpha\odot\beta$ as in Proposition \ref{prop.bel.F}. Then,%
\[
h_{m}\left. \prec\right. f=\sum_{\alpha\in\operatorname*{Comp}}\left(
-1\right) ^{\left\vert \alpha\right\vert }F_{\alpha\odot\left( m\right)
}R_{\omega\left( \alpha\right) }^{\perp}f.
\]
(Here, the sum on the right hand side converges, because all but finitely many
compositions $\alpha$ satisfy $R_{\omega\left( \alpha\right) }^{\perp}f=0$
for degree reasons.)
\end{theorem}
The proof is based on the following simple lemma:
\begin{lemma}
\label{lem.hmDless.lem}Let $a\in\operatorname*{QSym}$ and $f\in
\operatorname*{QSym}$. Then,%
\[
\sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert }\left( F_{\alpha} \bel a\right) R_{\omega\left(
\alpha\right) }^{\perp}f=a\left. \prec\right. f.
\]
\end{lemma}
\begin{proof}
[Proof of Lemma \ref{lem.hmDless.lem}.]The basis $\left( F_{\alpha}\right)
_{\alpha\in\operatorname*{Comp}}$ of $\operatorname*{QSym}$ and the basis
$\left( R_{\alpha}\right) _{\alpha\in\operatorname*{Comp}}$ of
$\operatorname*{NSym}$ are dual bases. Thus,%
\begin{equation}
\sum_{\alpha\in\operatorname*{Comp}}F_{\alpha}\left( R_{\alpha},g\right)
=g\ \ \ \ \ \ \ \ \ \ \text{for every }g\in\operatorname*{QSym}.
\label{pf.lem.hmDless.lem.dualbases}%
\end{equation}
Let us use Sweedler's notation. The map $\operatorname*{Comp}\rightarrow
\operatorname*{Comp},\ \alpha\mapsto\omega\left( \alpha\right) $ is a
bijection (since $\omega\left( \omega\left( \alpha\right) \right) =\alpha$
for any composition $\alpha$). Hence, we can substitute $\omega\left(
\alpha\right) $ for $\alpha$ in the sum $\sum_{\alpha\in\operatorname*{Comp}%
}\left( -1\right) ^{\left\vert \alpha\right\vert }\left( F_{\alpha}
\bel a\right) R_{\omega\left( \alpha\right) }^{\perp}f$. We thus obtain%
\begin{align*}
& \sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert }\left( F_{\alpha} \bel a\right) R_{\omega\left(
\alpha\right) }^{\perp}f\\
& =\sum_{\alpha\in\operatorname*{Comp}}\underbrace{\left( -1\right)
^{\left\vert \omega\left( \alpha\right) \right\vert }}_{\substack{=\left(
-1\right) ^{\left\vert \alpha\right\vert }\\\text{(since }\left\vert
\omega\left( \alpha\right) \right\vert =\left\vert \alpha\right\vert
\text{)}}}\left( F_{\omega\left( \alpha\right) } \bel a\right)
\underbrace{R_{\omega\left( \omega\left( \alpha\right) \right) }^{\perp}%
}_{\substack{=R_{\alpha}^{\perp}\\\text{(since }\omega\left( \omega\left(
\alpha\right) \right) =\alpha\text{)}}}f\\
& =\sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert }\left( F_{\omega\left( \alpha\right) } \bel a\right)
\underbrace{R_{\alpha}^{\perp}f}_{\substack{=\sum_{\left( f\right) }\left(
R_{\alpha},f_{\left( 1\right) }\right) f_{\left( 2\right) }\\\text{(by
(\ref{pf.hmDless.1}))}}}\\
& =\sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert }\left( F_{\omega\left( \alpha\right) } \bel a\right)
\sum_{\left( f\right) }\left( R_{\alpha},f_{\left( 1\right) }\right)
f_{\left( 2\right) }\\
& =\sum_{\left( f\right) }\sum_{\alpha\in\operatorname*{Comp}}\left(
-1\right) ^{\left\vert \alpha\right\vert }\left( F_{\omega\left(
\alpha\right) } \bel a\right) \left( R_{\alpha},f_{\left( 1\right)
}\right) f_{\left( 2\right) }\\
& =\sum_{\left( f\right) }\left( \left( \sum_{\alpha\in
\operatorname*{Comp}}\underbrace{\left( -1\right) ^{\left\vert
\alpha\right\vert }F_{\omega\left( \alpha\right) }}_{\substack{=S\left(
F_{\alpha}\right) \\\text{(by Proposition \ref{prop.S.F})}}}\left(
R_{\alpha},f_{\left( 1\right) }\right) \right) \bel a\right) f_{\left(
2\right) }\\
& =\sum_{\left( f\right) }\left( \left( \sum_{\alpha\in
\operatorname*{Comp}}S\left( F_{\alpha}\right) \left( R_{\alpha},f_{\left(
1\right) }\right) \right) \bel a\right) f_{\left( 2\right) }\\
& =\sum_{\left( f\right) }\left( S\left( \underbrace{\sum_{\alpha
\in\operatorname*{Comp}}F_{\alpha}\left( R_{\alpha},f_{\left( 1\right)
}\right) }_{\substack{_{\substack{=f_{\left( 1\right) }}}\\\text{(by
(\ref{pf.lem.hmDless.lem.dualbases}), applied to }g=f_{\left( 1\right)
}\text{)}}}\right) \bel a\right) f_{\left( 2\right) }=\sum_{\left(
f\right) }\left( S\left( f_{\left( 1\right) }\right) \bel a\right)
f_{\left( 2\right) }=a\left. \prec\right. f
\end{align*}
(by Theorem \ref{thm.beldend}, applied to $b=f$). This proves Lemma
\ref{lem.hmDless.lem}.
\end{proof}
\begin{proof}
[Proof of Theorem \ref{thm.hmDless}.]We have%
\begin{align*}
& \sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert }\underbrace{F_{\alpha\odot\left( m\right) }%
}_{\substack{=F_{\alpha} \bel h_{m}\\\text{(by (\ref{pf.hmDless.2}))}%
}}R_{\omega\left( \alpha\right) }^{\perp}f\\
& =\sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert }\left( F_{\alpha} \bel h_{m}\right) R_{\omega\left(
\alpha\right) }^{\perp}f=h_{m}\left. \prec\right. f
\end{align*}
(by Lemma \ref{lem.hmDless.lem}, applied to $a=h_{m}$). This proves Theorem
\ref{thm.hmDless}.
\end{proof}
As a consequence, we obtain the following result, conjectured by Mike Zabrocki
(private correspondence):
\begin{corollary}
\label{cor.zabrocki}For every positive integer $m$, define a $\mathbf{k}%
$-linear operator $\mathbf{W}_{m}:\operatorname*{QSym}\rightarrow
\operatorname*{QSym}$ by%
\[
\mathbf{W}_{m}=\sum_{\alpha\in\operatorname*{Comp}}\left( -1\right)
^{\left\vert \alpha\right\vert }F_{\alpha\odot\left( m\right) }%
R_{\omega\left( \alpha\right) }^{\perp}%
\]
(where $F_{\alpha\odot\left( m\right) }$ means left multiplication by
$F_{\alpha\odot\left( m\right) }$). Then, every composition $\alpha=\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) $ satisfies%
\[
\mathfrak{S}_{\alpha}^{\ast}=\left( \mathbf{W}_{\alpha_{1}}\circ
\mathbf{W}_{\alpha_{2}}\circ\cdots\circ\mathbf{W}_{\alpha_{\ell}}\right)
\left( 1\right) .
\]
\end{corollary}
\begin{proof}
[Proof of Corollary \ref{cor.zabrocki}.]For every positive integer $m$ and
every $f\in\operatorname*{QSym}$, we have%
\[
\mathbf{W}_{m}f=\sum_{\alpha\in\operatorname*{Comp}}\left( -1\right)
^{\left\vert \alpha\right\vert }F_{\alpha\odot\left( m\right) }%
R_{\omega\left( \alpha\right) }^{\perp}f=h_{m}\left. \prec\right.
f\ \ \ \ \ \ \ \ \ \ \left( \text{by Theorem \ref{thm.hmDless}}\right) .
\]
Hence, by induction, for every composition $\alpha=\left( \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right) $, we have%
\[
\mathbf{W}_{\alpha_{1}}\left( \mathbf{W}_{\alpha_{2}}\left( \cdots\left(
\mathbf{W}_{\alpha_{\ell}}\left( 1\right) \right) \cdots\right) \right)
=h_{\alpha_{1}}\left. \prec\right. \left( h_{\alpha_{2}}\left.
\prec\right. \left( \cdots\left. \prec\right. \left( h_{\alpha_{\ell}%
}\left. \prec\right. 1\right) \cdots\right) \right) =\mathfrak{S}%
_{\alpha}^{\ast}%
\]
(by Corollary \ref{cor.dualImm.dend.explicit}). In other words,%
\[
\mathfrak{S}_{\alpha}^{\ast}=\mathbf{W}_{\alpha_{1}}\left( \mathbf{W}%
_{\alpha_{2}}\left( \cdots\left( \mathbf{W}_{\alpha_{\ell}}\left( 1\right)
\right) \cdots\right) \right) =\left( \mathbf{W}_{\alpha_{1}}%
\circ\mathbf{W}_{\alpha_{2}}\circ\cdots\circ\mathbf{W}_{\alpha_{\ell}}\right)
\left( 1\right) .
\]
This proves Corollary \ref{cor.zabrocki}.
\end{proof}
Let us finish this section with two curiosities: two analogues of Theorem
\ref{thm.hmDless}, one of which can be viewed as an \textquotedblleft$m=0$
version\textquotedblright\ and the other as a \textquotedblleft negative $m$
version\textquotedblright. We begin with the \textquotedblleft$m=0$
one\textquotedblright, as it is the easier one to state:
\begin{proposition}
\label{prop.hmDless.analogue0}Let $f\in\operatorname*{QSym}$. Then,%
\[
\varepsilon\left( f\right) =\sum_{\alpha\in\operatorname*{Comp}}\left(
-1\right) ^{\left\vert \alpha\right\vert }F_{\alpha}R_{\omega\left(
\alpha\right) }^{\perp}f.
\]
\end{proposition}
\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.hmDless.analogue0}.]This proof can be found in
the detailed version of this note; it is similar to the proof of Theorem
\ref{thm.hmDless}.
\end{proof}
\end{vershort}
\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.hmDless.analogue0}.]Let us use Sweedler's
notation. The map
\[
\operatorname*{Comp}\rightarrow\operatorname*{Comp},\ \alpha\mapsto
\omega\left( \alpha\right)
\]
is a bijection (since $\omega\left( \omega\left( \alpha\right) \right)
=\alpha$ for any composition $\alpha$). Hence, we can substitute
$\omega\left( \alpha\right) $ for $\alpha$ in the sum $\sum_{\alpha
\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert \alpha\right\vert
}F_{\alpha}R_{\omega\left( \alpha\right) }^{\perp}f$. We thus obtain%
\begin{align*}
& \sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert }F_{\alpha}R_{\omega\left( \alpha\right) }^{\perp}f\\
& =\sum_{\alpha\in\operatorname*{Comp}}\underbrace{\left( -1\right)
^{\left\vert \omega\left( \alpha\right) \right\vert }}_{\substack{=\left(
-1\right) ^{\left\vert \alpha\right\vert }\\\text{(since }\left\vert
\omega\left( \alpha\right) \right\vert =\left\vert \alpha\right\vert
\text{)}}}F_{\omega\left( \alpha\right) }\underbrace{R_{\omega\left(
\omega\left( \alpha\right) \right) }^{\perp}}_{\substack{=R_{\alpha}%
^{\perp}\\\text{(since }\omega\left( \omega\left( \alpha\right) \right)
=\alpha\text{)}}}f\\
& =\sum_{\alpha\in\operatorname*{Comp}}\underbrace{\left( -1\right)
^{\left\vert \alpha\right\vert }F_{\omega\left( \alpha\right) }%
}_{\substack{=S\left( F_{\alpha}\right) \\\text{(by Proposition
\ref{prop.S.F})}}}\underbrace{R_{\alpha}^{\perp}f}_{\substack{=\sum_{\left(
f\right) }\left( R_{\alpha},f_{\left( 1\right) }\right) f_{\left(
2\right) }\\\text{(by (\ref{pf.hmDless.1}))}}}\\
& =\sum_{\alpha\in\operatorname*{Comp}}S\left( F_{\alpha}\right)
\sum_{\left( f\right) }\left( R_{\alpha},f_{\left( 1\right) }\right)
f_{\left( 2\right) }=\sum_{\alpha\in\operatorname*{Comp}}\sum_{\left(
f\right) }S\left( F_{\alpha}\right) \left( R_{\alpha},f_{\left( 1\right)
}\right) f_{\left( 2\right) }\\
& =\sum_{\left( f\right) }\left( \sum_{\alpha\in\operatorname*{Comp}%
}S\left( F_{\alpha}\right) \left( R_{\alpha},f_{\left( 1\right) }\right)
\right) f_{\left( 2\right) }=\sum_{\left( f\right) }S\left(
\underbrace{\sum_{\alpha\in\operatorname*{Comp}}F_{\alpha}\left( R_{\alpha
},f_{\left( 1\right) }\right) }_{\substack{_{\substack{=f_{\left(
1\right) }}}\\\text{(by (\ref{pf.lem.hmDless.lem.dualbases}), applied to
}g=f_{\left( 1\right) }\text{)}}}\right) f_{\left( 2\right) }\\
& =\sum_{\left( f\right) }S\left( f_{\left( 1\right) }\right)
f_{\left( 2\right) }=\varepsilon\left( f\right)
\end{align*}
(by one of the defining properties of the antipode). This proves Proposition
\ref{prop.hmDless.analogue0}.
\end{proof}
\end{verlong}
The \textquotedblleft negative $m$\textquotedblright\ analogue is less
obvious:\footnote{Proposition \ref{prop.hmDless.analogue-} does not literally
involve a negative $m$, but it involves an element $F_{\alpha}^{\setminus m}$
which can be viewed as \textquotedblleft something like $F_{\left(
\alpha\right) \odot\left( -m\right) }$\textquotedblright.}
\begin{proposition}
\label{prop.hmDless.analogue-}Let $f\in\operatorname*{QSym}$ and let $m$ be a
positive integer. For any composition $\alpha=\left( \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right) $, we define an element $F_{\alpha
}^{\setminus m}$ of $\operatorname*{QSym}$ as follows:
\begin{itemize}
\item If $\ell=0$ or $\alpha_{\ell}m$, then $F_{\alpha}^{\setminus m}=F_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1},\alpha_{\ell}-m\right) }$.
\end{itemize}
(Here, any equality or inequality in which $\alpha_{\ell}$ is mentioned is
understood to include the statement that $\ell>0$.)
Then,%
\[
\left( -1\right) ^{m}\sum_{\alpha\in\operatorname*{Comp}}\left( -1\right)
^{\left\vert \alpha\right\vert }F_{\alpha}^{\setminus m}R_{\omega\left(
\alpha\right) }^{\perp}f=\varepsilon\left( R_{\left( 1^{m}\right) }%
^{\perp}f\right) .
\]
Here, $\left( 1^{m}\right) $ denotes the composition $\left(
\underbrace{1,1,\ldots,1}_{m\text{ times}}\right) $.
\end{proposition}
\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.hmDless.analogue-}.]See the detailed version
of this note.
\end{proof}
\end{vershort}
\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.hmDless.analogue-}.]Let us first make some
auxiliary observations.
Any two elements $a$ and $b$ of $\operatorname*{NSym}$ satisfy
\begin{equation}
\left( ab\right) ^{\perp}=b^{\perp}\circ a^{\perp}
\label{pf.prop.hmDless.analogue-.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.hmDless.analogue-.1}):} Let $a$ and
$b$ be two elements of $\operatorname*{NSym}$. Let $c\in\operatorname*{QSym}$.
Then,%
\begin{align*}
\left( ab\right) ^{\perp}c & =\sum_{\left( c\right) }\underbrace{\left(
ab,c_{\left( 1\right) }\right) }_{\substack{=\sum_{\left( c_{\left(
1\right) }\right) }\left( a,\left( c_{\left( 1\right) }\right)
_{\left( 1\right) }\right) \left( b,\left( c_{\left( 1\right) }\right)
_{\left( 2\right) }\right) \\\text{(by (\ref{eq.(ab,c)}), applied to
}c_{\left( 1\right) }\text{ instead of }c\text{)}}}c_{\left( 2\right)
}\ \ \ \ \ \ \ \ \ \ \left( \text{by (\ref{pf.hmDless.1}), applied to
}g=ab\text{ and }f=c\right) \\
& =\sum_{\left( c\right) }\sum_{\left( c_{\left( 1\right) }\right)
}\left( a,\left( c_{\left( 1\right) }\right) _{\left( 1\right)
}\right) \left( b,\left( c_{\left( 1\right) }\right) _{\left( 2\right)
}\right) c_{\left( 2\right) }=\sum_{\left( c\right) }\sum_{\left(
c_{\left( 2\right) }\right) }\left( a,c_{\left( 1\right) }\right)
\left( b,\left( c_{\left( 2\right) }\right) _{\left( 1\right) }\right)
\left( c_{\left( 2\right) }\right) _{\left( 2\right) }\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the coassociativity of }\Delta\text{ yields}\\
\sum_{\left( c\right) }\sum_{\left( c_{\left( 1\right) }\right) }\left(
c_{\left( 1\right) }\right) _{\left( 1\right) }\otimes\left( c_{\left(
1\right) }\right) _{\left( 2\right) }\otimes c_{\left( 2\right) }%
=\sum_{\left( c\right) }\sum_{\left( c_{\left( 2\right) }\right)
}c_{\left( 1\right) }\otimes\left( c_{\left( 2\right) }\right) _{\left(
1\right) }\otimes\left( c_{\left( 2\right) }\right) _{\left( 2\right) }%
\end{array}
\right) \\
& =\sum_{\left( c\right) }\left( a,c_{\left( 1\right) }\right)
\sum_{\left( c_{\left( 2\right) }\right) }\left( b,\left( c_{\left(
2\right) }\right) _{\left( 1\right) }\right) \left( c_{\left( 2\right)
}\right) _{\left( 2\right) }.
\end{align*}
Compared with%
\begin{align*}
\left( b^{\perp}\circ a^{\perp}\right) \left( c\right) & =b^{\perp
}\left( \underbrace{a^{\perp}c}_{\substack{=\sum_{\left( c\right) }\left(
a,c_{\left( 1\right) }\right) c_{\left( 2\right) }\\\text{(by
(\ref{pf.hmDless.1}), applied to }g=a\text{ and }f=c\text{)}}}\right)
=b^{\perp}\left( \sum_{\left( c\right) }\left( a,c_{\left( 1\right)
}\right) c_{\left( 2\right) }\right) \\
& =\sum_{\left( c\right) }\left( a,c_{\left( 1\right) }\right)
\underbrace{b^{\perp}\left( c_{\left( 2\right) }\right) }_{\substack{=\sum
_{\left( c_{\left( 2\right) }\right) }\left( b,\left( c_{\left(
2\right) }\right) _{\left( 1\right) }\right) \left( c_{\left( 2\right)
}\right) _{\left( 2\right) }\\\text{(by (\ref{pf.hmDless.1}), applied to
}g=b\text{ and }f=c_{\left( 2\right) }\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the map }b^{\perp}\text{ is }\mathbf{k}\text{-linear}\right) \\
& =\sum_{\left( c\right) }\left( a,c_{\left( 1\right) }\right)
\sum_{\left( c_{\left( 2\right) }\right) }\left( b,\left( c_{\left(
2\right) }\right) _{\left( 1\right) }\right) \left( c_{\left( 2\right)
}\right) _{\left( 2\right) },
\end{align*}
this yields $\left( ab\right) ^{\perp}c=\left( b^{\perp}\circ a^{\perp
}\right) \left( c\right) $.
\par
Now, let us forget that we fixed $c$. We thus have shown that $\left(
ab\right) ^{\perp}c=\left( b^{\perp}\circ a^{\perp}\right) \left(
c\right) $ for every $c\in\operatorname*{QSym}$. In other words, $\left(
ab\right) ^{\perp}=b^{\perp}\circ a^{\perp}$. This proves
(\ref{pf.prop.hmDless.analogue-.1}).}.
For every two compositions $\alpha$ and $\beta$, we define a composition
$\left[ \alpha,\beta\right] $ by $\left[ \alpha,\beta\right] =\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell},\beta_{1},\beta_{2},\ldots
,\beta_{m}\right) $, where $\alpha$ and $\beta$ are written as $\alpha
=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) $ and
$\beta=\left( \beta_{1},\beta_{2},\ldots,\beta_{m}\right) $. We further
define a composition $\alpha\odot\beta$ as in Proposition \ref{prop.bel.F}.
Then, every two nonempty compositions $\alpha$ and $\beta$ satisfy%
\begin{equation}
R_{\alpha}R_{\beta}=R_{\left[ \alpha,\beta\right] }+R_{\alpha\odot\beta}.
\label{pf.prop.hmDless.analogue-.2}%
\end{equation}
(This is part of \cite[Theorem 5.4.10(c)]{Reiner}.) Now it is easy to see that%
\begin{equation}
R_{\omega\left( \left[ \alpha,\left( m\right) \right] \right)
}+R_{\omega\left( \alpha\odot\left( m\right) \right) }=R_{\left(
1^{m}\right) }R_{\omega\left( \alpha\right) }
\label{pf.prop.hmDless.analogue-.3}%
\end{equation}
for every nonempty composition $\alpha$\ \ \ \ \footnote{\textit{Proof of
(\ref{pf.prop.hmDless.analogue-.3}):} Let $\alpha$ be a nonempty composition.
Proposition \ref{prop.omega.odot} \textbf{(a)} shows that $\omega\left(
\left[ \alpha,\beta\right] \right) =\omega\left( \beta\right) \odot
\omega\left( \alpha\right) $ for every nonempty composition $\beta$.
Applying this to $\beta=\left( m\right) $, we obtain $\omega\left( \left[
\alpha,\left( m\right) \right] \right) =\underbrace{\omega\left( \left(
m\right) \right) }_{=\left( 1^{m}\right) }\odot\omega\left(
\alpha\right) =\left( 1^{m}\right) \odot\omega\left( \alpha\right) $. But
Proposition \ref{prop.omega.odot}\textbf{(b)} shows that $\omega\left(
\alpha\odot\beta\right) =\left[ \omega\left( \beta\right) ,\omega\left(
\alpha\right) \right] $ for every nonempty composition $\beta$. Applying
this to $\beta=\left( m\right) $, we obtain $\omega\left( \alpha
\odot\left( m\right) \right) =\left[ \underbrace{\omega\left( \left(
m\right) \right) }_{=\left( 1^{m}\right) },\omega\left( \alpha\right)
\right] =\left[ \left( 1^{m}\right) ,\omega\left( \alpha\right) \right]
$. Now,%
\begin{align*}
R_{\omega\left( \left[ \alpha,\left( m\right) \right] \right)
}+R_{\omega\left( \alpha\odot\left( m\right) \right) } & =R_{\omega
\left( \alpha\odot\left( m\right) \right) }+R_{\omega\left( \left[
\alpha,\left( m\right) \right] \right) }=R_{\left[ \left( 1^{m}\right)
,\omega\left( \alpha\right) \right] }+R_{\left( 1^{m}\right) \odot
\omega\left( \alpha\right) }\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since }\omega\left( \alpha\odot\left(
m\right) \right) =\left[ \left( 1^{m}\right) ,\omega\left(
\alpha\right) \right] \text{ and }\omega\left( \left[ \alpha,\left(
m\right) \right] \right) =\left( 1^{m}\right) \odot\omega\left(
\alpha\right) \right) \\
& =R_{\left( 1^{m}\right) }R_{\omega\left( \alpha\right) }%
\end{align*}
(since (\ref{pf.prop.hmDless.analogue-.2}) (applied to $\left( 1^{m}\right)
$ and $\omega\left( \alpha\right) $ instead of $\alpha$ and $\beta$) shows
that $R_{\left( 1^{m}\right) }R_{\omega\left( \alpha\right) }=R_{\left[
\left( 1^{m}\right) ,\omega\left( \alpha\right) \right] }+R_{\left(
1^{m}\right) \odot\omega\left( \alpha\right) }$). This proves
(\ref{pf.prop.hmDless.analogue-.3}).}. Hence, for every nonempty composition
$\alpha$, we have%
\begin{equation}
\left( \underbrace{R_{\omega\left( \left[ \alpha,\left( m\right) \right]
\right) }+R_{\omega\left( \alpha\odot\left( m\right) \right) }%
}_{=R_{\left( 1^{m}\right) }R_{\omega\left( \alpha\right) }}\right)
^{\perp}=\left( R_{\left( 1^{m}\right) }R_{\omega\left( \alpha\right)
}\right) ^{\perp}=R_{\omega\left( \alpha\right) }^{\perp}\circ R_{\left(
1^{m}\right) }^{\perp} \label{pf.prop.hmDless.analogue-.4}%
\end{equation}
(by (\ref{pf.prop.hmDless.analogue-.1}), applied to $a=R_{\left(
1^{m}\right) }$ and $b=R_{\omega\left( \alpha\right) }$).
We furthermore notice that $\omega\left( \varnothing\right) =\varnothing$
and thus $R_{\omega\left( \varnothing\right) }^{\perp}=R_{\varnothing
}^{\perp}=\operatorname*{id}$ (since $R_{\varnothing}=1$).
Now,%
\begin{align}
& \sum_{\substack{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp};\\\alpha_{\ell}=m}}\underbrace{\left( -1\right)
^{\left\vert \left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\right\vert }}_{\substack{=\left( -1\right) ^{\left\vert \left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell-1},m\right) \right\vert }\\\text{(since
}\alpha_{\ell}=m\text{)}}}\underbrace{F_{\left( \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right) }^{\setminus m}}_{\substack{=F_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right) }\\\text{(since }%
\alpha_{\ell}=m\text{)}}}\underbrace{R_{\omega\left( \left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \right) }^{\perp}}%
_{\substack{=R_{\omega\left( \left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1},m\right) \right) }^{\perp}\\\text{(since }\alpha_{\ell
}=m\text{)}}}f\nonumber\\
& =\underbrace{\sum_{\substack{\left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right) \in\operatorname*{Comp};\\\alpha_{\ell}=m}}}%
_{=\sum_{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)
\in\operatorname*{Comp}}}\underbrace{\left( -1\right) ^{\left\vert \left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1},m\right) \right\vert }%
}_{=\left( -1\right) ^{\left\vert \left( \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell-1}\right) \right\vert +m}}F_{\left( \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell-1}\right) }\underbrace{R_{\omega\left(
\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1},m\right) \right)
}^{\perp}}_{\substack{=R_{\omega\left( \left[ \left( \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell-1}\right) ,\left( m\right) \right] \right)
}^{\perp}\\\text{(since }\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1},m\right) \\=\left[ \left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell
-1}\right) ,\left( m\right) \right] \text{)}}}f\nonumber\\
& =\sum_{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right)
\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert \left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right) \right\vert +m}F_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right) }R_{\omega\left( \left[
\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right) ,\left(
m\right) \right] \right) }^{\perp}f\nonumber\\
& =\sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert +m}F_{\alpha}R_{\omega\left( \left[ \alpha,\left(
m\right) \right] \right) }^{\perp}f\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{here, we have substituted }\alpha\text{
for }\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1}\right) \text{ in
the sum}\right) \nonumber\\
& =\left( -1\right) ^{\left\vert \varnothing\right\vert +m}F_{\varnothing
}R_{\omega\left( \left[ \varnothing,\left( m\right) \right] \right)
}^{\perp}f+\sum_{\substack{\alpha\in\operatorname*{Comp};\\\alpha\text{ is
nonempty}}}\left( -1\right) ^{\left\vert \alpha\right\vert +m}F_{\alpha
}R_{\omega\left( \left[ \alpha,\left( m\right) \right] \right) }^{\perp
}f \label{pf.prop.hmDless.analogue-.11}%
\end{align}
(here, we have split off the addend for $\alpha=\varnothing$ from the sum). On
the other hand,%
\begin{align}
& \sum_{\substack{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp};\\\alpha_{\ell}>m}}\left( -1\right) ^{\left\vert
\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \right\vert
}\underbrace{F_{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
}^{\setminus m}}_{\substack{=F_{\left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell-1},\alpha_{\ell}-m\right) }\\\text{(since }\alpha_{\ell
}>m\text{)}}}R_{\omega\left( \left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right) \right) }^{\perp}f\nonumber\\
& =\sum_{\substack{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp};\\\alpha_{\ell}>m}}\left( -1\right) ^{\left\vert
\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \right\vert
}F_{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1},\alpha_{\ell
}-m\right) }R_{\omega\left( \left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right) \right) }^{\perp}f\nonumber\\
& =\sum_{\substack{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp};\\\ell>0}}\underbrace{\left( -1\right) ^{\left\vert
\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1},\alpha_{\ell}+m\right)
\right\vert }}_{=\left( -1\right) ^{\left\vert \left( \alpha_{1},\alpha
_{2},\ldots,\alpha_{\ell}\right) \right\vert +m}}F_{\left( \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right) }\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \underbrace{R_{\omega\left( \left( \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell-1},\alpha_{\ell}+m\right) \right) }^{\perp}%
}_{\substack{=R_{\omega\left( \left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right) \odot\left( m\right) \right) }^{\perp}\\\text{(since
}\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1},\alpha_{\ell}+m\right)
=\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \odot\left(
m\right) \text{)}}}f\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }\left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right) \\
\text{for }\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell-1},\alpha_{\ell
}-m\right) \text{ in the sum}%
\end{array}
\right) \nonumber\\
& =\sum_{\substack{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp};\\\ell>0}}\left( -1\right) ^{\left\vert \left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \right\vert +m}F_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) }R_{\omega\left( \left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \odot\left( m\right)
\right) }^{\perp}f\nonumber\\
& =\sum_{\substack{\alpha\in\operatorname*{Comp};\\\alpha\text{ is nonempty}%
}}\left( -1\right) ^{\left\vert \alpha\right\vert +m}F_{\alpha}%
R_{\omega\left( \alpha\odot\left( m\right) \right) }^{\perp}f
\label{pf.prop.hmDless.analogue-.16}%
\end{align}
(here, we have substituted $\alpha$ for $\left( \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{\ell}\right) $ in the sum).
But%
\begin{align*}
& \sum_{\alpha\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert
\alpha\right\vert }F_{\alpha}^{\setminus m}R_{\omega\left( \alpha\right)
}^{\perp}f\\
& =\sum_{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp}}\left( -1\right) ^{\left\vert \left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \right\vert }F_{\left(
\alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) }^{\setminus m}%
R_{\omega\left( \left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\right) }^{\perp}f\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{here, we have renamed the summation index
}\alpha\text{ as }\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\right) \\
& =\sum_{\substack{\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right)
\in\operatorname*{Comp};\\\ell=0\text{ or }\alpha_{\ell}m}}\left( -1\right) ^{\left\vert \left( \alpha_{1}%
,\alpha_{2},\ldots,\alpha_{\ell}\right) \right\vert }F_{\left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) }^{\setminus m}R_{\omega\left(
\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \right) }^{\perp
}f}_{\substack{=\sum_{\substack{\alpha\in\operatorname*{Comp};\\\alpha\text{
is nonempty}}}\left( -1\right) ^{\left\vert \alpha\right\vert +m}F_{\alpha
}R_{\omega\left( \alpha\odot\left( m\right) \right) }^{\perp}f\\\text{(by
(\ref{pf.prop.hmDless.analogue-.16}))}}}\\
& =\underbrace{\sum_{\substack{\left( \alpha_{1},\alpha_{2},\ldots
,\alpha_{\ell}\right) \in\operatorname*{Comp};\\\ell=0\text{ or }\alpha
_{\ell}0$. For any two
words $u=\left( X_{i_{1}},X_{i_{2}},\ldots,X_{i_{n}}\right) $ and $v=\left(
X_{j_{1}},X_{j_{2}},\ldots,X_{j_{m}}\right) $, we define the concatenation
$uv$ as the word \newline$\left( X_{i_{1}},X_{i_{2}},\ldots,X_{i_{n}%
},X_{j_{1}},X_{j_{2}},\ldots,X_{j_{m}}\right) $. Concatenation is an
associative operation and the empty word $1$ is a neutral element for it;
thus, the words form a monoid. We let $\operatorname*{Wrd}$ denote this
monoid. This monoid is the free monoid on the set $\left\{ X_{1},X_{2}%
,X_{3},\ldots\right\} $. Concatenation allows us to rewrite any word $\left(
X_{i_{1}},X_{i_{2}},\ldots,X_{i_{n}}\right) $ in the shorter form $X_{i_{1}%
}X_{i_{2}}\cdots X_{i_{n}}$.
Notice that $\operatorname*{Mon}$ (the set of all monomials) is also a monoid
under multiplication. We can thus define a monoid homomorphism $\pi
:\operatorname*{Wrd}\rightarrow\operatorname*{Mon}$ by $\pi\left(
X_{i}\right) =x_{i}$ for all $i\in\left\{ 1,2,3,\ldots\right\} $. This
homomorphism $\pi$ is surjective.
We define $\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle
\right\rangle $ to be the $\mathbf{k}$-module $\mathbf{k}^{\operatorname*{Wrd}%
}$; its elements are all families $\left( \lambda_{w}\right) _{w\in
\operatorname*{Wrd}}\in\mathbf{k}^{\operatorname*{Wrd}}$. We define a
multiplication on\textbf{ }$\mathbf{k}\left\langle \left\langle \mathbf{X}%
\right\rangle \right\rangle $ by
\begin{equation}
\left( \lambda_{w}\right) _{w\in\operatorname*{Wrd}}\cdot\left( \mu
_{w}\right) _{w\in\operatorname*{Wrd}}=\left( \sum_{\left( u,v\right)
\in\operatorname*{Wrd}\nolimits^{2};\ uv=w}\lambda_{u}\mu_{v}\right)
_{w\in\operatorname*{Wrd}}. \label{eq.WQSym.powerseries-mul}%
\end{equation}
This makes $\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle
\right\rangle $ into a $\mathbf{k}$-algebra, with unity $\left( \delta
_{w,1}\right) _{w\in\operatorname*{Wrd}}$. This $\mathbf{k}$-algebra is
called the $\mathbf{k}$\textit{-algebra of noncommutative power series in
}$X_{1},X_{2},X_{3},\ldots$. For every $u\in\operatorname*{Wrd}$, we identify
the word $u$ with the element $\left( \delta_{w,u}\right) _{w\in
\operatorname*{Wrd}}$ of $\mathbf{k}\left\langle \left\langle \mathbf{X}%
\right\rangle \right\rangle $\ \ \ \ \footnote{This identification is
harmless, since the map $\operatorname*{Wrd}\rightarrow\mathbf{k}\left\langle
\left\langle \mathbf{X}\right\rangle \right\rangle ,\ u\mapsto\left(
\delta_{w,u}\right) _{w\in\operatorname*{Wrd}}$ is a monoid homomorphism from
$\operatorname*{Wrd}$ to $\left( \mathbf{k}\left\langle \left\langle
\mathbf{X}\right\rangle \right\rangle ,\cdot\right) $. (However, it fails to
be injective if $\mathbf{k}=0$.)}. The $\mathbf{k}$-algebra $\mathbf{k}%
\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle $ becomes a
topological $\mathbf{k}$-algebra via the product topology (recalling that
$\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle
=\mathbf{k}^{\operatorname*{Wrd}}$ as sets). Thus, every element $\left(
\lambda_{w}\right) _{w\in\operatorname*{Wrd}}$ of $\mathbf{k}\left\langle
\left\langle \mathbf{X}\right\rangle \right\rangle $ can be rewritten in the
form $\sum_{w\in\operatorname*{Wrd}}\lambda_{w}w$. This turns the equality
(\ref{eq.WQSym.powerseries-mul}) into a distributive law (for infinite sums),
and explains why we refer to elements of $\mathbf{k}\left\langle \left\langle
\mathbf{X}\right\rangle \right\rangle $ as \textquotedblleft noncommutative
power series\textquotedblright. We think of words as noncommutative analogues
of monomials.
The \textit{degree} of a word $w$ will mean its length (i.e., the integer $n$
for which $w$ is an $n$-tuple). Let $\mathbf{k}\left\langle \left\langle
\mathbf{X}\right\rangle \right\rangle _{\operatorname*{bdd}}$ denote the
$\mathbf{k}$-subalgebra of $\mathbf{k}\left\langle \left\langle \mathbf{X}%
\right\rangle \right\rangle $ formed by the \textit{bounded-degree
noncommutative power series}\footnote{A noncommutative power series $\left(
\lambda_{w}\right) _{w\in\operatorname*{Wrd}}\in\mathbf{k}\left\langle
\left\langle \mathbf{X}\right\rangle \right\rangle $ is said to be
\textit{bounded-degree} if there is an $N\in\mathbb{N}$ such that every word
$w$ of length $>N$ satisfies $\lambda_{w}=0$.} in $\mathbf{k}\left\langle
\left\langle \mathbf{X}\right\rangle \right\rangle $. The surjective monoid
homomorphism $\pi:\operatorname*{Wrd}\rightarrow\operatorname*{Mon}$
canonically gives rise to surjective $\mathbf{k}$-algebra homomorphisms
$\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle
\rightarrow\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right]
$ and $\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle
\right\rangle _{\operatorname*{bdd}}\rightarrow\mathbf{k}\left[ \left[
x_{1},x_{2},x_{3},\ldots\right] \right] _{\operatorname*{bdd}}$, which we
also denote by $\pi$. Notice that the $\mathbf{k}$-algebra $\mathbf{k}%
\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle
_{\operatorname*{bdd}}$ is denoted $R\left\langle \mathbf{X}\right\rangle $ in
\cite[Section 8.1]{Reiner}.
If $w$ is a word, then we denote by $\operatorname*{Supp}w$ the subset%
\[
\left\{ i\in\left\{ 1,2,3,\ldots\right\} \ \mid\ \text{the symbol }%
X_{i}\text{ is an entry of }w\right\}
\]
of $\left\{ 1,2,3,\ldots\right\} $. Notice that $\operatorname*{Supp}%
w=\operatorname*{Supp}\left( \pi\left( w\right) \right) $ is a finite set.
A word $w$ is said to be \textit{packed} if there exists an $\ell\in
\mathbb{N}$ such that $\operatorname*{Supp}w=\left\{ 1,2,\ldots,\ell\right\}
$.
For each word $w$, we define a packed word $\operatorname*{pack}w$ as follows:
Replace the smallest letter\footnote{We use the total ordering on the set
$\left\{ X_{1},X_{2},X_{3},\ldots\right\} $ given by $X_{1}\min\left(
\operatorname*{Supp}v\right) ;\\
0, & \text{if }\min\left( \operatorname*{Supp}u\right) \leq\min\left(
\operatorname*{Supp}v\right)
\end{cases}
\]
for any two words $u$ and $v$.
\textbf{(d)} We define a binary operation $\bel :\mathbf{k}\left\langle
\left\langle \mathbf{X}\right\rangle \right\rangle \times\mathbf{k}%
\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle \rightarrow
\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle $
(written in infix notation) by the requirements that it be $\mathbf{k}%
$-bilinear and continuous with respect to the topology on $\mathbf{k}%
\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle $ and that it
satisfy%
\[
u \bel v=
\begin{cases}
uv, & \text{if }\max\left( \operatorname*{Supp}u\right) \leq\min\left(
\operatorname*{Supp}v\right) ;\\
0, & \text{if }\max\left( \operatorname*{Supp}u\right) >\min\left(
\operatorname*{Supp}v\right)
\end{cases}
\]
for any two words $u$ and $v$.
\textbf{(e)} We define a binary operation $\tvi :\mathbf{k}\left\langle
\left\langle \mathbf{X}\right\rangle \right\rangle \times\mathbf{k}%
\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle \rightarrow
\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle $
(written in infix notation) by the requirements that it be $\mathbf{k}%
$-bilinear and continuous with respect to the topology on $\mathbf{k}%
\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle $ and that it
satisfy%
\[
u \tvi v=
\begin{cases}
uv, & \text{if }\max\left( \operatorname*{Supp}u\right) <\min\left(
\operatorname*{Supp}v\right) ;\\
0, & \text{if }\max\left( \operatorname*{Supp}u\right) \geq\min\left(
\operatorname*{Supp}v\right)
\end{cases}
\]
for any two words $u$ and $v$.
\end{definition}
The first three of these five operations are closely related to those defined
by Novelli and Thibon in \cite{NoThi05}; the main difference is the use of
minima instead of maxima in our definitions.
The operations $\left. \prec\right. $, $\bel $ and $\tvi $ on
$\operatorname*{WQSym}$ lift the operations $\left. \prec\right. $, $\bel $
and $\tvi $ on $\operatorname*{QSym}$. More precisely, any $a\in
\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle $
and $b\in\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle
\right\rangle $ satisfy%
\begin{align*}
\pi\left( a\right) \left. \prec\right. \pi\left( b\right) &
=\pi\left( a\left. \prec\right. b\right) =\pi\left( b\left.
\succ\right. a\right) ;\\
\pi\left( a\right) \bel \pi\left( b\right) & =\pi\left( a
\bel b\right) ;\\
\pi\left( a\right) \tvi \pi\left( b\right) & =\pi\left( a
\tvi b\right)
\end{align*}
(and similar formulas would hold for $\circ$ and $\left. \succ\right. $ had
we bothered to define such operations on $\operatorname*{QSym}$). Also, using
the operation $\left. \succeq\right. $ defined in Remark \ref{rmk.dendri},
we have%
\[
\pi\left( a\right) \left. \succeq\right. \pi\left( b\right) =\pi\left(
a\left. \succ\right. b+a\circ b\right) \ \ \ \ \ \ \ \ \ \ \text{for any
}a\in\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle
\text{ and }b\in\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle
\right\rangle .
\]
We now have the following analogue of Proposition \ref{prop.QSym.closed}:
\begin{proposition}
\label{prop.WQSym.closed}Every $a\in\operatorname*{WQSym}$ and $b\in
\operatorname*{WQSym}$ satisfy $a\left. \prec\right. b\in
\operatorname*{WQSym}$, $a\circ b\in\operatorname*{WQSym}$, $a\left.
\succ\right. b\in\operatorname*{WQSym}$, $a \bel b\in\operatorname*{WQSym}$
and $a \tvi b\in\operatorname*{WQSym}$.
\end{proposition}
The proof of Proposition \ref{prop.WQSym.closed} is easier than that of
Proposition \ref{prop.QSym.closed}; we omit it here. In analogy to Remark
\ref{rmk.QSym.closed} and to (\ref{eq.WQSym.prod}), let us give explicit
formulas for these five operations on the basis $\left( \mathbf{M}%
_{u}\right) _{u\text{ is a packed word}}$ of $\operatorname*{WQSym}$:
\begin{remark}
\label{rmk.WQSym.closed}Let $u$ and $v$ be two packed words. Let $\ell$ be the
length of $u$. Then:
\textbf{(a)} We have%
\[
\mathbf{M}_{u}\left. \prec\right. \mathbf{M}_{v}=\sum_{\substack{w\text{ is
a packed word;}\\\operatorname*{pack}\left( w\left[ :\ell\right] \right)
=u;\ \operatorname*{pack}\left( w\left[ \ell:\right] \right)
=v;\\\min\left( \operatorname*{Supp}\left( w\left[ :\ell\right] \right)
\right) <\min\left( \operatorname*{Supp}\left( w\left[ \ell:\right]
\right) \right) }}\mathbf{M}_{w}.
\]
\textbf{(b)} We have%
\[
\mathbf{M}_{u}\circ\mathbf{M}_{v}=\sum_{\substack{w\text{ is a packed
word;}\\\operatorname*{pack}\left( w\left[ :\ell\right] \right)
=u;\ \operatorname*{pack}\left( w\left[ \ell:\right] \right)
=v;\\\min\left( \operatorname*{Supp}\left( w\left[ :\ell\right] \right)
\right) =\min\left( \operatorname*{Supp}\left( w\left[ \ell:\right]
\right) \right) }}\mathbf{M}_{w}.
\]
\textbf{(c)} We have%
\[
\mathbf{M}_{u}\left. \succ\right. \mathbf{M}_{v}=\sum_{\substack{w\text{ is
a packed word;}\\\operatorname*{pack}\left( w\left[ :\ell\right] \right)
=u;\ \operatorname*{pack}\left( w\left[ \ell:\right] \right)
=v;\\\min\left( \operatorname*{Supp}\left( w\left[ :\ell\right] \right)
\right) >\min\left( \operatorname*{Supp}\left( w\left[ \ell:\right]
\right) \right) }}\mathbf{M}_{w}.
\]
\textbf{(d)} We have%
\[
\mathbf{M}_{u} \bel \mathbf{M}_{v}=\sum_{\substack{w\text{ is a packed
word;}\\\operatorname*{pack}\left( w\left[ :\ell\right] \right)
=u;\ \operatorname*{pack}\left( w\left[ \ell:\right] \right)
=v;\\\max\left( \operatorname*{Supp}\left( w\left[ :\ell\right] \right)
\right) \leq\min\left( \operatorname*{Supp}\left( w\left[ \ell:\right]
\right) \right) }}\mathbf{M}_{w}.
\]
The sum on the right hand side consists of two addends (unless $u$ or $v$ is
empty), namely $\mathbf{M}_{uv^{+h-1}}$ and $\mathbf{M}_{uv^{+h}}$, where
$h=\max\left( \operatorname*{Supp}u\right) $, and where $v^{+j}$ denotes the
word obtained by replacing every letter $X_{k}$ in $v$ by $X_{k+j}$.
\textbf{(e)} We have%
\[
\mathbf{M}_{u} \tvi \mathbf{M}_{v}=\sum_{\substack{w\text{ is a packed
word;}\\\operatorname*{pack}\left( w\left[ :\ell\right] \right)
=u;\ \operatorname*{pack}\left( w\left[ \ell:\right] \right)
=v;\\\max\left( \operatorname*{Supp}\left( w\left[ :\ell\right] \right)
\right) <\min\left( \operatorname*{Supp}\left( w\left[ \ell:\right]
\right) \right) }}\mathbf{M}_{w}.
\]
The sum on the right hand side consists of one addend only, namely
$\mathbf{M}_{uv^{+h}}$.
\end{remark}
Let us now move on to the combinatorial Hopf algebra $\operatorname*{FQSym}$,
which is known as the \textit{Malvenuto-Reutenauer Hopf algebra} or the
\textit{Hopf algebra of free quasi-symmetric functions}. We shall define it as
a Hopf subalgebra of $\operatorname*{WQSym}$. This is not identical to the
definition in \cite[Section 8.1]{Reiner}, but equivalent to it.
For every $n\in\mathbb{N}$, we let $\mathfrak{S}_{n}$ be the symmetric group
on the set $\left\{ 1,2,\ldots,n\right\} $. (This notation is identical with
that in \cite{Reiner}. It has nothing to do with the $\mathfrak{S}_{\alpha}$
from \cite{BBSSZ}.) We let $\mathfrak{S}$ denote the disjoint union
$\bigsqcup_{n\in\mathbb{N}}\mathfrak{S}_{n}$. We identify permutations in
$\mathfrak{S}$ with certain words -- namely, every permutation $\pi
\in\mathfrak{S}$ is identified with the word $\left( X_{\pi\left( 1\right)
},X_{\pi\left( 2\right) },\ldots,X_{\pi\left( n\right) }\right) $, where
$n$ is such that $\pi\in\mathfrak{S}_{n}$. The words thus identified with
permutations in $\mathfrak{S}$ are precisely the packed words which do not
have repeated elements.
For every word $w$, we define a word $\operatorname*{std}w\in\mathfrak{S}$ as
follows: Write $w$ in the form $\left( X_{i_{1}},X_{i_{2}},\ldots,X_{i_{n}%
}\right) $. Then, $\operatorname*{std}w$ shall be the unique permutation
$\pi\in\mathfrak{S}_{n}$ such that, whenever $u$ and $v$ are two elements of
$\left\{ 1,2,\ldots,n\right\} $ satisfying $u\min\left( \operatorname*{Supp}\left( \pi\left[
\ell:\right] \right) \right) }}\mathbf{G}_{\pi}.
\]
\textbf{(b)} We have%
\[
\mathbf{G}_{\sigma} \bel \mathbf{G}_{\tau}=\sum_{\substack{\pi\in
\mathfrak{S};\\\operatorname*{std}\left( \pi\left[ :\ell\right] \right)
=\sigma;\ \operatorname*{std}\left( \pi\left[ \ell:\right] \right)
=\tau;\\\max\left( \operatorname*{Supp}\left( \pi\left[ :\ell\right]
\right) \right) \leq\min\left( \operatorname*{Supp}\left( \pi\left[
\ell:\right] \right) \right) }}\mathbf{G}_{\pi}.
\]
The sum on the right hand side consists of one addend only, namely
$\mathbf{G}_{\sigma\tau^{+\ell}}$.
\end{remark}
The statements of Remark \ref{rmk.FQSym.closed} can be easily derived from
Remark \ref{rmk.WQSym.closed}. The proof for \textbf{(a)} rests on the
following simple observations:
\begin{itemize}
\item Every word $w$ satisfies $\operatorname*{std}\left(
\operatorname*{pack}w\right) =\operatorname*{std}w$.
\item Every $n\in\mathbb{N}$, every word $w$ of length $n$ and every $\ell
\in\left\{ 0,1,\ldots,n\right\} $ satisfy
\[
\operatorname*{std}\left( \left( \operatorname*{std}w\right) \left[
:\ell\right] \right) =\operatorname*{std}\left( w\left[ :\ell\right]
\right) \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \operatorname*{std}%
\left( \left( \operatorname*{std}w\right) \left[ \ell:\right] \right)
=\operatorname*{std}\left( w\left[ \ell:\right] \right) .
\]
\item Every $n\in\mathbb{N}$, every word $w$ of length $n$ and every $\ell
\in\left\{ 0,1,\ldots,n\right\} $ satisfy the equivalence%
\begin{align*}
& \ \left( \min\left( \operatorname*{Supp}\left( w\left[ :\ell\right]
\right) \right) >\min\left( \operatorname*{Supp}\left( w\left[
\ell:\right] \right) \right) \right) \\
& \Longleftrightarrow\ \left( \min\left( \operatorname*{Supp}\left(
\left( \operatorname*{std}w\right) \left[ :\ell\right] \right) \right)
>\min\left( \operatorname*{Supp}\left( \left( \operatorname*{std}w\right)
\left[ \ell:\right] \right) \right) \right) .
\end{align*}
\end{itemize}
The third of these three observations would fail if the greater sign were to
be replaced by a smaller sign; this is essentially why $\operatorname*{FQSym}%
\subseteq\operatorname*{WQSym}$ is not closed under $\left. \prec\right. $.
The operation $\left. \succ\right. $ on $\operatorname*{FQSym}$ defined
above is closely related to the operation $\left. \succ\right. $ on
$\operatorname*{FQSym}$ introduced by Foissy in \cite[Section 4.2]{Foissy07}.
Indeed, the latter differs from the former in the use of $\max$ instead of
$\min$.
\section{\label{sect.epilogue}Epilogue}
We have introduced five binary operations $\left. \prec\right. $, $\circ$,
$\left. \succ\right. $, $\bel $, and $\tvi $ on $\mathbf{k}\left[ \left[
x_{1},x_{2},x_{3},\ldots\right] \right] $ and their restrictions to
$\operatorname*{QSym}$; we have further introduced five analogous operations
on $\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle \right\rangle
$ and their restrictions to $\operatorname*{WQSym}$ (as well as the
restrictions of two of them to $\operatorname*{FQSym}$). We have used these
operations (specifically, $\left. \prec\right. $ and $\bel $) to prove a
formula (Corollary \ref{cor.zabrocki}) for the dual immaculate functions
$\mathfrak{S}_{\alpha}^{\ast}$. Along the way, we have found that the
$\mathfrak{S}_{\alpha}^{\ast}$ can be obtained by repeated application of the
operation $\left. \prec\right. $ (Corollary \ref{cor.dualImm.dend.explicit}%
). A similar (but much more obvious) result can be obtained for the
fundamental quasisymmetric functions: For every $\alpha=\left( \alpha
_{1},\alpha_{2},\ldots,\alpha_{\ell}\right) \in\operatorname*{Comp}$, we have%
\[
F_{\alpha}=h_{\alpha_{1}} \tvi h_{\alpha_{2}} \tvi \cdots\tvi h_{\alpha_{\ell
}} \tvi 1
\]
(we do not use parentheses here, since $\tvi $ is associative). This shows
that the $\mathbf{k}$-algebra $\left( \operatorname*{QSym}, \tvi \right) $
is free. Moreover,%
\[
F_{\omega\left( \alpha\right) }=e_{\alpha_{\ell}} \bel e_{\alpha_{\ell-1}}
\bel \cdots\bel e_{\alpha_{1}} \bel 1,
\]
where $e_{m}$ stands for the $m$-th elementary symmetric function; thus, the
$\mathbf{k}$-algebra $\left( \operatorname*{QSym}, \bel \right) $ is also
free.\footnote{We owe these two observations to the referee.} (Incidentally,
this shows that $S\left( a \tvi b\right) =S\left( b\right) \bel S\left(
a\right) $ for any $a,b\in\operatorname*{QSym}$. But this does not hold for
$a,b\in\operatorname*{WQSym}$.)
One might wonder what \textquotedblleft functions\textquotedblright\ can be
similarly constructed using the operations $\left. \prec\right. $, $\circ$,
$\left. \succ\right. $, $\bel $, and $\tvi $ in $\operatorname*{WQSym}$,
using the noncommutative analogues $H_{m}=\sum_{i_{1}\leq i_{2}\leq\cdots\leq
i_{m}}X_{i_{1}}X_{i_{2}}\cdots X_{i_{m}}=\mathbf{G}_{\left( 1,2,\ldots
,m\right) }$ and $E_{m}=\sum_{i_{1}>i_{2}>\cdots>i_{m}}X_{i_{1}}X_{i_{2}%
}\cdots X_{i_{m}}=\mathbf{G}_{\left( m,m-1,\ldots,1\right) }$ of $h_{m}$ and
$e_{m}$. (These analogues actually live in $\operatorname*{NSym}$, where
$\operatorname*{NSym}$ is embedded into $\operatorname*{FQSym}$ as in
\cite[Corollary 8.1.14(b)]{Reiner}; but the operations do not preserve
$\operatorname*{NSym}$, and only two of them preserve $\operatorname*{FQSym}%
$.) However, it seems somewhat tricky to ask the right questions here; for
instance, the $\mathbf{k}$-linear span of the $\left. \succ\right. $-closure
of $\left\{ H_{m}\ \mid\ m\geq0\right\} $ is not a $\mathbf{k}$-subalgebra
of $\operatorname*{FQSym}$ (since $H_{2}H_{1}$ is not a $\mathbf{k}$-linear
combination of $H_{3}$, $H_{1}\left. \succ\right. \left( H_{1}\left.
\succ\right. H_{1}\right) $, $\left( H_{1}\left. \succ\right.
H_{1}\right) \left. \succ\right. H_{1}$, $H_{1}\left. \succ\right. H_{2}$
and $H_{2}\left. \succ\right. H_{1}$).
On the other hand, one might also try to write down the set of identities
satisfied by the operations $\cdot$, $\left. \prec\right. $, $\circ$,
$\left. \succeq\right. $, $\bel $ and $\tvi $ on the various spaces
($\mathbf{k}\left[ \left[ x_{1},x_{2},x_{3},\ldots\right] \right] $,
$\operatorname*{QSym}$, $\mathbf{k}\left\langle \left\langle \mathbf{X}%
\right\rangle \right\rangle $, $\operatorname*{WQSym}$ and
$\operatorname*{FQSym}$), or by subsets of these operations; these identities
could then be used to define new operads, i.e., algebraic structures
comprising a $\mathbf{k}$-module and some operations on it that imitate (some
of) the operations $\cdot$, $\left. \prec\right. $, $\circ$, $\left.
\succeq\right. $, $\bel $ and $\tvi $. For instance, apart from being
associative, the operations $\bel $ and $\tvi $ on $\mathbf{k}\left\langle
\left\langle \mathbf{X}\right\rangle \right\rangle $ satisfy the identity%
\begin{equation}
\left( a \bel b\right) \tvi c+\left( a \tvi b\right) \bel c=a \bel \left(
b \tvi c\right) +a \tvi \left( b \bel c\right) \label{eq.epilogue.bel-tvi}%
\end{equation}
for all $a,b,c\in\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle
\right\rangle $. This follows from the (easily verified) identities%
\begin{align*}
\left( a \bel b\right) \tvi c-a \bel \left( b \tvi c\right) &
=\varepsilon\left( b\right) \left( a \tvi c-a \bel c\right) ;\\
\left( a \tvi b\right) \bel c-a \tvi \left( b \bel c\right) &
=\varepsilon\left( b\right) \left( a \bel c-a \tvi c\right) ,
\end{align*}
where $\varepsilon:\mathbf{k}\left\langle \left\langle \mathbf{X}\right\rangle
\right\rangle \rightarrow\mathbf{k}$ is the map which sends every
noncommutative power series to its constant term. The equality
(\ref{eq.epilogue.bel-tvi}) (along with the associativity of $\bel $ and
$\tvi $) makes $\left( \mathbf{k}\left\langle \left\langle \mathbf{X}%
\right\rangle \right\rangle , \bel , \tvi \right) $ into what is called an
$As^{\left\langle 2\right\rangle }$\textit{-algebra} (see \cite[p.
39]{Zinbie10}). Is $\operatorname*{QSym}$ or $\operatorname*{WQSym}$ a free
$As^{\left\langle 2\right\rangle }$-algebra? What if we add the existence of a
common neutral element for the operations $\bel $ and $\tvi $ to the axioms of
this operad?
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\end{document}