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\begin{document}

\title{Project: Annihilating polynomials of Hopf algebra endomorphisms}
\author{Darij Grinberg}
\date{
%TCIMACRO{\TeXButton{today}{\today}}%
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\today
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\maketitle

This is an overview of some interrelated questions I'm interested in answering
or seeing answered. All of them concern connected graded Hopf algebras over a
commutative ring $\mathbf{k}$. I will use standard notations for Hopf algebras
(\cite[Chapter 1]{GriRei}). In particular:

\begin{itemize}
\item We fix a commutative ring $\mathbf{k}$. All tensor products, algebras,
Hom spaces, etc. will be over $\mathbf{k}$.

\item The comultiplication, the counit and the antipode of a Hopf algebra are
denoted by $\Delta$, $\epsilon$ and $S$. Sometimes, for disambiguation, we
denote them by $\Delta_{H}$, $\epsilon_{H}$ and $S_{H}$, where $H$ is the
relevant Hopf algebra.

\item \textquotedblleft Graded\textquotedblright\ (as in \textquotedblleft
graded $\mathbf{k}$-modules\textquotedblright, \textquotedblleft graded
$\mathbf{k}$-bialgebras, etc.\textquotedblright) will always mean
\textquotedblleft$\mathbb{N}$-graded\textquotedblright. We do \textbf{not}
twist our tensor products by the grading (i.e., we do not do any superalgebra here).

\item The $n$-th graded component of a graded $\mathbf{k}$-module $V$ will be
called $V_{n}$. If $n<0$, then this is the zero submodule $0$.

\item We let $\mathbb{N}=\left\{  0,1,2,\ldots\right\}  $.
\end{itemize}

The overarching thread of the first two questions is to find polynomials that
annihilate certain operators defined on any connected graded Hopf algebra. The
third question is an attempt to make questions like this brute-forceable (for
any given degree).

\section{Specific operators}

\subsection{Annihilating polynomials for $\operatorname*{id}-S^{2}$}

In \cite[Proposition 7]{AguLau14}, Aguiar and Lauve showed that
\[
\left(  \operatorname*{id}-S^{2}\right)  ^{n}\left(  H_{n}\right)  =0
\]
whenever $H$ is a connected graded Hopf algebra over a field and $n$ is a
positive integer. Later, Aguiar \cite[Lemma 12.50]{Aguiar17} strengthened this
equality to%
\[
\left(  \left(  \operatorname*{id}+S\right)  \circ\left(  \operatorname*{id}%
-S^{2}\right)  ^{n-1}\right)  \left(  H_{n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for each }n>0.
\]
For specific combinatorially interesting Hopf algebras, even stronger results
hold; in particular,
\[
\left(  \operatorname*{id}-S^{2}\right)  ^{n-1}\left(  H_{n}\right)
=0\ \ \ \ \ \ \ \ \ \ \text{holds for each }n>1
\]
when $H$ is the Malvenuto--Reutenauer Hopf algebra (see \cite[Example
8]{AguLau14}).

In \cite{antipode-squared}, I generalize this all to connected filtered
(rather than graded) Hopf algebras over commutative rings (rather than
fields), and even further. Here I am not interested in the filtered case, so
let me only state my results for graded Hopf algebras (\cite[\S 2.4]%
{antipode-squared}):\footnote{If $V$ is a graded $\mathbf{k}$-module, and if
$n\in\mathbb{N}$, then $V_{\leq n}$ denotes the $\mathbf{k}$-submodule
$V_{0}\oplus V_{1}\oplus\cdots\oplus V_{n}$ of $V$.}

\begin{corollary}
\label{cor.id-S2.gr1}Let $H$ be a connected graded $\mathbf{k}$-Hopf algebra
with antipode $S$. Then, for any positive integer $u$, we have%
\begin{equation}
\left(  \operatorname*{id}-S^{2}\right)  ^{u-1}\left(  H_{u}\right)
\subseteq\operatorname*{Prim}H \label{eq.cor.id-S2.gr1.claim1}%
\end{equation}
and%
\begin{equation}
\left(  \left(  \operatorname*{id}+S\right)  \circ\left(  \operatorname*{id}%
-S^{2}\right)  ^{u-1}\right)  \left(  H_{u}\right)  =0
\label{eq.cor.id-S2.gr1.claim3}%
\end{equation}
and%
\begin{equation}
\left(  \operatorname*{id}-S^{2}\right)  ^{u}\left(  H_{u}\right)  =0.
\label{eq.cor.id-S2.gr1.claim2}%
\end{equation}

\end{corollary}

\begin{corollary}
\label{cor.id-S2.grp}Let $H$ be a connected graded $\mathbf{k}$-Hopf algebra
with antipode $S$.

Let $p$ be a positive integer such that all $i\in\left\{  2,3,\ldots
,p\right\}  $ satisfy%
\begin{equation}
\left(  \operatorname*{id}-S^{2}\right)  \left(  H_{i}\right)  =0.
\label{eq.cor.id-S2.grp.ass-ann}%
\end{equation}


Then:

\textbf{(a)} For any integer $u>p$, we have%
\begin{equation}
\left(  \operatorname*{id}-S^{2}\right)  ^{u-p}\left(  H_{\leq u}\right)
\subseteq\operatorname*{Prim}H \label{eq.cor.id-S2.grp.claim1}%
\end{equation}
and%
\begin{equation}
\left(  \left(  \operatorname*{id}+S\right)  \circ\left(  \operatorname*{id}%
-S^{2}\right)  ^{u-p}\right)  \left(  H_{\leq u}\right)  =0.
\label{eq.cor.id-S2.grp.claim3}%
\end{equation}


\textbf{(b)} For any integer $u\geq p$, we have%
\begin{equation}
\left(  \operatorname*{id}-S^{2}\right)  ^{u-p+1}\left(  H_{\leq u}\right)
=0. \label{eq.cor.id-S2.grp.claim2}%
\end{equation}

\end{corollary}

\begin{corollary}
\label{cor.id-S2.gr2}Let $H$ be a connected graded $\mathbf{k}$-Hopf algebra
with antipode $S$. Assume that%
\begin{equation}
ab=ba\ \ \ \ \ \ \ \ \ \ \text{for every }a,b\in H_{1}.
\label{eq.cor.id-S2.gr2.commH1}%
\end{equation}
Then:

\textbf{(a)} We have
\[
\left(  \operatorname*{id}-S^{2}\right)  \left(  H_{2}\right)  =0.
\]


\textbf{(b)} For any integer $u>2$, we have%
\begin{equation}
\left(  \operatorname*{id}-S^{2}\right)  ^{u-2}\left(  H_{\leq u}\right)
\subseteq\operatorname*{Prim}H \label{eq.cor.id-S2.gr2.b.claim1}%
\end{equation}
and%
\begin{equation}
\left(  \left(  \operatorname*{id}+S\right)  \circ\left(  \operatorname*{id}%
-S^{2}\right)  ^{u-2}\right)  \left(  H_{\leq u}\right)  =0.
\label{eq.cor.id-S2.gr2.b.claim3}%
\end{equation}


\textbf{(c)} For any integer $u>1$, we have%
\begin{equation}
\left(  \operatorname*{id}-S^{2}\right)  ^{u-1}\left(  H_{\leq u}\right)  =0.
\label{eq.cor.id-S2.gr2.b.claim2}%
\end{equation}

\end{corollary}

\subsection{The random-to-top operator}

Now, fix a connected graded Hopf algebra $H$ over the commutative ring
$\mathbf{k}$.

For each $n\in\mathbb{N}$, let $p_{n}:H\rightarrow H$ be the projection onto
the $n$-th graded component $H_{n}$ of $H$ (viewed as an endomorphism of $H$).
Note that $p_{0}\left(  x\right)  =\epsilon\left(  x\right)  \cdot1_{H}$ for
each $x\in H$, since $H$ is connected. Also, each element of $H_{1}$ is
primitive (again since $H$ is connected).

Let $f:H\rightarrow H$ denote the map $p_{1}\star\operatorname*{id}$, where
the asterisk \textquotedblleft$\star$\textquotedblright\ means convolution of
linear maps (so that we have $f\left(  x\right)  =\sum_{\left(  x\right)
}p_{1}\left(  x_{\left(  1\right)  }\right)  x_{\left(  2\right)  }$ for any
$x\in H$, using Sweedler notation). The map $f$ can be viewed as a generalized
random-to-top or top-to-random shuffling operator: Indeed, the latter two
operators are obtained by specializing $H$ to be either the tensor algebra or
the shuffle algebra of a $\mathbf{k}$-module.

We claim the following:\footnote{In the following, the $\prod$ sign means a
product with respect to composition (not convolution). Also, the expression
\textquotedblleft$f-k$\textquotedblright\ for an integer $k$ always means the
map $f-k\cdot\operatorname*{id}\in\operatorname*{End}H$.}

\begin{theorem}
\label{thm.f.1}\textbf{(a)} We have $f=0$ on $H_{0}$, and we have
$f=\operatorname*{id}$ on $H_{1}$.

\textbf{(b)} For each $n\geq2$, we have%
\[
\left(  f-n\right)  \circ\left(  f-\left(  n-2\right)  \right)  ^{2}\circ
\prod_{i=0}^{n-3}\left(  f-i\right)  ^{n-1-i}=0\ \ \ \ \ \ \ \ \ \ \text{on
}H_{n}.
\]

\end{theorem}

For example, Theorem \ref{thm.f.1} \textbf{(b)} entails that%
\begin{align*}
\left(  f-2\right)  \circ f^{2}  &  =0\ \ \ \ \ \ \ \ \ \ \text{on }%
H_{2},\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(  f-3\right)  \circ\left(  f-1\right)  ^{2}\circ f^{2}  &
=0\ \ \ \ \ \ \ \ \ \ \text{on }H_{3},\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(  f-4\right)  \circ\left(  f-2\right)  ^{2}\circ\left(  f-1\right)
^{2}\circ f^{3}  &  =0\ \ \ \ \ \ \ \ \ \ \text{on }H_{4}.
\end{align*}


Theorem \ref{thm.f.1} should be compared with the classical fact (see, e.g.,
\cite[Theorem 3]{Grinberg-MO} and references therein) that $\left(
f-n\right)  \circ\prod_{i=0}^{n-2}\left(  f-i\right)  =0$ on the $n$-th graded
component of a tensor algebra. When passing to a general connected graded Hopf
algebra, we have only been able to salvage this equality \textquotedblleft up
to exponents\textquotedblright.

To prove Theorem \ref{thm.f.1}, I will introduce some notations:

\begin{itemize}
\item We set $\left[  p\right]  :=\left\{  1,2,\ldots,p\right\}  $ for any
$p\in\mathbb{N}$.

\item Given two integers $k$ and $\ell$, we define a $\left(  k,\ell\right)
$\emph{-primitive product} to be a product $a_{1}a_{2}\cdots a_{p}$ of
finitely many homogeneous primitive elements of $H$ such that $\deg\left(
a_{1}\right)  +\deg\left(  a_{2}\right)  +\cdots+\deg\left(  a_{p}\right)  =k$
and such that there are precisely $\ell$ many $i\in\left[  p\right]  $
satisfying $\deg\left(  a_{i}\right)  =1$.

For example, if $a,b,c,d$ are four homogeneous primitive elements of $H$ such
that $\deg a=2$ and $\deg b=1$ and $\deg c=3$ and $\deg d=1$, then $abcd$ is a
$\left(  7,2\right)  $-primitive product.

\item Given two integers $k$ and $\ell$, we let $P_{k,\ell}$ be the
$\mathbf{k}$-linear span of all $\left(  k,\ell\right)  $-primitive products
in $H$. This is a $\mathbf{k}$-submodule of $H_{k}$.

Note that
\[
P_{0,0}=\operatorname*{span}\left\{  1_{H}\right\}  =H_{0}.
\]
Furthermore,%
\[
P_{k,\ell}=0\ \ \ \ \ \ \ \ \ \ \text{whenever }\ell>k.
\]
Moreover, since $k-\left(  k-1\right)  =1$, we have
\[
P_{k,k-1}=0\ \ \ \ \ \ \ \ \ \ \text{for any }k\in\mathbb{Z}.
\]
Also, obviously,%
\[
P_{k,\ell}=0\ \ \ \ \ \ \ \ \ \ \text{whenever }\ell<0.
\]


\item If $U$ and $V$ are two $\mathbf{k}$-submodules of $H$, then $UV$ shall
mean the $\mathbf{k}$-linear span of all products $uv$ with $u\in U$ and $v\in
V$.
\end{itemize}

We will prove Theorem \ref{thm.f.1} by constructing a filtration of $H_{n}$
and a sequence of maps $g_{0},g_{1},\ldots,g_{n-2},g_{n}$ (sic!) that map
larger to smaller parts in this filtration:

\begin{theorem}
\label{thm.f.2}Let $n\in\mathbb{N}$. For each $k\in\mathbb{N}$, we define the
$\mathbf{k}$-linear map%
\[
g_{k}:=\left(  f-k\right)  \circ\prod_{i=0}^{k-2}\left(  f-i\right)
:H\rightarrow H.
\]


For each $k\in\mathbb{N}$, we set
\[
R_{k}:=\sum_{\ell\in\mathbb{N}}P_{k,\ell}H_{n-k};
\]
this is clearly a $\mathbf{k}$-submodule of $H_{k}H_{n-k}\subseteq H_{n}$.

For each $k\in\mathbb{N}$, we set%
\[
Q_{k}:=R_{k}+R_{k+1}+R_{k+2}+\cdots=\sum_{\ell\geq k}R_{\ell};
\]
again, this is a $\mathbf{k}$-submodule of $H_{n}$.

The following holds:

\textbf{(a)} We have $g_{k}\left(  Q_{k}\right)  \subseteq Q_{k+1}$ for each
$k\in\mathbb{N}$.

\textbf{(b)} We have $Q_{n-1}=Q_{n}$.

\textbf{(c)} We have $Q_{n+1}=0$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.f.2} (sketched).]This is mostly an aide-memoire for
myself.\footnote{Note to self: In the final version, use the notation
$b\rightharpoonup a$ for $\sum_{\left(  b\right)  }p_{1}\left(  b_{\left(
1\right)  }\right)  ab_{\left(  2\right)  }$. This should simplify the
argument below, possibly even getting rid of Sweedler notation altogether.}

We shall use Sweedler notation. First, observe that the map $p_{1}$ is an
$\left(  \epsilon,\epsilon\right)  $-derivation -- i.e., that we have
\begin{equation}
p_{1}\left(  ab\right)  =\epsilon\left(  a\right)  p_{1}\left(  b\right)
+\epsilon\left(  b\right)  p_{1}\left(  a\right)
\label{pf.thm.f.2.p1-prodrule}%
\end{equation}
for all $a,b\in H$. (This is straightforward to check.) Using this fact, it is
easy to see that
\begin{equation}
f\left(  ab\right)  =\sum_{\left(  b\right)  }p_{1}\left(  b_{\left(
1\right)  }\right)  ab_{\left(  2\right)  }+f\left(  a\right)  b
\label{pf.thm.f.2.f-prodrule}%
\end{equation}
for all $a,b\in H$. (Indeed, if $a,b\in H$, then%
\begin{align*}
f\left(  ab\right)   &  =\left(  p_{1}\star\operatorname*{id}\right)  \left(
ab\right)  =\sum_{\left(  a\right)  \left(  b\right)  }\underbrace{p_{1}%
\left(  a_{\left(  1\right)  }b_{\left(  1\right)  }\right)  }%
_{\substack{=\epsilon\left(  a_{\left(  1\right)  }\right)  p_{1}\left(
b_{\left(  1\right)  }\right)  +\epsilon\left(  b_{\left(  1\right)  }\right)
p_{1}\left(  a_{\left(  2\right)  }\right)  \\\text{(by
(\ref{pf.thm.f.2.p1-prodrule}))}}}a_{\left(  2\right)  }b_{\left(  2\right)
}\\
&  =\sum_{\left(  a\right)  \left(  b\right)  }\left(  \epsilon\left(
a_{\left(  1\right)  }\right)  p_{1}\left(  b_{\left(  1\right)  }\right)
+\epsilon\left(  b_{\left(  1\right)  }\right)  p_{1}\left(  a_{\left(
1\right)  }\right)  \right)  a_{\left(  2\right)  }b_{\left(  2\right)  }\\
&  =\underbrace{\sum_{\left(  a\right)  \left(  b\right)  }\epsilon\left(
a_{\left(  1\right)  }\right)  p_{1}\left(  b_{\left(  1\right)  }\right)
a_{\left(  2\right)  }b_{\left(  2\right)  }}_{=\sum_{\left(  b\right)  }%
p_{1}\left(  b_{\left(  1\right)  }\right)  ab_{\left(  2\right)  }%
}+\underbrace{\sum_{\left(  a\right)  \left(  b\right)  }\epsilon\left(
b_{\left(  1\right)  }\right)  p_{1}\left(  a_{\left(  1\right)  }\right)
a_{\left(  2\right)  }b_{\left(  2\right)  }}_{=\sum_{\left(  a\right)  }%
p_{1}\left(  a_{\left(  1\right)  }\right)  a_{\left(  2\right)  }b}\\
&  =\sum_{\left(  b\right)  }p_{1}\left(  b_{\left(  1\right)  }\right)
ab_{\left(  2\right)  }+\underbrace{\sum_{\left(  a\right)  }p_{1}\left(
a_{\left(  1\right)  }\right)  a_{\left(  2\right)  }}_{=\left(  p_{1}%
\star\operatorname*{id}\right)  \left(  a\right)  =f\left(  a\right)  }%
b=\sum_{\left(  b\right)  }p_{1}\left(  b_{\left(  1\right)  }\right)
ab_{\left(  2\right)  }+f\left(  a\right)  b,
\end{align*}
so that (\ref{pf.thm.f.2.f-prodrule}) is proven.)

Using (\ref{pf.thm.f.2.f-prodrule}) (and induction on $p$), we can easily
prove the following: If $a_{1},a_{2},\ldots,a_{p}$ are finitely many primitive
elements of $H$, then%
\begin{equation}
f\left(  a_{1}a_{2}\cdots a_{p}\right)  =\sum_{i=1}^{p}p_{1}\left(
a_{i}\right)  \cdot a_{1}a_{2}\cdots\widehat{a_{i}}\cdots a_{p}
\label{pf.thm.f.2.fa1ap}%
\end{equation}
(where the hat over the $a_{i}$ means \textquotedblleft omit $a_{i}$ from this
product\textquotedblright). Using this equality and
(\ref{pf.thm.f.2.f-prodrule}) again, we can conclude the following: If
$a_{1},a_{2},\ldots,a_{p}$ are finitely many primitive elements of $H$, and if
$b\in H$ is arbitrary, then%
\begin{align}
f\left(  a_{1}a_{2}\cdots a_{p}b\right)   &  =\sum_{\left(  b\right)  }%
p_{1}\left(  b_{\left(  1\right)  }\right)  a_{1}a_{2}\cdots a_{p}b_{\left(
2\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{i=1}^{p}p_{1}\left(  a_{i}\right)  \cdot
a_{1}a_{2}\cdots\widehat{a_{i}}\cdots a_{p}b \label{pf.thm.f.2.fa1apb1}%
\end{align}
(where the hat over the $a_{i}$ means \textquotedblleft omit $a_{i}$ from this
product\textquotedblright). Hence, in particular, if $a_{1},a_{2},\ldots
,a_{p}$ are finitely many homogeneous primitive elements of $H$, and if $b\in
H$ is arbitrary, then%
\begin{align}
f\left(  a_{1}a_{2}\cdots a_{p}b\right)   &  =\sum_{\left(  b\right)  }%
p_{1}\left(  b_{\left(  1\right)  }\right)  a_{1}a_{2}\cdots a_{p}b_{\left(
2\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{i\in\left[  p\right]  ;\\\deg\left(
a_{i}\right)  =1}}a_{i}\cdot a_{1}a_{2}\cdots\widehat{a_{i}}\cdots a_{p}b.
\label{pf.thm.f.2.fa1apb2}%
\end{align}


Next, we claim the following: For each $k,\ell\in\mathbb{Z}$, we have%
\begin{equation}
\left(  f-\ell\right)  \left(  P_{k,\ell}H_{n-k}\right)  \subseteq
P_{k+1,\ell+1}H_{n-k-1}+\left(  P_{k,\ell-1}+P_{k,\ell-2}\right)  H_{n-k}.
\label{pf.thm.f.2.mainrec}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.f.2.mainrec}):} Let $k,\ell\in\mathbb{Z}$. We
must prove (\ref{pf.thm.f.2.mainrec}). We know that $P_{k,\ell}H_{n-k}$ is the
$\mathbf{k}$-linear span of all products of the form $a_{1}a_{2}\cdots a_{p}%
b$, where $b\in H_{n-k}$ and where $a_{1},a_{2},\ldots,a_{p}$ are finitely
many homogeneous primitive elements of $H$ such that $\deg\left(
a_{1}\right)  +\deg\left(  a_{2}\right)  +\cdots+\deg\left(  a_{p}\right)  =k$
and such that there are precisely $\ell$ many $i\in\left[  p\right]  $
satisfying $\deg\left(  a_{i}\right)  =1$. Thus, it will suffice to show that
\begin{equation}
\left(  f-\ell\right)  \left(  a_{1}a_{2}\cdots a_{p}b\right)  \in
P_{k+1,\ell+1}H_{n-k-1}+\left(  P_{k,\ell-1}+P_{k,\ell-2}\right)  H_{n-k}
\label{pf.thm.f.2.mainrec.pf.goal}%
\end{equation}
for any such $b$ and such $a_{1},a_{2},\ldots,a_{p}$.

So let us fix such $b$ and such $a_{1},a_{2},\ldots,a_{p}$, and try to prove
(\ref{pf.thm.f.2.mainrec.pf.goal}). From (\ref{pf.thm.f.2.fa1apb2}), we see
that%
\begin{align*}
f\left(  a_{1}a_{2}\cdots a_{p}b\right)   &  =\sum_{\left(  b\right)  }%
p_{1}\left(  b_{\left(  1\right)  }\right)  a_{1}a_{2}\cdots a_{p}b_{\left(
2\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\sum_{\substack{i\in\left[  p\right]
;\\\deg\left(  a_{i}\right)  =1}}a_{i}\cdot a_{1}a_{2}\cdots\widehat{a_{i}%
}\cdots a_{p}b}_{\text{This sum has }\ell\text{ addends}},
\end{align*}
so that\footnote{We use the notation $\left[  x,y\right]  $ for the commutator
$xy-yx$ of two elements $x,y\in H$.}
\begin{align*}
&  \left(  f-\ell\right)  \left(  a_{1}a_{2}\cdots a_{p}b\right) \\
&  =\sum_{\left(  b\right)  }p_{1}\left(  b_{\left(  1\right)  }\right)
a_{1}a_{2}\cdots a_{p}b_{\left(  2\right)  }+\sum_{\substack{i\in\left[
p\right]  ;\\\deg\left(  a_{i}\right)  =1}}\underbrace{\left(  a_{i}\cdot
a_{1}a_{2}\cdots\widehat{a_{i}}\cdots a_{p}b-a_{1}a_{2}\cdots a_{p}b\right)
}_{\substack{=\sum_{j=1}^{i-1}a_{1}a_{2}\cdots a_{j-1}\left[  a_{i}%
,a_{j}\right]  a_{j+1}a_{j+2}\cdots\widehat{a_{i}}\cdots a_{p}b\\\text{(by a
standard manipulation with commutators)}}}\\
&  =\sum_{\left(  b\right)  }\underbrace{p_{1}\left(  b_{\left(  1\right)
}\right)  a_{1}a_{2}\cdots a_{p}b_{\left(  2\right)  }}_{\substack{\in
P_{k+1,\ell+1}H_{n-k-1}\\\text{(since }p_{1}\left(  b_{\left(  1\right)
}\right)  \text{ is a homogeneous}\\\text{primitive element of degree
}1\text{)}}}+\sum_{\substack{i\in\left[  p\right]  ;\\\deg\left(
a_{i}\right)  =1}}\ \ \sum_{j=1}^{i-1}\underbrace{a_{1}a_{2}\cdots
a_{j-1}\left[  a_{i},a_{j}\right]  a_{j+1}a_{j+2}\cdots\widehat{a_{i}}\cdots
a_{p}b}_{\substack{\in\left(  P_{k,\ell-1}+P_{k,\ell-2}\right)  H_{n-k}%
\\\text{(since }a_{1},a_{2},\ldots,a_{j-1},\left[  a_{i},a_{j}\right]
,a_{j+1},a_{j+2},\ldots,\widehat{a_{i}},\ldots,a_{p}\\\text{are homogeneous
primitive elements,}\\\text{and since }\deg\left[  a_{i},a_{j}\right]
>1\text{)}}}\\
&  \in P_{k+1,\ell+1}H_{n-k-1}+\left(  P_{k,\ell-1}+P_{k,\ell-2}\right)
H_{n-k}.
\end{align*}
This proves (\ref{pf.thm.f.2.mainrec.pf.goal}) and therefore
(\ref{pf.thm.f.2.mainrec}).]

Now, it is not hard to check all parts of Theorem \ref{thm.f.2}:

\textbf{(a)} Using (\ref{pf.thm.f.2.mainrec}), it is easy to see that
$f\left(  R_{k}\right)  \subseteq R_{k}+R_{k+1}\subseteq Q_{k}$ for each
$k\in\mathbb{N}$. Hence, it follows easily that%
\begin{equation}
f\left(  Q_{k}\right)  \subseteq Q_{k}\ \ \ \ \ \ \ \ \ \ \text{for each }%
k\in\mathbb{N}. \label{pf.thm.f.2.fQk}%
\end{equation}


Fix $k\in\mathbb{N}$. Then, (\ref{pf.thm.f.2.fQk}) yields $f\left(
Q_{k+1}\right)  \subseteq Q_{k+1}$. Hence, $Q_{k+1}$ is not just a
$\mathbf{k}$-module, but also a $\mathbf{k}\left[  f\right]  $-module (where
$f$ acts in the obvious way).

Next, we claim that each integer $\ell\geq-1$ satisfies%
\begin{equation}
\left(  \prod_{i=0}^{\ell}\left(  f-i\right)  \right)  \left(  P_{k,\ell
}H_{n-k}\right)  \subseteq Q_{k+1}. \label{pf.thm.f.2.a.rec1}%
\end{equation}
Indeed, (\ref{pf.thm.f.2.a.rec1}) easily follows by strong induction on $\ell$
(using $P_{k,-1}=0$ for the case $\ell=-1$, and using
(\ref{pf.thm.f.2.mainrec}) for the induction step\footnote{Here is the
\textit{induction step}: Let $\ell\geq0$. Assume (as the induction hypothesis)
that
\[
\left(  \prod_{i=0}^{\ell^{\prime}}\left(  f-i\right)  \right)  \left(
P_{k,\ell^{\prime}}H_{n-k}\right)  \subseteq Q_{k+1}%
\]
for every $\ell^{\prime}<\ell$. Hence, in particular,%
\begin{align*}
\left(  \prod_{i=0}^{\ell-1}\left(  f-i\right)  \right)  \left(  P_{k,\ell
-1}H_{n-k}\right)   &  \subseteq Q_{k+1}\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(  \prod_{i=0}^{\ell-2}\left(  f-i\right)  \right)  \left(  P_{k,\ell
-2}H_{n-k}\right)   &  \subseteq Q_{k+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{if
}\ell\geq1\right)  .
\end{align*}
The latter of these two relations quickly entails%
\[
\left(  \prod_{i=0}^{\ell-1}\left(  f-i\right)  \right)  \left(  P_{k,\ell
-2}H_{n-k}\right)  \subseteq Q_{k+1}%
\]
(in fact, this is obvious if $\ell<2$, and otherwise follows from $\left(
\prod_{i=0}^{\ell-1}\left(  f-i\right)  \right)  \left(  P_{k,\ell-2}%
H_{n-k}\right)  =\left(  f-\left(  \ell-1\right)  \right)  \underbrace{\left(
\left(  \prod_{i=0}^{\ell-2}\left(  f-i\right)  \right)  \left(  P_{k,\ell
-2}H_{n-k}\right)  \right)  }_{\subseteq Q_{k+1}}\subseteq\left(  f-\left(
\ell-1\right)  \right)  \left(  Q_{k+1}\right)  \subseteq Q_{k+1}$ (since
$Q_{k+1}$ is a $\mathbf{k}\left[  f\right]  $-module)). Now,%
\begin{align*}
&  \left(  \prod_{i=0}^{\ell}\left(  f-i\right)  \right)  \left(  P_{k,\ell
}H_{n-k}\right) \\
&  =\left(  \prod_{i=0}^{\ell-1}\left(  f-i\right)  \right)
\underbrace{\left(  \left(  f-\ell\right)  \left(  P_{k,\ell}H_{n-k}\right)
\right)  }_{\substack{\subseteq P_{k+1,\ell+1}H_{n-k-1}+\left(  P_{k,\ell
-1}+P_{k,\ell-2}\right)  H_{n-k}\\\text{(by (\ref{pf.thm.f.2.mainrec}))}}}\\
&  \subseteq\left(  \prod_{i=0}^{\ell-1}\left(  f-i\right)  \right)  \left(
\underbrace{P_{k+1,\ell+1}H_{n-k-1}}_{\subseteq R_{k+1}\subseteq Q_{k+1}%
}+\underbrace{\left(  P_{k,\ell-1}+P_{k,\ell-2}\right)  H_{n-k}}%
_{=P_{k,\ell-1}H_{n-k}+P_{k,\ell-2}H_{n-k}}\right) \\
&  \subseteq\underbrace{\left(  \prod_{i=0}^{\ell-1}\left(  f-i\right)
\right)  \left(  Q_{k+1}\right)  }_{\substack{\subseteq Q_{k+1}\\\text{(since
}Q_{k+1}\text{ is a }\mathbf{k}\left[  f\right]  \text{-module)}%
}}+\underbrace{\left(  \prod_{i=0}^{\ell-1}\left(  f-i\right)  \right)
\left(  P_{k,\ell-1}H_{n-k}\right)  }_{\substack{\subseteq Q_{k+1}\\\text{(as
shown above)}}}+\underbrace{\left(  \prod_{i=0}^{\ell-1}\left(  f-i\right)
\right)  \left(  P_{k,\ell-2}H_{n-k}\right)  }_{\substack{\subseteq
Q_{k+1}\\\text{(as shown above)}}}\\
&  \subseteq Q_{k+1}+Q_{k+1}+Q_{k+1}\subseteq Q_{k+1}.
\end{align*}
This completes the induction step.}). Thus, each integer $\ell\in\left\{
-1,0,\ldots,k-2\right\}  $ satisfies%
\begin{equation}
g_{k}\left(  P_{k,\ell}H_{n-k}\right)  \subseteq Q_{k+1}
\label{pf.thm.f.2.a.rec1a}%
\end{equation}
(since $g_{k}=\left(  f-k\right)  \circ\prod_{i=0}^{k-2}\left(  f-i\right)  $
can be written as a composition $u\circ\prod_{i=0}^{\ell}\left(  f-i\right)  $
for some $u\in\mathbf{k}\left[  f\right]  $, and since $Q_{k+1}$ is a
$\mathbf{k}\left[  f\right]  $-module).

Moreover,
\begin{equation}
g_{k}\left(  P_{k,k}H_{n-k}\right)  \subseteq Q_{k+1}
\label{pf.thm.f.2.a.rec1k}%
\end{equation}
(this is proved rather similarly to (\ref{pf.thm.f.2.a.rec1}), but
additionally using the fact that $P_{k,k-1}=0$).

Now,%
\begin{align*}
R_{k}  &  =\sum_{\ell\in\mathbb{N}}P_{k,\ell}H_{n-k}\\
&  =\sum_{\ell=0}^{k}P_{k,\ell}H_{n-k}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}P_{k,\ell}=0\text{ whenever }\ell>k\right) \\
&  =\sum_{\ell=0}^{k-2}P_{k,\ell}H_{n-k}+\underbrace{P_{k,k-1}}_{=0}%
H_{n-k}+P_{k,k}H_{n-k}\\
&  =\sum_{\ell=0}^{k-2}P_{k,\ell}H_{n-k}+P_{k,k}H_{n-k}.
\end{align*}
Applying the linear map $g_{k}$ to both sides of this equality, we obtain%
\[
g_{k}\left(  R_{k}\right)  =\sum_{\ell=0}^{k-2}g_{k}\left(  P_{k,\ell}%
H_{n-k}\right)  +g_{k}\left(  P_{k,k}H_{n-k}\right)  \subseteq Q_{k+1}%
\]
(by (\ref{pf.thm.f.2.a.rec1a}) and (\ref{pf.thm.f.2.a.rec1k})). On the other
hand, $g_{k}\left(  Q_{k+1}\right)  \subseteq Q_{k+1}$ (since $Q_{k+1}$ is a
$\mathbf{k}\left[  f\right]  $-module, whereas $g_{k}\in\mathbf{k}\left[
f\right]  $). Combining these two facts, we obtain $g_{k}\left(  Q_{k}\right)
\subseteq Q_{k+1}$ (since $Q_{k}=R_{k}+Q_{k+1}$). This proves Theorem
\ref{thm.f.2} \textbf{(a)}.

\textbf{(b)} We have $P_{n-1,\ell}H_{n-\left(  n-1\right)  }\subseteq
P_{n,\ell+1}$ for each $\ell\in\mathbb{Z}$, since $H_{n-\left(  n-1\right)
}=H_{1}$ consists of homogeneous primitive elements of degree $1$. Thus,
$R_{n-1}\subseteq R_{n}$ follows from the definitions of $R_{n-1}$ and $R_{n}%
$. This entails the claim of Theorem \ref{thm.f.2} \textbf{(b)}.

\textbf{(c)} Each $k>n$ satisfies $R_{k}=\sum_{\ell\in\mathbb{N}}P_{k,\ell
}\underbrace{H_{n-k}}_{\substack{=0\\\text{(since }k>n\text{)}}}=0$. Thus,
$Q_{n+1}=0$. This proves Theorem \ref{thm.f.2} \textbf{(c)}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.f.1} (sketched).]\textbf{(a)} Easy and left to the reader.

\textbf{(b)} Let $n\geq2$. Define the maps $g_{k}$ and the $\mathbf{k}%
$-submodules $R_{k}$ and $Q_{k}$ as in Theorem \ref{thm.f.2}. Then, it is easy
to see (by counting factors) that%
\[
\left(  f-n\right)  \circ\left(  f-\left(  n-2\right)  \right)  ^{2}\circ
\prod_{i=0}^{n-3}\left(  f-i\right)  ^{n-1-i}=g_{n}\circ\prod_{k=0}^{n-2}g_{k}%
\]
(note that all maps in this equality commute, since they all belong to the
commutative ring $\mathbf{k}\left[  f\right]  $). Hence, it remains to show
that%
\begin{equation}
g_{n}\circ\prod_{k=0}^{n-2}g_{k}=0\ \ \ \ \ \ \ \ \ \ \text{on }H_{n}.
\label{pf.thm.f.1.b.goal}%
\end{equation}


To do so, we observe that $1_{H}\in P_{0,0}$, so that $H_{n}\subseteq
P_{0,0}H_{n}=P_{0,0}H_{n-0}\subseteq R_{0}\subseteq Q_{0}$. Therefore,%
\begin{align*}
&  \left(  g_{n}\circ\prod_{k=0}^{n-2}g_{k}\right)  \left(  H_{n}\right) \\
&  \subseteq\left(  g_{n}\circ\prod_{k=0}^{n-2}g_{k}\right)  \left(
Q_{0}\right) \\
&  =g_{n}\left(  g_{n-2}\left(  g_{n-3}\left(  g_{n-4}\left(  \cdots\left(
g_{2}\left(  g_{1}\left(  \underbrace{g_{0}\left(  Q_{0}\right)
}_{\substack{\subseteq Q_{1}\\\text{(by Theorem \ref{thm.f.2} \textbf{(a)})}%
}}\right)  \right)  \right)  \right)  \right)  \right)  \right) \\
&  \subseteq g_{n}\left(  g_{n-2}\left(  g_{n-3}\left(  g_{n-4}\left(
\cdots\left(  g_{2}\left(  \underbrace{g_{1}\left(  Q_{1}\right)
}_{\substack{\subseteq Q_{2}\\\text{(by Theorem \ref{thm.f.2} \textbf{(a)})}%
}}\right)  \right)  \right)  \right)  \right)  \right) \\
&  \subseteq g_{n}\left(  g_{n-2}\left(  g_{n-3}\left(  g_{n-4}\left(
\cdots\left(  \underbrace{g_{2}\left(  Q_{2}\right)  }_{\substack{\subseteq
Q_{3}\\\text{(by Theorem \ref{thm.f.2} \textbf{(a)})}}}\right)  \right)
\right)  \right)  \right) \\
&  \subseteq\cdots\\
&  \subseteq g_{n}\left(  Q_{n-1}\right)  =g_{n}\left(  Q_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.f.2} \textbf{(b)}%
}\right) \\
&  \subseteq Q_{n+1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.f.2}
\textbf{(a)}}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.f.2} \textbf{(c)}%
}\right)  .
\end{align*}
This proves (\ref{pf.thm.f.1.b.goal}) and thus Theorem \ref{thm.f.1}
\textbf{(b)}.
\end{proof}

\begin{question}
Can the exponents in Theorem \ref{thm.f.1} \textbf{(b)} be improved (i.e.,
made smaller)?
\end{question}

It is easy to see that they cannot be improved for $n=2$ (consider the free
algebra with three generators $x,y_{1},z_{1}$ of degrees $2,1,1$,
respectively, with $\Delta\left(  x\right)  =x\otimes1+1\otimes x+y_{1}\otimes
y_{2}$). Some more complicated computations (using the Hopf algebra $H$ from
the proof of Theorem \ref{thm.pas-lin-ind} below) show that they cannot be
improved for $n=3$ and for $n=4$ either. I don't know about higher values of
$n$, but I wouldn't be surprised if the exponents are optimal.

It sounds like a good idea to test this on well-known combinatorial Hopf
algebras such as $\operatorname*{FQSym}$ and $\operatorname*{WQSym}$ (and in
fact, the $f$ for $H=\operatorname*{FQSym}$ was studied in \cite[\S 3.4.1]%
{Pang19} and \cite[Theorem 6.6?]{Pang18}). However, they are too well-behaved
to be of much use in this question. Indeed, if $H=\operatorname*{FQSym}$ or
$H=\operatorname*{WQSym}$, then $\left(  f-n\right)  \circ\prod_{i=0}%
^{n-2}\left(  f-i\right)  =0$ on $H_{n}$. More generally:

\begin{theorem}
\label{thm.f.1comm}Assume that every two elements of $H_{1}$ commute. Then,
each $n\in\mathbb{N}$ satisfies%
\[
\left(  f-n\right)  \circ\prod_{i=0}^{n-2}\left(  f-i\right)
=0\ \ \ \ \ \ \ \ \ \ \text{on }H_{n}.
\]

\end{theorem}

Note that this entails that $f$ is diagonalizable (assuming that every two
elements of $H_{1}$ commute) when $\mathbf{k}$ is a field of characteristic
$0$.

Here is a way of proving Theorem \ref{thm.f.1comm}. We define minor variations
of primitive elements and $\left(  k,\ell\right)  $-primitive products:

\begin{itemize}
\item We consider the Lie subalgebra of $H$ generated by $H_{1}$ (via the
commutator). Its elements are $\mathbf{k}$-linear combinations of nested
commutators of elements of $H_{1}$ (including the depth-$1$ commutators, which
are just the elements of $H_{1}$ themselves). The elements of this Lie
subalgebra will be called \emph{ultraprimitive} elements of $H$. It is
well-known (and easy to check) that all ultraprimitive elements of $H$ are primitive.

If every two elements of $H_{1}$ commute, then the ultraprimitive elements of
$H$ are precisely the elements of $H_{1}$.

\item Given two integers $k$ and $\ell$, we define a $\left(  k,\ell\right)
$\emph{-ultraprimitive product} to be a product $a_{1}a_{2}\cdots a_{p}$ of
finitely many homogeneous ultraprimitive elements of $H$ such that
$\deg\left(  a_{1}\right)  +\deg\left(  a_{2}\right)  +\cdots+\deg\left(
a_{p}\right)  =k$ and such that there are precisely $\ell$ many $i\in\left[
p\right]  $ satisfying $\deg\left(  a_{i}\right)  =1$.

\item Given two integers $k$ and $\ell$, we let $P_{k,\ell}^{\prime}$ be the
$\mathbf{k}$-linear span of all $\left(  k,\ell\right)  $-ultraprimitive
products in $H$. This is a $\mathbf{k}$-submodule of $H_{k}$.

Note that if every two elements of $H_{1}$ commute, then $P_{k,\ell}^{\prime
}=0$ whenever $k\neq\ell$ (because all nonzero ultraprimitive elements have
degree $1$ in this case).
\end{itemize}

Now, the following analogue of Theorem \ref{thm.f.2} holds:

\begin{theorem}
\label{thm.f.2ultra}Let $n\in\mathbb{N}$. For each $k\in\mathbb{N}$, we set
\[
R_{k}^{\prime}:=\sum_{\ell\in\mathbb{N}}P_{k,\ell}^{\prime}H_{n-k};
\]
this is clearly a $\mathbf{k}$-submodule of $H_{k}H_{n-k}\subseteq H_{n}$.

For each $k\in\mathbb{N}$, we set%
\[
Q_{k}^{\prime}:=R_{k}^{\prime}+R_{k+1}^{\prime}+R_{k+2}^{\prime}+\cdots
=\sum_{\ell\geq k}R_{\ell}^{\prime};
\]
again, this is a $\mathbf{k}$-submodule of $H_{n}$.

The following holds:

\textbf{(a)} If every two elements of $H_{1}$ commute, then $\left(
f-k\right)  \left(  Q_{k}^{\prime}\right)  \subseteq Q_{k+1}^{\prime}$ for
each $k\in\mathbb{N}$.

\textbf{(b)} We have $Q_{n-1}^{\prime}=Q_{n}^{\prime}$.

\textbf{(c)} We have $Q_{n+1}^{\prime}=0$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.f.2ultra} (sketched).]\textbf{(a)} Similar to the
proof of Theorem \ref{thm.f.2} \textbf{(a)}. More precisely: Use an argument
analogous to the proof of (\ref{pf.thm.f.2.mainrec}) to prove the equality%
\begin{equation}
\left(  f-\ell\right)  \left(  P_{k,\ell}^{\prime}H_{n-k}\right)  \subseteq
P_{k+1,\ell+1}^{\prime}H_{n-k-1}+\left(  P_{k,\ell-1}^{\prime}+P_{k,\ell
-2}^{\prime}\right)  H_{n-k} \label{pf.thm.f.2ultra.a.mainrec}%
\end{equation}
for each $k,\ell\in\mathbb{Z}$. Applying this to $\ell=k$, we obtain%
\begin{equation}
\left(  f-k\right)  \left(  P_{k,k}^{\prime}H_{n-k}\right)  \subseteq
P_{k+1,k+1}^{\prime}H_{n-k-1}+\left(  P_{k,k-1}^{\prime}+P_{k,k-2}^{\prime
}\right)  H_{n-k}. \label{pf.thm.f.2ultra.a.mainrec-k}%
\end{equation}


However, if every two elements of $H_{1}$ commute, then we have $P_{k,\ell
}^{\prime}=0$ whenever $k\neq\ell$, and thus we can simplify
(\ref{pf.thm.f.2ultra.a.mainrec-k}) to%
\[
\left(  f-k\right)  \left(  P_{k,k}^{\prime}H_{n-k}\right)  \subseteq
P_{k+1,k+1}^{\prime}H_{n-k-1}.
\]
This rewrites further as $\left(  f-k\right)  \left(  Q_{k}^{\prime}\right)
\subseteq Q_{k+1}^{\prime}$, since we have $Q_{k}^{\prime}=P_{k,k}^{\prime
}H_{n-k}$ (again because we have $P_{k,\ell}^{\prime}=0$ whenever $k\neq\ell$)
and $Q_{k+1}^{\prime}=P_{k+1,k+1}^{\prime}H_{n-k-1}$ (similarly). Thus,
Theorem \ref{thm.f.2ultra} \textbf{(a)} is proved.

\textbf{(b)}, \textbf{(c)} Similar to the proof of Theorem \ref{thm.f.2}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.f.1comm} (sketched).]This follows easily from
Theorem \ref{thm.f.2ultra} (similarly to how we proved Theorem \ref{thm.f.1}
using Theorem \ref{thm.f.2}).
\end{proof}

We remark the following:

\begin{theorem}
\label{thm.f.1comm-explicit}Assume that every two elements of $H_{1}$ commute.
Then, each $x\in H$ and $k\in\mathbb{N}$ satisfy%
\[
\left(  \prod_{i=0}^{k-1}\left(  f-i\right)  \right)  \left(  x\right)
=\sum_{\left(  x\right)  }p_{1}\left(  x_{\left(  k\right)  }\right)
p_{1}\left(  x_{\left(  k-1\right)  }\right)  \cdots p_{1}\left(  x_{\left(
1\right)  }\right)  \cdot x_{\left(  k+1\right)  }%
\]
(where we use Sweedler notation).
\end{theorem}

This theorem (which is not hard to prove by induction on $k$) can be used to
prove Theorem \ref{thm.f.1comm} again.

\begin{question}
\label{quest.fk.annpol}What happens if we generalize $f=p_{1}\star
\operatorname*{id}$ to $f_{k}=p_{k}\star\operatorname*{id}$ ? Is $f_{k}$ still
annihilated by a reasonably nice polynomial on each $H_{n}$ (in analogy to
Theorem \ref{thm.f.1})?
\end{question}

Here is a partial answer for $n=4$ and $k=2$. If $a\in H_{4}$, then%
\[
f_{2}\left(  a\right)  =\sum_{\left(  a\right)  }p_{2}\left(  a_{\left(
1\right)  }\right)  a_{\left(  2\right)  }\in H_{2}H_{2}.
\]
Hence, $f_{2}\left(  H_{4}\right)  \subseteq H_{2}H_{2}$. Furthermore, for any
$a,b\in H$, we have%
\begin{align*}
f_{2}\left(  ab\right)   &  =\sum_{\left(  a\right)  ,\left(  b\right)  }%
p_{2}\left(  a_{\left(  1\right)  }b_{\left(  1\right)  }\right)  a_{\left(
2\right)  }b_{\left(  2\right)  }\\
&  =\sum_{\left(  b\right)  }p_{2}\left(  b_{\left(  1\right)  }\right)
ab_{\left(  2\right)  }+\sum_{\left(  a\right)  ,\left(  b\right)  }%
p_{1}\left(  a_{\left(  1\right)  }\right)  p_{1}\left(  b_{\left(  1\right)
}\right)  a_{\left(  2\right)  }b_{\left(  2\right)  }+\sum_{\left(  a\right)
}p_{2}\left(  a_{\left(  1\right)  }\right)  a_{\left(  2\right)  }b.
\end{align*}
Thus, if $a,b\in H_{2}$, then%
\begin{align}
f_{2}\left(  ab\right)   &  =\underbrace{\sum_{\left(  b\right)  }p_{2}\left(
b_{\left(  1\right)  }\right)  ab_{\left(  2\right)  }}_{=ba}+\underbrace{\sum
_{\left(  a\right)  ,\left(  b\right)  }p_{1}\left(  a_{\left(  1\right)
}\right)  p_{1}\left(  b_{\left(  1\right)  }\right)  a_{\left(  2\right)
}b_{\left(  2\right)  }}_{\in H_{1}^{4}}+\underbrace{\sum_{\left(  a\right)
}p_{2}\left(  a_{\left(  1\right)  }\right)  a_{\left(  2\right)  }}%
_{=a}b\nonumber\\
&  \in ba+H_{1}^{4}+ab, \label{sol.quest.fk.annpol.24.4}%
\end{align}
so that
\[
\left(  f_{2}-2\right)  \left(  ab\right)  \in\underbrace{ab-ba}_{=\left[
a,b\right]  \in\left[  H_{2},H_{2}\right]  }+H_{1}^{4}\subseteq\left[
H_{2},H_{2}\right]  +H_{1}^{4}.
\]
Thus, $\left(  f_{2}-2\right)  \left(  H_{2}H_{2}\right)  \subseteq\left[
H_{2},H_{2}\right]  +H_{1}^{4}$. Furthermore, if $a,b\in H_{2}$, then%
\[
f_{2}\left(  \left[  a,b\right]  \right)  \in H_{1}^{4}%
\]
(by applying (\ref{sol.quest.fk.annpol.24.4}) to $b$ and $a$ instead of $a$
and $b$, and subtracting the result from the original
(\ref{sol.quest.fk.annpol.24.4})). Thus, $f_{2}\left(  \left[  H_{2}%
,H_{2}\right]  \right)  \subseteq H_{1}^{4}$. Finally, on $H_{1}^{4}$, the map
$f_{2}$ acts as the sum of all $\left(  2,2\right)  $-shuffles, so its
eigenvalues are (a subset of) $0,2,6$ (this is a particular case of
Proposition \ref{prop.sum-shuffle.evals} below). Thus, $\left(  f_{2}\left(
f_{2}-2\right)  \left(  f_{2}-6\right)  \right)  \left(  H_{1}^{4}\right)
=0$. Combining this all, we see that%
\[
\left(  f_{2}\left(  f_{2}-2\right)  \left(  f_{2}-6\right)  \right)  \circ
f_{2}\circ\left(  f_{2}-2\right)  \circ f_{2}=0.
\]


The last part of this argument (the part after falling down into $H_{1}^{4}$)
is easy to generalize, and the result is known:

\begin{proposition}
\label{prop.sum-shuffle.evals}Let $n\in\mathbb{N}$. In the group ring
$\mathbb{Z}\left[  \mathfrak{S}_{n}\right]  $, let $\operatorname*{sh}%
\nolimits_{k}$ denote the sum of all permutations $\sigma\in\mathfrak{S}_{n}$
that satisfy $\sigma\left(  1\right)  <\sigma\left(  2\right)  <\cdots
<\sigma\left(  k\right)  $ and $\sigma\left(  k+1\right)  <\sigma\left(
k+2\right)  <\cdots<\sigma\left(  n\right)  $. For each partition
$\lambda=\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{q}\right)  $ of $n$
(with all $i$ satisfying $\lambda_{i}>0$), we define the \emph{index} of
$\lambda$ to be the number of all subsets $T$ of $\left[  q\right]  $ that
satisfy $\sum_{i\in T}\lambda_{i}=k$. (Roughly speaking, the index of
$\lambda$ is thus the number of ways to choose a subsequence of $\lambda$ that
sums to $k$.) Let $G\left(  n,k\right)  $ denote the set of the indices of all
partitions of $n$. Then,%
\[
\prod_{i\in G\left(  n,k\right)  }\left(  \operatorname*{sh}\nolimits_{k}%
-i\right)  =0.
\]

\end{proposition}

This is a particular case of \cite[Theorem 3.5]{Pang18} (since
$\operatorname*{sh}\nolimits_{k}$ acting on a tensor algebra is exactly the
$f_{k}$ of that tensor algebra), but it also follows from \cite[(4.2) and
Theorem 4.1]{Schocker}.

\begin{question}
\label{quest.prop.sum-shuffle.evals}Can we prove Proposition
\ref{prop.sum-shuffle.evals} again using our filtrations?
\end{question}

The next question is surprisingly doable (apparently easier than Question
\ref{quest.fk.annpol}):

\begin{question}
\label{quest.fk.annpol-comm}Is there an analogue of Theorem \ref{thm.f.1comm}
for $f_{k}$ ? For instance, is $f_{k}$ diagonalizable (in characteristic $0$)
under the assumption that every two elements of $H_{1}+H_{2}+\cdots+H_{k}$ commute?
\end{question}

This has a positive and rather explicit answer:

For each $q\in\mathbb{Z}$, we let $\left[  q\right]  $ denote the set
$\left\{  1,2,\ldots,q\right\}  $. (This set is empty when $q\leq0$.)

Fix a positive integer $k$.

We define an $\left(  n,k\right)  $\textit{-snail} to be a tuple $\left(
a_{1},a_{2},\ldots,a_{q}\right)  $ (with arbitrary $q\in\mathbb{N}$) of
homogeneous elements of $H$ such that%
\[
0<\deg\left(  a_{i}\right)  \ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left[
q\right]
\]
and%
\[
\deg\left(  a_{i}\right)  \leq k\ \ \ \ \ \ \ \ \ \ \text{for each }%
i\in\left[  q-1\right]
\]
and%
\[
\deg\left(  a_{1}\right)  +\deg\left(  a_{2}\right)  +\cdots+\deg\left(
a_{q}\right)  =n.
\]
(Note that $\deg\left(  a_{q}\right)  $ can be any positive integer if $q>0$.)

The \textit{index} of an $\left(  n,k\right)  $-snail $\left(  a_{1}%
,a_{2},\ldots,a_{q}\right)  $ is defined as the number of all subsets $T$ of
$\left[  q\right]  $ satisfying $\sum_{i\in T}\deg\left(  a_{i}\right)  =k$.
(For example, if $k=1$, then the index of any $\left(  n,k\right)  $-snail
$\left(  a_{1},a_{2},\ldots,a_{q}\right)  $ is either $q-1$ or $q$, since the
only subsets $T\subseteq\left[  q\right]  $ satisfying $\sum_{i\in T}%
\deg\left(  a_{i}\right)  =1$ are $\left\{  1\right\}  ,\left\{  2\right\}
,\ldots,\left\{  q-1\right\}  $ and possibly $\left\{  q\right\}  $.)

Intuitively, you can think of the index of an $\left(  n,k\right)  $-snail
$\left(  a_{1},a_{2},\ldots,a_{q}\right)  $ as the number of all ways to
assemble a degree-$k$ product by striking some factors from the product
$a_{1}a_{2}\cdots a_{q}$.

We say that an integer $i\in\mathbb{Z}$ is $\left(  n,k\right)  $%
\textit{-friendly} if there is an $\left(  n,k\right)  $-snail $\left(
a_{1},a_{2},\ldots,a_{q}\right)  $ whose index is $i$.

Let $F\left(  n,k\right)  $ denote the set of all $\left(  n,k\right)
$-friendly integers.

Clearly, this set $F\left(  n,k\right)  $ is finite (since the number of all
possible degree profiles of $\left(  n,k\right)  $-snails is finite). Now I claim:

\begin{theorem}
\label{thm.fk.kcomm}Let $k$ be a positive integer. Assume that every two
elements of $H_{1}+H_{2}+\cdots+H_{k}$ commute. Let $f_{k}=p_{k}%
\star\operatorname*{id}$. Let $n$ be a positive integer. Then,%
\[
\prod_{i\in F\left(  n,k\right)  }\left(  f_{k}-i\right)
=0\ \ \ \ \ \ \ \ \ \ \text{on }H_{n}.
\]

\end{theorem}

\begin{example}
Let us see what this says for $n=4$ and $k=2$. In this case, the $\left(
n,k\right)  $-snails have the following forms:

\begin{itemize}
\item $\left(  a_{1}\right)  $ with $\deg a_{1}=4$; this has index $0$.

\item $\left(  a_{1},a_{2}\right)  $ with $\deg a_{1}=1$ and $\deg a_{2}=3$;
this has index $0$.

\item $\left(  a_{1},a_{2}\right)  $ with $\deg a_{1}=2$ and $\deg a_{2}=2$;
this has index $2$.

\item $\left(  a_{1},a_{2},a_{3}\right)  $ with $\deg a_{1}=1$ and $\deg
a_{2}=1$ and $\deg a_{3}=2$; this has index $2$.

\item $\left(  a_{1},a_{2},a_{3}\right)  $ with $\deg a_{1}=1$ and $\deg
a_{2}=2$ and $\deg a_{3}=1$; this has index $2$.

\item $\left(  a_{1},a_{2},a_{3}\right)  $ with $\deg a_{1}=2$ and $\deg
a_{2}=1$ and $\deg a_{3}=1$; this has index $2$.

\item $\left(  a_{1},a_{2},a_{3},a_{4}\right)  $ with $\deg a_{1}=1$ and $\deg
a_{2}=1$ and $\deg a_{3}=1$ and $\deg a_{4}=1$; this has index $6$.
\end{itemize}

Thus, $F\left(  n,k\right)  =\left\{  0,2,6\right\}  $ here. Hence, Theorem
\ref{thm.fk.kcomm} yields%
\[
f_{2}\left(  f_{2}-2\right)  \left(  f_{2}-6\right)
=0\ \ \ \ \ \ \ \ \ \ \text{on }H_{4}.
\]

\end{example}

We notice that Theorem \ref{thm.f.1comm} can easily be obtained by applying
Theorem \ref{thm.fk.kcomm} to $k=1$ (since $F\left(  n,1\right)  =\left\{
0,1,\ldots,n-2,n\right\}  $).

\begin{proof}
[Proof of Theorem \ref{thm.fk.kcomm} (sketched).]Very rough outline so far:

For each $i\in\mathbb{Z}$, we define $V_{i}$ to be the $\mathbf{k}$-linear
span of the products of all $\left(  n,k\right)  $-snails with index $\geq i$.
Thus, $V_{0}=H_{n}$, since each $h\in H_{n}$ is the product of the $\left(
n,k\right)  $-snail $\left(  h\right)  $ whose index is $\geq0$. On the other
hand, $V_{i}=0$ for all sufficiently large $i$. Furthermore, $V_{i}=V_{i+1}$
for any $i\in\mathbb{N}$ that is not $\left(  n,k\right)  $-friendly.

Now, we claim that
\begin{equation}
\left(  f_{k}-i\right)  \left(  V_{i}\right)  \subseteq V_{i+1}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\mathbb{N}.
\label{pf.thm.fk.kcomm.main}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.fk.kcomm.main}):} Here is just the main idea.

Let $i\in\mathbb{N}$. The typical generator of $V_{i}$ is a product
$a_{1}a_{2}\cdots a_{q}$ of an $\left(  n,k\right)  $-snail $\left(
a_{1},a_{2},\ldots,a_{q}\right)  $ with index $\geq i$. Consider this $\left(
n,k\right)  $-snail. We have%
\begin{align*}
&  f_{k}\left(  a_{1}a_{2}\cdots a_{q}\right) \\
&  =\left(  p_{k}\star\operatorname*{id}\right)  \left(  a_{1}a_{2}\cdots
a_{q}\right) \\
&  =\sum_{\left(  a_{1}\right)  ,\left(  a_{2}\right)  ,\ldots,\left(
a_{q}\right)  }p_{k}\left(  \left(  a_{1}\right)  _{\left(  1\right)  }\left(
a_{2}\right)  _{\left(  1\right)  }\cdots\left(  a_{q}\right)  _{\left(
1\right)  }\right)  \left(  a_{1}\right)  _{\left(  2\right)  }\left(
a_{2}\right)  _{\left(  2\right)  }\cdots\left(  a_{q}\right)  _{\left(
2\right)  }\\
&  =\sum_{j_{1}+j_{2}+\cdots+j_{q}=k}\ \ \sum_{\left(  a_{1}\right)  ,\left(
a_{2}\right)  ,\ldots,\left(  a_{q}\right)  }p_{j_{1}}\left(  \left(
a_{1}\right)  _{\left(  1\right)  }\right)  p_{j_{2}}\left(  \left(
a_{2}\right)  _{\left(  1\right)  }\right)  \cdots p_{j_{q}}\left(  \left(
a_{q}\right)  _{\left(  1\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  a_{1}\right)  _{\left(  2\right)  }\left(
a_{2}\right)  _{\left(  2\right)  }\cdots\left(  a_{q}\right)  _{\left(
2\right)  }%
\end{align*}
(since $p_{k}\left(  u_{1}u_{2}\cdots u_{q}\right)  =\sum_{j_{1}+j_{2}%
+\cdots+j_{q}=k}p_{j_{1}}\left(  u_{1}\right)  p_{j_{2}}\left(  u_{2}\right)
\cdots p_{j_{q}}\left(  u_{q}\right)  $ for any $u_{1},u_{2},\ldots,u_{q}$ in
any graded algebra).

The addends on the right hand side of this equality are again products of
$\left(  n,k\right)  $-snails\footnote{Here we are using the fact that
$p_{j_{q}}\left(  \left(  a_{q}\right)  _{\left(  1\right)  }\right)  $ is
homogeneous of degree $j_{q}\leq j_{1}+j_{2}+\cdots+j_{q}=k$.}. We must prove that

\begin{itemize}
\item \textbf{(a)} their indices are $\geq i$, and

\item \textbf{(b)} when their indices are $i$, the original snail $\left(
a_{1},a_{2},\ldots,a_{q}\right)  $ must itself have index $i$, and furthermore
these addends all equal $a_{1}a_{2}\cdots a_{q}$ and there are $i$ many of them.
\end{itemize}

\noindent This is not particularly hard, but annoying and finicky. Each addend
on the right hand side is a product of the form%
\[
p_{j_{1}}\left(  \left(  a_{1}\right)  _{\left(  1\right)  }\right)  p_{j_{2}%
}\left(  \left(  a_{2}\right)  _{\left(  1\right)  }\right)  \cdots p_{j_{q}%
}\left(  \left(  a_{q}\right)  _{\left(  1\right)  }\right)  \left(
a_{1}\right)  _{\left(  2\right)  }\left(  a_{2}\right)  _{\left(  2\right)
}\cdots\left(  a_{q}\right)  _{\left(  2\right)  }.
\]
After we remove the degree-$0$ factors from this product, we are left with a
product of $\leq2q$ factors. For each way of assembling a degree-$k$ product
from the original $a_{1}a_{2}\cdots a_{q}$ product, there is a way to do so
from the new product (just replace each $a_{m}$ by $p_{j_{m}}\left(  \left(
a_{m}\right)  _{\left(  1\right)  }\right)  \left(  a_{m}\right)  _{\left(
2\right)  }$). Moreover, there is at least one new way of assembling a
degree-$k$ product, namely by taking $p_{j_{1}}\left(  \left(  a_{1}\right)
_{\left(  1\right)  }\right)  p_{j_{2}}\left(  \left(  a_{2}\right)  _{\left(
1\right)  }\right)  \cdots p_{j_{q}}\left(  \left(  a_{q}\right)  _{\left(
1\right)  }\right)  $. This way is indeed new, unless each of the $q$ indices
$j_{s}$ equals either $0$ or $\deg a_{s}$\ \ \ \ \ \footnote{Why? Well, assume
that some $s$ satisfies $j_{s}\neq0$ and $j_{s}\neq\deg a_{s}$. We must show
that the product%
\[
p_{j_{1}}\left(  \left(  a_{1}\right)  _{\left(  1\right)  }\right)  p_{j_{2}%
}\left(  \left(  a_{2}\right)  _{\left(  1\right)  }\right)  \cdots p_{j_{q}%
}\left(  \left(  a_{q}\right)  _{\left(  1\right)  }\right)
\]
equals none of the $\prod_{t\in T}p_{j_{t}}\left(  \left(  a_{t}\right)
_{\left(  1\right)  }\right)  \left(  a_{t}\right)  _{\left(  2\right)  }$
products (as formal product, not just as elements of $H$). But this is easy:
This product contains the non-constant factor $p_{j_{s}}\left(  \left(
a_{s}\right)  _{\left(  1\right)  }\right)  $, but does not contain the
non-constant factor $\left(  a_{s}\right)  _{\left(  2\right)  }$; however,
any product of the form $\prod_{t\in T}p_{j_{t}}\left(  \left(  a_{t}\right)
_{\left(  1\right)  }\right)  \left(  a_{t}\right)  _{\left(  2\right)  }$
must contain either none or both of these two factors.}. So the index of our
addend is greater or equal to the index of $\left(  a_{1},a_{2},\ldots
,a_{q}\right)  $, and is strictly greater unless each of the $q$ indices
$j_{s}$ equals either $0$ or $\deg a_{s}$. This proves part \textbf{(a)}, and
more or less proves part \textbf{(b)}. All that remains to be done for part
\textbf{(b)} is showing that if each $j_{s}$ equals either $0$ or $\deg a_{s}%
$, then
\begin{align*}
&  p_{j_{1}}\left(  \left(  a_{1}\right)  _{\left(  1\right)  }\right)
p_{j_{2}}\left(  \left(  a_{2}\right)  _{\left(  1\right)  }\right)  \cdots
p_{j_{q}}\left(  \left(  a_{q}\right)  _{\left(  1\right)  }\right)  \left(
a_{1}\right)  _{\left(  2\right)  }\left(  a_{2}\right)  _{\left(  2\right)
}\cdots\left(  a_{q}\right)  _{\left(  2\right)  }\\
&  =a_{1}a_{2}\cdots a_{q}.
\end{align*}
This is easy: All but perhaps the last factor $\left(  a_{q}\right)  _{\left(
2\right)  }$ in the product belong to $H_{1}+H_{2}+\cdots+H_{k}$, and thus
commute with each other.]
\end{proof}

\begin{fineprint}
\begin{proof}
[Some thoughts on Question \ref{quest.fk.annpol}.]What follows is handwaving
that probably makes no sense. I am just storing it here so I don't forget it.

Again, argue by%
\[
f_{k}\left(  abcd\right)  =\sum_{u+v+w+x=k}\ \ \sum_{\left(  a\right)
,\ \left(  b\right)  ,\ \left(  c\right)  ,\ \left(  d\right)  }f_{u}\left(
a_{\left(  1\right)  }\right)  f_{v}\left(  b_{\left(  1\right)  }\right)
f_{w}\left(  c_{\left(  1\right)  }\right)  f_{x}\left(  d_{\left(  1\right)
}\right)  a_{\left(  2\right)  }b_{\left(  2\right)  }c_{\left(  2\right)
}d_{\left(  2\right)  }.
\]
Thus, factors of degree $>k$ get cut into pieces. Other factors get either cut
or moved around. Either at least one factor gets cut, or all factors get moved
(i.e., the factors get permuted). Thus, we obtain a filtration of $H_{n}$ by
the number of factors. Better, we obtain a \textquotedblleft poset
filtration\textquotedblright\ by the partitions of $n$ (partially ordered by
length). For each composition $\alpha=\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{p}\right)  $ of $n$, we let%
\[
W_{\alpha}:=H_{\alpha_{1}}H_{\alpha_{2}}\cdots H_{\alpha_{p}}\subseteq H_{n}.
\]
For each partition $\lambda$ of $n$, we let
\[
V_{\lambda}:=\sum_{\substack{\alpha\in\operatorname*{Comp}\nolimits_{n}%
;\\\lambda=\operatorname*{sort}\alpha}}W_{\alpha}%
\]
where $\operatorname*{Comp}\nolimits_{n}$ is the set of all compositions of
$n$ and where $\operatorname*{sort}\alpha$ means the result of sorting
$\alpha$ in weakly decreasing order. Then, $W_{\alpha}\subseteq W_{\beta}$
whenever $\alpha$ is a refinement of $\beta$. Moreover, $f_{k}$ preserves
$V_{\geq\lambda}:=V_{\lambda}+\sum_{\substack{\mu\in\operatorname*{Par}%
\nolimits_{n};\\\ell\left(  \mu\right)  >\ell\left(  \lambda\right)  }}V_{\mu
}$, and in fact acts by permutation of factors on $V_{\geq\lambda}%
/V_{>\lambda}$. This permutation action is some sort of Schocker sum like the
$\operatorname*{sh}\nolimits_{k}$ in Proposition \ref{prop.sum-shuffle.evals}.

Better yet, we can replace $f_{k}$ by any $p_{\alpha,\sigma}$ (see below) and
the map still preserves the filtration and acts on each factor as a quotient
of an $\mathbb{Z}\left[  S_{n}\right]  $-element.
\end{proof}
\end{fineprint}

Another question, less fitting into this project but related (possibly the
above results can be of some use):

\begin{question}
Can we find $\operatorname*{Ker}f$ if $\mathbf{k}$ has characteristic $0$ ?
This would generalize \cite[Theorem 7.15]{Grinbe16}.
\end{question}

To be fair, there are two ways to generalize \cite[Theorem 7.15]{Grinbe16},
and it is not clear which one is better. Indeed, if $H$ is the tensor algebra
of a $\mathbf{k}$-module $L$, then the map $\mathbf{t}^{\prime}$ from
\cite[\S 7]{Grinbe16} can be viewed as either $f=p_{1}\star\operatorname*{id}$
or $\left(  p_{1}\otimes\operatorname*{id}\right)  \circ\Delta$; the two
operators have the same kernel since the multiplication map $H\otimes
H\rightarrow H$ is injective on each single $H_{i}\otimes H_{j}$. More
generally, I believe I can show that $\operatorname*{Ker}f=\operatorname*{Ker}%
\left(  \left(  p_{1}\otimes\operatorname*{id}\right)  \circ\Delta\right)  $
whenever $H$ is cocommutative and $\operatorname*{char}\mathbf{k}=0$. However,
if $H$ is not cocommutative (e.g., the shuffle algebra) or if
$\operatorname*{char}\mathbf{k}\neq0$, then this is generally false.

\section{A Hopf algebra of natural Hopf algebra endomorphisms}

\bigskip

\textbf{This section is obsolete!} A more up-to-date version can be found in
\cite{solhopf}.

\bigskip



Let us now take a step back and look at both the Aguiar--Lauve $\left(
\operatorname*{id}-S^{2}\right)  ^{n}\left(  H_{n}\right)  =0$ result and
Theorem \ref{thm.f.1} from a bird's eye view: Both of them are statements of
the form \textquotedblleft A universally defined linear map on any connected
graded Hopf algebra is $0$ on the $n$-th homogeneous
component\textquotedblright. In the Aguiar--Lauve case, this map is $\left(
\operatorname*{id}-S^{2}\right)  ^{n}$, while in Theorem \ref{thm.f.1} it is
$\left(  f-n\right)  \circ\left(  f-\left(  n-2\right)  \right)  ^{2}%
\circ\prod_{i=0}^{n-3}\left(  f-i\right)  ^{n-1-i}$ (for $n\geq2$). Many other
such maps can be identified, such as the Adams operators $\operatorname*{id}%
\nolimits^{\star k}$ (thanks, Amy, for reminding me about them), or
composition powers $S^{k}$ of the antipode, or various compositions or
convolutions of projections $p_{0},p_{1},p_{2},\ldots$.

This makes the following question rather natural:

\begin{question}
Is there a mechanical way to prove such statements?
\end{question}

Let me try to concretize this, following the lead of Patras and Reutenauer
\cite{PatReu98}, which have studied the case of a connected graded
cocommutative Hopf algebra.

\subsection{The maps $p_{\alpha,\sigma}$ for a graded bialgebra $H$}

Consider a graded bialgebra $H$ over a commutative ring $\mathbf{k}$. The
$\mathbf{k}$-module $\operatorname*{End}\nolimits_{\operatorname*{gr}}H$ of
graded $\mathbf{k}$-module endomorphisms of $H$ is itself a $\mathbf{k}%
$-algebra in two different ways: It has an \textquotedblleft internal
multiplication\textquotedblright\ that corresponds to composition of
endomorphisms, and an \textquotedblleft external
multiplication\textquotedblright\ that corresponds to convolution of
endomorphisms. We consider the $\mathbf{k}$-submodule $\mathbf{E}\left(
H\right)  $ of $\operatorname*{End}\nolimits_{\operatorname*{gr}}H$ that
consists only of those $f\in\operatorname*{End}\nolimits_{\operatorname*{gr}%
}H$ that annihilate all but finitely many graded components of $H$ (that is,
that satisfy $f\left(  H_{n}\right)  =0$ for all sufficiently high $n$). This
submodule $\mathbf{E}\left(  H\right)  $ is itself graded, with the $n$-th
graded component being canonically isomorphic to $\operatorname*{End}%
\nolimits_{\mathbf{k}}\left(  H_{n}\right)  $. This submodule $\mathbf{E}%
\left(  H\right)  $ is preserved under both internal and external multiplication.

Unlike Hazewinkel in \cite{Hazewi04}, we shall however not consider
$\mathbf{E}\left(  H\right)  $ for a specific $H$, but we shall use it as
inspiration to study the functorial endomorphisms of $H$ defined for all
connected graded Hopf algebras. Such endomorphisms include

\begin{itemize}
\item the projections $p_{0},p_{1},p_{2},\ldots$;

\item their convolutions $p_{\left(  i_{1},i_{2},\ldots,i_{k}\right)
}:=p_{i_{1}}\star p_{i_{2}}\star\cdots\star p_{i_{k}}$ with $i_{1}%
,i_{2},\ldots,i_{k}\geq0$;

\item the compositions $p_{\alpha}\circ p_{\beta}\circ\cdots\circ p_{\kappa}$
of such convolutions.
\end{itemize}

However, we can define a broader class of such endomorphisms. To do so, we
need some notations:

\begin{definition}
\label{def.Sk-action}Let $H$ be any $\mathbf{k}$-module. Then, we let the
group $\mathfrak{S}_{k}$ act on $H^{\otimes k}$ from the left by permuting
tensorands, according to the rule%
\begin{align*}
\sigma\cdot\left(  h_{1}\otimes h_{2}\otimes\cdots\otimes h_{k}\right)   &
=h_{\sigma^{-1}\left(  1\right)  }\otimes h_{\sigma^{-1}\left(  2\right)
}\otimes\cdots\otimes h_{\sigma^{-1}\left(  k\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }\sigma\in\mathfrak{S}_{k}\text{ and
}h_{1},h_{2},\ldots,h_{k}\in H.
\end{align*}

\end{definition}

\begin{definition}
\label{def.iterated}Let $H$ be a $\mathbf{k}$-bialgebra. For any
$k\in\mathbb{N}$, we let $m^{\left[  k\right]  }:H^{\otimes k}\rightarrow H$
and $\Delta^{\left[  k\right]  }:H\rightarrow H^{\otimes k}$ be the iterated
multiplication and iterated comultiplication maps (denoted $m^{\left(
k-1\right)  }$ and $\Delta^{\left(  k-1\right)  }$ in \cite{GriRei}). Note
that $m^{\left[  k\right]  }$ sends each pure tensor $a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{k}$ to the product $a_{1}a_{2}\cdots a_{k}$,
whereas $\Delta^{\left[  k\right]  }$ sends each element $x\in H$ to
$\sum_{\left(  x\right)  }x_{\left(  1\right)  }\otimes x_{\left(  2\right)
}\otimes\cdots\otimes x_{\left(  k\right)  }$ (using Sweedler notation).
\end{definition}

\begin{definition}
\label{def.wc}\textbf{(a)} A \emph{weak composition} means a finite tuple of
nonnegative integers.

\textbf{(b)} If $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha
_{k}\right)  $ is a weak composition, then its \emph{size} $\left\vert
\alpha\right\vert $ is defined to be the number $\alpha_{1}+\alpha_{2}%
+\cdots+\alpha_{k}\in\mathbb{N}$.
\end{definition}

\begin{definition}
\label{def.Palpha}Let $H$ be a graded $\mathbf{k}$-module.

For any weak composition $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{k}\right)  $, we define the projection map $P_{\alpha}:H^{\otimes
k}\rightarrow H^{\otimes k}$ to be the tensor product $p_{\alpha_{1}}\otimes
p_{\alpha_{2}}\otimes\cdots\otimes p_{\alpha_{k}}$. (Thus, if we regard
$H^{\otimes k}$ as an $\mathbb{N}^{k}$-graded $\mathbf{k}$-module, then
$P_{\alpha}$ is its projection on its $\left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{k}\right)  $-degree component.)
\end{definition}

\begin{definition}
Let $H$ be a graded $\mathbf{k}$-bialgebra.

For any weak composition $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{k}\right)  $ and any permutation $\sigma\in\mathfrak{S}_{k}$, we
define a map $p_{\alpha,\sigma}:H\rightarrow H$ as follows:

\begin{itemize}
\item We can define it by the formula%
\[
p_{\alpha,\sigma}\left(  x\right)  =\sum_{\left(  x\right)  }p_{\alpha_{1}%
}\left(  x_{\left(  \sigma\left(  1\right)  \right)  }\right)  p_{\alpha_{2}%
}\left(  x_{\left(  \sigma\left(  2\right)  \right)  }\right)  \cdots
p_{\alpha_{k}}\left(  x_{\left(  \sigma\left(  k\right)  \right)  }\right)
\]
for every $x\in H$, where we are using the Sweedler notation $\sum_{\left(
x\right)  }x_{\left(  1\right)  }\otimes x_{\left(  2\right)  }\otimes
\cdots\otimes x_{\left(  k\right)  }$ for the iterated coproduct
$\Delta^{\left[  k\right]  }\left(  x\right)  \in H^{\otimes k}$.

\item More formally, we can define it as follows:%
\begin{equation}
p_{\alpha,\sigma}:=m^{\left[  k\right]  }\circ P_{\alpha}\circ\sigma^{-1}%
\circ\Delta^{\left[  k\right]  }, \label{eq.pas.formal-def}%
\end{equation}
where the $\sigma^{-1}$ really means the action of $\sigma^{-1}\in
\mathfrak{S}_{k}$ on $H^{\otimes k}$ (as in Definition \ref{def.Sk-action}).
\end{itemize}

This map $p_{\alpha,\sigma}$ is a graded $\mathbf{k}$-module endomorphism of
$H$ that sends $H_{\left\vert \alpha\right\vert }$ to $H_{\left\vert
\alpha\right\vert }$ and sends each other $H_{n}$ to $0$. Thus, $p_{\alpha
,\sigma}$ lies in the $\left\vert \alpha\right\vert $-th graded component of
$\mathbf{E}\left(  H\right)  $.
\end{definition}

If $H$ is commutative or cocommutative, then we can bring $p_{\alpha,\sigma}$
to the form $p_{\beta}$ for some weak composition $\beta$; indeed, we have%
\[
p_{\alpha,\sigma}=p_{\sigma\cdot\alpha}\ \ \ \ \ \ \ \ \ \ \text{if }H\text{
is commutative,}%
\]
where $\sigma\cdot\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{k}\right)
:=\left(  \alpha_{\sigma^{-1}\left(  1\right)  },\alpha_{\sigma^{-1}\left(
2\right)  },\ldots,\alpha_{\sigma^{-1}\left(  k\right)  }\right)  $, and we
have%
\[
p_{\alpha,\sigma}=p_{\alpha}\ \ \ \ \ \ \ \ \ \ \text{if }H\text{ is
cocommutative.}%
\]
However, in general, when $H$ is neither commutative nor cocommutative, we
cannot \textquotedblleft simplify\textquotedblright\ $p_{\alpha,\sigma}$.

Furthermore, when $H$ is connected, each $p_{\alpha,\sigma}$ for a weak
composition $\alpha$ and a permutation $\sigma$ can be rewritten in the form
$p_{\beta,\tau}$ for a composition (not just weak composition) $\beta$ and a
permutation $\tau$. (Indeed, since $H$ is connected, we can remove all
$p_{0}\left(  x_{\left(  i\right)  }\right)  $ factors from the product in the
definition of $p_{\alpha,\sigma}$.) See Proposition \ref{prop.mopis.reduce}
below for an explicit statement of this claim.

It is easy to see that any convolution of two maps of the form $p_{\alpha
,\sigma}$ is again a map of such form:

\begin{proposition}
\label{prop.sol-mac-0}Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{k}\right)  $ be a weak composition, and let $\sigma\in\mathfrak{S}%
_{k}$ be a permutation.

Let $\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{\ell}\right)  $ be a weak
composition, and let $\tau\in\mathfrak{S}_{\ell}$ be a permutation.

Then,
\begin{equation}
p_{\alpha,\sigma}\star p_{\beta,\tau}=p_{\alpha\beta,\sigma\oplus\tau},
\label{eq.pas-conv}%
\end{equation}
where $\alpha\beta$ is the concatenation of $\alpha$ and $\beta$ (that is, the
weak composition $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{k},\beta
_{1},\beta_{2},\ldots,\beta_{\ell}\right)  $), whereas $\sigma\oplus\tau$ is
the image of $\left(  \sigma,\tau\right)  $ under the obvious map
$\mathfrak{S}_{k}\times\mathfrak{S}_{\ell}\rightarrow\mathfrak{S}_{k+\ell}$.
\end{proposition}

It is more interesting to compute the composition of two maps of the form
$p_{\alpha,\sigma}$. It turns out that this is again a $\mathbf{k}$-linear
combination of maps of such form, and the explicit formula is similar to
Solomon's Mackey formula for the descent algebra (or, even more closely
related, \cite[Theorem 9.2 and \S 9.5.1]{Reuten93}):

\begin{theorem}
\label{thm.sol-mac-1}Let $\alpha=\left(  \alpha_{1},\alpha_{2},\ldots
,\alpha_{k}\right)  $ be a weak composition, and let $\sigma\in\mathfrak{S}%
_{k}$ be a permutation.

Let $\beta=\left(  \beta_{1},\beta_{2},\ldots,\beta_{\ell}\right)  $ be a weak
composition, and let $\tau\in\mathfrak{S}_{\ell}$ be a permutation.

Then,%
\[
p_{\alpha,\sigma}\circ p_{\beta,\tau}=\sum_{\substack{\gamma_{i,j}%
\in\mathbb{N}\text{ for all }i\in\left[  k\right]  \text{ and }j\in\left[
\ell\right]  ;\\\gamma_{i,1}+\gamma_{i,2}+\cdots+\gamma_{i,\ell}=\alpha
_{i}\text{ for all }i\in\left[  k\right]  ;\\\gamma_{1,j}+\gamma_{2,j}%
+\cdots+\gamma_{k,j}=\beta_{j}\text{ for all }j\in\left[  \ell\right]
}}p_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)
,\tau\left[  \sigma\right]  },
\]
where $\tau\left[  \sigma\right]  \in\mathfrak{S}_{k\ell}$ is the permutation
that sends each $\ell\left(  i-1\right)  +j$ (with $i\in\left[  k\right]  $
and $j\in\left[  \ell\right]  $) to $k\left(  \tau\left(  j\right)  -1\right)
+\sigma\left(  i\right)  $.
\end{theorem}

\begin{example}
For example, let us try to compute $p_{\left(  a,b\right)  ,s_{1}}\circ
p_{\left(  c,d\right)  ,s_{1}}$, where $s_{1}\in\mathfrak{S}_{2}$ is the
transposition that swaps $1$ with $2$. We have (using sumfree Sweedler
notation)%
\begin{align*}
&  \left(  p_{\left(  a,b\right)  ,s_{1}}\circ p_{\left(  c,d\right)  ,s_{1}%
}\right)  \left(  x\right) \\
&  =p_{\left(  a,b\right)  ,s_{1}}\underbrace{\left(  p_{\left(  c,d\right)
,s_{1}}\left(  x\right)  \right)  }_{=p_{c}\left(  x_{\left(  2\right)
}\right)  p_{d}\left(  x_{\left(  1\right)  }\right)  }\\
&  =p_{\left(  a,b\right)  ,s_{1}}\left(  p_{c}\left(  x_{\left(  2\right)
}\right)  p_{d}\left(  x_{\left(  1\right)  }\right)  \right) \\
&  =p_{a}\left(  \left(  p_{c}\left(  x_{\left(  2\right)  }\right)
p_{d}\left(  x_{\left(  1\right)  }\right)  \right)  _{\left(  2\right)
}\right)  \cdot p_{b}\left(  \left(  p_{c}\left(  x_{\left(  2\right)
}\right)  p_{d}\left(  x_{\left(  1\right)  }\right)  \right)  _{\left(
1\right)  }\right) \\
&  =p_{a}\left(  \left(  p_{c}\left(  x_{\left(  2\right)  }\right)  \right)
_{\left(  2\right)  }\left(  p_{d}\left(  x_{\left(  1\right)  }\right)
\right)  _{\left(  2\right)  }\right)  \cdot p_{b}\left(  \left(  p_{c}\left(
x_{\left(  2\right)  }\right)  \right)  _{\left(  1\right)  }\left(
p_{d}\left(  x_{\left(  1\right)  }\right)  \right)  _{\left(  1\right)
}\right) \\
&  =\sum_{\substack{c_{1}+c_{2}=c;\\d_{1}+d_{2}=d}}p_{a}\left(  p_{c_{2}%
}\left(  \left(  x_{\left(  2\right)  }\right)  _{\left(  2\right)  }\right)
p_{d_{2}}\left(  \left(  x_{\left(  1\right)  }\right)  _{\left(  2\right)
}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot p_{b}\left(  p_{c_{1}}\left(  \left(  x_{\left(
2\right)  }\right)  _{\left(  1\right)  }\right)  p_{d_{1}}\left(  \left(
x_{\left(  1\right)  }\right)  _{\left(  1\right)  }\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\circ
p_{c}=\left(  \sum_{c_{1}+c_{2}=c}p_{c_{1}}\otimes p_{c_{2}}\right)
\circ\Delta\text{ and likewise for }p_{d}\right) \\
&  =\sum_{\substack{c_{1}+c_{2}=c;\\d_{1}+d_{2}=d;\\c_{2}+d_{2}=a;\\c_{1}%
+d_{1}=b}}p_{c_{2}}\left(  \left(  x_{\left(  2\right)  }\right)  _{\left(
2\right)  }\right)  p_{d_{2}}\left(  \left(  x_{\left(  1\right)  }\right)
_{\left(  2\right)  }\right)  \cdot p_{c_{1}}\left(  \left(  x_{\left(
2\right)  }\right)  _{\left(  1\right)  }\right)  p_{d_{1}}\left(  \left(
x_{\left(  1\right)  }\right)  _{\left(  1\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }p_{a}\left(
p_{i}\left(  y\right)  p_{j}\left(  z\right)  \right)  =%
\begin{cases}
0, & \text{if }i+j\neq a;\\
p_{i}\left(  y\right)  p_{j}\left(  z\right)  , & \text{if }i+j=a
\end{cases}
\right) \\
&  =\sum_{\substack{c_{1}+c_{2}=c;\\d_{1}+d_{2}=d;\\c_{2}+d_{2}=a;\\c_{1}%
+d_{1}=b}}p_{c_{2}}\left(  x_{\left(  4\right)  }\right)  p_{d_{2}}\left(
x_{\left(  2\right)  }\right)  \cdot p_{c_{1}}\left(  x_{\left(  3\right)
}\right)  p_{d_{1}}\left(  x_{\left(  1\right)  }\right)  =\sum
_{\substack{c_{1}+c_{2}=c;\\d_{1}+d_{2}=d;\\c_{2}+d_{2}=a;\\c_{1}+d_{1}%
=b}}p_{\left(  c_{2},d_{2},c_{1},d_{1}\right)  ,\sigma}\left(  x\right)  ,
\end{align*}
where $\sigma\in\mathfrak{S}_{4}$ is the permutation that sends $1,2,3,4$ to
$4,2,3,1$, respectively. This is again a sum of $p_{\alpha,\sigma}$'s, with
$\ell\left(  \alpha\right)  $ varying between $2$ and $4$ (keep in mind that
some of $c_{1},c_{2},d_{1},d_{2}$ can be $0$, in which case we need to remove
them from our compositions).
\end{example}

\begin{proof}
[Proof of Theorem \ref{thm.sol-mac-1} (sketched).]The formal definition
(\ref{eq.pas.formal-def}) yields%
\begin{align}
p_{\alpha,\sigma}  &  =m^{\left[  k\right]  }\circ P_{\alpha}\circ\sigma
^{-1}\circ\Delta^{\left[  k\right]  }\ \ \ \ \ \ \ \ \ \ \text{and}%
\label{pf.thm.sol-mac-1.pas}\\
p_{\beta,\tau}  &  =m^{\left[  \ell\right]  }\circ P_{\beta}\circ\tau
^{-1}\circ\Delta^{\left[  \ell\right]  }. \label{pf.thm.sol-mac-1.pbt}%
\end{align}
We shall next state some rules for \textquotedblleft
commuting\textquotedblright\ the operators in these equalities past each other.

Let $\zeta\in\mathfrak{S}_{k\ell}$ be the permutation that sends each
$k\left(  j-1\right)  +i$ (with $i\in\left[  k\right]  $ and $j\in\left[
\ell\right]  $) to $\ell\left(  i-1\right)  +j$. This is called the
\emph{Zolotarev shuffle} (and appears, e.g., as $\nu^{-1}\mu$ in
\cite{Rousse94}).

From \cite[Exercise 1.4.22(c)]{GriRei} (applied to $k-1$ and $\ell-1$ instead
of $k$ and $\ell$), we obtain%
\begin{equation}
m_{H^{\otimes k}}^{\left[  \ell\right]  }\circ\left(  \Delta^{\left[
k\right]  }\right)  ^{\otimes\ell}=\Delta^{\left[  k\right]  }\circ m^{\left[
\ell\right]  }, \label{pf.thm.sol-mac-1.mD1}%
\end{equation}
where $m_{H^{\otimes k}}^{\left[  \ell\right]  }$ is the map defined just as
$m^{\left[  \ell\right]  }$ but for the algebra $H^{\otimes k}$ instead of
$H$. However, it is easy to see (and should be known) that
\begin{equation}
m_{H^{\otimes k}}^{\left[  \ell\right]  }=\left(  m^{\left[  \ell\right]
}\right)  ^{\otimes k}\circ\zeta. \label{pf.thm.sol-mac-1.mz}%
\end{equation}
Indeed, this can be checked on pure tensors, using the combinatorial
observation that%
\begin{align}
&  \zeta\left(  h_{1,1}\otimes h_{1,2}\otimes\cdots\otimes h_{1,k}\right.
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \otimes\ h_{2,1}\otimes h_{2,2}\otimes
\cdots\otimes h_{2,k}\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \otimes\ \cdots\right. \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \otimes\ h_{\ell,1}\otimes h_{\ell,2}%
\otimes\cdots\otimes h_{\ell,k}\right) \nonumber\\
&  =h_{1,1}\otimes h_{2,1}\otimes\cdots\otimes h_{\ell,1}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \otimes h_{1,2}\otimes h_{2,2}\otimes\cdots\otimes
h_{\ell,2}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \otimes\cdots\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \otimes h_{1,k}\otimes h_{2,k}\otimes\cdots\otimes
h_{\ell,k} \label{pf.thm.sol-mac-1.z-prod}%
\end{align}
for all matrices $\left(  h_{j,i}\right)  _{i\in\left[  k\right]
,\ j\in\left[  \ell\right]  }\in H^{k\times\ell}$ (sorry for the idiotic
indexing)\footnote{Let me sketch a \textit{proof of
(\ref{pf.thm.sol-mac-1.z-prod})} because it is far too easy to get such things
wrong.
\par
Define two maps $\lambda$ and $\rho$ from $\left[  k\right]  \times\left[
\ell\right]  $ to $\left[  k\ell\right]  $ by setting%
\[
\lambda\left(  i,j\right)  :=k\left(  j-1\right)  +i\in\left[  k\ell\right]
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ \rho\left(  i,j\right)
:=\ell\left(  i-1\right)  +j\in\left[  k\ell\right]
\]
for all $\left(  i,j\right)  \in\left[  k\right]  \times\left[  \ell\right]
$. These maps $\lambda$ and $\rho$ are bijections.
\par
Let $\left(  h_{j,i}\right)  _{i\in\left[  k\right]  ,\ j\in\left[
\ell\right]  }\in H^{k\times\ell}$ be a matrix. Set $g_{\left(  i,j\right)
}:=h_{j,i}$ for all $\left(  i,j\right)  \in\left[  k\right]  \times\left[
\ell\right]  $. Then,%
\begin{align*}
g_{\lambda^{-1}\left(  1\right)  }\otimes g_{\lambda^{-1}\left(  2\right)
}\otimes\cdots\otimes g_{\lambda^{-1}\left(  k\ell\right)  }  &
=h_{1,1}\otimes h_{1,2}\otimes\cdots\otimes h_{1,k}\\
&  \ \ \ \ \ \ \ \ \ \ \otimes h_{2,1}\otimes h_{2,2}\otimes\cdots\otimes
h_{2,k}\\
&  \ \ \ \ \ \ \ \ \ \ \otimes\cdots\\
&  \ \ \ \ \ \ \ \ \ \ \otimes h_{\ell,1}\otimes h_{\ell,2}\otimes
\cdots\otimes h_{\ell,k}%
\end{align*}
and%
\begin{align*}
g_{\rho^{-1}\left(  1\right)  }\otimes g_{\rho^{-1}\left(  2\right)  }%
\otimes\cdots\otimes g_{\rho^{-1}\left(  k\ell\right)  }  &  =h_{1,1}\otimes
h_{2,1}\otimes\cdots\otimes h_{\ell,1}\\
&  \ \ \ \ \ \ \ \ \ \ \otimes h_{1,2}\otimes h_{2,2}\otimes\cdots\otimes
h_{\ell,2}\\
&  \ \ \ \ \ \ \ \ \ \ \otimes\cdots\\
&  \ \ \ \ \ \ \ \ \ \ \otimes h_{1,k}\otimes h_{2,k}\otimes\cdots\otimes
h_{\ell,k}.
\end{align*}
Hence, we must show that%
\[
\zeta\left(  g_{\lambda^{-1}\left(  1\right)  }\otimes g_{\lambda^{-1}\left(
2\right)  }\otimes\cdots\otimes g_{\lambda^{-1}\left(  k\ell\right)  }\right)
=g_{\rho^{-1}\left(  1\right)  }\otimes g_{\rho^{-1}\left(  2\right)  }%
\otimes\cdots\otimes g_{\rho^{-1}\left(  k\ell\right)  }.
\]
Since%
\[
\zeta\left(  g_{\lambda^{-1}\left(  1\right)  }\otimes g_{\lambda^{-1}\left(
2\right)  }\otimes\cdots\otimes g_{\lambda^{-1}\left(  k\ell\right)  }\right)
=g_{\lambda^{-1}\left(  \zeta^{-1}\left(  1\right)  \right)  }\otimes
g_{\lambda^{-1}\left(  \zeta^{-1}\left(  2\right)  \right)  }\otimes
\cdots\otimes g_{\lambda^{-1}\left(  \zeta^{-1}\left(  k\ell\right)  \right)
},
\]
this is equivalent to showing that
\[
g_{\lambda^{-1}\left(  \zeta^{-1}\left(  1\right)  \right)  }\otimes
g_{\lambda^{-1}\left(  \zeta^{-1}\left(  2\right)  \right)  }\otimes
\cdots\otimes g_{\lambda^{-1}\left(  \zeta^{-1}\left(  k\ell\right)  \right)
}=g_{\rho^{-1}\left(  1\right)  }\otimes g_{\rho^{-1}\left(  2\right)
}\otimes\cdots\otimes g_{\rho^{-1}\left(  k\ell\right)  }.
\]
Thus, we need to check that $\lambda^{-1}\left(  \zeta^{-1}\left(  q\right)
\right)  =\rho^{-1}\left(  q\right)  $ for each $q\in\left[  k\ell\right]  $.
In other words, we need to check that $\lambda^{-1}\circ\zeta^{-1}=\rho^{-1}$.
In other words, we need to check that $\zeta\circ\lambda=\rho$. However, this
is easy: The definition of $\zeta$ yields $\zeta\left(  k\left(  j-1\right)
+i\right)  =\ell\left(  i-1\right)  +j$ for all $i,j$; but this rewrites as
$\zeta\left(  \lambda\left(  i,j\right)  \right)  =\rho\left(  i,j\right)  $.
So (\ref{pf.thm.sol-mac-1.z-prod}) is proven.}.

Using (\ref{pf.thm.sol-mac-1.mz}), we can rewrite (\ref{pf.thm.sol-mac-1.mD1})
as%
\begin{equation}
\left(  m^{\left[  \ell\right]  }\right)  ^{\otimes k}\circ\zeta\circ\left(
\Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}=\Delta^{\left[  k\right]
}\circ m^{\left[  \ell\right]  }. \label{pf.thm.sol-mac-1.mD2}%
\end{equation}


On the other hand, $\mathbb{N}^{k}$ is the set of all weak compositions of
length $k$. The symmetric group $\mathfrak{S}_{k}$ acts on the right on this
set $\mathbb{N}^{k}$ by permuting the entries: For any $\left(  \gamma
_{1},\gamma_{2},\ldots,\gamma_{k}\right)  \in\mathbb{N}^{k}$ and $\pi
\in\mathfrak{S}_{k}$, we have%
\[
\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{k}\right)  \cdot\pi=\left(
\gamma_{\pi\left(  1\right)  },\gamma_{\pi\left(  2\right)  },\ldots
,\gamma_{\pi\left(  k\right)  }\right)  .
\]
This action has the property that%
\begin{equation}
P_{\gamma}\circ\pi=\pi\circ P_{\gamma\cdot\pi} \label{pf.thm.sol-mac-1.Ppi}%
\end{equation}
for any $\pi\in\mathfrak{S}_{k}$ and $\gamma\in\mathbb{N}^{k}$. (This is easy
to check.) Hence,%
\begin{equation}
\pi\circ P_{\gamma}=P_{\gamma\cdot\pi^{-1}}\circ\pi
\label{pf.thm.sol-mac-1.piP}%
\end{equation}
for any $\pi\in\mathfrak{S}_{k}$ and $\gamma\in\mathbb{N}^{k}$.

The map $m^{\left[  \ell\right]  }:H^{\otimes\ell}\rightarrow H$ is graded;
thus, each $i\in\mathbb{N}$ satisfies%
\[
p_{i}\circ m^{\left[  \ell\right]  }=\sum_{\substack{\left(  i_{1}%
,i_{2},\ldots,i_{\ell}\right)  \in\mathbb{N}^{\ell};\\i_{1}+i_{2}%
+\cdots+i_{\ell}=i}}m^{\left[  \ell\right]  }\circ\left(  p_{i_{1}}\otimes
p_{i_{2}}\otimes\cdots\otimes p_{i_{\ell}}\right)  .
\]
By tensoring together $k$ such equalities, we obtain%
\begin{align}
&  P_{\gamma}\circ\left(  m^{\left[  \ell\right]  }\right)  ^{\otimes
k}\nonumber\\
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\gamma_{i}\text{ for all }i\in\left[  k\right]
}}\left(  m^{\left[  \ell\right]  }\right)  ^{\otimes k}\circ P_{\left(
\gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }
\label{pf.thm.sol-mac-1.Pm}%
\end{align}
for any $\gamma=\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{k}\right)
\in\mathbb{N}^{k}$. A dual argument shows that%
\begin{align}
&  \left(  \Delta^{\left[  \ell\right]  }\right)  ^{\otimes k}\circ P_{\gamma
}\nonumber\\
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\gamma_{i}\text{ for all }i\in\left[  k\right]
}}P_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }%
\circ\left(  \Delta^{\left[  \ell\right]  }\right)  ^{\otimes k}
\label{pf.thm.sol-mac-1.dP}%
\end{align}
for any $\gamma=\left(  \gamma_{1},\gamma_{2},\ldots,\gamma_{k}\right)
\in\mathbb{N}^{k}$. Switching the roles of $k$ and $\ell$ in this equality, we
obtain the following: For any $\gamma=\left(  \gamma_{1},\gamma_{2}%
,\ldots,\gamma_{\ell}\right)  \in\mathbb{N}^{\ell}$, we have%
\begin{align}
&  \left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}\circ P_{\gamma
}\nonumber\\
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
\ell\right]  \text{ and }j\in\left[  k\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,k}=\gamma_{i}\text{ for all }i\in\left[  \ell\right]
}}P_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{\ell,k}\right)  }%
\circ\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}\nonumber\\
&  =\sum_{\substack{\delta_{j,i}\in\mathbb{N}\text{ for all }i\in\left[
\ell\right]  \text{ and }j\in\left[  k\right]  ;\\\delta_{1,i}+\delta
_{2,i}+\cdots+\delta_{k,i}=\gamma_{i}\text{ for all }i\in\left[  \ell\right]
}}P_{\left(  \delta_{1,1},\delta_{2,1},\ldots,\delta_{k,\ell}\right)  }%
\circ\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed
the }\gamma_{i,j}\text{ as }\delta_{j,i}\right) \nonumber\\
&  =\sum_{\substack{\delta_{i,j}\in\mathbb{N}\text{ for all }j\in\left[
\ell\right]  \text{ and }i\in\left[  k\right]  ;\\\delta_{1,j}+\delta
_{2,j}+\cdots+\delta_{k,j}=\gamma_{j}\text{ for all }j\in\left[  \ell\right]
}}P_{\left(  \delta_{1,1},\delta_{2,1},\ldots,\delta_{k,\ell}\right)  }%
\circ\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed
the indices }i\text{ and }j\text{ as }j\text{ and }i\right) \nonumber\\
&  =\sum_{\substack{\delta_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\delta_{1,j}+\delta
_{2,j}+\cdots+\delta_{k,j}=\gamma_{j}\text{ for all }j\in\left[  \ell\right]
}}P_{\left(  \delta_{1,1},\delta_{2,1},\ldots,\delta_{k,\ell}\right)  }%
\circ\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}
\label{pf.thm.sol-mac-1.dP2}%
\end{align}
(here, we have just rewritten the \textquotedblleft$\delta_{i,j}\in\mathbb{N}$
for all $j\in\left[  \ell\right]  $ and $i\in\left[  k\right]  $%
\textquotedblright\ under the summation sign as \textquotedblleft$\delta
_{i,j}\in\mathbb{N}$ for all $i\in\left[  k\right]  $ and $j\in\left[
\ell\right]  $\textquotedblright).

Any two weak compositions $\gamma$ and $\delta$ of length $k\ell$ satisfy%
\begin{equation}
P_{\gamma}\circ P_{\delta}=%
\begin{cases}
P_{\gamma}, & \text{if }\gamma=\delta;\\
0, & \text{if }\gamma\neq\delta.
\end{cases}
\label{pf.thm.sol-mac-1.PP}%
\end{equation}


It is easy to see that%
\begin{equation}
\pi\circ\left(  m^{\left[  \ell\right]  }\right)  ^{\otimes k}=\left(
m^{\left[  \ell\right]  }\right)  ^{\otimes k}\circ\pi^{\times\ell}
\label{pf.thm.sol-mac-1.pim}%
\end{equation}
for each $\pi\in\mathfrak{S}_{k}$, where $\pi^{\times\ell}$ denotes the
permutation in $\mathfrak{S}_{k\ell}$ that sends each $\ell\left(  i-1\right)
+j$ (with $i\in\left[  k\right]  $ and $j\in\left[  \ell\right]  $) to
$\ell\left(  \pi\left(  i\right)  -1\right)  +j$\ \ \ \ \footnote{Actually,
this equality has nothing to do with $m^{\left[  \ell\right]  }$. More
generally, any permutation $\pi\in\mathfrak{S}_{k}$ and any $\mathbf{k}%
$-linear map $f:H^{\otimes\ell}\rightarrow H$ satisfy
\[
\pi\circ f^{\otimes k}=f^{\otimes k}\circ\pi^{\times\ell}.
\]
Applying this to $f=m^{\left[  \ell\right]  }$, we obtain
(\ref{pf.thm.sol-mac-1.pim}).}. Dually, it is easy to see that%
\begin{equation}
\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}\circ\pi
=\pi^{k\times}\circ\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}
\label{pf.thm.sol-mac-1.Delpi}%
\end{equation}
for each $\pi\in\mathfrak{S}_{\ell}$, where $\pi^{k\times}$ denotes the
permutation in $\mathfrak{S}_{k\ell}$ that sends each $k\left(  j-1\right)
+i$ (with $i\in\left[  k\right]  $ and $j\in\left[  \ell\right]  $) to
$k\left(  \pi\left(  j\right)  -1\right)  +i$.

Now, (\ref{pf.thm.sol-mac-1.pas}) and (\ref{pf.thm.sol-mac-1.pbt}) yield%
\begin{align*}
&  p_{\alpha,\sigma}\circ p_{\beta,\tau}\\
&  =m^{\left[  k\right]  }\circ P_{\alpha}\circ\sigma^{-1}\circ
\underbrace{\Delta^{\left[  k\right]  }\circ m^{\left[  \ell\right]  }%
}_{\substack{=\left(  m^{\left[  \ell\right]  }\right)  ^{\otimes k}\circ
\zeta\circ\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell
}\\\text{(by (\ref{pf.thm.sol-mac-1.mD2}))}}}\circ P_{\beta}\circ\tau
^{-1}\circ\Delta^{\left[  \ell\right]  }\\
&  =m^{\left[  k\right]  }\circ P_{\alpha}\circ\underbrace{\sigma^{-1}%
\circ\left(  m^{\left[  \ell\right]  }\right)  ^{\otimes k}}%
_{\substack{=\left(  m^{\left[  \ell\right]  }\right)  ^{\otimes k}%
\circ\left(  \sigma^{-1}\right)  ^{\times\ell}\\\text{(by
(\ref{pf.thm.sol-mac-1.pim}))}}}\circ\zeta\\
&  \ \ \ \ \ \ \ \ \ \ \circ\underbrace{\left(  \Delta^{\left[  k\right]
}\right)  ^{\otimes\ell}\circ P_{\beta}}_{\substack{=\sum_{\substack{\delta
_{i,j}\in\mathbb{N}\text{ for all }i\in\left[  k\right]  \text{ and }%
j\in\left[  \ell\right]  ;\\\delta_{1,j}+\delta_{2,j}+\cdots+\delta
_{k,j}=\beta_{j}\text{ for all }j\in\left[  \ell\right]  }}P_{\left(
\delta_{1,1},\delta_{2,1},\ldots,\delta_{k,\ell}\right)  }\circ\left(
\Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}\\\text{(by
(\ref{pf.thm.sol-mac-1.dP2}))}}}\circ\,\tau^{-1}\circ\Delta^{\left[
\ell\right]  }\\
&  =\sum_{\substack{\delta_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\delta_{1,j}+\delta
_{2,j}+\cdots+\delta_{k,j}=\beta_{j}\text{ for all }j\in\left[  \ell\right]
}}m^{\left[  k\right]  }\circ\underbrace{P_{\alpha}\circ\left(  m^{\left[
\ell\right]  }\right)  ^{\otimes k}}_{\substack{=\sum_{\substack{\gamma
_{i,j}\in\mathbb{N}\text{ for all }i\in\left[  k\right]  \text{ and }%
j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma_{i,2}+\cdots+\gamma_{i,\ell
}=\alpha_{i}\text{ for all }i\in\left[  k\right]  }}\left(  m^{\left[
\ell\right]  }\right)  ^{\otimes k}\circ P_{\left(  \gamma_{1,1},\gamma
_{1,2},\ldots,\gamma_{k,\ell}\right)  }\\\text{(by (\ref{pf.thm.sol-mac-1.Pm}%
))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \sigma^{-1}\right)  ^{\times\ell}%
\circ\zeta\circ P_{\left(  \delta_{1,1},\delta_{2,1},\ldots,\delta_{k,\ell
}\right)  }\circ\underbrace{\left(  \Delta^{\left[  k\right]  }\right)
^{\otimes\ell}\circ\tau^{-1}}_{\substack{=\left(  \tau^{-1}\right)  ^{k\times
}\circ\left(  \Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}\\\text{(by
(\ref{pf.thm.sol-mac-1.Delpi}))}}}\circ\,\Delta^{\left[  \ell\right]  }\\
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\alpha_{i}\text{ for all }i\in\left[  k\right]
}}\ \ \sum_{\substack{\delta_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\delta_{1,j}+\delta
_{2,j}+\cdots+\delta_{k,j}=\beta_{j}\text{ for all }j\in\left[  \ell\right]
}}\\
&  \ \ \ \ \ \ \ \ \ \ \underbrace{m^{\left[  k\right]  }\circ\left(
m^{\left[  \ell\right]  }\right)  ^{\otimes k}}_{=m^{\left[  k\ell\right]  }%
}\circ\,P_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)
}\circ\left(  \sigma^{-1}\right)  ^{\times\ell}\\
&  \ \ \ \ \ \ \ \ \ \ \circ\underbrace{\zeta\circ P_{\left(  \delta
_{1,1},\delta_{2,1},\ldots,\delta_{k,\ell}\right)  }}_{\substack{=P_{\left(
\delta_{1,1},\delta_{1,2},\ldots,\delta_{k,\ell}\right)  }\circ\zeta
\\\text{(by (\ref{pf.thm.sol-mac-1.Ppi}),}\\\text{since }\left(  \delta
_{1,1},\delta_{2,1},\ldots,\delta_{k,\ell}\right)  \cdot\zeta^{-1}\\=\left(
\delta_{1,1},\delta_{1,2},\ldots,\delta_{k,\ell}\right)  \text{)}}%
}\circ\left(  \tau^{-1}\right)  ^{k\times}\circ\underbrace{\left(
\Delta^{\left[  k\right]  }\right)  ^{\otimes\ell}\circ\Delta^{\left[
\ell\right]  }}_{=\Delta^{\left[  k\ell\right]  }}%
\end{align*}%
\begin{align*}
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\alpha_{i}\text{ for all }i\in\left[  k\right]
}}\ \ \sum_{\substack{\delta_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\delta_{1,j}+\delta
_{2,j}+\cdots+\delta_{k,j}=\beta_{j}\text{ for all }j\in\left[  \ell\right]
}}\\
&  \ \ \ \ \ \ \ \ \ \ m^{\left[  k\ell\right]  }\circ P_{\left(  \gamma
_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }\circ\underbrace{\left(
\sigma^{-1}\right)  ^{\times\ell}\circ P_{\left(  \delta_{1,1},\delta
_{1,2},\ldots,\delta_{k,\ell}\right)  }}_{\substack{=P_{\left(  \delta
_{\sigma\left(  1\right)  ,1},\delta_{\sigma\left(  1\right)  ,2}%
,\ldots,\delta_{\sigma\left(  k\right)  ,\ell}\right)  }\circ\left(
\sigma^{-1}\right)  ^{\times\ell}\\\text{(by (\ref{pf.thm.sol-mac-1.Ppi}%
),}\\\text{since }\left(  \delta_{1,1},\delta_{1,2},\ldots,\delta_{k,\ell
}\right)  \cdot\left(  \left(  \sigma^{-1}\right)  ^{\times\ell}\right)
^{-1}\\=\left(  \delta_{1,1},\delta_{1,2},\ldots,\delta_{k,\ell}\right)
\cdot\sigma^{\times\ell}\\=\left(  \delta_{\sigma\left(  1\right)  ,1}%
,\delta_{\sigma\left(  1\right)  ,2},\ldots,\delta_{\sigma\left(  k\right)
,\ell}\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \circ\zeta\circ\left(  \tau^{-1}\right)  ^{k\times
}\circ\Delta^{\left[  k\ell\right]  }\\
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\alpha_{i}\text{ for all }i\in\left[  k\right]
}}\ \ \sum_{\substack{\delta_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\delta_{1,j}+\delta
_{2,j}+\cdots+\delta_{k,j}=\beta_{j}\text{ for all }j\in\left[  \ell\right]
}}\\
&  \ \ \ \ \ \ \ \ \ \ m^{\left[  k\ell\right]  }\circ\underbrace{P_{\left(
\gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }\circ P_{\left(
\delta_{\sigma\left(  1\right)  ,1},\delta_{\sigma\left(  1\right)  ,2}%
,\ldots,\delta_{\sigma\left(  k\right)  ,\ell}\right)  }}_{\substack{=%
\begin{cases}
P_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }, &
\text{if }\gamma_{i,j}=\delta_{\sigma\left(  i\right)  ,j}\text{ for all
}i,j;\\
0, & \text{otherwise}%
\end{cases}
\\\text{(by (\ref{pf.thm.sol-mac-1.PP}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \circ\underbrace{\left(  \sigma^{-1}\right)
^{\times\ell}\circ\zeta\circ\left(  \tau^{-1}\right)  ^{k\times}}_{=\left(
\tau\left[  \sigma\right]  \right)  ^{-1}}\circ\,\Delta^{\left[  k\ell\right]
}\\
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\alpha_{i}\text{ for all }i\in\left[  k\right]
}}\ \ \sum_{\substack{\delta_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\delta_{1,j}+\delta
_{2,j}+\cdots+\delta_{k,j}=\beta_{j}\text{ for all }j\in\left[  \ell\right]
}}\\
&  \ \ \ \ \ \ \ \ \ \ m^{\left[  k\ell\right]  }\circ%
\begin{cases}
P_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }, &
\text{if }\gamma_{i,j}=\delta_{\sigma\left(  i\right)  ,j}\text{ for all
}i,j;\\
0, & \text{otherwise}%
\end{cases}
\\
&  \ \ \ \ \ \ \ \ \ \ \circ\left(  \tau\left[  \sigma\right]  \right)
^{-1}\circ\Delta^{\left[  k\ell\right]  }%
\end{align*}%
\begin{align*}
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\alpha_{i}\text{ for all }i\in\left[  k\right]
;\\\gamma_{\sigma^{-1}\left(  1\right)  ,j}+\gamma_{\sigma^{-1}\left(
2\right)  ,j}+\cdots+\gamma_{\sigma^{-1}\left(  k\right)  ,j}=\beta_{j}\text{
for all }j\in\left[  \ell\right]  }}m^{\left[  k\ell\right]  }\circ P_{\left(
\gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }\circ\left(
\tau\left[  \sigma\right]  \right)  ^{-1}\circ\Delta^{\left[  k\ell\right]
}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have discarded all the addends that do \textbf{not}}\\
\text{satisfy }\left(  \gamma_{i,j}=\delta_{\sigma\left(  i\right)  ,j}\text{
for all }i,j\right)  \text{, because these addends}\\
\text{are }0\text{; the remaining entries have been combined}\\
\text{into a single sum, with }\delta_{i,j}\text{ rewritten as }\gamma
_{\sigma^{-1}\left(  i\right)  ,j}%
\end{array}
\right) \\
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\alpha_{i}\text{ for all }i\in\left[  k\right]
;\\\gamma_{1,j}+\gamma_{2,j}+\cdots+\gamma_{k,j}=\beta_{j}\text{ for all }%
j\in\left[  \ell\right]  }}\underbrace{m^{\left[  k\ell\right]  }\circ
P_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }%
\circ\left(  \tau\left[  \sigma\right]  \right)  ^{-1}\circ\Delta^{\left[
k\ell\right]  }}_{\substack{=p_{\left(  \gamma_{1,1},\gamma_{1,2}%
,\ldots,\gamma_{k,\ell}\right)  ,\tau\left[  \sigma\right]  }\\\text{(by
(\ref{eq.pas.formal-def}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have rewritten the \textquotedblleft}\gamma_{\sigma^{-1}\left(
1\right)  ,j}+\gamma_{\sigma^{-1}\left(  2\right)  ,j}+\cdots+\gamma
_{\sigma^{-1}\left(  k\right)  ,j}\text{\textquotedblright}\\
\text{under the summation sign as \textquotedblleft}\gamma_{1,j}+\gamma
_{2,j}+\cdots+\gamma_{k,j}\text{\textquotedblright}%
\end{array}
\right) \\
&  =\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for all }i\in\left[
k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma_{i,1}+\gamma
_{i,2}+\cdots+\gamma_{i,\ell}=\alpha_{i}\text{ for all }i\in\left[  k\right]
;\\\gamma_{1,j}+\gamma_{2,j}+\cdots+\gamma_{k,j}=\beta_{j}\text{ for all }%
j\in\left[  \ell\right]  }}p_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots
,\gamma_{k,\ell}\right)  ,\tau\left[  \sigma\right]  }.
\end{align*}
This proves Theorem \ref{thm.sol-mac-1}.
\end{proof}

Note that both Theorem \ref{thm.sol-mac-1} and Proposition
\ref{prop.sol-mac-0} work for arbitrary graded (not necessarily connected) bialgebras.

Using Theorem \ref{thm.sol-mac-1} and (\ref{eq.pas-conv}), we can expand any
nested convolution-and-composition of $p_{\alpha,\sigma}$'s as a $\mathbf{k}%
$-linear combination of single $p_{\alpha,\sigma}$'s. This allows us to
mechnically prove equalities for $p_{\alpha,\sigma}$'s that involve only
convolution, composition and $\mathbf{k}$-linear combination and that are
supposed to be valid for any connected graded Hopf algebra $H$. The reason why
this works is the following \textquotedblleft generic linear
independence\textquotedblright\ theorem:

\begin{theorem}
\label{thm.pas-lin-ind}\textbf{(a)} There is a connected graded Hopf algebra
$H$ such that the family%
\[
\left(  p_{\alpha,\sigma}\right)  _{\substack{k\in\mathbb{N}\text{;}%
\\\alpha\text{ is a composition of length }k\text{;}\\\sigma\in\mathfrak{S}%
_{k}}}
\]
(of endomorphisms of $H$) is $\mathbf{k}$-linearly independent.

\textbf{(b)} Let $n\in\mathbb{N}$. Then, there is a connected graded Hopf
algebra $H$ such that the family%
\[
\left(  p_{\alpha,\sigma}\right)  _{\substack{k\in\mathbb{N}\text{;}%
\\\alpha\text{ is a composition of length }k\text{ and size }<n\text{;}%
\\\sigma\in\mathfrak{S}_{k}}}
\]
(of endomorphisms of $H$) is $\mathbf{k}$-linearly independent, and such that
each $H_{n}$ is a free $\mathbf{k}$-module with a finite basis.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.pas-lin-ind} (sketched).]\textbf{(b)} Let $H$ be
the free $\mathbf{k}$-algebra with generators
\[
x_{i,j}\ \ \ \ \ \ \ \ \ \ \text{with }i,j\in\mathbb{Z}\text{ satisfying
}1\leq i<j\leq n.
\]
We also set%
\[
x_{k,k}:=1_{H}\ \ \ \ \ \ \ \ \ \ \text{for each }k\in\left\{  1,2,\ldots
,n\right\}  .
\]
We make the $\mathbf{k}$-algebra $H$ graded by declaring each $x_{i,j}$ to be
homogeneous of degree $j-i$. We define a comultiplication $\Delta:H\rightarrow
H\otimes H$ on $H$ to be the $\mathbf{k}$-algebra homomorphism that satisfies%
\[
\Delta\left(  x_{i,j}\right)  =\sum_{k=i}^{j}x_{i,k}\otimes x_{k,j}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i,j\in\mathbb{Z}\text{ satisfying }1\leq
i<j\leq n.
\]
We define a counit $\epsilon:H\rightarrow\mathbf{k}$ in the obvious way to
preserve the grading (so that $\epsilon\left(  x_{i,j}\right)  =0$ whenever
$i<j$). It is easy to see that $H$ is a connected graded $\mathbf{k}%
$-bialgebra, thus a connected graded Hopf algebra.

(You can think of $H$ either as a noncommutative version of a reduced
incidence algebra of the chain poset with $n$ elements, or as a noncommutative
variant of the subalgebra of the Schur algebra corresponding to the
upper-triangular matrices with equal numbers on the diagonal. But we won't
need all this intuition.)

It is easy to see that%
\[
\Delta^{\left(  k-1\right)  }\left(  x_{i,j}\right)  =\sum_{i=u_{0}\leq
u_{1}\leq\cdots\leq u_{k}=j}x_{u_{0},u_{1}}\otimes x_{u_{1},u_{2}}%
\otimes\cdots\otimes x_{u_{k-1},u_{k}}%
\]
for any $i\leq j$ and any $k\in\mathbb{N}$. Thus, for any weak composition
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{k}\right)  $ and any
$\sigma\in\mathfrak{S}_{k}$ and any $1\leq i<j\leq n$ satisfying
$j-i=\left\vert \alpha\right\vert $, we have
\[
p_{\alpha,\sigma}\left(  x_{i,j}\right)  =x_{u_{\sigma\left(  1\right)
-1},u_{\sigma\left(  1\right)  }}x_{u_{\sigma\left(  2\right)  -1}%
,u_{\sigma\left(  2\right)  }}\cdots x_{u_{\sigma\left(  k\right)
-1},u_{\sigma\left(  k\right)  }},
\]
where $\left(  u_{0}\leq u_{1}\leq\cdots\leq u_{k}\right)  $ is the unique
weakly increasing sequence of integers satisfying $u_{0}=i$ and $u_{k}=j$ and
$u_{\sigma\left(  i\right)  }-u_{\sigma\left(  i\right)  -1}=\alpha_{i}$ for
all $i\in\left\{  1,2,\ldots,k\right\}  $. Hence, for any choice of $1\leq
i<j\leq n$, the images $p_{\alpha,\sigma}\left(  x_{i,j}\right)  $ as $\alpha$
runs over all compositions of $j-i$ and $\sigma$ runs over all permutations of
$\left[  \ell\left(  \alpha\right)  \right]  $ are distinct monomials and
therefore are $\mathbf{k}$-linearly independent. This yields the $\mathbf{k}%
$-linear independence of the family%
\[
\left(  p_{\alpha,\sigma}\right)  _{\substack{k\in\mathbb{N}\text{;}%
\\\alpha\text{ is a composition of length }k\text{ and size }m\text{;}%
\\\sigma\in\mathfrak{S}_{k}}}
\]
for any given $m\in\left\{  0,1,\ldots,n-1\right\}  $. Since each
$p_{\alpha,\sigma}$ lies in $\operatorname*{End}\nolimits_{\mathbf{k}}\left(
H_{\left\vert \alpha\right\vert }\right)  $, we thus obtain the $\mathbf{k}%
$-linear independence of the entire family%
\[
\left(  p_{\alpha,\sigma}\right)  _{\substack{k\in\mathbb{N}\text{;}%
\\\alpha\text{ is a composition of length }k\text{ and size }<n\text{;}%
\\\sigma\in\mathfrak{S}_{k}}}
\]
(since the sum $\sum_{m=0}^{n-1}\operatorname*{End}\nolimits_{\mathbf{k}%
}\left(  H_{m}\right)  $ is a direct sum). This proves Theorem
\ref{thm.pas-lin-ind} \textbf{(b)}.

\textbf{(a)} As for part \textbf{(b)}, but remove \textquotedblleft$\leq
n$\textquotedblright\ everywhere.
\end{proof}

Theorem \ref{thm.sol-mac-1} suggests that we should consider the algebra of
$p_{\alpha,\sigma}$'s as some twisted (Hopf-monoid-like?) version of
$\operatorname*{NSym}$, and the twist appears rather simple since all addends
on the right hand side have the same permutation $\tau\left[  \sigma\right]
$. However, one should keep in mind that some of the $\gamma_{i,j}$ can be
$0$, in which case some entries have to be \textquotedblleft
collapsed\textquotedblright\ to normalize the right hand side.

How all-encompassing are our $p_{\alpha,\sigma}$'s? Do they span all the
natural endomorphisms?

\begin{question}
Let $g$ be a natural graded $\mathbf{k}$-module endomorphism on the category
of connected graded $\mathbf{k}$-Hopf algebras. (That is, for each connected
graded Hopf algebra $H$, we have a graded $\mathbf{k}$-module endomorphism
$g_{H}$, and each graded Hopf algebra morphism $\varphi:H\rightarrow
H^{\prime}$ gives a commutative diagram.) Is it true that $g$ is an infinite
$\mathbf{k}$-linear combination of $p_{\alpha,\sigma}$'s?
\end{question}

We observe one more property of $p_{\alpha,\sigma}$'s:

\begin{proposition}
\label{prop.pas-on-tensors}Let $H$ and $G$ be two graded bialgebras. Let
$\alpha$ be a weak composition of length $k$, and let $\sigma\in
\mathfrak{S}_{k}$ be a permutation. Then,%
\[
\left(  p_{\alpha,\sigma}\text{ for }H\otimes G\right)  =\sum_{\substack{\beta
,\gamma\text{ weak compositions;}\\\text{entrywise sum }\beta+\gamma=\alpha
}}\left(  p_{\beta,\sigma}\text{ for }H\right)  \otimes\left(  p_{\gamma
,\sigma}\text{ for }G\right)
\]
as endomorphisms of $H\otimes G$.
\end{proposition}

\begin{proof}
Straightforward.
\end{proof}

\subsection{A combinatorial Hopf algebra of \textquotedblleft formal
$p_{\alpha,\sigma}$\textquotedblright s}

Next, let us try to find a combinatorial algebra in which the $p_{\alpha
,\sigma}$'s themselves live. This requires us to make some implicit things
explicit and introduce some more notation:

\begin{definition}
A \emph{mopiscotion} (please find a better name for this!) is a pair $\left(
\alpha,\sigma\right)  $, where $\alpha$ is a composition of length $k$ (for
some $k\in\mathbb{N}$) and $\sigma$ is a permutation in $\mathfrak{S}_{k}$.
\end{definition}

\begin{definition}
A \emph{weak mopiscotion} is a pair $\left(  \alpha,\sigma\right)  $, where
$\alpha$ is a weak composition of length $k$ (for some $k\in\mathbb{N}$) and
$\sigma$ is a permutation in $\mathfrak{S}_{k}$.
\end{definition}

\begin{definition}
Let $\left(  \alpha,\sigma\right)  $ be a weak mopiscotion, with
$\alpha=\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{k}\right)  $ and
$\sigma\in\mathfrak{S}_{k}$. Let $\left(  j_{1}<j_{2}<\cdots<j_{h}\right)  $
be the list of all elements $i$ of $\left\{  1,2,\ldots,k\right\}  $
satisfying $\alpha_{i}\neq0$, in increasing order. Let $\tau\in\mathfrak{S}%
_{h}$ be the standardization of the list $\left(  \sigma\left(  j_{1}\right)
,\sigma\left(  j_{2}\right)  ,\ldots,\sigma\left(  j_{h}\right)  \right)  $.
(See \cite[Definition 5.3.3]{GriRei} for the meaning of \textquotedblleft
standardization\textquotedblright.) Let $\alpha^{\operatorname*{red}}$ denote
the composition $\left(  \alpha_{j_{1}},\alpha_{j_{2}},\ldots,\alpha_{j_{h}%
}\right)  $ (which consists of all nonzero entries of $\alpha$). Then, we
define $\left(  \alpha,\sigma\right)  ^{\operatorname*{red}}$ to be the
mopiscotion $\left(  \alpha^{\operatorname*{red}},\tau\right)  $.
\end{definition}

For example,%
\[
\left(  \left(  3,0,1,2,0\right)  ,\ \left[  4,5,1,3,2\right]  \right)
^{\operatorname*{red}}=\left(  \left(  3,1,2\right)  ,\ \left[  3,1,2\right]
\right)  ,
\]
where the square brackets indicate a permutation written in one-line notation.

Clearly, if $\left(  \alpha,\sigma\right)  $ is a mopiscotion, then $\left(
\alpha,\sigma\right)  ^{\operatorname*{red}}=\left(  \alpha,\sigma\right)  $.

The following is easy to see:

\begin{proposition}
\label{prop.mopis.reduce}Let $H$ be a connected graded bialgebra. Let $\left(
\alpha,\sigma\right)  $ be a weak mopiscotion, and let $\left(  \beta
,\tau\right)  =\left(  \alpha,\sigma\right)  ^{\operatorname*{red}}$. Then,%
\[
p_{\alpha,\sigma}=p_{\beta,\tau}.
\]

\end{proposition}

\begin{definition}
\label{def.PNSym.ops}Let $\operatorname*{PNSym}$ be the free $\mathbf{k}%
$-module with basis $\left(  F_{\alpha,\sigma}\right)  _{\left(  \alpha
,\sigma\right)  \text{ is a mopiscotion}}$.

For any weak mopiscotion $\left(  \alpha,\sigma\right)  $, we set%
\[
F_{\alpha,\sigma}:=F_{\beta,\tau},
\]
where $\left(  \beta,\tau\right)  =\left(  \alpha,\sigma\right)
^{\operatorname*{red}}$.

Define two multiplications on $\operatorname*{PNSym}$: one \textquotedblleft
external multiplication\textquotedblright\ (which mirrors convolution of
$p_{\alpha,\sigma}$'s as expressed in Proposition \ref{prop.sol-mac-0}) given
by%
\[
F_{\alpha,\sigma}\cdot F_{\beta,\tau}=F_{\alpha\beta,\sigma\oplus\tau};
\]
and another \textquotedblleft internal multiplication\textquotedblright%
\ (which mirrors composition of $p_{\alpha,\sigma}$'s as expressed in Theorem
\ref{thm.sol-mac-1}) given by%
\[
F_{\alpha,\sigma}\ast F_{\beta,\tau}=\sum_{\substack{\gamma_{i,j}\in
\mathbb{N}\text{ for all }i\in\left[  k\right]  \text{ and }j\in\left[
\ell\right]  ;\\\gamma_{i,1}+\gamma_{i,2}+\cdots+\gamma_{i,\ell}=\alpha
_{i}\text{ for all }i\in\left[  k\right]  ;\\\gamma_{1,j}+\gamma_{2,j}%
+\cdots+\gamma_{k,j}=\beta_{j}\text{ for all }j\in\left[  \ell\right]
}}F_{\left(  \gamma_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)
,\tau\left[  \sigma\right]  }%
\]
(where $\alpha\in\mathbb{N}^{k}$ and $\beta\in\mathbb{N}^{\ell}$). Also, we
define a comultiplication $\Delta:\operatorname*{PNSym}\rightarrow
\operatorname*{PNSym}\otimes\operatorname*{PNSym}$ on $\operatorname*{PNSym}$
by%
\[
\Delta\left(  F_{\alpha,\sigma}\right)  =\sum_{\substack{\beta,\gamma\text{
weak compositions;}\\\text{entrywise sum }\beta+\gamma=\alpha}}F_{\beta
,\sigma}\otimes F_{\gamma,\sigma}%
\]
(mirroring the formula from Proposition \ref{prop.pas-on-tensors}).

We also equip the $\mathbf{k}$-module $\operatorname*{PNSym}$ with a grading
by letting each $F_{\alpha,\sigma}$ be homogeneous of degree $\left\vert
\alpha\right\vert $.
\end{definition}

If I have not made any mistakes, then:

\begin{theorem}
\label{thm.PNSym.alg}The $\mathbf{k}$-module $\operatorname*{PNSym}$ becomes a
connected graded Hopf algebra when equipped with the external multiplication
$\cdot$, and a (non-graded) non-unital bialgebra when equipped with the
internal multiplication $\ast$. In particular, both multiplications are associative.
\end{theorem}

There are two ways to prove this. I shall very briefly outline both:

\begin{proof}
[First proof idea for Theorem \ref{thm.PNSym.alg}.]Most claims can be derived
from properties of the operators $p_{\alpha,\sigma}$, using the $H$ from
Theorem \ref{thm.pas-lin-ind} \textbf{(a)} as a faithful representation.

For an example, let us prove that the internal multiplication $\ast$ on
$\operatorname*{PNSym}$ is associative.

Let $H$ be any connected graded $\mathbf{k}$-bialgebra. Let
$\operatorname*{ev}\nolimits_{H}:\operatorname*{PNSym}\rightarrow
\operatorname*{End}H$ be the $\mathbf{k}$-linear map that sends any
$F_{\alpha,\sigma}$ to the operator $p_{\alpha,\sigma}\in\operatorname*{End}H$
for any mopiscotion $\left(  \alpha,\sigma\right)  $. Note that
\begin{equation}
\operatorname*{ev}\nolimits_{H}\left(  F_{\alpha,\sigma}\right)
=p_{\alpha,\sigma} \label{pf.thm.PNSym.alg.1st.evF=p}%
\end{equation}
is true not only for all mopiscotions $\left(  \alpha,\sigma\right)  $, but
also for all weak mopiscotions $\left(  \alpha,\sigma\right)  $ (because if
$\left(  \alpha,\sigma\right)  $ is any weak mopiscotion, and if $\left(
\beta,\tau\right)  =\left(  \alpha,\sigma\right)  ^{\operatorname*{red}}$,
then $p_{\alpha,\sigma}=p_{\beta,\tau}$ and $F_{\alpha,\sigma}=F_{\beta,\tau}$).

Now, let $H$ be the connected graded Hopf algebra $H$ from Theorem
\ref{thm.pas-lin-ind} \textbf{(a)}. Then, Theorem \ref{thm.pas-lin-ind}
\textbf{(a)} says that the family $\left(  p_{\alpha,\sigma}\right)  _{\left(
\alpha,\sigma\right)  \text{ is a mopiscotion}}$ is $\mathbf{k}$-linearly
independent. Hence, the linear map $\operatorname*{ev}\nolimits_{H}$ is injective.

The formula for $F_{\alpha,\sigma}\ast F_{\beta,\tau}$ that we used to define
the internal multiplication $\ast$ is very similar to the formula for
$p_{\alpha,\sigma}\circ p_{\beta,\tau}$ in Theorem \ref{thm.sol-mac-1}. In
view of (\ref{pf.thm.PNSym.alg.1st.evF=p}), this entails that%
\[
\operatorname*{ev}\nolimits_{H}\left(  F_{\alpha,\sigma}\ast F_{\beta,\tau
}\right)  =p_{\alpha,\sigma}\circ p_{\beta,\tau}=\operatorname*{ev}%
\nolimits_{H}\left(  F_{\alpha,\sigma}\right)  \circ\operatorname*{ev}%
\nolimits_{H}\left(  F_{\beta,\tau}\right)
\]
for any two mopiscotions $\left(  \alpha,\sigma\right)  $ and $\left(
\beta,\tau\right)  $. By bilinearity, this entails that%
\[
\operatorname*{ev}\nolimits_{H}\left(  f\ast g\right)  =\left(
\operatorname*{ev}\nolimits_{H}f\right)  \circ\left(  \operatorname*{ev}%
\nolimits_{H}g\right)
\]
for any $f,g\in\operatorname*{PNSym}$. Thus, the injective $\mathbf{k}$-linear
map $\operatorname*{ev}\nolimits_{H}:\operatorname*{PNSym}\rightarrow
\operatorname*{End}H$ embeds the $\mathbf{k}$-module $\operatorname*{PNSym}$
with its binary operation $\ast$ into the algebra $\operatorname*{End}H$ with
its binary operation $\circ$. Since the latter operation $\circ$ is
associative, it thus follows that the former operation $\ast$ is associative
as well.

Similarly, we can show that the operation $\cdot$ on $\operatorname*{PNSym}$
is associative and unital with the unity $1=F_{\varnothing,\varnothing}$
(although this is pretty obvious).

It is very easy to see that the cooperation $\Delta$ is coassociative and
counital. It is also clear that both the multiplication $\cdot$ and the
comultiplication $\Delta$ on $\operatorname*{PNSym}$ are graded.

The next difficulty is to prove that $\Delta$ is a $\mathbf{k}$-algebra
homomorphism, i.e., that $\Delta\left(  fg\right)  =\Delta\left(  f\right)
\cdot\Delta\left(  g\right)  $ for all $f,g\in\operatorname*{PNSym}$ (where
the \textquotedblleft$\cdot$\textquotedblright\ on the right hand side is the
extension of the external multiplication $\cdot$ to $\operatorname*{PNSym}%
\otimes\operatorname*{PNSym}$). Here, we can argue as above, using the fact (a
consequence of Theorem \ref{prop.pas-on-tensors}) that%
\begin{align*}
\operatorname*{ev}\nolimits_{H\otimes H}\left(  f\right)   &  =\left(
\operatorname*{ev}\nolimits_{H}\otimes\operatorname*{ev}\nolimits_{H}\right)
\left(  \Delta\left(  f\right)  \right)  \in\operatorname*{End}\left(
H\otimes H\right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }f\in\operatorname*{PNSym}%
\end{align*}
to make sense of $\Delta$), and using the fact that the map
\[
\operatorname*{ev}\nolimits_{H}\otimes\operatorname*{ev}\nolimits_{H}%
:\operatorname*{PNSym}\otimes\operatorname*{PNSym}\rightarrow
\operatorname*{End}H\otimes\operatorname*{End}H\rightarrow\operatorname*{End}%
\left(  H\otimes H\right)
\]
is injective (this is not hard to show using the argument used in the proof of
Theorem \ref{thm.pas-lin-ind}).

What we have shown so far yields that $\operatorname*{PNSym}$ (equipped with
$\cdot$ and $\Delta$) is a connected graded $\mathbf{k}$-bialgebra. Thus,
$\operatorname*{PNSym}$ is a Hopf algebra (since any connected graded
$\mathbf{k}$-bialgebra is a Hopf algebra).

It remains to show that $\operatorname*{PNSym}$ (equipped with $\ast$ and
$\Delta$) is a non-unital $\mathbf{k}$-bialgebra. Having already verified that
$\ast$ is associative, we only need to show that $\Delta\left(  f\ast
g\right)  =\Delta\left(  f\right)  \ast\Delta\left(  g\right)  $ for all
$f,g\in\operatorname*{PNSym}$. But this is similar to the proof of
$\Delta\left(  fg\right)  =\Delta\left(  f\right)  \cdot\Delta\left(
g\right)  $ above. Thus, the proof of Theorem \ref{thm.PNSym.alg} is complete.
\end{proof}

\begin{proof}
[Second proof idea for Theorem \ref{thm.PNSym.alg}.]There is also a more
direct combinatorial approach to this theorem. First, we shall define two
smaller bialgebras $\operatorname*{NNSym}$ and $\operatorname*{Perm}$, and
then present $\operatorname*{PNSym}$ as a quotient of their tensor product
$\operatorname*{NNSym}\otimes\operatorname*{Perm}$.

Here are some details:

We define $\operatorname*{NNSym}$ to be the free $\mathbf{k}$-module with
basis $\left(  C_{\alpha}\right)  _{\alpha\text{ is a weak composition}}$. We
equip this $\mathbf{k}$-module $\operatorname*{NNSym}$ with an
\textquotedblleft external multiplication\textquotedblright\ defined by%
\[
C_{\alpha}\cdot C_{\beta}=C_{\alpha\beta}%
\]
(where $\alpha\beta$ is the concatenation of $\alpha$ and $\beta$), and an
\textquotedblleft internal multiplication\textquotedblright\ defined by%
\[
C_{\alpha}\ast C_{\beta}=\sum_{\substack{\gamma_{i,j}\in\mathbb{N}\text{ for
all }i\in\left[  k\right]  \text{ and }j\in\left[  \ell\right]  ;\\\gamma
_{i,1}+\gamma_{i,2}+\cdots+\gamma_{i,\ell}=\alpha_{i}\text{ for all }%
i\in\left[  k\right]  ;\\\gamma_{1,j}+\gamma_{2,j}+\cdots+\gamma_{k,j}%
=\beta_{j}\text{ for all }j\in\left[  \ell\right]  }}C_{\left(  \gamma
_{1,1},\gamma_{1,2},\ldots,\gamma_{k,\ell}\right)  }%
\]
(where $\alpha\in\mathbb{N}^{k}$ and $\beta\in\mathbb{N}^{\ell}$), and a
comultiplication $\Delta:\operatorname*{NNSym}\rightarrow\operatorname*{NNSym}%
\otimes\operatorname*{NNSym}$ defined by%
\[
\Delta\left(  C_{\alpha}\right)  =\sum_{\substack{\beta,\gamma\in
\mathbb{N}^{k}\text{;}\\\text{entrywise sum }\beta+\gamma=\alpha}}C_{\beta
}\otimes C_{\gamma}\ \ \ \ \ \ \ \ \ \ \text{for any }\alpha\in\mathbb{N}%
^{k}.
\]
It is not too hard to show that $\operatorname*{NNSym}$ thus becomes a graded
(but not connected!) $\mathbf{k}$-bialgebra when equipped with the external
multiplication $\cdot$, and a (non-graded) non-unital bialgebra when equipped
with the internal multiplication $\ast$. (Indeed, this $\operatorname*{NNSym}$
is a mild variation on the Hopf algebra $\operatorname*{NSym}$ of
noncommutative symmetric functions, which is studied (e.g.) in \cite{ncsf1} or
\cite[\S 5.4]{GriRei}; the only difference is that compositions have been
replaced by weak compositions.)

We define $\mathfrak{S}$ to be the disjoint union $\bigsqcup\limits_{k\in
\mathbb{N}}\mathfrak{S}_{k}$ of all symmetric groups $\mathfrak{S}_{k}$ for
all $k\in\mathbb{N}$. We define $\operatorname*{Perm}$ to be the free
$\mathbf{k}$-module with basis $\left(  P_{\sigma}\right)  _{\sigma
\in\mathfrak{S}}$. We equip this $\mathbf{k}$-module $\operatorname*{Perm}$
with an \textquotedblleft external multiplication\textquotedblright\ defined
by%
\[
P_{\sigma}\cdot P_{\tau}=P_{\sigma\oplus\tau},
\]
and an \textquotedblleft internal multiplication\textquotedblright\ defined by%
\[
P_{\sigma}\ast P_{\tau}=P_{\tau\left[  \sigma\right]  },
\]
and a comultiplication $\Delta:\operatorname*{Perm}\rightarrow
\operatorname*{Perm}\otimes\operatorname*{Perm}$ defined by%
\[
\Delta\left(  P_{\sigma}\right)  =P_{\sigma}\otimes P_{\sigma}.
\]
It is not too hard to show that $\operatorname*{Perm}$ becomes a $\mathbf{k}%
$-bialgebra when equipped with either of the two multiplications (but not a
graded one). Indeed, in both cases, it becomes the monoid algebra of an
appropriate monoid on the set $\operatorname*{Perm}$. (Checking the
associativity of the internal multiplication is a neat exercise in
combinatorial Yang--Baxter matrices.)

Now, let $\operatorname*{PNNSym}$ be the tensor product $\operatorname*{NNSym}%
\otimes\operatorname*{Perm}$. We equip this tensor product
$\operatorname*{PNNSym}$ with an \textquotedblleft external
multiplication\textquotedblright\ (obtained by tensoring the external
multiplications of $\operatorname*{NNSym}$ and of $\operatorname*{Perm}$), an
\textquotedblleft internal multiplication\textquotedblright\ (similarly) and a
comultiplication (similarly). We furthermore set%
\[
\widehat{F}_{\alpha,\sigma}:=C_{\alpha}\otimes P_{\sigma}\in
\operatorname*{PNNSym}\ \ \ \ \ \ \ \ \ \ \text{for any weak mopiscotion
}\left(  \alpha,\sigma\right)  .
\]
Then, $\left(  \widehat{F}_{\alpha,\sigma}\right)  _{\left(  \alpha
,\sigma\right)  \text{ is a weak mopiscotion}}$ is a basis of the $\mathbf{k}%
$-module $\operatorname*{PNNSym}$, and our operations $\cdot$, $\ast$ and
$\Delta$ on $\operatorname*{PNNSym}$ satisfy the same relations for this basis
as the analogous operations on $\operatorname*{PNSym}$ do for the basis
$\left(  F_{\alpha,\sigma}\right)  _{\left(  \alpha,\sigma\right)  \text{ is a
mopiscotion}}$ (we just have to replace each \textquotedblleft$F$%
\textquotedblright\ by \textquotedblleft$\widehat{F}$\textquotedblright). It
is thus easy to show that $\operatorname*{PNNSym}$ is a graded (but not
connected) $\mathbf{k}$-bialgebra when equipped with $\cdot$, and a
(non-graded) non-unital $\mathbf{k}$-bialgebra when equipped with $\ast$.
(Here we use the facts that the tensor product of two bialgebras is a
bialgebra, and that the tensor product of two non-unital bialgebras is a
non-unital bialgebra.)

As we said, $\operatorname*{PNNSym}$ is almost the $\operatorname*{PNSym}$
that we care about. But $\operatorname*{PNNSym}$ does not satisfy the rule%
\[
F_{\alpha,\sigma}=F_{\beta,\tau}\ \ \ \ \ \ \ \ \ \ \text{for }\left(
\beta,\tau\right)  =\left(  \alpha,\sigma\right)  ^{\operatorname*{red}}%
\]
that is fundamental to the definition of $\operatorname*{PNSym}$. Hence,
$\operatorname*{PNSym}$ is not quite $\operatorname*{PNNSym}$ but rather a
quotient of $\operatorname*{PNNSym}$. To be specific, we define a $\mathbf{k}%
$-submodule $I_{\operatorname*{red}}$ of $\operatorname*{PNNSym}$ by
\begin{align*}
I_{\operatorname*{red}}:=  &  \ \operatorname*{span}\nolimits_{\mathbf{k}%
}\left(  \widehat{F}_{\alpha,\sigma}-\widehat{F}_{\beta,\tau}\ \mid\ \left(
\beta,\tau\right)  =\left(  \alpha,\sigma\right)  ^{\operatorname*{red}%
}\right) \\
=  &  \ \operatorname*{span}\nolimits_{\mathbf{k}}\left(  \widehat{F}%
_{\alpha,\sigma}-\widehat{F}_{\beta,\tau}\ \mid\ \left(  \beta,\tau\right)
^{\operatorname*{red}}=\left(  \alpha,\sigma\right)  ^{\operatorname*{red}%
}\right)  .
\end{align*}
It is not too hard to show that this $I_{\operatorname*{red}}$ is an ideal of
$\operatorname*{PNNSym}$ with respect to both $\cdot$ and $\ast$ and a coideal
with respect to $\Delta$. Hence, the quotient $\operatorname*{PNNSym}%
/I_{\operatorname*{red}}$ inherits all operations of $\operatorname*{PNNSym}$,
thus becoming a graded bialgebra under $\cdot$ and $\Delta$ and a non-unital
bialgebra under $\ast$ and $\Delta$. Moreover, the graded bialgebra
$\operatorname*{PNNSym}/I_{\operatorname*{red}}$ is connected (since $\left(
\alpha,\sigma\right)  ^{\operatorname*{red}}=\left(  \varnothing
,\varnothing\right)  $ whenever $\left\vert \alpha\right\vert =0$), and thus
is a Hopf algebra. This completes the proof of Theorem \ref{thm.PNSym.alg} again.
\end{proof}

\begin{theorem}
\label{thm.PNSym.mods}Let $\operatorname*{PNSym}\nolimits^{\left(  2\right)
}$ be the non-unital algebra $\operatorname*{PNSym}$ with multiplication
$\ast$. Then, every connected graded bialgebra $H$ becomes a
$\operatorname*{PNSym}\nolimits^{\left(  2\right)  }$-module, with
$F_{\alpha,\sigma}$ acting as $p_{\alpha,\sigma}$.
\end{theorem}

\begin{proof}
[Proof idea.]This follows from Theorem \ref{thm.sol-mac-1}.
\end{proof}

There is also an analogue of the \textquotedblleft splitting
formula\textquotedblright\ (\cite[Proposition 5.2]{ncsf1}) for
$\operatorname*{PNSym}$, connecting the two products (internal and external)
with the comultiplication:

\begin{theorem}
\label{thm.PNSym.splitting}Let $f,g,h\in\operatorname*{PNSym}$. Write
$\Delta\left(  h\right)  $ as $\Delta\left(  h\right)  =\sum_{\left(
h\right)  }h_{\left(  1\right)  }\otimes h_{\left(  2\right)  }$ (using
Sweedler notation). Then,%
\[
\left(  fg\right)  \ast h=\sum_{\left(  h\right)  }\left(  f\ast h_{\left(
1\right)  }\right)  \left(  g\ast h_{\left(  2\right)  }\right)  .
\]

\end{theorem}

(To be more precise, the analogue of \cite[Proposition 5.2]{ncsf1} would be
the generalization of this formula to iterated coproducts $\Delta^{\left[
r\right]  }\left(  h\right)  $, but this generalization follows from Theorem
\ref{thm.PNSym.splitting} by a straightforward induction on $r$.)

\begin{proof}
[Proof idea for Theorem \ref{thm.PNSym.splitting}.]Fairly easy using
Definition \ref{def.PNSym.ops}. (An important first step is to show that the
formulas for $F_{\alpha,\sigma}\cdot F_{\beta,\tau}$ and $F_{\alpha,\sigma
}\ast F_{\beta,\tau}$ hold not only for mopiscotions $\left(  \alpha
,\sigma\right)  $ and $\left(  \beta,\tau\right)  $ but also for weak
mopiscotions $\left(  \alpha,\sigma\right)  $ and $\left(  \beta,\tau\right)
$.)
\end{proof}

\begin{question}
What is the combinatorial meaning of $\operatorname*{PNSym}$ ?
\end{question}

\begin{question}
Is there a cancellation-free formula for the antipode of
$\operatorname*{PNSym}$ ?
\end{question}

\begin{question}
Does $\operatorname*{PNSym}$ embed into any kind of noncommutative formal
power series?
\end{question}

\begin{remark}
An analogue of $\operatorname*{PNSym}\nolimits^{\left(  2\right)  }$ for Hopf
monoids is the Janus algebra of Aguiar and Mahajan \cite[\S 11.3]{AguMah20}.
Is there a way to translate results? Could $\operatorname*{PNSym}$ be the
result of transforming $\operatorname*{NSym}$ into a Hopf monoid using one of
Aguiar's functors, and then transforming it back into an algebra using
another? My impression is that this procedure could give us the permutations
$\sigma$ in the $F_{\alpha,\sigma}$'s.
\end{remark}

Note that there is a canonical bijection between mopiscotions and set
compositions into intervals. Specifically, if $\left(  \alpha,\sigma\right)  $
is a mopiscotion with $\left\vert \alpha\right\vert =n$, then we can consider
the set composition of the set $\left\{  1,2,\ldots,n\right\}  $ into the
intervals whose lengths are the parts of $\alpha$ and whose order is
determined by $\sigma$. (Unfortunately, there are some choices involved here,
and I'm not sure which is the right one.) This might be helpful for embedding
$\operatorname*{PNSym}$ in some Hopf algebra of set compositions, such as (the
dual of) $\operatorname*{WQSym}$.

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\end{thebibliography}


\end{document}