\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}%
\usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage}
\usepackage[all,cmtip]{xy}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{framed}
\usepackage{amsmath}
\usepackage{comment}
\usepackage{color}
\usepackage{hyperref}
\usepackage[sc]{mathpazo}
\usepackage[T1]{fontenc}
\usepackage{amsthm}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{LastRevised=Saturday, June 06, 2026 14:46:47}
%TCIDATA{SuppressPackageManagement}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\theoremstyle{definition}
\newtheorem{theo}{Theorem}[section]
\newenvironment{theorem}[1][]
{\begin{theo}[#1]\begin{leftbar}}
{\end{leftbar}\end{theo}}
\newtheorem{lem}[theo]{Lemma}
\newenvironment{lemma}[1][]
{\begin{lem}[#1]\begin{leftbar}}
{\end{leftbar}\end{lem}}
\newtheorem{prop}[theo]{Proposition}
\newenvironment{proposition}[1][]
{\begin{prop}[#1]\begin{leftbar}}
{\end{leftbar}\end{prop}}
\newtheorem{defi}[theo]{Definition}
\newenvironment{definition}[1][]
{\begin{defi}[#1]\begin{leftbar}}
{\end{leftbar}\end{defi}}
\newtheorem{remk}[theo]{Remark}
\newenvironment{remark}[1][]
{\begin{remk}[#1]\begin{leftbar}}
{\end{leftbar}\end{remk}}
\newtheorem{coro}[theo]{Corollary}
\newenvironment{corollary}[1][]
{\begin{coro}[#1]\begin{leftbar}}
{\end{leftbar}\end{coro}}
\newtheorem{conv}[theo]{Convention}
\newenvironment{condition}[1][]
{\begin{conv}[#1]\begin{leftbar}}
{\end{leftbar}\end{conv}}
\newtheorem{quest}[theo]{Question}
\newenvironment{algorithm}[1][]
{\begin{quest}[#1]\begin{leftbar}}
{\end{leftbar}\end{quest}}
\newtheorem{warn}[theo]{Warning}
\newenvironment{conclusion}[1][]
{\begin{warn}[#1]\begin{leftbar}}
{\end{leftbar}\end{warn}}
\newtheorem{conj}[theo]{Conjecture}
\newenvironment{conjecture}[1][]
{\begin{conj}[#1]\begin{leftbar}}
{\end{leftbar}\end{conj}}
\newtheorem{exmp}[theo]{Example}
\newenvironment{example}[1][]
{\begin{exmp}[#1]\begin{leftbar}}
{\end{leftbar}\end{exmp}}
\newenvironment{statement}{\begin{quote}}{\end{quote}}
\iffalse
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\fi
\newenvironment{verlong}{}{}
\newenvironment{vershort}{}{}
\newenvironment{noncompile}{}{}
\excludecomment{verlong}
\includecomment{vershort}
\excludecomment{noncompile}
\newcommand{\kk}{\mathbf{k}}
\newcommand{\id}{\operatorname{id}}
\newcommand{\ev}{\operatorname{ev}}
\newcommand{\Comp}{\operatorname{Comp}}
\newcommand{\bk}{\mathbf{k}}
\newcommand{\Nplus}{\mathbb{N}_{+}}
\newcommand{\NN}{\mathbb{N}}
\newcommand{\are}{\ar@{-}}
\let\sumnonlimits\sum
\let\prodnonlimits\prod
\let\bigcupnonlimits\bigcup
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
\renewcommand{\bigcup}{\bigcupnonlimits\limits}
\setlength\textheight{22.5cm}
\setlength\textwidth{15cm}
\ihead{Errata to ``Two interacting Hopf algebras''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Two interacting Hopf algebras of trees}

\textit{Damien Calaque, Kurusch Ebrahimi-Fard, and Dominique Manchon}

\href{https://arxiv.org/abs/0806.2238v4}{arXiv:0806.2238v4} [math.CO] 17 Dec 2011

\textbf{Errata and questions}
\end{center}

\setcounter{section}{13}

\appendix


\section{Corrections}

\begin{itemize}
\item \textbf{Page 3, two lines above (2):} You write: \textquotedblleft the
group $G_{0}=e+\mathfrak{g}_{0}$ of linear maps $\gamma$ that send the
bialgebra unit to the algebra unit, $\alpha\left(  1\right)  =1_{\mathcal{A}}%
$\textquotedblright. You mean $\gamma\left(  1\right)  $, not $\alpha\left(
1\right)  $ here.

\item \textbf{Page 3, equation (3):} Add a condition $\rho\left(  1\right)
=1_{\mathcal{A}}$ here (otherwise, the constant zero map $0$ would be a character...).

\item \textbf{Page 3, equation (5):} Remove the comma at the end of this equation.

\item \textbf{Page 3, section 3:} You write: \textquotedblleft Recall, that a
--non-planar-- \textit{rooted tree} is either the empty set, or
[...]\textquotedblright. I doubt that you really want the empty set to count
as a tree. If you do, then the definition of a rooted forest (page 4) allows
for \textquotedblleft invisible children\textquotedblright\ and infinitely
many different trees with one vertex only, which is in conflict with all your examples.

\item \textbf{Page 4, first line of the page:} Typo: \textquotedblleft
egde\textquotedblright\ should be \textquotedblleft edge\textquotedblright.

\item \textbf{Page 4, second paragraph of the page:} You write:
\textquotedblleft A \textit{rooted forest} is a finite collection $s=\left(
t_{1},t_{2},...,t_{n}\right)  $ of rooted trees\textquotedblright. There is
nothing really wrong with this, but I find the notation $\left(  t_{1}%
,t_{2},...,t_{n}\right)  $ not really appropriate for a collection where the
order of the elements doesn't matter (i.e., for a multiset): it conflicts with
the standard notation for the $n$-tuple $\left(  t_{1},t_{2},...,t_{n}\right)
$ (in which the order of the elements does matter). There doesn't seem to be a
(universally agreed upon) standard notation for multisets, though.

\item \textbf{Page 4, \S 4.1:} As I said above, the empty set does not really
belong into the class of trees. Thus you don't need to say \textquotedblleft%
\textit{excluding the empty tree}\textquotedblright. (The empty
\textbf{forest} makes sense, but this is just the empty collection $\left(
{}\right)  $ of rooted trees.)

\item \textbf{Page 4, equation (7):} It is worth saying that this formula can
be rewritten as%
\begin{equation}
\Delta\left(  t\right)  =\sum_{F\subseteq E\left(  t\right)  }\left(  t\mid
F\right)  \otimes\left(  t\diagup F\right)  , \tag{7'}\label{7'}%
\end{equation}
where

\begin{itemize}
\item $E\left(  t\right)  $ is the set of all edges of $t$;

\item $t\mid F$ means the result of removing all edges $e\notin F$ from $t$;

\item $t\diagup F$ means the result of contracting each edge $f\in F$ to a
point in $t$.
\end{itemize}

\item \textbf{Page 4, \S 4.1:} It is worth pointing out that the formula (7)
holds not only when $t$ is a tree, but also when $t$ is a forest (since the
right hand side is clearly multiplicative in $t$). The same applies to the
formula (\ref{7'}). This is used tacitly in the proof of Theorem 8. (More
precisely, the proof of Theorem 8 uses the analogous formulas for the map
$\Phi$, which lifts the map $\Delta$ to an algebra morphism from
$\mathcal{H}_{\operatorname*{CK}}$ to $\mathcal{H}\otimes\mathcal{H}%
_{\operatorname*{CK}}$.)

\item \textbf{Page 5, \S 4.2:} On the first line of \S 4.2, you write
\textquotedblleft$\widetilde{\mathcal{H}}=S\left(  T\right)  $%
\textquotedblright. The $S$ here should be a calligraphic $\mathcal{S}$.

\item \textbf{Page 5, \S 4.3:} You write: \textquotedblleft For any tree $t$
the corresponding $Z_{t}$ is an infinitesimal character of $\mathcal{H}%
$\textquotedblright. This holds only for trees $t\neq\bullet$.

\item \textbf{Page 5, \S 4.3:} In the formula%
\[
\left(  Z_{t}\star Z_{u}-Z_{u}\star Z_{t}\right)  \left(  v\right)  =\sum
_{s}Z_{t}\left(  s\right)  Z_{u}\left(  v\diagup s\right)  -\sum_{s}%
Z_{u}\left(  s\right)  Z_{t}\left(  v\diagup s\right)  ,
\]
it is worth explaining that the sums range over all subforests of $v$.

\item \textbf{Page 5, \S 4.3:} The formula%
\[
t\rhd u=\sum\limits_{v,\text{ }t\subset v\text{ and }v\diagup t=u}N\left(
t,u,v\right)  v
\]
needs some explanations to be understood correctly (I believe).

First of all, the sum $\sum\limits_{v,\text{ }t\subset v\text{ and }v\diagup
t=u}$ is a sum over all \textit{trees} $v$ satisfying $t\subset v$ and
$v\diagup t=u$.

Second, \textquotedblleft$t\subset v$ and $v\diagup t=u$\textquotedblright\ is
actually a shorthand notation for \textquotedblleft there is a subtree $s$ of
$v$ isomorphic to $t$ and satisfying $v\diagup s=u$ (with the $=$ sign
denoting isomorphism of trees!)\textquotedblright.

It would be better to rewrite the above formula in the following less
confusing way:%
\begin{equation}
t\rhd u=\sum\limits_{v\text{ is a tree}}N\left(  t,u,v\right)  v, \tag{Nab}%
\end{equation}
where $N\left(  t,u,v\right)  $ is the number of subforests $s$ of $v$ that
are isomorphic to $t$ and satisfy $v\diagup s\cong u$ (note that I don't like
speaking of $v/t$, since $t$ is itself not a subtree of $v$ but merely
isomorphic to a such, but the isomorphism class of $v/t$ does not depend on
$v$ and $t$ alone). Of course, the sum can be restricted to range only over
those $v$ for which $N\left(  t,u,v\right)  \neq0$ (in particular, we only
need to sum over the trees $v$ with at most $e\left(  t\right)  +e\left(
u\right)  $ many edges).

\begin{noncompile}
(There are two differences between summing over all forests and summing over
all subforests of $t$: The obvious difference is that not all forests are
subforests of $t$, but the more subtle difference is that when summing over
all forests, isomorphic forests do not count as distinct (because
\textquotedblleft forests\textquotedblright\ means \textquotedblleft forests
up to isomorphism\textquotedblright), but when summing over all subforests of
$t$, any two subforests of $t$ with distinct edge sets count as distinct (even
if they are isomorphic).)
\end{noncompile}

\item \textbf{Page 5, \S 4.3:} You write: \textquotedblleft where
$\mathfrak{g}=\operatorname*{Prim}\mathcal{H}^{\circ}$ is the Lie algebra
spanned by the $Z_{t}$'s for rooted trees $t$\textquotedblright. You mean
\textquotedblleft\lbrack...] for rooted trees $t\neq\bullet$\textquotedblright%
\ here, since $Z_{\bullet}$ is not primitive.

\item \textbf{Page 5, \S 4.3:} You write: \textquotedblleft The product $\rhd$
satisfies the left pre-Lie relation (5).\textquotedblright\ This is true, but
not at all obvious. In the appendix below (Appendix \ref{apx.pre-lie-proof}),
I give a proof (actually, this is the wonderful proof that you sent me in an email).

\item \textbf{Page 6, \S 4:} \textquotedblleft See [24] for more on the
combinatorics of rooted trees and Hopf algebras.\textquotedblright: The
reference [24] says nothing about trees. Probably a mis-reference.

\item \textbf{Page 6, Corollary 2:} It should be said that the summation
indices $k_{1},k_{2},\ldots,k_{r}$ are required to be positive.

\item \textbf{Page 6, proof of Proposition 3:} I don't understand what
\textquotedblleft any class of subforests contains only one
element\textquotedblright\ means and why this is needed.

\item \textbf{Page 7, \S 5:} \textquotedblleft perharps\textquotedblright%
\ $\rightarrow$ \textquotedblleft perhaps\textquotedblright.

\item \textbf{Page 7, (13):} Replace \textquotedblleft$\sum_{j\geq0}p_{2j}%
$\textquotedblright\ by \textquotedblleft$\sum_{j\geq1}p_{2j}$%
\textquotedblright, since there is no $p_{0}$.

\item \textbf{Page 8, Proposition 5:} I'm not sure this is true. For example,
let $t$ be the tree%
\[%
%TCIMACRO{\TeXButton{7-vertex tree}{\xymatrix{
%\bullet\are[d]^a & \bullet\are[d]^b & \bullet\are[d]^c \\
%\bullet\are[dr]_d & \bullet\are[d]^e & \bullet\are[dl]^f \\
%& \bullet}}}%
%BeginExpansion
\xymatrix{
\bullet\are[d]^a & \bullet\are[d]^b & \bullet\are[d]^c \\
\bullet\are[dr]_d & \bullet\are[d]^e & \bullet\are[dl]^f \\
& \bullet}%
%EndExpansion
\ \ .
\]
Then, $\Delta\left(  t\right)  $ contains an addend corresponding to the
contraction of $s=\left\{  a,c\right\}  $. But there are (at least) three
\textquotedblleft floor\textquotedblright\ functions $\mathbf{fl}:E\left(
t\right)  \rightarrow\mathbb{N}$ that correspond to this addend:

\begin{itemize}
\item one that sends $d,e,f$ to $0$ and $a$ to $1$ and $b$ to $2$ and $c$ to
$3$;

\item one that sends $d,e,f$ to $0$ and $a$ to $3$ and $b$ to $2$ and $c$ to
$1$;

\item one that sends $d,e,f,b$ to $0$ and $a,c$ to $1$.
\end{itemize}

This can be fixed by changing Definition 1 as follows: Instead of requiring
the image of $\mathbf{fl}$ to be an interval, we require that any two edges
$e,f$ with a common endpoint satisfy $\left\vert \mathbf{fl}\left(  e\right)
-\mathbf{fl}\left(  f\right)  \right\vert \leq1$. Then, it is easy to see that
the map%
\begin{align*}
\left\{  \text{floored tree structures on }t\right\}   &  \rightarrow\left\{
\text{subsets of }E\left(  t\right)  \right\}  ,\\
\mathbf{fl}  &  \mapsto\left\{  e\in E\left(  t\right)  \ \mid\ \mathbf{fl}%
\left(  e\right)  \text{ is even}\right\}
\end{align*}
is a bijection\footnote{Indeed, the inverse of this map sends any subset $F$
of $E\left(  t\right)  $ to the floored tree structure $\mathbf{fl}$ defined
as follows: For any $e\in E\left(  t\right)  $, we consider the path from the
top vertex of $e$ down to the root of $t$. Color each edge of this path black
if it belongs to $F$ and white if it does not. Also append an extra black edge
at the end of this path (after the root of $t$). Then, $\mathbf{fl}\left(
e\right)  $ is the number of times that the color of the edge changes as we
walk this path from one end to the other.}. This bijection furthermore has the
property that if it sends a floored tree structure $\mathbf{fl}$ (with
corresponding floored tree $\widetilde{t}$) to a subset $F$ of $E\left(
t\right)  $, then $\prod_{j\geq1}s_{2j-1}\left(  \widetilde{t}\right)  =t\mid
F$ and $t/\prod_{j\geq1}s_{2j-1}\left(  \widetilde{t}\right)  =t/F$. Hence,
the claim of Proposition 5 follows from the formula (\ref{7'}) by reindexing
the sum using this bijection. (I hope this was correct...)

Also, in (14), the summation sign \textquotedblleft$\sum_{r=1}^{h\left(
t\right)  }$\textquotedblright\ should be \textquotedblleft$\sum
_{r=1}^{h\left(  t\right)  +1}$\textquotedblright\ or just \textquotedblleft%
$\sum_{r\geq1}$\textquotedblright.

\item \textbf{Page 8, \S 8:} In the first displayed equation of \S 8, replace
\textquotedblleft$E_{\sigma}$\textquotedblright\ by \textquotedblleft%
$E_{\sigma}\left(  t\right)  $\textquotedblright.

While at that, it is worth saying that $E_{\sigma}\left(  t\right)  $ also
equals $\dfrac{1}{t!\sigma\left(  t\right)  }$ (by the definition of
$\operatorname*{CM}\left(  t\right)  $).

\item \textbf{Page 8, \S 8:} In all \textquotedblleft$\sum_{k_{1}+\cdots
+k_{r}=n}$\textquotedblright\ sums, the summation indices $k_{1},k_{2}%
,\ldots,k_{r}$ must be required to be positive.

\item \textbf{Page 9, \S 9.1:} \textquotedblleft spanned by nonempty rooted
trees\textquotedblright\ $\rightarrow$ \textquotedblleft spanned by rooted
trees\textquotedblright, since the empty graph is not a tree.

\item \textbf{Page 9, \S 9.1:} After (17), replace \textquotedblleft Here the
notation $V<W$ means that $x<y$ for any vertex $x$ of $v$ and any vertex $y$
of $w$ such that $x$ and $y$ are comparable\textquotedblright\ by
\textquotedblleft Here the notation $W<V$ means that $x<y$ for any vertex $x$
of $w$ and any vertex $y$ of $v$ such that $x$ and $y$ are
comparable\textquotedblright. This way, the order of $W$ and $V$ matches the
one in (17) and also in the next sentence.

\item \textbf{Page 9, (17):} The \textquotedblleft$t\otimes\mathbf{1}%
$\textquotedblright\ and \textquotedblleft$\mathbf{1}\otimes t$%
\textquotedblright\ addends should be removed from the right hand side.
Indeed, as you explain two sentences further on, you do count the empty cut
and the total cut as admissible cuts; thus, these two addends are already
included in the sum $\sum_{c\in\operatorname*{Adm}\left(  t\right)  }%
P^{c}\left(  t\right)  \otimes R^{c}\left(  t\right)  $.

\item \textbf{Page 10:} On the first line of this page, \textquotedblleft Here
we denote by $\operatorname*{Adm}\left(  t\right)  $ the set of admissible
cuts of a forest, i.e. the set of collections of edges such that any path from
the root to a leaf contains at most one edge of the
collection.\textquotedblright. Even with the clarification given in the next
sentence, this is only true when $t$ is a tree, not when $t$ is a forest. For
a forest, admissible cuts need to be defined as sets of vertices, not of edges.

\item \textbf{Page 10:} \textquotedblleft the subforest formed by the edges
above the cut $c\in\operatorname*{Adm}\left(  t\right)  $ (resp. the subforest
formed by the edges under the cut)\textquotedblright: In both cases, you want
not just the edges but also the vertices.

\item \textbf{Page 10:} After \textquotedblleft Note that the trunk of a tree
is a tree\textquotedblright, add \textquotedblleft or empty\textquotedblright.

\item \textbf{Page 10, equation (19):} The forest $u$ must be assumed to be
nonempty here.

\item \textbf{Page 10, equation (20):} The \textquotedblleft$V_{n}%
<\cdots<V_{1}$\textquotedblright\ under the summation sign should be
understood as saying that $V_{j}<V_{i}$ for all $i<j$ (not only that
$V_{i+1}<V_{i}$ for all $i$, which would be a weaker requirement).

\item \textbf{Page 10, definition of pre-Lie algebra structure:} Replace
\textquotedblleft$\mathcal{H}^{\circ}$\textquotedblright\ by \textquotedblleft%
$\mathcal{H}_{\operatorname*{CK}}^{\circ}$\textquotedblright\ twice (in
\textquotedblleft it is a primitive element of $\mathcal{H}^{\circ}%
$\textquotedblright\ and in \textquotedblleft the (convolution) product of
$\mathcal{H}^{\circ}$\textquotedblright).

\item \textbf{Page 10, definition of pre-Lie algebra structure:} The formula
\textquotedblleft$t\rightarrow u=\sum_{v}N^{\prime}\left(  t,u,v\right)
v$\textquotedblright\ seems to be meant in the sense that all of $t,u,v$ are
assumed to be trees. Otherwise, it would be pretty clear that $\delta_{t}%
\ast\delta_{u}=\delta_{t\rightarrow u}$, which makes the formula $\delta
_{t}\ast\delta_{u}-\delta_{u}\ast\delta_{t}=\delta_{t\rightarrow
u-u\rightarrow t}$ a trivial corollary.

\item \textbf{Page 10, definition of pre-Lie algebra structure:} In
\textquotedblleft where $N^{\prime}\left(  t,u,v\right)  $ is the number of
partitions $V\left(  t\right)  =V\ \amalg\ W$\textquotedblright, replace
\textquotedblleft$V\left(  t\right)  $\textquotedblright\ by \textquotedblleft%
$V\left(  v\right)  $\textquotedblright.

\item \textbf{Page 11, proof of Theorem 8:} This proof uses the formula%
\[
\Phi\left(  t\right)  =\sum_{\substack{s\text{ is a subforest}\\\text{of }%
t}}s\otimes t\diagup s\ \ \ \ \ \ \ \ \ \ \left(  \text{in }\mathcal{H}%
\otimes\mathcal{H}_{\operatorname*{CK}}\right)
\]
not just for all trees $t$ but also for all forests $t$.

\item \textbf{Page 11, proof of Theorem 8:} In the first display, the
parentheses around the \textquotedblleft$t$\textquotedblright\ in
\textquotedblleft$\Phi\left(  t\right)  $\textquotedblright\ have different sizes.

\item \textbf{Page 12, proof of Theorem 8:} In the first display, the $\Big/$
signs are misleadingly tall, looking as if they span the $\otimes$ signs. It
would probably be better to make them normally-sized but put parentheses
around \textquotedblleft$s\cap P^{c}\left(  t\right)  $\textquotedblright\ and
around \textquotedblleft$s\cap R^{c}\left(  t\right)  $\textquotedblright.

\item \textbf{Page 13, Corollary 12:} \textquotedblleft le $\alpha
$\textquotedblright\ should be \textquotedblleft let $\alpha$%
\textquotedblright.

\item \textbf{Page 13, Corollary 12:} \textquotedblleft linear maps
form\textquotedblright\ should be \textquotedblleft linear maps
from\textquotedblright.

\item \textbf{Page 15, (31):} Replace the comma in \textquotedblleft$B\left(
\alpha\ast\beta,a\right)  $\textquotedblright\ by a semicolon.

\item \textbf{Page 17, \S 11.1:} \textquotedblleft Let $\diamond$ the
product\textquotedblright\ $\rightarrow$ \textquotedblleft Let $\diamond$ be
the product\textquotedblright.

\item \textbf{Page 17, \S 11.1:} Replace \textquotedblleft connected graded
Hopf algebra\textquotedblright\ by \textquotedblleft connected filtered Hopf
algebra\textquotedblright. (I am also not sure if [23] is the right reference
for this.)

\item \textbf{Page 17, \S 11.1:} In the displayed equation just above Remark
18, the sign \textquotedblleft$\sum_{r=0}^{s}$\textquotedblright\ should be
\textquotedblleft$\sum_{r=0}^{k}$\textquotedblright.

\item \textbf{Page 19, proof of Theorem 20:} In the first display, replace
\textquotedblleft$\mathcal{H}_{\operatorname*{CK}}^{\left(  n\right)  }%
$\textquotedblright\ by \textquotedblleft$\mathcal{H}_{\operatorname*{CK}%
}^{\left(  k\right)  }$\textquotedblright.

\item \textbf{Page 19, Lemma 21:} \textquotedblleft Let $\widetilde{\delta
}:\mathcal{A}\rightarrow k$ the linear map\textquotedblright\ $\rightarrow$
\textquotedblleft Let $\widetilde{\delta}:\mathcal{A}\rightarrow k$ be the
linear map\textquotedblright.

\item \textbf{Page 19, proof of Lemma 21:} \textquotedblleft bigger than
$\sup\left(  l\left(  u\right)  ,l\left(  v\right)  \right)  $%
\textquotedblright\ should be \textquotedblleft$\geq\sup\left(  l\left(
u\right)  ,l\left(  v\right)  \right)  $\textquotedblright.

\item \textbf{Page 20, Corollary 23:} \textquotedblleft the
number\textquotedblright\ $\rightarrow$ \textquotedblleft be the
number\textquotedblright.

\item \textbf{Page 20, \S 11.4:} In the first displayed equation of \S 11.4,
replace \textquotedblleft$\sum_{k=0}^{n}$\textquotedblright\ by
\textquotedblleft$\sum_{r=0}^{k}$\textquotedblright.

\item \textbf{Page 20, \S 11.4:} \textquotedblleft the choice of a subset $E$
of $\left\{  1,\ldots l\right\}  $\textquotedblright\ should be
\textquotedblleft the choice of a subset $E$ of $\left\{  1,\ldots,k\right\}
$\textquotedblright.

\item \textbf{Page 20, \S 11.4:} \textquotedblleft
mutinomial\textquotedblright\ $\rightarrow$ \textquotedblleft
multinomial\textquotedblright.

\item \textbf{Page 21, definition of the generalized corolla }$\mathcal{C}%
_{k_{1},\ldots,k_{n}}$\textbf{:} What is called $E_{j}$ here was previously
denoted $E_{j-1}$.
\end{itemize}

\section{\label{apx.pre-lie-proof}Appendix: Proof of the pre-Lie relation for
the product $\rhd$ defined in \S 4.3}

\textit{Proof that the product }$\rhd$ \textit{defined in \S 4.3 satisfies the
left pre-Lie relation (5):}

First, some preparations.

For any assertion $\mathcal{A}$, we are going to denote by $\left[
\mathcal{A}\right]  $ the truth value of this assertion $\mathcal{A}$ (defined
by $\left[  \mathcal{A}\right]  =%
\begin{cases}
1, & \text{if }\mathcal{A}\text{ is true;}\\
0, & \text{if }\mathcal{A}\text{ is false}%
\end{cases}
\ \ $). Then, clearly, $\left[  \mathcal{A}\right]  \cdot\left[
\mathcal{B}\right]  =\left[  \mathcal{A}\text{ and }\mathcal{B}\right]  $ for
any two assertions $\mathcal{A}$ and $\mathcal{B}$.

\begin{lemma}
\label{lem.A1}Any two trees $t\neq\bullet$ and $u\neq\bullet$ satisfy%
\begin{equation}
Z_{t}\star Z_{u}=\left(  1+\left[  t=u\right]  \right)  Z_{tu}+Z_{t\rhd u}.
\tag{A1}%
\end{equation}

\end{lemma}

\begin{proof}
Let $t\neq\bullet$ and $u\neq\bullet$ be two trees. For any forest $w$, we
have%
\begin{align*}
&  \left(  Z_{t}\star Z_{u}\right)  \left(  w\right) \\
&  =\sum_{s\subseteq w}\underbrace{Z_{t}\left(  s\right)  }_{=\left[  s\cong
t\right]  }\underbrace{Z_{u}\left(  w\diagup s\right)  }_{=\left[  w\diagup
s\cong u\right]  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\left(
w\right)  =\sum_{s\subseteq w}s\otimes\left(  w\diagup s\right)  \text{ by
(7)}\right) \\
&  =\sum_{s\subseteq w}\underbrace{\left[  s\cong t\right]  \left[  w\diagup
s\cong u\right]  }_{\substack{=\left[  s\cong t\text{ and }w\diagup s\cong
u\right]  \\\text{(since }\left[  \mathcal{A}\right]  \cdot\left[
\mathcal{B}\right]  =\left[  \mathcal{A}\text{ and }\mathcal{B}\right]
\\\text{for any two assertions }\mathcal{A}\text{ and }\mathcal{B}\text{)}%
}}=\sum_{s\subseteq w}\left[  s\cong t\text{ and }w\diagup s\cong u\right] \\
&  =\sum_{\substack{s\subseteq w;\\s\cong t\text{ and }w\diagup s\cong
u}}\underbrace{\left[  s\cong t\text{ and }w\diagup s\cong u\right]
}_{\substack{=1\\\text{(since }\left(  s\cong t\text{ and }w\diagup s\cong
u\right)  \text{)}}}+\sum_{\substack{s\subseteq w;\\\text{not }\left(  s\cong
t\text{ and }w\diagup s\cong u\right)  }}\underbrace{\left[  s\cong t\text{
and }w\diagup s\cong u\right]  }_{\substack{=0\\\text{(since not }\left(
s\cong t\text{ and }w\diagup s\cong u\right)  \text{)}}}\\
&  =\underbrace{\sum_{\substack{s\subseteq w;\\s\cong t\text{ and }w\diagup
s\cong u}}1}_{\substack{=\left(  \text{number of subforests }s\subseteq
w\text{ such that }s\cong t\text{ and }w\diagup s\cong u\right)
\cdot1\\=\left(  \text{number of subforests }s\subseteq w\text{ such that
}s\cong t\text{ and }w\diagup s\cong u\right)  }}+\underbrace{\sum
_{\substack{s\subseteq w;\\\text{not }\left(  s\cong t\text{ and }w\diagup
s\cong u\right)  }}0}_{=0}\\
&  =\left(  \text{number of subforests }s\subseteq w\text{ such that }s\cong
t\text{ and }w\diagup s\cong u\right) \\
&  =\left(  \text{number of subtrees }s\subseteq w\text{ such that }s\cong
t\text{ and }w\diagup s\cong u\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since all subforests }s\subseteq w\text{ such that }s\cong t\text{ are
actually subtrees of }w\\
\text{(because if }s\cong t\text{, then }s\text{ is a tree (since }t\text{ is
a tree))}%
\end{array}
\right) \\
&  =N\left(  t,u,w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of }N\left(  t,u,w\right)  \right)  .
\end{align*}
Since the forests form a basis of $\mathcal{H}$, this yields that%
\begin{align}
&  Z_{t}\star Z_{u}\nonumber\\
&  =\sum_{v\text{ forest}}N\left(  t,u,v\right)  Z_{v}\nonumber\\
&  =\underbrace{\sum_{\substack{v\text{ forest;}\\v\text{ is a tree}}}N\left(
t,u,v\right)  Z_{v}}_{\substack{=\sum\limits_{v\text{ tree}}N\left(
t,u,v\right)  Z_{v}\\=\sum\limits_{v=\bullet}N\left(  t,u,v\right)  Z_{v}%
+\sum\limits_{v\text{ tree }\neq\bullet}N\left(  t,u,v\right)  Z_{v}%
}}+\underbrace{\sum_{\substack{v\text{ forest;}\\v\text{ is the forest }%
tu}}N\left(  t,u,v\right)  Z_{v}}_{=N\left(  t,u,tu\right)  Z_{tu}}%
+\sum_{\substack{v\text{ forest;}\\v\text{ is neither a tree}\\\text{nor the
forest }tu}}N\left(  t,u,v\right)  Z_{v}\nonumber\\
&  =\sum\limits_{v=\bullet}N\left(  t,u,v\right)  Z_{v}+\sum\limits_{v\text{
tree }\neq\bullet}N\left(  t,u,v\right)  Z_{v}+N\left(  t,u,tu\right)
Z_{tu}+\sum_{\substack{v\text{ forest;}\\v\text{ is neither a tree}\\\text{nor
the forest }tu}}N\left(  t,u,v\right)  Z_{v}. \tag{A2}%
\end{align}


Now, we recall that for every forest $v$, the number $N\left(  t,u,v\right)  $
is defined as the number of subtrees $s\subseteq v$ such that $s\cong t$ and
$v\diagup s\cong u$. As a consequence, we have:

\begin{itemize}
\item If $v$ is the tree $\bullet$, then $N\left(  t,u,v\right)  =0$ (since,
in this case, there are no subtrees $s\subseteq v$ such that $s\cong t$ and
$v\diagup s\cong u$ (because $t\neq\bullet$)).

\item If $v$ is neither a tree (not even $\bullet$) nor the forest $tu$, then
$N\left(  t,u,v\right)  =0$ (since, in this case, there are no subtrees
$s\subseteq v$ such that $s\cong t$ and $v\diagup s\cong u$%
\ \ \ \ \footnote{\textit{Proof.} Let $v$ be neither a tree (not even
$\bullet$) nor the forest $tu$. Assume that there exists some subtree
$s\subseteq v$ such that $s\cong t$ and $v\diagup s\cong u$. Consider this
subtree $s$.
\par
Since $v\diagup s\cong u$, it is clear that the forest $v\diagup s$ has only
one connected component (since $u$ is a tree and thus has only one connected
component). Here, when we say "connected component", we are \textit{not}
counting isolated vertices as extra components (because the one-vertex tree
$\bullet$ is the unity of our algebra $\mathcal{H}$, so we identify every
forest $f$ with the forest $f\bullet$).
\par
Since $v$ is not a tree, it is clear that $v$ is a forest with at least two
connected components. The subtree $s$ must be a subset of one of them (since
it is a tree, thus connected). If $s$ were a \textit{proper} subset of one of
these connected components, then the forest $v\diagup s$ would still have at
least two connected components (because when passing from $v$ to $v\diagup s$,
the connected component containing $s$ would not completely disappear, and the
other connected components would stay unchanged), which would contradict the
fact that $v\diagup s$ has only one connected component. Hence, $s$ is a
subset but not a proper subset of one of these connected components. In other
words, $s$ is one of these connected components. Hence, $v\diagup s$ is the
union of all the connected components other than $s$. Since $v\diagup s$ has
only one connected component, this means that there is only one connected
component other than $s$, and this connected component is isomorphic to
$v\diagup s$. Hence, our forest $v$ consists of two connected components, one
of them being $s\cong t$ and the other being $\cong v\diagup s\cong u$. Hence,
our forest $v$ is $tu$. But this contradicts the assumption $v\neq tu$.
\par
This contradiction shows that our assumption (that there exists some subtree
$s\subseteq v$ such that $s\cong t$ and $v\diagup s\cong u$) was wrong. Hence,
there are no subtrees $s\subseteq v$ such that $s\cong t$ and $v\diagup s\cong
u$, qed.}).

\item If $v$ is the forest $tu$, then $N\left(  t,u,v\right)  =1+\left[
t=u\right]  $\ \ \ \ \footnote{\textit{Proof.} Let $v$ be the forest $tu$. Let
$s\subseteq v$ be a subtree such that $s\cong t$ and $v\diagup s\cong u$.
\par
Since $v\diagup s\cong u$, it is clear that the forest $v\diagup s$ has only
one connected component (since $u$ is a tree and thus has only one connected
component). Here, when we say "connected component", we are \textit{not}
counting isolated vertices as extra components (because the one-vertex tree
$\bullet$ is the unity of our algebra $\mathcal{H}$, so we identify every
forest $f$ with the forest $f\bullet$).
\par
Since $v=tu$, it is clear that $v$ is a forest with exactly two connected
components $t$ and $u$. The subtree $s$ must be a subset of one of them (since
it is a tree, thus connected). If $s$ were a \textit{proper} subset of one of
these connected components, then the forest $v\diagup s$ would still have at
least two connected components (because when passing from $v$ to $v\diagup s$,
the connected component containing $s$ would not completely disappear, and the
other connected components would stay unchanged), which would contradict the
fact that $v\diagup s$ has only one connected component. Hence, $s$ is a
subset but not a proper subset of one of these connected components. In other
words, $s$ is one of these connected components.
\par
Now we recall that the two connected components of $v$ are $t$ and $u$. If
$t\neq u$, then this forces $s$ to be the component $t$ (because $s$ is one of
these two connected components, but it cannot be $u$ since $s\cong t\neq u$).
\par
Now forget that we fixed $s$. We have thus shown that every subtree
$s\subseteq v$ such that $s\cong t$ and $v\diagup s\cong u$ must be the
connected component $t$ of $v$. And conversely, $t\subseteq v$ is a subtree
such that $t\cong t$ and $v\diagup t\cong u$ (since $v=tu$). Hence, if $t\neq
u$, then there is exactly one subtree $s\subseteq v$ such that $s\cong t$ and
$v\diagup s\cong u$. In other words, if $t\neq u$, then $N\left(
t,u,v\right)  =1$ (since $N\left(  t,u,v\right)  $ is defined as the number of
subtrees $s\subseteq v$ such that $s\cong t$ and $v\diagup s\cong u$).
\par
By a similar argument, we see that if $t=u$, then $N\left(  t,u,v\right)  =2$
(in fact, if $t=u$, then every subtree $s\subseteq v$ such that $s\cong t$ and
$v\diagup s\cong u$ must be one of the two isomorphic connected components $t$
and $u$ of $v$, and each of these two components can be chosen to be $s$).
\par
So we know that $N\left(  t,u,v\right)  $ equals $1$ if $t\neq u$ and equals
$2$ if $t=u$. Thus,%
\[
N\left(  t,u,v\right)  =%
\begin{cases}
2, & \text{if }t=u;\\
1, & \text{if }t\neq u
\end{cases}
=1+\underbrace{%
\begin{cases}
1, & \text{if }t=u;\\
0, & \text{if }t\neq u
\end{cases}
}_{=\left[  t=u\right]  }=1+\left[  t=u\right]  ,
\]
qed.}. In other words, $N\left(  t,u,tu\right)  =1+\left[  t=u\right]  $.
\end{itemize}

Using these three observations, we see that (A2) becomes%
\begin{align}
&  Z_{t}\star Z_{u}\nonumber\\
&  =\sum\limits_{v=\bullet}\underbrace{N\left(  t,u,v\right)  }%
_{\substack{=0\\\text{(since }v\text{ is}\\\text{the tree }\bullet\text{)}%
}}Z_{v}+\sum\limits_{v\text{ tree }\neq\bullet}N\left(  t,u,v\right)
Z_{v}+\underbrace{N\left(  t,u,tu\right)  }_{=1+\left[  t=u\right]  }%
Z_{tu}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\sum_{\substack{v\text{
forest;}\\v\text{ is neither a tree}\\\text{nor the forest }tu}%
}\ \ \underbrace{N\left(  t,u,v\right)  }_{\substack{=0\\\text{(since }v\text{
is neither a }\\\text{tree nor the forest }tu\text{)}}}Z_{v}\nonumber\\
&  =\underbrace{\sum\limits_{v=\bullet}0Z_{v}}_{=0}+\sum\limits_{v\text{ tree
}\neq\bullet}N\left(  t,u,v\right)  Z_{v}+\left(  1+\left[  t=u\right]
\right)  Z_{tu}+\underbrace{\sum_{\substack{v\text{ forest;}\\v\text{ is
neither a tree}\\\text{nor the forest }tu}}0Z_{v}}_{=0}\nonumber\\
&  =\sum\limits_{v\text{ tree }\neq\bullet}N\left(  t,u,v\right)
Z_{v}+\left(  1+\left[  t=u\right]  \right)  Z_{tu}. \tag{A3}%
\end{align}


On the other hand, the equality (Nab) above says that%
\begin{align}
t\rhd u  &  =\sum\limits_{v\text{ tree}}N\left(  t,u,v\right)  v\nonumber\\
&  =\underbrace{N\left(  t,u,\bullet\right)  }_{\substack{=0\\\text{(since
there is no subforest }s\text{ of }\bullet\\\text{isomorphic to }t\text{
(because }t\neq\bullet\text{))}}}\bullet\,+\sum\limits_{v\text{ tree }%
\neq\bullet}N\left(  t,u,v\right)  v\nonumber\\
&  =\sum\limits_{v\text{ tree }\neq\bullet}N\left(  t,u,v\right)  v.\nonumber
\end{align}
In other words,%
\begin{equation}
\sum\limits_{v\text{ tree }\neq\bullet}N\left(  t,u,v\right)  v=t\rhd u.
\tag{A4}%
\end{equation}


\begin{noncompile}
OLD: On the other hand, the definition of $t\rhd u$ says that%
\[
t\rhd u=\sum\limits_{\substack{v\text{ tree }\neq\bullet;\\t\subseteq v\text{
and }v\diagup t\cong u}}N\left(  t,u,v\right)  v.
\]
But whenever $v$ is a tree $\neq\bullet$ which does \textit{not} satisfy
$\left(  t\subseteq v\text{ and }v\diagup t\cong u\right)  $, we have
$N\left(  t,u,v\right)  =0$\ \ \ \ \footnote{\textit{Proof.} Let $v$ be a tree
$\neq\bullet$ which does \textit{not} satisfy $\left(  t\subseteq v\text{ and
}v\diagup t\cong u\right)  $. Then, there are no subtrees $s\subseteq v$ such
that $s\cong t$ and $v\diagup s\cong u$. Hence, $N\left(  t,u,v\right)  =0$
(because $N\left(  t,u,v\right)  $ is the number of subtrees $s\subseteq v$
such that $s\cong t$ and $v\diagup s\cong u$), qed.}. Hence, $\sum
\limits_{\substack{v\text{ tree }\neq\bullet;\\\text{not }\left(  t\subseteq
v\text{ and }v\diagup t\cong u\right)  }}N\left(  t,u,v\right)  v=\sum
\limits_{\substack{v\text{ tree }\neq\bullet;\\\text{not }\left(  t\subseteq
v\text{ and }v\diagup t\cong u\right)  }}0v=0$.

Now,%
\begin{align}
\sum\limits_{v\text{ tree }\neq\bullet}N\left(  t,u,v\right)  v  &
=\underbrace{\sum\limits_{\substack{v\text{ tree }\neq\bullet;\\t\subseteq
v\text{ and }v\diagup t\cong u}}N\left(  t,u,v\right)  v}_{=t\rhd
u}+\underbrace{\sum\limits_{\substack{v\text{ tree }\neq\bullet;\\\text{not
}\left(  t\subseteq v\text{ and }v\diagup t\cong u\right)  }}N\left(
t,u,v\right)  v}_{=0}\nonumber\\
&  =t\rhd u. \tag{A4}%
\end{align}

\end{noncompile}

Now, (A3) becomes%
\begin{align*}
Z_{t}\star Z_{u}  &  =\underbrace{\sum\limits_{v\text{ tree }\neq\bullet
}N\left(  t,u,v\right)  Z_{v}}_{\substack{=Z_{\sum\limits_{v\text{ tree }%
\neq\bullet}N\left(  t,u,v\right)  v}=Z_{t\rhd u}\\\text{(by (A4))}}}+\left(
1+\left[  t=u\right]  \right)  Z_{tu}\\
&  =Z_{t\rhd u}+\left(  1+\left[  t=u\right]  \right)  Z_{tu}=\left(
1+\left[  t=u\right]  \right)  Z_{tu}+Z_{t\rhd u}.
\end{align*}
This proves Lemma~\ref{lem.A1}.
\end{proof}

Next let us notice an easy consequence of (A1):

\begin{lemma}
\label{lem.A9} Any two trees $t\neq\bullet$ and $u\neq\bullet$ satisfy%
\begin{equation}
Z_{t}\star Z_{u}-Z_{u}\star Z_{t}=Z_{t\rhd u-u\rhd t}. \tag{A9}%
\end{equation}

\end{lemma}

\begin{proof}
Let $t\neq\bullet$ and $u\neq\bullet$ be two trees. Applying (A1) to $u$ and
$t$ instead of $t$ and $u$ gives us $Z_{u}\star Z_{t}=\left(
1+\underbrace{\left[  u=t\right]  }_{=\left[  t=u\right]  }\right)
\underbrace{Z_{ut}}_{=Z_{tu}}+\,Z_{u\rhd t}=\left(  1+\left[  t=u\right]
\right)  Z_{tu}+Z_{u\rhd t}$. But applying (A1) directly gives us $Z_{t}\star
Z_{u}=\left(  1+\left[  t=u\right]  \right)  Z_{tu}+Z_{t\rhd u}$. Subtracting
the first of these two equations from the second, we obtain%
\begin{align*}
Z_{t}\star Z_{u}-Z_{u}\star Z_{t}  &  =\left(  \left(  1+\left[  t=u\right]
\right)  Z_{tu}+Z_{t\rhd u}\right)  -\left(  \left(  1+\left[  t=u\right]
\right)  Z_{tu}+Z_{u\rhd t}\right) \\
&  =Z_{t\rhd u}-Z_{u\rhd t}=Z_{t\rhd u-u\rhd t}.
\end{align*}
This proves Lemma~\ref{lem.A9}.
\end{proof}

Now recall that $\mathcal{H}$ is the symmetric algebra of $T^{\prime}$. Hence,
we can regard $k\bullet$ and $T^{\prime}$ as $k$-vector subspaces of
$\mathcal{H}$, respectively, where $k\bullet$ is contained in the $0$-th
graded component of $\mathcal{H}$ while $T^{\prime}$ is contained in the
direct sum of the other graded components. Hence, $T=T^{\prime}\oplus
k\bullet$ becomes a $k$-vector subspace of $\mathcal{H}$ as well. We notice
that $\Delta\left(  k\bullet\right)  \subseteq\mathcal{H}\otimes T$ (since
$\Delta\left(  \bullet\right)  =\underbrace{\bullet}_{\in\mathcal{H}}%
\otimes\underbrace{\bullet}_{\in T}\in\mathcal{H}\otimes T$) and
$\Delta\left(  T^{\prime}\right)  \subseteq\mathcal{H}\otimes T$ (since every
tree $t\in T^{\prime}$ satisfies%
\begin{align*}
\Delta\left(  t\right)   &  =\sum\limits_{s\subseteq t}\underbrace{s}%
_{\in\mathcal{H}}\otimes\underbrace{\left(  t\diagup s\right)  }%
_{\substack{\in T\\\text{(indeed, }t\diagup s\text{ is a quotient of a
tree,}\\\text{thus a tree, hence an element of }T\text{)}}}\\
&  \in\sum\limits_{s\subseteq t}\mathcal{H}\otimes T\subseteq\mathcal{H}%
\otimes T\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathcal{H}\otimes T\text{
is a vector space}\right)
\end{align*}
). Hence,
\[
\Delta\left(  \underbrace{T}_{=T^{\prime}\oplus k\bullet=T^{\prime}+k\bullet
}\right)  =\Delta\left(  T^{\prime}+k\bullet\right)  =\underbrace{\Delta
\left(  T^{\prime}\right)  }_{\subseteq\mathcal{H}\otimes T}%
+\underbrace{\Delta\left(  k\bullet\right)  }_{\subseteq\mathcal{H}\otimes
T}\subseteq\mathcal{H}\otimes T+\mathcal{H}\otimes T\subseteq\mathcal{H}%
\otimes T
\]
(since $\mathcal{H}\otimes T$ is a vector space). In other words, $T$ is a
left coideal of $\mathcal{H}$.

Now, for every subspace $U$ of $\mathcal{H}$, let $U^{\perp}$ denote the
subspace $\left\{  f\in\mathcal{H}^{\circ}\ \mid\ f\left(  U\right)
=0\right\}  $ of $\mathcal{H}^{\circ}$. By well-known properties of graded
duals of coalgebras, we know that whenever $U$ is a left coideal of
$\mathcal{H}$, the subspace $U^{\perp}$ is a left ideal of $\mathcal{H}%
^{\circ}$. Applied to $U=T$, this yields that $T^{\perp}$ is a left ideal of
$\mathcal{H}^{\circ}$.

Another easy consequence of (A1): Any two trees $t\neq\bullet$ and
$u\neq\bullet$ satisfy%
\begin{equation}
Z_{t}\star Z_{u}\equiv Z_{t\rhd u}\operatorname{mod}T^{\perp}. \tag{A11}%
\end{equation}
\footnote{\textit{Proof of (A11).} Let $t\neq\bullet$ and $u\neq\bullet$ be
two trees. Every tree $v\in T$ satisfies $v\neq tu$ (since $v$ is a tree,
whereas $tu$ is not a tree). Hence, every tree $v\in T$ satisfies
$Z_{tu}\left(  v\right)  =\left[  v=tu\right]  =0$ (since $v=tu$ is false
(because $v\neq tu$)). Thus, $Z_{tu}\left(  T\right)  =0$, so that $Z_{tu}%
\in\left\{  f\in\mathcal{H}^{\circ}\ \mid\ f\left(  T\right)  =0\right\}
=T^{\perp}$ (by the definition of $T^{\perp}$). But (A1) yields%
\begin{equation}
Z_{t}\star Z_{u}-Z_{t\rhd u}=\left(  1+\left[  t=u\right]  \right)
\underbrace{Z_{tu}}_{\in T^{\perp}}\in T^{\perp}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }T^{\perp}\text{ is a vector space}\right)  .\nonumber
\end{equation}
In other words, $Z_{t}\star Z_{u}\equiv Z_{t\rhd u}\operatorname{mod}T^{\perp
}$. This proves (A11).} By linearity, this yields the following: Any two
elements $x\in T^{\prime}$ and $y\in T^{\prime}$ satisfy%
\begin{equation}
Z_{x}\star Z_{y}\equiv Z_{x\rhd y}\operatorname{mod}T^{\perp}. \tag{A12}%
\end{equation}
(In fact, (A12) follows from (A11) because the term $Z_{x}\star Z_{y}-Z_{x\rhd
y}$ is bilinear with respect to $x$ and $y$, because the vector space
$T^{\prime}$ is generated by trees $\neq\bullet$, and because $T^{\perp}$ is a
vector space.)

Finally, let us notice that%
\begin{equation}
\text{if }x\in T^{\prime}\text{ and }y\in T^{\prime}\text{ are two elements
satisfying }Z_{x}\equiv Z_{y}\operatorname{mod}T^{\perp}\text{, then }x=y.
\tag{A15}%
\end{equation}
\footnote{\textit{Proof of (A15).} Let $x\in T^{\prime}$ and $y\in T^{\prime}$
be two elements satisfying $Z_{x}\equiv Z_{y}\operatorname{mod}T^{\perp}$.
Since $x\in T^{\prime}$, we can write $x$ in the form $x=\sum\limits_{t\text{
tree }\neq\bullet}\lambda_{t}t$ for some $\lambda_{t}\in k$ (because
$T^{\prime}$ is the $k$-vector space spanned by rooted trees $\neq\bullet$).
Consider these $\lambda_{t}\in k$. Since $y\in T^{\prime}$, we can write $y$
in the form $y=\sum\limits_{t\text{ tree }\neq\bullet}\mu_{t}t$ for some
$\mu_{t}\in k$ (because $T^{\prime}$ is the $k$-vector space spanned by rooted
trees $\neq\bullet$). Consider these $\mu_{t}\in k$. Every tree $s\neq\bullet$
satisfies $Z_{x}\left(  s\right)  =\lambda_{s}$ (since $x=\sum\limits_{t\text{
tree }\neq\bullet}\lambda_{t}t$ and thus
\begin{align*}
Z_{x}\left(  s\right)   &  =Z_{\sum\limits_{t\text{ tree }\neq\bullet}%
\lambda_{t}t}\left(  s\right)  =\sum\limits_{t\text{ tree }\neq\bullet}%
\lambda_{t}\underbrace{Z_{t}\left(  s\right)  }_{=\left[  s=t\right]  }%
=\sum\limits_{t\text{ tree }\neq\bullet}\lambda_{t}\left[  s=t\right] \\
&  =\sum\limits_{\substack{t\text{ tree }\neq\bullet;\\s=t}}\lambda
_{t}\underbrace{\left[  s=t\right]  }_{=1\text{ (since }s=t\text{)}}%
+\sum\limits_{\substack{t\text{ tree }\neq\bullet;\\s\neq t}}\lambda
_{t}\underbrace{\left[  s=t\right]  }_{=0\text{ (since }s\neq t\text{)}%
}=\underbrace{\sum\limits_{\substack{t\text{ tree }\neq\bullet;\\s=t}%
}\lambda_{t}1}_{=\lambda_{s}1=\lambda_{s}}+\underbrace{\sum
\limits_{\substack{t\text{ tree }\neq\bullet;\\s\neq t}}\lambda_{t}0}%
_{=0}=\lambda_{s}%
\end{align*}
) and $Z_{y}\left(  s\right)  =\mu_{s}$ (similarly). But $Z_{x}\equiv
Z_{y}\operatorname{mod}T^{\perp}$, so that $Z_{x}-Z_{y}\in T^{\perp}=\left\{
f\in\mathcal{H}^{\circ}\ \mid\ f\left(  T\right)  =0\right\}  $ (by the
definition of $T^{\perp}$), so that $\left(  Z_{x}-Z_{y}\right)  \left(
T\right)  =0$. Hence, every tree $s\neq\bullet$ satisfies $\left(  Z_{x}%
-Z_{y}\right)  \left(  s\right)  =0$ (since $s\in T^{\prime}\subseteq T$).
Thus, every tree $s\neq\bullet$ satisfies%
\[
0=\left(  Z_{x}-Z_{y}\right)  \left(  s\right)  =\underbrace{Z_{x}\left(
s\right)  }_{=\lambda_{s}}-\underbrace{Z_{y}\left(  s\right)  }_{=\mu_{s}%
}=\lambda_{s}-\mu_{s}.
\]
In other words, every tree $s\neq\bullet$ satisfies $\lambda_{s}=\mu_{s}$.
Renaming $s$ as $t$, we thus conclude: Every tree $t\neq\bullet$ satisfies
$\lambda_{t}=\mu_{t}$. Hence, $\sum\limits_{t\text{ tree }\neq\bullet}%
\lambda_{t}t=\sum\limits_{t\text{ tree }\neq\bullet}\mu_{t}t$. So we have
$x=\sum\limits_{t\text{ tree }\neq\bullet}\lambda_{t}t=\sum\limits_{t\text{
tree }\neq\bullet}\mu_{t}t=y$. This proves (A15).}

Now, let us finally prove that the product $\rhd$ defined in \S 4.3 satisfies
the left pre-Lie relation (5). In fact, let $a$, $b$ and $c$ be three trees in
$T^{\prime}$. Then, (A9) (applied to $t=a$ and $u=b$) yields%
\[
Z_{a}\star Z_{b}-Z_{b}\star Z_{a}=Z_{a\rhd b-b\rhd a}.
\]
Thus,%
\begin{align}
\left(  Z_{a}\star Z_{b}-Z_{b}\star Z_{a}\right)  \star Z_{c}  &  =Z_{a\rhd
b-b\rhd a}\star Z_{c}\nonumber\\
&  \equiv Z_{\left(  a\rhd b-b\rhd a\right)  \rhd c}\operatorname{mod}%
T^{\perp} \tag{A16}%
\end{align}
(by (A12), applied to $x=a\rhd b-b\rhd a$ and $y=c$). On the other hand, (A12)
(applied to $x=b$ and $y=c$) yields $Z_{b}\star Z_{c}\equiv Z_{b\rhd
c}\operatorname{mod}T^{\perp}$, so that $Z_{b}\star Z_{c}-Z_{b\rhd c}\in
T^{\perp}$. This yields $Z_{a}\star\left(  Z_{b}\star Z_{c}-Z_{b\rhd
c}\right)  \in T^{\perp}$ (since $T^{\perp}$ is a left ideal). This rewrites
as $Z_{a}\star\left(  Z_{b}\star Z_{c}\right)  -Z_{a}\star Z_{b\rhd c}\in
T^{\perp}$. In other words, $Z_{a}\star\left(  Z_{b}\star Z_{c}\right)  \equiv
Z_{a}\star Z_{b\rhd c}\operatorname{mod}T^{\perp}$. Now,%
\begin{align*}
\left(  Z_{a}\star Z_{b}\right)  \star Z_{c}  &  =Z_{a}\star\left(  Z_{b}\star
Z_{c}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\star\text{ is
associative}\right) \\
&  \equiv Z_{a}\star Z_{b\rhd c}\\
&  \equiv Z_{a\rhd\left(  b\rhd c\right)  }\operatorname{mod}T^{\perp
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (A12), applied to }x=a\text{ and
}y=b\rhd c\right)
\end{align*}
and similarly%
\[
\left(  Z_{b}\star Z_{a}\right)  \star Z_{c}\equiv Z_{b\rhd\left(  a\rhd
c\right)  }\operatorname{mod}T^{\perp}.
\]


Now, (A16) yields%
\begin{align*}
Z_{\left(  a\rhd b-b\rhd a\right)  \rhd c}  &  \equiv\left(  Z_{a}\star
Z_{b}-Z_{b}\star Z_{a}\right)  \star Z_{c}=\underbrace{\left(  Z_{a}\star
Z_{b}\right)  \star Z_{c}}_{\equiv Z_{a\rhd\left(  b\rhd c\right)
}\operatorname{mod}T^{\perp}}-\underbrace{\left(  Z_{b}\star Z_{a}\right)
\star Z_{c}}_{\equiv Z_{b\rhd\left(  a\rhd c\right)  }\operatorname{mod}%
T^{\perp}}\\
&  \equiv Z_{a\rhd\left(  b\rhd c\right)  }-Z_{b\rhd\left(  a\rhd c\right)
}=Z_{a\rhd\left(  b\rhd c\right)  -b\rhd\left(  a\rhd c\right)  }%
\operatorname{mod}T^{\perp}.
\end{align*}
By (A15) (applied to $x=\left(  a\rhd b-b\rhd a\right)  \rhd c$ and
$y=a\rhd\left(  b\rhd c\right)  -b\rhd\left(  a\rhd c\right)  $), this yields
that%
\[
\left(  a\rhd b-b\rhd a\right)  \rhd c=a\rhd\left(  b\rhd c\right)
-b\rhd\left(  a\rhd c\right)  .
\]
Since $\left(  a\rhd b-b\rhd a\right)  \rhd c=\left(  a\rhd b\right)  \rhd
c-\left(  b\rhd a\right)  \rhd c$, this rewrites as%
\[
\left(  a\rhd b\right)  \rhd c-\left(  b\rhd a\right)  \rhd c=a\rhd\left(
b\rhd c\right)  -b\rhd\left(  a\rhd c\right)  .
\]
In other words,%
\begin{equation}
\left(  a\rhd b\right)  \rhd c-a\rhd\left(  b\rhd c\right)  =\left(  b\rhd
a\right)  \rhd c-b\rhd\left(  a\rhd c\right)  . \tag{A20}%
\end{equation}


We have thus proven that any three trees $a$, $b$ and $c$ in $T^{\prime}$
satisfy (A20). But since the equation (A20) is multilinear in $a$, $b$ and
$c$, and since $T^{\prime}$ is generated (as a $k$-vector space) by the trees
in $T^{\prime}$, this yields that any three elements $a$, $b$ and $c$ in
$T^{\prime}$ (not necessarily trees) satisfy (A20). In other words, we have
shown that the product $\rhd$ defined in \S 4.3 satisfies the left pre-Lie
relation (5). Qed.


\end{document}