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\ihead{On PBW for pre-Lie algebras}
\ohead{\today}
\begin{document}

\title{On the PBW theorem for pre-Lie algebras}
\author{Darij Grinberg}
\date{\today}
\maketitle
\tableofcontents

\section*{Introduction}

In this note I shall explore the surroundings of Guin's and Oudom's
Poincar\'{e}-Birkhoff-Witt theorem for Lie algebras obtained from pre-Lie
algebras (\cite[Th\'{e}or\`{e}me 3.5]{GuiOud04}, \cite[Theorem 3.14]%
{GuiOud08}, \cite[Theorem 1.1]{Manchon11}, \cite[Corollary 1.3.1]{Schedler})
and, as a consequence, answer a MathOverflow question that I have asked in
2012 \cite{MO-norord} about the normal ordered product on differential
operators. The results proven here are elementary (and so are their proofs),
and few of them are new; yet (in my opinion) there are arguments and remarks
worthy of circulation among them (in particular, the question \cite{MO-norord}
feels natural to me, yet I have to find it in the existing literature).

The paper is long, due to the bulk of straightforward computations and
detailed proofs it contains. Since most of the proofs are relatively
straightforward (Theorem \ref{thm.nord.box} \textbf{(h)} is probably the only
tricky one), the reader is encouraged to skip them and regard the theorems as
exercises. I have also tried to be maximally explicit and technically correct
(e.g., I avoid identifying a Lie algebra $\mathfrak{g}$ with its image in its
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $ because the
canonical map $\mathfrak{g}\rightarrow U\left(  \mathfrak{g}\right)  $ is not
always injective over a ring); this, too, is responsible for some of the
length of this note.

In Section \ref{sect.action-thm}, we state (after fixing notations and
reminding the reader of basic terminology) an elementary and simple theorem
(Theorem \ref{thm.gen1}) about Lie algebra actions. This theorem states that
if $\mathfrak{g}$ is a Lie algebra over a commutative ring $\mathbf{k}$, if
$C$ is a $\mathbf{k}$-algebra (which, for us, means an associative
$\mathbf{k}$-algebra), if $K:\mathfrak{g}\rightarrow\operatorname*{Der}C$ is a
Lie algebra homomorphism from $\mathfrak{g}$ to the Lie algebra of derivations
of $C$, and if $f:\mathfrak{g}\rightarrow C$ is a $\mathbf{k}$-linear map
satisfying
\[
f\left(  \left[  a,b\right]  \right)  =\left[  f\left(  a\right)  ,f\left(
b\right)  \right]  +\left(  K\left(  a\right)  \right)  \left(  f\left(
b\right)  \right)  -\left(  K\left(  b\right)  \right)  \left(  f\left(
a\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in\mathfrak{g},
\]
then $C$ becomes a $\mathfrak{g}$-module via%
\[
a\rightharpoonup u=f\left(  a\right)  \cdot u+\left(  K\left(  a\right)
\right)  \left(  u\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a\in
\mathfrak{g}\text{ and }u\in C
\]
(where $a\rightharpoonup u$ is our notation for the action of $a$ on $u$). We
then show some additional properties of this action, the most significant of
which is Theorem \ref{thm.gen3}. This setting is rather general; we will only
end up using a particular case of it in the later sections.

In Section \ref{sect.prelude}, we present the setting of the
Poincar\'{e}-Birkhoff-Witt theorem(s): a Lie algebra $\mathfrak{g}$, its
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $, its symmetric
algebra $\operatorname*{Sym}\mathfrak{g}$, and various homomorphisms between
these modules (and their associated graded modules). Everything in this
section is well-known, but I found it worthwhile to explicitly state all
definitions and basic results, and even prove some of them (Proposition
\ref{prop.PBW-map} and Lemma \ref{lem.PBW-by-inverse-map} are probably the
most important ones), since the available literature leaves too much to the
reader and occasionally lacks precision. As a result, this section has become
rather long; a reader familiar with the Poincar\'{e}-Birkhoff-Witt theorems
will probably not be hurt by skipping it entirely.

Section \ref{sect.pre-lie} is the heart of this note. Here I first define the
notions of left and right pre-Lie algebras, and state their basic properties:
viz., that any (associative) algebra is both a left and a right pre-Lie
algebra, and that from any left or right pre-Lie algebra $A$ one can construct
a Lie algebra $A^{-}$. Then, we state the main properties of the Guin-Oudom
isomorphism (Theorem \ref{thm.manchon}). This is a $\mathbf{k}$-coalgebra
isomorphism $U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$ defined
for every left pre-Lie algebra $A$. Unlike Guin and Oudom (\cite{GuiOud04} and
\cite{GuiOud08}), we construct this isomorphism using Theorem \ref{thm.gen1}
and Lemma \ref{lem.PBW-by-inverse-map} rather than by extending the binary
operation of the pre-Lie algebra $A$ to its symmetric algebra
$\operatorname*{Sym}A$. As a consequence of this isomorphism, the
Poincar\'{e}-Birkhoff-Witt theorem holds for every Lie algebra of the form
$A^{-}$, where $A$ is a pre-Lie algebra (Theorem \ref{thm.manchon}
\textbf{(k)}); this stands in contrast to the usual versions of the
Poincar\'{e}-Birkhoff-Witt theorem, which make assumptions on the $\mathbf{k}%
$-module structure of the Lie algebra. We then use this isomorphism to define
a commutative multiplication $\boxdot$ on $U\left(  A^{-}\right)  $ (Corollary
\ref{cor.manchon.box}); it is defined by transporting the multiplication from
$\operatorname*{Sym}A$ to $U\left(  A^{-}\right)  $ via the Guin-Oudom
isomorphism $U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$.

In Section \ref{sect.nord}, we apply the above to the Lie algebra
$\mathfrak{gl}_{n}$ (which has the form $A^{-}$ for the pre-Lie algebra
$A=\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $), and relate it
to differential operators with polynomial coefficients. Part of the result
that we obtain states the following: Assume that $\mathbf{k}$ is a commutative
$\mathbb{Q}$-algebra. Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $G$
denote the $nm$-element set $\left\{  1,2,\ldots,n\right\}  \times\left\{
1,2,\ldots,m\right\}  $. Let $\mathcal{A}$ be the polynomial ring
$\mathbf{k}\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in G\right]  $ in the
$nm$ (commuting) indeterminates $x_{i,j}$. Let $\mathcal{D}$ denote the
$\mathbf{k}$-subalgebra of $\operatorname*{End}\mathcal{A}$ generated by the
multiplication operators $x_{i,j}$ (for $\left(  i,j\right)  \in
G$)\ \ \ \ \footnote{When writing $x_{i,j}$ here, we mean the linear map
$\mathcal{A}\rightarrow\mathcal{A},\ f\mapsto x_{i,j}\cdot f$.} and the
differential operators $\dfrac{\partial}{\partial x_{i,j}}$ (for $\left(
i,j\right)  \in G$). (Thus, $\mathcal{D}$ is the $\mathbf{k}$-algebra of
differential operators in the $x_{i,j}$ with polynomial coefficients.) Let
$\mathcal{A}^{\prime}$ be the polynomial ring $\mathbf{k}\left[
\partial_{i,j}\ \mid\ \left(  i,j\right)  \in G\right]  $ in the $nm$
(commuting) indeterminates $\partial_{i,j}$. The $\mathbf{k}$-linear map
\[
\xi:\mathcal{A}\otimes\mathcal{A}^{\prime}\rightarrow\mathcal{D}%
,\ \ \ \ \ \ \ \ \ \ P\otimes Q\mapsto P\cdot Q\left(  \left(  \dfrac
{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)
\]
is a $\mathbf{k}$-module isomorphism, but not (in general) a $\mathbf{k}%
$-algebra homomorphism. However, we can use this $\mathbf{k}$-module
isomorphism to transport the (commutative) multiplication on $\mathcal{A}%
\otimes\mathcal{A}^{\prime}$ to $\mathcal{D}$. We denote the resulting
multiplication by $\boxdot$ (explicitly, it is given by $A\boxdot B=\xi\left(
\xi^{-1}\left(  A\right)  \cdot\xi^{-1}\left(  B\right)  \right)  $). Next, we
define a $\mathbf{k}$-linear map $\omega:\mathfrak{gl}_{n}\rightarrow
\mathcal{D}^{-}$ by%
\[
\omega\left(  E_{i,j}\right)  =\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial
x_{j,k}}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}.
\]
This $\omega$ is actually a Lie algebra homomorphism (Proposition
\ref{prop.nord.l}). By the universal property of the universal enveloping
algebra $U\left(  \mathfrak{gl}_{n}\right)  $, it thus gives rise to a
$\mathbf{k}$-algebra homomorphism $\Omega:U\left(  \mathfrak{gl}_{n}\right)
\rightarrow\mathcal{D}$. It turns out that the image $\Omega\left(  U\left(
\mathfrak{gl}_{n}\right)  \right)  $ of this homomorphism is a $\mathbf{k}%
$-subalgebra of the $\mathbf{k}$-algebra $\left(  \mathcal{D},\boxdot\right)
$ (Theorem \ref{thm.nord.box-exists} \textbf{(a)}). Moreover, there exists a
commutative multiplication $\boxdot$ on the $\mathbf{k}$-module $U\left(
\mathfrak{gl}_{n}\right)  $, independent on $m$, such that $\Omega$ is a
$\mathbf{k}$-algebra homomorphism $\left(  U\left(  \mathfrak{gl}_{n}\right)
,\boxdot\right)  \rightarrow\left(  \mathcal{D},\boxdot\right)  $ (Theorem
\ref{thm.nord.box-exists} \textbf{(b)}). This multiplication $\boxdot$ is
obtained by an application of Corollary \ref{cor.manchon.box} (which explains
why we are using the notation $\boxdot$ for two seemingly unrelated
multiplications). This answers my MathOverflow question \cite{MO-norord} in
the affirmative.

In the final Section \ref{sect.bimod}, we shall address an obvious question of
symmetry. Namely, the Guin-Oudom isomorphism can be defined for a left pre-Lie
algebra, or (similarly) for a right pre-Lie algebra. An (associative)
$\mathbf{k}$-algebra can be viewed as a left and a right pre-Lie algebra at
the same time; thus it has two Guin-Oudom isomorphisms. How are these two
isomorphisms related? We shall show (Theorem \ref{thm.manchon.bi}) that they
are identical.

\subsection{Acknowledgments}

This note owes a significant part of its inspiration to Alexander Chervov's
comment at \cite{MO-norord} and Fr\'{e}d\'{e}ric Chapoton's answer at
\cite{MO-heisen}, the former of which suggested a crucial generalization that
made my question \cite{MO-norord} a lot more tractable, while the latter
referred me (on a related question) to the concept of pre-Lie algebras and
Manchon's beautiful exposition \cite{Manchon11} thereof.

\section{\label{sect.action-thm}A simple theorem on Lie algebra actions}

We begin with a rather general setting. While I am not aware of any
applications of this setting other than the properties of pre-Lie algebras to
which it is applied below, I prefer to start with the general and then move on
to the particular case, not least because the general case is more
\textquotedblleft classical\textquotedblright\ (for example, it does not
involve pre-Lie algebras) and has less complexity.

\subsection{Notations}

Let us first fix some notations:

\begin{condition}
In the following, all rings are associative and with unity.

Fix a commutative ring $\mathbf{k}$ (once and for all). In the following, all
$\mathbf{k}$-algebras are associative, unital and central. (\textquotedblleft
Central\textquotedblright\ means that $\lambda a=\left(  \lambda\cdot
1_{A}\right)  a=a\left(  \lambda\cdot1_{A}\right)  $ for every $\lambda
\in\mathbf{k}$ and every $a$ in the algebra.) All Lie algebras are defined
over $\mathbf{k}$. All tensor product signs, all \textquotedblleft%
$\operatorname*{Hom}$\textquotedblright\ signs, and all \textquotedblleft%
$\operatorname*{End}$\textquotedblright\ signs are understood to be defined
over $\mathbf{k}$ unless stated otherwise. All \textquotedblleft%
$\operatorname*{Hom}$\textquotedblright\ signs and all \textquotedblleft%
$\operatorname*{End}$\textquotedblright\ signs refer to homomorphisms
(respectively, endomorphisms) of $\mathbf{k}$-modules.

If $U$ and $V$ are two $\mathbf{k}$-submodules of a $\mathbf{k}$-algebra $A$,
then $UV$ denotes the $\mathbf{k}$-submodule of $A$ spanned by $\left\{
uv\ \mid\ \left(  u,v\right)  \in U\times V\right\}  $.

If $u$ and $v$ are two elements of a Lie algebra $\mathfrak{g}$, then $\left[
u,v\right]  $ denotes the Lie bracket of $\mathfrak{g}$ evaluated at $\left(
u,v\right)  $.

If $A$ is a $\mathbf{k}$-algebra, then \textquotedblleft$A$%
-module\textquotedblright\ means \textquotedblleft left $A$%
-module\textquotedblright\ unless explicitly stated otherwise.

The notation $\mathbb{N}$ stands for $\left\{  0,1,2,\ldots\right\}  $ (not
$\left\{  1,2,3,\ldots\right\}  $).

If $X$, $Y$ and $Z$ are three sets, and $f:X\rightarrow Y$ and $g:Y\rightarrow
Z$ are two maps, then $g\circ f$ denotes the map $X\rightarrow Z$ which sends
every $x\in X$ to $g\left(  f\left(  x\right)  \right)  $.
\end{condition}

Next, we recall the definition of a $\mathfrak{g}$-module (where
$\mathfrak{g}$ is a Lie algebra):

\begin{definition}
\label{def.liemod.left}Let $\mathfrak{g}$ be a Lie algebra. Let $V$ be a
$\mathbf{k}$-module. Let $\mu:\mathfrak{g}\times V\rightarrow V$ be a
$\mathbf{k}$-bilinear map. We say that $\left(  V,\mu\right)  $ is a
$\mathfrak{g}$\textit{-module} if and only if
\begin{equation}
\left(  \mu\left(  \left[  a,b\right]  ,v\right)  =\mu\left(  a,\mu\left(
b,v\right)  \right)  -\mu\left(  b,\mu\left(  a,v\right)  \right)  \text{ for
every }a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in V\right)  .
\label{eq.def.liemod.left.axiom}%
\end{equation}
If $\left(  V,\mu\right)  $ is a $\mathfrak{g}$-module, then the $\mathbf{k}%
$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow V$ is called the
\textit{Lie action} of the $\mathfrak{g}$-module $V$. If $\left(
V,\mu\right)  $ is a $\mathfrak{g}$-module, then the $\mathbf{k}$-module $V$
is called the \textit{underlying }$\mathbf{k}$\textit{-module} of $\left(
V,\mu\right)  $.
\end{definition}

The following conventions will simplify our life somewhat:

\begin{condition}
\label{conv.liemod.left}Let $\mathfrak{g}$ be a Lie algebra. Let $\left(
V,\mu\right)  $ be a $\mathfrak{g}$-module.

\textbf{(a)} For any $a\in\mathfrak{g}$ and $v\in V$, we shall abbreviate the
term $\mu\left(  a,v\right)  $ by $a\rightharpoonup v$, provided that the map
$\mu$ is obvious from the context. Using this notation, the relation
(\ref{eq.def.liemod.left.axiom}) rewrites as%
\begin{equation}
\left(  \left[  a,b\right]  \rightharpoonup v=a\rightharpoonup\left(
b\rightharpoonup v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)
\text{ for every }a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in
V\right)  . \label{eq.conv.liemod}%
\end{equation}


\textbf{Convention on the precedence of the }$\rightharpoonup$\textbf{ sign:
}The symbol $\rightharpoonup$ is understood to have the same precedence as the
multiplication sign (i.e., it binds as strongly as the multiplication sign).
Thus, $a\rightharpoonup v+w$ means $\left(  a\rightharpoonup v\right)  +w$
rather than $a\rightharpoonup\left(  v+w\right)  $, but $a\rightharpoonup
v\cdot w$ is undefined (it could mean both $\left(  a\rightharpoonup v\right)
\cdot w$ and $a\rightharpoonup\left(  v\cdot w\right)  $). Application of
functions will be supposed to bind more strongly than the $\rightharpoonup$
sign, so that $f\left(  v\right)  \rightharpoonup g\left(  w\right)  $ will
mean $\left(  f\left(  v\right)  \right)  \rightharpoonup\left(  g\left(
w\right)  \right)  $ (rather than $f\left(  v\rightharpoonup g\left(
w\right)  \right)  $ or $\left(  f\left(  v\rightharpoonup g\right)  \right)
\left(  w\right)  $ or anything else), but we will often use parentheses in
this case to make the correct interpretation of the formula even more obvious.

\textbf{(b)} We shall often refer to the $\mathfrak{g}$-module $\left(
V,\mu\right)  $ as \textquotedblleft the $\mathbf{k}$-module $V$, endowed with
the $\mathfrak{g}$-action $\mu$\textquotedblright\ (or by similar
formulations). We shall regard $\left(  V,\mu\right)  $ as the $\mathbf{k}%
$-module $V$ equipped with the additional data of the map $\mu$. Thus, when we
speak of \textquotedblleft elements of $\left(  V,\mu\right)  $%
\textquotedblright\ or \textquotedblleft maps to $\left(  V,\mu\right)
$\textquotedblright\ or \textquotedblleft$\mathbf{k}$-submodules of $\left(
V,\mu\right)  $\textquotedblright, we shall mean (respectively)
\textquotedblleft elements of $V$\textquotedblright, \textquotedblleft maps to
$V$\textquotedblright, or \textquotedblleft$\mathbf{k}$-submodules of
$V$\textquotedblright.

\textbf{(c)} By abuse of notation, we shall write \textquotedblleft$V$ is a
$\mathfrak{g}$-module\textquotedblright\ instead of \textquotedblleft$\left(
V,\mu\right)  $ is a $\mathfrak{g}$-module\textquotedblright\ when the map
$\mu$ is clear from the context or has not been introduced yet. (For instance,
when we say \textquotedblleft Let $V$ be a $\mathfrak{g}$%
-module\textquotedblright, we really mean to say \textquotedblleft Let
$\left(  V,\mu\right)  $ be a $\mathfrak{g}$-module\textquotedblright, where
the \textquotedblleft$\mu$\textquotedblright\ is some unused symbol. We will
mostly be able to avoid referring to this $\mu$, because our notation
$a\rightharpoonup v$ for $\mu\left(  a,v\right)  $ makes it possible to talk
about values of the Lie action $\mu$ without ever mentioning $\mu$.)
\end{condition}

\begin{definition}
Let $\mathfrak{g}$ be a Lie algebra. Let $V$ be a $\mathbf{k}$-module. A
$\mathfrak{g}$\textit{-module structure on }$V$ means a map $\mu
:\mathfrak{g}\times V\rightarrow V$ such that $\left(  V,\mu\right)  $ is a
$\mathfrak{g}$-module. In other words, a $\mathfrak{g}$-module structure on
$V$ means a $\mathbf{k}$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow V$
such that we have%
\[
\left(  \left[  a,b\right]  \rightharpoonup v=a\rightharpoonup\left(
b\rightharpoonup v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)
\text{ for every }a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in
V\right)  ,
\]
where we denote $\mu\left(  a,m\right)  $ by $a\rightharpoonup m$ for every
$a\in\mathfrak{g}$ and $m\in V$. Thus, a $\mathfrak{g}$-module is the same as
a $\mathbf{k}$-module endowed with a $\mathfrak{g}$-module structure.
\end{definition}

\begin{definition}
Let $\mathfrak{g}$ be a Lie algebra. Let $V$ be a $\mathfrak{g}$-module.

\textbf{(a)} A $\mathfrak{g}$\textit{-submodule} of $V$ means a $\mathfrak{g}%
$-module $V^{\prime}$ such that the $\mathbf{k}$-module $V^{\prime}$ is a
submodule of the $\mathbf{k}$-module $V$, and such that the Lie action of
$V^{\prime}$ is the restriction of the Lie action of $V$ to $\mathfrak{g}%
\times V^{\prime}$.

\textbf{(b)} Let $W$ be a further $\mathfrak{g}$-module. A $\mathfrak{g}%
$\textit{-module homomorphism from }$V$ \textit{to }$W$ means a $\mathbf{k}%
$-linear map $f:V\rightarrow W$ satisfying%
\[
\left(  a\rightharpoonup f\left(  v\right)  =f\left(  a\rightharpoonup
v\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and
}v\in V\right)  .
\]
A $\mathfrak{g}$\textit{-module isomorphism from }$V$ \textit{to }$W$ means an
invertible $\mathfrak{g}$-module homomorphism from $V$ to $W$ whose inverse is
also a $\mathfrak{g}$-module homomorphism. It is easy to show that any
invertible $\mathfrak{g}$-module homomorphism is a $\mathfrak{g}$-module isomorphism.

Thus, we can define a category of $\mathfrak{g}$-modules: Its objects are
$\mathfrak{g}$-modules, and its morphisms are $\mathfrak{g}$-module homomorphisms.
\end{definition}

\begin{definition}
\label{def.A-}If $A$ is a $\mathbf{k}$-algebra, then the $\mathbf{k}$-module
$A$ can be endowed with a Lie bracket defined by%
\[
\left(  \left[  a,b\right]  =ab-ba\text{ for every }a\in A\text{ and }b\in
A\right)  .
\]
The $\mathbf{k}$-module $A$ thus becomes a Lie algebra. This Lie algebra will
be denoted by $A^{-}$.
\end{definition}

\begin{definition}
Let $\mathfrak{g}$ be a Lie algebra. Then, $U\left(  \mathfrak{g}\right)  $
will denote the universal enveloping algebra of $\mathfrak{g}$. (See
Definition \ref{def.U(g)} for the definition of $U\left(  \mathfrak{g}\right)
$.) We denote by $\iota_{U,\mathfrak{g}}$ the canonical map $\mathfrak{g}%
\rightarrow U\left(  \mathfrak{g}\right)  $. (This map $\iota_{U,\mathfrak{g}%
}$ is also defined in Definition \ref{def.U(g)} further below.) This map
$\iota_{U,\mathfrak{g}}$ is a Lie algebra homomorphism from $\mathfrak{g}$ to
$\left(  U\left(  \mathfrak{g}\right)  \right)  ^{-}$. By abuse of notation,
some authors write $a$ for the image of an element $a\in\mathfrak{g}$ under
the map $\iota_{U,\mathfrak{g}}$; we will not do so, since the map
$\iota_{U,\mathfrak{g}}$ is not always injective (although the
Poincar\'{e}-Birkhoff-Witt theorem shows that it is if $\mathfrak{g}$ is a
free $\mathbf{k}$-module or if $\mathbf{k}$ is a $\mathbb{Q}$-algebra).
\end{definition}

The classical universal property of the universal enveloping algebra states
the following:

\begin{theorem}
\label{thm.U.univ}Let $\mathfrak{g}$ be a Lie algebra. Let $A$ be a
$\mathbf{k}$-algebra. Let $f:\mathfrak{g}\rightarrow A^{-}$ be a Lie algebra
homomorphism. Then, there exists a unique $\mathbf{k}$-algebra homomorphism
$F:U\left(  \mathfrak{g}\right)  \rightarrow A$ such that $f=F\circ
\iota_{U,\mathfrak{g}}$.
\end{theorem}

Using this universal property, we can construct a 1-to-1 correspondence
between $\mathfrak{g}$-modules and $U\left(  \mathfrak{g}\right)  $-modules:

\begin{definition}
\label{def.U.mod}Let $\mathfrak{g}$ be a Lie algebra.

\textbf{(a)} Every $U\left(  \mathfrak{g}\right)  $-module $M$ canonically
becomes a $\mathfrak{g}$-module by setting%
\[
\left(  a\rightharpoonup m=\iota_{U,\mathfrak{g}}\left(  a\right)
m\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathfrak{g}\text{ and }m\in M\right)
.
\]
Moreover, any $U\left(  \mathfrak{g}\right)  $-module homomorphism between two
$U\left(  \mathfrak{g}\right)  $-modules becomes a $\mathfrak{g}$-module
homomorphism if we regard these $U\left(  \mathfrak{g}\right)  $-modules as
$\mathfrak{g}$-modules. Thus, we obtain a functor from the category of
$U\left(  \mathfrak{g}\right)  $-modules to the category of $\mathfrak{g}$-modules.

\textbf{(b)} Every $\mathfrak{g}$-module $M$ canonically becomes a $U\left(
\mathfrak{g}\right)  $-module. To define the $U\left(  \mathfrak{g}\right)
$-module structure on $M$, we proceed as follows: Define a map $\varphi
:\mathfrak{g}\rightarrow\operatorname*{End}M$ by%
\[
\left(  \left(  \varphi\left(  a\right)  \right)  \left(  m\right)
=a\rightharpoonup m\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathfrak{g}\text{
and }m\in M\right)  .
\]
It is easy to see that this map $\varphi$ is a Lie algebra homomorphism from
$\mathfrak{g}$ to $\left(  \operatorname*{End}M\right)  ^{-}$. (Indeed, this
is a restatement of the axioms of a $\mathfrak{g}$-module; the fact that
$\varphi\left(  \left[  a,b\right]  \right)  =\left[  \varphi\left(  a\right)
,\varphi\left(  b\right)  \right]  $ for all $a,b\in\mathfrak{g}$ is
equivalent to the relation (\ref{eq.conv.liemod}).) Now, Theorem
\ref{thm.U.univ} (applied to $A=\operatorname*{End}M$ and $f=\varphi$) shows
that there exists a unique $\mathbf{k}$-algebra homomorphism $F:U\left(
\mathfrak{g}\right)  \rightarrow\operatorname*{End}M$ such that $\varphi
=F\circ\iota_{U,\mathfrak{g}}$. Consider this $F$. Now, we define a $U\left(
\mathfrak{g}\right)  $-module structure on $M$ by%
\[
\left(  pm=\left(  F\left(  p\right)  \right)  \left(  m\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in U\left(  \mathfrak{g}\right)  \text{
and }m\in M\right)  .
\]
Thus, every $\mathfrak{g}$-module canonically becomes a $U\left(
\mathfrak{g}\right)  $-module. Moreover, any $\mathfrak{g}$-module
homomorphism between two $\mathfrak{g}$-modules becomes a $U\left(
\mathfrak{g}\right)  $-module homomorphism if we regard these $\mathfrak{g}%
$-modules as $U\left(  \mathfrak{g}\right)  $-modules. Hence, we obtain a
functor from the category of $\mathfrak{g}$-modules to the category of
$U\left(  \mathfrak{g}\right)  $-modules.

\textbf{(c)} In Definition \ref{def.U.mod} \textbf{(a)}, we have constructed a
functor from the category of $U\left(  \mathfrak{g}\right)  $-modules to the
category of $\mathfrak{g}$-modules. In Definition \ref{def.U.mod}
\textbf{(b)}, we have constructed a functor from the category of
$\mathfrak{g}$-modules to the category of $U\left(  \mathfrak{g}\right)
$-modules. These two functors are mutually inverse. In particular, if $M$ is a
$\mathfrak{g}$-module, then the $U\left(  \mathfrak{g}\right)  $-module
structure on $M$ obtained according to Definition \ref{def.U.mod} \textbf{(b)}
satisfies
\[
\iota_{U,\mathfrak{g}}\left(  a\right)  m=a\rightharpoonup
m\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }m\in M.
\]


\textbf{(d)} According to Definition \ref{def.U.mod} \textbf{(a)}, every
$U\left(  \mathfrak{g}\right)  $-module canonically becomes a $\mathfrak{g}%
$-module. In particular, $U\left(  \mathfrak{g}\right)  $ itself becomes a
$\mathfrak{g}$-module (because $U\left(  \mathfrak{g}\right)  $ is a left
$U\left(  \mathfrak{g}\right)  $-module). This is the $\mathfrak{g}$-module
structure on $U\left(  \mathfrak{g}\right)  $ \textquotedblleft given by left
multiplication\textquotedblright\ (because it satisfies $x\rightharpoonup
u=\iota_{U,\mathfrak{g}}\left(  x\right)  u$ for every $x\in\mathfrak{g}$ and
$u\in U\left(  \mathfrak{g}\right)  $). Other canonical $\mathfrak{g}$-module
structures on $U\left(  \mathfrak{g}\right)  $ exist as well, but we shall not
use them for the time being.
\end{definition}

\begin{noncompile}
Let me first recall some basic definitions, in order to fix notations:

\begin{condition}
In the following, all rings are associative and with unity.

Fix a commutative ring $\mathbf{k}$ (once and for all). In the following, all
$\mathbf{k}$-algebras are associative, unital and central.
\end{condition}

\begin{definition}
\label{def.liealg}A \textit{Lie algebra} means a $\mathbf{k}$-module
$\mathfrak{g}$ endowed with a $\mathbf{k}$-bilinear map $\beta:\mathfrak{g}%
\times\mathfrak{g}\rightarrow\mathfrak{g}$ which satisfies the conditions%
\begin{align}
&  \left(  \beta\left(  v,v\right)  =0\text{ for every }v\in\mathfrak{g}%
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\label{eq.def.liealg.beta-1}\\
&  \left(  \beta\left(  u,\beta\left(  v,w\right)  \right)  +\beta\left(
v,\beta\left(  w,u\right)  \right)  +\beta\left(  w,\beta\left(  u,v\right)
\right)  =0\text{ for every }u\in\mathfrak{g}\text{, }v\in\mathfrak{g}\text{
and }w\in\mathfrak{g}\right)  . \label{eq.def.liealg.beta-2}%
\end{align}


By abuse of notation, we will often speak of the \textquotedblleft Lie algebra
$\mathfrak{g}$\textquotedblright\ without mentioning the map $\beta$ when the
map $\beta$ can be inferred from the context.

This $\mathbf{k}$-bilinear map $\beta$ is called the \textit{Lie bracket} of
the Lie algebra $\mathfrak{g}$. We shall often use the \textquotedblleft
square-brackets notation\textquotedblright\ for $\beta$; this means that we
shall write $\beta\left(  u,v\right)  $ as $\left[  u,v\right]  $ for any
$v\in\mathfrak{g}$ and $w\in\mathfrak{g}$. Using this notation, the axioms
(\ref{eq.def.liealg.beta-1}) and (\ref{eq.def.liealg.beta-2}) rewrite as%
\begin{align}
&  \left(  \left[  v,v\right]  =0\text{ for every }v\in\mathfrak{g}\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\label{eq.def.liealg.[]-1}\\
&  \left(  \left[  u,\left[  v,w\right]  \right]  +\left[  v,\left[
w,u\right]  \right]  +\left[  w,\left[  u,v\right]  \right]  =0\text{ for
every }u\in\mathfrak{g}\text{, }v\in\mathfrak{g}\text{ and }w\in
\mathfrak{g}\right)  . \label{eq.def.liealg.[]-2}%
\end{align}
\newline The equation (\ref{eq.def.liealg.beta-2}) (or its equivalent version
(\ref{eq.def.liealg.[]-2})) is called the \textit{Jacobi identity}.
\end{definition}

\begin{condition}
\label{conv.[]}We are going to use the notation $\left[  v,w\right]  $ as a
universal notation for the Lie bracket of two elements $v$ and $w$ in a Lie
algebra. This means that whenever we have some Lie algebra $\mathfrak{g}$, and
we are given two elements $v$ and $w$ of $\mathfrak{g}$, we will denote by
$\left[  v,w\right]  $ the Lie bracket of $\mathfrak{g}$ applied to $\left(
v,w\right)  $ (unless we explicitly stated that the notation $\left[
v,w\right]  $ means something different). (Of course, the Lie algebra and its
elements need not actually be called $\mathfrak{g}$, $v$ and $w$ for this
convention to apply. The labels $\mathfrak{g}$, $v$ and $w$ are local to this convention.)
\end{condition}

\begin{definition}
\label{def.liemod}Let $\mathfrak{g}$ be a Lie algebra. Let $V$ be a
$\mathbf{k}$-module. Let $\mu:\mathfrak{g}\times V\rightarrow V$ be a
$\mathbf{k}$-bilinear map. We say that $\left(  V,\mu\right)  $ is a
$\mathfrak{g}$\textit{-module} if and only if
\begin{equation}
\left(  \mu\left(  \left[  a,b\right]  ,v\right)  =\mu\left(  a,\mu\left(
b,v\right)  \right)  -\mu\left(  b,\mu\left(  a,v\right)  \right)  \text{ for
every }a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in V\right)  .
\label{eq.def.liemod.L-mod}%
\end{equation}
If $\left(  V,\mu\right)  $ is a $\mathfrak{g}$-module, then the $\mathbf{k}%
$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow V$ is called the
\textit{Lie action} of the $\mathfrak{g}$-module $V$.

Often, when the map $\mu$ is obvious from the context, we abbreviate the term
$\mu\left(  a,v\right)  $ by $a\rightharpoonup v$ for any $a\in\mathfrak{g}$
and $v\in V$. Using this notation, the relation (\ref{eq.def.liemod.L-mod})
rewrites as%
\begin{equation}
\left(  \left[  a,b\right]  \rightharpoonup v=a\rightharpoonup\left(
b\rightharpoonup v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)
\text{ for every }a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in
V\right)  . \label{eq.def.liemod.[L-mod]}%
\end{equation}
\newline Also, an abuse of notation allows us to write \textquotedblleft$V$ is
a $\mathfrak{g}$-module\textquotedblright\ instead of \textquotedblleft%
$\left(  V,\mu\right)  $ is a $\mathfrak{g}$-module\textquotedblright\ if the
map $\mu$ is clear from the context or has not been introduced yet.\newline
Besides, when $\left(  V,\mu\right)  $ is a $\mathfrak{g}$-module, we will say
that $\mu$ is a $\mathfrak{g}$\textit{-module structure on }$V$. In other
words, if $V$ is a $\mathbf{k}$-module, then a $\mathfrak{g}$\textit{-module
structure on }$V$ means a map $\mu:\mathfrak{g}\times V\rightarrow V$ such
that $\left(  V,\mu\right)  $ is a $\mathfrak{g}$-module.
\end{definition}

\begin{condition}
\label{conv.->}We are going to use the notation $a\rightharpoonup v$ as a
universal notation for the Lie action of a $\mathfrak{g}$-module. This means
that whenever we have some Lie algebra $\mathfrak{g}$ and some $\mathfrak{g}%
$-module $V$, and we are given two elements $a\in\mathfrak{g}$ and $v\in V$,
we will denote by $a\rightharpoonup v$ the Lie action of $V$ applied to
$\left(  a,v\right)  $ (unless we explicitly stated that the notation
$a\rightharpoonup v$ means something different). (Of course, the notations
$\mathfrak{g}$, $V$, $a$ and $v$ are local to this convention.)\newline%
\textbf{Convention on the precedence of the }$\rightharpoonup$\textbf{ sign:
}When we use the notation $a\rightharpoonup v$, the $\rightharpoonup$ sign is
supposed to have the same precedence as the multiplication sign (i. e. bind as
strongly as the multiplication sign). Thus, $a\rightharpoonup v+w$ means
$\left(  a\rightharpoonup v\right)  +w$ rather than $a\rightharpoonup\left(
v+w\right)  $, but $a\rightharpoonup v\cdot w$ is undefined (it may mean both
$\left(  a\rightharpoonup v\right)  \cdot w$ and $a\rightharpoonup\left(
v\cdot w\right)  $). Application of functions will be supposed to bind more
strongly than the $\rightharpoonup$ sign, so that $f\left(  v\right)
\rightharpoonup g\left(  w\right)  $ will mean $\left(  f\left(  v\right)
\right)  \rightharpoonup\left(  g\left(  w\right)  \right)  $ (rather than
$f\left(  v\rightharpoonup g\left(  w\right)  \right)  $ or $\left(  f\left(
v\rightharpoonup g\right)  \right)  \left(  w\right)  $ or anything else), but
we will often use parentheses in this case to make the correct interpretation
of the formula even more obvious.
\end{condition}

\begin{definition}
\textbf{(a)} If $A$ is a $\mathbf{k}$-algebra, then the $\mathbf{k}$-module
$A$ can be endowed with a Lie bracket defined by%
\[
\left(  \left[  a,b\right]  =ab-ba\text{ for every }a\in A\text{ and }b\in
A\right)  .
\]
The $\mathbf{k}$-algebra $A$ thus becomes a Lie algebra. This Lie algebra will
be denoted by $A^{-}$.

\textbf{(b)} If $A$ is a $\mathbf{k}$-algebra and $M$ is an $A$-module, then
$M$ canonically becomes an $A^{-}$-module. Indeed, the $A^{-}$-module
structure on $M$ is given by%
\[
\left(  a\rightharpoonup m=am\text{ for every }a\in A\text{ and }m\in
M\right)  .
\]


\textbf{(c)} Let $M$ be a $\mathbf{k}$-module. Then, $\operatorname*{End}M$ is
a $\mathbf{k}$-algebra, and thus $\left(  \operatorname*{End}M\right)  ^{-}$
is a Lie algebra. Furthermore, $M$ is an $\operatorname*{End}M$-module, and
thus becomes an $\left(  \operatorname*{End}M\right)  ^{-}$-module.
\end{definition}

\begin{todo}
Define Lie subalgebras, Lie algebra homomorphisms, submodule, module homomorphisms.

Define coalgebras, bialgebras, primitive elements.

Define $U$.
\end{todo}
\end{noncompile}

Next, let us define the notion of a derivation:

\begin{definition}
\label{def.Der}Let $C$ be a $\mathbf{k}$-algebra. A $\mathbf{k}$-linear map
$d:C\rightarrow C$ is said to be a \textit{derivation} of $C$ if and only if
it satisfies%
\begin{equation}
\left(  d\left(  ab\right)  =ad\left(  b\right)  +d\left(  a\right)  b\text{
for every }a\in C\text{ and }b\in C\right)  . \label{eq.def.Der.defder}%
\end{equation}
We let $\operatorname*{Der}C$ denote the set of all derivations of $C$.
\end{definition}

We state a few simple properties of derivations:

\begin{proposition}
\label{prop.Der.basics}Let $C$ be a $\mathbf{k}$-algebra.

\textbf{(a)} The set $\operatorname*{Der}C$ is a Lie subalgebra of $\left(
\operatorname*{End}C\right)  ^{-}$.

\textbf{(b)} Let $f\in\operatorname*{Der}C$. Let $n\in\mathbb{N}$ and let
$a_{1},a_{2},\ldots,a_{n}\in C$. Then,%
\[
f\left(  a_{1}a_{2}\cdots a_{n}\right)  =\sum_{i=1}^{n}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{n}.
\]


\textbf{(c)} Let $f\in\operatorname*{Der}C$. Then, $f\left(  1\right)  =0$.
\end{proposition}

Proposition \ref{prop.Der.basics} \textbf{(a)} is proven in \cite[Remark
1.26]{derivat}, and Proposition \ref{prop.Der.basics} \textbf{(b)} is a
particular case of \cite[Theorem 1.14]{derivat} (for $A=C$ and $M=C$).
Proposition \ref{prop.Der.basics} \textbf{(c)} is proven in \cite[Theorem
1.12]{derivat}.

We shall also use the following three facts:

\begin{lemma}
\label{lem.der.tensid}Let $A$ be a $\mathbf{k}$-algebra. Let $f:A\rightarrow
A$ be a derivation. Let $B$ be a $\mathbf{k}$-algebra.

\textbf{(a)} The map $f\otimes\operatorname*{id}\nolimits_{B}:A\otimes
B\rightarrow A\otimes B$ is a derivation.

\textbf{(b)} The map $\operatorname*{id}\nolimits_{B}\otimes f:B\otimes
A\rightarrow B\otimes A$ is a derivation.
\end{lemma}

\begin{lemma}
\label{lem.derivation.unique}Let $A$ be a $\mathbf{k}$-algebra. Let
$d:A\rightarrow A$ and $e:A\rightarrow A$ be two derivations. Let $S$ be a
subset of $A$ which generates $A$ as a $\mathbf{k}$-algebra. Assume that
$d\mid_{S}=e\mid_{S}$. Then, $d=e$.
\end{lemma}

\begin{lemma}
\label{lem.Der.fdef}Let $A$ and $B$ be two $\mathbf{k}$-algebras. Let
$f:A\rightarrow B$ be a $\mathbf{k}$-algebra homomorphism. Let $d:A\rightarrow
A$ and $e:B\rightarrow B$ be two derivations. Let $S$ be a subset of $A$ which
generates $A$ as a $\mathbf{k}$-algebra. Assume that $\left(  f\circ d\right)
\mid_{S}=\left(  e\circ f\right)  \mid_{S}$. Then, $f\circ d=e\circ f$.
\end{lemma}

These three lemmas are elementary and the reader should have no trouble
proving them.\footnote{Complete proofs can be found in \cite{derivat}. More
precisely: Lemma \ref{lem.der.tensid} is a particular case of
\cite[Proposition 1.44]{derivat} (for $M=A$). Lemma
\ref{lem.derivation.unique} is a particular case of \cite[Proposition
1.29]{derivat} (for $M=A$). Lemma \ref{lem.Der.fdef} is \cite[Corollary
1.45]{derivat}.}

\subsection{A Lie algebra action}

\begin{theorem}
\label{thm.gen1}Let $\mathfrak{g}$ be a Lie algebra. Let $C$ be a $\mathbf{k}%
$-algebra. Let $K:\mathfrak{g}\rightarrow\operatorname*{Der}C$ be a Lie
algebra homomorphism. Let $f:\mathfrak{g}\rightarrow C$ be a $\mathbf{k}%
$-linear map. Assume that%
\begin{equation}
f\left(  \left[  a,b\right]  \right)  =\left[  f\left(  a\right)  ,f\left(
b\right)  \right]  +\left(  K\left(  a\right)  \right)  \left(  f\left(
b\right)  \right)  -\left(  K\left(  b\right)  \right)  \left(  f\left(
a\right)  \right)  \label{eq.thm.gen1.axiom1}%
\end{equation}
for every $a\in\mathfrak{g}$ and $b\in\mathfrak{g}$ (where the Lie bracket
$\left[  f\left(  a\right)  ,f\left(  b\right)  \right]  $ is computed in the
Lie algebra $C^{-}$).

\textbf{(a)} Then, we can define a $\mathfrak{g}$-module structure on $C$ by
setting%
\begin{equation}
\left(  a\rightharpoonup u=f\left(  a\right)  \cdot u+\left(  K\left(
a\right)  \right)  \left(  u\right)  \text{ for all }a\in\mathfrak{g}\text{
and }u\in C\right)  . \label{eq.thm.gen1.a}%
\end{equation}


In the following, we will regard $C$ as a $\mathfrak{g}$-module by means of
this $\mathfrak{g}$-module structure.

\textbf{(b)} Being a $\mathfrak{g}$-module, $C$ becomes a $U\left(
\mathfrak{g}\right)  $-module. Define a map $\eta:U\left(  \mathfrak{g}%
\right)  \rightarrow C$ by%
\[
\eta\left(  u\right)  =u1_{C}\ \ \ \ \ \ \ \ \ \ \text{for every }u\in
U\left(  \mathfrak{g}\right)  .
\]
Then, $\eta$ is a $\mathfrak{g}$-module homomorphism.

\textbf{(c)} For every $a\in\mathfrak{g}$, $b\in C$ and $c\in C$, we have%
\[
a\rightharpoonup\left(  bc\right)  -b\cdot\left(  a\rightharpoonup c\right)
=\left(  K\left(  a\right)  \right)  \left(  b\right)  \cdot c+\left[
f\left(  a\right)  ,b\right]  c.
\]
(Here, again, the Lie bracket $\left[  f\left(  a\right)  ,b\right]  $ is
computed in the Lie algebra $C^{-}$.)
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.gen1}.]\textbf{(a)} We define a $\mathbf{k}%
$-bilinear map $\mu:\mathfrak{g}\times C\rightarrow C$ by%
\[
\left(  \mu\left(  a,u\right)  =f\left(  a\right)  \cdot u+\left(  K\left(
a\right)  \right)  \left(  u\right)  \text{ for all }a\in\mathfrak{g}\text{
and }u\in C\right)  .
\]
We write $a\rightharpoonup u$ for $\mu\left(  a,u\right)  $ whenever
$a\in\mathfrak{g}$ and $u\in C$. Thus, (\ref{eq.thm.gen1.a}) holds.

We now have to show that $\mu$ is a $\mathfrak{g}$-module structure on $C$. In
other words, we need to show that%
\begin{equation}
\left[  a,b\right]  \rightharpoonup v=a\rightharpoonup\left(  b\rightharpoonup
v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)  \text{ for every
}a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in C.
\label{pf.thm.gen1.a.goal}%
\end{equation}


So let us fix $a\in\mathfrak{g}$, $b\in\mathfrak{g}$ and $v\in C$. Recall that
$K:\mathfrak{g}\rightarrow\operatorname*{Der}C$ is a Lie algebra homomorphism;
thus,
\[
K\left(  \left[  a,b\right]  \right)  =\left[  K\left(  a\right)  ,K\left(
b\right)  \right]  =K\left(  a\right)  \circ K\left(  b\right)  -K\left(
b\right)  \circ K\left(  a\right)  .
\]
Now, (\ref{eq.thm.gen1.a}) (applied to $\left[  a,b\right]  $ and $v$ instead
of $a$ and $u$) shows that%
\begin{align}
\left[  a,b\right]  \rightharpoonup v  &  =\underbrace{f\left(  \left[
a,b\right]  \right)  }_{\substack{=\left[  f\left(  a\right)  ,f\left(
b\right)  \right]  +\left(  K\left(  a\right)  \right)  \left(  f\left(
b\right)  \right)  -\left(  K\left(  b\right)  \right)  \left(  f\left(
a\right)  \right)  \\\text{(by (\ref{eq.thm.gen1.axiom1}))}}}\cdot
v+\underbrace{\left(  K\left(  \left[  a,b\right]  \right)  \right)
}_{=K\left(  a\right)  \circ K\left(  b\right)  -K\left(  b\right)  \circ
K\left(  a\right)  }\left(  v\right) \nonumber\\
&  =\underbrace{\left(  \left[  f\left(  a\right)  ,f\left(  b\right)
\right]  +\left(  K\left(  a\right)  \right)  \left(  f\left(  b\right)
\right)  -\left(  K\left(  b\right)  \right)  \left(  f\left(  a\right)
\right)  \right)  \cdot v}_{=\left[  f\left(  a\right)  ,f\left(  b\right)
\right]  \cdot v+\left(  K\left(  a\right)  \right)  \left(  f\left(
b\right)  \right)  \cdot v-\left(  K\left(  b\right)  \right)  \left(
f\left(  a\right)  \right)  \cdot v}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  K\left(  a\right)  \circ K\left(
b\right)  -K\left(  b\right)  \circ K\left(  a\right)  \right)  \left(
v\right)  }_{=\left(  K\left(  a\right)  \circ K\left(  b\right)  \right)
\left(  v\right)  -\left(  K\left(  b\right)  \circ K\left(  a\right)
\right)  \left(  v\right)  }\nonumber\\
&  =\left[  f\left(  a\right)  ,f\left(  b\right)  \right]  \cdot v+\left(
K\left(  a\right)  \right)  \left(  f\left(  b\right)  \right)  \cdot
v-\left(  K\left(  b\right)  \right)  \left(  f\left(  a\right)  \right)
\cdot v\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  K\left(  a\right)  \circ K\left(  b\right)
\right)  \left(  v\right)  -\left(  K\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  v\right)  . \label{pf.thm.gen1.a.0}%
\end{align}


On the other hand, (\ref{eq.thm.gen1.a}) (applied to $b$ and $v$ instead of
$a$ and $u$) yields $b\rightharpoonup v=f\left(  b\right)  \cdot v+\left(
K\left(  b\right)  \right)  \left(  v\right)  $. But $K\left(  \underbrace{a}%
_{\in\mathfrak{g}}\right)  \in K\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Der}C$; in other words, $K\left(  a\right)  :C\rightarrow C$ is
a derivation. Now, (\ref{eq.thm.gen1.a}) (applied to $b\rightharpoonup v$
instead of $u$) yields%
\begin{align}
&  a\rightharpoonup\left(  b\rightharpoonup v\right) \nonumber\\
&  =f\left(  a\right)  \cdot\underbrace{\left(  b\rightharpoonup v\right)
}_{=f\left(  b\right)  \cdot v+\left(  K\left(  b\right)  \right)  \left(
v\right)  }+K\left(  a\right)  \left(  \underbrace{b\rightharpoonup
v}_{=\left(  f\left(  b\right)  \cdot v+\left(  K\left(  b\right)  \right)
\left(  v\right)  \right)  }\right) \nonumber\\
&  =\underbrace{f\left(  a\right)  \cdot\left(  \left(  f\left(  b\right)
\cdot v+\left(  K\left(  b\right)  \right)  \left(  v\right)  \right)
\right)  }_{=f\left(  a\right)  \cdot f\left(  b\right)  \cdot v+f\left(
a\right)  \cdot\left(  K\left(  b\right)  \right)  \left(  v\right)
}+\underbrace{K\left(  a\right)  \left(  f\left(  b\right)  \cdot v+\left(
K\left(  b\right)  \right)  \left(  v\right)  \right)  }_{=K\left(  a\right)
\left(  f\left(  b\right)  \cdot v\right)  +\left(  K\left(  a\right)
\right)  \left(  \left(  K\left(  b\right)  \right)  \left(  v\right)
\right)  }\nonumber\\
&  =f\left(  a\right)  \cdot f\left(  b\right)  \cdot v+f\left(  a\right)
\cdot\left(  K\left(  b\right)  \right)  \left(  v\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  K\left(  a\right)  \right)
\left(  f\left(  b\right)  \cdot v\right)  }_{\substack{=f\left(  b\right)
\cdot\left(  K\left(  a\right)  \right)  \left(  v\right)  +\left(  K\left(
a\right)  \right)  \left(  f\left(  b\right)  \right)  \cdot v\\\text{(since
}K\left(  a\right)  \text{ is a derivation)}}}+\underbrace{\left(  K\left(
a\right)  \right)  \left(  \left(  K\left(  b\right)  \right)  \left(
v\right)  \right)  }_{=\left(  K\left(  a\right)  \circ K\left(  b\right)
\right)  \left(  v\right)  }\nonumber\\
&  =f\left(  a\right)  \cdot f\left(  b\right)  \cdot v+f\left(  a\right)
\cdot\left(  K\left(  b\right)  \right)  \left(  v\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +f\left(  b\right)  \cdot\left(  K\left(  a\right)
\right)  \left(  v\right)  +\left(  K\left(  a\right)  \right)  \left(
f\left(  b\right)  \right)  \cdot v+\left(  K\left(  a\right)  \circ K\left(
b\right)  \right)  \left(  v\right)  . \label{pf.thm.gen1.a.1}%
\end{align}
The same argument (applied to $b$ and $a$ instead of $a$ and $b$) shows that%
\begin{align*}
&  b\rightharpoonup\left(  a\rightharpoonup v\right) \\
&  =f\left(  b\right)  \cdot f\left(  a\right)  \cdot v+f\left(  b\right)
\cdot\left(  K\left(  a\right)  \right)  \left(  v\right) \\
&  \ \ \ \ \ \ \ \ \ \ +f\left(  a\right)  \cdot\left(  K\left(  b\right)
\right)  \left(  v\right)  +\left(  K\left(  b\right)  \right)  \left(
f\left(  a\right)  \right)  \cdot v+\left(  K\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  v\right)  .
\end{align*}
Subtracting this equality from (\ref{pf.thm.gen1.a.1}), we obtain%
\begin{align*}
&  a\rightharpoonup\left(  b\rightharpoonup v\right)  -b\rightharpoonup\left(
a\rightharpoonup v\right) \\
&  =\left(  f\left(  a\right)  \cdot f\left(  b\right)  \cdot v+f\left(
a\right)  \cdot\left(  K\left(  b\right)  \right)  \left(  v\right)  \right.
\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  +f\left(  b\right)
\cdot\left(  K\left(  a\right)  \right)  \left(  v\right)  +\left(  K\left(
a\right)  \right)  \left(  f\left(  b\right)  \right)  \cdot v+\left(
K\left(  a\right)  \circ K\left(  b\right)  \right)  \left(  v\right)  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ -\left(  f\left(  b\right)  \cdot f\left(  a\right)
\cdot v+f\left(  b\right)  \cdot\left(  K\left(  a\right)  \right)  \left(
v\right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  +f\left(  a\right)
\cdot\left(  K\left(  b\right)  \right)  \left(  v\right)  +\left(  K\left(
b\right)  \right)  \left(  f\left(  a\right)  \right)  \cdot v+\left(
K\left(  b\right)  \circ K\left(  a\right)  \right)  \left(  v\right)  \right)
\\
&  =\underbrace{f\left(  a\right)  \cdot f\left(  b\right)  \cdot v-f\left(
b\right)  \cdot f\left(  a\right)  \cdot v}_{\substack{=\left(  f\left(
a\right)  \cdot f\left(  b\right)  -f\left(  b\right)  \cdot f\left(
a\right)  \right)  \left(  v\right)  \\=\left[  f\left(  a\right)  ,f\left(
b\right)  \right]  \cdot v\\\text{(since }f\left(  a\right)  \cdot f\left(
b\right)  -f\left(  b\right)  \cdot f\left(  a\right)  =\left[  f\left(
a\right)  ,f\left(  b\right)  \right]  \text{)}}}+\left(  K\left(  a\right)
\right)  \left(  f\left(  b\right)  \right)  \cdot v-\left(  K\left(
b\right)  \right)  \left(  f\left(  a\right)  \right)  \cdot v\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  K\left(  a\right)  \circ K\left(  b\right)
\right)  \left(  v\right)  -\left(  K\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  v\right) \\
&  =\left[  f\left(  a\right)  ,f\left(  b\right)  \right]  \cdot v+\left(
K\left(  a\right)  \right)  \left(  f\left(  b\right)  \right)  \cdot
v-\left(  K\left(  b\right)  \right)  \left(  f\left(  a\right)  \right)
\cdot v\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  K\left(  a\right)  \circ K\left(  b\right)
\right)  \left(  v\right)  -\left(  K\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  v\right)  .
\end{align*}
Comparing this with (\ref{pf.thm.gen1.a.0}), we obtain $\left[  a,b\right]
\rightharpoonup v=a\rightharpoonup\left(  b\rightharpoonup v\right)
-b\rightharpoonup\left(  a\rightharpoonup v\right)  $. Thus,
(\ref{pf.thm.gen1.a.goal}) is proven. Hence, $\mu$ is a $\mathfrak{g}$-module
structure on $C$. Theorem \ref{thm.gen1} \textbf{(a)} is proven.

\textbf{(b)} For every $a\in\mathfrak{g}$ and $u\in U\left(  \mathfrak{g}%
\right)  $, we have%
\begin{align*}
&  \eta\left(  \underbrace{a\rightharpoonup u}_{\substack{=au\\\text{(by the
definition of the}\\\mathfrak{g}\text{-module structure on }U\left(
\mathfrak{g}\right)  \text{)}}}\right) \\
&  =\eta\left(  au\right)  =au1_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\eta\right) \\
&  =\iota_{U,\mathfrak{g}}\left(  a\right)  u1_{C}\ \ \ \ \ \ \ \ \ \ \left(
\text{since we use }a\text{ as a shorthand for }\iota_{U,\mathfrak{g}}\left(
a\right)  \right) \\
&  =a\rightharpoonup\underbrace{\left(  u1_{C}\right)  }_{\substack{=\eta
\left(  u\right)  \\\text{(since }\eta\left(  u\right)  =u1_{C}\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a\rightharpoonup\left(
u1_{C}\right)  =\iota_{U,\mathfrak{g}}\left(  a\right)  u1_{C}\right) \\
&  =a\rightharpoonup\eta\left(  u\right)  .
\end{align*}
In other words, $\eta$ is a $\mathfrak{g}$-module homomorphism. This proves
Theorem \ref{thm.gen1} \textbf{(b)}.

\textbf{(c)} Let $a\in\mathfrak{g}$, $b\in C$ and $c\in C$. The map $K$ has
target $\operatorname*{Der}C$. Hence, $K\left(  a\right)  \in
\operatorname*{Der}C$. In other words, $K\left(  a\right)  :C\rightarrow C$ is
a derivation.

The definition of the Lie bracket of $C^{-}$ shows that $\left[  f\left(
a\right)  ,b\right]  =f\left(  a\right)  \cdot b-b\cdot f\left(  a\right)  $.

Now, the definition of the $\mathfrak{g}$-module structure on $C$ shows that%
\begin{align*}
&  a\rightharpoonup\left(  bc\right) \\
&  =f\left(  a\right)  \cdot bc+\underbrace{\left(  K\left(  a\right)
\right)  \left(  bc\right)  }_{\substack{=b\cdot\left(  K\left(  a\right)
\right)  \left(  c\right)  +\left(  K\left(  a\right)  \right)  \left(
b\right)  \cdot c\\\text{(since }K\left(  a\right)  \text{ is a derivation)}%
}}\\
&  =f\left(  a\right)  \cdot bc+b\cdot\left(  K\left(  a\right)  \right)
\left(  c\right)  +\left(  K\left(  a\right)  \right)  \left(  b\right)  \cdot
c.
\end{align*}
On the other hand, the definition of the $\mathfrak{g}$-module structure on
$C$ shows that $a\rightharpoonup c=f\left(  a\right)  \cdot c+\left(  K\left(
a\right)  \right)  \left(  c\right)  $. Hence,%
\[
b\cdot\underbrace{\left(  a\rightharpoonup c\right)  }_{=f\left(  a\right)
\cdot c+\left(  K\left(  a\right)  \right)  \left(  c\right)  }=b\cdot\left(
f\left(  a\right)  \cdot c+\left(  K\left(  a\right)  \right)  \left(
c\right)  \right)  =b\cdot f\left(  a\right)  \cdot c+b\cdot\left(  K\left(
a\right)  \right)  \left(  c\right)  .
\]
Thus,%
\begin{align*}
&  \underbrace{a\rightharpoonup\left(  bc\right)  }_{=f\left(  a\right)  \cdot
bc+b\cdot\left(  K\left(  a\right)  \right)  \left(  c\right)  +\left(
K\left(  a\right)  \right)  \left(  b\right)  \cdot c}-\underbrace{b\cdot
\left(  a\rightharpoonup c\right)  }_{=b\cdot f\left(  a\right)  \cdot
c+b\cdot\left(  K\left(  a\right)  \right)  \left(  c\right)  }\\
&  =\left(  f\left(  a\right)  \cdot bc+b\cdot\left(  K\left(  a\right)
\right)  \left(  c\right)  +\left(  K\left(  a\right)  \right)  \left(
b\right)  \cdot c\right)  -\left(  b\cdot f\left(  a\right)  \cdot
c+b\cdot\left(  K\left(  a\right)  \right)  \left(  c\right)  \right) \\
&  =f\left(  a\right)  \cdot bc+\left(  K\left(  a\right)  \right)  \left(
b\right)  \cdot c-b\cdot f\left(  a\right)  \cdot c\\
&  =\left(  K\left(  a\right)  \right)  \left(  b\right)  \cdot
c+\underbrace{f\left(  a\right)  \cdot bc-b\cdot f\left(  a\right)  \cdot
c}_{=\left(  f\left(  a\right)  \cdot b-b\cdot f\left(  a\right)  \right)
\cdot c}\\
&  =\left(  K\left(  a\right)  \right)  \left(  b\right)  \cdot
c+\underbrace{\left(  f\left(  a\right)  \cdot b-b\cdot f\left(  a\right)
\right)  }_{\substack{=\left[  f\left(  a\right)  ,b\right]  \\\text{(since
}\left[  f\left(  a\right)  ,b\right]  =f\left(  a\right)  \cdot b-b\cdot
f\left(  a\right)  \text{)}}}\cdot c\\
&  =\left(  K\left(  a\right)  \right)  \left(  b\right)  \cdot c+\left[
f\left(  a\right)  ,b\right]  c.
\end{align*}
This proves Theorem \ref{thm.gen1} \textbf{(c)}.
\end{proof}

\subsection{The bialgebra case}

\begin{condition}
We shall use the notions of $\mathbf{k}$-coalgebras and $\mathbf{k}%
$-bialgebras. See, for example, \cite[\S 1]{GriRei15} for an introduction to
these notions. We always assume $\mathbf{k}$-coalgebras to be counital and
coassociative. We will use the notations $\Delta$ and $\epsilon$ for the
comultiplication and the counit of a $\mathbf{k}$-coalgebra.

If $\mathfrak{g}$ is a Lie algebra, then the universal enveloping algebra
$U\left(  \mathfrak{g}\right)  $ comes equipped with a canonical Hopf algebra
structure (obtained by projecting the Hopf algebra structure on the tensor
algebra of $\mathfrak{g}$).
\end{condition}

\begin{definition}
Let $C$ be a $\mathbf{k}$-coalgebra. Then, a \textit{coderivation} of $C$
means a $\mathbf{k}$-linear map $f:C\rightarrow C$ such that $\Delta\circ
f=\left(  f\otimes\operatorname*{id}+\operatorname*{id}\otimes f\right)
\circ\Delta$. We let $\operatorname*{Coder}C$ denote the set of all
coderivations of a $\mathbf{k}$-coalgebra $C$. It is known that
$\operatorname*{Coder}C$ is a Lie subalgebra of $\left(  \operatorname*{End}%
C\right)  ^{-}$; however, we shall only be interested in
$\operatorname*{Coder}C$ as a set.
\end{definition}

\begin{proposition}
\label{prop.coder.epsi}Let $C$ be a $\mathbf{k}$-coalgebra. Let $f$ be a
coderivation of $C$. Then, $\epsilon\circ f=0$.
\end{proposition}

Proposition \ref{prop.coder.epsi}, of course, is the \textquotedblleft
dual\textquotedblright\ of Proposition \ref{prop.Der.basics} \textbf{(c)}
(that is, of the well-known fact that any derivation of a $\mathbf{k}$-algebra
sends $1$ to $0$).

\begin{proof}
[First proof of Proposition \ref{prop.coder.epsi}.]Let $\kappa_{1}%
:C\rightarrow C\otimes\mathbf{k}$ and $\kappa_{2}:C\rightarrow\mathbf{k}%
\otimes C$ be the canonical $\mathbf{k}$-module isomorphisms. Let
$\kappa:\mathbf{k}\rightarrow\mathbf{k}\otimes\mathbf{k}$ be the canonical
$\mathbf{k}$-module isomorphism. Then, two of the axioms of a $\mathbf{k}%
$-coalgebra (applied to the $\mathbf{k}$-coalgebra $C$) yield $\left(
\operatorname*{id}\otimes\epsilon\right)  \circ\Delta=\kappa_{1}$ and $\left(
\epsilon\otimes\operatorname*{id}\right)  \circ\Delta=\kappa_{2}$. Now,
\[
\underbrace{\left(  \epsilon\otimes\epsilon\right)  }_{=\left(  \epsilon
\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes\epsilon\right)  }\circ\Delta=\left(  \epsilon\otimes\operatorname*{id}%
\right)  \circ\underbrace{\left(  \operatorname*{id}\otimes\epsilon\right)
\circ\Delta}_{=\kappa_{1}}=\left(  \epsilon\otimes\operatorname*{id}\right)
\circ\kappa_{1}=\kappa\circ\epsilon.
\]
Therefore,%
\begin{align*}
\underbrace{\kappa\circ\epsilon}_{=\left(  \epsilon\otimes\epsilon\right)
\circ\Delta}\circ f  &  =\left(  \epsilon\otimes\epsilon\right)
\circ\underbrace{\Delta\circ f}_{\substack{=\left(  f\otimes\operatorname*{id}%
+\operatorname*{id}\otimes f\right)  \circ\Delta\\\text{(since }f\text{ is a
coderivation)}}}=\underbrace{\left(  \epsilon\otimes\epsilon\right)
\circ\left(  f\otimes\operatorname*{id}+\operatorname*{id}\otimes f\right)
}_{=\left(  \epsilon\circ f\right)  \otimes\left(  \epsilon\circ
\operatorname*{id}\right)  +\left(  \epsilon\circ\operatorname*{id}\right)
\otimes\left(  \epsilon\circ f\right)  }\circ\Delta\\
&  =\left(  \left(  \epsilon\circ f\right)  \otimes\underbrace{\left(
\epsilon\circ\operatorname*{id}\right)  }_{=\epsilon}+\underbrace{\left(
\epsilon\circ\operatorname*{id}\right)  }_{=\epsilon}\otimes\left(
\epsilon\circ f\right)  \right)  \circ\Delta=\left(  \left(  \epsilon\circ
f\right)  \otimes\epsilon+\epsilon\otimes\left(  \epsilon\circ f\right)
\right)  \circ\Delta\\
&  =\underbrace{\left(  \left(  \epsilon\circ f\right)  \otimes\epsilon
\right)  }_{=\left(  \left(  \epsilon\circ f\right)  \otimes\operatorname*{id}%
\right)  \circ\left(  \operatorname*{id}\otimes\epsilon\right)  }\circ
\Delta+\underbrace{\left(  \epsilon\otimes\left(  \epsilon\circ f\right)
\right)  }_{=\left(  \operatorname*{id}\otimes\left(  \epsilon\circ f\right)
\right)  \circ\left(  \epsilon\otimes\operatorname*{id}\right)  }\circ\Delta\\
&  =\left(  \left(  \epsilon\circ f\right)  \otimes\operatorname*{id}\right)
\circ\underbrace{\left(  \operatorname*{id}\otimes\epsilon\right)  \circ
\Delta}_{=\kappa_{1}}+\left(  \operatorname*{id}\otimes\left(  \epsilon\circ
f\right)  \right)  \circ\underbrace{\left(  \epsilon\otimes\operatorname*{id}%
\right)  \circ\Delta}_{=\kappa_{2}}\\
&  =\underbrace{\left(  \left(  \epsilon\circ f\right)  \otimes
\operatorname*{id}\right)  \circ\kappa_{1}}_{=\kappa\circ\left(  \epsilon\circ
f\right)  }+\underbrace{\left(  \operatorname*{id}\otimes\left(  \epsilon\circ
f\right)  \right)  \circ\kappa_{2}}_{=\kappa\circ\left(  \epsilon\circ
f\right)  }\\
&  =\kappa\circ\left(  \epsilon\circ f\right)  +\kappa\circ\left(
\epsilon\circ f\right)  =2\kappa\circ\left(  \epsilon\circ f\right)
=2\kappa\circ\epsilon\circ f.
\end{align*}
Subtracting $\kappa\circ\epsilon\circ f$ from this equality, we obtain
$0=\kappa\circ\epsilon\circ f$. Since $\kappa$ is an isomorphism, we can
cancel $\kappa$ from this equality. As a result, we obtain $0=\epsilon\circ
f$. This proves Proposition \ref{prop.coder.epsi}.
\end{proof}

\begin{proof}
[Second proof of Proposition \ref{prop.coder.epsi}.]Recall that the dual
$\mathbf{k}$-module $C^{\ast}=\operatorname*{Hom}\nolimits_{\mathbf{k}}\left(
C,\mathbf{k}\right)  $ (this is the $\mathbf{k}$-module of all $\mathbf{k}%
$-linear maps from $C$ to $\mathbf{k}$) canonically becomes a $\mathbf{k}%
$-algebra. The counit $\epsilon:C\rightarrow\mathbf{k}$ of $C$ is the unity of
this $\mathbf{k}$-algebra; in other words, $\epsilon=1_{C^{\ast}}$.

Now, the adjoint map $f^{\ast}:C^{\ast}\rightarrow C^{\ast}$ of the
coderivation $f:C\rightarrow C$ is a derivation of the $\mathbf{k}$-algebra
$C^{\ast}$\ \ \ \ \footnote{\textit{Proof.} Let $a\in C^{\ast}$ and $b\in
C^{\ast}$. We must prove that $f^{\ast}\left(  ab\right)  =f^{\ast}\left(
a\right)  \cdot b+a\cdot f^{\ast}\left(  b\right)  $.
\par
Let $\mu_{\mathbf{k}}$ be the canonical $\mathbf{k}$-algebra homomorphism
$\mathbf{k}\otimes\mathbf{k}\rightarrow\mathbf{k},\ \lambda\otimes\mu
\mapsto\lambda\mu$. (This is, of course, an isomorphism.) Then, the definition
of the multiplication on $C^{\ast}$ shows that $ab=\mu_{\mathbf{k}}%
\circ\left(  a\otimes b\right)  \circ\Delta$. Now, the definition of $f^{\ast
}$ shows that%
\begin{align*}
f^{\ast}\left(  ab\right)   &  =\underbrace{\left(  ab\right)  }%
_{=\mu_{\mathbf{k}}\circ\left(  a\otimes b\right)  \circ\Delta}\circ
f=\mu_{\mathbf{k}}\circ\left(  a\otimes b\right)  \circ\underbrace{\Delta\circ
f}_{\substack{=\left(  f\otimes\operatorname*{id}+\operatorname*{id}\otimes
f\right)  \circ\Delta\\\text{(since }f\text{ is a coderivation)}}%
}=\mu_{\mathbf{k}}\circ\underbrace{\left(  a\otimes b\right)  \circ\left(
f\otimes\operatorname*{id}+\operatorname*{id}\otimes f\right)  }_{=\left(
a\otimes b\right)  \circ\left(  f\otimes\operatorname*{id}\right)  +\left(
a\otimes b\right)  \circ\left(  \operatorname*{id}\otimes f\right)  }%
\circ\Delta\\
&  =\mu_{\mathbf{k}}\circ\left(  \underbrace{\left(  a\otimes b\right)
\circ\left(  f\otimes\operatorname*{id}\right)  }_{=\left(  a\circ f\right)
\otimes b}+\underbrace{\left(  a\otimes b\right)  \circ\left(
\operatorname*{id}\otimes f\right)  }_{=a\otimes\left(  b\circ f\right)
}\right)  \circ\Delta\\
&  =\mu_{\mathbf{k}}\circ\left(  \left(  a\circ f\right)  \otimes
b+a\otimes\left(  b\circ f\right)  \right)  \circ\Delta=\mu_{\mathbf{k}}%
\circ\left(  \left(  a\circ f\right)  \otimes b\right)  \circ\Delta
+\mu_{\mathbf{k}}\circ\left(  a\otimes\left(  b\circ f\right)  \right)
\circ\Delta.
\end{align*}
Compared with%
\begin{align*}
\underbrace{f^{\ast}\left(  a\right)  }_{=a\circ f}\cdot b+a\cdot
\underbrace{f^{\ast}\left(  b\right)  }_{=b\circ f}  &  =\underbrace{\left(
a\circ f\right)  \cdot b}_{\substack{=\mu_{\mathbf{k}}\circ\left(  \left(
a\circ f\right)  \otimes b\right)  \circ\Delta\\\text{(by the definition of
the}\\\text{multiplication on }C^{\ast}\text{)}}}+\underbrace{a\cdot\left(
b\circ f\right)  }_{\substack{=\mu_{\mathbf{k}}\circ\left(  a\otimes\left(
b\circ f\right)  \right)  \circ\Delta\\\text{(by the definition of
the}\\\text{multiplication on }C^{\ast}\text{)}}}\\
&  =\mu_{\mathbf{k}}\circ\left(  \left(  a\circ f\right)  \otimes b\right)
\circ\Delta+\mu_{\mathbf{k}}\circ\left(  a\otimes\left(  b\circ f\right)
\right)  \circ\Delta,
\end{align*}
this shows that $f^{\ast}\left(  ab\right)  =f^{\ast}\left(  a\right)  \cdot
b+a\cdot f^{\ast}\left(  b\right)  $.
\par
Thus, we have proven that $f^{\ast}\left(  ab\right)  =f^{\ast}\left(
a\right)  \cdot b+a\cdot f^{\ast}\left(  b\right)  $ for every $a\in C^{\ast}$
and $b\in C^{\ast}$. In other words, $f^{\ast}$ is a derivation of $C^{\ast}$,
qed.}. Therefore, $f^{\ast}$ sends $1$ to $0$ (since a well-known result
states that any derivation of a $\mathbf{k}$-algebra sends $1$ to $0$). In
other words, $f^{\ast}\left(  1_{C^{\ast}}\right)  =0$. Since $1_{C^{\ast}%
}=\epsilon$, this rewrites as $f^{\ast}\left(  \epsilon\right)  =0$. Since
$f^{\ast}\left(  \epsilon\right)  =\epsilon\circ f$, this rewrites as
$\epsilon\circ f=0$. Proposition \ref{prop.coder.epsi} thus is proven again.
\end{proof}

\begin{definition}
A \textit{primitive} element of a $\mathbf{k}$-bialgebra $H$ means an element
$x\in H$ satisfying $\Delta\left(  x\right)  =x\otimes1+1\otimes x$. We let
$\operatorname*{Prim}H$ denote the set of all primitive elements of a
$\mathbf{k}$-bialgebra $H$. It is well-known that $\operatorname*{Prim}H$ is a
Lie subalgebra of $H^{-}$.
\end{definition}

\begin{theorem}
\label{thm.gen3}Let $\mathfrak{g}$ be a Lie algebra. Let $C$ be a $\mathbf{k}%
$-bialgebra. Let $K:\mathfrak{g}\rightarrow\operatorname*{Der}C$ be a Lie
algebra homomorphism such that $K\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Coder}C$. Let $f:\mathfrak{g}\rightarrow C$ be a $\mathbf{k}%
$-linear map such that $f\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Prim}C$. Assume that (\ref{eq.thm.gen1.axiom1}) holds for every
$a\in\mathfrak{g}$ and $b\in\mathfrak{g}$.

Consider the $\mathfrak{g}$-module structure on $C$ defined in Theorem
\ref{thm.gen1} \textbf{(a)}, and the map $\eta:U\left(  \mathfrak{g}\right)
\rightarrow C$ defined in Theorem \ref{thm.gen1} \textbf{(b)}.

\textbf{(a)} For every $a\in\mathfrak{g}$, the map $C\rightarrow C,\ c\mapsto
a\rightharpoonup c$ is a coderivation of $C$.

\textbf{(b)} The map $\eta:U\left(  \mathfrak{g}\right)  \rightarrow C$ is a
$\mathbf{k}$-coalgebra homomorphism.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.gen3}.]\textbf{(a)} Let $a\in\mathfrak{g}$. Let
$\zeta$ be the map $C\rightarrow C,\ c\mapsto a\rightharpoonup c$. We must
prove that this map $\zeta$ is a coderivation of $C$. In other words, we must
prove that $\Delta\circ\zeta=\left(  \zeta\otimes\operatorname*{id}%
+\operatorname*{id}\otimes\zeta\right)  \circ\Delta$.

We have $K\left(  a\right)  \in K\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Coder}C$. In other words, $K\left(  a\right)  $ is a
coderivation of $C$.

Let $u\in C$. Then, by the definition of $\zeta$, we have $\zeta\left(
u\right)  =a\rightharpoonup u=f\left(  a\right)  \cdot u+\left(  K\left(
a\right)  \right)  \left(  u\right)  $ (by (\ref{eq.thm.gen1.a})). Hence,
\begin{align}
\left(  \Delta\circ\zeta\right)  \left(  u\right)   &  =\Delta\left(
\underbrace{\zeta\left(  u\right)  }_{=f\left(  a\right)  \cdot u+\left(
K\left(  a\right)  \right)  \left(  u\right)  }\right)  =\Delta\left(
f\left(  a\right)  \cdot u+\left(  K\left(  a\right)  \right)  \left(
u\right)  \right) \nonumber\\
&  =\underbrace{\Delta\left(  f\left(  a\right)  \right)  }%
_{\substack{=f\left(  a\right)  \otimes1+1\otimes f\left(  a\right)
\\\text{(since }f\left(  a\right)  \in f\left(  \mathfrak{g}\right)
\subseteq\operatorname*{Prim}C\text{)}}}\cdot\Delta\left(  u\right)
+\underbrace{\Delta\left(  \left(  K\left(  a\right)  \right)  \left(
u\right)  \right)  }_{=\left(  \Delta\circ K\left(  a\right)  \right)  \left(
u\right)  }\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\text{ is a }\mathbf{k}%
\text{-algebra homomorphism}\right) \nonumber\\
&  =\left(  f\left(  a\right)  \otimes1+1\otimes f\left(  a\right)  \right)
\cdot\Delta\left(  u\right)  +\underbrace{\left(  \Delta\circ K\left(
a\right)  \right)  }_{\substack{=\left(  K\left(  a\right)  \otimes
\operatorname*{id}+\operatorname*{id}\otimes K\left(  a\right)  \right)
\circ\Delta\\\text{(since }K\left(  a\right)  \text{ is a coderivation of
}C\text{)}}}\left(  u\right) \nonumber\\
&  =\left(  f\left(  a\right)  \otimes1+1\otimes f\left(  a\right)  \right)
\cdot\Delta\left(  u\right)  +\underbrace{\left(  \left(  K\left(  a\right)
\otimes\operatorname*{id}+\operatorname*{id}\otimes K\left(  a\right)
\right)  \circ\Delta\right)  \left(  u\right)  }_{=\left(  K\left(  a\right)
\otimes\operatorname*{id}+\operatorname*{id}\otimes K\left(  a\right)
\right)  \left(  \Delta\left(  u\right)  \right)  }\nonumber\\
&  =\left(  f\left(  a\right)  \otimes1+1\otimes f\left(  a\right)  \right)
\cdot\Delta\left(  u\right)  +\left(  K\left(  a\right)  \otimes
\operatorname*{id}+\operatorname*{id}\otimes K\left(  a\right)  \right)
\left(  \Delta\left(  u\right)  \right)  . \label{pf.thm.gen3.a.1}%
\end{align}
But every $p\in C\otimes C$ satisfies%
\begin{equation}
\left(  \zeta\otimes\operatorname*{id}+\operatorname*{id}\otimes\zeta\right)
\left(  p\right)  =\left(  f\left(  a\right)  \otimes1+1\otimes f\left(
a\right)  \right)  \cdot p+\left(  K\left(  a\right)  \otimes
\operatorname*{id}+\operatorname*{id}\otimes K\left(  a\right)  \right)
\left(  p\right)  \label{pf.thm.gen3.a.2}%
\end{equation}
\footnote{\textit{Proof.} Let $p\in C\otimes C$. We need to prove the equality
(\ref{pf.thm.gen3.a.2}). Since this equality is $\mathbf{k}$-linear in $p$, we
can WLOG assume that $p$ is a pure tensor (since the pure tensors span the
$\mathbf{k}$-module $C\otimes C$). Assume this. Thus, $p=q\otimes r$ for some
$q\in C$ and $r\in C$. Consider these $q$ and $r$. We have%
\begin{align*}
&  \left(  \zeta\otimes\operatorname*{id}+\operatorname*{id}\otimes
\zeta\right)  \left(  \underbrace{p}_{=q\otimes r}\right) \\
&  =\left(  \zeta\otimes\operatorname*{id}+\operatorname*{id}\otimes
\zeta\right)  \left(  q\otimes r\right) \\
&  =\underbrace{\zeta\left(  q\right)  }_{\substack{=a\rightharpoonup
q\\\text{(by the definition of }\zeta\text{)}}}\otimes
\underbrace{\operatorname*{id}\left(  r\right)  }_{=r}%
+\underbrace{\operatorname*{id}\left(  q\right)  }_{=q}\otimes
\underbrace{\zeta\left(  r\right)  }_{\substack{=a\rightharpoonup r\\\text{(by
the definition of }\zeta\text{)}}}\\
&  =\underbrace{\left(  a\rightharpoonup q\right)  }_{\substack{=f\left(
a\right)  \cdot q+\left(  K\left(  a\right)  \right)  \left(  q\right)
\\\text{(by (\ref{eq.thm.gen1.a}))}}}\otimes r+q\otimes\underbrace{\left(
a\rightharpoonup r\right)  }_{\substack{=f\left(  a\right)  \cdot r+\left(
K\left(  a\right)  \right)  \left(  r\right)  \\\text{(by (\ref{eq.thm.gen1.a}%
))}}}\\
&  =\left(  f\left(  a\right)  \cdot q+\left(  K\left(  a\right)  \right)
\left(  q\right)  \right)  \otimes r+q\otimes\left(  f\left(  a\right)  \cdot
r+\left(  K\left(  a\right)  \right)  \left(  r\right)  \right) \\
&  =f\left(  a\right)  \cdot q\otimes r+\left(  K\left(  a\right)  \right)
\left(  q\right)  \otimes r+q\otimes f\left(  a\right)  \cdot r+q\otimes
\left(  K\left(  a\right)  \right)  \left(  r\right)  .
\end{align*}
Compared with%
\begin{align*}
&  \left(  f\left(  a\right)  \otimes1+1\otimes f\left(  a\right)  \right)
\cdot\underbrace{p}_{=q\otimes r}+\left(  K\left(  a\right)  \otimes
\operatorname*{id}+\operatorname*{id}\otimes K\left(  a\right)  \right)
\left(  \underbrace{p}_{=q\otimes r}\right) \\
&  =\left(  f\left(  a\right)  \otimes1+1\otimes f\left(  a\right)  \right)
\cdot\left(  q\otimes r\right)  +\left(  K\left(  a\right)  \otimes
\operatorname*{id}+\operatorname*{id}\otimes K\left(  a\right)  \right)
\left(  q\otimes r\right) \\
&  =\underbrace{\left(  f\left(  a\right)  \otimes1\right)  \cdot\left(
q\otimes r\right)  }_{=f\left(  a\right)  \cdot q\otimes r}%
+\underbrace{\left(  1\otimes f\left(  a\right)  \right)  \cdot\left(
q\otimes r\right)  }_{=q\otimes f\left(  a\right)  \cdot r}%
+\underbrace{\left(  K\left(  a\right)  \otimes\operatorname*{id}\right)
\left(  q\otimes r\right)  }_{\substack{=\left(  K\left(  a\right)  \right)
\left(  q\right)  \otimes\operatorname*{id}\left(  r\right)  \\=\left(
K\left(  a\right)  \right)  \left(  q\right)  \otimes r}}+\underbrace{\left(
\operatorname*{id}\otimes K\left(  a\right)  \right)  \left(  q\otimes
r\right)  }_{\substack{=\operatorname*{id}\left(  q\right)  \otimes\left(
K\left(  a\right)  \right)  \left(  r\right)  \\=q\otimes\left(  K\left(
a\right)  \right)  \left(  r\right)  }}\\
&  =f\left(  a\right)  \cdot q\otimes r+q\otimes f\left(  a\right)  \cdot
r+\left(  K\left(  a\right)  \right)  \left(  q\right)  \otimes r+q\otimes
\left(  K\left(  a\right)  \right)  \left(  r\right) \\
&  =f\left(  a\right)  \cdot q\otimes r+\left(  K\left(  a\right)  \right)
\left(  q\right)  \otimes r+q\otimes f\left(  a\right)  \cdot r+q\otimes
\left(  K\left(  a\right)  \right)  \left(  r\right)  ,
\end{align*}
this yields $\left(  \zeta\otimes\operatorname*{id}+\operatorname*{id}%
\otimes\zeta\right)  \left(  p\right)  =\left(  f\left(  a\right)
\otimes1+1\otimes f\left(  a\right)  \right)  \cdot p+\left(  K\left(
a\right)  \otimes\operatorname*{id}+\operatorname*{id}\otimes K\left(
a\right)  \right)  \left(  p\right)  $. This proves (\ref{pf.thm.gen3.a.2}).}.
Applying this to $p=\Delta\left(  u\right)  $, we obtain%
\begin{align*}
&  \left(  \zeta\otimes\operatorname*{id}+\operatorname*{id}\otimes
\zeta\right)  \left(  \Delta\left(  u\right)  \right) \\
&  =\left(  f\left(  a\right)  \otimes1+1\otimes f\left(  a\right)  \right)
\cdot\Delta\left(  u\right)  +\left(  K\left(  a\right)  \otimes
\operatorname*{id}+\operatorname*{id}\otimes K\left(  a\right)  \right)
\left(  \Delta\left(  u\right)  \right)  .
\end{align*}
Compared with (\ref{pf.thm.gen3.a.1}), this shows that%
\begin{equation}
\left(  \Delta\circ\zeta\right)  \left(  u\right)  =\left(  \zeta
\otimes\operatorname*{id}+\operatorname*{id}\otimes\zeta\right)  \left(
\Delta\left(  u\right)  \right)  =\left(  \left(  \zeta\otimes
\operatorname*{id}+\operatorname*{id}\otimes\zeta\right)  \circ\Delta\right)
\left(  u\right)  . \label{pf.thm.gen3.a.5}%
\end{equation}


Now, let us forget that we fixed $u$. We thus have proven
(\ref{pf.thm.gen3.a.5}) for every $u\in C$. In other words, $\Delta\circ
\zeta=\left(  \zeta\otimes\operatorname*{id}+\operatorname*{id}\otimes
\zeta\right)  \circ\Delta$. In other words, $\zeta$ is a coderivation of $C$.
This completes our proof of Theorem \ref{thm.gen3} \textbf{(a)}.

\textbf{(b)} We need to prove that $\eta$ is a $\mathbf{k}$-coalgebra
homomorphism. In other words, we need to prove that $\Delta\circ\eta=\left(
\eta\otimes\eta\right)  \circ\Delta$ and $\epsilon\circ\eta=\epsilon$. We
shall prove $\Delta\circ\eta=\left(  \eta\otimes\eta\right)  \circ\Delta$ first.

We define a map $\Xi:U\left(  \mathfrak{g}\right)  \rightarrow
\operatorname*{End}C$ by setting%
\[
\left(  \left(  \Xi\left(  p\right)  \right)  \left(  c\right)
=pc\ \ \ \ \ \ \ \ \ \ \text{for every }p\in U\left(  \mathfrak{g}\right)
\text{ and }c\in C\right)  .
\]
Thus, $\Xi$ is a $\mathbf{k}$-algebra homomorphism. (More precisely, $\Xi$ is
the $\mathbf{k}$-algebra homomorphism $U\left(  \mathfrak{g}\right)
\rightarrow\operatorname*{End}C$ which describes the left $U\left(
\mathfrak{g}\right)  $-module $C$.) Hence, $\Xi\otimes\Xi:U\left(
\mathfrak{g}\right)  \otimes U\left(  \mathfrak{g}\right)  \rightarrow
\operatorname*{End}C\otimes\operatorname*{End}C$ is a $\mathbf{k}$-algebra
homomorphism as well.

We let $\mathbf{z}:\operatorname*{End}C\otimes\operatorname*{End}%
C\rightarrow\operatorname*{End}\left(  C\otimes C\right)  $ be the
$\mathbf{k}$-linear map which sends every $f\otimes g\in\operatorname*{End}%
C\otimes\operatorname*{End}C$ to the endomorphism $f\otimes g$ of $C\otimes
C$. This $\mathbf{z}$ is a $\mathbf{k}$-algebra homomorphism as well.

We define a $\mathbf{k}$-linear map $\Xi^{\prime}:U\left(  \mathfrak{g}%
\right)  \rightarrow\operatorname*{End}\left(  C\otimes C\right)  $ by
$\Xi^{\prime}=\mathbf{z}\circ\left(  \Xi\otimes\Xi\right)  \circ\Delta$. This
$\Xi^{\prime}$ is a $\mathbf{k}$-algebra homomorphism (since it is a
composition of three $\mathbf{k}$-algebra homomorphisms).

Now, let $\mathfrak{P}$ be the subset%
\[
\left\{  p\in U\left(  \mathfrak{g}\right)  \ \mid\ \Delta\circ\Xi\left(
p\right)  =\Xi^{\prime}\left(  p\right)  \circ\Delta\right\}
\]
\footnote{Keep in mind that $\Delta\circ\Xi\left(  p\right)  $ means
$\Delta\circ\left(  \Xi\left(  p\right)  \right)  $ rather than $\left(
\Delta\circ\Xi\right)  \left(  p\right)  $.} of $U\left(  \mathfrak{g}\right)
$. Then, $\mathfrak{P}$ is a $\mathbf{k}$-subalgebra of $U\left(
\mathfrak{g}\right)  $\ \ \ \ \footnote{\textit{Proof.} The set $\mathfrak{P}$
is the set of all $p\in U\left(  \mathfrak{g}\right)  $ satisfying the
equation $\Delta\circ\Xi\left(  p\right)  =\Xi^{\prime}\left(  p\right)
\circ\Delta$. Since this equation is $\mathbf{k}$-linear in $p$, this shows
that $\mathfrak{P}$ is a $\mathbf{k}$-submodule of $U\left(  \mathfrak{g}%
\right)  $.
\par
Since $\Xi$ and $\Xi^{\prime}$ are $\mathbf{k}$-algebra homomorphisms, we have
$\Xi\left(  1\right)  =\operatorname*{id}$ and $\Xi^{\prime}\left(  1\right)
=\operatorname*{id}$. Hence, $\Delta\circ\underbrace{\Xi\left(  1\right)
}_{=\operatorname*{id}}=\Delta$ and $\underbrace{\Xi^{\prime}\left(  1\right)
}_{=\operatorname*{id}}\circ\Delta=\Delta$, so that $\Delta\circ\Xi\left(
1\right)  =\Delta=\Xi^{\prime}\left(  1\right)  \circ\Delta$. In other words,
$1\in\mathfrak{P}$ (by the definition of $\mathfrak{P}$).
\par
Now, let $a\in\mathfrak{P}$ and $b\in\mathfrak{P}$. We shall prove that
$ab\in\mathfrak{P}$.
\par
Indeed, we have $a\in\mathfrak{P}$. In other words, $\Delta\circ\Xi\left(
a\right)  =\Xi^{\prime}\left(  a\right)  \circ\Delta$ (by the definition of
$\mathfrak{P}$). Similarly, from $b\in\mathfrak{P}$, we obtain $\Delta\circ
\Xi\left(  b\right)  =\Xi^{\prime}\left(  b\right)  \circ\Delta$. Now,
$\Xi\left(  ab\right)  =\Xi\left(  a\right)  \circ\Xi\left(  b\right)  $
(since $\Xi$ is a $\mathbf{k}$-algebra homomorphism) and $\Xi^{\prime}\left(
ab\right)  =\Xi^{\prime}\left(  a\right)  \circ\Xi^{\prime}\left(  b\right)  $
(since $\Xi^{\prime}$ is a $\mathbf{k}$-algebra homomorphism). Now,%
\[
\Delta\circ\underbrace{\Xi\left(  ab\right)  }_{=\Xi\left(  a\right)  \circ
\Xi\left(  b\right)  }=\underbrace{\Delta\circ\Xi\left(  a\right)  }%
_{=\Xi^{\prime}\left(  a\right)  \circ\Delta}\circ\Xi\left(  b\right)
=\Xi^{\prime}\left(  a\right)  \circ\underbrace{\Delta\circ\Xi\left(
b\right)  }_{=\Xi^{\prime}\left(  b\right)  \circ\Delta}=\underbrace{\Xi
^{\prime}\left(  a\right)  \circ\Xi^{\prime}\left(  b\right)  }_{=\Xi^{\prime
}\left(  ab\right)  }\circ\Delta=\Xi^{\prime}\left(  ab\right)  \circ\Delta.
\]
In other words, $ab\in\mathfrak{P}$ (by the definition of $\mathfrak{P}$).
\par
Now, let us forget that we fixed $a$ and $b$. We thus have proven that
$ab\in\mathfrak{P}$ for every $a\in\mathfrak{P}$ and $b\in\mathfrak{P}$.
Combining this with the facts that $\mathfrak{P}$ is a $\mathbf{k}$-submodule
of $U\left(  \mathfrak{g}\right)  $ and that we have $1\in\mathfrak{P}$, we
conclude that $\mathfrak{P}$ is a $\mathbf{k}$-subalgebra of $U\left(
\mathfrak{g}\right)  $. Qed.} and satisfies $\iota_{U,\mathfrak{g}}\left(
\mathfrak{g}\right)  \subseteq\mathfrak{P}$\ \ \ \ \footnote{\textit{Proof.}
Let $a\in\mathfrak{g}$. We shall show that $\iota_{U,\mathfrak{g}}\left(
a\right)  \in\mathfrak{P}$.
\par
Indeed, let $\zeta$ be the map $C\rightarrow C,\ c\mapsto a\rightharpoonup c$.
Then, this map $\zeta$ is a coderivation of $C$ (by Theorem \ref{thm.gen3}
\textbf{(a)}). In other words, $\Delta\circ\zeta=\left(  \zeta\otimes
\operatorname*{id}+\operatorname*{id}\otimes\zeta\right)  \circ\Delta$.
\par
Now, every $c\in C$ satisfies%
\[
\left(  \Xi\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)  \right)
\left(  c\right)  =\iota_{U,\mathfrak{g}}\left(  a\right)  c=a\rightharpoonup
c=\zeta\left(  c\right)
\]
(since $\zeta\left(  c\right)  $ is defined to be $a\rightharpoonup c$). In
other words, $\Xi\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)
=\zeta$.
\par
On the other hand, $\Xi$ is a $\mathbf{k}$-algebra homomorphism, and thus
$\Xi\left(  1\right)  =\operatorname*{id}$.
\par
We have $\Delta\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)
=\iota_{U,\mathfrak{g}}\left(  a\right)  \otimes1+1\otimes\iota
_{U,\mathfrak{g}}\left(  a\right)  $ (by the definition of the coalgebra
structure on $U\left(  \mathfrak{g}\right)  $) and thus
\begin{align*}
\underbrace{\Xi^{\prime}}_{=\mathbf{z}\circ\left(  \Xi\otimes\Xi\right)
\circ\Delta}\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)   &
=\left(  \mathbf{z}\circ\left(  \Xi\otimes\Xi\right)  \circ\Delta\right)
\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right) \\
&  =\mathbf{z}\left(  \left(  \Xi\otimes\Xi\right)  \underbrace{\left(
\Delta\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)  \right)
}_{=\iota_{U,\mathfrak{g}}\left(  a\right)  \otimes1+1\otimes\iota
_{U,\mathfrak{g}}\left(  a\right)  }\right) \\
&  =\mathbf{z}\left(  \underbrace{\left(  \Xi\otimes\Xi\right)  \left(
\iota_{U,\mathfrak{g}}\left(  a\right)  \otimes1+1\otimes\iota_{U,\mathfrak{g}%
}\left(  a\right)  \right)  }_{=\Xi\left(  \iota_{U,\mathfrak{g}}\left(
a\right)  \right)  \otimes\Xi\left(  1\right)  +\Xi\left(  1\right)
\otimes\Xi\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)  }\right) \\
&  =\mathbf{z}\left(  \underbrace{\Xi\left(  \iota_{U,\mathfrak{g}}\left(
a\right)  \right)  }_{=\zeta}\otimes\underbrace{\Xi\left(  1\right)
}_{=\operatorname*{id}}+\underbrace{\Xi\left(  1\right)  }%
_{=\operatorname*{id}}\otimes\underbrace{\Xi\left(  \iota_{U,\mathfrak{g}%
}\left(  a\right)  \right)  }_{=\zeta}\right) \\
&  =\mathbf{z}\left(  \zeta\otimes\operatorname*{id}+\operatorname*{id}%
\otimes\zeta\right)  =\zeta\otimes\operatorname*{id}+\operatorname*{id}%
\otimes\zeta
\end{align*}
(by the definition of the map $\mathbf{z}$). Now,
\[
\Delta\circ\underbrace{\Xi\left(  \iota_{U,\mathfrak{g}}\left(  a\right)
\right)  }_{=\zeta}=\Delta\circ\zeta=\underbrace{\left(  \zeta\otimes
\operatorname*{id}+\operatorname*{id}\otimes\zeta\right)  }_{=\Xi^{\prime
}\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)  }\circ\Delta
=\Xi^{\prime}\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)
\circ\Delta.
\]
In other words, $\iota_{U,\mathfrak{g}}\left(  a\right)  \in\mathfrak{P}$ (by
the definition of $\mathfrak{P}$).
\par
Now let us forget that we fixed $a$. We thus have proven that $\iota
_{U,\mathfrak{g}}\left(  a\right)  \in\mathfrak{P}$ for every $a\in
\mathfrak{g}$. In other words, $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}%
\right)  \subseteq\mathfrak{P}$, qed.}.

But the subset $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $ generates
the $\mathbf{k}$-algebra $U\left(  \mathfrak{g}\right)  $. In other words,
every $\mathbf{k}$-subalgebra $\mathfrak{B}$ of $U\left(  \mathfrak{g}\right)
$ satisfying $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)
\subseteq\mathfrak{B}$ must satisfy $\mathfrak{B}=U\left(  \mathfrak{g}%
\right)  $. Applying this to $\mathfrak{B}=\mathfrak{P}$, we thus obtain
$\mathfrak{P}=U\left(  \mathfrak{g}\right)  $ (since $\mathfrak{P}$ is a
$\mathbf{k}$-subalgebra of $U\left(  \mathfrak{g}\right)  $ and satisfies
$\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)  \subseteq\mathfrak{P}$).
Thus,%
\[
U\left(  \mathfrak{g}\right)  =\mathfrak{P}=\left\{  p\in U\left(
\mathfrak{g}\right)  \ \mid\ \Delta\circ\Xi\left(  p\right)  =\Xi^{\prime
}\left(  p\right)  \circ\Delta\right\}  .
\]
In other words, every $p\in U\left(  \mathfrak{g}\right)  $ satisfies%
\begin{equation}
\Delta\circ\Xi\left(  p\right)  =\Xi^{\prime}\left(  p\right)  \circ\Delta.
\label{pf.thm.gen3.b.6}%
\end{equation}


But every $q\in U\left(  \mathfrak{g}\right)  \otimes U\left(  \mathfrak{g}%
\right)  $ satisfies
\begin{equation}
\left(  \left(  \mathbf{z}\circ\left(  \Xi\otimes\Xi\right)  \right)  \left(
q\right)  \right)  \left(  1_{C}\otimes1_{C}\right)  =\left(  \eta\otimes
\eta\right)  \left(  q\right)  \label{pf.thm.gen3.b.7}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.gen3.b.7}):} Let $q\in U\left(
\mathfrak{g}\right)  \otimes U\left(  \mathfrak{g}\right)  $. We must prove
the equality (\ref{pf.thm.gen3.b.7}). This equality is $\mathbf{k}$-linear in
$q$. Hence, we can WLOG assume that $q$ is a pure tensor (since the pure
tensors span the $\mathbf{k}$-module $U\left(  \mathfrak{g}\right)  \otimes
U\left(  \mathfrak{g}\right)  $). Assume this. In other words, $q=u\otimes v$
for some $u\in U\left(  \mathfrak{g}\right)  $ and $v\in U\left(
\mathfrak{g}\right)  $. Consider these $u$ and $v$.
\par
The definition of $\Xi$ shows that $\left(  \Xi\left(  u\right)  \right)
\left(  1_{C}\right)  =u1_{C}=\eta\left(  u\right)  $ (since $\eta\left(
u\right)  $ is defined to be $u1_{C}$). Similarly, $\left(  \Xi\left(
v\right)  \right)  \left(  1_{C}\right)  =\eta\left(  v\right)  $. Now,%
\begin{align*}
\left(  \mathbf{z}\circ\left(  \Xi\otimes\Xi\right)  \right)  \left(
\underbrace{q}_{=u\otimes v}\right)   &  =\left(  \mathbf{z}\circ\left(
\Xi\otimes\Xi\right)  \right)  \left(  u\otimes v\right)  =\mathbf{z}\left(
\underbrace{\left(  \Xi\otimes\Xi\right)  \left(  u\otimes v\right)  }%
_{=\Xi\left(  u\right)  \otimes\Xi\left(  v\right)  }\right) \\
&  =\mathbf{z}\left(  \Xi\left(  u\right)  \otimes\Xi\left(  v\right)
\right)  =\Xi\left(  u\right)  \otimes\Xi\left(  v\right)
\end{align*}
(by the definition of $\mathbf{z}$). Hence,%
\begin{align*}
\left(  \underbrace{\left(  \mathbf{z}\circ\left(  \Xi\otimes\Xi\right)
\right)  \left(  q\right)  }_{=\Xi\left(  u\right)  \otimes\Xi\left(
v\right)  }\right)  \left(  1_{C}\otimes1_{C}\right)   &  =\left(  \Xi\left(
u\right)  \otimes\Xi\left(  v\right)  \right)  \left(  1_{C}\otimes
1_{C}\right)  =\underbrace{\left(  \Xi\left(  u\right)  \right)  \left(
1_{C}\right)  }_{=\eta\left(  u\right)  }\otimes\underbrace{\left(  \Xi\left(
v\right)  \right)  \left(  1_{C}\right)  }_{=\eta\left(  v\right)  }\\
&  =\eta\left(  u\right)  \otimes\eta\left(  v\right)  =\left(  \eta
\otimes\eta\right)  \left(  \underbrace{u\otimes v}_{=q}\right)  =\left(
\eta\otimes\eta\right)  \left(  q\right)  .
\end{align*}
This proves (\ref{pf.thm.gen3.b.7}).}.

Now, let $p\in U\left(  \mathfrak{g}\right)  $. We shall prove that $\left(
\Delta\circ\eta\right)  \left(  p\right)  =\left(  \left(  \eta\otimes
\eta\right)  \circ\Delta\right)  \left(  p\right)  $.

The definition of $\Xi$ shows that $\left(  \Xi\left(  p\right)  \right)
\left(  1_{C}\right)  =p1_{C}=\eta\left(  p\right)  $ (since $\eta\left(
p\right)  $ is defined to be $p1_{C}$). Thus,%
\begin{align*}
\left(  \Delta\circ\eta\right)  \left(  p\right)   &  =\Delta\left(
\underbrace{\eta\left(  p\right)  }_{=\left(  \Xi\left(  p\right)  \right)
\left(  1_{C}\right)  }\right)  =\Delta\left(  \left(  \Xi\left(  p\right)
\right)  \left(  1_{C}\right)  \right)  =\underbrace{\left(  \Delta\circ
\Xi\left(  p\right)  \right)  }_{\substack{=\Xi^{\prime}\left(  p\right)
\circ\Delta\\\text{(by (\ref{pf.thm.gen3.b.6}))}}}\left(  1_{C}\right) \\
&  =\left(  \Xi^{\prime}\left(  p\right)  \circ\Delta\right)  \left(
1_{C}\right)  =\left(  \underbrace{\Xi^{\prime}}_{=\mathbf{z}\circ\left(
\Xi\otimes\Xi\right)  \circ\Delta}\left(  p\right)  \right)  \left(
\underbrace{\Delta\left(  1_{C}\right)  }_{=1_{C}\otimes1_{C}}\right) \\
&  =\underbrace{\left(  \left(  \mathbf{z}\circ\left(  \Xi\otimes\Xi\right)
\circ\Delta\right)  \left(  p\right)  \right)  }_{=\left(  \mathbf{z}%
\circ\left(  \Xi\otimes\Xi\right)  \right)  \left(  \Delta\left(  p\right)
\right)  }\left(  1_{C}\otimes1_{C}\right)  =\left(  \left(  \mathbf{z}%
\circ\left(  \Xi\otimes\Xi\right)  \right)  \left(  \Delta\left(  p\right)
\right)  \right)  \left(  1_{C}\otimes1_{C}\right) \\
&  =\left(  \eta\otimes\eta\right)  \left(  \Delta\left(  p\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.gen3.b.7}), applied to
}q=\Delta\left(  p\right)  \right) \\
&  =\left(  \left(  \eta\otimes\eta\right)  \circ\Delta\right)  \left(
p\right)  .
\end{align*}


Now, let us forget that we fixed $p$. We thus have proven that $\left(
\Delta\circ\eta\right)  \left(  p\right)  =\left(  \left(  \eta\otimes
\eta\right)  \circ\Delta\right)  \left(  p\right)  $ for every $p\in U\left(
\mathfrak{g}\right)  $. In other words, $\Delta\circ\eta=\left(  \eta
\otimes\eta\right)  \circ\Delta$.

It remains to prove that $\epsilon\circ\eta=\epsilon$.

We let $\mathfrak{R}$ be the subset%
\[
\left\{  p\in U\left(  \mathfrak{g}\right)  \ \mid\ \epsilon\circ\Xi\left(
p\right)  =\epsilon\left(  p\right)  \cdot\epsilon\right\}
\]
of $U\left(  \mathfrak{g}\right)  $. Then, $\mathfrak{R}$ is a $\mathbf{k}%
$-subalgebra of $U\left(  \mathfrak{g}\right)  $%
\ \ \ \ \footnote{\textit{Proof.} The set $\mathfrak{R}$ is the set of all
$p\in U\left(  \mathfrak{g}\right)  $ satisfying the equation $\epsilon
\circ\Xi\left(  p\right)  =\epsilon\left(  p\right)  \cdot\epsilon$. Since
this equation is $\mathbf{k}$-linear in $p$, this shows that $\mathfrak{R}$ is
a $\mathbf{k}$-submodule of $U\left(  \mathfrak{g}\right)  $.
\par
Since $\Xi$ and $\epsilon$ are $\mathbf{k}$-algebra homomorphisms, we have
$\Xi\left(  1\right)  =\operatorname*{id}$ and $\epsilon\left(  1\right)  =1$.
Hence, $\epsilon\circ\underbrace{\Xi\left(  1\right)  }_{=\operatorname*{id}%
}=\epsilon$ and $\underbrace{\epsilon\left(  1\right)  }_{=1}\cdot
\epsilon=\epsilon$, so that $\epsilon\circ\Xi\left(  1\right)  =\epsilon
=\epsilon\left(  1\right)  \cdot\epsilon$. In other words, $1\in\mathfrak{R}$
(by the definition of $\mathfrak{R}$).
\par
Now, let $a\in\mathfrak{R}$ and $b\in\mathfrak{R}$. We shall prove that
$ab\in\mathfrak{R}$.
\par
Indeed, we have $a\in\mathfrak{R}$. In other words, $\epsilon\circ\Xi\left(
a\right)  =\epsilon\left(  a\right)  \cdot\epsilon$ (by the definition of
$\mathfrak{R}$). Similarly, from $b\in\mathfrak{R}$, we obtain $\epsilon
\circ\Xi\left(  b\right)  =\epsilon\left(  b\right)  \cdot\epsilon$. Now,
$\Xi\left(  ab\right)  =\Xi\left(  a\right)  \circ\Xi\left(  b\right)  $
(since $\Xi$ is a $\mathbf{k}$-algebra homomorphism) and $\epsilon\left(
ab\right)  =\epsilon\left(  a\right)  \cdot\epsilon\left(  b\right)  $ (since
$\epsilon$ is a $\mathbf{k}$-algebra homomorphism). Now,%
\[
\epsilon\circ\underbrace{\Xi\left(  ab\right)  }_{=\Xi\left(  a\right)
\circ\Xi\left(  b\right)  }=\underbrace{\epsilon\circ\Xi\left(  a\right)
}_{=\epsilon\left(  a\right)  \cdot\epsilon}\circ\Xi\left(  b\right)
=\epsilon\left(  a\right)  \cdot\underbrace{\epsilon\circ\Xi\left(  b\right)
}_{=\epsilon\left(  b\right)  \cdot\epsilon}=\underbrace{\epsilon\left(
a\right)  \cdot\epsilon\left(  b\right)  }_{=\epsilon\left(  ab\right)  }%
\cdot\epsilon=\epsilon\left(  ab\right)  \cdot\epsilon.
\]
In other words, $ab\in\mathfrak{R}$ (by the definition of $\mathfrak{R}$).
\par
Now, let us forget that we fixed $a$ and $b$. We thus have proven that
$ab\in\mathfrak{R}$ for every $a\in\mathfrak{R}$ and $b\in\mathfrak{R}$.
Combining this with the facts that $\mathfrak{R}$ is a $\mathbf{k}$-submodule
of $U\left(  \mathfrak{g}\right)  $ and that we have $1\in\mathfrak{R}$, we
conclude that $\mathfrak{R}$ is a $\mathbf{k}$-subalgebra of $U\left(
\mathfrak{g}\right)  $. Qed.} and satisfies $\iota_{U,\mathfrak{g}}\left(
\mathfrak{g}\right)  \subseteq\mathfrak{R}$\ \ \ \ \footnote{\textit{Proof.}
Let $a\in\mathfrak{g}$. We shall show that $\iota_{U,\mathfrak{g}}\left(
a\right)  \in\mathfrak{R}$.
\par
Indeed, let $\zeta$ be the map $C\rightarrow C,\ c\mapsto a\rightharpoonup c$.
Then, this map $\zeta$ is a coderivation of $C$ (by Theorem \ref{thm.gen3}
\textbf{(a)}). Hence, $\epsilon\circ\zeta=0$ (by Proposition
\ref{prop.coder.epsi}, applied to $f=\zeta$).
\par
Now, every $c\in C$ satisfies%
\[
\left(  \Xi\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)  \right)
\left(  c\right)  =\iota_{U,\mathfrak{g}}\left(  a\right)  c=a\rightharpoonup
c=\zeta\left(  c\right)
\]
(since $\zeta\left(  c\right)  $ is defined to be $a\rightharpoonup c$). In
other words, $\Xi\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)
=\zeta$.
\par
We have $\epsilon\left(  \iota_{U,\mathfrak{g}}\left(  a\right)  \right)  =0$
(by the definition of the coalgebra structure on $U\left(  \mathfrak{g}%
\right)  $) and thus $\underbrace{\epsilon\left(  \iota_{U,\mathfrak{g}%
}\left(  a\right)  \right)  }_{=0}\cdot\epsilon=0$. Now,
\[
\epsilon\circ\underbrace{\Xi\left(  \iota_{U,\mathfrak{g}}\left(  a\right)
\right)  }_{=\zeta}=\epsilon\circ\zeta=0=\epsilon\left(  \iota_{U,\mathfrak{g}%
}\left(  a\right)  \right)  \cdot\epsilon.
\]
In other words, $\iota_{U,\mathfrak{g}}\left(  a\right)  \in\mathfrak{R}$ (by
the definition of $\mathfrak{R}$).
\par
Now let us forget that we fixed $a$. We thus have proven that $\iota
_{U,\mathfrak{g}}\left(  a\right)  \in\mathfrak{R}$ for every $a\in
\mathfrak{g}$. In other words, $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}%
\right)  \subseteq\mathfrak{R}$, qed.}.

But the subset $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $ generates
the $\mathbf{k}$-algebra $U\left(  \mathfrak{g}\right)  $. In other words,
every $\mathbf{k}$-subalgebra $\mathfrak{B}$ of $U\left(  \mathfrak{g}\right)
$ satisfying $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)
\subseteq\mathfrak{B}$ must satisfy $\mathfrak{B}=U\left(  \mathfrak{g}%
\right)  $. Applying this to $\mathfrak{B}=\mathfrak{R}$, we thus obtain
$\mathfrak{R}=U\left(  \mathfrak{g}\right)  $ (since $\mathfrak{R}$ is a
$\mathbf{k}$-subalgebra of $U\left(  \mathfrak{g}\right)  $ and satisfies
$\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)  \subseteq\mathfrak{R}$).
Thus,%
\[
U\left(  \mathfrak{g}\right)  =\mathfrak{R}=\left\{  p\in U\left(
\mathfrak{g}\right)  \ \mid\ \epsilon\circ\Xi\left(  p\right)  =\epsilon
\left(  p\right)  \cdot\epsilon\right\}  .
\]
In other words, every $p\in U\left(  \mathfrak{g}\right)  $ satisfies%
\begin{equation}
\epsilon\circ\Xi\left(  p\right)  =\epsilon\left(  p\right)  \cdot\epsilon.
\label{pf.thm.gen3.b.16}%
\end{equation}


Now, let $p\in U\left(  \mathfrak{g}\right)  $. We shall prove that $\left(
\epsilon\circ\eta\right)  \left(  p\right)  =\epsilon\left(  p\right)  $.

The definition of $\Xi$ shows that $\left(  \Xi\left(  p\right)  \right)
\left(  1_{C}\right)  =p1_{C}=\eta\left(  p\right)  $ (since $\eta\left(
p\right)  $ is defined to be $p1_{C}$). Thus,%
\begin{align*}
\left(  \epsilon\circ\eta\right)  \left(  p\right)   &  =\epsilon\left(
\underbrace{\eta\left(  p\right)  }_{=\left(  \Xi\left(  p\right)  \right)
\left(  1_{C}\right)  }\right)  =\epsilon\left(  \left(  \Xi\left(  p\right)
\right)  \left(  1_{C}\right)  \right)  =\underbrace{\left(  \epsilon\circ
\Xi\left(  p\right)  \right)  }_{\substack{=\epsilon\left(  p\right)
\cdot\epsilon\\\text{(by (\ref{pf.thm.gen3.b.16}))}}}\left(  1_{C}\right) \\
&  =\left(  \epsilon\left(  p\right)  \cdot\epsilon\right)  \left(
1_{C}\right)  =\epsilon\left(  p\right)  \cdot\underbrace{\epsilon\left(
1_{C}\right)  }_{=1}=\epsilon\left(  p\right)  .
\end{align*}


Now, let us forget that we fixed $p$. We thus have proven that $\left(
\epsilon\circ\eta\right)  \left(  p\right)  =\epsilon\left(  p\right)  $ for
every $p\in U\left(  \mathfrak{g}\right)  $. In other words, $\epsilon
\circ\eta=\epsilon$. As explained above, this concludes the proof of Theorem
\ref{thm.gen3} \textbf{(b)}.
\end{proof}

\section{\label{sect.prelude}Filtrations, gradings and the prelude to PBW}

In this section, we shall recall some fundamental definitions and facts
pertaining to filtered and graded $\mathbf{k}$-modules and $\mathbf{k}%
$-algebras, and specifically to the (standard) filtration on the universal
enveloping algebra $U\left(  \mathfrak{g}\right)  $ and its relation to the
symmetric algebra $\operatorname*{Sym}\mathfrak{g}$. All of what is said and
proved in this section is shallow and well-known (and so the section can
probably be skimmed at a high speed); we nevertheless prefer to be maximally
explicit and detailed (although we do not prove the most fundamental results,
whose proofs are straightforward and should be in basic abstract algebra
texts) in order to fix the terminology (which, unfortunately, is not standard
across different texts and authors).

\subsection{Definitions on filtered and graded $\mathbf{k}$-modules}

We shall now make some definitions concerned with filtered and graded
algebras. These definitions are all classical, but the notations are not
standardized across the literature (as many authors use the words
\textquotedblleft filtered\textquotedblright\ and \textquotedblleft
graded\textquotedblright\ in different meanings).

\begin{definition}
\label{def.filtered-mod}\textbf{(a)} Let $V$ be a $\mathbf{k}$-module. A
$\mathbf{k}$\textit{-module filtration of }$V$ will mean a sequence $\left(
V_{n}\right)  _{n\geq0}$ of $\mathbf{k}$-submodules of $V$ such that
$\bigcup\limits_{n\geq0}V_{n}=V$ and $V_{0}\subseteq V_{1}\subseteq
V_{2}\subseteq\cdots$.

\textbf{(b)} A \textit{filtered }$\mathbf{k}$\textit{-module} means a
$\mathbf{k}$-module $V$ equipped with a $\mathbf{k}$-module filtration of $V$.
This $\mathbf{k}$-module filtration will be referred to as the
\textit{filtration of }$V$. By abuse of notation, we will often use the same
letter $V$ for this filtered $\mathbf{k}$-module and for its underlying
$\mathbf{k}$-module $V$, when the filtration is clear from the context.

\textbf{(c)} Let $V$ and $W$ be two filtered $\mathbf{k}$-modules. Let
$\left(  V_{n}\right)  _{n\geq0}$ and $\left(  W_{n}\right)  _{n\geq0}$ be the
filtrations of $V$ and $W$, respectively. Let $f:V\rightarrow W$ be a
$\mathbf{k}$-linear map. We say that the map $f$ is \textit{filtered} if and
only if every $n\in\mathbb{N}$ satisfies $f\left(  V_{n}\right)  \subseteq
W_{n}$.

(Instead of saying that \textquotedblleft the map $f$ is
filtered\textquotedblright, some authors say that \textquotedblleft the map
$f$ respects the filtrations $\left(  V_{n}\right)  _{n\geq0}$ and $\left(
W_{n}\right)  _{n\geq0}$\textquotedblright\ or that \textquotedblleft the map
$f$ preserves the filtration\textquotedblright.)
\end{definition}

The filtered $\mathbf{k}$-modules form a category, whose morphisms are the
filtered $\mathbf{k}$-linear maps.\footnote{Nota bene: An isomorphism in this
category is an invertible filtered $\mathbf{k}$-linear map whose inverse is
again filtered. The \textquotedblleft whose inverse is again
filtered\textquotedblright\ requirement cannot be dropped: Not every
invertible filtered $\mathbf{k}$-linear map is an isomorphism in this
category!}

\begin{definition}
\label{def.graded-mod}\textbf{(a)} Let $V$ be a $\mathbf{k}$-module. A
$\mathbf{k}$\textit{-module grading} on $V$ will mean a sequence $\left(
V_{n}\right)  _{n\geq0}$ of $\mathbf{k}$-submodules of $V$ such that
$\bigoplus\limits_{n\geq0}V_{n}=V$. (Here, the direct sum $\bigoplus
\limits_{n\geq0}V_{n}$ is internal.)

\textbf{(b)} A \textit{graded }$\mathbf{k}$\textit{-module} means a
$\mathbf{k}$-module $V$ equipped with a $\mathbf{k}$-module grading on $V$.
This $\mathbf{k}$-module grading will be referred to as the \textit{grading of
}$V$. By abuse of notation, we will often use the same letter $V$ for this
graded $\mathbf{k}$-module and for its underlying $\mathbf{k}$-module $V$,
when the grading is clear from the context.

\textbf{(c)} Let $V$ and $W$ be two graded $\mathbf{k}$-modules. Let $\left(
V_{n}\right)  _{n\geq0}$ and $\left(  W_{n}\right)  _{n\geq0}$ be the gradings
of $V$ and $W$, respectively. Let $f:V\rightarrow W$ be a $\mathbf{k}$-linear
map. We say that the map $f$ is \textit{graded} if and only if every
$n\in\mathbb{N}$ satisfies $f\left(  V_{n}\right)  \subseteq W_{n}$.

(Instead of saying that \textquotedblleft the map $f$ is
graded\textquotedblright, some authors say that \textquotedblleft the map $f$
respects the gradings $\left(  V_{n}\right)  _{n\geq0}$ and $\left(
W_{n}\right)  _{n\geq0}$\textquotedblright\ or that \textquotedblleft the map
$f$ preserves the degree\textquotedblright.)
\end{definition}

The graded $\mathbf{k}$-modules form a category, whose morphisms are the
graded $\mathbf{k}$-linear maps.\footnote{This time, isomorphisms in this
category are easy to characterize: They are the invertible graded $\mathbf{k}%
$-linear maps. Thus, the inverse of any invertible graded $\mathbf{k}$-linear
map is again graded. (This is not hard to check.)}

\begin{definition}
\label{def.filtered-from-graded}Let $V$ be a graded $\mathbf{k}$-module. Let
$\left(  V_{n}\right)  _{n\geq0}$ be the grading of $V$. Thus, $\bigoplus
\limits_{n\geq0}V_{n}=V$. For every $m\in\mathbb{N}$, define a $\mathbf{k}%
$-submodule $V_{\leq m}$ of $V$ by%
\[
V_{\leq m}=\bigoplus\limits_{n=0}^{m}V_{n}.
\]
(This is well-defined, because $\bigoplus\limits_{n=0}^{m}V_{n}$ is a subsum
of the direct sum $\bigoplus\limits_{n\geq0}V_{n}$.) Then, $\left(  V_{\leq
n}\right)  _{n\geq0}$ is a filtration of the $\mathbf{k}$-module $V$. Equipped
with this filtration, $V$ becomes a filtered $\mathbf{k}$-module. We denote
this filtered $\mathbf{k}$-module again by $V$, since it is defined
canonically in terms of $V$. Thus, every graded $\mathbf{k}$-module $V$
canonically becomes a filtered $\mathbf{k}$-module.
\end{definition}

\begin{remark}
\label{rmk.filtered-from-graded.map}Let $V$ and $W$ be two graded $\mathbf{k}%
$-modules. Thus, both $V$ and $W$ canonically become filtered $\mathbf{k}%
$-modules (according to Definition \ref{def.filtered-from-graded}). Every
graded $\mathbf{k}$-linear map $f:V\rightarrow W$ between the graded
$\mathbf{k}$-modules $V$ and $W$ is automatically a filtered $\mathbf{k}%
$-linear map between the filtered $\mathbf{k}$-modules $V$ and $W$.
\end{remark}

\begin{definition}
\label{def.gr}Let $V$ be a filtered $\mathbf{k}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be the filtration of $V$. We set $V_{-1}$ to be the
$\mathbf{k}$-submodule $0$ of $V$; thus, $V_{-1}\subseteq V_{0}\subseteq
V_{1}\subseteq\cdots$. For every $m\in\mathbb{N}$, let $\operatorname*{gr}%
\nolimits_{m}V$ denote the quotient module $V_{m}/V_{m-1}$. Let
$\operatorname*{gr}V$ denote the graded $\mathbf{k}$-module $\bigoplus
\limits_{m\geq0}\operatorname*{gr}\nolimits_{m}V$, equipped with the grading
$\left(  \operatorname*{gr}\nolimits_{m}V\right)  _{m\in\mathbb{N}}$. (Here,
of course, we identify each $\operatorname*{gr}\nolimits_{m}V$ with the
corresponding submodule of the direct sum $\bigoplus\limits_{m\geq
0}\operatorname*{gr}\nolimits_{m}V$.) This graded $\mathbf{k}$-module
$\operatorname*{gr}V$ is known as the \textit{associated graded module} of the
filtered $\mathbf{k}$-module $V$.

For every $m\in\mathbb{N}$ and every $v\in V_{m}$, we let $\left[  v\right]
_{m}$ denote the residue class $v+V_{m-1}$ of $v\in V_{m}$ in $V_{m}%
/V_{m-1}=\operatorname*{gr}\nolimits_{m}V$.
\end{definition}

\begin{definition}
\label{def.gr.map}Let $V$ and $W$ be two filtered $\mathbf{k}$-modules. Then,
two graded $\mathbf{k}$-modules $\operatorname*{gr}V$ and $\operatorname*{gr}%
W$ are defined according to Definition \ref{def.gr}.

Let $f:V\rightarrow W$ be a filtered $\mathbf{k}$-linear map. Then, for every
$m\in\mathbb{N}$, we define a $\mathbf{k}$-linear map $\operatorname*{gr}%
\nolimits_{m}f:\operatorname*{gr}\nolimits_{m}V\rightarrow\operatorname*{gr}%
\nolimits_{m}W$ as follows:

Let $\left(  V_{n}\right)  _{n\geq0}$ and $\left(  W_{n}\right)  _{n\geq0}$ be
the filtrations of $V$ and $W$, respectively. Then, $f\left(  V_{m}\right)
\subseteq W_{m}$ (since the map $f$ is filtered). Hence, the map $f$ restricts
to a $\mathbf{k}$-linear map $V_{m}\rightarrow W_{m}$. This $\mathbf{k}%
$-linear map sends $V_{m-1}$ to $W_{m-1}$ (since $f\left(  V_{m-1}\right)
\subseteq W_{m-1}$ (again because $f$ is filtered, and since $V_{-1}=0$)), and
thus gives rise to a $\mathbf{k}$-linear map%
\[
\varphi:V_{m}/V_{m-1}\rightarrow W_{m}/W_{m-1},\ \ \ \ \ \ \ \ \ \ \left[
v\right]  _{m}\mapsto\left[  f\left(  v\right)  \right]  _{m}.
\]
We denote this $\mathbf{k}$-linear map $\varphi$ by $\operatorname*{gr}%
\nolimits_{m}f$. Thus, $\operatorname*{gr}\nolimits_{m}f$ is a $\mathbf{k}%
$-linear map $V_{m}/V_{m-1}\rightarrow W_{m}/W_{m-1}$. In other words,
$\operatorname*{gr}\nolimits_{m}f$ is a $\mathbf{k}$-linear map
$\operatorname*{gr}\nolimits_{m}V\rightarrow\operatorname*{gr}\nolimits_{m}W$
(since $\operatorname*{gr}\nolimits_{m}V=V_{m}/V_{m-1}$ and
$\operatorname*{gr}\nolimits_{m}W=W_{m}/W_{m-1}$).

Now, we have defined a $\mathbf{k}$-linear map $\operatorname*{gr}%
\nolimits_{m}f:\operatorname*{gr}\nolimits_{m}V\rightarrow\operatorname*{gr}%
\nolimits_{m}W$ for every $m\in\mathbb{N}$. This $\operatorname*{gr}%
\nolimits_{m}f$ satisfies%
\[
\left(  \operatorname*{gr}\nolimits_{m}f\right)  \left(  \left[  v\right]
_{m}\right)  =\left[  f\left(  v\right)  \right]  _{m}%
\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}\text{ and }v\in V_{m}.
\]


We finally define a $\mathbf{k}$-linear map $\operatorname*{gr}f:\bigoplus
\limits_{m\geq0}\operatorname*{gr}\nolimits_{m}V\rightarrow\bigoplus
\limits_{m\geq0}\operatorname*{gr}\nolimits_{m}W$ by $\operatorname*{gr}%
f=\bigoplus\limits_{m\geq0}\operatorname*{gr}\nolimits_{m}f$. Thus,
$\operatorname*{gr}f$ is a graded $\mathbf{k}$-linear map from
$\operatorname*{gr}V$ to $\operatorname*{gr}W$ (because $\bigoplus
\limits_{m\geq0}\operatorname*{gr}\nolimits_{m}V=\operatorname*{gr}V$ and
$\bigoplus\limits_{m\geq0}\operatorname*{gr}\nolimits_{m}W=\operatorname*{gr}%
W$). It satisfies%
\[
\left(  \operatorname*{gr}f\right)  \left(  \left[  v\right]  _{m}\right)
=\left[  f\left(  v\right)  \right]  _{m}\ \ \ \ \ \ \ \ \ \ \text{for every
}m\in\mathbb{N}\text{ and }v\in V_{m}.
\]

\end{definition}

Definition \ref{def.filtered-from-graded} shows a way to interpret every
graded $\mathbf{k}$-module as a filtered $\mathbf{k}$-module (though the
grading cannot be reconstructed from the filtration, and thus we lose
information when we replace the former by the latter). Together with Remark
\ref{rmk.filtered-from-graded.map} (which shows what happens to graded
$\mathbf{k}$-linear maps under this interpretation), it therefore defines a
functor from the category of graded $\mathbf{k}$-modules (and graded
$\mathbf{k}$-linear maps) to the category of filtered $\mathbf{k}$-modules
(and filtered $\mathbf{k}$-linear maps). This functor can be regarded as a
forgetful functor, since it preserves the underlying set of the graded
$\mathbf{k}$-module (although it is not a forgetful functor in the strict
meaning of this word, since it replace the grading by a filtration, which is
\textbf{not} the grading).

Definition \ref{def.gr} shows how to construct a graded $\mathbf{k}$-module
from any filtered $\mathbf{k}$-module (although not on the same underlying
set, and again, at the cost of losing information). Together with Definition
\ref{def.gr.map} (which constructs a graded $\mathbf{k}$-linear map from any
filtered $\mathbf{k}$-linear map), it thus defines a functor from the category
of filtered $\mathbf{k}$-modules (and filtered $\mathbf{k}$-linear maps) to
the category of graded $\mathbf{k}$-modules (and graded $\mathbf{k}$-linear
maps). This functor does not preserve the underlying set, and it often has the
effect of making the underlying set \textquotedblleft
coarser\textquotedblright\ in an appropriate informal sense of this word.

So we have found two functors: one from the category of graded $\mathbf{k}%
$-modules to the category of filtered $\mathbf{k}$-modules, and one in the
opposite direction. Neither of these two functors is an isomorphism; however,
applying the latter after the former gives a functor naturally isomorphic to
the identity:

\begin{definition}
\label{def.grad}Let $V$ be a graded $\mathbf{k}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be the grading of $V$. Consider the $\mathbf{k}%
$-submodules $V_{\leq m}$ for all $m\in\mathbb{N}$ (defined as in Definition
\ref{def.filtered-from-graded}). Thus, $V$ becomes a filtered $\mathbf{k}%
$-module (equipped with the filtration $\left(  V_{\leq n}\right)  _{n\geq0}%
$). Hence, the graded $\mathbf{k}$-module $\operatorname*{gr}V$ is
well-defined (according to Definition \ref{def.gr}). For every $m\in
\mathbb{N}$, we define a $\mathbf{k}$-linear map%
\[
\operatorname*{grad}\nolimits_{m,V}:V_{m}\rightarrow V_{\leq m}/V_{\leq
m-1},\ \ \ \ \ \ \ \ \ \ v\mapsto\left[  v\right]  _{m}%
\]
(making use of the fact that $v\in V_{m}\subseteq V_{\leq m}$ for every $v\in
V_{m}$). It is easy to see that this map $\operatorname*{grad}\nolimits_{m,V}$
is a $\mathbf{k}$-module isomorphism.

Thus, for every $m\in\mathbb{N}$, the map $\operatorname*{grad}\nolimits_{m,V}%
$ is a $\mathbf{k}$-module isomorphism from $V_{m}$ to $V_{\leq m}/V_{\leq
m-1}=\operatorname*{gr}\nolimits_{m}V$. Hence, $\bigoplus\limits_{m\geq
0}\operatorname*{grad}\nolimits_{m,V}$ is a graded $\mathbf{k}$-module
isomorphism from $V$ to $\operatorname*{gr}V$ (since $V=\bigoplus
\limits_{m\geq0}V_{m}$ with grading $\left(  V_{m}\right)  _{m\geq0}$, and
since $\operatorname*{gr}V=\bigoplus\limits_{m\geq0}\operatorname*{gr}%
\nolimits_{m}V$ with grading $\left(  \operatorname*{gr}\nolimits_{m}V\right)
_{m\geq0}$). We denote this graded $\mathbf{k}$-module isomorphism by
$\operatorname*{grad}\nolimits_{V}$.

Thus, $\operatorname*{grad}\nolimits_{V}$ is a graded $\mathbf{k}$-module
isomorphism from $V$ to $\operatorname*{gr}V$. It satisfies%
\[
\operatorname*{grad}\nolimits_{V}\left(  v\right)  =\left[  v\right]
_{m}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}\text{ and }v\in
V_{m}.
\]
The inverse $\operatorname*{grad}\nolimits_{V}^{-1}$ of $\operatorname*{grad}%
\nolimits_{V}$ is also a graded $\mathbf{k}$-module isomorphism.
\end{definition}

\begin{remark}
\label{rmk.grad.map}Let $V$ and $W$ be two graded $\mathbf{k}$-modules. Let
$f:V\rightarrow W$ be a graded $\mathbf{k}$-linear map. Then, the diagram%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix@C=5pc{
%V \ar[r]^f \ar[d]_{\operatorname{grad}_V} & W \ar[d]^{\operatorname{grad}_W}
%\\
%\operatorname{gr} V \ar[r]_{\operatorname{gr} f} & \operatorname{gr} W
%}}}%
%BeginExpansion
\xymatrix@C=5pc{
V \ar[r]^f \ar[d]_{\operatorname{grad}_V} & W \ar[d]^{\operatorname{grad}_W}
\\
\operatorname{gr} V \ar[r]_{\operatorname{gr} f} & \operatorname{gr} W
}%
%EndExpansion
\]
is commutative.
\end{remark}

The following (rather well-known) theorem is a basic property of filtered
maps; it allows us to use filtrations to prove that certain maps are invertible:

\begin{theorem}
\label{thm.filtered.isocrit}Let $V$ and $W$ be two filtered $\mathbf{k}%
$-modules. Let $f:V\rightarrow W$ be a filtered $\mathbf{k}$-linear map.
Assume that the map $\operatorname*{gr}f:\operatorname*{gr}V\rightarrow
\operatorname*{gr}W$ is invertible. Then, the map $f$ is invertible, and its
inverse $f^{-1}$ is again a filtered $\mathbf{k}$-linear map.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.filtered.isocrit}.]The map $\operatorname*{gr}%
f:\operatorname*{gr}V\rightarrow\operatorname*{gr}W$ is graded, $\mathbf{k}%
$-linear and invertible. Hence, its inverse $\left(  \operatorname*{gr}%
f\right)  ^{-1}$ also is a graded invertible $\mathbf{k}$-linear map.

Let $\left(  V_{n}\right)  _{n\geq0}$ be the filtration on $V$. Let $\left(
W_{n}\right)  _{n\geq0}$ be the filtration on $W$. Thus,%
\begin{equation}
f\left(  V_{n}\right)  \subseteq W_{n}\ \ \ \ \ \ \ \ \ \ \text{for every
}n\in\mathbb{N} \label{pf.thm.filtered.isocrit.filt}%
\end{equation}
(since the map $f$ is filtered). As usual, for every negative integer $n$, we
set $V_{n}=0$ and $W_{n}=0$.

Now,%
\begin{equation}
W_{n}=f\left(  V_{n}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
n\in\mathbb{N}\cup\left\{  -1\right\}  \label{pf.thm.filtered.isocrit.surj.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.filtered.isocrit.surj.1}):} We shall
prove (\ref{pf.thm.filtered.isocrit.surj.1}) by induction over $n$:
\par
\textit{Induction base:} We have $W_{-1}=0$. Compared with $f\left(
\underbrace{V_{-1}}_{=0}\right)  =f\left(  0\right)  =0$, this yields
$W_{-1}=f\left(  V_{-1}\right)  $. In other words,
(\ref{pf.thm.filtered.isocrit.surj.1}) holds for $n=-1$. This completes the
induction base.
\par
\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
(\ref{pf.thm.filtered.isocrit.surj.1}) holds for $n=N-1$. We must prove that
(\ref{pf.thm.filtered.isocrit.surj.1}) holds for $n=N$.
\par
Let $w\in W_{N}$. Thus, $\left[  w\right]  _{N}\in\operatorname*{gr}%
\nolimits_{N}W$ is well-defined. Since the map $\left(  \operatorname*{gr}%
f\right)  ^{-1}$ is graded, we have $\left(  \operatorname*{gr}f\right)
^{-1}\left(  \left[  w\right]  _{N}\right)  \in\operatorname*{gr}%
\nolimits_{N}V$ (since $\left[  w\right]  _{N}\in\operatorname*{gr}%
\nolimits_{N}W$). Thus, $\left(  \operatorname*{gr}f\right)  ^{-1}\left(
\left[  w\right]  _{N}\right)  \in\operatorname*{gr}\nolimits_{N}%
V=V_{N}/V_{N-1}$. In other words, there exists some $v\in V_{N}$ such that
$\left(  \operatorname*{gr}f\right)  ^{-1}\left(  \left[  w\right]
_{N}\right)  =\left[  v\right]  _{N}$. Consider this $v$.
\par
From $\left(  \operatorname*{gr}f\right)  ^{-1}\left(  \left[  w\right]
_{N}\right)  =\left[  v\right]  _{N}$, we obtain $\left[  w\right]
_{N}=\left(  \operatorname*{gr}f\right)  \left(  \left[  v\right]
_{N}\right)  =\left[  f\left(  v\right)  \right]  _{N}$ (by the definition of
$\operatorname*{gr}f$). In other words, $w\equiv f\left(  v\right)
\operatorname{mod}W_{N-1}$. In other words, $w-f\left(  v\right)  \in W_{N-1}%
$.
\par
We assumed that (\ref{pf.thm.filtered.isocrit.surj.1}) holds for $n=N-1$. In
other words, $W_{N-1}=f\left(  V_{N-1}\right)  $. Hence, $w-f\left(  v\right)
\in W_{N-1}=f\left(  V_{N-1}\right)  $. Thus,%
\[
w\in f\left(  \underbrace{v}_{\in V_{N}}\right)  +f\left(  \underbrace{V_{N-1}%
}_{\subseteq V_{N}}\right)  \subseteq f\left(  V_{N}\right)  +f\left(
V_{N}\right)  \subseteq f\left(  V_{N}\right)
\]
(since $f\left(  V_{N}\right)  $ is a $\mathbf{k}$-module).
\par
Let us now forget that we fixed $w$. We thus have shown that $w\in f\left(
V_{N}\right)  $ for every $w\in W_{N}$. In other words, $W_{N}\subseteq
f\left(  V_{N}\right)  $. Combined with $f\left(  V_{N}\right)  \subseteq
W_{N}$ (which follows from (\ref{pf.thm.filtered.isocrit.filt}), applied to
$n=N$), this yields $W_{N}=f\left(  V_{N}\right)  $. In other words,
(\ref{pf.thm.filtered.isocrit.surj.1}) holds for $n=N$. This completes the
induction step.
\par
Thus, (\ref{pf.thm.filtered.isocrit.surj.1}) is proven by induction.}. Hence,
\[
W=\bigcup_{n\in\mathbb{N}}\underbrace{W_{n}}_{\substack{=f\left(
V_{n}\right)  \\\text{(by (\ref{pf.thm.filtered.isocrit.surj.1}))}}%
}=\bigcup_{n\in\mathbb{N}}f\left(  V_{n}\right)  =f\left(  \underbrace{\bigcup
_{n\in\mathbb{N}}V_{n}}_{=V}\right)  =f\left(  V\right)  .
\]
In other words, the map $f$ is surjective.

Furthermore,%
\begin{equation}
\left(  \operatorname*{Ker}f\right)  \cap V_{n}=0\ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N}\cup\left\{  -1\right\}
\label{pf.thm.filtered.isocrit.inj.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.filtered.isocrit.inj.1}):} We shall
prove (\ref{pf.thm.filtered.isocrit.inj.1}) by induction over $n$:
\par
\textit{Induction base:} We have $\left(  \operatorname*{Ker}f\right)
\cap\underbrace{V_{-1}}_{=0}=\left(  \operatorname*{Ker}f\right)  \cap0=0$. In
other words, (\ref{pf.thm.filtered.isocrit.inj.1}) holds for $n=-1$. This
completes the induction base.
\par
\textit{Induction step:} Let $N\in\mathbb{N}$. Assume that
(\ref{pf.thm.filtered.isocrit.inj.1}) holds for $n=N-1$. We must prove that
(\ref{pf.thm.filtered.isocrit.inj.1}) holds for $n=N$.
\par
We assumed that (\ref{pf.thm.filtered.isocrit.inj.1}) holds for $n=N-1$. In
other words, $\left(  \operatorname*{Ker}f\right)  \cap V_{N-1}=0$.
\par
Let $v\in\left(  \operatorname*{Ker}f\right)  \cap V_{N}$. Thus, $v\in\left(
\operatorname*{Ker}f\right)  \cap V_{N}\subseteq\operatorname*{Ker}f$, so that
$f\left(  v\right)  =0$. Also, $v\in\left(  \operatorname*{Ker}f\right)  \cap
V_{N}\subseteq V_{N}$, and thus $\left[  v\right]  _{N}\in\operatorname*{gr}%
\nolimits_{N}V$ is well-defined. The definition of $\operatorname*{gr}f$ shows
that $\left(  \operatorname*{gr}f\right)  \left(  \left[  v\right]
_{N}\right)  =\left[  \underbrace{f\left(  v\right)  }_{=0}\right]
_{N}=\left[  0\right]  _{N}=0$. Since the map $\operatorname*{gr}f$ is
injective (because $\operatorname*{gr}f$ is invertible), this shows that
$\left[  v\right]  _{N}=0$. In other words, $v\in V_{N-1}$. Combined with
$v\in\operatorname*{Ker}f$, this shows that $v\in\left(  \operatorname*{Ker}%
f\right)  \cap V_{N-1}=0$. In other words, $v=0$.
\par
Let us now forget that we fixed $v$. We thus have proven that $v=0$ for every
$v\in\left(  \operatorname*{Ker}f\right)  \cap V_{N}$. In other words,
$\left(  \operatorname*{Ker}f\right)  \cap V_{N}=0$. In other words,
(\ref{pf.thm.filtered.isocrit.inj.1}) holds for $n=N$. This completes the
induction step.
\par
Thus, (\ref{pf.thm.filtered.isocrit.inj.1}) is proven by induction.}. Now,
$\operatorname*{Ker}f\subseteq V$, so that $\left(  \operatorname*{Ker}%
f\right)  \cap V=\operatorname*{Ker}f$. Hence,%
\begin{align*}
\operatorname*{Ker}f  &  =\left(  \operatorname*{Ker}f\right)  \cap
\underbrace{V}_{=\bigcup_{n\in\mathbb{N}}V_{n}}=\left(  \operatorname*{Ker}%
f\right)  \cap\left(  \bigcup_{n\in\mathbb{N}}V_{n}\right) \\
&  =\bigcup_{n\in\mathbb{N}}\underbrace{\left(  \left(  \operatorname*{Ker}%
f\right)  \cap V_{n}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.thm.filtered.isocrit.inj.1}))}}}=\bigcup_{n\in\mathbb{N}}0=0.
\end{align*}
In other words, the map $f$ is injective. Thus, the map $f$ is both injective
and surjective. Hence, the map $f$ is bijective, i.e., invertible. Clearly,
its inverse $f^{-1}$ is a $\mathbf{k}$-linear map.

It suffices to prove that this map $f^{-1}$ is filtered. For every
$n\in\mathbb{N}$, we have%
\[
f^{-1}\left(  \underbrace{W_{n}}_{\substack{=f\left(  V_{n}\right)
\\\text{(by (\ref{pf.thm.filtered.isocrit.surj.1}))}}}\right)  =f^{-1}\left(
f\left(  V_{n}\right)  \right)  =V_{n}%
\]
(since the map $f$ is invertible). Hence, the map $f^{-1}$ is filtered. This
completes the proof of Theorem \ref{thm.filtered.isocrit}.
\end{proof}

\subsection{Definitions on filtered and graded $\mathbf{k}$-algebras}

Recall that $\mathbf{k}$-algebras are $\mathbf{k}$-modules. When a
$\mathbf{k}$-algebra has a filtration, it may and may not be a filtered
$\mathbf{k}$-algebra, depending on whether the $\mathbf{k}$-algebra structure
\textquotedblleft plays well\textquotedblright\ with respect to the
filtration. Here is what this means in detail:

\begin{definition}
\label{def.filtered-alg}Let $A$ be a $\mathbf{k}$-algebra. Let $\left(
A_{n}\right)  _{n\geq0}$ be a filtration on the $\mathbf{k}$-module $A$. Thus,
$A$ becomes a filtered $\mathbf{k}$-module. We say that the $\mathbf{k}%
$-module $A$ (endowed with the given $\mathbf{k}$-algebra structure on $A$ and
with the filtration $\left(  A_{n}\right)  _{n\geq0}$) is a \textit{filtered
}$\mathbf{k}$\textit{-algebra} if and only if we have%
\begin{align*}
1_{A}  &  \in A_{0}\ \ \ \ \ \ \ \ \ \ \text{and}\\
&  \left(  A_{n}A_{m}\subseteq A_{n+m}\ \ \ \ \ \ \ \ \ \ \text{for every
}n\in\mathbb{N}\text{ and }m\in\mathbb{N}\right)  .
\end{align*}
Thus, a filtered $\mathbf{k}$-algebra automatically is a filtered $\mathbf{k}%
$-module and (at the same time) a $\mathbf{k}$-algebra.
\end{definition}

We define a similar notation for gradings:

\begin{definition}
\label{def.graded-alg}Let $A$ be a $\mathbf{k}$-algebra. Let $\left(
A_{n}\right)  _{n\geq0}$ be a grading on the $\mathbf{k}$-module $A$. Thus,
$A$ becomes a graded $\mathbf{k}$-module. We say that the $\mathbf{k}$-module
$A$ (endowed with the given $\mathbf{k}$-algebra structure on $A$ and with the
grading $\left(  A_{n}\right)  _{n\geq0}$) is a \textit{graded }$\mathbf{k}%
$\textit{-algebra} if and only if we have%
\begin{align*}
1_{A}  &  \in A_{0}\ \ \ \ \ \ \ \ \ \ \text{and}\\
&  \left(  A_{n}A_{m}\subseteq A_{n+m}\ \ \ \ \ \ \ \ \ \ \text{for every
}n\in\mathbb{N}\text{ and }m\in\mathbb{N}\right)  .
\end{align*}
Thus, a graded $\mathbf{k}$-algebra automatically is a graded $\mathbf{k}%
$-module and (at the same time) a $\mathbf{k}$-algebra.
\end{definition}

\begin{remark}
\label{rmk.graded-alg.is-filtered}Let $A$ be a graded $\mathbf{k}$-algebra.
Consider the $\mathbf{k}$-submodules $A_{\leq m}$ for all $m\in\mathbb{N}$
(defined as in Definition \ref{def.filtered-from-graded}). Then, $A$ (endowed
with the given $\mathbf{k}$-algebra structure on $A$ and with the filtration
$\left(  A_{\leq n}\right)  _{n\geq0}$) is a filtered $\mathbf{k}$-algebra.
\end{remark}

We notice that the requirement $1_{A}\in A_{0}$ in Definition
\ref{def.graded-alg} could be dropped (it follows from the other
requirements); however, the requirement $1_{A}\in A_{0}$ in Definition
\ref{def.filtered-alg} cannot be dropped.

\begin{definition}
\label{def.gr.alg}Let $A$ be a filtered $\mathbf{k}$-algebra. Then, we can
define a $\mathbf{k}$-algebra structure on the graded $\mathbf{k}$-module
$\operatorname*{gr}A$ as follows:

Let $\left(  A_{n}\right)  _{n\geq0}$ be a filtration of the filtered
$\mathbf{k}$-module $A$. Then, there exists a unique binary operation $\ast$
on $\operatorname*{gr}A$ which satisfies%
\[
\left(  \left[  v\right]  _{n}\ast\left[  w\right]  _{m}=\left[  vw\right]
_{n+m}\ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{, }m\in
\mathbb{N}\text{, }v\in A_{n}\text{ and }w\in A_{m}\right)  .
\]
The $\mathbf{k}$-module $\operatorname*{gr}A$, equipped with this binary
operation $\ast$, becomes a $\mathbf{k}$-algebra with unity
$1_{\operatorname*{gr}A}=\left[  1_{A}\right]  _{0}\in\operatorname*{gr}%
\nolimits_{0}A\subseteq\operatorname*{gr}A$. This $\mathbf{k}$-algebra
$\operatorname*{gr}A$ furthermore becomes a graded $\mathbf{k}$-algebra (when
combined with the grading on $\operatorname*{gr}A$). This graded $\mathbf{k}%
$-algebra $\operatorname*{gr}A$ is known as the \textit{associated graded
algebra} of the filtered $\mathbf{k}$-algebra $A$. We shall write the binary
operation $\ast$ as a common multiplication; i.e., we shall write $\alpha
\beta$ or $\alpha\cdot\beta$ for $\alpha\ast\beta$.
\end{definition}

\begin{remark}
\label{rmk.gr.alg.map}Let $A$ and $B$ be two filtered $\mathbf{k}$-algebras.
Let $f:A\rightarrow B$ be a filtered $\mathbf{k}$-algebra homomorphism. Then,
$\operatorname*{gr}A$ and $\operatorname*{gr}B$ are graded $\mathbf{k}%
$-algebras (according to Definition \ref{def.gr.map}). The map
$\operatorname*{gr}f:\operatorname*{gr}A\rightarrow\operatorname*{gr}B$
(defined according to Definition \ref{def.gr.map}) is a graded $\mathbf{k}%
$-algebra homomorphism.
\end{remark}

\begin{remark}
\label{rmk.grad.alg}Let $A$ be a graded $\mathbf{k}$-algebra. Then, the map
$\operatorname*{grad}\nolimits_{A}:A\rightarrow\operatorname*{gr}A$ (defined
according to Definition \ref{def.grad}) is a graded $\mathbf{k}$-algebra isomorphism.
\end{remark}

Let us record a simple addendum to Theorem \ref{thm.gen1}:

\begin{theorem}
\label{thm.gen1filt}Let $\mathfrak{g}$ be a Lie algebra. Let $C$ be a filtered
$\mathbf{k}$-algebra. Let $\left(  C_{n}\right)  _{n\geq0}$ be the filtration
of $C$. Set $C_{-1}=0$.

Let $K:\mathfrak{g}\rightarrow\operatorname*{Der}C$ be a Lie algebra
homomorphism. Let $f:\mathfrak{g}\rightarrow C$ be a $\mathbf{k}$-linear map.
Assume that (\ref{eq.thm.gen1.axiom1}) holds for every $a\in\mathfrak{g}$ and
$b\in\mathfrak{g}$ (where the Lie bracket $\left[  f\left(  a\right)
,f\left(  b\right)  \right]  $ is computed in the Lie algebra $C^{-}$). Define
a $\mathfrak{g}$-module structure on $C$ as in Theorem \ref{thm.gen1}
\textbf{(a)}. Define a map $\eta:U\left(  \mathfrak{g}\right)  \rightarrow C$
as in Theorem \ref{thm.gen1} \textbf{(b)}.

Assume furthermore that $f\left(  \mathfrak{g}\right)  \subseteq C_{1}$. Also,
assume that the map $K\left(  a\right)  :C\rightarrow C$ is filtered for every
$a\in\mathfrak{g}$.

Then,%
\begin{equation}
\eta\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)
\right)  \in f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{n}\right)  +C_{n-1} \label{eq.thm.gen1filt.claim}%
\end{equation}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.gen1filt}.]We shall prove
(\ref{eq.thm.gen1filt.claim}) by induction over $n$:

\textit{Induction base:} The equality (\ref{eq.thm.gen1filt.claim}) holds in
the case when $n=0$\ \ \ \ \footnote{\textit{Proof:} If $a_{1},a_{2}%
,\ldots,a_{0}$ are $0$ elements of $\mathfrak{g}$, then%
\begin{align*}
&  \eta\left(  \underbrace{\iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{0}\right)  }_{=\left(  \text{empty product}\right)  =1}\right) \\
&  =\eta\left(  1\right)  =1\cdot1_{C}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\eta\right) \\
&  =1_{C}=f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{0}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\left(  a_{1}\right)
f\left(  a_{2}\right)  \cdots f\left(  a_{0}\right)  =\left(  \text{empty
product}\right)  =1_{C}\right) \\
&  \in f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{0}\right)  +C_{0-1}%
\end{align*}
In other words, (\ref{eq.thm.gen1filt.claim}) holds in the case when $n=0$.
Qed.}. This completes the induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that
(\ref{eq.thm.gen1filt.claim}) holds in the case when $n=N-1$. We now must show
that (\ref{eq.thm.gen1filt.claim}) holds in the case when $n=N$.

We have assumed that (\ref{eq.thm.gen1filt.claim}) holds in the case when
$n=N-1$. In other words, we have%
\begin{equation}
\eta\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{N-1}\right)
\right)  \in f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{N-1}\right)  +C_{\left(  N-1\right)  -1} \label{pf.thm.gen1filt.indhyp}%
\end{equation}
for every $a_{1},a_{2},\ldots,a_{N-1}\in\mathfrak{g}$.

Now, let $a_{1},a_{2},\ldots,a_{N}\in\mathfrak{g}$.

We have assumed that the map $K\left(  a\right)  :C\rightarrow C$ is filtered
for every $a\in\mathfrak{g}$. Applying this to $a=a_{1}$, we conclude that the
map $K\left(  a_{1}\right)  :C\rightarrow C$ is filtered.

For every $p\in\left\{  2,3,\ldots,N\right\}  $, we have $f\left(
\underbrace{a_{p}}_{\in\mathfrak{g}}\right)  \in f\left(  \mathfrak{g}\right)
\subseteq C_{1}$. Multiplying these relations for all $p\in\left\{
2,3,\ldots,N\right\}  $, we obtain%
\[
f\left(  a_{2}\right)  f\left(  a_{3}\right)  \cdots f\left(  a_{N}\right)
\in\underbrace{C_{1}C_{1}\cdots C_{1}}_{N-1\text{ factors}}\subseteq C_{N-1}%
\]
(since $C$ is a filtered $\mathbf{k}$-algebra). Also, $f\left(
\underbrace{a_{1}}_{\in\mathfrak{g}}\right)  \in f\left(  \mathfrak{g}\right)
\subseteq C_{1}$.

Let $g$ denote the element $\eta\left(  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \iota_{U,\mathfrak{g}}\left(  a_{3}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{N}\right)  \right)  $ of $C$. Then,%
\begin{equation}
g=\eta\left(  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \iota
_{U,\mathfrak{g}}\left(  a_{3}\right)  \cdots\iota_{U,\mathfrak{g}}\left(
a_{N}\right)  \right)  \in f\left(  a_{2}\right)  f\left(  a_{3}\right)
\cdots f\left(  a_{N}\right)  +C_{\left(  N-1\right)  -1}
\label{pf.thm.gen1filt.indhyp-applied}%
\end{equation}
(by (\ref{pf.thm.gen1filt.indhyp}), applied to $a_{2},a_{3},\ldots,a_{N-1}$
instead of $a_{1},a_{2},\ldots,a_{N-1}$). Hence,%
\[
g\in\underbrace{f\left(  a_{2}\right)  f\left(  a_{3}\right)  \cdots f\left(
a_{N}\right)  }_{\in C_{N-1}}+\underbrace{C_{\left(  N-1\right)  -1}%
}_{\substack{\subseteq C_{N-1}\\\text{(since }C\text{ is filtered)}}}\subseteq
C_{N-1}+C_{N-1}\subseteq C_{N-1}%
\]
(since $C_{N-1}$ is a $\mathbf{k}$-module). Applying the map $K\left(
a_{1}\right)  $ to both sides of this relation, we obtain%
\begin{equation}
\left(  K\left(  a_{1}\right)  \right)  \left(  g\right)  \in\left(  K\left(
a_{1}\right)  \right)  \left(  C_{N-1}\right)  \subseteq C_{N-1}
\label{pf.thm.gen1filt.La1g}%
\end{equation}
(since the map $K\left(  a_{1}\right)  $ is filtered).

Also, multiplying both sides of the relation
(\ref{pf.thm.gen1filt.indhyp-applied}) with $f\left(  a_{1}\right)  $ from the
left, we obtain%
\begin{align}
f\left(  a_{1}\right)  \cdot g  &  \in f\left(  a_{1}\right)  \cdot\left(
f\left(  a_{2}\right)  f\left(  a_{3}\right)  \cdots f\left(  a_{N}\right)
+C_{\left(  N-1\right)  -1}\right) \nonumber\\
&  \subseteq\underbrace{f\left(  a_{1}\right)  \cdot f\left(  a_{2}\right)
f\left(  a_{3}\right)  \cdots f\left(  a_{N}\right)  }_{=f\left(
a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{N}\right)
}+\underbrace{f\left(  a_{1}\right)  }_{\in C_{1}}\cdot C_{\left(  N-1\right)
-1}\nonumber\\
&  \subseteq f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{N}\right)  +\underbrace{C_{1}\cdot C_{\left(  N-1\right)  -1}%
}_{\substack{\subseteq C_{1+\left(  \left(  N-1\right)  -1\right)
}\\\text{(since }C\text{ is a filtered }\mathbf{k}\text{-algebra)}%
}}\nonumber\\
&  \subseteq f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{N}\right)  +\underbrace{C_{1+\left(  \left(  N-1\right)  -1\right)  }%
}_{=C_{N-1}}\nonumber\\
&  =f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{N}\right)  +C_{N-1}. \label{pf.thm.gen1filt.a1g}%
\end{align}


But%
\begin{equation}
g=\eta\left(  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \iota
_{U,\mathfrak{g}}\left(  a_{3}\right)  \cdots\iota_{U,\mathfrak{g}}\left(
a_{N}\right)  \right)  =\left(  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)
\iota_{U,\mathfrak{g}}\left(  a_{3}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{N}\right)  \right)  1_{C} \label{pf.thm.gen1filt.5}%
\end{equation}
(by the definition of $\eta$). Now, the definition of $\eta$ yields%
\begin{align*}
&  \eta\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota
_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(
a_{N}\right)  \right) \\
&  =\underbrace{\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{N}\right)  \right)  }_{=\left(  \iota_{U,\mathfrak{g}}\left(
a_{1}\right)  \right)  \cdot\left(  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \iota_{U,\mathfrak{g}}\left(  a_{3}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{N}\right)  \right)  }1_{C}\\
&  =\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \right)
\cdot\underbrace{\left(  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)
\iota_{U,\mathfrak{g}}\left(  a_{3}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{N}\right)  \right)  1_{C}}_{\substack{=g\\\text{(by
(\ref{pf.thm.gen1filt.5}))}}}=\left(  \iota_{U,\mathfrak{g}}\left(
a_{1}\right)  \right)  \cdot g\\
&  =a_{1}\rightharpoonup g\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition
of the }U\left(  \mathfrak{g}\right)  \text{-module structure on }C\right) \\
&  =\underbrace{f\left(  a_{1}\right)  \cdot g}_{\substack{\in f\left(
a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(  a_{N}\right)
+C_{N-1}\\\text{(by (\ref{pf.thm.gen1filt.a1g}))}}}+\underbrace{\left(
K\left(  a_{1}\right)  \right)  \left(  g\right)  }_{\substack{\in
C_{N-1}\\\text{(by (\ref{pf.thm.gen1filt.La1g}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the }\mathfrak{g}%
\text{-module structure on }C\right) \\
&  \in f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{N}\right)  +\underbrace{C_{N-1}+C_{N-1}}_{\substack{\subseteq
C_{N-1}\\\text{(since }C_{N-1}\text{ is a }\mathbf{k}\text{-module)}}}\\
&  \subseteq f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{N}\right)  +C_{N-1}.
\end{align*}


Let us now forget that we fixed $a_{1},a_{2},\ldots,a_{N}\in\mathfrak{g}$. We
thus have shown that%
\[
\eta\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{N}\right)
\right)  \in f\left(  a_{1}\right)  f\left(  a_{2}\right)  \cdots f\left(
a_{N}\right)  +C_{N-1}%
\]
for every $a_{1},a_{2},\ldots,a_{N}\in\mathfrak{g}$. In other words,
(\ref{eq.thm.gen1filt.claim}) holds in the case when $n=N$. This completes the
induction step. The induction proof of (\ref{eq.thm.gen1filt.claim}) is thus
complete. Thus, Theorem \ref{thm.gen1filt} is proven.
\end{proof}

\subsection{$\operatorname*{Sym}V$ as a graded $\mathbf{k}$-algebra}

\begin{definition}
\label{def.Sym}Let $V$ be a $\mathbf{k}$-module. For every $n\in\mathbb{N}$,
we let $V^{\otimes n}$ denote the $n$-th tensor power of $V$ (that is, the
$\mathbf{k}$-module $\underbrace{V\otimes V\otimes\cdots\otimes V}_{n\text{
times }V}$). We let $T\left(  V\right)  $ denote the tensor algebra of $V$.
This is the graded $\mathbf{k}$-algebra whose underlying $\mathbf{k}$-module
is $\bigoplus\limits_{n\geq0}V^{\otimes n}$, whose grading is $\left(
V^{\otimes n}\right)  _{n\geq0}$, and whose multiplication is given by%
\begin{align*}
&  \left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  \cdot\left(
b_{1}\otimes b_{2}\otimes\cdots\otimes b_{m}\right) \\
&  =a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\otimes b_{1}\otimes
b_{2}\otimes\cdots\otimes b_{m}\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{for every }n\in\mathbb{N}\text{, }%
m\in\mathbb{N}\text{, }a_{1},a_{2},\ldots,a_{n}\in V\text{ and }b_{1}%
,b_{2},\ldots,b_{m}\in V\right.  .
\end{align*}


We let $\operatorname*{Sym}V$ denote the quotient algebra of the $\mathbf{k}%
$-algebra $T\left(  V\right)  $ by its ideal generated by the tensors of the
form $v\otimes w-w\otimes v$ with $v\in V$ and $w\in V$. We let $\pi
_{\operatorname*{Sym},V}$ be the canonical projection from $T\left(  V\right)
$ to its quotient $\mathbf{k}$-algebra $\operatorname*{Sym}V$. It is
well-known that $\operatorname*{Sym}V$ (endowed with the grading $\left(
\pi_{\operatorname*{Sym},V}\left(  V^{\otimes n}\right)  \right)  _{n\geq0}$)
is a graded $\mathbf{k}$-algebra (since the ideal of $T\left(  V\right)  $
generated by the products of the form $v\otimes w-w\otimes v$ with $v\in V$
and $w\in V$ is a graded $\mathbf{k}$-submodule of $T\left(  V\right)  $).
This graded $\mathbf{k}$-algebra $\operatorname*{Sym}V$ is known as the
\textit{symmetric algebra} of $V$. For every $n\in\mathbb{N}$, we write
$\operatorname*{Sym}\nolimits^{n}V$ for the $\mathbf{k}$-submodule
$\pi_{\operatorname*{Sym},V}\left(  V^{\otimes n}\right)  $ of
$\operatorname*{Sym}V$. Thus, $\operatorname*{Sym}V=\bigoplus\limits_{n\geq
0}\operatorname*{Sym}\nolimits^{n}V$, and the grading of $\operatorname*{Sym}%
V$ is $\left(  \operatorname*{Sym}\nolimits^{n}V\right)  _{n\geq0}$.

Both $T\left(  V\right)  $ and $\operatorname*{Sym}V$ are graded $\mathbf{k}%
$-algebras, and thus canonically become filtered $\mathbf{k}$-algebras. It is
well-known that the $\mathbf{k}$-algebra $\operatorname*{Sym}V$ is commutative.

We let $\iota_{T,V}$ be the canonical inclusion map of $V$ into $T\left(
V\right)  $. This map is the composition $V\overset{\cong}{\longrightarrow
}V^{\otimes1}\overset{\text{inclusion}}{\longrightarrow}\bigoplus
\limits_{n\geq0}V^{\otimes n}=T\left(  V\right)  $. (Here, the
\textquotedblleft$T$\textquotedblright\ in \textquotedblleft$\iota_{T,V}%
$\textquotedblright\ is not a variable, but stands for the letter
\textquotedblleft t\textquotedblright\ in \textquotedblleft\textbf{t}ensor
algebra\textquotedblright.) We have $\iota_{T,V}\left(  V\right)
=V^{\otimes1}$. We will usually (but not always) identify $V$ with the
$\mathbf{k}$-submodule $V^{\otimes1}$ of $T\left(  V\right)  $ via this map
$\iota_{T,V}$; this identification is harmless since $\iota_{T,V}$ is
injective. Thus, for every $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots,a_{n}\in
V$, we have $a_{1}a_{2}\cdots a_{n}=a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}$ in the $\mathbf{k}$-algebra $T\left(  V\right)  $.

We let $\iota_{\operatorname*{Sym},V}$ denote the composition $\pi
_{\operatorname*{Sym},V}\circ\iota_{T,V}$. This composition $\iota
_{\operatorname*{Sym},V}$ is a $\mathbf{k}$-linear map $V\rightarrow
\operatorname*{Sym}V$. It is well-known that this map $\iota
_{\operatorname*{Sym},V}$ is injective and satisfies $\iota
_{\operatorname*{Sym},V}\left(  V\right)  =\operatorname*{Sym}\nolimits^{1}V$.

The $\mathbf{k}$-algebra $T\left(  V\right)  $ becomes a $\mathbf{k}%
$-bialgebra as follows: We define the coproduct $\Delta$ of $T\left(
V\right)  $ as the unique $\mathbf{k}$-algebra homomorphism $\Delta:T\left(
V\right)  \rightarrow T\left(  V\right)  \otimes T\left(  V\right)  $
satisfying%
\[
\left(  \Delta\left(  \iota_{T,V}\left(  v\right)  \right)  =\iota
_{T,V}\left(  v\right)  \otimes1+1\otimes\iota_{T,V}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V\right)  .
\]
We define the counit $\epsilon$ of $T\left(  V\right)  $ as the unique
$\mathbf{k}$-algebra homomorphism $\epsilon:T\left(  V\right)  \rightarrow
\mathbf{k}$ satisfying%
\[
\left(  \epsilon\left(  \iota_{T,V}\left(  v\right)  \right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V\right)  .
\]


Explicitly, $\Delta$ and $\epsilon$ are given by the following formulas:%
\begin{align*}
\Delta\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)   &
=\sum_{I\subseteq\left\{  1,2,\ldots,n\right\}  }a_{I}\otimes a_{\left\{
1,2,\ldots,n\right\}  \setminus I};\\
\epsilon\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)   &  =%
\begin{cases}
1, & \text{if }n=0;\\
0, & \text{if }n>0
\end{cases}
\end{align*}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in V$, where
for every subset $J=\left\{  j_{1}<j_{2}<\cdots<j_{k}\right\}  \subseteq
\left\{  1,2,\ldots,n\right\}  $, we define $a_{J}$ to be the tensor
$a_{j_{1}}\otimes a_{j_{2}}\otimes\cdots\otimes a_{j_{k}}$. The $\mathbf{k}%
$-bialgebra $T\left(  V\right)  $ is actually a Hopf algebra.

It is easy to see that the elements of $\iota_{T,V}\left(  V\right)  $ are
primitive elements of $T\left(  V\right)  $. In other words, $\iota
_{T,V}\left(  V\right)  \subseteq\operatorname*{Prim}\left(  T\left(
V\right)  \right)  $.

The $\mathbf{k}$-algebra $\operatorname*{Sym}V$ also becomes a $\mathbf{k}%
$-bialgebra, being a quotient of the $\mathbf{k}$-bialgebra $T\left(
V\right)  $ by a biideal. It is easy to see that $\iota_{\operatorname*{Sym}%
,V}\left(  V\right)  \subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}%
V\right)  $.
\end{definition}

\begin{remark}
\label{rmk.Sym.span}Let $V$ be a $\mathbf{k}$-module. Let $n\in\mathbb{N}$.

\textbf{(a)} We have $\pi_{\operatorname*{Sym},V}\left(  a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}\right)  =\iota_{\operatorname*{Sym},V}\left(
a_{1}\right)  \iota_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},V}\left(  a_{n}\right)  $ for any $a_{1}%
,a_{2},\ldots,a_{n}\in V$.

\textbf{(b)} The $\mathbf{k}$-module $\operatorname*{Sym}\nolimits^{n}V$ is
spanned by the elements $\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},V}\left(  a_{n}\right)  $ with $a_{1},a_{2},\ldots
,a_{n}\in V$.

\textbf{(c)} Assume that $n$ is positive. Then, $\operatorname*{Sym}%
\nolimits^{n}V=\left(  \operatorname*{Sym}\nolimits^{1}V\right)  \left(
\operatorname*{Sym}\nolimits^{n-1}V\right)  $.
\end{remark}

\begin{proof}
[Proof of Remark \ref{rmk.Sym.span}.]\textbf{(a)} Let $a_{1},a_{2}%
,\ldots,a_{n}\in V$. For any $p\in\left\{  1,2,\ldots,n\right\}  $, we have%
\[
\underbrace{\iota_{\operatorname*{Sym},V}}_{=\pi_{\operatorname*{Sym},V}%
\circ\iota_{T,V}}\left(  a_{p}\right)  =\left(  \pi_{\operatorname*{Sym}%
,V}\circ\iota_{T,V}\right)  \left(  a_{p}\right)  =\pi_{\operatorname*{Sym}%
,V}\left(  \iota_{T,V}\left(  a_{p}\right)  \right)  .
\]
Thus,%
\begin{align*}
&  \iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,V}\left(  a_{n}\right) \\
&  =\pi_{\operatorname*{Sym},V}\left(  \iota_{T,V}\left(  a_{1}\right)
\right)  \pi_{\operatorname*{Sym},V}\left(  \iota_{T,V}\left(  a_{2}\right)
\right)  \cdots\pi_{\operatorname*{Sym},V}\left(  \iota_{T,V}\left(
a_{n}\right)  \right) \\
&  =\pi_{\operatorname*{Sym},V}\left(  \underbrace{\iota_{T,V}\left(
a_{1}\right)  \iota_{T,V}\left(  a_{2}\right)  \cdots\iota_{T,V}\left(
a_{n}\right)  }_{\substack{=a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\\\text{(by the definition of the product in }T\left(  V\right)
\text{)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi_{\operatorname*{Sym},V}\text{
is a }\mathbf{k}\text{-algebra homomorphism}\right) \\
&  =\pi_{\operatorname*{Sym},V}\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\right)  .
\end{align*}
This proves Remark \ref{rmk.Sym.span} \textbf{(a)}.

\textbf{(b)} The $\mathbf{k}$-module $V^{\otimes n}$ is spanned by pure
tensors. In other words, the $\mathbf{k}$-module $V^{\otimes n}$ is spanned by
the elements $a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}$ with $a_{1}%
,a_{2},\ldots,a_{n}\in V$. Thus, the $\mathbf{k}$-module $\pi
_{\operatorname*{Sym},V}\left(  V^{\otimes n}\right)  $ is spanned by the
elements $\pi_{\operatorname*{Sym},V}\left(  a_{1}\otimes a_{2}\otimes
\cdots\otimes a_{n}\right)  $ with $a_{1},a_{2},\ldots,a_{n}\in V$ (since the
map $\pi_{\operatorname*{Sym},V}$ is $\mathbf{k}$-linear). In other words, the
$\mathbf{k}$-module $\operatorname*{Sym}\nolimits^{n}V$ is spanned by the
elements $\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,V}\left(  a_{n}\right)  $ with $a_{1},a_{2},\ldots,a_{n}\in V$ (because
$\pi_{\operatorname*{Sym},V}\left(  V^{\otimes n}\right)  =\operatorname*{Sym}%
\nolimits^{n}V$ and because every $a_{1},a_{2},\ldots,a_{n}\in V$ satisfy
$\pi_{\operatorname*{Sym},V}\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\right)  =\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},V}\left(  a_{n}\right)  $). This proves Remark
\ref{rmk.Sym.span} \textbf{(b)}.

\textbf{(c)} Clearly, $\left(  \operatorname*{Sym}\nolimits^{1}V\right)
\left(  \operatorname*{Sym}\nolimits^{n-1}V\right)  \subseteq
\operatorname*{Sym}\nolimits^{n}V$ (since the $\mathbf{k}$-algebra
$\operatorname*{Sym}V$ is graded).

Now, let $v\in\operatorname*{Sym}\nolimits^{n}V$. We shall show that
$v\in\left(  \operatorname*{Sym}\nolimits^{1}V\right)  \left(
\operatorname*{Sym}\nolimits^{n-1}V\right)  $.

We have $v\in\operatorname*{Sym}\nolimits^{n}V$. Thus, $v$ is a $\mathbf{k}%
$-linear combination of elements of the form $\iota_{\operatorname*{Sym}%
,V}\left(  a_{1}\right)  \iota_{\operatorname*{Sym},V}\left(  a_{2}\right)
\cdots\iota_{\operatorname*{Sym},V}\left(  a_{n}\right)  $ with $a_{1}%
,a_{2},\ldots,a_{n}\in V$ (due to Remark \ref{rmk.Sym.span} \textbf{(b)}).

We need to prove the relation $v\in\left(  \operatorname*{Sym}\nolimits^{1}%
V\right)  \left(  \operatorname*{Sym}\nolimits^{n-1}V\right)  $. This relation
is clearly $\mathbf{k}$-linear in $v$. Hence, we can WLOG assume that $v$ has
the form $\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,V}\left(  a_{n}\right)  $ with $a_{1},a_{2},\ldots,a_{n}\in V$ (because $v$
is a $\mathbf{k}$-linear combination of elements of the form $\iota
_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota_{\operatorname*{Sym}%
,V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym},V}\left(
a_{n}\right)  $ with $a_{1},a_{2},\ldots,a_{n}\in V$). Assume this. Thus,
$v=\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,V}\left(  a_{n}\right)  $ for some $a_{1},a_{2},\ldots,a_{n}\in V$. Consider
these $a_{1},a_{2},\ldots,a_{n}$. We have%
\begin{align*}
v  &  =\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,V}\left(  a_{n}\right) \\
&  =\underbrace{\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  }%
_{\in\operatorname*{Sym}\nolimits^{1}V}\cdot\underbrace{\left(  \iota
_{\operatorname*{Sym},V}\left(  a_{2}\right)  \iota_{\operatorname*{Sym}%
,V}\left(  a_{3}\right)  \cdots\iota_{\operatorname*{Sym},V}\left(
a_{n}\right)  \right)  }_{\substack{\in\left(  \operatorname*{Sym}%
\nolimits^{1}V\right)  ^{n-1}\\\text{(since }\iota_{\operatorname*{Sym}%
,V}\left(  a_{p}\right)  \in\operatorname*{Sym}\nolimits^{1}V\text{ for every
}p\in\left\{  2,3,\ldots,n\right\}  \text{)}}}\\
&  \in\left(  \operatorname*{Sym}\nolimits^{1}V\right)  \left(
\operatorname*{Sym}\nolimits^{n-1}V\right)  .
\end{align*}


Now, let us forget that we have fixed $v$. We thus have shown $v\in\left(
\operatorname*{Sym}\nolimits^{1}V\right)  \left(  \operatorname*{Sym}%
\nolimits^{n-1}V\right)  $ for every $v\in\operatorname*{Sym}\nolimits^{n}V$.
In other words, $\operatorname*{Sym}\nolimits^{n}V\subseteq\left(
\operatorname*{Sym}\nolimits^{1}V\right)  \left(  \operatorname*{Sym}%
\nolimits^{n-1}V\right)  $. Combined with $\left(  \operatorname*{Sym}%
\nolimits^{1}V\right)  \left(  \operatorname*{Sym}\nolimits^{n-1}V\right)
\subseteq\operatorname*{Sym}\nolimits^{n}V$, this shows that $\left(
\operatorname*{Sym}\nolimits^{1}V\right)  \left(  \operatorname*{Sym}%
\nolimits^{n-1}V\right)  =\operatorname*{Sym}\nolimits^{n}V$. Remark
\ref{rmk.Sym.span} \textbf{(c)} is proven.
\end{proof}

The following fact is well-known as the universal property of the symmetric algebra:

\begin{proposition}
\label{prop.Sym.univ}Let $V$ be a $\mathbf{k}$-module. Let $A$ be a
commutative $\mathbf{k}$-algebra. Let $f:V\rightarrow A$ be a $\mathbf{k}%
$-linear map. Then, there exists a unique $\mathbf{k}$-algebra homomorphism
$F:\operatorname*{Sym}V\rightarrow A$ such that $F\circ\iota
_{\operatorname*{Sym},V}=f$. (See Definition \ref{def.Sym} for the definition
of $\iota_{\operatorname*{Sym},V}$.)
\end{proposition}

Let us also state a \textquotedblleft derivational analogue\textquotedblright%
\ of this universal property:

\begin{proposition}
\label{prop.Sym.deruniv}Let $V$ be a $\mathbf{k}$-module. Let $f:V\rightarrow
\operatorname*{Sym}V$ be a $\mathbf{k}$-linear map. Then, there exists a
unique derivation $F:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$ such
that $F\circ\iota_{\operatorname*{Sym},V}=f$.
\end{proposition}

Proposition \ref{prop.Sym.deruniv} follows from \cite[Proposition
2.7]{derivat} (applied to $M=\operatorname*{Sym}V$). We shall use the
following fact, which strengthens Proposition \ref{prop.Sym.deruniv} in a
particular case:

\begin{proposition}
\label{prop.Sym.bider}Let $V$ be a $\mathbf{k}$-module. Let $f:V\rightarrow V$
be a $\mathbf{k}$-linear map.

\textbf{(a)} There exists a unique derivation $F:\operatorname*{Sym}%
V\rightarrow\operatorname*{Sym}V$ such that $F\circ\iota_{\operatorname*{Sym}%
,V}=\iota_{\operatorname*{Sym},V}\circ f$.

We shall denote this derivation $F$ by $\widetilde{f}$.

\textbf{(b)} The map $\widetilde{f}:\operatorname*{Sym}V\rightarrow
\operatorname*{Sym}V$ is furthermore a coderivation.

\textbf{(c)} The map $\widetilde{f}:\operatorname*{Sym}V\rightarrow
\operatorname*{Sym}V$ is graded.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.Sym.bider}.]\textbf{(a)} Proposition
\ref{prop.Sym.deruniv} (applied to $\iota_{\operatorname*{Sym},V}\circ f$
instead of $f$) shows that there exists a unique derivation
$F:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$ such that $F\circ
\iota_{\operatorname*{Sym},V}=\iota_{\operatorname*{Sym},V}\circ f$. This
proves Proposition \ref{prop.Sym.bider} \textbf{(a)}.

The map $\widetilde{f}$ was defined as the unique derivation
$F:\operatorname*{Sym}V\rightarrow\operatorname*{Sym}V$ such that $F\circ
\iota_{\operatorname*{Sym},V}=\iota_{\operatorname*{Sym},V}\circ f$. Thus,
$\widetilde{f}$ is a derivation $\operatorname*{Sym}V\rightarrow
\operatorname*{Sym}V$ and satisfies%
\[
\widetilde{f}\circ\iota_{\operatorname*{Sym},V}=\iota_{\operatorname*{Sym}%
,V}\circ f.
\]
We have $\widetilde{f}\in\operatorname*{Der}\left(  \operatorname*{Sym}%
V\right)  $ (since $\widetilde{f}$ is a derivation). For every $v\in V$, we
have%
\begin{align}
\widetilde{f}\left(  \iota_{\operatorname*{Sym},V}\left(  v\right)  \right)
&  =\left(  \underbrace{\widetilde{f}\circ\iota_{\operatorname*{Sym},V}%
}_{=\iota_{\operatorname*{Sym},V}\circ f}\right)  \left(  v\right)  =\left(
\iota_{\operatorname*{Sym},V}\circ f\right)  \left(  v\right)  =\iota
_{\operatorname*{Sym},V}\left(  \underbrace{f\left(  v\right)  }_{\in
V}\right) \nonumber\\
&  \in\iota_{\operatorname*{Sym},V}\left(  V\right)  =\operatorname*{Sym}%
\nolimits^{1}V. \label{pf.prop.Sym.bider.deg1}%
\end{align}


\textbf{(c)} Let $n\in\mathbb{N}$. Let $x\in\operatorname*{Sym}\nolimits^{n}%
V$. We are going to prove the relation $\widetilde{f}\left(  x\right)
\in\operatorname*{Sym}\nolimits^{n}V$.

This relation is $\mathbf{k}$-linear in $x$. Hence, we can WLOG assume that
$x$ is of the form $\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},V}\left(  a_{n}\right)  $ for some $a_{1},a_{2}%
,\ldots,a_{n}\in V$ (because the $\mathbf{k}$-module $\operatorname*{Sym}%
\nolimits^{n}V$ is spanned by the elements $\iota_{\operatorname*{Sym}%
,V}\left(  a_{1}\right)  \iota_{\operatorname*{Sym},V}\left(  a_{2}\right)
\cdots\iota_{\operatorname*{Sym},V}\left(  a_{n}\right)  $ with $a_{1}%
,a_{2},\ldots,a_{n}\in V$ (by Remark \ref{rmk.Sym.span} \textbf{(b)})). Assume
this, and consider these $a_{1},a_{2},\ldots,a_{n}$. Thus,%
\[
x=\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,V}\left(  a_{n}\right)  .
\]
Applying the map $\widetilde{f}$ to both sides of this equality, we obtain%
\begin{align*}
\widetilde{f}\left(  x\right)   &  =\widetilde{f}\left(  \iota
_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota_{\operatorname*{Sym}%
,V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym},V}\left(
a_{n}\right)  \right) \\
&  =\sum_{i=1}^{n}\underbrace{\iota_{\operatorname*{Sym},V}\left(
a_{1}\right)  \iota_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},V}\left(  a_{i-1}\right)  }_{\in\operatorname*{Sym}%
\nolimits^{i-1}V}\underbrace{\widetilde{f}\left(  \iota_{\operatorname*{Sym}%
,V}\left(  a_{i}\right)  \right)  }_{\substack{\in\operatorname*{Sym}%
\nolimits^{1}V\\\text{(by (\ref{pf.prop.Sym.bider.deg1}), applied}\\\text{to
}v=a_{i}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \underbrace{\iota_{\operatorname*{Sym},V}\left(
a_{i+1}\right)  \iota_{\operatorname*{Sym},V}\left(  a_{i+2}\right)
\cdots\iota_{\operatorname*{Sym},V}\left(  a_{n}\right)  }_{\in
\operatorname*{Sym}\nolimits^{n-i}V}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.Der.basics} \textbf{(b)}, applied to
}\operatorname*{Sym}V\text{ and }\iota_{\operatorname*{Sym},V}\left(
a_{i}\right) \\
\text{instead of }C\text{ and }a_{i}%
\end{array}
\right) \\
&  \in\sum_{i=1}^{n}\underbrace{\left(  \operatorname*{Sym}\nolimits^{i-1}%
V\right)  \left(  \operatorname*{Sym}\nolimits^{1}V\right)  \left(
\operatorname*{Sym}\nolimits^{n-i}V\right)  }_{\substack{\subseteq
\operatorname*{Sym}\nolimits^{\left(  i-1\right)  +1+\left(  n-i\right)
}V\\\text{(since }\operatorname*{Sym}V\text{ is a graded }\mathbf{k}%
\text{-algebra)}}}\subseteq\sum_{i=1}^{n}\underbrace{\operatorname*{Sym}%
\nolimits^{\left(  i-1\right)  +1+\left(  n-i\right)  }V}%
_{=\operatorname*{Sym}\nolimits^{n}V}\\
&  =\sum_{i=1}^{n}\operatorname*{Sym}\nolimits^{n}V\subseteq
\operatorname*{Sym}\nolimits^{n}V
\end{align*}
(since $\operatorname*{Sym}\nolimits^{n}V$ is a $\mathbf{k}$-module).

Now, let us forget that we fixed $x$. We thus have proven that $\widetilde{f}%
\left(  x\right)  \in\operatorname*{Sym}\nolimits^{n}V$ for every
$x\in\operatorname*{Sym}\nolimits^{n}V$. In other words, $\widetilde{f}\left(
\operatorname*{Sym}\nolimits^{n}V\right)  \subseteq\operatorname*{Sym}%
\nolimits^{n}V$.

Let us now forget that we fixed $n$. We thus have proven that $\widetilde{f}%
\left(  \operatorname*{Sym}\nolimits^{n}V\right)  \subseteq\operatorname*{Sym}%
\nolimits^{n}V$ for every $n\in\mathbb{N}$. In other words, the map
$\widetilde{f}$ is graded. This proves Proposition \ref{prop.Sym.bider}
\textbf{(c)}.

\textbf{(b)} Let $A$ be the $\mathbf{k}$-algebra $\operatorname*{Sym}V$. Then,
the comultiplication of the $\mathbf{k}$-coalgebra $\operatorname*{Sym}V$ is a
$\mathbf{k}$-linear map $\Delta:A\rightarrow A\otimes A$. This map $\Delta$ is
a $\mathbf{k}$-algebra homomorphism (since $\operatorname*{Sym}V$ is a
$\mathbf{k}$-bialgebra). The map $\widetilde{f}\otimes\operatorname*{id}%
\nolimits_{A}:A\otimes A\rightarrow A\otimes A$ is a derivation (by Lemma
\ref{lem.der.tensid} \textbf{(a)}, applied to $A$ and $\widetilde{f}$ instead
of $B$ and $f$). In other words, $\widetilde{f}\otimes\operatorname*{id}%
\nolimits_{A}\in\operatorname*{Der}\left(  A\otimes A\right)  $. The map
$\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}:A\otimes A\rightarrow
A\otimes A$ is a derivation (by Lemma \ref{lem.der.tensid} \textbf{(b)},
applied to $A$ and $\widetilde{f}$ instead of $B$ and $f$). In other words,
$\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\in\operatorname*{Der}%
\left(  A\otimes A\right)  $. But $\operatorname*{Der}\left(  A\otimes
A\right)  $ is a Lie subalgebra of $\left(  \operatorname*{End}\left(
A\otimes A\right)  \right)  ^{-}$ (by Proposition \ref{prop.Der.basics}
\textbf{(a)}, applied to $C=A\otimes A$), and thus a $\mathbf{k}$-module.
Hence, $\widetilde{f}\otimes\operatorname*{id}\nolimits_{A}+\operatorname*{id}%
\nolimits_{A}\otimes\widetilde{f}\in\operatorname*{Der}\left(  A\otimes
A\right)  $ (since both $\widetilde{f}\otimes\operatorname*{id}\nolimits_{A}$
and $\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}$ belong to
$\operatorname*{Der}\left(  A\otimes A\right)  $). In other words,
$\widetilde{f}\otimes\operatorname*{id}\nolimits_{A}+\operatorname*{id}%
\nolimits_{A}\otimes\widetilde{f}:A\otimes A\rightarrow A\otimes A$ is a derivation.

Let $S$ be the subset $\iota_{\operatorname*{Sym},V}\left(  V\right)  $ of
$\operatorname*{Sym}V=A$. Thus, $S=\iota_{\operatorname*{Sym},V}\left(
V\right)  =\operatorname*{Sym}\nolimits^{1}V$. Hence, $S$ generates
$\operatorname*{Sym}V=A$ as a $\mathbf{k}$-algebra.

We have $\widetilde{f}\in\operatorname*{Der}\left(  \operatorname*{Sym}%
V\right)  $. Thus, Proposition \ref{prop.Der.basics} \textbf{(c)} (applied to
$\operatorname*{Sym}V$ and $\widetilde{f}$ instead of $C$ and $f$) shows that
$\widetilde{f}\left(  1\right)  =0$.

We have
\begin{equation}
\left(  \Delta\circ\widetilde{f}\right)  \mid_{S}=\left(  \left(
\widetilde{f}\otimes\operatorname*{id}\nolimits_{A}+\operatorname*{id}%
\nolimits_{A}\otimes\widetilde{f}\right)  \circ\Delta\right)  \mid_{S}
\label{pf.prop.Sym.bider.b.Srestr1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.Sym.bider.b.Srestr1}):} Let $s\in S$.
We shall show that $\left(  \Delta\circ\widetilde{f}\right)  \left(  s\right)
=\left(  \left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
+\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\right)  \circ
\Delta\right)  \left(  s\right)  $.
\par
The definition of the $\mathbf{k}$-bialgebra structure on $\operatorname*{Sym}%
V$ shows that%
\begin{equation}
\Delta\left(  \iota_{\operatorname*{Sym},V}\left(  w\right)  \right)
=\iota_{\operatorname*{Sym},V}\left(  w\right)  \otimes1+1\otimes
\iota_{\operatorname*{Sym},V}\left(  w\right)
\label{pf.prop.Sym.bider.b.Srestr1.pf.1}%
\end{equation}
for every $w\in V$.
\par
Now, $s\in S=\iota_{\operatorname*{Sym},V}\left(  V\right)  $. Hence,
$s=\iota_{\operatorname*{Sym},V}\left(  v\right)  $ for some $v\in V$.
Consider this $v$. Now,%
\[
\widetilde{f}\left(  \underbrace{s}_{=\iota_{\operatorname*{Sym},V}\left(
v\right)  }\right)  =\widetilde{f}\left(  \iota_{\operatorname*{Sym},V}\left(
v\right)  \right)  =\underbrace{\left(  \widetilde{f}\circ\iota
_{\operatorname*{Sym},V}\right)  }_{=\iota_{\operatorname*{Sym},V}\circ
f}\left(  v\right)  =\left(  \iota_{\operatorname*{Sym},V}\circ f\right)
\left(  v\right)  =\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)
\right)  .
\]
Hence,%
\begin{align}
\left(  \Delta\circ\widetilde{f}\right)  \left(  s\right)   &  =\Delta\left(
\underbrace{\widetilde{f}\left(  s\right)  }_{=\iota_{\operatorname*{Sym}%
,V}\left(  f\left(  v\right)  \right)  }\right)  =\Delta\left(  \iota
_{\operatorname*{Sym},V}\left(  f\left(  v\right)  \right)  \right)
\nonumber\\
&  =\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)  \right)
\otimes1+1\otimes\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)
\right)  \label{pf.prop.Sym.bider.b.Srestr1.pf.3}%
\end{align}
(by (\ref{pf.prop.Sym.bider.b.Srestr1.pf.1}), applied to $w=f\left(  v\right)
$).
\par
On the other hand,%
\begin{align*}
&  \left(  \left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
+\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\right)  \circ
\Delta\right)  \left(  s\right) \\
&  =\left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
+\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\right)  \left(
\Delta\left(  \underbrace{s}_{=\iota_{\operatorname*{Sym},V}\left(  v\right)
}\right)  \right) \\
&  =\left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
+\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\right)  \left(
\underbrace{\Delta\left(  \iota_{\operatorname*{Sym},V}\left(  v\right)
\right)  }_{\substack{=\iota_{\operatorname*{Sym},V}\left(  v\right)
\otimes1+1\otimes\iota_{\operatorname*{Sym},V}\left(  v\right)  \\\text{(by
(\ref{pf.prop.Sym.bider.b.Srestr1.pf.1}), applied to }w=v\text{)}}}\right) \\
&  =\left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
+\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\right)  \left(
\iota_{\operatorname*{Sym},V}\left(  v\right)  \otimes1+1\otimes
\iota_{\operatorname*{Sym},V}\left(  v\right)  \right) \\
&  =\left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}\right)
\left(  \iota_{\operatorname*{Sym},V}\left(  v\right)  \otimes1+1\otimes
\iota_{\operatorname*{Sym},V}\left(  v\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  \operatorname*{id}\nolimits_{A}\otimes
\widetilde{f}\right)  \left(  \iota_{\operatorname*{Sym},V}\left(  v\right)
\otimes1+1\otimes\iota_{\operatorname*{Sym},V}\left(  v\right)  \right)  .
\end{align*}
Since%
\begin{align*}
&  \left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}\right)  \left(
\iota_{\operatorname*{Sym},V}\left(  v\right)  \otimes1+1\otimes
\iota_{\operatorname*{Sym},V}\left(  v\right)  \right) \\
&  =\underbrace{\left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
\right)  \left(  \iota_{\operatorname*{Sym},V}\left(  v\right)  \otimes
1\right)  }_{=\widetilde{f}\left(  \iota_{\operatorname*{Sym},V}\left(
v\right)  \right)  \otimes\operatorname*{id}\nolimits_{A}\left(  1\right)
}+\underbrace{\left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
\right)  \left(  1\otimes\iota_{\operatorname*{Sym},V}\left(  v\right)
\right)  }_{=\widetilde{f}\left(  1\right)  \otimes\operatorname*{id}%
\nolimits_{A}\left(  \iota_{\operatorname*{Sym},V}\left(  v\right)  \right)
}\\
&  =\underbrace{\widetilde{f}\left(  \iota_{\operatorname*{Sym},V}\left(
v\right)  \right)  }_{=\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)
\right)  }\otimes\underbrace{\operatorname*{id}\nolimits_{A}\left(  1\right)
}_{=1}+\underbrace{\widetilde{f}\left(  1\right)  }_{=0}\otimes
\operatorname*{id}\nolimits_{A}\left(  \iota_{\operatorname*{Sym},V}\left(
v\right)  \right) \\
&  =\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)  \right)
\otimes1+\underbrace{0\otimes\operatorname*{id}\nolimits_{A}\left(
\iota_{\operatorname*{Sym},V}\left(  v\right)  \right)  }_{=0}=\iota
_{\operatorname*{Sym},V}\left(  f\left(  v\right)  \right)  \otimes1
\end{align*}
and%
\begin{align*}
&  \left(  \operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\right)  \left(
\iota_{\operatorname*{Sym},V}\left(  v\right)  \otimes1+1\otimes
\iota_{\operatorname*{Sym},V}\left(  v\right)  \right) \\
&  =\underbrace{\left(  \operatorname*{id}\nolimits_{A}\otimes\widetilde{f}%
\right)  \left(  \iota_{\operatorname*{Sym},V}\left(  v\right)  \otimes
1\right)  }_{=\operatorname*{id}\nolimits_{A}\left(  \iota
_{\operatorname*{Sym},V}\left(  v\right)  \right)  \otimes\widetilde{f}\left(
1\right)  }+\underbrace{\left(  \operatorname*{id}\nolimits_{A}\otimes
\widetilde{f}\right)  \left(  1\otimes\iota_{\operatorname*{Sym},V}\left(
v\right)  \right)  }_{=\operatorname*{id}\nolimits_{A}\left(  1\right)
\otimes\widetilde{f}\left(  \iota_{\operatorname*{Sym},V}\left(  v\right)
\right)  }\\
&  =\operatorname*{id}\nolimits_{A}\left(  \iota_{\operatorname*{Sym}%
,V}\left(  v\right)  \right)  \otimes\underbrace{\widetilde{f}\left(
1\right)  }_{=0}+\underbrace{\operatorname*{id}\nolimits_{A}\left(  1\right)
}_{=1}\otimes\underbrace{\widetilde{f}\left(  \iota_{\operatorname*{Sym}%
,V}\left(  v\right)  \right)  }_{=\iota_{\operatorname*{Sym},V}\left(
f\left(  v\right)  \right)  }\\
&  =\underbrace{\operatorname*{id}\nolimits_{A}\left(  \iota
_{\operatorname*{Sym},V}\left(  v\right)  \right)  \otimes0}_{=0}%
+1\otimes\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)  \right)
=1\otimes\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)  \right)  ,
\end{align*}
this becomes%
\begin{align*}
&  \left(  \left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
+\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\right)  \circ
\Delta\right)  \left(  s\right) \\
&  =\underbrace{\left(  \widetilde{f}\otimes\operatorname*{id}\nolimits_{A}%
\right)  \left(  \iota_{\operatorname*{Sym},V}\left(  v\right)  \otimes
1+1\otimes\iota_{\operatorname*{Sym},V}\left(  v\right)  \right)  }%
_{=\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)  \right)  \otimes
1}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \operatorname*{id}\nolimits_{A}%
\otimes\widetilde{f}\right)  \left(  \iota_{\operatorname*{Sym},V}\left(
v\right)  \otimes1+1\otimes\iota_{\operatorname*{Sym},V}\left(  v\right)
\right)  }_{=1\otimes\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)
\right)  }\\
&  =\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)  \right)
\otimes1+1\otimes\iota_{\operatorname*{Sym},V}\left(  f\left(  v\right)
\right)  .
\end{align*}
Compared with (\ref{pf.prop.Sym.bider.b.Srestr1.pf.3}), this yields $\left(
\Delta\circ\widetilde{f}\right)  \left(  s\right)  =\left(  \left(
\widetilde{f}\otimes\operatorname*{id}\nolimits_{A}+\operatorname*{id}%
\nolimits_{A}\otimes\widetilde{f}\right)  \circ\Delta\right)  \left(
s\right)  $.
\par
Now, let us forget that we fixed $s$. We thus have shown that $\left(
\Delta\circ\widetilde{f}\right)  \left(  s\right)  =\left(  \left(
\widetilde{f}\otimes\operatorname*{id}\nolimits_{A}+\operatorname*{id}%
\nolimits_{A}\otimes\widetilde{f}\right)  \circ\Delta\right)  \left(
s\right)  $ for every $s\in S$. In other words, $\left(  \Delta\circ
\widetilde{f}\right)  \mid_{S}=\left(  \left(  \widetilde{f}\otimes
\operatorname*{id}\nolimits_{A}+\operatorname*{id}\nolimits_{A}\otimes
\widetilde{f}\right)  \circ\Delta\right)  \mid_{S}$. This proves
(\ref{pf.prop.Sym.bider.b.Srestr1}).}. Hence, Lemma \ref{lem.Der.fdef}
(applied to $A\otimes A$, $\Delta$, $\widetilde{f}$ and $\widetilde{f}%
\otimes\operatorname*{id}\nolimits_{A}+\operatorname*{id}\nolimits_{A}%
\otimes\widetilde{f}$ instead of $B$, $f$, $d$ and $e$) shows that
$\Delta\circ\widetilde{f}=\left(  \widetilde{f}\otimes\operatorname*{id}%
\nolimits_{A}+\operatorname*{id}\nolimits_{A}\otimes\widetilde{f}\right)
\circ\Delta$. In other words, $\Delta\circ\widetilde{f}=\left(  \widetilde{f}%
\otimes\operatorname*{id}+\operatorname*{id}\otimes\widetilde{f}\right)
\circ\Delta$. In other words, $\widetilde{f}$ is a coderivation (by the
definition of a \textquotedblleft coderivation\textquotedblright). This proves
Proposition \ref{prop.Sym.bider} \textbf{(b)}.
\end{proof}

\subsection{$U\left(  \mathfrak{g}\right)  $ as a filtered $\mathbf{k}%
$-algebra}

\begin{definition}
\label{def.U(g)}Let $\mathfrak{g}$ be a Lie algebra. We recall that the
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $ is defined as
the quotient algebra of the $\mathbf{k}$-algebra $T\left(  \mathfrak{g}%
\right)  $ by its ideal $J_{\mathfrak{g}}$, where $J_{\mathfrak{g}}$ is the
ideal of $T\left(  \mathfrak{g}\right)  $ generated by all tensors of the form%
\[
v\otimes w-w\otimes v-\left[  v,w\right]  \ \ \ \ \ \ \ \ \ \ \text{with }%
v\in\mathfrak{g}\text{ and }w\in\mathfrak{g}.
\]
We let $\pi_{U,\mathfrak{g}}$ be the canonical projection from $T\left(
\mathfrak{g}\right)  $ to its quotient $\mathbf{k}$-algebra $U\left(
\mathfrak{g}\right)  $. (Here, the \textquotedblleft$U$\textquotedblright\ in
\textquotedblleft$\pi_{U,\mathfrak{g}}$\textquotedblright\ is not a variable,
but stands for the letter \textquotedblleft u\textquotedblright\ in
\textquotedblleft\textbf{u}niversal enveloping algebra\textquotedblright.)

As we know, $T\left(  \mathfrak{g}\right)  $ is a graded $\mathbf{k}$-algebra,
and thus a filtered $\mathbf{k}$-algebra. Its filtration is $\left(
\mathfrak{g}^{\otimes\leq m}\right)  _{m\geq0}$, where we set%
\[
\mathfrak{g}^{\otimes\leq m}=\bigoplus\limits_{n=0}^{m}\mathfrak{g}^{\otimes
n}\ \ \ \ \ \ \ \ \ \ \text{for every }m\in\mathbb{N}.
\]
This filtration $\left(  \mathfrak{g}^{\otimes\leq m}\right)  _{m\geq0}$ on
$T\left(  \mathfrak{g}\right)  $ gives rise to a filtration $\left(
\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq m}\right)  \right)
_{m\geq0}$ on the quotient algebra $U\left(  \mathfrak{g}\right)  $. This
latter filtration will be denoted by $\left(  U_{\leq m}\left(  \mathfrak{g}%
\right)  \right)  _{m\geq0}$ (thus, we set $U_{\leq m}\left(  \mathfrak{g}%
\right)  =\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq m}\right)  $
for every $m\in\mathbb{N}$), and makes $U\left(  \mathfrak{g}\right)  $ into a
filtered $\mathbf{k}$-algebra.

If $\mathfrak{g}$ is an abelian Lie algebra (i.e., a Lie algebra satisfying
$\left[  v,w\right]  =0$ for all $v\in\mathfrak{g}$ and $w\in\mathfrak{g}$),
then $U\left(  \mathfrak{g}\right)  =\operatorname*{Sym}\mathfrak{g}$.
However, in general, $U\left(  \mathfrak{g}\right)  $ differs noticeably from
$\operatorname*{Sym}\mathfrak{g}$ (and is usually noncommutative, unlike
$\operatorname*{Sym}\mathfrak{g}$). Various results comparing $U\left(
\mathfrak{g}\right)  $ to $\operatorname*{Sym}\mathfrak{g}$ are collectively
known as \textquotedblleft Poincar\'{e}-Birkhoff-Witt
theorems\textquotedblright; usually, they state (under certain requirements)
either that $U\left(  \mathfrak{g}\right)  \cong\operatorname*{Sym}%
\mathfrak{g}$ as $\mathbf{k}$-modules (or, sometimes, as $\mathfrak{g}%
$-modules when equipped with appropriate $\mathfrak{g}$-module structures; but
almost never as $\mathbf{k}$-algebras), or that $\operatorname*{gr}\left(
U\left(  \mathfrak{g}\right)  \right)  \cong\operatorname*{Sym}\mathfrak{g}$
as $\mathbf{k}$-algebras. See \cite[Theorem 5.9]{Gri-thes} for a collection of
such \textquotedblleft Poincar\'{e}-Birkhoff-Witt theorems\textquotedblright.
In this note, we shall explore one such theorem, which has a condition saying
that the Lie algebra structure $\mathfrak{g}$ comes from a pre-Lie algebra
structure. (We shall introduce the latter notion in due time.)

Let $\iota_{U,\mathfrak{g}}$ denote the map $\pi_{U,\mathfrak{g}}\circ
\iota_{T,\mathfrak{g}}:\mathfrak{g}\rightarrow U\left(  \mathfrak{g}\right)
$. This map $\iota_{U,\mathfrak{g}}$ is often called the \textquotedblleft
canonical inclusion of $\mathfrak{g}$ into $U\left(  \mathfrak{g}\right)
$\textquotedblright, although in some (rather perverse) cases it fails to be
injective. For every $a\in\mathfrak{g}$, we have%
\begin{equation}
\underbrace{\iota_{U,\mathfrak{g}}}_{=\pi_{U,\mathfrak{g}}\circ\iota
_{T,\mathfrak{g}}}\left(  a\right)  =\left(  \pi_{U,\mathfrak{g}}\circ
\iota_{T,\mathfrak{g}}\right)  \left(  a\right)  =\pi_{U,\mathfrak{g}}\left(
\underbrace{\iota_{T,\mathfrak{g}}\left(  a\right)  }_{=a}\right)
=\pi_{U,\mathfrak{g}}\left(  a\right)  . \label{eq.def.U(g).iUg(a)}%
\end{equation}
It is common to write $a$ for this element $\iota_{U,\mathfrak{g}}\left(
a\right)  =\pi_{U,\mathfrak{g}}\left(  a\right)  $ of $U\left(  \mathfrak{g}%
\right)  $, but this notation is slightly abusive (since, as explained above,
$\iota_{U,\mathfrak{g}}$ is not always injective).

Let $\iota_{\operatorname*{gr}U,\mathfrak{g}}$ denote the map $\mathfrak{g}%
\rightarrow\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)  $
which sends every $v\in\mathfrak{g}$ to $\left[  \iota_{U,\mathfrak{g}}\left(
v\right)  \right]  _{1}\in\operatorname*{gr}\nolimits_{1}\left(  U\left(
\mathfrak{g}\right)  \right)  \subseteq\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  $.
\end{definition}

\begin{proposition}
\label{prop.PBW-map}Let $\mathfrak{g}$ be a Lie algebra.

\textbf{(a)} The $\mathbf{k}$-algebra $U\left(  \mathfrak{g}\right)  $ is
generated by its subset $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $.

\textbf{(b)} The $\mathbf{k}$-algebra $\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  $ is generated by its subset $\iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $.

\textbf{(c)} The $\mathbf{k}$-algebra $\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  $ is commutative.

\textbf{(d)} There exists a unique $\mathbf{k}$-algebra homomorphism
$\Phi:\operatorname*{Sym}\mathfrak{g}\rightarrow\operatorname*{gr}\left(
U\left(  \mathfrak{g}\right)  \right)  $ such that $\Phi\circ\iota
_{\operatorname*{Sym},\mathfrak{g}}=\iota_{\operatorname*{gr}U,\mathfrak{g}}$.
Let us denote this $\mathbf{k}$-algebra homomorphism $\Phi$ by
$\operatorname*{PBW}\nolimits_{\mathfrak{g}}$, and call it the \textit{PBW
homomorphism} of $\mathfrak{g}$. (As usual, \textquotedblleft
PBW\textquotedblright\ stands for \textquotedblleft
Poincar\'{e}-Birkhoff-Witt\textquotedblright.)

\textbf{(e)} The PBW homomorphism $\operatorname*{PBW}\nolimits_{\mathfrak{g}%
}$ is graded and surjective.

\textbf{(f)} Let $n\in\mathbb{N}$. The $\mathbf{k}$-module $\operatorname*{gr}%
\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)  $ is spanned by
the elements $\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{n}\right)  \right]  _{n}$ with $a_{1},a_{2},\ldots,a_{n}\in V$.
\end{proposition}

Proposition \ref{prop.PBW-map} is well-known, but let us give a proof for the
sake of completeness:

\begin{proof}
[Proof of Proposition \ref{prop.PBW-map}.]\textbf{(a)} The $\mathbf{k}%
$-algebra $T\left(  \mathfrak{g}\right)  $ is generated by its subset
$\iota_{T,\mathfrak{g}}\left(  \mathfrak{g}\right)  $ (since every tensor is a
$\mathbf{k}$-linear combination of pure tensors). Hence, the $\mathbf{k}%
$-algebra $U\left(  \mathfrak{g}\right)  =\pi_{U,\mathfrak{g}}\left(  T\left(
\mathfrak{g}\right)  \right)  $ is generated by its subset $\pi
_{U,\mathfrak{g}}\left(  \iota_{T,\mathfrak{g}}\left(  \mathfrak{g}\right)
\right)  =\underbrace{\left(  \pi_{U,\mathfrak{g}}\circ\iota_{T,\mathfrak{g}%
}\right)  }_{=\iota_{U,\mathfrak{g}}}\left(  \mathfrak{g}\right)
=\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $. This proves Proposition
\ref{prop.PBW-map} \textbf{(a)}.

\textbf{(b)} The projection $\pi_{U,\mathfrak{g}}:T\left(  \mathfrak{g}%
\right)  \rightarrow U\left(  \mathfrak{g}\right)  $ is filtered (due to the
definition of the filtration on $U\left(  \mathfrak{g}\right)  $). Hence, it
gives rise to a graded $\mathbf{k}$-linear map $\operatorname*{gr}\left(
\pi_{U,\mathfrak{g}}\right)  :\operatorname*{gr}\left(  T\left(
\mathfrak{g}\right)  \right)  \rightarrow\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  $. This $\mathbf{k}$-linear map
$\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  $ is a graded
$\mathbf{k}$-algebra homomorphism (because of Remark \ref{rmk.gr.alg.map}). It
is easy to see that%
\begin{equation}
\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)
=\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)
\left(  \operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}
\label{pf.prop.PBW-map.b.grn1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.PBW-map.b.grn1}):} Let $n\in
\mathbb{N}$. Let $p\in\operatorname*{gr}\nolimits_{n}\left(  U\left(
\mathfrak{g}\right)  \right)  $. We shall show that $p\in\left(
\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)  \left(
\operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)  \right)
\right)  $.
\par
We have $p\in\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}%
\right)  \right)  =\left(  \pi_{U,\mathfrak{g}}\left(  \mathfrak{g}%
^{\otimes\leq n}\right)  \right)  /\left(  \pi_{U,\mathfrak{g}}\left(
\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }\right)  \right)  $ (since the
filtration on $U\left(  \mathfrak{g}\right)  $ was defined to be $\left(
\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq m}\right)  \right)
_{m\geq0}$). Hence, we can write $p$ in the form $p=\left[  q\right]  _{n}$
for some $q\in\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq n}\right)
$. Consider this $q$. We have $q\in\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}%
^{\otimes\leq n}\right)  $. Thus, $q=\pi_{U,\mathfrak{g}}\left(  Q\right)  $
for some $Q\in\mathfrak{g}^{\otimes\leq n}$. Consider this $Q$.
\par
We have $Q\in\mathfrak{g}^{\otimes\leq n}=\mathfrak{g}^{\otimes n}%
+\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }$. Thus, $Q=Q_{1}+Q_{2}$ for
some $Q_{1}\in\mathfrak{g}^{\otimes n}$ and $Q_{2}\in\mathfrak{g}^{\otimes
\leq\left(  n-1\right)  }$. Consider these $Q_{1}$ and $Q_{2}$. We have%
\begin{align*}
q  &  =\pi_{U,\mathfrak{g}}\left(  \underbrace{Q}_{=Q_{1}+Q_{2}}\right)
=\pi_{U,\mathfrak{g}}\left(  Q_{1}+Q_{2}\right)  =\pi_{U,\mathfrak{g}}\left(
Q_{1}\right)  +\pi_{U,\mathfrak{g}}\left(  \underbrace{Q_{2}}_{\in
\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }}\right) \\
&  \in\pi_{U,\mathfrak{g}}\left(  Q_{1}\right)  +\pi_{U,\mathfrak{g}}\left(
\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }\right)  ,
\end{align*}
so that $q\equiv\pi_{U,\mathfrak{g}}\left(  Q_{1}\right)  \operatorname{mod}%
\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq\left(  n-1\right)
}\right)  $. In other words, $\left[  q\right]  _{n}=\left[  \pi
_{U,\mathfrak{g}}\left(  Q_{1}\right)  \right]  _{n}$.
\par
On the other hand, consider the element $\left[  Q_{1}\right]  _{n}$ of
$\mathfrak{g}^{\otimes\leq n}/\mathfrak{g}^{\otimes\leq\left(  n-1\right)
}=\operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)
\right)  $. By the definition of $\operatorname*{gr}\left(  \pi
_{U,\mathfrak{g}}\right)  $, we have $\left(  \operatorname*{gr}\left(
\pi_{U,\mathfrak{g}}\right)  \right)  \left(  \left[  Q_{1}\right]
_{n}\right)  =\left[  \pi_{U,\mathfrak{g}}\left(  Q_{1}\right)  \right]  _{n}%
$. Comparing this with $\left[  q\right]  _{n}=\left[  \pi_{U,\mathfrak{g}%
}\left(  Q_{1}\right)  \right]  _{n}$, we obtain $\left[  q\right]
_{n}=\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)
\left(  \left[  Q_{1}\right]  _{n}\right)  $. Hence, $p=\left[  q\right]
_{n}=\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)
\left(  \underbrace{\left[  Q_{1}\right]  _{n}}_{\in\operatorname*{gr}%
\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)  \right)  }\right)
\in\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)
\left(  \operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)
\right)  \right)  $.
\par
Let us now forget that we fixed $p$. We thus have proven that $p\in\left(
\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)  \left(
\operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)  \right)
\right)  $ for every $p\in\operatorname*{gr}\nolimits_{n}\left(  U\left(
\mathfrak{g}\right)  \right)  $. In other words, $\operatorname*{gr}%
\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)  \subseteq\left(
\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)  \left(
\operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)  \right)
\right)  $. Combined with the inclusion $\left(  \operatorname*{gr}\left(
\pi_{U,\mathfrak{g}}\right)  \right)  \left(  \operatorname*{gr}%
\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)  \right)  \right)
\subseteq\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)
\right)  $ (which is obvious, since the map $\operatorname*{gr}\left(
\pi_{U,\mathfrak{g}}\right)  $ is graded), this results in $\operatorname*{gr}%
\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)  =\left(
\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)  \left(
\operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)  \right)
\right)  $. This proves (\ref{pf.prop.PBW-map.b.grn1}).}.

On the other hand, the $\mathbf{k}$-algebra $T\left(  \mathfrak{g}\right)  $
is graded. Thus, $\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)
}:T\left(  \mathfrak{g}\right)  \rightarrow\operatorname*{gr}\left(  T\left(
\mathfrak{g}\right)  \right)  $ is a graded $\mathbf{k}$-algebra isomorphism
(by Remark \ref{rmk.grad.alg}). Define a $\mathbf{k}$-linear map
$\omega:T\left(  \mathfrak{g}\right)  \rightarrow\operatorname*{gr}\left(
U\left(  \mathfrak{g}\right)  \right)  $ by%
\[
\omega=\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \circ
\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }.
\]
Then, $\omega$ is a graded $\mathbf{k}$-algebra homomorphism\footnote{since it
is the composition of the graded $\mathbf{k}$-algebra homomorphisms
$\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  $ and
$\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }$}. For every
$n\in\mathbb{N}$ and $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$, we have
\begin{equation}
\omega\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  =\left[
\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}
\label{pf.prop.PBW-map.b.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.PBW-map.b.1}):} Let $n\in\mathbb{N}$
and $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$. Then, $a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}\in\mathfrak{g}^{\otimes n}$. Hence, the
definition of $\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }$
yields $\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }\left(
a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  =\left[  a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}\right]  _{n}$. Now, from $\omega
=\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \circ
\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }$, we obtain%
\begin{align}
\omega\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)   &
=\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)
\left(  \underbrace{\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}%
\right)  }\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)
}_{=\left[  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right]  _{n}}\right)
\nonumber\\
&  =\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)
\left(  \left[  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right]
_{n}\right)  =\left[  \pi_{U,\mathfrak{g}}\left(  a_{1}\otimes a_{2}%
\otimes\cdots\otimes a_{n}\right)  \right]  _{n}
\label{pf.prop.PBW-map.b.1.pf.1}%
\end{align}
(by the definition of $\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)
$).
\par
But (\ref{eq.def.U(g).iUg(a)}) (applied to $a=a_{i}$ for each $i\in\left\{
1,2,\ldots,n\right\}  $, and then multiplied) shows that%
\begin{align*}
\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)   &
=\pi_{U,\mathfrak{g}}\left(  a_{1}\right)  \pi_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\pi_{U,\mathfrak{g}}\left(  a_{n}\right) \\
&  =\pi_{U,\mathfrak{g}}\underbrace{\left(  a_{1}a_{2}\cdots a_{n}\right)
}_{=a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\pi_{U,\mathfrak{g}}\text{ is a }\mathbf{k}\text{-algebra
homomorphism}\right) \\
&  =\pi_{U,\mathfrak{g}}\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\right)  .
\end{align*}
Hence, (\ref{pf.prop.PBW-map.b.1.pf.1}) becomes%
\[
\omega\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  =\left[
\underbrace{\pi_{U,\mathfrak{g}}\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes
a_{n}\right)  }_{=\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota
_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(
a_{n}\right)  }\right]  _{n}=\left[  \iota_{U,\mathfrak{g}}\left(
a_{1}\right)  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}.
\]
This proves (\ref{pf.prop.PBW-map.b.1}).}. Furthermore, we have%
\begin{equation}
\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)
=\omega\left(  \mathfrak{g}^{\otimes n}\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N} \label{pf.prop.PBW-map.b.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.PBW-map.b.3}):} Let $n\in\mathbb{N}$.
The map $\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }$ is a
graded $\mathbf{k}$-algebra isomorphism. Thus, it is invertible, and its
inverse is again graded. Hence, $\operatorname*{grad}\nolimits_{T\left(
\mathfrak{g}\right)  }\left(  \mathfrak{g}^{\otimes n}\right)
=\operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)  \right)
$.
\par
Now, $\omega=\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)
\circ\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }$, so that%
\begin{align*}
\omega\left(  \mathfrak{g}^{\otimes n}\right)   &  =\left(  \operatorname*{gr}%
\left(  \pi_{U,\mathfrak{g}}\right)  \circ\operatorname*{grad}%
\nolimits_{T\left(  \mathfrak{g}\right)  }\right)  \left(  \mathfrak{g}%
^{\otimes n}\right)  =\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}%
}\right)  \right)  \left(  \underbrace{\operatorname*{grad}\nolimits_{T\left(
\mathfrak{g}\right)  }\left(  \mathfrak{g}^{\otimes n}\right)  }%
_{=\operatorname*{gr}\nolimits_{n}\left(  T\left(  \mathfrak{g}\right)
\right)  }\right)  =\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}%
}\right)  \right)  \left(  \operatorname*{gr}\nolimits_{n}\left(  T\left(
\mathfrak{g}\right)  \right)  \right) \\
&  =\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.PBW-map.b.grn1}%
)}\right)  .
\end{align*}
This proves (\ref{pf.prop.PBW-map.b.3}).}.

Let $Z$ be the $\mathbf{k}$-subalgebra of $\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  $ generated by $\iota_{\operatorname*{gr}%
U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $. Our goal is to prove that the
$\mathbf{k}$-algebra $\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)
\right)  $ is generated by its subset $\iota_{\operatorname*{gr}%
U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $. In other words, our goal is to
prove that $\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)
=Z$.

Clearly, $\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}\right)
\subseteq Z$ (since $Z$ is the $\mathbf{k}$-subalgebra of $\operatorname*{gr}%
\left(  U\left(  \mathfrak{g}\right)  \right)  $ generated by $\iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $).

Let $n\in\mathbb{N}$. For every $a\in\mathfrak{g}$, we have $\iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  a\right)  =\left[  \iota
_{U,\mathfrak{g}}\left(  a\right)  \right]  _{1}$ (by the definition of
$\iota_{\operatorname*{gr}U,\mathfrak{g}}$). Thus, for every $a_{1}%
,a_{2},\ldots,a_{n}\in\mathfrak{g}$, we have%
\begin{align}
&  \underbrace{\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  a_{1}\right)
}_{=\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \right]  _{1}%
}\underbrace{\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  a_{2}\right)
}_{=\left[  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \right]  _{1}}%
\cdots\underbrace{\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(
a_{n}\right)  }_{=\left[  \iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]
_{1}}\nonumber\\
&  =\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \right]  _{1}\left[
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \right]  _{1}\cdots\left[
\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{1}=\left[
\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]
_{n}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of multiplication on
}\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)  \right)
\nonumber\\
&  =\omega\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.PBW-map.b.1})}\right)  .
\label{pf.prop.PBW-map.b.14}%
\end{align}
Hence, for every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$, we have%
\begin{align}
\omega\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)   &
=\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  \underbrace{a_{1}}%
_{\in\mathfrak{g}}\right)  \iota_{\operatorname*{gr}U,\mathfrak{g}}\left(
\underbrace{a_{2}}_{\in\mathfrak{g}}\right)  \cdots\iota_{\operatorname*{gr}%
U,\mathfrak{g}}\left(  \underbrace{a_{n}}_{\in\mathfrak{g}}\right) \nonumber\\
&  \in\underbrace{\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(
\mathfrak{g}\right)  \iota_{\operatorname*{gr}U,\mathfrak{g}}\left(
\mathfrak{g}\right)  \cdots\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(
\mathfrak{g}\right)  }_{n\text{ times}}=\left(  \underbrace{\iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}\right)  }_{\subseteq
Z}\right)  ^{n}\subseteq Z^{n}\subseteq Z \label{pf.prop.PBW-map.b.15}%
\end{align}
(since $Z$ is a $\mathbf{k}$-subalgebra).

Now, the $\mathbf{k}$-module $\mathfrak{g}^{\otimes n}$ is spanned by the pure
tensors $a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}$ with $a_{1}%
,a_{2},\ldots,a_{n}\in\mathfrak{g}$. Hence, its image $\omega\left(
\mathfrak{g}^{\otimes n}\right)  $ under the map $\omega$ is spanned by the
elements $\omega\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  $
with $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$. Since all of these elements
$\omega\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  $ belong
to $Z$ (according to (\ref{pf.prop.PBW-map.b.15})), this shows that
$\omega\left(  \mathfrak{g}^{\otimes n}\right)  \subseteq Z$.

Now, let us forget that we fixed $n$. We thus have proven that $\omega\left(
\mathfrak{g}^{\otimes n}\right)  \subseteq Z$ for every $n\in\mathbb{N}$. Now,%
\[
\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)
=\bigoplus_{n\geq0}\operatorname*{gr}\nolimits_{n}\left(  U\left(
\mathfrak{g}\right)  \right)  =\sum_{n\geq0}\underbrace{\operatorname*{gr}%
\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)  }%
_{\substack{=\omega\left(  \mathfrak{g}^{\otimes n}\right)  \\\text{(by
(\ref{pf.prop.PBW-map.b.3}))}}}=\sum_{n\geq0}\underbrace{\omega\left(
\mathfrak{g}^{\otimes n}\right)  }_{\subseteq Z}\subseteq\sum_{n\geq
0}Z\subseteq Z.
\]
Combined with $Z\subseteq\operatorname*{gr}\left(  U\left(  \mathfrak{g}%
\right)  \right)  $ (which is obvious), this shows that $\operatorname*{gr}%
\left(  U\left(  \mathfrak{g}\right)  \right)  =Z$. This completes the proof
of Proposition \ref{prop.PBW-map} \textbf{(b)}.

\textbf{(c)} It is known that a $\mathbf{k}$-algebra is commutative if it has
a generating set whose elements commute pairwise.

Now, it is easy to see that the elements of the subset $\iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $ of
$\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)  $ commute
pairwise\footnote{\textit{Proof.} Let $x\in\iota_{\operatorname*{gr}%
U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $ and $y\in\iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $. We must
prove that $xy=yx$.
\par
We have $x\in\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}%
\right)  $. In other words, there exists $a\in\mathfrak{g}$ such that
$x=\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  a\right)  $. Consider this
$a$.
\par
We have $y\in\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}%
\right)  $. In other words, there exists $b\in\mathfrak{g}$ such that
$y=\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  b\right)  $. Consider this
$b$.
\par
\begin{itemize}
\item The element $a\otimes b-b\otimes a-\left[  a,b\right]  $ of $T\left(
\mathfrak{g}\right)  $ belongs to the ideal $J_{\mathfrak{g}}$ defined in
Definition \ref{def.U(g)}. Since $U\left(  \mathfrak{g}\right)  $ is the
quotient of $T\left(  \mathfrak{g}\right)  $ modulo this ideal, this shows
that $\pi_{U,\mathfrak{g}}\left(  a\otimes b-b\otimes a-\left[  a,b\right]
\right)  =0$. Hence,
\begin{align*}
0  &  =\pi_{U,\mathfrak{g}}\left(  \underbrace{a\otimes b}_{=ab}%
-\underbrace{b\otimes a}_{=ba}-\left[  a,b\right]  \right) \\
&  =\pi_{U,\mathfrak{g}}\left(  ab-ba-\left[  a,b\right]  \right)
=\underbrace{\pi_{U,\mathfrak{g}}\left(  a\right)  }_{=\iota_{U,\mathfrak{g}%
}\left(  a\right)  }\underbrace{\pi_{U,\mathfrak{g}}\left(  b\right)
}_{=\iota_{U,\mathfrak{g}}\left(  b\right)  }-\underbrace{\pi_{U,\mathfrak{g}%
}\left(  b\right)  }_{=\iota_{U,\mathfrak{g}}\left(  b\right)  }%
\underbrace{\pi_{U,\mathfrak{g}}\left(  a\right)  }_{=\iota_{U,\mathfrak{g}%
}\left(  a\right)  }-\pi_{U,\mathfrak{g}}\left(  \left[  a,b\right]  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi_{U,\mathfrak{g}}\text{ is a
}\mathbf{k}\text{-algebra homomorphism}\right) \\
&  =\iota_{U,\mathfrak{g}}\left(  a\right)  \iota_{U,\mathfrak{g}}\left(
b\right)  -\iota_{U,\mathfrak{g}}\left(  b\right)  \iota_{U,\mathfrak{g}%
}\left(  a\right)  -\pi_{U,\mathfrak{g}}\left(  \left[  a,b\right]  \right)  .
\end{align*}
Thus,
\begin{equation}
\iota_{U,\mathfrak{g}}\left(  a\right)  \iota_{U,\mathfrak{g}}\left(
b\right)  -\iota_{U,\mathfrak{g}}\left(  b\right)  \iota_{U,\mathfrak{g}%
}\left(  a\right)  =\pi_{U,\mathfrak{g}}\left(  \underbrace{\left[
a,b\right]  }_{\in\mathfrak{g}^{\otimes1}\subseteq\mathfrak{g}^{\otimes\leq1}%
}\right)  \in\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq1}\right)  .
\label{pf.prop.PBW-map.c.3}%
\end{equation}
\end{itemize}
\par
We have $x=\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  a\right)  =\left[
\iota_{U,\mathfrak{g}}\left(  a\right)  \right]  _{1}$ (by the definition of
$\iota_{\operatorname*{gr}U,\mathfrak{g}}$) and $y=\left[  \iota
_{U,\mathfrak{g}}\left(  b\right)  \right]  _{1}$ (similarly). Hence,%
\begin{align*}
&  \underbrace{x}_{=\left[  \iota_{U,\mathfrak{g}}\left(  a\right)  \right]
_{1}}\underbrace{y}_{=\left[  \iota_{U,\mathfrak{g}}\left(  b\right)  \right]
_{1}}-\underbrace{y}_{=\left[  \iota_{U,\mathfrak{g}}\left(  b\right)
\right]  _{1}}\underbrace{x}_{=\left[  \iota_{U,\mathfrak{g}}\left(  a\right)
\right]  _{1}}\\
&  =\underbrace{\left[  \iota_{U,\mathfrak{g}}\left(  a\right)  \right]
_{1}\left[  \iota_{U,\mathfrak{g}}\left(  b\right)  \right]  _{1}%
}_{\substack{=\left[  \iota_{U,\mathfrak{g}}\left(  a\right)  \iota
_{U,\mathfrak{g}}\left(  b\right)  \right]  _{2}\\\text{(by the definition of
the product in }\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)
\right)  \text{)}}}-\underbrace{\left[  \iota_{U,\mathfrak{g}}\left(
b\right)  \right]  _{1}\left[  \iota_{U,\mathfrak{g}}\left(  a\right)
\right]  _{1}}_{\substack{=\left[  \iota_{U,\mathfrak{g}}\left(  b\right)
\iota_{U,\mathfrak{g}}\left(  a\right)  \right]  _{2}\\\text{(by the
definition of the product in }\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  \text{)}}}\\
&  =\left[  \iota_{U,\mathfrak{g}}\left(  a\right)  \iota_{U,\mathfrak{g}%
}\left(  b\right)  \right]  _{2}-\left[  \iota_{U,\mathfrak{g}}\left(
b\right)  \iota_{U,\mathfrak{g}}\left(  a\right)  \right]  _{2}=\left[
\iota_{U,\mathfrak{g}}\left(  a\right)  \iota_{U,\mathfrak{g}}\left(
b\right)  -\iota_{U,\mathfrak{g}}\left(  b\right)  \iota_{U,\mathfrak{g}%
}\left(  a\right)  \right]  _{2}=0
\end{align*}
(by (\ref{pf.prop.PBW-map.c.3})). In other words, $xy=yx$, qed.}. Thus the
$\mathbf{k}$-algebra $\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)
\right)  $ has a generating set whose elements commute pairwise (namely, the
set $\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  \mathfrak{g}\right)  $,
because Proposition \ref{prop.PBW-map} \textbf{(b)} shows that this set is a
generating set for $\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)
\right)  $). Therefore, the $\mathbf{k}$-algebra $\operatorname*{gr}\left(
U\left(  \mathfrak{g}\right)  \right)  $ is commutative (since a $\mathbf{k}%
$-algebra is commutative if it has a generating set whose elements pairwise
commute). This proves Proposition \ref{prop.PBW-map} \textbf{(c)}.

\textbf{(d)} Proposition \ref{prop.PBW-map} \textbf{(c)} shows that the
$\mathbf{k}$-algebra $\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)
\right)  $ is commutative. Hence, Proposition \ref{prop.Sym.univ} (applied to
$V=\mathfrak{g}$, $A=\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)
\right)  $ and $f=\iota_{\operatorname*{gr}U,\mathfrak{g}}$) shows that there
exists a unique $\mathbf{k}$-algebra homomorphism $F:\operatorname*{Sym}%
\mathfrak{g}\rightarrow\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)
\right)  $ such that $F\circ\iota_{\operatorname*{Sym},\mathfrak{g}}%
=\iota_{\operatorname*{gr}U,\mathfrak{g}}$. Renaming $F$ as $\Phi$ in this
statement, we obtain precisely Proposition \ref{prop.PBW-map} \textbf{(d)}.

\textbf{(e)} The definition of $\operatorname*{PBW}\nolimits_{\mathfrak{g}}$
shows that $\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\iota
_{\operatorname*{Sym},\mathfrak{g}}=\iota_{\operatorname*{gr}U,\mathfrak{g}}$.
Now let $n\in\mathbb{N}$.

Consider the map $\omega:T\left(  \mathfrak{g}\right)  \rightarrow
\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)  $ defined in
our above proof of Proposition \ref{prop.PBW-map} \textbf{(c)}. We recall the
facts (which we have proven in the proof of Proposition \ref{prop.PBW-map}
\textbf{(c)}):

\begin{itemize}
\item We have $\omega=\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)
\circ\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }$.

\item The map $\omega$ is a graded $\mathbf{k}$-algebra homomorphism.

\item We have $\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}%
\right)  \right)  =\omega\left(  \mathfrak{g}^{\otimes n}\right)  $ for every
$n\in\mathbb{N}$.
\end{itemize}

We have $\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\pi
_{\operatorname*{Sym},\mathfrak{g}}=\omega$\ \ \ \ \footnote{\textit{Proof.}
It is well-known that the tensor algebra $T\left(  \mathfrak{g}\right)  $ is
generated by $\mathfrak{g}^{\otimes1}$ (as a $\mathbf{k}$-algebra).
\par
Now, $\operatorname*{PBW}\nolimits_{\mathfrak{g}}$ and $\pi
_{\operatorname*{Sym},\mathfrak{g}}$ are $\mathbf{k}$-algebra homomorphisms.
Thus, their composition $\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ
\pi_{\operatorname*{Sym},\mathfrak{g}}$ is a $\mathbf{k}$-algebra
homomorphism.
\par
Let $w\in\mathfrak{g}^{\otimes1}$. Then, $w\in\mathfrak{g}^{\otimes1}%
=\iota_{T,\mathfrak{g}}\left(  \mathfrak{g}\right)  $. Hence, $w=\iota
_{T,\mathfrak{g}}\left(  v\right)  $ for some $v\in\mathfrak{g}$. Consider
this $v$. Comparing%
\begin{align*}
&  \left(  \operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\pi
_{\operatorname*{Sym},\mathfrak{g}}\right)  \left(  \underbrace{w}%
_{=\iota_{T,\mathfrak{g}}\left(  v\right)  }\right) \\
&  =\left(  \operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\pi
_{\operatorname*{Sym},\mathfrak{g}}\right)  \left(  \iota_{T,\mathfrak{g}%
}\left(  v\right)  \right)  =\left(  \operatorname*{PBW}%
\nolimits_{\mathfrak{g}}\circ\underbrace{\pi_{\operatorname*{Sym}%
,\mathfrak{g}}\circ\iota_{T,\mathfrak{g}}}_{=\iota_{\operatorname*{Sym}%
,\mathfrak{g}}}\right)  \left(  v\right) \\
&  =\left(  \underbrace{\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ
\iota_{\operatorname*{Sym},\mathfrak{g}}}_{=\iota_{\operatorname*{gr}%
U,\mathfrak{g}}}\right)  \left(  v\right)  =\iota_{\operatorname*{gr}%
U,\mathfrak{g}}\left(  v\right)  =\left[  \underbrace{\iota_{U,\mathfrak{g}}%
}_{=\pi_{U,\mathfrak{g}}\circ\iota_{T,\mathfrak{g}}}\left(  v\right)  \right]
_{1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  v\right)  \right) \\
&  =\left[  \left(  \pi_{U,\mathfrak{g}}\circ\iota_{T,\mathfrak{g}}\right)
\left(  v\right)  \right]  _{1}=\left[  \pi_{U,\mathfrak{g}}\left(
\iota_{T,\mathfrak{g}}\left(  v\right)  \right)  \right]  _{1}%
\end{align*}
with%
\begin{align*}
\underbrace{\omega}_{=\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)
\circ\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }}\left(
\underbrace{w}_{=\iota_{T,\mathfrak{g}}\left(  v\right)  }\right)   &
=\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \circ
\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }\right)  \left(
\iota_{T,\mathfrak{g}}\left(  v\right)  \right)  =\left(  \operatorname*{gr}%
\left(  \pi_{U,\mathfrak{g}}\right)  \right)  \left(
\underbrace{\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)
}\left(  \iota_{T,\mathfrak{g}}\left(  v\right)  \right)  }%
_{\substack{=\left[  \iota_{T,\mathfrak{g}}\left(  v\right)  \right]
_{1}\\\text{(by the definition of }\operatorname*{grad}\nolimits_{T\left(
\mathfrak{g}\right)  }\text{,}\\\text{since }\iota_{T,\mathfrak{g}}\left(
v\right)  \in\mathfrak{g}^{\otimes1}\subseteq\mathfrak{g}^{\otimes\leq
1}\text{)}}}\right) \\
&  =\left(  \operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)  \right)
\left(  \left[  \iota_{T,\mathfrak{g}}\left(  v\right)  \right]  _{1}\right)
=\left[  \pi_{U,\mathfrak{g}}\left(  \iota_{T,\mathfrak{g}}\left(  v\right)
\right)  \right]  _{1},
\end{align*}
we obtain $\left(  \operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ
\pi_{\operatorname*{Sym},\mathfrak{g}}\right)  \left(  w\right)
=\omega\left(  w\right)  $.
\par
Now let us forget that we fixed $w$. We thus have shown that $\left(
\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\pi_{\operatorname*{Sym}%
,\mathfrak{g}}\right)  \left(  w\right)  =\omega\left(  w\right)  $ for every
$w\in\mathfrak{g}^{\otimes1}$. In other words, the two maps
$\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\pi_{\operatorname*{Sym}%
,\mathfrak{g}}$ and $\omega$ are equal to each other on every element of
$\mathfrak{g}^{\otimes1}$. Since the $\mathbf{k}$-algebra $T\left(
\mathfrak{g}\right)  $ is generated by $\mathfrak{g}^{\otimes1}$, this shows
that the two maps $\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ
\pi_{\operatorname*{Sym},\mathfrak{g}}$ and $\omega$ must be identical (since
these two maps are $\mathbf{k}$-algebra homomorphisms). In other words,
$\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\pi_{\operatorname*{Sym}%
,\mathfrak{g}}=\omega$, qed.}. Now, for every $n\in\mathbb{N}$, we have%
\begin{align*}
\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)
&  =\underbrace{\omega}_{=\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ
\pi_{\operatorname*{Sym},\mathfrak{g}}}\left(  \mathfrak{g}^{\otimes
n}\right)  =\left(  \operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ
\pi_{\operatorname*{Sym},\mathfrak{g}}\right)  \left(  \mathfrak{g}^{\otimes
n}\right) \\
&  =\operatorname*{PBW}\nolimits_{\mathfrak{g}}\underbrace{\left(
\pi_{\operatorname*{Sym},\mathfrak{g}}\left(  \mathfrak{g}^{\otimes n}\right)
\right)  }_{=\operatorname*{Sym}\nolimits^{n}\mathfrak{g}}=\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}\left(  \operatorname*{Sym}\nolimits^{n}%
\mathfrak{g}\right)  .
\end{align*}
Thus, in particular, we have $\operatorname*{PBW}\nolimits_{\mathfrak{g}%
}\left(  \operatorname*{Sym}\nolimits^{n}\mathfrak{g}\right)  \subseteq
\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)
$ for every $n\in\mathbb{N}$. Hence, the map $\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}$ is graded. Moreover,%
\begin{align*}
\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)   &
=\underbrace{\bigoplus_{n\in\mathbb{N}}}_{=\sum_{n\in\mathbb{N}}%
}\underbrace{\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}%
\right)  \right)  }_{=\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(
\operatorname*{Sym}\nolimits^{n}\mathfrak{g}\right)  }=\sum_{n\in\mathbb{N}%
}\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \operatorname*{Sym}%
\nolimits^{n}\mathfrak{g}\right) \\
&  =\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \underbrace{\sum
_{n\in\mathbb{N}}\operatorname*{Sym}\nolimits^{n}\mathfrak{g}}%
_{=\operatorname*{Sym}\mathfrak{g}}\right)  =\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}\left(  \operatorname*{Sym}\mathfrak{g}\right)  .
\end{align*}
Hence, the map $\operatorname*{PBW}\nolimits_{\mathfrak{g}}$ is surjective.
This proves Proposition \ref{prop.PBW-map} \textbf{(e)}.

\textbf{(f)} Consider the map $\omega:T\left(  \mathfrak{g}\right)
\rightarrow\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)  $
defined in our above proof of Proposition \ref{prop.PBW-map} \textbf{(c)}. We
recall the facts (which we have proven in the proof of Proposition
\ref{prop.PBW-map} \textbf{(c)}):

\begin{itemize}
\item We have $\omega=\operatorname*{gr}\left(  \pi_{U,\mathfrak{g}}\right)
\circ\operatorname*{grad}\nolimits_{T\left(  \mathfrak{g}\right)  }$.

\item The map $\omega$ is a graded $\mathbf{k}$-algebra homomorphism.

\item For every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$, we have%
\begin{equation}
\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  a_{1}\right)  \iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{gr}U,\mathfrak{g}}\left(  a_{n}\right)  =\omega\left(
a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  .
\label{pf.prop.PBW-map.f.1}%
\end{equation}


\item We have $\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}%
\right)  \right)  =\omega\left(  \mathfrak{g}^{\otimes n}\right)  $.
\end{itemize}

Now, the $\mathbf{k}$-module $\mathfrak{g}^{\otimes n}$ is spanned by pure
tensors. In other words, the $\mathbf{k}$-module $\mathfrak{g}^{\otimes n}$ is
spanned by the elements $a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}$ with
$a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$. Thus, the $\mathbf{k}$-module
$\omega\left(  \mathfrak{g}^{\otimes n}\right)  $ is spanned by the elements
$\omega\left(  a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}\right)  $ with
$a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$ (since the map $\omega$ is
$\mathbf{k}$-linear). In other words, the $\mathbf{k}$-module
$\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)
$ is spanned by the elements $\iota_{\operatorname*{Sym},\mathfrak{g}}\left(
a_{1}\right)  \iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)
\cdots\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  $ with
$a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$ (because $\omega\left(
\mathfrak{g}^{\otimes n}\right)  =\operatorname*{gr}\nolimits_{n}\left(
U\left(  \mathfrak{g}\right)  \right)  $ and because every $a_{1},a_{2}%
,\ldots,a_{n}\in\mathfrak{g}$ satisfy $\omega\left(  a_{1}\otimes a_{2}%
\otimes\cdots\otimes a_{n}\right)  =\iota_{\operatorname*{gr}U,\mathfrak{g}%
}\left(  a_{1}\right)  \iota_{\operatorname*{gr}U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(
a_{n}\right)  $). This proves Proposition \ref{prop.PBW-map} \textbf{(f)}.
\end{proof}

Nothing in Proposition \ref{prop.PBW-map} states that the map
$\operatorname*{PBW}\nolimits_{\mathfrak{g}}$ is bijective. This is, in fact,
not true in general; and when it is true, the proof is usually not trivial.
One case where it is true is when $\mathfrak{g}$ is a free $\mathbf{k}%
$-module; another is when $\mathbf{k}$ is a $\mathbb{Q}$-algebra. See
\cite[Theorem 5.9]{Gri-thes} for a list of such results; they are collectively
known as the \textquotedblleft Poincar\'{e}-Birkhoff-Witt
theorems\textquotedblright\ (although the most commonly encountered one is the
one where $\mathfrak{g}$ is assumed to be a free $\mathbf{k}$-algebra). We
shall derive another criterion for $\operatorname*{PBW}\nolimits_{\mathfrak{g}%
}$ to be bijective. First, we state a criterion which (on its own) is not very
interesting, but will serve as a lemma:

\begin{lemma}
\label{lem.PBW-by-inverse-map}Let $\mathfrak{g}$ be a Lie algebra. Consider
the graded $\mathbf{k}$-algebra homomorphism $\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}$ constructed in Proposition \ref{prop.PBW-map}.

Let $\phi:U\left(  \mathfrak{g}\right)  \rightarrow\operatorname*{Sym}%
\mathfrak{g}$ be a $\mathbf{k}$-linear map. Assume that
\begin{align}
&  \phi\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota
_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(
a_{n}\right)  \right) \nonumber\\
&  \in\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  +\sum
_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}\mathfrak{g}
\label{eq.lem.PBW-by-inverse-map.1}%
\end{align}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}%
$. Then:

\textbf{(a)} The map $\phi$ is a filtered $\mathbf{k}$-module isomorphism.

\textbf{(b)} The inverse $\phi^{-1}$ of the map $\phi$ is filtered.

\textbf{(c)} The map $\operatorname*{grad}\nolimits_{\operatorname*{Sym}%
\mathfrak{g}}^{-1}\circ\left(  \operatorname*{gr}\phi\right)
:\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)
\rightarrow\operatorname*{Sym}\mathfrak{g}$ is an inverse to the map
$\operatorname*{PBW}\nolimits_{\mathfrak{g}}$. In particular, the map
$\operatorname*{PBW}\nolimits_{\mathfrak{g}}$ is invertible.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.PBW-by-inverse-map}.]We have%
\begin{equation}
\phi\left(  \pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes n}\right)
\right)  \subseteq\sum\limits_{k=0}^{n}\operatorname*{Sym}\nolimits^{k}%
\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}
\label{pf.lem.PBW-by-inverse-map.a.filt.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.PBW-by-inverse-map.a.filt.1}):} Let
$n\in\mathbb{N}$. We must prove that $\phi\left(  \pi_{U,\mathfrak{g}}\left(
v\right)  \right)  \in\sum\limits_{k=0}^{n}\operatorname*{Sym}\nolimits^{k}%
\mathfrak{g}\ $for every $v\in\mathfrak{g}^{\otimes n}$. So let $v\in
\mathfrak{g}^{\otimes n}$. Since this claim is $\mathbf{k}$-linear in $v$, we
can WLOG assume that $v$ is a pure tensor (since the tensor power
$\mathfrak{g}^{\otimes n}$ is spanned by pure tensors). Assume this. Thus,
$v=a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}$ with $a_{1},a_{2}%
,\ldots,a_{n}\in\mathfrak{g}$. Consider these $a_{1},a_{2},\ldots,a_{n}$.
\par
The definition of the product on $T\left(  \mathfrak{g}\right)  $ shows that
the element $\iota_{T,\mathfrak{g}}\left(  a_{1}\right)  \iota_{T,\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{T,\mathfrak{g}}\left(  a_{n}\right)  $
equals the tensor $a_{1}\otimes a_{2}\otimes\cdots\otimes a_{n}$. Thus,
$\iota_{T,\mathfrak{g}}\left(  a_{1}\right)  \iota_{T,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{T,\mathfrak{g}}\left(  a_{n}\right)  =a_{1}\otimes
a_{2}\otimes\cdots\otimes a_{n}=v$. Hence,%
\begin{align*}
\pi_{U,\mathfrak{g}}\left(  \underbrace{v}_{=\iota_{T,\mathfrak{g}}\left(
a_{1}\right)  \iota_{T,\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{T,\mathfrak{g}}\left(  a_{n}\right)  }\right)   &  =\pi_{U,\mathfrak{g}%
}\left(  \iota_{T,\mathfrak{g}}\left(  a_{1}\right)  \iota_{T,\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{T,\mathfrak{g}}\left(  a_{n}\right)
\right) \\
&  =\pi_{U,\mathfrak{g}}\left(  \iota_{T,\mathfrak{g}}\left(  a_{1}\right)
\right)  \pi_{U,\mathfrak{g}}\left(  \iota_{T,\mathfrak{g}}\left(
a_{2}\right)  \right)  \cdots\pi_{U,\mathfrak{g}}\left(  \iota_{T,\mathfrak{g}%
}\left(  a_{n}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\pi_{U,\mathfrak{g}}\text{ is a
}\mathbf{k}\text{-algebra homomorphism}\right) \\
&  =\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)
\end{align*}
(because every $p\in\left\{  1,2,\ldots,n\right\}  $ satisfies $\pi
_{U,\mathfrak{g}}\left(  \iota_{T,\mathfrak{g}}\left(  a_{p}\right)  \right)
=\underbrace{\left(  \pi_{U,\mathfrak{g}}\circ\iota_{T,\mathfrak{g}}\right)
}_{=\iota_{U,\mathfrak{g}}}\left(  a_{p}\right)  =\iota_{U,\mathfrak{g}%
}\left(  a_{p}\right)  $). Applying $\phi$ to both sides of this equality, we
obtain%
\begin{align*}
\phi\left(  \pi_{U,\mathfrak{g}}\left(  v\right)  \right)   &  =\phi\left(
\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right) \\
&  \in\underbrace{\iota_{\operatorname*{Sym},\mathfrak{g}}\left(
a_{1}\right)  \iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)
\cdots\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)
}_{\substack{\in\left(  \operatorname*{Sym}\nolimits^{1}\mathfrak{g}\right)
^{n}\\\text{(since }\iota_{\operatorname*{Sym},\mathfrak{g}}\left(
a_{p}\right)  \in\operatorname*{Sym}\nolimits^{1}\mathfrak{g}\text{ for every
}p\in\left\{  1,2,\ldots,n\right\}  \text{)}}}+\sum_{k=0}^{n-1}%
\operatorname*{Sym}\nolimits^{k}\mathfrak{g} \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{eq.lem.PBW-by-inverse-map.1})}\right) \\
&  \subseteq\underbrace{\left(  \operatorname*{Sym}\nolimits^{1}%
\mathfrak{g}\right)  ^{n}}_{\substack{\subseteq\operatorname*{Sym}%
\nolimits^{n}\mathfrak{g}\\\text{(since }\operatorname*{Sym}\mathfrak{g}\text{
is a graded}\\\mathbf{k}\text{-algebra)}}}+\sum_{k=0}^{n-1}\operatorname*{Sym}%
\nolimits^{k}\mathfrak{g}=\operatorname*{Sym}\nolimits^{n}\mathfrak{g+}%
\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}\mathfrak{g} \subseteq
\sum_{k=0}^{n}\operatorname*{Sym}\nolimits^{k}\mathfrak{g.}%
\end{align*}
This is precisely what we wanted to prove. Thus,
(\ref{pf.lem.PBW-by-inverse-map.a.filt.1}) is proven.}. Therefore, the map
$\phi$ is filtered\footnote{\textit{Proof.} Let $m\in\mathbb{N}$. Then,
$\mathfrak{g}^{\otimes\leq m}=\sum_{n=0}^{m}\mathfrak{g}^{\otimes n}$ (by the
definition of $\mathfrak{g}^{\otimes\leq m}$), so that%
\begin{align*}
\phi\left(  \pi_{U,\mathfrak{g}}\left(  \underbrace{\mathfrak{g}^{\otimes\leq
m}}_{=\sum_{n=0}^{m}\mathfrak{g}^{\otimes n}}\right)  \right)   &
=\phi\left(  \pi_{U,\mathfrak{g}}\left(  \sum_{n=0}^{m}\mathfrak{g}^{\otimes
n}\right)  \right)  =\sum_{n=0}^{m}\underbrace{\phi\left(  \pi_{U,\mathfrak{g}%
}\left(  \mathfrak{g}^{\otimes n}\right)  \right)  }_{\substack{\subseteq
\sum\limits_{k=0}^{n}\operatorname*{Sym}\nolimits^{k}\mathfrak{g}\\\text{(by
(\ref{pf.lem.PBW-by-inverse-map.a.filt.1}))}}}\\
&  \subseteq\sum_{n=0}^{m}\sum\limits_{k=0}^{n}\operatorname*{Sym}%
\nolimits^{k}\mathfrak{g}=\sum\limits_{k=0}^{m}\underbrace{\sum_{n=k}%
^{m}\operatorname*{Sym}\nolimits^{k}\mathfrak{g}}_{\subseteq
\operatorname*{Sym}\nolimits^{k}\mathfrak{g}}\subseteq\sum\limits_{k=0}%
^{m}\operatorname*{Sym}\nolimits^{k}\mathfrak{g.}%
\end{align*}
\par
Now, let us forget that we fixed $m$. Thus we have shown that $\phi\left(
\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq m}\right)  \right)
\subseteq\sum\limits_{k=0}^{m}\operatorname*{Sym}\nolimits^{k}\mathfrak{g}$
for every $m\in\mathbb{N}$. In other words, the map $\phi$ is filtered (since
$\left(  \pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq m}\right)
\right)  _{m\geq0}$ is the filtration on $U\left(  \mathfrak{g}\right)  $, and
since $\left(  \sum\limits_{k=0}^{m}\operatorname*{Sym}\nolimits^{k}%
\mathfrak{g}\right)  _{m\geq0}$ is the filtration on $\operatorname*{Sym}%
\mathfrak{g}$). Qed.}. Hence, the graded $\mathbf{k}$-linear map
$\operatorname*{gr}\phi:\operatorname*{gr}\left(  U\left(  \mathfrak{g}%
\right)  \right)  \rightarrow\operatorname*{gr}\left(  \operatorname*{Sym}%
\mathfrak{g}\right)  $ is well-defined.

\textbf{(c)} Let $\eta$ denote the map $\operatorname*{grad}%
\nolimits_{\operatorname*{Sym}\mathfrak{g}}^{-1}\circ\left(
\operatorname*{gr}\phi\right)  :\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  \rightarrow\operatorname*{gr}\left(
\operatorname*{Sym}\mathfrak{g}\right)  $. Clearly, $\eta$ is a $\mathbf{k}%
$-linear map.

We have%
\begin{equation}
\left(  \operatorname*{gr}\phi\right)  \left(  \left[  \iota_{U,\mathfrak{g}%
}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}\right)  =\left[
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}
\label{pf.lem.PBW-by-inverse-map.c.1}%
\end{equation}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}%
$\ \ \ \ \footnote{\textit{Proof of (\ref{pf.lem.PBW-by-inverse-map.c.1}):}
Let $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$. We have
$\underbrace{\iota_{U,\mathfrak{g}}}_{=\pi_{U,\mathfrak{g}}\circ
\iota_{T,\mathfrak{g}}}\left(  a_{p}\right)  =\left(  \pi_{U,\mathfrak{g}%
}\circ\iota_{T,\mathfrak{g}}\right)  \left(  a_{p}\right)  =\pi
_{U,\mathfrak{g}}\left(  \underbrace{\iota_{T,\mathfrak{g}}\left(
a_{p}\right)  }_{\in\mathfrak{g}^{\otimes1}\subseteq\mathfrak{g}^{\otimes
\leq1}}\right)  \in\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq
1}\right)  $ for every $p\in\left\{  1,2,\ldots,n\right\}  $. Hence,
\[
\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \in\left(
\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq1}\right)  \right)
^{n}\subseteq\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq n}\right)
\]
(since $\left(  \pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq
m}\right)  \right)  _{m\geq0}$ is a filtration of the $\mathbf{k}$-algebra
$U\left(  \mathfrak{g}\right)  $). Thus, $\left[  \iota_{U,\mathfrak{g}%
}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}$ is a well-defined
element of $\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}%
\right)  \right)  $. The definition of $\operatorname*{gr}\phi$ therefore
yields%
\begin{align*}
&  \left(  \operatorname*{gr}\phi\right)  \left(  \left[  \iota
_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]
_{n}\right) \\
&  =\left[  \phi\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{n}\right)  \right)  \right]  _{n}=\left[  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}%
\end{align*}
(because of (\ref{eq.lem.PBW-by-inverse-map.1}), since $\left(  \sum
\limits_{k=0}^{m}\operatorname*{Sym}\nolimits^{k}\mathfrak{g}\right)
_{m\geq0}$ is the filtration on $\operatorname*{Sym}\mathfrak{g}$). This
proves (\ref{pf.lem.PBW-by-inverse-map.c.1}).}. Hence,%
\begin{equation}
\eta\left(  \left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota
_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(
a_{n}\right)  \right]  _{n}\right)  =\iota_{\operatorname*{Sym},\mathfrak{g}%
}\left(  a_{1}\right)  \iota_{\operatorname*{Sym},\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{\operatorname*{Sym},\mathfrak{g}}\left(
a_{n}\right)  \label{pf.lem.PBW-by-inverse-map.c.2}%
\end{equation}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}%
$\ \ \ \ \footnote{\textit{Proof of (\ref{pf.lem.PBW-by-inverse-map.c.2}):}
Let $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$. We have
$\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{p}\right)  \in
\operatorname*{Sym}\nolimits^{1}\mathfrak{g}$ for every $p\in\left\{
1,2,\ldots,n\right\}  $. Hence,
\begin{align*}
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)   &  \in\left(
\operatorname*{Sym}\nolimits^{1}\mathfrak{g}\right)  ^{n}\subseteq
\operatorname*{Sym}\nolimits^{n}\mathfrak{g}\ \ \ \ \ \ \ \ \ \ \left(
\text{since the }\mathbf{k}\text{-algebra }\operatorname*{Sym}\mathfrak{g}%
\text{ is graded}\right) \\
&  \subseteq\sum_{k=0}^{n}\operatorname*{Sym}\nolimits^{k}\mathfrak{g}.
\end{align*}
Thus, $\left[  \iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}$
is a well-defined element of $\operatorname*{gr}\nolimits_{n}\left(
\operatorname*{Sym}\left(  \mathfrak{g}\right)  \right)  $. The definition of
$\operatorname*{grad}\nolimits_{\operatorname{Sym}\left(  \mathfrak{g}\right)
}$ therefore yields%
\begin{equation}
\operatorname*{grad}\nolimits_{\operatorname{Sym}\left(  \mathfrak{g}\right)
}\left(  \iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right)
=\left[  \iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}.
\label{pf.lem.PBW-by-inverse-map.c.2.pf.1}%
\end{equation}
Now,
\begin{align*}
&  \underbrace{\eta}_{=\operatorname*{grad}\nolimits_{\operatorname{Sym}%
\left(  \mathfrak{g}\right)  }^{-1}\circ\left(  \operatorname*{gr}\phi\right)
}\left(  \left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota
_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(
a_{n}\right)  \right]  _{n}\right) \\
&  =\left(  \operatorname*{grad}\nolimits_{\operatorname{Sym}\left(
\mathfrak{g}\right)  }^{-1}\circ\left(  \operatorname*{gr}\phi\right)
\right)  \left(  \left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{n}\right)  \right]  _{n}\right) \\
&  =\operatorname*{grad}\nolimits_{\operatorname{Sym}\left(  \mathfrak{g}%
\right)  }^{-1}\left(  \underbrace{\left(  \operatorname*{gr}\phi\right)
\left(  \left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota
_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(
a_{n}\right)  \right]  _{n}\right)  }_{\substack{=\left[  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right]
_{n}\\\text{(by (\ref{pf.lem.PBW-by-inverse-map.c.1}))}}}\right) \\
&  =\operatorname*{grad}\nolimits_{\operatorname{Sym}\left(  \mathfrak{g}%
\right)  }^{-1}\left(  \left[  \iota_{\operatorname*{Sym},\mathfrak{g}}\left(
a_{1}\right)  \iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)
\cdots\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right]
_{n}\right)  =\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)
\end{align*}
(by (\ref{pf.lem.PBW-by-inverse-map.c.2.pf.1})), qed.}.

We have%
\begin{equation}
\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \iota_{\operatorname*{Sym}%
,\mathfrak{g}}\left(  a_{1}\right)  \iota_{\operatorname*{Sym},\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym},\mathfrak{g}}\left(
a_{n}\right)  \right)  =\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{n}\right)  \right]  _{n} \label{pf.lem.PBW-by-inverse-map.c.PBW}%
\end{equation}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}%
$\ \ \ \ \footnote{\textit{Proof of (\ref{pf.lem.PBW-by-inverse-map.c.PBW}):}
Let $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$. We have
$\underbrace{\iota_{U,\mathfrak{g}}}_{=\pi_{U,\mathfrak{g}}\circ
\iota_{T,\mathfrak{g}}}\left(  a_{p}\right)  =\left(  \pi_{U,\mathfrak{g}%
}\circ\iota_{T,\mathfrak{g}}\right)  \left(  a_{p}\right)  =\pi
_{U,\mathfrak{g}}\left(  \underbrace{\iota_{T,\mathfrak{g}}\left(
a_{p}\right)  }_{\in\mathfrak{g}^{\otimes1}\subseteq\mathfrak{g}^{\otimes
\leq1}}\right)  \in\pi_{U,\mathfrak{g}}\left(  \mathfrak{g}^{\otimes\leq
1}\right)  $ for every $p\in\left\{  1,2,\ldots,n\right\}  $. Hence, $\left[
\iota_{U,\mathfrak{g}}\left(  a_{p}\right)  \right]  _{1}$ is well-defined for
every $p\in\left\{  1,2,\ldots,n\right\}  $, and the definition of the product
in $\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)  $ yields%
\[
\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \right]  _{1}\left[
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \right]  _{1}\cdots\left[
\iota_{U,\mathfrak{g}}\left(  a_{p}\right)  \right]  _{1}=\left[
\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]
_{n}.
\]
\par
The definition of $\operatorname*{PBW}\nolimits_{\mathfrak{g}}$ shows that
$\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\iota_{\operatorname*{Sym}%
,\mathfrak{g}}=\iota_{\operatorname*{gr}U,\mathfrak{g}}$.
\par
On the other hand, recall that $\operatorname*{PBW}\nolimits_{\mathfrak{g}}$
is a $\mathbf{k}$-algebra homomorphism. Thus,%
\begin{align*}
&  \operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right) \\
&  =\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)  \right)
\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \iota_{\operatorname*{Sym}%
,\mathfrak{g}}\left(  a_{2}\right)  \right)  \cdots\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}\left(  \iota_{\operatorname*{Sym},\mathfrak{g}%
}\left(  a_{n}\right)  \right) \\
&  =\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \right]  _{1}\left[
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \right]  _{1}\cdots\left[
\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because for every }p\in\left\{  1,2,\ldots,n\right\}  \text{, we have}\\
\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \iota_{\operatorname*{Sym}%
,\mathfrak{g}}\left(  a_{p}\right)  \right)  =\underbrace{\left(
\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\iota_{\operatorname*{Sym}%
,\mathfrak{g}}\right)  }_{=\iota_{\operatorname*{gr}U,\mathfrak{g}}}\left(
a_{p}\right)  =\iota_{\operatorname*{gr}U,\mathfrak{g}}\left(  a_{p}\right)
=\left[  \iota_{U,\mathfrak{g}}\left(  a_{p}\right)  \right]  _{1}\\
\text{(by the definition of }\iota_{\operatorname*{gr}U,\mathfrak{g}}\text{)}%
\end{array}
\right) \\
&  =\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)
\right]  _{n}.
\end{align*}
This proves (\ref{pf.lem.PBW-by-inverse-map.c.PBW}).}.

Now, we have $\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\eta
=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.} Let $v\in
\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)  \right)  $. We are
going to prove the identity $\left(  \operatorname*{PBW}%
\nolimits_{\mathfrak{g}}\circ\eta\right)  \left(  v\right)  =v$.
\par
Indeed, this identity is $\mathbf{k}$-linear in $v$. Hence, we can WLOG assume
that $v$ is a homogeneous element of $\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  $ (because every element of $\operatorname*{gr}%
\left(  U\left(  \mathfrak{g}\right)  \right)  $ is a sum of homogeneous
elements). Assume this. Thus, there exists an $n\in\mathbb{N}$ such that
$v\in\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)
\right)  $. Consider this $v$.
\par
We must prove the identity $\left(  \operatorname*{PBW}\nolimits_{\mathfrak{g}%
}\circ\eta\right)  \left(  v\right)  =v$. Since this identity is $\mathbf{k}%
$-linear in $v$, we can WLOG assume that $v$ has the form $\left[
\iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}}\left(
a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}$
with $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$ (because the $\mathbf{k}%
$-module $\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)
\right)  $ is spanned by the elements $\left[  \iota_{U,\mathfrak{g}}\left(
a_{1}\right)  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}$ with $a_{1}%
,a_{2},\ldots,a_{n}\in\mathfrak{g}$ (due to Proposition \ref{prop.PBW-map}
\textbf{(f)})). Assume this. Thus, there exist $a_{1},a_{2},\ldots,a_{n}%
\in\mathfrak{g}$ such that $v=\left[  \iota_{U,\mathfrak{g}}\left(
a_{1}\right)  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}$. Consider these
$a_{1},a_{2},\ldots,a_{n}$. Then,%
\[
\eta\left(  \underbrace{v}_{=\left[  \iota_{U,\mathfrak{g}}\left(
a_{1}\right)  \iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{n}\right)  \right]  _{n}}\right)
=\eta\left(  \left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{n}\right)  \right]  _{n}\right)  =\iota_{\operatorname*{Sym}%
,\mathfrak{g}}\left(  a_{1}\right)  \iota_{\operatorname*{Sym},\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym},\mathfrak{g}}\left(
a_{n}\right)
\]
(by (\ref{pf.lem.PBW-by-inverse-map.c.2})). Applying the map
$\operatorname*{PBW}\nolimits_{\mathfrak{g}}$ to both sides of this equality,
we obtain%
\begin{align*}
\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \eta\left(  v\right)
\right)   &  =\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)  \right) \\
&  =\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)
\right]  _{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.PBW-by-inverse-map.c.PBW})}\right) \\
&  =v.
\end{align*}
Thus, $\left(  \operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\eta\right)
\left(  v\right)  =\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(
\eta\left(  v\right)  \right)  =v$.
\par
Now, let us forget that we have fixed $v$. Thus, we have shown that $\left(
\operatorname*{PBW}\nolimits_{\mathfrak{g}}\circ\eta\right)  \left(  v\right)
=v$ for every $v\in\operatorname*{gr}\left(  U\left(  \mathfrak{g}\right)
\right)  $. In other words, $\operatorname*{PBW}\nolimits_{\mathfrak{g}}%
\circ\eta=\operatorname*{id}$, qed.} and $\eta\circ\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}=\operatorname*{id}$\ \ \ \ \footnote{\textit{Proof.}
Let $v\in\operatorname*{Sym}\mathfrak{g}$. We are going to prove the identity
$\left(  \eta\circ\operatorname*{PBW}\nolimits_{\mathfrak{g}}\right)  \left(
v\right)  =v$.
\par
Indeed, this identity is $\mathbf{k}$-linear in $v$. Hence, we can WLOG assume
that $v$ is a homogeneous element of $\operatorname*{Sym}\mathfrak{g}$
(because every element of $\operatorname*{Sym}\mathfrak{g}$ is a sum of
homogeneous elements). Assume this. Thus, there exists an $n\in\mathbb{N}$
such that $v\in\operatorname*{Sym}\nolimits^{n}\mathfrak{g}$. Consider this
$v$.
\par
We must prove the identity $\left(  \eta\circ\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}\right)  \left(  v\right)  =v$. Since this identity is
$\mathbf{k}$-linear in $v$, we can WLOG assume that $v$ has the form
$\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota_{\operatorname*{Sym}%
,V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym},V}\left(
a_{n}\right)  $ with $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$ (because the
$\mathbf{k}$-module $\operatorname*{Sym}\nolimits^{n}\mathfrak{g}$ is spanned
by the elements $\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},V}\left(  a_{n}\right)  $ with $a_{1},a_{2},\ldots
,a_{n}\in\mathfrak{g}$ (due to Remark \ref{rmk.Sym.span})). Assume this. Thus,
there exist $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$ such that $v=\iota
_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota_{\operatorname*{Sym}%
,V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym},V}\left(
a_{n}\right)  $. Consider these $a_{1},a_{2},\ldots,a_{n}$. Then,%
\begin{align*}
\operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  \underbrace{v}%
_{=\iota_{\operatorname*{Sym},V}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,V}\left(  a_{n}\right)  }\right)   &  =\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}\left(  \iota_{\operatorname*{Sym},V}\left(
a_{1}\right)  \iota_{\operatorname*{Sym},V}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},V}\left(  a_{n}\right)  \right) \\
&  =\left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)  \iota_{U,\mathfrak{g}%
}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{n}\right)
\right]  _{n}%
\end{align*}
(by (\ref{pf.lem.PBW-by-inverse-map.c.PBW})). Applying the map $\eta$ to both
sides of this equality, we obtain%
\begin{align*}
\eta\left(  \operatorname*{PBW}\nolimits_{\mathfrak{g}}\left(  v\right)
\right)   &  =\eta\left(  \left[  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{n}\right)  \right]  _{n}\right) \\
&  =\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{2}\right)  \cdots
\iota_{\operatorname*{Sym},\mathfrak{g}}\left(  a_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.PBW-by-inverse-map.c.2}%
)}\right) \\
&  =v.
\end{align*}
Thus, $\left(  \eta\circ\operatorname*{PBW}\nolimits_{\mathfrak{g}}\right)
\left(  v\right)  =\eta\left(  \operatorname*{PBW}\nolimits_{\mathfrak{g}%
}\left(  v\right)  \right)  =v$.
\par
Now, let us forget that we have fixed $v$. Thus, we have shown that $\left(
\eta\circ\operatorname*{PBW}\nolimits_{\mathfrak{g}}\right)  \left(  v\right)
=v$ for every $v\in\operatorname*{Sym}\mathfrak{g}$. In other words,
$\eta\circ\operatorname*{PBW}\nolimits_{\mathfrak{g}}=\operatorname*{id}$,
qed.}. These two equalities show that the maps $\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}$ and $\eta$ are mutually inverse. In other words, the
map $\eta$ is an inverse to the map $\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}$. In other words, the map $\operatorname*{grad}%
\nolimits_{\operatorname*{Sym}\mathfrak{g}}^{-1}\circ\left(
\operatorname*{gr}\phi\right)  $ is an inverse to the map $\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}$ (since $\eta=\operatorname*{grad}%
\nolimits_{\operatorname*{Sym}\mathfrak{g}}^{-1}\circ\left(
\operatorname*{gr}\phi\right)  $). Thus, the map $\operatorname*{PBW}%
\nolimits_{\mathfrak{g}}$ is invertible. Lemma \ref{lem.PBW-by-inverse-map}
\textbf{(c)} is proven.

Lemma \ref{lem.PBW-by-inverse-map} \textbf{(c)} also shows that the map
$\operatorname*{grad}\nolimits_{\operatorname*{Sym}\mathfrak{g}}^{-1}%
\circ\left(  \operatorname*{gr}\phi\right)  $ is invertible (since it is an
inverse to the map $\operatorname*{PBW}\nolimits_{\mathfrak{g}}$). But
$\operatorname*{grad}\nolimits_{\operatorname*{Sym}\mathfrak{g}}$ is also an
invertible map (since $\operatorname*{grad}\nolimits_{\operatorname*{Sym}%
\mathfrak{g}}$ is a graded $\mathbf{k}$-module isomorphism). Hence, the map
\newline$\operatorname*{grad}\nolimits_{\operatorname*{Sym}\mathfrak{g}}%
\circ\left(  \operatorname*{grad}\nolimits_{\operatorname*{Sym}\mathfrak{g}%
}^{-1}\circ\left(  \operatorname*{gr}\phi\right)  \right)  $ is also
invertible (being the composition of the two invertible maps
$\operatorname*{grad}\nolimits_{\operatorname*{Sym}\mathfrak{g}}$ and
$\operatorname*{grad}\nolimits_{\operatorname*{Sym}\mathfrak{g}}^{-1}%
\circ\left(  \operatorname*{gr}\phi\right)  $). Since $\operatorname*{grad}%
\nolimits_{\operatorname*{Sym}\mathfrak{g}}\circ\left(  \operatorname*{grad}%
\nolimits_{\operatorname*{Sym}\mathfrak{g}}^{-1}\circ\left(
\operatorname*{gr}\phi\right)  \right)  =\underbrace{\operatorname*{grad}%
\nolimits_{\operatorname*{Sym}\mathfrak{g}}\circ\operatorname*{grad}%
\nolimits_{\operatorname*{Sym}\mathfrak{g}}^{-1}}_{=\operatorname*{id}}%
\circ\left(  \operatorname*{gr}\phi\right)  =\operatorname*{gr}\phi$, this
rewrites as follows: The map $\operatorname*{gr}\phi$ is invertible.
Therefore, Theorem \ref{thm.filtered.isocrit} (applied to $V=U\left(
\mathfrak{g}\right)  $, $W=\operatorname*{Sym}\mathfrak{g}$ and $f=\phi$)
shows that the map $\phi$ is invertible, and that its inverse $\phi^{-1}$ is
again a filtered $\mathbf{k}$-algebra map. This proves Lemma
\ref{lem.PBW-by-inverse-map} \textbf{(b)}.

The map $\phi$ is a filtered $\mathbf{k}$-module isomorphism (since it is
filtered, $\mathbf{k}$-linear and invertible). This proves Lemma
\ref{lem.PBW-by-inverse-map} \textbf{(a)}.
\end{proof}

\section{\label{sect.pre-lie}Pre-Lie algebras}

\subsection{Definitions}

We now introduce the main notions of this note:

\begin{definition}
\label{def.prelie.left}A \textit{left pre-Lie algebra} is a $\mathbf{k}%
$-module $A$, equipped with a $\mathbf{k}$-bilinear map $\mu:A\times
A\rightarrow A$, satisfying the relation%
\begin{equation}
\mu\left(  \mu\left(  a,b\right)  ,c\right)  -\mu\left(  a,\mu\left(
b,c\right)  \right)  =\mu\left(  \mu\left(  b,a\right)  ,c\right)  -\mu\left(
b,\mu\left(  a,c\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a,b,c\in
A. \label{eq.def.prelie.left.axiom.mu}%
\end{equation}


When $A$ is a left pre-Lie algebra, and when the map $\mu$ is clear from the
context, we use the name $\vartriangleright$ for the map $\mu$ and write it in
infix notation (i.e., we use the notation $a\vartriangleright b$ for
$\mu\left(  a,b\right)  $ when $a\in A$ and $b\in A$). Thus, the axiom
(\ref{eq.def.prelie.left.axiom.mu}) rewrites as%
\begin{equation}
\left(  a\vartriangleright b\right)  \vartriangleright c-a\vartriangleright
\left(  b\vartriangleright c\right)  =\left(  b\vartriangleright a\right)
\vartriangleright c-b\vartriangleright\left(  a\vartriangleright c\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a,b,c\in A.
\label{eq.def.prelie.left.axiom}%
\end{equation}

\end{definition}

\begin{definition}
\label{def.prelie.right}A \textit{right pre-Lie algebra} is a $\mathbf{k}%
$-module $A$, equipped with a $\mathbf{k}$-bilinear map $\mu:A\times
A\rightarrow A$, satisfying the relation%
\begin{equation}
\mu\left(  \mu\left(  a,b\right)  ,c\right)  -\mu\left(  a,\mu\left(
b,c\right)  \right)  =\mu\left(  \mu\left(  a,c\right)  ,b\right)  -\mu\left(
a,\mu\left(  c,b\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a,b,c\in
A. \label{eq.def.prelie.right.axiom.mu}%
\end{equation}


When $A$ is a right pre-Lie algebra, and when the map $\mu$ is clear from the
context, we use the name $\vartriangleleft$ for the map $\mu$ and write it in
infix notation (i.e., we use the notation $a\vartriangleleft b$ for
$\mu\left(  a,b\right)  $ when $a\in A$ and $b\in A$). Thus, the axiom
(\ref{eq.def.prelie.right.axiom.mu}) rewrites as%
\begin{equation}
\left(  a\vartriangleleft b\right)  \vartriangleleft c-a\vartriangleleft
\left(  b\vartriangleleft c\right)  =\left(  a\vartriangleleft c\right)
\vartriangleleft b-a\vartriangleleft\left(  c\vartriangleleft b\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a,b,c\in A.
\label{eq.def.prelie.right.axiom}%
\end{equation}

\end{definition}

Left pre-Lie algebras are also known as left-symmetric algebras, Vinberg
algebras, or quasi-associative algebras. Similar epithets exist for right
pre-Lie algebras. We shall prove some basic properties of these objects in
this section; more advanced (and more interesting) results and examples can be
found in the expositions \cite{Manchon11} and \cite{Burde11}.

First, let us observe that the notions of left pre-Lie algebras and of right
pre-Lie algebras are \textquotedblleft symmetric to each
other\textquotedblright, in the sense that we can translate between the two by
reversing the order of the arguments in the binary operation:

\begin{proposition}
\label{prop.prelie.left-right}Let $A$ be a left pre-Lie algebra. We define a
map $\vartriangleleft:A\times A\rightarrow A$ (written in infix notation) by%
\[
\left(  a\vartriangleleft b=b\vartriangleright a\ \ \ \ \ \ \ \ \ \ \text{for
all }a,b\in A\right)  .
\]
Then, $A$ (equipped with this map $\vartriangleleft$) is a right pre-Lie algebra.
\end{proposition}

\begin{proposition}
\label{prop.prelie.right-left}Let $A$ be a right pre-Lie algebra. We define a
map $\vartriangleright:A\times A\rightarrow A$ (written in infix notation) by%
\[
\left(  a\vartriangleright b=b\vartriangleleft a\ \ \ \ \ \ \ \ \ \ \text{for
all }a,b\in A\right)  .
\]
Then, $A$ (equipped with this map $\vartriangleright$) is a left pre-Lie algebra.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.prelie.left-right}.]The map $\vartriangleright
:A\times A\rightarrow A$ is $\mathbf{k}$-bilinear (since $A$ is a left pre-Lie
algebra). Thus, the element $b\vartriangleright a\in A$ (for $a,b\in A$)
depends $\mathbf{k}$-linearly on each of $a$ and $b$. In other words, the
element $a\vartriangleleft b$ (for $a,b\in A$) depends $\mathbf{k}$-linearly
on each of $a$ and $b$ (since $a\vartriangleleft b=b\vartriangleright a$). In
other words, the map $\vartriangleleft$ is $\mathbf{k}$-bilinear.

Also, (\ref{eq.def.prelie.left.axiom}) holds (since $A$ is a left pre-Lie algebra).

Now, we shall prove that $\left(  a\vartriangleleft b\right)  \vartriangleleft
c-a\vartriangleleft\left(  b\vartriangleleft c\right)  =\left(
a\vartriangleleft c\right)  \vartriangleleft b-a\vartriangleleft\left(
c\vartriangleleft b\right)  $ for all $a,b,c\in A$.

Indeed, let $a,b,c\in A$. We have%
\begin{align*}
&  \underbrace{\left(  a\vartriangleleft b\right)  }%
_{\substack{=b\vartriangleright a\\\text{(by the definition of }%
\vartriangleleft\text{)}}}\vartriangleleft c-a\vartriangleleft
\underbrace{\left(  b\vartriangleleft c\right)  }%
_{\substack{=c\vartriangleright b\\\text{(by the definition of }%
\vartriangleleft\text{)}}}\\
&  =\underbrace{\left(  b\vartriangleright a\right)  \vartriangleleft
c}_{\substack{=c\vartriangleright\left(  b\vartriangleright a\right)
\\\text{(by the definition of }\vartriangleleft\text{)}}%
}-\underbrace{a\vartriangleleft\left(  c\vartriangleright b\right)
}_{\substack{=\left(  c\vartriangleright b\right)  \vartriangleright
a\\\text{(by the definition of }\vartriangleleft\text{)}}}\\
&  =c\vartriangleright\left(  b\vartriangleright a\right)  -\left(
c\vartriangleright b\right)  \vartriangleright a=-\underbrace{\left(  \left(
c\vartriangleright b\right)  \vartriangleright a-c\vartriangleright\left(
b\vartriangleright a\right)  \right)  }_{\substack{=\left(  b\vartriangleright
c\right)  \vartriangleright a-b\vartriangleright\left(  c\vartriangleright
a\right)  \\\text{(by (\ref{eq.def.prelie.left.axiom}) (applied to}\\c\text{
and }a\text{ instead of }a\text{ and }c\text{))}}}\\
&  =-\left(  \left(  b\vartriangleright c\right)  \vartriangleright
a-b\vartriangleright\left(  c\vartriangleright a\right)  \right)
=b\vartriangleright\left(  c\vartriangleright a\right)  -\left(
b\vartriangleright c\right)  \vartriangleright a.
\end{align*}
Comparing this to%
\begin{align*}
&  \underbrace{\left(  a\vartriangleleft c\right)  }%
_{\substack{=c\vartriangleright a\\\text{(by the definition of }%
\vartriangleleft\text{)}}}\vartriangleleft b-a\vartriangleleft
\underbrace{\left(  c\vartriangleleft b\right)  }%
_{\substack{=b\vartriangleright c\\\text{(by the definition of }%
\vartriangleleft\text{)}}}\\
&  =\underbrace{\left(  c\vartriangleright a\right)  \vartriangleleft
b}_{\substack{=b\vartriangleright\left(  c\vartriangleright a\right)
\\\text{(by the definition of }\vartriangleleft\text{)}}%
}-\underbrace{a\vartriangleleft\left(  b\vartriangleright c\right)
}_{\substack{=\left(  b\vartriangleright c\right)  \vartriangleright
a\\\text{(by the definition of }\vartriangleleft\text{)}}}\\
&  =b\vartriangleright\left(  c\vartriangleright a\right)  -\left(
b\vartriangleright c\right)  \vartriangleright a,
\end{align*}
we obtain $\left(  a\vartriangleleft b\right)  \vartriangleleft
c-a\vartriangleleft\left(  b\vartriangleleft c\right)  =\left(
a\vartriangleleft c\right)  \vartriangleleft b-a\vartriangleleft\left(
c\vartriangleleft b\right)  $.

Let us now forget that we fixed $a,b,c$. We thus have proven that $\left(
a\vartriangleleft b\right)  \vartriangleleft c-a\vartriangleleft\left(
b\vartriangleleft c\right)  =\left(  a\vartriangleleft c\right)
\vartriangleleft b-a\vartriangleleft\left(  c\vartriangleleft b\right)  $ for
all $a,b,c\in A$. In other words, (\ref{eq.def.prelie.right.axiom}) holds.
This (combined with the fact that the map $\vartriangleleft$ is $\mathbf{k}%
$-bilinear) shows that $A$ (equipped with the map $\vartriangleleft$) is a
right pre-Lie algebra (by the definition of a \textquotedblleft right pre-Lie
algebra\textquotedblright). This proves Proposition
\ref{prop.prelie.left-right}.
\end{proof}

\begin{proof}
[Proof of Proposition \ref{prop.prelie.right-left}.]This proof is highly
similar to the proof of Proposition \ref{prop.prelie.left-right} above, and
thus is left to the reader.
\end{proof}

The following two propositions show that the notions of (left and right)
pre-Lie algebras are a middle ground between (associative) algebras and Lie algebras.

\begin{proposition}
\label{prop.prelie.assoc}Let $A$ be a $\mathbf{k}$-algebra. (Recall that
\textquotedblleft$\mathbf{k}$-algebra\textquotedblright\ means
\textquotedblleft associative unital $\mathbf{k}$-algebra\textquotedblright%
\ for us; that said, unitality is not important in this proposition.)

\textbf{(a)} The $\mathbf{k}$-module $A$ (equipped with the multiplication of
$A$) is a left pre-Lie algebra.

\textbf{(b)} The $\mathbf{k}$-module $A$ (equipped with the multiplication of
$A$) is a right pre-Lie algebra.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.prelie.assoc}.]\textbf{(a)} The multiplication
of $A$ is clearly $\mathbf{k}$-bilinear. It thus remains to prove that
(\ref{eq.def.prelie.left.axiom.mu}) holds when $\mu$ is taken to be the
multiplication of $A$. In other words, it remains to prove that
\[
\left(  ab\right)  c-a\left(  bc\right)  =\left(  ba\right)  c-b\left(
ac\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }a,b,c\in A.
\]
But this obviously follows from comparing $\left(  ab\right)  c-a\left(
bc\right)  =abc-abc=0$ with $\left(  ba\right)  c-b\left(  ac\right)
=bac-bac=0$. Thus, Proposition \ref{prop.prelie.assoc} \textbf{(a)} is proven.

\textbf{(b)} The proof of Proposition \ref{prop.prelie.assoc} \textbf{(b)} is
analogous to that of Proposition \ref{prop.prelie.assoc} \textbf{(a)}, and
left to the reader.
\end{proof}

\begin{proposition}
\label{prop.prelie.lie}Let $A$ be a left pre-Lie algebra. Define a
$\mathbf{k}$-bilinear map $\beta:A\times A\rightarrow A$ by%
\[
\beta\left(  a,b\right)  =a\vartriangleright b-b\vartriangleright
a\ \ \ \ \ \ \ \ \ \ \text{for all }a\in A\text{ and }b\in A\text{.}%
\]
Then, the $\mathbf{k}$-module $A$, endowed with this map $\beta$, is a Lie
algebra. This Lie algebra is denoted by $A^{-}$. (This clearly generalizes the
definition of the Lie algebra $A^{-}$ when $A$ is a $\mathbf{k}$-algebra.)
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.prelie.lie}.]The operation $\vartriangleright$
is $\mathbf{k}$-bilinear (since $A$ is a left pre-Lie algebra). Hence,
$a\vartriangleright b-b\vartriangleright a$ (for $a\in A$ and $b\in A$)
depends $\mathbf{k}$-linearly on each of $a$ and $b$. The $\mathbf{k}%
$-bilinear map $\beta$ is thus well-defined.

The axiom (\ref{eq.def.prelie.left.axiom}) holds, since $A$ is a left pre-Lie algebra.

We only need to prove that the $\mathbf{k}$-module $A$, endowed with the map
$\beta$, is a Lie algebra. To do this, it suffices to verify that%
\begin{equation}
\left(  \beta\left(  v,v\right)  =0\text{ for every }v\in A\right)
\label{pf.prop.prelie.lie.axi1}%
\end{equation}
and that%
\begin{equation}
\left(  \beta\left(  u,\beta\left(  v,w\right)  \right)  +\beta\left(
v,\beta\left(  w,u\right)  \right)  +\beta\left(  w,\beta\left(  u,v\right)
\right)  =0\text{ for every }u\in A\text{, }v\in A\text{ and }w\in A\right)
\label{pf.prop.prelie.lie.axi2}%
\end{equation}
(because it is clear that the map $\beta$ is $\mathbf{k}$-bilinear).

\textit{Proof of (\ref{pf.prop.prelie.lie.axi1}):} Let $v\in A$. Then, the
definition of $\beta$ shows that $\beta\left(  v,v\right)  =v\vartriangleright
v-v\vartriangleright v=0$. Now let us forget that we fixed $v$. We thus have
shown that $\beta\left(  v,v\right)  =0$ for every $v\in A$. In other words,
(\ref{pf.prop.prelie.lie.axi1}) holds.

\textit{Proof of (\ref{pf.prop.prelie.lie.axi2}):} Let $u\in A$, $v\in A$ and
$w\in A$. Now,%
\begin{align}
&  \beta\left(  u,\underbrace{\beta\left(  v,w\right)  }%
_{\substack{=v\vartriangleright w-w\vartriangleright v\\\text{(by the
definition of }\beta\text{)}}}\right) \nonumber\\
&  =\beta\left(  u,v\vartriangleright w-w\vartriangleright v\right)
=\underbrace{u\vartriangleright\left(  v\vartriangleright w-w\vartriangleright
v\right)  }_{\substack{=u\vartriangleright\left(  v\vartriangleright w\right)
-u\vartriangleright\left(  w\vartriangleright v\right)  \\\text{(since the
operation }\vartriangleright\text{ is }\mathbf{k}\text{-bilinear)}%
}}-\underbrace{\left(  v\vartriangleright w-w\vartriangleright v\right)
\vartriangleright u}_{\substack{=\left(  v\vartriangleright w\right)
\vartriangleright u-\left(  w\vartriangleright v\right)  \vartriangleright
u\\\text{(since the operation }\vartriangleright\text{ is }\mathbf{k}%
\text{-bilinear)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\beta\right)
\nonumber\\
&  =\left(  u\vartriangleright\left(  v\vartriangleright w\right)
-u\vartriangleright\left(  w\vartriangleright v\right)  \right)  -\left(
\left(  v\vartriangleright w\right)  \vartriangleright u-\left(
w\vartriangleright v\right)  \vartriangleright u\right) \nonumber\\
&  =u\vartriangleright\left(  v\vartriangleright w\right)  -u\vartriangleright
\left(  w\vartriangleright v\right)  -\left(  v\vartriangleright w\right)
\vartriangleright u+\left(  w\vartriangleright v\right)  \vartriangleright u.
\label{pf.prop.prelie.lie.1}%
\end{align}


Applying (\ref{pf.prop.prelie.lie.1}) to $v$, $w$ and $u$ instead of $u$, $v$
and $w$, we obtain%
\begin{equation}
\beta\left(  v,\beta\left(  w,u\right)  \right)  =v\vartriangleright\left(
w\vartriangleright u\right)  -v\vartriangleright\left(  u\vartriangleright
w\right)  -\left(  w\vartriangleright u\right)  \vartriangleright v+\left(
u\vartriangleright w\right)  \vartriangleright v. \label{pf.prop.prelie.lie.2}%
\end{equation}


Applying (\ref{pf.prop.prelie.lie.1}) to $w$, $u$ and $v$ instead of $u$, $v$
and $w$, we obtain%
\begin{equation}
\beta\left(  w,\beta\left(  u,v\right)  \right)  =w\vartriangleright\left(
u\vartriangleright v\right)  -w\vartriangleright\left(  v\vartriangleright
u\right)  -\left(  u\vartriangleright v\right)  \vartriangleright w+\left(
v\vartriangleright u\right)  \vartriangleright w. \label{pf.prop.prelie.lie.3}%
\end{equation}


Adding together the identities (\ref{pf.prop.prelie.lie.1}),
(\ref{pf.prop.prelie.lie.2}) and (\ref{pf.prop.prelie.lie.3}), we obtain%
\begin{align*}
&  \beta\left(  u,\beta\left(  v,w\right)  \right)  +\beta\left(
v,\beta\left(  w,u\right)  \right)  +\beta\left(  w,\beta\left(  u,v\right)
\right) \\
&  =u\vartriangleright\left(  v\vartriangleright w\right)  -u\vartriangleright
\left(  w\vartriangleright v\right)  -\left(  v\vartriangleright w\right)
\vartriangleright u+\left(  w\vartriangleright v\right)  \vartriangleright u\\
&  \ \ \ \ \ \ \ \ \ \ +v\vartriangleright\left(  w\vartriangleright u\right)
-v\vartriangleright\left(  u\vartriangleright w\right)  -\left(
w\vartriangleright u\right)  \vartriangleright v+\left(  u\vartriangleright
w\right)  \vartriangleright v\\
&  \ \ \ \ \ \ \ \ \ \ +w\vartriangleright\left(  u\vartriangleright v\right)
-w\vartriangleright\left(  v\vartriangleright u\right)  -\left(
u\vartriangleright v\right)  \vartriangleright w+\left(  v\vartriangleright
u\right)  \vartriangleright w\\
&  =\underbrace{\left(  \left(  w\vartriangleright v\right)  \vartriangleright
u-w\vartriangleright\left(  v\vartriangleright u\right)  \right)
}_{\substack{=\left(  v\vartriangleright w\right)  \vartriangleright
u-v\vartriangleright\left(  w\vartriangleright u\right)  \\\text{(by
(\ref{eq.def.prelie.left.axiom}), applied}\\\text{to }a=w\text{, }b=v\text{
and }c=u\text{)}}}+\underbrace{\left(  \left(  u\vartriangleright w\right)
\vartriangleright v-u\vartriangleright\left(  w\vartriangleright v\right)
\right)  }_{\substack{=\left(  w\vartriangleright u\right)  \vartriangleright
v-w\vartriangleright\left(  u\vartriangleright v\right)  \\\text{(by
(\ref{eq.def.prelie.left.axiom}), applied}\\\text{to }a=u\text{, }b=w\text{
and }c=v\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{\left(  \left(  v\vartriangleright
u\right)  \vartriangleright w-v\vartriangleright\left(  u\vartriangleright
w\right)  \right)  }_{\substack{=\left(  u\vartriangleright v\right)
\vartriangleright w-u\vartriangleright\left(  v\vartriangleright w\right)
\\\text{(by (\ref{eq.def.prelie.left.axiom}), applied}\\\text{to }a=v\text{,
}b=u\text{ and }c=w\text{)}}}+u\vartriangleright\left(  v\vartriangleright
w\right)  -\left(  v\vartriangleright w\right)  \vartriangleright u\\
&  \ \ \ \ \ \ \ \ \ \ +v\vartriangleright\left(  w\vartriangleright u\right)
-\left(  w\vartriangleright u\right)  \vartriangleright v+w\vartriangleright
\left(  u\vartriangleright v\right)  -\left(  u\vartriangleright v\right)
\vartriangleright w\\
&  =\left(  v\vartriangleright w\right)  \vartriangleright
u-v\vartriangleright\left(  w\vartriangleright u\right)  +\left(
w\vartriangleright u\right)  \vartriangleright v-w\vartriangleright\left(
u\vartriangleright v\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  u\vartriangleright v\right)  \vartriangleright
w-u\vartriangleright\left(  v\vartriangleright w\right)  +u\vartriangleright
\left(  v\vartriangleright w\right)  -\left(  v\vartriangleright w\right)
\vartriangleright u\\
&  \ \ \ \ \ \ \ \ \ \ +v\vartriangleright\left(  w\vartriangleright u\right)
-\left(  w\vartriangleright u\right)  \vartriangleright v+w\vartriangleright
\left(  u\vartriangleright v\right)  -\left(  u\vartriangleright v\right)
\vartriangleright w\\
&  =0.
\end{align*}
Now, let us forget that we fixed $u$, $v$ and $w$. We thus have shown that
$\beta\left(  u,\beta\left(  v,w\right)  \right)  +\beta\left(  v,\beta\left(
w,u\right)  \right)  +\beta\left(  w,\beta\left(  u,v\right)  \right)  =0$ for
every $u\in A$, $v\in A$ and $w\in A$. In other words,
(\ref{pf.prop.prelie.lie.axi2}) holds.

Now, we have shown that (\ref{pf.prop.prelie.lie.axi1}) and
(\ref{pf.prop.prelie.lie.axi2}) hold. Since the map $\beta$ is $\mathbf{k}%
$-bilinear, this shows that the $\mathbf{k}$-module $A$, endowed with this map
$\beta$, is a Lie algebra (by the definition of a \textquotedblleft Lie
algebra\textquotedblright). This proves Proposition \ref{prop.prelie.lie}.
\end{proof}

Let us state the analogue of Proposition \ref{prop.prelie.lie} for right
pre-Lie algebras:

\begin{proposition}
\label{prop.prelie.lie.right}Let $A$ be a right pre-Lie algebra. Define a
$\mathbf{k}$-bilinear map $\beta:A\times A\rightarrow A$ by%
\[
\beta\left(  a,b\right)  =a\vartriangleleft b-b\vartriangleleft
a\ \ \ \ \ \ \ \ \ \ \text{for all }a\in A\text{ and }b\in A\text{.}%
\]
Then, the $\mathbf{k}$-module $A$, endowed with this map $\beta$, is a Lie
algebra. This Lie algebra is denoted by $A^{-}$. (This clearly generalizes the
definition of the Lie algebra $A^{-}$ when $A$ is a $\mathbf{k}$-algebra.)
\end{proposition}

The proof of Proposition \ref{prop.prelie.lie.right} is highly similar to that
of Proposition \ref{prop.prelie.lie}; thus we leave it to the reader.

\begin{remark}
\label{rmk.prelie.A-}It is worth repeating one obvious observation. Namely,
let $A$ be a $\mathbf{k}$-algebra. Then, we can construct three Lie algebras,
all of which we denote by $A^{-}$:

\begin{itemize}
\item One Lie algebra denoted by $A^{-}$ has been defined in Definition
\ref{def.A-}.

\item Another Lie algebra denoted by $A^{-}$ can be obtained from Proposition
\ref{prop.prelie.lie}, after we first make $A$ into a left pre-Lie algebra
(according to Proposition \ref{prop.prelie.assoc} \textbf{(a)}).

\item A third Lie algebra denoted by $A^{-}$ can be obtained from Proposition
\ref{prop.prelie.lie.right}, after we first make $A$ into a right pre-Lie
algebra (according to Proposition \ref{prop.prelie.assoc} \textbf{(b)}).
\end{itemize}

These three Lie algebras denoted by $A^{-}$ are \textbf{all identical}; thus,
the notations do not conflict.
\end{remark}

\subsection{The Guin-Oudom isomorphism}

\begin{definition}
\label{def.manchon.danger}From now on, we shall employ the following abuse of
notation: If $V$ is a $\mathbf{k}$-module, then every $v\in V$ will be
identified with the element $\iota_{\operatorname*{Sym},V}\left(  v\right)  $
of $\operatorname*{Sym}V$ and also identified with the element $\iota
_{T,V}\left(  v\right)  $ of $T\left(  V\right)  $. This, of course, is
dangerous, because a product of the form $v_{1}v_{2}\cdots v_{k}$ (with
$k\in\mathbb{N}$ and $v_{1},v_{2},\ldots,v_{k}\in V$) can mean two different
things (namely, the tensor $v_{1}\otimes v_{2}\otimes\cdots\otimes v_{k}$ in
$T\left(  V\right)  $, and the projection $\pi_{\operatorname*{Sym},V}\left(
v_{1}\otimes v_{2}\otimes\cdots\otimes v_{k}\right)  $ of this tensor on
$\operatorname*{Sym}V$). (It becomes even more dangerous when $V$ itself has a
$\mathbf{k}$-algebra structure, because then $v_{1}v_{2}\cdots v_{k}$ might
also mean a product inside $V$.) However, we shall rely on the context to
clear up any ambiguities.

We observe that any $k$ elements $v_{1},v_{2},\ldots,v_{k}$ of a $\mathbf{k}%
$-module $V$ satisfy%
\begin{equation}
v_{1}v_{2}\cdots v_{k}=\pi_{\operatorname*{Sym},V}\left(  v_{1}\otimes
v_{2}\otimes\cdots\otimes v_{k}\right)  \label{eq.def.manchon.danger.eq}%
\end{equation}
in $\operatorname*{Sym}V$.
\end{definition}

The following result extends \cite[Theorem 1.1]{Manchon11}:

\begin{theorem}
\label{thm.manchon}Let $A$ be a left pre-Lie algebra. Recall that a Lie
algebra $A^{-}$ is defined (according to Proposition \ref{prop.prelie.lie}).

For any $a\in A$, let $\operatorname*{left}\nolimits_{a}$ be the $\mathbf{k}%
$-linear map $A\rightarrow A,\ b\mapsto a\vartriangleright b$. For any $a\in
A$, we define a derivation $L_{a}:\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$ as follows: Proposition \ref{prop.Sym.deruniv} (applied
to $V=A$, $M=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}%
\circ\operatorname*{left}\nolimits_{a}$) shows that there exists a unique
derivation $F:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ such that
$F\circ\iota_{\operatorname*{Sym},A}=\operatorname*{left}\nolimits_{a}%
\circ\iota_{\operatorname*{Sym},A}$. We let $L_{a}$ be this derivation.

In the following, we will identify every $a\in A$ with the element
$\iota_{\operatorname*{Sym},A}\left(  a\right)  $ of $\operatorname*{Sym}A$.
(This is a particular case of the abuse of notation introduced in Definition
\ref{def.manchon.danger}.) Thus, the injection $\iota_{\operatorname*{Sym}%
,A}:A\rightarrow\operatorname*{Sym}A$ becomes an inclusion, and the
$\mathbf{k}$-module $A$ becomes a $\mathbf{k}$-submodule of
$\operatorname*{Sym}A$ (although the left pre-Lie algebra $A$ does
\textbf{not} become a subalgebra of $\operatorname*{Sym}A$). Thus, products
such as $b_{1}b_{2}\cdots b_{n}$ (where $n\in\mathbb{N}$ and $b_{1}%
,b_{2},\ldots,b_{n}\in A$) will always mean products inside
$\operatorname*{Sym}A$. (This notation will not conflict with the binary
operation on the left pre-Lie algebra $A$, since the latter operation is
denoted by $\vartriangleright$.)

\textbf{(a)} If $a\in A$, $n\in\mathbb{N}$ and $b_{1},b_{2},\ldots,b_{n}\in
A$, then%
\[
L_{a}\left(  b_{1}b_{2}\cdots b_{n}\right)  =\sum_{k=1}^{n}b_{1}b_{2}\cdots
b_{k-1}\left(  a\vartriangleright b_{k}\right)  b_{k+1}b_{k+2}\cdots b_{n}.
\]


\textbf{(b)} For every $a\in A$, the map $L_{a}:\operatorname*{Sym}%
A\rightarrow\operatorname*{Sym}A$ is graded.

\textbf{(c)} For every $a\in A$, the map $L_{a}:\operatorname*{Sym}%
A\rightarrow\operatorname*{Sym}A$ is a coderivation.

\textbf{(d)} We define a map $K:A\rightarrow\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $ by
\[
\left(  K\left(  a\right)  =L_{a}\ \ \ \ \ \ \ \ \ \ \text{for every }a\in
A\right)  .
\]
Then, this map $K$ is a Lie algebra homomorphism from $A^{-}$ to
$\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $. (See Proposition
\ref{prop.Der.basics} \textbf{(a)} for the definition of the Lie algebra
$\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $.)

\textbf{(e)} We can define an $A^{-}$-module structure on $\operatorname*{Sym}%
A$ by setting%
\[
\left(  a\rightharpoonup u=au+L_{a}\left(  u\right)  \text{ for all }a\in
A^{-}\text{ and }u\in\operatorname*{Sym}A\right)  .
\]
In the following, we will regard $\operatorname*{Sym}A$ as an $A^{-}$-module
by means of this $A^{-}$-module structure.

\textbf{(f)} Being an $A^{-}$-module, $\operatorname*{Sym}A$ becomes a
$U\left(  A^{-}\right)  $-module. Define a map $\eta:U\left(  A^{-}\right)
\rightarrow\operatorname*{Sym}A$ by%
\[
\eta\left(  u\right)  =u1_{\operatorname*{Sym}A}\ \ \ \ \ \ \ \ \ \ \text{for
every }u\in U\left(  A^{-}\right)  .
\]
Then, $\eta$ is an $A^{-}$-module homomorphism.

\textbf{(g)} We have $\eta\left(  \iota_{U,A^{-}}\left(  a\right)  \right)
=a$ for every $a\in A^{-}$.

\textbf{(h)} The map $\eta$ is an $A^{-}$-module isomorphism.

\textbf{(i)} We have%
\[
\eta\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}}\left(
a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right)  \in
a_{1}a_{2}\cdots a_{n}+\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}A
\]
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in A^{-}$.

\textbf{(j)} The map $\eta$ and its inverse $\eta^{-1}$ are filtered (where
$U\left(  A^{-}\right)  $ is endowed with the usual filtration on a universal
enveloping algebra).

\textbf{(k)} The map $\operatorname*{grad}\nolimits_{\operatorname*{Sym}%
A}^{-1}\circ\left(  \operatorname*{gr}\eta\right)  :\operatorname*{gr}\left(
U\left(  A^{-}\right)  \right)  \rightarrow\operatorname*{Sym}A$ is an inverse
to the map $\operatorname*{PBW}\nolimits_{A^{-}}$. In particular, the map
$\operatorname*{PBW}\nolimits_{A^{-}}$ is invertible.

\textbf{(l)} The map $\eta:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ is a $\mathbf{k}$-coalgebra isomorphism.

\textbf{(m)} For every $a\in A$, $b\in A$ and $c\in\operatorname*{Sym}A$, we
have%
\[
a\rightharpoonup\left(  bc\right)  -b\cdot\left(  a\rightharpoonup c\right)
=\left(  a\vartriangleright b\right)  \cdot c.
\]

\end{theorem}

We suggest to call the $\mathbf{k}$-coalgebra isomorphism $\eta:U\left(
A^{-}\right)  \rightarrow\operatorname*{Sym}A$ in Theorem \ref{thm.manchon}
the \textit{Guin-Oudom isomorphism}.

The most striking part of Theorem \ref{thm.manchon} is part \textbf{(k)},
which gives a PBW theorem for any Lie algebra of the form $A^{-}$ for $A$
being a pre-Lie algebra. This result originates in Oudom's and Guin's
\cite[Th\'{e}or\`{e}me 3.5]{GuiOud04} and \cite[Theorem 3.14]{GuiOud08}%
\footnote{although Oudom and Guin work with right pre-Lie algebras (but this
does not change much, as there is a symmetry in the concepts) and approach the
theorem in a significantly different way (this actually matters)}. Parts
\textbf{(a)}, \textbf{(b)} and \textbf{(d)}--\textbf{(k)} of Theorem
\ref{thm.manchon} appear (explicitly or implicitly) in \cite[Theorem 1.1 and
its proof]{Manchon11} and \cite[Theorem 19 and its proof]{Manchon15}; parts
\textbf{(c)} and \textbf{(l)} appear in \cite[Remark 3.3 and Theorem
3.14]{GuiOud08}. Notice that \cite[Corollary 1.3.1]{Schedler} also follows
from Theorem \ref{thm.manchon} \textbf{(l)}\footnote{although here, again, one
has to translate from left to right pre-Lie algebras}. We shall mostly derive
Theorem \ref{thm.manchon} from what has already been shown.

\begin{proof}
[Proof of Theorem \ref{thm.manchon}.]For every $a\in A$, we know that
$L_{a}:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ is a derivation
and satisfies%
\begin{equation}
L_{a}\circ\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}%
\circ\operatorname*{left}\nolimits_{a} \label{pf.thm.manchon.La-def}%
\end{equation}
\footnote{\textit{Proof.} Let $a\in A$. Then, $L_{a}$ is the unique derivation
$F:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ such that $F\circ
\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}\circ
\operatorname*{left}\nolimits_{a}$. Hence, $L_{a}$ is a derivation
$\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ and satisfies
$L_{a}\circ\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}%
\circ\operatorname*{left}\nolimits_{a}$. Qed.}. Hence, for every $a\in A$ and
$b\in A$, we have%
\begin{align}
L_{a}\left(  \iota_{\operatorname*{Sym},A}\left(  b\right)  \right)   &
=\underbrace{\left(  L_{a}\circ\iota_{\operatorname*{Sym},A}\right)
}_{\substack{=\iota_{\operatorname*{Sym},A}\circ\operatorname*{left}%
\nolimits_{a}\\\text{(by (\ref{pf.thm.manchon.La-def}))}}}\left(  b\right)
=\iota_{\operatorname*{Sym},A}\left(  \underbrace{\operatorname*{left}%
\nolimits_{a}\left(  b\right)  }_{\substack{=a\vartriangleright b\\\text{(by
the definition}\\\text{of }\operatorname*{left}\nolimits_{a}\text{)}}}\right)
\nonumber\\
&  =\iota_{\operatorname*{Sym},A}\left(  a\vartriangleright b\right)  .
\label{pf.thm.manchon.La(iSymb)}%
\end{align}
In other words, for every $a\in A$ and $b\in A$, we have
\begin{equation}
L_{a}\left(  b\right)  =a\vartriangleright b \label{pf.thm.manchon.La(b)}%
\end{equation}
\footnote{since $\iota_{\operatorname*{Sym},A}\left(  b\right)  =b$ (because
we are identifying $\iota_{\operatorname*{Sym},A}\left(  x\right)  $ with $x$
for every $x\in A$) and $\iota_{\operatorname*{Sym},A}\left(
a\vartriangleright b\right)  =a\vartriangleright b$ (for the same reason)}.

\textbf{(a)} Let $a\in A$, $n\in\mathbb{N}$ and $b_{1},b_{2},\ldots,b_{n}\in
A$. The map $L_{a}$ is a derivation, thus an element of $\operatorname*{Der}%
\left(  \operatorname*{Sym}A\right)  $. Hence, Proposition
\ref{prop.Der.basics} \textbf{(b)} (applied to $C=\operatorname*{Sym}A$,
$f=L_{a}$ and $a_{i}=b_{i}$) yields
\begin{align*}
&  L_{a}\left(  b_{1}b_{2}\cdots b_{n}\right) \\
&  =\sum_{i=1}^{n}b_{1}b_{2}\cdots b_{i-1}\underbrace{L_{a}\left(
b_{i}\right)  }_{\substack{=a\vartriangleright b_{i}\\\text{(by
(\ref{pf.thm.manchon.La(b)}))}}}b_{i+1}b_{i+2}\cdots b_{n}\\
&  =\sum_{i=1}^{n}b_{1}b_{2}\cdots b_{i-1}\left(  a\vartriangleright
b_{i}\right)  b_{i+1}b_{i+2}\cdots b_{n}=\sum_{k=1}^{n}b_{1}b_{2}\cdots
b_{k-1}\left(  a\vartriangleright b_{k}\right)  b_{k+1}b_{k+2}\cdots b_{n}%
\end{align*}
(here, we have renamed the summation index $i$ as $k$). This proves Theorem
\ref{thm.manchon} \textbf{(a)}.

\textbf{(b)} Let $a\in A$.

In Proposition \ref{prop.Sym.bider} (applied to $V=A$ and
$f=\operatorname*{left}\nolimits_{a}$), a map $\widetilde{f}%
:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ was defined as the
unique derivation $F:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ such
that $F\circ\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}%
\circ\operatorname*{left}\nolimits_{a}$. This derivation $F$ must clearly be
our map $L_{a}$ (since we know that $L_{a}:\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$ is a derivation and satisfies $L_{a}\circ\iota
_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}\circ
\operatorname*{left}\nolimits_{a}$). Hence, the $\widetilde{f}$ defined in
Proposition \ref{prop.Sym.bider} (applied to $V=A$ and $f=\operatorname*{left}%
\nolimits_{a}$) is our map $L_{a}$. Thus, Proposition \ref{prop.Sym.bider} can
be applied to $V=A$, $f=\operatorname*{left}\nolimits_{a}$ and $\widetilde{f}%
=L_{a}$.

As a consequence, Proposition \ref{prop.Sym.bider} \textbf{(c)} (applied to
$V=A$, $f=\operatorname*{left}\nolimits_{a}$ and $\widetilde{f}=L_{a}$) shows
that the map $L_{a}:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ is
graded. This proves Theorem \ref{thm.manchon} \textbf{(b)}.

\textbf{(c)} Let $a\in A$. In our above proof of Theorem \ref{thm.manchon}
\textbf{(b)}, we have shown that Proposition \ref{prop.Sym.bider} can be
applied to $V=A$, $f=\operatorname*{left}\nolimits_{a}$ and $\widetilde{f}%
=L_{a}$. Hence, Proposition \ref{prop.Sym.bider} \textbf{(b)} (applied to
$V=A$, $f=\operatorname*{left}\nolimits_{a}$ and $\widetilde{f}=L_{a}$) shows
that the map $L_{a}:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ is a
coderivation. Theorem \ref{thm.manchon} \textbf{(c)} is thus proven.

\textbf{(d)} The map $K$ is a map from $A$ to $\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $, thus a map from $A^{-}$ to
$\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $ (since $A^{-}=A$ as sets).

We shall first prove that%
\begin{equation}
K\left(  \left[  a,b\right]  \right)  =\left[  K\left(  a\right)  ,K\left(
b\right)  \right]  \label{pf.thm.manchon.d.goal1}%
\end{equation}
for every $a\in A^{-}$ and $b\in A^{-}$.

\textit{Proof of (\ref{pf.thm.manchon.d.goal1}):} Recall that we are regarding
$A$ as a $\mathbf{k}$-submodule of $\operatorname*{Sym}A$ via the injection
$\iota_{\operatorname*{Sym},A}:A\rightarrow\operatorname*{Sym}A$. Thus,
$A=\iota_{\operatorname*{Sym},A}\left(  A\right)  =\operatorname*{Sym}%
\nolimits^{1}A$. Recall that the subset $\operatorname*{Sym}\nolimits^{1}A$ of
$\operatorname*{Sym}A$ generates the $\mathbf{k}$-algebra $\operatorname*{Sym}%
A$. In other words, the subset $A$ of $\operatorname*{Sym}A$ generates the
$\mathbf{k}$-algebra $\operatorname*{Sym}A$ (since $A=\operatorname*{Sym}%
\nolimits^{1}A$).

Let $a\in A^{-}$ and $b\in A^{-}$. Then, $K\left(  a\right)  $ and $K\left(
b\right)  $ are elements of $\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $. Thus $\left[  K\left(  a\right)  ,K\left(  b\right)  \right]  $
is an element of $\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $
(since $\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $ is a Lie
algebra), hence a derivation $\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$. Also, $K\left(  \left[  a,b\right]  \right)  $ is an
element of $\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $, hence a
derivation $\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$.

We have $\left[  K\left(  a\right)  ,K\left(  b\right)  \right]  \mid
_{A}=K\left(  \left[  a,b\right]  \right)  \mid_{A}$%
\ \ \ \ \footnote{\textit{Proof.} Let $c\in A$. The definition of $K$ yields
$K\left(  a\right)  =L_{a}$, $K\left(  b\right)  =L_{b}$ and $K\left(  \left[
a,b\right]  \right)  =L_{\left[  a,b\right]  }$. Now,
\begin{align*}
&  \left(  \left[  K\left(  a\right)  ,K\left(  b\right)  \right]  \mid
_{A}\right)  \left(  c\right) \\
&  =\underbrace{\left[  K\left(  a\right)  ,K\left(  b\right)  \right]
}_{=K\left(  a\right)  \circ K\left(  b\right)  -K\left(  b\right)  \circ
K\left(  a\right)  }\left(  c\right)  =\left(  \underbrace{K\left(  a\right)
}_{=L_{a}}\circ\underbrace{K\left(  b\right)  }_{=L_{b}}-\underbrace{K\left(
b\right)  }_{=L_{b}}\circ\underbrace{K\left(  a\right)  }_{=L_{a}}\right)
\left(  c\right) \\
&  =\left(  L_{a}\circ L_{b}-L_{b}\circ L_{a}\right)  \left(  c\right)
=L_{a}\left(  \underbrace{L_{b}\left(  c\right)  }%
_{\substack{=b\vartriangleright c\\\text{(by (\ref{pf.thm.manchon.La(b)}),
applied to }b\text{ and }c\\\text{instead of }a\text{ and }b\text{)}}}\right)
-L_{b}\left(  \underbrace{L_{a}\left(  c\right)  }%
_{\substack{=a\vartriangleright c\\\text{(by (\ref{pf.thm.manchon.La(b)}),
applied to }c\\\text{instead of }b\text{)}}}\right) \\
&  =\underbrace{L_{a}\left(  b\vartriangleright c\right)  }%
_{\substack{=a\vartriangleright\left(  b\vartriangleright c\right)
\\\text{(by (\ref{pf.thm.manchon.La(b)}), applied to }b\vartriangleright
c\\\text{instead of }b\text{)}}}-\underbrace{L_{b}\left(  a\vartriangleright
c\right)  }_{\substack{=b\vartriangleright\left(  a\vartriangleright c\right)
\\\text{(by (\ref{pf.thm.manchon.La(b)}), applied to }b\text{ and
}a\vartriangleright c\\\text{instead of }a\text{ and }b\text{)}}}\\
&  =a\vartriangleright\left(  b\vartriangleright c\right)  -b\vartriangleright
\left(  a\vartriangleright c\right)  .
\end{align*}
Comparing this with%
\begin{align*}
&  \left(  K\left(  \left[  a,b\right]  \right)  \mid_{A}\right)  \left(
c\right) \\
&  =\underbrace{\left(  K\left(  \left[  a,b\right]  \right)  \right)
}_{=L_{\left[  a,b\right]  }}\left(  c\right)  =L_{\left[  a,b\right]
}\left(  c\right)  =\underbrace{\left[  a,b\right]  }%
_{\substack{=a\vartriangleright b-b\vartriangleright a\\\text{(by the
definition}\\\text{of the Lie algebra }A^{-}\text{)}}}\vartriangleright
c\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{pf.thm.manchon.La(b)}), applied to }\left[  a,b\right]  \text{
and }c\\
\text{instead of }a\text{ and }b
\end{array}
\right) \\
&  =\left(  a\vartriangleright b-b\vartriangleright a\right)
\vartriangleright c=\left(  a\vartriangleright b\right)  \vartriangleright
c-\left(  b\vartriangleright a\right)  \vartriangleright c=a\vartriangleright
\left(  b\vartriangleright c\right)  -b\vartriangleright\left(
a\vartriangleright c\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  a\vartriangleright
b\right)  \vartriangleright c-a\vartriangleright\left(  b\vartriangleright
c\right)  =\left(  b\vartriangleright a\right)  \vartriangleright
c-b\vartriangleright\left(  a\vartriangleright c\right)  \text{ (by
(\ref{eq.def.prelie.left.axiom}))}\right)  ,
\end{align*}
we obtain $\left(  \left[  K\left(  a\right)  ,K\left(  b\right)  \right]
\mid_{A}\right)  \left(  c\right)  =\left(  K\left(  \left[  a,b\right]
\right)  \mid_{A}\right)  \left(  c\right)  $.
\par
Now, let us forget that we fixed $c$. We thus have shown that $\left(  \left[
K\left(  a\right)  ,K\left(  b\right)  \right]  \mid_{A}\right)  \left(
c\right)  =\left(  K\left(  \left[  a,b\right]  \right)  \mid_{A}\right)
\left(  c\right)  $ for every $c\in A$. In other words, $\left[  K\left(
a\right)  ,K\left(  b\right)  \right]  \mid_{A}=K\left(  \left[  a,b\right]
\right)  \mid_{A}$, qed.}. Hence, Lemma \ref{lem.derivation.unique} (applied
to $\operatorname*{Sym}A$, $\left[  K\left(  a\right)  ,K\left(  b\right)
\right]  $, $K\left(  \left[  a,b\right]  \right)  $ and $A$ instead of $A$,
$d$, $e$ and $S$) shows that $\left[  K\left(  a\right)  ,K\left(  b\right)
\right]  =K\left(  \left[  a,b\right]  \right)  $. This proves
(\ref{pf.thm.manchon.d.goal1}).

Next, we notice that%
\begin{equation}
K\left(  \lambda a+\mu b\right)  =\lambda K\left(  a\right)  +\mu K\left(
b\right)  \label{pf.thm.manchon.d.goal2}%
\end{equation}
for every $a\in A^{-}$ and $b\in A^{-}$ and every $\lambda\in\mathbf{k}$ and
$\mu\in\mathbf{k}$.

\textit{Proof of (\ref{pf.thm.manchon.d.goal2}):} The proof of
(\ref{pf.thm.manchon.d.goal2}) is similar to the above proof of
(\ref{pf.thm.manchon.d.goal1}), but even simpler (because instead of $\left[
a,b\right]  \vartriangleright c=a\vartriangleright\left(  b\vartriangleright
c\right)  -b\vartriangleright\left(  a\vartriangleright c\right)  $, we now
need to use the much simpler equality $\left(  \lambda a+\mu b\right)
\vartriangleright c=\lambda\left(  a\vartriangleright c\right)  +\mu\left(
b\vartriangleright c\right)  $). Hence, we leave it to the reader.

Now, the map $K:A^{-}\rightarrow\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $ is $\mathbf{k}$-linear (because of (\ref{pf.thm.manchon.d.goal2}))
and thus a Lie algebra homomorphism (because of (\ref{pf.thm.manchon.d.goal1}%
)). This proves Theorem \ref{thm.manchon} \textbf{(d)}.

\textbf{(e)} The map $K:A^{-}\rightarrow\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $ is a Lie algebra homomorphism (according to
Theorem \ref{thm.manchon} \textbf{(d)}). Moreover, we have%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K\left(  a\right)  \right)  \left(
\iota_{\operatorname*{Sym},A}\left(  b\right)  \right)  -\left(  K\left(
b\right)  \right)  \left(  \iota_{\operatorname*{Sym},A}\left(  a\right)
\right)
\]
for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  $ is computed in the Lie algebra $\left(
\operatorname*{Sym}A\right)  ^{-}$)\ \ \ \ \footnote{\textit{Proof.} Let $a\in
A^{-}$ and $b\in A^{-}$. Recall that we are identifying $\iota
_{\operatorname*{Sym},A}\left(  x\right)  $ with $x$ for every $x\in A$. Thus,
$\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
a,b\right]  $, $\iota_{\operatorname*{Sym},A}\left(  a\right)  =a$ and
$\iota_{\operatorname*{Sym},A}\left(  b\right)  =b$. Moreover, the
$\mathbf{k}$-algebra $\operatorname*{Sym}A$ is commutative, and thus the Lie
bracket of the Lie algebra $\left(  \operatorname*{Sym}A\right)  ^{-}$ is
identically $0$. Hence, $\left[  \iota_{\operatorname*{Sym},A}\left(
a\right)  ,\iota_{\operatorname*{Sym},A}\left(  b\right)  \right]  =0$. Now,%
\begin{align*}
&  \underbrace{\left[  \iota_{\operatorname*{Sym},A}\left(  a\right)
,\iota_{\operatorname*{Sym},A}\left(  b\right)  \right]  }_{=0}%
+\underbrace{\left(  K\left(  a\right)  \right)  }_{\substack{=L_{a}%
\\\text{(by the definition of }K\text{)}}}\left(  \underbrace{\iota
_{\operatorname*{Sym},A}\left(  b\right)  }_{=b}\right)  -\underbrace{\left(
K\left(  b\right)  \right)  }_{\substack{=L_{b}\\\text{(by the definition of
}K\text{)}}}\left(  \underbrace{\iota_{\operatorname*{Sym},A}\left(  a\right)
}_{=a}\right) \\
&  =\underbrace{L_{a}\left(  b\right)  }_{\substack{=a\vartriangleright
b\\\text{(by (\ref{pf.thm.manchon.La(b)}))}}}-\underbrace{L_{b}\left(
a\right)  }_{\substack{=b\vartriangleright a\\\text{(by
(\ref{pf.thm.manchon.La(b)}) (applied to}\\b\text{ and }a\text{ instead of
}a\text{ and }b\text{))}}}=a\vartriangleright b-b\vartriangleright a.
\end{align*}
Compared with%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
a,b\right]  =a\vartriangleright b-b\vartriangleright a
\]
(by the definition of the Lie bracket on $A^{-}$), this shows that
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K\left(  a\right)  \right)  \left(
\iota_{\operatorname*{Sym},A}\left(  b\right)  \right)  -\left(  K\left(
b\right)  \right)  \left(  \iota_{\operatorname*{Sym},A}\left(  a\right)
\right)  .
\]
Qed.}. Hence, we can apply Theorem \ref{thm.gen1} to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$.

Thus, Theorem \ref{thm.gen1} \textbf{(a)} (applied to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$) shows that we
can define an $A^{-}$-module structure on $\operatorname*{Sym}A$ by setting%
\[
\left(  a\rightharpoonup u=\iota_{\operatorname*{Sym},A}\left(  a\right)
\cdot u+\left(  K\left(  a\right)  \right)  \left(  u\right)  \text{ for all
}a\in A^{-}\text{ and }u\in\operatorname*{Sym}A\right)  .
\]
In other words, we can define an $A^{-}$-module structure on
$\operatorname*{Sym}A$ by setting%
\[
\left(  a\rightharpoonup u=au+L_{a}\left(  u\right)  \text{ for all }a\in
A^{-}\text{ and }u\in\operatorname*{Sym}A\right)
\]
(because every $a\in A^{-}$ and $u\in\operatorname*{Sym}A$ satisfy
$\underbrace{\iota_{\operatorname*{Sym},A}\left(  a\right)  }_{=a}\cdot
u+\underbrace{\left(  K\left(  a\right)  \right)  }_{\substack{=L_{a}%
\\\text{(by the definition of }K\text{)}}}\left(  u\right)  =au+L_{a}\left(
u\right)  $). This proves Theorem \ref{thm.manchon} \textbf{(e)}.

As a consequence of this proof of Theorem \ref{thm.manchon} \textbf{(e)}, we
see that the $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1} \textbf{(a)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$). We
notice further that the map $\eta:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ defined in Theorem \ref{thm.manchon} \textbf{(f)} is
precisely the map $\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}%
A$ that is constructed by Theorem \ref{thm.gen1} \textbf{(b)} (applied to
$\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota
_{\operatorname*{Sym},A}$) (since these two maps have the same definition).

\textbf{(f)} As we have seen in our proof of Theorem \ref{thm.manchon}
\textbf{(e)}, we can apply Theorem \ref{thm.gen1} to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$. Moreover, the
following holds:

\begin{itemize}
\item The $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1} \textbf{(a)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$).

\item The map $\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$
defined in Theorem \ref{thm.manchon} \textbf{(f)} is precisely the map
$\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$ that is
constructed by Theorem \ref{thm.gen1} \textbf{(b)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$).
\end{itemize}

Hence, Theorem \ref{thm.gen1} \textbf{(b)} (applied to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$) shows that
$\eta$ is an $A^{-}$-module homomorphism. This proves Theorem
\ref{thm.manchon} \textbf{(f)}.

\textbf{(g)} Let $a\in A^{-}$. The map $L_{a}$ is a derivation. In other
words, $L_{a}\in\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $.
Thus, $L_{a}\left(  1\right)  =0$ (by Proposition \ref{prop.Der.basics}
\textbf{(c)}, applied to $\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $ and $L_{a}$ instead of $C$ and $f$).

Now, the definition of $\eta$ shows that
\begin{align*}
&  \eta\left(  \iota_{U,A^{-}}\left(  a\right)  \right) \\
&  =\iota_{U,A^{-}}\left(  a\right)  1_{\operatorname*{Sym}A}=a\rightharpoonup
1_{\operatorname*{Sym}A}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the action of
}U\left(  A^{-}\right)  \text{ on }\operatorname*{Sym}A\right) \\
&  =\underbrace{a1_{\operatorname*{Sym}A}}_{=a}+\underbrace{L_{a}\left(
1_{\operatorname*{Sym}A}\right)  }_{=L_{a}\left(  1\right)  =0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the }A^{-}\text{-module
structure on }\operatorname*{Sym}A\right) \\
&  =a.
\end{align*}
This proves Theorem \ref{thm.manchon} \textbf{(g)}.

\textbf{(i)} We make the following observations:

\begin{itemize}
\item The map $K:A^{-}\rightarrow\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $ is a Lie algebra homomorphism (according to
Theorem \ref{thm.manchon} \textbf{(d)}).

\item We have%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K\left(  a\right)  \right)  \left(
\iota_{\operatorname*{Sym},A}\left(  b\right)  \right)  -\left(  K\left(
b\right)  \right)  \left(  \iota_{\operatorname*{Sym},A}\left(  a\right)
\right)
\]
for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  $ is computed in the Lie algebra $\left(
\operatorname*{Sym}A\right)  ^{-}$)\ \ \ \ \footnote{This was proven in our
above proof of Theorem \ref{thm.manchon} \textbf{(e)}.}.

\item The $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1} \textbf{(a)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}%
$).\ \ \ \ \footnote{This was proven above.}

\item The map $\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$
defined in Theorem \ref{thm.manchon} \textbf{(f)} is precisely the map
$\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$ that is
constructed by Theorem \ref{thm.gen1} \textbf{(b)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$).

\item The filtration of the filtered $\mathbf{k}$-algebra $\operatorname*{Sym}%
A$ is $\left(  \bigoplus_{k=0}^{m}\operatorname*{Sym}\nolimits^{k}A\right)
_{m\geq0}$.

\item We have $\iota_{\operatorname*{Sym},A}\left(  \underbrace{A^{-}}%
_{=A}\right)  =\iota_{\operatorname*{Sym},A}\left(  A\right)
=\operatorname*{Sym}\nolimits^{1}A\subseteq\bigoplus_{k=0}^{1}%
\operatorname*{Sym}\nolimits^{k}A$.

\item The map $K\left(  a\right)  :\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$ is filtered for every $a\in A^{-}$%
\ \ \ \ \footnote{\textit{Proof.} Let $a\in A^{-}$. Thus, the definition of
$K$ shows that $K\left(  a\right)  =L_{a}$. Now, the map $L_{a}$ is graded (by
Theorem \ref{thm.manchon} \textbf{(b)}), and thus filtered (since any graded
$\mathbf{k}$-linear map is filtered). In other words, the map $K\left(
a\right)  $ is filtered (since $K\left(  a\right)  =L_{a}$). Qed.}.
\end{itemize}

Hence, we can apply Theorem \ref{thm.gen1filt} to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$, $\left(  C_{m}\right)  _{m\geq0}=\left(
\bigoplus_{k=0}^{m}\operatorname*{Sym}\nolimits^{k}A\right)  _{m\geq0}$ and
$f=\iota_{\operatorname*{Sym},A}$. As a result, we conclude that
\begin{align}
&  \eta\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}}\left(
a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right) \nonumber\\
&  \in\iota_{\operatorname*{Sym},A}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},A}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,A}\left(  a_{n}\right)  +\bigoplus_{k=0}^{n-1}\operatorname*{Sym}%
\nolimits^{k}A \label{pf.thm.manchon.i.1}%
\end{align}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in A^{-}$.
Thus, for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in A^{-}%
$, we have%
\[
\eta\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}}\left(
a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right)  \in
a_{1}a_{2}\cdots a_{n}+\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}A
\]
\footnote{\textit{Proof.} Let $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots
,a_{n}\in A^{-}$.
\par
Recall that we are identifying every $x\in A$ with $\iota_{\operatorname*{Sym}%
,A}\left(  x\right)  $. Thus, every $x\in A$ satisfies $x=\iota
_{\operatorname*{Sym},A}\left(  x\right)  =\iota_{\operatorname*{Sym},A^{-}%
}\left(  x\right)  $ (since $\iota_{\operatorname*{Sym},A}=\iota
_{\operatorname*{Sym},A^{-}}$). In particular, every $i\in\left\{
1,2,\ldots,n\right\}  $ satisfies the identity $a_{i}=\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{i}\right)  $. Multiplying these
identities over all $i\in\left\{  1,2,\ldots,n\right\}  $, we obtain
$a_{1}a_{2}\cdots a_{n}=\iota_{\operatorname*{Sym},A^{-}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  $. Now,
(\ref{pf.thm.manchon.i.1}) becomes%
\begin{align*}
\eta\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}}\left(
a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right)   &
\in\underbrace{\iota_{\operatorname*{Sym},A}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},A}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A}\left(  a_{n}\right)  }_{=a_{1}a_{2}\cdots a_{n}%
}+\underbrace{\bigoplus_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}%
A}_{\substack{=\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}A\\\text{(since
direct sums}\\\text{are sums)}}}\\
&  =a_{1}a_{2}\cdots a_{n}+\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}A,
\end{align*}
qed.}. This proves Theorem \ref{thm.manchon} \textbf{(i)}.

\textbf{(h)} Recall that $\eta$ is a $\mathbf{k}$-linear map $U\left(
A^{-}\right)  \rightarrow\operatorname*{Sym}A$. In other words, $\eta$ is a
$\mathbf{k}$-linear map $U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}%
\left(  A^{-}\right)  $ (since $A^{-}=A$ as $\mathbf{k}$-modules).

For every $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots,a_{n}\in A^{-}$, we have%
\begin{align*}
&  \eta\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}}\left(
a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right) \\
&  \in\iota_{\operatorname*{Sym},A^{-}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  +\sum_{k=0}^{n-1}%
\operatorname*{Sym}\nolimits^{k}\left(  A^{-}\right)
\end{align*}
\footnote{\textit{Proof.} Let $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots
,a_{n}\in A^{-}$. We have $A^{-}=A$ as $\mathbf{k}$-modules. Hence,
$\iota_{\operatorname*{Sym},A^{-}}=\iota_{\operatorname*{Sym},A}$.
\par
Recall that we are identifying every $x\in A$ with $\iota_{\operatorname*{Sym}%
,A}\left(  x\right)  $. Thus, every $x\in A$ satisfies $x=\iota
_{\operatorname*{Sym},A}\left(  x\right)  =\iota_{\operatorname*{Sym},A^{-}%
}\left(  x\right)  $ (since $\iota_{\operatorname*{Sym},A}=\iota
_{\operatorname*{Sym},A^{-}}$). In particular, every $i\in\left\{
1,2,\ldots,n\right\}  $ satisfies the identity $a_{i}=\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{i}\right)  $. Multiplying these
identities over all $i\in\left\{  1,2,\ldots,n\right\}  $, we obtain
$a_{1}a_{2}\cdots a_{n}=\iota_{\operatorname*{Sym},A^{-}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  $.
\par
Now, Theorem \ref{thm.manchon} \textbf{(i)} shows that%
\begin{align*}
&  \eta\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}}\left(
a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right) \\
&  \in\underbrace{a_{1}a_{2}\cdots a_{n}}_{=\iota_{\operatorname*{Sym},A^{-}%
}\left(  a_{1}\right)  \iota_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)
\cdots\iota_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  }+\sum
_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}\underbrace{A}_{=A^{-}}\\
&  =\iota_{\operatorname*{Sym},A^{-}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  +\sum_{k=0}^{n-1}%
\operatorname*{Sym}\nolimits^{k}\left(  A^{-}\right)  ,
\end{align*}
qed.}. Thus, we can apply Lemma \ref{lem.PBW-by-inverse-map} to $\mathfrak{g}%
=A^{-}$ and $\phi=\eta$.

Lemma \ref{lem.PBW-by-inverse-map} \textbf{(a)} (applied to $\mathfrak{g}%
=A^{-}$ and $\phi=\eta$) shows that the map $\eta$ is a filtered $\mathbf{k}%
$-module isomorphism. In particular, the map $\eta$ is invertible.

But we know that $\eta$ is an $A^{-}$-module homomorphism (by Theorem
\ref{thm.manchon} \textbf{(f)}). Since $\eta$ is invertible, this shows that
$\eta$ is an $A^{-}$-module isomorphism. This proves Theorem \ref{thm.manchon}
\textbf{(h)}.

\textbf{(j)} In our above proof of Theorem \ref{thm.manchon} \textbf{(h)}, we
have shown that we can apply Lemma \ref{lem.PBW-by-inverse-map} to
$\mathfrak{g}=A^{-}$ and $\phi=\eta$. Hence, Lemma
\ref{lem.PBW-by-inverse-map} \textbf{(a)} (applied to $\mathfrak{g}=A^{-}$ and
$\phi=\eta$) shows that the map $\eta$ is a filtered $\mathbf{k}$-module
isomorphism. Also, Lemma \ref{lem.PBW-by-inverse-map} \textbf{(b)} (applied to
$\mathfrak{g}=A^{-}$ and $\phi=\eta$) shows that the inverse $\eta^{-1}$ of
the map $\eta$ is filtered. Thus, both maps $\eta$ and $\eta^{-1}$ are
filtered. This proves Theorem \ref{thm.manchon} \textbf{(j)}.

\textbf{(k)} In our above proof of Theorem \ref{thm.manchon} \textbf{(h)}, we
have shown that we can apply Lemma \ref{lem.PBW-by-inverse-map} to
$\mathfrak{g}=A^{-}$ and $\phi=\eta$. Hence, the first claim of Lemma
\ref{lem.PBW-by-inverse-map} \textbf{(c)} (applied to $\mathfrak{g}=A^{-}$ and
$\phi=\eta$) shows that the map $\operatorname*{grad}%
\nolimits_{\operatorname*{Sym}\left(  A^{-}\right)  }^{-1}\circ\left(
\operatorname*{gr}\eta\right)  :\operatorname*{gr}\left(  U\left(
A^{-}\right)  \right)  \rightarrow\operatorname*{Sym}\left(  A^{-}\right)  $
is an inverse to the map $\operatorname*{PBW}\nolimits_{A^{-}}$. Since
$A^{-}=A$ as $\mathbf{k}$-modules, this rewrites as follows: The map
$\operatorname*{grad}\nolimits_{\operatorname*{Sym}A}^{-1}\circ\left(
\operatorname*{gr}\eta\right)  :\operatorname*{gr}\left(  U\left(
A^{-}\right)  \right)  \rightarrow\operatorname*{Sym}A$ is an inverse to the
map $\operatorname*{PBW}\nolimits_{A^{-}}$. In particular, the map
$\operatorname*{PBW}\nolimits_{A^{-}}$ is invertible. This proves Theorem
\ref{thm.manchon} \textbf{(k)}.

\textbf{(l)} We first observe the following:

\begin{itemize}
\item The map $K:A^{-}\rightarrow\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $ is a Lie algebra homomorphism. (This follows
from Theorem \ref{thm.manchon} \textbf{(d)}.)

\item We have $K\left(  A^{-}\right)  \subseteq\operatorname*{Coder}\left(
\operatorname*{Sym}A\right)  $\ \ \ \ \footnote{\textit{Proof.} Let $g\in
K\left(  A^{-}\right)  $. Thus, there exists some $a\in A^{-}$ such that
$g=K\left(  a\right)  $. Consider this $a$. We have $a\in A^{-}=A$ and
$g=K\left(  a\right)  =L_{a}$ (by the definition of $K$). But $L_{a}%
:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ is a coderivation
(according to Theorem \ref{thm.manchon} \textbf{(c)}). In other words,
$L_{a}\in\operatorname*{Coder}\left(  \operatorname*{Sym}A\right)  $. Hence,
$g=L_{a}\in\operatorname*{Coder}\left(  \operatorname*{Sym}A\right)  $.
\par
Now, let us forget that we fixed $g$. We thus have shown that $g\in
\operatorname*{Coder}\left(  \operatorname*{Sym}A\right)  $ for every $g\in
K\left(  A^{-}\right)  $. In other words, $K\left(  A^{-}\right)
\subseteq\operatorname*{Coder}\left(  \operatorname*{Sym}A\right)  $, qed.}.

\item We have $\iota_{\operatorname*{Sym},A}\left(  A^{-}\right)
\subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}A\right)
$\ \ \ \ \footnote{\textit{Proof.} Recall that $\iota_{\operatorname*{Sym}%
,V}\left(  V\right)  \subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}%
V\right)  $ for every $\mathbf{k}$-module $V$. Applying this to $V=A$, we
obtain $\iota_{\operatorname*{Sym},A}\left(  A\right)  \subseteq
\operatorname*{Prim}\left(  \operatorname*{Sym}A\right)  $. Since $A^{-}=A$ as
$\mathbf{k}$-modules, this rewrites as $\iota_{\operatorname*{Sym},A}\left(
A^{-}\right)  \subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}%
A\right)  $. Qed.}.

\item The equality%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K\left(  a\right)  \right)  \left(
\iota_{\operatorname*{Sym},A}\left(  b\right)  \right)  -\left(  K\left(
b\right)  \right)  \left(  \iota_{\operatorname*{Sym},A}\left(  a\right)
\right)
\]
holds for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  $ is computed in the Lie algebra $\left(
\operatorname*{Sym}A\right)  ^{-}$). (This was proven during our above proof
of Theorem \ref{thm.manchon} \textbf{(e)}.)

\item The $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1} \textbf{(a)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$).
(This follows from our above proof of Theorem \ref{thm.manchon} \textbf{(e)}.)

\item The map $\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$
defined in Theorem \ref{thm.manchon} \textbf{(f)} is precisely the map
$\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$ that is
constructed by Theorem \ref{thm.gen1} \textbf{(b)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$).
(This is clear, because the definitions of these two maps are the same.)
\end{itemize}

Combining these observations, we see that Theorem \ref{thm.gen3} can be
applied to $\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota
_{\operatorname*{Sym},A}$. Consequently, Theorem \ref{thm.gen3} \textbf{(b)}
shows that the map $\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}%
A$ is a $\mathbf{k}$-coalgebra homomorphism. Since $\eta$ is invertible, this
shows that $\eta$ is a $\mathbf{k}$-coalgebra isomorphism. This finishes the
proof of Theorem \ref{thm.manchon} \textbf{(l)}.

\textbf{(m)} As we have seen in our proof of Theorem \ref{thm.manchon}
\textbf{(e)}, we can apply Theorem \ref{thm.gen1} to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$. Moreover, the
$A^{-}$-module structure on $\operatorname*{Sym}A$ defined in Theorem
\ref{thm.manchon} \textbf{(e)} is precisely the one that is constructed by
Theorem \ref{thm.gen1} \textbf{(a)} (applied to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$). Hence,
Theorem \ref{thm.gen1} \textbf{(c)} (applied to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$) shows that,
for every $a\in A^{-}$, $b\in\operatorname*{Sym}A$ and $c\in
\operatorname*{Sym}A$, we have%
\begin{equation}
a\rightharpoonup\left(  bc\right)  -b\cdot\left(  a\rightharpoonup c\right)
=\left(  K\left(  a\right)  \right)  \left(  b\right)  \cdot c+\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,b\right]  c.
\label{pf.thm.manchon.m.used}%
\end{equation}


Now, fix $a\in A$, $b\in A$ and $c\in\operatorname*{Sym}A$. We have $a\in
A=A^{-}$ and $b\in A\subseteq\operatorname*{Sym}A$ (since we regard $A$ as a
$\mathbf{k}$-submodule of $\operatorname*{Sym}A$). Moreover, the $\mathbf{k}%
$-algebra $\operatorname*{Sym}A$ is commutative, and thus the Lie bracket of
the Lie algebra $\left(  \operatorname*{Sym}A\right)  ^{-}$ is identically
$0$. Hence, $\left[  \iota_{\operatorname*{Sym},A}\left(  a\right)  ,b\right]
=0$. Furthermore, $K\left(  a\right)  =L_{a}$ (by the definition of $K$), so
that%
\[
\underbrace{\left(  K\left(  a\right)  \right)  }_{=L_{a}}\left(  b\right)
=L_{a}\left(  b\right)  =a\vartriangleright b\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.thm.manchon.La(b)})}\right)  .
\]


Now, (\ref{pf.thm.manchon.m.used}) shows that%
\[
a\rightharpoonup\left(  bc\right)  -b\cdot\left(  a\rightharpoonup c\right)
=\underbrace{\left(  K\left(  a\right)  \right)  \left(  b\right)
}_{=a\vartriangleright b}\cdot c+\underbrace{\left[  \iota
_{\operatorname*{Sym},A}\left(  a\right)  ,b\right]  }_{=0}c=\left(
a\vartriangleright b\right)  \cdot c+\underbrace{0c}_{=0}=\left(
a\vartriangleright b\right)  \cdot c.
\]
This proves Theorem \ref{thm.manchon} \textbf{(m)}.
\end{proof}

\begin{corollary}
\label{cor.manchon.box}Let $A$ be a left pre-Lie algebra. Define the map
$\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$ as in Theorem
\ref{thm.manchon} \textbf{(f)}. Theorem \ref{thm.manchon} \textbf{(h)} shows
that $\eta$ is an $A^{-}$-module isomorphism, thus a $\mathbf{k}$-module
isomorphism. Hence, we can define a new, \textbf{commutative} multiplication
$\boxdot$ on $U\left(  A^{-}\right)  $ (formally speaking, a $\mathbf{k}%
$-bilinear map $\boxdot:U\left(  A^{-}\right)  \times U\left(  A^{-}\right)
\rightarrow U\left(  A^{-}\right)  $, written in infix notation) by letting%
\[
x\boxdot y=\eta\left(  \eta^{-1}\left(  x\right)  \cdot\eta^{-1}\left(
y\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for all }x\in U\left(
A^{-}\right)  \text{ and }y\in U\left(  A^{-}\right)  .
\]
(In other words, this multiplication $\boxdot$ is the result of transporting
the multiplication of the $\mathbf{k}$-algebra $\operatorname*{Sym}A$ to
$U\left(  A^{-}\right)  $ along the isomorphism $\eta$.)

\textbf{(a)} Let us write $\left(  U\left(  A^{-}\right)  ,\boxdot\right)  $
for the $\mathbf{k}$-module $U\left(  A^{-}\right)  $ equipped with the
multiplication $\boxdot$. This is a commutative $\mathbf{k}$-algebra with
unity $1_{U\left(  A^{-}\right)  }$.

\textbf{(b)} Let $\Delta$ and $\epsilon$ denote the comultiplication and the
counit of the $\mathbf{k}$-bialgebra $U\left(  A^{-}\right)  $. The
$\mathbf{k}$-algebra $\left(  U\left(  A^{-}\right)  ,\boxdot\right)  $,
equipped with the comultiplication $\Delta$ and the counit $\epsilon$, is a
$\mathbf{k}$-bialgebra. We shall denote this $\mathbf{k}$-bialgebra by
$\left(  U\left(  A^{-}\right)  ,\boxdot\right)  $.

\textbf{(c)} The map $\eta$ is a $\mathbf{k}$-bialgebra isomorphism $\left(
U\left(  A^{-}\right)  ,\boxdot\right)  \rightarrow\operatorname*{Sym}A$.

\textbf{(d)} We have $1_{\left(  U\left(  A^{-}\right)  ,\boxdot\right)
}=1_{U\left(  A^{-}\right)  }$.

\textbf{(e)} For every $c\in A^{-}$, let $\overline{c}$ denote the element
$\iota_{U,A^{-}}\left(  c\right)  $ of $U\left(  A^{-}\right)  $. For every
$n\in\mathbb{N}$, every $a\in A^{-}$ and every $b_{1},b_{2},\ldots,b_{n}\in
A^{-}$, we have%
\begin{align*}
&  \overline{a}\boxdot\left(  \overline{b_{1}}\boxdot\overline{b_{2}}%
\boxdot\cdots\boxdot\overline{b_{n}}\right) \\
&  =\overline{a}\left(  \overline{b_{1}}\boxdot\overline{b_{2}}\boxdot
\cdots\boxdot\overline{b_{n}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\sum_{k=1}^{n}\overline{b_{1}}\boxdot\overline{b_{2}%
}\boxdot\cdots\boxdot\overline{b_{k-1}}\boxdot\left(  \overline
{a\vartriangleright b_{k}}\right)  \boxdot\overline{b_{k+1}}\boxdot
\overline{b_{k+2}}\boxdot\cdots\boxdot\overline{b_{n}}.
\end{align*}

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.manchon.box}.]\textbf{(a)} The definition of
$\eta$ yields $\eta\left(  1_{U\left(  A^{-}\right)  }\right)  =1_{U\left(
A^{-}\right)  }1_{\operatorname*{Sym}A}=1_{\operatorname*{Sym}A}$, so that
$\eta^{-1}\left(  1_{\operatorname*{Sym}A}\right)  =1_{U\left(  A^{-}\right)
}$.

Now, the multiplication of the $\mathbf{k}$-algebra $\operatorname*{Sym}A$ is
associative and commutative and has neutral element $1_{\operatorname*{Sym}A}%
$. Therefore, the binary operation $\boxdot$ is associative and commutative
and has neutral element $\eta^{-1}\left(  1_{\operatorname*{Sym}A}\right)  $
(because the binary operation $\boxdot$ is the result of transporting the
multiplication of the $\mathbf{k}$-algebra $\operatorname*{Sym}A$ to $U\left(
A^{-}\right)  $ along the isomorphism $\eta$). In other words, the binary
operation $\boxdot$ is associative and commutative and has neutral element
$1_{U\left(  A^{-}\right)  }$ (since $\eta^{-1}\left(  1_{\operatorname*{Sym}%
A}\right)  =1_{U\left(  A^{-}\right)  }$). In other words, the $\mathbf{k}%
$-module $U\left(  A^{-}\right)  $ equipped with the multiplication $\boxdot$
is a commutative $\mathbf{k}$-algebra with unity $1_{U\left(  A^{-}\right)  }%
$. This proves Corollary \ref{cor.manchon.box} \textbf{(a)}.

\textbf{(b)} Let us define a \textit{pre-bialgebra} to mean a $\mathbf{k}%
$-module $V$ equipped with a $\mathbf{k}$-bilinear map $\mu_{V}:V\times
V\rightarrow V$ called its \textit{multiplication}, an element $e_{V}\in V$
called its \textit{unity}, a $\mathbf{k}$-linear map $\Delta_{V}:V\rightarrow
V\otimes V$ called its \textit{comultiplication}, and a $\mathbf{k}$-linear
map $\epsilon_{V}:V\rightarrow\mathbf{k}$ called its \textit{counit}. These
maps $\mu_{V}$, $\Delta_{V}$ and $\epsilon_{V}$ and this element $e_{V}$ are
not required to satisfy any axioms (unlike for a $\mathbf{k}$-bialgebra).
Thus, a $\mathbf{k}$-bialgebra is a pre-bialgebra satisfying certain axioms.
We define a \textit{homomorphism of pre-bialgebras} to be a $\mathbf{k}%
$-linear map from one pre-bialgebra to another that respects the
multiplication, the unity, the comultiplication and the counit. As usual, we
define an \textit{isomorphism of pre-bialgebras} to be an invertible
homomorphism of pre-bialgebras whose inverse is also a homomorphism of
pre-bialgebras. Finally, we say that two pre-bialgebras are
\textit{isomorphic} if there exists an isomorphism of pre-bialgebras between them.

Now, let us make one trivial yet crucial observation: If a pre-bialgebra $V$
is isomorphic to a $\mathbf{k}$-bialgebra, then $V$ itself is a $\mathbf{k}%
$-bialgebra. More precisely, we have the following fact:

\textit{Observation 1:} If $V$ is a pre-bialgebra, if $W$ is a $\mathbf{k}%
$-bialgebra, and if $\varphi:V\rightarrow W$ is an isomorphism of
pre-bialgebras, then $V$ is a $\mathbf{k}$-bialgebra, and $\varphi$ is a
$\mathbf{k}$-bialgebra isomorphism.

Now, the $\mathbf{k}$-module $U\left(  A^{-}\right)  $, equipped with the
multiplication $\boxdot$, the unity $1_{U\left(  A^{-}\right)  }$, the
comultiplication $\Delta$ and the counit $\epsilon$ is a pre-bialgebra. The
map $\eta$ is an isomorphism of pre-bialgebras from this pre-bialgebra to the
$\mathbf{k}$-bialgebra $\operatorname*{Sym}A$\ \ \ \ \footnote{\textit{Proof.}
We must show the following five statements:
\par
\begin{enumerate}
\item The map $\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$ is
an isomorphism of $\mathbf{k}$-modules.
\par
\item The map $\eta$ preserves the multiplication (i.e., we have $\eta\left(
x\boxdot y\right)  =\eta\left(  x\right)  \cdot\eta\left(  y\right)  $ for all
$x\in U\left(  A^{-}\right)  $ and $y\in U\left(  A^{-}\right)  $).
\par
\item The map $\eta$ preserves the unity (i.e., we have $\eta\left(
1_{U\left(  A^{-}\right)  }\right)  =1_{\operatorname*{Sym}A}$).
\par
\item The map $\eta$ preserves the comultiplication (i.e., we have $\left(
\eta\otimes\eta\right)  \circ\Delta=\Delta_{\operatorname*{Sym}A}\circ\eta$).
\par
\item The map $\eta$ preserves the counit (i.e., we have $\epsilon
=\epsilon_{\operatorname*{Sym}A}\circ\eta$).
\end{enumerate}
\par
The first and the third of these five statements have already been proven. The
second follows from the fact that the binary operation $\boxdot$ is the result
of transporting the multiplication of the $\mathbf{k}$-algebra
$\operatorname*{Sym}A$ to $U\left(  A^{-}\right)  $ along the isomorphism
$\eta$. The fourth and fifth statements follow from the fact that $\eta$ is a
coalgebra isomorphism $U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$
(by Theorem \ref{thm.manchon} \textbf{(l)}). Thus, all five statements are
proven, qed.}. Thus, Observation 1 (applied to $V=U\left(  A^{-}\right)  $
(equipped with the multiplication $\boxdot$, the unity $1_{U\left(
A^{-}\right)  }$, the comultiplication $\Delta$ and the counit $\epsilon$),
$W=\operatorname*{Sym}A$ and $\varphi=\eta$) shows that $U\left(
A^{-}\right)  $ (equipped with the multiplication $\boxdot$, the unity
$1_{U\left(  A^{-}\right)  }$, the comultiplication $\Delta$ and the counit
$\epsilon$) is a $\mathbf{k}$-bialgebra, and $\eta$ is a $\mathbf{k}%
$-bialgebra isomorphism.

In particular, $U\left(  A^{-}\right)  $ (equipped with the multiplication
$\boxdot$, the unity $1_{U\left(  A^{-}\right)  }$, the comultiplication
$\Delta$ and the counit $\epsilon$) is a $\mathbf{k}$-bialgebra. In other
words, the $\mathbf{k}$-algebra $\left(  U\left(  A^{-}\right)  ,\boxdot
\right)  $, equipped with the comultiplication $\Delta$ and the counit
$\epsilon$, is a $\mathbf{k}$-bialgebra. This proves Corollary
\ref{cor.manchon.box} \textbf{(b)}.

\textbf{(c)} In our proof of Corollary \ref{cor.manchon.box} \textbf{(b)}, we
have shown that $\eta$ is a $\mathbf{k}$-bialgebra isomorphism. This proves
Corollary \ref{cor.manchon.box} \textbf{(c)}.

\textbf{(d)} Corollary \ref{cor.manchon.box} \textbf{(c)} shows that the map
$\eta$ is a $\mathbf{k}$-bialgebra isomorphism $\left(  U\left(  A^{-}\right)
,\boxdot\right)  \rightarrow\operatorname*{Sym}A$. In particular, the map
$\eta$ is a $\mathbf{k}$-algebra isomorphism $\left(  U\left(  A^{-}\right)
,\boxdot\right)  \rightarrow\operatorname*{Sym}A$. In particular, the map
$\eta$ is a $\mathbf{k}$-algebra homomorphism $\left(  U\left(  A^{-}\right)
,\boxdot\right)  \rightarrow\operatorname*{Sym}A$. Hence, $\eta\left(
1_{\left(  U\left(  A^{-}\right)  ,\boxdot\right)  }\right)
=1_{\operatorname*{Sym}A}$.

But the definition of $\eta$ shows that $\eta\left(  1_{U\left(  A^{-}\right)
}\right)  =1_{U\left(  A^{-}\right)  }1_{\operatorname*{Sym}A}%
=1_{\operatorname*{Sym}A}$. Since $\eta$ is invertible (because $\eta$ is an
isomorphism), this shows that $1_{U\left(  A^{-}\right)  }=\eta^{-1}\left(
1_{\operatorname*{Sym}A}\right)  =1_{\left(  U\left(  A^{-}\right)
,\boxdot\right)  }$ (since $\eta\left(  1_{\left(  U\left(  A^{-}\right)
,\boxdot\right)  }\right)  =1_{\operatorname*{Sym}A}$). This proves Corollary
\ref{cor.manchon.box} \textbf{(d)}.

\textbf{(e)} In the following, we will identify every $a\in A$ with the
element $\iota_{\operatorname*{Sym},A}\left(  a\right)  $ of
$\operatorname*{Sym}A$ (as it was done in Theorem \ref{thm.manchon}).

The map $\eta$ is a $\mathbf{k}$-algebra isomorphism $\left(  U\left(
A^{-}\right)  ,\boxdot\right)  \rightarrow\operatorname*{Sym}A$ (as we have
shown in the proof of Corollary \ref{cor.manchon.box} \textbf{(d)}). Thus, its
inverse $\eta^{-1}$ is a $\mathbf{k}$-algebra isomorphism $\operatorname*{Sym}%
A\rightarrow\left(  U\left(  A^{-}\right)  ,\boxdot\right)  $.

On the other hand, the map $\eta$ is an $A^{-}$-module isomorphism. Hence, its
inverse $\eta^{-1}$ is an $A^{-}$-module isomorphism as well.

For every $c\in A^{-}$, we have%
\begin{equation}
\eta^{-1}\left(  c\right)  =\overline{c} \label{pf.cor.manchon.box.e.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.manchon.box.e.1}):} Let $c\in A^{-}$.
Then, $\overline{c}=\iota_{U,A^{-}}\left(  c\right)  $ (by the definition of
$\overline{c}$). Hence, $\eta\left(  \underbrace{\overline{c}}_{=\iota
_{U,A^{-}}\left(  c\right)  }\right)  =\eta\left(  \iota_{U,A^{-}}\left(
c\right)  \right)  =c$ (by Theorem \ref{thm.manchon} \textbf{(g)}, applied to
$a=c$). Hence, $\eta^{-1}\left(  c\right)  =\overline{c}$. This proves
(\ref{pf.cor.manchon.box.e.1}).}.

Let $n\in\mathbb{N}$. Let $a\in A^{-}$ and $b_{1},b_{2},\ldots,b_{n}\in A^{-}%
$. Then, $\overline{a}=\iota_{U,A^{-}}\left(  a\right)  $ (by the definition
of $\overline{a}$).

The definition of the $A^{-}$-module structure on $\operatorname*{Sym}A$ shows
that%
\begin{align*}
&  a\rightharpoonup\left(  b_{1}b_{2}\cdots b_{n}\right)  =a\left(  b_{1}%
b_{2}\cdots b_{n}\right)  +\underbrace{L_{a}\left(  b_{1}b_{2}\cdots
b_{n}\right)  }_{\substack{=\sum_{k=1}^{n}b_{1}b_{2}\cdots b_{k-1}\left(
a\vartriangleright b_{k}\right)  b_{k+1}b_{k+2}\cdots b_{n}\\\text{(by Theorem
\ref{thm.manchon} \textbf{(a)})}}}\\
&  =a\left(  b_{1}b_{2}\cdots b_{n}\right)  +\sum_{k=1}^{n}b_{1}b_{2}\cdots
b_{k-1}\left(  a\vartriangleright b_{k}\right)  b_{k+1}b_{k+2}\cdots b_{n}.
\end{align*}
Applying the map $\eta^{-1}$ to this equality, we obtain%
\begin{align*}
&  \eta^{-1}\left(  a\rightharpoonup\left(  b_{1}b_{2}\cdots b_{n}\right)
\right) \\
&  =\eta^{-1}\left(  a\left(  b_{1}b_{2}\cdots b_{n}\right)  +\sum_{k=1}%
^{n}b_{1}b_{2}\cdots b_{k-1}\left(  a\vartriangleright b_{k}\right)
b_{k+1}b_{k+2}\cdots b_{n}\right) \\
&  =\underbrace{\eta^{-1}\left(  a\right)  }_{\substack{=\overline
{a}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}}}\boxdot\left(
\underbrace{\eta^{-1}\left(  b_{1}\right)  }_{\substack{=\overline{b_{1}%
}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}}}\boxdot\underbrace{\eta
^{-1}\left(  b_{2}\right)  }_{\substack{=\overline{b_{2}}\\\text{(by
(\ref{pf.cor.manchon.box.e.1}))}}}\boxdot\cdots\boxdot\underbrace{\eta
^{-1}\left(  b_{n}\right)  }_{\substack{=\overline{b_{n}}\\\text{(by
(\ref{pf.cor.manchon.box.e.1}))}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{k=1}^{n}\underbrace{\eta^{-1}\left(
b_{1}\right)  }_{\substack{=\overline{b_{1}}\\\text{(by
(\ref{pf.cor.manchon.box.e.1}))}}}\boxdot\underbrace{\eta^{-1}\left(
b_{2}\right)  }_{\substack{=\overline{b_{2}}\\\text{(by
(\ref{pf.cor.manchon.box.e.1}))}}}\boxdot\cdots\boxdot\underbrace{\eta
^{-1}\left(  b_{k-1}\right)  }_{\substack{=\overline{b_{k-1}}\\\text{(by
(\ref{pf.cor.manchon.box.e.1}))}}}\boxdot\underbrace{\eta^{-1}\left(
a\vartriangleright b_{k}\right)  }_{\substack{=\overline{a\vartriangleright
b_{k}}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \boxdot\underbrace{\eta^{-1}\left(  b_{k+1}\right)
}_{\substack{=\overline{b_{k+1}}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}%
}}\boxdot\underbrace{\eta^{-1}\left(  b_{k+2}\right)  }_{\substack{=\overline
{b_{k+2}}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}}}\boxdot\cdots
\boxdot\underbrace{\eta\left(  b_{n}\right)  }_{\substack{=\overline{b_{n}%
}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta^{-1}\text{ is a }%
\mathbf{k}\text{-algebra isomorphism }\operatorname*{Sym}A\rightarrow\left(
U\left(  A^{-}\right)  ,\boxdot\right)  \right) \\
&  =\overline{a}\boxdot\left(  \overline{b_{1}}\boxdot\overline{b_{2}}%
\boxdot\cdots\boxdot\overline{b_{n}}\right)  +\sum_{k=1}^{n}\overline{b_{1}%
}\boxdot\overline{b_{2}}\boxdot\cdots\boxdot\overline{b_{k-1}}\boxdot\left(
\overline{a\vartriangleright b_{k}}\right)  \boxdot\overline{b_{k+1}}%
\boxdot\overline{b_{k+2}}\boxdot\cdots\boxdot\overline{b_{n}}.
\end{align*}
Comparing this with%
\begin{align*}
&  \eta^{-1}\left(  a\rightharpoonup\left(  b_{1}b_{2}\cdots b_{n}\right)
\right) \\
&  =a\rightharpoonup\underbrace{\eta^{-1}\left(  b_{1}b_{2}\cdots
b_{n}\right)  }_{\substack{=\eta^{-1}\left(  b_{1}\right)  \boxdot\eta
^{-1}\left(  b_{2}\right)  \boxdot\cdots\boxdot\eta^{-1}\left(  b_{n}\right)
\\\text{(since }\eta^{-1}\text{ is a }\mathbf{k}\text{-algebra isomorphism
}\operatorname*{Sym}A\rightarrow\left(  U\left(  A^{-}\right)  ,\boxdot
\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta^{-1}\text{ is an }%
A^{-}\text{-module isomorphism}\right) \\
&  =a\rightharpoonup\left(  \underbrace{\eta^{-1}\left(  b_{1}\right)
}_{\substack{=\overline{b_{1}}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}%
}}\boxdot\underbrace{\eta^{-1}\left(  b_{2}\right)  }_{\substack{=\overline
{b_{2}}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}}}\boxdot\cdots
\boxdot\underbrace{\eta^{-1}\left(  b_{n}\right)  }_{\substack{=\overline
{b_{n}}\\\text{(by (\ref{pf.cor.manchon.box.e.1}))}}}\right) \\
&  =a\rightharpoonup\left(  \overline{b_{1}}\boxdot\overline{b_{2}}%
\boxdot\cdots\boxdot\overline{b_{n}}\right)  =\underbrace{\iota_{U,A^{-}%
}\left(  a\right)  }_{=\overline{a}}\left(  \overline{b_{1}}\boxdot
\overline{b_{2}}\boxdot\cdots\boxdot\overline{b_{n}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the }A^{-}%
\text{-module structure on }U\left(  A^{-}\right)  \right) \\
&  =\overline{a}\left(  \overline{b_{1}}\boxdot\overline{b_{2}}\boxdot
\cdots\boxdot\overline{b_{n}}\right)  ,
\end{align*}
we obtain%
\begin{align*}
&  \overline{a}\boxdot\left(  \overline{b_{1}}\boxdot\overline{b_{2}}%
\boxdot\cdots\boxdot\overline{b_{n}}\right)  +\sum_{k=1}^{n}\overline{b_{1}%
}\boxdot\overline{b_{2}}\boxdot\cdots\boxdot\overline{b_{k-1}}\boxdot\left(
\overline{a\vartriangleright b_{k}}\right)  \boxdot\overline{b_{k+1}}%
\boxdot\overline{b_{k+2}}\boxdot\cdots\boxdot\overline{b_{n}}\\
&  =\overline{a}\left(  \overline{b_{1}}\boxdot\overline{b_{2}}\boxdot
\cdots\boxdot\overline{b_{n}}\right)  .
\end{align*}
In other words,%
\begin{align*}
&  \overline{a}\boxdot\left(  \overline{b_{1}}\boxdot\overline{b_{2}}%
\boxdot\cdots\boxdot\overline{b_{n}}\right) \\
&  =\overline{a}\left(  \overline{b_{1}}\boxdot\overline{b_{2}}\boxdot
\cdots\boxdot\overline{b_{n}}\right)  -\sum_{k=1}^{n}\overline{b_{1}}%
\boxdot\overline{b_{2}}\boxdot\cdots\boxdot\overline{b_{k-1}}\boxdot\left(
\overline{a\vartriangleright b_{k}}\right)  \boxdot\overline{b_{k+1}}%
\boxdot\overline{b_{k+2}}\boxdot\cdots\boxdot\overline{b_{n}}.
\end{align*}
This proves Corollary \ref{cor.manchon.box} \textbf{(e)}.
\end{proof}

\section{\label{sect.nord}The normal ordered product of differential
operators}

In this section, we shall explore an application of Theorem \ref{thm.manchon}
to the calculus of differential operators, answering the question posed in
\cite{MO-norord}.

\begin{definition}
\label{def.nord.setup1}For the rest of Section \ref{sect.nord}, we shall be
working in the following setup:

Assume that $\mathbf{k}$ is a commutative $\mathbb{Q}$-algebra (so that
$1,2,3,\ldots$ are invertible in $\mathbf{k}$). Let $n\in\mathbb{N}$ and
$m\in\mathbb{N}$. Let $G$ denote the $nm$-element set $\left\{  1,2,\ldots
,n\right\}  \times\left\{  1,2,\ldots,m\right\}  $. Let $\mathcal{A}$ be the
polynomial ring $\mathbf{k}\left[  x_{i,j}\ \mid\ \left(  i,j\right)  \in
G\right]  $ in the $nm$ (commuting) indeterminates $x_{i,j}$. For every
$a\in\mathcal{A}$, we shall denote the map $\mathcal{A}\rightarrow
\mathcal{A},\ u\mapsto au$ (the \textquotedblleft multiplication by
$a$\textquotedblright\ operator) by $a$ (by abuse of notation). Such an
operator will be called a \textit{multiplication operator}.

The abuse of notation that we have just introduced makes certain notations
ambiguous. Namely, if $f:\mathcal{A}\rightarrow\mathcal{A}$ is a map, and if
$a\in\mathcal{A}$ is element, then the expression \textquotedblleft%
$fa$\textquotedblright\ (or \textquotedblleft$f\left(  a\right)
$\textquotedblright) can mean two different things: It can mean the image of
$a$ under the map $f$, or it can mean the composition $f\circ a$ of the map
$f$ with the \textquotedblleft multiplication by $a$\textquotedblright%
\ operator (which we have just agreed to denote by $a$). In order to
disambiguate such expressions, let us agree to denote the image of $a$ under
the map $f$ by $f\left[  a\right]  $ (instead of $fa$ or $f\left(  a\right)
$). Thus, for example, $\dfrac{\partial}{\partial x_{i,j}}\left[
x_{i,j}x_{u,v}\right]  =x_{u,v}$ (if $\left(  u,v\right)  \neq\left(
i,j\right)  $), whereas $\dfrac{\partial}{\partial x_{i,j}}\left(
x_{i,j}x_{u,v}\right)  $ is the endomorphism of $\mathcal{A}$ defined as the
composition $\dfrac{\partial}{\partial x_{i,j}}\circ\left(  x_{i,j}%
x_{u,v}\right)  $.

Let $\mathcal{D}$ denote the $\mathbf{k}$-subalgebra of $\operatorname*{End}%
\mathcal{A}$ generated by the multiplication operators $x_{i,j}$ (for $\left(
i,j\right)  \in G$) and the differential operators $\dfrac{\partial}{\partial
x_{i,j}}$ (for $\left(  i,j\right)  \in G$). This $\mathbf{k}$-algebra
$\mathcal{D}$ is known as the $\mathbf{k}$\textit{-algebra of differential
operators} in the $nm$ variables $x_{i,j}$ \textit{with polynomial
coefficients}. It is well-known that
\begin{equation}
\left(  M\left(  \left(  x_{i,j}\right)  _{\left(  i,j\right)  \in G}\right)
\cdot N\left(  \left(  \dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(
i,j\right)  \in G}\right)  \right)  _{M\text{ and }N\text{ are two monomials}}
\label{eq.def.nord.setup1.basis-of-D}%
\end{equation}
is a basis of the $\mathbf{k}$-module $\mathcal{D}$ (where \textquotedblleft
monomial\textquotedblright\ means \textquotedblleft monomial in the variables
$x_{i,j}$ for $\left(  i,j\right)  \in G$\textquotedblright).

On the other hand, let $\mathcal{A}^{\prime}$ be the polynomial ring
$\mathbf{k}\left[  \partial_{i,j}\ \mid\ \left(  i,j\right)  \in G\right]  $
in the $nm$ (commuting) indeterminates $\partial_{i,j}$. Let $\xi
:\mathcal{A}\otimes\mathcal{A}^{\prime}\rightarrow\mathcal{D}$ be the
$\mathbf{k}$-linear map defined by%
\[
\xi\left(  P\otimes Q\right)  =P\cdot Q\left(  \left(  \dfrac{\partial
}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }P\in\mathcal{A}\text{ and }Q\in
\mathcal{A}^{\prime}.
\]


Clearly, the family $\left(  M\left(  \left(  x_{i,j}\right)  _{\left(
i,j\right)  \in G}\right)  \otimes N\left(  \left(  \partial_{i,j}\right)
_{\left(  i,j\right)  \in G}\right)  \right)  _{M\text{ and }N\text{ are two
monomials}}$ is a basis of the $\mathbf{k}$-module $\mathcal{A}\otimes
\mathcal{A}^{\prime}$. The $\mathbf{k}$-linear map $\xi$ takes this basis to
the family (\ref{eq.def.nord.setup1.basis-of-D}); but this latter family is a
basis of the $\mathbf{k}$-module $\mathcal{D}$. Hence, the $\mathbf{k}$-linear
map $\xi$ is a $\mathbf{k}$-module isomorphism.

The map $\xi$ is not (in general) a $\mathbf{k}$-algebra homomorphism. Indeed,
the $\mathbf{k}$-algebra $\mathcal{A}\otimes\mathcal{A}^{\prime}$ is
commutative, while the $\mathbf{k}$-algebra $\mathcal{D}$ usually is not.
However, we can define a new, \textbf{commutative} multiplication $\boxdot$ on
$\mathcal{D}$ (formally speaking, a $\mathbf{k}$-bilinear map $\boxdot
:\mathcal{D}\times\mathcal{D}\rightarrow\mathcal{D}$, written in infix
notation) by letting%
\[
A\boxdot B=\xi\left(  \xi^{-1}\left(  A\right)  \cdot\xi^{-1}\left(  B\right)
\right)  \ \ \ \ \ \ \ \ \ \ \text{for all }A\in\mathcal{D}\text{ and }%
B\in\mathcal{D}.
\]
(In other words, this multiplication $\boxdot$ is the result of transporting
the multiplication of the $\mathbf{k}$-algebra $\mathcal{A}\otimes
\mathcal{A}^{\prime}$ to $\mathcal{D}$ along the isomorphism $\xi$.)

The multiplication $\boxdot$ is called the \textit{normal ordered product} (or
\textit{normally ordered product}) on $\mathcal{D}$. Let us write $\left(
\mathcal{D},\boxdot\right)  $ for the $\mathbf{k}$-module $\mathcal{D}$
equipped with the multiplication $\boxdot$. This is a commutative $\mathbf{k}$-algebra.

(The multiplication $\boxdot$ so far has nothing to do with the multiplication
$\boxdot$ in Corollary \ref{cor.manchon.box}; but we will relate them soon enough.)

Now, let $\mathfrak{gl}_{n}$ denote the Lie algebra $\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  ^{-}$ (where
$\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  =\mathbf{k}^{n\times
n}$ is the ring of $n\times n$-matrices over $\mathbf{k}$). The $\mathbf{k}%
$-module $\mathfrak{gl}_{n}$ has a basis $\left(  E_{i,j}\right)  _{\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}$ consisting of the
elementary matrices (i.e., for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$, the matrix $E_{i,j}$ is the $n\times n$-matrix
whose $\left(  i,j\right)  $-th entry is $1$ and whose all other entries are
$0$). We define a $\mathbf{k}$-linear map $\omega:\mathfrak{gl}_{n}%
\rightarrow\mathcal{D}^{-}$ by
\[
\omega\left(  E_{i,j}\right)  =\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial
x_{j,k}}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}.
\]
Then, this map $\omega:\mathfrak{gl}_{n}\rightarrow\mathcal{D}^{-}$ is a Lie
algebra homomorphism (this will be proven in Proposition \ref{prop.nord.l}
below). Thus, Theorem \ref{thm.U.univ} (applied to $\mathfrak{gl}_{n}$,
$\mathcal{D}$ and $\omega$ instead of $\mathfrak{g}$, $A$ and $f$) shows that
there exists a unique $\mathbf{k}$-algebra homomorphism $F:U\left(
\mathfrak{gl}_{n}\right)  \rightarrow\mathcal{D}$ such that $\omega
=F\circ\iota_{U,\mathfrak{gl}_{n}}$. Denote this $F$ by $\Omega$. Thus,
$\Omega$ is a $\mathbf{k}$-algebra homomorphism $U\left(  \mathfrak{gl}%
_{n}\right)  \rightarrow\mathcal{D}$ such that $\omega=\Omega\circ
\iota_{U,\mathfrak{gl}_{n}}$.
\end{definition}

Notice that we have required $\mathbf{k}$ to be a $\mathbb{Q}$-algebra in
order for $\xi$ to be an isomorphism. We could get rid of this requirement by
changing the definition of $\mathcal{D}$ (in such a way that the elements of
$\mathcal{D}$ become \textquotedblleft formal differential
operators\textquotedblright\ rather than actual endomorphisms of $\mathcal{A}%
$; this would then require defining the product of the $\mathbf{k}$-algebra
$\mathcal{D}$ independently of composition of differential operators).
Alternatively, it might be possible to define the multiplication $\boxdot$
without reference to $\xi$, though this would be significantly more difficult
(if possible).

\begin{proposition}
\label{prop.nord.l}We use the setup of Definition \ref{def.nord.setup1}. The
map $\omega:\mathfrak{gl}_{n}\rightarrow\mathcal{D}^{-}$ is a Lie algebra homomorphism.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.nord.l}.]We shall show that%
\begin{equation}
\omega\left(  \left[  x,y\right]  \right)  =\left[  \omega\left(  x\right)
,\omega\left(  y\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }%
x\in\mathfrak{gl}_{n}\text{ and }y\in\mathfrak{gl}_{n}.
\label{pf.prop.nord.l.1}%
\end{equation}


\textit{Proof of (\ref{pf.prop.nord.l.1}):} Let $x\in\mathfrak{gl}_{n}$ and
$y\in\mathfrak{gl}_{n}$. We must prove the equality (\ref{pf.prop.nord.l.1}).
This equality is $\mathbf{k}$-linear in each of $x$ and $y$. Hence, we can
WLOG assume that both $x$ and $y$ belong to the basis $\left(  E_{i,j}\right)
_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}$ of the
$\mathbf{k}$-module $\mathfrak{gl}_{n}$. Assume this. Thus, $x=E_{i,j}$ and
$y=E_{u,v}$ for some $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}$ and $\left(  u,v\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$.
Consider these $\left(  i,j\right)  $ and $\left(  u,v\right)  $.

We shall use the notation $\delta_{p,q}$ for the integer $%
\begin{cases}
1, & \text{if }p=q;\\
0, & \text{if }p\neq q
\end{cases}
$ whenever $p$ and $q$ are two objects.

It is well-known that the Lie bracket of $\mathfrak{gl}_{n}$ satisfies%
\[
\left[  E_{i,j},E_{u,v}\right]  =\delta_{j,u}E_{i,v}-\delta_{v,i}E_{u,j}.
\]
Hence, $\left[  \underbrace{x}_{=E_{i,j}},\underbrace{y}_{=E_{u,v}}\right]
=\left[  E_{i,j},E_{u,v}\right]  =\delta_{j,u}E_{i,v}-\delta_{v,i}E_{u,j}$.
Applying the map $\omega$ to both sides of this equality, we obtain%
\begin{align}
\omega\left(  \left[  x,y\right]  \right)   &  =\omega\left(  \delta
_{j,u}E_{i,v}-\delta_{v,i}E_{u,j}\right)  =\delta_{j,u}\underbrace{\omega
\left(  E_{i,v}\right)  }_{\substack{=\sum_{k=1}^{m}x_{i,k}\dfrac{\partial
}{\partial x_{v,k}}\\\text{(by the definition of }\omega\text{)}}%
}-\delta_{v,i}\underbrace{\omega\left(  E_{u,j}\right)  }_{\substack{=\sum
_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{j,k}}\\\text{(by the definition
of }\omega\text{)}}}\nonumber\\
&  =\delta_{j,u}\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{v,k}}%
-\delta_{v,i}\sum_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{j,k}}.
\label{pf.prop.nord.l.1.pf.LHS}%
\end{align}


On the other hand, it is known that every $\left(  a,b\right)  \in G$ and
$\left(  c,d\right)  \in G$ satisfy%
\begin{align}
x_{a,b}x_{c,d}  &  =x_{c,d}x_{a,b};\label{pf.prop.nord.l.1.pf.1}\\
\dfrac{\partial}{\partial x_{a,b}}\dfrac{\partial}{\partial x_{c,d}}  &
=\dfrac{\partial}{\partial x_{c,d}}\dfrac{\partial}{\partial x_{a,b}%
};\label{pf.prop.nord.l.1.pf.2}\\
\dfrac{\partial}{\partial x_{a,b}}\cdot x_{c,d}  &  =x_{c,d}\dfrac{\partial
}{\partial x_{a,b}}+\delta_{a,c}\delta_{b,d}. \label{pf.prop.nord.l.1.pf.3}%
\end{align}


Now,
\[
\omega\left(  \underbrace{x}_{=E_{i,j}}\right)  =\omega\left(  E_{i,j}\right)
=\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right)
\]
and%
\begin{align*}
\omega\left(  \underbrace{y}_{=E_{u,v}}\right)   &  =\omega\left(
E_{u,v}\right)  =\sum_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{v,k}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right) \\
&  =\sum_{h=1}^{m}x_{u,h}\dfrac{\partial}{\partial x_{v,h}}.
\end{align*}
Hence,%
\begin{align}
&  \left[  \underbrace{\omega\left(  x\right)  }_{=\sum_{k=1}^{m}x_{i,k}%
\dfrac{\partial}{\partial x_{j,k}}},\underbrace{\omega\left(  y\right)
}_{=\sum_{h=1}^{m}x_{u,h}\dfrac{\partial}{\partial x_{v,h}}}\right]
\nonumber\\
&  =\left[  \sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}},\sum
_{h=1}^{m}x_{u,h}\dfrac{\partial}{\partial x_{v,h}}\right]  =\sum_{k=1}%
^{m}\sum_{h=1}^{m}\left[  x_{i,k}\dfrac{\partial}{\partial x_{j,k}}%
,x_{u,h}\dfrac{\partial}{\partial x_{v,h}}\right]  . \label{pf.prop.nord.l.8}%
\end{align}
But every $k\in\left\{  1,2,\ldots,m\right\}  $ and $h\in\left\{
1,2,\ldots,m\right\}  $ satisfy%
\begin{align*}
&  \left[  x_{i,k}\dfrac{\partial}{\partial x_{j,k}},x_{u,h}\dfrac{\partial
}{\partial x_{v,h}}\right] \\
&  =x_{i,k}\underbrace{\dfrac{\partial}{\partial x_{j,k}}\cdot x_{u,h}%
}_{\substack{=x_{u,h}\dfrac{\partial}{\partial x_{j,k}}+\delta_{j,u}%
\delta_{k,h}\\\text{(by (\ref{pf.prop.nord.l.1.pf.3}), applied to}\\\left(
a,b\right)  =\left(  j,k\right)  \text{ and }\left(  c,d\right)  =\left(
u,h\right)  \text{)}}}\dfrac{\partial}{\partial x_{v,h}}-x_{u,h}%
\underbrace{\dfrac{\partial}{\partial x_{v,h}}\cdot x_{i,k}}%
_{\substack{=x_{i,k}\dfrac{\partial}{\partial x_{v,h}}+\delta_{v,i}%
\delta_{h,k}\\\text{(by (\ref{pf.prop.nord.l.1.pf.3}), applied to}\\\left(
a,b\right)  =\left(  v,h\right)  \text{ and }\left(  c,d\right)  =\left(
i,k\right)  \text{)}}}\dfrac{\partial}{\partial x_{j,k}}\\
&  =x_{i,k}\left(  x_{u,h}\dfrac{\partial}{\partial x_{j,k}}+\delta
_{j,u}\delta_{k,h}\right)  \dfrac{\partial}{\partial x_{v,h}}-x_{u,h}\left(
x_{i,k}\dfrac{\partial}{\partial x_{v,h}}+\delta_{v,i}\delta_{h,k}\right)
\dfrac{\partial}{\partial x_{j,k}}\\
&  =\underbrace{x_{i,k}x_{u,h}}_{\substack{=x_{u,h}x_{i,k}\\\text{(by
(\ref{pf.prop.nord.l.1.pf.1}), applied to}\\\left(  a,b\right)  =\left(
i,k\right)  \text{ and }\left(  c,d\right)  =\left(  u,h\right)  \text{)}%
}}\underbrace{\dfrac{\partial}{\partial x_{j,k}}\dfrac{\partial}{\partial
x_{v,h}}}_{\substack{=\dfrac{\partial}{\partial x_{v,h}}\dfrac{\partial
}{\partial x_{j,k}}\\\text{(by (\ref{pf.prop.nord.l.1.pf.2}), applied
to}\\\left(  a,b\right)  =\left(  j,k\right)  \text{ and }\left(  c,d\right)
=\left(  v,h\right)  \text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +x_{i,k}\delta_{j,u}\delta_{k,h}\dfrac{\partial
}{\partial x_{v,h}}-x_{u,h}x_{i,k}\dfrac{\partial}{\partial x_{v,h}}%
\dfrac{\partial}{\partial x_{j,k}}-x_{u,h}\delta_{v,i}\underbrace{\delta
_{h,k}}_{=\delta_{k,h}}\dfrac{\partial}{\partial x_{j,k}}\\
&  =x_{u,h}x_{i,k}\dfrac{\partial}{\partial x_{v,h}}\dfrac{\partial}{\partial
x_{j,k}}+x_{i,k}\delta_{j,u}\delta_{k,h}\dfrac{\partial}{\partial x_{v,h}%
}-x_{u,h}x_{i,k}\dfrac{\partial}{\partial x_{v,h}}\dfrac{\partial}{\partial
x_{j,k}}-x_{u,h}\delta_{v,i}\delta_{k,h}\dfrac{\partial}{\partial x_{j,k}}\\
&  =x_{i,k}\delta_{j,u}\delta_{k,h}\dfrac{\partial}{\partial x_{v,h}}%
-x_{u,h}\delta_{v,i}\delta_{k,h}\dfrac{\partial}{\partial x_{j,k}}%
=\delta_{k,h}\left(  x_{i,k}\delta_{j,u}\dfrac{\partial}{\partial x_{v,h}%
}-x_{u,h}\delta_{v,i}\dfrac{\partial}{\partial x_{j,k}}\right)  .
\end{align*}
Thus, (\ref{pf.prop.nord.l.8}) becomes%
\begin{align*}
&  \left[  \omega\left(  x\right)  ,\omega\left(  y\right)  \right] \\
&  =\sum_{k=1}^{m}\sum_{h=1}^{m}\underbrace{\left[  x_{i,k}\dfrac{\partial
}{\partial x_{j,k}},x_{u,h}\dfrac{\partial}{\partial x_{v,h}}\right]
}_{=\delta_{k,h}\left(  x_{i,k}\delta_{j,u}\dfrac{\partial}{\partial x_{v,h}%
}-x_{u,h}\delta_{v,i}\dfrac{\partial}{\partial x_{j,k}}\right)  }\\
&  =\sum_{k=1}^{m}\underbrace{\sum_{h=1}^{m}\delta_{k,h}\left(  x_{i,k}%
\delta_{j,u}\dfrac{\partial}{\partial x_{v,h}}-x_{u,h}\delta_{v,i}%
\dfrac{\partial}{\partial x_{j,k}}\right)  }_{\substack{=x_{i,k}\delta
_{j,u}\dfrac{\partial}{\partial x_{v,k}}-x_{u,k}\delta_{v,i}\dfrac{\partial
}{\partial x_{j,k}}\\\text{(because the presence of the }\delta_{k,h}\text{
factor in the sum}\\\text{ensures that only the addend for }h=k\text{ can be
nonzero)}}}\\
&  =\sum_{k=1}^{m}\left(  x_{i,k}\delta_{j,u}\dfrac{\partial}{\partial
x_{v,k}}-x_{u,k}\delta_{v,i}\dfrac{\partial}{\partial x_{j,k}}\right) \\
&  =\delta_{j,u}\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{v,k}}%
-\delta_{v,i}\sum_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{j,k}}%
=\omega\left(  \left[  x,y\right]  \right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.prop.nord.l.1.pf.LHS})}\right)  .
\end{align*}
This proves (\ref{pf.prop.nord.l.1}).

The map $\omega:\mathfrak{gl}_{n}\rightarrow\mathcal{D}^{-}$ is $\mathbf{k}%
$-linear. Thus, (\ref{pf.prop.nord.l.1}) shows that $\omega$ is a Lie algebra
homomorphism. This proves Proposition \ref{prop.nord.l}.
\end{proof}

\begin{corollary}
\label{cor.nord.l}We use the setup of Definition \ref{def.nord.setup1}.

\textbf{(a)} We can define a $\mathfrak{gl}_{n}$-module structure on the
$\mathbf{k}$-module $\mathcal{D}$ by setting%
\[
a\rightharpoonup F=\omega\left(  a\right)  \cdot
F\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{gl}_{n}\text{ and }%
F\in\mathcal{D}%
\]
(where we are using the fact that every $a\in\mathfrak{gl}_{n}$ satisfies
$\omega\left(  a\right)  \in\mathcal{D}^{-}=\mathcal{D}$). Let us make
$\mathcal{D}$ into a $\mathfrak{gl}_{n}$-module using this $\mathfrak{gl}_{n}%
$-module structure.

\textbf{(b)} The map $\Omega:U\left(  \mathfrak{gl}_{n}\right)  \rightarrow
\mathcal{D}$ is a $\mathfrak{gl}_{n}$-module homomorphism.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.nord.l}.]\textbf{(a)} Define a map $\mu
:\mathfrak{gl}_{n}\times\mathcal{D}\rightarrow\mathcal{D}$ by%
\[
\left(  \mu\left(  a,F\right)  =\omega\left(  a\right)  \cdot
F\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{gl}_{n}\text{ and }%
f\in\mathcal{D}\right)  .
\]
Clearly, this map $\mu$ is $\mathbf{k}$-bilinear (since the map $\omega$ is
$\mathbf{k}$-linear). Let us denote $\mu\left(  a,v\right)  $ by
$a\rightharpoonup v$ for every $a\in\mathfrak{gl}_{n}$ and $v\in\mathcal{D}$.
Thus, we have%
\begin{equation}
a\rightharpoonup v=\mu\left(  a,v\right)  =\omega\left(  a\right)  \cdot
v\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\mu\right)
\label{pf.cor.nord.l.a.1}%
\end{equation}
for every $a\in\mathfrak{gl}_{n}$ and $v\in\mathcal{D}$. Then, we have%
\[
\left[  a,b\right]  \rightharpoonup v=a\rightharpoonup\left(  b\rightharpoonup
v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)  \text{ for every
}a\in\mathfrak{gl}_{n}\text{, }b\in\mathfrak{gl}_{n}\text{ and }%
v\in\mathcal{D}%
\]
\footnote{\textit{Proof.} Let $a\in\mathfrak{gl}_{n}$, $b\in\mathfrak{gl}_{n}$
and $v\in\mathcal{D}$. Then, $\omega$ is a Lie algebra homomorphism from
$\mathfrak{gl}_{n}$ to $\mathcal{D}^{-}$ (by Proposition \ref{prop.nord.l}).
Thus,%
\[
\omega\left(  \left[  a,b\right]  \right)  =\left[  \omega\left(  a\right)
,\omega\left(  b\right)  \right]  =\omega\left(  a\right)  \cdot\omega\left(
b\right)  -\omega\left(  b\right)  \cdot\omega\left(  a\right)
\]
(by the definition of the Lie bracket on $\mathcal{D}^{-}$). Now,%
\begin{align*}
&  \left[  a,b\right]  \rightharpoonup v\\
&  =\underbrace{\omega\left(  \left[  a,b\right]  \right)  }_{=\omega\left(
a\right)  \cdot\omega\left(  b\right)  -\omega\left(  b\right)  \cdot
\omega\left(  a\right)  }\cdot v\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.cor.nord.l.a.1}), applied to }\left[  a,b\right]  \text{ instead of
}a\right) \\
&  =\left(  \omega\left(  a\right)  \cdot\omega\left(  b\right)
-\omega\left(  b\right)  \cdot\omega\left(  a\right)  \right)  \cdot v\\
&  =\omega\left(  a\right)  \cdot\omega\left(  b\right)  \cdot v-\omega\left(
b\right)  \cdot\omega\left(  a\right)  \cdot v.
\end{align*}
Comparing this with%
\begin{align*}
&  a\rightharpoonup\underbrace{\left(  b\rightharpoonup v\right)
}_{\substack{=\omega\left(  b\right)  \cdot v\\\text{(by
(\ref{pf.cor.nord.l.a.1}), applied to}\\b\text{ instead of }a\text{)}%
}}-b\rightharpoonup\underbrace{\left(  a\rightharpoonup v\right)
}_{\substack{=\omega\left(  a\right)  \cdot v\\\text{(by
(\ref{pf.cor.nord.l.a.1}))}}}\\
&  =\underbrace{a\rightharpoonup\left(  \omega\left(  b\right)  \cdot
v\right)  }_{\substack{=\omega\left(  a\right)  \cdot\omega\left(  b\right)
\cdot v\\\text{(by (\ref{pf.cor.nord.l.a.1}), applied to}\\\omega\left(
b\right)  \cdot v\text{ instead of }v\text{)}}}-\underbrace{b\rightharpoonup
\left(  \omega\left(  a\right)  \cdot v\right)  }_{\substack{=\omega\left(
b\right)  \cdot\omega\left(  a\right)  \cdot v\\\text{(by
(\ref{pf.cor.nord.l.a.1}), applied to}\\b\text{ and }\omega\left(  a\right)
\cdot v\text{ instead of }a\text{ and }v\text{)}}}\\
&  =\omega\left(  a\right)  \cdot\omega\left(  b\right)  \cdot v-\omega\left(
b\right)  \cdot\omega\left(  a\right)  \cdot v,
\end{align*}
we obtain $\left[  a,b\right]  \rightharpoonup v=a\rightharpoonup\left(
b\rightharpoonup v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)
$. Qed.}. Hence, the map $\mu$ is a $\mathfrak{gl}_{n}$-module structure on
the $\mathbf{k}$-module $\mathcal{D}$ (since the map $\mu$ is $\mathbf{k}%
$-bilinear). In other words, we can define a $\mathfrak{gl}_{n}$-module
structure on the $\mathbf{k}$-module $\mathcal{D}$ by setting%
\[
a\rightharpoonup F=\omega\left(  a\right)  \cdot
F\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{gl}_{n}\text{ and }%
F\in\mathcal{D}%
\]
(because this definition yields exactly the $\mathfrak{gl}_{n}$-module
structure $\mu$). This proves Corollary \ref{cor.nord.l} \textbf{(a)}.

\textbf{(b)} Let $a\in\mathfrak{gl}_{n}$ and $z\in U\left(  \mathfrak{gl}%
_{n}\right)  $. We shall show that $\Omega\left(  a\rightharpoonup z\right)
=a\rightharpoonup\Omega\left(  z\right)  $.

Indeed, the definition of the $\mathfrak{gl}_{n}$-module structure on
$U\left(  \mathfrak{gl}_{n}\right)  $ shows that $a\rightharpoonup
z=\iota_{U,\mathfrak{gl}_{n}}\left(  a\right)  \cdot z$.

On the other hand, the definition of the $\mathfrak{gl}_{n}$-structure on
$\mathcal{D}$ shows that
\begin{align*}
a\rightharpoonup\Omega\left(  z\right)   &  =\underbrace{\omega}_{=\Omega
\circ\iota_{U,\mathfrak{gl}_{n}}}\left(  a\right)  \cdot\Omega\left(
z\right)  =\underbrace{\left(  \Omega\circ\iota_{U,\mathfrak{gl}_{n}}\right)
\left(  a\right)  }_{=\Omega\left(  \iota_{U,\mathfrak{gl}_{n}}\left(
a\right)  \right)  }\cdot\Omega\left(  z\right) \\
&  =\Omega\left(  \iota_{U,\mathfrak{gl}_{n}}\left(  a\right)  \right)
\cdot\Omega\left(  z\right)  =\Omega\left(  \iota_{U,\mathfrak{gl}_{n}}\left(
a\right)  \cdot z\right)
\end{align*}
(since $\Omega$ is a $\mathbf{k}$-algebra homomorphism). Comparing this with%
\[
\Omega\left(  \underbrace{a\rightharpoonup z}_{=\iota_{U,\mathfrak{gl}_{n}%
}\left(  a\right)  \cdot z}\right)  =\Omega\left(  \iota_{U,\mathfrak{gl}_{n}%
}\left(  a\right)  \cdot z\right)  ,
\]
we obtain $\Omega\left(  a\rightharpoonup z\right)  =a\rightharpoonup
\Omega\left(  z\right)  $.

Let us now forget that we fixed $a$ and $z$. We thus have shown that
$\Omega\left(  a\rightharpoonup z\right)  =a\rightharpoonup\Omega\left(
z\right)  $ for every $a\in\mathfrak{gl}_{n}$ and $z\in U\left(
\mathfrak{gl}_{n}\right)  $. In other words, the map $\Omega:U\left(
\mathfrak{gl}_{n}\right)  \rightarrow\mathcal{D}$ is a $\mathfrak{gl}_{n}%
$-module homomorphism. This proves Corollary \ref{cor.nord.l} \textbf{(b)}.
\end{proof}

\begin{proposition}
\label{prop.nord.derivs}We use the setup of Definition \ref{def.nord.setup1}.
Then, $\omega\left(  \mathfrak{gl}_{n}\right)  \subseteq\operatorname*{Der}%
\mathcal{A}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.nord.derivs}.]We want to prove that
$\omega\left(  \mathfrak{gl}_{n}\right)  \subseteq\operatorname*{Der}%
\mathcal{A}$. In other words, we want to prove that $\omega\left(  b\right)
\in\operatorname*{Der}\mathcal{A}$ for every $b\in\mathfrak{gl}_{n}$. So let
us fix $b\in\mathfrak{gl}_{n}$.

We must prove the relation $\omega\left(  b\right)  \in\operatorname*{Der}%
\mathcal{A}$. This relation is $\mathbf{k}$-linear in $b$. Hence, we can WLOG
assume that $b$ belongs to the basis $\left(  E_{i,j}\right)  _{\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}$ of the $\mathbf{k}%
$-module $\mathfrak{gl}_{n}$. Assume this. Thus, $b=E_{i,j}$ for some $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$. Consider this $\left(
i,j\right)  $. Hence,
\begin{equation}
\omega\left(  \underbrace{b}_{=E_{i,j}}\right)  =\omega\left(  E_{i,j}\right)
=\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right)  .
\label{pf.prop.nord.derivs.omegax}%
\end{equation}


Now, let $p\in\mathcal{A}$ and $q\in\mathcal{A}$.

Let $k\in\left\{  1,2,\ldots,m\right\}  $. Then, $\dfrac{\partial}{\partial
x_{j,k}}\in\operatorname*{End}\mathcal{A}$ is a derivation of $\mathcal{A}$
(since $\dfrac{\partial}{\partial x_{j,k}}$ is a partial derivative operator).
Hence,
\begin{equation}
\dfrac{\partial}{\partial x_{j,k}}\left[  pq\right]  =p\dfrac{\partial
}{\partial x_{j,k}}\left[  q\right]  +\dfrac{\partial}{\partial x_{j,k}%
}\left[  p\right]  \cdot q. \label{pf.prop.nord.derivs.1}%
\end{equation}
(Recall that we are using the notation $f\left[  a\right]  $ for the image of
an element $a\in\mathcal{A}$ under a map $f:\mathcal{A}\rightarrow\mathcal{A}$.)

Now, let us forget that we fixed $k$. We thus have proven
(\ref{pf.prop.nord.derivs.1}) for every $k\in\left\{  1,2,\ldots,m\right\}  $.
Now,%
\begin{align*}
&  \underbrace{\left(  \omega\left(  b\right)  \right)  }_{\substack{=\sum
_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}}\\\text{(by
(\ref{pf.prop.nord.derivs.omegax}))}}}\left[  pq\right] \\
&  =\left(  \sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}}\right)
\left[  pq\right]  =\sum_{k=1}^{m}x_{i,k}\underbrace{\dfrac{\partial}{\partial
x_{j,k}}\left[  pq\right]  }_{\substack{=p\dfrac{\partial}{\partial x_{j,k}%
}\left[  q\right]  +\dfrac{\partial}{\partial x_{j,k}}\left[  p\right]  \cdot
q\\\text{(by (\ref{pf.prop.nord.derivs.1}))}}}\\
&  =\sum_{k=1}^{m}x_{i,k}\left(  p\dfrac{\partial}{\partial x_{j,k}}\left[
q\right]  +\dfrac{\partial}{\partial x_{j,k}}\left[  p\right]  \cdot q\right)
\\
&  =\sum_{k=1}^{m}\underbrace{x_{i,k}p}_{=px_{i,k}}\dfrac{\partial}{\partial
x_{j,k}}\left[  q\right]  +\underbrace{\sum_{k=1}^{m}x_{i,k}\dfrac{\partial
}{\partial x_{j,k}}\left[  p\right]  \cdot q}_{=\left(  \sum_{k=1}^{m}%
x_{i,k}\dfrac{\partial}{\partial x_{j,k}}\left[  p\right]  \right)  \cdot q}\\
&  =\underbrace{\sum_{k=1}^{m}px_{i,k}\dfrac{\partial}{\partial x_{j,k}%
}\left[  q\right]  }_{\substack{=p\sum_{k=1}^{m}x_{i,k}\dfrac{\partial
}{\partial x_{j,k}}\left[  q\right]  \\=p\left(  \sum_{k=1}^{m}x_{i,k}%
\dfrac{\partial}{\partial x_{j,k}}\right)  \left[  q\right]  }%
}+\underbrace{\left(  \sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}%
}\left[  p\right]  \right)  }_{=\left(  \sum_{k=1}^{m}x_{i,k}\dfrac{\partial
}{\partial x_{j,k}}\right)  \left[  p\right]  }\cdot q\\
&  =p\underbrace{\left(  \sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial
x_{j,k}}\right)  }_{\substack{=\omega\left(  b\right)  \\\text{(by
(\ref{pf.prop.nord.derivs.omegax}))}}}\left[  q\right]  +\underbrace{\left(
\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}}\right)  }%
_{\substack{=\omega\left(  b\right)  \\\text{(by
(\ref{pf.prop.nord.derivs.omegax}))}}}\left[  p\right]  \cdot q\\
&  =p\left(  \omega\left(  b\right)  \right)  \left[  q\right]  +\left(
\omega\left(  b\right)  \right)  \left[  p\right]  \cdot q.
\end{align*}


Now let us forget that we fixed $p$ and $q$. We thus have shown that $\left(
\omega\left(  b\right)  \right)  \left[  pq\right]  =p\left(  \omega\left(
b\right)  \right)  \left[  q\right]  +\left(  \omega\left(  b\right)  \right)
\left[  p\right]  \cdot q$ for every $p\in\mathcal{A}$ and $q\in\mathcal{A}$.
In other words, the map $\omega\left(  b\right)  $ is a derivation. In other
words, $\omega\left(  b\right)  \in\operatorname*{Der}\mathcal{A}$. This
proves Proposition \ref{prop.nord.derivs}.
\end{proof}

\begin{proposition}
\label{prop.nord.Dunity}We use the setup of Definition \ref{def.nord.setup1}.

\textbf{(a)} The map $\xi$ is a $\mathbf{k}$-algebra isomorphism from
$\mathcal{A}\otimes\mathcal{A}^{\prime}$ to the $\mathbf{k}$-algebra $\left(
\mathcal{D},\boxdot\right)  $.

\textbf{(b)} We have $1_{\left(  \mathcal{D},\boxdot\right)  }=1_{\mathcal{D}%
}$.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.nord.Dunity}.]\textbf{(a)} The map $\xi$ is a
$\mathbf{k}$-module isomorphism $\mathcal{A}\otimes\mathcal{A}^{\prime
}\rightarrow\mathcal{D}$, and thus is invertible. Moreover, it sends products
in $\mathcal{A}\otimes\mathcal{A}^{\prime}$ to products inside the
$\mathbf{k}$-algebra $\left(  \mathcal{D},\boxdot\right)  $ (i.e., it
satisfies $\xi\left(  ab\right)  =\xi\left(  a\right)  \boxdot\xi\left(
b\right)  $ for every $a\in\mathcal{A}\otimes\mathcal{A}^{\prime}$ and
$b\in\mathcal{A}\otimes\mathcal{A}^{\prime}$), because the multiplication
$\boxdot$ on $\mathcal{D}$ is the result of transporting the multiplication of
the $\mathbf{k}$-algebra $\mathcal{A}\otimes\mathcal{A}^{\prime}$ to
$\mathcal{D}$ along the isomorphism $\xi$. Hence, this map $\xi$ is a
$\mathbf{k}$-algebra isomorphism from $\mathcal{A}\otimes\mathcal{A}^{\prime}$
to the $\mathbf{k}$-algebra $\left(  \mathcal{D},\boxdot\right)  $ (because
any invertible $\mathbf{k}$-linear map between two $\mathbf{k}$-algebras which
sends products in its domain to products inside its target must be a
$\mathbf{k}$-algebra isomorphism). This proves Proposition
\ref{prop.nord.Dunity} \textbf{(a)}.

\textbf{(b)} Proposition \ref{prop.nord.Dunity} \textbf{(a)} shows that the
map $\xi$ is a $\mathbf{k}$-algebra isomorphism from $\mathcal{A}%
\otimes\mathcal{A}^{\prime}$ to the $\mathbf{k}$-algebra $\left(
\mathcal{D},\boxdot\right)  $. Hence, $\xi$ sends the unity of the
$\mathbf{k}$-algebra $\mathcal{A}\otimes\mathcal{A}^{\prime}$ to the unity of
the $\mathbf{k}$-algebra $\left(  \mathcal{D},\boxdot\right)  $. In other
words, $\xi\left(  1_{\mathcal{A}\otimes\mathcal{A}^{\prime}}\right)
=1_{\left(  \mathcal{D},\boxdot\right)  }$. Thus,%
\begin{align*}
1_{\left(  \mathcal{D},\boxdot\right)  }  &  =\xi\left(
\underbrace{1_{\mathcal{A}\otimes\mathcal{A}^{\prime}}}_{=1_{\mathcal{A}%
}\otimes1_{\mathcal{A}^{\prime}}}\right)  =\xi\left(  1_{\mathcal{A}}%
\otimes1_{\mathcal{A}^{\prime}}\right) \\
&  =\underbrace{1_{\mathcal{A}}}_{=1}\cdot\underbrace{1_{\mathcal{A}^{\prime}%
}\left(  \left(  \dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(
i,j\right)  \in G}\right)  }_{=1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }\xi\right) \\
&  =1\cdot1=1=1_{\mathcal{D}}.
\end{align*}
This proves Proposition \ref{prop.nord.Dunity} \textbf{(b)}.
\end{proof}

The following theorem states a positive answer to MathOverflow question
\cite{MO-norord}:

\begin{theorem}
\label{thm.nord.box-exists}We use the setup of Definition
\ref{def.nord.setup1}.

\textbf{(a)} The set $\Omega\left(  U\left(  \mathfrak{gl}_{n}\right)
\right)  $ is a $\mathbf{k}$-subalgebra of the $\mathbf{k}$-algebra $\left(
\mathcal{D},\boxdot\right)  $.

\textbf{(b)} There exists a commutative multiplication $\boxdot$ on the
$\mathbf{k}$-module $U\left(  \mathfrak{gl}_{n}\right)  $ such that $\Omega$
is a $\mathbf{k}$-algebra homomorphism $\left(  U\left(  \mathfrak{gl}%
_{n}\right)  ,\boxdot\right)  \rightarrow\left(  \mathcal{D},\boxdot\right)
$. Moreover, this multiplication $\boxdot$ can be chosen to be independent on
$m$.
\end{theorem}

We shall prove Theorem \ref{thm.nord.box-exists} by constructing the
multiplication $\boxdot$ whose existence is alleged in Theorem
\ref{thm.nord.box-exists} \textbf{(b)} using Theorem \ref{thm.manchon}:

\begin{theorem}
\label{thm.nord.box}We use the setup of Definition \ref{def.nord.setup1}.

The $\mathbf{k}$-module $\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}%
\right)  $ (equipped with the multiplication of $\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $) is a left pre-Lie algebra (by
Proposition \ref{prop.prelie.assoc} \textbf{(a)}). Hence, we can construct a
map $\eta:U\left(  \left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  ^{-}\right)  \rightarrow\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ according
to Theorem \ref{thm.manchon} \textbf{(f)} (applied to $A=\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $). The map $\eta$ is a map from
$U\left(  \left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)
\right)  ^{-}\right)  $ to $\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $, therefore a map from
$U\left(  \mathfrak{gl}_{n}\right)  $ to $\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ (since
$\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)
^{-}=\mathfrak{gl}_{n}$).

\textbf{(a)} The map $\eta$ is a $\mathbf{k}$-module isomorphism. Hence, we
can define a new, \textbf{commutative} multiplication $\boxdot$ on $U\left(
\mathfrak{gl}_{n}\right)  $ (formally speaking, a $\mathbf{k}$-bilinear map
$\boxdot:U\left(  \mathfrak{gl}_{n}\right)  \times U\left(  \mathfrak{gl}%
_{n}\right)  \rightarrow U\left(  \mathfrak{gl}_{n}\right)  $, written in
infix notation) by letting%
\[
x\boxdot y=\eta\left(  \eta^{-1}\left(  x\right)  \cdot\eta^{-1}\left(
y\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for all }x\in U\left(
\mathfrak{gl}_{n}\right)  \text{ and }y\in U\left(  \mathfrak{gl}_{n}\right)
.
\]
(In other words, this multiplication $\boxdot$ is the result of transporting
the multiplication of the $\mathbf{k}$-algebra $\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ to
$U\left(  \mathfrak{gl}_{n}\right)  $ along the isomorphism $\eta$.)

\textbf{(b)} Let us write $\left(  U\left(  \mathfrak{gl}_{n}\right)
,\boxdot\right)  $ for the $\mathbf{k}$-module $U\left(  \mathfrak{gl}%
_{n}\right)  $ equipped with the multiplication $\boxdot$. This is a
commutative $\mathbf{k}$-algebra with unity $1_{U\left(  \mathfrak{gl}%
_{n}\right)  }$.

\textbf{(c)} Let $\Delta$ and $\epsilon$ denote the comultiplication and the
counit of the $\mathbf{k}$-bialgebra $U\left(  \mathfrak{gl}_{n}\right)  $.
The $\mathbf{k}$-algebra $\left(  U\left(  \mathfrak{gl}_{n}\right)
,\boxdot\right)  $, equipped with the comultiplication $\Delta$ and the counit
$\epsilon$, is a $\mathbf{k}$-bialgebra. We shall denote this $\mathbf{k}%
$-bialgebra by $\left(  U\left(  \mathfrak{gl}_{n}\right)  ,\boxdot\right)  $.

\textbf{(d)} The map $\eta$ is a $\mathbf{k}$-bialgebra isomorphism $\left(
U\left(  \mathfrak{gl}_{n}\right)  ,\boxdot\right)  \rightarrow
\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $.

\textbf{(e)} For every $p\in\mathcal{A}$, $q\in\mathcal{A}^{\prime}$ and
$h\in\mathcal{D}$, we have%
\[
h\boxdot\xi\left(  p\otimes q\right)  =p\cdot h\cdot q\left(  \left(
\dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  .
\]


\textbf{(f)} For every $p\in\mathcal{A}$, $q\in\mathcal{A}^{\prime}$ and
$a\in\mathfrak{gl}_{n}$, we have%
\[
\omega\left(  a\right)  \cdot\xi\left(  p\otimes q\right)  =\omega\left(
a\right)  \boxdot\xi\left(  p\otimes q\right)  +\xi\left(  \left(
\omega\left(  a\right)  \right)  \left[  p\right]  \otimes q\right)  .
\]
(Recall that we are using the notation $f\left[  a\right]  $ for the image of
an element $a\in\mathcal{A}$ under a map $f:\mathcal{A}\rightarrow\mathcal{A}$.)

\textbf{(g)} For every $a\in\mathfrak{gl}_{n}$, $b\in\mathfrak{gl}_{n}$ and
$h\in\mathcal{D}$, we have%
\[
\omega\left(  a\right)  \cdot\left(  \omega\left(  b\right)  \boxdot h\right)
=\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \cdot
h\right)  +\omega\left(  a\vartriangleright b\right)  \boxdot h.
\]


\textbf{(h)} The map $\Omega$ is a $\mathbf{k}$-algebra homomorphism $\left(
U\left(  \mathfrak{gl}_{n}\right)  ,\boxdot\right)  \rightarrow\left(
\mathcal{D},\boxdot\right)  $.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.nord.box}.]Define a $\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  ^{-}$-module structure on
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $ as in Theorem \ref{thm.manchon} \textbf{(e)}
(applied to $A=\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $).
Thus, $\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $ is an $\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  ^{-}$-module. In other words,
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $ is a $\mathfrak{gl}_{n}$-module (since
$\mathfrak{gl}_{n}=\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $).

\textbf{(a)} Theorem \ref{thm.manchon} \textbf{(h)} (applied to
$A=\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $) shows that the
map $\eta$ is an $\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}%
\right)  \right)  ^{-}$-module isomorphism. Thus, in particular, the map
$\eta$ is a $\mathbf{k}$-module isomorphism. This proves Theorem
\ref{thm.nord.box} \textbf{(a)} (since the remaining claims of Theorem
\ref{thm.nord.box} \textbf{(a)} are obvious).

\textbf{(b)} Theorem \ref{thm.nord.box} \textbf{(b)} follows from Corollary
\ref{cor.manchon.box} \textbf{(a)} (applied to $A=\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $), upon realizing that $\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)
^{-}=\mathfrak{gl}_{n}$.

\textbf{(c)} Theorem \ref{thm.nord.box} \textbf{(c)} follows from Corollary
\ref{cor.manchon.box} \textbf{(b)} (applied to $A=\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $), upon realizing that $\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)
^{-}=\mathfrak{gl}_{n}$.

\textbf{(d)} Theorem \ref{thm.nord.box} \textbf{(d)} follows from Corollary
\ref{cor.manchon.box} \textbf{(c)} (applied to $A=\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $), upon realizing that $\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)
^{-}=\mathfrak{gl}_{n}$.

\textbf{(e)} The map $\xi$ is a $\mathbf{k}$-algebra isomorphism from
$\mathcal{A}\otimes\mathcal{A}^{\prime}$ to the $\mathbf{k}$-algebra $\left(
\mathcal{D},\boxdot\right)  $ (by Proposition \ref{prop.nord.Dunity}
\textbf{(a)}), therefore also a $\mathbf{k}$-algebra homomorphism from
$\mathcal{A}\otimes\mathcal{A}^{\prime}$ to the $\mathbf{k}$-algebra $\left(
\mathcal{D},\boxdot\right)  $.

Let $p\in\mathcal{A}$ and $q\in\mathcal{A}^{\prime}$. We are first going to
prove that%
\begin{equation}
\xi\left(  u\right)  \boxdot\xi\left(  p\otimes q\right)  =p\cdot\xi\left(
u\right)  \cdot q\left(  \left(  \dfrac{\partial}{\partial x_{i,j}}\right)
_{\left(  i,j\right)  \in G}\right)  \label{pf.thm.nord.box.e.xi-claim}%
\end{equation}
for every $u\in\mathcal{A}\otimes\mathcal{A}^{\prime}$.

\textit{Proof of (\ref{pf.thm.nord.box.e.xi-claim}):} Let $u\in\mathcal{A}%
\otimes\mathcal{A}^{\prime}$. We need to prove the equality
(\ref{pf.thm.nord.box.e.xi-claim}). This equality is $\mathbf{k}$-linear in
$u$. Hence, we can WLOG assume that $u$ is a pure tensor (since the
$\mathbf{k}$-module $\mathcal{A}\otimes\mathcal{A}^{\prime}$ is spanned by the
pure tensors). Assume this. Thus, $u=p^{\prime}\otimes q^{\prime}$ for some
$p^{\prime}\in\mathcal{A}$ and $q^{\prime}\in\mathcal{A}^{\prime}$. Consider
these $p^{\prime}$ and $q^{\prime}$. (The notations $p^{\prime}$ and
$q^{\prime}$ have nothing to do with derivatives.) We have%
\begin{equation}
\xi\left(  \underbrace{u}_{=p^{\prime}\otimes q^{\prime}}\right)  =\xi\left(
p^{\prime}\otimes q^{\prime}\right)  =p^{\prime}\cdot q^{\prime}\left(
\left(  \dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  \label{pf.thm.nord.box.e.xi-claim.pf.1}%
\end{equation}
(by the definition of $\xi$). On the other hand, recall that the map $\xi$ is
a $\mathbf{k}$-algebra homomorphism from $\mathcal{A}\otimes\mathcal{A}%
^{\prime}$ to the $\mathbf{k}$-algebra $\left(  \mathcal{D},\boxdot\right)  $.
Hence,%
\begin{align*}
\xi\left(  u\right)  \boxdot\xi\left(  p\otimes q\right)   &  =\xi\left(
\underbrace{u}_{=p^{\prime}\otimes q^{\prime}}\cdot\left(  p\otimes q\right)
\right)  =\xi\left(  \underbrace{\left(  p^{\prime}\otimes q^{\prime}\right)
\cdot\left(  p\otimes q\right)  }_{=p^{\prime}p\otimes q^{\prime}q}\right) \\
&  =\xi\left(  \underbrace{p^{\prime}p}_{\substack{=pp^{\prime}\\\text{(since
the }\mathbf{k}\text{-algebra }\mathcal{A}\\\text{is commutative)}}}\otimes
q^{\prime}q\right)  =\xi\left(  pp^{\prime}\otimes q^{\prime}q\right) \\
&  =\underbrace{\left(  pp^{\prime}\right)  }_{=p\cdot p^{\prime}}%
\cdot\underbrace{\left(  q^{\prime}q\right)  \left(  \left(  \dfrac{\partial
}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)  }%
_{=q^{\prime}\left(  \left(  \dfrac{\partial}{\partial x_{i,j}}\right)
_{\left(  i,j\right)  \in G}\right)  \cdot q\left(  \left(  \dfrac{\partial
}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\xi\right) \\
&  =p\cdot\underbrace{p^{\prime}\cdot q^{\prime}\left(  \left(  \dfrac
{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)
}_{\substack{=\xi\left(  u\right)  \\\text{(by
(\ref{pf.thm.nord.box.e.xi-claim.pf.1}))}}}\cdot q\left(  \left(
\dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)
\\
&  =p\cdot\xi\left(  u\right)  \cdot q\left(  \left(  \dfrac{\partial
}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)  .
\end{align*}
This proves (\ref{pf.thm.nord.box.e.xi-claim.pf.1}).

Now, let $h\in\mathcal{D}$. Let $u=\xi^{-1}\left(  h\right)  $. This $u$ is a
well-defined element of $\mathcal{A}\otimes\mathcal{A}^{\prime}$ (since $\xi$
is a $\mathbf{k}$-module isomorphism). Now, $\xi\left(  u\right)  =h$ (since
$u=\xi^{-1}\left(  h\right)  $), so that $h=\xi\left(  u\right)  $. Thus,%
\begin{align*}
\underbrace{h}_{=\xi\left(  u\right)  }\boxdot\xi\left(  p\otimes q\right)
&  =\xi\left(  u\right)  \boxdot\xi\left(  p\otimes q\right)  =p\cdot
\underbrace{\xi\left(  u\right)  }_{=h}\cdot q\left(  \left(  \dfrac{\partial
}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.nord.box.e.xi-claim.pf.1}%
)}\right) \\
&  =p\cdot h\cdot q\left(  \left(  \dfrac{\partial}{\partial x_{i,j}}\right)
_{\left(  i,j\right)  \in G}\right)  .
\end{align*}
This proves Theorem \ref{thm.nord.box} \textbf{(e)}.

\textbf{(f)} Let $p\in\mathcal{A}$, $q\in\mathcal{A}^{\prime}$ and
$a\in\mathfrak{gl}_{n}$.

Let $h$ denote the element $\omega\left(  a\right)  $ of $\mathcal{D}$. Then,
$h=\omega\left(  \underbrace{a}_{\in\mathfrak{gl}_{n}}\right)  \in
\omega\left(  \mathfrak{gl}_{n}\right)  \subseteq\operatorname*{Der}%
\mathcal{A}$ (by Proposition \ref{prop.nord.derivs}). In other words, $h$ is a
derivation of $\mathcal{A}$. Thus, we have%
\begin{equation}
h\left[  s_{1}s_{2}\right]  =s_{1}\cdot h\left[  s_{2}\right]  +h\left[
s_{1}\right]  \cdot s_{2} \label{pf.thm.nord.box.f.deriv}%
\end{equation}
for every $s_{1}\in\mathcal{A}$ and $s_{2}\in\mathcal{A}$.

Let $q_{\partial}$ denote the element $q\left(  \left(  \dfrac{\partial
}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)
\in\mathcal{D}$. Thus, $q\left(  \left(  \dfrac{\partial}{\partial x_{i,j}%
}\right)  _{\left(  i,j\right)  \in G}\right)  =q_{\partial}$.

Let $r\in\mathcal{A}$.

The definition of $\xi$ shows that%
\[
\xi\left(  p\otimes q\right)  =p\cdot\underbrace{q\left(  \left(
\dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  }_{=q_{\partial}}=p\cdot q_{\partial}%
\]
and%
\[
\xi\left(  h\left[  p\right]  \otimes q\right)  =h\left[  p\right]
\cdot\underbrace{q\left(  \left(  \dfrac{\partial}{\partial x_{i,j}}\right)
_{\left(  i,j\right)  \in G}\right)  }_{=q_{\partial}}=h\left[  p\right]
\cdot q_{\partial}.
\]


But Theorem \ref{thm.nord.box} \textbf{(e)} shows that%
\[
h\boxdot\xi\left(  p\otimes q\right)  =p\cdot h\cdot\underbrace{q\left(
\left(  \dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  }_{=q_{\partial}}=p\cdot h\cdot q_{\partial},
\]
and thus%
\begin{equation}
\underbrace{\left(  h\boxdot\xi\left(  p\otimes q\right)  \right)  }_{=p\cdot
h\cdot q_{\partial}}\left[  r\right]  =\left(  p\cdot h\cdot q_{\partial
}\right)  \left[  r\right]  =p\cdot\underbrace{\left(  h\cdot q_{\partial
}\right)  \left[  r\right]  }_{=h\left[  q_{\partial}\left[  r\right]
\right]  }=p\cdot h\left[  q_{\partial}\left[  r\right]  \right]  .
\label{pf.thm.nord.box.f.4}%
\end{equation}
Moreover,
\begin{equation}
\underbrace{\left(  \xi\left(  h\left[  p\right]  \otimes q\right)  \right)
}_{=h\left[  p\right]  \cdot q_{\partial}}\left[  r\right]  =\left(  h\left[
p\right]  \cdot q_{\partial}\right)  \left[  r\right]  =h\left[  p\right]
\cdot q_{\partial}\left[  r\right]  . \label{pf.thm.nord.box.f.5}%
\end{equation}


Also,
\begin{align*}
\left(  h\cdot\underbrace{\xi\left(  p\otimes q\right)  }_{=p\cdot
q_{\partial}}\right)  \left[  r\right]   &  =\left(  h\cdot p\cdot
q_{\partial}\right)  \left[  r\right]  =h\left[  \underbrace{\left(  p\cdot
q_{\partial}\right)  \left[  r\right]  }_{=p\cdot q_{\partial}\left[
r\right]  }\right]  =h\left[  p\cdot q_{\partial}\left[  r\right]  \right] \\
&  =\underbrace{p\cdot h\left[  q_{\partial}\left[  r\right]  \right]
}_{\substack{=\left(  h\boxdot\xi\left(  p\otimes q\right)  \right)  \left[
r\right]  \\\text{(by (\ref{pf.thm.nord.box.f.4}))}}}+\underbrace{h\left[
p\right]  \cdot q_{\partial}\left[  r\right]  }_{\substack{=\left(  \xi\left(
h\left[  p\right]  \otimes q\right)  \right)  \left[  r\right]  \\\text{(by
(\ref{pf.thm.nord.box.f.5}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.nord.box.f.deriv}),
applied to }s_{1}=p\text{ and }s_{2}=q_{\partial}\left[  r\right]  \right) \\
&  =\left(  h\boxdot\xi\left(  p\otimes q\right)  \right)  \left[  r\right]
+\left(  \xi\left(  h\left[  p\right]  \otimes q\right)  \right)  \left[
r\right] \\
&  =\left(  h\boxdot\xi\left(  p\otimes q\right)  +\xi\left(  h\left[
p\right]  \otimes q\right)  \right)  \left[  r\right]  .
\end{align*}


Now, let us forget that we fixed $r$. We thus have shown that%
\[
\left(  h\cdot\xi\left(  p\otimes q\right)  \right)  \left[  r\right]
=\left(  h\boxdot\xi\left(  p\otimes q\right)  +\xi\left(  h\left[  p\right]
\otimes q\right)  \right)  \left[  r\right]
\]
for every $r\in\mathcal{A}$. In other words,%
\[
h\cdot\xi\left(  p\otimes q\right)  =h\boxdot\xi\left(  p\otimes q\right)
+\xi\left(  h\left[  p\right]  \otimes q\right)  .
\]
Since $h=\omega\left(  a\right)  $, this rewrites as%
\[
\omega\left(  a\right)  \cdot\xi\left(  p\otimes q\right)  =\omega\left(
a\right)  \boxdot\xi\left(  p\otimes q\right)  +\xi\left(  \left(
\omega\left(  a\right)  \right)  \left[  p\right]  \otimes q\right)  .
\]
This proves Theorem \ref{thm.nord.box} \textbf{(f)}.

\textbf{(g)} Recall that the $\mathbf{k}$-module $\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $ (equipped with the multiplication of
$\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $) is a left pre-Lie
algebra. Hence, if $a\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}%
\right)  $ and $b\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $
are two matrices, then $a\vartriangleright b$ is the product of the matrices
$a$ and $b$ inside $\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)
$. Hence, for every $\left(  u,v\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}$ and $\left(  y,z\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$, we
have%
\begin{align}
E_{u,v}\vartriangleright E_{y,z}  &  =\left(  \text{the product of the
matrices }E_{u,v}\text{ and }E_{y,z}\text{ inside }\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right) \nonumber\\
&  =\delta_{v,y}E_{u,z} \label{pf.thm.nord.box.g.EuvEij}%
\end{align}
(by the standard multiplication rule for elementary matrices).

Next, let us observe that%
\begin{equation}
\omega\left(  E_{u,v}\right)  \boxdot\xi\left(  p\otimes q\right)  =\sum
_{k=1}^{m}\xi\left(  x_{u,k}p\otimes\partial_{v,k}q\right)
\label{pf.thm.nord.box.g.omega-acting-on-xi}%
\end{equation}
for every $\left(  u,v\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$,
$p\in\mathcal{A}$ and $q\in\mathcal{A}^{\prime}$.

\textit{Proof of (\ref{pf.thm.nord.box.g.omega-acting-on-xi}):} Let $\left(
u,v\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$, $p\in\mathcal{A}$ and
$q\in\mathcal{A}^{\prime}$. We have%
\begin{align*}
&  \omega\left(  E_{u,v}\right)  \boxdot\xi\left(  p\otimes q\right)  \\
&  =p\cdot\underbrace{\omega\left(  E_{u,v}\right)  }_{\substack{=\sum
_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{v,k}}\\\text{(by the definition
of }\omega\text{)}}}\cdot q\left(  \left(  \dfrac{\partial}{\partial x_{i,j}%
}\right)  _{\left(  i,j\right)  \in G}\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.nord.box}
\textbf{(e)} (applied to }h=\omega\left(  E_{u,v}\right)  \text{)}\right)  \\
&  =p\cdot\left(  \sum_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{v,k}%
}\right)  \cdot q\left(  \left(  \dfrac{\partial}{\partial x_{i,j}}\right)
_{\left(  i,j\right)  \in G}\right)  \\
&  =\sum_{k=1}^{m}px_{u,k}\dfrac{\partial}{\partial x_{v,k}}\cdot q\left(
\left(  \dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  .
\end{align*}
Comparing this with%
\begin{align*}
&  \sum_{k=1}^{m}\underbrace{\xi\left(  x_{u,k}p\otimes\partial_{v,k}q\right)
}_{\substack{=\left(  x_{u,k}p\right)  \cdot\left(  \partial_{v,k}q\right)
\left(  \left(  \dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(
i,j\right)  \in G}\right)  \\\text{(by the definition of }\xi\text{)}}}\\
&  =\sum_{k=1}^{m}\underbrace{\left(  x_{u,k}p\right)  }_{=px_{u,k}}%
\cdot\underbrace{\left(  \partial_{v,k}q\right)  \left(  \left(
\dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  }_{=\partial_{v,k}\left(  \left(  \dfrac{\partial}{\partial
x_{i,j}}\right)  _{\left(  i,j\right)  \in G}\right)  \cdot q\left(  \left(
\dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  }\\
&  =\sum_{k=1}^{m}px_{u,k}\cdot\underbrace{\partial_{v,k}\left(  \left(
\dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  }_{=\dfrac{\partial}{\partial x_{v,k}}}\cdot q\left(  \left(
\dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  \\
&  =\sum_{k=1}^{m}px_{u,k}\dfrac{\partial}{\partial x_{v,k}}\cdot q\left(
\left(  \dfrac{\partial}{\partial x_{i,j}}\right)  _{\left(  i,j\right)  \in
G}\right)  ,
\end{align*}
this yields $\omega\left(  E_{u,v}\right)  \boxdot\xi\left(  p\otimes
q\right)  =\sum_{k=1}^{m}\xi\left(  x_{u,k}p\otimes\partial_{v,k}q\right)  $.
This proves (\ref{pf.thm.nord.box.g.omega-acting-on-xi}).

We are now going to show that%
\begin{equation}
\omega\left(  a\right)  \cdot\left(  \omega\left(  b\right)  \boxdot\xi\left(
s\right)  \right)  =\omega\left(  b\right)  \boxdot\left(  \omega\left(
a\right)  \cdot\xi\left(  s\right)  \right)  +\omega\left(  a\vartriangleright
b\right)  \boxdot\xi\left(  s\right)  \label{pf.thm.nord.box.g.xi-claim}%
\end{equation}
for every $a\in\mathfrak{gl}_{n}$, $b\in\mathfrak{gl}_{n}$ and $s\in
\mathcal{A}\otimes\mathcal{A}^{\prime}$.

\textit{Proof of (\ref{pf.thm.nord.box.e.xi-claim}):} Let $a\in\mathfrak{gl}%
_{n}$, $b\in\mathfrak{gl}_{n}$ and $s\in\mathcal{A}\otimes\mathcal{A}^{\prime
}$.

We want to prove the equality (\ref{pf.thm.nord.box.e.xi-claim}). This
equality is $\mathbf{k}$-linear in $s$. Hence, we can WLOG assume that $s$ is
a pure tensor (since the $\mathbf{k}$-module $\mathcal{A}\otimes
\mathcal{A}^{\prime}$ is spanned by the pure tensors). Assume this. Thus,
$s=p\otimes q$ for some $p\in\mathcal{A}$ and $q\in\mathcal{A}$. Consider
these $p$ and $q$.

We want to prove the equality (\ref{pf.thm.nord.box.e.xi-claim}). This
equality is $\mathbf{k}$-linear in each of $a$ and $b$. Hence, we can WLOG
assume that both $a$ and $b$ belong to the basis $\left(  E_{i,j}\right)
_{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}$ of the
$\mathbf{k}$-module $\mathfrak{gl}_{n}$. Assume this. Thus, $a=E_{u,v}$ and
$b=E_{y,z}$ for some $\left(  u,v\right)  \in\left\{  1,2,\ldots,n\right\}
^{2}$ and some $\left(  y,z\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$.
Consider this $\left(  u,v\right)  $ and this $\left(  y,z\right)  $.

Let $h$ denote the element $\omega\left(  a\right)  $ of $\mathcal{D}$. Then,
$h=\omega\left(  \underbrace{a}_{\in\mathfrak{gl}_{n}}\right)  \in
\omega\left(  \mathfrak{gl}_{n}\right)  \subseteq\operatorname*{Der}%
\mathcal{A}$ (by Proposition \ref{prop.nord.derivs}). In other words, $h$ is a
derivation of $\mathcal{A}$. Thus, we have%
\begin{equation}
h\left[  s_{1}s_{2}\right]  =s_{1}\cdot h\left[  s_{2}\right]  +h\left[
s_{1}\right]  \cdot s_{2} \label{pf.thm.nord.box.g.xi-claim.pf.h-der}%
\end{equation}
for every $s_{1}\in\mathcal{A}$ and $s_{2}\in\mathcal{A}$.

We have
\begin{equation}
\underbrace{a}_{=E_{u,v}}\vartriangleright\underbrace{b}_{=E_{y,z}}%
=E_{u,v}\vartriangleright E_{y,z}=\delta_{v,y}E_{u,z}
\label{pf.thm.nord.box.g.xi-claim.pf.1}%
\end{equation}
(by (\ref{pf.thm.nord.box.g.EuvEij})). Hence,%
\begin{equation}
h\left[  x_{y,\ell}\right]  =\delta_{v,y}x_{u,\ell}
\label{pf.thm.nord.box.g.xi-claim.pf.1b}%
\end{equation}
for every $\ell\in\left\{  1,2,\ldots,m\right\}  $%
\ \ \ \ \footnote{\textit{Proof of (\ref{pf.thm.nord.box.g.xi-claim.pf.1b}):}
Let $\ell\in\left\{  1,2,\ldots,m\right\}  $. Then, $h=\omega\left(
\underbrace{a}_{=E_{u,v}}\right)  =\omega\left(  E_{u,v}\right)  =\sum
_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{v,k}}$ (by the definition of
$\omega$). Hence,
\begin{align*}
&  \underbrace{h}_{=\sum_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{v,k}}%
}\left[  x_{y,\ell}\right] \\
&  =\left(  \sum_{k=1}^{m}x_{u,k}\dfrac{\partial}{\partial x_{v,k}}\right)
\left[  x_{y,\ell}\right]  =\underbrace{\sum_{k=1}^{m}}_{=\sum_{k\in\left\{
1,2,\ldots,m\right\}  }}x_{u,k}\underbrace{\dfrac{\partial}{\partial x_{v,k}%
}\left[  x_{y,\ell}\right]  }_{=\delta_{\left(  v,k\right)  ,\left(
y,\ell\right)  }=\delta_{v,y}\delta_{k,\ell}}=\sum_{k\in\left\{
1,2,\ldots,m\right\}  }x_{u,k}\delta_{v,y}\delta_{k,\ell}\\
&  =x_{u,\ell}\delta_{v,y}\underbrace{\delta_{\ell,\ell}}_{=1}+\sum
_{\substack{k\in\left\{  1,2,\ldots,m\right\}  ;\\k\neq\ell}}x_{u,k}%
\delta_{v,y}\underbrace{\delta_{k,\ell}}_{\substack{=0\\\text{(since }%
k\neq\ell\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=\ell\text{ from the sum}\right) \\
&  =x_{u,\ell}\delta_{v,y}+\underbrace{\sum_{\substack{k\in\left\{
1,2,\ldots,m\right\}  ;\\k\neq\ell}}x_{u,k}\delta_{v,y}0}_{=0}=x_{u,\ell
}\delta_{v,y}=\delta_{v,y}x_{u,\ell}.
\end{align*}
This proves (\ref{pf.thm.nord.box.g.xi-claim.pf.1b}).}.

We have%
\begin{equation}
\omega\left(  \underbrace{b}_{=E_{y,z}}\right)  =\omega\left(  E_{y,z}\right)
=\sum_{k=1}^{m}x_{y,k}\dfrac{\partial}{\partial x_{z,k}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right)  .
\label{pf.thm.nord.box.g.xi-claim.pf.3}%
\end{equation}


On the other hand,%
\begin{align*}
\omega\left(  a\right)  \cdot\xi\left(  \underbrace{s}_{=p\otimes q}\right)
&  =\omega\left(  a\right)  \cdot\xi\left(  p\otimes q\right)  =\omega\left(
a\right)  \boxdot\xi\left(  \underbrace{p\otimes q}_{=s}\right)  +\xi\left(
\underbrace{\left(  \omega\left(  a\right)  \right)  }_{=h}\left[  p\right]
\otimes q\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.nord.box}
\textbf{(f)}}\right) \\
&  =\omega\left(  a\right)  \boxdot\xi\left(  s\right)  +\xi\left(  h\left[
p\right]  \otimes q\right)  .
\end{align*}
Hence,%
\begin{align}
&  \omega\left(  b\right)  \boxdot\underbrace{\left(  \omega\left(  a\right)
\cdot\xi\left(  s\right)  \right)  }_{=\omega\left(  a\right)  \boxdot
\xi\left(  s\right)  +\xi\left(  h\left[  p\right]  \otimes q\right)
}\nonumber\\
&  =\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \boxdot
\xi\left(  s\right)  +\xi\left(  h\left[  p\right]  \otimes q\right)  \right)
\nonumber\\
&  =\underbrace{\omega\left(  b\right)  \boxdot\omega\left(  a\right)
}_{\substack{=\omega\left(  a\right)  \boxdot\omega\left(  b\right)
\\\text{(since the }\mathbf{k}\text{-algebra }\left(  \mathcal{D}%
,\boxdot\right)  \\\text{is commutative)}}}\boxdot\xi\left(  s\right)
+\omega\left(  \underbrace{b}_{=E_{y,z}}\right)  \boxdot\xi\left(  h\left[
p\right]  \otimes q\right) \nonumber\\
&  =\omega\left(  a\right)  \boxdot\omega\left(  b\right)  \boxdot\xi\left(
s\right)  +\underbrace{\omega\left(  E_{u,z}\right)  \boxdot\xi\left(
h\left[  p\right]  \otimes q\right)  }_{\substack{=\sum_{k=1}^{m}\xi\left(
x_{y,k}\cdot h\left[  p\right]  \otimes\partial_{z,k}q\right)  \\\text{(by
(\ref{pf.thm.nord.box.g.omega-acting-on-xi}), applied to}\\h\left[  p\right]
\text{ and }\left(  y,z\right)  \text{ instead of }p\text{ and }\left(
u,v\right)  \text{)}}}\nonumber\\
&  =\omega\left(  a\right)  \boxdot\omega\left(  b\right)  \boxdot\xi\left(
s\right)  +\sum_{k=1}^{m}\xi\left(  x_{y,k}\cdot h\left[  p\right]
\otimes\partial_{z,k}q\right)  . \label{pf.thm.nord.box.g.xi-claim.pf.RHS-1.1}%
\end{align}


On the other hand,%
\begin{equation}
\omega\left(  \underbrace{b}_{=E_{y,z}}\right)  \boxdot\xi\left(
\underbrace{s}_{=p\otimes q}\right)  =\omega\left(  E_{y,z}\right)  \boxdot
\xi\left(  p\otimes q\right)  =\sum_{k=1}^{m}\xi\left(  x_{y,k}p\otimes
\partial_{z,k}q\right)  \label{pf.thm.nord.box.g.xi-claim.pf.LHS-1.a}%
\end{equation}
(by (\ref{pf.thm.nord.box.g.omega-acting-on-xi}) (applied to $\left(
y,z\right)  $ instead of $\left(  u,v\right)  $)). Hence,%
\begin{align}
&  \omega\left(  a\right)  \cdot\underbrace{\left(  \omega\left(  b\right)
\boxdot\xi\left(  s\right)  \right)  }_{=\sum_{k=1}^{m}\xi\left(
x_{y,k}p\otimes\partial_{z,k}q\right)  }\nonumber\\
&  =\omega\left(  a\right)  \cdot\sum_{k=1}^{m}\xi\left(  x_{y,k}%
p\otimes\partial_{z,k}q\right)  =\sum_{k=1}^{m}\underbrace{\omega\left(
a\right)  \cdot\xi\left(  x_{y,k}p\otimes\partial_{z,k}q\right)
}_{\substack{=\omega\left(  a\right)  \boxdot\xi\left(  x_{y,k}p\otimes
\partial_{z,k}q\right)  +\xi\left(  \left(  \omega\left(  a\right)  \right)
\left[  x_{y,k}p\right]  \otimes\partial_{z,k}q\right)  \\\text{(by Theorem
\ref{thm.nord.box} \textbf{(f)} (applied to }x_{y,k}p\text{ and }%
\partial_{z,k}q\\\text{instead of }p\text{ and }q\text{))}}}\nonumber\\
&  =\sum_{k=1}^{m}\left(  \omega\left(  a\right)  \boxdot\xi\left(
x_{y,k}p\otimes\partial_{z,k}q\right)  +\xi\left(  \underbrace{\left(
\omega\left(  a\right)  \right)  }_{=h}\left[  x_{y,k}p\right]  \otimes
\partial_{z,k}q\right)  \right) \nonumber\\
&  =\underbrace{\sum_{k=1}^{m}\omega\left(  a\right)  \boxdot\xi\left(
x_{y,k}p\otimes\partial_{z,k}q\right)  }_{=\omega\left(  a\right)  \boxdot
\sum_{k=1}^{m}\xi\left(  x_{y,k}p\otimes\partial_{z,k}q\right)  }+\sum
_{k=1}^{m}\xi\left(  h\left[  x_{y,k}p\right]  \otimes\partial_{z,k}q\right)
\nonumber\\
&  =\omega\left(  a\right)  \boxdot\underbrace{\sum_{k=1}^{m}\xi\left(
x_{y,k}p\otimes\partial_{z,k}q\right)  }_{\substack{=\omega\left(  b\right)
\boxdot\xi\left(  s\right)  \\\text{(by
(\ref{pf.thm.nord.box.g.xi-claim.pf.LHS-1.a}))}}}+\sum_{k=1}^{m}\xi\left(
h\left[  x_{y,k}p\right]  \otimes\partial_{z,k}q\right) \nonumber\\
&  =\omega\left(  a\right)  \boxdot\omega\left(  b\right)  \boxdot\xi\left(
s\right)  +\sum_{k=1}^{m}\xi\left(  h\left[  x_{y,k}p\right]  \otimes
\partial_{z,k}q\right)  . \label{pf.thm.nord.box.g.xi-claim.pf.LHS-1.1}%
\end{align}
But for every $k\in\left\{  1,2,\ldots,m\right\}  $, we have%
\begin{align*}
&  \xi\left(  \underbrace{h\left[  x_{y,k}p\right]  }_{\substack{=x_{y,k}\cdot
h\left[  p\right]  +h\left[  x_{y,k}\right]  \cdot p\\\text{(by
(\ref{pf.thm.nord.box.g.xi-claim.pf.h-der}), applied to}\\s_{1}=x_{y,k}\text{
and }s_{2}=p\text{)}}}\otimes\partial_{z,k}q\right) \\
&  =\xi\left(  \underbrace{\left(  x_{y,k}\cdot h\left[  p\right]  +h\left[
x_{y,k}\right]  \cdot p\right)  \otimes\partial_{z,k}q}_{=x_{y,k}\cdot
h\left[  p\right]  \otimes\partial_{z,k}q+h\left[  x_{y,k}\right]  \cdot
p\otimes\partial_{z,k}q}\right) \\
&  =\xi\left(  x_{y,k}\cdot h\left[  p\right]  \otimes\partial_{z,k}q+h\left[
x_{y,k}\right]  \cdot p\otimes\partial_{z,k}q\right) \\
&  =\xi\left(  x_{y,k}\cdot h\left[  p\right]  \otimes\partial_{z,k}q\right)
+\xi\left(  h\left[  x_{y,k}\right]  \cdot p\otimes\partial_{z,k}q\right)  .
\end{align*}
Thus, (\ref{pf.thm.nord.box.g.xi-claim.pf.LHS-1.1}) becomes%
\begin{align}
&  \omega\left(  a\right)  \cdot\left(  \omega\left(  b\right)  \boxdot
\xi\left(  s\right)  \right) \nonumber\\
&  =\omega\left(  a\right)  \boxdot\omega\left(  b\right)  \boxdot\xi\left(
s\right)  +\sum_{k=1}^{m}\underbrace{\xi\left(  h\left[  x_{y,k}p\right]
\otimes\partial_{z,k}q\right)  }_{=\xi\left(  x_{y,k}\cdot h\left[  p\right]
\otimes\partial_{z,k}q\right)  +\xi\left(  h\left[  x_{y,k}\right]  \cdot
p\otimes\partial_{z,k}q\right)  }\nonumber\\
&  =\omega\left(  a\right)  \boxdot\omega\left(  b\right)  \boxdot\xi\left(
s\right)  +\underbrace{\sum_{k=1}^{m}\left(  \xi\left(  x_{y,k}\cdot h\left[
p\right]  \otimes\partial_{z,k}q\right)  +\xi\left(  h\left[  x_{y,k}\right]
\cdot p\otimes\partial_{z,k}q\right)  \right)  }_{=\sum_{k=1}^{m}\xi\left(
x_{y,k}\cdot h\left[  p\right]  \otimes\partial_{z,k}q\right)  +\sum_{k=1}%
^{m}\xi\left(  h\left[  x_{y,k}\right]  \cdot p\otimes\partial_{z,k}q\right)
}\nonumber\\
&  =\underbrace{\omega\left(  a\right)  \boxdot\omega\left(  b\right)
\boxdot\xi\left(  s\right)  +\sum_{k=1}^{m}\xi\left(  x_{y,k}\cdot h\left[
p\right]  \otimes\partial_{z,k}q\right)  }_{\substack{=\omega\left(  b\right)
\boxdot\left(  \omega\left(  a\right)  \cdot\xi\left(  s\right)  \right)
\\\text{(by (\ref{pf.thm.nord.box.g.xi-claim.pf.RHS-1.1}))}}}+\sum_{k=1}%
^{m}\xi\left(  \underbrace{h\left[  x_{y,k}\right]  }_{\substack{=\delta
_{v,y}x_{u,k}\\\text{(by (\ref{pf.thm.nord.box.g.xi-claim.pf.1b}%
)}\\\text{(applied to }\ell=k\text{))}}}\cdot p\otimes\partial_{z,k}q\right)
\nonumber\\
&  =\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \cdot
\xi\left(  s\right)  \right)  +\sum_{k=1}^{m}\underbrace{\xi\left(
\delta_{v,y}x_{u,k}\cdot p\otimes\partial_{z,k}q\right)  }_{\substack{=\delta
_{v,y}\xi\left(  x_{u,k}\cdot p\otimes\partial_{z,k}q\right)  \\=\delta
_{v,y}\xi\left(  x_{u,k}p\otimes\partial_{z,k}q\right)  }}\nonumber\\
&  =\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \cdot
\xi\left(  s\right)  \right)  +\sum_{k=1}^{m}\delta_{v,y}\xi\left(
x_{u,k}p\otimes\partial_{z,k}q\right)  .
\label{pf.thm.nord.box.g.xi-claim.pf.LHS-1.3}%
\end{align}


On the other hand, applying the map $\omega$ to the equality
(\ref{pf.thm.nord.box.g.xi-claim.pf.1}), we obtain%
\[
\omega\left(  a\vartriangleright b\right)  =\omega\left(  \delta_{v,y}%
E_{u,z}\right)  =\delta_{v,y}\omega\left(  E_{u,z}\right)  .
\]
Hence,%
\begin{align}
\underbrace{\omega\left(  a\vartriangleright b\right)  }_{=\delta_{v,y}%
\omega\left(  E_{u,z}\right)  }\boxdot\xi\left(  \underbrace{s}_{=p\otimes
q}\right)   &  =\delta_{v,y}\underbrace{\omega\left(  E_{u,z}\right)
\boxdot\xi\left(  p\otimes q\right)  }_{\substack{=\sum_{k=1}^{m}\xi\left(
x_{u,k}p\otimes\partial_{z,k}q\right)  \\\text{(by
(\ref{pf.thm.nord.box.g.omega-acting-on-xi}) (applied to }\left(  u,z\right)
\\\text{instead of }\left(  u,v\right)  \text{))}}}\nonumber\\
&  =\delta_{v,y}\sum_{k=1}^{m}\xi\left(  x_{u,k}p\otimes\partial
_{z,k}q\right)  =\sum_{k=1}^{m}\delta_{v,y}\xi\left(  x_{u,k}p\otimes
\partial_{z,k}q\right)  . \label{pf.thm.nord.box.g.xi-claim.pf.RHS-2.1}%
\end{align}
Hence, (\ref{pf.thm.nord.box.g.xi-claim.pf.LHS-1.3}) becomes%
\begin{align*}
&  \omega\left(  a\right)  \cdot\left(  \omega\left(  b\right)  \boxdot
\xi\left(  s\right)  \right) \\
&  =\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \cdot
\xi\left(  s\right)  \right)  +\underbrace{\sum_{k=1}^{m}\delta_{v,y}%
\xi\left(  x_{u,k}p\otimes\partial_{z,k}q\right)  }_{\substack{=\omega\left(
a\vartriangleright b\right)  \boxdot\xi\left(  s\right)  \\\text{(by
(\ref{pf.thm.nord.box.g.xi-claim.pf.RHS-2.1}))}}}\\
&  =\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \cdot
\xi\left(  s\right)  \right)  +\omega\left(  a\vartriangleright b\right)
\boxdot\xi\left(  s\right)  .
\end{align*}
Thus, (\ref{pf.thm.nord.box.g.xi-claim}) is proven.

Now, fix $a\in\mathfrak{gl}_{n}$, $b\in\mathfrak{gl}_{n}$ and $h\in
\mathcal{D}$. Let $s=\xi^{-1}\left(  h\right)  $. This $s$ is a well-defined
element of $\mathcal{A}\otimes\mathcal{A}^{\prime}$ (since $\xi$ is a
$\mathbf{k}$-module isomorphism). Now, $\xi\left(  s\right)  =h$ (since
$s=\xi^{-1}\left(  h\right)  $), so that $h=\xi\left(  s\right)  $. Thus,%
\begin{align*}
&  \omega\left(  a\right)  \cdot\left(  \omega\left(  b\right)  \boxdot
\underbrace{h}_{=\xi\left(  s\right)  }\right) \\
&  =\omega\left(  a\right)  \cdot\left(  \omega\left(  b\right)  \boxdot
\xi\left(  s\right)  \right)  =\omega\left(  b\right)  \boxdot\left(
\omega\left(  a\right)  \cdot\underbrace{\xi\left(  s\right)  }_{=h}\right)
+\omega\left(  a\vartriangleright b\right)  \boxdot\underbrace{\xi\left(
s\right)  }_{=h}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.nord.box.g.xi-claim})}\right) \\
&  =\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \cdot
h\right)  +\omega\left(  a\vartriangleright b\right)  \boxdot h.
\end{align*}
This proves Theorem \ref{thm.nord.box} \textbf{(g)}.

\textbf{(h)} This proof is going to be somewhat complicated, so we subdivide
it into several steps for the reader's convenience.

\textit{Step 1:} Recall that $\left(  E_{i,j}\right)  _{\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}}$ is a basis of the $\mathbf{k}$-module
$\mathfrak{gl}_{n}=\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $.
Hence, we can define a $\mathbf{k}$-linear map $\phi:\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \rightarrow\mathcal{A}\otimes
\mathcal{A}^{\prime}$ by%
\[
\phi\left(  E_{i,j}\right)  =\sum_{k=1}^{m}x_{i,k}\otimes\partial
_{j,k}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}.
\]
Consider this map $\phi$. Proposition \ref{prop.Sym.univ} (applied to
$V=\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $, $A=\mathcal{A}%
\otimes\mathcal{A}^{\prime}$ and $f=\phi$) shows that there exists a unique
$\mathbf{k}$-algebra homomorphism $F:\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)
\rightarrow\mathcal{A}\otimes\mathcal{A}^{\prime}$ such that $F\circ
\iota_{\operatorname*{Sym},\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  }=\phi$. Let us denote this $F$ by $\Phi$. Thus, $\Phi$ is
a $\mathbf{k}$-algebra homomorphism $\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)
\rightarrow\mathcal{A}\otimes\mathcal{A}^{\prime}$ such that $\Phi\circ
\iota_{\operatorname*{Sym},\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  }=\phi$.

\textit{Step 2:} Let us make a convention. Namely, we shall employ the abuse
of notation introduced in Definition \ref{def.manchon.danger} (applied to
$V=\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $). Thus, we will
identify every $a\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $
with the element $\iota_{\operatorname*{Sym},\operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  }\left(  a\right)  $ of $\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $. As a
consequence, products such as $b_{1}b_{2}\cdots b_{p}$ (where $p\in\mathbb{N}$
and $b_{1},b_{2},\ldots,b_{p}\in\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  $) will always mean products inside $\operatorname*{Sym}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $
(and not products inside $\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}%
\right)  $). In particular, we are \textbf{not} going to denote the product of
two elements $a$ and $b$ of the matrix ring $\operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  $ by $ab$, because the notation $ab$ will mean the
product of the elements $a$ and $b$ of $\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $.

Fortunately, we have a different notation for products inside
$\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $. Namely, recall
that the $\mathbf{k}$-module $\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  $ (equipped with the multiplication of $\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $) is a left pre-Lie algebra. Hence,
if $a\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $ and
$b\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $ are two
matrices, then $a\vartriangleright b$ is the product of the matrices $a$ and
$b$ inside $\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $. Hence,
for every $\left(  u,v\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ and
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$, we have%
\begin{align}
E_{u,v}\vartriangleright E_{i,j}  &  =\left(  \text{the product of the
matrices }E_{u,v}\text{ and }E_{i,j}\text{ inside }\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right) \nonumber\\
&  =\delta_{v,i}E_{u,j} \label{pf.thm.nord.box.e.EijEuv}%
\end{align}
(by the standard multiplication rule for elementary matrices).

We identify every $a\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}%
\right)  $ with the element $\iota_{\operatorname*{Sym},\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  }\left(  a\right)  $ of
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $. Thus, $\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  =\iota_{\operatorname*{Sym},\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  }\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  =\operatorname*{Sym}%
\nolimits^{1}\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)
\right)  $.

\textit{Step 3:} The map $\xi$ is a $\mathbf{k}$-algebra isomorphism from
$\mathcal{A}\otimes\mathcal{A}^{\prime}$ to the $\mathbf{k}$-algebra $\left(
\mathcal{D},\boxdot\right)  $ (by Proposition \ref{prop.nord.Dunity}
\textbf{(a)}), therefore also a $\mathbf{k}$-algebra homomorphism from
$\mathcal{A}\otimes\mathcal{A}^{\prime}$ to the $\mathbf{k}$-algebra $\left(
\mathcal{D},\boxdot\right)  $.

\textit{Step 4:} Now, define a map $\psi:\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)
\rightarrow\mathcal{D}$ by%
\[
\psi=\xi\circ\Phi.
\]
Then, $\psi$ is the composition of the two $\mathbf{k}$-algebra homomorphisms
$\xi:\mathcal{A}\otimes\mathcal{A}^{\prime}\rightarrow\left(  \mathcal{D}%
,\boxdot\right)  $ and $\Phi:\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  \rightarrow\mathcal{A}%
\otimes\mathcal{A}^{\prime}$. Hence, $\psi$ is a $\mathbf{k}$-algebra
homomorphism from $\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  $ to $\left(  \mathcal{D},\boxdot\right)
$.

The following diagram shows the most important maps that we have so far
introduced:%
\[%
%TCIMACRO{\TeXButton{x}{\xymatrix{
%\left(U\left(\mathfrak{gl}_n\right), \boxdot\right) \ar[r]^{\eta}
%\ar[d]_{\Omega} & \Sym\left(\operatorname{M}_n\left(\kk\right)\right
%) \ar[d]^{\Phi} \ar[dl]^{\psi} \\
%\left(\mathcal{D}, \boxdot\right) & \mathcal{A} \otimes\mathcal{A}^\prime
%\ar[l]^{\xi}
%}}}%
%BeginExpansion
\xymatrix{
\left(U\left(\mathfrak{gl}_n\right), \boxdot\right) \ar[r]^{\eta}
\ar[d]_{\Omega} & \Sym\left(\operatorname{M}_n\left(\kk\right)\right
) \ar[d]^{\Phi} \ar[dl]^{\psi} \\
\left(\mathcal{D}, \boxdot\right) & \mathcal{A} \otimes\mathcal{A}^\prime
\ar[l]^{\xi}
}%
%EndExpansion
.
\]
We do not yet know whether the left triangle of this diagram is commutative
(although we will later prove this, in Step 10). The right triangle is
commutative by definition of $\psi$. All arrows in this diagram except for
$\Omega$ are known to be $\mathbf{k}$-algebra homomorphisms; proving this for
$\Omega$ is our goal.

\textit{Step 5:} For every $\left(  i,j\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{2}$, we have%
\begin{equation}
\Phi\left(  E_{i,j}\right)  =\sum_{k=1}^{m}x_{i,k}\otimes\partial_{j,k}
\label{pf.thm.nord.box.e.Phi-formula.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.nord.box.e.Phi-formula.1}):} Let
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$. We have
$\Phi\circ\iota_{\operatorname*{Sym},\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  }=\phi$. Hence, $\left(  \Phi\circ\iota
_{\operatorname*{Sym},\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)
}\right)  \left(  E_{i,j}\right)  =\phi\left(  E_{i,j}\right)  =\sum_{k=1}%
^{m}x_{i,k}\otimes\partial_{j,k}$ (by the definition of $\phi$). Compared with%
\[
\left(  \Phi\circ\iota_{\operatorname*{Sym},\operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  }\right)  \left(  E_{i,j}\right)  =\Phi\left(
\underbrace{\iota_{\operatorname*{Sym},\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  }\left(  E_{i,j}\right)  }_{\substack{=E_{i,j}\\\text{(by
our abuse of notation)}}}\right)  =\Phi\left(  E_{i,j}\right)  ,
\]
this shows that $\Phi\left(  E_{i,j}\right)  =\sum_{k=1}^{m}x_{i,k}%
\otimes\partial_{j,k}$. This proves (\ref{pf.thm.nord.box.e.Phi-formula.1}).}
and%
\begin{equation}
\psi\left(  E_{i,j}\right)  =\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial
x_{j,k}} \label{pf.thm.nord.box.e.psi-formula-1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.nord.box.e.psi-formula-1}):} Every
$u\in\left\{  1,2,\ldots,n\right\}  $, $v\in\left\{  1,2,\ldots,n\right\}  $
and $k\in\left\{  1,2,\ldots,m\right\}  $ satisfy%
\begin{align}
\xi\left(  x_{u,k}\otimes\partial_{v,k}\right)   &  =x_{u,k}\cdot
\underbrace{\partial_{v,k}\left(  \left(  \dfrac{\partial}{\partial x_{i,j}%
}\right)  _{\left(  i,j\right)  \in G}\right)  }_{=\dfrac{\partial}{\partial
x_{v,k}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\xi\right)
\nonumber\\
&  =x_{u,k}\dfrac{\partial}{\partial x_{v,k}}.
\label{pf.thm.nord.box.e.psi-formula-1.pf.1}%
\end{align}
\par
Now, let $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$. Then,%
\begin{align*}
\underbrace{\psi}_{=\xi\circ\Phi}\left(  E_{i,j}\right)   &  =\left(  \xi
\circ\Phi\right)  \left(  E_{i,j}\right)  =\xi\left(  \underbrace{\Phi\left(
E_{i,j}\right)  }_{\substack{=\sum_{k=1}^{m}x_{i,k}\otimes\partial
_{j,k}\\\text{(by (\ref{pf.thm.nord.box.e.Phi-formula.1}))}}}\right)
=\xi\left(  \sum_{k=1}^{m}x_{i,k}\otimes\partial_{j,k}\right) \\
&  =\sum_{k=1}^{m}\underbrace{\xi\left(  x_{i,k}\otimes\partial_{j,k}\right)
}_{\substack{=x_{i,k}\dfrac{\partial}{\partial x_{j,k}}\\\text{(by
(\ref{pf.thm.nord.box.e.psi-formula-1.pf.1}), applied to }u=i\text{ and
}v=j\text{)}}}=\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}}.
\end{align*}
This proves (\ref{pf.thm.nord.box.e.psi-formula-1}).}. Hence,%
\begin{equation}
\psi\left(  a\right)  =\omega\left(  a\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{gl}_{n} \label{pf.thm.nord.box.g.psi-formula-2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.nord.box.g.psi-formula-2}):} Let
$a\in\mathfrak{gl}_{n}$. We must prove the equality $\psi\left(  a\right)
=\omega\left(  a\right)  $. This equality is $\mathbf{k}$-linear in $a$.
Hence, we can WLOG assume that $a$ belongs to the basis $\left(
E_{i,j}\right)  _{\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}}$
of the $\mathbf{k}$-module $\mathfrak{gl}_{n}$. Assume this. Thus, $a=E_{i,j}$
for some $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$.
Consider this $\left(  i,j\right)  $. Comparing%
\[
\psi\left(  \underbrace{a}_{=E_{i,j}}\right)  =\psi\left(  E_{i,j}\right)
=\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.nord.box.e.psi-formula-1}%
)}\right)
\]
with%
\[
\omega\left(  \underbrace{a}_{=E_{i,j}}\right)  =\omega\left(  E_{i,j}\right)
=\sum_{k=1}^{m}x_{i,k}\dfrac{\partial}{\partial x_{j,k}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\omega\right)  ,
\]
we obtain $\psi\left(  a\right)  =\omega\left(  a\right)  $. This proves
(\ref{pf.thm.nord.box.g.psi-formula-2}).}.

\textit{Step 6:} Recall that we have defined a $\mathfrak{gl}_{n}$-module
structure on the $\mathbf{k}$-module $\mathcal{D}$ in Corollary
\ref{cor.nord.l}. Let us make $\mathcal{D}$ into a $\mathfrak{gl}_{n}$-module
using this $\mathfrak{gl}_{n}$-module structure.

On the other hand, recall that $\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ is a $\mathfrak{gl}_{n}%
$-module. This $\mathfrak{gl}_{n}$-module satisfies%
\begin{equation}
a\rightharpoonup1_{\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  }=a\ \ \ \ \ \ \ \ \ \ \text{for every
}a\in\mathfrak{gl}_{n} \label{pf.thm.nord.box.g.trivact}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.nord.box.g.trivact}):} For every
$a\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $, we define a
derivation $L_{a}:\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  \rightarrow\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ as in
Theorem \ref{thm.manchon} (applied to $A=\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  $).
\par
Let $a\in\mathfrak{gl}_{n}$. Thus, $a\in\mathfrak{gl}_{n}=\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $. Thus, $L_{a}$ is a derivation
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  \rightarrow\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $. In other
words, $L_{a}\in\operatorname*{Der}\left(  \operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  \right)  $.
Hence, Proposition \ref{prop.Der.basics} \textbf{(c)} (applied to
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $ and $L_{a}$ instead of $C$ and $f$) shows that
$L_{a}\left(  1\right)  =0$.
\par
Now, the definition of the $\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  ^{-}$-module structure on $\operatorname*{Sym}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $
shows that%
\[
a\rightharpoonup1_{\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  }=\underbrace{a\cdot1_{\operatorname*{Sym}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }%
}_{=a}+L_{a}\left(  \underbrace{1_{\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }}%
_{=1}\right)  =a+\underbrace{L_{a}\left(  1\right)  }_{=0}=a.
\]
This proves (\ref{pf.thm.nord.box.g.trivact}).}.

\textit{Step 7:} We shall now show that%
\begin{equation}
\psi\left(  a\rightharpoonup z\right)  =a\rightharpoonup\psi\left(  z\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }N\in\mathbb{N}\text{, }a\in
\mathfrak{gl}_{n}\text{ and }z\in\operatorname*{Sym}\nolimits^{N}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  .
\label{pf.thm.nord.box.g.inductor}%
\end{equation}


\textit{Proof of (\ref{pf.thm.nord.box.g.inductor}):} We shall prove
(\ref{pf.thm.nord.box.g.inductor}) by induction over $N$:

\textit{Induction base:} We have $\psi\left(  a\rightharpoonup z\right)
=a\rightharpoonup\psi\left(  z\right)  $ for every $a\in\mathfrak{gl}_{n}$ and
$z\in\operatorname*{Sym}\nolimits^{0}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$a\in\mathfrak{gl}_{n}$ and $z\in\operatorname*{Sym}\nolimits^{0}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $. Recall
that $\operatorname*{Sym}\nolimits^{0}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  $ is a free $\mathbf{k}$-module with basis
$\left(  1_{\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  }\right)  $.
\par
We need to prove the equality $\psi\left(  a\rightharpoonup z\right)
=a\rightharpoonup\psi\left(  z\right)  $. This equality is $\mathbf{k}$-linear
in $z$. Hence, we can WLOG assume that $z$ belongs to the basis $\left(
1_{\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  }\right)  $ of the $\mathbf{k}$-module
$\operatorname*{Sym}\nolimits^{0}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $. Assume this. Thus, $z=1_{\operatorname*{Sym}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }$.
Now, $\psi\left(  \underbrace{z}_{=1_{\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }}\right)
=\psi\left(  1_{\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  }\right)  =1_{\left(  \mathcal{D}%
,\boxdot\right)  }$ (since $\psi$ is a $\mathbf{k}$-algebra homomorphism from
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $ to $\left(  \mathcal{D},\boxdot\right)  $), so
that $\psi\left(  1_{\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }\right)  =1_{\left(
\mathcal{D},\boxdot\right)  }=1_{\mathcal{D}}$ (by Proposition
\ref{prop.nord.Dunity} \textbf{(b)}). Now,%
\begin{align*}
&  a\rightharpoonup\psi\left(  z\right)  =\omega\left(  a\right)
\cdot\underbrace{\psi\left(  z\right)  }_{=1_{\mathcal{D}}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the }\mathfrak{gl}%
_{n}\text{-structure on }\mathcal{D}\right) \\
&  =\omega\left(  a\right)  \cdot1_{\mathcal{D}}=\omega\left(  a\right)  .
\end{align*}
\par
But%
\[
a\rightharpoonup\underbrace{z}_{=1_{\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }%
}=a\rightharpoonup1_{\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }%
=a\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.nord.box.g.trivact}%
)}\right)  .
\]
Applying the map $\psi$ to this equality, we obtain $\psi\left(
a\rightharpoonup z\right)  =\psi\left(  a\right)  =\omega\left(  a\right)  $
(by (\ref{pf.thm.nord.box.g.psi-formula-2})). Comparing this with
$a\rightharpoonup\psi\left(  z\right)  =\omega\left(  a\right)  $, we obtain
$\psi\left(  a\rightharpoonup z\right)  =a\rightharpoonup\psi\left(  z\right)
$. Qed.}. In other words, (\ref{pf.thm.nord.box.g.inductor}) holds for $N=0$.
This completes the induction base.

\textit{Induction step:} Let $M$ be a positive integer. Assume that
(\ref{pf.thm.nord.box.g.inductor}) holds for $N=M-1$. We must prove that
(\ref{pf.thm.nord.box.g.inductor}) holds for $N=M$.

We have assumed that (\ref{pf.thm.nord.box.g.inductor}) holds for $n=N-1$. In
other words, we have%
\begin{equation}
\psi\left(  a\rightharpoonup z\right)  =a\rightharpoonup\psi\left(  z\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{gl}_{n}\text{ and }%
z\in\operatorname*{Sym}\nolimits^{M-1}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  .
\label{pf.thm.nord.box.g.inductor.pf.indhyp}%
\end{equation}


Now, let $a\in\mathfrak{gl}_{n}$ and $z\in\operatorname*{Sym}\nolimits^{M}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $.
We are going to show that $\psi\left(  a\rightharpoonup z\right)
=a\rightharpoonup\psi\left(  z\right)  $.

We have $z\in\operatorname*{Sym}\nolimits^{M}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  =\left(  \operatorname*{Sym}%
\nolimits^{1}\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)
\right)  \right)  \left(  \operatorname*{Sym}\nolimits^{M-1}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  \right)  $
(by Remark \ref{rmk.Sym.span} \textbf{(c)} (applied to $\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $ and $M$ instead of $V$ and $n$)).
Thus, $z$ is a $\mathbf{k}$-linear combination of elements of the form $bu$
with $b\in\operatorname*{Sym}\nolimits^{1}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ and $u\in
\operatorname*{Sym}\nolimits^{M-1}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  $.

Now, we must prove the equality $\psi\left(  a\rightharpoonup z\right)
=a\rightharpoonup\psi\left(  z\right)  $. This equality is $\mathbf{k}$-linear
in $z$. Hence, we can WLOG assume that $z$ has the form $bu$ with
$b\in\operatorname*{Sym}\nolimits^{1}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  $ and $u\in\operatorname*{Sym}%
\nolimits^{M-1}\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}%
\right)  \right)  $ (because $z$ is a $\mathbf{k}$-linear combination of
elements of the form $bu$ with $b\in\operatorname*{Sym}\nolimits^{1}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ and
$u\in\operatorname*{Sym}\nolimits^{M-1}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $). Assume this. Hence,
$z=bu$ for some $b\in\operatorname*{Sym}\nolimits^{1}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ and $u\in
\operatorname*{Sym}\nolimits^{M-1}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  $. Consider this $b$ and $u$.

We have $b\in\operatorname*{Sym}\nolimits^{1}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  =\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  =\mathfrak{gl}_{n}$. Hence,
(\ref{pf.thm.nord.box.g.psi-formula-2}) (applied to $b$ instead of $a$) shows
that $\psi\left(  b\right)  =\omega\left(  b\right)  $.

Now, $a\in\mathfrak{gl}_{n}=\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  $ and $b\in\operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  $. Hence, $a\vartriangleright b\in\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $. Hence,
(\ref{pf.thm.nord.box.g.psi-formula-2}) (applied to $a\vartriangleright b$
instead of $a$) shows that $\psi\left(  a\vartriangleright b\right)
=\omega\left(  a\vartriangleright b\right)  $.

Applying (\ref{pf.thm.nord.box.g.inductor.pf.indhyp}) to $u$ instead of $z$,
we see that%
\begin{equation}
\psi\left(  a\rightharpoonup u\right)  =a\rightharpoonup\psi\left(  u\right)
=\omega\left(  a\right)  \cdot\psi\left(  u\right)
\label{pf.thm.nord.box.g.inductor.pf.2}%
\end{equation}
(by the definition of the $\mathfrak{gl}_{n}$-module structure on
$\mathcal{D}$). On the other hand,%
\begin{equation}
a\rightharpoonup\psi\left(  \underbrace{z}_{=bu}\right)  =a\rightharpoonup
\psi\left(  bu\right)  =\omega\left(  a\right)  \cdot\psi\left(  bu\right)
\label{pf.thm.nord.box.g.inductor.pf.3}%
\end{equation}
(by the definition of the $\mathfrak{gl}_{n}$-module structure on
$\mathcal{D}$).

But recall that $\psi$ is a $\mathbf{k}$-algebra homomorphism from
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $ to $\left(  \mathcal{D},\boxdot\right)  $.
Hence, $\psi\left(  bu\right)  =\underbrace{\psi\left(  b\right)  }%
_{=\omega\left(  b\right)  }\boxdot\psi\left(  u\right)  =\omega\left(
b\right)  \boxdot\psi\left(  u\right)  $. Hence,
(\ref{pf.thm.nord.box.g.inductor.pf.3}) becomes%
\begin{align}
a\rightharpoonup\psi\left(  z\right)   &  =\omega\left(  a\right)
\cdot\underbrace{\psi\left(  bu\right)  }_{=\omega\left(  b\right)
\boxdot\psi\left(  u\right)  }=\omega\left(  a\right)  \cdot\left(
\omega\left(  b\right)  \boxdot\psi\left(  u\right)  \right) \nonumber\\
&  =\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \cdot
\psi\left(  u\right)  \right)  +\omega\left(  a\vartriangleright b\right)
\boxdot\psi\left(  u\right) \label{pf.thm.nord.box.g.inductor.pf.4}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.nord.box}
\textbf{(g)} (applied to }h=\psi\left(  u\right)  \text{)}\right)  .\nonumber
\end{align}


But $a\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $ and
$b\in\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $. Hence,
Theorem \ref{thm.manchon} \textbf{(m)} (applied to $A=\operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  $ and $c=u$) shows that%
\[
a\rightharpoonup\left(  bu\right)  -b\cdot\left(  a\rightharpoonup u\right)
=\left(  a\vartriangleright b\right)  \cdot u.
\]
Hence, $a\rightharpoonup\left(  bu\right)  =b\cdot\left(  a\rightharpoonup
u\right)  +\left(  a\vartriangleright b\right)  \cdot u$. Now,%
\[
a\rightharpoonup\underbrace{z}_{=bu}=a\rightharpoonup\left(  bu\right)
=b\cdot\left(  a\rightharpoonup u\right)  +\left(  a\vartriangleright
b\right)  \cdot u.
\]
Applying the map $\psi$ to this equality, we obtain%
\begin{align*}
\psi\left(  a\rightharpoonup z\right)   &  =\psi\left(  b\cdot\left(
a\rightharpoonup u\right)  +\left(  a\vartriangleright b\right)  \cdot
u\right)  =\underbrace{\psi\left(  b\right)  }_{=\omega\left(  b\right)
}\boxdot\underbrace{\psi\left(  a\rightharpoonup u\right)  }%
_{\substack{=\omega\left(  a\right)  \cdot\psi\left(  u\right)  \\\text{(by
(\ref{pf.thm.nord.box.g.inductor.pf.2}))}}}+\underbrace{\psi\left(
a\vartriangleright b\right)  }_{=\omega\left(  a\vartriangleright b\right)
}\boxdot\psi\left(  u\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\psi\text{ is a }\mathbf{k}\text{-algebra homomorphism}\\
\text{from }\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  \text{ to }\left(  \mathcal{D},\boxdot\right)
\end{array}
\right) \\
&  =\omega\left(  b\right)  \boxdot\left(  \omega\left(  a\right)  \cdot
\psi\left(  u\right)  \right)  +\omega\left(  a\vartriangleright b\right)
\boxdot\psi\left(  u\right)  .
\end{align*}
Comparing this with (\ref{pf.thm.nord.box.g.inductor.pf.4}), we obtain
$\psi\left(  a\rightharpoonup z\right)  =a\rightharpoonup\psi\left(  z\right)
$.

Let us now forget that we fixed $a$ and $z$. We thus have shown that%
\[
\psi\left(  a\rightharpoonup z\right)  =a\rightharpoonup\psi\left(  z\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{gl}_{n}\text{ and }%
z\in\operatorname*{Sym}\nolimits^{M}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  .
\]
In other words, (\ref{pf.thm.nord.box.g.inductor}) holds for $N=M$. This
completes the induction step. Hence, the induction proof of
(\ref{pf.thm.nord.box.g.inductor}) is complete.

\textit{Step 8:} We will now show that%
\begin{equation}
\psi\text{ is a }\mathfrak{gl}_{n}\text{-module homomorphism from
}\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  \text{ to }\mathcal{D}.
\label{pf.thm.nord.box.g.glnmodhom}%
\end{equation}


\textit{Proof of (\ref{pf.thm.nord.box.g.glnmodhom}):} We want to prove that
$\psi$ is a $\mathfrak{gl}_{n}$-module homomorphism from $\operatorname*{Sym}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ to
$\mathcal{D}$. In other words, we want to prove that $\psi\left(
a\rightharpoonup z\right)  =a\rightharpoonup\psi\left(  z\right)  $ for every
$a\in\mathfrak{gl}_{n}$ and $z\in\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $. So let us fix
$a\in\mathfrak{gl}_{n}$ and $z\in\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $.

We want to prove the equality $\psi\left(  a\rightharpoonup z\right)
=a\rightharpoonup\psi\left(  z\right)  $. This equality is $\mathbf{k}$-linear
in $z$. Hence, we can WLOG assume that $z$ is a homogeneous element of
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $ (since every element of $\operatorname*{Sym}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ is
a $\mathbf{k}$-linear combination of homogeneous elements). Assume this. Thus,
$z\in\operatorname*{Sym}\nolimits^{N}\left(  \operatorname*{M}\nolimits_{n}%
\left(  \mathbf{k}\right)  \right)  $ for some $N\in\mathbb{N}$. Consider this
$N$. Now, (\ref{pf.thm.nord.box.g.inductor}) shows that $\psi\left(
a\rightharpoonup z\right)  =a\rightharpoonup\psi\left(  z\right)  $.

Now, let us forget that we fixed $a$ and $z$. We thus have shown that
$\psi\left(  a\rightharpoonup z\right)  =a\rightharpoonup\psi\left(  z\right)
$ for every $a\in\mathfrak{gl}_{n}$ and $z\in\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $. In other
words, $\psi$ is a $\mathfrak{gl}_{n}$-module homomorphism from
$\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $ to $\mathcal{D}$. This proves
(\ref{pf.thm.nord.box.g.glnmodhom}).

\textit{Step 9:} The definition of the map $\eta$ shows that $\eta\left(
1_{U\left(  \mathfrak{gl}_{n}\right)  }\right)  =1_{U\left(  \mathfrak{gl}%
_{n}\right)  }\cdot1_{\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }=1_{\operatorname*{Sym}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }$.
Hence,%
\begin{align*}
\left(  \psi\circ\eta\right)  \left(  1_{U\left(  \mathfrak{gl}_{n}\right)
}\right)   &  =\psi\left(  \underbrace{\eta\left(  1_{U\left(  \mathfrak{gl}%
_{n}\right)  }\right)  }_{=1_{\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  }}\right)  =\psi\left(
1_{\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  }\right)  =1_{\left(  \mathcal{D},\boxdot\right)
}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\psi\text{ is a }\mathbf{k}\text{-algebra homomorphism}\\
\text{from }\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  \text{ to }\left(  \mathcal{D},\boxdot\right)
\end{array}
\right) \\
&  =1_{\mathcal{D}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.nord.Dunity} \textbf{(b)}}\right)  .
\end{align*}
On the other hand, $\Omega\left(  1_{U\left(  \mathfrak{gl}_{n}\right)
}\right)  =1_{\mathcal{D}}$ (since $\Omega$ is a $\mathbf{k}$-algebra
homomorphism from $U\left(  \mathfrak{gl}_{n}\right)  $ to $\mathcal{D}$).

\textit{Step 10:} Theorem \ref{thm.manchon} \textbf{(h)} (applied to
$A=\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  $) shows that the
map $\eta$ is an $\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}%
\right)  \right)  ^{-}$-module isomorphism. In other words, the map $\eta$ is
an $\mathfrak{gl}_{n}$-module isomorphism (since $\mathfrak{gl}_{n}=\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  ^{-}$).

But (\ref{pf.thm.nord.box.g.glnmodhom}) shows that the map $\psi$ is a
$\mathfrak{gl}_{n}$-module homomorphism.

The map $\psi\circ\eta:U\left(  \mathfrak{gl}_{n}\right)  \rightarrow
\mathcal{D}$ is a $\mathfrak{gl}_{n}$-module homomorphism (since it is the
composition of the $\mathfrak{gl}_{n}$-module homomorphisms $\eta:U\left(
\mathfrak{gl}_{n}\right)  \rightarrow\operatorname*{Sym}\left(
\operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ and
$\psi:\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  \rightarrow\mathcal{D}$). The map $\Omega$ is also
a $\mathfrak{gl}_{n}$-module homomorphism (by Corollary \ref{cor.nord.l}
\textbf{(b)}).

Now, we are going to prove that $\psi\circ\eta=\Omega$.

Indeed, let $z\in U\left(  \mathfrak{gl}_{n}\right)  $. We shall show that
$\left(  \psi\circ\eta\right)  \left(  z\right)  =\Omega\left(  z\right)  $.

The $\mathbf{k}$-modules $U\left(  \mathfrak{gl}_{n}\right)  $ and
$\mathcal{D}$ are $\mathfrak{gl}_{n}$-modules, and thus left $U\left(
\mathfrak{gl}_{n}\right)  $-modules (since every $\mathfrak{gl}_{n}$-module is
a left $U\left(  \mathfrak{gl}_{n}\right)  $-module). The maps $\psi\circ\eta$
and $\Omega$ are $\mathfrak{gl}_{n}$-module homomorphisms, and thus left
$U\left(  \mathfrak{gl}_{n}\right)  $-module homomorphisms (since every
$\mathfrak{gl}_{n}$-module homomorphism is a left $U\left(  \mathfrak{gl}%
_{n}\right)  $-module homomorphism).

The left $U\left(  \mathfrak{gl}_{n}\right)  $-module structure on $U\left(
\mathfrak{gl}_{n}\right)  $ is given by left multiplication. Hence,
$z\cdot1_{U\left(  \mathfrak{gl}_{n}\right)  }=z$. Now, $\psi\circ\eta$ is a
left $U\left(  \mathfrak{gl}_{n}\right)  $-module homomorphism. Hence,%
\[
\left(  \psi\circ\eta\right)  \left(  z\cdot1_{U\left(  \mathfrak{gl}%
_{n}\right)  }\right)  =z\rightharpoonup\underbrace{\left(  \psi\circ
\eta\right)  \left(  1_{U\left(  \mathfrak{gl}_{n}\right)  }\right)
}_{=1_{\mathcal{D}}}=z\rightharpoonup1_{\mathcal{D}}.
\]
Since $z\cdot1_{U\left(  \mathfrak{gl}_{n}\right)  }=z$, this rewrites as
$\left(  \psi\circ\eta\right)  \left(  z\right)  =z\rightharpoonup
1_{\mathcal{D}}$.

On the other hand, $\Omega$ is a left $U\left(  \mathfrak{gl}_{n}\right)
$-module homomorphism. Hence,
\[
\Omega\left(  z\cdot1_{U\left(  \mathfrak{gl}_{n}\right)  }\right)
=z\rightharpoonup\underbrace{\Omega\left(  1_{U\left(  \mathfrak{gl}%
_{n}\right)  }\right)  }_{=1_{\mathcal{D}}}=z\rightharpoonup1_{\mathcal{D}}.
\]
Since $z\cdot1_{U\left(  \mathfrak{gl}_{n}\right)  }=z$, this rewrites as
$\Omega\left(  z\right)  =z\rightharpoonup1_{\mathcal{D}}$. Comparing this
with $\left(  \psi\circ\eta\right)  \left(  z\right)  =z\rightharpoonup
1_{\mathcal{D}}$, we obtain $\left(  \psi\circ\eta\right)  \left(  z\right)
=\Omega\left(  z\right)  $.

Let us now forget that we fixed $z$. We thus have shown that $\left(
\psi\circ\eta\right)  \left(  z\right)  =\Omega\left(  z\right)  $ for every
$z\in U\left(  \mathfrak{gl}_{n}\right)  $. In other words, $\psi\circ
\eta=\Omega$.

\textit{Step 11:} Theorem \ref{thm.nord.box} \textbf{(b)} shows that the map
$\eta$ is a $\mathbf{k}$-bialgebra isomorphism $\left(  U\left(
\mathfrak{gl}_{n}\right)  ,\boxdot\right)  \rightarrow\operatorname*{Sym}%
\left(  \operatorname*{M}\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $.
In particular, the map $\eta$ is a $\mathbf{k}$-algebra isomorphism $\left(
U\left(  \mathfrak{gl}_{n}\right)  ,\boxdot\right)  \rightarrow
\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $. Hence, $\eta$ is a $\mathbf{k}$-algebra
homomorphism $\left(  U\left(  \mathfrak{gl}_{n}\right)  ,\boxdot\right)
\rightarrow\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  $.

The map $\psi\circ\eta$ is a $\mathbf{k}$-algebra homomorphism $\left(
U\left(  \mathfrak{gl}_{n}\right)  ,\boxdot\right)  \rightarrow\left(
\mathcal{D},\boxdot\right)  $ (since it is the composition of the $\mathbf{k}%
$-algebra homomorphisms $\eta:\left(  U\left(  \mathfrak{gl}_{n}\right)
,\boxdot\right)  \rightarrow\operatorname*{Sym}\left(  \operatorname*{M}%
\nolimits_{n}\left(  \mathbf{k}\right)  \right)  $ and $\psi
:\operatorname*{Sym}\left(  \operatorname*{M}\nolimits_{n}\left(
\mathbf{k}\right)  \right)  \rightarrow\left(  \mathcal{D},\boxdot\right)  $).
Since $\psi\circ\eta=\Omega$, we can rewrite this as follows: The map $\Omega$
is a $\mathbf{k}$-algebra homomorphism $\left(  U\left(  \mathfrak{gl}%
_{n}\right)  ,\boxdot\right)  \rightarrow\left(  \mathcal{D},\boxdot\right)
$. This proves Theorem \ref{thm.nord.box} \textbf{(h)}.
\end{proof}

Now that Theorem \ref{thm.nord.box} is proven, we can finally prove Theorem
\ref{thm.nord.box-exists}:

\begin{proof}
[Proof of Theorem \ref{thm.nord.box-exists}.]\textbf{(b)} Consider the
multiplication $\boxdot$ on the $\mathbf{k}$-module $U\left(  \mathfrak{gl}%
_{n}\right)  $ defined in Theorem \ref{thm.nord.box} \textbf{(a)}. This
multiplication is commutative (according to Theorem \ref{thm.nord.box}
\textbf{(a)}) and has the property that $\Omega$ is a $\mathbf{k}$-algebra
homomorphism $\left(  U\left(  \mathfrak{gl}_{n}\right)  ,\boxdot\right)
\rightarrow\left(  \mathcal{D},\boxdot\right)  $ (according to Theorem
\ref{thm.nord.box} \textbf{(h)}). Moreover, it is independent on $m$ (due to
its construction). Thus, Theorem \ref{thm.nord.box-exists} \textbf{(b)} is proven.

\textbf{(a)} Consider the multiplication $\boxdot$ on the $\mathbf{k}$-module
$U\left(  \mathfrak{gl}_{n}\right)  $ defined in Theorem \ref{thm.nord.box}
\textbf{(a)}. Theorem \ref{thm.nord.box} \textbf{(h)} shows that $\Omega$ is a
$\mathbf{k}$-algebra homomorphism $\left(  U\left(  \mathfrak{gl}_{n}\right)
,\boxdot\right)  \rightarrow\left(  \mathcal{D},\boxdot\right)  $. Hence,
$\Omega\left(  U\left(  \mathfrak{gl}_{n}\right)  \right)  $ is the image of
the $\mathbf{k}$-algebra $\left(  U\left(  \mathfrak{gl}_{n}\right)
,\boxdot\right)  $ under a $\mathbf{k}$-algebra homomorphism, and thus itself
a $\mathbf{k}$-subalgebra of $\left(  \mathcal{D},\boxdot\right)  $. This
proves Theorem \ref{thm.nord.box-exists} \textbf{(a)}.
\end{proof}

\section{\label{sect.bimod}Commutation of left and right actions}

Many of the results we have stated above were not \textquotedblleft left-right
symmetric\textquotedblright: For instance, Definition \ref{def.liemod.left}
involves a map $\mu:\mathfrak{g}\times V\rightarrow V$, whereas we could just
as well consider $\mathbf{k}$-modules $V$ equipped with a map $\mu
:V\times\mathfrak{g}\rightarrow\mathfrak{g}$. Theorem \ref{thm.gen1}
constructed a left $U\left(  \mathfrak{g}\right)  $-module structure on $C$,
but one could also ask for a right $U\left(  \mathfrak{g}\right)  $-module
structure. Theorem \ref{thm.manchon} concerns left pre-Lie algebras, but we
could just as well state an analogous result for right pre-Lie
algebras\footnote{Right pre-Lie algebras are a \textquotedblleft
symmetric\textquotedblright\ notion to that of left pre-Lie algebras; see
Proposition \ref{prop.prelie.left-right} and Proposition
\ref{prop.prelie.right-left}.}. And so on.

Stating and proving these analogues is a rather unexciting and straightforward
task; one essentially needs to \textquotedblleft turn all products
around\textquotedblright\ (i.e., replacing $ab$ by $ba$, replacing $\mu\left(
a,v\right)  $ by $\mu\left(  v,a\right)  $, replacing $a\vartriangleright b$
by $b\vartriangleleft a$, etc.). What is more interesting is to combine these
analogues with the original results and to see how they interact. As an
example of such an interaction, we can recall that an (associative)
$\mathbf{k}$-algebra $A$ can be regarded both as a left pre-Lie algebra
(according to Proposition \ref{prop.prelie.assoc} \textbf{(a)}) and as a right
pre-Lie algebra (according to Proposition \ref{prop.prelie.assoc}
\textbf{(b)}). The left pre-Lie algebra structure on $A$ allows us to apply
Theorem \ref{thm.manchon} to this $A$, thus obtaining a map $\eta:U\left(
A^{-}\right)  \rightarrow\operatorname*{Sym}A$. The right pre-Lie algebra
structure on $A$ allows us to apply Theorem \ref{thm.manchon.right} (the
straightforward analogue of Theorem \ref{thm.manchon} for right pre-Lie
algebras) to this $A$, thus obtaining a second map $\eta^{\prime}:U\left(
A^{-}\right)  \rightarrow\operatorname*{Sym}A$\ \ \ \ \footnote{The $A^{-}$
appearing in this map $\eta^{\prime}$ is the same as the $A^{-}$ appearing in
the map $\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$, because
of Remark \ref{rmk.prelie.A-}.}. How are these two maps $\eta$ and
$\eta^{\prime}$ related? We will see (in Theorem \ref{thm.manchon.bi} below)
that they are identical.

\subsection{Right $\mathfrak{g}$-modules}

Let us, however, return to a more general setting. In Definition
\ref{def.liemod.left}, we have defined the notion of \textquotedblleft%
$\mathfrak{g}$-modules\textquotedblright; we shall sometimes refer to them as
\textquotedblleft left $\mathfrak{g}$-modules\textquotedblright\ from now on.
Let us now define a notion of \textquotedblleft right $\mathfrak{g}%
$-modules\textquotedblright, which is a \textquotedblleft
right\textquotedblright\ analogue to the notion of $\mathfrak{g}$-modules
introduced in Definition \ref{def.liemod.left}:

\begin{definition}
\label{def.liemod.right}Let $\mathfrak{g}$ be a Lie algebra. Let $V$ be a
$\mathbf{k}$-module. Let $\mu:V\times\mathfrak{g}\rightarrow V$ be a
$\mathbf{k}$-bilinear map. We say that $\left(  V,\mu\right)  $ is a
\textit{right }$\mathfrak{g}$\textit{-module} if and only if
\begin{equation}
\left(  \mu\left(  v,\left[  a,b\right]  \right)  =\mu\left(  \mu\left(
v,a\right)  ,b\right)  -\mu\left(  \mu\left(  v,b\right)  ,a\right)  \text{
for every }a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in V\right)
. \label{eq.def.liemod.right.axiom}%
\end{equation}
If $\left(  V,\mu\right)  $ is a right $\mathfrak{g}$-module, then the
$\mathbf{k}$-bilinear map $\mu:V\times\mathfrak{g}\rightarrow V$ is called the
\textit{Lie action} of the right $\mathfrak{g}$-module $V$. If $\left(
V,\mu\right)  $ is a right $\mathfrak{g}$-module, then the $\mathbf{k}$-module
$V$ is called the \textit{underlying }$\mathbf{k}$\textit{-module} of $\left(
V,\mu\right)  $.
\end{definition}

Similarly to Convention \ref{conv.liemod.left}, we now make the following convention:

\begin{condition}
\label{conv.liemod.right}Let $\mathfrak{g}$ be a Lie algebra. Let $\left(
V,\mu\right)  $ be a right $\mathfrak{g}$-module.

\textbf{(a)} For any $a\in\mathfrak{g}$ and $v\in V$, we shall abbreviate the
term $\mu\left(  v,a\right)  $ by $v\leftharpoonup a$, provided that the map
$\mu$ is obvious from the context. Using this notation, the relation
(\ref{eq.def.liemod.right.axiom}) rewrites as%
\begin{equation}
\left(  v\leftharpoonup\left[  a,b\right]  =\left(  v\leftharpoonup a\right)
\leftharpoonup b-\left(  v\leftharpoonup b\right)  \leftharpoonup a\text{ for
every }a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in V\right)  .
\label{eq.conv.liemod.right}%
\end{equation}


\textbf{Convention on the precedence of the }$\leftharpoonup$\textbf{ sign:
}The symbol $\leftharpoonup$ is understood to have the same precedence as the
multiplication sign (i.e., it binds as strongly as the multiplication sign).
Thus, $v+w\leftharpoonup a$ means $v+\left(  w\leftharpoonup a\right)  $
rather than $\left(  v+w\right)  \leftharpoonup a$, but $v\cdot
w\leftharpoonup a$ is undefined (it could mean both $v\cdot\left(
w\leftharpoonup a\right)  $ and $\left(  v\cdot w\right)  \leftharpoonup a$).
Application of functions will be supposed to bind more strongly than the
$\rightharpoonup$ sign, so that $f\left(  v\right)  \leftharpoonup g\left(
w\right)  $ will mean $\left(  f\left(  v\right)  \right)  \leftharpoonup
\left(  g\left(  w\right)  \right)  $ (rather than $f\left(  v\leftharpoonup
g\left(  w\right)  \right)  $ or $\left(  f\left(  v\leftharpoonup g\right)
\right)  \left(  w\right)  $ or anything else), but we will often use
parentheses in this case to make the correct interpretation of the formula
even more obvious.

\textbf{(b)} We shall often refer to the right $\mathfrak{g}$-module $\left(
V,\mu\right)  $ as \textquotedblleft the $\mathbf{k}$-module $V$, endowed with
the right $\mathfrak{g}$-action $\mu$\textquotedblright\ (or by similar
formulations). We shall regard $\left(  V,\mu\right)  $ as the $\mathbf{k}%
$-module $V$ equipped with the additional data of the map $\mu$. Thus, when we
speak of \textquotedblleft elements of $\left(  V,\mu\right)  $%
\textquotedblright\ or \textquotedblleft maps to $\left(  V,\mu\right)
$\textquotedblright\ or \textquotedblleft$\mathbf{k}$-submodules of $\left(
V,\mu\right)  $\textquotedblright, we shall mean (respectively)
\textquotedblleft elements of $V$\textquotedblright, \textquotedblleft maps to
$V$\textquotedblright, or \textquotedblleft$\mathbf{k}$-submodules of
$V$\textquotedblright.

\textbf{(c)} By abuse of notation, we shall write \textquotedblleft$V$ is a
right $\mathfrak{g}$-module\textquotedblright\ instead of \textquotedblleft%
$\left(  V,\mu\right)  $ is a right $\mathfrak{g}$-module\textquotedblright%
\ when the map $\mu$ is clear from the context or has not been introduced yet.
\end{condition}

\begin{definition}
Let $\mathfrak{g}$ be a Lie algebra. Let $V$ be a $\mathbf{k}$-module. A
\textit{right }$\mathfrak{g}$\textit{-module structure on }$V$ means a map
$\mu:V\times\mathfrak{g}\rightarrow V$ such that $\left(  V,\mu\right)  $ is a
right $\mathfrak{g}$-module. In other words, a right $\mathfrak{g}$-module
structure on $V$ means a $\mathbf{k}$-bilinear map $\mu:V\times\mathfrak{g}%
\rightarrow V$ such that we have%
\[
\left(  v\leftharpoonup\left[  a,b\right]  =\left(  v\leftharpoonup a\right)
\leftharpoonup b-\left(  v\leftharpoonup b\right)  \leftharpoonup a\text{ for
every }a\in\mathfrak{g}\text{, }b\in\mathfrak{g}\text{ and }v\in V\right)  ,
\]
where we denote $\mu\left(  m,a\right)  $ by $m\leftharpoonup a$ for every
$a\in\mathfrak{g}$ and $m\in V$. Thus, a right $\mathfrak{g}$-module is the
same as a $\mathbf{k}$-module endowed with a right $\mathfrak{g}$-module structure.
\end{definition}

\begin{definition}
Let $\mathfrak{g}$ be a Lie algebra. Let $V$ be a right $\mathfrak{g}$-module.

\textbf{(a)} A \textit{right }$\mathfrak{g}$\textit{-submodule} of $V$ means a
right $\mathfrak{g}$-module $V^{\prime}$ such that the $\mathbf{k}$-module
$V^{\prime}$ is a submodule of the $\mathbf{k}$-module $V$, and such that the
Lie action of $V^{\prime}$ is the restriction of the Lie action of $V$ to
$V^{\prime}\times\mathfrak{g}$.

\textbf{(b)} Let $W$ be a further right $\mathfrak{g}$-module. A \textit{right
}$\mathfrak{g}$\textit{-module homomorphism from }$V$ \textit{to }$W$ means a
$\mathbf{k}$-linear map $f:V\rightarrow W$ satisfying%
\[
\left(  f\left(  v\right)  \leftharpoonup a=f\left(  v\leftharpoonup a\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }v\in
V\right)  .
\]
A \textit{right }$\mathfrak{g}$\textit{-module isomorphism from }$V$
\textit{to }$W$ means an invertible right $\mathfrak{g}$-module homomorphism
from $V$ to $W$ whose inverse is also a right $\mathfrak{g}$-module
homomorphism. It is easy to show that any invertible right $\mathfrak{g}%
$-module homomorphism is a $\mathfrak{g}$-module isomorphism.

Thus, we can define a category of right $\mathfrak{g}$-modules: Its objects
are right $\mathfrak{g}$-modules, and its morphisms are right $\mathfrak{g}%
$-module homomorphisms.
\end{definition}

You will not encounter right $\mathfrak{g}$-modules in the literature too
often (unlike right $A$-modules over a $\mathbf{k}$-algebra $A$). This has a
rather simple reason: They are in 1-to-1 correspondence with (left)
$\mathfrak{g}$-modules. Namely, the following holds:

\begin{remark}
\label{rmk.liemod.left-right}Let $\mathfrak{g}$ be a Lie algebra.

\textbf{(a)} If $V$ is a left $\mathfrak{g}$-module, then we can define a
right $\mathfrak{g}$-module structure on $V$ by%
\[
\left(  v\leftharpoonup a=-a\rightharpoonup v\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{ and }v\in V\right)  .
\]
Thus, every left $\mathfrak{g}$-module becomes a right $\mathfrak{g}$-module.

\textbf{(b)} If $V$ is a right $\mathfrak{g}$-module, then we can define a
right $\mathfrak{g}$-module structure on $V$ by%
\[
\left(  a\rightharpoonup v=-v\leftharpoonup a\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{ and }v\in V\right)  .
\]
Thus, every right $\mathfrak{g}$-module becomes a left $\mathfrak{g}$-module.

\textbf{(c)} If $V$ and $W$ are two left $\mathfrak{g}$-modules, and if
$f:V\rightarrow W$ is a left $\mathfrak{g}$-module homomorphism, then
$f:V\rightarrow W$ is also a right $\mathfrak{g}$-module homomorphism (where
$V$ and $W$ become right $\mathfrak{g}$-modules according to Remark
\ref{rmk.liemod.left-right} \textbf{(a)}). Thus, we obtain a functor from the
category of left $\mathfrak{g}$-modules to the category of right
$\mathfrak{g}$-modules.

\textbf{(d)} Similarly, we obtain a functor from the category of right
$\mathfrak{g}$-modules to the category of left $\mathfrak{g}$-modules.

\textbf{(e)} These two functors are mutually inverse.
\end{remark}

We leave the proof of Remark \ref{rmk.liemod.left-right} to the reader; it is
an easy consequence of the fact that any two elements $a$ and $b$ of a Lie
algebra $\mathfrak{g}$ satisfy $\left[  a,b\right]  =-\left[  b,a\right]  $.

Remark \ref{rmk.liemod.left-right} shows that (roughly speaking) everything
that can be said using the notion of right $\mathfrak{g}$-modules can also be
said without it. Nevertheless, we find this notion useful, because when we
will state analogues of our results above (e.g., of Theorem \ref{thm.manchon}%
), it will help to make the analogy obvious.

\subsection{Opposite algebras}

Next, we shall define some concepts which boil down to \textquotedblleft
turning around the product\textquotedblright\ in a $\mathbf{k}$-algebra, in a
Lie algebra and in a pre-Lie algebra:

\begin{definition}
\label{def.op.alg}Let $A$ be a $\mathbf{k}$-algebra. Then, we define
$A^{\operatorname*{op}}$ to be the $\mathbf{k}$-module $A$, equipped with the
multiplication $\ast:A\times A\rightarrow A$ (written in infix notation) which
is defined by%
\[
\left(  a\ast b=ba\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }b\in
A\right)  .
\]
It is easy to see that $A^{\operatorname*{op}}$ is again a $\mathbf{k}%
$-algebra, with the same unity as $A$. (Roughly speaking,
$A^{\operatorname*{op}}$ is the same $\mathbf{k}$-algebra as $A$, except that
the order of the factors in its multiplication has been turned around.) We
call $A^{\operatorname*{op}}$ the \textit{opposite algebra} of $A$.
\end{definition}

\begin{definition}
\label{def.op.lie}Let $\mathfrak{g}$ be a Lie algebra. Then, we define
$\mathfrak{g}^{\operatorname*{op}}$ to be the $\mathbf{k}$-module
$\mathfrak{g}$, equipped with the Lie bracket $\lambda:\mathfrak{g}%
\times\mathfrak{g}\rightarrow\mathfrak{g}$ which is defined by%
\[
\left(  \lambda\left(  a,b\right)  =\left[  b,a\right]
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }%
b\in\mathfrak{g}\right)  .
\]
It is easy to see that $\mathfrak{g}^{\operatorname*{op}}$ is again a Lie
algebra. (Roughly speaking, $\mathfrak{g}^{\operatorname*{op}}$ is the same
Lie algebra as $\mathfrak{g}$, except that the order of the arguments in its
Lie bracket has been turned around.) We call $\mathfrak{g}^{\operatorname*{op}%
}$ the \textit{opposite Lie algebra} of $\mathfrak{g}$.
\end{definition}

\begin{definition}
\label{def.op.prelie.left}Let $A$ be a left pre-Lie algebra. Then, we define
$A^{\operatorname*{op}}$ to be the $\mathbf{k}$-module $A$, equipped with the
map $\vartriangleleft:A\times A\rightarrow A$ (written in infix notation)
which is defined by%
\[
\left(  a\vartriangleleft b=b\vartriangleright a\ \ \ \ \ \ \ \ \ \ \text{for
all }a,b\in A\right)  .
\]
Proposition \ref{prop.prelie.left-right} shows that $A^{\operatorname*{op}}$
is a right pre-Lie algebra. We call $A^{\operatorname*{op}}$ the
\textit{opposite pre-Lie algebra} of $A$.
\end{definition}

\begin{definition}
\label{def.op.prelie.right}Let $A$ be a right pre-Lie algebra. Then, we define
$A^{\operatorname*{op}}$ to be the $\mathbf{k}$-module $A$, equipped with the
map $\vartriangleright:A\times A\rightarrow A$ (written in infix notation)
which is defined by%
\[
\left(  a\vartriangleright b=b\vartriangleleft a\ \ \ \ \ \ \ \ \ \ \text{for
all }a,b\in A\right)  .
\]
Proposition \ref{prop.prelie.right-left} shows that $A^{\operatorname*{op}}$
is a left pre-Lie algebra. We call $A^{\operatorname*{op}}$ the
\textit{opposite pre-Lie algebra} of $A$.
\end{definition}

The four definitions that we just made are similar to each other: Each of them
transforms an algebraic structure by switching the order of arguments in its
binary operation. (Similar operations exist for groups, monoids, etc..)
However, their properties differ. For example, every Lie algebra
$\mathfrak{g}$ is isomorphic to its opposite Lie algebra $\mathfrak{g}%
^{\operatorname*{op}}$:

\begin{remark}
\label{rmk.op.lie.iso}Let $\mathfrak{g}$ be a Lie algebra. Then, the map%
\[
\mathfrak{g}\rightarrow\mathfrak{g}^{\operatorname*{op}}%
,\ \ \ \ \ \ \ \ \ \ x\mapsto-x
\]
is a Lie algebra isomorphism.
\end{remark}

However, in general, a $\mathbf{k}$-algebra $A$ is not isomorphic to
$A^{\operatorname*{op}}$, and for pre-Lie algebras, such a statement would not
even make sense. Also, the opposite Lie algebra $\mathfrak{g}%
^{\operatorname*{op}}$ of a Lie algebra $\mathfrak{g}$ can be simply obtained
by multiplying the Lie bracket of $\mathfrak{g}$ by $-1$.

\begin{remark}
Let $A$ be a $\mathbf{k}$-algebra. Then, $\left(  A^{\operatorname*{op}%
}\right)  ^{-}=\left(  A^{-}\right)  ^{\operatorname*{op}}$.
\end{remark}

\begin{definition}
Let $V$ be a $\mathbf{k}$-module. We let $\operatorname*{End}%
\nolimits^{\operatorname*{op}}V$ be the $\mathbf{k}$-module
$\operatorname*{End}V$, equipped with the multiplication $\ast:\left(
\operatorname*{End}V\right)  \times\left(  \operatorname*{End}V\right)
\rightarrow\operatorname*{End}V$ (written in infix notation) which is defined
by%
\[
\left(  a\ast b=b\circ a\ \ \ \ \ \ \ \ \ \ \text{for every }a\in
\operatorname*{End}V\text{ and }b\in\operatorname*{End}V\right)  .
\]
This $\operatorname*{End}\nolimits^{\operatorname*{op}}V$ is a $\mathbf{k}%
$-algebra. Actually, $\operatorname*{End}\nolimits^{\operatorname*{op}%
}V=\left(  \operatorname*{End}V\right)  ^{\operatorname*{op}}$ as $\mathbf{k}$-algebras.
\end{definition}

We have introduced the special notation $\operatorname*{End}%
\nolimits^{\operatorname*{op}}V$ (instead of using $\left(
\operatorname*{End}V\right)  ^{\operatorname*{op}}$) because we want to stress
that $\operatorname*{End}\nolimits^{\operatorname*{op}}V$ is an analogue of
$\operatorname*{End}V$ (rather than just the opposite algebra of
$\operatorname*{End}V$). More precisely, $\operatorname*{End}%
\nolimits^{\operatorname*{op}}V$ is defined in the same way as
$\operatorname*{End}V$, with the only difference that the multiplication now
corresponds to the composition of maps \textquotedblleft in the opposite
order\textquotedblright. The $\mathbf{k}$-algebra $\operatorname*{End}%
\nolimits^{\operatorname*{op}}V$ plays the same role in regard to right
modules as the $\mathbf{k}$-algebra $\operatorname*{End}V$ plays in regard to
left modules; more precisely, we have the following two analogous propositions:

\begin{proposition}
\label{prop.EndV.leftmod}Let $V$ be a $\mathbf{k}$-module. Let $A$ be a
$\mathbf{k}$-algebra. Left $A$-module structures on $V$ are in a 1-to-1
correspondence with $\mathbf{k}$-algebra homomorphisms $A\rightarrow
\operatorname*{End}V$; this correspondence is defined as follows: Given a left
$A$-module structure on $V$, we can define a $\mathbf{k}$-algebra homomorphism
$\Phi:A\rightarrow\operatorname*{End}V$ by setting
\[
\left(  \left(  \Phi\left(  a\right)  \right)  \left(  v\right)
=av\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }v\in V\right)  .
\]
Conversely, given a $\mathbf{k}$-algebra homomorphism $\Phi:A\rightarrow
\operatorname*{End}V$, we can define a left $A$-module structure on $V$ by
setting%
\[
\left(  av=\left(  \Phi\left(  a\right)  \right)  \left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }v\in V\right)  .
\]

\end{proposition}

\begin{proposition}
\label{prop.EndV.rightmod}Let $V$ be a $\mathbf{k}$-module. Let $A$ be a
$\mathbf{k}$-algebra. Right $A$-module structures on $V$ are in a 1-to-1
correspondence with $\mathbf{k}$-algebra homomorphisms $A\rightarrow
\operatorname*{End}\nolimits^{\operatorname*{op}}V$; this correspondence is
defined as follows: Given a right $A$-module structure on $V$, we can define a
$\mathbf{k}$-algebra homomorphism $\Phi:A\rightarrow\operatorname*{End}%
\nolimits^{\operatorname*{op}}V$ by setting
\[
\left(  \left(  \Phi\left(  a\right)  \right)  \left(  v\right)
=va\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }v\in V\right)  .
\]
Conversely, given a $\mathbf{k}$-algebra homomorphism $\Phi:A\rightarrow
\operatorname*{End}\nolimits^{\operatorname*{op}}V$, we can define a right
$A$-module structure on $V$ by setting%
\[
\left(  va=\left(  \Phi\left(  a\right)  \right)  \left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in A\text{ and }v\in V\right)  .
\]

\end{proposition}

The analogy between Proposition \ref{prop.EndV.leftmod} and Proposition
\ref{prop.EndV.rightmod} would become even more obvious if we would write the
action of $\Phi\left(  a\right)  $ on the right of the $v$ in Proposition
\ref{prop.EndV.rightmod} (that is, if we would write $v\left(  \Phi\left(
a\right)  \right)  $ instead of $\left(  \Phi\left(  a\right)  \right)
\left(  v\right)  $); but we prefer to keep to the more standard notations.

We can now state the analogue of Definition \ref{def.U.mod} for right
$\mathfrak{g}$-modules:

\begin{definition}
\label{def.U.rightmod}Let $\mathfrak{g}$ be a Lie algebra.

\textbf{(a)} Every right $U\left(  \mathfrak{g}\right)  $-module $M$
canonically becomes a right $\mathfrak{g}$-module by setting%
\[
\left(  m\leftharpoonup a=m\iota_{U,\mathfrak{g}}\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathfrak{g}\text{ and }m\in M\right)
.
\]
Moreover, any right $U\left(  \mathfrak{g}\right)  $-module homomorphism
between two right $U\left(  \mathfrak{g}\right)  $-modules becomes a right
$\mathfrak{g}$-module homomorphism if we regard these right $U\left(
\mathfrak{g}\right)  $-modules as right $\mathfrak{g}$-modules. Thus, we
obtain a functor from the category of right $U\left(  \mathfrak{g}\right)
$-modules to the category of right $\mathfrak{g}$-modules.

\textbf{(b)} Every right $\mathfrak{g}$-module $M$ canonically becomes a right
$U\left(  \mathfrak{g}\right)  $-module. To define the right $U\left(
\mathfrak{g}\right)  $-module structure on $M$, we proceed as follows: Define
a map $\varphi:\mathfrak{g}\rightarrow\operatorname*{End}%
\nolimits^{\operatorname*{op}}M$ by%
\[
\left(  \left(  \varphi\left(  a\right)  \right)  \left(  m\right)
=m\leftharpoonup a\ \ \ \ \ \ \ \ \ \ \text{for all }a\in\mathfrak{g}\text{
and }m\in M\right)  .
\]
It is easy to see that this map $\varphi$ is a Lie algebra homomorphism from
$\mathfrak{g}$ to $\left(  \operatorname*{End}\nolimits^{\operatorname*{op}%
}M\right)  ^{-}$. (Indeed, this is a restatement of the axioms of a right
$\mathfrak{g}$-module; the fact that $\varphi\left(  \left[  a,b\right]
\right)  =\left[  \varphi\left(  a\right)  ,\varphi\left(  b\right)  \right]
$ for all $a,b\in\mathfrak{g}$ is equivalent to the relation
(\ref{eq.conv.liemod.right}).) Now, Theorem \ref{thm.U.univ} (applied to
$A=\operatorname*{End}\nolimits^{\operatorname*{op}}M$ and $f=\varphi$) shows
that there exists a unique $\mathbf{k}$-algebra homomorphism $F:U\left(
\mathfrak{g}\right)  \rightarrow\operatorname*{End}%
\nolimits^{\operatorname*{op}}M$ such that $\varphi=F\circ\iota
_{U,\mathfrak{g}}$. Consider this $F$. Now, we define a right $U\left(
\mathfrak{g}\right)  $-module structure on $M$ by%
\[
\left(  mp=\left(  F\left(  p\right)  \right)  \left(  m\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }p\in U\left(  \mathfrak{g}\right)  \text{
and }m\in M\right)  .
\]
Thus, every right $\mathfrak{g}$-module canonically becomes a right $U\left(
\mathfrak{g}\right)  $-module. Moreover, any right $\mathfrak{g}$-module
homomorphism between two right $\mathfrak{g}$-modules becomes a right
$U\left(  \mathfrak{g}\right)  $-module homomorphism if we regard these right
$\mathfrak{g}$-modules as right $U\left(  \mathfrak{g}\right)  $-modules.
Hence, we obtain a functor from the category of right $\mathfrak{g}$-modules
to the category of right $U\left(  \mathfrak{g}\right)  $-modules.

\textbf{(c)} In Definition \ref{def.U.rightmod} \textbf{(a)}, we have
constructed a functor from the category of right $U\left(  \mathfrak{g}%
\right)  $-modules to the category of right $\mathfrak{g}$-modules. In
Definition \ref{def.U.rightmod} \textbf{(b)}, we have constructed a functor
from the category of right $\mathfrak{g}$-modules to the category of right
$U\left(  \mathfrak{g}\right)  $-modules. These two functors are mutually
inverse. In particular, if $M$ is a right $\mathfrak{g}$-module, then the
right $U\left(  \mathfrak{g}\right)  $-module structure on $M$ obtained
according to Definition \ref{def.U.rightmod} \textbf{(b)} satisfies
\[
m\iota_{U,\mathfrak{g}}\left(  a\right)  =m\leftharpoonup
a\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }m\in M.
\]


\textbf{(d)} According to Definition \ref{def.U.rightmod} \textbf{(a)}, every
right $U\left(  \mathfrak{g}\right)  $-module canonically becomes a right
$\mathfrak{g}$-module. In particular, $U\left(  \mathfrak{g}\right)  $ itself
becomes a right $\mathfrak{g}$-module (because $U\left(  \mathfrak{g}\right)
$ is a right $U\left(  \mathfrak{g}\right)  $-module). This is the right
$\mathfrak{g}$-module structure on $U\left(  \mathfrak{g}\right)  $
\textquotedblleft given by right multiplication\textquotedblright\ (because it
satisfies $u\leftharpoonup x=u\iota_{U,\mathfrak{g}}\left(  x\right)  $ for
every $x\in\mathfrak{g}$ and $u\in U\left(  \mathfrak{g}\right)  $). Other
canonical right $\mathfrak{g}$-module structures on $U\left(  \mathfrak{g}%
\right)  $ exist as well, but we shall not use them for the time being.
\end{definition}

We furthermore define a \textquotedblleft right analogue\textquotedblright\ of
$\operatorname*{Der}C$:

\begin{definition}
\label{def.Der.right}Let $C$ be a $\mathbf{k}$-algebra. We can regard
$\operatorname*{Der}C$ not only as a subset of $\operatorname*{End}C$, but
also as a subset of $\operatorname*{End}\nolimits^{\operatorname*{op}}C$
(since $\operatorname*{End}\nolimits^{\operatorname*{op}}C=\operatorname*{End}%
C$ as sets). Then, $\operatorname*{Der}C$ is a Lie subalgebra of $\left(
\operatorname*{End}\nolimits^{\operatorname*{op}}C\right)  ^{-}$. (This is an
analogue of Proposition \ref{prop.Der.basics} \textbf{(a)}.) We define
$\operatorname*{Der}\nolimits^{\operatorname*{op}}C$ to be the Lie subalgebra
$\operatorname*{Der}C$ of $\left(  \operatorname*{End}%
\nolimits^{\operatorname*{op}}C\right)  ^{-}$. (Thus, $\operatorname*{Der}%
\nolimits^{\operatorname*{op}}C=\operatorname*{Der}C$ as sets and as
$\mathbf{k}$-modules, but $\operatorname*{Der}\nolimits^{\operatorname*{op}%
}C=\left(  \operatorname*{Der}C\right)  ^{\operatorname*{op}}$ as Lie algebras.)
\end{definition}

\subsection{Analogues of results on Lie algebra actions}

We can now state the analogue of Theorem \ref{thm.gen1} for right
$\mathfrak{g}$-modules:

\begin{theorem}
\label{thm.gen1.right}Let $\mathfrak{g}$ be a Lie algebra. Let $C$ be a
$\mathbf{k}$-algebra. Let $K^{\prime}:\mathfrak{g}\rightarrow
\operatorname*{Der}\nolimits^{\operatorname*{op}}C$ be a Lie algebra
homomorphism. Let $f^{\prime}:\mathfrak{g}\rightarrow C$ be a $\mathbf{k}%
$-linear map. Assume that%
\begin{equation}
f^{\prime}\left(  \left[  a,b\right]  \right)  =\left[  f^{\prime}\left(
a\right)  ,f^{\prime}\left(  b\right)  \right]  +\left(  K^{\prime}\left(
b\right)  \right)  \left(  f^{\prime}\left(  a\right)  \right)  -\left(
K^{\prime}\left(  a\right)  \right)  \left(  f^{\prime}\left(  b\right)
\right)  \label{eq.thm.gen1.right.axiom1}%
\end{equation}
for every $a\in\mathfrak{g}$ and $b\in\mathfrak{g}$ (where the Lie bracket
$\left[  f^{\prime}\left(  a\right)  ,f^{\prime}\left(  b\right)  \right]  $
is computed in the Lie algebra $C^{-}$).

\textbf{(a)} Then, we can define a right $\mathfrak{g}$-module structure on
$C$ by setting%
\begin{equation}
\left(  u\leftharpoonup a=u\cdot f^{\prime}\left(  a\right)  +\left(
K^{\prime}\left(  a\right)  \right)  \left(  u\right)  \text{ for all }%
a\in\mathfrak{g}\text{ and }u\in C\right)  . \label{eq.thm.gen1.right.a}%
\end{equation}


In the following, we will regard $C$ as a right $\mathfrak{g}$-module by means
of this right $\mathfrak{g}$-module structure.

\textbf{(b)} Being a right $\mathfrak{g}$-module, $C$ becomes a right
$U\left(  \mathfrak{g}\right)  $-module. Define a map $\eta^{\prime}:U\left(
\mathfrak{g}\right)  \rightarrow C$ by%
\[
\eta^{\prime}\left(  u\right)  =1_{C}u\ \ \ \ \ \ \ \ \ \ \text{for every
}u\in U\left(  \mathfrak{g}\right)  .
\]
Then, $\eta^{\prime}$ is a right $\mathfrak{g}$-module homomorphism.

\textbf{(c)} For every $a\in\mathfrak{g}$, $b\in C$ and $c\in C$, we have%
\[
\left(  cb\right)  \leftharpoonup a-\left(  c\leftharpoonup a\right)  \cdot
b=c\cdot\left(  K^{\prime}\left(  a\right)  \right)  \left(  b\right)
-c\left[  f^{\prime}\left(  a\right)  ,b\right]  .
\]
(Here, again, the Lie bracket $\left[  f^{\prime}\left(  a\right)  ,b\right]
$ is computed in the Lie algebra $C^{-}$.)
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.gen1.right}.]In order to prove Theorem
\ref{thm.gen1.right}, it suffices to follow the proof of Theorem
\ref{thm.gen1}, making only some minor changes. For example, \textquotedblleft%
$f$\textquotedblright, \textquotedblleft$K$\textquotedblright,
\textquotedblleft$\operatorname*{Der}C$\textquotedblright\ and
\textquotedblleft$\eta$\textquotedblright\ have to be replaced by
\textquotedblleft$f^{\prime}$\textquotedblright, \textquotedblleft$K^{\prime}%
$\textquotedblright, \textquotedblleft$\operatorname*{Der}%
\nolimits^{\operatorname*{op}}C$\textquotedblright\ and \textquotedblleft%
$\eta^{\prime}$\textquotedblright; some plus signs have to be changed to minus
signs and vice versa; and in various products the factors have to be
reordered. Thus, for example:

\begin{itemize}
\item The map $\mu:\mathfrak{g}\times C\rightarrow C$ in the proof of Theorem
\ref{thm.gen1} has to be replaced by a map $\mu:C\times\mathfrak{g}\rightarrow
C$.

\item The equality
\[
K\left(  \left[  a,b\right]  \right)  =\left[  K\left(  a\right)  ,K\left(
b\right)  \right]  =K\left(  a\right)  \circ K\left(  b\right)  -K\left(
b\right)  \circ K\left(  a\right)
\]
has to be replaced by%
\[
K^{\prime}\left(  \left[  a,b\right]  \right)  =\left[  K^{\prime}\left(
a\right)  ,K^{\prime}\left(  b\right)  \right]  =K^{\prime}\left(  b\right)
\circ K^{\prime}\left(  a\right)  -K^{\prime}\left(  a\right)  \circ
K^{\prime}\left(  b\right)
\]
(since $\left[  K^{\prime}\left(  a\right)  ,K^{\prime}\left(  b\right)
\right]  $ now has to be understood as a Lie bracket inside
$\operatorname*{Der}\nolimits^{\operatorname*{op}}C$ and not inside
$\operatorname*{Der}C$).
\end{itemize}
\end{proof}

Next, let us state the analogue of Theorem \ref{thm.gen3} for right
$\mathfrak{g}$-modules:

\begin{theorem}
\label{thm.gen3.right}Let $\mathfrak{g}$ be a Lie algebra. Let $C$ be a
$\mathbf{k}$-bialgebra. Let $K^{\prime}:\mathfrak{g}\rightarrow
\operatorname*{Der}\nolimits^{\operatorname*{op}}C$ be a Lie algebra
homomorphism such that $K^{\prime}\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Coder}C$. (We are regarding $\operatorname*{Coder}C$ as a set
here, whence we do not have to replace it by its \textquotedblleft right
analogue\textquotedblright.) Let $f^{\prime}:\mathfrak{g}\rightarrow C$ be a
$\mathbf{k}$-linear map such that $f^{\prime}\left(  \mathfrak{g}\right)
\subseteq\operatorname*{Prim}C$. Assume that (\ref{eq.thm.gen1.right.axiom1})
holds for every $a\in\mathfrak{g}$ and $b\in\mathfrak{g}$.

Consider the right $\mathfrak{g}$-module structure on $C$ defined in Theorem
\ref{thm.gen1.right} \textbf{(a)}, and the map $\eta^{\prime}:U\left(
\mathfrak{g}\right)  \rightarrow C$ defined in Theorem \ref{thm.gen1.right}
\textbf{(b)}.

\textbf{(a)} For every $a\in\mathfrak{g}$, the map $C\rightarrow C,\ c\mapsto
c\leftharpoonup a$ is a coderivation of $C$.

\textbf{(b)} The map $\eta^{\prime}:U\left(  \mathfrak{g}\right)  \rightarrow
C$ is a $\mathbf{k}$-coalgebra homomorphism.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.gen3.right}.]In order to prove Theorem
\ref{thm.gen3.right}, it suffices to follow the proof of Theorem
\ref{thm.gen3}, making only some minor changes. For example, \textquotedblleft%
$f$\textquotedblright, \textquotedblleft$K$\textquotedblright,
\textquotedblleft$\operatorname*{Der}C$\textquotedblright\ and
\textquotedblleft$\eta$\textquotedblright\ have to be replaced by
\textquotedblleft$f^{\prime}$\textquotedblright, \textquotedblleft$K^{\prime}%
$\textquotedblright, \textquotedblleft$\operatorname*{Der}%
\nolimits^{\operatorname*{op}}C$\textquotedblright\ and \textquotedblleft%
$\eta^{\prime}$\textquotedblright; some plus signs have to be changed to minus
signs and vice versa; and in various products the factors have to be
reordered. Here are some concrete examples of what these changes mean:

\begin{itemize}
\item The map $\zeta$, which was defined as the map $C\rightarrow C,\ c\mapsto
a\rightharpoonup c$, must now be defined as the map $C\rightarrow C,\ c\mapsto
c\leftharpoonup a$ instead.

\item The equality (\ref{pf.thm.gen3.a.1}) has to be replaced by%
\[
\left(  \Delta\circ\zeta\right)  \left(  u\right)  =\Delta\left(  u\right)
\cdot\left(  f^{\prime}\left(  a\right)  \otimes1+1\otimes f^{\prime}\left(
a\right)  \right)  +\left(  K^{\prime}\left(  a\right)  \otimes
\operatorname*{id}+\operatorname*{id}\otimes K^{\prime}\left(  a\right)
\right)  \left(  \Delta\left(  u\right)  \right)  .
\]


\item The equality (\ref{pf.thm.gen3.a.2}) has to be replaced by%
\[
\left(  \zeta\otimes\operatorname*{id}+\operatorname*{id}\otimes\zeta\right)
\left(  p\right)  =p\cdot\left(  f^{\prime}\left(  a\right)  \otimes1+1\otimes
f^{\prime}\left(  a\right)  \right)  +\left(  K^{\prime}\left(  a\right)
\otimes\operatorname*{id}+\operatorname*{id}\otimes K^{\prime}\left(
a\right)  \right)  \left(  p\right)  .
\]


\item The $\mathbf{k}$-algebra homomorphism $\Xi:U\left(  \mathfrak{g}\right)
\rightarrow\operatorname*{End}C$ has to be replaced by a $\mathbf{k}$-algebra
homomorphism $\Xi:U\left(  \mathfrak{g}\right)  \rightarrow\operatorname*{End}%
\nolimits^{\operatorname*{op}}C$ defined by setting%
\begin{equation}
\left(  \left(  \Xi\left(  p\right)  \right)  \left(  c\right)
=cp\ \ \ \ \ \ \ \ \ \ \text{for every }p\in U\left(  \mathfrak{g}\right)
\text{ and }c\in C\right)  . \label{pf.thm.gen3.right.Xi}%
\end{equation}


\item The map $\mathbf{z}$ needs not be changed, but it now has to be
considered as a $\mathbf{k}$-algebra homomorphism from $\operatorname*{End}%
\nolimits^{\operatorname*{op}}C\otimes\operatorname*{End}%
\nolimits^{\operatorname*{op}}C$ to $\operatorname*{End}%
\nolimits^{\operatorname*{op}}\left(  C\otimes C\right)  $.

\item The map $\Xi^{\prime}$ is defined in the same manner as in the proof of
Theorem \ref{thm.gen3}, but now using the new map $\Xi$ (defined by
(\ref{pf.thm.gen3.right.Xi})); thus, this map $\Xi^{\prime}$ will be a
$\mathbf{k}$-algebra homomorphism $U\left(  \mathfrak{g}\right)
\rightarrow\operatorname*{End}\nolimits^{\operatorname*{op}}\left(  C\otimes
C\right)  $.

\item The proof of the fact that $\mathfrak{P}$ is a $\mathbf{k}$-subalgebra
of $U\left(  \mathfrak{g}\right)  $ must be modified: For example,
\textquotedblleft$\Xi\left(  ab\right)  =\Xi\left(  a\right)  \circ\Xi\left(
b\right)  $\textquotedblright\ has to be replaced by \textquotedblleft%
$\Xi\left(  ab\right)  =\Xi\left(  b\right)  \circ\Xi\left(  a\right)
$\textquotedblright\ (since $\Xi$ is no longer a $\mathbf{k}$-algebra
homomorphism from $U\left(  \mathfrak{g}\right)  $ to $\operatorname*{End}C$,
but now is a $\mathbf{k}$-algebra homomorphism from $U\left(  \mathfrak{g}%
\right)  $ to $\operatorname*{End}\nolimits^{\operatorname*{op}}C$); similarly
for $\Xi^{\prime}$.

\item The proof of $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)
\subseteq\mathfrak{P}$ has to be modified as well: For example,
\textquotedblleft$\left(  \Xi\left(  \iota_{U,\mathfrak{g}}\left(  a\right)
\right)  \right)  \left(  c\right)  =\iota_{U,\mathfrak{g}}\left(  a\right)
c=a\rightharpoonup c=\zeta\left(  c\right)  $\textquotedblright\ now must be
replaced by \textquotedblleft$\left(  \Xi\left(  \iota_{U,\mathfrak{g}}\left(
a\right)  \right)  \right)  \left(  c\right)  =c\iota_{U,\mathfrak{g}}\left(
a\right)  =c\leftharpoonup a=\zeta\left(  c\right)  $\textquotedblright.

\item The proof of the fact that $\mathfrak{R}$ is a $\mathbf{k}$-subalgebra
of $U\left(  \mathfrak{g}\right)  $ must be modified (similarly to the proof
of the fact that $\mathfrak{P}$ is a $\mathbf{k}$-subalgebra of $U\left(
\mathfrak{g}\right)  $).

\item The proof of $\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)
\subseteq\mathfrak{R}$ must be modified (similarly to the proof of
$\iota_{U,\mathfrak{g}}\left(  \mathfrak{g}\right)  \subseteq\mathfrak{P}$).
\end{itemize}

The reader should be able to do all these changes without trouble.
\end{proof}

Furthermore, let us state the analogue of Theorem \ref{thm.gen1filt}:

\begin{theorem}
\label{thm.gen1filt.right}Let $\mathfrak{g}$ be a Lie algebra. Let $C$ be a
filtered $\mathbf{k}$-algebra. Let $\left(  C_{n}\right)  _{n\geq0}$ be the
filtration of $C$. Set $C_{-1}=0$.

Let $K^{\prime}:\mathfrak{g}\rightarrow\operatorname*{Der}%
\nolimits^{\operatorname*{op}}C$ be a Lie algebra homomorphism. Let
$f^{\prime}:\mathfrak{g}\rightarrow C$ be a $\mathbf{k}$-linear map. Assume
that (\ref{eq.thm.gen1.right.axiom1}) holds for every $a\in\mathfrak{g}$ and
$b\in\mathfrak{g}$ (where the Lie bracket $\left[  f^{\prime}\left(  a\right)
,f^{\prime}\left(  b\right)  \right]  $ is computed in the Lie algebra $C^{-}%
$). Define a right $\mathfrak{g}$-module structure on $C$ as in Theorem
\ref{thm.gen1.right} \textbf{(a)}. Define a map $\eta^{\prime}:U\left(
\mathfrak{g}\right)  \rightarrow C$ as in Theorem \ref{thm.gen1.right}
\textbf{(b)}.

Assume furthermore that $f^{\prime}\left(  \mathfrak{g}\right)  \subseteq
C_{1}$. Also, assume that the map $K^{\prime}\left(  a\right)  :C\rightarrow
C$ is filtered for every $a\in\mathfrak{g}$.

Then,%
\begin{equation}
\eta^{\prime}\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{n}\right)  \right)  \in f^{\prime}\left(  a_{1}\right)  f^{\prime
}\left(  a_{2}\right)  \cdots f^{\prime}\left(  a_{n}\right)  +C_{n-1}
\label{eq.thm.gen1filt.right.claim}%
\end{equation}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.gen1filt.right}.]We shall first show a slightly
restated version of the claim of Theorem \ref{thm.gen1filt.right}:

\begin{statement}
\textit{Claim 1:} We have
\[
\eta^{\prime}\left(  \iota_{U,\mathfrak{g}}\left(  a_{n}\right)
\iota_{U,\mathfrak{g}}\left(  a_{n-1}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{1}\right)  \right)  \in f^{\prime}\left(  a_{n}\right)  f^{\prime
}\left(  a_{n-1}\right)  \cdots f^{\prime}\left(  a_{1}\right)  +C_{n-1}%
\]
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$.
\end{statement}

\textit{Proof of Claim 1.} In order to prove Claim 1, it suffices to follow
the proof of Theorem \ref{thm.gen1filt}, making only some minor changes. For
example, \textquotedblleft$f$\textquotedblright, \textquotedblleft%
$K$\textquotedblright, \textquotedblleft$\operatorname*{Der}C$%
\textquotedblright\ and \textquotedblleft$\eta$\textquotedblright\ have to be
replaced by \textquotedblleft$f^{\prime}$\textquotedblright, \textquotedblleft%
$K^{\prime}$\textquotedblright, \textquotedblleft$\operatorname*{Der}%
\nolimits^{\operatorname*{op}}C$\textquotedblright\ and \textquotedblleft%
$\eta^{\prime}$\textquotedblright; some plus signs have to be changed to minus
signs and vice versa; and in various products the factors have to be
reordered. Here are some concrete examples of what these changes mean:

\begin{itemize}
\item The equality (\ref{pf.thm.gen1filt.indhyp}) has to be replaced by%
\begin{equation}
\eta^{\prime}\left(  \iota_{U,\mathfrak{g}}\left(  a_{N-1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{N-2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{1}\right)  \right)  \in f^{\prime}\left(  a_{N-1}\right)  f\left(
a_{N-2}\right)  \cdots f\left(  a_{1}\right)  +C_{\left(  N-1\right)  -1}.
\label{pf.thm.gen1filt.right.indhyp-applied}%
\end{equation}


\item The element $g$ has to be replaced by the element $\eta^{\prime}\left(
\iota_{U,\mathfrak{g}}\left(  a_{N}\right)  \iota_{U,\mathfrak{g}}\left(
a_{N-1}\right)  \cdots\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \right)  $.

\item The product $f\left(  a_{2}\right)  f\left(  a_{3}\right)  \cdots
f\left(  a_{N}\right)  $ has to be replaced by $f^{\prime}\left(
a_{N}\right)  f^{\prime}\left(  a_{N-1}\right)  \cdots f^{\prime}\left(
a_{2}\right)  $.

\item Instead of multiplying both sides of the relation
(\ref{pf.thm.gen1filt.indhyp-applied}) with $f\left(  a_{1}\right)  $ from the
left, we need to multiply both sides of the relation
(\ref{pf.thm.gen1filt.right.indhyp-applied}) with $f\left(  a_{1}\right)  $
from the right.

\item The equality (\ref{pf.thm.gen1filt.5}) has to be replaced by%
\[
g=\eta^{\prime}\left(  \iota_{U,\mathfrak{g}}\left(  a_{N}\right)
\iota_{U,\mathfrak{g}}\left(  a_{N-1}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{2}\right)  \right)  =1_{C}\left(  \iota_{U,\mathfrak{g}}\left(
a_{N}\right)  \iota_{U,\mathfrak{g}}\left(  a_{N-1}\right)  \cdots
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \right)  .
\]

\end{itemize}

Thus, Claim 1 is proven.

Now, let $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots,a_{n}\in\mathfrak{g}$. Then,
Claim 1 (applied to $a_{n},a_{n-1},\ldots,a_{1}$ instead of $a_{1}%
,a_{2},\ldots,a_{n}$) shows that%
\[
\eta^{\prime}\left(  \iota_{U,\mathfrak{g}}\left(  a_{1}\right)
\iota_{U,\mathfrak{g}}\left(  a_{2}\right)  \cdots\iota_{U,\mathfrak{g}%
}\left(  a_{n}\right)  \right)  \in f^{\prime}\left(  a_{1}\right)  f^{\prime
}\left(  a_{2}\right)  \cdots f^{\prime}\left(  a_{n}\right)  +C_{n-1}.
\]
This proves Theorem \ref{thm.gen1filt.right}.
\end{proof}

\subsection{A theorem connecting left and right actions}

As we have said, what we have done so far in Section \ref{sect.bimod} was
rather unexciting: We have just stated analogues of previous results; the
proofs of these analogues were mostly analogous to the proofs of those latter
results. Now, however, we shall connect the analogues with the original
results, to obtain something new:

\begin{theorem}
\label{thm.gen5}Let $\mathfrak{g}$ be a Lie algebra. Let $C$ be a $\mathbf{k}$-algebra.

Let $K:\mathfrak{g}\rightarrow\operatorname*{Der}C$ be a Lie algebra
homomorphism. Let $f:\mathfrak{g}\rightarrow C$ be a $\mathbf{k}$-linear map.
Assume that (\ref{eq.thm.gen1.axiom1}) holds for every $a\in\mathfrak{g}$ and
$b\in\mathfrak{g}$ (where the Lie bracket $\left[  f\left(  a\right)
,f\left(  b\right)  \right]  $ is computed in the Lie algebra $C^{-}$).
Theorem \ref{thm.gen1} \textbf{(a)} shows that we can define a $\mathfrak{g}%
$-module structure on $C$ by setting%
\[
\left(  a\rightharpoonup u=f\left(  a\right)  \cdot u+\left(  K\left(
a\right)  \right)  \left(  u\right)  \text{ for all }a\in\mathfrak{g}\text{
and }u\in C\right)  .
\]
In the following, we will regard $C$ as a $\mathfrak{g}$-module by means of
this $\mathfrak{g}$-module structure.

Let $K^{\prime}:\mathfrak{g}\rightarrow\operatorname*{Der}%
\nolimits^{\operatorname*{op}}C$ be a Lie algebra homomorphism. Let
$f^{\prime}:\mathfrak{g}\rightarrow C$ be a $\mathbf{k}$-linear map. Assume
that (\ref{eq.thm.gen1.right.axiom1}) holds for every $a\in\mathfrak{g}$ and
$b\in\mathfrak{g}$ (where the Lie bracket $\left[  f^{\prime}\left(  a\right)
,f^{\prime}\left(  b\right)  \right]  $ is computed in the Lie algebra $C^{-}%
$). Theorem \ref{thm.gen1.right} \textbf{(a)} shows that we can define a right
$\mathfrak{g}$-module structure on $C$ by setting%
\[
\left(  u\leftharpoonup a=u\cdot f^{\prime}\left(  a\right)  +\left(
K^{\prime}\left(  a\right)  \right)  \left(  u\right)  \text{ for all }%
a\in\mathfrak{g}\text{ and }u\in C\right)  .
\]
In the following, we will regard $C$ as a right $\mathfrak{g}$-module by means
of this right $\mathfrak{g}$-module structure.

Furthermore, assume that%
\begin{equation}
\left(  K\left(  a\right)  \circ K^{\prime}\left(  b\right)  -K^{\prime
}\left(  b\right)  \circ K\left(  a\right)  \right)  \left(  u\right)
=\left(  K^{\prime}\left(  b\right)  \right)  \left(  f\left(  a\right)
\right)  \cdot u-u\cdot\left(  K\left(  a\right)  \right)  \left(  f^{\prime
}\left(  b\right)  \right)  \label{eq.thm.gen5.cond1}%
\end{equation}
for all $a\in\mathfrak{g}$, $b\in\mathfrak{g}$ and $u\in C$. (Here,
\textquotedblleft$K\left(  a\right)  \circ K^{\prime}\left(  b\right)
$\textquotedblright\ has to be read as $\left(  K\left(  a\right)  \right)
\circ\left(  K^{\prime}\left(  b\right)  \right)  $, and similarly for the
expression \textquotedblleft$K^{\prime}\left(  b\right)  \circ K\left(
a\right)  $\textquotedblright.)

\textbf{(a)} The left $\mathfrak{g}$-module structure and the right
$\mathfrak{g}$-module structure defined on $C$ satisfy%
\[
a\rightharpoonup\left(  u\leftharpoonup b\right)  =\left(  a\rightharpoonup
u\right)  \leftharpoonup b
\]
for all $a\in\mathfrak{g}$, $b\in\mathfrak{g}$ and $u\in C$.

\textbf{(b)} The $\mathbf{k}$-module $C$ is a left $\mathfrak{g}$-module, and
thus becomes a left $U\left(  \mathfrak{g}\right)  $-module. Also, the
$\mathbf{k}$-module $C$ is a right $\mathfrak{g}$-module, and thus becomes a
right $U\left(  \mathfrak{g}\right)  $-module.

The left $U\left(  \mathfrak{g}\right)  $-module structure on $C$ and the
right $U\left(  \mathfrak{g}\right)  $-module structure on $C$ can be combined
to form a $\left(  U\left(  \mathfrak{g}\right)  ,U\left(  \mathfrak{g}%
\right)  \right)  $-bimodule structure on $C$.

\textbf{(c)} Assume further that $f=f^{\prime}$.

Consider the map $\eta:U\left(  \mathfrak{g}\right)  \rightarrow C$ defined in
Theorem \ref{thm.gen1} \textbf{(b)}. Consider the map $\eta^{\prime}:U\left(
\mathfrak{g}\right)  \rightarrow C$ defined in Theorem \ref{thm.gen1.right}
\textbf{(b)}. Then, $\eta=\eta^{\prime}$.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.gen5}.]\textbf{(a)} Let $a\in\mathfrak{g}$,
$b\in\mathfrak{g}$ and $u\in C$. We have $K\left(  \underbrace{a}%
_{\in\mathfrak{g}}\right)  \in K\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Der}C$. In other words, $K\left(  a\right)  :C\rightarrow C$ is
a derivation. Also, $K^{\prime}\left(  \underbrace{b}_{\in\mathfrak{g}%
}\right)  \in K^{\prime}\left(  \mathfrak{g}\right)  \subseteq
\operatorname*{Der}\nolimits^{\operatorname*{op}}C=\operatorname*{Der}C$. In
other words, $K^{\prime}\left(  b\right)  :C\rightarrow C$ is a derivation.
Now, subtracting the equality%
\begin{align*}
&  a\rightharpoonup\underbrace{\left(  u\leftharpoonup b\right)
}_{\substack{=u\cdot f^{\prime}\left(  b\right)  +\left(  K^{\prime}\left(
b\right)  \right)  \left(  u\right)  \\\text{(by the definition of the
right}\\\mathfrak{g}\text{-module structure on }C\text{)}}}\\
&  =a\rightharpoonup\left(  u\cdot f^{\prime}\left(  b\right)  +\left(
K^{\prime}\left(  b\right)  \right)  \left(  u\right)  \right) \\
&  =\underbrace{a\rightharpoonup\left(  u\cdot f^{\prime}\left(  b\right)
\right)  }_{\substack{=f\left(  a\right)  \cdot u\cdot f^{\prime}\left(
b\right)  +\left(  K\left(  a\right)  \right)  \left(  u\cdot f^{\prime
}\left(  b\right)  \right)  \\\text{(by the definition of the left}%
\\\mathfrak{g}\text{-module structure on }C\text{)}}%
}+\underbrace{a\rightharpoonup\left(  K^{\prime}\left(  b\right)  \right)
\left(  u\right)  }_{\substack{=f\left(  a\right)  \cdot\left(  K^{\prime
}\left(  b\right)  \right)  \left(  u\right)  +\left(  K\left(  a\right)
\right)  \left(  \left(  K^{\prime}\left(  b\right)  \right)  \left(
u\right)  \right)  \\\text{(by the definition of the left}\\\mathfrak{g}%
\text{-module structure on }C\text{)}}}\\
&  =f\left(  a\right)  \cdot u\cdot f^{\prime}\left(  b\right)
+\underbrace{\left(  K\left(  a\right)  \right)  \left(  u\cdot f^{\prime
}\left(  b\right)  \right)  }_{\substack{=u\cdot\left(  K\left(  a\right)
\right)  \left(  f^{\prime}\left(  b\right)  \right)  +\left(  K\left(
a\right)  \right)  \left(  u\right)  \cdot f^{\prime}\left(  b\right)
\\\text{(since }K\left(  a\right)  \text{ is a derivation)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +f\left(  a\right)  \cdot\left(  K^{\prime}\left(
b\right)  \right)  \left(  u\right)  +\underbrace{\left(  K\left(  a\right)
\right)  \left(  \left(  K^{\prime}\left(  b\right)  \right)  \left(
u\right)  \right)  }_{=\left(  K\left(  a\right)  \circ K^{\prime}\left(
b\right)  \right)  \left(  u\right)  }\\
&  =f\left(  a\right)  \cdot u\cdot f^{\prime}\left(  b\right)  +u\cdot\left(
K\left(  a\right)  \right)  \left(  f^{\prime}\left(  b\right)  \right)
+\left(  K\left(  a\right)  \right)  \left(  u\right)  \cdot f^{\prime}\left(
b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +f\left(  a\right)  \cdot\left(  K^{\prime}\left(
b\right)  \right)  \left(  u\right)  +\left(  K\left(  a\right)  \circ
K^{\prime}\left(  b\right)  \right)  \left(  u\right)
\end{align*}
from the equality%
\begin{align*}
&  \underbrace{\left(  a\rightharpoonup u\right)  }_{\substack{=f\left(
a\right)  \cdot u+\left(  K\left(  a\right)  \right)  \left(  u\right)
\\\text{(by the definition of the left}\\\mathfrak{g}\text{-module structure
on }C\text{)}}}\leftharpoonup b\\
&  =\left(  f\left(  a\right)  \cdot u+\left(  K\left(  a\right)  \right)
\left(  u\right)  \right)  \leftharpoonup b\\
&  =\underbrace{\left(  f\left(  a\right)  \cdot u\right)  \leftharpoonup
b}_{\substack{=f\left(  a\right)  \cdot u\cdot f^{\prime}\left(  b\right)
+\left(  K^{\prime}\left(  b\right)  \right)  \left(  f\left(  a\right)  \cdot
u\right)  \\\text{(by the definition of the right}\\\mathfrak{g}\text{-module
structure on }C\text{)}}}+\underbrace{\left(  K\left(  a\right)  \right)
\left(  u\right)  \leftharpoonup b}_{\substack{=\left(  K\left(  a\right)
\right)  \left(  u\right)  \cdot f^{\prime}\left(  b\right)  +\left(
K^{\prime}\left(  b\right)  \right)  \left(  \left(  K\left(  a\right)
\right)  \left(  u\right)  \right)  \\\text{(by the definition of the
right}\\\mathfrak{g}\text{-module structure on }C\text{)}}}\\
&  =f\left(  a\right)  \cdot u\cdot f^{\prime}\left(  b\right)
+\underbrace{\left(  K^{\prime}\left(  b\right)  \right)  \left(  f\left(
a\right)  \cdot u\right)  }_{\substack{=f\left(  a\right)  \cdot\left(
K^{\prime}\left(  b\right)  \right)  \left(  u\right)  +\left(  K^{\prime
}\left(  b\right)  \right)  \left(  f\left(  a\right)  \right)  \cdot
u\\\text{(since }K^{\prime}\left(  b\right)  \text{ is a derivation)}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  K\left(  a\right)  \right)  \left(  u\right)
\cdot f^{\prime}\left(  b\right)  +\underbrace{\left(  K^{\prime}\left(
b\right)  \right)  \left(  \left(  K\left(  a\right)  \right)  \left(
u\right)  \right)  }_{=\left(  K^{\prime}\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  u\right)  }\\
&  =f\left(  a\right)  \cdot u\cdot f^{\prime}\left(  b\right)  +f\left(
a\right)  \cdot\left(  K^{\prime}\left(  b\right)  \right)  \left(  u\right)
+\left(  K^{\prime}\left(  b\right)  \right)  \left(  f\left(  a\right)
\right)  \cdot u\\
&  \ \ \ \ \ \ \ \ \ \ +\left(  K\left(  a\right)  \right)  \left(  u\right)
\cdot f^{\prime}\left(  b\right)  +\left(  K^{\prime}\left(  b\right)  \circ
K\left(  a\right)  \right)  \left(  u\right)  ,
\end{align*}
we obtain%
\begin{align*}
&  \left(  a\rightharpoonup u\right)  \leftharpoonup b-a\rightharpoonup\left(
u\leftharpoonup b\right) \\
&  =\left(  f\left(  a\right)  \cdot u\cdot f^{\prime}\left(  b\right)
+f\left(  a\right)  \cdot\left(  K^{\prime}\left(  b\right)  \right)  \left(
u\right)  +\left(  K^{\prime}\left(  b\right)  \right)  \left(  f\left(
a\right)  \right)  \cdot u\right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  +\left(  K\left(  a\right)
\right)  \left(  u\right)  \cdot f^{\prime}\left(  b\right)  +\left(
K^{\prime}\left(  b\right)  \circ K\left(  a\right)  \right)  \left(
u\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  f\left(  a\right)  \cdot u\cdot f^{\prime
}\left(  b\right)  +u\cdot\left(  K\left(  a\right)  \right)  \left(
f^{\prime}\left(  b\right)  \right)  +\left(  K\left(  a\right)  \right)
\left(  u\right)  \cdot f^{\prime}\left(  b\right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  +f\left(  a\right)
\cdot\left(  K^{\prime}\left(  b\right)  \right)  \left(  u\right)  +\left(
K\left(  a\right)  \circ K^{\prime}\left(  b\right)  \right)  \left(
u\right)  \right) \\
&  =\left(  K^{\prime}\left(  b\right)  \right)  \left(  f\left(  a\right)
\right)  \cdot u+\left(  K^{\prime}\left(  b\right)  \circ K\left(  a\right)
\right)  \left(  u\right)  -u\cdot\left(  K\left(  a\right)  \right)  \left(
f^{\prime}\left(  b\right)  \right)  -\left(  K\left(  a\right)  \circ
K^{\prime}\left(  b\right)  \right)  \left(  u\right) \\
&  =\underbrace{\left(  K^{\prime}\left(  b\right)  \right)  \left(  f\left(
a\right)  \right)  \cdot u-u\cdot\left(  K\left(  a\right)  \right)  \left(
f^{\prime}\left(  b\right)  \right)  }_{\substack{=\left(  K\left(  a\right)
\circ K^{\prime}\left(  b\right)  -K^{\prime}\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  u\right)  \\\text{(by (\ref{eq.thm.gen5.cond1}))}%
}}+\left(  K^{\prime}\left(  b\right)  \circ K\left(  a\right)  \right)
\left(  u\right)  -\left(  K\left(  a\right)  \circ K^{\prime}\left(
b\right)  \right)  \left(  u\right) \\
&  =\underbrace{\left(  K\left(  a\right)  \circ K^{\prime}\left(  b\right)
-K^{\prime}\left(  b\right)  \circ K\left(  a\right)  \right)  \left(
u\right)  }_{=\left(  K\left(  a\right)  \circ K^{\prime}\left(  b\right)
\right)  \left(  u\right)  -\left(  K^{\prime}\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  u\right)  }+\left(  K^{\prime}\left(  b\right)
\circ K\left(  a\right)  \right)  \left(  u\right)  -\left(  K\left(
a\right)  \circ K^{\prime}\left(  b\right)  \right)  \left(  u\right) \\
&  =\left(  K\left(  a\right)  \circ K^{\prime}\left(  b\right)  \right)
\left(  u\right)  -\left(  K^{\prime}\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  u\right)  +\left(  K^{\prime}\left(  b\right)
\circ K\left(  a\right)  \right)  \left(  u\right)  -\left(  K\left(
a\right)  \circ K^{\prime}\left(  b\right)  \right)  \left(  u\right) \\
&  =0.
\end{align*}
In other words, $a\rightharpoonup\left(  u\leftharpoonup b\right)  =\left(
a\rightharpoonup u\right)  \leftharpoonup b$. This proves Theorem
\ref{thm.gen5} \textbf{(a)}.

\textbf{(b)} Fix $a\in\mathfrak{g}$. Let $\lambda_{a}$ be the map
$C\rightarrow C,\ u\mapsto a\rightharpoonup u$. Then, the map $\lambda_{a}$ is
a right $\mathfrak{g}$-module homomorphism\footnote{\textit{Proof.} Let
$b\in\mathfrak{g}$ and $u\in C$. Then, the definition of $\lambda_{a}$ shows
that $\lambda_{a}\left(  u\leftharpoonup b\right)  =a\rightharpoonup\left(
u\leftharpoonup b\right)  =\left(  a\rightharpoonup u\right)  \leftharpoonup
b$ (by Theorem \ref{thm.gen5} \textbf{(a)}). Also, $\lambda_{a}\left(
u\right)  =a\rightharpoonup u$ (by the definition of $\lambda_{a}$), so that
$\underbrace{\lambda_{a}\left(  u\right)  }_{=a\rightharpoonup u}%
\leftharpoonup b=\left(  a\rightharpoonup u\right)  \leftharpoonup b$.
Comparing this with $\lambda_{a}\left(  u\leftharpoonup b\right)  =\left(
a\rightharpoonup u\right)  \leftharpoonup b$, we obtain $\lambda_{a}\left(
u\leftharpoonup b\right)  =\lambda_{a}\left(  u\right)  \leftharpoonup b$.
\par
Now, let us forget that we fixed $b$ and $u$. We thus have proven that
$\lambda_{a}\left(  u\leftharpoonup b\right)  =\lambda_{a}\left(  u\right)
\leftharpoonup b$ for every $b\in\mathfrak{g}$ and $u\in C$. In other words,
$\lambda_{a}$ is a right $\mathfrak{g}$-module homomorphism. Qed.}, and thus
is a right $U\left(  \mathfrak{g}\right)  $-module homomorphism (since every
right $\mathfrak{g}$-module homomorphism is a right $U\left(  \mathfrak{g}%
\right)  $-module homomorphism). Thus, we have%
\begin{equation}
a\rightharpoonup\left(  uq\right)  =\left(  a\rightharpoonup u\right)
q\ \ \ \ \ \ \ \ \ \ \text{for every }u\in C\text{ and }q\in U\left(
\mathfrak{g}\right)  \label{pf.thm.gen5.b.step1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.gen5.b.step1}):} Let $u\in C$ and
$q\in U\left(  \mathfrak{g}\right)  $. The definition of $\lambda_{a}$ shows
that $\lambda_{a}\left(  uq\right)  =a\rightharpoonup\left(  uq\right)  $. But
$\lambda_{a}$ is a right $U\left(  \mathfrak{g}\right)  $-module homomorphism.
Hence, $\lambda_{a}\left(  uq\right)  =\underbrace{\lambda_{a}\left(
u\right)  }_{\substack{=a\rightharpoonup u\\\text{(by the definition}%
\\\text{of }\lambda_{a}\text{)}}}q=\left(  a\rightharpoonup u\right)  q$.
Comparing this with $\lambda_{a}\left(  uq\right)  =a\rightharpoonup\left(
uq\right)  $, we obtain $a\rightharpoonup\left(  uq\right)  =\left(
a\rightharpoonup u\right)  q$. This proves (\ref{pf.thm.gen5.b.step1}).}.

Let us now forget that we fixed $a$. We thus have proven
(\ref{pf.thm.gen5.b.step1}) for every $a\in\mathfrak{g}$.

Let us now fix $q\in U\left(  \mathfrak{g}\right)  $. Let $\rho_{q}$ be the
map $C\rightarrow C,\ u\mapsto uq$. Then, the map $\rho_{q}$ is a left
$\mathfrak{g}$-module homomorphism\footnote{\textit{Proof.} Let $a\in
\mathfrak{g}$ and $u\in C$. The definition of $\rho_{q}$ shows that $\rho
_{q}\left(  u\right)  =uq$, so that $a\rightharpoonup\underbrace{\rho
_{q}\left(  u\right)  }_{=uq}=a\rightharpoonup\left(  uq\right)  =\left(
a\rightharpoonup u\right)  q$ (by (\ref{pf.thm.gen5.b.step1})). On the other
hand, the definition of $\rho_{q}$ shows that $\rho_{q}\left(
a\rightharpoonup u\right)  =\left(  a\rightharpoonup u\right)  q$. Comparing
this with $a\rightharpoonup\rho_{q}\left(  u\right)  =\left(  a\rightharpoonup
u\right)  q$, we obtain $\rho_{q}\left(  a\rightharpoonup u\right)
=a\rightharpoonup\rho_{q}\left(  u\right)  $.
\par
Now, let us forget that we fixed $a$ and $u$. We thus have proven that
$\rho_{q}\left(  a\rightharpoonup u\right)  =a\rightharpoonup\rho_{q}\left(
u\right)  $ for every $a\in\mathfrak{g}$ and $u\in C$. In other words, the map
$\rho_{q}$ is a left $\mathfrak{g}$-module homomorphism. Qed.}, and thus is a
left $U\left(  \mathfrak{g}\right)  $-module homomorphism (since every left
$\mathfrak{g}$-module homomorphism is a left $U\left(  \mathfrak{g}\right)
$-module homomorphism). Thus, we have%
\begin{equation}
p\left(  uq\right)  =\left(  pu\right)  q\ \ \ \ \ \ \ \ \ \ \text{for every
}u\in C\text{ and }p\in U\left(  \mathfrak{g}\right)
\label{pf.thm.gen5.b.step2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.gen5.b.step2}):} Let $u\in C$ and
$p\in U\left(  \mathfrak{g}\right)  $. The definition of $\rho_{q}$ shows that
$\rho_{q}\left(  pu\right)  =\left(  pu\right)  q$. But $\rho_{q}$ is a left
$U\left(  \mathfrak{g}\right)  $-module homomorphism. Hence, $\rho_{q}\left(
pu\right)  =p\underbrace{\rho_{q}\left(  u\right)  }_{\substack{=uq\\\text{(by
the definition}\\\text{of }\rho_{q}\text{)}}}=p\left(  uq\right)  $. Comparing
this with $\rho_{q}\left(  pu\right)  =\left(  pu\right)  q$, we obtain
$p\left(  uq\right)  =\left(  pu\right)  q$. This proves
(\ref{pf.thm.gen5.b.step2}).}.

Let us now forget that we fixed $q$. We thus have proven
(\ref{pf.thm.gen5.b.step1}) for every $q\in U\left(  \mathfrak{g}\right)  $.
In other words, we have $p\left(  uq\right)  =\left(  pu\right)  q$ for every
$u\in C$, $p\in U\left(  \mathfrak{g}\right)  $ and $q\in U\left(
\mathfrak{g}\right)  $. In other words, the left $U\left(  \mathfrak{g}%
\right)  $-module structure on $C$ and the right $U\left(  \mathfrak{g}%
\right)  $-module structure on $C$ can be combined to form a $\left(  U\left(
\mathfrak{g}\right)  ,U\left(  \mathfrak{g}\right)  \right)  $-bimodule
structure on $C$. This proves Theorem \ref{thm.gen5} \textbf{(b)}.

\textbf{(c)} The map $\eta^{\prime}:U\left(  \mathfrak{g}\right)  \rightarrow
C$ is a left $\mathfrak{g}$-module homomorphism\footnote{\textit{Proof.} Let
$a\in\mathfrak{g}$ and $u\in U\left(  \mathfrak{g}\right)  $.
\par
We have $K^{\prime}\left(  \underbrace{a}_{\in\mathfrak{g}}\right)  \in
K^{\prime}\left(  \mathfrak{g}\right)  \subseteq\operatorname*{Der}%
\nolimits^{\operatorname*{op}}C=\operatorname*{Der}C$. Hence, Proposition
\ref{prop.Der.basics} \textbf{(c)} (applied to $K^{\prime}\left(  a\right)  $
instead of $f$) shows that $\left(  K^{\prime}\left(  a\right)  \right)
\left(  1\right)  =0$. Thus, $\left(  K^{\prime}\left(  a\right)  \right)
\left(  \underbrace{1_{C}}_{=1}\right)  =\left(  K^{\prime}\left(  a\right)
\right)  \left(  1\right)  =0$.
\par
We have $K\left(  \underbrace{a}_{\in\mathfrak{g}}\right)  \in K\left(
\mathfrak{g}\right)  \subseteq\operatorname*{Der}C$. Hence, Proposition
\ref{prop.Der.basics} \textbf{(c)} (applied to $K\left(  a\right)  $ instead
of $f$) shows that $\left(  K\left(  a\right)  \right)  \left(  1\right)  =0$.
Thus, $\left(  K\left(  a\right)  \right)  \left(  \underbrace{1_{C}}%
_{=1}\right)  =\left(  K\left(  a\right)  \right)  \left(  1\right)  =0$.
\par
The definition of the right $U\left(  \mathfrak{g}\right)  $-module structure
on $C$ shows that%
\begin{align*}
1_{C}\iota_{U,\mathfrak{g}}\left(  a\right)   &  =1_{C}\leftharpoonup
a=1_{C}\cdot f^{\prime}\left(  a\right)  +\underbrace{\left(  K^{\prime
}\left(  a\right)  \right)  \left(  1_{C}\right)  }_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the right
}\mathfrak{g}\text{-module structure on }C\right) \\
&  =1_{C}\cdot f^{\prime}\left(  a\right)  =f^{\prime}\left(  a\right)  .
\end{align*}
Comparing this with%
\begin{align*}
a\rightharpoonup1_{C}  &  =f\left(  a\right)  \cdot1_{C}+\underbrace{\left(
K\left(  a\right)  \right)  \left(  1_{C}\right)  }_{=0}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the left
}\mathfrak{g}\text{-module structure on }C\right) \\
&  =f\left(  a\right)  \cdot1_{C}=\underbrace{f}_{=f^{\prime}}\left(
a\right)  =f^{\prime}\left(  a\right)  ,
\end{align*}
we obtain $1_{C}\iota_{U,\mathfrak{g}}\left(  a\right)  =a\rightharpoonup
1_{C}$.
\par
The definition of $\eta^{\prime}$ shows that $\eta^{\prime}\left(  u\right)
=1_{C}u$. Hence,%
\[
a\rightharpoonup\underbrace{\eta^{\prime}\left(  u\right)  }_{=1_{C}%
u}=a\rightharpoonup\left(  1_{C}u\right)  =\left(  a\rightharpoonup
1_{C}\right)  u
\]
(by (\ref{pf.thm.gen5.b.step1}) (applied to $1_{C}$ and $u$ instead of $u$ and
$q$)). But the definition of $\eta^{\prime}$ also shows that
\begin{align*}
\eta^{\prime}\left(  a\rightharpoonup u\right)   &  =1_{C}\underbrace{\left(
a\rightharpoonup u\right)  }_{\substack{=\iota_{U,\mathfrak{g}}\left(
a\right)  u\\\text{(by the definition of the left}\\\mathfrak{g}\text{-module
structure on }U\left(  \mathfrak{g}\right)  \text{)}}}=1_{C}\left(
\iota_{U,\mathfrak{g}}\left(  a\right)  u\right) \\
&  =\underbrace{\left(  1_{C}\iota_{U,\mathfrak{g}}\left(  a\right)  \right)
}_{=a\rightharpoonup1_{C}}u=\left(  a\rightharpoonup1_{C}\right)  u.
\end{align*}
Comparing this with $a\rightharpoonup\eta^{\prime}\left(  u\right)  =\left(
a\rightharpoonup1_{C}\right)  u$, we obtain $\eta^{\prime}\left(
a\rightharpoonup u\right)  =a\rightharpoonup\eta^{\prime}\left(  u\right)  $.
\par
Now, let us forget that we fixed $a$ and $u$. We thus have shown that
$\eta^{\prime}\left(  a\rightharpoonup u\right)  =a\rightharpoonup\eta
^{\prime}\left(  u\right)  $ for every $a\in\mathfrak{g}$ and $u\in U\left(
\mathfrak{g}\right)  $. In other words, $\eta^{\prime}:U\left(  \mathfrak{g}%
\right)  \rightarrow C$ is a left $\mathfrak{g}$-module homomorphism, qed.},
and therefore is a left $U\left(  \mathfrak{g}\right)  $-module homomorphism.

Now, every $u\in U\left(  \mathfrak{g}\right)  $ satisfies%
\begin{align*}
\eta^{\prime}\left(  \underbrace{u}_{=u\cdot1}\right)   &  =\eta^{\prime
}\left(  u\cdot1\right)  =u\cdot\underbrace{\eta^{\prime}\left(  1\right)
}_{\substack{=1\cdot1_{C}\\\text{(by the definition of }\eta^{\prime}\text{)}%
}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta^{\prime}\text{ is a left
}U\left(  \mathfrak{g}\right)  \text{-module homomorphism}\right) \\
&  =u\cdot1\cdot1_{C}=u1_{C}\\
&  =\eta\left(  u\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta\left(
u\right)  =u1_{C}\text{ (by the definition of }\eta\text{)}\right)  .
\end{align*}
In other words, $\eta^{\prime}=\eta$. This proves Theorem \ref{thm.gen5}
\textbf{(c)}.
\end{proof}

\subsection{The right Guin-Oudom isomorphism}

We can now prove the analogue of Theorem \ref{thm.manchon} for right pre-Lie algebras:

\begin{theorem}
\label{thm.manchon.right}Let $A$ be a right pre-Lie algebra. Recall that a Lie
algebra $A^{-}$ is defined (according to Proposition
\ref{prop.prelie.lie.right}).

For any $a\in A$, let $\operatorname*{right}\nolimits_{a}$ be the $\mathbf{k}%
$-linear map $A\rightarrow A,\ b\mapsto b\vartriangleleft a$. For any $a\in
A$, we define a derivation $R_{a}:\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$ as follows: Proposition \ref{prop.Sym.deruniv} (applied
to $V=A$, $M=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}%
\circ\operatorname*{right}\nolimits_{a}$) shows that there exists a unique
derivation $F:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ such that
$F\circ\iota_{\operatorname*{Sym},A}=\operatorname*{right}\nolimits_{a}%
\circ\iota_{\operatorname*{Sym},A}$. We let $R_{a}$ be this derivation.

In the following, we will identify every $a\in A$ with the element
$\iota_{\operatorname*{Sym},A}\left(  a\right)  $ of $\operatorname*{Sym}A$.
(This is a particular case of the abuse of notation introduced in Definition
\ref{def.manchon.danger}.) Thus, the injection $\iota_{\operatorname*{Sym}%
,A}:A\rightarrow\operatorname*{Sym}A$ becomes an inclusion, and the
$\mathbf{k}$-module $A$ becomes a $\mathbf{k}$-submodule of
$\operatorname*{Sym}A$ (although the right pre-Lie algebra $A$ does
\textbf{not} become a subalgebra of $\operatorname*{Sym}A$). Thus, products
such as $b_{1}b_{2}\cdots b_{n}$ (where $n\in\mathbb{N}$ and $b_{1}%
,b_{2},\ldots,b_{n}\in A$) will always mean products inside
$\operatorname*{Sym}A$. (This notation will not conflict with the binary
operation on the right pre-Lie algebra $A$, since the latter operation is
denoted by $\vartriangleleft$.)

\textbf{(a)} If $a\in A$, $n\in\mathbb{N}$ and $b_{1},b_{2},\ldots,b_{n}\in
A$, then%
\[
R_{a}\left(  b_{1}b_{2}\cdots b_{n}\right)  =\sum_{k=1}^{n}b_{1}b_{2}\cdots
b_{k-1}\left(  b_{k}\vartriangleleft a\right)  b_{k+1}b_{k+2}\cdots b_{n}.
\]


\textbf{(b)} For every $a\in A$, the map $R_{a}:\operatorname*{Sym}%
A\rightarrow\operatorname*{Sym}A$ is graded.

\textbf{(c)} For every $a\in A$, the map $R_{a}:\operatorname*{Sym}%
A\rightarrow\operatorname*{Sym}A$ is a coderivation.

\textbf{(d)} We define a map $K^{\prime}:A\rightarrow\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $ by
\[
\left(  K^{\prime}\left(  a\right)  =R_{a}\ \ \ \ \ \ \ \ \ \ \text{for every
}a\in A\right)  .
\]
Then, this map $K^{\prime}$ is a Lie algebra homomorphism from $A^{-}$ to
$\operatorname*{Der}\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}%
A\right)  $. (See Definition \ref{def.Der.right} for the definition of the Lie
algebra $\operatorname*{Der}\nolimits^{\operatorname*{op}}\left(
\operatorname*{Sym}A\right)  $.)

\textbf{(e)} We can define a right $A^{-}$-module structure on
$\operatorname*{Sym}A$ by setting%
\[
\left(  u\leftharpoonup a=ua+R_{a}\left(  u\right)  \text{ for all }a\in
A^{-}\text{ and }u\in\operatorname*{Sym}A\right)  .
\]
In the following, we will regard $\operatorname*{Sym}A$ as a right $A^{-}%
$-module by means of this right $A^{-}$-module structure.

\textbf{(f)} Being a right $A^{-}$-module, $\operatorname*{Sym}A$ becomes a
right $U\left(  A^{-}\right)  $-module. Define a map $\eta^{\prime}:U\left(
A^{-}\right)  \rightarrow\operatorname*{Sym}A$ by%
\[
\eta^{\prime}\left(  u\right)  =1_{\operatorname*{Sym}A}%
u\ \ \ \ \ \ \ \ \ \ \text{for every }u\in U\left(  A^{-}\right)  .
\]
Then, $\eta^{\prime}$ is a right $A^{-}$-module homomorphism.

\textbf{(g)} We have $\eta^{\prime}\left(  \iota_{U,A^{-}}\left(  a\right)
\right)  =a$ for every $a\in A^{-}$.

\textbf{(h)} The map $\eta^{\prime}$ is a right $A^{-}$-module isomorphism.

\textbf{(i)} We have%
\[
\eta^{\prime}\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}%
}\left(  a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right)  \in
a_{1}a_{2}\cdots a_{n}+\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}A
\]
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in A^{-}$.

\textbf{(j)} The map $\eta^{\prime}$ and its inverse $\left(  \eta^{\prime
}\right)  ^{-1}$ are filtered (where $U\left(  A^{-}\right)  $ is endowed with
the usual filtration on a universal enveloping algebra).

\textbf{(k)} The map $\operatorname*{grad}\nolimits_{\operatorname*{Sym}%
A}^{-1}\circ\left(  \operatorname*{gr}\left(  \eta^{\prime}\right)  \right)
:\operatorname*{gr}\left(  U\left(  A^{-}\right)  \right)  \rightarrow
\operatorname*{Sym}A$ is an inverse to the map $\operatorname*{PBW}%
\nolimits_{A^{-}}$. In particular, the map $\operatorname*{PBW}%
\nolimits_{A^{-}}$ is invertible.

\textbf{(l)} The map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ is a $\mathbf{k}$-coalgebra isomorphism.

\textbf{(m)} For every $a\in A$, $b\in A$ and $c\in\operatorname*{Sym}A$, we
have%
\[
\left(  cb\right)  \leftharpoonup a-\left(  c\leftharpoonup a\right)  \cdot
b=c\cdot\left(  b\vartriangleleft a\right)  .
\]

\end{theorem}

The proof of Theorem \ref{thm.manchon.right} is analogous to that of Theorem
\ref{thm.manchon}. I shall, however, spell it out in full instead of just
pointing out the necessary changes, since this is the only way I can make sure
that I don't make mistakes.

\begin{proof}
[Proof of Theorem \ref{thm.manchon.right}.]For every $a\in A$, we know that
$R_{a}:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ is a derivation
and satisfies%
\begin{equation}
R_{a}\circ\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}%
\circ\operatorname*{left}\nolimits_{a} \label{pf.thm.manchon.right.La-def}%
\end{equation}
\footnote{\textit{Proof.} Let $a\in A$. Then, $R_{a}$ is the unique derivation
$F:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ such that $F\circ
\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}\circ
\operatorname*{right}\nolimits_{a}$. Hence, $R_{a}$ is a derivation
$\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ and satisfies
$L_{a}\circ\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}%
\circ\operatorname*{right}\nolimits_{a}$. Qed.}. Hence, for every $a\in A$ and
$b\in A$, we have%
\begin{align}
R_{a}\left(  \iota_{\operatorname*{Sym},A}\left(  b\right)  \right)   &
=\underbrace{\left(  R_{a}\circ\iota_{\operatorname*{Sym},A}\right)
}_{\substack{=\iota_{\operatorname*{Sym},A}\circ\operatorname*{right}%
\nolimits_{a}\\\text{(by (\ref{pf.thm.manchon.right.La-def}))}}}\left(
b\right)  =\iota_{\operatorname*{Sym},A}\left(
\underbrace{\operatorname*{right}\nolimits_{a}\left(  b\right)  }%
_{\substack{=b\vartriangleleft a\\\text{(by the definition}\\\text{of
}\operatorname*{right}\nolimits_{a}\text{)}}}\right) \nonumber\\
&  =\iota_{\operatorname*{Sym},A}\left(  b\vartriangleleft a\right)  .
\label{pf.thm.manchon.right.La(iSymb)}%
\end{align}
In other words, for every $a\in A$ and $b\in A$, we have
\begin{equation}
R_{a}\left(  b\right)  =b\vartriangleleft a \label{pf.thm.manchon.right.La(b)}%
\end{equation}
\footnote{since $\iota_{\operatorname*{Sym},A}\left(  b\right)  =b$ (because
we are identifying $\iota_{\operatorname*{Sym},A}\left(  x\right)  $ with $x$
for every $x\in A$) and $\iota_{\operatorname*{Sym},A}\left(
b\vartriangleleft a\right)  =b\vartriangleleft a$ (for the same reason)}.

\textbf{(a)} Let $a\in A$, $n\in\mathbb{N}$ and $b_{1},b_{2},\ldots,b_{n}\in
A$. The map $R_{a}$ is a derivation, thus an element of $\operatorname*{Der}%
\left(  \operatorname*{Sym}A\right)  $. Hence, Proposition
\ref{prop.Der.basics} \textbf{(b)} (applied to $C=\operatorname*{Sym}A$,
$f=R_{a}$ and $a_{i}=b_{i}$) yields
\begin{align*}
&  R_{a}\left(  b_{1}b_{2}\cdots b_{n}\right) \\
&  =\sum_{i=1}^{n}b_{1}b_{2}\cdots b_{i-1}\underbrace{R_{a}\left(
b_{i}\right)  }_{\substack{=b_{i}\vartriangleleft a\\\text{(by
(\ref{pf.thm.manchon.right.La(b)}))}}}b_{i+1}b_{i+2}\cdots b_{n}\\
&  =\sum_{i=1}^{n}b_{1}b_{2}\cdots b_{i-1}\left(  b_{i}\vartriangleleft
a\right)  b_{i+1}b_{i+2}\cdots b_{n}=\sum_{k=1}^{n}b_{1}b_{2}\cdots
b_{k-1}\left(  b_{k}\vartriangleleft a\right)  b_{k+1}b_{k+2}\cdots b_{n}%
\end{align*}
(here, we have renamed the summation index $i$ as $k$). This proves Theorem
\ref{thm.manchon.right} \textbf{(a)}.

\textbf{(b)} Let $a\in A$.

In Proposition \ref{prop.Sym.bider} (applied to $V=A$ and
$f=\operatorname*{right}\nolimits_{a}$), a map $\widetilde{f}%
:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ was defined as the
unique derivation $F:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ such
that $F\circ\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}%
\circ\operatorname*{right}\nolimits_{a}$. This derivation $F$ must clearly be
our map $R_{a}$ (since we know that $R_{a}:\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$ is a derivation and satisfies $R_{a}\circ\iota
_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}\circ
\operatorname*{right}\nolimits_{a}$). Hence, the $\widetilde{f}$ defined in
Proposition \ref{prop.Sym.bider} (applied to $V=A$ and
$f=\operatorname*{right}\nolimits_{a}$) is our map $R_{a}$. Thus, Proposition
\ref{prop.Sym.bider} can be applied to $V=A$, $f=\operatorname*{right}%
\nolimits_{a}$ and $\widetilde{f}=R_{a}$.

As a consequence, Proposition \ref{prop.Sym.bider} \textbf{(c)} (applied to
$V=A$, $f=\operatorname*{right}\nolimits_{a}$ and $\widetilde{f}=R_{a}$) shows
that the map $R_{a}:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ is
graded. This proves Theorem \ref{thm.manchon.right} \textbf{(b)}.

\textbf{(c)} Let $a\in A$. In our above proof of Theorem
\ref{thm.manchon.right} \textbf{(b)}, we have shown that Proposition
\ref{prop.Sym.bider} can be applied to $V=A$, $f=\operatorname*{right}%
\nolimits_{a}$ and $\widetilde{f}=R_{a}$. Hence, Proposition
\ref{prop.Sym.bider} \textbf{(b)} (applied to $V=A$, $f=\operatorname*{right}%
\nolimits_{a}$ and $\widetilde{f}=R_{a}$) shows that the map $R_{a}%
:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ is a coderivation.
Theorem \ref{thm.manchon.right} \textbf{(c)} is thus proven.

\textbf{(d)} The map $K^{\prime}$ is a map from $A$ to $\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $, thus a
map from $A^{-}$ to $\operatorname*{Der}\nolimits^{\operatorname*{op}}\left(
\operatorname*{Sym}A\right)  $ (since $A^{-}=A$ as sets).

We shall first prove that%
\begin{equation}
K^{\prime}\left(  \left[  a,b\right]  \right)  =\left[  K^{\prime}\left(
a\right)  ,K^{\prime}\left(  b\right)  \right]
\label{pf.thm.manchon.right.d.goal1}%
\end{equation}
for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
K^{\prime}\left(  a\right)  ,K^{\prime}\left(  b\right)  \right]  $ is
computed in $\operatorname*{Der}\nolimits^{\operatorname*{op}}\left(
\operatorname*{Sym}A\right)  $).

\textit{Proof of (\ref{pf.thm.manchon.right.d.goal1}):} Recall that we are
regarding $A$ as a $\mathbf{k}$-submodule of $\operatorname*{Sym}A$ via the
injection $\iota_{\operatorname*{Sym},A}:A\rightarrow\operatorname*{Sym}A$.
Thus, $A=\iota_{\operatorname*{Sym},A}\left(  A\right)  =\operatorname*{Sym}%
\nolimits^{1}A$. Recall that the subset $\operatorname*{Sym}\nolimits^{1}A$ of
$\operatorname*{Sym}A$ generates the $\mathbf{k}$-algebra $\operatorname*{Sym}%
A$. In other words, the subset $A$ of $\operatorname*{Sym}A$ generates the
$\mathbf{k}$-algebra $\operatorname*{Sym}A$ (since $A=\operatorname*{Sym}%
\nolimits^{1}A$).

Let $a\in A^{-}$ and $b\in A^{-}$. Then, $K^{\prime}\left(  a\right)  $ and
$K^{\prime}\left(  b\right)  $ are elements of $\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $. Thus
$\left[  K^{\prime}\left(  a\right)  ,K^{\prime}\left(  b\right)  \right]  $
is an element of $\operatorname*{Der}\nolimits^{\operatorname*{op}}\left(
\operatorname*{Sym}A\right)  $ (since $\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $ is a Lie
algebra), hence an element of $\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $ (since $\operatorname*{Der}\nolimits^{\operatorname*{op}}\left(
\operatorname*{Sym}A\right)  =\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $ as sets), hence a derivation $\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$. Also, $K^{\prime}\left(  \left[  a,b\right]  \right)  $
is an element of $\operatorname*{Der}\nolimits^{\operatorname*{op}}\left(
\operatorname*{Sym}A\right)  =\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $, hence a derivation $\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$.

We have $\left[  K^{\prime}\left(  a\right)  ,K^{\prime}\left(  b\right)
\right]  \mid_{A}=K^{\prime}\left(  \left[  a,b\right]  \right)  \mid_{A}%
$\ \ \ \ \footnote{\textit{Proof.} Let $c\in A$. The definition of $K^{\prime
}$ yields $K^{\prime}\left(  a\right)  =R_{a}$, $K^{\prime}\left(  b\right)
=R_{b}$ and $K^{\prime}\left(  \left[  a,b\right]  \right)  =R_{\left[
a,b\right]  }$. Now,
\begin{align*}
&  \left(  \left[  K^{\prime}\left(  a\right)  ,K^{\prime}\left(  b\right)
\right]  \mid_{A}\right)  \left(  c\right) \\
&  =\underbrace{\left[  K^{\prime}\left(  a\right)  ,K^{\prime}\left(
b\right)  \right]  }_{\substack{=K^{\prime}\left(  b\right)  \circ K^{\prime
}\left(  a\right)  -K^{\prime}\left(  a\right)  \circ K^{\prime}\left(
b\right)  \\\text{(since the Lie bracket used here}\\\text{is that of
}\operatorname*{Der}\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}%
A\right)  \text{)}}}\left(  c\right)  =\left(  \underbrace{K^{\prime}\left(
b\right)  }_{=R_{b}}\circ\underbrace{K^{\prime}\left(  a\right)  }_{=R_{a}%
}-\underbrace{K^{\prime}\left(  a\right)  }_{=R_{a}}\circ\underbrace{K^{\prime
}\left(  b\right)  }_{=R_{b}}\right)  \left(  c\right) \\
&  =\left(  R_{b}\circ R_{a}-R_{a}\circ R_{b}\right)  \left(  c\right)
=R_{b}\left(  \underbrace{R_{a}\left(  c\right)  }%
_{\substack{=c\vartriangleleft a\\\text{(by (\ref{pf.thm.manchon.right.La(b)}%
), applied to }c\\\text{instead of }b\text{)}}}\right)  -R_{a}\left(
\underbrace{R_{b}\left(  c\right)  }_{\substack{=c\vartriangleleft
b\\\text{(by (\ref{pf.thm.manchon.right.La(b)}), applied to }b\text{ and
}c\\\text{instead of }a\text{ and }b\text{)}}}\right) \\
&  =\underbrace{R_{b}\left(  c\vartriangleleft a\right)  }_{\substack{=\left(
c\vartriangleleft a\right)  \vartriangleleft b\\\text{(by
(\ref{pf.thm.manchon.right.La(b)}), applied to }b\text{ and }c\vartriangleleft
a\\\text{instead of }a\text{ and }b\text{)}}}-\underbrace{R_{a}\left(
c\vartriangleleft b\right)  }_{\substack{=\left(  c\vartriangleleft b\right)
\vartriangleleft a\\\text{(by (\ref{pf.thm.manchon.right.La(b)}), applied to
}c\vartriangleleft b\\\text{instead of }b\text{)}}}\\
&  =\left(  c\vartriangleleft a\right)  \vartriangleleft b-\left(
c\vartriangleleft b\right)  \vartriangleleft a.
\end{align*}
Comparing this with%
\begin{align*}
&  \left(  K^{\prime}\left(  \left[  a,b\right]  \right)  \mid_{A}\right)
\left(  c\right) \\
&  =\underbrace{\left(  K^{\prime}\left(  \left[  a,b\right]  \right)
\right)  }_{=R_{\left[  a,b\right]  }}\left(  c\right)  =R_{\left[
a,b\right]  }\left(  c\right)  =c\vartriangleleft\underbrace{\left[
a,b\right]  }_{\substack{=a\vartriangleleft b-b\vartriangleleft a\\\text{(by
the definition}\\\text{of the Lie algebra }A^{-}\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{pf.thm.manchon.right.La(b)}), applied to }\left[  a,b\right]
\text{ and }c\\
\text{instead of }a\text{ and }b
\end{array}
\right) \\
&  =c\vartriangleleft\left(  a\vartriangleleft b-b\vartriangleleft a\right)
=c\vartriangleleft\left(  a\vartriangleleft b\right)  -c\vartriangleleft
\left(  b\vartriangleleft a\right)  =\left(  c\vartriangleleft a\right)
\vartriangleleft b-\left(  c\vartriangleleft b\right)  \vartriangleleft a\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\left(  c\vartriangleleft b\right)  \vartriangleleft
a-c\vartriangleleft\left(  b\vartriangleleft a\right)  =\left(
c\vartriangleleft a\right)  \vartriangleleft b-c\vartriangleleft\left(
a\vartriangleleft b\right) \\
\text{(by (\ref{eq.def.prelie.right.axiom}), applied to }c\text{ and }a\text{
instead of }a\text{ and }c\text{)}%
\end{array}
\right)  ,
\end{align*}
we obtain $\left(  \left[  K^{\prime}\left(  a\right)  ,K^{\prime}\left(
b\right)  \right]  \mid_{A}\right)  \left(  c\right)  =\left(  K^{\prime
}\left(  \left[  a,b\right]  \right)  \mid_{A}\right)  \left(  c\right)  $.
\par
Now, let us forget that we fixed $c$. We thus have shown that $\left(  \left[
K^{\prime}\left(  a\right)  ,K^{\prime}\left(  b\right)  \right]  \mid
_{A}\right)  \left(  c\right)  =\left(  K^{\prime}\left(  \left[  a,b\right]
\right)  \mid_{A}\right)  \left(  c\right)  $ for every $c\in A$. In other
words, $\left[  K^{\prime}\left(  a\right)  ,K^{\prime}\left(  b\right)
\right]  \mid_{A}=K^{\prime}\left(  \left[  a,b\right]  \right)  \mid_{A}$,
qed.}. Hence, Lemma \ref{lem.derivation.unique} (applied to
$\operatorname*{Sym}A$, $\left[  K^{\prime}\left(  a\right)  ,K^{\prime
}\left(  b\right)  \right]  $, $K^{\prime}\left(  \left[  a,b\right]  \right)
$ and $A$ instead of $A$, $d$, $e$ and $S$) shows that $\left[  K^{\prime
}\left(  a\right)  ,K^{\prime}\left(  b\right)  \right]  =K^{\prime}\left(
\left[  a,b\right]  \right)  $. This proves
(\ref{pf.thm.manchon.right.d.goal1}).

Next, we notice that%
\begin{equation}
K^{\prime}\left(  \lambda a+\mu b\right)  =\lambda K^{\prime}\left(  a\right)
+\mu K^{\prime}\left(  b\right)  \label{pf.thm.manchon.right.d.goal2}%
\end{equation}
for every $a\in A^{-}$ and $b\in A^{-}$ and every $\lambda\in\mathbf{k}$ and
$\mu\in\mathbf{k}$.

\textit{Proof of (\ref{pf.thm.manchon.right.d.goal2}):} The proof of
(\ref{pf.thm.manchon.right.d.goal2}) is similar to the above proof of
(\ref{pf.thm.manchon.right.d.goal1}), but even simpler (because instead of
$c\vartriangleleft\left[  a,b\right]  =\left(  c\vartriangleleft a\right)
\vartriangleleft b-\left(  c\vartriangleleft b\right)  \vartriangleleft a$, we
now need to use the much simpler equality $c\vartriangleleft\left(  \lambda
a+\mu b\right)  =\lambda\left(  c\vartriangleleft a\right)  +\mu\left(
c\vartriangleleft b\right)  $). Hence, we leave it to the reader.

Now, the map $K^{\prime}:A^{-}\rightarrow\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $ is
$\mathbf{k}$-linear (because of (\ref{pf.thm.manchon.right.d.goal2})) and thus
a Lie algebra homomorphism (because of (\ref{pf.thm.manchon.right.d.goal1})).
This proves Theorem \ref{thm.manchon.right} \textbf{(d)}.

\textbf{(e)} The map $K^{\prime}:A^{-}\rightarrow\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $ is a Lie
algebra homomorphism (according to Theorem \ref{thm.manchon.right}
\textbf{(d)}). Moreover, we have%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K^{\prime}\left(  b\right)  \right)
\left(  \iota_{\operatorname*{Sym},A}\left(  a\right)  \right)  -\left(
K^{\prime}\left(  a\right)  \right)  \left(  \iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right)
\]
for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  $ is computed in the Lie algebra $\left(
\operatorname*{Sym}A\right)  ^{-}$)\ \ \ \ \footnote{\textit{Proof.} Let $a\in
A^{-}$ and $b\in A^{-}$. Recall that we are identifying $\iota
_{\operatorname*{Sym},A}\left(  x\right)  $ with $x$ for every $x\in A$. Thus,
$\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
a,b\right]  $, $\iota_{\operatorname*{Sym},A}\left(  a\right)  =a$ and
$\iota_{\operatorname*{Sym},A}\left(  b\right)  =b$. Moreover, the
$\mathbf{k}$-algebra $\operatorname*{Sym}A$ is commutative, and thus the Lie
bracket of the Lie algebra $\left(  \operatorname*{Sym}A\right)  ^{-}$ is
identically $0$. Hence, $\left[  \iota_{\operatorname*{Sym},A}\left(
a\right)  ,\iota_{\operatorname*{Sym},A}\left(  b\right)  \right]  =0$. Now,%
\begin{align*}
&  \underbrace{\left[  \iota_{\operatorname*{Sym},A}\left(  a\right)
,\iota_{\operatorname*{Sym},A}\left(  b\right)  \right]  }_{=0}%
+\underbrace{\left(  K^{\prime}\left(  b\right)  \right)  }_{\substack{=R_{b}%
\\\text{(by the definition of }K^{\prime}\text{)}}}\left(  \underbrace{\iota
_{\operatorname*{Sym},A}\left(  a\right)  }_{=a}\right)  -\underbrace{\left(
K^{\prime}\left(  a\right)  \right)  }_{\substack{=R_{a}\\\text{(by the
definition of }K^{\prime}\text{)}}}\left(  \underbrace{\iota
_{\operatorname*{Sym},A}\left(  b\right)  }_{=b}\right) \\
&  =\underbrace{R_{b}\left(  a\right)  }_{\substack{=a\vartriangleleft
b\\\text{(by (\ref{pf.thm.manchon.right.La(b)}) (applied to}\\b\text{ and
}a\text{ instead of }a\text{ and }b\text{))}}}-\underbrace{R_{a}\left(
b\right)  }_{\substack{=b\vartriangleleft a\\\text{(by
(\ref{pf.thm.manchon.right.La(b)}))}}}=a\vartriangleleft b-b\vartriangleleft
a.
\end{align*}
Compared with%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
a,b\right]  =a\vartriangleleft b-b\vartriangleleft a
\]
(by the definition of the Lie bracket on $A^{-}$), this shows that
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K^{\prime}\left(  b\right)  \right)
\left(  \iota_{\operatorname*{Sym},A}\left(  a\right)  \right)  -\left(
K^{\prime}\left(  a\right)  \right)  \left(  \iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right)  .
\]
Qed.}. Hence, we can apply Theorem \ref{thm.gen1.right} to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota_{\operatorname*{Sym}%
,A}$.

Thus, Theorem \ref{thm.gen1.right} \textbf{(a)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota_{\operatorname*{Sym}%
,A}$) shows that we can define a right $A^{-}$-module structure on
$\operatorname*{Sym}A$ by setting%
\[
\left(  u\leftharpoonup a=u\cdot\iota_{\operatorname*{Sym},A}\left(  a\right)
+\left(  K^{\prime}\left(  a\right)  \right)  \left(  u\right)  \text{ for all
}a\in A^{-}\text{ and }u\in\operatorname*{Sym}A\right)  .
\]
In other words, we can define a right $A^{-}$-module structure on
$\operatorname*{Sym}A$ by setting%
\[
\left(  u\leftharpoonup a=ua+R_{a}\left(  u\right)  \text{ for all }a\in
A^{-}\text{ and }u\in\operatorname*{Sym}A\right)
\]
(because every $a\in A^{-}$ and $u\in\operatorname*{Sym}A$ satisfy
$u\cdot\underbrace{\iota_{\operatorname*{Sym},A}\left(  a\right)  }%
_{=a}+\underbrace{\left(  K^{\prime}\left(  a\right)  \right)  }%
_{\substack{=R_{a}\\\text{(by the definition of }K^{\prime}\text{)}}}\left(
u\right)  =ua+R_{a}\left(  u\right)  $). This proves Theorem
\ref{thm.manchon.right} \textbf{(e)}.

As a consequence of this proof of Theorem \ref{thm.manchon.right}
\textbf{(e)}, we see that the $A^{-}$-module structure on $\operatorname*{Sym}%
A$ defined in Theorem \ref{thm.manchon.right} \textbf{(e)} is precisely the
one that is constructed by Theorem \ref{thm.gen1.right} \textbf{(a)} (applied
to $\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota
_{\operatorname*{Sym},A}$). We notice further that the map $\eta:U\left(
A^{-}\right)  \rightarrow\operatorname*{Sym}A$ defined in Theorem
\ref{thm.manchon.right} \textbf{(f)} is precisely the map $\eta:U\left(
A^{-}\right)  \rightarrow\operatorname*{Sym}A$ that is constructed by Theorem
\ref{thm.gen1.right} \textbf{(b)} (applied to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$) (since these
two maps have the same definition).

\textbf{(f)} As we have seen in our proof of Theorem \ref{thm.manchon.right}
\textbf{(e)}, we can apply Theorem \ref{thm.gen1.right} to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota_{\operatorname*{Sym}%
,A}$. Moreover, the following holds:

\begin{itemize}
\item The right $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon.right} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1.right} \textbf{(a)} (applied to
$\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota
_{\operatorname*{Sym},A}$).

\item The map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ defined in Theorem \ref{thm.manchon.right} \textbf{(f)}
is precisely the map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ that is constructed by Theorem \ref{thm.gen1.right}
\textbf{(b)} (applied to $\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and
$f^{\prime}=\iota_{\operatorname*{Sym},A}$).
\end{itemize}

Hence, Theorem \ref{thm.gen1.right} \textbf{(b)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota_{\operatorname*{Sym}%
,A}$) shows that $\eta^{\prime}$ is a right $A^{-}$-module homomorphism. This
proves Theorem \ref{thm.manchon.right} \textbf{(f)}.

\textbf{(g)} Let $a\in A^{-}$. The map $R_{a}$ is a derivation. In other
words, $R_{a}\in\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $.
Thus, $R_{a}\left(  1\right)  =0$ (by Proposition \ref{prop.Der.basics}
\textbf{(c)}, applied to $\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $ and $R_{a}$ instead of $C$ and $f$).

Now, the definition of $\eta^{\prime}$ shows that
\begin{align*}
&  \eta^{\prime}\left(  \iota_{U,A^{-}}\left(  a\right)  \right) \\
&  =1_{\operatorname*{Sym}A}\iota_{U,A^{-}}\left(  a\right)
=1_{\operatorname*{Sym}A}\leftharpoonup a\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the right action of
}U\left(  A^{-}\right)  \text{ on }\operatorname*{Sym}A\right) \\
&  =\underbrace{1_{\operatorname*{Sym}A}a}_{=a}+\underbrace{R_{a}\left(
1_{\operatorname*{Sym}A}\right)  }_{=R_{a}\left(  1\right)  =0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the right }%
A^{-}\text{-module structure on }\operatorname*{Sym}A\right) \\
&  =a.
\end{align*}
This proves Theorem \ref{thm.manchon.right} \textbf{(g)}.

\textbf{(i)} We make the following observations:

\begin{itemize}
\item The map $K^{\prime}:A^{-}\rightarrow\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $ is a Lie
algebra homomorphism (according to Theorem \ref{thm.manchon.right}
\textbf{(d)}).

\item We have%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K^{\prime}\left(  b\right)  \right)
\left(  \iota_{\operatorname*{Sym},A}\left(  a\right)  \right)  -\left(
K^{\prime}\left(  a\right)  \right)  \left(  \iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right)
\]
for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  $ is computed in the Lie algebra $\left(
\operatorname*{Sym}A\right)  ^{-}$)\ \ \ \ \footnote{This was proven in our
above proof of Theorem \ref{thm.manchon.right} \textbf{(e)}.}.

\item The right $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon.right} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1.right} \textbf{(a)} (applied to
$\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota
_{\operatorname*{Sym},A}$).\ \ \ \ \footnote{This was proven above.}

\item The map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ defined in Theorem \ref{thm.manchon.right} \textbf{(f)}
is precisely the map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ that is constructed by Theorem \ref{thm.gen1.right}
\textbf{(b)} (applied to $\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and
$f^{\prime}=\iota_{\operatorname*{Sym},A}$).

\item The filtration of the filtered $\mathbf{k}$-algebra $\operatorname*{Sym}%
A$ is $\left(  \bigoplus_{k=0}^{m}\operatorname*{Sym}\nolimits^{k}A\right)
_{m\geq0}$.

\item We have $\iota_{\operatorname*{Sym},A}\left(  \underbrace{A^{-}}%
_{=A}\right)  =\iota_{\operatorname*{Sym},A}\left(  A\right)
=\operatorname*{Sym}\nolimits^{1}A\subseteq\bigoplus_{k=0}^{1}%
\operatorname*{Sym}\nolimits^{k}A$.

\item The map $K^{\prime}\left(  a\right)  :\operatorname*{Sym}A\rightarrow
\operatorname*{Sym}A$ is filtered for every $a\in A^{-}$%
\ \ \ \ \footnote{\textit{Proof.} Let $a\in A^{-}$. Thus, the definition of
$K^{\prime}$ shows that $K^{\prime}\left(  a\right)  =R_{a}$. Now, the map
$R_{a}$ is graded (by Theorem \ref{thm.manchon.right} \textbf{(b)}), and thus
filtered (since any graded $\mathbf{k}$-linear map is filtered). In other
words, the map $K^{\prime}\left(  a\right)  $ is filtered (since $K^{\prime
}\left(  a\right)  =R_{a}$). Qed.}.
\end{itemize}

Hence, we can apply Theorem \ref{thm.gen1filt.right} to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$, $\left(  C_{m}\right)  _{m\geq0}=\left(
\bigoplus_{k=0}^{m}\operatorname*{Sym}\nolimits^{k}A\right)  _{m\geq0}$ and
$f^{\prime}=\iota_{\operatorname*{Sym},A}$. As a result, we conclude that
\begin{align}
&  \eta\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}}\left(
a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right) \nonumber\\
&  \in\iota_{\operatorname*{Sym},A}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},A}\left(  a_{2}\right)  \cdots\iota_{\operatorname*{Sym}%
,A}\left(  a_{n}\right)  +\bigoplus_{k=0}^{n-1}\operatorname*{Sym}%
\nolimits^{k}A \label{pf.thm.manchon.right.i.1}%
\end{align}
for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in A^{-}$.
Thus, for every $n\in\mathbb{N}$ and every $a_{1},a_{2},\ldots,a_{n}\in A^{-}%
$, we have%
\[
\eta\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}}\left(
a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right)  \in
a_{1}a_{2}\cdots a_{n}+\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}A
\]
\footnote{\textit{Proof.} Let $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots
,a_{n}\in A^{-}$.
\par
Recall that we are identifying every $x\in A$ with $\iota_{\operatorname*{Sym}%
,A}\left(  x\right)  $. Thus, every $x\in A$ satisfies $x=\iota
_{\operatorname*{Sym},A}\left(  x\right)  =\iota_{\operatorname*{Sym},A^{-}%
}\left(  x\right)  $ (since $\iota_{\operatorname*{Sym},A}=\iota
_{\operatorname*{Sym},A^{-}}$). In particular, every $i\in\left\{
1,2,\ldots,n\right\}  $ satisfies the identity $a_{i}=\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{i}\right)  $. Multiplying these
identities over all $i\in\left\{  1,2,\ldots,n\right\}  $, we obtain
$a_{1}a_{2}\cdots a_{n}=\iota_{\operatorname*{Sym},A^{-}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  $. Now,
(\ref{pf.thm.manchon.right.i.1}) becomes%
\begin{align*}
\eta^{\prime}\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}%
}\left(  a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right)   &
\in\underbrace{\iota_{\operatorname*{Sym},A}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},A}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A}\left(  a_{n}\right)  }_{=a_{1}a_{2}\cdots a_{n}%
}+\underbrace{\bigoplus_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}%
A}_{\substack{=\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}A\\\text{(since
direct sums}\\\text{are sums)}}}\\
&  =a_{1}a_{2}\cdots a_{n}+\sum_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}A,
\end{align*}
qed.}. This proves Theorem \ref{thm.manchon.right} \textbf{(i)}.

\textbf{(h)} Recall that $\eta^{\prime}$ is a $\mathbf{k}$-linear map
$U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$. In other words,
$\eta^{\prime}$ is a $\mathbf{k}$-linear map $U\left(  A^{-}\right)
\rightarrow\operatorname*{Sym}\left(  A^{-}\right)  $ (since $A^{-}=A$ as
$\mathbf{k}$-modules).

For every $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots,a_{n}\in A^{-}$, we have%
\begin{align*}
&  \eta^{\prime}\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}%
}\left(  a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right) \\
&  \in\iota_{\operatorname*{Sym},A^{-}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  +\sum_{k=0}^{n-1}%
\operatorname*{Sym}\nolimits^{k}\left(  A^{-}\right)
\end{align*}
\footnote{\textit{Proof.} Let $n\in\mathbb{N}$ and $a_{1},a_{2},\ldots
,a_{n}\in A^{-}$. We have $A^{-}=A$ as $\mathbf{k}$-modules. Hence,
$\iota_{\operatorname*{Sym},A^{-}}=\iota_{\operatorname*{Sym},A}$.
\par
Recall that we are identifying every $x\in A$ with $\iota_{\operatorname*{Sym}%
,A}\left(  x\right)  $. Thus, every $x\in A$ satisfies $x=\iota
_{\operatorname*{Sym},A}\left(  x\right)  =\iota_{\operatorname*{Sym},A^{-}%
}\left(  x\right)  $ (since $\iota_{\operatorname*{Sym},A}=\iota
_{\operatorname*{Sym},A^{-}}$). In particular, every $i\in\left\{
1,2,\ldots,n\right\}  $ satisfies the identity $a_{i}=\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{i}\right)  $. Multiplying these
identities over all $i\in\left\{  1,2,\ldots,n\right\}  $, we obtain
$a_{1}a_{2}\cdots a_{n}=\iota_{\operatorname*{Sym},A^{-}}\left(  a_{1}\right)
\iota_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  $.
\par
Now, Theorem \ref{thm.manchon.right} \textbf{(i)} shows that%
\begin{align*}
&  \eta^{\prime}\left(  \iota_{U,A^{-}}\left(  a_{1}\right)  \iota_{U,A^{-}%
}\left(  a_{2}\right)  \cdots\iota_{U,A^{-}}\left(  a_{n}\right)  \right) \\
&  \in\underbrace{a_{1}a_{2}\cdots a_{n}}_{=\iota_{\operatorname*{Sym},A^{-}%
}\left(  a_{1}\right)  \iota_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)
\cdots\iota_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  }+\sum
_{k=0}^{n-1}\operatorname*{Sym}\nolimits^{k}\underbrace{A}_{=A^{-}}\\
&  =\iota_{\operatorname*{Sym},A^{-}}\left(  a_{1}\right)  \iota
_{\operatorname*{Sym},A^{-}}\left(  a_{2}\right)  \cdots\iota
_{\operatorname*{Sym},A^{-}}\left(  a_{n}\right)  +\sum_{k=0}^{n-1}%
\operatorname*{Sym}\nolimits^{k}\left(  A^{-}\right)  ,
\end{align*}
qed.}. Thus, we can apply Lemma \ref{lem.PBW-by-inverse-map} to $\mathfrak{g}%
=A^{-}$ and $\phi=\eta^{\prime}$.

Lemma \ref{lem.PBW-by-inverse-map} \textbf{(a)} (applied to $\mathfrak{g}%
=A^{-}$ and $\phi=\eta^{\prime}$) shows that the map $\eta^{\prime}$ is a
filtered $\mathbf{k}$-module isomorphism. In particular, the map $\eta
^{\prime}$ is invertible.

But we know that $\eta^{\prime}$ is a right $A^{-}$-module homomorphism (by
Theorem \ref{thm.manchon.right} \textbf{(f)}). Since $\eta^{\prime}$ is
invertible, this shows that $\eta^{\prime}$ is a right $A^{-}$-module
isomorphism. This proves Theorem \ref{thm.manchon.right} \textbf{(h)}.

\textbf{(j)} In our above proof of Theorem \ref{thm.manchon.right}
\textbf{(h)}, we have shown that we can apply Lemma
\ref{lem.PBW-by-inverse-map} to $\mathfrak{g}=A^{-}$ and $\phi=\eta^{\prime}$.
Hence, Lemma \ref{lem.PBW-by-inverse-map} \textbf{(a)} (applied to
$\mathfrak{g}=A^{-}$ and $\phi=\eta^{\prime}$) shows that the map
$\eta^{\prime}$ is a filtered $\mathbf{k}$-module isomorphism. Also, Lemma
\ref{lem.PBW-by-inverse-map} \textbf{(b)} (applied to $\mathfrak{g}=A^{-}$ and
$\phi=\eta^{\prime}$) shows that the inverse $\left(  \eta^{\prime}\right)
^{-1}$ of the map $\eta^{\prime}$ is filtered. Thus, both maps $\eta^{\prime}$
and $\left(  \eta^{\prime}\right)  ^{-1}$ are filtered. This proves Theorem
\ref{thm.manchon.right} \textbf{(j)}.

\textbf{(k)} In our above proof of Theorem \ref{thm.manchon.right}
\textbf{(h)}, we have shown that we can apply Lemma
\ref{lem.PBW-by-inverse-map} to $\mathfrak{g}=A^{-}$ and $\phi=\eta^{\prime}$.
Hence, the first claim of Lemma \ref{lem.PBW-by-inverse-map} \textbf{(c)}
(applied to $\mathfrak{g}=A^{-}$ and $\phi=\eta^{\prime}$) shows that the map
$\operatorname*{grad}\nolimits_{\operatorname*{Sym}\left(  A^{-}\right)
}^{-1}\circ\left(  \operatorname*{gr}\left(  \eta^{\prime}\right)  \right)
:\operatorname*{gr}\left(  U\left(  A^{-}\right)  \right)  \rightarrow
\operatorname*{Sym}\left(  A^{-}\right)  $ is an inverse to the map
$\operatorname*{PBW}\nolimits_{A^{-}}$. Since $A^{-}=A$ as $\mathbf{k}%
$-modules, this rewrites as follows: The map $\operatorname*{grad}%
\nolimits_{\operatorname*{Sym}A}^{-1}\circ\left(  \operatorname*{gr}\left(
\eta^{\prime}\right)  \right)  :\operatorname*{gr}\left(  U\left(
A^{-}\right)  \right)  \rightarrow\operatorname*{Sym}A$ is an inverse to the
map $\operatorname*{PBW}\nolimits_{A^{-}}$. In particular, the map
$\operatorname*{PBW}\nolimits_{A^{-}}$ is invertible. This proves Theorem
\ref{thm.manchon.right} \textbf{(k)}.

\textbf{(l)} We first observe the following:

\begin{itemize}
\item The map $K^{\prime}:A^{-}\rightarrow\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $ is a Lie algebra homomorphism. (This follows
from Theorem \ref{thm.manchon.right} \textbf{(d)}.)

\item We have $K^{\prime}\left(  A^{-}\right)  \subseteq\operatorname*{Coder}%
\left(  \operatorname*{Sym}A\right)  $\ \ \ \ \footnote{\textit{Proof.} Let
$g\in K^{\prime}\left(  A^{-}\right)  $. Thus, there exists some $a\in A^{-}$
such that $g=K^{\prime}\left(  a\right)  $. Consider this $a$. We have $a\in
A^{-}=A$ and $g=K^{\prime}\left(  a\right)  =R_{a}$ (by the definition of
$K^{\prime}$). But $R_{a}:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$
is a coderivation (according to Theorem \ref{thm.manchon.right} \textbf{(c)}).
In other words, $R_{a}\in\operatorname*{Coder}\left(  \operatorname*{Sym}%
A\right)  $. Hence, $g=R_{a}\in\operatorname*{Coder}\left(
\operatorname*{Sym}A\right)  $.
\par
Now, let us forget that we fixed $g$. We thus have shown that $g\in
\operatorname*{Coder}\left(  \operatorname*{Sym}A\right)  $ for every $g\in
K^{\prime}\left(  A^{-}\right)  $. In other words, $K^{\prime}\left(
A^{-}\right)  \subseteq\operatorname*{Coder}\left(  \operatorname*{Sym}%
A\right)  $, qed.}.

\item We have $\iota_{\operatorname*{Sym},A}\left(  A^{-}\right)
\subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}A\right)
$\ \ \ \ \footnote{\textit{Proof.} Recall that $\iota_{\operatorname*{Sym}%
,V}\left(  V\right)  \subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}%
V\right)  $ for every $\mathbf{k}$-module $V$. Applying this to $V=A$, we
obtain $\iota_{\operatorname*{Sym},A}\left(  A\right)  \subseteq
\operatorname*{Prim}\left(  \operatorname*{Sym}A\right)  $. Since $A^{-}=A$ as
$\mathbf{k}$-modules, this rewrites as $\iota_{\operatorname*{Sym},A}\left(
A^{-}\right)  \subseteq\operatorname*{Prim}\left(  \operatorname*{Sym}%
A\right)  $. Qed.}.

\item The equality%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K^{\prime}\left(  b\right)  \right)
\left(  \iota_{\operatorname*{Sym},A}\left(  a\right)  \right)  -\left(
K^{\prime}\left(  a\right)  \right)  \left(  \iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right)
\]
holds for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  $ is computed in the Lie algebra $\left(
\operatorname*{Sym}A\right)  ^{-}$). (This was proven during our above proof
of Theorem \ref{thm.manchon.right} \textbf{(e)}.)

\item The right $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon.right} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1.right} \textbf{(a)} (applied to
$\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota
_{\operatorname*{Sym},A}$). (This follows from our above proof of Theorem
\ref{thm.manchon.right} \textbf{(e)}.)

\item The map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ defined in Theorem \ref{thm.manchon.right} \textbf{(f)}
is precisely the map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ that is constructed by Theorem \ref{thm.gen1.right}
\textbf{(b)} (applied to $\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and
$f^{\prime}=\iota_{\operatorname*{Sym},A}$). (This is clear, because the
definitions of these two maps are the same.)
\end{itemize}

Combining these observations, we see that Theorem \ref{thm.gen3.right} can be
applied to $\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime
}=\iota_{\operatorname*{Sym},A}$. Consequently, Theorem \ref{thm.gen3.right}
\textbf{(b)} shows that the map $\eta^{\prime}:U\left(  A^{-}\right)
\rightarrow\operatorname*{Sym}A$ is a $\mathbf{k}$-coalgebra homomorphism.
Since $\eta^{\prime}$ is invertible, this shows that $\eta^{\prime}$ is a
$\mathbf{k}$-coalgebra isomorphism. This finishes the proof of Theorem
\ref{thm.manchon.right} \textbf{(l)}.

\textbf{(m)} As we have seen in our proof of Theorem \ref{thm.manchon.right}
\textbf{(e)}, we can apply Theorem \ref{thm.gen1.right} to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota_{\operatorname*{Sym}%
,A}$. Moreover, the right $A^{-}$-module structure on $\operatorname*{Sym}A$
defined in Theorem \ref{thm.manchon.right} \textbf{(e)} is precisely the one
that is constructed by Theorem \ref{thm.gen1.right} \textbf{(a)} (applied to
$\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota
_{\operatorname*{Sym},A}$). Hence, Theorem \ref{thm.gen1.right} \textbf{(c)}
(applied to $\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime
}=\iota_{\operatorname*{Sym},A}$) shows that, for every $a\in A^{-}$,
$b\in\operatorname*{Sym}A$ and $c\in\operatorname*{Sym}A$, we have%
\begin{equation}
\left(  cb\right)  \leftharpoonup a-\left(  c\leftharpoonup a\right)  \cdot
b=c\cdot\left(  K^{\prime}\left(  a\right)  \right)  \left(  b\right)
-c\left[  \iota_{\operatorname*{Sym},A}\left(  a\right)  ,b\right]  .
\label{pf.thm.manchon.right.m.used}%
\end{equation}


Now, fix $a\in A$, $b\in A$ and $c\in\operatorname*{Sym}A$. We have $a\in
A=A^{-}$ and $b\in A\subseteq\operatorname*{Sym}A$ (since we regard $A$ as a
$\mathbf{k}$-submodule of $\operatorname*{Sym}A$). Moreover, the $\mathbf{k}%
$-algebra $\operatorname*{Sym}A$ is commutative, and thus the Lie bracket of
the Lie algebra $\left(  \operatorname*{Sym}A\right)  ^{-}$ is identically
$0$. Hence, $\left[  \iota_{\operatorname*{Sym},A}\left(  a\right)  ,b\right]
=0$. Furthermore, $K^{\prime}\left(  a\right)  =R_{a}$ (by the definition of
$K^{\prime}$), so that%
\[
\underbrace{\left(  K^{\prime}\left(  a\right)  \right)  }_{=R_{a}}\left(
b\right)  =R_{a}\left(  b\right)  =b\vartriangleleft
a\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.manchon.right.La(b)}%
)}\right)  .
\]


Now, (\ref{pf.thm.manchon.right.m.used}) shows that%
\[
\left(  cb\right)  \leftharpoonup a-\left(  c\leftharpoonup a\right)  \cdot
b=c\cdot\underbrace{\left(  K^{\prime}\left(  a\right)  \right)  \left(
b\right)  }_{=b\vartriangleleft a}-c\underbrace{\left[  \iota
_{\operatorname*{Sym},A}\left(  a\right)  ,b\right]  }_{=0}=c\cdot\left(
b\vartriangleleft a\right)  -\underbrace{c0}_{=0}=c\cdot\left(
b\vartriangleleft a\right)  .
\]
This proves Theorem \ref{thm.manchon.right} \textbf{(m)}.
\end{proof}

\subsection{The case of an associative algebra}

Now we can see Theorem \ref{thm.manchon} with Theorem \ref{thm.manchon.right} interact:

\begin{theorem}
\label{thm.manchon.bi}Let $A$ be a $\mathbf{k}$-algebra. Thus, $A$ becomes a
left pre-Lie algebra (according to Proposition \ref{prop.prelie.assoc}
\textbf{(a)}) and a right pre-Lie algebra (according to Proposition
\ref{prop.prelie.assoc} \textbf{(b)}). Consider the map $\eta:U\left(
A^{-}\right)  \rightarrow\operatorname*{Sym}A$ defined in Theorem
\ref{thm.manchon} \textbf{(f)} (using the left pre-Lie algebra structure on
$A$). Also, consider the map $\eta^{\prime}:U\left(  A^{-}\right)
\rightarrow\operatorname*{Sym}A$ defined in Theorem \ref{thm.manchon.right}
\textbf{(f)} (using the right pre-Lie algebra structure on $A$).

Then, $\eta=\eta^{\prime}$.
\end{theorem}

Before we prove this theorem, let us state a trivial lemma:

\begin{lemma}
\label{lem.manchon.bi.triv}Let $A$ be a $\mathbf{k}$-algebra. Thus, $A$
becomes a left pre-Lie algebra (according to Proposition
\ref{prop.prelie.assoc} \textbf{(a)}) and a right pre-Lie algebra (according
to Proposition \ref{prop.prelie.assoc} \textbf{(b)}).

\textbf{(a)} We have $a\vartriangleright b=a\vartriangleleft b$ for any $a\in
A$ and $b\in A$.

\textbf{(b)} We have $a\vartriangleright\left(  c\vartriangleleft b\right)
=\left(  a\vartriangleright c\right)  \vartriangleleft b$ for any $a\in A$,
$b\in A$ and $c\in A$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.manchon.bi.triv}.]\textbf{(a)} Let $a\in A$ and $b\in
A$. We have $a\vartriangleright b=ab$ (by the definition of the left pre-Lie
algebra structure on $A$). We also have $a\vartriangleleft b=ab$ (by the
definition of the right pre-Lie algebra structure on $A$). Comparing this with
$a\vartriangleright b=ab$, we obtain $a\vartriangleright b=a\vartriangleleft
b$. This proves Lemma \ref{lem.manchon.bi.triv} \textbf{(a)}.

\textbf{(b)} Let $a\in A$, $b\in A$ and $c\in A$. We have $c\vartriangleleft
b=cb$ (by the definition of the right pre-Lie algebra structure on $A$). We
also have $a\vartriangleright c=ac$ (by the definition of the left pre-Lie
algebra structure on $A$). Now, the definition of the left pre-Lie algebra
structure on $A$ shows that%
\[
a\vartriangleright\left(  c\vartriangleleft b\right)  =a\underbrace{\left(
c\vartriangleleft b\right)  }_{=cb}=acb.
\]
On the other hand, the definition of the right pre-Lie algebra structure on
$A$ shows that%
\[
\left(  a\vartriangleright c\right)  \vartriangleleft b=\underbrace{\left(
a\vartriangleright c\right)  }_{=ac}b=acb.
\]
Comparing this with $a\vartriangleright\left(  c\vartriangleleft b\right)
=acb$, we obtain $a\vartriangleright\left(  c\vartriangleleft b\right)
=\left(  a\vartriangleright c\right)  \vartriangleleft b$. This proves Lemma
\ref{lem.manchon.bi.triv} \textbf{(b)}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.manchon.bi}.]We shall use all notations introduced
in Theorem \ref{thm.manchon} (in particular, the map $K:A\rightarrow
\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $ and the left $A^{-}%
$-module structure on $\operatorname*{Sym}A$). We shall also use all notations
introduced in Theorem \ref{thm.manchon.right} (in particular, the map
$K^{\prime}:A\rightarrow\operatorname*{Der}\nolimits^{\operatorname*{op}%
}\left(  \operatorname*{Sym}A\right)  $ and the right $A^{-}$-module structure
on $\operatorname*{Sym}A$).

Our setup causes a notational ambiguity: When $a$ and $b$ are two elements of
$A$, then the expression \textquotedblleft$ab$\textquotedblright\ might mean
the product of $a$ and $b$ in the $\mathbf{k}$-algebra $A$, but can also mean
the product $ab$ of the elements $a$ and $b$ of the symmetric algebra
$\operatorname*{Sym}A$. We resolve this ambiguity by agreeing \textbf{not} to
use the notation $ab$ with the former meaning. Thus, in this proof, a product
such as $ab$ (with $a\in A$ and $b\in A$) shall always mean the product of $a$
and $b$ regarded as elements of the symmetric algebra $\operatorname*{Sym}A$,
rather than the product of $a$ and $b$ regarded as elements of $A$.

For every $a\in A^{-}$ and $b\in A^{-}$, we have%
\begin{equation}
K\left(  a\right)  \circ K^{\prime}\left(  b\right)  =K^{\prime}\left(
b\right)  \circ K\left(  a\right)  \label{pf.thm.manchon.bi.KK'}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.manchon.bi.KK'}):} Let $a\in A^{-}$
and $b\in A^{-}$.
\par
Recall that we are regarding $A$ as a $\mathbf{k}$-submodule of
$\operatorname*{Sym}A$ via the injection $\iota_{\operatorname*{Sym}%
,A}:A\rightarrow\operatorname*{Sym}A$. Thus, $A=\iota_{\operatorname*{Sym}%
,A}\left(  A\right)  =\operatorname*{Sym}\nolimits^{1}A$. Recall that the
subset $\operatorname*{Sym}\nolimits^{1}A$ of $\operatorname*{Sym}A$ generates
the $\mathbf{k}$-algebra $\operatorname*{Sym}A$. In other words, the subset
$A$ of $\operatorname*{Sym}A$ generates the $\mathbf{k}$-algebra
$\operatorname*{Sym}A$ (since $A=\operatorname*{Sym}\nolimits^{1}A$).
\par
The target of the map $K^{\prime}$ is $\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)
=\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $. Thus, $K^{\prime
}\left(  b\right)  \in\operatorname*{Der}\left(  \operatorname*{Sym}A\right)
$. Also, $K\left(  a\right)  \in\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $ (since the target of the map $K$ is $\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $).
\par
Now, $K\left(  a\right)  $ and $K^{\prime}\left(  b\right)  $ both are
elements of the Lie algebra $\operatorname*{Der}\left(  \operatorname*{Sym}%
A\right)  $. Hence, $\left[  K\left(  a\right)  ,K^{\prime}\left(  b\right)
\right]  \in\operatorname*{Der}\left(  \operatorname*{Sym}A\right)  $, where
we are using the Lie bracket of the Lie algebra $\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $ (not of the Lie algebra $\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $). In
other words, $\left[  K\left(  a\right)  ,K^{\prime}\left(  b\right)  \right]
$ is a derivation $\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$.
\par
The map $0:\operatorname*{Sym}A\rightarrow\operatorname*{Sym}A$ is also a
derivation.
\par
Now, we shall show that $\left[  K\left(  a\right)  ,K^{\prime}\left(
b\right)  \right]  \mid_{A}=0\mid_{A}$.
\par
Indeed, let $c\in A$. Lemma \ref{lem.manchon.bi.triv} \textbf{(b)} shows that
$a\vartriangleright\left(  c\vartriangleleft b\right)  =\left(
a\vartriangleright c\right)  \vartriangleleft b$.
\par
The definition of $K$ yields $K\left(  a\right)  =L_{a}$. The definition of
$K^{\prime}$ yields $K^{\prime}\left(  b\right)  =R_{b}$. Now,
\begin{align*}
&  \left(  \left[  K\left(  a\right)  ,K^{\prime}\left(  b\right)  \right]
\mid_{A}\right)  \left(  c\right) \\
&  =\underbrace{\left[  K\left(  a\right)  ,K^{\prime}\left(  b\right)
\right]  }_{\substack{=K\left(  a\right)  \circ K^{\prime}\left(  b\right)
-K^{\prime}\left(  b\right)  \circ K\left(  a\right)  \\\text{(since the Lie
bracket used here}\\\text{is that of }\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  \text{)}}}\left(  c\right)  =\left(
\underbrace{K\left(  a\right)  }_{=L_{a}}\circ\underbrace{K^{\prime}\left(
b\right)  }_{=R_{b}}-\underbrace{K^{\prime}\left(  b\right)  }_{=R_{b}}%
\circ\underbrace{K\left(  a\right)  }_{=L_{a}}\right)  \left(  c\right) \\
&  =\left(  L_{a}\circ R_{b}-R_{b}\circ L_{a}\right)  \left(  c\right)
=L_{a}\left(  \underbrace{R_{b}\left(  c\right)  }%
_{\substack{=c\vartriangleleft b\\\text{(by (\ref{pf.thm.manchon.right.La(b)}%
), applied to }b\text{ and }c\\\text{instead of }a\text{ and }b\text{)}%
}}\right)  -R_{b}\left(  \underbrace{L_{a}\left(  c\right)  }%
_{\substack{=a\vartriangleright c\\\text{(by (\ref{pf.thm.manchon.La(b)}),
applied to }c\\\text{instead of }b\text{)}}}\right) \\
&  =\underbrace{L_{a}\left(  c\vartriangleleft b\right)  }%
_{\substack{=a\vartriangleright\left(  c\vartriangleleft b\right)  \\\text{(by
(\ref{pf.thm.manchon.La(b)}), applied to }c\vartriangleleft b\\\text{instead
of }b\text{)}}}-\underbrace{R_{b}\left(  a\vartriangleright c\right)
}_{\substack{=\left(  a\vartriangleright c\right)  \vartriangleleft
b\\\text{(by (\ref{pf.thm.manchon.right.La(b)}), applied to }b\text{ and
}a\vartriangleright c\\\text{instead of }a\text{ and }b\text{)}}}\\
&  =\underbrace{a\vartriangleright\left(  c\vartriangleleft b\right)
}_{\substack{=\left(  a\vartriangleright c\right)  \vartriangleleft
b}}-\left(  a\vartriangleright c\right)  \vartriangleleft b=\left(
a\vartriangleright c\right)  \vartriangleleft b-\left(  a\vartriangleright
c\right)  \vartriangleleft b=0.
\end{align*}
Comparing this with $\left(  0\mid_{A}\right)  \left(  c\right)  =0\left(
c\right)  =0$, we obtain $\left(  \left[  K\left(  a\right)  ,K^{\prime
}\left(  b\right)  \right]  \mid_{A}\right)  \left(  c\right)  =\left(
0\mid_{A}\right)  \left(  c\right)  $.
\par
Now, let us forget that we fixed $c$. We thus have shown that $\left(  \left[
K\left(  a\right)  ,K^{\prime}\left(  b\right)  \right]  \mid_{A}\right)
\left(  c\right)  =\left(  0\mid_{A}\right)  \left(  c\right)  $ for every
$c\in A$. In other words, $\left[  K\left(  a\right)  ,K^{\prime}\left(
b\right)  \right]  \mid_{A}=0\mid_{A}$. Hence, Lemma
\ref{lem.derivation.unique} (applied to $\operatorname*{Sym}A$, $\left[
K\left(  a\right)  ,K^{\prime}\left(  b\right)  \right]  $, $0$ and $A$
instead of $A$, $d$, $e$ and $S$) shows that $\left[  K\left(  a\right)
,K^{\prime}\left(  b\right)  \right]  =0$. Since $\left[  K\left(  a\right)
,K^{\prime}\left(  b\right)  \right]  =K\left(  a\right)  \circ K^{\prime
}\left(  b\right)  -K^{\prime}\left(  b\right)  \circ K\left(  a\right)  $,
this rewrites as $K\left(  a\right)  \circ K^{\prime}\left(  b\right)
-K^{\prime}\left(  b\right)  \circ K\left(  a\right)  =0$. In other words,
$K\left(  a\right)  \circ K^{\prime}\left(  b\right)  =K^{\prime}\left(
b\right)  \circ K\left(  a\right)  $. This proves (\ref{pf.thm.manchon.bi.KK'}%
).}.

In the proof of Theorem \ref{thm.manchon} \textbf{(e)}, we have shown that the
following holds:

\begin{itemize}
\item The map $K:A^{-}\rightarrow\operatorname*{Der}\left(
\operatorname*{Sym}A\right)  $ is a Lie algebra homomorphism.

\item We have%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K\left(  a\right)  \right)  \left(
\iota_{\operatorname*{Sym},A}\left(  b\right)  \right)  -\left(  K\left(
b\right)  \right)  \left(  \iota_{\operatorname*{Sym},A}\left(  a\right)
\right)
\]
for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  $ is computed in the Lie algebra $\left(
\operatorname*{Sym}A\right)  ^{-}$).

\item We can apply Theorem \ref{thm.gen1} to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$.

\item The $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1} \textbf{(a)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$).

\item The map $\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$
defined in Theorem \ref{thm.manchon} \textbf{(f)} is precisely the map
$\eta:U\left(  A^{-}\right)  \rightarrow\operatorname*{Sym}A$ that is
constructed by Theorem \ref{thm.gen1} \textbf{(b)} (applied to $\mathfrak{g}%
=A^{-}$, $C=\operatorname*{Sym}A$ and $f=\iota_{\operatorname*{Sym},A}$).
\end{itemize}

Similarly, in the proof of Theorem \ref{thm.manchon.right} \textbf{(e)}, we
have shown that the following holds:

\begin{itemize}
\item The map $K^{\prime}:A^{-}\rightarrow\operatorname*{Der}%
\nolimits^{\operatorname*{op}}\left(  \operatorname*{Sym}A\right)  $ is a Lie
algebra homomorphism.

\item We have%
\[
\iota_{\operatorname*{Sym},A}\left(  \left[  a,b\right]  \right)  =\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  +\left(  K^{\prime}\left(  b\right)  \right)
\left(  \iota_{\operatorname*{Sym},A}\left(  a\right)  \right)  -\left(
K^{\prime}\left(  a\right)  \right)  \left(  \iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right)
\]
for every $a\in A^{-}$ and $b\in A^{-}$ (where the Lie bracket $\left[
\iota_{\operatorname*{Sym},A}\left(  a\right)  ,\iota_{\operatorname*{Sym}%
,A}\left(  b\right)  \right]  $ is computed in the Lie algebra $\left(
\operatorname*{Sym}A\right)  ^{-}$).

\item We can apply Theorem \ref{thm.gen1.right} to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$ and $f^{\prime}=\iota_{\operatorname*{Sym},A}$.

\item The right $A^{-}$-module structure on $\operatorname*{Sym}A$ defined in
Theorem \ref{thm.manchon.right} \textbf{(e)} is precisely the one that is
constructed by Theorem \ref{thm.gen1.right} \textbf{(a)} (applied to
$\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and $f^{\prime}=\iota
_{\operatorname*{Sym},A}$).

\item The map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ defined in Theorem \ref{thm.manchon.right} \textbf{(f)}
is precisely the map $\eta^{\prime}:U\left(  A^{-}\right)  \rightarrow
\operatorname*{Sym}A$ that is constructed by Theorem \ref{thm.gen1.right}
\textbf{(b)} (applied to $\mathfrak{g}=A^{-}$, $C=\operatorname*{Sym}A$ and
$f^{\prime}=\iota_{\operatorname*{Sym},A}$).
\end{itemize}

Furthermore, we observe the following:

\begin{itemize}
\item We have%
\[
\left(  K\left(  a\right)  \circ K^{\prime}\left(  b\right)  -K^{\prime
}\left(  b\right)  \circ K\left(  a\right)  \right)  \left(  u\right)
=\left(  K^{\prime}\left(  b\right)  \right)  \left(  \iota
_{\operatorname*{Sym},A}\left(  a\right)  \right)  \cdot u-u\cdot\left(
K\left(  a\right)  \right)  \left(  \iota_{\operatorname*{Sym},A}\left(
b\right)  \right)
\]
for all $a\in A^{-}$, $b\in A^{-}$ and $u\in\operatorname*{Sym}A$%
\ \ \ \ \footnote{\textit{Proof.} Let $a\in A^{-}$, $b\in A^{-}$ and
$u\in\operatorname*{Sym}A$.
\par
Recall that we are identifying $\iota_{\operatorname*{Sym},A}\left(  x\right)
$ with $x$ for every $x\in A$. Thus, $\iota_{\operatorname*{Sym},A}\left(
a\right)  =a$ and $\iota_{\operatorname*{Sym},A}\left(  b\right)  =b$.
\par
The definition of $K$ yields $K\left(  a\right)  =L_{a}$. Hence,
\begin{align*}
\underbrace{\left(  K\left(  a\right)  \right)  }_{=L_{a}}\left(
\underbrace{\iota_{\operatorname*{Sym},A}\left(  b\right)  }_{=b}\right)   &
=L_{a}\left(  b\right)  =a\vartriangleright b\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.thm.manchon.La(b)})}\right) \\
&  =a\vartriangleleft b\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma
\ref{lem.manchon.bi.triv} \textbf{(a)}}\right)  .
\end{align*}
On the other hand, the definition of $K^{\prime}$ yields $K^{\prime}\left(
b\right)  =R_{b}$. Now,%
\[
\underbrace{\left(  K^{\prime}\left(  b\right)  \right)  }_{=R_{b}}\left(
\underbrace{\iota_{\operatorname*{Sym},A}\left(  a\right)  }_{=a}\right)
=R_{b}\left(  a\right)  =a\vartriangleleft b
\]
(by (\ref{pf.thm.manchon.right.La(b)}), applied to $b$ and $a$ instead of $a$
and $b$).
\par
Now,%
\begin{align*}
&  \left(  \underbrace{K\left(  a\right)  \circ K^{\prime}\left(  b\right)
}_{\substack{=K^{\prime}\left(  b\right)  \circ K\left(  a\right)  \\\text{(by
(\ref{pf.thm.manchon.bi.KK'}))}}}-K^{\prime}\left(  b\right)  \circ K\left(
a\right)  \right)  \left(  u\right) \\
&  =\underbrace{\left(  K^{\prime}\left(  b\right)  \circ K\left(  a\right)
-K^{\prime}\left(  b\right)  \circ K\left(  a\right)  \right)  }_{=0}\left(
u\right)  =0\left(  u\right)  =0.
\end{align*}
Comparing this with%
\begin{align*}
\underbrace{\left(  K^{\prime}\left(  b\right)  \right)  \left(
\iota_{\operatorname*{Sym},A}\left(  a\right)  \right)  }_{=a\vartriangleleft
b}\cdot u-u\cdot\underbrace{\left(  K\left(  a\right)  \right)  \left(
\iota_{\operatorname*{Sym},A}\left(  b\right)  \right)  }_{=a\vartriangleleft
b}  &  =\left(  a\vartriangleleft b\right)  \cdot u-\underbrace{u\cdot\left(
a\vartriangleleft b\right)  }_{\substack{=\left(  a\vartriangleleft b\right)
\cdot u\\\text{(since the }\mathbf{k}\text{-algebra }\operatorname*{Sym}%
A\\\text{is commutative)}}}\\
&  =\left(  a\vartriangleleft b\right)  \cdot u-\left(  a\vartriangleleft
b\right)  \cdot u=0,
\end{align*}
we obtain%
\[
\left(  K\left(  a\right)  \circ K^{\prime}\left(  b\right)  -K^{\prime
}\left(  b\right)  \circ K\left(  a\right)  \right)  \left(  u\right)
=\left(  K^{\prime}\left(  b\right)  \right)  \left(  \iota
_{\operatorname*{Sym},A}\left(  a\right)  \right)  \cdot u-u\cdot\left(
K\left(  a\right)  \right)  \left(  \iota_{\operatorname*{Sym},A}\left(
b\right)  \right)  ,
\]
qed.}.

\item We have $\iota_{\operatorname*{Sym},A}=\iota_{\operatorname*{Sym},A}$.
\end{itemize}

Combining all the observations in the bullet points made above, we conclude
that we can apply Theorem \ref{thm.gen5} \textbf{(c)} to $\mathfrak{g}=A^{-}$,
$C=\operatorname*{Sym}A$, $f=\iota_{\operatorname*{Sym},A}$ and $f^{\prime
}=\iota_{\operatorname*{Sym},A}$. As a result, we obtain $\eta=\eta^{\prime}$.
This proves Theorem \ref{thm.manchon.bi}.
\end{proof}

\begin{thebibliography}{999999999}                                                                                        %


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\end{thebibliography}


\end{document}