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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}